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Regularity for Certain Nonlinear Parabolic SystemsHyeong-Ohk Bae ab Hi Jun Choe ca Department of Mathematics Ajou University Suwon Republic of Koreab Department of Mathematics Ajou University Suwon Republic of Korea Republicof Koreac Department of Mathematics Yonsei University Seoul Republic of Korea

Online Publication Date 01 May 2004To cite this Article Bae Hyeong-Ohk and Choe Hi Jun (2004) Regularity for CertainNonlinear Parabolic Systems Communications in Partial Differential Equations295 611 - 645To link to this article DOI 101081PDE-120037327

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COMMUNICATIONS IN PARTIAL DIFFERENTIAL EQUATIONSVol 29 Nos 5 amp 6 pp 611ndash645 2004

Regularity for Certain Nonlinear Parabolic Systems

Hyeong-Ohk Bae1 and Hi Jun Choe2

1Department of Mathematics Ajou UniversitySuwon Republic of Korea

2Department of Mathematics Yonsei UniversitySeoul Republic of Korea

ABSTRACT

By means of an inequality of Poincareacute type a weak Harnack inequality forthe gradient of a solution and an integral inequality of Campanato type it isshown that a solution to certain degenerate parabolic system is locally Houmlldercontinuous The system is a generalization of p-Laplacian system Using adifference quotient method and Moser type iteration it is then proved that thegradient of a solution is locally bounded Finally using the iteration and scalingit is shown that the gradient of the solution satisfies a Campanato type integralinequality and is locally Houmllder continuous

Key Words p-Laplacian system Houmllder continuous Poincareacute inequality

lowastCorrespondence Hyeong-Ohk Bae Department of Mathematics Ajou University Suwon443-749 Republic of Korea E-mail hobaeajouackr

611

DOI 101081PDE-120037327 0360-5302 (Print) 1532-4133 (Online)Copyright copy 2004 by Marcel Dekker Inc wwwdekkercom

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612 Bae and Choe

1 INTRODUCTION

In this paper we study the regularity of second order parabolic systems ofgeneral growth We may consider these systems as generalizations for evolutionaryp-Laplacian functionals Choe (1992a) established interior Houmllder regularity for thegradient of the bounded minimizers of

min Iu = minintFu (11)

under general growth conditions on F such that

Qpminus22 le FQiQ

jQi

j le 1

(Qpminus2 + Qqminus2)2 (12)

for all Q isin nN and some gt 0 where 1 lt p le q lt p+ 1 Thus it is naturalto ask regularity questions for parabolic cases Here we prove Houmllder continuityof weak solutions and their gradients for parabolic systems under general growthconditions The theory for the parabolic cases are rather complicated because ofscaling

Now we define our problem Suppose sub n is a bounded domain andwe define T = times 0 T for positive T Let 1 lt p le q lt p+ 1 and we assumeF rarr is a 3 function satisfying (12) for all Q isin nN and some gt 0 and

F primeprimeprimes le cspminus3 + sqminus3 (13)

for some c and for all s gt 0 From the ellipticity condition (12) we note that

spminus2 le F primeprimes le 1spminus2 + sqminus2 (14)

for all s gt 0 We define AiQ

def= FQiQ for notational simplicity

We say u isin 00 T L2 cap L20 TW 1p is a solution to

uit minus

(Ai

u)x= 0 i = 1 N (15)

if u satisfiesintT

minusui13it + Ai

u13ixdx dt = 0

for all 13 isin 0 T We prove that any solution u of (15) is Houmllder continuous and

the gradient of u is Houmllder continuousFor stationary cases ie for elliptic systems a number of authors have

considered regularity questions for the case p = q Uhlenbeck (1977) did show u isHoumllder continuous when p ge 2 The 1 regularity was extended for all p isin 1when u is a scalar valued function by various authors (Lewis 1983 Lieberman1991 ) When 1 lt p lt 2 and u is a vector valued function Tolksdorff (1983)showed that u is Houmllder continuous On the other hand systems with nonstandard

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Regularity for Certain Nonlinear Parabolic Systems 613

growth conditions have been interested by many authors Indeed Fusco andSbordone (1990) showed a higher integrability result of u under the assumptionthat 1 lt p le q lt npnminus p and F satisfies 2 condition Also Marcellini (1989)considered various anisotropic cases Choe (1992a) showed that if u is a boundedminimizer of (11) then u is bounded under the assumption (12) Finally weremark that Lieberman (1991) generalized the structure of Ladyzhenskaya andUralrsquotzeva and Uraltseva (1968) also showed the C-regularity

For the evolutionary case a number of authors proved 0 regularity for thesolutions of degenerate parabolic equations that is for the case p = q DiBenedetto(1986) proved 0 regularity for evolutionary p-Laplacian equations when p isin2 His method was extended for all p isin 1 by Chen and DiBenedetto(1988) Also considering a reflection principle Chen and DiBenedetto (1989)gave another proof for Houmllder continuity of solutions to systems up to theboundary when p isin 2nn+ 2 Lieberman (1993) considered boundary behaviorof degenerate parabolic equations We also remark that Choe (1992b) provedHoumllder continuity of solutions to systems for all p isin 1 in a rather simpleway by showing a Poincareacute inequality and weak Harnack inequalities for uFurthermore we note that DiBenedetto and Friedman (1985) proved the gradientsof solutions of parabolic systems are Houmllder continuous when p isin 2nn+ 2Independently Wiegner (1986) proved the gradients of solutions of parabolicsystems are Houmllder continuous when p isin 2 Choe (1991) also proved that thegradient of solutions of parabolic systems are Houmllder continuous for all p isin 1

as long as u lies in a highly integrable space To be more specific when u isin Lqloc

q gt N2minus pp then u is Houmllder continuous for all p isin 1In the next section we show that solutions of (15) are locally bounded using a

Moser type iteration which is suitable for systems with general growth conditionsNamely we show that u isin L

loc if u isin Lr0loc for some r0 (see Theorem 21) The

restriction of the integrability of u is necessary to start the iteration In fact ifp isin 2nn+ 2 and p le q le p2+ 1 or p2+ 1 le q lt pn+ 3+ 4n+ 4then r0 le 2 and there is no restriction on the integrability of u Once u is boundedwe prove an inequality of Poincareacute type which is known for evolutionary p-Laplacian systems (see Choe 1992b) We then obtain a weak Harnack inequality fora solution of equation using a scaling argument as in Choe (1992b) and an iterationmethod as in Choe (1991) Consequently a Campanato type growth condition for u

is obtained by combining Poincareacutersquos inequality a weak Harnack inequality for u

and a Caccioppoli type inequality Hence Houmllder continuity of u follows from theisomorphism theorem of Da Prato (1965)

Following Choe (1991) that is using a difference quotient method we showthat u has space derivatives Once u has space derivatives we find an integralidentity which is suitable for Moser iteration Indeed we use integration by partsand a Caccioppoli type inequality for the second derivatives of u (see Choe andLewis 1991 for elliptic cases) For evolutionary p-Laplacian we also recall thatit is necessary to show u isin Ls for all s isin p when p isin 1 2 in proving u

is bounded Once u is bounded we construct an iteration scheme to obtain aCampanato growth condition on u There we use a perturbation method tooTherefore Houmllder continuity of u follows from the isomorphism theorem

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614 Bae and Choe

We define symbols and notations as follows

BRx0 = x isin n x minus x0 lt R Rt0 = t isin t0 minus Rp lt t lt t0

QRx0 t0 = BRx0timesRt0 SRx0 t0 = BRx0times t0 minus R2 t0

pQR is the parabolic boundary of QR z = x t

ndashintAu dz = 1

AintAu dz uR = ndash

intQRx0t0

u dz uRt = ndashintBRx0

ux tdx

We write z0 = x0 t0 = 0 0 as a generic point On the other hand if there is noconfusion we simply drop x0 t0 from various symbols and notations Finally wewrite c as a constant depending only on exterior data and as an arbitrarily smallconstant depending only on exterior data

2 L ESTIMATE FOR u

In this section we prove that u is bounded To begin with Moser iteration weneed that u is highly integrable

Theorem 21 Suppose that u isin Lr0locT for some r0 and 1 lt p lt then we have

u isin Lloc Furthermore suppose SR0

sub T then u satisfies

supS R0

2

u le c

[ndashintSR0

us0+s1 dz

] 1s0+s2 + c (21)

for any number s0 gt minuss2 and some c independent of R0 wheres1 = p

p+1minusq s2 = 2minusn+2qminusp

p+1minusqif q ge p

2 + 1

s1 = 2 s2 = n+ 2minus 2np

if q le p

2 + 1

Here r0 = s0 + s1

Proof Let 0 lt lt R be numbers Let be a standard cutoff function such that = 1 for x t isin S = 0 for x t isin pSR 0 le le 1 t le cRminus p lecRminus for some c We take usuip as a test function to (15) and we obtain

1s + 2

intSR

d

dt

(us+2p)dzminus p

s + 2

intSR

us+2pminus1t dz

+intSR

Aiu

(usuix+ susminus2uiujuj

x

)p dz = minusp

intSR

Aiuusuix

pminus1 dz

(22)

Owing to (14) one has that there is c gt 0 such that

pminus 1spminus1 minus c le F primes le 1

(spminus1

pminus 1+ sqminus1

q minus 1

)+ c

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Regularity for Certain Nonlinear Parabolic Systems 615

for all s gt 0 We note that Fuixu = F primeuui

xu so we have that

pminus 1up minus cu le Ai

uuixle 1

( uppminus 1

+ uqq minus 1

)+ cu (23)

We also note that

Aiuu

iujujxge

pminus 1upminus2ui

xujxuiuj minus c

uixujxuiuj

u ge minuscu2u (24)

and that

Aiuus+1pminus1 le cupminus1 + uqminus1us+1pminus1 (25)

Combining (23) (24) and (25) and using Houmllderrsquos inequality we obtain from (22)that

supt

intBR

us+2p dx +intSR

upusp dz

le cintSR

us+2pminus1tdz+ cintSR

(us+pp + u p

p+1minusq +s pp+1minusq

)dz

+ cint

usup dz

for some c and

supt

intBR

us+2pdx +intSR

∣∣∣(u p+sp

)∣∣∣p dzle c

intSR

us+2pminus1tdz+ cintSR

(us+pp + u p

p+1minusq +s pp+1minusq

)dz

+ cint

pus dz

By the integrability assumption on u the right side is bounded and thereforethe left side is also bounded Therefore from Sobolevrsquos embedding theorem andHoumllderrsquos inequality we prove that for p lt n

intS

up+s+s+2 pn dz

leint [ int

us+2p dx] p

n[ int

us+p nnminusp

npnminusp dx

] nminuspn

dt

le[sup

t

intus+2p dx

] pnint ∣∣∣(u s+p

p )∣∣∣p dz

le c

[1

Rminus p

p+1minusq

intSR

(us+2 + us+p + u pp+1minusq +s + us)dz]1+ p

n

(26)

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616 Bae and Choe

We assume that n gt p ge 2 then we see that pp+ 1minus q gt p ge 2 Hence fromYoungrsquos inequality we can write (26) as

intS

us+p+s+2 pn dx le c

[1

Rminus p

p+1minusq

intSR

u pp+1minusq +s dz+ SR

Rminus p

p+1minusq

]1+ pn

(27)

for some c We set si by si+1 + pp+ 1minus q = si + p+ si + 2pn that is forsufficiently large s0

si = i

(s0 +

2minus n+ 2q minus p

p+ 1minus q

)+ n+ 2q minus pminus 2

p+ 1minus q

where def= 1+ pn Define Ri

def= R021+ 2minusi Now we take s = si = Ri+1 andR = Ri in (27) Define

i

def= ndashintSRi

usi+ pp+1minusq dz

then (27) can be written as

Rn+2i+1 i+1 le c

(2i+2

R0

) pp+1minusq

Rn+2i

[

i + 1

]

and

i+1 le c

(R0

2

)n+2 pnminus p

p+1minusq

2i+1 pp+1minusq minusn+2

[

i + 1

]

and hence

i+1 le cii + ci (28)

for some c We also have that

i+1 le ciciminus1iminus1 + ciminus1 + ci le ciciminus1

2

iminus1 + ciciminus1 + ci

le ciciminus1ciminus223

iminus2 + ciciminus1ciminus22 + ciciminus1 + ci

le ci+iminus1+iminus22+iminus33+middotmiddotmiddot+2iminus2+iminus1

i+1

0 + ci + ci+iminus1

+ ci+iminus1+iminus22 + middot middot middot + ci+iminus1+middotmiddotmiddot+2iminus2+iminus1

le cn2

p2iminus1minus n

p ii+1

0 + icn2

p2iminus1minus n

p i

le ci+1

i+1

0 + ici+1

If s0 gt n+ 2q minus pminus 2p+ 1minus q then s0 + 2minus n+ 2q minus pp+ 1minus q gt0 Hence we have that

limi

si = and limi

sii

= s0 +2minus n+ 2q minus p

p+ 1minus q

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Regularity for Certain Nonlinear Parabolic Systems 617

Iterating (28) with some large s0 we have

supS R0

2

u le c

[ndashintSR0

u pp+1minusq +s0 dz

] 1

s0+ 2minusn+2qminuspp+1minusq + c

and this completes the proof for the case n gt p ge 2Now we assume that 1 lt p le 2 If q ge p2+ 1 then pp+ 1minus q ge 2

In this case applying Youngrsquos inequality on the last term of (26) we have (27)Hence following the argument for the case p ge 2 we prove (21) and this completesthe proof for the case q ge p2+ 1

Now we assume that p lt n and

p

2+ 1 gt q ge p gt 1

In this case s + 2 gt s + pp+ 1minus q and we have from (26)

intS

us+p+s+2 pn dx le c

[1

Rminus 2

intSR

us+2 dz+ SRRminus 2

]1+ pn

(29)

for some c We set si by sidef= is0 + n+ 2minus 2npi minus 1 Hence if s0 gt 2np

minus nminus 2 then

limi

si = and limi

sii

= s0 + n+ 2minus 2np

We take s = si = Ri+1 and R = Ri in (29) Iterating (29) we obtain

supS R0

2

u le c

[ndashintSR0

us0+2 dz

] 1s0+n+2minus 2n

p + c

for some c independent of R0 and this completes the proof for the case p2+ 1 gtq ge p gt 1

We now assume that n = p Then from Sobolevrsquos embedding theorem andHoumllderrsquos inequality we obtain that for p = n

intS

u s+22 +s+p dz le

int [ intus+2p dx

] 12[ int

u2s+p2p dx] 1

2dt

le[sup

t

intus+2p dx

] 12[ int (

u s+pp

)2pdx

] 12dt

le[sup

t

intus+2p dx

] 12int ∣∣∣(u s+p

p )∣∣∣p dz

le c

[1

Rminus p

intSR

(us+2 + us+p + u pp+1minusq +s + us)dz] 3

2

for some c The remaining things can be done similarly as in the case 2 le p lt n

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618 Bae and Choe

Now we consider the case p gt n Then from Sobolevrsquos embedding theorem andHoumllderrsquos inequality we obtain that for p gt n

intS

us+2+ p+s2 dz le

int [ intus+2p dx

](sup

x

u p+sp

) p2

dt

le(sup

t

intus+2p dx

) int ( int ∣∣∣(u s+pp

)∣∣∣p dx)12

dt

le c

[1

Rminus p

intSR

(us+2 + us+p + u pp+1minusq +s + us)dz] 3

2

The remaining things can be done in a similar way as before

The restriction of the integrability of u is necessary to start Moser typeiteration In fact if p isin 2nn+ 2 and p le q le p2+ 1 or p2+ 1 le q ltpn+ 3+ 4n+ 4 then r0 le 2 and there is no restriction on the integrability of u

3 L ESTIMATE FOR u

In this section we prove that u is bounded To begin with Moser iterationwe need that u is highly integrable Indeed the higher integrability of u followsfrom Caccioppoli type inequality for the second derivative of u and integration byparts technique Here the boundedness of u is crucial This method has been appliedin proving for the regularity questions for singular parabolic equations (see Choe1991) By approximation we assume that u has second derivative with respect to xHence differentiating (15) with respect to x we get

(uix

)tminus (

Ai

Qj

uujxx

)x= 0 (31)

Lemma 31 Let SR0sub T Suppose that u is bounded then for all s

intS R0

2

us dz le cintSR0

up dz+ cR0 s

for some c

Proof Let be a standard cutoff function such that = 1 for x t isin S = 0 forx t isin pSR 0 le le 1 t le cRminus 2 le cRminus for some c

From integration by parts we have

intBR

us+2k dx =intBR

usuixuixk dx = minus

intBR

u div(usuk

)dx

le csuL

intBR

us2uk dx + csuL

intBR

us+1kminus1 dx

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Regularity for Certain Nonlinear Parabolic Systems 619

and applying Youngrsquos inequality we obtain

intBR

us+2k dx le cintBR

usminus22u2k dx + cintBR

us2 dx

for some c depending on uL and s Integrating with respect to t we obtain

intSR

us+2k dz le cintSR

usminus22u2k dz+ cintSR

us2 dz (32)

for some c depending on uL and s Now we take uixusminuspk as a test function

to (31) and we get

intSR

d

dt

(us+2minuspk)dz+

intSR

usminus22u2k dz

le cintSR

us+2minuspkminus1tdz+ cintSR

us+2qminusp2 dz+ cintSR

us2 dz(33)

for some c Therefore combining (32) and (33) we obtain

intSR

us+2k dz le cintSR

us2 dz+ cintSR

us+2minusptdz

+ cintSR

us+2qminusp2 dz (34)

Assume p ge 2 Since s + 2q minus p ge s + 2minus p we get from (34) that

intS

us+2 dz le c

Rminus 2

intSR

us+2qminusp dz+ cSR

Rminus 2(35)

for some c Since we are assuming 2 le p le q lt p+ 1 2q minus p lt 2 Define si bysi+1 + 2q minus p = si + 2 that is si = s0 + i2+ 2pminus 2q Note that limirarr si = Set s0 = 3pminus 2q then s0 + 2minus p gt 0 Define Ri = R021+ 2minusi i = 0 1 2 Now we set s = si = Ri+1 and R = Ri Hence defining i =

intSRi

usi+2qminusp dz wecan rewrite (35) as

i+1 le cii + ciBR0 (36)

Therefore iterating (36) we conclude that

intS R

2

us dz le cintSR

up dz+ c (37)

for all s and this completes the proof for the case p ge 2Now we consider the case that 1 lt p le 2 Suppose that s + 2minus p le s

+ 2q minus p that is p+ 22 le q lt p+ 1 In this case we prove (37) in the

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620 Bae and Choe

same way for the case p ge 2 Finally we assume that p le q lt p+ 22 that iss + 2minus p ge s + 2q minus p Considering Youngrsquos inequality in (34) we get

intS

us+2 dz le c

Rminus 2

intSR

us+2minusp dz+ cSR

Rminus 2 (38)

Taking = Ri+1 R = Ri and si = s0 + ip with s0 = 2pminus 2 we obtain from (38)

intSRi+1

usi+1+2minusp dz le ci

R2

intSRi

usi+2minusp dz+ ci SRR2

for some c Therefore iterating this we obtain (37)

Once we have shown that u is in a high Ls space we can apply Moser iteration

Theorem 32 We have u isin L Furthermore suppose SR0sub T then u satisfies

supS R0

2

u le c[ndashintSR0

us0+s1 dz] 1

s0+s2 + c (39)

for some large s0 and some c independent of R0 where

s1 = 2q minus p s2 = 2minus nq minus p if q ge p

2 + 1

s1 = 2 s2 = np

2 + 2minus n if q le p

2 + 1

Proof Let be a standard cutoff function We take usuix2 as a test function to

(31) and we obtain

1s+2

int d

dt

(us+2)2 dz+

intAi

Qj

uujxx

(usuixx

+ susminus2uixukxukxx

)2 dz

= minus2int

Ai

Qj

uujxx

usuixx

dz

Considering the ellipticity condition

Qpminus22 le Ai

Qj

Qi

j le c

(Qpminus2 + Qqminus2)2

we see that

supt

intus+22 dx +

int ∣∣(u s+p2

)∣∣2 dzle c

intus+2tdz+ c

int (us+p + us+2qminusp)2 dz

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Regularity for Certain Nonlinear Parabolic Systems 621

for some c Therefore from Sobolevrsquos embedding theorem and Houmllderrsquos inequalitywe prove that

intS

us+p+s+2 2n dx le

int [ intus+22 dx

] 2n[ int

us+p nnminus2

2nnminus2dx

] nminus2n

dt

le[sup

t

intus+22 dx

] 2nint ∣∣∣(u s+p

2 )∣∣∣2 dz

le c

[1

Rminus 2

intSR

(us+2 + us+p + us+2qminusp)dz

]1+ 2n

(310)

We assume that p ge 2 then we see that s + 2q minus p ge s + 2 Hence fromYoungrsquos inequality we can write (310) as

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2qminusp dz+ SRRminus 2

]1+ 2n

(311)

for some c We set si by si+1 + 2q minus p = si + p+ si + 22n that is for

sufficiently large s0 si = is0 +(2minus nq minus p

)i minus 1 where

def= 1+ 2n Define

Ri

def= R021+ 2minusi Now we take s = si = Ri+1 and R = Ri in (311) Define

i

def= ndashintSRi

usi+2qminusp dz then (311) can be written as

i+1 le cii + ci (312)

If s0 gt nq minus pminus 2 then s0 + 2minus nq minus p gt 0 Hence we have that

limi

si = and limi

sii

= s0 + 2minus nq minus p

From Lemma 31 we can take s0 as large as possible Iterating (312) with some larges0 we have

supS R0

2

u le c[ndashintSR0

us0+2qminusp dz] 1

s0+2minusnqminusp + c

and this completes the proof for the case p ge 2Now we assume that 1 lt p le 2 If q ge p2+ 1 then s + 2q minus p ge s + 2 In this

case applying Youngrsquos inequality on the last term of (310) we have

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2qminusp dz+ SRRminus 2

]1+ 2n

for some c Hence following the argument for the case p ge 2 we prove (39) andthis completes the proof for the case q ge p2+ 1

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622 Bae and Choe

Now we assume that p2+ 1 gt q ge p gt 1 In this case s + 2 gt s + 2q minus p andwe have from (310)

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2 dz+ SRRminus 2

]1+ 2n

(313)

for some c We set si by sidef= is0 + np2+ 2minus ni minus 1 Hence if s0 gt nminus 2

minus np2 then

limi

si = and limi

sii

= s0 + 2minus n+ np

2

We take s = si = Ri+1 and R = Ri in (313) Iterating (313) we obtain

supS R0

2

u le c

[ndashintSR0

us0+2 dz

] 1s0+2+ np

2 minusn + c

for some c independent of R0 and this completes the proof for the casep2+ 1 gt q ge p gt 1

4 HOumlLDER CONTINUITY OF u

In this section we prove that u is Houmllder continuous

Theorem 41 Define r0 by the constant appearing in Theorem 21 Suppose that uis L

r0locT then u is Houmllder continuous Moreover for all Q2R sub T with R lt 1 u

satisfies

ndashintQR

uminus uR2 dz le cRs

for some c independent of R where we define

s =2 when p ge 2

ppminus 1 when p isin 1 2

We introduce a nonnegative cutoff function isin 0 BR such that = 1 in

BR2 0 le le 1 le cR We define minus

R

def= t0 minus 2Rp t0 minus Rp and QminusR = BR timesminus

R

Also we define uRt

def= 1BRintBR

ux tdx and uR

def= 1QRintQR

ux tdz Firstwe prove a lemma which is essential for a Poincareacute inequality for solutions of adegenerate parabolic system

Lemma 42 Suppose Q2R sub T then u satisfies the following inequality

suptisinR

intminus

R

dsintBR

2ux tminus uRs2 dx

le cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz+ cintQminus

R

uminus uRt2 dz

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Regularity for Certain Nonlinear Parabolic Systems 623

for all R lt R0 where c depends only on nN and p and

s1 = p s2 = p and s3 = pqminus1

pminus1 when p ge 2

s1 = p2minus p s2 = ppminus 1 and s3 = pq minus 1 when 1 lt p lt 2

Proof Since u isin 0(0 T L2

) there exists ts isin R for all s isin minus

R such that

intBR

pux t minus uRs2dx = suptisinR

intBR

pux tminus uRs2 dx

We take uminus uRspxst as a test function to (15) where st is the characteristic

function such that

st =1 for all isin s t

0 for all isin s t

Hence we have

intut middot uminus uRs

pxst dz+int

Au middot uminus uRspst dz = 0 (41)

Now we assume p isin 2 Considering the structure condition (12) on A (23)and (25) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

leint

uminus uRs2x sp dx + cintBRtimesst

upminus1uminus uRspminus1dz

+ cintBRtimesst

uqminus1uminus uRspminus1dz (42)

for some c independent of R From Youngrsquos inequality we have

intBRtimesst

upminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

up dz (43)

for some c and small Similarly the last term in (42) can be estimated as follows

intBRtimesst

uqminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

u pqminus1pminus1 dz (44)

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for some c and small Combining (42) (43) and (44) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

le cint

uminus uRs2x sp dx +

Rp

intBRtimesst

uminus uRspp dz

+ cintBRtimesst

up dz+ cintBRtimesst

u pqminus1pminus1 dz (45)

for some c and small Integrating (45) with respect to s from t0 minus 2Rp to t0 minus Rpwe have

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRs2p dz+ c

Rp

intminus

R

dsintBRtimesst

uminus uRs2p dz

+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

By the choice of t we have that for small

intminus

R

dsintBR

uminusuRs2x t pdx

le cintQminus

R

uminusuRs2p dz+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

So we proved Lemma 42 when p ge 2In case p isin 1 2 we estimate the second term of (41) as follows

intuppst dz+ p

intAu middot uminus uRs

pminus1st dz

leint

Au middot uminus uRspst dz (46)

and using the fact that u is bounded and ppminus 1 gt 2 we have

∣∣∣ int Au middot uminusuRspminus1st dz

∣∣∣le c

intBRtimesst

upminus1uminusuRspminus1st dz

+ cintBRtimesst

uqminus1pminus1uminusuRsst dz

le

Rp

intuminusuRs

ppminus1 pst dz+

c

Rp2minusp

intuppminus1 dz+ c

Rp2minusp

intupqminus1 dz

(47)

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Regularity for Certain Nonlinear Parabolic Systems 625

Combining (41) (46) and (47) we have

intBR

uminus uRs2x t p dx

le cu2minusppminus1L

Rp

intBRtimesst

uminus uRs2pst dz+c

Rp2minusp

intBRtimesst

uppminus1 dz

+ c

Rp2minusp

intBRtimesst

upqminus1 dz+intBR

uminus uRs2x sp dx (48)

for small Integrating (48) with respect to s and considering the choice of t weprove Lemma 42 when 1 lt p lt 2

Now we prove a Poincareacute inequality

Theorem 43 Suppose Q2R sub T with R lt 1 then u satisfies the following inequality

intQR

2

uminus uR22 dz le cR2

intQ2R

u2 dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

where c is independent of R

Proof Using Lemma 42 we have

intQR

2

uminus uR22 dz le

intQR

uminus uRs2p dz

le c

Rp

intminus

R

dsintQR

uminus uRs2p dz le c suptisinR

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRt2p dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

le cR2intQ2R

u2 dz+ cRs1

intQ2R

us2z+ cRs1

intQ2R

us3 dz

where we used a Poincareacute type inequality for x variables only that is

intBR

uminus uRs2p dx le cR2intBR

u2 dx

Proof of Theorem 41 Since u is bounded owing to Theorem 32 we have fromTheorem 43

intQR

uminus uRs2 dz lecRn+p+2 when p isin 2

cRn+p+ppminus1 when p isin 1 2

where c is independent of R Hence by isomorphism theorem of Da Prato (1965) weprove Theorem 41

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626 Bae and Choe

5 HOumlLDER CONTINUITY FOR u

In this section we prove that u is Houmllder continuous Because of degeneracyof the pde we need scaled cylinders (see DiBenedetto and Friedman 1985)Let R0 gt 0 be fixed and consider a cylinder

SR0

def= BR0times t0 minus R2minus

0 t0

for small For any 0 lt R le R02 we introduce a scaled cylinder

SRdef= BR times t0 minus R22minusp t0

for fixed ge supSR0u

We assume that pminus2 gt R0 if p gt 2 Hence we note that SR sub S

R0and

ge supSRu

The following theorem is our main theorem in this section

Theorem 51 Suppose that SR sub T then u satisfies for all r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR 2 dz+ cR

for some gt 0

For the proof of Theorem 51 we follow a usual program for degenerateequations such that if for some isin 0 1∣∣z isin SR u2 le 1minus 2

∣∣ gt ∣∣SR

∣∣ (51)

then

supS R

2

u le (52)

for some small isin 0 1 and otherwise that is (51) fails then for all 0 lt r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR2 dz (53)

for some gt 0 Combining (52) and (53) we prove Theorem 51First we prove that (52) under the assumption (51) A similar argument for

the case p = q ge 2 appeared in DiBenedetto and Friedman (1985) We assume that(51) is true Set v = u2 For each h gt 0 we set AhRt

def= x isin BR v gt h and

BhR

def= z isin SR v gt h From the mean value theorem we estimate AhRt

Lemma 52 For any isin 0 there exists isin t0 minus 2minuspR20 t0 minus 2minuspR2

0 such that

A1minus2R0 le 1minus

1minus BR0

where is independent of

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Regularity for Certain Nonlinear Parabolic Systems 627

Proof Suppose that the assertion is not true then

B1minus2R0 ge

int t0minus2minuspR20

t0minus2minuspR20

A1minus2R0d gt 1minus BR0

2minuspR20 = 1minus SR0

and this contradicts the assumption (51)

The following lemma is useful in estimating sup u Indeed when p = q ge 2DiBenedetto and Friedman proved a similar theorem (see Lemma 62 inDiBenedetto and Friedman 1985) Now choose = 2 and in Lemma 52We define St

R

def= BR times t

Lemma 53 There exists a positive integer r independent of such that

∣∣A1minus 2r

2R0t

∣∣ le (1minus

(2

)2)BR0

for all t isin t0 minus

22minuspR2

0 t0

Proof Consider the function

v = log+[

2

2minus vminus 1minus 2+ + 2

2r

]

where r is a positive integer This function was used in Choe (1991) and DiBenedettoand Friedman (1985) We take 13 = ui

x2vprime 2 as a test function for (31) where

isin 0 is a standard cutoff function such that B1minusR for small decided later is

contained in support Integrating from to t we obtain

12

intStR

vt2prime 2 dz+

intStR

Ai

Qj

uujxx

uixx

2prime 2 dz

+ 2intStR

Ai

Qj

uujxx

uixukxx

ukx2primeprime 2 dz

+ 2intStR

Ai

Qj

uujxx

uix x

2prime dz = 0

Considering the fact 2primeprime = 21+ prime2 we have

intBR

2vx t 2xdx +intStR

upminus2v21+ prime2 2 dz

leintBR

2vx 2xdx + cintStR

upminus2v prime dz

+ cintStR

uqminus2v prime dz

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628 Bae and Choe

Applying Youngrsquos inequality on the last two terms we getintBR

2vx t 2xdx

leintBR

2vx 2xdx + cintStR

upminus2 2 dz+ cintStR

u2qminuspminus2 2 dz

for some c Since le r log 2 and u is bounded we getintBR

2vx t 2xdx leintBR

2vx 2xdx + crBR

From Lemma 52 we getintBR

2vx t 2xdx le r log 221minus

1minus

2

BR

From definition of we haveintB1minusRcapvgt1minus

2r 22vx tdx ge r minus 1 log 22A1minus

2r 21minusRt

and

A1minus 2r

21minusRt le[( r

r minus 1

)2 1minus

1minus

2

+ c

r

]BR

Hence we obtain

A(1minus

2r

)2R

t le A1minus 2r

21minusRt + BRB1minusR

le[( r

r minus 1

)2 1minus

1minus

2

+ c

r+ n

]BR

Taking r so large that( r

r minus 1

)2 le(1minus

2

)1+

c

rle 3

82

and = 382n we obtain

A1minus 2r

2Rt le(1minus

(2

)2)BR

We take 13 = uixvminus h+ 2 as a test function in (31) where v = u2 and is

a cutoff function such that = 1 in Sr = 0 on pSR le cRminus r and t le cpminus2Rminus r2 Considering the ellipticity condition (12) we have

supt

int ∣∣vminus h+∣∣2 2 dx +

intupminus2vminus h+2 2 dz

le c

Rminus r2

intSR

upminus2vminus h+2 dz

+ c

Rminus r2

intSR

u2qminuspminus2vminus h+2 dz+ cintSR

vminus h+2 tdz (54)

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Regularity for Certain Nonlinear Parabolic Systems 629

for some c Now choose h ge 12 For such a choice of h we haveint

upminus2vminus h+2 2 dz ge cpminus2int

vminus h+2 2 dz

Hence from (54) we obtain

supt

intvminus h+2 2 dx + pminus2

intvminus h+2 2 dz

le[

2qminuspminus2

Rminus r2+ pminus2

Rminus r2

] intSR

vminus h+2 dz (55)

for some c independent of Observe that is bounded above by some fixednumber We make change of variables

= pminus2t minus t0 vx = vx t0 + 2minusp

and we write (55) as

supt

intBR

vminus h+2 2 dx +intSR

vminus h+2 2 dz le c

Rminus r2

intSR

vminus h+dz

Once we have this Caccioppoli type inequality we can apply truncation argumentTherefore following to similar argument in DiBenedetto and Friedman (1984 or1985) we have a maximum principle Now the proof for the next lemma is almostthe same as in DiBenedetto and Friedman (1984 or 1985) we omit the proof

Lemma 54 If for some and isin 0 1∣∣x t isin S1+R u2 lt 1minus 2∣∣ ge S1+R (56)

then there exists a number isin 0 1 independent of and R such that

supS R

2

ux t le

Now we assume that (56) fails that is∣∣x t isin S1+R u2 lt 1minus 2∣∣ lt S1+R (57)

Following a comparison argument similar to Choe (1991) or DiBenedetto andFriedman (1984) we prove that u satisfies a Campanato type integral inequalityLet V isin nN be a constant vector such that 1

2 le V le Taking uxkminus Vk

2 as atest function to (31) we have the following lemma

Lemma 55 There exists a constant c independent of such that

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2tdx +intS R

2

upminus22u2 dz

le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

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630 Bae and Choe

Now suppose that v is a solution to

vit minus

(Ai

Qj

Vvjx

)x

= 0 (58)

with v = u on pSR2 Since 1

2 le V le le c for some fixed c we see that fromthe structure condition of F

2 le 2minuspAi

Qj

Vi

j le c1+ qminusp2

Hence we see that v satisfies a Campanato type integral inequality

Lemma 56 Suppose that 0 lt r le R2 There exists a constant c independent ofVR and such that

intSr

vminus vr 2 dz le c( r

R

)n+4 intS R

2

uminus V 2 dz (59)

Proof We write (58) as

vit minus pminus2

Ai

Qj

V

pminus2vjx

x

= 0

We change of variable = pminus2t minus t0 then the function vx = vx 2minusp+ t0satisfies a uniformly parabolic system

vi minus

Ai

Qj

V

pminus2vjx

x

= 0

Thus from results of Campanato (1966) we prove

intS

vminus v2 dz le c(

R

)n+4 intS R

2

uminus V 2 dz

and scaling back prove (59)

We estimate the L2 norm of uminus v in SR2

Lemma 57 For each lt R2 there exists a constant c independent of such that

intS

uminus v2 dz le c(1+ qminusp 4

n+2)(

minus2 ndashintSR

uminus V 2 dz) int

SRuminus V 2 dz

for = min(12

2n

)

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Regularity for Certain Nonlinear Parabolic Systems 631

Proof Subtracting(Ai

Qj

Vujx

)x

from both sides of (15) we get

uit minus

(Ai

Qj

Vujx

)x= (

Aiuminus Ai

Qj

Vujx

)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(510)

From (58) and (510) we see that uminus v satisfies

ui minus vit minus(Ai

Qj

Vuj minus vjx)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(511)

Simply applying uminus v to (511) we get

pminus2intS R

2uminusv2dzle

intS R

2

∣∣∣ int 1

0Ai

Qj

V+suminus VminusAi

Qj

Vds∣∣∣

times uminusV uminusvdz

and it follows thatintS R

2

uminus v2 dz

le minus2pminus2intS R

2

∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣2

uminus V 2 dz (512)

Note that AiV = F primeV V i

V and

Ai

Qj

V = F primeprimeV ViV

j

V 2 + F primeV (ij

V minus V iV

j

V 3)

where ij and are Kronecker delta functions Also we recall the structurecondition (13) F primeprimeprimeh le chpminus3 + hqminus3 We have that

∣∣∣ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds∣∣∣

le∣∣∣∣int 1

0F primeprime(V + suminus V)

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2

minusF primeprimeV ViV

j

V 2 ds

∣∣∣∣+int 1

0

∣∣∣∣F primeV + suminus VV + suminus V minus F primeV

V ∣∣∣∣ds

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632 Bae and Choe

+∣∣∣∣int 1

0F primeV + suminus V

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V3

minusF primeV ViV

j

V 3 ds

∣∣∣∣= I + II + III

The first term I is estimates as follows

I leint 1

0

∣∣∣F primeprimeV + suminus Vminus F primeprimeV ∣∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2∣∣∣∣∣ ds

+int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

(513)

By the mean value theorem we have that

int 1

0

∣∣∣F primeprimeV+suminusVminusF primeprimeV ∣∣∣ds le c

int 1

0

int 1

0

∣∣F primeprimeprimeV+shuminusV∣∣ds dhuminusV (514)

for some c and

int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

le cF primeprimeV int 1

0

int 1

0

ds dh∣∣V + shuminus V∣∣ uminus V (515)

If V ge 2uminus V then for all s isin 0 1 12 V le V + suminus V le 2V and from

(513) (514) and (515) we have I le c(V pminus3 + V qminus3

)uminus V If V le 2uminus V and p ge 2 then

∣∣F primeprimeV + suminus V∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le cuminus V pminus2 + cuminus V qminus2

and

∣∣F primeprimeV ∣∣ le cV pminus2 + cV qminus2 le cuminus V pminus2 + cuminus V qminus2

Hence we have that

I le cint 1

0

∣∣F primeprimeV + suminus V∣∣+ ∣∣F primeprimeV ∣∣ds le c1+ pminusquminus V pminus2

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Regularity for Certain Nonlinear Parabolic Systems 633

for some c independent of If V le 2uminus V and 1 lt p lt 2 then∣∣F primeprime(V + suminus V)∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le c1+ pminusqV + suminus Vpminus2

for some c We note thatint 1

0V + suminus Vpminus2 ds le uminus V pminus2

int 1

0

∣∣∣∣ V uminus V minus s

∣∣∣∣pminus2

ds le cuminus V pminus2

Therefore combining all these together we have

I le cV pminus3 uminus V (516)

for some c independent of For II and III we observe that

F primeV V le cV pminus2 + cV qminus2

∣∣∣∣FprimeV V i

Vj

V 3∣∣∣∣ le cV pminus2 + cV qminus2

∣∣∣∣(F primeV V

)∣∣∣∣ le cV pminus3 + cV qminus3

∣∣∣∣(F primeV V i

Vj

V 3)∣∣∣∣ le cV pminus3 + cV qminus3

Hence following the same argument for I we conclude that

II + III le cV pminus3 uminus V (517)

for some c independent of Combining (516) and (517) we obtain that∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣ le cV pminus3 uminus V (518)

and considering (518) in (512)intS R

2

uminus V 2 dz le cminus2pminus2V 2pminus3intS R

2

uminus V 4 dz

le cminus2intS R

2

uminus V 4 dz (519)

From Houmllderrsquos inequality we have thatintS R

2

uminus V 4 dz

=intS R

2

uminus V 8n+ 4nminus8n dz

le[

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx] 2

n int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

(520)

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From Lemma 55 we have

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx le c2 supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2 dx

le cp + 2qminusp

R2

intSR

uminus V 2 dz

for some c independent of Now recalling that 12 le V le we obtain that

int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

le c22minuspnminus2

n

int t0

t0minus 2minuspR24

[ intB R

2

(u + V )2pminus2uminus V 4 dx] nminus2

n

dt

le c4nminuspnminus8

n

int t0

t0minus 2minuspR24

[ intB R

2

∣∣u pminus22 uminus V pminus2

2 V∣∣ 2n

nminus2 dx] nminus2

n

dt

le c4nminuspnminus8

n

intS R

2

upminus2 2u2 dz+ c2nminuspnminus8

n pminus2intSR

uminus V 2 dz (521)

for some c independent of where we have used Sobolevrsquos inequality FromLemma 55 we haveint

S R2

upminus2 2u2 dz le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

Thus combining (519) (520) and (521) we have that

intS R

2uminus V 2 dz le c

(1+ qminusp 4

n+2)[

minus2ndashintSR

uminus V 2 dz] 2

n intSR

uminus V 2 dz

and this completes the proof

Now we prove an iteration lemma which says that if mean oscillation of unear some fixed V is small then u satisfies a Campanato type integral inequality

Lemma 58 Fix 0 lt 0 lt 1 then there exist small positive numbers independentof such that if V0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists V1 isin nN satisfying(12minusradic

0

) le V1 le

(1+radic

0

) (522)

ndashintSR

uminus V12 dz le 0 ndashintSR

uminus V02 dz (523)

ndashintSR

uminus V12 dz le 2

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Regularity for Certain Nonlinear Parabolic Systems 635

Proof Since V0 satisfies the assumption of Lemma 57 we apply Lemma 57 withV = V0 and obtainint

S R2

vminus u2 dz le cintSR

uminus V02 dz

So from the usual comparison argument we haveintSR

uminus uR2 dz le cintSR

vminus vR2 dz+ cintSR

uminus v2 dz

le cn+4intS R

2

uminus u R22 dz+ c

intS R

2

uminus v2 dz

le c(n+4 +

) intSR

uminus V02 dz

where c is independent of Choosing and isin 0 12 small we prove (523)

with V1 = uR Inequality ndashintSR

uminus V12 dz le 2 follows from (523) andassumption ndash

intSR

uminus V02 dz le 2 Now we also estimate

V1 minus V0 =∣∣∣∣∣ ndashintSR

vminus V0 dz

∣∣∣∣le c

[ndashintSR

vminus u2 dz] 1

2

+ c

[ndashintSR

uminus V02 dz] 1

2

le cminus1minusn2(1+

120

)12

Therefore using the same technique of DiBenedetto (1986) Lemma 44 and if wetake very small compared to (522) holds and this completes the proof

Lemma 59 There exist small positive numbers 0 independent of such that ifV0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists a sequence of vectors Vi i = 1 2 3 sub nN satisfying

14 le Vi le 4 ndash

intSiR

uminus Vi2 dz le 2

ndashintSi+1R

uminus Vi+12 dz le 0 ndashintSiR

uminus Vi2 dz

Proof For the proof of the lemma we need to show that 14 le Vi le 4 But this

follows from that

Vi+1 minus Vi2 le c(0 + minusnminus2

)∣∣∣ ndashintSiR

uminus Vi dz∣∣∣

le ci0

i∣∣∣ ndashint

SRuminus V02 dz

∣∣∣ le ci0

i2

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636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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ORDER REPRINTS

642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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t 200

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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612 Bae and Choe

1 INTRODUCTION

In this paper we study the regularity of second order parabolic systems ofgeneral growth We may consider these systems as generalizations for evolutionaryp-Laplacian functionals Choe (1992a) established interior Houmllder regularity for thegradient of the bounded minimizers of

min Iu = minintFu (11)

under general growth conditions on F such that

Qpminus22 le FQiQ

jQi

j le 1

(Qpminus2 + Qqminus2)2 (12)

for all Q isin nN and some gt 0 where 1 lt p le q lt p+ 1 Thus it is naturalto ask regularity questions for parabolic cases Here we prove Houmllder continuityof weak solutions and their gradients for parabolic systems under general growthconditions The theory for the parabolic cases are rather complicated because ofscaling

Now we define our problem Suppose sub n is a bounded domain andwe define T = times 0 T for positive T Let 1 lt p le q lt p+ 1 and we assumeF rarr is a 3 function satisfying (12) for all Q isin nN and some gt 0 and

F primeprimeprimes le cspminus3 + sqminus3 (13)

for some c and for all s gt 0 From the ellipticity condition (12) we note that

spminus2 le F primeprimes le 1spminus2 + sqminus2 (14)

for all s gt 0 We define AiQ

def= FQiQ for notational simplicity

We say u isin 00 T L2 cap L20 TW 1p is a solution to

uit minus

(Ai

u)x= 0 i = 1 N (15)

if u satisfiesintT

minusui13it + Ai

u13ixdx dt = 0

for all 13 isin 0 T We prove that any solution u of (15) is Houmllder continuous and

the gradient of u is Houmllder continuousFor stationary cases ie for elliptic systems a number of authors have

considered regularity questions for the case p = q Uhlenbeck (1977) did show u isHoumllder continuous when p ge 2 The 1 regularity was extended for all p isin 1when u is a scalar valued function by various authors (Lewis 1983 Lieberman1991 ) When 1 lt p lt 2 and u is a vector valued function Tolksdorff (1983)showed that u is Houmllder continuous On the other hand systems with nonstandard

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growth conditions have been interested by many authors Indeed Fusco andSbordone (1990) showed a higher integrability result of u under the assumptionthat 1 lt p le q lt npnminus p and F satisfies 2 condition Also Marcellini (1989)considered various anisotropic cases Choe (1992a) showed that if u is a boundedminimizer of (11) then u is bounded under the assumption (12) Finally weremark that Lieberman (1991) generalized the structure of Ladyzhenskaya andUralrsquotzeva and Uraltseva (1968) also showed the C-regularity

For the evolutionary case a number of authors proved 0 regularity for thesolutions of degenerate parabolic equations that is for the case p = q DiBenedetto(1986) proved 0 regularity for evolutionary p-Laplacian equations when p isin2 His method was extended for all p isin 1 by Chen and DiBenedetto(1988) Also considering a reflection principle Chen and DiBenedetto (1989)gave another proof for Houmllder continuity of solutions to systems up to theboundary when p isin 2nn+ 2 Lieberman (1993) considered boundary behaviorof degenerate parabolic equations We also remark that Choe (1992b) provedHoumllder continuity of solutions to systems for all p isin 1 in a rather simpleway by showing a Poincareacute inequality and weak Harnack inequalities for uFurthermore we note that DiBenedetto and Friedman (1985) proved the gradientsof solutions of parabolic systems are Houmllder continuous when p isin 2nn+ 2Independently Wiegner (1986) proved the gradients of solutions of parabolicsystems are Houmllder continuous when p isin 2 Choe (1991) also proved that thegradient of solutions of parabolic systems are Houmllder continuous for all p isin 1

as long as u lies in a highly integrable space To be more specific when u isin Lqloc

q gt N2minus pp then u is Houmllder continuous for all p isin 1In the next section we show that solutions of (15) are locally bounded using a

Moser type iteration which is suitable for systems with general growth conditionsNamely we show that u isin L

loc if u isin Lr0loc for some r0 (see Theorem 21) The

restriction of the integrability of u is necessary to start the iteration In fact ifp isin 2nn+ 2 and p le q le p2+ 1 or p2+ 1 le q lt pn+ 3+ 4n+ 4then r0 le 2 and there is no restriction on the integrability of u Once u is boundedwe prove an inequality of Poincareacute type which is known for evolutionary p-Laplacian systems (see Choe 1992b) We then obtain a weak Harnack inequality fora solution of equation using a scaling argument as in Choe (1992b) and an iterationmethod as in Choe (1991) Consequently a Campanato type growth condition for u

is obtained by combining Poincareacutersquos inequality a weak Harnack inequality for u

and a Caccioppoli type inequality Hence Houmllder continuity of u follows from theisomorphism theorem of Da Prato (1965)

Following Choe (1991) that is using a difference quotient method we showthat u has space derivatives Once u has space derivatives we find an integralidentity which is suitable for Moser iteration Indeed we use integration by partsand a Caccioppoli type inequality for the second derivatives of u (see Choe andLewis 1991 for elliptic cases) For evolutionary p-Laplacian we also recall thatit is necessary to show u isin Ls for all s isin p when p isin 1 2 in proving u

is bounded Once u is bounded we construct an iteration scheme to obtain aCampanato growth condition on u There we use a perturbation method tooTherefore Houmllder continuity of u follows from the isomorphism theorem

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We define symbols and notations as follows

BRx0 = x isin n x minus x0 lt R Rt0 = t isin t0 minus Rp lt t lt t0

QRx0 t0 = BRx0timesRt0 SRx0 t0 = BRx0times t0 minus R2 t0

pQR is the parabolic boundary of QR z = x t

ndashintAu dz = 1

AintAu dz uR = ndash

intQRx0t0

u dz uRt = ndashintBRx0

ux tdx

We write z0 = x0 t0 = 0 0 as a generic point On the other hand if there is noconfusion we simply drop x0 t0 from various symbols and notations Finally wewrite c as a constant depending only on exterior data and as an arbitrarily smallconstant depending only on exterior data

2 L ESTIMATE FOR u

In this section we prove that u is bounded To begin with Moser iteration weneed that u is highly integrable

Theorem 21 Suppose that u isin Lr0locT for some r0 and 1 lt p lt then we have

u isin Lloc Furthermore suppose SR0

sub T then u satisfies

supS R0

2

u le c

[ndashintSR0

us0+s1 dz

] 1s0+s2 + c (21)

for any number s0 gt minuss2 and some c independent of R0 wheres1 = p

p+1minusq s2 = 2minusn+2qminusp

p+1minusqif q ge p

2 + 1

s1 = 2 s2 = n+ 2minus 2np

if q le p

2 + 1

Here r0 = s0 + s1

Proof Let 0 lt lt R be numbers Let be a standard cutoff function such that = 1 for x t isin S = 0 for x t isin pSR 0 le le 1 t le cRminus p lecRminus for some c We take usuip as a test function to (15) and we obtain

1s + 2

intSR

d

dt

(us+2p)dzminus p

s + 2

intSR

us+2pminus1t dz

+intSR

Aiu

(usuix+ susminus2uiujuj

x

)p dz = minusp

intSR

Aiuusuix

pminus1 dz

(22)

Owing to (14) one has that there is c gt 0 such that

pminus 1spminus1 minus c le F primes le 1

(spminus1

pminus 1+ sqminus1

q minus 1

)+ c

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Regularity for Certain Nonlinear Parabolic Systems 615

for all s gt 0 We note that Fuixu = F primeuui

xu so we have that

pminus 1up minus cu le Ai

uuixle 1

( uppminus 1

+ uqq minus 1

)+ cu (23)

We also note that

Aiuu

iujujxge

pminus 1upminus2ui

xujxuiuj minus c

uixujxuiuj

u ge minuscu2u (24)

and that

Aiuus+1pminus1 le cupminus1 + uqminus1us+1pminus1 (25)

Combining (23) (24) and (25) and using Houmllderrsquos inequality we obtain from (22)that

supt

intBR

us+2p dx +intSR

upusp dz

le cintSR

us+2pminus1tdz+ cintSR

(us+pp + u p

p+1minusq +s pp+1minusq

)dz

+ cint

usup dz

for some c and

supt

intBR

us+2pdx +intSR

∣∣∣(u p+sp

)∣∣∣p dzle c

intSR

us+2pminus1tdz+ cintSR

(us+pp + u p

p+1minusq +s pp+1minusq

)dz

+ cint

pus dz

By the integrability assumption on u the right side is bounded and thereforethe left side is also bounded Therefore from Sobolevrsquos embedding theorem andHoumllderrsquos inequality we prove that for p lt n

intS

up+s+s+2 pn dz

leint [ int

us+2p dx] p

n[ int

us+p nnminusp

npnminusp dx

] nminuspn

dt

le[sup

t

intus+2p dx

] pnint ∣∣∣(u s+p

p )∣∣∣p dz

le c

[1

Rminus p

p+1minusq

intSR

(us+2 + us+p + u pp+1minusq +s + us)dz]1+ p

n

(26)

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We assume that n gt p ge 2 then we see that pp+ 1minus q gt p ge 2 Hence fromYoungrsquos inequality we can write (26) as

intS

us+p+s+2 pn dx le c

[1

Rminus p

p+1minusq

intSR

u pp+1minusq +s dz+ SR

Rminus p

p+1minusq

]1+ pn

(27)

for some c We set si by si+1 + pp+ 1minus q = si + p+ si + 2pn that is forsufficiently large s0

si = i

(s0 +

2minus n+ 2q minus p

p+ 1minus q

)+ n+ 2q minus pminus 2

p+ 1minus q

where def= 1+ pn Define Ri

def= R021+ 2minusi Now we take s = si = Ri+1 andR = Ri in (27) Define

i

def= ndashintSRi

usi+ pp+1minusq dz

then (27) can be written as

Rn+2i+1 i+1 le c

(2i+2

R0

) pp+1minusq

Rn+2i

[

i + 1

]

and

i+1 le c

(R0

2

)n+2 pnminus p

p+1minusq

2i+1 pp+1minusq minusn+2

[

i + 1

]

and hence

i+1 le cii + ci (28)

for some c We also have that

i+1 le ciciminus1iminus1 + ciminus1 + ci le ciciminus1

2

iminus1 + ciciminus1 + ci

le ciciminus1ciminus223

iminus2 + ciciminus1ciminus22 + ciciminus1 + ci

le ci+iminus1+iminus22+iminus33+middotmiddotmiddot+2iminus2+iminus1

i+1

0 + ci + ci+iminus1

+ ci+iminus1+iminus22 + middot middot middot + ci+iminus1+middotmiddotmiddot+2iminus2+iminus1

le cn2

p2iminus1minus n

p ii+1

0 + icn2

p2iminus1minus n

p i

le ci+1

i+1

0 + ici+1

If s0 gt n+ 2q minus pminus 2p+ 1minus q then s0 + 2minus n+ 2q minus pp+ 1minus q gt0 Hence we have that

limi

si = and limi

sii

= s0 +2minus n+ 2q minus p

p+ 1minus q

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Iterating (28) with some large s0 we have

supS R0

2

u le c

[ndashintSR0

u pp+1minusq +s0 dz

] 1

s0+ 2minusn+2qminuspp+1minusq + c

and this completes the proof for the case n gt p ge 2Now we assume that 1 lt p le 2 If q ge p2+ 1 then pp+ 1minus q ge 2

In this case applying Youngrsquos inequality on the last term of (26) we have (27)Hence following the argument for the case p ge 2 we prove (21) and this completesthe proof for the case q ge p2+ 1

Now we assume that p lt n and

p

2+ 1 gt q ge p gt 1

In this case s + 2 gt s + pp+ 1minus q and we have from (26)

intS

us+p+s+2 pn dx le c

[1

Rminus 2

intSR

us+2 dz+ SRRminus 2

]1+ pn

(29)

for some c We set si by sidef= is0 + n+ 2minus 2npi minus 1 Hence if s0 gt 2np

minus nminus 2 then

limi

si = and limi

sii

= s0 + n+ 2minus 2np

We take s = si = Ri+1 and R = Ri in (29) Iterating (29) we obtain

supS R0

2

u le c

[ndashintSR0

us0+2 dz

] 1s0+n+2minus 2n

p + c

for some c independent of R0 and this completes the proof for the case p2+ 1 gtq ge p gt 1

We now assume that n = p Then from Sobolevrsquos embedding theorem andHoumllderrsquos inequality we obtain that for p = n

intS

u s+22 +s+p dz le

int [ intus+2p dx

] 12[ int

u2s+p2p dx] 1

2dt

le[sup

t

intus+2p dx

] 12[ int (

u s+pp

)2pdx

] 12dt

le[sup

t

intus+2p dx

] 12int ∣∣∣(u s+p

p )∣∣∣p dz

le c

[1

Rminus p

intSR

(us+2 + us+p + u pp+1minusq +s + us)dz] 3

2

for some c The remaining things can be done similarly as in the case 2 le p lt n

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Now we consider the case p gt n Then from Sobolevrsquos embedding theorem andHoumllderrsquos inequality we obtain that for p gt n

intS

us+2+ p+s2 dz le

int [ intus+2p dx

](sup

x

u p+sp

) p2

dt

le(sup

t

intus+2p dx

) int ( int ∣∣∣(u s+pp

)∣∣∣p dx)12

dt

le c

[1

Rminus p

intSR

(us+2 + us+p + u pp+1minusq +s + us)dz] 3

2

The remaining things can be done in a similar way as before

The restriction of the integrability of u is necessary to start Moser typeiteration In fact if p isin 2nn+ 2 and p le q le p2+ 1 or p2+ 1 le q ltpn+ 3+ 4n+ 4 then r0 le 2 and there is no restriction on the integrability of u

3 L ESTIMATE FOR u

In this section we prove that u is bounded To begin with Moser iterationwe need that u is highly integrable Indeed the higher integrability of u followsfrom Caccioppoli type inequality for the second derivative of u and integration byparts technique Here the boundedness of u is crucial This method has been appliedin proving for the regularity questions for singular parabolic equations (see Choe1991) By approximation we assume that u has second derivative with respect to xHence differentiating (15) with respect to x we get

(uix

)tminus (

Ai

Qj

uujxx

)x= 0 (31)

Lemma 31 Let SR0sub T Suppose that u is bounded then for all s

intS R0

2

us dz le cintSR0

up dz+ cR0 s

for some c

Proof Let be a standard cutoff function such that = 1 for x t isin S = 0 forx t isin pSR 0 le le 1 t le cRminus 2 le cRminus for some c

From integration by parts we have

intBR

us+2k dx =intBR

usuixuixk dx = minus

intBR

u div(usuk

)dx

le csuL

intBR

us2uk dx + csuL

intBR

us+1kminus1 dx

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Regularity for Certain Nonlinear Parabolic Systems 619

and applying Youngrsquos inequality we obtain

intBR

us+2k dx le cintBR

usminus22u2k dx + cintBR

us2 dx

for some c depending on uL and s Integrating with respect to t we obtain

intSR

us+2k dz le cintSR

usminus22u2k dz+ cintSR

us2 dz (32)

for some c depending on uL and s Now we take uixusminuspk as a test function

to (31) and we get

intSR

d

dt

(us+2minuspk)dz+

intSR

usminus22u2k dz

le cintSR

us+2minuspkminus1tdz+ cintSR

us+2qminusp2 dz+ cintSR

us2 dz(33)

for some c Therefore combining (32) and (33) we obtain

intSR

us+2k dz le cintSR

us2 dz+ cintSR

us+2minusptdz

+ cintSR

us+2qminusp2 dz (34)

Assume p ge 2 Since s + 2q minus p ge s + 2minus p we get from (34) that

intS

us+2 dz le c

Rminus 2

intSR

us+2qminusp dz+ cSR

Rminus 2(35)

for some c Since we are assuming 2 le p le q lt p+ 1 2q minus p lt 2 Define si bysi+1 + 2q minus p = si + 2 that is si = s0 + i2+ 2pminus 2q Note that limirarr si = Set s0 = 3pminus 2q then s0 + 2minus p gt 0 Define Ri = R021+ 2minusi i = 0 1 2 Now we set s = si = Ri+1 and R = Ri Hence defining i =

intSRi

usi+2qminusp dz wecan rewrite (35) as

i+1 le cii + ciBR0 (36)

Therefore iterating (36) we conclude that

intS R

2

us dz le cintSR

up dz+ c (37)

for all s and this completes the proof for the case p ge 2Now we consider the case that 1 lt p le 2 Suppose that s + 2minus p le s

+ 2q minus p that is p+ 22 le q lt p+ 1 In this case we prove (37) in the

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same way for the case p ge 2 Finally we assume that p le q lt p+ 22 that iss + 2minus p ge s + 2q minus p Considering Youngrsquos inequality in (34) we get

intS

us+2 dz le c

Rminus 2

intSR

us+2minusp dz+ cSR

Rminus 2 (38)

Taking = Ri+1 R = Ri and si = s0 + ip with s0 = 2pminus 2 we obtain from (38)

intSRi+1

usi+1+2minusp dz le ci

R2

intSRi

usi+2minusp dz+ ci SRR2

for some c Therefore iterating this we obtain (37)

Once we have shown that u is in a high Ls space we can apply Moser iteration

Theorem 32 We have u isin L Furthermore suppose SR0sub T then u satisfies

supS R0

2

u le c[ndashintSR0

us0+s1 dz] 1

s0+s2 + c (39)

for some large s0 and some c independent of R0 where

s1 = 2q minus p s2 = 2minus nq minus p if q ge p

2 + 1

s1 = 2 s2 = np

2 + 2minus n if q le p

2 + 1

Proof Let be a standard cutoff function We take usuix2 as a test function to

(31) and we obtain

1s+2

int d

dt

(us+2)2 dz+

intAi

Qj

uujxx

(usuixx

+ susminus2uixukxukxx

)2 dz

= minus2int

Ai

Qj

uujxx

usuixx

dz

Considering the ellipticity condition

Qpminus22 le Ai

Qj

Qi

j le c

(Qpminus2 + Qqminus2)2

we see that

supt

intus+22 dx +

int ∣∣(u s+p2

)∣∣2 dzle c

intus+2tdz+ c

int (us+p + us+2qminusp)2 dz

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Regularity for Certain Nonlinear Parabolic Systems 621

for some c Therefore from Sobolevrsquos embedding theorem and Houmllderrsquos inequalitywe prove that

intS

us+p+s+2 2n dx le

int [ intus+22 dx

] 2n[ int

us+p nnminus2

2nnminus2dx

] nminus2n

dt

le[sup

t

intus+22 dx

] 2nint ∣∣∣(u s+p

2 )∣∣∣2 dz

le c

[1

Rminus 2

intSR

(us+2 + us+p + us+2qminusp)dz

]1+ 2n

(310)

We assume that p ge 2 then we see that s + 2q minus p ge s + 2 Hence fromYoungrsquos inequality we can write (310) as

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2qminusp dz+ SRRminus 2

]1+ 2n

(311)

for some c We set si by si+1 + 2q minus p = si + p+ si + 22n that is for

sufficiently large s0 si = is0 +(2minus nq minus p

)i minus 1 where

def= 1+ 2n Define

Ri

def= R021+ 2minusi Now we take s = si = Ri+1 and R = Ri in (311) Define

i

def= ndashintSRi

usi+2qminusp dz then (311) can be written as

i+1 le cii + ci (312)

If s0 gt nq minus pminus 2 then s0 + 2minus nq minus p gt 0 Hence we have that

limi

si = and limi

sii

= s0 + 2minus nq minus p

From Lemma 31 we can take s0 as large as possible Iterating (312) with some larges0 we have

supS R0

2

u le c[ndashintSR0

us0+2qminusp dz] 1

s0+2minusnqminusp + c

and this completes the proof for the case p ge 2Now we assume that 1 lt p le 2 If q ge p2+ 1 then s + 2q minus p ge s + 2 In this

case applying Youngrsquos inequality on the last term of (310) we have

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2qminusp dz+ SRRminus 2

]1+ 2n

for some c Hence following the argument for the case p ge 2 we prove (39) andthis completes the proof for the case q ge p2+ 1

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Now we assume that p2+ 1 gt q ge p gt 1 In this case s + 2 gt s + 2q minus p andwe have from (310)

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2 dz+ SRRminus 2

]1+ 2n

(313)

for some c We set si by sidef= is0 + np2+ 2minus ni minus 1 Hence if s0 gt nminus 2

minus np2 then

limi

si = and limi

sii

= s0 + 2minus n+ np

2

We take s = si = Ri+1 and R = Ri in (313) Iterating (313) we obtain

supS R0

2

u le c

[ndashintSR0

us0+2 dz

] 1s0+2+ np

2 minusn + c

for some c independent of R0 and this completes the proof for the casep2+ 1 gt q ge p gt 1

4 HOumlLDER CONTINUITY OF u

In this section we prove that u is Houmllder continuous

Theorem 41 Define r0 by the constant appearing in Theorem 21 Suppose that uis L

r0locT then u is Houmllder continuous Moreover for all Q2R sub T with R lt 1 u

satisfies

ndashintQR

uminus uR2 dz le cRs

for some c independent of R where we define

s =2 when p ge 2

ppminus 1 when p isin 1 2

We introduce a nonnegative cutoff function isin 0 BR such that = 1 in

BR2 0 le le 1 le cR We define minus

R

def= t0 minus 2Rp t0 minus Rp and QminusR = BR timesminus

R

Also we define uRt

def= 1BRintBR

ux tdx and uR

def= 1QRintQR

ux tdz Firstwe prove a lemma which is essential for a Poincareacute inequality for solutions of adegenerate parabolic system

Lemma 42 Suppose Q2R sub T then u satisfies the following inequality

suptisinR

intminus

R

dsintBR

2ux tminus uRs2 dx

le cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz+ cintQminus

R

uminus uRt2 dz

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Regularity for Certain Nonlinear Parabolic Systems 623

for all R lt R0 where c depends only on nN and p and

s1 = p s2 = p and s3 = pqminus1

pminus1 when p ge 2

s1 = p2minus p s2 = ppminus 1 and s3 = pq minus 1 when 1 lt p lt 2

Proof Since u isin 0(0 T L2

) there exists ts isin R for all s isin minus

R such that

intBR

pux t minus uRs2dx = suptisinR

intBR

pux tminus uRs2 dx

We take uminus uRspxst as a test function to (15) where st is the characteristic

function such that

st =1 for all isin s t

0 for all isin s t

Hence we have

intut middot uminus uRs

pxst dz+int

Au middot uminus uRspst dz = 0 (41)

Now we assume p isin 2 Considering the structure condition (12) on A (23)and (25) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

leint

uminus uRs2x sp dx + cintBRtimesst

upminus1uminus uRspminus1dz

+ cintBRtimesst

uqminus1uminus uRspminus1dz (42)

for some c independent of R From Youngrsquos inequality we have

intBRtimesst

upminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

up dz (43)

for some c and small Similarly the last term in (42) can be estimated as follows

intBRtimesst

uqminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

u pqminus1pminus1 dz (44)

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624 Bae and Choe

for some c and small Combining (42) (43) and (44) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

le cint

uminus uRs2x sp dx +

Rp

intBRtimesst

uminus uRspp dz

+ cintBRtimesst

up dz+ cintBRtimesst

u pqminus1pminus1 dz (45)

for some c and small Integrating (45) with respect to s from t0 minus 2Rp to t0 minus Rpwe have

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRs2p dz+ c

Rp

intminus

R

dsintBRtimesst

uminus uRs2p dz

+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

By the choice of t we have that for small

intminus

R

dsintBR

uminusuRs2x t pdx

le cintQminus

R

uminusuRs2p dz+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

So we proved Lemma 42 when p ge 2In case p isin 1 2 we estimate the second term of (41) as follows

intuppst dz+ p

intAu middot uminus uRs

pminus1st dz

leint

Au middot uminus uRspst dz (46)

and using the fact that u is bounded and ppminus 1 gt 2 we have

∣∣∣ int Au middot uminusuRspminus1st dz

∣∣∣le c

intBRtimesst

upminus1uminusuRspminus1st dz

+ cintBRtimesst

uqminus1pminus1uminusuRsst dz

le

Rp

intuminusuRs

ppminus1 pst dz+

c

Rp2minusp

intuppminus1 dz+ c

Rp2minusp

intupqminus1 dz

(47)

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Regularity for Certain Nonlinear Parabolic Systems 625

Combining (41) (46) and (47) we have

intBR

uminus uRs2x t p dx

le cu2minusppminus1L

Rp

intBRtimesst

uminus uRs2pst dz+c

Rp2minusp

intBRtimesst

uppminus1 dz

+ c

Rp2minusp

intBRtimesst

upqminus1 dz+intBR

uminus uRs2x sp dx (48)

for small Integrating (48) with respect to s and considering the choice of t weprove Lemma 42 when 1 lt p lt 2

Now we prove a Poincareacute inequality

Theorem 43 Suppose Q2R sub T with R lt 1 then u satisfies the following inequality

intQR

2

uminus uR22 dz le cR2

intQ2R

u2 dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

where c is independent of R

Proof Using Lemma 42 we have

intQR

2

uminus uR22 dz le

intQR

uminus uRs2p dz

le c

Rp

intminus

R

dsintQR

uminus uRs2p dz le c suptisinR

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRt2p dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

le cR2intQ2R

u2 dz+ cRs1

intQ2R

us2z+ cRs1

intQ2R

us3 dz

where we used a Poincareacute type inequality for x variables only that is

intBR

uminus uRs2p dx le cR2intBR

u2 dx

Proof of Theorem 41 Since u is bounded owing to Theorem 32 we have fromTheorem 43

intQR

uminus uRs2 dz lecRn+p+2 when p isin 2

cRn+p+ppminus1 when p isin 1 2

where c is independent of R Hence by isomorphism theorem of Da Prato (1965) weprove Theorem 41

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626 Bae and Choe

5 HOumlLDER CONTINUITY FOR u

In this section we prove that u is Houmllder continuous Because of degeneracyof the pde we need scaled cylinders (see DiBenedetto and Friedman 1985)Let R0 gt 0 be fixed and consider a cylinder

SR0

def= BR0times t0 minus R2minus

0 t0

for small For any 0 lt R le R02 we introduce a scaled cylinder

SRdef= BR times t0 minus R22minusp t0

for fixed ge supSR0u

We assume that pminus2 gt R0 if p gt 2 Hence we note that SR sub S

R0and

ge supSRu

The following theorem is our main theorem in this section

Theorem 51 Suppose that SR sub T then u satisfies for all r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR 2 dz+ cR

for some gt 0

For the proof of Theorem 51 we follow a usual program for degenerateequations such that if for some isin 0 1∣∣z isin SR u2 le 1minus 2

∣∣ gt ∣∣SR

∣∣ (51)

then

supS R

2

u le (52)

for some small isin 0 1 and otherwise that is (51) fails then for all 0 lt r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR2 dz (53)

for some gt 0 Combining (52) and (53) we prove Theorem 51First we prove that (52) under the assumption (51) A similar argument for

the case p = q ge 2 appeared in DiBenedetto and Friedman (1985) We assume that(51) is true Set v = u2 For each h gt 0 we set AhRt

def= x isin BR v gt h and

BhR

def= z isin SR v gt h From the mean value theorem we estimate AhRt

Lemma 52 For any isin 0 there exists isin t0 minus 2minuspR20 t0 minus 2minuspR2

0 such that

A1minus2R0 le 1minus

1minus BR0

where is independent of

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Regularity for Certain Nonlinear Parabolic Systems 627

Proof Suppose that the assertion is not true then

B1minus2R0 ge

int t0minus2minuspR20

t0minus2minuspR20

A1minus2R0d gt 1minus BR0

2minuspR20 = 1minus SR0

and this contradicts the assumption (51)

The following lemma is useful in estimating sup u Indeed when p = q ge 2DiBenedetto and Friedman proved a similar theorem (see Lemma 62 inDiBenedetto and Friedman 1985) Now choose = 2 and in Lemma 52We define St

R

def= BR times t

Lemma 53 There exists a positive integer r independent of such that

∣∣A1minus 2r

2R0t

∣∣ le (1minus

(2

)2)BR0

for all t isin t0 minus

22minuspR2

0 t0

Proof Consider the function

v = log+[

2

2minus vminus 1minus 2+ + 2

2r

]

where r is a positive integer This function was used in Choe (1991) and DiBenedettoand Friedman (1985) We take 13 = ui

x2vprime 2 as a test function for (31) where

isin 0 is a standard cutoff function such that B1minusR for small decided later is

contained in support Integrating from to t we obtain

12

intStR

vt2prime 2 dz+

intStR

Ai

Qj

uujxx

uixx

2prime 2 dz

+ 2intStR

Ai

Qj

uujxx

uixukxx

ukx2primeprime 2 dz

+ 2intStR

Ai

Qj

uujxx

uix x

2prime dz = 0

Considering the fact 2primeprime = 21+ prime2 we have

intBR

2vx t 2xdx +intStR

upminus2v21+ prime2 2 dz

leintBR

2vx 2xdx + cintStR

upminus2v prime dz

+ cintStR

uqminus2v prime dz

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628 Bae and Choe

Applying Youngrsquos inequality on the last two terms we getintBR

2vx t 2xdx

leintBR

2vx 2xdx + cintStR

upminus2 2 dz+ cintStR

u2qminuspminus2 2 dz

for some c Since le r log 2 and u is bounded we getintBR

2vx t 2xdx leintBR

2vx 2xdx + crBR

From Lemma 52 we getintBR

2vx t 2xdx le r log 221minus

1minus

2

BR

From definition of we haveintB1minusRcapvgt1minus

2r 22vx tdx ge r minus 1 log 22A1minus

2r 21minusRt

and

A1minus 2r

21minusRt le[( r

r minus 1

)2 1minus

1minus

2

+ c

r

]BR

Hence we obtain

A(1minus

2r

)2R

t le A1minus 2r

21minusRt + BRB1minusR

le[( r

r minus 1

)2 1minus

1minus

2

+ c

r+ n

]BR

Taking r so large that( r

r minus 1

)2 le(1minus

2

)1+

c

rle 3

82

and = 382n we obtain

A1minus 2r

2Rt le(1minus

(2

)2)BR

We take 13 = uixvminus h+ 2 as a test function in (31) where v = u2 and is

a cutoff function such that = 1 in Sr = 0 on pSR le cRminus r and t le cpminus2Rminus r2 Considering the ellipticity condition (12) we have

supt

int ∣∣vminus h+∣∣2 2 dx +

intupminus2vminus h+2 2 dz

le c

Rminus r2

intSR

upminus2vminus h+2 dz

+ c

Rminus r2

intSR

u2qminuspminus2vminus h+2 dz+ cintSR

vminus h+2 tdz (54)

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Regularity for Certain Nonlinear Parabolic Systems 629

for some c Now choose h ge 12 For such a choice of h we haveint

upminus2vminus h+2 2 dz ge cpminus2int

vminus h+2 2 dz

Hence from (54) we obtain

supt

intvminus h+2 2 dx + pminus2

intvminus h+2 2 dz

le[

2qminuspminus2

Rminus r2+ pminus2

Rminus r2

] intSR

vminus h+2 dz (55)

for some c independent of Observe that is bounded above by some fixednumber We make change of variables

= pminus2t minus t0 vx = vx t0 + 2minusp

and we write (55) as

supt

intBR

vminus h+2 2 dx +intSR

vminus h+2 2 dz le c

Rminus r2

intSR

vminus h+dz

Once we have this Caccioppoli type inequality we can apply truncation argumentTherefore following to similar argument in DiBenedetto and Friedman (1984 or1985) we have a maximum principle Now the proof for the next lemma is almostthe same as in DiBenedetto and Friedman (1984 or 1985) we omit the proof

Lemma 54 If for some and isin 0 1∣∣x t isin S1+R u2 lt 1minus 2∣∣ ge S1+R (56)

then there exists a number isin 0 1 independent of and R such that

supS R

2

ux t le

Now we assume that (56) fails that is∣∣x t isin S1+R u2 lt 1minus 2∣∣ lt S1+R (57)

Following a comparison argument similar to Choe (1991) or DiBenedetto andFriedman (1984) we prove that u satisfies a Campanato type integral inequalityLet V isin nN be a constant vector such that 1

2 le V le Taking uxkminus Vk

2 as atest function to (31) we have the following lemma

Lemma 55 There exists a constant c independent of such that

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2tdx +intS R

2

upminus22u2 dz

le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

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630 Bae and Choe

Now suppose that v is a solution to

vit minus

(Ai

Qj

Vvjx

)x

= 0 (58)

with v = u on pSR2 Since 1

2 le V le le c for some fixed c we see that fromthe structure condition of F

2 le 2minuspAi

Qj

Vi

j le c1+ qminusp2

Hence we see that v satisfies a Campanato type integral inequality

Lemma 56 Suppose that 0 lt r le R2 There exists a constant c independent ofVR and such that

intSr

vminus vr 2 dz le c( r

R

)n+4 intS R

2

uminus V 2 dz (59)

Proof We write (58) as

vit minus pminus2

Ai

Qj

V

pminus2vjx

x

= 0

We change of variable = pminus2t minus t0 then the function vx = vx 2minusp+ t0satisfies a uniformly parabolic system

vi minus

Ai

Qj

V

pminus2vjx

x

= 0

Thus from results of Campanato (1966) we prove

intS

vminus v2 dz le c(

R

)n+4 intS R

2

uminus V 2 dz

and scaling back prove (59)

We estimate the L2 norm of uminus v in SR2

Lemma 57 For each lt R2 there exists a constant c independent of such that

intS

uminus v2 dz le c(1+ qminusp 4

n+2)(

minus2 ndashintSR

uminus V 2 dz) int

SRuminus V 2 dz

for = min(12

2n

)

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Regularity for Certain Nonlinear Parabolic Systems 631

Proof Subtracting(Ai

Qj

Vujx

)x

from both sides of (15) we get

uit minus

(Ai

Qj

Vujx

)x= (

Aiuminus Ai

Qj

Vujx

)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(510)

From (58) and (510) we see that uminus v satisfies

ui minus vit minus(Ai

Qj

Vuj minus vjx)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(511)

Simply applying uminus v to (511) we get

pminus2intS R

2uminusv2dzle

intS R

2

∣∣∣ int 1

0Ai

Qj

V+suminus VminusAi

Qj

Vds∣∣∣

times uminusV uminusvdz

and it follows thatintS R

2

uminus v2 dz

le minus2pminus2intS R

2

∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣2

uminus V 2 dz (512)

Note that AiV = F primeV V i

V and

Ai

Qj

V = F primeprimeV ViV

j

V 2 + F primeV (ij

V minus V iV

j

V 3)

where ij and are Kronecker delta functions Also we recall the structurecondition (13) F primeprimeprimeh le chpminus3 + hqminus3 We have that

∣∣∣ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds∣∣∣

le∣∣∣∣int 1

0F primeprime(V + suminus V)

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2

minusF primeprimeV ViV

j

V 2 ds

∣∣∣∣+int 1

0

∣∣∣∣F primeV + suminus VV + suminus V minus F primeV

V ∣∣∣∣ds

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632 Bae and Choe

+∣∣∣∣int 1

0F primeV + suminus V

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V3

minusF primeV ViV

j

V 3 ds

∣∣∣∣= I + II + III

The first term I is estimates as follows

I leint 1

0

∣∣∣F primeprimeV + suminus Vminus F primeprimeV ∣∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2∣∣∣∣∣ ds

+int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

(513)

By the mean value theorem we have that

int 1

0

∣∣∣F primeprimeV+suminusVminusF primeprimeV ∣∣∣ds le c

int 1

0

int 1

0

∣∣F primeprimeprimeV+shuminusV∣∣ds dhuminusV (514)

for some c and

int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

le cF primeprimeV int 1

0

int 1

0

ds dh∣∣V + shuminus V∣∣ uminus V (515)

If V ge 2uminus V then for all s isin 0 1 12 V le V + suminus V le 2V and from

(513) (514) and (515) we have I le c(V pminus3 + V qminus3

)uminus V If V le 2uminus V and p ge 2 then

∣∣F primeprimeV + suminus V∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le cuminus V pminus2 + cuminus V qminus2

and

∣∣F primeprimeV ∣∣ le cV pminus2 + cV qminus2 le cuminus V pminus2 + cuminus V qminus2

Hence we have that

I le cint 1

0

∣∣F primeprimeV + suminus V∣∣+ ∣∣F primeprimeV ∣∣ds le c1+ pminusquminus V pminus2

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Regularity for Certain Nonlinear Parabolic Systems 633

for some c independent of If V le 2uminus V and 1 lt p lt 2 then∣∣F primeprime(V + suminus V)∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le c1+ pminusqV + suminus Vpminus2

for some c We note thatint 1

0V + suminus Vpminus2 ds le uminus V pminus2

int 1

0

∣∣∣∣ V uminus V minus s

∣∣∣∣pminus2

ds le cuminus V pminus2

Therefore combining all these together we have

I le cV pminus3 uminus V (516)

for some c independent of For II and III we observe that

F primeV V le cV pminus2 + cV qminus2

∣∣∣∣FprimeV V i

Vj

V 3∣∣∣∣ le cV pminus2 + cV qminus2

∣∣∣∣(F primeV V

)∣∣∣∣ le cV pminus3 + cV qminus3

∣∣∣∣(F primeV V i

Vj

V 3)∣∣∣∣ le cV pminus3 + cV qminus3

Hence following the same argument for I we conclude that

II + III le cV pminus3 uminus V (517)

for some c independent of Combining (516) and (517) we obtain that∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣ le cV pminus3 uminus V (518)

and considering (518) in (512)intS R

2

uminus V 2 dz le cminus2pminus2V 2pminus3intS R

2

uminus V 4 dz

le cminus2intS R

2

uminus V 4 dz (519)

From Houmllderrsquos inequality we have thatintS R

2

uminus V 4 dz

=intS R

2

uminus V 8n+ 4nminus8n dz

le[

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx] 2

n int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

(520)

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634 Bae and Choe

From Lemma 55 we have

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx le c2 supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2 dx

le cp + 2qminusp

R2

intSR

uminus V 2 dz

for some c independent of Now recalling that 12 le V le we obtain that

int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

le c22minuspnminus2

n

int t0

t0minus 2minuspR24

[ intB R

2

(u + V )2pminus2uminus V 4 dx] nminus2

n

dt

le c4nminuspnminus8

n

int t0

t0minus 2minuspR24

[ intB R

2

∣∣u pminus22 uminus V pminus2

2 V∣∣ 2n

nminus2 dx] nminus2

n

dt

le c4nminuspnminus8

n

intS R

2

upminus2 2u2 dz+ c2nminuspnminus8

n pminus2intSR

uminus V 2 dz (521)

for some c independent of where we have used Sobolevrsquos inequality FromLemma 55 we haveint

S R2

upminus2 2u2 dz le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

Thus combining (519) (520) and (521) we have that

intS R

2uminus V 2 dz le c

(1+ qminusp 4

n+2)[

minus2ndashintSR

uminus V 2 dz] 2

n intSR

uminus V 2 dz

and this completes the proof

Now we prove an iteration lemma which says that if mean oscillation of unear some fixed V is small then u satisfies a Campanato type integral inequality

Lemma 58 Fix 0 lt 0 lt 1 then there exist small positive numbers independentof such that if V0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists V1 isin nN satisfying(12minusradic

0

) le V1 le

(1+radic

0

) (522)

ndashintSR

uminus V12 dz le 0 ndashintSR

uminus V02 dz (523)

ndashintSR

uminus V12 dz le 2

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Regularity for Certain Nonlinear Parabolic Systems 635

Proof Since V0 satisfies the assumption of Lemma 57 we apply Lemma 57 withV = V0 and obtainint

S R2

vminus u2 dz le cintSR

uminus V02 dz

So from the usual comparison argument we haveintSR

uminus uR2 dz le cintSR

vminus vR2 dz+ cintSR

uminus v2 dz

le cn+4intS R

2

uminus u R22 dz+ c

intS R

2

uminus v2 dz

le c(n+4 +

) intSR

uminus V02 dz

where c is independent of Choosing and isin 0 12 small we prove (523)

with V1 = uR Inequality ndashintSR

uminus V12 dz le 2 follows from (523) andassumption ndash

intSR

uminus V02 dz le 2 Now we also estimate

V1 minus V0 =∣∣∣∣∣ ndashintSR

vminus V0 dz

∣∣∣∣le c

[ndashintSR

vminus u2 dz] 1

2

+ c

[ndashintSR

uminus V02 dz] 1

2

le cminus1minusn2(1+

120

)12

Therefore using the same technique of DiBenedetto (1986) Lemma 44 and if wetake very small compared to (522) holds and this completes the proof

Lemma 59 There exist small positive numbers 0 independent of such that ifV0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists a sequence of vectors Vi i = 1 2 3 sub nN satisfying

14 le Vi le 4 ndash

intSiR

uminus Vi2 dz le 2

ndashintSi+1R

uminus Vi+12 dz le 0 ndashintSiR

uminus Vi2 dz

Proof For the proof of the lemma we need to show that 14 le Vi le 4 But this

follows from that

Vi+1 minus Vi2 le c(0 + minusnminus2

)∣∣∣ ndashintSiR

uminus Vi dz∣∣∣

le ci0

i∣∣∣ ndashint

SRuminus V02 dz

∣∣∣ le ci0

i2

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636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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ORDER REPRINTS

642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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Regularity for Certain Nonlinear Parabolic Systems 613

growth conditions have been interested by many authors Indeed Fusco andSbordone (1990) showed a higher integrability result of u under the assumptionthat 1 lt p le q lt npnminus p and F satisfies 2 condition Also Marcellini (1989)considered various anisotropic cases Choe (1992a) showed that if u is a boundedminimizer of (11) then u is bounded under the assumption (12) Finally weremark that Lieberman (1991) generalized the structure of Ladyzhenskaya andUralrsquotzeva and Uraltseva (1968) also showed the C-regularity

For the evolutionary case a number of authors proved 0 regularity for thesolutions of degenerate parabolic equations that is for the case p = q DiBenedetto(1986) proved 0 regularity for evolutionary p-Laplacian equations when p isin2 His method was extended for all p isin 1 by Chen and DiBenedetto(1988) Also considering a reflection principle Chen and DiBenedetto (1989)gave another proof for Houmllder continuity of solutions to systems up to theboundary when p isin 2nn+ 2 Lieberman (1993) considered boundary behaviorof degenerate parabolic equations We also remark that Choe (1992b) provedHoumllder continuity of solutions to systems for all p isin 1 in a rather simpleway by showing a Poincareacute inequality and weak Harnack inequalities for uFurthermore we note that DiBenedetto and Friedman (1985) proved the gradientsof solutions of parabolic systems are Houmllder continuous when p isin 2nn+ 2Independently Wiegner (1986) proved the gradients of solutions of parabolicsystems are Houmllder continuous when p isin 2 Choe (1991) also proved that thegradient of solutions of parabolic systems are Houmllder continuous for all p isin 1

as long as u lies in a highly integrable space To be more specific when u isin Lqloc

q gt N2minus pp then u is Houmllder continuous for all p isin 1In the next section we show that solutions of (15) are locally bounded using a

Moser type iteration which is suitable for systems with general growth conditionsNamely we show that u isin L

loc if u isin Lr0loc for some r0 (see Theorem 21) The

restriction of the integrability of u is necessary to start the iteration In fact ifp isin 2nn+ 2 and p le q le p2+ 1 or p2+ 1 le q lt pn+ 3+ 4n+ 4then r0 le 2 and there is no restriction on the integrability of u Once u is boundedwe prove an inequality of Poincareacute type which is known for evolutionary p-Laplacian systems (see Choe 1992b) We then obtain a weak Harnack inequality fora solution of equation using a scaling argument as in Choe (1992b) and an iterationmethod as in Choe (1991) Consequently a Campanato type growth condition for u

is obtained by combining Poincareacutersquos inequality a weak Harnack inequality for u

and a Caccioppoli type inequality Hence Houmllder continuity of u follows from theisomorphism theorem of Da Prato (1965)

Following Choe (1991) that is using a difference quotient method we showthat u has space derivatives Once u has space derivatives we find an integralidentity which is suitable for Moser iteration Indeed we use integration by partsand a Caccioppoli type inequality for the second derivatives of u (see Choe andLewis 1991 for elliptic cases) For evolutionary p-Laplacian we also recall thatit is necessary to show u isin Ls for all s isin p when p isin 1 2 in proving u

is bounded Once u is bounded we construct an iteration scheme to obtain aCampanato growth condition on u There we use a perturbation method tooTherefore Houmllder continuity of u follows from the isomorphism theorem

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614 Bae and Choe

We define symbols and notations as follows

BRx0 = x isin n x minus x0 lt R Rt0 = t isin t0 minus Rp lt t lt t0

QRx0 t0 = BRx0timesRt0 SRx0 t0 = BRx0times t0 minus R2 t0

pQR is the parabolic boundary of QR z = x t

ndashintAu dz = 1

AintAu dz uR = ndash

intQRx0t0

u dz uRt = ndashintBRx0

ux tdx

We write z0 = x0 t0 = 0 0 as a generic point On the other hand if there is noconfusion we simply drop x0 t0 from various symbols and notations Finally wewrite c as a constant depending only on exterior data and as an arbitrarily smallconstant depending only on exterior data

2 L ESTIMATE FOR u

In this section we prove that u is bounded To begin with Moser iteration weneed that u is highly integrable

Theorem 21 Suppose that u isin Lr0locT for some r0 and 1 lt p lt then we have

u isin Lloc Furthermore suppose SR0

sub T then u satisfies

supS R0

2

u le c

[ndashintSR0

us0+s1 dz

] 1s0+s2 + c (21)

for any number s0 gt minuss2 and some c independent of R0 wheres1 = p

p+1minusq s2 = 2minusn+2qminusp

p+1minusqif q ge p

2 + 1

s1 = 2 s2 = n+ 2minus 2np

if q le p

2 + 1

Here r0 = s0 + s1

Proof Let 0 lt lt R be numbers Let be a standard cutoff function such that = 1 for x t isin S = 0 for x t isin pSR 0 le le 1 t le cRminus p lecRminus for some c We take usuip as a test function to (15) and we obtain

1s + 2

intSR

d

dt

(us+2p)dzminus p

s + 2

intSR

us+2pminus1t dz

+intSR

Aiu

(usuix+ susminus2uiujuj

x

)p dz = minusp

intSR

Aiuusuix

pminus1 dz

(22)

Owing to (14) one has that there is c gt 0 such that

pminus 1spminus1 minus c le F primes le 1

(spminus1

pminus 1+ sqminus1

q minus 1

)+ c

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for all s gt 0 We note that Fuixu = F primeuui

xu so we have that

pminus 1up minus cu le Ai

uuixle 1

( uppminus 1

+ uqq minus 1

)+ cu (23)

We also note that

Aiuu

iujujxge

pminus 1upminus2ui

xujxuiuj minus c

uixujxuiuj

u ge minuscu2u (24)

and that

Aiuus+1pminus1 le cupminus1 + uqminus1us+1pminus1 (25)

Combining (23) (24) and (25) and using Houmllderrsquos inequality we obtain from (22)that

supt

intBR

us+2p dx +intSR

upusp dz

le cintSR

us+2pminus1tdz+ cintSR

(us+pp + u p

p+1minusq +s pp+1minusq

)dz

+ cint

usup dz

for some c and

supt

intBR

us+2pdx +intSR

∣∣∣(u p+sp

)∣∣∣p dzle c

intSR

us+2pminus1tdz+ cintSR

(us+pp + u p

p+1minusq +s pp+1minusq

)dz

+ cint

pus dz

By the integrability assumption on u the right side is bounded and thereforethe left side is also bounded Therefore from Sobolevrsquos embedding theorem andHoumllderrsquos inequality we prove that for p lt n

intS

up+s+s+2 pn dz

leint [ int

us+2p dx] p

n[ int

us+p nnminusp

npnminusp dx

] nminuspn

dt

le[sup

t

intus+2p dx

] pnint ∣∣∣(u s+p

p )∣∣∣p dz

le c

[1

Rminus p

p+1minusq

intSR

(us+2 + us+p + u pp+1minusq +s + us)dz]1+ p

n

(26)

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We assume that n gt p ge 2 then we see that pp+ 1minus q gt p ge 2 Hence fromYoungrsquos inequality we can write (26) as

intS

us+p+s+2 pn dx le c

[1

Rminus p

p+1minusq

intSR

u pp+1minusq +s dz+ SR

Rminus p

p+1minusq

]1+ pn

(27)

for some c We set si by si+1 + pp+ 1minus q = si + p+ si + 2pn that is forsufficiently large s0

si = i

(s0 +

2minus n+ 2q minus p

p+ 1minus q

)+ n+ 2q minus pminus 2

p+ 1minus q

where def= 1+ pn Define Ri

def= R021+ 2minusi Now we take s = si = Ri+1 andR = Ri in (27) Define

i

def= ndashintSRi

usi+ pp+1minusq dz

then (27) can be written as

Rn+2i+1 i+1 le c

(2i+2

R0

) pp+1minusq

Rn+2i

[

i + 1

]

and

i+1 le c

(R0

2

)n+2 pnminus p

p+1minusq

2i+1 pp+1minusq minusn+2

[

i + 1

]

and hence

i+1 le cii + ci (28)

for some c We also have that

i+1 le ciciminus1iminus1 + ciminus1 + ci le ciciminus1

2

iminus1 + ciciminus1 + ci

le ciciminus1ciminus223

iminus2 + ciciminus1ciminus22 + ciciminus1 + ci

le ci+iminus1+iminus22+iminus33+middotmiddotmiddot+2iminus2+iminus1

i+1

0 + ci + ci+iminus1

+ ci+iminus1+iminus22 + middot middot middot + ci+iminus1+middotmiddotmiddot+2iminus2+iminus1

le cn2

p2iminus1minus n

p ii+1

0 + icn2

p2iminus1minus n

p i

le ci+1

i+1

0 + ici+1

If s0 gt n+ 2q minus pminus 2p+ 1minus q then s0 + 2minus n+ 2q minus pp+ 1minus q gt0 Hence we have that

limi

si = and limi

sii

= s0 +2minus n+ 2q minus p

p+ 1minus q

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Iterating (28) with some large s0 we have

supS R0

2

u le c

[ndashintSR0

u pp+1minusq +s0 dz

] 1

s0+ 2minusn+2qminuspp+1minusq + c

and this completes the proof for the case n gt p ge 2Now we assume that 1 lt p le 2 If q ge p2+ 1 then pp+ 1minus q ge 2

In this case applying Youngrsquos inequality on the last term of (26) we have (27)Hence following the argument for the case p ge 2 we prove (21) and this completesthe proof for the case q ge p2+ 1

Now we assume that p lt n and

p

2+ 1 gt q ge p gt 1

In this case s + 2 gt s + pp+ 1minus q and we have from (26)

intS

us+p+s+2 pn dx le c

[1

Rminus 2

intSR

us+2 dz+ SRRminus 2

]1+ pn

(29)

for some c We set si by sidef= is0 + n+ 2minus 2npi minus 1 Hence if s0 gt 2np

minus nminus 2 then

limi

si = and limi

sii

= s0 + n+ 2minus 2np

We take s = si = Ri+1 and R = Ri in (29) Iterating (29) we obtain

supS R0

2

u le c

[ndashintSR0

us0+2 dz

] 1s0+n+2minus 2n

p + c

for some c independent of R0 and this completes the proof for the case p2+ 1 gtq ge p gt 1

We now assume that n = p Then from Sobolevrsquos embedding theorem andHoumllderrsquos inequality we obtain that for p = n

intS

u s+22 +s+p dz le

int [ intus+2p dx

] 12[ int

u2s+p2p dx] 1

2dt

le[sup

t

intus+2p dx

] 12[ int (

u s+pp

)2pdx

] 12dt

le[sup

t

intus+2p dx

] 12int ∣∣∣(u s+p

p )∣∣∣p dz

le c

[1

Rminus p

intSR

(us+2 + us+p + u pp+1minusq +s + us)dz] 3

2

for some c The remaining things can be done similarly as in the case 2 le p lt n

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Now we consider the case p gt n Then from Sobolevrsquos embedding theorem andHoumllderrsquos inequality we obtain that for p gt n

intS

us+2+ p+s2 dz le

int [ intus+2p dx

](sup

x

u p+sp

) p2

dt

le(sup

t

intus+2p dx

) int ( int ∣∣∣(u s+pp

)∣∣∣p dx)12

dt

le c

[1

Rminus p

intSR

(us+2 + us+p + u pp+1minusq +s + us)dz] 3

2

The remaining things can be done in a similar way as before

The restriction of the integrability of u is necessary to start Moser typeiteration In fact if p isin 2nn+ 2 and p le q le p2+ 1 or p2+ 1 le q ltpn+ 3+ 4n+ 4 then r0 le 2 and there is no restriction on the integrability of u

3 L ESTIMATE FOR u

In this section we prove that u is bounded To begin with Moser iterationwe need that u is highly integrable Indeed the higher integrability of u followsfrom Caccioppoli type inequality for the second derivative of u and integration byparts technique Here the boundedness of u is crucial This method has been appliedin proving for the regularity questions for singular parabolic equations (see Choe1991) By approximation we assume that u has second derivative with respect to xHence differentiating (15) with respect to x we get

(uix

)tminus (

Ai

Qj

uujxx

)x= 0 (31)

Lemma 31 Let SR0sub T Suppose that u is bounded then for all s

intS R0

2

us dz le cintSR0

up dz+ cR0 s

for some c

Proof Let be a standard cutoff function such that = 1 for x t isin S = 0 forx t isin pSR 0 le le 1 t le cRminus 2 le cRminus for some c

From integration by parts we have

intBR

us+2k dx =intBR

usuixuixk dx = minus

intBR

u div(usuk

)dx

le csuL

intBR

us2uk dx + csuL

intBR

us+1kminus1 dx

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and applying Youngrsquos inequality we obtain

intBR

us+2k dx le cintBR

usminus22u2k dx + cintBR

us2 dx

for some c depending on uL and s Integrating with respect to t we obtain

intSR

us+2k dz le cintSR

usminus22u2k dz+ cintSR

us2 dz (32)

for some c depending on uL and s Now we take uixusminuspk as a test function

to (31) and we get

intSR

d

dt

(us+2minuspk)dz+

intSR

usminus22u2k dz

le cintSR

us+2minuspkminus1tdz+ cintSR

us+2qminusp2 dz+ cintSR

us2 dz(33)

for some c Therefore combining (32) and (33) we obtain

intSR

us+2k dz le cintSR

us2 dz+ cintSR

us+2minusptdz

+ cintSR

us+2qminusp2 dz (34)

Assume p ge 2 Since s + 2q minus p ge s + 2minus p we get from (34) that

intS

us+2 dz le c

Rminus 2

intSR

us+2qminusp dz+ cSR

Rminus 2(35)

for some c Since we are assuming 2 le p le q lt p+ 1 2q minus p lt 2 Define si bysi+1 + 2q minus p = si + 2 that is si = s0 + i2+ 2pminus 2q Note that limirarr si = Set s0 = 3pminus 2q then s0 + 2minus p gt 0 Define Ri = R021+ 2minusi i = 0 1 2 Now we set s = si = Ri+1 and R = Ri Hence defining i =

intSRi

usi+2qminusp dz wecan rewrite (35) as

i+1 le cii + ciBR0 (36)

Therefore iterating (36) we conclude that

intS R

2

us dz le cintSR

up dz+ c (37)

for all s and this completes the proof for the case p ge 2Now we consider the case that 1 lt p le 2 Suppose that s + 2minus p le s

+ 2q minus p that is p+ 22 le q lt p+ 1 In this case we prove (37) in the

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same way for the case p ge 2 Finally we assume that p le q lt p+ 22 that iss + 2minus p ge s + 2q minus p Considering Youngrsquos inequality in (34) we get

intS

us+2 dz le c

Rminus 2

intSR

us+2minusp dz+ cSR

Rminus 2 (38)

Taking = Ri+1 R = Ri and si = s0 + ip with s0 = 2pminus 2 we obtain from (38)

intSRi+1

usi+1+2minusp dz le ci

R2

intSRi

usi+2minusp dz+ ci SRR2

for some c Therefore iterating this we obtain (37)

Once we have shown that u is in a high Ls space we can apply Moser iteration

Theorem 32 We have u isin L Furthermore suppose SR0sub T then u satisfies

supS R0

2

u le c[ndashintSR0

us0+s1 dz] 1

s0+s2 + c (39)

for some large s0 and some c independent of R0 where

s1 = 2q minus p s2 = 2minus nq minus p if q ge p

2 + 1

s1 = 2 s2 = np

2 + 2minus n if q le p

2 + 1

Proof Let be a standard cutoff function We take usuix2 as a test function to

(31) and we obtain

1s+2

int d

dt

(us+2)2 dz+

intAi

Qj

uujxx

(usuixx

+ susminus2uixukxukxx

)2 dz

= minus2int

Ai

Qj

uujxx

usuixx

dz

Considering the ellipticity condition

Qpminus22 le Ai

Qj

Qi

j le c

(Qpminus2 + Qqminus2)2

we see that

supt

intus+22 dx +

int ∣∣(u s+p2

)∣∣2 dzle c

intus+2tdz+ c

int (us+p + us+2qminusp)2 dz

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for some c Therefore from Sobolevrsquos embedding theorem and Houmllderrsquos inequalitywe prove that

intS

us+p+s+2 2n dx le

int [ intus+22 dx

] 2n[ int

us+p nnminus2

2nnminus2dx

] nminus2n

dt

le[sup

t

intus+22 dx

] 2nint ∣∣∣(u s+p

2 )∣∣∣2 dz

le c

[1

Rminus 2

intSR

(us+2 + us+p + us+2qminusp)dz

]1+ 2n

(310)

We assume that p ge 2 then we see that s + 2q minus p ge s + 2 Hence fromYoungrsquos inequality we can write (310) as

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2qminusp dz+ SRRminus 2

]1+ 2n

(311)

for some c We set si by si+1 + 2q minus p = si + p+ si + 22n that is for

sufficiently large s0 si = is0 +(2minus nq minus p

)i minus 1 where

def= 1+ 2n Define

Ri

def= R021+ 2minusi Now we take s = si = Ri+1 and R = Ri in (311) Define

i

def= ndashintSRi

usi+2qminusp dz then (311) can be written as

i+1 le cii + ci (312)

If s0 gt nq minus pminus 2 then s0 + 2minus nq minus p gt 0 Hence we have that

limi

si = and limi

sii

= s0 + 2minus nq minus p

From Lemma 31 we can take s0 as large as possible Iterating (312) with some larges0 we have

supS R0

2

u le c[ndashintSR0

us0+2qminusp dz] 1

s0+2minusnqminusp + c

and this completes the proof for the case p ge 2Now we assume that 1 lt p le 2 If q ge p2+ 1 then s + 2q minus p ge s + 2 In this

case applying Youngrsquos inequality on the last term of (310) we have

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2qminusp dz+ SRRminus 2

]1+ 2n

for some c Hence following the argument for the case p ge 2 we prove (39) andthis completes the proof for the case q ge p2+ 1

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Now we assume that p2+ 1 gt q ge p gt 1 In this case s + 2 gt s + 2q minus p andwe have from (310)

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2 dz+ SRRminus 2

]1+ 2n

(313)

for some c We set si by sidef= is0 + np2+ 2minus ni minus 1 Hence if s0 gt nminus 2

minus np2 then

limi

si = and limi

sii

= s0 + 2minus n+ np

2

We take s = si = Ri+1 and R = Ri in (313) Iterating (313) we obtain

supS R0

2

u le c

[ndashintSR0

us0+2 dz

] 1s0+2+ np

2 minusn + c

for some c independent of R0 and this completes the proof for the casep2+ 1 gt q ge p gt 1

4 HOumlLDER CONTINUITY OF u

In this section we prove that u is Houmllder continuous

Theorem 41 Define r0 by the constant appearing in Theorem 21 Suppose that uis L

r0locT then u is Houmllder continuous Moreover for all Q2R sub T with R lt 1 u

satisfies

ndashintQR

uminus uR2 dz le cRs

for some c independent of R where we define

s =2 when p ge 2

ppminus 1 when p isin 1 2

We introduce a nonnegative cutoff function isin 0 BR such that = 1 in

BR2 0 le le 1 le cR We define minus

R

def= t0 minus 2Rp t0 minus Rp and QminusR = BR timesminus

R

Also we define uRt

def= 1BRintBR

ux tdx and uR

def= 1QRintQR

ux tdz Firstwe prove a lemma which is essential for a Poincareacute inequality for solutions of adegenerate parabolic system

Lemma 42 Suppose Q2R sub T then u satisfies the following inequality

suptisinR

intminus

R

dsintBR

2ux tminus uRs2 dx

le cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz+ cintQminus

R

uminus uRt2 dz

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for all R lt R0 where c depends only on nN and p and

s1 = p s2 = p and s3 = pqminus1

pminus1 when p ge 2

s1 = p2minus p s2 = ppminus 1 and s3 = pq minus 1 when 1 lt p lt 2

Proof Since u isin 0(0 T L2

) there exists ts isin R for all s isin minus

R such that

intBR

pux t minus uRs2dx = suptisinR

intBR

pux tminus uRs2 dx

We take uminus uRspxst as a test function to (15) where st is the characteristic

function such that

st =1 for all isin s t

0 for all isin s t

Hence we have

intut middot uminus uRs

pxst dz+int

Au middot uminus uRspst dz = 0 (41)

Now we assume p isin 2 Considering the structure condition (12) on A (23)and (25) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

leint

uminus uRs2x sp dx + cintBRtimesst

upminus1uminus uRspminus1dz

+ cintBRtimesst

uqminus1uminus uRspminus1dz (42)

for some c independent of R From Youngrsquos inequality we have

intBRtimesst

upminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

up dz (43)

for some c and small Similarly the last term in (42) can be estimated as follows

intBRtimesst

uqminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

u pqminus1pminus1 dz (44)

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for some c and small Combining (42) (43) and (44) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

le cint

uminus uRs2x sp dx +

Rp

intBRtimesst

uminus uRspp dz

+ cintBRtimesst

up dz+ cintBRtimesst

u pqminus1pminus1 dz (45)

for some c and small Integrating (45) with respect to s from t0 minus 2Rp to t0 minus Rpwe have

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRs2p dz+ c

Rp

intminus

R

dsintBRtimesst

uminus uRs2p dz

+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

By the choice of t we have that for small

intminus

R

dsintBR

uminusuRs2x t pdx

le cintQminus

R

uminusuRs2p dz+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

So we proved Lemma 42 when p ge 2In case p isin 1 2 we estimate the second term of (41) as follows

intuppst dz+ p

intAu middot uminus uRs

pminus1st dz

leint

Au middot uminus uRspst dz (46)

and using the fact that u is bounded and ppminus 1 gt 2 we have

∣∣∣ int Au middot uminusuRspminus1st dz

∣∣∣le c

intBRtimesst

upminus1uminusuRspminus1st dz

+ cintBRtimesst

uqminus1pminus1uminusuRsst dz

le

Rp

intuminusuRs

ppminus1 pst dz+

c

Rp2minusp

intuppminus1 dz+ c

Rp2minusp

intupqminus1 dz

(47)

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Combining (41) (46) and (47) we have

intBR

uminus uRs2x t p dx

le cu2minusppminus1L

Rp

intBRtimesst

uminus uRs2pst dz+c

Rp2minusp

intBRtimesst

uppminus1 dz

+ c

Rp2minusp

intBRtimesst

upqminus1 dz+intBR

uminus uRs2x sp dx (48)

for small Integrating (48) with respect to s and considering the choice of t weprove Lemma 42 when 1 lt p lt 2

Now we prove a Poincareacute inequality

Theorem 43 Suppose Q2R sub T with R lt 1 then u satisfies the following inequality

intQR

2

uminus uR22 dz le cR2

intQ2R

u2 dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

where c is independent of R

Proof Using Lemma 42 we have

intQR

2

uminus uR22 dz le

intQR

uminus uRs2p dz

le c

Rp

intminus

R

dsintQR

uminus uRs2p dz le c suptisinR

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRt2p dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

le cR2intQ2R

u2 dz+ cRs1

intQ2R

us2z+ cRs1

intQ2R

us3 dz

where we used a Poincareacute type inequality for x variables only that is

intBR

uminus uRs2p dx le cR2intBR

u2 dx

Proof of Theorem 41 Since u is bounded owing to Theorem 32 we have fromTheorem 43

intQR

uminus uRs2 dz lecRn+p+2 when p isin 2

cRn+p+ppminus1 when p isin 1 2

where c is independent of R Hence by isomorphism theorem of Da Prato (1965) weprove Theorem 41

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5 HOumlLDER CONTINUITY FOR u

In this section we prove that u is Houmllder continuous Because of degeneracyof the pde we need scaled cylinders (see DiBenedetto and Friedman 1985)Let R0 gt 0 be fixed and consider a cylinder

SR0

def= BR0times t0 minus R2minus

0 t0

for small For any 0 lt R le R02 we introduce a scaled cylinder

SRdef= BR times t0 minus R22minusp t0

for fixed ge supSR0u

We assume that pminus2 gt R0 if p gt 2 Hence we note that SR sub S

R0and

ge supSRu

The following theorem is our main theorem in this section

Theorem 51 Suppose that SR sub T then u satisfies for all r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR 2 dz+ cR

for some gt 0

For the proof of Theorem 51 we follow a usual program for degenerateequations such that if for some isin 0 1∣∣z isin SR u2 le 1minus 2

∣∣ gt ∣∣SR

∣∣ (51)

then

supS R

2

u le (52)

for some small isin 0 1 and otherwise that is (51) fails then for all 0 lt r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR2 dz (53)

for some gt 0 Combining (52) and (53) we prove Theorem 51First we prove that (52) under the assumption (51) A similar argument for

the case p = q ge 2 appeared in DiBenedetto and Friedman (1985) We assume that(51) is true Set v = u2 For each h gt 0 we set AhRt

def= x isin BR v gt h and

BhR

def= z isin SR v gt h From the mean value theorem we estimate AhRt

Lemma 52 For any isin 0 there exists isin t0 minus 2minuspR20 t0 minus 2minuspR2

0 such that

A1minus2R0 le 1minus

1minus BR0

where is independent of

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Regularity for Certain Nonlinear Parabolic Systems 627

Proof Suppose that the assertion is not true then

B1minus2R0 ge

int t0minus2minuspR20

t0minus2minuspR20

A1minus2R0d gt 1minus BR0

2minuspR20 = 1minus SR0

and this contradicts the assumption (51)

The following lemma is useful in estimating sup u Indeed when p = q ge 2DiBenedetto and Friedman proved a similar theorem (see Lemma 62 inDiBenedetto and Friedman 1985) Now choose = 2 and in Lemma 52We define St

R

def= BR times t

Lemma 53 There exists a positive integer r independent of such that

∣∣A1minus 2r

2R0t

∣∣ le (1minus

(2

)2)BR0

for all t isin t0 minus

22minuspR2

0 t0

Proof Consider the function

v = log+[

2

2minus vminus 1minus 2+ + 2

2r

]

where r is a positive integer This function was used in Choe (1991) and DiBenedettoand Friedman (1985) We take 13 = ui

x2vprime 2 as a test function for (31) where

isin 0 is a standard cutoff function such that B1minusR for small decided later is

contained in support Integrating from to t we obtain

12

intStR

vt2prime 2 dz+

intStR

Ai

Qj

uujxx

uixx

2prime 2 dz

+ 2intStR

Ai

Qj

uujxx

uixukxx

ukx2primeprime 2 dz

+ 2intStR

Ai

Qj

uujxx

uix x

2prime dz = 0

Considering the fact 2primeprime = 21+ prime2 we have

intBR

2vx t 2xdx +intStR

upminus2v21+ prime2 2 dz

leintBR

2vx 2xdx + cintStR

upminus2v prime dz

+ cintStR

uqminus2v prime dz

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628 Bae and Choe

Applying Youngrsquos inequality on the last two terms we getintBR

2vx t 2xdx

leintBR

2vx 2xdx + cintStR

upminus2 2 dz+ cintStR

u2qminuspminus2 2 dz

for some c Since le r log 2 and u is bounded we getintBR

2vx t 2xdx leintBR

2vx 2xdx + crBR

From Lemma 52 we getintBR

2vx t 2xdx le r log 221minus

1minus

2

BR

From definition of we haveintB1minusRcapvgt1minus

2r 22vx tdx ge r minus 1 log 22A1minus

2r 21minusRt

and

A1minus 2r

21minusRt le[( r

r minus 1

)2 1minus

1minus

2

+ c

r

]BR

Hence we obtain

A(1minus

2r

)2R

t le A1minus 2r

21minusRt + BRB1minusR

le[( r

r minus 1

)2 1minus

1minus

2

+ c

r+ n

]BR

Taking r so large that( r

r minus 1

)2 le(1minus

2

)1+

c

rle 3

82

and = 382n we obtain

A1minus 2r

2Rt le(1minus

(2

)2)BR

We take 13 = uixvminus h+ 2 as a test function in (31) where v = u2 and is

a cutoff function such that = 1 in Sr = 0 on pSR le cRminus r and t le cpminus2Rminus r2 Considering the ellipticity condition (12) we have

supt

int ∣∣vminus h+∣∣2 2 dx +

intupminus2vminus h+2 2 dz

le c

Rminus r2

intSR

upminus2vminus h+2 dz

+ c

Rminus r2

intSR

u2qminuspminus2vminus h+2 dz+ cintSR

vminus h+2 tdz (54)

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Regularity for Certain Nonlinear Parabolic Systems 629

for some c Now choose h ge 12 For such a choice of h we haveint

upminus2vminus h+2 2 dz ge cpminus2int

vminus h+2 2 dz

Hence from (54) we obtain

supt

intvminus h+2 2 dx + pminus2

intvminus h+2 2 dz

le[

2qminuspminus2

Rminus r2+ pminus2

Rminus r2

] intSR

vminus h+2 dz (55)

for some c independent of Observe that is bounded above by some fixednumber We make change of variables

= pminus2t minus t0 vx = vx t0 + 2minusp

and we write (55) as

supt

intBR

vminus h+2 2 dx +intSR

vminus h+2 2 dz le c

Rminus r2

intSR

vminus h+dz

Once we have this Caccioppoli type inequality we can apply truncation argumentTherefore following to similar argument in DiBenedetto and Friedman (1984 or1985) we have a maximum principle Now the proof for the next lemma is almostthe same as in DiBenedetto and Friedman (1984 or 1985) we omit the proof

Lemma 54 If for some and isin 0 1∣∣x t isin S1+R u2 lt 1minus 2∣∣ ge S1+R (56)

then there exists a number isin 0 1 independent of and R such that

supS R

2

ux t le

Now we assume that (56) fails that is∣∣x t isin S1+R u2 lt 1minus 2∣∣ lt S1+R (57)

Following a comparison argument similar to Choe (1991) or DiBenedetto andFriedman (1984) we prove that u satisfies a Campanato type integral inequalityLet V isin nN be a constant vector such that 1

2 le V le Taking uxkminus Vk

2 as atest function to (31) we have the following lemma

Lemma 55 There exists a constant c independent of such that

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2tdx +intS R

2

upminus22u2 dz

le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

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630 Bae and Choe

Now suppose that v is a solution to

vit minus

(Ai

Qj

Vvjx

)x

= 0 (58)

with v = u on pSR2 Since 1

2 le V le le c for some fixed c we see that fromthe structure condition of F

2 le 2minuspAi

Qj

Vi

j le c1+ qminusp2

Hence we see that v satisfies a Campanato type integral inequality

Lemma 56 Suppose that 0 lt r le R2 There exists a constant c independent ofVR and such that

intSr

vminus vr 2 dz le c( r

R

)n+4 intS R

2

uminus V 2 dz (59)

Proof We write (58) as

vit minus pminus2

Ai

Qj

V

pminus2vjx

x

= 0

We change of variable = pminus2t minus t0 then the function vx = vx 2minusp+ t0satisfies a uniformly parabolic system

vi minus

Ai

Qj

V

pminus2vjx

x

= 0

Thus from results of Campanato (1966) we prove

intS

vminus v2 dz le c(

R

)n+4 intS R

2

uminus V 2 dz

and scaling back prove (59)

We estimate the L2 norm of uminus v in SR2

Lemma 57 For each lt R2 there exists a constant c independent of such that

intS

uminus v2 dz le c(1+ qminusp 4

n+2)(

minus2 ndashintSR

uminus V 2 dz) int

SRuminus V 2 dz

for = min(12

2n

)

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Regularity for Certain Nonlinear Parabolic Systems 631

Proof Subtracting(Ai

Qj

Vujx

)x

from both sides of (15) we get

uit minus

(Ai

Qj

Vujx

)x= (

Aiuminus Ai

Qj

Vujx

)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(510)

From (58) and (510) we see that uminus v satisfies

ui minus vit minus(Ai

Qj

Vuj minus vjx)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(511)

Simply applying uminus v to (511) we get

pminus2intS R

2uminusv2dzle

intS R

2

∣∣∣ int 1

0Ai

Qj

V+suminus VminusAi

Qj

Vds∣∣∣

times uminusV uminusvdz

and it follows thatintS R

2

uminus v2 dz

le minus2pminus2intS R

2

∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣2

uminus V 2 dz (512)

Note that AiV = F primeV V i

V and

Ai

Qj

V = F primeprimeV ViV

j

V 2 + F primeV (ij

V minus V iV

j

V 3)

where ij and are Kronecker delta functions Also we recall the structurecondition (13) F primeprimeprimeh le chpminus3 + hqminus3 We have that

∣∣∣ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds∣∣∣

le∣∣∣∣int 1

0F primeprime(V + suminus V)

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2

minusF primeprimeV ViV

j

V 2 ds

∣∣∣∣+int 1

0

∣∣∣∣F primeV + suminus VV + suminus V minus F primeV

V ∣∣∣∣ds

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632 Bae and Choe

+∣∣∣∣int 1

0F primeV + suminus V

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V3

minusF primeV ViV

j

V 3 ds

∣∣∣∣= I + II + III

The first term I is estimates as follows

I leint 1

0

∣∣∣F primeprimeV + suminus Vminus F primeprimeV ∣∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2∣∣∣∣∣ ds

+int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

(513)

By the mean value theorem we have that

int 1

0

∣∣∣F primeprimeV+suminusVminusF primeprimeV ∣∣∣ds le c

int 1

0

int 1

0

∣∣F primeprimeprimeV+shuminusV∣∣ds dhuminusV (514)

for some c and

int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

le cF primeprimeV int 1

0

int 1

0

ds dh∣∣V + shuminus V∣∣ uminus V (515)

If V ge 2uminus V then for all s isin 0 1 12 V le V + suminus V le 2V and from

(513) (514) and (515) we have I le c(V pminus3 + V qminus3

)uminus V If V le 2uminus V and p ge 2 then

∣∣F primeprimeV + suminus V∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le cuminus V pminus2 + cuminus V qminus2

and

∣∣F primeprimeV ∣∣ le cV pminus2 + cV qminus2 le cuminus V pminus2 + cuminus V qminus2

Hence we have that

I le cint 1

0

∣∣F primeprimeV + suminus V∣∣+ ∣∣F primeprimeV ∣∣ds le c1+ pminusquminus V pminus2

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Regularity for Certain Nonlinear Parabolic Systems 633

for some c independent of If V le 2uminus V and 1 lt p lt 2 then∣∣F primeprime(V + suminus V)∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le c1+ pminusqV + suminus Vpminus2

for some c We note thatint 1

0V + suminus Vpminus2 ds le uminus V pminus2

int 1

0

∣∣∣∣ V uminus V minus s

∣∣∣∣pminus2

ds le cuminus V pminus2

Therefore combining all these together we have

I le cV pminus3 uminus V (516)

for some c independent of For II and III we observe that

F primeV V le cV pminus2 + cV qminus2

∣∣∣∣FprimeV V i

Vj

V 3∣∣∣∣ le cV pminus2 + cV qminus2

∣∣∣∣(F primeV V

)∣∣∣∣ le cV pminus3 + cV qminus3

∣∣∣∣(F primeV V i

Vj

V 3)∣∣∣∣ le cV pminus3 + cV qminus3

Hence following the same argument for I we conclude that

II + III le cV pminus3 uminus V (517)

for some c independent of Combining (516) and (517) we obtain that∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣ le cV pminus3 uminus V (518)

and considering (518) in (512)intS R

2

uminus V 2 dz le cminus2pminus2V 2pminus3intS R

2

uminus V 4 dz

le cminus2intS R

2

uminus V 4 dz (519)

From Houmllderrsquos inequality we have thatintS R

2

uminus V 4 dz

=intS R

2

uminus V 8n+ 4nminus8n dz

le[

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx] 2

n int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

(520)

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634 Bae and Choe

From Lemma 55 we have

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx le c2 supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2 dx

le cp + 2qminusp

R2

intSR

uminus V 2 dz

for some c independent of Now recalling that 12 le V le we obtain that

int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

le c22minuspnminus2

n

int t0

t0minus 2minuspR24

[ intB R

2

(u + V )2pminus2uminus V 4 dx] nminus2

n

dt

le c4nminuspnminus8

n

int t0

t0minus 2minuspR24

[ intB R

2

∣∣u pminus22 uminus V pminus2

2 V∣∣ 2n

nminus2 dx] nminus2

n

dt

le c4nminuspnminus8

n

intS R

2

upminus2 2u2 dz+ c2nminuspnminus8

n pminus2intSR

uminus V 2 dz (521)

for some c independent of where we have used Sobolevrsquos inequality FromLemma 55 we haveint

S R2

upminus2 2u2 dz le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

Thus combining (519) (520) and (521) we have that

intS R

2uminus V 2 dz le c

(1+ qminusp 4

n+2)[

minus2ndashintSR

uminus V 2 dz] 2

n intSR

uminus V 2 dz

and this completes the proof

Now we prove an iteration lemma which says that if mean oscillation of unear some fixed V is small then u satisfies a Campanato type integral inequality

Lemma 58 Fix 0 lt 0 lt 1 then there exist small positive numbers independentof such that if V0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists V1 isin nN satisfying(12minusradic

0

) le V1 le

(1+radic

0

) (522)

ndashintSR

uminus V12 dz le 0 ndashintSR

uminus V02 dz (523)

ndashintSR

uminus V12 dz le 2

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Regularity for Certain Nonlinear Parabolic Systems 635

Proof Since V0 satisfies the assumption of Lemma 57 we apply Lemma 57 withV = V0 and obtainint

S R2

vminus u2 dz le cintSR

uminus V02 dz

So from the usual comparison argument we haveintSR

uminus uR2 dz le cintSR

vminus vR2 dz+ cintSR

uminus v2 dz

le cn+4intS R

2

uminus u R22 dz+ c

intS R

2

uminus v2 dz

le c(n+4 +

) intSR

uminus V02 dz

where c is independent of Choosing and isin 0 12 small we prove (523)

with V1 = uR Inequality ndashintSR

uminus V12 dz le 2 follows from (523) andassumption ndash

intSR

uminus V02 dz le 2 Now we also estimate

V1 minus V0 =∣∣∣∣∣ ndashintSR

vminus V0 dz

∣∣∣∣le c

[ndashintSR

vminus u2 dz] 1

2

+ c

[ndashintSR

uminus V02 dz] 1

2

le cminus1minusn2(1+

120

)12

Therefore using the same technique of DiBenedetto (1986) Lemma 44 and if wetake very small compared to (522) holds and this completes the proof

Lemma 59 There exist small positive numbers 0 independent of such that ifV0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists a sequence of vectors Vi i = 1 2 3 sub nN satisfying

14 le Vi le 4 ndash

intSiR

uminus Vi2 dz le 2

ndashintSi+1R

uminus Vi+12 dz le 0 ndashintSiR

uminus Vi2 dz

Proof For the proof of the lemma we need to show that 14 le Vi le 4 But this

follows from that

Vi+1 minus Vi2 le c(0 + minusnminus2

)∣∣∣ ndashintSiR

uminus Vi dz∣∣∣

le ci0

i∣∣∣ ndashint

SRuminus V02 dz

∣∣∣ le ci0

i2

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636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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t 200

7

ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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614 Bae and Choe

We define symbols and notations as follows

BRx0 = x isin n x minus x0 lt R Rt0 = t isin t0 minus Rp lt t lt t0

QRx0 t0 = BRx0timesRt0 SRx0 t0 = BRx0times t0 minus R2 t0

pQR is the parabolic boundary of QR z = x t

ndashintAu dz = 1

AintAu dz uR = ndash

intQRx0t0

u dz uRt = ndashintBRx0

ux tdx

We write z0 = x0 t0 = 0 0 as a generic point On the other hand if there is noconfusion we simply drop x0 t0 from various symbols and notations Finally wewrite c as a constant depending only on exterior data and as an arbitrarily smallconstant depending only on exterior data

2 L ESTIMATE FOR u

In this section we prove that u is bounded To begin with Moser iteration weneed that u is highly integrable

Theorem 21 Suppose that u isin Lr0locT for some r0 and 1 lt p lt then we have

u isin Lloc Furthermore suppose SR0

sub T then u satisfies

supS R0

2

u le c

[ndashintSR0

us0+s1 dz

] 1s0+s2 + c (21)

for any number s0 gt minuss2 and some c independent of R0 wheres1 = p

p+1minusq s2 = 2minusn+2qminusp

p+1minusqif q ge p

2 + 1

s1 = 2 s2 = n+ 2minus 2np

if q le p

2 + 1

Here r0 = s0 + s1

Proof Let 0 lt lt R be numbers Let be a standard cutoff function such that = 1 for x t isin S = 0 for x t isin pSR 0 le le 1 t le cRminus p lecRminus for some c We take usuip as a test function to (15) and we obtain

1s + 2

intSR

d

dt

(us+2p)dzminus p

s + 2

intSR

us+2pminus1t dz

+intSR

Aiu

(usuix+ susminus2uiujuj

x

)p dz = minusp

intSR

Aiuusuix

pminus1 dz

(22)

Owing to (14) one has that there is c gt 0 such that

pminus 1spminus1 minus c le F primes le 1

(spminus1

pminus 1+ sqminus1

q minus 1

)+ c

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Regularity for Certain Nonlinear Parabolic Systems 615

for all s gt 0 We note that Fuixu = F primeuui

xu so we have that

pminus 1up minus cu le Ai

uuixle 1

( uppminus 1

+ uqq minus 1

)+ cu (23)

We also note that

Aiuu

iujujxge

pminus 1upminus2ui

xujxuiuj minus c

uixujxuiuj

u ge minuscu2u (24)

and that

Aiuus+1pminus1 le cupminus1 + uqminus1us+1pminus1 (25)

Combining (23) (24) and (25) and using Houmllderrsquos inequality we obtain from (22)that

supt

intBR

us+2p dx +intSR

upusp dz

le cintSR

us+2pminus1tdz+ cintSR

(us+pp + u p

p+1minusq +s pp+1minusq

)dz

+ cint

usup dz

for some c and

supt

intBR

us+2pdx +intSR

∣∣∣(u p+sp

)∣∣∣p dzle c

intSR

us+2pminus1tdz+ cintSR

(us+pp + u p

p+1minusq +s pp+1minusq

)dz

+ cint

pus dz

By the integrability assumption on u the right side is bounded and thereforethe left side is also bounded Therefore from Sobolevrsquos embedding theorem andHoumllderrsquos inequality we prove that for p lt n

intS

up+s+s+2 pn dz

leint [ int

us+2p dx] p

n[ int

us+p nnminusp

npnminusp dx

] nminuspn

dt

le[sup

t

intus+2p dx

] pnint ∣∣∣(u s+p

p )∣∣∣p dz

le c

[1

Rminus p

p+1minusq

intSR

(us+2 + us+p + u pp+1minusq +s + us)dz]1+ p

n

(26)

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616 Bae and Choe

We assume that n gt p ge 2 then we see that pp+ 1minus q gt p ge 2 Hence fromYoungrsquos inequality we can write (26) as

intS

us+p+s+2 pn dx le c

[1

Rminus p

p+1minusq

intSR

u pp+1minusq +s dz+ SR

Rminus p

p+1minusq

]1+ pn

(27)

for some c We set si by si+1 + pp+ 1minus q = si + p+ si + 2pn that is forsufficiently large s0

si = i

(s0 +

2minus n+ 2q minus p

p+ 1minus q

)+ n+ 2q minus pminus 2

p+ 1minus q

where def= 1+ pn Define Ri

def= R021+ 2minusi Now we take s = si = Ri+1 andR = Ri in (27) Define

i

def= ndashintSRi

usi+ pp+1minusq dz

then (27) can be written as

Rn+2i+1 i+1 le c

(2i+2

R0

) pp+1minusq

Rn+2i

[

i + 1

]

and

i+1 le c

(R0

2

)n+2 pnminus p

p+1minusq

2i+1 pp+1minusq minusn+2

[

i + 1

]

and hence

i+1 le cii + ci (28)

for some c We also have that

i+1 le ciciminus1iminus1 + ciminus1 + ci le ciciminus1

2

iminus1 + ciciminus1 + ci

le ciciminus1ciminus223

iminus2 + ciciminus1ciminus22 + ciciminus1 + ci

le ci+iminus1+iminus22+iminus33+middotmiddotmiddot+2iminus2+iminus1

i+1

0 + ci + ci+iminus1

+ ci+iminus1+iminus22 + middot middot middot + ci+iminus1+middotmiddotmiddot+2iminus2+iminus1

le cn2

p2iminus1minus n

p ii+1

0 + icn2

p2iminus1minus n

p i

le ci+1

i+1

0 + ici+1

If s0 gt n+ 2q minus pminus 2p+ 1minus q then s0 + 2minus n+ 2q minus pp+ 1minus q gt0 Hence we have that

limi

si = and limi

sii

= s0 +2minus n+ 2q minus p

p+ 1minus q

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Regularity for Certain Nonlinear Parabolic Systems 617

Iterating (28) with some large s0 we have

supS R0

2

u le c

[ndashintSR0

u pp+1minusq +s0 dz

] 1

s0+ 2minusn+2qminuspp+1minusq + c

and this completes the proof for the case n gt p ge 2Now we assume that 1 lt p le 2 If q ge p2+ 1 then pp+ 1minus q ge 2

In this case applying Youngrsquos inequality on the last term of (26) we have (27)Hence following the argument for the case p ge 2 we prove (21) and this completesthe proof for the case q ge p2+ 1

Now we assume that p lt n and

p

2+ 1 gt q ge p gt 1

In this case s + 2 gt s + pp+ 1minus q and we have from (26)

intS

us+p+s+2 pn dx le c

[1

Rminus 2

intSR

us+2 dz+ SRRminus 2

]1+ pn

(29)

for some c We set si by sidef= is0 + n+ 2minus 2npi minus 1 Hence if s0 gt 2np

minus nminus 2 then

limi

si = and limi

sii

= s0 + n+ 2minus 2np

We take s = si = Ri+1 and R = Ri in (29) Iterating (29) we obtain

supS R0

2

u le c

[ndashintSR0

us0+2 dz

] 1s0+n+2minus 2n

p + c

for some c independent of R0 and this completes the proof for the case p2+ 1 gtq ge p gt 1

We now assume that n = p Then from Sobolevrsquos embedding theorem andHoumllderrsquos inequality we obtain that for p = n

intS

u s+22 +s+p dz le

int [ intus+2p dx

] 12[ int

u2s+p2p dx] 1

2dt

le[sup

t

intus+2p dx

] 12[ int (

u s+pp

)2pdx

] 12dt

le[sup

t

intus+2p dx

] 12int ∣∣∣(u s+p

p )∣∣∣p dz

le c

[1

Rminus p

intSR

(us+2 + us+p + u pp+1minusq +s + us)dz] 3

2

for some c The remaining things can be done similarly as in the case 2 le p lt n

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618 Bae and Choe

Now we consider the case p gt n Then from Sobolevrsquos embedding theorem andHoumllderrsquos inequality we obtain that for p gt n

intS

us+2+ p+s2 dz le

int [ intus+2p dx

](sup

x

u p+sp

) p2

dt

le(sup

t

intus+2p dx

) int ( int ∣∣∣(u s+pp

)∣∣∣p dx)12

dt

le c

[1

Rminus p

intSR

(us+2 + us+p + u pp+1minusq +s + us)dz] 3

2

The remaining things can be done in a similar way as before

The restriction of the integrability of u is necessary to start Moser typeiteration In fact if p isin 2nn+ 2 and p le q le p2+ 1 or p2+ 1 le q ltpn+ 3+ 4n+ 4 then r0 le 2 and there is no restriction on the integrability of u

3 L ESTIMATE FOR u

In this section we prove that u is bounded To begin with Moser iterationwe need that u is highly integrable Indeed the higher integrability of u followsfrom Caccioppoli type inequality for the second derivative of u and integration byparts technique Here the boundedness of u is crucial This method has been appliedin proving for the regularity questions for singular parabolic equations (see Choe1991) By approximation we assume that u has second derivative with respect to xHence differentiating (15) with respect to x we get

(uix

)tminus (

Ai

Qj

uujxx

)x= 0 (31)

Lemma 31 Let SR0sub T Suppose that u is bounded then for all s

intS R0

2

us dz le cintSR0

up dz+ cR0 s

for some c

Proof Let be a standard cutoff function such that = 1 for x t isin S = 0 forx t isin pSR 0 le le 1 t le cRminus 2 le cRminus for some c

From integration by parts we have

intBR

us+2k dx =intBR

usuixuixk dx = minus

intBR

u div(usuk

)dx

le csuL

intBR

us2uk dx + csuL

intBR

us+1kminus1 dx

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and applying Youngrsquos inequality we obtain

intBR

us+2k dx le cintBR

usminus22u2k dx + cintBR

us2 dx

for some c depending on uL and s Integrating with respect to t we obtain

intSR

us+2k dz le cintSR

usminus22u2k dz+ cintSR

us2 dz (32)

for some c depending on uL and s Now we take uixusminuspk as a test function

to (31) and we get

intSR

d

dt

(us+2minuspk)dz+

intSR

usminus22u2k dz

le cintSR

us+2minuspkminus1tdz+ cintSR

us+2qminusp2 dz+ cintSR

us2 dz(33)

for some c Therefore combining (32) and (33) we obtain

intSR

us+2k dz le cintSR

us2 dz+ cintSR

us+2minusptdz

+ cintSR

us+2qminusp2 dz (34)

Assume p ge 2 Since s + 2q minus p ge s + 2minus p we get from (34) that

intS

us+2 dz le c

Rminus 2

intSR

us+2qminusp dz+ cSR

Rminus 2(35)

for some c Since we are assuming 2 le p le q lt p+ 1 2q minus p lt 2 Define si bysi+1 + 2q minus p = si + 2 that is si = s0 + i2+ 2pminus 2q Note that limirarr si = Set s0 = 3pminus 2q then s0 + 2minus p gt 0 Define Ri = R021+ 2minusi i = 0 1 2 Now we set s = si = Ri+1 and R = Ri Hence defining i =

intSRi

usi+2qminusp dz wecan rewrite (35) as

i+1 le cii + ciBR0 (36)

Therefore iterating (36) we conclude that

intS R

2

us dz le cintSR

up dz+ c (37)

for all s and this completes the proof for the case p ge 2Now we consider the case that 1 lt p le 2 Suppose that s + 2minus p le s

+ 2q minus p that is p+ 22 le q lt p+ 1 In this case we prove (37) in the

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same way for the case p ge 2 Finally we assume that p le q lt p+ 22 that iss + 2minus p ge s + 2q minus p Considering Youngrsquos inequality in (34) we get

intS

us+2 dz le c

Rminus 2

intSR

us+2minusp dz+ cSR

Rminus 2 (38)

Taking = Ri+1 R = Ri and si = s0 + ip with s0 = 2pminus 2 we obtain from (38)

intSRi+1

usi+1+2minusp dz le ci

R2

intSRi

usi+2minusp dz+ ci SRR2

for some c Therefore iterating this we obtain (37)

Once we have shown that u is in a high Ls space we can apply Moser iteration

Theorem 32 We have u isin L Furthermore suppose SR0sub T then u satisfies

supS R0

2

u le c[ndashintSR0

us0+s1 dz] 1

s0+s2 + c (39)

for some large s0 and some c independent of R0 where

s1 = 2q minus p s2 = 2minus nq minus p if q ge p

2 + 1

s1 = 2 s2 = np

2 + 2minus n if q le p

2 + 1

Proof Let be a standard cutoff function We take usuix2 as a test function to

(31) and we obtain

1s+2

int d

dt

(us+2)2 dz+

intAi

Qj

uujxx

(usuixx

+ susminus2uixukxukxx

)2 dz

= minus2int

Ai

Qj

uujxx

usuixx

dz

Considering the ellipticity condition

Qpminus22 le Ai

Qj

Qi

j le c

(Qpminus2 + Qqminus2)2

we see that

supt

intus+22 dx +

int ∣∣(u s+p2

)∣∣2 dzle c

intus+2tdz+ c

int (us+p + us+2qminusp)2 dz

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for some c Therefore from Sobolevrsquos embedding theorem and Houmllderrsquos inequalitywe prove that

intS

us+p+s+2 2n dx le

int [ intus+22 dx

] 2n[ int

us+p nnminus2

2nnminus2dx

] nminus2n

dt

le[sup

t

intus+22 dx

] 2nint ∣∣∣(u s+p

2 )∣∣∣2 dz

le c

[1

Rminus 2

intSR

(us+2 + us+p + us+2qminusp)dz

]1+ 2n

(310)

We assume that p ge 2 then we see that s + 2q minus p ge s + 2 Hence fromYoungrsquos inequality we can write (310) as

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2qminusp dz+ SRRminus 2

]1+ 2n

(311)

for some c We set si by si+1 + 2q minus p = si + p+ si + 22n that is for

sufficiently large s0 si = is0 +(2minus nq minus p

)i minus 1 where

def= 1+ 2n Define

Ri

def= R021+ 2minusi Now we take s = si = Ri+1 and R = Ri in (311) Define

i

def= ndashintSRi

usi+2qminusp dz then (311) can be written as

i+1 le cii + ci (312)

If s0 gt nq minus pminus 2 then s0 + 2minus nq minus p gt 0 Hence we have that

limi

si = and limi

sii

= s0 + 2minus nq minus p

From Lemma 31 we can take s0 as large as possible Iterating (312) with some larges0 we have

supS R0

2

u le c[ndashintSR0

us0+2qminusp dz] 1

s0+2minusnqminusp + c

and this completes the proof for the case p ge 2Now we assume that 1 lt p le 2 If q ge p2+ 1 then s + 2q minus p ge s + 2 In this

case applying Youngrsquos inequality on the last term of (310) we have

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2qminusp dz+ SRRminus 2

]1+ 2n

for some c Hence following the argument for the case p ge 2 we prove (39) andthis completes the proof for the case q ge p2+ 1

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Now we assume that p2+ 1 gt q ge p gt 1 In this case s + 2 gt s + 2q minus p andwe have from (310)

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2 dz+ SRRminus 2

]1+ 2n

(313)

for some c We set si by sidef= is0 + np2+ 2minus ni minus 1 Hence if s0 gt nminus 2

minus np2 then

limi

si = and limi

sii

= s0 + 2minus n+ np

2

We take s = si = Ri+1 and R = Ri in (313) Iterating (313) we obtain

supS R0

2

u le c

[ndashintSR0

us0+2 dz

] 1s0+2+ np

2 minusn + c

for some c independent of R0 and this completes the proof for the casep2+ 1 gt q ge p gt 1

4 HOumlLDER CONTINUITY OF u

In this section we prove that u is Houmllder continuous

Theorem 41 Define r0 by the constant appearing in Theorem 21 Suppose that uis L

r0locT then u is Houmllder continuous Moreover for all Q2R sub T with R lt 1 u

satisfies

ndashintQR

uminus uR2 dz le cRs

for some c independent of R where we define

s =2 when p ge 2

ppminus 1 when p isin 1 2

We introduce a nonnegative cutoff function isin 0 BR such that = 1 in

BR2 0 le le 1 le cR We define minus

R

def= t0 minus 2Rp t0 minus Rp and QminusR = BR timesminus

R

Also we define uRt

def= 1BRintBR

ux tdx and uR

def= 1QRintQR

ux tdz Firstwe prove a lemma which is essential for a Poincareacute inequality for solutions of adegenerate parabolic system

Lemma 42 Suppose Q2R sub T then u satisfies the following inequality

suptisinR

intminus

R

dsintBR

2ux tminus uRs2 dx

le cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz+ cintQminus

R

uminus uRt2 dz

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for all R lt R0 where c depends only on nN and p and

s1 = p s2 = p and s3 = pqminus1

pminus1 when p ge 2

s1 = p2minus p s2 = ppminus 1 and s3 = pq minus 1 when 1 lt p lt 2

Proof Since u isin 0(0 T L2

) there exists ts isin R for all s isin minus

R such that

intBR

pux t minus uRs2dx = suptisinR

intBR

pux tminus uRs2 dx

We take uminus uRspxst as a test function to (15) where st is the characteristic

function such that

st =1 for all isin s t

0 for all isin s t

Hence we have

intut middot uminus uRs

pxst dz+int

Au middot uminus uRspst dz = 0 (41)

Now we assume p isin 2 Considering the structure condition (12) on A (23)and (25) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

leint

uminus uRs2x sp dx + cintBRtimesst

upminus1uminus uRspminus1dz

+ cintBRtimesst

uqminus1uminus uRspminus1dz (42)

for some c independent of R From Youngrsquos inequality we have

intBRtimesst

upminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

up dz (43)

for some c and small Similarly the last term in (42) can be estimated as follows

intBRtimesst

uqminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

u pqminus1pminus1 dz (44)

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for some c and small Combining (42) (43) and (44) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

le cint

uminus uRs2x sp dx +

Rp

intBRtimesst

uminus uRspp dz

+ cintBRtimesst

up dz+ cintBRtimesst

u pqminus1pminus1 dz (45)

for some c and small Integrating (45) with respect to s from t0 minus 2Rp to t0 minus Rpwe have

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRs2p dz+ c

Rp

intminus

R

dsintBRtimesst

uminus uRs2p dz

+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

By the choice of t we have that for small

intminus

R

dsintBR

uminusuRs2x t pdx

le cintQminus

R

uminusuRs2p dz+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

So we proved Lemma 42 when p ge 2In case p isin 1 2 we estimate the second term of (41) as follows

intuppst dz+ p

intAu middot uminus uRs

pminus1st dz

leint

Au middot uminus uRspst dz (46)

and using the fact that u is bounded and ppminus 1 gt 2 we have

∣∣∣ int Au middot uminusuRspminus1st dz

∣∣∣le c

intBRtimesst

upminus1uminusuRspminus1st dz

+ cintBRtimesst

uqminus1pminus1uminusuRsst dz

le

Rp

intuminusuRs

ppminus1 pst dz+

c

Rp2minusp

intuppminus1 dz+ c

Rp2minusp

intupqminus1 dz

(47)

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Regularity for Certain Nonlinear Parabolic Systems 625

Combining (41) (46) and (47) we have

intBR

uminus uRs2x t p dx

le cu2minusppminus1L

Rp

intBRtimesst

uminus uRs2pst dz+c

Rp2minusp

intBRtimesst

uppminus1 dz

+ c

Rp2minusp

intBRtimesst

upqminus1 dz+intBR

uminus uRs2x sp dx (48)

for small Integrating (48) with respect to s and considering the choice of t weprove Lemma 42 when 1 lt p lt 2

Now we prove a Poincareacute inequality

Theorem 43 Suppose Q2R sub T with R lt 1 then u satisfies the following inequality

intQR

2

uminus uR22 dz le cR2

intQ2R

u2 dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

where c is independent of R

Proof Using Lemma 42 we have

intQR

2

uminus uR22 dz le

intQR

uminus uRs2p dz

le c

Rp

intminus

R

dsintQR

uminus uRs2p dz le c suptisinR

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRt2p dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

le cR2intQ2R

u2 dz+ cRs1

intQ2R

us2z+ cRs1

intQ2R

us3 dz

where we used a Poincareacute type inequality for x variables only that is

intBR

uminus uRs2p dx le cR2intBR

u2 dx

Proof of Theorem 41 Since u is bounded owing to Theorem 32 we have fromTheorem 43

intQR

uminus uRs2 dz lecRn+p+2 when p isin 2

cRn+p+ppminus1 when p isin 1 2

where c is independent of R Hence by isomorphism theorem of Da Prato (1965) weprove Theorem 41

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5 HOumlLDER CONTINUITY FOR u

In this section we prove that u is Houmllder continuous Because of degeneracyof the pde we need scaled cylinders (see DiBenedetto and Friedman 1985)Let R0 gt 0 be fixed and consider a cylinder

SR0

def= BR0times t0 minus R2minus

0 t0

for small For any 0 lt R le R02 we introduce a scaled cylinder

SRdef= BR times t0 minus R22minusp t0

for fixed ge supSR0u

We assume that pminus2 gt R0 if p gt 2 Hence we note that SR sub S

R0and

ge supSRu

The following theorem is our main theorem in this section

Theorem 51 Suppose that SR sub T then u satisfies for all r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR 2 dz+ cR

for some gt 0

For the proof of Theorem 51 we follow a usual program for degenerateequations such that if for some isin 0 1∣∣z isin SR u2 le 1minus 2

∣∣ gt ∣∣SR

∣∣ (51)

then

supS R

2

u le (52)

for some small isin 0 1 and otherwise that is (51) fails then for all 0 lt r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR2 dz (53)

for some gt 0 Combining (52) and (53) we prove Theorem 51First we prove that (52) under the assumption (51) A similar argument for

the case p = q ge 2 appeared in DiBenedetto and Friedman (1985) We assume that(51) is true Set v = u2 For each h gt 0 we set AhRt

def= x isin BR v gt h and

BhR

def= z isin SR v gt h From the mean value theorem we estimate AhRt

Lemma 52 For any isin 0 there exists isin t0 minus 2minuspR20 t0 minus 2minuspR2

0 such that

A1minus2R0 le 1minus

1minus BR0

where is independent of

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Regularity for Certain Nonlinear Parabolic Systems 627

Proof Suppose that the assertion is not true then

B1minus2R0 ge

int t0minus2minuspR20

t0minus2minuspR20

A1minus2R0d gt 1minus BR0

2minuspR20 = 1minus SR0

and this contradicts the assumption (51)

The following lemma is useful in estimating sup u Indeed when p = q ge 2DiBenedetto and Friedman proved a similar theorem (see Lemma 62 inDiBenedetto and Friedman 1985) Now choose = 2 and in Lemma 52We define St

R

def= BR times t

Lemma 53 There exists a positive integer r independent of such that

∣∣A1minus 2r

2R0t

∣∣ le (1minus

(2

)2)BR0

for all t isin t0 minus

22minuspR2

0 t0

Proof Consider the function

v = log+[

2

2minus vminus 1minus 2+ + 2

2r

]

where r is a positive integer This function was used in Choe (1991) and DiBenedettoand Friedman (1985) We take 13 = ui

x2vprime 2 as a test function for (31) where

isin 0 is a standard cutoff function such that B1minusR for small decided later is

contained in support Integrating from to t we obtain

12

intStR

vt2prime 2 dz+

intStR

Ai

Qj

uujxx

uixx

2prime 2 dz

+ 2intStR

Ai

Qj

uujxx

uixukxx

ukx2primeprime 2 dz

+ 2intStR

Ai

Qj

uujxx

uix x

2prime dz = 0

Considering the fact 2primeprime = 21+ prime2 we have

intBR

2vx t 2xdx +intStR

upminus2v21+ prime2 2 dz

leintBR

2vx 2xdx + cintStR

upminus2v prime dz

+ cintStR

uqminus2v prime dz

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Applying Youngrsquos inequality on the last two terms we getintBR

2vx t 2xdx

leintBR

2vx 2xdx + cintStR

upminus2 2 dz+ cintStR

u2qminuspminus2 2 dz

for some c Since le r log 2 and u is bounded we getintBR

2vx t 2xdx leintBR

2vx 2xdx + crBR

From Lemma 52 we getintBR

2vx t 2xdx le r log 221minus

1minus

2

BR

From definition of we haveintB1minusRcapvgt1minus

2r 22vx tdx ge r minus 1 log 22A1minus

2r 21minusRt

and

A1minus 2r

21minusRt le[( r

r minus 1

)2 1minus

1minus

2

+ c

r

]BR

Hence we obtain

A(1minus

2r

)2R

t le A1minus 2r

21minusRt + BRB1minusR

le[( r

r minus 1

)2 1minus

1minus

2

+ c

r+ n

]BR

Taking r so large that( r

r minus 1

)2 le(1minus

2

)1+

c

rle 3

82

and = 382n we obtain

A1minus 2r

2Rt le(1minus

(2

)2)BR

We take 13 = uixvminus h+ 2 as a test function in (31) where v = u2 and is

a cutoff function such that = 1 in Sr = 0 on pSR le cRminus r and t le cpminus2Rminus r2 Considering the ellipticity condition (12) we have

supt

int ∣∣vminus h+∣∣2 2 dx +

intupminus2vminus h+2 2 dz

le c

Rminus r2

intSR

upminus2vminus h+2 dz

+ c

Rminus r2

intSR

u2qminuspminus2vminus h+2 dz+ cintSR

vminus h+2 tdz (54)

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Regularity for Certain Nonlinear Parabolic Systems 629

for some c Now choose h ge 12 For such a choice of h we haveint

upminus2vminus h+2 2 dz ge cpminus2int

vminus h+2 2 dz

Hence from (54) we obtain

supt

intvminus h+2 2 dx + pminus2

intvminus h+2 2 dz

le[

2qminuspminus2

Rminus r2+ pminus2

Rminus r2

] intSR

vminus h+2 dz (55)

for some c independent of Observe that is bounded above by some fixednumber We make change of variables

= pminus2t minus t0 vx = vx t0 + 2minusp

and we write (55) as

supt

intBR

vminus h+2 2 dx +intSR

vminus h+2 2 dz le c

Rminus r2

intSR

vminus h+dz

Once we have this Caccioppoli type inequality we can apply truncation argumentTherefore following to similar argument in DiBenedetto and Friedman (1984 or1985) we have a maximum principle Now the proof for the next lemma is almostthe same as in DiBenedetto and Friedman (1984 or 1985) we omit the proof

Lemma 54 If for some and isin 0 1∣∣x t isin S1+R u2 lt 1minus 2∣∣ ge S1+R (56)

then there exists a number isin 0 1 independent of and R such that

supS R

2

ux t le

Now we assume that (56) fails that is∣∣x t isin S1+R u2 lt 1minus 2∣∣ lt S1+R (57)

Following a comparison argument similar to Choe (1991) or DiBenedetto andFriedman (1984) we prove that u satisfies a Campanato type integral inequalityLet V isin nN be a constant vector such that 1

2 le V le Taking uxkminus Vk

2 as atest function to (31) we have the following lemma

Lemma 55 There exists a constant c independent of such that

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2tdx +intS R

2

upminus22u2 dz

le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

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Now suppose that v is a solution to

vit minus

(Ai

Qj

Vvjx

)x

= 0 (58)

with v = u on pSR2 Since 1

2 le V le le c for some fixed c we see that fromthe structure condition of F

2 le 2minuspAi

Qj

Vi

j le c1+ qminusp2

Hence we see that v satisfies a Campanato type integral inequality

Lemma 56 Suppose that 0 lt r le R2 There exists a constant c independent ofVR and such that

intSr

vminus vr 2 dz le c( r

R

)n+4 intS R

2

uminus V 2 dz (59)

Proof We write (58) as

vit minus pminus2

Ai

Qj

V

pminus2vjx

x

= 0

We change of variable = pminus2t minus t0 then the function vx = vx 2minusp+ t0satisfies a uniformly parabolic system

vi minus

Ai

Qj

V

pminus2vjx

x

= 0

Thus from results of Campanato (1966) we prove

intS

vminus v2 dz le c(

R

)n+4 intS R

2

uminus V 2 dz

and scaling back prove (59)

We estimate the L2 norm of uminus v in SR2

Lemma 57 For each lt R2 there exists a constant c independent of such that

intS

uminus v2 dz le c(1+ qminusp 4

n+2)(

minus2 ndashintSR

uminus V 2 dz) int

SRuminus V 2 dz

for = min(12

2n

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Regularity for Certain Nonlinear Parabolic Systems 631

Proof Subtracting(Ai

Qj

Vujx

)x

from both sides of (15) we get

uit minus

(Ai

Qj

Vujx

)x= (

Aiuminus Ai

Qj

Vujx

)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(510)

From (58) and (510) we see that uminus v satisfies

ui minus vit minus(Ai

Qj

Vuj minus vjx)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(511)

Simply applying uminus v to (511) we get

pminus2intS R

2uminusv2dzle

intS R

2

∣∣∣ int 1

0Ai

Qj

V+suminus VminusAi

Qj

Vds∣∣∣

times uminusV uminusvdz

and it follows thatintS R

2

uminus v2 dz

le minus2pminus2intS R

2

∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣2

uminus V 2 dz (512)

Note that AiV = F primeV V i

V and

Ai

Qj

V = F primeprimeV ViV

j

V 2 + F primeV (ij

V minus V iV

j

V 3)

where ij and are Kronecker delta functions Also we recall the structurecondition (13) F primeprimeprimeh le chpminus3 + hqminus3 We have that

∣∣∣ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds∣∣∣

le∣∣∣∣int 1

0F primeprime(V + suminus V)

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2

minusF primeprimeV ViV

j

V 2 ds

∣∣∣∣+int 1

0

∣∣∣∣F primeV + suminus VV + suminus V minus F primeV

V ∣∣∣∣ds

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632 Bae and Choe

+∣∣∣∣int 1

0F primeV + suminus V

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V3

minusF primeV ViV

j

V 3 ds

∣∣∣∣= I + II + III

The first term I is estimates as follows

I leint 1

0

∣∣∣F primeprimeV + suminus Vminus F primeprimeV ∣∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2∣∣∣∣∣ ds

+int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

(513)

By the mean value theorem we have that

int 1

0

∣∣∣F primeprimeV+suminusVminusF primeprimeV ∣∣∣ds le c

int 1

0

int 1

0

∣∣F primeprimeprimeV+shuminusV∣∣ds dhuminusV (514)

for some c and

int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

le cF primeprimeV int 1

0

int 1

0

ds dh∣∣V + shuminus V∣∣ uminus V (515)

If V ge 2uminus V then for all s isin 0 1 12 V le V + suminus V le 2V and from

(513) (514) and (515) we have I le c(V pminus3 + V qminus3

)uminus V If V le 2uminus V and p ge 2 then

∣∣F primeprimeV + suminus V∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le cuminus V pminus2 + cuminus V qminus2

and

∣∣F primeprimeV ∣∣ le cV pminus2 + cV qminus2 le cuminus V pminus2 + cuminus V qminus2

Hence we have that

I le cint 1

0

∣∣F primeprimeV + suminus V∣∣+ ∣∣F primeprimeV ∣∣ds le c1+ pminusquminus V pminus2

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Regularity for Certain Nonlinear Parabolic Systems 633

for some c independent of If V le 2uminus V and 1 lt p lt 2 then∣∣F primeprime(V + suminus V)∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le c1+ pminusqV + suminus Vpminus2

for some c We note thatint 1

0V + suminus Vpminus2 ds le uminus V pminus2

int 1

0

∣∣∣∣ V uminus V minus s

∣∣∣∣pminus2

ds le cuminus V pminus2

Therefore combining all these together we have

I le cV pminus3 uminus V (516)

for some c independent of For II and III we observe that

F primeV V le cV pminus2 + cV qminus2

∣∣∣∣FprimeV V i

Vj

V 3∣∣∣∣ le cV pminus2 + cV qminus2

∣∣∣∣(F primeV V

)∣∣∣∣ le cV pminus3 + cV qminus3

∣∣∣∣(F primeV V i

Vj

V 3)∣∣∣∣ le cV pminus3 + cV qminus3

Hence following the same argument for I we conclude that

II + III le cV pminus3 uminus V (517)

for some c independent of Combining (516) and (517) we obtain that∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣ le cV pminus3 uminus V (518)

and considering (518) in (512)intS R

2

uminus V 2 dz le cminus2pminus2V 2pminus3intS R

2

uminus V 4 dz

le cminus2intS R

2

uminus V 4 dz (519)

From Houmllderrsquos inequality we have thatintS R

2

uminus V 4 dz

=intS R

2

uminus V 8n+ 4nminus8n dz

le[

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx] 2

n int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

(520)

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634 Bae and Choe

From Lemma 55 we have

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx le c2 supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2 dx

le cp + 2qminusp

R2

intSR

uminus V 2 dz

for some c independent of Now recalling that 12 le V le we obtain that

int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

le c22minuspnminus2

n

int t0

t0minus 2minuspR24

[ intB R

2

(u + V )2pminus2uminus V 4 dx] nminus2

n

dt

le c4nminuspnminus8

n

int t0

t0minus 2minuspR24

[ intB R

2

∣∣u pminus22 uminus V pminus2

2 V∣∣ 2n

nminus2 dx] nminus2

n

dt

le c4nminuspnminus8

n

intS R

2

upminus2 2u2 dz+ c2nminuspnminus8

n pminus2intSR

uminus V 2 dz (521)

for some c independent of where we have used Sobolevrsquos inequality FromLemma 55 we haveint

S R2

upminus2 2u2 dz le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

Thus combining (519) (520) and (521) we have that

intS R

2uminus V 2 dz le c

(1+ qminusp 4

n+2)[

minus2ndashintSR

uminus V 2 dz] 2

n intSR

uminus V 2 dz

and this completes the proof

Now we prove an iteration lemma which says that if mean oscillation of unear some fixed V is small then u satisfies a Campanato type integral inequality

Lemma 58 Fix 0 lt 0 lt 1 then there exist small positive numbers independentof such that if V0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists V1 isin nN satisfying(12minusradic

0

) le V1 le

(1+radic

0

) (522)

ndashintSR

uminus V12 dz le 0 ndashintSR

uminus V02 dz (523)

ndashintSR

uminus V12 dz le 2

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Regularity for Certain Nonlinear Parabolic Systems 635

Proof Since V0 satisfies the assumption of Lemma 57 we apply Lemma 57 withV = V0 and obtainint

S R2

vminus u2 dz le cintSR

uminus V02 dz

So from the usual comparison argument we haveintSR

uminus uR2 dz le cintSR

vminus vR2 dz+ cintSR

uminus v2 dz

le cn+4intS R

2

uminus u R22 dz+ c

intS R

2

uminus v2 dz

le c(n+4 +

) intSR

uminus V02 dz

where c is independent of Choosing and isin 0 12 small we prove (523)

with V1 = uR Inequality ndashintSR

uminus V12 dz le 2 follows from (523) andassumption ndash

intSR

uminus V02 dz le 2 Now we also estimate

V1 minus V0 =∣∣∣∣∣ ndashintSR

vminus V0 dz

∣∣∣∣le c

[ndashintSR

vminus u2 dz] 1

2

+ c

[ndashintSR

uminus V02 dz] 1

2

le cminus1minusn2(1+

120

)12

Therefore using the same technique of DiBenedetto (1986) Lemma 44 and if wetake very small compared to (522) holds and this completes the proof

Lemma 59 There exist small positive numbers 0 independent of such that ifV0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists a sequence of vectors Vi i = 1 2 3 sub nN satisfying

14 le Vi le 4 ndash

intSiR

uminus Vi2 dz le 2

ndashintSi+1R

uminus Vi+12 dz le 0 ndashintSiR

uminus Vi2 dz

Proof For the proof of the lemma we need to show that 14 le Vi le 4 But this

follows from that

Vi+1 minus Vi2 le c(0 + minusnminus2

)∣∣∣ ndashintSiR

uminus Vi dz∣∣∣

le ci0

i∣∣∣ ndashint

SRuminus V02 dz

∣∣∣ le ci0

i2

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636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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Regularity for Certain Nonlinear Parabolic Systems 615

for all s gt 0 We note that Fuixu = F primeuui

xu so we have that

pminus 1up minus cu le Ai

uuixle 1

( uppminus 1

+ uqq minus 1

)+ cu (23)

We also note that

Aiuu

iujujxge

pminus 1upminus2ui

xujxuiuj minus c

uixujxuiuj

u ge minuscu2u (24)

and that

Aiuus+1pminus1 le cupminus1 + uqminus1us+1pminus1 (25)

Combining (23) (24) and (25) and using Houmllderrsquos inequality we obtain from (22)that

supt

intBR

us+2p dx +intSR

upusp dz

le cintSR

us+2pminus1tdz+ cintSR

(us+pp + u p

p+1minusq +s pp+1minusq

)dz

+ cint

usup dz

for some c and

supt

intBR

us+2pdx +intSR

∣∣∣(u p+sp

)∣∣∣p dzle c

intSR

us+2pminus1tdz+ cintSR

(us+pp + u p

p+1minusq +s pp+1minusq

)dz

+ cint

pus dz

By the integrability assumption on u the right side is bounded and thereforethe left side is also bounded Therefore from Sobolevrsquos embedding theorem andHoumllderrsquos inequality we prove that for p lt n

intS

up+s+s+2 pn dz

leint [ int

us+2p dx] p

n[ int

us+p nnminusp

npnminusp dx

] nminuspn

dt

le[sup

t

intus+2p dx

] pnint ∣∣∣(u s+p

p )∣∣∣p dz

le c

[1

Rminus p

p+1minusq

intSR

(us+2 + us+p + u pp+1minusq +s + us)dz]1+ p

n

(26)

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616 Bae and Choe

We assume that n gt p ge 2 then we see that pp+ 1minus q gt p ge 2 Hence fromYoungrsquos inequality we can write (26) as

intS

us+p+s+2 pn dx le c

[1

Rminus p

p+1minusq

intSR

u pp+1minusq +s dz+ SR

Rminus p

p+1minusq

]1+ pn

(27)

for some c We set si by si+1 + pp+ 1minus q = si + p+ si + 2pn that is forsufficiently large s0

si = i

(s0 +

2minus n+ 2q minus p

p+ 1minus q

)+ n+ 2q minus pminus 2

p+ 1minus q

where def= 1+ pn Define Ri

def= R021+ 2minusi Now we take s = si = Ri+1 andR = Ri in (27) Define

i

def= ndashintSRi

usi+ pp+1minusq dz

then (27) can be written as

Rn+2i+1 i+1 le c

(2i+2

R0

) pp+1minusq

Rn+2i

[

i + 1

]

and

i+1 le c

(R0

2

)n+2 pnminus p

p+1minusq

2i+1 pp+1minusq minusn+2

[

i + 1

]

and hence

i+1 le cii + ci (28)

for some c We also have that

i+1 le ciciminus1iminus1 + ciminus1 + ci le ciciminus1

2

iminus1 + ciciminus1 + ci

le ciciminus1ciminus223

iminus2 + ciciminus1ciminus22 + ciciminus1 + ci

le ci+iminus1+iminus22+iminus33+middotmiddotmiddot+2iminus2+iminus1

i+1

0 + ci + ci+iminus1

+ ci+iminus1+iminus22 + middot middot middot + ci+iminus1+middotmiddotmiddot+2iminus2+iminus1

le cn2

p2iminus1minus n

p ii+1

0 + icn2

p2iminus1minus n

p i

le ci+1

i+1

0 + ici+1

If s0 gt n+ 2q minus pminus 2p+ 1minus q then s0 + 2minus n+ 2q minus pp+ 1minus q gt0 Hence we have that

limi

si = and limi

sii

= s0 +2minus n+ 2q minus p

p+ 1minus q

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Regularity for Certain Nonlinear Parabolic Systems 617

Iterating (28) with some large s0 we have

supS R0

2

u le c

[ndashintSR0

u pp+1minusq +s0 dz

] 1

s0+ 2minusn+2qminuspp+1minusq + c

and this completes the proof for the case n gt p ge 2Now we assume that 1 lt p le 2 If q ge p2+ 1 then pp+ 1minus q ge 2

In this case applying Youngrsquos inequality on the last term of (26) we have (27)Hence following the argument for the case p ge 2 we prove (21) and this completesthe proof for the case q ge p2+ 1

Now we assume that p lt n and

p

2+ 1 gt q ge p gt 1

In this case s + 2 gt s + pp+ 1minus q and we have from (26)

intS

us+p+s+2 pn dx le c

[1

Rminus 2

intSR

us+2 dz+ SRRminus 2

]1+ pn

(29)

for some c We set si by sidef= is0 + n+ 2minus 2npi minus 1 Hence if s0 gt 2np

minus nminus 2 then

limi

si = and limi

sii

= s0 + n+ 2minus 2np

We take s = si = Ri+1 and R = Ri in (29) Iterating (29) we obtain

supS R0

2

u le c

[ndashintSR0

us0+2 dz

] 1s0+n+2minus 2n

p + c

for some c independent of R0 and this completes the proof for the case p2+ 1 gtq ge p gt 1

We now assume that n = p Then from Sobolevrsquos embedding theorem andHoumllderrsquos inequality we obtain that for p = n

intS

u s+22 +s+p dz le

int [ intus+2p dx

] 12[ int

u2s+p2p dx] 1

2dt

le[sup

t

intus+2p dx

] 12[ int (

u s+pp

)2pdx

] 12dt

le[sup

t

intus+2p dx

] 12int ∣∣∣(u s+p

p )∣∣∣p dz

le c

[1

Rminus p

intSR

(us+2 + us+p + u pp+1minusq +s + us)dz] 3

2

for some c The remaining things can be done similarly as in the case 2 le p lt n

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618 Bae and Choe

Now we consider the case p gt n Then from Sobolevrsquos embedding theorem andHoumllderrsquos inequality we obtain that for p gt n

intS

us+2+ p+s2 dz le

int [ intus+2p dx

](sup

x

u p+sp

) p2

dt

le(sup

t

intus+2p dx

) int ( int ∣∣∣(u s+pp

)∣∣∣p dx)12

dt

le c

[1

Rminus p

intSR

(us+2 + us+p + u pp+1minusq +s + us)dz] 3

2

The remaining things can be done in a similar way as before

The restriction of the integrability of u is necessary to start Moser typeiteration In fact if p isin 2nn+ 2 and p le q le p2+ 1 or p2+ 1 le q ltpn+ 3+ 4n+ 4 then r0 le 2 and there is no restriction on the integrability of u

3 L ESTIMATE FOR u

In this section we prove that u is bounded To begin with Moser iterationwe need that u is highly integrable Indeed the higher integrability of u followsfrom Caccioppoli type inequality for the second derivative of u and integration byparts technique Here the boundedness of u is crucial This method has been appliedin proving for the regularity questions for singular parabolic equations (see Choe1991) By approximation we assume that u has second derivative with respect to xHence differentiating (15) with respect to x we get

(uix

)tminus (

Ai

Qj

uujxx

)x= 0 (31)

Lemma 31 Let SR0sub T Suppose that u is bounded then for all s

intS R0

2

us dz le cintSR0

up dz+ cR0 s

for some c

Proof Let be a standard cutoff function such that = 1 for x t isin S = 0 forx t isin pSR 0 le le 1 t le cRminus 2 le cRminus for some c

From integration by parts we have

intBR

us+2k dx =intBR

usuixuixk dx = minus

intBR

u div(usuk

)dx

le csuL

intBR

us2uk dx + csuL

intBR

us+1kminus1 dx

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Regularity for Certain Nonlinear Parabolic Systems 619

and applying Youngrsquos inequality we obtain

intBR

us+2k dx le cintBR

usminus22u2k dx + cintBR

us2 dx

for some c depending on uL and s Integrating with respect to t we obtain

intSR

us+2k dz le cintSR

usminus22u2k dz+ cintSR

us2 dz (32)

for some c depending on uL and s Now we take uixusminuspk as a test function

to (31) and we get

intSR

d

dt

(us+2minuspk)dz+

intSR

usminus22u2k dz

le cintSR

us+2minuspkminus1tdz+ cintSR

us+2qminusp2 dz+ cintSR

us2 dz(33)

for some c Therefore combining (32) and (33) we obtain

intSR

us+2k dz le cintSR

us2 dz+ cintSR

us+2minusptdz

+ cintSR

us+2qminusp2 dz (34)

Assume p ge 2 Since s + 2q minus p ge s + 2minus p we get from (34) that

intS

us+2 dz le c

Rminus 2

intSR

us+2qminusp dz+ cSR

Rminus 2(35)

for some c Since we are assuming 2 le p le q lt p+ 1 2q minus p lt 2 Define si bysi+1 + 2q minus p = si + 2 that is si = s0 + i2+ 2pminus 2q Note that limirarr si = Set s0 = 3pminus 2q then s0 + 2minus p gt 0 Define Ri = R021+ 2minusi i = 0 1 2 Now we set s = si = Ri+1 and R = Ri Hence defining i =

intSRi

usi+2qminusp dz wecan rewrite (35) as

i+1 le cii + ciBR0 (36)

Therefore iterating (36) we conclude that

intS R

2

us dz le cintSR

up dz+ c (37)

for all s and this completes the proof for the case p ge 2Now we consider the case that 1 lt p le 2 Suppose that s + 2minus p le s

+ 2q minus p that is p+ 22 le q lt p+ 1 In this case we prove (37) in the

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620 Bae and Choe

same way for the case p ge 2 Finally we assume that p le q lt p+ 22 that iss + 2minus p ge s + 2q minus p Considering Youngrsquos inequality in (34) we get

intS

us+2 dz le c

Rminus 2

intSR

us+2minusp dz+ cSR

Rminus 2 (38)

Taking = Ri+1 R = Ri and si = s0 + ip with s0 = 2pminus 2 we obtain from (38)

intSRi+1

usi+1+2minusp dz le ci

R2

intSRi

usi+2minusp dz+ ci SRR2

for some c Therefore iterating this we obtain (37)

Once we have shown that u is in a high Ls space we can apply Moser iteration

Theorem 32 We have u isin L Furthermore suppose SR0sub T then u satisfies

supS R0

2

u le c[ndashintSR0

us0+s1 dz] 1

s0+s2 + c (39)

for some large s0 and some c independent of R0 where

s1 = 2q minus p s2 = 2minus nq minus p if q ge p

2 + 1

s1 = 2 s2 = np

2 + 2minus n if q le p

2 + 1

Proof Let be a standard cutoff function We take usuix2 as a test function to

(31) and we obtain

1s+2

int d

dt

(us+2)2 dz+

intAi

Qj

uujxx

(usuixx

+ susminus2uixukxukxx

)2 dz

= minus2int

Ai

Qj

uujxx

usuixx

dz

Considering the ellipticity condition

Qpminus22 le Ai

Qj

Qi

j le c

(Qpminus2 + Qqminus2)2

we see that

supt

intus+22 dx +

int ∣∣(u s+p2

)∣∣2 dzle c

intus+2tdz+ c

int (us+p + us+2qminusp)2 dz

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Regularity for Certain Nonlinear Parabolic Systems 621

for some c Therefore from Sobolevrsquos embedding theorem and Houmllderrsquos inequalitywe prove that

intS

us+p+s+2 2n dx le

int [ intus+22 dx

] 2n[ int

us+p nnminus2

2nnminus2dx

] nminus2n

dt

le[sup

t

intus+22 dx

] 2nint ∣∣∣(u s+p

2 )∣∣∣2 dz

le c

[1

Rminus 2

intSR

(us+2 + us+p + us+2qminusp)dz

]1+ 2n

(310)

We assume that p ge 2 then we see that s + 2q minus p ge s + 2 Hence fromYoungrsquos inequality we can write (310) as

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2qminusp dz+ SRRminus 2

]1+ 2n

(311)

for some c We set si by si+1 + 2q minus p = si + p+ si + 22n that is for

sufficiently large s0 si = is0 +(2minus nq minus p

)i minus 1 where

def= 1+ 2n Define

Ri

def= R021+ 2minusi Now we take s = si = Ri+1 and R = Ri in (311) Define

i

def= ndashintSRi

usi+2qminusp dz then (311) can be written as

i+1 le cii + ci (312)

If s0 gt nq minus pminus 2 then s0 + 2minus nq minus p gt 0 Hence we have that

limi

si = and limi

sii

= s0 + 2minus nq minus p

From Lemma 31 we can take s0 as large as possible Iterating (312) with some larges0 we have

supS R0

2

u le c[ndashintSR0

us0+2qminusp dz] 1

s0+2minusnqminusp + c

and this completes the proof for the case p ge 2Now we assume that 1 lt p le 2 If q ge p2+ 1 then s + 2q minus p ge s + 2 In this

case applying Youngrsquos inequality on the last term of (310) we have

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2qminusp dz+ SRRminus 2

]1+ 2n

for some c Hence following the argument for the case p ge 2 we prove (39) andthis completes the proof for the case q ge p2+ 1

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622 Bae and Choe

Now we assume that p2+ 1 gt q ge p gt 1 In this case s + 2 gt s + 2q minus p andwe have from (310)

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2 dz+ SRRminus 2

]1+ 2n

(313)

for some c We set si by sidef= is0 + np2+ 2minus ni minus 1 Hence if s0 gt nminus 2

minus np2 then

limi

si = and limi

sii

= s0 + 2minus n+ np

2

We take s = si = Ri+1 and R = Ri in (313) Iterating (313) we obtain

supS R0

2

u le c

[ndashintSR0

us0+2 dz

] 1s0+2+ np

2 minusn + c

for some c independent of R0 and this completes the proof for the casep2+ 1 gt q ge p gt 1

4 HOumlLDER CONTINUITY OF u

In this section we prove that u is Houmllder continuous

Theorem 41 Define r0 by the constant appearing in Theorem 21 Suppose that uis L

r0locT then u is Houmllder continuous Moreover for all Q2R sub T with R lt 1 u

satisfies

ndashintQR

uminus uR2 dz le cRs

for some c independent of R where we define

s =2 when p ge 2

ppminus 1 when p isin 1 2

We introduce a nonnegative cutoff function isin 0 BR such that = 1 in

BR2 0 le le 1 le cR We define minus

R

def= t0 minus 2Rp t0 minus Rp and QminusR = BR timesminus

R

Also we define uRt

def= 1BRintBR

ux tdx and uR

def= 1QRintQR

ux tdz Firstwe prove a lemma which is essential for a Poincareacute inequality for solutions of adegenerate parabolic system

Lemma 42 Suppose Q2R sub T then u satisfies the following inequality

suptisinR

intminus

R

dsintBR

2ux tminus uRs2 dx

le cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz+ cintQminus

R

uminus uRt2 dz

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Regularity for Certain Nonlinear Parabolic Systems 623

for all R lt R0 where c depends only on nN and p and

s1 = p s2 = p and s3 = pqminus1

pminus1 when p ge 2

s1 = p2minus p s2 = ppminus 1 and s3 = pq minus 1 when 1 lt p lt 2

Proof Since u isin 0(0 T L2

) there exists ts isin R for all s isin minus

R such that

intBR

pux t minus uRs2dx = suptisinR

intBR

pux tminus uRs2 dx

We take uminus uRspxst as a test function to (15) where st is the characteristic

function such that

st =1 for all isin s t

0 for all isin s t

Hence we have

intut middot uminus uRs

pxst dz+int

Au middot uminus uRspst dz = 0 (41)

Now we assume p isin 2 Considering the structure condition (12) on A (23)and (25) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

leint

uminus uRs2x sp dx + cintBRtimesst

upminus1uminus uRspminus1dz

+ cintBRtimesst

uqminus1uminus uRspminus1dz (42)

for some c independent of R From Youngrsquos inequality we have

intBRtimesst

upminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

up dz (43)

for some c and small Similarly the last term in (42) can be estimated as follows

intBRtimesst

uqminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

u pqminus1pminus1 dz (44)

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624 Bae and Choe

for some c and small Combining (42) (43) and (44) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

le cint

uminus uRs2x sp dx +

Rp

intBRtimesst

uminus uRspp dz

+ cintBRtimesst

up dz+ cintBRtimesst

u pqminus1pminus1 dz (45)

for some c and small Integrating (45) with respect to s from t0 minus 2Rp to t0 minus Rpwe have

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRs2p dz+ c

Rp

intminus

R

dsintBRtimesst

uminus uRs2p dz

+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

By the choice of t we have that for small

intminus

R

dsintBR

uminusuRs2x t pdx

le cintQminus

R

uminusuRs2p dz+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

So we proved Lemma 42 when p ge 2In case p isin 1 2 we estimate the second term of (41) as follows

intuppst dz+ p

intAu middot uminus uRs

pminus1st dz

leint

Au middot uminus uRspst dz (46)

and using the fact that u is bounded and ppminus 1 gt 2 we have

∣∣∣ int Au middot uminusuRspminus1st dz

∣∣∣le c

intBRtimesst

upminus1uminusuRspminus1st dz

+ cintBRtimesst

uqminus1pminus1uminusuRsst dz

le

Rp

intuminusuRs

ppminus1 pst dz+

c

Rp2minusp

intuppminus1 dz+ c

Rp2minusp

intupqminus1 dz

(47)

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Regularity for Certain Nonlinear Parabolic Systems 625

Combining (41) (46) and (47) we have

intBR

uminus uRs2x t p dx

le cu2minusppminus1L

Rp

intBRtimesst

uminus uRs2pst dz+c

Rp2minusp

intBRtimesst

uppminus1 dz

+ c

Rp2minusp

intBRtimesst

upqminus1 dz+intBR

uminus uRs2x sp dx (48)

for small Integrating (48) with respect to s and considering the choice of t weprove Lemma 42 when 1 lt p lt 2

Now we prove a Poincareacute inequality

Theorem 43 Suppose Q2R sub T with R lt 1 then u satisfies the following inequality

intQR

2

uminus uR22 dz le cR2

intQ2R

u2 dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

where c is independent of R

Proof Using Lemma 42 we have

intQR

2

uminus uR22 dz le

intQR

uminus uRs2p dz

le c

Rp

intminus

R

dsintQR

uminus uRs2p dz le c suptisinR

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRt2p dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

le cR2intQ2R

u2 dz+ cRs1

intQ2R

us2z+ cRs1

intQ2R

us3 dz

where we used a Poincareacute type inequality for x variables only that is

intBR

uminus uRs2p dx le cR2intBR

u2 dx

Proof of Theorem 41 Since u is bounded owing to Theorem 32 we have fromTheorem 43

intQR

uminus uRs2 dz lecRn+p+2 when p isin 2

cRn+p+ppminus1 when p isin 1 2

where c is independent of R Hence by isomorphism theorem of Da Prato (1965) weprove Theorem 41

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5 HOumlLDER CONTINUITY FOR u

In this section we prove that u is Houmllder continuous Because of degeneracyof the pde we need scaled cylinders (see DiBenedetto and Friedman 1985)Let R0 gt 0 be fixed and consider a cylinder

SR0

def= BR0times t0 minus R2minus

0 t0

for small For any 0 lt R le R02 we introduce a scaled cylinder

SRdef= BR times t0 minus R22minusp t0

for fixed ge supSR0u

We assume that pminus2 gt R0 if p gt 2 Hence we note that SR sub S

R0and

ge supSRu

The following theorem is our main theorem in this section

Theorem 51 Suppose that SR sub T then u satisfies for all r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR 2 dz+ cR

for some gt 0

For the proof of Theorem 51 we follow a usual program for degenerateequations such that if for some isin 0 1∣∣z isin SR u2 le 1minus 2

∣∣ gt ∣∣SR

∣∣ (51)

then

supS R

2

u le (52)

for some small isin 0 1 and otherwise that is (51) fails then for all 0 lt r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR2 dz (53)

for some gt 0 Combining (52) and (53) we prove Theorem 51First we prove that (52) under the assumption (51) A similar argument for

the case p = q ge 2 appeared in DiBenedetto and Friedman (1985) We assume that(51) is true Set v = u2 For each h gt 0 we set AhRt

def= x isin BR v gt h and

BhR

def= z isin SR v gt h From the mean value theorem we estimate AhRt

Lemma 52 For any isin 0 there exists isin t0 minus 2minuspR20 t0 minus 2minuspR2

0 such that

A1minus2R0 le 1minus

1minus BR0

where is independent of

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Regularity for Certain Nonlinear Parabolic Systems 627

Proof Suppose that the assertion is not true then

B1minus2R0 ge

int t0minus2minuspR20

t0minus2minuspR20

A1minus2R0d gt 1minus BR0

2minuspR20 = 1minus SR0

and this contradicts the assumption (51)

The following lemma is useful in estimating sup u Indeed when p = q ge 2DiBenedetto and Friedman proved a similar theorem (see Lemma 62 inDiBenedetto and Friedman 1985) Now choose = 2 and in Lemma 52We define St

R

def= BR times t

Lemma 53 There exists a positive integer r independent of such that

∣∣A1minus 2r

2R0t

∣∣ le (1minus

(2

)2)BR0

for all t isin t0 minus

22minuspR2

0 t0

Proof Consider the function

v = log+[

2

2minus vminus 1minus 2+ + 2

2r

]

where r is a positive integer This function was used in Choe (1991) and DiBenedettoand Friedman (1985) We take 13 = ui

x2vprime 2 as a test function for (31) where

isin 0 is a standard cutoff function such that B1minusR for small decided later is

contained in support Integrating from to t we obtain

12

intStR

vt2prime 2 dz+

intStR

Ai

Qj

uujxx

uixx

2prime 2 dz

+ 2intStR

Ai

Qj

uujxx

uixukxx

ukx2primeprime 2 dz

+ 2intStR

Ai

Qj

uujxx

uix x

2prime dz = 0

Considering the fact 2primeprime = 21+ prime2 we have

intBR

2vx t 2xdx +intStR

upminus2v21+ prime2 2 dz

leintBR

2vx 2xdx + cintStR

upminus2v prime dz

+ cintStR

uqminus2v prime dz

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628 Bae and Choe

Applying Youngrsquos inequality on the last two terms we getintBR

2vx t 2xdx

leintBR

2vx 2xdx + cintStR

upminus2 2 dz+ cintStR

u2qminuspminus2 2 dz

for some c Since le r log 2 and u is bounded we getintBR

2vx t 2xdx leintBR

2vx 2xdx + crBR

From Lemma 52 we getintBR

2vx t 2xdx le r log 221minus

1minus

2

BR

From definition of we haveintB1minusRcapvgt1minus

2r 22vx tdx ge r minus 1 log 22A1minus

2r 21minusRt

and

A1minus 2r

21minusRt le[( r

r minus 1

)2 1minus

1minus

2

+ c

r

]BR

Hence we obtain

A(1minus

2r

)2R

t le A1minus 2r

21minusRt + BRB1minusR

le[( r

r minus 1

)2 1minus

1minus

2

+ c

r+ n

]BR

Taking r so large that( r

r minus 1

)2 le(1minus

2

)1+

c

rle 3

82

and = 382n we obtain

A1minus 2r

2Rt le(1minus

(2

)2)BR

We take 13 = uixvminus h+ 2 as a test function in (31) where v = u2 and is

a cutoff function such that = 1 in Sr = 0 on pSR le cRminus r and t le cpminus2Rminus r2 Considering the ellipticity condition (12) we have

supt

int ∣∣vminus h+∣∣2 2 dx +

intupminus2vminus h+2 2 dz

le c

Rminus r2

intSR

upminus2vminus h+2 dz

+ c

Rminus r2

intSR

u2qminuspminus2vminus h+2 dz+ cintSR

vminus h+2 tdz (54)

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Regularity for Certain Nonlinear Parabolic Systems 629

for some c Now choose h ge 12 For such a choice of h we haveint

upminus2vminus h+2 2 dz ge cpminus2int

vminus h+2 2 dz

Hence from (54) we obtain

supt

intvminus h+2 2 dx + pminus2

intvminus h+2 2 dz

le[

2qminuspminus2

Rminus r2+ pminus2

Rminus r2

] intSR

vminus h+2 dz (55)

for some c independent of Observe that is bounded above by some fixednumber We make change of variables

= pminus2t minus t0 vx = vx t0 + 2minusp

and we write (55) as

supt

intBR

vminus h+2 2 dx +intSR

vminus h+2 2 dz le c

Rminus r2

intSR

vminus h+dz

Once we have this Caccioppoli type inequality we can apply truncation argumentTherefore following to similar argument in DiBenedetto and Friedman (1984 or1985) we have a maximum principle Now the proof for the next lemma is almostthe same as in DiBenedetto and Friedman (1984 or 1985) we omit the proof

Lemma 54 If for some and isin 0 1∣∣x t isin S1+R u2 lt 1minus 2∣∣ ge S1+R (56)

then there exists a number isin 0 1 independent of and R such that

supS R

2

ux t le

Now we assume that (56) fails that is∣∣x t isin S1+R u2 lt 1minus 2∣∣ lt S1+R (57)

Following a comparison argument similar to Choe (1991) or DiBenedetto andFriedman (1984) we prove that u satisfies a Campanato type integral inequalityLet V isin nN be a constant vector such that 1

2 le V le Taking uxkminus Vk

2 as atest function to (31) we have the following lemma

Lemma 55 There exists a constant c independent of such that

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2tdx +intS R

2

upminus22u2 dz

le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

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630 Bae and Choe

Now suppose that v is a solution to

vit minus

(Ai

Qj

Vvjx

)x

= 0 (58)

with v = u on pSR2 Since 1

2 le V le le c for some fixed c we see that fromthe structure condition of F

2 le 2minuspAi

Qj

Vi

j le c1+ qminusp2

Hence we see that v satisfies a Campanato type integral inequality

Lemma 56 Suppose that 0 lt r le R2 There exists a constant c independent ofVR and such that

intSr

vminus vr 2 dz le c( r

R

)n+4 intS R

2

uminus V 2 dz (59)

Proof We write (58) as

vit minus pminus2

Ai

Qj

V

pminus2vjx

x

= 0

We change of variable = pminus2t minus t0 then the function vx = vx 2minusp+ t0satisfies a uniformly parabolic system

vi minus

Ai

Qj

V

pminus2vjx

x

= 0

Thus from results of Campanato (1966) we prove

intS

vminus v2 dz le c(

R

)n+4 intS R

2

uminus V 2 dz

and scaling back prove (59)

We estimate the L2 norm of uminus v in SR2

Lemma 57 For each lt R2 there exists a constant c independent of such that

intS

uminus v2 dz le c(1+ qminusp 4

n+2)(

minus2 ndashintSR

uminus V 2 dz) int

SRuminus V 2 dz

for = min(12

2n

)

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Regularity for Certain Nonlinear Parabolic Systems 631

Proof Subtracting(Ai

Qj

Vujx

)x

from both sides of (15) we get

uit minus

(Ai

Qj

Vujx

)x= (

Aiuminus Ai

Qj

Vujx

)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(510)

From (58) and (510) we see that uminus v satisfies

ui minus vit minus(Ai

Qj

Vuj minus vjx)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(511)

Simply applying uminus v to (511) we get

pminus2intS R

2uminusv2dzle

intS R

2

∣∣∣ int 1

0Ai

Qj

V+suminus VminusAi

Qj

Vds∣∣∣

times uminusV uminusvdz

and it follows thatintS R

2

uminus v2 dz

le minus2pminus2intS R

2

∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣2

uminus V 2 dz (512)

Note that AiV = F primeV V i

V and

Ai

Qj

V = F primeprimeV ViV

j

V 2 + F primeV (ij

V minus V iV

j

V 3)

where ij and are Kronecker delta functions Also we recall the structurecondition (13) F primeprimeprimeh le chpminus3 + hqminus3 We have that

∣∣∣ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds∣∣∣

le∣∣∣∣int 1

0F primeprime(V + suminus V)

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2

minusF primeprimeV ViV

j

V 2 ds

∣∣∣∣+int 1

0

∣∣∣∣F primeV + suminus VV + suminus V minus F primeV

V ∣∣∣∣ds

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632 Bae and Choe

+∣∣∣∣int 1

0F primeV + suminus V

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V3

minusF primeV ViV

j

V 3 ds

∣∣∣∣= I + II + III

The first term I is estimates as follows

I leint 1

0

∣∣∣F primeprimeV + suminus Vminus F primeprimeV ∣∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2∣∣∣∣∣ ds

+int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

(513)

By the mean value theorem we have that

int 1

0

∣∣∣F primeprimeV+suminusVminusF primeprimeV ∣∣∣ds le c

int 1

0

int 1

0

∣∣F primeprimeprimeV+shuminusV∣∣ds dhuminusV (514)

for some c and

int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

le cF primeprimeV int 1

0

int 1

0

ds dh∣∣V + shuminus V∣∣ uminus V (515)

If V ge 2uminus V then for all s isin 0 1 12 V le V + suminus V le 2V and from

(513) (514) and (515) we have I le c(V pminus3 + V qminus3

)uminus V If V le 2uminus V and p ge 2 then

∣∣F primeprimeV + suminus V∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le cuminus V pminus2 + cuminus V qminus2

and

∣∣F primeprimeV ∣∣ le cV pminus2 + cV qminus2 le cuminus V pminus2 + cuminus V qminus2

Hence we have that

I le cint 1

0

∣∣F primeprimeV + suminus V∣∣+ ∣∣F primeprimeV ∣∣ds le c1+ pminusquminus V pminus2

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Regularity for Certain Nonlinear Parabolic Systems 633

for some c independent of If V le 2uminus V and 1 lt p lt 2 then∣∣F primeprime(V + suminus V)∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le c1+ pminusqV + suminus Vpminus2

for some c We note thatint 1

0V + suminus Vpminus2 ds le uminus V pminus2

int 1

0

∣∣∣∣ V uminus V minus s

∣∣∣∣pminus2

ds le cuminus V pminus2

Therefore combining all these together we have

I le cV pminus3 uminus V (516)

for some c independent of For II and III we observe that

F primeV V le cV pminus2 + cV qminus2

∣∣∣∣FprimeV V i

Vj

V 3∣∣∣∣ le cV pminus2 + cV qminus2

∣∣∣∣(F primeV V

)∣∣∣∣ le cV pminus3 + cV qminus3

∣∣∣∣(F primeV V i

Vj

V 3)∣∣∣∣ le cV pminus3 + cV qminus3

Hence following the same argument for I we conclude that

II + III le cV pminus3 uminus V (517)

for some c independent of Combining (516) and (517) we obtain that∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣ le cV pminus3 uminus V (518)

and considering (518) in (512)intS R

2

uminus V 2 dz le cminus2pminus2V 2pminus3intS R

2

uminus V 4 dz

le cminus2intS R

2

uminus V 4 dz (519)

From Houmllderrsquos inequality we have thatintS R

2

uminus V 4 dz

=intS R

2

uminus V 8n+ 4nminus8n dz

le[

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx] 2

n int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

(520)

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634 Bae and Choe

From Lemma 55 we have

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx le c2 supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2 dx

le cp + 2qminusp

R2

intSR

uminus V 2 dz

for some c independent of Now recalling that 12 le V le we obtain that

int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

le c22minuspnminus2

n

int t0

t0minus 2minuspR24

[ intB R

2

(u + V )2pminus2uminus V 4 dx] nminus2

n

dt

le c4nminuspnminus8

n

int t0

t0minus 2minuspR24

[ intB R

2

∣∣u pminus22 uminus V pminus2

2 V∣∣ 2n

nminus2 dx] nminus2

n

dt

le c4nminuspnminus8

n

intS R

2

upminus2 2u2 dz+ c2nminuspnminus8

n pminus2intSR

uminus V 2 dz (521)

for some c independent of where we have used Sobolevrsquos inequality FromLemma 55 we haveint

S R2

upminus2 2u2 dz le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

Thus combining (519) (520) and (521) we have that

intS R

2uminus V 2 dz le c

(1+ qminusp 4

n+2)[

minus2ndashintSR

uminus V 2 dz] 2

n intSR

uminus V 2 dz

and this completes the proof

Now we prove an iteration lemma which says that if mean oscillation of unear some fixed V is small then u satisfies a Campanato type integral inequality

Lemma 58 Fix 0 lt 0 lt 1 then there exist small positive numbers independentof such that if V0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists V1 isin nN satisfying(12minusradic

0

) le V1 le

(1+radic

0

) (522)

ndashintSR

uminus V12 dz le 0 ndashintSR

uminus V02 dz (523)

ndashintSR

uminus V12 dz le 2

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Regularity for Certain Nonlinear Parabolic Systems 635

Proof Since V0 satisfies the assumption of Lemma 57 we apply Lemma 57 withV = V0 and obtainint

S R2

vminus u2 dz le cintSR

uminus V02 dz

So from the usual comparison argument we haveintSR

uminus uR2 dz le cintSR

vminus vR2 dz+ cintSR

uminus v2 dz

le cn+4intS R

2

uminus u R22 dz+ c

intS R

2

uminus v2 dz

le c(n+4 +

) intSR

uminus V02 dz

where c is independent of Choosing and isin 0 12 small we prove (523)

with V1 = uR Inequality ndashintSR

uminus V12 dz le 2 follows from (523) andassumption ndash

intSR

uminus V02 dz le 2 Now we also estimate

V1 minus V0 =∣∣∣∣∣ ndashintSR

vminus V0 dz

∣∣∣∣le c

[ndashintSR

vminus u2 dz] 1

2

+ c

[ndashintSR

uminus V02 dz] 1

2

le cminus1minusn2(1+

120

)12

Therefore using the same technique of DiBenedetto (1986) Lemma 44 and if wetake very small compared to (522) holds and this completes the proof

Lemma 59 There exist small positive numbers 0 independent of such that ifV0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists a sequence of vectors Vi i = 1 2 3 sub nN satisfying

14 le Vi le 4 ndash

intSiR

uminus Vi2 dz le 2

ndashintSi+1R

uminus Vi+12 dz le 0 ndashintSiR

uminus Vi2 dz

Proof For the proof of the lemma we need to show that 14 le Vi le 4 But this

follows from that

Vi+1 minus Vi2 le c(0 + minusnminus2

)∣∣∣ ndashintSiR

uminus Vi dz∣∣∣

le ci0

i∣∣∣ ndashint

SRuminus V02 dz

∣∣∣ le ci0

i2

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636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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ORDER REPRINTS

642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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t 200

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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616 Bae and Choe

We assume that n gt p ge 2 then we see that pp+ 1minus q gt p ge 2 Hence fromYoungrsquos inequality we can write (26) as

intS

us+p+s+2 pn dx le c

[1

Rminus p

p+1minusq

intSR

u pp+1minusq +s dz+ SR

Rminus p

p+1minusq

]1+ pn

(27)

for some c We set si by si+1 + pp+ 1minus q = si + p+ si + 2pn that is forsufficiently large s0

si = i

(s0 +

2minus n+ 2q minus p

p+ 1minus q

)+ n+ 2q minus pminus 2

p+ 1minus q

where def= 1+ pn Define Ri

def= R021+ 2minusi Now we take s = si = Ri+1 andR = Ri in (27) Define

i

def= ndashintSRi

usi+ pp+1minusq dz

then (27) can be written as

Rn+2i+1 i+1 le c

(2i+2

R0

) pp+1minusq

Rn+2i

[

i + 1

]

and

i+1 le c

(R0

2

)n+2 pnminus p

p+1minusq

2i+1 pp+1minusq minusn+2

[

i + 1

]

and hence

i+1 le cii + ci (28)

for some c We also have that

i+1 le ciciminus1iminus1 + ciminus1 + ci le ciciminus1

2

iminus1 + ciciminus1 + ci

le ciciminus1ciminus223

iminus2 + ciciminus1ciminus22 + ciciminus1 + ci

le ci+iminus1+iminus22+iminus33+middotmiddotmiddot+2iminus2+iminus1

i+1

0 + ci + ci+iminus1

+ ci+iminus1+iminus22 + middot middot middot + ci+iminus1+middotmiddotmiddot+2iminus2+iminus1

le cn2

p2iminus1minus n

p ii+1

0 + icn2

p2iminus1minus n

p i

le ci+1

i+1

0 + ici+1

If s0 gt n+ 2q minus pminus 2p+ 1minus q then s0 + 2minus n+ 2q minus pp+ 1minus q gt0 Hence we have that

limi

si = and limi

sii

= s0 +2minus n+ 2q minus p

p+ 1minus q

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Iterating (28) with some large s0 we have

supS R0

2

u le c

[ndashintSR0

u pp+1minusq +s0 dz

] 1

s0+ 2minusn+2qminuspp+1minusq + c

and this completes the proof for the case n gt p ge 2Now we assume that 1 lt p le 2 If q ge p2+ 1 then pp+ 1minus q ge 2

In this case applying Youngrsquos inequality on the last term of (26) we have (27)Hence following the argument for the case p ge 2 we prove (21) and this completesthe proof for the case q ge p2+ 1

Now we assume that p lt n and

p

2+ 1 gt q ge p gt 1

In this case s + 2 gt s + pp+ 1minus q and we have from (26)

intS

us+p+s+2 pn dx le c

[1

Rminus 2

intSR

us+2 dz+ SRRminus 2

]1+ pn

(29)

for some c We set si by sidef= is0 + n+ 2minus 2npi minus 1 Hence if s0 gt 2np

minus nminus 2 then

limi

si = and limi

sii

= s0 + n+ 2minus 2np

We take s = si = Ri+1 and R = Ri in (29) Iterating (29) we obtain

supS R0

2

u le c

[ndashintSR0

us0+2 dz

] 1s0+n+2minus 2n

p + c

for some c independent of R0 and this completes the proof for the case p2+ 1 gtq ge p gt 1

We now assume that n = p Then from Sobolevrsquos embedding theorem andHoumllderrsquos inequality we obtain that for p = n

intS

u s+22 +s+p dz le

int [ intus+2p dx

] 12[ int

u2s+p2p dx] 1

2dt

le[sup

t

intus+2p dx

] 12[ int (

u s+pp

)2pdx

] 12dt

le[sup

t

intus+2p dx

] 12int ∣∣∣(u s+p

p )∣∣∣p dz

le c

[1

Rminus p

intSR

(us+2 + us+p + u pp+1minusq +s + us)dz] 3

2

for some c The remaining things can be done similarly as in the case 2 le p lt n

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Now we consider the case p gt n Then from Sobolevrsquos embedding theorem andHoumllderrsquos inequality we obtain that for p gt n

intS

us+2+ p+s2 dz le

int [ intus+2p dx

](sup

x

u p+sp

) p2

dt

le(sup

t

intus+2p dx

) int ( int ∣∣∣(u s+pp

)∣∣∣p dx)12

dt

le c

[1

Rminus p

intSR

(us+2 + us+p + u pp+1minusq +s + us)dz] 3

2

The remaining things can be done in a similar way as before

The restriction of the integrability of u is necessary to start Moser typeiteration In fact if p isin 2nn+ 2 and p le q le p2+ 1 or p2+ 1 le q ltpn+ 3+ 4n+ 4 then r0 le 2 and there is no restriction on the integrability of u

3 L ESTIMATE FOR u

In this section we prove that u is bounded To begin with Moser iterationwe need that u is highly integrable Indeed the higher integrability of u followsfrom Caccioppoli type inequality for the second derivative of u and integration byparts technique Here the boundedness of u is crucial This method has been appliedin proving for the regularity questions for singular parabolic equations (see Choe1991) By approximation we assume that u has second derivative with respect to xHence differentiating (15) with respect to x we get

(uix

)tminus (

Ai

Qj

uujxx

)x= 0 (31)

Lemma 31 Let SR0sub T Suppose that u is bounded then for all s

intS R0

2

us dz le cintSR0

up dz+ cR0 s

for some c

Proof Let be a standard cutoff function such that = 1 for x t isin S = 0 forx t isin pSR 0 le le 1 t le cRminus 2 le cRminus for some c

From integration by parts we have

intBR

us+2k dx =intBR

usuixuixk dx = minus

intBR

u div(usuk

)dx

le csuL

intBR

us2uk dx + csuL

intBR

us+1kminus1 dx

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and applying Youngrsquos inequality we obtain

intBR

us+2k dx le cintBR

usminus22u2k dx + cintBR

us2 dx

for some c depending on uL and s Integrating with respect to t we obtain

intSR

us+2k dz le cintSR

usminus22u2k dz+ cintSR

us2 dz (32)

for some c depending on uL and s Now we take uixusminuspk as a test function

to (31) and we get

intSR

d

dt

(us+2minuspk)dz+

intSR

usminus22u2k dz

le cintSR

us+2minuspkminus1tdz+ cintSR

us+2qminusp2 dz+ cintSR

us2 dz(33)

for some c Therefore combining (32) and (33) we obtain

intSR

us+2k dz le cintSR

us2 dz+ cintSR

us+2minusptdz

+ cintSR

us+2qminusp2 dz (34)

Assume p ge 2 Since s + 2q minus p ge s + 2minus p we get from (34) that

intS

us+2 dz le c

Rminus 2

intSR

us+2qminusp dz+ cSR

Rminus 2(35)

for some c Since we are assuming 2 le p le q lt p+ 1 2q minus p lt 2 Define si bysi+1 + 2q minus p = si + 2 that is si = s0 + i2+ 2pminus 2q Note that limirarr si = Set s0 = 3pminus 2q then s0 + 2minus p gt 0 Define Ri = R021+ 2minusi i = 0 1 2 Now we set s = si = Ri+1 and R = Ri Hence defining i =

intSRi

usi+2qminusp dz wecan rewrite (35) as

i+1 le cii + ciBR0 (36)

Therefore iterating (36) we conclude that

intS R

2

us dz le cintSR

up dz+ c (37)

for all s and this completes the proof for the case p ge 2Now we consider the case that 1 lt p le 2 Suppose that s + 2minus p le s

+ 2q minus p that is p+ 22 le q lt p+ 1 In this case we prove (37) in the

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same way for the case p ge 2 Finally we assume that p le q lt p+ 22 that iss + 2minus p ge s + 2q minus p Considering Youngrsquos inequality in (34) we get

intS

us+2 dz le c

Rminus 2

intSR

us+2minusp dz+ cSR

Rminus 2 (38)

Taking = Ri+1 R = Ri and si = s0 + ip with s0 = 2pminus 2 we obtain from (38)

intSRi+1

usi+1+2minusp dz le ci

R2

intSRi

usi+2minusp dz+ ci SRR2

for some c Therefore iterating this we obtain (37)

Once we have shown that u is in a high Ls space we can apply Moser iteration

Theorem 32 We have u isin L Furthermore suppose SR0sub T then u satisfies

supS R0

2

u le c[ndashintSR0

us0+s1 dz] 1

s0+s2 + c (39)

for some large s0 and some c independent of R0 where

s1 = 2q minus p s2 = 2minus nq minus p if q ge p

2 + 1

s1 = 2 s2 = np

2 + 2minus n if q le p

2 + 1

Proof Let be a standard cutoff function We take usuix2 as a test function to

(31) and we obtain

1s+2

int d

dt

(us+2)2 dz+

intAi

Qj

uujxx

(usuixx

+ susminus2uixukxukxx

)2 dz

= minus2int

Ai

Qj

uujxx

usuixx

dz

Considering the ellipticity condition

Qpminus22 le Ai

Qj

Qi

j le c

(Qpminus2 + Qqminus2)2

we see that

supt

intus+22 dx +

int ∣∣(u s+p2

)∣∣2 dzle c

intus+2tdz+ c

int (us+p + us+2qminusp)2 dz

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for some c Therefore from Sobolevrsquos embedding theorem and Houmllderrsquos inequalitywe prove that

intS

us+p+s+2 2n dx le

int [ intus+22 dx

] 2n[ int

us+p nnminus2

2nnminus2dx

] nminus2n

dt

le[sup

t

intus+22 dx

] 2nint ∣∣∣(u s+p

2 )∣∣∣2 dz

le c

[1

Rminus 2

intSR

(us+2 + us+p + us+2qminusp)dz

]1+ 2n

(310)

We assume that p ge 2 then we see that s + 2q minus p ge s + 2 Hence fromYoungrsquos inequality we can write (310) as

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2qminusp dz+ SRRminus 2

]1+ 2n

(311)

for some c We set si by si+1 + 2q minus p = si + p+ si + 22n that is for

sufficiently large s0 si = is0 +(2minus nq minus p

)i minus 1 where

def= 1+ 2n Define

Ri

def= R021+ 2minusi Now we take s = si = Ri+1 and R = Ri in (311) Define

i

def= ndashintSRi

usi+2qminusp dz then (311) can be written as

i+1 le cii + ci (312)

If s0 gt nq minus pminus 2 then s0 + 2minus nq minus p gt 0 Hence we have that

limi

si = and limi

sii

= s0 + 2minus nq minus p

From Lemma 31 we can take s0 as large as possible Iterating (312) with some larges0 we have

supS R0

2

u le c[ndashintSR0

us0+2qminusp dz] 1

s0+2minusnqminusp + c

and this completes the proof for the case p ge 2Now we assume that 1 lt p le 2 If q ge p2+ 1 then s + 2q minus p ge s + 2 In this

case applying Youngrsquos inequality on the last term of (310) we have

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2qminusp dz+ SRRminus 2

]1+ 2n

for some c Hence following the argument for the case p ge 2 we prove (39) andthis completes the proof for the case q ge p2+ 1

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Now we assume that p2+ 1 gt q ge p gt 1 In this case s + 2 gt s + 2q minus p andwe have from (310)

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2 dz+ SRRminus 2

]1+ 2n

(313)

for some c We set si by sidef= is0 + np2+ 2minus ni minus 1 Hence if s0 gt nminus 2

minus np2 then

limi

si = and limi

sii

= s0 + 2minus n+ np

2

We take s = si = Ri+1 and R = Ri in (313) Iterating (313) we obtain

supS R0

2

u le c

[ndashintSR0

us0+2 dz

] 1s0+2+ np

2 minusn + c

for some c independent of R0 and this completes the proof for the casep2+ 1 gt q ge p gt 1

4 HOumlLDER CONTINUITY OF u

In this section we prove that u is Houmllder continuous

Theorem 41 Define r0 by the constant appearing in Theorem 21 Suppose that uis L

r0locT then u is Houmllder continuous Moreover for all Q2R sub T with R lt 1 u

satisfies

ndashintQR

uminus uR2 dz le cRs

for some c independent of R where we define

s =2 when p ge 2

ppminus 1 when p isin 1 2

We introduce a nonnegative cutoff function isin 0 BR such that = 1 in

BR2 0 le le 1 le cR We define minus

R

def= t0 minus 2Rp t0 minus Rp and QminusR = BR timesminus

R

Also we define uRt

def= 1BRintBR

ux tdx and uR

def= 1QRintQR

ux tdz Firstwe prove a lemma which is essential for a Poincareacute inequality for solutions of adegenerate parabolic system

Lemma 42 Suppose Q2R sub T then u satisfies the following inequality

suptisinR

intminus

R

dsintBR

2ux tminus uRs2 dx

le cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz+ cintQminus

R

uminus uRt2 dz

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Regularity for Certain Nonlinear Parabolic Systems 623

for all R lt R0 where c depends only on nN and p and

s1 = p s2 = p and s3 = pqminus1

pminus1 when p ge 2

s1 = p2minus p s2 = ppminus 1 and s3 = pq minus 1 when 1 lt p lt 2

Proof Since u isin 0(0 T L2

) there exists ts isin R for all s isin minus

R such that

intBR

pux t minus uRs2dx = suptisinR

intBR

pux tminus uRs2 dx

We take uminus uRspxst as a test function to (15) where st is the characteristic

function such that

st =1 for all isin s t

0 for all isin s t

Hence we have

intut middot uminus uRs

pxst dz+int

Au middot uminus uRspst dz = 0 (41)

Now we assume p isin 2 Considering the structure condition (12) on A (23)and (25) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

leint

uminus uRs2x sp dx + cintBRtimesst

upminus1uminus uRspminus1dz

+ cintBRtimesst

uqminus1uminus uRspminus1dz (42)

for some c independent of R From Youngrsquos inequality we have

intBRtimesst

upminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

up dz (43)

for some c and small Similarly the last term in (42) can be estimated as follows

intBRtimesst

uqminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

u pqminus1pminus1 dz (44)

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for some c and small Combining (42) (43) and (44) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

le cint

uminus uRs2x sp dx +

Rp

intBRtimesst

uminus uRspp dz

+ cintBRtimesst

up dz+ cintBRtimesst

u pqminus1pminus1 dz (45)

for some c and small Integrating (45) with respect to s from t0 minus 2Rp to t0 minus Rpwe have

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRs2p dz+ c

Rp

intminus

R

dsintBRtimesst

uminus uRs2p dz

+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

By the choice of t we have that for small

intminus

R

dsintBR

uminusuRs2x t pdx

le cintQminus

R

uminusuRs2p dz+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

So we proved Lemma 42 when p ge 2In case p isin 1 2 we estimate the second term of (41) as follows

intuppst dz+ p

intAu middot uminus uRs

pminus1st dz

leint

Au middot uminus uRspst dz (46)

and using the fact that u is bounded and ppminus 1 gt 2 we have

∣∣∣ int Au middot uminusuRspminus1st dz

∣∣∣le c

intBRtimesst

upminus1uminusuRspminus1st dz

+ cintBRtimesst

uqminus1pminus1uminusuRsst dz

le

Rp

intuminusuRs

ppminus1 pst dz+

c

Rp2minusp

intuppminus1 dz+ c

Rp2minusp

intupqminus1 dz

(47)

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Regularity for Certain Nonlinear Parabolic Systems 625

Combining (41) (46) and (47) we have

intBR

uminus uRs2x t p dx

le cu2minusppminus1L

Rp

intBRtimesst

uminus uRs2pst dz+c

Rp2minusp

intBRtimesst

uppminus1 dz

+ c

Rp2minusp

intBRtimesst

upqminus1 dz+intBR

uminus uRs2x sp dx (48)

for small Integrating (48) with respect to s and considering the choice of t weprove Lemma 42 when 1 lt p lt 2

Now we prove a Poincareacute inequality

Theorem 43 Suppose Q2R sub T with R lt 1 then u satisfies the following inequality

intQR

2

uminus uR22 dz le cR2

intQ2R

u2 dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

where c is independent of R

Proof Using Lemma 42 we have

intQR

2

uminus uR22 dz le

intQR

uminus uRs2p dz

le c

Rp

intminus

R

dsintQR

uminus uRs2p dz le c suptisinR

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRt2p dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

le cR2intQ2R

u2 dz+ cRs1

intQ2R

us2z+ cRs1

intQ2R

us3 dz

where we used a Poincareacute type inequality for x variables only that is

intBR

uminus uRs2p dx le cR2intBR

u2 dx

Proof of Theorem 41 Since u is bounded owing to Theorem 32 we have fromTheorem 43

intQR

uminus uRs2 dz lecRn+p+2 when p isin 2

cRn+p+ppminus1 when p isin 1 2

where c is independent of R Hence by isomorphism theorem of Da Prato (1965) weprove Theorem 41

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5 HOumlLDER CONTINUITY FOR u

In this section we prove that u is Houmllder continuous Because of degeneracyof the pde we need scaled cylinders (see DiBenedetto and Friedman 1985)Let R0 gt 0 be fixed and consider a cylinder

SR0

def= BR0times t0 minus R2minus

0 t0

for small For any 0 lt R le R02 we introduce a scaled cylinder

SRdef= BR times t0 minus R22minusp t0

for fixed ge supSR0u

We assume that pminus2 gt R0 if p gt 2 Hence we note that SR sub S

R0and

ge supSRu

The following theorem is our main theorem in this section

Theorem 51 Suppose that SR sub T then u satisfies for all r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR 2 dz+ cR

for some gt 0

For the proof of Theorem 51 we follow a usual program for degenerateequations such that if for some isin 0 1∣∣z isin SR u2 le 1minus 2

∣∣ gt ∣∣SR

∣∣ (51)

then

supS R

2

u le (52)

for some small isin 0 1 and otherwise that is (51) fails then for all 0 lt r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR2 dz (53)

for some gt 0 Combining (52) and (53) we prove Theorem 51First we prove that (52) under the assumption (51) A similar argument for

the case p = q ge 2 appeared in DiBenedetto and Friedman (1985) We assume that(51) is true Set v = u2 For each h gt 0 we set AhRt

def= x isin BR v gt h and

BhR

def= z isin SR v gt h From the mean value theorem we estimate AhRt

Lemma 52 For any isin 0 there exists isin t0 minus 2minuspR20 t0 minus 2minuspR2

0 such that

A1minus2R0 le 1minus

1minus BR0

where is independent of

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Regularity for Certain Nonlinear Parabolic Systems 627

Proof Suppose that the assertion is not true then

B1minus2R0 ge

int t0minus2minuspR20

t0minus2minuspR20

A1minus2R0d gt 1minus BR0

2minuspR20 = 1minus SR0

and this contradicts the assumption (51)

The following lemma is useful in estimating sup u Indeed when p = q ge 2DiBenedetto and Friedman proved a similar theorem (see Lemma 62 inDiBenedetto and Friedman 1985) Now choose = 2 and in Lemma 52We define St

R

def= BR times t

Lemma 53 There exists a positive integer r independent of such that

∣∣A1minus 2r

2R0t

∣∣ le (1minus

(2

)2)BR0

for all t isin t0 minus

22minuspR2

0 t0

Proof Consider the function

v = log+[

2

2minus vminus 1minus 2+ + 2

2r

]

where r is a positive integer This function was used in Choe (1991) and DiBenedettoand Friedman (1985) We take 13 = ui

x2vprime 2 as a test function for (31) where

isin 0 is a standard cutoff function such that B1minusR for small decided later is

contained in support Integrating from to t we obtain

12

intStR

vt2prime 2 dz+

intStR

Ai

Qj

uujxx

uixx

2prime 2 dz

+ 2intStR

Ai

Qj

uujxx

uixukxx

ukx2primeprime 2 dz

+ 2intStR

Ai

Qj

uujxx

uix x

2prime dz = 0

Considering the fact 2primeprime = 21+ prime2 we have

intBR

2vx t 2xdx +intStR

upminus2v21+ prime2 2 dz

leintBR

2vx 2xdx + cintStR

upminus2v prime dz

+ cintStR

uqminus2v prime dz

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Applying Youngrsquos inequality on the last two terms we getintBR

2vx t 2xdx

leintBR

2vx 2xdx + cintStR

upminus2 2 dz+ cintStR

u2qminuspminus2 2 dz

for some c Since le r log 2 and u is bounded we getintBR

2vx t 2xdx leintBR

2vx 2xdx + crBR

From Lemma 52 we getintBR

2vx t 2xdx le r log 221minus

1minus

2

BR

From definition of we haveintB1minusRcapvgt1minus

2r 22vx tdx ge r minus 1 log 22A1minus

2r 21minusRt

and

A1minus 2r

21minusRt le[( r

r minus 1

)2 1minus

1minus

2

+ c

r

]BR

Hence we obtain

A(1minus

2r

)2R

t le A1minus 2r

21minusRt + BRB1minusR

le[( r

r minus 1

)2 1minus

1minus

2

+ c

r+ n

]BR

Taking r so large that( r

r minus 1

)2 le(1minus

2

)1+

c

rle 3

82

and = 382n we obtain

A1minus 2r

2Rt le(1minus

(2

)2)BR

We take 13 = uixvminus h+ 2 as a test function in (31) where v = u2 and is

a cutoff function such that = 1 in Sr = 0 on pSR le cRminus r and t le cpminus2Rminus r2 Considering the ellipticity condition (12) we have

supt

int ∣∣vminus h+∣∣2 2 dx +

intupminus2vminus h+2 2 dz

le c

Rminus r2

intSR

upminus2vminus h+2 dz

+ c

Rminus r2

intSR

u2qminuspminus2vminus h+2 dz+ cintSR

vminus h+2 tdz (54)

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Regularity for Certain Nonlinear Parabolic Systems 629

for some c Now choose h ge 12 For such a choice of h we haveint

upminus2vminus h+2 2 dz ge cpminus2int

vminus h+2 2 dz

Hence from (54) we obtain

supt

intvminus h+2 2 dx + pminus2

intvminus h+2 2 dz

le[

2qminuspminus2

Rminus r2+ pminus2

Rminus r2

] intSR

vminus h+2 dz (55)

for some c independent of Observe that is bounded above by some fixednumber We make change of variables

= pminus2t minus t0 vx = vx t0 + 2minusp

and we write (55) as

supt

intBR

vminus h+2 2 dx +intSR

vminus h+2 2 dz le c

Rminus r2

intSR

vminus h+dz

Once we have this Caccioppoli type inequality we can apply truncation argumentTherefore following to similar argument in DiBenedetto and Friedman (1984 or1985) we have a maximum principle Now the proof for the next lemma is almostthe same as in DiBenedetto and Friedman (1984 or 1985) we omit the proof

Lemma 54 If for some and isin 0 1∣∣x t isin S1+R u2 lt 1minus 2∣∣ ge S1+R (56)

then there exists a number isin 0 1 independent of and R such that

supS R

2

ux t le

Now we assume that (56) fails that is∣∣x t isin S1+R u2 lt 1minus 2∣∣ lt S1+R (57)

Following a comparison argument similar to Choe (1991) or DiBenedetto andFriedman (1984) we prove that u satisfies a Campanato type integral inequalityLet V isin nN be a constant vector such that 1

2 le V le Taking uxkminus Vk

2 as atest function to (31) we have the following lemma

Lemma 55 There exists a constant c independent of such that

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2tdx +intS R

2

upminus22u2 dz

le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

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Now suppose that v is a solution to

vit minus

(Ai

Qj

Vvjx

)x

= 0 (58)

with v = u on pSR2 Since 1

2 le V le le c for some fixed c we see that fromthe structure condition of F

2 le 2minuspAi

Qj

Vi

j le c1+ qminusp2

Hence we see that v satisfies a Campanato type integral inequality

Lemma 56 Suppose that 0 lt r le R2 There exists a constant c independent ofVR and such that

intSr

vminus vr 2 dz le c( r

R

)n+4 intS R

2

uminus V 2 dz (59)

Proof We write (58) as

vit minus pminus2

Ai

Qj

V

pminus2vjx

x

= 0

We change of variable = pminus2t minus t0 then the function vx = vx 2minusp+ t0satisfies a uniformly parabolic system

vi minus

Ai

Qj

V

pminus2vjx

x

= 0

Thus from results of Campanato (1966) we prove

intS

vminus v2 dz le c(

R

)n+4 intS R

2

uminus V 2 dz

and scaling back prove (59)

We estimate the L2 norm of uminus v in SR2

Lemma 57 For each lt R2 there exists a constant c independent of such that

intS

uminus v2 dz le c(1+ qminusp 4

n+2)(

minus2 ndashintSR

uminus V 2 dz) int

SRuminus V 2 dz

for = min(12

2n

)

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Regularity for Certain Nonlinear Parabolic Systems 631

Proof Subtracting(Ai

Qj

Vujx

)x

from both sides of (15) we get

uit minus

(Ai

Qj

Vujx

)x= (

Aiuminus Ai

Qj

Vujx

)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(510)

From (58) and (510) we see that uminus v satisfies

ui minus vit minus(Ai

Qj

Vuj minus vjx)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(511)

Simply applying uminus v to (511) we get

pminus2intS R

2uminusv2dzle

intS R

2

∣∣∣ int 1

0Ai

Qj

V+suminus VminusAi

Qj

Vds∣∣∣

times uminusV uminusvdz

and it follows thatintS R

2

uminus v2 dz

le minus2pminus2intS R

2

∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣2

uminus V 2 dz (512)

Note that AiV = F primeV V i

V and

Ai

Qj

V = F primeprimeV ViV

j

V 2 + F primeV (ij

V minus V iV

j

V 3)

where ij and are Kronecker delta functions Also we recall the structurecondition (13) F primeprimeprimeh le chpminus3 + hqminus3 We have that

∣∣∣ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds∣∣∣

le∣∣∣∣int 1

0F primeprime(V + suminus V)

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2

minusF primeprimeV ViV

j

V 2 ds

∣∣∣∣+int 1

0

∣∣∣∣F primeV + suminus VV + suminus V minus F primeV

V ∣∣∣∣ds

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632 Bae and Choe

+∣∣∣∣int 1

0F primeV + suminus V

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V3

minusF primeV ViV

j

V 3 ds

∣∣∣∣= I + II + III

The first term I is estimates as follows

I leint 1

0

∣∣∣F primeprimeV + suminus Vminus F primeprimeV ∣∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2∣∣∣∣∣ ds

+int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

(513)

By the mean value theorem we have that

int 1

0

∣∣∣F primeprimeV+suminusVminusF primeprimeV ∣∣∣ds le c

int 1

0

int 1

0

∣∣F primeprimeprimeV+shuminusV∣∣ds dhuminusV (514)

for some c and

int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

le cF primeprimeV int 1

0

int 1

0

ds dh∣∣V + shuminus V∣∣ uminus V (515)

If V ge 2uminus V then for all s isin 0 1 12 V le V + suminus V le 2V and from

(513) (514) and (515) we have I le c(V pminus3 + V qminus3

)uminus V If V le 2uminus V and p ge 2 then

∣∣F primeprimeV + suminus V∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le cuminus V pminus2 + cuminus V qminus2

and

∣∣F primeprimeV ∣∣ le cV pminus2 + cV qminus2 le cuminus V pminus2 + cuminus V qminus2

Hence we have that

I le cint 1

0

∣∣F primeprimeV + suminus V∣∣+ ∣∣F primeprimeV ∣∣ds le c1+ pminusquminus V pminus2

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Regularity for Certain Nonlinear Parabolic Systems 633

for some c independent of If V le 2uminus V and 1 lt p lt 2 then∣∣F primeprime(V + suminus V)∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le c1+ pminusqV + suminus Vpminus2

for some c We note thatint 1

0V + suminus Vpminus2 ds le uminus V pminus2

int 1

0

∣∣∣∣ V uminus V minus s

∣∣∣∣pminus2

ds le cuminus V pminus2

Therefore combining all these together we have

I le cV pminus3 uminus V (516)

for some c independent of For II and III we observe that

F primeV V le cV pminus2 + cV qminus2

∣∣∣∣FprimeV V i

Vj

V 3∣∣∣∣ le cV pminus2 + cV qminus2

∣∣∣∣(F primeV V

)∣∣∣∣ le cV pminus3 + cV qminus3

∣∣∣∣(F primeV V i

Vj

V 3)∣∣∣∣ le cV pminus3 + cV qminus3

Hence following the same argument for I we conclude that

II + III le cV pminus3 uminus V (517)

for some c independent of Combining (516) and (517) we obtain that∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣ le cV pminus3 uminus V (518)

and considering (518) in (512)intS R

2

uminus V 2 dz le cminus2pminus2V 2pminus3intS R

2

uminus V 4 dz

le cminus2intS R

2

uminus V 4 dz (519)

From Houmllderrsquos inequality we have thatintS R

2

uminus V 4 dz

=intS R

2

uminus V 8n+ 4nminus8n dz

le[

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx] 2

n int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

(520)

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634 Bae and Choe

From Lemma 55 we have

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx le c2 supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2 dx

le cp + 2qminusp

R2

intSR

uminus V 2 dz

for some c independent of Now recalling that 12 le V le we obtain that

int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

le c22minuspnminus2

n

int t0

t0minus 2minuspR24

[ intB R

2

(u + V )2pminus2uminus V 4 dx] nminus2

n

dt

le c4nminuspnminus8

n

int t0

t0minus 2minuspR24

[ intB R

2

∣∣u pminus22 uminus V pminus2

2 V∣∣ 2n

nminus2 dx] nminus2

n

dt

le c4nminuspnminus8

n

intS R

2

upminus2 2u2 dz+ c2nminuspnminus8

n pminus2intSR

uminus V 2 dz (521)

for some c independent of where we have used Sobolevrsquos inequality FromLemma 55 we haveint

S R2

upminus2 2u2 dz le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

Thus combining (519) (520) and (521) we have that

intS R

2uminus V 2 dz le c

(1+ qminusp 4

n+2)[

minus2ndashintSR

uminus V 2 dz] 2

n intSR

uminus V 2 dz

and this completes the proof

Now we prove an iteration lemma which says that if mean oscillation of unear some fixed V is small then u satisfies a Campanato type integral inequality

Lemma 58 Fix 0 lt 0 lt 1 then there exist small positive numbers independentof such that if V0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists V1 isin nN satisfying(12minusradic

0

) le V1 le

(1+radic

0

) (522)

ndashintSR

uminus V12 dz le 0 ndashintSR

uminus V02 dz (523)

ndashintSR

uminus V12 dz le 2

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Regularity for Certain Nonlinear Parabolic Systems 635

Proof Since V0 satisfies the assumption of Lemma 57 we apply Lemma 57 withV = V0 and obtainint

S R2

vminus u2 dz le cintSR

uminus V02 dz

So from the usual comparison argument we haveintSR

uminus uR2 dz le cintSR

vminus vR2 dz+ cintSR

uminus v2 dz

le cn+4intS R

2

uminus u R22 dz+ c

intS R

2

uminus v2 dz

le c(n+4 +

) intSR

uminus V02 dz

where c is independent of Choosing and isin 0 12 small we prove (523)

with V1 = uR Inequality ndashintSR

uminus V12 dz le 2 follows from (523) andassumption ndash

intSR

uminus V02 dz le 2 Now we also estimate

V1 minus V0 =∣∣∣∣∣ ndashintSR

vminus V0 dz

∣∣∣∣le c

[ndashintSR

vminus u2 dz] 1

2

+ c

[ndashintSR

uminus V02 dz] 1

2

le cminus1minusn2(1+

120

)12

Therefore using the same technique of DiBenedetto (1986) Lemma 44 and if wetake very small compared to (522) holds and this completes the proof

Lemma 59 There exist small positive numbers 0 independent of such that ifV0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists a sequence of vectors Vi i = 1 2 3 sub nN satisfying

14 le Vi le 4 ndash

intSiR

uminus Vi2 dz le 2

ndashintSi+1R

uminus Vi+12 dz le 0 ndashintSiR

uminus Vi2 dz

Proof For the proof of the lemma we need to show that 14 le Vi le 4 But this

follows from that

Vi+1 minus Vi2 le c(0 + minusnminus2

)∣∣∣ ndashintSiR

uminus Vi dz∣∣∣

le ci0

i∣∣∣ ndashint

SRuminus V02 dz

∣∣∣ le ci0

i2

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636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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t 02

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t 200

7

ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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Regularity for Certain Nonlinear Parabolic Systems 617

Iterating (28) with some large s0 we have

supS R0

2

u le c

[ndashintSR0

u pp+1minusq +s0 dz

] 1

s0+ 2minusn+2qminuspp+1minusq + c

and this completes the proof for the case n gt p ge 2Now we assume that 1 lt p le 2 If q ge p2+ 1 then pp+ 1minus q ge 2

In this case applying Youngrsquos inequality on the last term of (26) we have (27)Hence following the argument for the case p ge 2 we prove (21) and this completesthe proof for the case q ge p2+ 1

Now we assume that p lt n and

p

2+ 1 gt q ge p gt 1

In this case s + 2 gt s + pp+ 1minus q and we have from (26)

intS

us+p+s+2 pn dx le c

[1

Rminus 2

intSR

us+2 dz+ SRRminus 2

]1+ pn

(29)

for some c We set si by sidef= is0 + n+ 2minus 2npi minus 1 Hence if s0 gt 2np

minus nminus 2 then

limi

si = and limi

sii

= s0 + n+ 2minus 2np

We take s = si = Ri+1 and R = Ri in (29) Iterating (29) we obtain

supS R0

2

u le c

[ndashintSR0

us0+2 dz

] 1s0+n+2minus 2n

p + c

for some c independent of R0 and this completes the proof for the case p2+ 1 gtq ge p gt 1

We now assume that n = p Then from Sobolevrsquos embedding theorem andHoumllderrsquos inequality we obtain that for p = n

intS

u s+22 +s+p dz le

int [ intus+2p dx

] 12[ int

u2s+p2p dx] 1

2dt

le[sup

t

intus+2p dx

] 12[ int (

u s+pp

)2pdx

] 12dt

le[sup

t

intus+2p dx

] 12int ∣∣∣(u s+p

p )∣∣∣p dz

le c

[1

Rminus p

intSR

(us+2 + us+p + u pp+1minusq +s + us)dz] 3

2

for some c The remaining things can be done similarly as in the case 2 le p lt n

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618 Bae and Choe

Now we consider the case p gt n Then from Sobolevrsquos embedding theorem andHoumllderrsquos inequality we obtain that for p gt n

intS

us+2+ p+s2 dz le

int [ intus+2p dx

](sup

x

u p+sp

) p2

dt

le(sup

t

intus+2p dx

) int ( int ∣∣∣(u s+pp

)∣∣∣p dx)12

dt

le c

[1

Rminus p

intSR

(us+2 + us+p + u pp+1minusq +s + us)dz] 3

2

The remaining things can be done in a similar way as before

The restriction of the integrability of u is necessary to start Moser typeiteration In fact if p isin 2nn+ 2 and p le q le p2+ 1 or p2+ 1 le q ltpn+ 3+ 4n+ 4 then r0 le 2 and there is no restriction on the integrability of u

3 L ESTIMATE FOR u

In this section we prove that u is bounded To begin with Moser iterationwe need that u is highly integrable Indeed the higher integrability of u followsfrom Caccioppoli type inequality for the second derivative of u and integration byparts technique Here the boundedness of u is crucial This method has been appliedin proving for the regularity questions for singular parabolic equations (see Choe1991) By approximation we assume that u has second derivative with respect to xHence differentiating (15) with respect to x we get

(uix

)tminus (

Ai

Qj

uujxx

)x= 0 (31)

Lemma 31 Let SR0sub T Suppose that u is bounded then for all s

intS R0

2

us dz le cintSR0

up dz+ cR0 s

for some c

Proof Let be a standard cutoff function such that = 1 for x t isin S = 0 forx t isin pSR 0 le le 1 t le cRminus 2 le cRminus for some c

From integration by parts we have

intBR

us+2k dx =intBR

usuixuixk dx = minus

intBR

u div(usuk

)dx

le csuL

intBR

us2uk dx + csuL

intBR

us+1kminus1 dx

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Regularity for Certain Nonlinear Parabolic Systems 619

and applying Youngrsquos inequality we obtain

intBR

us+2k dx le cintBR

usminus22u2k dx + cintBR

us2 dx

for some c depending on uL and s Integrating with respect to t we obtain

intSR

us+2k dz le cintSR

usminus22u2k dz+ cintSR

us2 dz (32)

for some c depending on uL and s Now we take uixusminuspk as a test function

to (31) and we get

intSR

d

dt

(us+2minuspk)dz+

intSR

usminus22u2k dz

le cintSR

us+2minuspkminus1tdz+ cintSR

us+2qminusp2 dz+ cintSR

us2 dz(33)

for some c Therefore combining (32) and (33) we obtain

intSR

us+2k dz le cintSR

us2 dz+ cintSR

us+2minusptdz

+ cintSR

us+2qminusp2 dz (34)

Assume p ge 2 Since s + 2q minus p ge s + 2minus p we get from (34) that

intS

us+2 dz le c

Rminus 2

intSR

us+2qminusp dz+ cSR

Rminus 2(35)

for some c Since we are assuming 2 le p le q lt p+ 1 2q minus p lt 2 Define si bysi+1 + 2q minus p = si + 2 that is si = s0 + i2+ 2pminus 2q Note that limirarr si = Set s0 = 3pminus 2q then s0 + 2minus p gt 0 Define Ri = R021+ 2minusi i = 0 1 2 Now we set s = si = Ri+1 and R = Ri Hence defining i =

intSRi

usi+2qminusp dz wecan rewrite (35) as

i+1 le cii + ciBR0 (36)

Therefore iterating (36) we conclude that

intS R

2

us dz le cintSR

up dz+ c (37)

for all s and this completes the proof for the case p ge 2Now we consider the case that 1 lt p le 2 Suppose that s + 2minus p le s

+ 2q minus p that is p+ 22 le q lt p+ 1 In this case we prove (37) in the

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620 Bae and Choe

same way for the case p ge 2 Finally we assume that p le q lt p+ 22 that iss + 2minus p ge s + 2q minus p Considering Youngrsquos inequality in (34) we get

intS

us+2 dz le c

Rminus 2

intSR

us+2minusp dz+ cSR

Rminus 2 (38)

Taking = Ri+1 R = Ri and si = s0 + ip with s0 = 2pminus 2 we obtain from (38)

intSRi+1

usi+1+2minusp dz le ci

R2

intSRi

usi+2minusp dz+ ci SRR2

for some c Therefore iterating this we obtain (37)

Once we have shown that u is in a high Ls space we can apply Moser iteration

Theorem 32 We have u isin L Furthermore suppose SR0sub T then u satisfies

supS R0

2

u le c[ndashintSR0

us0+s1 dz] 1

s0+s2 + c (39)

for some large s0 and some c independent of R0 where

s1 = 2q minus p s2 = 2minus nq minus p if q ge p

2 + 1

s1 = 2 s2 = np

2 + 2minus n if q le p

2 + 1

Proof Let be a standard cutoff function We take usuix2 as a test function to

(31) and we obtain

1s+2

int d

dt

(us+2)2 dz+

intAi

Qj

uujxx

(usuixx

+ susminus2uixukxukxx

)2 dz

= minus2int

Ai

Qj

uujxx

usuixx

dz

Considering the ellipticity condition

Qpminus22 le Ai

Qj

Qi

j le c

(Qpminus2 + Qqminus2)2

we see that

supt

intus+22 dx +

int ∣∣(u s+p2

)∣∣2 dzle c

intus+2tdz+ c

int (us+p + us+2qminusp)2 dz

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Regularity for Certain Nonlinear Parabolic Systems 621

for some c Therefore from Sobolevrsquos embedding theorem and Houmllderrsquos inequalitywe prove that

intS

us+p+s+2 2n dx le

int [ intus+22 dx

] 2n[ int

us+p nnminus2

2nnminus2dx

] nminus2n

dt

le[sup

t

intus+22 dx

] 2nint ∣∣∣(u s+p

2 )∣∣∣2 dz

le c

[1

Rminus 2

intSR

(us+2 + us+p + us+2qminusp)dz

]1+ 2n

(310)

We assume that p ge 2 then we see that s + 2q minus p ge s + 2 Hence fromYoungrsquos inequality we can write (310) as

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2qminusp dz+ SRRminus 2

]1+ 2n

(311)

for some c We set si by si+1 + 2q minus p = si + p+ si + 22n that is for

sufficiently large s0 si = is0 +(2minus nq minus p

)i minus 1 where

def= 1+ 2n Define

Ri

def= R021+ 2minusi Now we take s = si = Ri+1 and R = Ri in (311) Define

i

def= ndashintSRi

usi+2qminusp dz then (311) can be written as

i+1 le cii + ci (312)

If s0 gt nq minus pminus 2 then s0 + 2minus nq minus p gt 0 Hence we have that

limi

si = and limi

sii

= s0 + 2minus nq minus p

From Lemma 31 we can take s0 as large as possible Iterating (312) with some larges0 we have

supS R0

2

u le c[ndashintSR0

us0+2qminusp dz] 1

s0+2minusnqminusp + c

and this completes the proof for the case p ge 2Now we assume that 1 lt p le 2 If q ge p2+ 1 then s + 2q minus p ge s + 2 In this

case applying Youngrsquos inequality on the last term of (310) we have

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2qminusp dz+ SRRminus 2

]1+ 2n

for some c Hence following the argument for the case p ge 2 we prove (39) andthis completes the proof for the case q ge p2+ 1

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622 Bae and Choe

Now we assume that p2+ 1 gt q ge p gt 1 In this case s + 2 gt s + 2q minus p andwe have from (310)

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2 dz+ SRRminus 2

]1+ 2n

(313)

for some c We set si by sidef= is0 + np2+ 2minus ni minus 1 Hence if s0 gt nminus 2

minus np2 then

limi

si = and limi

sii

= s0 + 2minus n+ np

2

We take s = si = Ri+1 and R = Ri in (313) Iterating (313) we obtain

supS R0

2

u le c

[ndashintSR0

us0+2 dz

] 1s0+2+ np

2 minusn + c

for some c independent of R0 and this completes the proof for the casep2+ 1 gt q ge p gt 1

4 HOumlLDER CONTINUITY OF u

In this section we prove that u is Houmllder continuous

Theorem 41 Define r0 by the constant appearing in Theorem 21 Suppose that uis L

r0locT then u is Houmllder continuous Moreover for all Q2R sub T with R lt 1 u

satisfies

ndashintQR

uminus uR2 dz le cRs

for some c independent of R where we define

s =2 when p ge 2

ppminus 1 when p isin 1 2

We introduce a nonnegative cutoff function isin 0 BR such that = 1 in

BR2 0 le le 1 le cR We define minus

R

def= t0 minus 2Rp t0 minus Rp and QminusR = BR timesminus

R

Also we define uRt

def= 1BRintBR

ux tdx and uR

def= 1QRintQR

ux tdz Firstwe prove a lemma which is essential for a Poincareacute inequality for solutions of adegenerate parabolic system

Lemma 42 Suppose Q2R sub T then u satisfies the following inequality

suptisinR

intminus

R

dsintBR

2ux tminus uRs2 dx

le cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz+ cintQminus

R

uminus uRt2 dz

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Regularity for Certain Nonlinear Parabolic Systems 623

for all R lt R0 where c depends only on nN and p and

s1 = p s2 = p and s3 = pqminus1

pminus1 when p ge 2

s1 = p2minus p s2 = ppminus 1 and s3 = pq minus 1 when 1 lt p lt 2

Proof Since u isin 0(0 T L2

) there exists ts isin R for all s isin minus

R such that

intBR

pux t minus uRs2dx = suptisinR

intBR

pux tminus uRs2 dx

We take uminus uRspxst as a test function to (15) where st is the characteristic

function such that

st =1 for all isin s t

0 for all isin s t

Hence we have

intut middot uminus uRs

pxst dz+int

Au middot uminus uRspst dz = 0 (41)

Now we assume p isin 2 Considering the structure condition (12) on A (23)and (25) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

leint

uminus uRs2x sp dx + cintBRtimesst

upminus1uminus uRspminus1dz

+ cintBRtimesst

uqminus1uminus uRspminus1dz (42)

for some c independent of R From Youngrsquos inequality we have

intBRtimesst

upminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

up dz (43)

for some c and small Similarly the last term in (42) can be estimated as follows

intBRtimesst

uqminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

u pqminus1pminus1 dz (44)

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ORDER REPRINTS

624 Bae and Choe

for some c and small Combining (42) (43) and (44) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

le cint

uminus uRs2x sp dx +

Rp

intBRtimesst

uminus uRspp dz

+ cintBRtimesst

up dz+ cintBRtimesst

u pqminus1pminus1 dz (45)

for some c and small Integrating (45) with respect to s from t0 minus 2Rp to t0 minus Rpwe have

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRs2p dz+ c

Rp

intminus

R

dsintBRtimesst

uminus uRs2p dz

+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

By the choice of t we have that for small

intminus

R

dsintBR

uminusuRs2x t pdx

le cintQminus

R

uminusuRs2p dz+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

So we proved Lemma 42 when p ge 2In case p isin 1 2 we estimate the second term of (41) as follows

intuppst dz+ p

intAu middot uminus uRs

pminus1st dz

leint

Au middot uminus uRspst dz (46)

and using the fact that u is bounded and ppminus 1 gt 2 we have

∣∣∣ int Au middot uminusuRspminus1st dz

∣∣∣le c

intBRtimesst

upminus1uminusuRspminus1st dz

+ cintBRtimesst

uqminus1pminus1uminusuRsst dz

le

Rp

intuminusuRs

ppminus1 pst dz+

c

Rp2minusp

intuppminus1 dz+ c

Rp2minusp

intupqminus1 dz

(47)

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Regularity for Certain Nonlinear Parabolic Systems 625

Combining (41) (46) and (47) we have

intBR

uminus uRs2x t p dx

le cu2minusppminus1L

Rp

intBRtimesst

uminus uRs2pst dz+c

Rp2minusp

intBRtimesst

uppminus1 dz

+ c

Rp2minusp

intBRtimesst

upqminus1 dz+intBR

uminus uRs2x sp dx (48)

for small Integrating (48) with respect to s and considering the choice of t weprove Lemma 42 when 1 lt p lt 2

Now we prove a Poincareacute inequality

Theorem 43 Suppose Q2R sub T with R lt 1 then u satisfies the following inequality

intQR

2

uminus uR22 dz le cR2

intQ2R

u2 dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

where c is independent of R

Proof Using Lemma 42 we have

intQR

2

uminus uR22 dz le

intQR

uminus uRs2p dz

le c

Rp

intminus

R

dsintQR

uminus uRs2p dz le c suptisinR

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRt2p dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

le cR2intQ2R

u2 dz+ cRs1

intQ2R

us2z+ cRs1

intQ2R

us3 dz

where we used a Poincareacute type inequality for x variables only that is

intBR

uminus uRs2p dx le cR2intBR

u2 dx

Proof of Theorem 41 Since u is bounded owing to Theorem 32 we have fromTheorem 43

intQR

uminus uRs2 dz lecRn+p+2 when p isin 2

cRn+p+ppminus1 when p isin 1 2

where c is independent of R Hence by isomorphism theorem of Da Prato (1965) weprove Theorem 41

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5 HOumlLDER CONTINUITY FOR u

In this section we prove that u is Houmllder continuous Because of degeneracyof the pde we need scaled cylinders (see DiBenedetto and Friedman 1985)Let R0 gt 0 be fixed and consider a cylinder

SR0

def= BR0times t0 minus R2minus

0 t0

for small For any 0 lt R le R02 we introduce a scaled cylinder

SRdef= BR times t0 minus R22minusp t0

for fixed ge supSR0u

We assume that pminus2 gt R0 if p gt 2 Hence we note that SR sub S

R0and

ge supSRu

The following theorem is our main theorem in this section

Theorem 51 Suppose that SR sub T then u satisfies for all r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR 2 dz+ cR

for some gt 0

For the proof of Theorem 51 we follow a usual program for degenerateequations such that if for some isin 0 1∣∣z isin SR u2 le 1minus 2

∣∣ gt ∣∣SR

∣∣ (51)

then

supS R

2

u le (52)

for some small isin 0 1 and otherwise that is (51) fails then for all 0 lt r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR2 dz (53)

for some gt 0 Combining (52) and (53) we prove Theorem 51First we prove that (52) under the assumption (51) A similar argument for

the case p = q ge 2 appeared in DiBenedetto and Friedman (1985) We assume that(51) is true Set v = u2 For each h gt 0 we set AhRt

def= x isin BR v gt h and

BhR

def= z isin SR v gt h From the mean value theorem we estimate AhRt

Lemma 52 For any isin 0 there exists isin t0 minus 2minuspR20 t0 minus 2minuspR2

0 such that

A1minus2R0 le 1minus

1minus BR0

where is independent of

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Regularity for Certain Nonlinear Parabolic Systems 627

Proof Suppose that the assertion is not true then

B1minus2R0 ge

int t0minus2minuspR20

t0minus2minuspR20

A1minus2R0d gt 1minus BR0

2minuspR20 = 1minus SR0

and this contradicts the assumption (51)

The following lemma is useful in estimating sup u Indeed when p = q ge 2DiBenedetto and Friedman proved a similar theorem (see Lemma 62 inDiBenedetto and Friedman 1985) Now choose = 2 and in Lemma 52We define St

R

def= BR times t

Lemma 53 There exists a positive integer r independent of such that

∣∣A1minus 2r

2R0t

∣∣ le (1minus

(2

)2)BR0

for all t isin t0 minus

22minuspR2

0 t0

Proof Consider the function

v = log+[

2

2minus vminus 1minus 2+ + 2

2r

]

where r is a positive integer This function was used in Choe (1991) and DiBenedettoand Friedman (1985) We take 13 = ui

x2vprime 2 as a test function for (31) where

isin 0 is a standard cutoff function such that B1minusR for small decided later is

contained in support Integrating from to t we obtain

12

intStR

vt2prime 2 dz+

intStR

Ai

Qj

uujxx

uixx

2prime 2 dz

+ 2intStR

Ai

Qj

uujxx

uixukxx

ukx2primeprime 2 dz

+ 2intStR

Ai

Qj

uujxx

uix x

2prime dz = 0

Considering the fact 2primeprime = 21+ prime2 we have

intBR

2vx t 2xdx +intStR

upminus2v21+ prime2 2 dz

leintBR

2vx 2xdx + cintStR

upminus2v prime dz

+ cintStR

uqminus2v prime dz

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628 Bae and Choe

Applying Youngrsquos inequality on the last two terms we getintBR

2vx t 2xdx

leintBR

2vx 2xdx + cintStR

upminus2 2 dz+ cintStR

u2qminuspminus2 2 dz

for some c Since le r log 2 and u is bounded we getintBR

2vx t 2xdx leintBR

2vx 2xdx + crBR

From Lemma 52 we getintBR

2vx t 2xdx le r log 221minus

1minus

2

BR

From definition of we haveintB1minusRcapvgt1minus

2r 22vx tdx ge r minus 1 log 22A1minus

2r 21minusRt

and

A1minus 2r

21minusRt le[( r

r minus 1

)2 1minus

1minus

2

+ c

r

]BR

Hence we obtain

A(1minus

2r

)2R

t le A1minus 2r

21minusRt + BRB1minusR

le[( r

r minus 1

)2 1minus

1minus

2

+ c

r+ n

]BR

Taking r so large that( r

r minus 1

)2 le(1minus

2

)1+

c

rle 3

82

and = 382n we obtain

A1minus 2r

2Rt le(1minus

(2

)2)BR

We take 13 = uixvminus h+ 2 as a test function in (31) where v = u2 and is

a cutoff function such that = 1 in Sr = 0 on pSR le cRminus r and t le cpminus2Rminus r2 Considering the ellipticity condition (12) we have

supt

int ∣∣vminus h+∣∣2 2 dx +

intupminus2vminus h+2 2 dz

le c

Rminus r2

intSR

upminus2vminus h+2 dz

+ c

Rminus r2

intSR

u2qminuspminus2vminus h+2 dz+ cintSR

vminus h+2 tdz (54)

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Regularity for Certain Nonlinear Parabolic Systems 629

for some c Now choose h ge 12 For such a choice of h we haveint

upminus2vminus h+2 2 dz ge cpminus2int

vminus h+2 2 dz

Hence from (54) we obtain

supt

intvminus h+2 2 dx + pminus2

intvminus h+2 2 dz

le[

2qminuspminus2

Rminus r2+ pminus2

Rminus r2

] intSR

vminus h+2 dz (55)

for some c independent of Observe that is bounded above by some fixednumber We make change of variables

= pminus2t minus t0 vx = vx t0 + 2minusp

and we write (55) as

supt

intBR

vminus h+2 2 dx +intSR

vminus h+2 2 dz le c

Rminus r2

intSR

vminus h+dz

Once we have this Caccioppoli type inequality we can apply truncation argumentTherefore following to similar argument in DiBenedetto and Friedman (1984 or1985) we have a maximum principle Now the proof for the next lemma is almostthe same as in DiBenedetto and Friedman (1984 or 1985) we omit the proof

Lemma 54 If for some and isin 0 1∣∣x t isin S1+R u2 lt 1minus 2∣∣ ge S1+R (56)

then there exists a number isin 0 1 independent of and R such that

supS R

2

ux t le

Now we assume that (56) fails that is∣∣x t isin S1+R u2 lt 1minus 2∣∣ lt S1+R (57)

Following a comparison argument similar to Choe (1991) or DiBenedetto andFriedman (1984) we prove that u satisfies a Campanato type integral inequalityLet V isin nN be a constant vector such that 1

2 le V le Taking uxkminus Vk

2 as atest function to (31) we have the following lemma

Lemma 55 There exists a constant c independent of such that

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2tdx +intS R

2

upminus22u2 dz

le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

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630 Bae and Choe

Now suppose that v is a solution to

vit minus

(Ai

Qj

Vvjx

)x

= 0 (58)

with v = u on pSR2 Since 1

2 le V le le c for some fixed c we see that fromthe structure condition of F

2 le 2minuspAi

Qj

Vi

j le c1+ qminusp2

Hence we see that v satisfies a Campanato type integral inequality

Lemma 56 Suppose that 0 lt r le R2 There exists a constant c independent ofVR and such that

intSr

vminus vr 2 dz le c( r

R

)n+4 intS R

2

uminus V 2 dz (59)

Proof We write (58) as

vit minus pminus2

Ai

Qj

V

pminus2vjx

x

= 0

We change of variable = pminus2t minus t0 then the function vx = vx 2minusp+ t0satisfies a uniformly parabolic system

vi minus

Ai

Qj

V

pminus2vjx

x

= 0

Thus from results of Campanato (1966) we prove

intS

vminus v2 dz le c(

R

)n+4 intS R

2

uminus V 2 dz

and scaling back prove (59)

We estimate the L2 norm of uminus v in SR2

Lemma 57 For each lt R2 there exists a constant c independent of such that

intS

uminus v2 dz le c(1+ qminusp 4

n+2)(

minus2 ndashintSR

uminus V 2 dz) int

SRuminus V 2 dz

for = min(12

2n

)

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Regularity for Certain Nonlinear Parabolic Systems 631

Proof Subtracting(Ai

Qj

Vujx

)x

from both sides of (15) we get

uit minus

(Ai

Qj

Vujx

)x= (

Aiuminus Ai

Qj

Vujx

)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(510)

From (58) and (510) we see that uminus v satisfies

ui minus vit minus(Ai

Qj

Vuj minus vjx)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(511)

Simply applying uminus v to (511) we get

pminus2intS R

2uminusv2dzle

intS R

2

∣∣∣ int 1

0Ai

Qj

V+suminus VminusAi

Qj

Vds∣∣∣

times uminusV uminusvdz

and it follows thatintS R

2

uminus v2 dz

le minus2pminus2intS R

2

∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣2

uminus V 2 dz (512)

Note that AiV = F primeV V i

V and

Ai

Qj

V = F primeprimeV ViV

j

V 2 + F primeV (ij

V minus V iV

j

V 3)

where ij and are Kronecker delta functions Also we recall the structurecondition (13) F primeprimeprimeh le chpminus3 + hqminus3 We have that

∣∣∣ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds∣∣∣

le∣∣∣∣int 1

0F primeprime(V + suminus V)

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2

minusF primeprimeV ViV

j

V 2 ds

∣∣∣∣+int 1

0

∣∣∣∣F primeV + suminus VV + suminus V minus F primeV

V ∣∣∣∣ds

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632 Bae and Choe

+∣∣∣∣int 1

0F primeV + suminus V

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V3

minusF primeV ViV

j

V 3 ds

∣∣∣∣= I + II + III

The first term I is estimates as follows

I leint 1

0

∣∣∣F primeprimeV + suminus Vminus F primeprimeV ∣∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2∣∣∣∣∣ ds

+int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

(513)

By the mean value theorem we have that

int 1

0

∣∣∣F primeprimeV+suminusVminusF primeprimeV ∣∣∣ds le c

int 1

0

int 1

0

∣∣F primeprimeprimeV+shuminusV∣∣ds dhuminusV (514)

for some c and

int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

le cF primeprimeV int 1

0

int 1

0

ds dh∣∣V + shuminus V∣∣ uminus V (515)

If V ge 2uminus V then for all s isin 0 1 12 V le V + suminus V le 2V and from

(513) (514) and (515) we have I le c(V pminus3 + V qminus3

)uminus V If V le 2uminus V and p ge 2 then

∣∣F primeprimeV + suminus V∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le cuminus V pminus2 + cuminus V qminus2

and

∣∣F primeprimeV ∣∣ le cV pminus2 + cV qminus2 le cuminus V pminus2 + cuminus V qminus2

Hence we have that

I le cint 1

0

∣∣F primeprimeV + suminus V∣∣+ ∣∣F primeprimeV ∣∣ds le c1+ pminusquminus V pminus2

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Regularity for Certain Nonlinear Parabolic Systems 633

for some c independent of If V le 2uminus V and 1 lt p lt 2 then∣∣F primeprime(V + suminus V)∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le c1+ pminusqV + suminus Vpminus2

for some c We note thatint 1

0V + suminus Vpminus2 ds le uminus V pminus2

int 1

0

∣∣∣∣ V uminus V minus s

∣∣∣∣pminus2

ds le cuminus V pminus2

Therefore combining all these together we have

I le cV pminus3 uminus V (516)

for some c independent of For II and III we observe that

F primeV V le cV pminus2 + cV qminus2

∣∣∣∣FprimeV V i

Vj

V 3∣∣∣∣ le cV pminus2 + cV qminus2

∣∣∣∣(F primeV V

)∣∣∣∣ le cV pminus3 + cV qminus3

∣∣∣∣(F primeV V i

Vj

V 3)∣∣∣∣ le cV pminus3 + cV qminus3

Hence following the same argument for I we conclude that

II + III le cV pminus3 uminus V (517)

for some c independent of Combining (516) and (517) we obtain that∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣ le cV pminus3 uminus V (518)

and considering (518) in (512)intS R

2

uminus V 2 dz le cminus2pminus2V 2pminus3intS R

2

uminus V 4 dz

le cminus2intS R

2

uminus V 4 dz (519)

From Houmllderrsquos inequality we have thatintS R

2

uminus V 4 dz

=intS R

2

uminus V 8n+ 4nminus8n dz

le[

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx] 2

n int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

(520)

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634 Bae and Choe

From Lemma 55 we have

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx le c2 supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2 dx

le cp + 2qminusp

R2

intSR

uminus V 2 dz

for some c independent of Now recalling that 12 le V le we obtain that

int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

le c22minuspnminus2

n

int t0

t0minus 2minuspR24

[ intB R

2

(u + V )2pminus2uminus V 4 dx] nminus2

n

dt

le c4nminuspnminus8

n

int t0

t0minus 2minuspR24

[ intB R

2

∣∣u pminus22 uminus V pminus2

2 V∣∣ 2n

nminus2 dx] nminus2

n

dt

le c4nminuspnminus8

n

intS R

2

upminus2 2u2 dz+ c2nminuspnminus8

n pminus2intSR

uminus V 2 dz (521)

for some c independent of where we have used Sobolevrsquos inequality FromLemma 55 we haveint

S R2

upminus2 2u2 dz le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

Thus combining (519) (520) and (521) we have that

intS R

2uminus V 2 dz le c

(1+ qminusp 4

n+2)[

minus2ndashintSR

uminus V 2 dz] 2

n intSR

uminus V 2 dz

and this completes the proof

Now we prove an iteration lemma which says that if mean oscillation of unear some fixed V is small then u satisfies a Campanato type integral inequality

Lemma 58 Fix 0 lt 0 lt 1 then there exist small positive numbers independentof such that if V0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists V1 isin nN satisfying(12minusradic

0

) le V1 le

(1+radic

0

) (522)

ndashintSR

uminus V12 dz le 0 ndashintSR

uminus V02 dz (523)

ndashintSR

uminus V12 dz le 2

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Regularity for Certain Nonlinear Parabolic Systems 635

Proof Since V0 satisfies the assumption of Lemma 57 we apply Lemma 57 withV = V0 and obtainint

S R2

vminus u2 dz le cintSR

uminus V02 dz

So from the usual comparison argument we haveintSR

uminus uR2 dz le cintSR

vminus vR2 dz+ cintSR

uminus v2 dz

le cn+4intS R

2

uminus u R22 dz+ c

intS R

2

uminus v2 dz

le c(n+4 +

) intSR

uminus V02 dz

where c is independent of Choosing and isin 0 12 small we prove (523)

with V1 = uR Inequality ndashintSR

uminus V12 dz le 2 follows from (523) andassumption ndash

intSR

uminus V02 dz le 2 Now we also estimate

V1 minus V0 =∣∣∣∣∣ ndashintSR

vminus V0 dz

∣∣∣∣le c

[ndashintSR

vminus u2 dz] 1

2

+ c

[ndashintSR

uminus V02 dz] 1

2

le cminus1minusn2(1+

120

)12

Therefore using the same technique of DiBenedetto (1986) Lemma 44 and if wetake very small compared to (522) holds and this completes the proof

Lemma 59 There exist small positive numbers 0 independent of such that ifV0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists a sequence of vectors Vi i = 1 2 3 sub nN satisfying

14 le Vi le 4 ndash

intSiR

uminus Vi2 dz le 2

ndashintSi+1R

uminus Vi+12 dz le 0 ndashintSiR

uminus Vi2 dz

Proof For the proof of the lemma we need to show that 14 le Vi le 4 But this

follows from that

Vi+1 minus Vi2 le c(0 + minusnminus2

)∣∣∣ ndashintSiR

uminus Vi dz∣∣∣

le ci0

i∣∣∣ ndashint

SRuminus V02 dz

∣∣∣ le ci0

i2

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636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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t 200

7

ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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618 Bae and Choe

Now we consider the case p gt n Then from Sobolevrsquos embedding theorem andHoumllderrsquos inequality we obtain that for p gt n

intS

us+2+ p+s2 dz le

int [ intus+2p dx

](sup

x

u p+sp

) p2

dt

le(sup

t

intus+2p dx

) int ( int ∣∣∣(u s+pp

)∣∣∣p dx)12

dt

le c

[1

Rminus p

intSR

(us+2 + us+p + u pp+1minusq +s + us)dz] 3

2

The remaining things can be done in a similar way as before

The restriction of the integrability of u is necessary to start Moser typeiteration In fact if p isin 2nn+ 2 and p le q le p2+ 1 or p2+ 1 le q ltpn+ 3+ 4n+ 4 then r0 le 2 and there is no restriction on the integrability of u

3 L ESTIMATE FOR u

In this section we prove that u is bounded To begin with Moser iterationwe need that u is highly integrable Indeed the higher integrability of u followsfrom Caccioppoli type inequality for the second derivative of u and integration byparts technique Here the boundedness of u is crucial This method has been appliedin proving for the regularity questions for singular parabolic equations (see Choe1991) By approximation we assume that u has second derivative with respect to xHence differentiating (15) with respect to x we get

(uix

)tminus (

Ai

Qj

uujxx

)x= 0 (31)

Lemma 31 Let SR0sub T Suppose that u is bounded then for all s

intS R0

2

us dz le cintSR0

up dz+ cR0 s

for some c

Proof Let be a standard cutoff function such that = 1 for x t isin S = 0 forx t isin pSR 0 le le 1 t le cRminus 2 le cRminus for some c

From integration by parts we have

intBR

us+2k dx =intBR

usuixuixk dx = minus

intBR

u div(usuk

)dx

le csuL

intBR

us2uk dx + csuL

intBR

us+1kminus1 dx

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Regularity for Certain Nonlinear Parabolic Systems 619

and applying Youngrsquos inequality we obtain

intBR

us+2k dx le cintBR

usminus22u2k dx + cintBR

us2 dx

for some c depending on uL and s Integrating with respect to t we obtain

intSR

us+2k dz le cintSR

usminus22u2k dz+ cintSR

us2 dz (32)

for some c depending on uL and s Now we take uixusminuspk as a test function

to (31) and we get

intSR

d

dt

(us+2minuspk)dz+

intSR

usminus22u2k dz

le cintSR

us+2minuspkminus1tdz+ cintSR

us+2qminusp2 dz+ cintSR

us2 dz(33)

for some c Therefore combining (32) and (33) we obtain

intSR

us+2k dz le cintSR

us2 dz+ cintSR

us+2minusptdz

+ cintSR

us+2qminusp2 dz (34)

Assume p ge 2 Since s + 2q minus p ge s + 2minus p we get from (34) that

intS

us+2 dz le c

Rminus 2

intSR

us+2qminusp dz+ cSR

Rminus 2(35)

for some c Since we are assuming 2 le p le q lt p+ 1 2q minus p lt 2 Define si bysi+1 + 2q minus p = si + 2 that is si = s0 + i2+ 2pminus 2q Note that limirarr si = Set s0 = 3pminus 2q then s0 + 2minus p gt 0 Define Ri = R021+ 2minusi i = 0 1 2 Now we set s = si = Ri+1 and R = Ri Hence defining i =

intSRi

usi+2qminusp dz wecan rewrite (35) as

i+1 le cii + ciBR0 (36)

Therefore iterating (36) we conclude that

intS R

2

us dz le cintSR

up dz+ c (37)

for all s and this completes the proof for the case p ge 2Now we consider the case that 1 lt p le 2 Suppose that s + 2minus p le s

+ 2q minus p that is p+ 22 le q lt p+ 1 In this case we prove (37) in the

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620 Bae and Choe

same way for the case p ge 2 Finally we assume that p le q lt p+ 22 that iss + 2minus p ge s + 2q minus p Considering Youngrsquos inequality in (34) we get

intS

us+2 dz le c

Rminus 2

intSR

us+2minusp dz+ cSR

Rminus 2 (38)

Taking = Ri+1 R = Ri and si = s0 + ip with s0 = 2pminus 2 we obtain from (38)

intSRi+1

usi+1+2minusp dz le ci

R2

intSRi

usi+2minusp dz+ ci SRR2

for some c Therefore iterating this we obtain (37)

Once we have shown that u is in a high Ls space we can apply Moser iteration

Theorem 32 We have u isin L Furthermore suppose SR0sub T then u satisfies

supS R0

2

u le c[ndashintSR0

us0+s1 dz] 1

s0+s2 + c (39)

for some large s0 and some c independent of R0 where

s1 = 2q minus p s2 = 2minus nq minus p if q ge p

2 + 1

s1 = 2 s2 = np

2 + 2minus n if q le p

2 + 1

Proof Let be a standard cutoff function We take usuix2 as a test function to

(31) and we obtain

1s+2

int d

dt

(us+2)2 dz+

intAi

Qj

uujxx

(usuixx

+ susminus2uixukxukxx

)2 dz

= minus2int

Ai

Qj

uujxx

usuixx

dz

Considering the ellipticity condition

Qpminus22 le Ai

Qj

Qi

j le c

(Qpminus2 + Qqminus2)2

we see that

supt

intus+22 dx +

int ∣∣(u s+p2

)∣∣2 dzle c

intus+2tdz+ c

int (us+p + us+2qminusp)2 dz

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Regularity for Certain Nonlinear Parabolic Systems 621

for some c Therefore from Sobolevrsquos embedding theorem and Houmllderrsquos inequalitywe prove that

intS

us+p+s+2 2n dx le

int [ intus+22 dx

] 2n[ int

us+p nnminus2

2nnminus2dx

] nminus2n

dt

le[sup

t

intus+22 dx

] 2nint ∣∣∣(u s+p

2 )∣∣∣2 dz

le c

[1

Rminus 2

intSR

(us+2 + us+p + us+2qminusp)dz

]1+ 2n

(310)

We assume that p ge 2 then we see that s + 2q minus p ge s + 2 Hence fromYoungrsquos inequality we can write (310) as

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2qminusp dz+ SRRminus 2

]1+ 2n

(311)

for some c We set si by si+1 + 2q minus p = si + p+ si + 22n that is for

sufficiently large s0 si = is0 +(2minus nq minus p

)i minus 1 where

def= 1+ 2n Define

Ri

def= R021+ 2minusi Now we take s = si = Ri+1 and R = Ri in (311) Define

i

def= ndashintSRi

usi+2qminusp dz then (311) can be written as

i+1 le cii + ci (312)

If s0 gt nq minus pminus 2 then s0 + 2minus nq minus p gt 0 Hence we have that

limi

si = and limi

sii

= s0 + 2minus nq minus p

From Lemma 31 we can take s0 as large as possible Iterating (312) with some larges0 we have

supS R0

2

u le c[ndashintSR0

us0+2qminusp dz] 1

s0+2minusnqminusp + c

and this completes the proof for the case p ge 2Now we assume that 1 lt p le 2 If q ge p2+ 1 then s + 2q minus p ge s + 2 In this

case applying Youngrsquos inequality on the last term of (310) we have

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2qminusp dz+ SRRminus 2

]1+ 2n

for some c Hence following the argument for the case p ge 2 we prove (39) andthis completes the proof for the case q ge p2+ 1

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Now we assume that p2+ 1 gt q ge p gt 1 In this case s + 2 gt s + 2q minus p andwe have from (310)

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2 dz+ SRRminus 2

]1+ 2n

(313)

for some c We set si by sidef= is0 + np2+ 2minus ni minus 1 Hence if s0 gt nminus 2

minus np2 then

limi

si = and limi

sii

= s0 + 2minus n+ np

2

We take s = si = Ri+1 and R = Ri in (313) Iterating (313) we obtain

supS R0

2

u le c

[ndashintSR0

us0+2 dz

] 1s0+2+ np

2 minusn + c

for some c independent of R0 and this completes the proof for the casep2+ 1 gt q ge p gt 1

4 HOumlLDER CONTINUITY OF u

In this section we prove that u is Houmllder continuous

Theorem 41 Define r0 by the constant appearing in Theorem 21 Suppose that uis L

r0locT then u is Houmllder continuous Moreover for all Q2R sub T with R lt 1 u

satisfies

ndashintQR

uminus uR2 dz le cRs

for some c independent of R where we define

s =2 when p ge 2

ppminus 1 when p isin 1 2

We introduce a nonnegative cutoff function isin 0 BR such that = 1 in

BR2 0 le le 1 le cR We define minus

R

def= t0 minus 2Rp t0 minus Rp and QminusR = BR timesminus

R

Also we define uRt

def= 1BRintBR

ux tdx and uR

def= 1QRintQR

ux tdz Firstwe prove a lemma which is essential for a Poincareacute inequality for solutions of adegenerate parabolic system

Lemma 42 Suppose Q2R sub T then u satisfies the following inequality

suptisinR

intminus

R

dsintBR

2ux tminus uRs2 dx

le cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz+ cintQminus

R

uminus uRt2 dz

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Regularity for Certain Nonlinear Parabolic Systems 623

for all R lt R0 where c depends only on nN and p and

s1 = p s2 = p and s3 = pqminus1

pminus1 when p ge 2

s1 = p2minus p s2 = ppminus 1 and s3 = pq minus 1 when 1 lt p lt 2

Proof Since u isin 0(0 T L2

) there exists ts isin R for all s isin minus

R such that

intBR

pux t minus uRs2dx = suptisinR

intBR

pux tminus uRs2 dx

We take uminus uRspxst as a test function to (15) where st is the characteristic

function such that

st =1 for all isin s t

0 for all isin s t

Hence we have

intut middot uminus uRs

pxst dz+int

Au middot uminus uRspst dz = 0 (41)

Now we assume p isin 2 Considering the structure condition (12) on A (23)and (25) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

leint

uminus uRs2x sp dx + cintBRtimesst

upminus1uminus uRspminus1dz

+ cintBRtimesst

uqminus1uminus uRspminus1dz (42)

for some c independent of R From Youngrsquos inequality we have

intBRtimesst

upminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

up dz (43)

for some c and small Similarly the last term in (42) can be estimated as follows

intBRtimesst

uqminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

u pqminus1pminus1 dz (44)

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for some c and small Combining (42) (43) and (44) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

le cint

uminus uRs2x sp dx +

Rp

intBRtimesst

uminus uRspp dz

+ cintBRtimesst

up dz+ cintBRtimesst

u pqminus1pminus1 dz (45)

for some c and small Integrating (45) with respect to s from t0 minus 2Rp to t0 minus Rpwe have

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRs2p dz+ c

Rp

intminus

R

dsintBRtimesst

uminus uRs2p dz

+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

By the choice of t we have that for small

intminus

R

dsintBR

uminusuRs2x t pdx

le cintQminus

R

uminusuRs2p dz+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

So we proved Lemma 42 when p ge 2In case p isin 1 2 we estimate the second term of (41) as follows

intuppst dz+ p

intAu middot uminus uRs

pminus1st dz

leint

Au middot uminus uRspst dz (46)

and using the fact that u is bounded and ppminus 1 gt 2 we have

∣∣∣ int Au middot uminusuRspminus1st dz

∣∣∣le c

intBRtimesst

upminus1uminusuRspminus1st dz

+ cintBRtimesst

uqminus1pminus1uminusuRsst dz

le

Rp

intuminusuRs

ppminus1 pst dz+

c

Rp2minusp

intuppminus1 dz+ c

Rp2minusp

intupqminus1 dz

(47)

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Regularity for Certain Nonlinear Parabolic Systems 625

Combining (41) (46) and (47) we have

intBR

uminus uRs2x t p dx

le cu2minusppminus1L

Rp

intBRtimesst

uminus uRs2pst dz+c

Rp2minusp

intBRtimesst

uppminus1 dz

+ c

Rp2minusp

intBRtimesst

upqminus1 dz+intBR

uminus uRs2x sp dx (48)

for small Integrating (48) with respect to s and considering the choice of t weprove Lemma 42 when 1 lt p lt 2

Now we prove a Poincareacute inequality

Theorem 43 Suppose Q2R sub T with R lt 1 then u satisfies the following inequality

intQR

2

uminus uR22 dz le cR2

intQ2R

u2 dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

where c is independent of R

Proof Using Lemma 42 we have

intQR

2

uminus uR22 dz le

intQR

uminus uRs2p dz

le c

Rp

intminus

R

dsintQR

uminus uRs2p dz le c suptisinR

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRt2p dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

le cR2intQ2R

u2 dz+ cRs1

intQ2R

us2z+ cRs1

intQ2R

us3 dz

where we used a Poincareacute type inequality for x variables only that is

intBR

uminus uRs2p dx le cR2intBR

u2 dx

Proof of Theorem 41 Since u is bounded owing to Theorem 32 we have fromTheorem 43

intQR

uminus uRs2 dz lecRn+p+2 when p isin 2

cRn+p+ppminus1 when p isin 1 2

where c is independent of R Hence by isomorphism theorem of Da Prato (1965) weprove Theorem 41

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5 HOumlLDER CONTINUITY FOR u

In this section we prove that u is Houmllder continuous Because of degeneracyof the pde we need scaled cylinders (see DiBenedetto and Friedman 1985)Let R0 gt 0 be fixed and consider a cylinder

SR0

def= BR0times t0 minus R2minus

0 t0

for small For any 0 lt R le R02 we introduce a scaled cylinder

SRdef= BR times t0 minus R22minusp t0

for fixed ge supSR0u

We assume that pminus2 gt R0 if p gt 2 Hence we note that SR sub S

R0and

ge supSRu

The following theorem is our main theorem in this section

Theorem 51 Suppose that SR sub T then u satisfies for all r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR 2 dz+ cR

for some gt 0

For the proof of Theorem 51 we follow a usual program for degenerateequations such that if for some isin 0 1∣∣z isin SR u2 le 1minus 2

∣∣ gt ∣∣SR

∣∣ (51)

then

supS R

2

u le (52)

for some small isin 0 1 and otherwise that is (51) fails then for all 0 lt r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR2 dz (53)

for some gt 0 Combining (52) and (53) we prove Theorem 51First we prove that (52) under the assumption (51) A similar argument for

the case p = q ge 2 appeared in DiBenedetto and Friedman (1985) We assume that(51) is true Set v = u2 For each h gt 0 we set AhRt

def= x isin BR v gt h and

BhR

def= z isin SR v gt h From the mean value theorem we estimate AhRt

Lemma 52 For any isin 0 there exists isin t0 minus 2minuspR20 t0 minus 2minuspR2

0 such that

A1minus2R0 le 1minus

1minus BR0

where is independent of

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Regularity for Certain Nonlinear Parabolic Systems 627

Proof Suppose that the assertion is not true then

B1minus2R0 ge

int t0minus2minuspR20

t0minus2minuspR20

A1minus2R0d gt 1minus BR0

2minuspR20 = 1minus SR0

and this contradicts the assumption (51)

The following lemma is useful in estimating sup u Indeed when p = q ge 2DiBenedetto and Friedman proved a similar theorem (see Lemma 62 inDiBenedetto and Friedman 1985) Now choose = 2 and in Lemma 52We define St

R

def= BR times t

Lemma 53 There exists a positive integer r independent of such that

∣∣A1minus 2r

2R0t

∣∣ le (1minus

(2

)2)BR0

for all t isin t0 minus

22minuspR2

0 t0

Proof Consider the function

v = log+[

2

2minus vminus 1minus 2+ + 2

2r

]

where r is a positive integer This function was used in Choe (1991) and DiBenedettoand Friedman (1985) We take 13 = ui

x2vprime 2 as a test function for (31) where

isin 0 is a standard cutoff function such that B1minusR for small decided later is

contained in support Integrating from to t we obtain

12

intStR

vt2prime 2 dz+

intStR

Ai

Qj

uujxx

uixx

2prime 2 dz

+ 2intStR

Ai

Qj

uujxx

uixukxx

ukx2primeprime 2 dz

+ 2intStR

Ai

Qj

uujxx

uix x

2prime dz = 0

Considering the fact 2primeprime = 21+ prime2 we have

intBR

2vx t 2xdx +intStR

upminus2v21+ prime2 2 dz

leintBR

2vx 2xdx + cintStR

upminus2v prime dz

+ cintStR

uqminus2v prime dz

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628 Bae and Choe

Applying Youngrsquos inequality on the last two terms we getintBR

2vx t 2xdx

leintBR

2vx 2xdx + cintStR

upminus2 2 dz+ cintStR

u2qminuspminus2 2 dz

for some c Since le r log 2 and u is bounded we getintBR

2vx t 2xdx leintBR

2vx 2xdx + crBR

From Lemma 52 we getintBR

2vx t 2xdx le r log 221minus

1minus

2

BR

From definition of we haveintB1minusRcapvgt1minus

2r 22vx tdx ge r minus 1 log 22A1minus

2r 21minusRt

and

A1minus 2r

21minusRt le[( r

r minus 1

)2 1minus

1minus

2

+ c

r

]BR

Hence we obtain

A(1minus

2r

)2R

t le A1minus 2r

21minusRt + BRB1minusR

le[( r

r minus 1

)2 1minus

1minus

2

+ c

r+ n

]BR

Taking r so large that( r

r minus 1

)2 le(1minus

2

)1+

c

rle 3

82

and = 382n we obtain

A1minus 2r

2Rt le(1minus

(2

)2)BR

We take 13 = uixvminus h+ 2 as a test function in (31) where v = u2 and is

a cutoff function such that = 1 in Sr = 0 on pSR le cRminus r and t le cpminus2Rminus r2 Considering the ellipticity condition (12) we have

supt

int ∣∣vminus h+∣∣2 2 dx +

intupminus2vminus h+2 2 dz

le c

Rminus r2

intSR

upminus2vminus h+2 dz

+ c

Rminus r2

intSR

u2qminuspminus2vminus h+2 dz+ cintSR

vminus h+2 tdz (54)

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Regularity for Certain Nonlinear Parabolic Systems 629

for some c Now choose h ge 12 For such a choice of h we haveint

upminus2vminus h+2 2 dz ge cpminus2int

vminus h+2 2 dz

Hence from (54) we obtain

supt

intvminus h+2 2 dx + pminus2

intvminus h+2 2 dz

le[

2qminuspminus2

Rminus r2+ pminus2

Rminus r2

] intSR

vminus h+2 dz (55)

for some c independent of Observe that is bounded above by some fixednumber We make change of variables

= pminus2t minus t0 vx = vx t0 + 2minusp

and we write (55) as

supt

intBR

vminus h+2 2 dx +intSR

vminus h+2 2 dz le c

Rminus r2

intSR

vminus h+dz

Once we have this Caccioppoli type inequality we can apply truncation argumentTherefore following to similar argument in DiBenedetto and Friedman (1984 or1985) we have a maximum principle Now the proof for the next lemma is almostthe same as in DiBenedetto and Friedman (1984 or 1985) we omit the proof

Lemma 54 If for some and isin 0 1∣∣x t isin S1+R u2 lt 1minus 2∣∣ ge S1+R (56)

then there exists a number isin 0 1 independent of and R such that

supS R

2

ux t le

Now we assume that (56) fails that is∣∣x t isin S1+R u2 lt 1minus 2∣∣ lt S1+R (57)

Following a comparison argument similar to Choe (1991) or DiBenedetto andFriedman (1984) we prove that u satisfies a Campanato type integral inequalityLet V isin nN be a constant vector such that 1

2 le V le Taking uxkminus Vk

2 as atest function to (31) we have the following lemma

Lemma 55 There exists a constant c independent of such that

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2tdx +intS R

2

upminus22u2 dz

le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

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Now suppose that v is a solution to

vit minus

(Ai

Qj

Vvjx

)x

= 0 (58)

with v = u on pSR2 Since 1

2 le V le le c for some fixed c we see that fromthe structure condition of F

2 le 2minuspAi

Qj

Vi

j le c1+ qminusp2

Hence we see that v satisfies a Campanato type integral inequality

Lemma 56 Suppose that 0 lt r le R2 There exists a constant c independent ofVR and such that

intSr

vminus vr 2 dz le c( r

R

)n+4 intS R

2

uminus V 2 dz (59)

Proof We write (58) as

vit minus pminus2

Ai

Qj

V

pminus2vjx

x

= 0

We change of variable = pminus2t minus t0 then the function vx = vx 2minusp+ t0satisfies a uniformly parabolic system

vi minus

Ai

Qj

V

pminus2vjx

x

= 0

Thus from results of Campanato (1966) we prove

intS

vminus v2 dz le c(

R

)n+4 intS R

2

uminus V 2 dz

and scaling back prove (59)

We estimate the L2 norm of uminus v in SR2

Lemma 57 For each lt R2 there exists a constant c independent of such that

intS

uminus v2 dz le c(1+ qminusp 4

n+2)(

minus2 ndashintSR

uminus V 2 dz) int

SRuminus V 2 dz

for = min(12

2n

)

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Regularity for Certain Nonlinear Parabolic Systems 631

Proof Subtracting(Ai

Qj

Vujx

)x

from both sides of (15) we get

uit minus

(Ai

Qj

Vujx

)x= (

Aiuminus Ai

Qj

Vujx

)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(510)

From (58) and (510) we see that uminus v satisfies

ui minus vit minus(Ai

Qj

Vuj minus vjx)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(511)

Simply applying uminus v to (511) we get

pminus2intS R

2uminusv2dzle

intS R

2

∣∣∣ int 1

0Ai

Qj

V+suminus VminusAi

Qj

Vds∣∣∣

times uminusV uminusvdz

and it follows thatintS R

2

uminus v2 dz

le minus2pminus2intS R

2

∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣2

uminus V 2 dz (512)

Note that AiV = F primeV V i

V and

Ai

Qj

V = F primeprimeV ViV

j

V 2 + F primeV (ij

V minus V iV

j

V 3)

where ij and are Kronecker delta functions Also we recall the structurecondition (13) F primeprimeprimeh le chpminus3 + hqminus3 We have that

∣∣∣ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds∣∣∣

le∣∣∣∣int 1

0F primeprime(V + suminus V)

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2

minusF primeprimeV ViV

j

V 2 ds

∣∣∣∣+int 1

0

∣∣∣∣F primeV + suminus VV + suminus V minus F primeV

V ∣∣∣∣ds

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+∣∣∣∣int 1

0F primeV + suminus V

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V3

minusF primeV ViV

j

V 3 ds

∣∣∣∣= I + II + III

The first term I is estimates as follows

I leint 1

0

∣∣∣F primeprimeV + suminus Vminus F primeprimeV ∣∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2∣∣∣∣∣ ds

+int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

(513)

By the mean value theorem we have that

int 1

0

∣∣∣F primeprimeV+suminusVminusF primeprimeV ∣∣∣ds le c

int 1

0

int 1

0

∣∣F primeprimeprimeV+shuminusV∣∣ds dhuminusV (514)

for some c and

int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

le cF primeprimeV int 1

0

int 1

0

ds dh∣∣V + shuminus V∣∣ uminus V (515)

If V ge 2uminus V then for all s isin 0 1 12 V le V + suminus V le 2V and from

(513) (514) and (515) we have I le c(V pminus3 + V qminus3

)uminus V If V le 2uminus V and p ge 2 then

∣∣F primeprimeV + suminus V∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le cuminus V pminus2 + cuminus V qminus2

and

∣∣F primeprimeV ∣∣ le cV pminus2 + cV qminus2 le cuminus V pminus2 + cuminus V qminus2

Hence we have that

I le cint 1

0

∣∣F primeprimeV + suminus V∣∣+ ∣∣F primeprimeV ∣∣ds le c1+ pminusquminus V pminus2

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Regularity for Certain Nonlinear Parabolic Systems 633

for some c independent of If V le 2uminus V and 1 lt p lt 2 then∣∣F primeprime(V + suminus V)∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le c1+ pminusqV + suminus Vpminus2

for some c We note thatint 1

0V + suminus Vpminus2 ds le uminus V pminus2

int 1

0

∣∣∣∣ V uminus V minus s

∣∣∣∣pminus2

ds le cuminus V pminus2

Therefore combining all these together we have

I le cV pminus3 uminus V (516)

for some c independent of For II and III we observe that

F primeV V le cV pminus2 + cV qminus2

∣∣∣∣FprimeV V i

Vj

V 3∣∣∣∣ le cV pminus2 + cV qminus2

∣∣∣∣(F primeV V

)∣∣∣∣ le cV pminus3 + cV qminus3

∣∣∣∣(F primeV V i

Vj

V 3)∣∣∣∣ le cV pminus3 + cV qminus3

Hence following the same argument for I we conclude that

II + III le cV pminus3 uminus V (517)

for some c independent of Combining (516) and (517) we obtain that∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣ le cV pminus3 uminus V (518)

and considering (518) in (512)intS R

2

uminus V 2 dz le cminus2pminus2V 2pminus3intS R

2

uminus V 4 dz

le cminus2intS R

2

uminus V 4 dz (519)

From Houmllderrsquos inequality we have thatintS R

2

uminus V 4 dz

=intS R

2

uminus V 8n+ 4nminus8n dz

le[

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx] 2

n int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

(520)

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634 Bae and Choe

From Lemma 55 we have

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx le c2 supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2 dx

le cp + 2qminusp

R2

intSR

uminus V 2 dz

for some c independent of Now recalling that 12 le V le we obtain that

int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

le c22minuspnminus2

n

int t0

t0minus 2minuspR24

[ intB R

2

(u + V )2pminus2uminus V 4 dx] nminus2

n

dt

le c4nminuspnminus8

n

int t0

t0minus 2minuspR24

[ intB R

2

∣∣u pminus22 uminus V pminus2

2 V∣∣ 2n

nminus2 dx] nminus2

n

dt

le c4nminuspnminus8

n

intS R

2

upminus2 2u2 dz+ c2nminuspnminus8

n pminus2intSR

uminus V 2 dz (521)

for some c independent of where we have used Sobolevrsquos inequality FromLemma 55 we haveint

S R2

upminus2 2u2 dz le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

Thus combining (519) (520) and (521) we have that

intS R

2uminus V 2 dz le c

(1+ qminusp 4

n+2)[

minus2ndashintSR

uminus V 2 dz] 2

n intSR

uminus V 2 dz

and this completes the proof

Now we prove an iteration lemma which says that if mean oscillation of unear some fixed V is small then u satisfies a Campanato type integral inequality

Lemma 58 Fix 0 lt 0 lt 1 then there exist small positive numbers independentof such that if V0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists V1 isin nN satisfying(12minusradic

0

) le V1 le

(1+radic

0

) (522)

ndashintSR

uminus V12 dz le 0 ndashintSR

uminus V02 dz (523)

ndashintSR

uminus V12 dz le 2

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Regularity for Certain Nonlinear Parabolic Systems 635

Proof Since V0 satisfies the assumption of Lemma 57 we apply Lemma 57 withV = V0 and obtainint

S R2

vminus u2 dz le cintSR

uminus V02 dz

So from the usual comparison argument we haveintSR

uminus uR2 dz le cintSR

vminus vR2 dz+ cintSR

uminus v2 dz

le cn+4intS R

2

uminus u R22 dz+ c

intS R

2

uminus v2 dz

le c(n+4 +

) intSR

uminus V02 dz

where c is independent of Choosing and isin 0 12 small we prove (523)

with V1 = uR Inequality ndashintSR

uminus V12 dz le 2 follows from (523) andassumption ndash

intSR

uminus V02 dz le 2 Now we also estimate

V1 minus V0 =∣∣∣∣∣ ndashintSR

vminus V0 dz

∣∣∣∣le c

[ndashintSR

vminus u2 dz] 1

2

+ c

[ndashintSR

uminus V02 dz] 1

2

le cminus1minusn2(1+

120

)12

Therefore using the same technique of DiBenedetto (1986) Lemma 44 and if wetake very small compared to (522) holds and this completes the proof

Lemma 59 There exist small positive numbers 0 independent of such that ifV0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists a sequence of vectors Vi i = 1 2 3 sub nN satisfying

14 le Vi le 4 ndash

intSiR

uminus Vi2 dz le 2

ndashintSi+1R

uminus Vi+12 dz le 0 ndashintSiR

uminus Vi2 dz

Proof For the proof of the lemma we need to show that 14 le Vi le 4 But this

follows from that

Vi+1 minus Vi2 le c(0 + minusnminus2

)∣∣∣ ndashintSiR

uminus Vi dz∣∣∣

le ci0

i∣∣∣ ndashint

SRuminus V02 dz

∣∣∣ le ci0

i2

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636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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Regularity for Certain Nonlinear Parabolic Systems 619

and applying Youngrsquos inequality we obtain

intBR

us+2k dx le cintBR

usminus22u2k dx + cintBR

us2 dx

for some c depending on uL and s Integrating with respect to t we obtain

intSR

us+2k dz le cintSR

usminus22u2k dz+ cintSR

us2 dz (32)

for some c depending on uL and s Now we take uixusminuspk as a test function

to (31) and we get

intSR

d

dt

(us+2minuspk)dz+

intSR

usminus22u2k dz

le cintSR

us+2minuspkminus1tdz+ cintSR

us+2qminusp2 dz+ cintSR

us2 dz(33)

for some c Therefore combining (32) and (33) we obtain

intSR

us+2k dz le cintSR

us2 dz+ cintSR

us+2minusptdz

+ cintSR

us+2qminusp2 dz (34)

Assume p ge 2 Since s + 2q minus p ge s + 2minus p we get from (34) that

intS

us+2 dz le c

Rminus 2

intSR

us+2qminusp dz+ cSR

Rminus 2(35)

for some c Since we are assuming 2 le p le q lt p+ 1 2q minus p lt 2 Define si bysi+1 + 2q minus p = si + 2 that is si = s0 + i2+ 2pminus 2q Note that limirarr si = Set s0 = 3pminus 2q then s0 + 2minus p gt 0 Define Ri = R021+ 2minusi i = 0 1 2 Now we set s = si = Ri+1 and R = Ri Hence defining i =

intSRi

usi+2qminusp dz wecan rewrite (35) as

i+1 le cii + ciBR0 (36)

Therefore iterating (36) we conclude that

intS R

2

us dz le cintSR

up dz+ c (37)

for all s and this completes the proof for the case p ge 2Now we consider the case that 1 lt p le 2 Suppose that s + 2minus p le s

+ 2q minus p that is p+ 22 le q lt p+ 1 In this case we prove (37) in the

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620 Bae and Choe

same way for the case p ge 2 Finally we assume that p le q lt p+ 22 that iss + 2minus p ge s + 2q minus p Considering Youngrsquos inequality in (34) we get

intS

us+2 dz le c

Rminus 2

intSR

us+2minusp dz+ cSR

Rminus 2 (38)

Taking = Ri+1 R = Ri and si = s0 + ip with s0 = 2pminus 2 we obtain from (38)

intSRi+1

usi+1+2minusp dz le ci

R2

intSRi

usi+2minusp dz+ ci SRR2

for some c Therefore iterating this we obtain (37)

Once we have shown that u is in a high Ls space we can apply Moser iteration

Theorem 32 We have u isin L Furthermore suppose SR0sub T then u satisfies

supS R0

2

u le c[ndashintSR0

us0+s1 dz] 1

s0+s2 + c (39)

for some large s0 and some c independent of R0 where

s1 = 2q minus p s2 = 2minus nq minus p if q ge p

2 + 1

s1 = 2 s2 = np

2 + 2minus n if q le p

2 + 1

Proof Let be a standard cutoff function We take usuix2 as a test function to

(31) and we obtain

1s+2

int d

dt

(us+2)2 dz+

intAi

Qj

uujxx

(usuixx

+ susminus2uixukxukxx

)2 dz

= minus2int

Ai

Qj

uujxx

usuixx

dz

Considering the ellipticity condition

Qpminus22 le Ai

Qj

Qi

j le c

(Qpminus2 + Qqminus2)2

we see that

supt

intus+22 dx +

int ∣∣(u s+p2

)∣∣2 dzle c

intus+2tdz+ c

int (us+p + us+2qminusp)2 dz

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Regularity for Certain Nonlinear Parabolic Systems 621

for some c Therefore from Sobolevrsquos embedding theorem and Houmllderrsquos inequalitywe prove that

intS

us+p+s+2 2n dx le

int [ intus+22 dx

] 2n[ int

us+p nnminus2

2nnminus2dx

] nminus2n

dt

le[sup

t

intus+22 dx

] 2nint ∣∣∣(u s+p

2 )∣∣∣2 dz

le c

[1

Rminus 2

intSR

(us+2 + us+p + us+2qminusp)dz

]1+ 2n

(310)

We assume that p ge 2 then we see that s + 2q minus p ge s + 2 Hence fromYoungrsquos inequality we can write (310) as

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2qminusp dz+ SRRminus 2

]1+ 2n

(311)

for some c We set si by si+1 + 2q minus p = si + p+ si + 22n that is for

sufficiently large s0 si = is0 +(2minus nq minus p

)i minus 1 where

def= 1+ 2n Define

Ri

def= R021+ 2minusi Now we take s = si = Ri+1 and R = Ri in (311) Define

i

def= ndashintSRi

usi+2qminusp dz then (311) can be written as

i+1 le cii + ci (312)

If s0 gt nq minus pminus 2 then s0 + 2minus nq minus p gt 0 Hence we have that

limi

si = and limi

sii

= s0 + 2minus nq minus p

From Lemma 31 we can take s0 as large as possible Iterating (312) with some larges0 we have

supS R0

2

u le c[ndashintSR0

us0+2qminusp dz] 1

s0+2minusnqminusp + c

and this completes the proof for the case p ge 2Now we assume that 1 lt p le 2 If q ge p2+ 1 then s + 2q minus p ge s + 2 In this

case applying Youngrsquos inequality on the last term of (310) we have

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2qminusp dz+ SRRminus 2

]1+ 2n

for some c Hence following the argument for the case p ge 2 we prove (39) andthis completes the proof for the case q ge p2+ 1

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622 Bae and Choe

Now we assume that p2+ 1 gt q ge p gt 1 In this case s + 2 gt s + 2q minus p andwe have from (310)

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2 dz+ SRRminus 2

]1+ 2n

(313)

for some c We set si by sidef= is0 + np2+ 2minus ni minus 1 Hence if s0 gt nminus 2

minus np2 then

limi

si = and limi

sii

= s0 + 2minus n+ np

2

We take s = si = Ri+1 and R = Ri in (313) Iterating (313) we obtain

supS R0

2

u le c

[ndashintSR0

us0+2 dz

] 1s0+2+ np

2 minusn + c

for some c independent of R0 and this completes the proof for the casep2+ 1 gt q ge p gt 1

4 HOumlLDER CONTINUITY OF u

In this section we prove that u is Houmllder continuous

Theorem 41 Define r0 by the constant appearing in Theorem 21 Suppose that uis L

r0locT then u is Houmllder continuous Moreover for all Q2R sub T with R lt 1 u

satisfies

ndashintQR

uminus uR2 dz le cRs

for some c independent of R where we define

s =2 when p ge 2

ppminus 1 when p isin 1 2

We introduce a nonnegative cutoff function isin 0 BR such that = 1 in

BR2 0 le le 1 le cR We define minus

R

def= t0 minus 2Rp t0 minus Rp and QminusR = BR timesminus

R

Also we define uRt

def= 1BRintBR

ux tdx and uR

def= 1QRintQR

ux tdz Firstwe prove a lemma which is essential for a Poincareacute inequality for solutions of adegenerate parabolic system

Lemma 42 Suppose Q2R sub T then u satisfies the following inequality

suptisinR

intminus

R

dsintBR

2ux tminus uRs2 dx

le cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz+ cintQminus

R

uminus uRt2 dz

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Regularity for Certain Nonlinear Parabolic Systems 623

for all R lt R0 where c depends only on nN and p and

s1 = p s2 = p and s3 = pqminus1

pminus1 when p ge 2

s1 = p2minus p s2 = ppminus 1 and s3 = pq minus 1 when 1 lt p lt 2

Proof Since u isin 0(0 T L2

) there exists ts isin R for all s isin minus

R such that

intBR

pux t minus uRs2dx = suptisinR

intBR

pux tminus uRs2 dx

We take uminus uRspxst as a test function to (15) where st is the characteristic

function such that

st =1 for all isin s t

0 for all isin s t

Hence we have

intut middot uminus uRs

pxst dz+int

Au middot uminus uRspst dz = 0 (41)

Now we assume p isin 2 Considering the structure condition (12) on A (23)and (25) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

leint

uminus uRs2x sp dx + cintBRtimesst

upminus1uminus uRspminus1dz

+ cintBRtimesst

uqminus1uminus uRspminus1dz (42)

for some c independent of R From Youngrsquos inequality we have

intBRtimesst

upminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

up dz (43)

for some c and small Similarly the last term in (42) can be estimated as follows

intBRtimesst

uqminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

u pqminus1pminus1 dz (44)

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624 Bae and Choe

for some c and small Combining (42) (43) and (44) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

le cint

uminus uRs2x sp dx +

Rp

intBRtimesst

uminus uRspp dz

+ cintBRtimesst

up dz+ cintBRtimesst

u pqminus1pminus1 dz (45)

for some c and small Integrating (45) with respect to s from t0 minus 2Rp to t0 minus Rpwe have

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRs2p dz+ c

Rp

intminus

R

dsintBRtimesst

uminus uRs2p dz

+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

By the choice of t we have that for small

intminus

R

dsintBR

uminusuRs2x t pdx

le cintQminus

R

uminusuRs2p dz+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

So we proved Lemma 42 when p ge 2In case p isin 1 2 we estimate the second term of (41) as follows

intuppst dz+ p

intAu middot uminus uRs

pminus1st dz

leint

Au middot uminus uRspst dz (46)

and using the fact that u is bounded and ppminus 1 gt 2 we have

∣∣∣ int Au middot uminusuRspminus1st dz

∣∣∣le c

intBRtimesst

upminus1uminusuRspminus1st dz

+ cintBRtimesst

uqminus1pminus1uminusuRsst dz

le

Rp

intuminusuRs

ppminus1 pst dz+

c

Rp2minusp

intuppminus1 dz+ c

Rp2minusp

intupqminus1 dz

(47)

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Regularity for Certain Nonlinear Parabolic Systems 625

Combining (41) (46) and (47) we have

intBR

uminus uRs2x t p dx

le cu2minusppminus1L

Rp

intBRtimesst

uminus uRs2pst dz+c

Rp2minusp

intBRtimesst

uppminus1 dz

+ c

Rp2minusp

intBRtimesst

upqminus1 dz+intBR

uminus uRs2x sp dx (48)

for small Integrating (48) with respect to s and considering the choice of t weprove Lemma 42 when 1 lt p lt 2

Now we prove a Poincareacute inequality

Theorem 43 Suppose Q2R sub T with R lt 1 then u satisfies the following inequality

intQR

2

uminus uR22 dz le cR2

intQ2R

u2 dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

where c is independent of R

Proof Using Lemma 42 we have

intQR

2

uminus uR22 dz le

intQR

uminus uRs2p dz

le c

Rp

intminus

R

dsintQR

uminus uRs2p dz le c suptisinR

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRt2p dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

le cR2intQ2R

u2 dz+ cRs1

intQ2R

us2z+ cRs1

intQ2R

us3 dz

where we used a Poincareacute type inequality for x variables only that is

intBR

uminus uRs2p dx le cR2intBR

u2 dx

Proof of Theorem 41 Since u is bounded owing to Theorem 32 we have fromTheorem 43

intQR

uminus uRs2 dz lecRn+p+2 when p isin 2

cRn+p+ppminus1 when p isin 1 2

where c is independent of R Hence by isomorphism theorem of Da Prato (1965) weprove Theorem 41

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626 Bae and Choe

5 HOumlLDER CONTINUITY FOR u

In this section we prove that u is Houmllder continuous Because of degeneracyof the pde we need scaled cylinders (see DiBenedetto and Friedman 1985)Let R0 gt 0 be fixed and consider a cylinder

SR0

def= BR0times t0 minus R2minus

0 t0

for small For any 0 lt R le R02 we introduce a scaled cylinder

SRdef= BR times t0 minus R22minusp t0

for fixed ge supSR0u

We assume that pminus2 gt R0 if p gt 2 Hence we note that SR sub S

R0and

ge supSRu

The following theorem is our main theorem in this section

Theorem 51 Suppose that SR sub T then u satisfies for all r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR 2 dz+ cR

for some gt 0

For the proof of Theorem 51 we follow a usual program for degenerateequations such that if for some isin 0 1∣∣z isin SR u2 le 1minus 2

∣∣ gt ∣∣SR

∣∣ (51)

then

supS R

2

u le (52)

for some small isin 0 1 and otherwise that is (51) fails then for all 0 lt r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR2 dz (53)

for some gt 0 Combining (52) and (53) we prove Theorem 51First we prove that (52) under the assumption (51) A similar argument for

the case p = q ge 2 appeared in DiBenedetto and Friedman (1985) We assume that(51) is true Set v = u2 For each h gt 0 we set AhRt

def= x isin BR v gt h and

BhR

def= z isin SR v gt h From the mean value theorem we estimate AhRt

Lemma 52 For any isin 0 there exists isin t0 minus 2minuspR20 t0 minus 2minuspR2

0 such that

A1minus2R0 le 1minus

1minus BR0

where is independent of

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Regularity for Certain Nonlinear Parabolic Systems 627

Proof Suppose that the assertion is not true then

B1minus2R0 ge

int t0minus2minuspR20

t0minus2minuspR20

A1minus2R0d gt 1minus BR0

2minuspR20 = 1minus SR0

and this contradicts the assumption (51)

The following lemma is useful in estimating sup u Indeed when p = q ge 2DiBenedetto and Friedman proved a similar theorem (see Lemma 62 inDiBenedetto and Friedman 1985) Now choose = 2 and in Lemma 52We define St

R

def= BR times t

Lemma 53 There exists a positive integer r independent of such that

∣∣A1minus 2r

2R0t

∣∣ le (1minus

(2

)2)BR0

for all t isin t0 minus

22minuspR2

0 t0

Proof Consider the function

v = log+[

2

2minus vminus 1minus 2+ + 2

2r

]

where r is a positive integer This function was used in Choe (1991) and DiBenedettoand Friedman (1985) We take 13 = ui

x2vprime 2 as a test function for (31) where

isin 0 is a standard cutoff function such that B1minusR for small decided later is

contained in support Integrating from to t we obtain

12

intStR

vt2prime 2 dz+

intStR

Ai

Qj

uujxx

uixx

2prime 2 dz

+ 2intStR

Ai

Qj

uujxx

uixukxx

ukx2primeprime 2 dz

+ 2intStR

Ai

Qj

uujxx

uix x

2prime dz = 0

Considering the fact 2primeprime = 21+ prime2 we have

intBR

2vx t 2xdx +intStR

upminus2v21+ prime2 2 dz

leintBR

2vx 2xdx + cintStR

upminus2v prime dz

+ cintStR

uqminus2v prime dz

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628 Bae and Choe

Applying Youngrsquos inequality on the last two terms we getintBR

2vx t 2xdx

leintBR

2vx 2xdx + cintStR

upminus2 2 dz+ cintStR

u2qminuspminus2 2 dz

for some c Since le r log 2 and u is bounded we getintBR

2vx t 2xdx leintBR

2vx 2xdx + crBR

From Lemma 52 we getintBR

2vx t 2xdx le r log 221minus

1minus

2

BR

From definition of we haveintB1minusRcapvgt1minus

2r 22vx tdx ge r minus 1 log 22A1minus

2r 21minusRt

and

A1minus 2r

21minusRt le[( r

r minus 1

)2 1minus

1minus

2

+ c

r

]BR

Hence we obtain

A(1minus

2r

)2R

t le A1minus 2r

21minusRt + BRB1minusR

le[( r

r minus 1

)2 1minus

1minus

2

+ c

r+ n

]BR

Taking r so large that( r

r minus 1

)2 le(1minus

2

)1+

c

rle 3

82

and = 382n we obtain

A1minus 2r

2Rt le(1minus

(2

)2)BR

We take 13 = uixvminus h+ 2 as a test function in (31) where v = u2 and is

a cutoff function such that = 1 in Sr = 0 on pSR le cRminus r and t le cpminus2Rminus r2 Considering the ellipticity condition (12) we have

supt

int ∣∣vminus h+∣∣2 2 dx +

intupminus2vminus h+2 2 dz

le c

Rminus r2

intSR

upminus2vminus h+2 dz

+ c

Rminus r2

intSR

u2qminuspminus2vminus h+2 dz+ cintSR

vminus h+2 tdz (54)

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Regularity for Certain Nonlinear Parabolic Systems 629

for some c Now choose h ge 12 For such a choice of h we haveint

upminus2vminus h+2 2 dz ge cpminus2int

vminus h+2 2 dz

Hence from (54) we obtain

supt

intvminus h+2 2 dx + pminus2

intvminus h+2 2 dz

le[

2qminuspminus2

Rminus r2+ pminus2

Rminus r2

] intSR

vminus h+2 dz (55)

for some c independent of Observe that is bounded above by some fixednumber We make change of variables

= pminus2t minus t0 vx = vx t0 + 2minusp

and we write (55) as

supt

intBR

vminus h+2 2 dx +intSR

vminus h+2 2 dz le c

Rminus r2

intSR

vminus h+dz

Once we have this Caccioppoli type inequality we can apply truncation argumentTherefore following to similar argument in DiBenedetto and Friedman (1984 or1985) we have a maximum principle Now the proof for the next lemma is almostthe same as in DiBenedetto and Friedman (1984 or 1985) we omit the proof

Lemma 54 If for some and isin 0 1∣∣x t isin S1+R u2 lt 1minus 2∣∣ ge S1+R (56)

then there exists a number isin 0 1 independent of and R such that

supS R

2

ux t le

Now we assume that (56) fails that is∣∣x t isin S1+R u2 lt 1minus 2∣∣ lt S1+R (57)

Following a comparison argument similar to Choe (1991) or DiBenedetto andFriedman (1984) we prove that u satisfies a Campanato type integral inequalityLet V isin nN be a constant vector such that 1

2 le V le Taking uxkminus Vk

2 as atest function to (31) we have the following lemma

Lemma 55 There exists a constant c independent of such that

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2tdx +intS R

2

upminus22u2 dz

le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

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630 Bae and Choe

Now suppose that v is a solution to

vit minus

(Ai

Qj

Vvjx

)x

= 0 (58)

with v = u on pSR2 Since 1

2 le V le le c for some fixed c we see that fromthe structure condition of F

2 le 2minuspAi

Qj

Vi

j le c1+ qminusp2

Hence we see that v satisfies a Campanato type integral inequality

Lemma 56 Suppose that 0 lt r le R2 There exists a constant c independent ofVR and such that

intSr

vminus vr 2 dz le c( r

R

)n+4 intS R

2

uminus V 2 dz (59)

Proof We write (58) as

vit minus pminus2

Ai

Qj

V

pminus2vjx

x

= 0

We change of variable = pminus2t minus t0 then the function vx = vx 2minusp+ t0satisfies a uniformly parabolic system

vi minus

Ai

Qj

V

pminus2vjx

x

= 0

Thus from results of Campanato (1966) we prove

intS

vminus v2 dz le c(

R

)n+4 intS R

2

uminus V 2 dz

and scaling back prove (59)

We estimate the L2 norm of uminus v in SR2

Lemma 57 For each lt R2 there exists a constant c independent of such that

intS

uminus v2 dz le c(1+ qminusp 4

n+2)(

minus2 ndashintSR

uminus V 2 dz) int

SRuminus V 2 dz

for = min(12

2n

)

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Regularity for Certain Nonlinear Parabolic Systems 631

Proof Subtracting(Ai

Qj

Vujx

)x

from both sides of (15) we get

uit minus

(Ai

Qj

Vujx

)x= (

Aiuminus Ai

Qj

Vujx

)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(510)

From (58) and (510) we see that uminus v satisfies

ui minus vit minus(Ai

Qj

Vuj minus vjx)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(511)

Simply applying uminus v to (511) we get

pminus2intS R

2uminusv2dzle

intS R

2

∣∣∣ int 1

0Ai

Qj

V+suminus VminusAi

Qj

Vds∣∣∣

times uminusV uminusvdz

and it follows thatintS R

2

uminus v2 dz

le minus2pminus2intS R

2

∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣2

uminus V 2 dz (512)

Note that AiV = F primeV V i

V and

Ai

Qj

V = F primeprimeV ViV

j

V 2 + F primeV (ij

V minus V iV

j

V 3)

where ij and are Kronecker delta functions Also we recall the structurecondition (13) F primeprimeprimeh le chpminus3 + hqminus3 We have that

∣∣∣ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds∣∣∣

le∣∣∣∣int 1

0F primeprime(V + suminus V)

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2

minusF primeprimeV ViV

j

V 2 ds

∣∣∣∣+int 1

0

∣∣∣∣F primeV + suminus VV + suminus V minus F primeV

V ∣∣∣∣ds

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632 Bae and Choe

+∣∣∣∣int 1

0F primeV + suminus V

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V3

minusF primeV ViV

j

V 3 ds

∣∣∣∣= I + II + III

The first term I is estimates as follows

I leint 1

0

∣∣∣F primeprimeV + suminus Vminus F primeprimeV ∣∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2∣∣∣∣∣ ds

+int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

(513)

By the mean value theorem we have that

int 1

0

∣∣∣F primeprimeV+suminusVminusF primeprimeV ∣∣∣ds le c

int 1

0

int 1

0

∣∣F primeprimeprimeV+shuminusV∣∣ds dhuminusV (514)

for some c and

int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

le cF primeprimeV int 1

0

int 1

0

ds dh∣∣V + shuminus V∣∣ uminus V (515)

If V ge 2uminus V then for all s isin 0 1 12 V le V + suminus V le 2V and from

(513) (514) and (515) we have I le c(V pminus3 + V qminus3

)uminus V If V le 2uminus V and p ge 2 then

∣∣F primeprimeV + suminus V∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le cuminus V pminus2 + cuminus V qminus2

and

∣∣F primeprimeV ∣∣ le cV pminus2 + cV qminus2 le cuminus V pminus2 + cuminus V qminus2

Hence we have that

I le cint 1

0

∣∣F primeprimeV + suminus V∣∣+ ∣∣F primeprimeV ∣∣ds le c1+ pminusquminus V pminus2

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Regularity for Certain Nonlinear Parabolic Systems 633

for some c independent of If V le 2uminus V and 1 lt p lt 2 then∣∣F primeprime(V + suminus V)∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le c1+ pminusqV + suminus Vpminus2

for some c We note thatint 1

0V + suminus Vpminus2 ds le uminus V pminus2

int 1

0

∣∣∣∣ V uminus V minus s

∣∣∣∣pminus2

ds le cuminus V pminus2

Therefore combining all these together we have

I le cV pminus3 uminus V (516)

for some c independent of For II and III we observe that

F primeV V le cV pminus2 + cV qminus2

∣∣∣∣FprimeV V i

Vj

V 3∣∣∣∣ le cV pminus2 + cV qminus2

∣∣∣∣(F primeV V

)∣∣∣∣ le cV pminus3 + cV qminus3

∣∣∣∣(F primeV V i

Vj

V 3)∣∣∣∣ le cV pminus3 + cV qminus3

Hence following the same argument for I we conclude that

II + III le cV pminus3 uminus V (517)

for some c independent of Combining (516) and (517) we obtain that∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣ le cV pminus3 uminus V (518)

and considering (518) in (512)intS R

2

uminus V 2 dz le cminus2pminus2V 2pminus3intS R

2

uminus V 4 dz

le cminus2intS R

2

uminus V 4 dz (519)

From Houmllderrsquos inequality we have thatintS R

2

uminus V 4 dz

=intS R

2

uminus V 8n+ 4nminus8n dz

le[

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx] 2

n int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

(520)

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634 Bae and Choe

From Lemma 55 we have

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx le c2 supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2 dx

le cp + 2qminusp

R2

intSR

uminus V 2 dz

for some c independent of Now recalling that 12 le V le we obtain that

int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

le c22minuspnminus2

n

int t0

t0minus 2minuspR24

[ intB R

2

(u + V )2pminus2uminus V 4 dx] nminus2

n

dt

le c4nminuspnminus8

n

int t0

t0minus 2minuspR24

[ intB R

2

∣∣u pminus22 uminus V pminus2

2 V∣∣ 2n

nminus2 dx] nminus2

n

dt

le c4nminuspnminus8

n

intS R

2

upminus2 2u2 dz+ c2nminuspnminus8

n pminus2intSR

uminus V 2 dz (521)

for some c independent of where we have used Sobolevrsquos inequality FromLemma 55 we haveint

S R2

upminus2 2u2 dz le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

Thus combining (519) (520) and (521) we have that

intS R

2uminus V 2 dz le c

(1+ qminusp 4

n+2)[

minus2ndashintSR

uminus V 2 dz] 2

n intSR

uminus V 2 dz

and this completes the proof

Now we prove an iteration lemma which says that if mean oscillation of unear some fixed V is small then u satisfies a Campanato type integral inequality

Lemma 58 Fix 0 lt 0 lt 1 then there exist small positive numbers independentof such that if V0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists V1 isin nN satisfying(12minusradic

0

) le V1 le

(1+radic

0

) (522)

ndashintSR

uminus V12 dz le 0 ndashintSR

uminus V02 dz (523)

ndashintSR

uminus V12 dz le 2

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Regularity for Certain Nonlinear Parabolic Systems 635

Proof Since V0 satisfies the assumption of Lemma 57 we apply Lemma 57 withV = V0 and obtainint

S R2

vminus u2 dz le cintSR

uminus V02 dz

So from the usual comparison argument we haveintSR

uminus uR2 dz le cintSR

vminus vR2 dz+ cintSR

uminus v2 dz

le cn+4intS R

2

uminus u R22 dz+ c

intS R

2

uminus v2 dz

le c(n+4 +

) intSR

uminus V02 dz

where c is independent of Choosing and isin 0 12 small we prove (523)

with V1 = uR Inequality ndashintSR

uminus V12 dz le 2 follows from (523) andassumption ndash

intSR

uminus V02 dz le 2 Now we also estimate

V1 minus V0 =∣∣∣∣∣ ndashintSR

vminus V0 dz

∣∣∣∣le c

[ndashintSR

vminus u2 dz] 1

2

+ c

[ndashintSR

uminus V02 dz] 1

2

le cminus1minusn2(1+

120

)12

Therefore using the same technique of DiBenedetto (1986) Lemma 44 and if wetake very small compared to (522) holds and this completes the proof

Lemma 59 There exist small positive numbers 0 independent of such that ifV0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists a sequence of vectors Vi i = 1 2 3 sub nN satisfying

14 le Vi le 4 ndash

intSiR

uminus Vi2 dz le 2

ndashintSi+1R

uminus Vi+12 dz le 0 ndashintSiR

uminus Vi2 dz

Proof For the proof of the lemma we need to show that 14 le Vi le 4 But this

follows from that

Vi+1 minus Vi2 le c(0 + minusnminus2

)∣∣∣ ndashintSiR

uminus Vi dz∣∣∣

le ci0

i∣∣∣ ndashint

SRuminus V02 dz

∣∣∣ le ci0

i2

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636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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t 02

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t 200

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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620 Bae and Choe

same way for the case p ge 2 Finally we assume that p le q lt p+ 22 that iss + 2minus p ge s + 2q minus p Considering Youngrsquos inequality in (34) we get

intS

us+2 dz le c

Rminus 2

intSR

us+2minusp dz+ cSR

Rminus 2 (38)

Taking = Ri+1 R = Ri and si = s0 + ip with s0 = 2pminus 2 we obtain from (38)

intSRi+1

usi+1+2minusp dz le ci

R2

intSRi

usi+2minusp dz+ ci SRR2

for some c Therefore iterating this we obtain (37)

Once we have shown that u is in a high Ls space we can apply Moser iteration

Theorem 32 We have u isin L Furthermore suppose SR0sub T then u satisfies

supS R0

2

u le c[ndashintSR0

us0+s1 dz] 1

s0+s2 + c (39)

for some large s0 and some c independent of R0 where

s1 = 2q minus p s2 = 2minus nq minus p if q ge p

2 + 1

s1 = 2 s2 = np

2 + 2minus n if q le p

2 + 1

Proof Let be a standard cutoff function We take usuix2 as a test function to

(31) and we obtain

1s+2

int d

dt

(us+2)2 dz+

intAi

Qj

uujxx

(usuixx

+ susminus2uixukxukxx

)2 dz

= minus2int

Ai

Qj

uujxx

usuixx

dz

Considering the ellipticity condition

Qpminus22 le Ai

Qj

Qi

j le c

(Qpminus2 + Qqminus2)2

we see that

supt

intus+22 dx +

int ∣∣(u s+p2

)∣∣2 dzle c

intus+2tdz+ c

int (us+p + us+2qminusp)2 dz

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Regularity for Certain Nonlinear Parabolic Systems 621

for some c Therefore from Sobolevrsquos embedding theorem and Houmllderrsquos inequalitywe prove that

intS

us+p+s+2 2n dx le

int [ intus+22 dx

] 2n[ int

us+p nnminus2

2nnminus2dx

] nminus2n

dt

le[sup

t

intus+22 dx

] 2nint ∣∣∣(u s+p

2 )∣∣∣2 dz

le c

[1

Rminus 2

intSR

(us+2 + us+p + us+2qminusp)dz

]1+ 2n

(310)

We assume that p ge 2 then we see that s + 2q minus p ge s + 2 Hence fromYoungrsquos inequality we can write (310) as

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2qminusp dz+ SRRminus 2

]1+ 2n

(311)

for some c We set si by si+1 + 2q minus p = si + p+ si + 22n that is for

sufficiently large s0 si = is0 +(2minus nq minus p

)i minus 1 where

def= 1+ 2n Define

Ri

def= R021+ 2minusi Now we take s = si = Ri+1 and R = Ri in (311) Define

i

def= ndashintSRi

usi+2qminusp dz then (311) can be written as

i+1 le cii + ci (312)

If s0 gt nq minus pminus 2 then s0 + 2minus nq minus p gt 0 Hence we have that

limi

si = and limi

sii

= s0 + 2minus nq minus p

From Lemma 31 we can take s0 as large as possible Iterating (312) with some larges0 we have

supS R0

2

u le c[ndashintSR0

us0+2qminusp dz] 1

s0+2minusnqminusp + c

and this completes the proof for the case p ge 2Now we assume that 1 lt p le 2 If q ge p2+ 1 then s + 2q minus p ge s + 2 In this

case applying Youngrsquos inequality on the last term of (310) we have

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2qminusp dz+ SRRminus 2

]1+ 2n

for some c Hence following the argument for the case p ge 2 we prove (39) andthis completes the proof for the case q ge p2+ 1

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622 Bae and Choe

Now we assume that p2+ 1 gt q ge p gt 1 In this case s + 2 gt s + 2q minus p andwe have from (310)

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2 dz+ SRRminus 2

]1+ 2n

(313)

for some c We set si by sidef= is0 + np2+ 2minus ni minus 1 Hence if s0 gt nminus 2

minus np2 then

limi

si = and limi

sii

= s0 + 2minus n+ np

2

We take s = si = Ri+1 and R = Ri in (313) Iterating (313) we obtain

supS R0

2

u le c

[ndashintSR0

us0+2 dz

] 1s0+2+ np

2 minusn + c

for some c independent of R0 and this completes the proof for the casep2+ 1 gt q ge p gt 1

4 HOumlLDER CONTINUITY OF u

In this section we prove that u is Houmllder continuous

Theorem 41 Define r0 by the constant appearing in Theorem 21 Suppose that uis L

r0locT then u is Houmllder continuous Moreover for all Q2R sub T with R lt 1 u

satisfies

ndashintQR

uminus uR2 dz le cRs

for some c independent of R where we define

s =2 when p ge 2

ppminus 1 when p isin 1 2

We introduce a nonnegative cutoff function isin 0 BR such that = 1 in

BR2 0 le le 1 le cR We define minus

R

def= t0 minus 2Rp t0 minus Rp and QminusR = BR timesminus

R

Also we define uRt

def= 1BRintBR

ux tdx and uR

def= 1QRintQR

ux tdz Firstwe prove a lemma which is essential for a Poincareacute inequality for solutions of adegenerate parabolic system

Lemma 42 Suppose Q2R sub T then u satisfies the following inequality

suptisinR

intminus

R

dsintBR

2ux tminus uRs2 dx

le cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz+ cintQminus

R

uminus uRt2 dz

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Regularity for Certain Nonlinear Parabolic Systems 623

for all R lt R0 where c depends only on nN and p and

s1 = p s2 = p and s3 = pqminus1

pminus1 when p ge 2

s1 = p2minus p s2 = ppminus 1 and s3 = pq minus 1 when 1 lt p lt 2

Proof Since u isin 0(0 T L2

) there exists ts isin R for all s isin minus

R such that

intBR

pux t minus uRs2dx = suptisinR

intBR

pux tminus uRs2 dx

We take uminus uRspxst as a test function to (15) where st is the characteristic

function such that

st =1 for all isin s t

0 for all isin s t

Hence we have

intut middot uminus uRs

pxst dz+int

Au middot uminus uRspst dz = 0 (41)

Now we assume p isin 2 Considering the structure condition (12) on A (23)and (25) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

leint

uminus uRs2x sp dx + cintBRtimesst

upminus1uminus uRspminus1dz

+ cintBRtimesst

uqminus1uminus uRspminus1dz (42)

for some c independent of R From Youngrsquos inequality we have

intBRtimesst

upminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

up dz (43)

for some c and small Similarly the last term in (42) can be estimated as follows

intBRtimesst

uqminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

u pqminus1pminus1 dz (44)

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624 Bae and Choe

for some c and small Combining (42) (43) and (44) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

le cint

uminus uRs2x sp dx +

Rp

intBRtimesst

uminus uRspp dz

+ cintBRtimesst

up dz+ cintBRtimesst

u pqminus1pminus1 dz (45)

for some c and small Integrating (45) with respect to s from t0 minus 2Rp to t0 minus Rpwe have

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRs2p dz+ c

Rp

intminus

R

dsintBRtimesst

uminus uRs2p dz

+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

By the choice of t we have that for small

intminus

R

dsintBR

uminusuRs2x t pdx

le cintQminus

R

uminusuRs2p dz+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

So we proved Lemma 42 when p ge 2In case p isin 1 2 we estimate the second term of (41) as follows

intuppst dz+ p

intAu middot uminus uRs

pminus1st dz

leint

Au middot uminus uRspst dz (46)

and using the fact that u is bounded and ppminus 1 gt 2 we have

∣∣∣ int Au middot uminusuRspminus1st dz

∣∣∣le c

intBRtimesst

upminus1uminusuRspminus1st dz

+ cintBRtimesst

uqminus1pminus1uminusuRsst dz

le

Rp

intuminusuRs

ppminus1 pst dz+

c

Rp2minusp

intuppminus1 dz+ c

Rp2minusp

intupqminus1 dz

(47)

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Regularity for Certain Nonlinear Parabolic Systems 625

Combining (41) (46) and (47) we have

intBR

uminus uRs2x t p dx

le cu2minusppminus1L

Rp

intBRtimesst

uminus uRs2pst dz+c

Rp2minusp

intBRtimesst

uppminus1 dz

+ c

Rp2minusp

intBRtimesst

upqminus1 dz+intBR

uminus uRs2x sp dx (48)

for small Integrating (48) with respect to s and considering the choice of t weprove Lemma 42 when 1 lt p lt 2

Now we prove a Poincareacute inequality

Theorem 43 Suppose Q2R sub T with R lt 1 then u satisfies the following inequality

intQR

2

uminus uR22 dz le cR2

intQ2R

u2 dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

where c is independent of R

Proof Using Lemma 42 we have

intQR

2

uminus uR22 dz le

intQR

uminus uRs2p dz

le c

Rp

intminus

R

dsintQR

uminus uRs2p dz le c suptisinR

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRt2p dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

le cR2intQ2R

u2 dz+ cRs1

intQ2R

us2z+ cRs1

intQ2R

us3 dz

where we used a Poincareacute type inequality for x variables only that is

intBR

uminus uRs2p dx le cR2intBR

u2 dx

Proof of Theorem 41 Since u is bounded owing to Theorem 32 we have fromTheorem 43

intQR

uminus uRs2 dz lecRn+p+2 when p isin 2

cRn+p+ppminus1 when p isin 1 2

where c is independent of R Hence by isomorphism theorem of Da Prato (1965) weprove Theorem 41

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626 Bae and Choe

5 HOumlLDER CONTINUITY FOR u

In this section we prove that u is Houmllder continuous Because of degeneracyof the pde we need scaled cylinders (see DiBenedetto and Friedman 1985)Let R0 gt 0 be fixed and consider a cylinder

SR0

def= BR0times t0 minus R2minus

0 t0

for small For any 0 lt R le R02 we introduce a scaled cylinder

SRdef= BR times t0 minus R22minusp t0

for fixed ge supSR0u

We assume that pminus2 gt R0 if p gt 2 Hence we note that SR sub S

R0and

ge supSRu

The following theorem is our main theorem in this section

Theorem 51 Suppose that SR sub T then u satisfies for all r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR 2 dz+ cR

for some gt 0

For the proof of Theorem 51 we follow a usual program for degenerateequations such that if for some isin 0 1∣∣z isin SR u2 le 1minus 2

∣∣ gt ∣∣SR

∣∣ (51)

then

supS R

2

u le (52)

for some small isin 0 1 and otherwise that is (51) fails then for all 0 lt r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR2 dz (53)

for some gt 0 Combining (52) and (53) we prove Theorem 51First we prove that (52) under the assumption (51) A similar argument for

the case p = q ge 2 appeared in DiBenedetto and Friedman (1985) We assume that(51) is true Set v = u2 For each h gt 0 we set AhRt

def= x isin BR v gt h and

BhR

def= z isin SR v gt h From the mean value theorem we estimate AhRt

Lemma 52 For any isin 0 there exists isin t0 minus 2minuspR20 t0 minus 2minuspR2

0 such that

A1minus2R0 le 1minus

1minus BR0

where is independent of

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Regularity for Certain Nonlinear Parabolic Systems 627

Proof Suppose that the assertion is not true then

B1minus2R0 ge

int t0minus2minuspR20

t0minus2minuspR20

A1minus2R0d gt 1minus BR0

2minuspR20 = 1minus SR0

and this contradicts the assumption (51)

The following lemma is useful in estimating sup u Indeed when p = q ge 2DiBenedetto and Friedman proved a similar theorem (see Lemma 62 inDiBenedetto and Friedman 1985) Now choose = 2 and in Lemma 52We define St

R

def= BR times t

Lemma 53 There exists a positive integer r independent of such that

∣∣A1minus 2r

2R0t

∣∣ le (1minus

(2

)2)BR0

for all t isin t0 minus

22minuspR2

0 t0

Proof Consider the function

v = log+[

2

2minus vminus 1minus 2+ + 2

2r

]

where r is a positive integer This function was used in Choe (1991) and DiBenedettoand Friedman (1985) We take 13 = ui

x2vprime 2 as a test function for (31) where

isin 0 is a standard cutoff function such that B1minusR for small decided later is

contained in support Integrating from to t we obtain

12

intStR

vt2prime 2 dz+

intStR

Ai

Qj

uujxx

uixx

2prime 2 dz

+ 2intStR

Ai

Qj

uujxx

uixukxx

ukx2primeprime 2 dz

+ 2intStR

Ai

Qj

uujxx

uix x

2prime dz = 0

Considering the fact 2primeprime = 21+ prime2 we have

intBR

2vx t 2xdx +intStR

upminus2v21+ prime2 2 dz

leintBR

2vx 2xdx + cintStR

upminus2v prime dz

+ cintStR

uqminus2v prime dz

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628 Bae and Choe

Applying Youngrsquos inequality on the last two terms we getintBR

2vx t 2xdx

leintBR

2vx 2xdx + cintStR

upminus2 2 dz+ cintStR

u2qminuspminus2 2 dz

for some c Since le r log 2 and u is bounded we getintBR

2vx t 2xdx leintBR

2vx 2xdx + crBR

From Lemma 52 we getintBR

2vx t 2xdx le r log 221minus

1minus

2

BR

From definition of we haveintB1minusRcapvgt1minus

2r 22vx tdx ge r minus 1 log 22A1minus

2r 21minusRt

and

A1minus 2r

21minusRt le[( r

r minus 1

)2 1minus

1minus

2

+ c

r

]BR

Hence we obtain

A(1minus

2r

)2R

t le A1minus 2r

21minusRt + BRB1minusR

le[( r

r minus 1

)2 1minus

1minus

2

+ c

r+ n

]BR

Taking r so large that( r

r minus 1

)2 le(1minus

2

)1+

c

rle 3

82

and = 382n we obtain

A1minus 2r

2Rt le(1minus

(2

)2)BR

We take 13 = uixvminus h+ 2 as a test function in (31) where v = u2 and is

a cutoff function such that = 1 in Sr = 0 on pSR le cRminus r and t le cpminus2Rminus r2 Considering the ellipticity condition (12) we have

supt

int ∣∣vminus h+∣∣2 2 dx +

intupminus2vminus h+2 2 dz

le c

Rminus r2

intSR

upminus2vminus h+2 dz

+ c

Rminus r2

intSR

u2qminuspminus2vminus h+2 dz+ cintSR

vminus h+2 tdz (54)

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Regularity for Certain Nonlinear Parabolic Systems 629

for some c Now choose h ge 12 For such a choice of h we haveint

upminus2vminus h+2 2 dz ge cpminus2int

vminus h+2 2 dz

Hence from (54) we obtain

supt

intvminus h+2 2 dx + pminus2

intvminus h+2 2 dz

le[

2qminuspminus2

Rminus r2+ pminus2

Rminus r2

] intSR

vminus h+2 dz (55)

for some c independent of Observe that is bounded above by some fixednumber We make change of variables

= pminus2t minus t0 vx = vx t0 + 2minusp

and we write (55) as

supt

intBR

vminus h+2 2 dx +intSR

vminus h+2 2 dz le c

Rminus r2

intSR

vminus h+dz

Once we have this Caccioppoli type inequality we can apply truncation argumentTherefore following to similar argument in DiBenedetto and Friedman (1984 or1985) we have a maximum principle Now the proof for the next lemma is almostthe same as in DiBenedetto and Friedman (1984 or 1985) we omit the proof

Lemma 54 If for some and isin 0 1∣∣x t isin S1+R u2 lt 1minus 2∣∣ ge S1+R (56)

then there exists a number isin 0 1 independent of and R such that

supS R

2

ux t le

Now we assume that (56) fails that is∣∣x t isin S1+R u2 lt 1minus 2∣∣ lt S1+R (57)

Following a comparison argument similar to Choe (1991) or DiBenedetto andFriedman (1984) we prove that u satisfies a Campanato type integral inequalityLet V isin nN be a constant vector such that 1

2 le V le Taking uxkminus Vk

2 as atest function to (31) we have the following lemma

Lemma 55 There exists a constant c independent of such that

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2tdx +intS R

2

upminus22u2 dz

le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

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630 Bae and Choe

Now suppose that v is a solution to

vit minus

(Ai

Qj

Vvjx

)x

= 0 (58)

with v = u on pSR2 Since 1

2 le V le le c for some fixed c we see that fromthe structure condition of F

2 le 2minuspAi

Qj

Vi

j le c1+ qminusp2

Hence we see that v satisfies a Campanato type integral inequality

Lemma 56 Suppose that 0 lt r le R2 There exists a constant c independent ofVR and such that

intSr

vminus vr 2 dz le c( r

R

)n+4 intS R

2

uminus V 2 dz (59)

Proof We write (58) as

vit minus pminus2

Ai

Qj

V

pminus2vjx

x

= 0

We change of variable = pminus2t minus t0 then the function vx = vx 2minusp+ t0satisfies a uniformly parabolic system

vi minus

Ai

Qj

V

pminus2vjx

x

= 0

Thus from results of Campanato (1966) we prove

intS

vminus v2 dz le c(

R

)n+4 intS R

2

uminus V 2 dz

and scaling back prove (59)

We estimate the L2 norm of uminus v in SR2

Lemma 57 For each lt R2 there exists a constant c independent of such that

intS

uminus v2 dz le c(1+ qminusp 4

n+2)(

minus2 ndashintSR

uminus V 2 dz) int

SRuminus V 2 dz

for = min(12

2n

)

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Regularity for Certain Nonlinear Parabolic Systems 631

Proof Subtracting(Ai

Qj

Vujx

)x

from both sides of (15) we get

uit minus

(Ai

Qj

Vujx

)x= (

Aiuminus Ai

Qj

Vujx

)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(510)

From (58) and (510) we see that uminus v satisfies

ui minus vit minus(Ai

Qj

Vuj minus vjx)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(511)

Simply applying uminus v to (511) we get

pminus2intS R

2uminusv2dzle

intS R

2

∣∣∣ int 1

0Ai

Qj

V+suminus VminusAi

Qj

Vds∣∣∣

times uminusV uminusvdz

and it follows thatintS R

2

uminus v2 dz

le minus2pminus2intS R

2

∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣2

uminus V 2 dz (512)

Note that AiV = F primeV V i

V and

Ai

Qj

V = F primeprimeV ViV

j

V 2 + F primeV (ij

V minus V iV

j

V 3)

where ij and are Kronecker delta functions Also we recall the structurecondition (13) F primeprimeprimeh le chpminus3 + hqminus3 We have that

∣∣∣ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds∣∣∣

le∣∣∣∣int 1

0F primeprime(V + suminus V)

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2

minusF primeprimeV ViV

j

V 2 ds

∣∣∣∣+int 1

0

∣∣∣∣F primeV + suminus VV + suminus V minus F primeV

V ∣∣∣∣ds

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632 Bae and Choe

+∣∣∣∣int 1

0F primeV + suminus V

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V3

minusF primeV ViV

j

V 3 ds

∣∣∣∣= I + II + III

The first term I is estimates as follows

I leint 1

0

∣∣∣F primeprimeV + suminus Vminus F primeprimeV ∣∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2∣∣∣∣∣ ds

+int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

(513)

By the mean value theorem we have that

int 1

0

∣∣∣F primeprimeV+suminusVminusF primeprimeV ∣∣∣ds le c

int 1

0

int 1

0

∣∣F primeprimeprimeV+shuminusV∣∣ds dhuminusV (514)

for some c and

int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

le cF primeprimeV int 1

0

int 1

0

ds dh∣∣V + shuminus V∣∣ uminus V (515)

If V ge 2uminus V then for all s isin 0 1 12 V le V + suminus V le 2V and from

(513) (514) and (515) we have I le c(V pminus3 + V qminus3

)uminus V If V le 2uminus V and p ge 2 then

∣∣F primeprimeV + suminus V∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le cuminus V pminus2 + cuminus V qminus2

and

∣∣F primeprimeV ∣∣ le cV pminus2 + cV qminus2 le cuminus V pminus2 + cuminus V qminus2

Hence we have that

I le cint 1

0

∣∣F primeprimeV + suminus V∣∣+ ∣∣F primeprimeV ∣∣ds le c1+ pminusquminus V pminus2

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Regularity for Certain Nonlinear Parabolic Systems 633

for some c independent of If V le 2uminus V and 1 lt p lt 2 then∣∣F primeprime(V + suminus V)∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le c1+ pminusqV + suminus Vpminus2

for some c We note thatint 1

0V + suminus Vpminus2 ds le uminus V pminus2

int 1

0

∣∣∣∣ V uminus V minus s

∣∣∣∣pminus2

ds le cuminus V pminus2

Therefore combining all these together we have

I le cV pminus3 uminus V (516)

for some c independent of For II and III we observe that

F primeV V le cV pminus2 + cV qminus2

∣∣∣∣FprimeV V i

Vj

V 3∣∣∣∣ le cV pminus2 + cV qminus2

∣∣∣∣(F primeV V

)∣∣∣∣ le cV pminus3 + cV qminus3

∣∣∣∣(F primeV V i

Vj

V 3)∣∣∣∣ le cV pminus3 + cV qminus3

Hence following the same argument for I we conclude that

II + III le cV pminus3 uminus V (517)

for some c independent of Combining (516) and (517) we obtain that∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣ le cV pminus3 uminus V (518)

and considering (518) in (512)intS R

2

uminus V 2 dz le cminus2pminus2V 2pminus3intS R

2

uminus V 4 dz

le cminus2intS R

2

uminus V 4 dz (519)

From Houmllderrsquos inequality we have thatintS R

2

uminus V 4 dz

=intS R

2

uminus V 8n+ 4nminus8n dz

le[

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx] 2

n int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

(520)

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634 Bae and Choe

From Lemma 55 we have

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx le c2 supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2 dx

le cp + 2qminusp

R2

intSR

uminus V 2 dz

for some c independent of Now recalling that 12 le V le we obtain that

int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

le c22minuspnminus2

n

int t0

t0minus 2minuspR24

[ intB R

2

(u + V )2pminus2uminus V 4 dx] nminus2

n

dt

le c4nminuspnminus8

n

int t0

t0minus 2minuspR24

[ intB R

2

∣∣u pminus22 uminus V pminus2

2 V∣∣ 2n

nminus2 dx] nminus2

n

dt

le c4nminuspnminus8

n

intS R

2

upminus2 2u2 dz+ c2nminuspnminus8

n pminus2intSR

uminus V 2 dz (521)

for some c independent of where we have used Sobolevrsquos inequality FromLemma 55 we haveint

S R2

upminus2 2u2 dz le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

Thus combining (519) (520) and (521) we have that

intS R

2uminus V 2 dz le c

(1+ qminusp 4

n+2)[

minus2ndashintSR

uminus V 2 dz] 2

n intSR

uminus V 2 dz

and this completes the proof

Now we prove an iteration lemma which says that if mean oscillation of unear some fixed V is small then u satisfies a Campanato type integral inequality

Lemma 58 Fix 0 lt 0 lt 1 then there exist small positive numbers independentof such that if V0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists V1 isin nN satisfying(12minusradic

0

) le V1 le

(1+radic

0

) (522)

ndashintSR

uminus V12 dz le 0 ndashintSR

uminus V02 dz (523)

ndashintSR

uminus V12 dz le 2

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Regularity for Certain Nonlinear Parabolic Systems 635

Proof Since V0 satisfies the assumption of Lemma 57 we apply Lemma 57 withV = V0 and obtainint

S R2

vminus u2 dz le cintSR

uminus V02 dz

So from the usual comparison argument we haveintSR

uminus uR2 dz le cintSR

vminus vR2 dz+ cintSR

uminus v2 dz

le cn+4intS R

2

uminus u R22 dz+ c

intS R

2

uminus v2 dz

le c(n+4 +

) intSR

uminus V02 dz

where c is independent of Choosing and isin 0 12 small we prove (523)

with V1 = uR Inequality ndashintSR

uminus V12 dz le 2 follows from (523) andassumption ndash

intSR

uminus V02 dz le 2 Now we also estimate

V1 minus V0 =∣∣∣∣∣ ndashintSR

vminus V0 dz

∣∣∣∣le c

[ndashintSR

vminus u2 dz] 1

2

+ c

[ndashintSR

uminus V02 dz] 1

2

le cminus1minusn2(1+

120

)12

Therefore using the same technique of DiBenedetto (1986) Lemma 44 and if wetake very small compared to (522) holds and this completes the proof

Lemma 59 There exist small positive numbers 0 independent of such that ifV0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists a sequence of vectors Vi i = 1 2 3 sub nN satisfying

14 le Vi le 4 ndash

intSiR

uminus Vi2 dz le 2

ndashintSi+1R

uminus Vi+12 dz le 0 ndashintSiR

uminus Vi2 dz

Proof For the proof of the lemma we need to show that 14 le Vi le 4 But this

follows from that

Vi+1 minus Vi2 le c(0 + minusnminus2

)∣∣∣ ndashintSiR

uminus Vi dz∣∣∣

le ci0

i∣∣∣ ndashint

SRuminus V02 dz

∣∣∣ le ci0

i2

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636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

Dow

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t 200

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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Regularity for Certain Nonlinear Parabolic Systems 621

for some c Therefore from Sobolevrsquos embedding theorem and Houmllderrsquos inequalitywe prove that

intS

us+p+s+2 2n dx le

int [ intus+22 dx

] 2n[ int

us+p nnminus2

2nnminus2dx

] nminus2n

dt

le[sup

t

intus+22 dx

] 2nint ∣∣∣(u s+p

2 )∣∣∣2 dz

le c

[1

Rminus 2

intSR

(us+2 + us+p + us+2qminusp)dz

]1+ 2n

(310)

We assume that p ge 2 then we see that s + 2q minus p ge s + 2 Hence fromYoungrsquos inequality we can write (310) as

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2qminusp dz+ SRRminus 2

]1+ 2n

(311)

for some c We set si by si+1 + 2q minus p = si + p+ si + 22n that is for

sufficiently large s0 si = is0 +(2minus nq minus p

)i minus 1 where

def= 1+ 2n Define

Ri

def= R021+ 2minusi Now we take s = si = Ri+1 and R = Ri in (311) Define

i

def= ndashintSRi

usi+2qminusp dz then (311) can be written as

i+1 le cii + ci (312)

If s0 gt nq minus pminus 2 then s0 + 2minus nq minus p gt 0 Hence we have that

limi

si = and limi

sii

= s0 + 2minus nq minus p

From Lemma 31 we can take s0 as large as possible Iterating (312) with some larges0 we have

supS R0

2

u le c[ndashintSR0

us0+2qminusp dz] 1

s0+2minusnqminusp + c

and this completes the proof for the case p ge 2Now we assume that 1 lt p le 2 If q ge p2+ 1 then s + 2q minus p ge s + 2 In this

case applying Youngrsquos inequality on the last term of (310) we have

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2qminusp dz+ SRRminus 2

]1+ 2n

for some c Hence following the argument for the case p ge 2 we prove (39) andthis completes the proof for the case q ge p2+ 1

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622 Bae and Choe

Now we assume that p2+ 1 gt q ge p gt 1 In this case s + 2 gt s + 2q minus p andwe have from (310)

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2 dz+ SRRminus 2

]1+ 2n

(313)

for some c We set si by sidef= is0 + np2+ 2minus ni minus 1 Hence if s0 gt nminus 2

minus np2 then

limi

si = and limi

sii

= s0 + 2minus n+ np

2

We take s = si = Ri+1 and R = Ri in (313) Iterating (313) we obtain

supS R0

2

u le c

[ndashintSR0

us0+2 dz

] 1s0+2+ np

2 minusn + c

for some c independent of R0 and this completes the proof for the casep2+ 1 gt q ge p gt 1

4 HOumlLDER CONTINUITY OF u

In this section we prove that u is Houmllder continuous

Theorem 41 Define r0 by the constant appearing in Theorem 21 Suppose that uis L

r0locT then u is Houmllder continuous Moreover for all Q2R sub T with R lt 1 u

satisfies

ndashintQR

uminus uR2 dz le cRs

for some c independent of R where we define

s =2 when p ge 2

ppminus 1 when p isin 1 2

We introduce a nonnegative cutoff function isin 0 BR such that = 1 in

BR2 0 le le 1 le cR We define minus

R

def= t0 minus 2Rp t0 minus Rp and QminusR = BR timesminus

R

Also we define uRt

def= 1BRintBR

ux tdx and uR

def= 1QRintQR

ux tdz Firstwe prove a lemma which is essential for a Poincareacute inequality for solutions of adegenerate parabolic system

Lemma 42 Suppose Q2R sub T then u satisfies the following inequality

suptisinR

intminus

R

dsintBR

2ux tminus uRs2 dx

le cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz+ cintQminus

R

uminus uRt2 dz

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Regularity for Certain Nonlinear Parabolic Systems 623

for all R lt R0 where c depends only on nN and p and

s1 = p s2 = p and s3 = pqminus1

pminus1 when p ge 2

s1 = p2minus p s2 = ppminus 1 and s3 = pq minus 1 when 1 lt p lt 2

Proof Since u isin 0(0 T L2

) there exists ts isin R for all s isin minus

R such that

intBR

pux t minus uRs2dx = suptisinR

intBR

pux tminus uRs2 dx

We take uminus uRspxst as a test function to (15) where st is the characteristic

function such that

st =1 for all isin s t

0 for all isin s t

Hence we have

intut middot uminus uRs

pxst dz+int

Au middot uminus uRspst dz = 0 (41)

Now we assume p isin 2 Considering the structure condition (12) on A (23)and (25) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

leint

uminus uRs2x sp dx + cintBRtimesst

upminus1uminus uRspminus1dz

+ cintBRtimesst

uqminus1uminus uRspminus1dz (42)

for some c independent of R From Youngrsquos inequality we have

intBRtimesst

upminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

up dz (43)

for some c and small Similarly the last term in (42) can be estimated as follows

intBRtimesst

uqminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

u pqminus1pminus1 dz (44)

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624 Bae and Choe

for some c and small Combining (42) (43) and (44) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

le cint

uminus uRs2x sp dx +

Rp

intBRtimesst

uminus uRspp dz

+ cintBRtimesst

up dz+ cintBRtimesst

u pqminus1pminus1 dz (45)

for some c and small Integrating (45) with respect to s from t0 minus 2Rp to t0 minus Rpwe have

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRs2p dz+ c

Rp

intminus

R

dsintBRtimesst

uminus uRs2p dz

+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

By the choice of t we have that for small

intminus

R

dsintBR

uminusuRs2x t pdx

le cintQminus

R

uminusuRs2p dz+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

So we proved Lemma 42 when p ge 2In case p isin 1 2 we estimate the second term of (41) as follows

intuppst dz+ p

intAu middot uminus uRs

pminus1st dz

leint

Au middot uminus uRspst dz (46)

and using the fact that u is bounded and ppminus 1 gt 2 we have

∣∣∣ int Au middot uminusuRspminus1st dz

∣∣∣le c

intBRtimesst

upminus1uminusuRspminus1st dz

+ cintBRtimesst

uqminus1pminus1uminusuRsst dz

le

Rp

intuminusuRs

ppminus1 pst dz+

c

Rp2minusp

intuppminus1 dz+ c

Rp2minusp

intupqminus1 dz

(47)

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Regularity for Certain Nonlinear Parabolic Systems 625

Combining (41) (46) and (47) we have

intBR

uminus uRs2x t p dx

le cu2minusppminus1L

Rp

intBRtimesst

uminus uRs2pst dz+c

Rp2minusp

intBRtimesst

uppminus1 dz

+ c

Rp2minusp

intBRtimesst

upqminus1 dz+intBR

uminus uRs2x sp dx (48)

for small Integrating (48) with respect to s and considering the choice of t weprove Lemma 42 when 1 lt p lt 2

Now we prove a Poincareacute inequality

Theorem 43 Suppose Q2R sub T with R lt 1 then u satisfies the following inequality

intQR

2

uminus uR22 dz le cR2

intQ2R

u2 dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

where c is independent of R

Proof Using Lemma 42 we have

intQR

2

uminus uR22 dz le

intQR

uminus uRs2p dz

le c

Rp

intminus

R

dsintQR

uminus uRs2p dz le c suptisinR

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRt2p dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

le cR2intQ2R

u2 dz+ cRs1

intQ2R

us2z+ cRs1

intQ2R

us3 dz

where we used a Poincareacute type inequality for x variables only that is

intBR

uminus uRs2p dx le cR2intBR

u2 dx

Proof of Theorem 41 Since u is bounded owing to Theorem 32 we have fromTheorem 43

intQR

uminus uRs2 dz lecRn+p+2 when p isin 2

cRn+p+ppminus1 when p isin 1 2

where c is independent of R Hence by isomorphism theorem of Da Prato (1965) weprove Theorem 41

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626 Bae and Choe

5 HOumlLDER CONTINUITY FOR u

In this section we prove that u is Houmllder continuous Because of degeneracyof the pde we need scaled cylinders (see DiBenedetto and Friedman 1985)Let R0 gt 0 be fixed and consider a cylinder

SR0

def= BR0times t0 minus R2minus

0 t0

for small For any 0 lt R le R02 we introduce a scaled cylinder

SRdef= BR times t0 minus R22minusp t0

for fixed ge supSR0u

We assume that pminus2 gt R0 if p gt 2 Hence we note that SR sub S

R0and

ge supSRu

The following theorem is our main theorem in this section

Theorem 51 Suppose that SR sub T then u satisfies for all r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR 2 dz+ cR

for some gt 0

For the proof of Theorem 51 we follow a usual program for degenerateequations such that if for some isin 0 1∣∣z isin SR u2 le 1minus 2

∣∣ gt ∣∣SR

∣∣ (51)

then

supS R

2

u le (52)

for some small isin 0 1 and otherwise that is (51) fails then for all 0 lt r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR2 dz (53)

for some gt 0 Combining (52) and (53) we prove Theorem 51First we prove that (52) under the assumption (51) A similar argument for

the case p = q ge 2 appeared in DiBenedetto and Friedman (1985) We assume that(51) is true Set v = u2 For each h gt 0 we set AhRt

def= x isin BR v gt h and

BhR

def= z isin SR v gt h From the mean value theorem we estimate AhRt

Lemma 52 For any isin 0 there exists isin t0 minus 2minuspR20 t0 minus 2minuspR2

0 such that

A1minus2R0 le 1minus

1minus BR0

where is independent of

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Regularity for Certain Nonlinear Parabolic Systems 627

Proof Suppose that the assertion is not true then

B1minus2R0 ge

int t0minus2minuspR20

t0minus2minuspR20

A1minus2R0d gt 1minus BR0

2minuspR20 = 1minus SR0

and this contradicts the assumption (51)

The following lemma is useful in estimating sup u Indeed when p = q ge 2DiBenedetto and Friedman proved a similar theorem (see Lemma 62 inDiBenedetto and Friedman 1985) Now choose = 2 and in Lemma 52We define St

R

def= BR times t

Lemma 53 There exists a positive integer r independent of such that

∣∣A1minus 2r

2R0t

∣∣ le (1minus

(2

)2)BR0

for all t isin t0 minus

22minuspR2

0 t0

Proof Consider the function

v = log+[

2

2minus vminus 1minus 2+ + 2

2r

]

where r is a positive integer This function was used in Choe (1991) and DiBenedettoand Friedman (1985) We take 13 = ui

x2vprime 2 as a test function for (31) where

isin 0 is a standard cutoff function such that B1minusR for small decided later is

contained in support Integrating from to t we obtain

12

intStR

vt2prime 2 dz+

intStR

Ai

Qj

uujxx

uixx

2prime 2 dz

+ 2intStR

Ai

Qj

uujxx

uixukxx

ukx2primeprime 2 dz

+ 2intStR

Ai

Qj

uujxx

uix x

2prime dz = 0

Considering the fact 2primeprime = 21+ prime2 we have

intBR

2vx t 2xdx +intStR

upminus2v21+ prime2 2 dz

leintBR

2vx 2xdx + cintStR

upminus2v prime dz

+ cintStR

uqminus2v prime dz

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628 Bae and Choe

Applying Youngrsquos inequality on the last two terms we getintBR

2vx t 2xdx

leintBR

2vx 2xdx + cintStR

upminus2 2 dz+ cintStR

u2qminuspminus2 2 dz

for some c Since le r log 2 and u is bounded we getintBR

2vx t 2xdx leintBR

2vx 2xdx + crBR

From Lemma 52 we getintBR

2vx t 2xdx le r log 221minus

1minus

2

BR

From definition of we haveintB1minusRcapvgt1minus

2r 22vx tdx ge r minus 1 log 22A1minus

2r 21minusRt

and

A1minus 2r

21minusRt le[( r

r minus 1

)2 1minus

1minus

2

+ c

r

]BR

Hence we obtain

A(1minus

2r

)2R

t le A1minus 2r

21minusRt + BRB1minusR

le[( r

r minus 1

)2 1minus

1minus

2

+ c

r+ n

]BR

Taking r so large that( r

r minus 1

)2 le(1minus

2

)1+

c

rle 3

82

and = 382n we obtain

A1minus 2r

2Rt le(1minus

(2

)2)BR

We take 13 = uixvminus h+ 2 as a test function in (31) where v = u2 and is

a cutoff function such that = 1 in Sr = 0 on pSR le cRminus r and t le cpminus2Rminus r2 Considering the ellipticity condition (12) we have

supt

int ∣∣vminus h+∣∣2 2 dx +

intupminus2vminus h+2 2 dz

le c

Rminus r2

intSR

upminus2vminus h+2 dz

+ c

Rminus r2

intSR

u2qminuspminus2vminus h+2 dz+ cintSR

vminus h+2 tdz (54)

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Regularity for Certain Nonlinear Parabolic Systems 629

for some c Now choose h ge 12 For such a choice of h we haveint

upminus2vminus h+2 2 dz ge cpminus2int

vminus h+2 2 dz

Hence from (54) we obtain

supt

intvminus h+2 2 dx + pminus2

intvminus h+2 2 dz

le[

2qminuspminus2

Rminus r2+ pminus2

Rminus r2

] intSR

vminus h+2 dz (55)

for some c independent of Observe that is bounded above by some fixednumber We make change of variables

= pminus2t minus t0 vx = vx t0 + 2minusp

and we write (55) as

supt

intBR

vminus h+2 2 dx +intSR

vminus h+2 2 dz le c

Rminus r2

intSR

vminus h+dz

Once we have this Caccioppoli type inequality we can apply truncation argumentTherefore following to similar argument in DiBenedetto and Friedman (1984 or1985) we have a maximum principle Now the proof for the next lemma is almostthe same as in DiBenedetto and Friedman (1984 or 1985) we omit the proof

Lemma 54 If for some and isin 0 1∣∣x t isin S1+R u2 lt 1minus 2∣∣ ge S1+R (56)

then there exists a number isin 0 1 independent of and R such that

supS R

2

ux t le

Now we assume that (56) fails that is∣∣x t isin S1+R u2 lt 1minus 2∣∣ lt S1+R (57)

Following a comparison argument similar to Choe (1991) or DiBenedetto andFriedman (1984) we prove that u satisfies a Campanato type integral inequalityLet V isin nN be a constant vector such that 1

2 le V le Taking uxkminus Vk

2 as atest function to (31) we have the following lemma

Lemma 55 There exists a constant c independent of such that

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2tdx +intS R

2

upminus22u2 dz

le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

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630 Bae and Choe

Now suppose that v is a solution to

vit minus

(Ai

Qj

Vvjx

)x

= 0 (58)

with v = u on pSR2 Since 1

2 le V le le c for some fixed c we see that fromthe structure condition of F

2 le 2minuspAi

Qj

Vi

j le c1+ qminusp2

Hence we see that v satisfies a Campanato type integral inequality

Lemma 56 Suppose that 0 lt r le R2 There exists a constant c independent ofVR and such that

intSr

vminus vr 2 dz le c( r

R

)n+4 intS R

2

uminus V 2 dz (59)

Proof We write (58) as

vit minus pminus2

Ai

Qj

V

pminus2vjx

x

= 0

We change of variable = pminus2t minus t0 then the function vx = vx 2minusp+ t0satisfies a uniformly parabolic system

vi minus

Ai

Qj

V

pminus2vjx

x

= 0

Thus from results of Campanato (1966) we prove

intS

vminus v2 dz le c(

R

)n+4 intS R

2

uminus V 2 dz

and scaling back prove (59)

We estimate the L2 norm of uminus v in SR2

Lemma 57 For each lt R2 there exists a constant c independent of such that

intS

uminus v2 dz le c(1+ qminusp 4

n+2)(

minus2 ndashintSR

uminus V 2 dz) int

SRuminus V 2 dz

for = min(12

2n

)

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Regularity for Certain Nonlinear Parabolic Systems 631

Proof Subtracting(Ai

Qj

Vujx

)x

from both sides of (15) we get

uit minus

(Ai

Qj

Vujx

)x= (

Aiuminus Ai

Qj

Vujx

)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(510)

From (58) and (510) we see that uminus v satisfies

ui minus vit minus(Ai

Qj

Vuj minus vjx)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(511)

Simply applying uminus v to (511) we get

pminus2intS R

2uminusv2dzle

intS R

2

∣∣∣ int 1

0Ai

Qj

V+suminus VminusAi

Qj

Vds∣∣∣

times uminusV uminusvdz

and it follows thatintS R

2

uminus v2 dz

le minus2pminus2intS R

2

∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣2

uminus V 2 dz (512)

Note that AiV = F primeV V i

V and

Ai

Qj

V = F primeprimeV ViV

j

V 2 + F primeV (ij

V minus V iV

j

V 3)

where ij and are Kronecker delta functions Also we recall the structurecondition (13) F primeprimeprimeh le chpminus3 + hqminus3 We have that

∣∣∣ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds∣∣∣

le∣∣∣∣int 1

0F primeprime(V + suminus V)

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2

minusF primeprimeV ViV

j

V 2 ds

∣∣∣∣+int 1

0

∣∣∣∣F primeV + suminus VV + suminus V minus F primeV

V ∣∣∣∣ds

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632 Bae and Choe

+∣∣∣∣int 1

0F primeV + suminus V

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V3

minusF primeV ViV

j

V 3 ds

∣∣∣∣= I + II + III

The first term I is estimates as follows

I leint 1

0

∣∣∣F primeprimeV + suminus Vminus F primeprimeV ∣∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2∣∣∣∣∣ ds

+int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

(513)

By the mean value theorem we have that

int 1

0

∣∣∣F primeprimeV+suminusVminusF primeprimeV ∣∣∣ds le c

int 1

0

int 1

0

∣∣F primeprimeprimeV+shuminusV∣∣ds dhuminusV (514)

for some c and

int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

le cF primeprimeV int 1

0

int 1

0

ds dh∣∣V + shuminus V∣∣ uminus V (515)

If V ge 2uminus V then for all s isin 0 1 12 V le V + suminus V le 2V and from

(513) (514) and (515) we have I le c(V pminus3 + V qminus3

)uminus V If V le 2uminus V and p ge 2 then

∣∣F primeprimeV + suminus V∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le cuminus V pminus2 + cuminus V qminus2

and

∣∣F primeprimeV ∣∣ le cV pminus2 + cV qminus2 le cuminus V pminus2 + cuminus V qminus2

Hence we have that

I le cint 1

0

∣∣F primeprimeV + suminus V∣∣+ ∣∣F primeprimeV ∣∣ds le c1+ pminusquminus V pminus2

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Regularity for Certain Nonlinear Parabolic Systems 633

for some c independent of If V le 2uminus V and 1 lt p lt 2 then∣∣F primeprime(V + suminus V)∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le c1+ pminusqV + suminus Vpminus2

for some c We note thatint 1

0V + suminus Vpminus2 ds le uminus V pminus2

int 1

0

∣∣∣∣ V uminus V minus s

∣∣∣∣pminus2

ds le cuminus V pminus2

Therefore combining all these together we have

I le cV pminus3 uminus V (516)

for some c independent of For II and III we observe that

F primeV V le cV pminus2 + cV qminus2

∣∣∣∣FprimeV V i

Vj

V 3∣∣∣∣ le cV pminus2 + cV qminus2

∣∣∣∣(F primeV V

)∣∣∣∣ le cV pminus3 + cV qminus3

∣∣∣∣(F primeV V i

Vj

V 3)∣∣∣∣ le cV pminus3 + cV qminus3

Hence following the same argument for I we conclude that

II + III le cV pminus3 uminus V (517)

for some c independent of Combining (516) and (517) we obtain that∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣ le cV pminus3 uminus V (518)

and considering (518) in (512)intS R

2

uminus V 2 dz le cminus2pminus2V 2pminus3intS R

2

uminus V 4 dz

le cminus2intS R

2

uminus V 4 dz (519)

From Houmllderrsquos inequality we have thatintS R

2

uminus V 4 dz

=intS R

2

uminus V 8n+ 4nminus8n dz

le[

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx] 2

n int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

(520)

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634 Bae and Choe

From Lemma 55 we have

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx le c2 supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2 dx

le cp + 2qminusp

R2

intSR

uminus V 2 dz

for some c independent of Now recalling that 12 le V le we obtain that

int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

le c22minuspnminus2

n

int t0

t0minus 2minuspR24

[ intB R

2

(u + V )2pminus2uminus V 4 dx] nminus2

n

dt

le c4nminuspnminus8

n

int t0

t0minus 2minuspR24

[ intB R

2

∣∣u pminus22 uminus V pminus2

2 V∣∣ 2n

nminus2 dx] nminus2

n

dt

le c4nminuspnminus8

n

intS R

2

upminus2 2u2 dz+ c2nminuspnminus8

n pminus2intSR

uminus V 2 dz (521)

for some c independent of where we have used Sobolevrsquos inequality FromLemma 55 we haveint

S R2

upminus2 2u2 dz le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

Thus combining (519) (520) and (521) we have that

intS R

2uminus V 2 dz le c

(1+ qminusp 4

n+2)[

minus2ndashintSR

uminus V 2 dz] 2

n intSR

uminus V 2 dz

and this completes the proof

Now we prove an iteration lemma which says that if mean oscillation of unear some fixed V is small then u satisfies a Campanato type integral inequality

Lemma 58 Fix 0 lt 0 lt 1 then there exist small positive numbers independentof such that if V0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists V1 isin nN satisfying(12minusradic

0

) le V1 le

(1+radic

0

) (522)

ndashintSR

uminus V12 dz le 0 ndashintSR

uminus V02 dz (523)

ndashintSR

uminus V12 dz le 2

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Regularity for Certain Nonlinear Parabolic Systems 635

Proof Since V0 satisfies the assumption of Lemma 57 we apply Lemma 57 withV = V0 and obtainint

S R2

vminus u2 dz le cintSR

uminus V02 dz

So from the usual comparison argument we haveintSR

uminus uR2 dz le cintSR

vminus vR2 dz+ cintSR

uminus v2 dz

le cn+4intS R

2

uminus u R22 dz+ c

intS R

2

uminus v2 dz

le c(n+4 +

) intSR

uminus V02 dz

where c is independent of Choosing and isin 0 12 small we prove (523)

with V1 = uR Inequality ndashintSR

uminus V12 dz le 2 follows from (523) andassumption ndash

intSR

uminus V02 dz le 2 Now we also estimate

V1 minus V0 =∣∣∣∣∣ ndashintSR

vminus V0 dz

∣∣∣∣le c

[ndashintSR

vminus u2 dz] 1

2

+ c

[ndashintSR

uminus V02 dz] 1

2

le cminus1minusn2(1+

120

)12

Therefore using the same technique of DiBenedetto (1986) Lemma 44 and if wetake very small compared to (522) holds and this completes the proof

Lemma 59 There exist small positive numbers 0 independent of such that ifV0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists a sequence of vectors Vi i = 1 2 3 sub nN satisfying

14 le Vi le 4 ndash

intSiR

uminus Vi2 dz le 2

ndashintSi+1R

uminus Vi+12 dz le 0 ndashintSiR

uminus Vi2 dz

Proof For the proof of the lemma we need to show that 14 le Vi le 4 But this

follows from that

Vi+1 minus Vi2 le c(0 + minusnminus2

)∣∣∣ ndashintSiR

uminus Vi dz∣∣∣

le ci0

i∣∣∣ ndashint

SRuminus V02 dz

∣∣∣ le ci0

i2

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636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

Dow

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t 200

7

ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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622 Bae and Choe

Now we assume that p2+ 1 gt q ge p gt 1 In this case s + 2 gt s + 2q minus p andwe have from (310)

intS

us+p+s+2 2n dx le c

[1

Rminus 2

intSR

us+2 dz+ SRRminus 2

]1+ 2n

(313)

for some c We set si by sidef= is0 + np2+ 2minus ni minus 1 Hence if s0 gt nminus 2

minus np2 then

limi

si = and limi

sii

= s0 + 2minus n+ np

2

We take s = si = Ri+1 and R = Ri in (313) Iterating (313) we obtain

supS R0

2

u le c

[ndashintSR0

us0+2 dz

] 1s0+2+ np

2 minusn + c

for some c independent of R0 and this completes the proof for the casep2+ 1 gt q ge p gt 1

4 HOumlLDER CONTINUITY OF u

In this section we prove that u is Houmllder continuous

Theorem 41 Define r0 by the constant appearing in Theorem 21 Suppose that uis L

r0locT then u is Houmllder continuous Moreover for all Q2R sub T with R lt 1 u

satisfies

ndashintQR

uminus uR2 dz le cRs

for some c independent of R where we define

s =2 when p ge 2

ppminus 1 when p isin 1 2

We introduce a nonnegative cutoff function isin 0 BR such that = 1 in

BR2 0 le le 1 le cR We define minus

R

def= t0 minus 2Rp t0 minus Rp and QminusR = BR timesminus

R

Also we define uRt

def= 1BRintBR

ux tdx and uR

def= 1QRintQR

ux tdz Firstwe prove a lemma which is essential for a Poincareacute inequality for solutions of adegenerate parabolic system

Lemma 42 Suppose Q2R sub T then u satisfies the following inequality

suptisinR

intminus

R

dsintBR

2ux tminus uRs2 dx

le cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz+ cintQminus

R

uminus uRt2 dz

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Regularity for Certain Nonlinear Parabolic Systems 623

for all R lt R0 where c depends only on nN and p and

s1 = p s2 = p and s3 = pqminus1

pminus1 when p ge 2

s1 = p2minus p s2 = ppminus 1 and s3 = pq minus 1 when 1 lt p lt 2

Proof Since u isin 0(0 T L2

) there exists ts isin R for all s isin minus

R such that

intBR

pux t minus uRs2dx = suptisinR

intBR

pux tminus uRs2 dx

We take uminus uRspxst as a test function to (15) where st is the characteristic

function such that

st =1 for all isin s t

0 for all isin s t

Hence we have

intut middot uminus uRs

pxst dz+int

Au middot uminus uRspst dz = 0 (41)

Now we assume p isin 2 Considering the structure condition (12) on A (23)and (25) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

leint

uminus uRs2x sp dx + cintBRtimesst

upminus1uminus uRspminus1dz

+ cintBRtimesst

uqminus1uminus uRspminus1dz (42)

for some c independent of R From Youngrsquos inequality we have

intBRtimesst

upminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

up dz (43)

for some c and small Similarly the last term in (42) can be estimated as follows

intBRtimesst

uqminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

u pqminus1pminus1 dz (44)

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624 Bae and Choe

for some c and small Combining (42) (43) and (44) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

le cint

uminus uRs2x sp dx +

Rp

intBRtimesst

uminus uRspp dz

+ cintBRtimesst

up dz+ cintBRtimesst

u pqminus1pminus1 dz (45)

for some c and small Integrating (45) with respect to s from t0 minus 2Rp to t0 minus Rpwe have

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRs2p dz+ c

Rp

intminus

R

dsintBRtimesst

uminus uRs2p dz

+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

By the choice of t we have that for small

intminus

R

dsintBR

uminusuRs2x t pdx

le cintQminus

R

uminusuRs2p dz+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

So we proved Lemma 42 when p ge 2In case p isin 1 2 we estimate the second term of (41) as follows

intuppst dz+ p

intAu middot uminus uRs

pminus1st dz

leint

Au middot uminus uRspst dz (46)

and using the fact that u is bounded and ppminus 1 gt 2 we have

∣∣∣ int Au middot uminusuRspminus1st dz

∣∣∣le c

intBRtimesst

upminus1uminusuRspminus1st dz

+ cintBRtimesst

uqminus1pminus1uminusuRsst dz

le

Rp

intuminusuRs

ppminus1 pst dz+

c

Rp2minusp

intuppminus1 dz+ c

Rp2minusp

intupqminus1 dz

(47)

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Regularity for Certain Nonlinear Parabolic Systems 625

Combining (41) (46) and (47) we have

intBR

uminus uRs2x t p dx

le cu2minusppminus1L

Rp

intBRtimesst

uminus uRs2pst dz+c

Rp2minusp

intBRtimesst

uppminus1 dz

+ c

Rp2minusp

intBRtimesst

upqminus1 dz+intBR

uminus uRs2x sp dx (48)

for small Integrating (48) with respect to s and considering the choice of t weprove Lemma 42 when 1 lt p lt 2

Now we prove a Poincareacute inequality

Theorem 43 Suppose Q2R sub T with R lt 1 then u satisfies the following inequality

intQR

2

uminus uR22 dz le cR2

intQ2R

u2 dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

where c is independent of R

Proof Using Lemma 42 we have

intQR

2

uminus uR22 dz le

intQR

uminus uRs2p dz

le c

Rp

intminus

R

dsintQR

uminus uRs2p dz le c suptisinR

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRt2p dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

le cR2intQ2R

u2 dz+ cRs1

intQ2R

us2z+ cRs1

intQ2R

us3 dz

where we used a Poincareacute type inequality for x variables only that is

intBR

uminus uRs2p dx le cR2intBR

u2 dx

Proof of Theorem 41 Since u is bounded owing to Theorem 32 we have fromTheorem 43

intQR

uminus uRs2 dz lecRn+p+2 when p isin 2

cRn+p+ppminus1 when p isin 1 2

where c is independent of R Hence by isomorphism theorem of Da Prato (1965) weprove Theorem 41

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626 Bae and Choe

5 HOumlLDER CONTINUITY FOR u

In this section we prove that u is Houmllder continuous Because of degeneracyof the pde we need scaled cylinders (see DiBenedetto and Friedman 1985)Let R0 gt 0 be fixed and consider a cylinder

SR0

def= BR0times t0 minus R2minus

0 t0

for small For any 0 lt R le R02 we introduce a scaled cylinder

SRdef= BR times t0 minus R22minusp t0

for fixed ge supSR0u

We assume that pminus2 gt R0 if p gt 2 Hence we note that SR sub S

R0and

ge supSRu

The following theorem is our main theorem in this section

Theorem 51 Suppose that SR sub T then u satisfies for all r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR 2 dz+ cR

for some gt 0

For the proof of Theorem 51 we follow a usual program for degenerateequations such that if for some isin 0 1∣∣z isin SR u2 le 1minus 2

∣∣ gt ∣∣SR

∣∣ (51)

then

supS R

2

u le (52)

for some small isin 0 1 and otherwise that is (51) fails then for all 0 lt r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR2 dz (53)

for some gt 0 Combining (52) and (53) we prove Theorem 51First we prove that (52) under the assumption (51) A similar argument for

the case p = q ge 2 appeared in DiBenedetto and Friedman (1985) We assume that(51) is true Set v = u2 For each h gt 0 we set AhRt

def= x isin BR v gt h and

BhR

def= z isin SR v gt h From the mean value theorem we estimate AhRt

Lemma 52 For any isin 0 there exists isin t0 minus 2minuspR20 t0 minus 2minuspR2

0 such that

A1minus2R0 le 1minus

1minus BR0

where is independent of

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Regularity for Certain Nonlinear Parabolic Systems 627

Proof Suppose that the assertion is not true then

B1minus2R0 ge

int t0minus2minuspR20

t0minus2minuspR20

A1minus2R0d gt 1minus BR0

2minuspR20 = 1minus SR0

and this contradicts the assumption (51)

The following lemma is useful in estimating sup u Indeed when p = q ge 2DiBenedetto and Friedman proved a similar theorem (see Lemma 62 inDiBenedetto and Friedman 1985) Now choose = 2 and in Lemma 52We define St

R

def= BR times t

Lemma 53 There exists a positive integer r independent of such that

∣∣A1minus 2r

2R0t

∣∣ le (1minus

(2

)2)BR0

for all t isin t0 minus

22minuspR2

0 t0

Proof Consider the function

v = log+[

2

2minus vminus 1minus 2+ + 2

2r

]

where r is a positive integer This function was used in Choe (1991) and DiBenedettoand Friedman (1985) We take 13 = ui

x2vprime 2 as a test function for (31) where

isin 0 is a standard cutoff function such that B1minusR for small decided later is

contained in support Integrating from to t we obtain

12

intStR

vt2prime 2 dz+

intStR

Ai

Qj

uujxx

uixx

2prime 2 dz

+ 2intStR

Ai

Qj

uujxx

uixukxx

ukx2primeprime 2 dz

+ 2intStR

Ai

Qj

uujxx

uix x

2prime dz = 0

Considering the fact 2primeprime = 21+ prime2 we have

intBR

2vx t 2xdx +intStR

upminus2v21+ prime2 2 dz

leintBR

2vx 2xdx + cintStR

upminus2v prime dz

+ cintStR

uqminus2v prime dz

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628 Bae and Choe

Applying Youngrsquos inequality on the last two terms we getintBR

2vx t 2xdx

leintBR

2vx 2xdx + cintStR

upminus2 2 dz+ cintStR

u2qminuspminus2 2 dz

for some c Since le r log 2 and u is bounded we getintBR

2vx t 2xdx leintBR

2vx 2xdx + crBR

From Lemma 52 we getintBR

2vx t 2xdx le r log 221minus

1minus

2

BR

From definition of we haveintB1minusRcapvgt1minus

2r 22vx tdx ge r minus 1 log 22A1minus

2r 21minusRt

and

A1minus 2r

21minusRt le[( r

r minus 1

)2 1minus

1minus

2

+ c

r

]BR

Hence we obtain

A(1minus

2r

)2R

t le A1minus 2r

21minusRt + BRB1minusR

le[( r

r minus 1

)2 1minus

1minus

2

+ c

r+ n

]BR

Taking r so large that( r

r minus 1

)2 le(1minus

2

)1+

c

rle 3

82

and = 382n we obtain

A1minus 2r

2Rt le(1minus

(2

)2)BR

We take 13 = uixvminus h+ 2 as a test function in (31) where v = u2 and is

a cutoff function such that = 1 in Sr = 0 on pSR le cRminus r and t le cpminus2Rminus r2 Considering the ellipticity condition (12) we have

supt

int ∣∣vminus h+∣∣2 2 dx +

intupminus2vminus h+2 2 dz

le c

Rminus r2

intSR

upminus2vminus h+2 dz

+ c

Rminus r2

intSR

u2qminuspminus2vminus h+2 dz+ cintSR

vminus h+2 tdz (54)

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Regularity for Certain Nonlinear Parabolic Systems 629

for some c Now choose h ge 12 For such a choice of h we haveint

upminus2vminus h+2 2 dz ge cpminus2int

vminus h+2 2 dz

Hence from (54) we obtain

supt

intvminus h+2 2 dx + pminus2

intvminus h+2 2 dz

le[

2qminuspminus2

Rminus r2+ pminus2

Rminus r2

] intSR

vminus h+2 dz (55)

for some c independent of Observe that is bounded above by some fixednumber We make change of variables

= pminus2t minus t0 vx = vx t0 + 2minusp

and we write (55) as

supt

intBR

vminus h+2 2 dx +intSR

vminus h+2 2 dz le c

Rminus r2

intSR

vminus h+dz

Once we have this Caccioppoli type inequality we can apply truncation argumentTherefore following to similar argument in DiBenedetto and Friedman (1984 or1985) we have a maximum principle Now the proof for the next lemma is almostthe same as in DiBenedetto and Friedman (1984 or 1985) we omit the proof

Lemma 54 If for some and isin 0 1∣∣x t isin S1+R u2 lt 1minus 2∣∣ ge S1+R (56)

then there exists a number isin 0 1 independent of and R such that

supS R

2

ux t le

Now we assume that (56) fails that is∣∣x t isin S1+R u2 lt 1minus 2∣∣ lt S1+R (57)

Following a comparison argument similar to Choe (1991) or DiBenedetto andFriedman (1984) we prove that u satisfies a Campanato type integral inequalityLet V isin nN be a constant vector such that 1

2 le V le Taking uxkminus Vk

2 as atest function to (31) we have the following lemma

Lemma 55 There exists a constant c independent of such that

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2tdx +intS R

2

upminus22u2 dz

le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

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630 Bae and Choe

Now suppose that v is a solution to

vit minus

(Ai

Qj

Vvjx

)x

= 0 (58)

with v = u on pSR2 Since 1

2 le V le le c for some fixed c we see that fromthe structure condition of F

2 le 2minuspAi

Qj

Vi

j le c1+ qminusp2

Hence we see that v satisfies a Campanato type integral inequality

Lemma 56 Suppose that 0 lt r le R2 There exists a constant c independent ofVR and such that

intSr

vminus vr 2 dz le c( r

R

)n+4 intS R

2

uminus V 2 dz (59)

Proof We write (58) as

vit minus pminus2

Ai

Qj

V

pminus2vjx

x

= 0

We change of variable = pminus2t minus t0 then the function vx = vx 2minusp+ t0satisfies a uniformly parabolic system

vi minus

Ai

Qj

V

pminus2vjx

x

= 0

Thus from results of Campanato (1966) we prove

intS

vminus v2 dz le c(

R

)n+4 intS R

2

uminus V 2 dz

and scaling back prove (59)

We estimate the L2 norm of uminus v in SR2

Lemma 57 For each lt R2 there exists a constant c independent of such that

intS

uminus v2 dz le c(1+ qminusp 4

n+2)(

minus2 ndashintSR

uminus V 2 dz) int

SRuminus V 2 dz

for = min(12

2n

)

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Regularity for Certain Nonlinear Parabolic Systems 631

Proof Subtracting(Ai

Qj

Vujx

)x

from both sides of (15) we get

uit minus

(Ai

Qj

Vujx

)x= (

Aiuminus Ai

Qj

Vujx

)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(510)

From (58) and (510) we see that uminus v satisfies

ui minus vit minus(Ai

Qj

Vuj minus vjx)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(511)

Simply applying uminus v to (511) we get

pminus2intS R

2uminusv2dzle

intS R

2

∣∣∣ int 1

0Ai

Qj

V+suminus VminusAi

Qj

Vds∣∣∣

times uminusV uminusvdz

and it follows thatintS R

2

uminus v2 dz

le minus2pminus2intS R

2

∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣2

uminus V 2 dz (512)

Note that AiV = F primeV V i

V and

Ai

Qj

V = F primeprimeV ViV

j

V 2 + F primeV (ij

V minus V iV

j

V 3)

where ij and are Kronecker delta functions Also we recall the structurecondition (13) F primeprimeprimeh le chpminus3 + hqminus3 We have that

∣∣∣ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds∣∣∣

le∣∣∣∣int 1

0F primeprime(V + suminus V)

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2

minusF primeprimeV ViV

j

V 2 ds

∣∣∣∣+int 1

0

∣∣∣∣F primeV + suminus VV + suminus V minus F primeV

V ∣∣∣∣ds

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632 Bae and Choe

+∣∣∣∣int 1

0F primeV + suminus V

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V3

minusF primeV ViV

j

V 3 ds

∣∣∣∣= I + II + III

The first term I is estimates as follows

I leint 1

0

∣∣∣F primeprimeV + suminus Vminus F primeprimeV ∣∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2∣∣∣∣∣ ds

+int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

(513)

By the mean value theorem we have that

int 1

0

∣∣∣F primeprimeV+suminusVminusF primeprimeV ∣∣∣ds le c

int 1

0

int 1

0

∣∣F primeprimeprimeV+shuminusV∣∣ds dhuminusV (514)

for some c and

int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

le cF primeprimeV int 1

0

int 1

0

ds dh∣∣V + shuminus V∣∣ uminus V (515)

If V ge 2uminus V then for all s isin 0 1 12 V le V + suminus V le 2V and from

(513) (514) and (515) we have I le c(V pminus3 + V qminus3

)uminus V If V le 2uminus V and p ge 2 then

∣∣F primeprimeV + suminus V∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le cuminus V pminus2 + cuminus V qminus2

and

∣∣F primeprimeV ∣∣ le cV pminus2 + cV qminus2 le cuminus V pminus2 + cuminus V qminus2

Hence we have that

I le cint 1

0

∣∣F primeprimeV + suminus V∣∣+ ∣∣F primeprimeV ∣∣ds le c1+ pminusquminus V pminus2

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Regularity for Certain Nonlinear Parabolic Systems 633

for some c independent of If V le 2uminus V and 1 lt p lt 2 then∣∣F primeprime(V + suminus V)∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le c1+ pminusqV + suminus Vpminus2

for some c We note thatint 1

0V + suminus Vpminus2 ds le uminus V pminus2

int 1

0

∣∣∣∣ V uminus V minus s

∣∣∣∣pminus2

ds le cuminus V pminus2

Therefore combining all these together we have

I le cV pminus3 uminus V (516)

for some c independent of For II and III we observe that

F primeV V le cV pminus2 + cV qminus2

∣∣∣∣FprimeV V i

Vj

V 3∣∣∣∣ le cV pminus2 + cV qminus2

∣∣∣∣(F primeV V

)∣∣∣∣ le cV pminus3 + cV qminus3

∣∣∣∣(F primeV V i

Vj

V 3)∣∣∣∣ le cV pminus3 + cV qminus3

Hence following the same argument for I we conclude that

II + III le cV pminus3 uminus V (517)

for some c independent of Combining (516) and (517) we obtain that∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣ le cV pminus3 uminus V (518)

and considering (518) in (512)intS R

2

uminus V 2 dz le cminus2pminus2V 2pminus3intS R

2

uminus V 4 dz

le cminus2intS R

2

uminus V 4 dz (519)

From Houmllderrsquos inequality we have thatintS R

2

uminus V 4 dz

=intS R

2

uminus V 8n+ 4nminus8n dz

le[

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx] 2

n int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

(520)

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634 Bae and Choe

From Lemma 55 we have

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx le c2 supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2 dx

le cp + 2qminusp

R2

intSR

uminus V 2 dz

for some c independent of Now recalling that 12 le V le we obtain that

int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

le c22minuspnminus2

n

int t0

t0minus 2minuspR24

[ intB R

2

(u + V )2pminus2uminus V 4 dx] nminus2

n

dt

le c4nminuspnminus8

n

int t0

t0minus 2minuspR24

[ intB R

2

∣∣u pminus22 uminus V pminus2

2 V∣∣ 2n

nminus2 dx] nminus2

n

dt

le c4nminuspnminus8

n

intS R

2

upminus2 2u2 dz+ c2nminuspnminus8

n pminus2intSR

uminus V 2 dz (521)

for some c independent of where we have used Sobolevrsquos inequality FromLemma 55 we haveint

S R2

upminus2 2u2 dz le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

Thus combining (519) (520) and (521) we have that

intS R

2uminus V 2 dz le c

(1+ qminusp 4

n+2)[

minus2ndashintSR

uminus V 2 dz] 2

n intSR

uminus V 2 dz

and this completes the proof

Now we prove an iteration lemma which says that if mean oscillation of unear some fixed V is small then u satisfies a Campanato type integral inequality

Lemma 58 Fix 0 lt 0 lt 1 then there exist small positive numbers independentof such that if V0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists V1 isin nN satisfying(12minusradic

0

) le V1 le

(1+radic

0

) (522)

ndashintSR

uminus V12 dz le 0 ndashintSR

uminus V02 dz (523)

ndashintSR

uminus V12 dz le 2

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Regularity for Certain Nonlinear Parabolic Systems 635

Proof Since V0 satisfies the assumption of Lemma 57 we apply Lemma 57 withV = V0 and obtainint

S R2

vminus u2 dz le cintSR

uminus V02 dz

So from the usual comparison argument we haveintSR

uminus uR2 dz le cintSR

vminus vR2 dz+ cintSR

uminus v2 dz

le cn+4intS R

2

uminus u R22 dz+ c

intS R

2

uminus v2 dz

le c(n+4 +

) intSR

uminus V02 dz

where c is independent of Choosing and isin 0 12 small we prove (523)

with V1 = uR Inequality ndashintSR

uminus V12 dz le 2 follows from (523) andassumption ndash

intSR

uminus V02 dz le 2 Now we also estimate

V1 minus V0 =∣∣∣∣∣ ndashintSR

vminus V0 dz

∣∣∣∣le c

[ndashintSR

vminus u2 dz] 1

2

+ c

[ndashintSR

uminus V02 dz] 1

2

le cminus1minusn2(1+

120

)12

Therefore using the same technique of DiBenedetto (1986) Lemma 44 and if wetake very small compared to (522) holds and this completes the proof

Lemma 59 There exist small positive numbers 0 independent of such that ifV0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists a sequence of vectors Vi i = 1 2 3 sub nN satisfying

14 le Vi le 4 ndash

intSiR

uminus Vi2 dz le 2

ndashintSi+1R

uminus Vi+12 dz le 0 ndashintSiR

uminus Vi2 dz

Proof For the proof of the lemma we need to show that 14 le Vi le 4 But this

follows from that

Vi+1 minus Vi2 le c(0 + minusnminus2

)∣∣∣ ndashintSiR

uminus Vi dz∣∣∣

le ci0

i∣∣∣ ndashint

SRuminus V02 dz

∣∣∣ le ci0

i2

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636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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t 200

7

ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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Regularity for Certain Nonlinear Parabolic Systems 623

for all R lt R0 where c depends only on nN and p and

s1 = p s2 = p and s3 = pqminus1

pminus1 when p ge 2

s1 = p2minus p s2 = ppminus 1 and s3 = pq minus 1 when 1 lt p lt 2

Proof Since u isin 0(0 T L2

) there exists ts isin R for all s isin minus

R such that

intBR

pux t minus uRs2dx = suptisinR

intBR

pux tminus uRs2 dx

We take uminus uRspxst as a test function to (15) where st is the characteristic

function such that

st =1 for all isin s t

0 for all isin s t

Hence we have

intut middot uminus uRs

pxst dz+int

Au middot uminus uRspst dz = 0 (41)

Now we assume p isin 2 Considering the structure condition (12) on A (23)and (25) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

leint

uminus uRs2x sp dx + cintBRtimesst

upminus1uminus uRspminus1dz

+ cintBRtimesst

uqminus1uminus uRspminus1dz (42)

for some c independent of R From Youngrsquos inequality we have

intBRtimesst

upminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

up dz (43)

for some c and small Similarly the last term in (42) can be estimated as follows

intBRtimesst

uqminus1uminus uRspminus1dz

le

Rp

intBRtimesst

uminus uRspp dz+ cintBRtimesst

u pqminus1pminus1 dz (44)

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624 Bae and Choe

for some c and small Combining (42) (43) and (44) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

le cint

uminus uRs2x sp dx +

Rp

intBRtimesst

uminus uRspp dz

+ cintBRtimesst

up dz+ cintBRtimesst

u pqminus1pminus1 dz (45)

for some c and small Integrating (45) with respect to s from t0 minus 2Rp to t0 minus Rpwe have

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRs2p dz+ c

Rp

intminus

R

dsintBRtimesst

uminus uRs2p dz

+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

By the choice of t we have that for small

intminus

R

dsintBR

uminusuRs2x t pdx

le cintQminus

R

uminusuRs2p dz+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

So we proved Lemma 42 when p ge 2In case p isin 1 2 we estimate the second term of (41) as follows

intuppst dz+ p

intAu middot uminus uRs

pminus1st dz

leint

Au middot uminus uRspst dz (46)

and using the fact that u is bounded and ppminus 1 gt 2 we have

∣∣∣ int Au middot uminusuRspminus1st dz

∣∣∣le c

intBRtimesst

upminus1uminusuRspminus1st dz

+ cintBRtimesst

uqminus1pminus1uminusuRsst dz

le

Rp

intuminusuRs

ppminus1 pst dz+

c

Rp2minusp

intuppminus1 dz+ c

Rp2minusp

intupqminus1 dz

(47)

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Regularity for Certain Nonlinear Parabolic Systems 625

Combining (41) (46) and (47) we have

intBR

uminus uRs2x t p dx

le cu2minusppminus1L

Rp

intBRtimesst

uminus uRs2pst dz+c

Rp2minusp

intBRtimesst

uppminus1 dz

+ c

Rp2minusp

intBRtimesst

upqminus1 dz+intBR

uminus uRs2x sp dx (48)

for small Integrating (48) with respect to s and considering the choice of t weprove Lemma 42 when 1 lt p lt 2

Now we prove a Poincareacute inequality

Theorem 43 Suppose Q2R sub T with R lt 1 then u satisfies the following inequality

intQR

2

uminus uR22 dz le cR2

intQ2R

u2 dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

where c is independent of R

Proof Using Lemma 42 we have

intQR

2

uminus uR22 dz le

intQR

uminus uRs2p dz

le c

Rp

intminus

R

dsintQR

uminus uRs2p dz le c suptisinR

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRt2p dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

le cR2intQ2R

u2 dz+ cRs1

intQ2R

us2z+ cRs1

intQ2R

us3 dz

where we used a Poincareacute type inequality for x variables only that is

intBR

uminus uRs2p dx le cR2intBR

u2 dx

Proof of Theorem 41 Since u is bounded owing to Theorem 32 we have fromTheorem 43

intQR

uminus uRs2 dz lecRn+p+2 when p isin 2

cRn+p+ppminus1 when p isin 1 2

where c is independent of R Hence by isomorphism theorem of Da Prato (1965) weprove Theorem 41

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626 Bae and Choe

5 HOumlLDER CONTINUITY FOR u

In this section we prove that u is Houmllder continuous Because of degeneracyof the pde we need scaled cylinders (see DiBenedetto and Friedman 1985)Let R0 gt 0 be fixed and consider a cylinder

SR0

def= BR0times t0 minus R2minus

0 t0

for small For any 0 lt R le R02 we introduce a scaled cylinder

SRdef= BR times t0 minus R22minusp t0

for fixed ge supSR0u

We assume that pminus2 gt R0 if p gt 2 Hence we note that SR sub S

R0and

ge supSRu

The following theorem is our main theorem in this section

Theorem 51 Suppose that SR sub T then u satisfies for all r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR 2 dz+ cR

for some gt 0

For the proof of Theorem 51 we follow a usual program for degenerateequations such that if for some isin 0 1∣∣z isin SR u2 le 1minus 2

∣∣ gt ∣∣SR

∣∣ (51)

then

supS R

2

u le (52)

for some small isin 0 1 and otherwise that is (51) fails then for all 0 lt r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR2 dz (53)

for some gt 0 Combining (52) and (53) we prove Theorem 51First we prove that (52) under the assumption (51) A similar argument for

the case p = q ge 2 appeared in DiBenedetto and Friedman (1985) We assume that(51) is true Set v = u2 For each h gt 0 we set AhRt

def= x isin BR v gt h and

BhR

def= z isin SR v gt h From the mean value theorem we estimate AhRt

Lemma 52 For any isin 0 there exists isin t0 minus 2minuspR20 t0 minus 2minuspR2

0 such that

A1minus2R0 le 1minus

1minus BR0

where is independent of

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Regularity for Certain Nonlinear Parabolic Systems 627

Proof Suppose that the assertion is not true then

B1minus2R0 ge

int t0minus2minuspR20

t0minus2minuspR20

A1minus2R0d gt 1minus BR0

2minuspR20 = 1minus SR0

and this contradicts the assumption (51)

The following lemma is useful in estimating sup u Indeed when p = q ge 2DiBenedetto and Friedman proved a similar theorem (see Lemma 62 inDiBenedetto and Friedman 1985) Now choose = 2 and in Lemma 52We define St

R

def= BR times t

Lemma 53 There exists a positive integer r independent of such that

∣∣A1minus 2r

2R0t

∣∣ le (1minus

(2

)2)BR0

for all t isin t0 minus

22minuspR2

0 t0

Proof Consider the function

v = log+[

2

2minus vminus 1minus 2+ + 2

2r

]

where r is a positive integer This function was used in Choe (1991) and DiBenedettoand Friedman (1985) We take 13 = ui

x2vprime 2 as a test function for (31) where

isin 0 is a standard cutoff function such that B1minusR for small decided later is

contained in support Integrating from to t we obtain

12

intStR

vt2prime 2 dz+

intStR

Ai

Qj

uujxx

uixx

2prime 2 dz

+ 2intStR

Ai

Qj

uujxx

uixukxx

ukx2primeprime 2 dz

+ 2intStR

Ai

Qj

uujxx

uix x

2prime dz = 0

Considering the fact 2primeprime = 21+ prime2 we have

intBR

2vx t 2xdx +intStR

upminus2v21+ prime2 2 dz

leintBR

2vx 2xdx + cintStR

upminus2v prime dz

+ cintStR

uqminus2v prime dz

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628 Bae and Choe

Applying Youngrsquos inequality on the last two terms we getintBR

2vx t 2xdx

leintBR

2vx 2xdx + cintStR

upminus2 2 dz+ cintStR

u2qminuspminus2 2 dz

for some c Since le r log 2 and u is bounded we getintBR

2vx t 2xdx leintBR

2vx 2xdx + crBR

From Lemma 52 we getintBR

2vx t 2xdx le r log 221minus

1minus

2

BR

From definition of we haveintB1minusRcapvgt1minus

2r 22vx tdx ge r minus 1 log 22A1minus

2r 21minusRt

and

A1minus 2r

21minusRt le[( r

r minus 1

)2 1minus

1minus

2

+ c

r

]BR

Hence we obtain

A(1minus

2r

)2R

t le A1minus 2r

21minusRt + BRB1minusR

le[( r

r minus 1

)2 1minus

1minus

2

+ c

r+ n

]BR

Taking r so large that( r

r minus 1

)2 le(1minus

2

)1+

c

rle 3

82

and = 382n we obtain

A1minus 2r

2Rt le(1minus

(2

)2)BR

We take 13 = uixvminus h+ 2 as a test function in (31) where v = u2 and is

a cutoff function such that = 1 in Sr = 0 on pSR le cRminus r and t le cpminus2Rminus r2 Considering the ellipticity condition (12) we have

supt

int ∣∣vminus h+∣∣2 2 dx +

intupminus2vminus h+2 2 dz

le c

Rminus r2

intSR

upminus2vminus h+2 dz

+ c

Rminus r2

intSR

u2qminuspminus2vminus h+2 dz+ cintSR

vminus h+2 tdz (54)

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Regularity for Certain Nonlinear Parabolic Systems 629

for some c Now choose h ge 12 For such a choice of h we haveint

upminus2vminus h+2 2 dz ge cpminus2int

vminus h+2 2 dz

Hence from (54) we obtain

supt

intvminus h+2 2 dx + pminus2

intvminus h+2 2 dz

le[

2qminuspminus2

Rminus r2+ pminus2

Rminus r2

] intSR

vminus h+2 dz (55)

for some c independent of Observe that is bounded above by some fixednumber We make change of variables

= pminus2t minus t0 vx = vx t0 + 2minusp

and we write (55) as

supt

intBR

vminus h+2 2 dx +intSR

vminus h+2 2 dz le c

Rminus r2

intSR

vminus h+dz

Once we have this Caccioppoli type inequality we can apply truncation argumentTherefore following to similar argument in DiBenedetto and Friedman (1984 or1985) we have a maximum principle Now the proof for the next lemma is almostthe same as in DiBenedetto and Friedman (1984 or 1985) we omit the proof

Lemma 54 If for some and isin 0 1∣∣x t isin S1+R u2 lt 1minus 2∣∣ ge S1+R (56)

then there exists a number isin 0 1 independent of and R such that

supS R

2

ux t le

Now we assume that (56) fails that is∣∣x t isin S1+R u2 lt 1minus 2∣∣ lt S1+R (57)

Following a comparison argument similar to Choe (1991) or DiBenedetto andFriedman (1984) we prove that u satisfies a Campanato type integral inequalityLet V isin nN be a constant vector such that 1

2 le V le Taking uxkminus Vk

2 as atest function to (31) we have the following lemma

Lemma 55 There exists a constant c independent of such that

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2tdx +intS R

2

upminus22u2 dz

le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

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630 Bae and Choe

Now suppose that v is a solution to

vit minus

(Ai

Qj

Vvjx

)x

= 0 (58)

with v = u on pSR2 Since 1

2 le V le le c for some fixed c we see that fromthe structure condition of F

2 le 2minuspAi

Qj

Vi

j le c1+ qminusp2

Hence we see that v satisfies a Campanato type integral inequality

Lemma 56 Suppose that 0 lt r le R2 There exists a constant c independent ofVR and such that

intSr

vminus vr 2 dz le c( r

R

)n+4 intS R

2

uminus V 2 dz (59)

Proof We write (58) as

vit minus pminus2

Ai

Qj

V

pminus2vjx

x

= 0

We change of variable = pminus2t minus t0 then the function vx = vx 2minusp+ t0satisfies a uniformly parabolic system

vi minus

Ai

Qj

V

pminus2vjx

x

= 0

Thus from results of Campanato (1966) we prove

intS

vminus v2 dz le c(

R

)n+4 intS R

2

uminus V 2 dz

and scaling back prove (59)

We estimate the L2 norm of uminus v in SR2

Lemma 57 For each lt R2 there exists a constant c independent of such that

intS

uminus v2 dz le c(1+ qminusp 4

n+2)(

minus2 ndashintSR

uminus V 2 dz) int

SRuminus V 2 dz

for = min(12

2n

)

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Regularity for Certain Nonlinear Parabolic Systems 631

Proof Subtracting(Ai

Qj

Vujx

)x

from both sides of (15) we get

uit minus

(Ai

Qj

Vujx

)x= (

Aiuminus Ai

Qj

Vujx

)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(510)

From (58) and (510) we see that uminus v satisfies

ui minus vit minus(Ai

Qj

Vuj minus vjx)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(511)

Simply applying uminus v to (511) we get

pminus2intS R

2uminusv2dzle

intS R

2

∣∣∣ int 1

0Ai

Qj

V+suminus VminusAi

Qj

Vds∣∣∣

times uminusV uminusvdz

and it follows thatintS R

2

uminus v2 dz

le minus2pminus2intS R

2

∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣2

uminus V 2 dz (512)

Note that AiV = F primeV V i

V and

Ai

Qj

V = F primeprimeV ViV

j

V 2 + F primeV (ij

V minus V iV

j

V 3)

where ij and are Kronecker delta functions Also we recall the structurecondition (13) F primeprimeprimeh le chpminus3 + hqminus3 We have that

∣∣∣ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds∣∣∣

le∣∣∣∣int 1

0F primeprime(V + suminus V)

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2

minusF primeprimeV ViV

j

V 2 ds

∣∣∣∣+int 1

0

∣∣∣∣F primeV + suminus VV + suminus V minus F primeV

V ∣∣∣∣ds

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632 Bae and Choe

+∣∣∣∣int 1

0F primeV + suminus V

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V3

minusF primeV ViV

j

V 3 ds

∣∣∣∣= I + II + III

The first term I is estimates as follows

I leint 1

0

∣∣∣F primeprimeV + suminus Vminus F primeprimeV ∣∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2∣∣∣∣∣ ds

+int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

(513)

By the mean value theorem we have that

int 1

0

∣∣∣F primeprimeV+suminusVminusF primeprimeV ∣∣∣ds le c

int 1

0

int 1

0

∣∣F primeprimeprimeV+shuminusV∣∣ds dhuminusV (514)

for some c and

int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

le cF primeprimeV int 1

0

int 1

0

ds dh∣∣V + shuminus V∣∣ uminus V (515)

If V ge 2uminus V then for all s isin 0 1 12 V le V + suminus V le 2V and from

(513) (514) and (515) we have I le c(V pminus3 + V qminus3

)uminus V If V le 2uminus V and p ge 2 then

∣∣F primeprimeV + suminus V∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le cuminus V pminus2 + cuminus V qminus2

and

∣∣F primeprimeV ∣∣ le cV pminus2 + cV qminus2 le cuminus V pminus2 + cuminus V qminus2

Hence we have that

I le cint 1

0

∣∣F primeprimeV + suminus V∣∣+ ∣∣F primeprimeV ∣∣ds le c1+ pminusquminus V pminus2

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Regularity for Certain Nonlinear Parabolic Systems 633

for some c independent of If V le 2uminus V and 1 lt p lt 2 then∣∣F primeprime(V + suminus V)∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le c1+ pminusqV + suminus Vpminus2

for some c We note thatint 1

0V + suminus Vpminus2 ds le uminus V pminus2

int 1

0

∣∣∣∣ V uminus V minus s

∣∣∣∣pminus2

ds le cuminus V pminus2

Therefore combining all these together we have

I le cV pminus3 uminus V (516)

for some c independent of For II and III we observe that

F primeV V le cV pminus2 + cV qminus2

∣∣∣∣FprimeV V i

Vj

V 3∣∣∣∣ le cV pminus2 + cV qminus2

∣∣∣∣(F primeV V

)∣∣∣∣ le cV pminus3 + cV qminus3

∣∣∣∣(F primeV V i

Vj

V 3)∣∣∣∣ le cV pminus3 + cV qminus3

Hence following the same argument for I we conclude that

II + III le cV pminus3 uminus V (517)

for some c independent of Combining (516) and (517) we obtain that∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣ le cV pminus3 uminus V (518)

and considering (518) in (512)intS R

2

uminus V 2 dz le cminus2pminus2V 2pminus3intS R

2

uminus V 4 dz

le cminus2intS R

2

uminus V 4 dz (519)

From Houmllderrsquos inequality we have thatintS R

2

uminus V 4 dz

=intS R

2

uminus V 8n+ 4nminus8n dz

le[

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx] 2

n int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

(520)

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634 Bae and Choe

From Lemma 55 we have

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx le c2 supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2 dx

le cp + 2qminusp

R2

intSR

uminus V 2 dz

for some c independent of Now recalling that 12 le V le we obtain that

int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

le c22minuspnminus2

n

int t0

t0minus 2minuspR24

[ intB R

2

(u + V )2pminus2uminus V 4 dx] nminus2

n

dt

le c4nminuspnminus8

n

int t0

t0minus 2minuspR24

[ intB R

2

∣∣u pminus22 uminus V pminus2

2 V∣∣ 2n

nminus2 dx] nminus2

n

dt

le c4nminuspnminus8

n

intS R

2

upminus2 2u2 dz+ c2nminuspnminus8

n pminus2intSR

uminus V 2 dz (521)

for some c independent of where we have used Sobolevrsquos inequality FromLemma 55 we haveint

S R2

upminus2 2u2 dz le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

Thus combining (519) (520) and (521) we have that

intS R

2uminus V 2 dz le c

(1+ qminusp 4

n+2)[

minus2ndashintSR

uminus V 2 dz] 2

n intSR

uminus V 2 dz

and this completes the proof

Now we prove an iteration lemma which says that if mean oscillation of unear some fixed V is small then u satisfies a Campanato type integral inequality

Lemma 58 Fix 0 lt 0 lt 1 then there exist small positive numbers independentof such that if V0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists V1 isin nN satisfying(12minusradic

0

) le V1 le

(1+radic

0

) (522)

ndashintSR

uminus V12 dz le 0 ndashintSR

uminus V02 dz (523)

ndashintSR

uminus V12 dz le 2

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Regularity for Certain Nonlinear Parabolic Systems 635

Proof Since V0 satisfies the assumption of Lemma 57 we apply Lemma 57 withV = V0 and obtainint

S R2

vminus u2 dz le cintSR

uminus V02 dz

So from the usual comparison argument we haveintSR

uminus uR2 dz le cintSR

vminus vR2 dz+ cintSR

uminus v2 dz

le cn+4intS R

2

uminus u R22 dz+ c

intS R

2

uminus v2 dz

le c(n+4 +

) intSR

uminus V02 dz

where c is independent of Choosing and isin 0 12 small we prove (523)

with V1 = uR Inequality ndashintSR

uminus V12 dz le 2 follows from (523) andassumption ndash

intSR

uminus V02 dz le 2 Now we also estimate

V1 minus V0 =∣∣∣∣∣ ndashintSR

vminus V0 dz

∣∣∣∣le c

[ndashintSR

vminus u2 dz] 1

2

+ c

[ndashintSR

uminus V02 dz] 1

2

le cminus1minusn2(1+

120

)12

Therefore using the same technique of DiBenedetto (1986) Lemma 44 and if wetake very small compared to (522) holds and this completes the proof

Lemma 59 There exist small positive numbers 0 independent of such that ifV0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists a sequence of vectors Vi i = 1 2 3 sub nN satisfying

14 le Vi le 4 ndash

intSiR

uminus Vi2 dz le 2

ndashintSi+1R

uminus Vi+12 dz le 0 ndashintSiR

uminus Vi2 dz

Proof For the proof of the lemma we need to show that 14 le Vi le 4 But this

follows from that

Vi+1 minus Vi2 le c(0 + minusnminus2

)∣∣∣ ndashintSiR

uminus Vi dz∣∣∣

le ci0

i∣∣∣ ndashint

SRuminus V02 dz

∣∣∣ le ci0

i2

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636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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t 200

7

ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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624 Bae and Choe

for some c and small Combining (42) (43) and (44) we have

intuminus uRs2x t p dx +

intBRtimesst

upp dz

le cint

uminus uRs2x sp dx +

Rp

intBRtimesst

uminus uRspp dz

+ cintBRtimesst

up dz+ cintBRtimesst

u pqminus1pminus1 dz (45)

for some c and small Integrating (45) with respect to s from t0 minus 2Rp to t0 minus Rpwe have

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRs2p dz+ c

Rp

intminus

R

dsintBRtimesst

uminus uRs2p dz

+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

By the choice of t we have that for small

intminus

R

dsintBR

uminusuRs2x t pdx

le cintQminus

R

uminusuRs2p dz+ cRpintQ2R

up dz+ cRpintQ2R

u pqminus1pminus1 dz

So we proved Lemma 42 when p ge 2In case p isin 1 2 we estimate the second term of (41) as follows

intuppst dz+ p

intAu middot uminus uRs

pminus1st dz

leint

Au middot uminus uRspst dz (46)

and using the fact that u is bounded and ppminus 1 gt 2 we have

∣∣∣ int Au middot uminusuRspminus1st dz

∣∣∣le c

intBRtimesst

upminus1uminusuRspminus1st dz

+ cintBRtimesst

uqminus1pminus1uminusuRsst dz

le

Rp

intuminusuRs

ppminus1 pst dz+

c

Rp2minusp

intuppminus1 dz+ c

Rp2minusp

intupqminus1 dz

(47)

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Regularity for Certain Nonlinear Parabolic Systems 625

Combining (41) (46) and (47) we have

intBR

uminus uRs2x t p dx

le cu2minusppminus1L

Rp

intBRtimesst

uminus uRs2pst dz+c

Rp2minusp

intBRtimesst

uppminus1 dz

+ c

Rp2minusp

intBRtimesst

upqminus1 dz+intBR

uminus uRs2x sp dx (48)

for small Integrating (48) with respect to s and considering the choice of t weprove Lemma 42 when 1 lt p lt 2

Now we prove a Poincareacute inequality

Theorem 43 Suppose Q2R sub T with R lt 1 then u satisfies the following inequality

intQR

2

uminus uR22 dz le cR2

intQ2R

u2 dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

where c is independent of R

Proof Using Lemma 42 we have

intQR

2

uminus uR22 dz le

intQR

uminus uRs2p dz

le c

Rp

intminus

R

dsintQR

uminus uRs2p dz le c suptisinR

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRt2p dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

le cR2intQ2R

u2 dz+ cRs1

intQ2R

us2z+ cRs1

intQ2R

us3 dz

where we used a Poincareacute type inequality for x variables only that is

intBR

uminus uRs2p dx le cR2intBR

u2 dx

Proof of Theorem 41 Since u is bounded owing to Theorem 32 we have fromTheorem 43

intQR

uminus uRs2 dz lecRn+p+2 when p isin 2

cRn+p+ppminus1 when p isin 1 2

where c is independent of R Hence by isomorphism theorem of Da Prato (1965) weprove Theorem 41

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626 Bae and Choe

5 HOumlLDER CONTINUITY FOR u

In this section we prove that u is Houmllder continuous Because of degeneracyof the pde we need scaled cylinders (see DiBenedetto and Friedman 1985)Let R0 gt 0 be fixed and consider a cylinder

SR0

def= BR0times t0 minus R2minus

0 t0

for small For any 0 lt R le R02 we introduce a scaled cylinder

SRdef= BR times t0 minus R22minusp t0

for fixed ge supSR0u

We assume that pminus2 gt R0 if p gt 2 Hence we note that SR sub S

R0and

ge supSRu

The following theorem is our main theorem in this section

Theorem 51 Suppose that SR sub T then u satisfies for all r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR 2 dz+ cR

for some gt 0

For the proof of Theorem 51 we follow a usual program for degenerateequations such that if for some isin 0 1∣∣z isin SR u2 le 1minus 2

∣∣ gt ∣∣SR

∣∣ (51)

then

supS R

2

u le (52)

for some small isin 0 1 and otherwise that is (51) fails then for all 0 lt r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR2 dz (53)

for some gt 0 Combining (52) and (53) we prove Theorem 51First we prove that (52) under the assumption (51) A similar argument for

the case p = q ge 2 appeared in DiBenedetto and Friedman (1985) We assume that(51) is true Set v = u2 For each h gt 0 we set AhRt

def= x isin BR v gt h and

BhR

def= z isin SR v gt h From the mean value theorem we estimate AhRt

Lemma 52 For any isin 0 there exists isin t0 minus 2minuspR20 t0 minus 2minuspR2

0 such that

A1minus2R0 le 1minus

1minus BR0

where is independent of

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Regularity for Certain Nonlinear Parabolic Systems 627

Proof Suppose that the assertion is not true then

B1minus2R0 ge

int t0minus2minuspR20

t0minus2minuspR20

A1minus2R0d gt 1minus BR0

2minuspR20 = 1minus SR0

and this contradicts the assumption (51)

The following lemma is useful in estimating sup u Indeed when p = q ge 2DiBenedetto and Friedman proved a similar theorem (see Lemma 62 inDiBenedetto and Friedman 1985) Now choose = 2 and in Lemma 52We define St

R

def= BR times t

Lemma 53 There exists a positive integer r independent of such that

∣∣A1minus 2r

2R0t

∣∣ le (1minus

(2

)2)BR0

for all t isin t0 minus

22minuspR2

0 t0

Proof Consider the function

v = log+[

2

2minus vminus 1minus 2+ + 2

2r

]

where r is a positive integer This function was used in Choe (1991) and DiBenedettoand Friedman (1985) We take 13 = ui

x2vprime 2 as a test function for (31) where

isin 0 is a standard cutoff function such that B1minusR for small decided later is

contained in support Integrating from to t we obtain

12

intStR

vt2prime 2 dz+

intStR

Ai

Qj

uujxx

uixx

2prime 2 dz

+ 2intStR

Ai

Qj

uujxx

uixukxx

ukx2primeprime 2 dz

+ 2intStR

Ai

Qj

uujxx

uix x

2prime dz = 0

Considering the fact 2primeprime = 21+ prime2 we have

intBR

2vx t 2xdx +intStR

upminus2v21+ prime2 2 dz

leintBR

2vx 2xdx + cintStR

upminus2v prime dz

+ cintStR

uqminus2v prime dz

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628 Bae and Choe

Applying Youngrsquos inequality on the last two terms we getintBR

2vx t 2xdx

leintBR

2vx 2xdx + cintStR

upminus2 2 dz+ cintStR

u2qminuspminus2 2 dz

for some c Since le r log 2 and u is bounded we getintBR

2vx t 2xdx leintBR

2vx 2xdx + crBR

From Lemma 52 we getintBR

2vx t 2xdx le r log 221minus

1minus

2

BR

From definition of we haveintB1minusRcapvgt1minus

2r 22vx tdx ge r minus 1 log 22A1minus

2r 21minusRt

and

A1minus 2r

21minusRt le[( r

r minus 1

)2 1minus

1minus

2

+ c

r

]BR

Hence we obtain

A(1minus

2r

)2R

t le A1minus 2r

21minusRt + BRB1minusR

le[( r

r minus 1

)2 1minus

1minus

2

+ c

r+ n

]BR

Taking r so large that( r

r minus 1

)2 le(1minus

2

)1+

c

rle 3

82

and = 382n we obtain

A1minus 2r

2Rt le(1minus

(2

)2)BR

We take 13 = uixvminus h+ 2 as a test function in (31) where v = u2 and is

a cutoff function such that = 1 in Sr = 0 on pSR le cRminus r and t le cpminus2Rminus r2 Considering the ellipticity condition (12) we have

supt

int ∣∣vminus h+∣∣2 2 dx +

intupminus2vminus h+2 2 dz

le c

Rminus r2

intSR

upminus2vminus h+2 dz

+ c

Rminus r2

intSR

u2qminuspminus2vminus h+2 dz+ cintSR

vminus h+2 tdz (54)

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Regularity for Certain Nonlinear Parabolic Systems 629

for some c Now choose h ge 12 For such a choice of h we haveint

upminus2vminus h+2 2 dz ge cpminus2int

vminus h+2 2 dz

Hence from (54) we obtain

supt

intvminus h+2 2 dx + pminus2

intvminus h+2 2 dz

le[

2qminuspminus2

Rminus r2+ pminus2

Rminus r2

] intSR

vminus h+2 dz (55)

for some c independent of Observe that is bounded above by some fixednumber We make change of variables

= pminus2t minus t0 vx = vx t0 + 2minusp

and we write (55) as

supt

intBR

vminus h+2 2 dx +intSR

vminus h+2 2 dz le c

Rminus r2

intSR

vminus h+dz

Once we have this Caccioppoli type inequality we can apply truncation argumentTherefore following to similar argument in DiBenedetto and Friedman (1984 or1985) we have a maximum principle Now the proof for the next lemma is almostthe same as in DiBenedetto and Friedman (1984 or 1985) we omit the proof

Lemma 54 If for some and isin 0 1∣∣x t isin S1+R u2 lt 1minus 2∣∣ ge S1+R (56)

then there exists a number isin 0 1 independent of and R such that

supS R

2

ux t le

Now we assume that (56) fails that is∣∣x t isin S1+R u2 lt 1minus 2∣∣ lt S1+R (57)

Following a comparison argument similar to Choe (1991) or DiBenedetto andFriedman (1984) we prove that u satisfies a Campanato type integral inequalityLet V isin nN be a constant vector such that 1

2 le V le Taking uxkminus Vk

2 as atest function to (31) we have the following lemma

Lemma 55 There exists a constant c independent of such that

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2tdx +intS R

2

upminus22u2 dz

le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

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630 Bae and Choe

Now suppose that v is a solution to

vit minus

(Ai

Qj

Vvjx

)x

= 0 (58)

with v = u on pSR2 Since 1

2 le V le le c for some fixed c we see that fromthe structure condition of F

2 le 2minuspAi

Qj

Vi

j le c1+ qminusp2

Hence we see that v satisfies a Campanato type integral inequality

Lemma 56 Suppose that 0 lt r le R2 There exists a constant c independent ofVR and such that

intSr

vminus vr 2 dz le c( r

R

)n+4 intS R

2

uminus V 2 dz (59)

Proof We write (58) as

vit minus pminus2

Ai

Qj

V

pminus2vjx

x

= 0

We change of variable = pminus2t minus t0 then the function vx = vx 2minusp+ t0satisfies a uniformly parabolic system

vi minus

Ai

Qj

V

pminus2vjx

x

= 0

Thus from results of Campanato (1966) we prove

intS

vminus v2 dz le c(

R

)n+4 intS R

2

uminus V 2 dz

and scaling back prove (59)

We estimate the L2 norm of uminus v in SR2

Lemma 57 For each lt R2 there exists a constant c independent of such that

intS

uminus v2 dz le c(1+ qminusp 4

n+2)(

minus2 ndashintSR

uminus V 2 dz) int

SRuminus V 2 dz

for = min(12

2n

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Regularity for Certain Nonlinear Parabolic Systems 631

Proof Subtracting(Ai

Qj

Vujx

)x

from both sides of (15) we get

uit minus

(Ai

Qj

Vujx

)x= (

Aiuminus Ai

Qj

Vujx

)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(510)

From (58) and (510) we see that uminus v satisfies

ui minus vit minus(Ai

Qj

Vuj minus vjx)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(511)

Simply applying uminus v to (511) we get

pminus2intS R

2uminusv2dzle

intS R

2

∣∣∣ int 1

0Ai

Qj

V+suminus VminusAi

Qj

Vds∣∣∣

times uminusV uminusvdz

and it follows thatintS R

2

uminus v2 dz

le minus2pminus2intS R

2

∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣2

uminus V 2 dz (512)

Note that AiV = F primeV V i

V and

Ai

Qj

V = F primeprimeV ViV

j

V 2 + F primeV (ij

V minus V iV

j

V 3)

where ij and are Kronecker delta functions Also we recall the structurecondition (13) F primeprimeprimeh le chpminus3 + hqminus3 We have that

∣∣∣ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds∣∣∣

le∣∣∣∣int 1

0F primeprime(V + suminus V)

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2

minusF primeprimeV ViV

j

V 2 ds

∣∣∣∣+int 1

0

∣∣∣∣F primeV + suminus VV + suminus V minus F primeV

V ∣∣∣∣ds

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632 Bae and Choe

+∣∣∣∣int 1

0F primeV + suminus V

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V3

minusF primeV ViV

j

V 3 ds

∣∣∣∣= I + II + III

The first term I is estimates as follows

I leint 1

0

∣∣∣F primeprimeV + suminus Vminus F primeprimeV ∣∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2∣∣∣∣∣ ds

+int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

(513)

By the mean value theorem we have that

int 1

0

∣∣∣F primeprimeV+suminusVminusF primeprimeV ∣∣∣ds le c

int 1

0

int 1

0

∣∣F primeprimeprimeV+shuminusV∣∣ds dhuminusV (514)

for some c and

int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

le cF primeprimeV int 1

0

int 1

0

ds dh∣∣V + shuminus V∣∣ uminus V (515)

If V ge 2uminus V then for all s isin 0 1 12 V le V + suminus V le 2V and from

(513) (514) and (515) we have I le c(V pminus3 + V qminus3

)uminus V If V le 2uminus V and p ge 2 then

∣∣F primeprimeV + suminus V∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le cuminus V pminus2 + cuminus V qminus2

and

∣∣F primeprimeV ∣∣ le cV pminus2 + cV qminus2 le cuminus V pminus2 + cuminus V qminus2

Hence we have that

I le cint 1

0

∣∣F primeprimeV + suminus V∣∣+ ∣∣F primeprimeV ∣∣ds le c1+ pminusquminus V pminus2

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Regularity for Certain Nonlinear Parabolic Systems 633

for some c independent of If V le 2uminus V and 1 lt p lt 2 then∣∣F primeprime(V + suminus V)∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le c1+ pminusqV + suminus Vpminus2

for some c We note thatint 1

0V + suminus Vpminus2 ds le uminus V pminus2

int 1

0

∣∣∣∣ V uminus V minus s

∣∣∣∣pminus2

ds le cuminus V pminus2

Therefore combining all these together we have

I le cV pminus3 uminus V (516)

for some c independent of For II and III we observe that

F primeV V le cV pminus2 + cV qminus2

∣∣∣∣FprimeV V i

Vj

V 3∣∣∣∣ le cV pminus2 + cV qminus2

∣∣∣∣(F primeV V

)∣∣∣∣ le cV pminus3 + cV qminus3

∣∣∣∣(F primeV V i

Vj

V 3)∣∣∣∣ le cV pminus3 + cV qminus3

Hence following the same argument for I we conclude that

II + III le cV pminus3 uminus V (517)

for some c independent of Combining (516) and (517) we obtain that∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣ le cV pminus3 uminus V (518)

and considering (518) in (512)intS R

2

uminus V 2 dz le cminus2pminus2V 2pminus3intS R

2

uminus V 4 dz

le cminus2intS R

2

uminus V 4 dz (519)

From Houmllderrsquos inequality we have thatintS R

2

uminus V 4 dz

=intS R

2

uminus V 8n+ 4nminus8n dz

le[

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx] 2

n int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

(520)

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634 Bae and Choe

From Lemma 55 we have

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx le c2 supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2 dx

le cp + 2qminusp

R2

intSR

uminus V 2 dz

for some c independent of Now recalling that 12 le V le we obtain that

int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

le c22minuspnminus2

n

int t0

t0minus 2minuspR24

[ intB R

2

(u + V )2pminus2uminus V 4 dx] nminus2

n

dt

le c4nminuspnminus8

n

int t0

t0minus 2minuspR24

[ intB R

2

∣∣u pminus22 uminus V pminus2

2 V∣∣ 2n

nminus2 dx] nminus2

n

dt

le c4nminuspnminus8

n

intS R

2

upminus2 2u2 dz+ c2nminuspnminus8

n pminus2intSR

uminus V 2 dz (521)

for some c independent of where we have used Sobolevrsquos inequality FromLemma 55 we haveint

S R2

upminus2 2u2 dz le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

Thus combining (519) (520) and (521) we have that

intS R

2uminus V 2 dz le c

(1+ qminusp 4

n+2)[

minus2ndashintSR

uminus V 2 dz] 2

n intSR

uminus V 2 dz

and this completes the proof

Now we prove an iteration lemma which says that if mean oscillation of unear some fixed V is small then u satisfies a Campanato type integral inequality

Lemma 58 Fix 0 lt 0 lt 1 then there exist small positive numbers independentof such that if V0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists V1 isin nN satisfying(12minusradic

0

) le V1 le

(1+radic

0

) (522)

ndashintSR

uminus V12 dz le 0 ndashintSR

uminus V02 dz (523)

ndashintSR

uminus V12 dz le 2

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Regularity for Certain Nonlinear Parabolic Systems 635

Proof Since V0 satisfies the assumption of Lemma 57 we apply Lemma 57 withV = V0 and obtainint

S R2

vminus u2 dz le cintSR

uminus V02 dz

So from the usual comparison argument we haveintSR

uminus uR2 dz le cintSR

vminus vR2 dz+ cintSR

uminus v2 dz

le cn+4intS R

2

uminus u R22 dz+ c

intS R

2

uminus v2 dz

le c(n+4 +

) intSR

uminus V02 dz

where c is independent of Choosing and isin 0 12 small we prove (523)

with V1 = uR Inequality ndashintSR

uminus V12 dz le 2 follows from (523) andassumption ndash

intSR

uminus V02 dz le 2 Now we also estimate

V1 minus V0 =∣∣∣∣∣ ndashintSR

vminus V0 dz

∣∣∣∣le c

[ndashintSR

vminus u2 dz] 1

2

+ c

[ndashintSR

uminus V02 dz] 1

2

le cminus1minusn2(1+

120

)12

Therefore using the same technique of DiBenedetto (1986) Lemma 44 and if wetake very small compared to (522) holds and this completes the proof

Lemma 59 There exist small positive numbers 0 independent of such that ifV0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists a sequence of vectors Vi i = 1 2 3 sub nN satisfying

14 le Vi le 4 ndash

intSiR

uminus Vi2 dz le 2

ndashintSi+1R

uminus Vi+12 dz le 0 ndashintSiR

uminus Vi2 dz

Proof For the proof of the lemma we need to show that 14 le Vi le 4 But this

follows from that

Vi+1 minus Vi2 le c(0 + minusnminus2

)∣∣∣ ndashintSiR

uminus Vi dz∣∣∣

le ci0

i∣∣∣ ndashint

SRuminus V02 dz

∣∣∣ le ci0

i2

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636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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Regularity for Certain Nonlinear Parabolic Systems 625

Combining (41) (46) and (47) we have

intBR

uminus uRs2x t p dx

le cu2minusppminus1L

Rp

intBRtimesst

uminus uRs2pst dz+c

Rp2minusp

intBRtimesst

uppminus1 dz

+ c

Rp2minusp

intBRtimesst

upqminus1 dz+intBR

uminus uRs2x sp dx (48)

for small Integrating (48) with respect to s and considering the choice of t weprove Lemma 42 when 1 lt p lt 2

Now we prove a Poincareacute inequality

Theorem 43 Suppose Q2R sub T with R lt 1 then u satisfies the following inequality

intQR

2

uminus uR22 dz le cR2

intQ2R

u2 dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

where c is independent of R

Proof Using Lemma 42 we have

intQR

2

uminus uR22 dz le

intQR

uminus uRs2p dz

le c

Rp

intminus

R

dsintQR

uminus uRs2p dz le c suptisinR

intminus

R

dsintBR

uminus uRs2x t p dx

le cintQminus

R

uminus uRt2p dz+ cRs1

intQ2R

us2 dz+ cRs1

intQ2R

us3 dz

le cR2intQ2R

u2 dz+ cRs1

intQ2R

us2z+ cRs1

intQ2R

us3 dz

where we used a Poincareacute type inequality for x variables only that is

intBR

uminus uRs2p dx le cR2intBR

u2 dx

Proof of Theorem 41 Since u is bounded owing to Theorem 32 we have fromTheorem 43

intQR

uminus uRs2 dz lecRn+p+2 when p isin 2

cRn+p+ppminus1 when p isin 1 2

where c is independent of R Hence by isomorphism theorem of Da Prato (1965) weprove Theorem 41

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626 Bae and Choe

5 HOumlLDER CONTINUITY FOR u

In this section we prove that u is Houmllder continuous Because of degeneracyof the pde we need scaled cylinders (see DiBenedetto and Friedman 1985)Let R0 gt 0 be fixed and consider a cylinder

SR0

def= BR0times t0 minus R2minus

0 t0

for small For any 0 lt R le R02 we introduce a scaled cylinder

SRdef= BR times t0 minus R22minusp t0

for fixed ge supSR0u

We assume that pminus2 gt R0 if p gt 2 Hence we note that SR sub S

R0and

ge supSRu

The following theorem is our main theorem in this section

Theorem 51 Suppose that SR sub T then u satisfies for all r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR 2 dz+ cR

for some gt 0

For the proof of Theorem 51 we follow a usual program for degenerateequations such that if for some isin 0 1∣∣z isin SR u2 le 1minus 2

∣∣ gt ∣∣SR

∣∣ (51)

then

supS R

2

u le (52)

for some small isin 0 1 and otherwise that is (51) fails then for all 0 lt r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR2 dz (53)

for some gt 0 Combining (52) and (53) we prove Theorem 51First we prove that (52) under the assumption (51) A similar argument for

the case p = q ge 2 appeared in DiBenedetto and Friedman (1985) We assume that(51) is true Set v = u2 For each h gt 0 we set AhRt

def= x isin BR v gt h and

BhR

def= z isin SR v gt h From the mean value theorem we estimate AhRt

Lemma 52 For any isin 0 there exists isin t0 minus 2minuspR20 t0 minus 2minuspR2

0 such that

A1minus2R0 le 1minus

1minus BR0

where is independent of

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Regularity for Certain Nonlinear Parabolic Systems 627

Proof Suppose that the assertion is not true then

B1minus2R0 ge

int t0minus2minuspR20

t0minus2minuspR20

A1minus2R0d gt 1minus BR0

2minuspR20 = 1minus SR0

and this contradicts the assumption (51)

The following lemma is useful in estimating sup u Indeed when p = q ge 2DiBenedetto and Friedman proved a similar theorem (see Lemma 62 inDiBenedetto and Friedman 1985) Now choose = 2 and in Lemma 52We define St

R

def= BR times t

Lemma 53 There exists a positive integer r independent of such that

∣∣A1minus 2r

2R0t

∣∣ le (1minus

(2

)2)BR0

for all t isin t0 minus

22minuspR2

0 t0

Proof Consider the function

v = log+[

2

2minus vminus 1minus 2+ + 2

2r

]

where r is a positive integer This function was used in Choe (1991) and DiBenedettoand Friedman (1985) We take 13 = ui

x2vprime 2 as a test function for (31) where

isin 0 is a standard cutoff function such that B1minusR for small decided later is

contained in support Integrating from to t we obtain

12

intStR

vt2prime 2 dz+

intStR

Ai

Qj

uujxx

uixx

2prime 2 dz

+ 2intStR

Ai

Qj

uujxx

uixukxx

ukx2primeprime 2 dz

+ 2intStR

Ai

Qj

uujxx

uix x

2prime dz = 0

Considering the fact 2primeprime = 21+ prime2 we have

intBR

2vx t 2xdx +intStR

upminus2v21+ prime2 2 dz

leintBR

2vx 2xdx + cintStR

upminus2v prime dz

+ cintStR

uqminus2v prime dz

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628 Bae and Choe

Applying Youngrsquos inequality on the last two terms we getintBR

2vx t 2xdx

leintBR

2vx 2xdx + cintStR

upminus2 2 dz+ cintStR

u2qminuspminus2 2 dz

for some c Since le r log 2 and u is bounded we getintBR

2vx t 2xdx leintBR

2vx 2xdx + crBR

From Lemma 52 we getintBR

2vx t 2xdx le r log 221minus

1minus

2

BR

From definition of we haveintB1minusRcapvgt1minus

2r 22vx tdx ge r minus 1 log 22A1minus

2r 21minusRt

and

A1minus 2r

21minusRt le[( r

r minus 1

)2 1minus

1minus

2

+ c

r

]BR

Hence we obtain

A(1minus

2r

)2R

t le A1minus 2r

21minusRt + BRB1minusR

le[( r

r minus 1

)2 1minus

1minus

2

+ c

r+ n

]BR

Taking r so large that( r

r minus 1

)2 le(1minus

2

)1+

c

rle 3

82

and = 382n we obtain

A1minus 2r

2Rt le(1minus

(2

)2)BR

We take 13 = uixvminus h+ 2 as a test function in (31) where v = u2 and is

a cutoff function such that = 1 in Sr = 0 on pSR le cRminus r and t le cpminus2Rminus r2 Considering the ellipticity condition (12) we have

supt

int ∣∣vminus h+∣∣2 2 dx +

intupminus2vminus h+2 2 dz

le c

Rminus r2

intSR

upminus2vminus h+2 dz

+ c

Rminus r2

intSR

u2qminuspminus2vminus h+2 dz+ cintSR

vminus h+2 tdz (54)

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Regularity for Certain Nonlinear Parabolic Systems 629

for some c Now choose h ge 12 For such a choice of h we haveint

upminus2vminus h+2 2 dz ge cpminus2int

vminus h+2 2 dz

Hence from (54) we obtain

supt

intvminus h+2 2 dx + pminus2

intvminus h+2 2 dz

le[

2qminuspminus2

Rminus r2+ pminus2

Rminus r2

] intSR

vminus h+2 dz (55)

for some c independent of Observe that is bounded above by some fixednumber We make change of variables

= pminus2t minus t0 vx = vx t0 + 2minusp

and we write (55) as

supt

intBR

vminus h+2 2 dx +intSR

vminus h+2 2 dz le c

Rminus r2

intSR

vminus h+dz

Once we have this Caccioppoli type inequality we can apply truncation argumentTherefore following to similar argument in DiBenedetto and Friedman (1984 or1985) we have a maximum principle Now the proof for the next lemma is almostthe same as in DiBenedetto and Friedman (1984 or 1985) we omit the proof

Lemma 54 If for some and isin 0 1∣∣x t isin S1+R u2 lt 1minus 2∣∣ ge S1+R (56)

then there exists a number isin 0 1 independent of and R such that

supS R

2

ux t le

Now we assume that (56) fails that is∣∣x t isin S1+R u2 lt 1minus 2∣∣ lt S1+R (57)

Following a comparison argument similar to Choe (1991) or DiBenedetto andFriedman (1984) we prove that u satisfies a Campanato type integral inequalityLet V isin nN be a constant vector such that 1

2 le V le Taking uxkminus Vk

2 as atest function to (31) we have the following lemma

Lemma 55 There exists a constant c independent of such that

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2tdx +intS R

2

upminus22u2 dz

le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

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630 Bae and Choe

Now suppose that v is a solution to

vit minus

(Ai

Qj

Vvjx

)x

= 0 (58)

with v = u on pSR2 Since 1

2 le V le le c for some fixed c we see that fromthe structure condition of F

2 le 2minuspAi

Qj

Vi

j le c1+ qminusp2

Hence we see that v satisfies a Campanato type integral inequality

Lemma 56 Suppose that 0 lt r le R2 There exists a constant c independent ofVR and such that

intSr

vminus vr 2 dz le c( r

R

)n+4 intS R

2

uminus V 2 dz (59)

Proof We write (58) as

vit minus pminus2

Ai

Qj

V

pminus2vjx

x

= 0

We change of variable = pminus2t minus t0 then the function vx = vx 2minusp+ t0satisfies a uniformly parabolic system

vi minus

Ai

Qj

V

pminus2vjx

x

= 0

Thus from results of Campanato (1966) we prove

intS

vminus v2 dz le c(

R

)n+4 intS R

2

uminus V 2 dz

and scaling back prove (59)

We estimate the L2 norm of uminus v in SR2

Lemma 57 For each lt R2 there exists a constant c independent of such that

intS

uminus v2 dz le c(1+ qminusp 4

n+2)(

minus2 ndashintSR

uminus V 2 dz) int

SRuminus V 2 dz

for = min(12

2n

)

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Regularity for Certain Nonlinear Parabolic Systems 631

Proof Subtracting(Ai

Qj

Vujx

)x

from both sides of (15) we get

uit minus

(Ai

Qj

Vujx

)x= (

Aiuminus Ai

Qj

Vujx

)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(510)

From (58) and (510) we see that uminus v satisfies

ui minus vit minus(Ai

Qj

Vuj minus vjx)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(511)

Simply applying uminus v to (511) we get

pminus2intS R

2uminusv2dzle

intS R

2

∣∣∣ int 1

0Ai

Qj

V+suminus VminusAi

Qj

Vds∣∣∣

times uminusV uminusvdz

and it follows thatintS R

2

uminus v2 dz

le minus2pminus2intS R

2

∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣2

uminus V 2 dz (512)

Note that AiV = F primeV V i

V and

Ai

Qj

V = F primeprimeV ViV

j

V 2 + F primeV (ij

V minus V iV

j

V 3)

where ij and are Kronecker delta functions Also we recall the structurecondition (13) F primeprimeprimeh le chpminus3 + hqminus3 We have that

∣∣∣ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds∣∣∣

le∣∣∣∣int 1

0F primeprime(V + suminus V)

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2

minusF primeprimeV ViV

j

V 2 ds

∣∣∣∣+int 1

0

∣∣∣∣F primeV + suminus VV + suminus V minus F primeV

V ∣∣∣∣ds

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ORDER REPRINTS

632 Bae and Choe

+∣∣∣∣int 1

0F primeV + suminus V

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V3

minusF primeV ViV

j

V 3 ds

∣∣∣∣= I + II + III

The first term I is estimates as follows

I leint 1

0

∣∣∣F primeprimeV + suminus Vminus F primeprimeV ∣∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2∣∣∣∣∣ ds

+int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

(513)

By the mean value theorem we have that

int 1

0

∣∣∣F primeprimeV+suminusVminusF primeprimeV ∣∣∣ds le c

int 1

0

int 1

0

∣∣F primeprimeprimeV+shuminusV∣∣ds dhuminusV (514)

for some c and

int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

le cF primeprimeV int 1

0

int 1

0

ds dh∣∣V + shuminus V∣∣ uminus V (515)

If V ge 2uminus V then for all s isin 0 1 12 V le V + suminus V le 2V and from

(513) (514) and (515) we have I le c(V pminus3 + V qminus3

)uminus V If V le 2uminus V and p ge 2 then

∣∣F primeprimeV + suminus V∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le cuminus V pminus2 + cuminus V qminus2

and

∣∣F primeprimeV ∣∣ le cV pminus2 + cV qminus2 le cuminus V pminus2 + cuminus V qminus2

Hence we have that

I le cint 1

0

∣∣F primeprimeV + suminus V∣∣+ ∣∣F primeprimeV ∣∣ds le c1+ pminusquminus V pminus2

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Regularity for Certain Nonlinear Parabolic Systems 633

for some c independent of If V le 2uminus V and 1 lt p lt 2 then∣∣F primeprime(V + suminus V)∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le c1+ pminusqV + suminus Vpminus2

for some c We note thatint 1

0V + suminus Vpminus2 ds le uminus V pminus2

int 1

0

∣∣∣∣ V uminus V minus s

∣∣∣∣pminus2

ds le cuminus V pminus2

Therefore combining all these together we have

I le cV pminus3 uminus V (516)

for some c independent of For II and III we observe that

F primeV V le cV pminus2 + cV qminus2

∣∣∣∣FprimeV V i

Vj

V 3∣∣∣∣ le cV pminus2 + cV qminus2

∣∣∣∣(F primeV V

)∣∣∣∣ le cV pminus3 + cV qminus3

∣∣∣∣(F primeV V i

Vj

V 3)∣∣∣∣ le cV pminus3 + cV qminus3

Hence following the same argument for I we conclude that

II + III le cV pminus3 uminus V (517)

for some c independent of Combining (516) and (517) we obtain that∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣ le cV pminus3 uminus V (518)

and considering (518) in (512)intS R

2

uminus V 2 dz le cminus2pminus2V 2pminus3intS R

2

uminus V 4 dz

le cminus2intS R

2

uminus V 4 dz (519)

From Houmllderrsquos inequality we have thatintS R

2

uminus V 4 dz

=intS R

2

uminus V 8n+ 4nminus8n dz

le[

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx] 2

n int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

(520)

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634 Bae and Choe

From Lemma 55 we have

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx le c2 supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2 dx

le cp + 2qminusp

R2

intSR

uminus V 2 dz

for some c independent of Now recalling that 12 le V le we obtain that

int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

le c22minuspnminus2

n

int t0

t0minus 2minuspR24

[ intB R

2

(u + V )2pminus2uminus V 4 dx] nminus2

n

dt

le c4nminuspnminus8

n

int t0

t0minus 2minuspR24

[ intB R

2

∣∣u pminus22 uminus V pminus2

2 V∣∣ 2n

nminus2 dx] nminus2

n

dt

le c4nminuspnminus8

n

intS R

2

upminus2 2u2 dz+ c2nminuspnminus8

n pminus2intSR

uminus V 2 dz (521)

for some c independent of where we have used Sobolevrsquos inequality FromLemma 55 we haveint

S R2

upminus2 2u2 dz le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

Thus combining (519) (520) and (521) we have that

intS R

2uminus V 2 dz le c

(1+ qminusp 4

n+2)[

minus2ndashintSR

uminus V 2 dz] 2

n intSR

uminus V 2 dz

and this completes the proof

Now we prove an iteration lemma which says that if mean oscillation of unear some fixed V is small then u satisfies a Campanato type integral inequality

Lemma 58 Fix 0 lt 0 lt 1 then there exist small positive numbers independentof such that if V0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists V1 isin nN satisfying(12minusradic

0

) le V1 le

(1+radic

0

) (522)

ndashintSR

uminus V12 dz le 0 ndashintSR

uminus V02 dz (523)

ndashintSR

uminus V12 dz le 2

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Regularity for Certain Nonlinear Parabolic Systems 635

Proof Since V0 satisfies the assumption of Lemma 57 we apply Lemma 57 withV = V0 and obtainint

S R2

vminus u2 dz le cintSR

uminus V02 dz

So from the usual comparison argument we haveintSR

uminus uR2 dz le cintSR

vminus vR2 dz+ cintSR

uminus v2 dz

le cn+4intS R

2

uminus u R22 dz+ c

intS R

2

uminus v2 dz

le c(n+4 +

) intSR

uminus V02 dz

where c is independent of Choosing and isin 0 12 small we prove (523)

with V1 = uR Inequality ndashintSR

uminus V12 dz le 2 follows from (523) andassumption ndash

intSR

uminus V02 dz le 2 Now we also estimate

V1 minus V0 =∣∣∣∣∣ ndashintSR

vminus V0 dz

∣∣∣∣le c

[ndashintSR

vminus u2 dz] 1

2

+ c

[ndashintSR

uminus V02 dz] 1

2

le cminus1minusn2(1+

120

)12

Therefore using the same technique of DiBenedetto (1986) Lemma 44 and if wetake very small compared to (522) holds and this completes the proof

Lemma 59 There exist small positive numbers 0 independent of such that ifV0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists a sequence of vectors Vi i = 1 2 3 sub nN satisfying

14 le Vi le 4 ndash

intSiR

uminus Vi2 dz le 2

ndashintSi+1R

uminus Vi+12 dz le 0 ndashintSiR

uminus Vi2 dz

Proof For the proof of the lemma we need to show that 14 le Vi le 4 But this

follows from that

Vi+1 minus Vi2 le c(0 + minusnminus2

)∣∣∣ ndashintSiR

uminus Vi dz∣∣∣

le ci0

i∣∣∣ ndashint

SRuminus V02 dz

∣∣∣ le ci0

i2

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636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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ORDER REPRINTS

642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

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Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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626 Bae and Choe

5 HOumlLDER CONTINUITY FOR u

In this section we prove that u is Houmllder continuous Because of degeneracyof the pde we need scaled cylinders (see DiBenedetto and Friedman 1985)Let R0 gt 0 be fixed and consider a cylinder

SR0

def= BR0times t0 minus R2minus

0 t0

for small For any 0 lt R le R02 we introduce a scaled cylinder

SRdef= BR times t0 minus R22minusp t0

for fixed ge supSR0u

We assume that pminus2 gt R0 if p gt 2 Hence we note that SR sub S

R0and

ge supSRu

The following theorem is our main theorem in this section

Theorem 51 Suppose that SR sub T then u satisfies for all r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR 2 dz+ cR

for some gt 0

For the proof of Theorem 51 we follow a usual program for degenerateequations such that if for some isin 0 1∣∣z isin SR u2 le 1minus 2

∣∣ gt ∣∣SR

∣∣ (51)

then

supS R

2

u le (52)

for some small isin 0 1 and otherwise that is (51) fails then for all 0 lt r le R2

ndashintSr

uminus uSr 2 dz le c( r

R

)

ndashintSR

uminus uSR2 dz (53)

for some gt 0 Combining (52) and (53) we prove Theorem 51First we prove that (52) under the assumption (51) A similar argument for

the case p = q ge 2 appeared in DiBenedetto and Friedman (1985) We assume that(51) is true Set v = u2 For each h gt 0 we set AhRt

def= x isin BR v gt h and

BhR

def= z isin SR v gt h From the mean value theorem we estimate AhRt

Lemma 52 For any isin 0 there exists isin t0 minus 2minuspR20 t0 minus 2minuspR2

0 such that

A1minus2R0 le 1minus

1minus BR0

where is independent of

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Regularity for Certain Nonlinear Parabolic Systems 627

Proof Suppose that the assertion is not true then

B1minus2R0 ge

int t0minus2minuspR20

t0minus2minuspR20

A1minus2R0d gt 1minus BR0

2minuspR20 = 1minus SR0

and this contradicts the assumption (51)

The following lemma is useful in estimating sup u Indeed when p = q ge 2DiBenedetto and Friedman proved a similar theorem (see Lemma 62 inDiBenedetto and Friedman 1985) Now choose = 2 and in Lemma 52We define St

R

def= BR times t

Lemma 53 There exists a positive integer r independent of such that

∣∣A1minus 2r

2R0t

∣∣ le (1minus

(2

)2)BR0

for all t isin t0 minus

22minuspR2

0 t0

Proof Consider the function

v = log+[

2

2minus vminus 1minus 2+ + 2

2r

]

where r is a positive integer This function was used in Choe (1991) and DiBenedettoand Friedman (1985) We take 13 = ui

x2vprime 2 as a test function for (31) where

isin 0 is a standard cutoff function such that B1minusR for small decided later is

contained in support Integrating from to t we obtain

12

intStR

vt2prime 2 dz+

intStR

Ai

Qj

uujxx

uixx

2prime 2 dz

+ 2intStR

Ai

Qj

uujxx

uixukxx

ukx2primeprime 2 dz

+ 2intStR

Ai

Qj

uujxx

uix x

2prime dz = 0

Considering the fact 2primeprime = 21+ prime2 we have

intBR

2vx t 2xdx +intStR

upminus2v21+ prime2 2 dz

leintBR

2vx 2xdx + cintStR

upminus2v prime dz

+ cintStR

uqminus2v prime dz

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628 Bae and Choe

Applying Youngrsquos inequality on the last two terms we getintBR

2vx t 2xdx

leintBR

2vx 2xdx + cintStR

upminus2 2 dz+ cintStR

u2qminuspminus2 2 dz

for some c Since le r log 2 and u is bounded we getintBR

2vx t 2xdx leintBR

2vx 2xdx + crBR

From Lemma 52 we getintBR

2vx t 2xdx le r log 221minus

1minus

2

BR

From definition of we haveintB1minusRcapvgt1minus

2r 22vx tdx ge r minus 1 log 22A1minus

2r 21minusRt

and

A1minus 2r

21minusRt le[( r

r minus 1

)2 1minus

1minus

2

+ c

r

]BR

Hence we obtain

A(1minus

2r

)2R

t le A1minus 2r

21minusRt + BRB1minusR

le[( r

r minus 1

)2 1minus

1minus

2

+ c

r+ n

]BR

Taking r so large that( r

r minus 1

)2 le(1minus

2

)1+

c

rle 3

82

and = 382n we obtain

A1minus 2r

2Rt le(1minus

(2

)2)BR

We take 13 = uixvminus h+ 2 as a test function in (31) where v = u2 and is

a cutoff function such that = 1 in Sr = 0 on pSR le cRminus r and t le cpminus2Rminus r2 Considering the ellipticity condition (12) we have

supt

int ∣∣vminus h+∣∣2 2 dx +

intupminus2vminus h+2 2 dz

le c

Rminus r2

intSR

upminus2vminus h+2 dz

+ c

Rminus r2

intSR

u2qminuspminus2vminus h+2 dz+ cintSR

vminus h+2 tdz (54)

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Regularity for Certain Nonlinear Parabolic Systems 629

for some c Now choose h ge 12 For such a choice of h we haveint

upminus2vminus h+2 2 dz ge cpminus2int

vminus h+2 2 dz

Hence from (54) we obtain

supt

intvminus h+2 2 dx + pminus2

intvminus h+2 2 dz

le[

2qminuspminus2

Rminus r2+ pminus2

Rminus r2

] intSR

vminus h+2 dz (55)

for some c independent of Observe that is bounded above by some fixednumber We make change of variables

= pminus2t minus t0 vx = vx t0 + 2minusp

and we write (55) as

supt

intBR

vminus h+2 2 dx +intSR

vminus h+2 2 dz le c

Rminus r2

intSR

vminus h+dz

Once we have this Caccioppoli type inequality we can apply truncation argumentTherefore following to similar argument in DiBenedetto and Friedman (1984 or1985) we have a maximum principle Now the proof for the next lemma is almostthe same as in DiBenedetto and Friedman (1984 or 1985) we omit the proof

Lemma 54 If for some and isin 0 1∣∣x t isin S1+R u2 lt 1minus 2∣∣ ge S1+R (56)

then there exists a number isin 0 1 independent of and R such that

supS R

2

ux t le

Now we assume that (56) fails that is∣∣x t isin S1+R u2 lt 1minus 2∣∣ lt S1+R (57)

Following a comparison argument similar to Choe (1991) or DiBenedetto andFriedman (1984) we prove that u satisfies a Campanato type integral inequalityLet V isin nN be a constant vector such that 1

2 le V le Taking uxkminus Vk

2 as atest function to (31) we have the following lemma

Lemma 55 There exists a constant c independent of such that

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2tdx +intS R

2

upminus22u2 dz

le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

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630 Bae and Choe

Now suppose that v is a solution to

vit minus

(Ai

Qj

Vvjx

)x

= 0 (58)

with v = u on pSR2 Since 1

2 le V le le c for some fixed c we see that fromthe structure condition of F

2 le 2minuspAi

Qj

Vi

j le c1+ qminusp2

Hence we see that v satisfies a Campanato type integral inequality

Lemma 56 Suppose that 0 lt r le R2 There exists a constant c independent ofVR and such that

intSr

vminus vr 2 dz le c( r

R

)n+4 intS R

2

uminus V 2 dz (59)

Proof We write (58) as

vit minus pminus2

Ai

Qj

V

pminus2vjx

x

= 0

We change of variable = pminus2t minus t0 then the function vx = vx 2minusp+ t0satisfies a uniformly parabolic system

vi minus

Ai

Qj

V

pminus2vjx

x

= 0

Thus from results of Campanato (1966) we prove

intS

vminus v2 dz le c(

R

)n+4 intS R

2

uminus V 2 dz

and scaling back prove (59)

We estimate the L2 norm of uminus v in SR2

Lemma 57 For each lt R2 there exists a constant c independent of such that

intS

uminus v2 dz le c(1+ qminusp 4

n+2)(

minus2 ndashintSR

uminus V 2 dz) int

SRuminus V 2 dz

for = min(12

2n

)

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Regularity for Certain Nonlinear Parabolic Systems 631

Proof Subtracting(Ai

Qj

Vujx

)x

from both sides of (15) we get

uit minus

(Ai

Qj

Vujx

)x= (

Aiuminus Ai

Qj

Vujx

)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(510)

From (58) and (510) we see that uminus v satisfies

ui minus vit minus(Ai

Qj

Vuj minus vjx)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(511)

Simply applying uminus v to (511) we get

pminus2intS R

2uminusv2dzle

intS R

2

∣∣∣ int 1

0Ai

Qj

V+suminus VminusAi

Qj

Vds∣∣∣

times uminusV uminusvdz

and it follows thatintS R

2

uminus v2 dz

le minus2pminus2intS R

2

∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣2

uminus V 2 dz (512)

Note that AiV = F primeV V i

V and

Ai

Qj

V = F primeprimeV ViV

j

V 2 + F primeV (ij

V minus V iV

j

V 3)

where ij and are Kronecker delta functions Also we recall the structurecondition (13) F primeprimeprimeh le chpminus3 + hqminus3 We have that

∣∣∣ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds∣∣∣

le∣∣∣∣int 1

0F primeprime(V + suminus V)

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2

minusF primeprimeV ViV

j

V 2 ds

∣∣∣∣+int 1

0

∣∣∣∣F primeV + suminus VV + suminus V minus F primeV

V ∣∣∣∣ds

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632 Bae and Choe

+∣∣∣∣int 1

0F primeV + suminus V

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V3

minusF primeV ViV

j

V 3 ds

∣∣∣∣= I + II + III

The first term I is estimates as follows

I leint 1

0

∣∣∣F primeprimeV + suminus Vminus F primeprimeV ∣∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2∣∣∣∣∣ ds

+int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

(513)

By the mean value theorem we have that

int 1

0

∣∣∣F primeprimeV+suminusVminusF primeprimeV ∣∣∣ds le c

int 1

0

int 1

0

∣∣F primeprimeprimeV+shuminusV∣∣ds dhuminusV (514)

for some c and

int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

le cF primeprimeV int 1

0

int 1

0

ds dh∣∣V + shuminus V∣∣ uminus V (515)

If V ge 2uminus V then for all s isin 0 1 12 V le V + suminus V le 2V and from

(513) (514) and (515) we have I le c(V pminus3 + V qminus3

)uminus V If V le 2uminus V and p ge 2 then

∣∣F primeprimeV + suminus V∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le cuminus V pminus2 + cuminus V qminus2

and

∣∣F primeprimeV ∣∣ le cV pminus2 + cV qminus2 le cuminus V pminus2 + cuminus V qminus2

Hence we have that

I le cint 1

0

∣∣F primeprimeV + suminus V∣∣+ ∣∣F primeprimeV ∣∣ds le c1+ pminusquminus V pminus2

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Regularity for Certain Nonlinear Parabolic Systems 633

for some c independent of If V le 2uminus V and 1 lt p lt 2 then∣∣F primeprime(V + suminus V)∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le c1+ pminusqV + suminus Vpminus2

for some c We note thatint 1

0V + suminus Vpminus2 ds le uminus V pminus2

int 1

0

∣∣∣∣ V uminus V minus s

∣∣∣∣pminus2

ds le cuminus V pminus2

Therefore combining all these together we have

I le cV pminus3 uminus V (516)

for some c independent of For II and III we observe that

F primeV V le cV pminus2 + cV qminus2

∣∣∣∣FprimeV V i

Vj

V 3∣∣∣∣ le cV pminus2 + cV qminus2

∣∣∣∣(F primeV V

)∣∣∣∣ le cV pminus3 + cV qminus3

∣∣∣∣(F primeV V i

Vj

V 3)∣∣∣∣ le cV pminus3 + cV qminus3

Hence following the same argument for I we conclude that

II + III le cV pminus3 uminus V (517)

for some c independent of Combining (516) and (517) we obtain that∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣ le cV pminus3 uminus V (518)

and considering (518) in (512)intS R

2

uminus V 2 dz le cminus2pminus2V 2pminus3intS R

2

uminus V 4 dz

le cminus2intS R

2

uminus V 4 dz (519)

From Houmllderrsquos inequality we have thatintS R

2

uminus V 4 dz

=intS R

2

uminus V 8n+ 4nminus8n dz

le[

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx] 2

n int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

(520)

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634 Bae and Choe

From Lemma 55 we have

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx le c2 supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2 dx

le cp + 2qminusp

R2

intSR

uminus V 2 dz

for some c independent of Now recalling that 12 le V le we obtain that

int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

le c22minuspnminus2

n

int t0

t0minus 2minuspR24

[ intB R

2

(u + V )2pminus2uminus V 4 dx] nminus2

n

dt

le c4nminuspnminus8

n

int t0

t0minus 2minuspR24

[ intB R

2

∣∣u pminus22 uminus V pminus2

2 V∣∣ 2n

nminus2 dx] nminus2

n

dt

le c4nminuspnminus8

n

intS R

2

upminus2 2u2 dz+ c2nminuspnminus8

n pminus2intSR

uminus V 2 dz (521)

for some c independent of where we have used Sobolevrsquos inequality FromLemma 55 we haveint

S R2

upminus2 2u2 dz le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

Thus combining (519) (520) and (521) we have that

intS R

2uminus V 2 dz le c

(1+ qminusp 4

n+2)[

minus2ndashintSR

uminus V 2 dz] 2

n intSR

uminus V 2 dz

and this completes the proof

Now we prove an iteration lemma which says that if mean oscillation of unear some fixed V is small then u satisfies a Campanato type integral inequality

Lemma 58 Fix 0 lt 0 lt 1 then there exist small positive numbers independentof such that if V0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists V1 isin nN satisfying(12minusradic

0

) le V1 le

(1+radic

0

) (522)

ndashintSR

uminus V12 dz le 0 ndashintSR

uminus V02 dz (523)

ndashintSR

uminus V12 dz le 2

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Regularity for Certain Nonlinear Parabolic Systems 635

Proof Since V0 satisfies the assumption of Lemma 57 we apply Lemma 57 withV = V0 and obtainint

S R2

vminus u2 dz le cintSR

uminus V02 dz

So from the usual comparison argument we haveintSR

uminus uR2 dz le cintSR

vminus vR2 dz+ cintSR

uminus v2 dz

le cn+4intS R

2

uminus u R22 dz+ c

intS R

2

uminus v2 dz

le c(n+4 +

) intSR

uminus V02 dz

where c is independent of Choosing and isin 0 12 small we prove (523)

with V1 = uR Inequality ndashintSR

uminus V12 dz le 2 follows from (523) andassumption ndash

intSR

uminus V02 dz le 2 Now we also estimate

V1 minus V0 =∣∣∣∣∣ ndashintSR

vminus V0 dz

∣∣∣∣le c

[ndashintSR

vminus u2 dz] 1

2

+ c

[ndashintSR

uminus V02 dz] 1

2

le cminus1minusn2(1+

120

)12

Therefore using the same technique of DiBenedetto (1986) Lemma 44 and if wetake very small compared to (522) holds and this completes the proof

Lemma 59 There exist small positive numbers 0 independent of such that ifV0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists a sequence of vectors Vi i = 1 2 3 sub nN satisfying

14 le Vi le 4 ndash

intSiR

uminus Vi2 dz le 2

ndashintSi+1R

uminus Vi+12 dz le 0 ndashintSiR

uminus Vi2 dz

Proof For the proof of the lemma we need to show that 14 le Vi le 4 But this

follows from that

Vi+1 minus Vi2 le c(0 + minusnminus2

)∣∣∣ ndashintSiR

uminus Vi dz∣∣∣

le ci0

i∣∣∣ ndashint

SRuminus V02 dz

∣∣∣ le ci0

i2

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636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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ORDER REPRINTS

642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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t 200

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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Regularity for Certain Nonlinear Parabolic Systems 627

Proof Suppose that the assertion is not true then

B1minus2R0 ge

int t0minus2minuspR20

t0minus2minuspR20

A1minus2R0d gt 1minus BR0

2minuspR20 = 1minus SR0

and this contradicts the assumption (51)

The following lemma is useful in estimating sup u Indeed when p = q ge 2DiBenedetto and Friedman proved a similar theorem (see Lemma 62 inDiBenedetto and Friedman 1985) Now choose = 2 and in Lemma 52We define St

R

def= BR times t

Lemma 53 There exists a positive integer r independent of such that

∣∣A1minus 2r

2R0t

∣∣ le (1minus

(2

)2)BR0

for all t isin t0 minus

22minuspR2

0 t0

Proof Consider the function

v = log+[

2

2minus vminus 1minus 2+ + 2

2r

]

where r is a positive integer This function was used in Choe (1991) and DiBenedettoand Friedman (1985) We take 13 = ui

x2vprime 2 as a test function for (31) where

isin 0 is a standard cutoff function such that B1minusR for small decided later is

contained in support Integrating from to t we obtain

12

intStR

vt2prime 2 dz+

intStR

Ai

Qj

uujxx

uixx

2prime 2 dz

+ 2intStR

Ai

Qj

uujxx

uixukxx

ukx2primeprime 2 dz

+ 2intStR

Ai

Qj

uujxx

uix x

2prime dz = 0

Considering the fact 2primeprime = 21+ prime2 we have

intBR

2vx t 2xdx +intStR

upminus2v21+ prime2 2 dz

leintBR

2vx 2xdx + cintStR

upminus2v prime dz

+ cintStR

uqminus2v prime dz

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ORDER REPRINTS

628 Bae and Choe

Applying Youngrsquos inequality on the last two terms we getintBR

2vx t 2xdx

leintBR

2vx 2xdx + cintStR

upminus2 2 dz+ cintStR

u2qminuspminus2 2 dz

for some c Since le r log 2 and u is bounded we getintBR

2vx t 2xdx leintBR

2vx 2xdx + crBR

From Lemma 52 we getintBR

2vx t 2xdx le r log 221minus

1minus

2

BR

From definition of we haveintB1minusRcapvgt1minus

2r 22vx tdx ge r minus 1 log 22A1minus

2r 21minusRt

and

A1minus 2r

21minusRt le[( r

r minus 1

)2 1minus

1minus

2

+ c

r

]BR

Hence we obtain

A(1minus

2r

)2R

t le A1minus 2r

21minusRt + BRB1minusR

le[( r

r minus 1

)2 1minus

1minus

2

+ c

r+ n

]BR

Taking r so large that( r

r minus 1

)2 le(1minus

2

)1+

c

rle 3

82

and = 382n we obtain

A1minus 2r

2Rt le(1minus

(2

)2)BR

We take 13 = uixvminus h+ 2 as a test function in (31) where v = u2 and is

a cutoff function such that = 1 in Sr = 0 on pSR le cRminus r and t le cpminus2Rminus r2 Considering the ellipticity condition (12) we have

supt

int ∣∣vminus h+∣∣2 2 dx +

intupminus2vminus h+2 2 dz

le c

Rminus r2

intSR

upminus2vminus h+2 dz

+ c

Rminus r2

intSR

u2qminuspminus2vminus h+2 dz+ cintSR

vminus h+2 tdz (54)

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Regularity for Certain Nonlinear Parabolic Systems 629

for some c Now choose h ge 12 For such a choice of h we haveint

upminus2vminus h+2 2 dz ge cpminus2int

vminus h+2 2 dz

Hence from (54) we obtain

supt

intvminus h+2 2 dx + pminus2

intvminus h+2 2 dz

le[

2qminuspminus2

Rminus r2+ pminus2

Rminus r2

] intSR

vminus h+2 dz (55)

for some c independent of Observe that is bounded above by some fixednumber We make change of variables

= pminus2t minus t0 vx = vx t0 + 2minusp

and we write (55) as

supt

intBR

vminus h+2 2 dx +intSR

vminus h+2 2 dz le c

Rminus r2

intSR

vminus h+dz

Once we have this Caccioppoli type inequality we can apply truncation argumentTherefore following to similar argument in DiBenedetto and Friedman (1984 or1985) we have a maximum principle Now the proof for the next lemma is almostthe same as in DiBenedetto and Friedman (1984 or 1985) we omit the proof

Lemma 54 If for some and isin 0 1∣∣x t isin S1+R u2 lt 1minus 2∣∣ ge S1+R (56)

then there exists a number isin 0 1 independent of and R such that

supS R

2

ux t le

Now we assume that (56) fails that is∣∣x t isin S1+R u2 lt 1minus 2∣∣ lt S1+R (57)

Following a comparison argument similar to Choe (1991) or DiBenedetto andFriedman (1984) we prove that u satisfies a Campanato type integral inequalityLet V isin nN be a constant vector such that 1

2 le V le Taking uxkminus Vk

2 as atest function to (31) we have the following lemma

Lemma 55 There exists a constant c independent of such that

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2tdx +intS R

2

upminus22u2 dz

le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

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630 Bae and Choe

Now suppose that v is a solution to

vit minus

(Ai

Qj

Vvjx

)x

= 0 (58)

with v = u on pSR2 Since 1

2 le V le le c for some fixed c we see that fromthe structure condition of F

2 le 2minuspAi

Qj

Vi

j le c1+ qminusp2

Hence we see that v satisfies a Campanato type integral inequality

Lemma 56 Suppose that 0 lt r le R2 There exists a constant c independent ofVR and such that

intSr

vminus vr 2 dz le c( r

R

)n+4 intS R

2

uminus V 2 dz (59)

Proof We write (58) as

vit minus pminus2

Ai

Qj

V

pminus2vjx

x

= 0

We change of variable = pminus2t minus t0 then the function vx = vx 2minusp+ t0satisfies a uniformly parabolic system

vi minus

Ai

Qj

V

pminus2vjx

x

= 0

Thus from results of Campanato (1966) we prove

intS

vminus v2 dz le c(

R

)n+4 intS R

2

uminus V 2 dz

and scaling back prove (59)

We estimate the L2 norm of uminus v in SR2

Lemma 57 For each lt R2 there exists a constant c independent of such that

intS

uminus v2 dz le c(1+ qminusp 4

n+2)(

minus2 ndashintSR

uminus V 2 dz) int

SRuminus V 2 dz

for = min(12

2n

)

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Regularity for Certain Nonlinear Parabolic Systems 631

Proof Subtracting(Ai

Qj

Vujx

)x

from both sides of (15) we get

uit minus

(Ai

Qj

Vujx

)x= (

Aiuminus Ai

Qj

Vujx

)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(510)

From (58) and (510) we see that uminus v satisfies

ui minus vit minus(Ai

Qj

Vuj minus vjx)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(511)

Simply applying uminus v to (511) we get

pminus2intS R

2uminusv2dzle

intS R

2

∣∣∣ int 1

0Ai

Qj

V+suminus VminusAi

Qj

Vds∣∣∣

times uminusV uminusvdz

and it follows thatintS R

2

uminus v2 dz

le minus2pminus2intS R

2

∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣2

uminus V 2 dz (512)

Note that AiV = F primeV V i

V and

Ai

Qj

V = F primeprimeV ViV

j

V 2 + F primeV (ij

V minus V iV

j

V 3)

where ij and are Kronecker delta functions Also we recall the structurecondition (13) F primeprimeprimeh le chpminus3 + hqminus3 We have that

∣∣∣ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds∣∣∣

le∣∣∣∣int 1

0F primeprime(V + suminus V)

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2

minusF primeprimeV ViV

j

V 2 ds

∣∣∣∣+int 1

0

∣∣∣∣F primeV + suminus VV + suminus V minus F primeV

V ∣∣∣∣ds

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632 Bae and Choe

+∣∣∣∣int 1

0F primeV + suminus V

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V3

minusF primeV ViV

j

V 3 ds

∣∣∣∣= I + II + III

The first term I is estimates as follows

I leint 1

0

∣∣∣F primeprimeV + suminus Vminus F primeprimeV ∣∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2∣∣∣∣∣ ds

+int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

(513)

By the mean value theorem we have that

int 1

0

∣∣∣F primeprimeV+suminusVminusF primeprimeV ∣∣∣ds le c

int 1

0

int 1

0

∣∣F primeprimeprimeV+shuminusV∣∣ds dhuminusV (514)

for some c and

int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

le cF primeprimeV int 1

0

int 1

0

ds dh∣∣V + shuminus V∣∣ uminus V (515)

If V ge 2uminus V then for all s isin 0 1 12 V le V + suminus V le 2V and from

(513) (514) and (515) we have I le c(V pminus3 + V qminus3

)uminus V If V le 2uminus V and p ge 2 then

∣∣F primeprimeV + suminus V∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le cuminus V pminus2 + cuminus V qminus2

and

∣∣F primeprimeV ∣∣ le cV pminus2 + cV qminus2 le cuminus V pminus2 + cuminus V qminus2

Hence we have that

I le cint 1

0

∣∣F primeprimeV + suminus V∣∣+ ∣∣F primeprimeV ∣∣ds le c1+ pminusquminus V pminus2

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Regularity for Certain Nonlinear Parabolic Systems 633

for some c independent of If V le 2uminus V and 1 lt p lt 2 then∣∣F primeprime(V + suminus V)∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le c1+ pminusqV + suminus Vpminus2

for some c We note thatint 1

0V + suminus Vpminus2 ds le uminus V pminus2

int 1

0

∣∣∣∣ V uminus V minus s

∣∣∣∣pminus2

ds le cuminus V pminus2

Therefore combining all these together we have

I le cV pminus3 uminus V (516)

for some c independent of For II and III we observe that

F primeV V le cV pminus2 + cV qminus2

∣∣∣∣FprimeV V i

Vj

V 3∣∣∣∣ le cV pminus2 + cV qminus2

∣∣∣∣(F primeV V

)∣∣∣∣ le cV pminus3 + cV qminus3

∣∣∣∣(F primeV V i

Vj

V 3)∣∣∣∣ le cV pminus3 + cV qminus3

Hence following the same argument for I we conclude that

II + III le cV pminus3 uminus V (517)

for some c independent of Combining (516) and (517) we obtain that∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣ le cV pminus3 uminus V (518)

and considering (518) in (512)intS R

2

uminus V 2 dz le cminus2pminus2V 2pminus3intS R

2

uminus V 4 dz

le cminus2intS R

2

uminus V 4 dz (519)

From Houmllderrsquos inequality we have thatintS R

2

uminus V 4 dz

=intS R

2

uminus V 8n+ 4nminus8n dz

le[

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx] 2

n int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

(520)

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634 Bae and Choe

From Lemma 55 we have

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx le c2 supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2 dx

le cp + 2qminusp

R2

intSR

uminus V 2 dz

for some c independent of Now recalling that 12 le V le we obtain that

int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

le c22minuspnminus2

n

int t0

t0minus 2minuspR24

[ intB R

2

(u + V )2pminus2uminus V 4 dx] nminus2

n

dt

le c4nminuspnminus8

n

int t0

t0minus 2minuspR24

[ intB R

2

∣∣u pminus22 uminus V pminus2

2 V∣∣ 2n

nminus2 dx] nminus2

n

dt

le c4nminuspnminus8

n

intS R

2

upminus2 2u2 dz+ c2nminuspnminus8

n pminus2intSR

uminus V 2 dz (521)

for some c independent of where we have used Sobolevrsquos inequality FromLemma 55 we haveint

S R2

upminus2 2u2 dz le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

Thus combining (519) (520) and (521) we have that

intS R

2uminus V 2 dz le c

(1+ qminusp 4

n+2)[

minus2ndashintSR

uminus V 2 dz] 2

n intSR

uminus V 2 dz

and this completes the proof

Now we prove an iteration lemma which says that if mean oscillation of unear some fixed V is small then u satisfies a Campanato type integral inequality

Lemma 58 Fix 0 lt 0 lt 1 then there exist small positive numbers independentof such that if V0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists V1 isin nN satisfying(12minusradic

0

) le V1 le

(1+radic

0

) (522)

ndashintSR

uminus V12 dz le 0 ndashintSR

uminus V02 dz (523)

ndashintSR

uminus V12 dz le 2

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Regularity for Certain Nonlinear Parabolic Systems 635

Proof Since V0 satisfies the assumption of Lemma 57 we apply Lemma 57 withV = V0 and obtainint

S R2

vminus u2 dz le cintSR

uminus V02 dz

So from the usual comparison argument we haveintSR

uminus uR2 dz le cintSR

vminus vR2 dz+ cintSR

uminus v2 dz

le cn+4intS R

2

uminus u R22 dz+ c

intS R

2

uminus v2 dz

le c(n+4 +

) intSR

uminus V02 dz

where c is independent of Choosing and isin 0 12 small we prove (523)

with V1 = uR Inequality ndashintSR

uminus V12 dz le 2 follows from (523) andassumption ndash

intSR

uminus V02 dz le 2 Now we also estimate

V1 minus V0 =∣∣∣∣∣ ndashintSR

vminus V0 dz

∣∣∣∣le c

[ndashintSR

vminus u2 dz] 1

2

+ c

[ndashintSR

uminus V02 dz] 1

2

le cminus1minusn2(1+

120

)12

Therefore using the same technique of DiBenedetto (1986) Lemma 44 and if wetake very small compared to (522) holds and this completes the proof

Lemma 59 There exist small positive numbers 0 independent of such that ifV0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists a sequence of vectors Vi i = 1 2 3 sub nN satisfying

14 le Vi le 4 ndash

intSiR

uminus Vi2 dz le 2

ndashintSi+1R

uminus Vi+12 dz le 0 ndashintSiR

uminus Vi2 dz

Proof For the proof of the lemma we need to show that 14 le Vi le 4 But this

follows from that

Vi+1 minus Vi2 le c(0 + minusnminus2

)∣∣∣ ndashintSiR

uminus Vi dz∣∣∣

le ci0

i∣∣∣ ndashint

SRuminus V02 dz

∣∣∣ le ci0

i2

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636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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628 Bae and Choe

Applying Youngrsquos inequality on the last two terms we getintBR

2vx t 2xdx

leintBR

2vx 2xdx + cintStR

upminus2 2 dz+ cintStR

u2qminuspminus2 2 dz

for some c Since le r log 2 and u is bounded we getintBR

2vx t 2xdx leintBR

2vx 2xdx + crBR

From Lemma 52 we getintBR

2vx t 2xdx le r log 221minus

1minus

2

BR

From definition of we haveintB1minusRcapvgt1minus

2r 22vx tdx ge r minus 1 log 22A1minus

2r 21minusRt

and

A1minus 2r

21minusRt le[( r

r minus 1

)2 1minus

1minus

2

+ c

r

]BR

Hence we obtain

A(1minus

2r

)2R

t le A1minus 2r

21minusRt + BRB1minusR

le[( r

r minus 1

)2 1minus

1minus

2

+ c

r+ n

]BR

Taking r so large that( r

r minus 1

)2 le(1minus

2

)1+

c

rle 3

82

and = 382n we obtain

A1minus 2r

2Rt le(1minus

(2

)2)BR

We take 13 = uixvminus h+ 2 as a test function in (31) where v = u2 and is

a cutoff function such that = 1 in Sr = 0 on pSR le cRminus r and t le cpminus2Rminus r2 Considering the ellipticity condition (12) we have

supt

int ∣∣vminus h+∣∣2 2 dx +

intupminus2vminus h+2 2 dz

le c

Rminus r2

intSR

upminus2vminus h+2 dz

+ c

Rminus r2

intSR

u2qminuspminus2vminus h+2 dz+ cintSR

vminus h+2 tdz (54)

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 629

for some c Now choose h ge 12 For such a choice of h we haveint

upminus2vminus h+2 2 dz ge cpminus2int

vminus h+2 2 dz

Hence from (54) we obtain

supt

intvminus h+2 2 dx + pminus2

intvminus h+2 2 dz

le[

2qminuspminus2

Rminus r2+ pminus2

Rminus r2

] intSR

vminus h+2 dz (55)

for some c independent of Observe that is bounded above by some fixednumber We make change of variables

= pminus2t minus t0 vx = vx t0 + 2minusp

and we write (55) as

supt

intBR

vminus h+2 2 dx +intSR

vminus h+2 2 dz le c

Rminus r2

intSR

vminus h+dz

Once we have this Caccioppoli type inequality we can apply truncation argumentTherefore following to similar argument in DiBenedetto and Friedman (1984 or1985) we have a maximum principle Now the proof for the next lemma is almostthe same as in DiBenedetto and Friedman (1984 or 1985) we omit the proof

Lemma 54 If for some and isin 0 1∣∣x t isin S1+R u2 lt 1minus 2∣∣ ge S1+R (56)

then there exists a number isin 0 1 independent of and R such that

supS R

2

ux t le

Now we assume that (56) fails that is∣∣x t isin S1+R u2 lt 1minus 2∣∣ lt S1+R (57)

Following a comparison argument similar to Choe (1991) or DiBenedetto andFriedman (1984) we prove that u satisfies a Campanato type integral inequalityLet V isin nN be a constant vector such that 1

2 le V le Taking uxkminus Vk

2 as atest function to (31) we have the following lemma

Lemma 55 There exists a constant c independent of such that

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2tdx +intS R

2

upminus22u2 dz

le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

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630 Bae and Choe

Now suppose that v is a solution to

vit minus

(Ai

Qj

Vvjx

)x

= 0 (58)

with v = u on pSR2 Since 1

2 le V le le c for some fixed c we see that fromthe structure condition of F

2 le 2minuspAi

Qj

Vi

j le c1+ qminusp2

Hence we see that v satisfies a Campanato type integral inequality

Lemma 56 Suppose that 0 lt r le R2 There exists a constant c independent ofVR and such that

intSr

vminus vr 2 dz le c( r

R

)n+4 intS R

2

uminus V 2 dz (59)

Proof We write (58) as

vit minus pminus2

Ai

Qj

V

pminus2vjx

x

= 0

We change of variable = pminus2t minus t0 then the function vx = vx 2minusp+ t0satisfies a uniformly parabolic system

vi minus

Ai

Qj

V

pminus2vjx

x

= 0

Thus from results of Campanato (1966) we prove

intS

vminus v2 dz le c(

R

)n+4 intS R

2

uminus V 2 dz

and scaling back prove (59)

We estimate the L2 norm of uminus v in SR2

Lemma 57 For each lt R2 there exists a constant c independent of such that

intS

uminus v2 dz le c(1+ qminusp 4

n+2)(

minus2 ndashintSR

uminus V 2 dz) int

SRuminus V 2 dz

for = min(12

2n

)

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Regularity for Certain Nonlinear Parabolic Systems 631

Proof Subtracting(Ai

Qj

Vujx

)x

from both sides of (15) we get

uit minus

(Ai

Qj

Vujx

)x= (

Aiuminus Ai

Qj

Vujx

)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(510)

From (58) and (510) we see that uminus v satisfies

ui minus vit minus(Ai

Qj

Vuj minus vjx)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(511)

Simply applying uminus v to (511) we get

pminus2intS R

2uminusv2dzle

intS R

2

∣∣∣ int 1

0Ai

Qj

V+suminus VminusAi

Qj

Vds∣∣∣

times uminusV uminusvdz

and it follows thatintS R

2

uminus v2 dz

le minus2pminus2intS R

2

∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣2

uminus V 2 dz (512)

Note that AiV = F primeV V i

V and

Ai

Qj

V = F primeprimeV ViV

j

V 2 + F primeV (ij

V minus V iV

j

V 3)

where ij and are Kronecker delta functions Also we recall the structurecondition (13) F primeprimeprimeh le chpminus3 + hqminus3 We have that

∣∣∣ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds∣∣∣

le∣∣∣∣int 1

0F primeprime(V + suminus V)

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2

minusF primeprimeV ViV

j

V 2 ds

∣∣∣∣+int 1

0

∣∣∣∣F primeV + suminus VV + suminus V minus F primeV

V ∣∣∣∣ds

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632 Bae and Choe

+∣∣∣∣int 1

0F primeV + suminus V

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V3

minusF primeV ViV

j

V 3 ds

∣∣∣∣= I + II + III

The first term I is estimates as follows

I leint 1

0

∣∣∣F primeprimeV + suminus Vminus F primeprimeV ∣∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2∣∣∣∣∣ ds

+int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

(513)

By the mean value theorem we have that

int 1

0

∣∣∣F primeprimeV+suminusVminusF primeprimeV ∣∣∣ds le c

int 1

0

int 1

0

∣∣F primeprimeprimeV+shuminusV∣∣ds dhuminusV (514)

for some c and

int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

le cF primeprimeV int 1

0

int 1

0

ds dh∣∣V + shuminus V∣∣ uminus V (515)

If V ge 2uminus V then for all s isin 0 1 12 V le V + suminus V le 2V and from

(513) (514) and (515) we have I le c(V pminus3 + V qminus3

)uminus V If V le 2uminus V and p ge 2 then

∣∣F primeprimeV + suminus V∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le cuminus V pminus2 + cuminus V qminus2

and

∣∣F primeprimeV ∣∣ le cV pminus2 + cV qminus2 le cuminus V pminus2 + cuminus V qminus2

Hence we have that

I le cint 1

0

∣∣F primeprimeV + suminus V∣∣+ ∣∣F primeprimeV ∣∣ds le c1+ pminusquminus V pminus2

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Regularity for Certain Nonlinear Parabolic Systems 633

for some c independent of If V le 2uminus V and 1 lt p lt 2 then∣∣F primeprime(V + suminus V)∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le c1+ pminusqV + suminus Vpminus2

for some c We note thatint 1

0V + suminus Vpminus2 ds le uminus V pminus2

int 1

0

∣∣∣∣ V uminus V minus s

∣∣∣∣pminus2

ds le cuminus V pminus2

Therefore combining all these together we have

I le cV pminus3 uminus V (516)

for some c independent of For II and III we observe that

F primeV V le cV pminus2 + cV qminus2

∣∣∣∣FprimeV V i

Vj

V 3∣∣∣∣ le cV pminus2 + cV qminus2

∣∣∣∣(F primeV V

)∣∣∣∣ le cV pminus3 + cV qminus3

∣∣∣∣(F primeV V i

Vj

V 3)∣∣∣∣ le cV pminus3 + cV qminus3

Hence following the same argument for I we conclude that

II + III le cV pminus3 uminus V (517)

for some c independent of Combining (516) and (517) we obtain that∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣ le cV pminus3 uminus V (518)

and considering (518) in (512)intS R

2

uminus V 2 dz le cminus2pminus2V 2pminus3intS R

2

uminus V 4 dz

le cminus2intS R

2

uminus V 4 dz (519)

From Houmllderrsquos inequality we have thatintS R

2

uminus V 4 dz

=intS R

2

uminus V 8n+ 4nminus8n dz

le[

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx] 2

n int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

(520)

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634 Bae and Choe

From Lemma 55 we have

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx le c2 supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2 dx

le cp + 2qminusp

R2

intSR

uminus V 2 dz

for some c independent of Now recalling that 12 le V le we obtain that

int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

le c22minuspnminus2

n

int t0

t0minus 2minuspR24

[ intB R

2

(u + V )2pminus2uminus V 4 dx] nminus2

n

dt

le c4nminuspnminus8

n

int t0

t0minus 2minuspR24

[ intB R

2

∣∣u pminus22 uminus V pminus2

2 V∣∣ 2n

nminus2 dx] nminus2

n

dt

le c4nminuspnminus8

n

intS R

2

upminus2 2u2 dz+ c2nminuspnminus8

n pminus2intSR

uminus V 2 dz (521)

for some c independent of where we have used Sobolevrsquos inequality FromLemma 55 we haveint

S R2

upminus2 2u2 dz le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

Thus combining (519) (520) and (521) we have that

intS R

2uminus V 2 dz le c

(1+ qminusp 4

n+2)[

minus2ndashintSR

uminus V 2 dz] 2

n intSR

uminus V 2 dz

and this completes the proof

Now we prove an iteration lemma which says that if mean oscillation of unear some fixed V is small then u satisfies a Campanato type integral inequality

Lemma 58 Fix 0 lt 0 lt 1 then there exist small positive numbers independentof such that if V0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists V1 isin nN satisfying(12minusradic

0

) le V1 le

(1+radic

0

) (522)

ndashintSR

uminus V12 dz le 0 ndashintSR

uminus V02 dz (523)

ndashintSR

uminus V12 dz le 2

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Regularity for Certain Nonlinear Parabolic Systems 635

Proof Since V0 satisfies the assumption of Lemma 57 we apply Lemma 57 withV = V0 and obtainint

S R2

vminus u2 dz le cintSR

uminus V02 dz

So from the usual comparison argument we haveintSR

uminus uR2 dz le cintSR

vminus vR2 dz+ cintSR

uminus v2 dz

le cn+4intS R

2

uminus u R22 dz+ c

intS R

2

uminus v2 dz

le c(n+4 +

) intSR

uminus V02 dz

where c is independent of Choosing and isin 0 12 small we prove (523)

with V1 = uR Inequality ndashintSR

uminus V12 dz le 2 follows from (523) andassumption ndash

intSR

uminus V02 dz le 2 Now we also estimate

V1 minus V0 =∣∣∣∣∣ ndashintSR

vminus V0 dz

∣∣∣∣le c

[ndashintSR

vminus u2 dz] 1

2

+ c

[ndashintSR

uminus V02 dz] 1

2

le cminus1minusn2(1+

120

)12

Therefore using the same technique of DiBenedetto (1986) Lemma 44 and if wetake very small compared to (522) holds and this completes the proof

Lemma 59 There exist small positive numbers 0 independent of such that ifV0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists a sequence of vectors Vi i = 1 2 3 sub nN satisfying

14 le Vi le 4 ndash

intSiR

uminus Vi2 dz le 2

ndashintSi+1R

uminus Vi+12 dz le 0 ndashintSiR

uminus Vi2 dz

Proof For the proof of the lemma we need to show that 14 le Vi le 4 But this

follows from that

Vi+1 minus Vi2 le c(0 + minusnminus2

)∣∣∣ ndashintSiR

uminus Vi dz∣∣∣

le ci0

i∣∣∣ ndashint

SRuminus V02 dz

∣∣∣ le ci0

i2

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636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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t 200

7

ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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Regularity for Certain Nonlinear Parabolic Systems 629

for some c Now choose h ge 12 For such a choice of h we haveint

upminus2vminus h+2 2 dz ge cpminus2int

vminus h+2 2 dz

Hence from (54) we obtain

supt

intvminus h+2 2 dx + pminus2

intvminus h+2 2 dz

le[

2qminuspminus2

Rminus r2+ pminus2

Rminus r2

] intSR

vminus h+2 dz (55)

for some c independent of Observe that is bounded above by some fixednumber We make change of variables

= pminus2t minus t0 vx = vx t0 + 2minusp

and we write (55) as

supt

intBR

vminus h+2 2 dx +intSR

vminus h+2 2 dz le c

Rminus r2

intSR

vminus h+dz

Once we have this Caccioppoli type inequality we can apply truncation argumentTherefore following to similar argument in DiBenedetto and Friedman (1984 or1985) we have a maximum principle Now the proof for the next lemma is almostthe same as in DiBenedetto and Friedman (1984 or 1985) we omit the proof

Lemma 54 If for some and isin 0 1∣∣x t isin S1+R u2 lt 1minus 2∣∣ ge S1+R (56)

then there exists a number isin 0 1 independent of and R such that

supS R

2

ux t le

Now we assume that (56) fails that is∣∣x t isin S1+R u2 lt 1minus 2∣∣ lt S1+R (57)

Following a comparison argument similar to Choe (1991) or DiBenedetto andFriedman (1984) we prove that u satisfies a Campanato type integral inequalityLet V isin nN be a constant vector such that 1

2 le V le Taking uxkminus Vk

2 as atest function to (31) we have the following lemma

Lemma 55 There exists a constant c independent of such that

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2tdx +intS R

2

upminus22u2 dz

le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

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630 Bae and Choe

Now suppose that v is a solution to

vit minus

(Ai

Qj

Vvjx

)x

= 0 (58)

with v = u on pSR2 Since 1

2 le V le le c for some fixed c we see that fromthe structure condition of F

2 le 2minuspAi

Qj

Vi

j le c1+ qminusp2

Hence we see that v satisfies a Campanato type integral inequality

Lemma 56 Suppose that 0 lt r le R2 There exists a constant c independent ofVR and such that

intSr

vminus vr 2 dz le c( r

R

)n+4 intS R

2

uminus V 2 dz (59)

Proof We write (58) as

vit minus pminus2

Ai

Qj

V

pminus2vjx

x

= 0

We change of variable = pminus2t minus t0 then the function vx = vx 2minusp+ t0satisfies a uniformly parabolic system

vi minus

Ai

Qj

V

pminus2vjx

x

= 0

Thus from results of Campanato (1966) we prove

intS

vminus v2 dz le c(

R

)n+4 intS R

2

uminus V 2 dz

and scaling back prove (59)

We estimate the L2 norm of uminus v in SR2

Lemma 57 For each lt R2 there exists a constant c independent of such that

intS

uminus v2 dz le c(1+ qminusp 4

n+2)(

minus2 ndashintSR

uminus V 2 dz) int

SRuminus V 2 dz

for = min(12

2n

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Regularity for Certain Nonlinear Parabolic Systems 631

Proof Subtracting(Ai

Qj

Vujx

)x

from both sides of (15) we get

uit minus

(Ai

Qj

Vujx

)x= (

Aiuminus Ai

Qj

Vujx

)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(510)

From (58) and (510) we see that uminus v satisfies

ui minus vit minus(Ai

Qj

Vuj minus vjx)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(511)

Simply applying uminus v to (511) we get

pminus2intS R

2uminusv2dzle

intS R

2

∣∣∣ int 1

0Ai

Qj

V+suminus VminusAi

Qj

Vds∣∣∣

times uminusV uminusvdz

and it follows thatintS R

2

uminus v2 dz

le minus2pminus2intS R

2

∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣2

uminus V 2 dz (512)

Note that AiV = F primeV V i

V and

Ai

Qj

V = F primeprimeV ViV

j

V 2 + F primeV (ij

V minus V iV

j

V 3)

where ij and are Kronecker delta functions Also we recall the structurecondition (13) F primeprimeprimeh le chpminus3 + hqminus3 We have that

∣∣∣ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds∣∣∣

le∣∣∣∣int 1

0F primeprime(V + suminus V)

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2

minusF primeprimeV ViV

j

V 2 ds

∣∣∣∣+int 1

0

∣∣∣∣F primeV + suminus VV + suminus V minus F primeV

V ∣∣∣∣ds

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632 Bae and Choe

+∣∣∣∣int 1

0F primeV + suminus V

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V3

minusF primeV ViV

j

V 3 ds

∣∣∣∣= I + II + III

The first term I is estimates as follows

I leint 1

0

∣∣∣F primeprimeV + suminus Vminus F primeprimeV ∣∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2∣∣∣∣∣ ds

+int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

(513)

By the mean value theorem we have that

int 1

0

∣∣∣F primeprimeV+suminusVminusF primeprimeV ∣∣∣ds le c

int 1

0

int 1

0

∣∣F primeprimeprimeV+shuminusV∣∣ds dhuminusV (514)

for some c and

int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

le cF primeprimeV int 1

0

int 1

0

ds dh∣∣V + shuminus V∣∣ uminus V (515)

If V ge 2uminus V then for all s isin 0 1 12 V le V + suminus V le 2V and from

(513) (514) and (515) we have I le c(V pminus3 + V qminus3

)uminus V If V le 2uminus V and p ge 2 then

∣∣F primeprimeV + suminus V∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le cuminus V pminus2 + cuminus V qminus2

and

∣∣F primeprimeV ∣∣ le cV pminus2 + cV qminus2 le cuminus V pminus2 + cuminus V qminus2

Hence we have that

I le cint 1

0

∣∣F primeprimeV + suminus V∣∣+ ∣∣F primeprimeV ∣∣ds le c1+ pminusquminus V pminus2

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Regularity for Certain Nonlinear Parabolic Systems 633

for some c independent of If V le 2uminus V and 1 lt p lt 2 then∣∣F primeprime(V + suminus V)∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le c1+ pminusqV + suminus Vpminus2

for some c We note thatint 1

0V + suminus Vpminus2 ds le uminus V pminus2

int 1

0

∣∣∣∣ V uminus V minus s

∣∣∣∣pminus2

ds le cuminus V pminus2

Therefore combining all these together we have

I le cV pminus3 uminus V (516)

for some c independent of For II and III we observe that

F primeV V le cV pminus2 + cV qminus2

∣∣∣∣FprimeV V i

Vj

V 3∣∣∣∣ le cV pminus2 + cV qminus2

∣∣∣∣(F primeV V

)∣∣∣∣ le cV pminus3 + cV qminus3

∣∣∣∣(F primeV V i

Vj

V 3)∣∣∣∣ le cV pminus3 + cV qminus3

Hence following the same argument for I we conclude that

II + III le cV pminus3 uminus V (517)

for some c independent of Combining (516) and (517) we obtain that∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣ le cV pminus3 uminus V (518)

and considering (518) in (512)intS R

2

uminus V 2 dz le cminus2pminus2V 2pminus3intS R

2

uminus V 4 dz

le cminus2intS R

2

uminus V 4 dz (519)

From Houmllderrsquos inequality we have thatintS R

2

uminus V 4 dz

=intS R

2

uminus V 8n+ 4nminus8n dz

le[

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx] 2

n int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

(520)

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634 Bae and Choe

From Lemma 55 we have

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx le c2 supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2 dx

le cp + 2qminusp

R2

intSR

uminus V 2 dz

for some c independent of Now recalling that 12 le V le we obtain that

int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

le c22minuspnminus2

n

int t0

t0minus 2minuspR24

[ intB R

2

(u + V )2pminus2uminus V 4 dx] nminus2

n

dt

le c4nminuspnminus8

n

int t0

t0minus 2minuspR24

[ intB R

2

∣∣u pminus22 uminus V pminus2

2 V∣∣ 2n

nminus2 dx] nminus2

n

dt

le c4nminuspnminus8

n

intS R

2

upminus2 2u2 dz+ c2nminuspnminus8

n pminus2intSR

uminus V 2 dz (521)

for some c independent of where we have used Sobolevrsquos inequality FromLemma 55 we haveint

S R2

upminus2 2u2 dz le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

Thus combining (519) (520) and (521) we have that

intS R

2uminus V 2 dz le c

(1+ qminusp 4

n+2)[

minus2ndashintSR

uminus V 2 dz] 2

n intSR

uminus V 2 dz

and this completes the proof

Now we prove an iteration lemma which says that if mean oscillation of unear some fixed V is small then u satisfies a Campanato type integral inequality

Lemma 58 Fix 0 lt 0 lt 1 then there exist small positive numbers independentof such that if V0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists V1 isin nN satisfying(12minusradic

0

) le V1 le

(1+radic

0

) (522)

ndashintSR

uminus V12 dz le 0 ndashintSR

uminus V02 dz (523)

ndashintSR

uminus V12 dz le 2

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Regularity for Certain Nonlinear Parabolic Systems 635

Proof Since V0 satisfies the assumption of Lemma 57 we apply Lemma 57 withV = V0 and obtainint

S R2

vminus u2 dz le cintSR

uminus V02 dz

So from the usual comparison argument we haveintSR

uminus uR2 dz le cintSR

vminus vR2 dz+ cintSR

uminus v2 dz

le cn+4intS R

2

uminus u R22 dz+ c

intS R

2

uminus v2 dz

le c(n+4 +

) intSR

uminus V02 dz

where c is independent of Choosing and isin 0 12 small we prove (523)

with V1 = uR Inequality ndashintSR

uminus V12 dz le 2 follows from (523) andassumption ndash

intSR

uminus V02 dz le 2 Now we also estimate

V1 minus V0 =∣∣∣∣∣ ndashintSR

vminus V0 dz

∣∣∣∣le c

[ndashintSR

vminus u2 dz] 1

2

+ c

[ndashintSR

uminus V02 dz] 1

2

le cminus1minusn2(1+

120

)12

Therefore using the same technique of DiBenedetto (1986) Lemma 44 and if wetake very small compared to (522) holds and this completes the proof

Lemma 59 There exist small positive numbers 0 independent of such that ifV0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists a sequence of vectors Vi i = 1 2 3 sub nN satisfying

14 le Vi le 4 ndash

intSiR

uminus Vi2 dz le 2

ndashintSi+1R

uminus Vi+12 dz le 0 ndashintSiR

uminus Vi2 dz

Proof For the proof of the lemma we need to show that 14 le Vi le 4 But this

follows from that

Vi+1 minus Vi2 le c(0 + minusnminus2

)∣∣∣ ndashintSiR

uminus Vi dz∣∣∣

le ci0

i∣∣∣ ndashint

SRuminus V02 dz

∣∣∣ le ci0

i2

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636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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t 200

7

ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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630 Bae and Choe

Now suppose that v is a solution to

vit minus

(Ai

Qj

Vvjx

)x

= 0 (58)

with v = u on pSR2 Since 1

2 le V le le c for some fixed c we see that fromthe structure condition of F

2 le 2minuspAi

Qj

Vi

j le c1+ qminusp2

Hence we see that v satisfies a Campanato type integral inequality

Lemma 56 Suppose that 0 lt r le R2 There exists a constant c independent ofVR and such that

intSr

vminus vr 2 dz le c( r

R

)n+4 intS R

2

uminus V 2 dz (59)

Proof We write (58) as

vit minus pminus2

Ai

Qj

V

pminus2vjx

x

= 0

We change of variable = pminus2t minus t0 then the function vx = vx 2minusp+ t0satisfies a uniformly parabolic system

vi minus

Ai

Qj

V

pminus2vjx

x

= 0

Thus from results of Campanato (1966) we prove

intS

vminus v2 dz le c(

R

)n+4 intS R

2

uminus V 2 dz

and scaling back prove (59)

We estimate the L2 norm of uminus v in SR2

Lemma 57 For each lt R2 there exists a constant c independent of such that

intS

uminus v2 dz le c(1+ qminusp 4

n+2)(

minus2 ndashintSR

uminus V 2 dz) int

SRuminus V 2 dz

for = min(12

2n

)

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Regularity for Certain Nonlinear Parabolic Systems 631

Proof Subtracting(Ai

Qj

Vujx

)x

from both sides of (15) we get

uit minus

(Ai

Qj

Vujx

)x= (

Aiuminus Ai

Qj

Vujx

)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(510)

From (58) and (510) we see that uminus v satisfies

ui minus vit minus(Ai

Qj

Vuj minus vjx)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(511)

Simply applying uminus v to (511) we get

pminus2intS R

2uminusv2dzle

intS R

2

∣∣∣ int 1

0Ai

Qj

V+suminus VminusAi

Qj

Vds∣∣∣

times uminusV uminusvdz

and it follows thatintS R

2

uminus v2 dz

le minus2pminus2intS R

2

∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣2

uminus V 2 dz (512)

Note that AiV = F primeV V i

V and

Ai

Qj

V = F primeprimeV ViV

j

V 2 + F primeV (ij

V minus V iV

j

V 3)

where ij and are Kronecker delta functions Also we recall the structurecondition (13) F primeprimeprimeh le chpminus3 + hqminus3 We have that

∣∣∣ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds∣∣∣

le∣∣∣∣int 1

0F primeprime(V + suminus V)

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2

minusF primeprimeV ViV

j

V 2 ds

∣∣∣∣+int 1

0

∣∣∣∣F primeV + suminus VV + suminus V minus F primeV

V ∣∣∣∣ds

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ORDER REPRINTS

632 Bae and Choe

+∣∣∣∣int 1

0F primeV + suminus V

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V3

minusF primeV ViV

j

V 3 ds

∣∣∣∣= I + II + III

The first term I is estimates as follows

I leint 1

0

∣∣∣F primeprimeV + suminus Vminus F primeprimeV ∣∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2∣∣∣∣∣ ds

+int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

(513)

By the mean value theorem we have that

int 1

0

∣∣∣F primeprimeV+suminusVminusF primeprimeV ∣∣∣ds le c

int 1

0

int 1

0

∣∣F primeprimeprimeV+shuminusV∣∣ds dhuminusV (514)

for some c and

int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

le cF primeprimeV int 1

0

int 1

0

ds dh∣∣V + shuminus V∣∣ uminus V (515)

If V ge 2uminus V then for all s isin 0 1 12 V le V + suminus V le 2V and from

(513) (514) and (515) we have I le c(V pminus3 + V qminus3

)uminus V If V le 2uminus V and p ge 2 then

∣∣F primeprimeV + suminus V∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le cuminus V pminus2 + cuminus V qminus2

and

∣∣F primeprimeV ∣∣ le cV pminus2 + cV qminus2 le cuminus V pminus2 + cuminus V qminus2

Hence we have that

I le cint 1

0

∣∣F primeprimeV + suminus V∣∣+ ∣∣F primeprimeV ∣∣ds le c1+ pminusquminus V pminus2

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Regularity for Certain Nonlinear Parabolic Systems 633

for some c independent of If V le 2uminus V and 1 lt p lt 2 then∣∣F primeprime(V + suminus V)∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le c1+ pminusqV + suminus Vpminus2

for some c We note thatint 1

0V + suminus Vpminus2 ds le uminus V pminus2

int 1

0

∣∣∣∣ V uminus V minus s

∣∣∣∣pminus2

ds le cuminus V pminus2

Therefore combining all these together we have

I le cV pminus3 uminus V (516)

for some c independent of For II and III we observe that

F primeV V le cV pminus2 + cV qminus2

∣∣∣∣FprimeV V i

Vj

V 3∣∣∣∣ le cV pminus2 + cV qminus2

∣∣∣∣(F primeV V

)∣∣∣∣ le cV pminus3 + cV qminus3

∣∣∣∣(F primeV V i

Vj

V 3)∣∣∣∣ le cV pminus3 + cV qminus3

Hence following the same argument for I we conclude that

II + III le cV pminus3 uminus V (517)

for some c independent of Combining (516) and (517) we obtain that∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣ le cV pminus3 uminus V (518)

and considering (518) in (512)intS R

2

uminus V 2 dz le cminus2pminus2V 2pminus3intS R

2

uminus V 4 dz

le cminus2intS R

2

uminus V 4 dz (519)

From Houmllderrsquos inequality we have thatintS R

2

uminus V 4 dz

=intS R

2

uminus V 8n+ 4nminus8n dz

le[

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx] 2

n int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

(520)

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634 Bae and Choe

From Lemma 55 we have

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx le c2 supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2 dx

le cp + 2qminusp

R2

intSR

uminus V 2 dz

for some c independent of Now recalling that 12 le V le we obtain that

int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

le c22minuspnminus2

n

int t0

t0minus 2minuspR24

[ intB R

2

(u + V )2pminus2uminus V 4 dx] nminus2

n

dt

le c4nminuspnminus8

n

int t0

t0minus 2minuspR24

[ intB R

2

∣∣u pminus22 uminus V pminus2

2 V∣∣ 2n

nminus2 dx] nminus2

n

dt

le c4nminuspnminus8

n

intS R

2

upminus2 2u2 dz+ c2nminuspnminus8

n pminus2intSR

uminus V 2 dz (521)

for some c independent of where we have used Sobolevrsquos inequality FromLemma 55 we haveint

S R2

upminus2 2u2 dz le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

Thus combining (519) (520) and (521) we have that

intS R

2uminus V 2 dz le c

(1+ qminusp 4

n+2)[

minus2ndashintSR

uminus V 2 dz] 2

n intSR

uminus V 2 dz

and this completes the proof

Now we prove an iteration lemma which says that if mean oscillation of unear some fixed V is small then u satisfies a Campanato type integral inequality

Lemma 58 Fix 0 lt 0 lt 1 then there exist small positive numbers independentof such that if V0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists V1 isin nN satisfying(12minusradic

0

) le V1 le

(1+radic

0

) (522)

ndashintSR

uminus V12 dz le 0 ndashintSR

uminus V02 dz (523)

ndashintSR

uminus V12 dz le 2

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Regularity for Certain Nonlinear Parabolic Systems 635

Proof Since V0 satisfies the assumption of Lemma 57 we apply Lemma 57 withV = V0 and obtainint

S R2

vminus u2 dz le cintSR

uminus V02 dz

So from the usual comparison argument we haveintSR

uminus uR2 dz le cintSR

vminus vR2 dz+ cintSR

uminus v2 dz

le cn+4intS R

2

uminus u R22 dz+ c

intS R

2

uminus v2 dz

le c(n+4 +

) intSR

uminus V02 dz

where c is independent of Choosing and isin 0 12 small we prove (523)

with V1 = uR Inequality ndashintSR

uminus V12 dz le 2 follows from (523) andassumption ndash

intSR

uminus V02 dz le 2 Now we also estimate

V1 minus V0 =∣∣∣∣∣ ndashintSR

vminus V0 dz

∣∣∣∣le c

[ndashintSR

vminus u2 dz] 1

2

+ c

[ndashintSR

uminus V02 dz] 1

2

le cminus1minusn2(1+

120

)12

Therefore using the same technique of DiBenedetto (1986) Lemma 44 and if wetake very small compared to (522) holds and this completes the proof

Lemma 59 There exist small positive numbers 0 independent of such that ifV0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists a sequence of vectors Vi i = 1 2 3 sub nN satisfying

14 le Vi le 4 ndash

intSiR

uminus Vi2 dz le 2

ndashintSi+1R

uminus Vi+12 dz le 0 ndashintSiR

uminus Vi2 dz

Proof For the proof of the lemma we need to show that 14 le Vi le 4 But this

follows from that

Vi+1 minus Vi2 le c(0 + minusnminus2

)∣∣∣ ndashintSiR

uminus Vi dz∣∣∣

le ci0

i∣∣∣ ndashint

SRuminus V02 dz

∣∣∣ le ci0

i2

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ORDER REPRINTS

636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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ORDER REPRINTS

642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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t 200

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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Regularity for Certain Nonlinear Parabolic Systems 631

Proof Subtracting(Ai

Qj

Vujx

)x

from both sides of (15) we get

uit minus

(Ai

Qj

Vujx

)x= (

Aiuminus Ai

Qj

Vujx

)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(510)

From (58) and (510) we see that uminus v satisfies

ui minus vit minus(Ai

Qj

Vuj minus vjx)x

=([ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

](ujxminus V

j

))x

(511)

Simply applying uminus v to (511) we get

pminus2intS R

2uminusv2dzle

intS R

2

∣∣∣ int 1

0Ai

Qj

V+suminus VminusAi

Qj

Vds∣∣∣

times uminusV uminusvdz

and it follows thatintS R

2

uminus v2 dz

le minus2pminus2intS R

2

∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣2

uminus V 2 dz (512)

Note that AiV = F primeV V i

V and

Ai

Qj

V = F primeprimeV ViV

j

V 2 + F primeV (ij

V minus V iV

j

V 3)

where ij and are Kronecker delta functions Also we recall the structurecondition (13) F primeprimeprimeh le chpminus3 + hqminus3 We have that

∣∣∣ int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds∣∣∣

le∣∣∣∣int 1

0F primeprime(V + suminus V)

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2

minusF primeprimeV ViV

j

V 2 ds

∣∣∣∣+int 1

0

∣∣∣∣F primeV + suminus VV + suminus V minus F primeV

V ∣∣∣∣ds

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632 Bae and Choe

+∣∣∣∣int 1

0F primeV + suminus V

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V3

minusF primeV ViV

j

V 3 ds

∣∣∣∣= I + II + III

The first term I is estimates as follows

I leint 1

0

∣∣∣F primeprimeV + suminus Vminus F primeprimeV ∣∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2∣∣∣∣∣ ds

+int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

(513)

By the mean value theorem we have that

int 1

0

∣∣∣F primeprimeV+suminusVminusF primeprimeV ∣∣∣ds le c

int 1

0

int 1

0

∣∣F primeprimeprimeV+shuminusV∣∣ds dhuminusV (514)

for some c and

int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

le cF primeprimeV int 1

0

int 1

0

ds dh∣∣V + shuminus V∣∣ uminus V (515)

If V ge 2uminus V then for all s isin 0 1 12 V le V + suminus V le 2V and from

(513) (514) and (515) we have I le c(V pminus3 + V qminus3

)uminus V If V le 2uminus V and p ge 2 then

∣∣F primeprimeV + suminus V∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le cuminus V pminus2 + cuminus V qminus2

and

∣∣F primeprimeV ∣∣ le cV pminus2 + cV qminus2 le cuminus V pminus2 + cuminus V qminus2

Hence we have that

I le cint 1

0

∣∣F primeprimeV + suminus V∣∣+ ∣∣F primeprimeV ∣∣ds le c1+ pminusquminus V pminus2

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Regularity for Certain Nonlinear Parabolic Systems 633

for some c independent of If V le 2uminus V and 1 lt p lt 2 then∣∣F primeprime(V + suminus V)∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le c1+ pminusqV + suminus Vpminus2

for some c We note thatint 1

0V + suminus Vpminus2 ds le uminus V pminus2

int 1

0

∣∣∣∣ V uminus V minus s

∣∣∣∣pminus2

ds le cuminus V pminus2

Therefore combining all these together we have

I le cV pminus3 uminus V (516)

for some c independent of For II and III we observe that

F primeV V le cV pminus2 + cV qminus2

∣∣∣∣FprimeV V i

Vj

V 3∣∣∣∣ le cV pminus2 + cV qminus2

∣∣∣∣(F primeV V

)∣∣∣∣ le cV pminus3 + cV qminus3

∣∣∣∣(F primeV V i

Vj

V 3)∣∣∣∣ le cV pminus3 + cV qminus3

Hence following the same argument for I we conclude that

II + III le cV pminus3 uminus V (517)

for some c independent of Combining (516) and (517) we obtain that∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣ le cV pminus3 uminus V (518)

and considering (518) in (512)intS R

2

uminus V 2 dz le cminus2pminus2V 2pminus3intS R

2

uminus V 4 dz

le cminus2intS R

2

uminus V 4 dz (519)

From Houmllderrsquos inequality we have thatintS R

2

uminus V 4 dz

=intS R

2

uminus V 8n+ 4nminus8n dz

le[

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx] 2

n int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

(520)

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634 Bae and Choe

From Lemma 55 we have

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx le c2 supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2 dx

le cp + 2qminusp

R2

intSR

uminus V 2 dz

for some c independent of Now recalling that 12 le V le we obtain that

int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

le c22minuspnminus2

n

int t0

t0minus 2minuspR24

[ intB R

2

(u + V )2pminus2uminus V 4 dx] nminus2

n

dt

le c4nminuspnminus8

n

int t0

t0minus 2minuspR24

[ intB R

2

∣∣u pminus22 uminus V pminus2

2 V∣∣ 2n

nminus2 dx] nminus2

n

dt

le c4nminuspnminus8

n

intS R

2

upminus2 2u2 dz+ c2nminuspnminus8

n pminus2intSR

uminus V 2 dz (521)

for some c independent of where we have used Sobolevrsquos inequality FromLemma 55 we haveint

S R2

upminus2 2u2 dz le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

Thus combining (519) (520) and (521) we have that

intS R

2uminus V 2 dz le c

(1+ qminusp 4

n+2)[

minus2ndashintSR

uminus V 2 dz] 2

n intSR

uminus V 2 dz

and this completes the proof

Now we prove an iteration lemma which says that if mean oscillation of unear some fixed V is small then u satisfies a Campanato type integral inequality

Lemma 58 Fix 0 lt 0 lt 1 then there exist small positive numbers independentof such that if V0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists V1 isin nN satisfying(12minusradic

0

) le V1 le

(1+radic

0

) (522)

ndashintSR

uminus V12 dz le 0 ndashintSR

uminus V02 dz (523)

ndashintSR

uminus V12 dz le 2

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Regularity for Certain Nonlinear Parabolic Systems 635

Proof Since V0 satisfies the assumption of Lemma 57 we apply Lemma 57 withV = V0 and obtainint

S R2

vminus u2 dz le cintSR

uminus V02 dz

So from the usual comparison argument we haveintSR

uminus uR2 dz le cintSR

vminus vR2 dz+ cintSR

uminus v2 dz

le cn+4intS R

2

uminus u R22 dz+ c

intS R

2

uminus v2 dz

le c(n+4 +

) intSR

uminus V02 dz

where c is independent of Choosing and isin 0 12 small we prove (523)

with V1 = uR Inequality ndashintSR

uminus V12 dz le 2 follows from (523) andassumption ndash

intSR

uminus V02 dz le 2 Now we also estimate

V1 minus V0 =∣∣∣∣∣ ndashintSR

vminus V0 dz

∣∣∣∣le c

[ndashintSR

vminus u2 dz] 1

2

+ c

[ndashintSR

uminus V02 dz] 1

2

le cminus1minusn2(1+

120

)12

Therefore using the same technique of DiBenedetto (1986) Lemma 44 and if wetake very small compared to (522) holds and this completes the proof

Lemma 59 There exist small positive numbers 0 independent of such that ifV0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists a sequence of vectors Vi i = 1 2 3 sub nN satisfying

14 le Vi le 4 ndash

intSiR

uminus Vi2 dz le 2

ndashintSi+1R

uminus Vi+12 dz le 0 ndashintSiR

uminus Vi2 dz

Proof For the proof of the lemma we need to show that 14 le Vi le 4 But this

follows from that

Vi+1 minus Vi2 le c(0 + minusnminus2

)∣∣∣ ndashintSiR

uminus Vi dz∣∣∣

le ci0

i∣∣∣ ndashint

SRuminus V02 dz

∣∣∣ le ci0

i2

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636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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ORDER REPRINTS

642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

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Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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t 200

7

ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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632 Bae and Choe

+∣∣∣∣int 1

0F primeV + suminus V

(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V3

minusF primeV ViV

j

V 3 ds

∣∣∣∣= I + II + III

The first term I is estimates as follows

I leint 1

0

∣∣∣F primeprimeV + suminus Vminus F primeprimeV ∣∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V

j + suj

xminus V

j)

V + suminus V2∣∣∣∣∣ ds

+int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

(513)

By the mean value theorem we have that

int 1

0

∣∣∣F primeprimeV+suminusVminusF primeprimeV ∣∣∣ds le c

int 1

0

int 1

0

∣∣F primeprimeprimeV+shuminusV∣∣ds dhuminusV (514)

for some c and

int 1

0

∣∣F primeprimeV ∣∣∣∣∣∣∣(V i

+ suixminus V i

)(V j

+ sujxminus V

j)

V + suminus V2 minus V iV

j

V 2∣∣∣∣∣ ds

le cF primeprimeV int 1

0

int 1

0

ds dh∣∣V + shuminus V∣∣ uminus V (515)

If V ge 2uminus V then for all s isin 0 1 12 V le V + suminus V le 2V and from

(513) (514) and (515) we have I le c(V pminus3 + V qminus3

)uminus V If V le 2uminus V and p ge 2 then

∣∣F primeprimeV + suminus V∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le cuminus V pminus2 + cuminus V qminus2

and

∣∣F primeprimeV ∣∣ le cV pminus2 + cV qminus2 le cuminus V pminus2 + cuminus V qminus2

Hence we have that

I le cint 1

0

∣∣F primeprimeV + suminus V∣∣+ ∣∣F primeprimeV ∣∣ds le c1+ pminusquminus V pminus2

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Regularity for Certain Nonlinear Parabolic Systems 633

for some c independent of If V le 2uminus V and 1 lt p lt 2 then∣∣F primeprime(V + suminus V)∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le c1+ pminusqV + suminus Vpminus2

for some c We note thatint 1

0V + suminus Vpminus2 ds le uminus V pminus2

int 1

0

∣∣∣∣ V uminus V minus s

∣∣∣∣pminus2

ds le cuminus V pminus2

Therefore combining all these together we have

I le cV pminus3 uminus V (516)

for some c independent of For II and III we observe that

F primeV V le cV pminus2 + cV qminus2

∣∣∣∣FprimeV V i

Vj

V 3∣∣∣∣ le cV pminus2 + cV qminus2

∣∣∣∣(F primeV V

)∣∣∣∣ le cV pminus3 + cV qminus3

∣∣∣∣(F primeV V i

Vj

V 3)∣∣∣∣ le cV pminus3 + cV qminus3

Hence following the same argument for I we conclude that

II + III le cV pminus3 uminus V (517)

for some c independent of Combining (516) and (517) we obtain that∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣ le cV pminus3 uminus V (518)

and considering (518) in (512)intS R

2

uminus V 2 dz le cminus2pminus2V 2pminus3intS R

2

uminus V 4 dz

le cminus2intS R

2

uminus V 4 dz (519)

From Houmllderrsquos inequality we have thatintS R

2

uminus V 4 dz

=intS R

2

uminus V 8n+ 4nminus8n dz

le[

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx] 2

n int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

(520)

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634 Bae and Choe

From Lemma 55 we have

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx le c2 supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2 dx

le cp + 2qminusp

R2

intSR

uminus V 2 dz

for some c independent of Now recalling that 12 le V le we obtain that

int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

le c22minuspnminus2

n

int t0

t0minus 2minuspR24

[ intB R

2

(u + V )2pminus2uminus V 4 dx] nminus2

n

dt

le c4nminuspnminus8

n

int t0

t0minus 2minuspR24

[ intB R

2

∣∣u pminus22 uminus V pminus2

2 V∣∣ 2n

nminus2 dx] nminus2

n

dt

le c4nminuspnminus8

n

intS R

2

upminus2 2u2 dz+ c2nminuspnminus8

n pminus2intSR

uminus V 2 dz (521)

for some c independent of where we have used Sobolevrsquos inequality FromLemma 55 we haveint

S R2

upminus2 2u2 dz le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

Thus combining (519) (520) and (521) we have that

intS R

2uminus V 2 dz le c

(1+ qminusp 4

n+2)[

minus2ndashintSR

uminus V 2 dz] 2

n intSR

uminus V 2 dz

and this completes the proof

Now we prove an iteration lemma which says that if mean oscillation of unear some fixed V is small then u satisfies a Campanato type integral inequality

Lemma 58 Fix 0 lt 0 lt 1 then there exist small positive numbers independentof such that if V0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists V1 isin nN satisfying(12minusradic

0

) le V1 le

(1+radic

0

) (522)

ndashintSR

uminus V12 dz le 0 ndashintSR

uminus V02 dz (523)

ndashintSR

uminus V12 dz le 2

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Regularity for Certain Nonlinear Parabolic Systems 635

Proof Since V0 satisfies the assumption of Lemma 57 we apply Lemma 57 withV = V0 and obtainint

S R2

vminus u2 dz le cintSR

uminus V02 dz

So from the usual comparison argument we haveintSR

uminus uR2 dz le cintSR

vminus vR2 dz+ cintSR

uminus v2 dz

le cn+4intS R

2

uminus u R22 dz+ c

intS R

2

uminus v2 dz

le c(n+4 +

) intSR

uminus V02 dz

where c is independent of Choosing and isin 0 12 small we prove (523)

with V1 = uR Inequality ndashintSR

uminus V12 dz le 2 follows from (523) andassumption ndash

intSR

uminus V02 dz le 2 Now we also estimate

V1 minus V0 =∣∣∣∣∣ ndashintSR

vminus V0 dz

∣∣∣∣le c

[ndashintSR

vminus u2 dz] 1

2

+ c

[ndashintSR

uminus V02 dz] 1

2

le cminus1minusn2(1+

120

)12

Therefore using the same technique of DiBenedetto (1986) Lemma 44 and if wetake very small compared to (522) holds and this completes the proof

Lemma 59 There exist small positive numbers 0 independent of such that ifV0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists a sequence of vectors Vi i = 1 2 3 sub nN satisfying

14 le Vi le 4 ndash

intSiR

uminus Vi2 dz le 2

ndashintSi+1R

uminus Vi+12 dz le 0 ndashintSiR

uminus Vi2 dz

Proof For the proof of the lemma we need to show that 14 le Vi le 4 But this

follows from that

Vi+1 minus Vi2 le c(0 + minusnminus2

)∣∣∣ ndashintSiR

uminus Vi dz∣∣∣

le ci0

i∣∣∣ ndashint

SRuminus V02 dz

∣∣∣ le ci0

i2

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636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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ORDER REPRINTS

642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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Regularity for Certain Nonlinear Parabolic Systems 633

for some c independent of If V le 2uminus V and 1 lt p lt 2 then∣∣F primeprime(V + suminus V)∣∣ le cV + suminus Vpminus2 + cV + suminus Vqminus2

le c1+ pminusqV + suminus Vpminus2

for some c We note thatint 1

0V + suminus Vpminus2 ds le uminus V pminus2

int 1

0

∣∣∣∣ V uminus V minus s

∣∣∣∣pminus2

ds le cuminus V pminus2

Therefore combining all these together we have

I le cV pminus3 uminus V (516)

for some c independent of For II and III we observe that

F primeV V le cV pminus2 + cV qminus2

∣∣∣∣FprimeV V i

Vj

V 3∣∣∣∣ le cV pminus2 + cV qminus2

∣∣∣∣(F primeV V

)∣∣∣∣ le cV pminus3 + cV qminus3

∣∣∣∣(F primeV V i

Vj

V 3)∣∣∣∣ le cV pminus3 + cV qminus3

Hence following the same argument for I we conclude that

II + III le cV pminus3 uminus V (517)

for some c independent of Combining (516) and (517) we obtain that∣∣∣∣int 1

0Ai

Qj

V + suminus Vminus Ai

Qj

Vds

∣∣∣∣ le cV pminus3 uminus V (518)

and considering (518) in (512)intS R

2

uminus V 2 dz le cminus2pminus2V 2pminus3intS R

2

uminus V 4 dz

le cminus2intS R

2

uminus V 4 dz (519)

From Houmllderrsquos inequality we have thatintS R

2

uminus V 4 dz

=intS R

2

uminus V 8n+ 4nminus8n dz

le[

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx] 2

n int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

(520)

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ORDER REPRINTS

634 Bae and Choe

From Lemma 55 we have

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx le c2 supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2 dx

le cp + 2qminusp

R2

intSR

uminus V 2 dz

for some c independent of Now recalling that 12 le V le we obtain that

int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

le c22minuspnminus2

n

int t0

t0minus 2minuspR24

[ intB R

2

(u + V )2pminus2uminus V 4 dx] nminus2

n

dt

le c4nminuspnminus8

n

int t0

t0minus 2minuspR24

[ intB R

2

∣∣u pminus22 uminus V pminus2

2 V∣∣ 2n

nminus2 dx] nminus2

n

dt

le c4nminuspnminus8

n

intS R

2

upminus2 2u2 dz+ c2nminuspnminus8

n pminus2intSR

uminus V 2 dz (521)

for some c independent of where we have used Sobolevrsquos inequality FromLemma 55 we haveint

S R2

upminus2 2u2 dz le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

Thus combining (519) (520) and (521) we have that

intS R

2uminus V 2 dz le c

(1+ qminusp 4

n+2)[

minus2ndashintSR

uminus V 2 dz] 2

n intSR

uminus V 2 dz

and this completes the proof

Now we prove an iteration lemma which says that if mean oscillation of unear some fixed V is small then u satisfies a Campanato type integral inequality

Lemma 58 Fix 0 lt 0 lt 1 then there exist small positive numbers independentof such that if V0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists V1 isin nN satisfying(12minusradic

0

) le V1 le

(1+radic

0

) (522)

ndashintSR

uminus V12 dz le 0 ndashintSR

uminus V02 dz (523)

ndashintSR

uminus V12 dz le 2

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Regularity for Certain Nonlinear Parabolic Systems 635

Proof Since V0 satisfies the assumption of Lemma 57 we apply Lemma 57 withV = V0 and obtainint

S R2

vminus u2 dz le cintSR

uminus V02 dz

So from the usual comparison argument we haveintSR

uminus uR2 dz le cintSR

vminus vR2 dz+ cintSR

uminus v2 dz

le cn+4intS R

2

uminus u R22 dz+ c

intS R

2

uminus v2 dz

le c(n+4 +

) intSR

uminus V02 dz

where c is independent of Choosing and isin 0 12 small we prove (523)

with V1 = uR Inequality ndashintSR

uminus V12 dz le 2 follows from (523) andassumption ndash

intSR

uminus V02 dz le 2 Now we also estimate

V1 minus V0 =∣∣∣∣∣ ndashintSR

vminus V0 dz

∣∣∣∣le c

[ndashintSR

vminus u2 dz] 1

2

+ c

[ndashintSR

uminus V02 dz] 1

2

le cminus1minusn2(1+

120

)12

Therefore using the same technique of DiBenedetto (1986) Lemma 44 and if wetake very small compared to (522) holds and this completes the proof

Lemma 59 There exist small positive numbers 0 independent of such that ifV0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists a sequence of vectors Vi i = 1 2 3 sub nN satisfying

14 le Vi le 4 ndash

intSiR

uminus Vi2 dz le 2

ndashintSi+1R

uminus Vi+12 dz le 0 ndashintSiR

uminus Vi2 dz

Proof For the proof of the lemma we need to show that 14 le Vi le 4 But this

follows from that

Vi+1 minus Vi2 le c(0 + minusnminus2

)∣∣∣ ndashintSiR

uminus Vi dz∣∣∣

le ci0

i∣∣∣ ndashint

SRuminus V02 dz

∣∣∣ le ci0

i2

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ORDER REPRINTS

636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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ORDER REPRINTS

638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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ORDER REPRINTS

640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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ORDER REPRINTS

642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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t 02

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t 200

7

ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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ORDER REPRINTS

634 Bae and Choe

From Lemma 55 we have

supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 4 dx le c2 supt0minus 2minuspR2

4 letlet0

intB R

2

uminus V 2 dx

le cp + 2qminusp

R2

intSR

uminus V 2 dz

for some c independent of Now recalling that 12 le V le we obtain that

int t0

t0minus 2minuspR24

[ intB R

2

uminus V 4 dx] nminus2

n

dt

le c22minuspnminus2

n

int t0

t0minus 2minuspR24

[ intB R

2

(u + V )2pminus2uminus V 4 dx] nminus2

n

dt

le c4nminuspnminus8

n

int t0

t0minus 2minuspR24

[ intB R

2

∣∣u pminus22 uminus V pminus2

2 V∣∣ 2n

nminus2 dx] nminus2

n

dt

le c4nminuspnminus8

n

intS R

2

upminus2 2u2 dz+ c2nminuspnminus8

n pminus2intSR

uminus V 2 dz (521)

for some c independent of where we have used Sobolevrsquos inequality FromLemma 55 we haveint

S R2

upminus2 2u2 dz le cpminus2 + 2qminuspminus2

R2

intSR

uminus V 2 dz

Thus combining (519) (520) and (521) we have that

intS R

2uminus V 2 dz le c

(1+ qminusp 4

n+2)[

minus2ndashintSR

uminus V 2 dz] 2

n intSR

uminus V 2 dz

and this completes the proof

Now we prove an iteration lemma which says that if mean oscillation of unear some fixed V is small then u satisfies a Campanato type integral inequality

Lemma 58 Fix 0 lt 0 lt 1 then there exist small positive numbers independentof such that if V0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists V1 isin nN satisfying(12minusradic

0

) le V1 le

(1+radic

0

) (522)

ndashintSR

uminus V12 dz le 0 ndashintSR

uminus V02 dz (523)

ndashintSR

uminus V12 dz le 2

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Regularity for Certain Nonlinear Parabolic Systems 635

Proof Since V0 satisfies the assumption of Lemma 57 we apply Lemma 57 withV = V0 and obtainint

S R2

vminus u2 dz le cintSR

uminus V02 dz

So from the usual comparison argument we haveintSR

uminus uR2 dz le cintSR

vminus vR2 dz+ cintSR

uminus v2 dz

le cn+4intS R

2

uminus u R22 dz+ c

intS R

2

uminus v2 dz

le c(n+4 +

) intSR

uminus V02 dz

where c is independent of Choosing and isin 0 12 small we prove (523)

with V1 = uR Inequality ndashintSR

uminus V12 dz le 2 follows from (523) andassumption ndash

intSR

uminus V02 dz le 2 Now we also estimate

V1 minus V0 =∣∣∣∣∣ ndashintSR

vminus V0 dz

∣∣∣∣le c

[ndashintSR

vminus u2 dz] 1

2

+ c

[ndashintSR

uminus V02 dz] 1

2

le cminus1minusn2(1+

120

)12

Therefore using the same technique of DiBenedetto (1986) Lemma 44 and if wetake very small compared to (522) holds and this completes the proof

Lemma 59 There exist small positive numbers 0 independent of such that ifV0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists a sequence of vectors Vi i = 1 2 3 sub nN satisfying

14 le Vi le 4 ndash

intSiR

uminus Vi2 dz le 2

ndashintSi+1R

uminus Vi+12 dz le 0 ndashintSiR

uminus Vi2 dz

Proof For the proof of the lemma we need to show that 14 le Vi le 4 But this

follows from that

Vi+1 minus Vi2 le c(0 + minusnminus2

)∣∣∣ ndashintSiR

uminus Vi dz∣∣∣

le ci0

i∣∣∣ ndashint

SRuminus V02 dz

∣∣∣ le ci0

i2

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ORDER REPRINTS

636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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ORDER REPRINTS

640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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ORDER REPRINTS

642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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t 02

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t 200

7

ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 635

Proof Since V0 satisfies the assumption of Lemma 57 we apply Lemma 57 withV = V0 and obtainint

S R2

vminus u2 dz le cintSR

uminus V02 dz

So from the usual comparison argument we haveintSR

uminus uR2 dz le cintSR

vminus vR2 dz+ cintSR

uminus v2 dz

le cn+4intS R

2

uminus u R22 dz+ c

intS R

2

uminus v2 dz

le c(n+4 +

) intSR

uminus V02 dz

where c is independent of Choosing and isin 0 12 small we prove (523)

with V1 = uR Inequality ndashintSR

uminus V12 dz le 2 follows from (523) andassumption ndash

intSR

uminus V02 dz le 2 Now we also estimate

V1 minus V0 =∣∣∣∣∣ ndashintSR

vminus V0 dz

∣∣∣∣le c

[ndashintSR

vminus u2 dz] 1

2

+ c

[ndashintSR

uminus V02 dz] 1

2

le cminus1minusn2(1+

120

)12

Therefore using the same technique of DiBenedetto (1986) Lemma 44 and if wetake very small compared to (522) holds and this completes the proof

Lemma 59 There exist small positive numbers 0 independent of such that ifV0 is any vector in nN satisfying 1

2 le V0 le and

ndashintSR

uminus V02 dz le 2

then there exists a sequence of vectors Vi i = 1 2 3 sub nN satisfying

14 le Vi le 4 ndash

intSiR

uminus Vi2 dz le 2

ndashintSi+1R

uminus Vi+12 dz le 0 ndashintSiR

uminus Vi2 dz

Proof For the proof of the lemma we need to show that 14 le Vi le 4 But this

follows from that

Vi+1 minus Vi2 le c(0 + minusnminus2

)∣∣∣ ndashintSiR

uminus Vi dz∣∣∣

le ci0

i∣∣∣ ndashint

SRuminus V02 dz

∣∣∣ le ci0

i2

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ORDER REPRINTS

636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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ORDER REPRINTS

640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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ORDER REPRINTS

642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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636 Bae and Choe

Thus Vi is a Cauchy sequence and the limit lies in the appropriate range as inLemma 58 Therefore we can apply Lemma 58 continuously and the proof iscomplete (Also see the proof of Lemma 45 in DiBenedetto 1986)

We estimate mean oscillation of u under the assumption (57) Indeed weuse a Caccioppoli type inequality for u The following lemma is of fundamentalin showing that mean oscillation is small that is u satisfies the assumptions ofLemma 58 We take in the assumption (57) We define R and minus

R by

Rdef= BR times

(t0 minus

122minuspR2 t0

)

minusR

def= BR times(t0 minus

122minuspR2 t0 minus

142minuspR2

)

Let isin 0 BR be a cutoff function such that u = 1 in BR

2

Lemma 510 Suppose that S2R sub T then u satisfies

suptisinR

intminus

R ds

intBR

uminus uRs22 dx le c1

n+1 4minuspRn+2 (524)

where

uRs

def= 1BRx0

intBRx0

ux sdx

Proof Let ts isin R be the time when

intBR

∣∣ux t minus uRs

∣∣22 dx = suptisinR

intBR

∣∣ux tminus uRs

∣∣22 dxNow we take

(uixminus ui

xRs

)2st as test function in (31) Hence we obtain

12

int (uix

)t

(uixminus ui

xRs

)2st dz+

intAi

Qj

ujx

uix

2st dz

+ 2int

Ai

Qj

ujx

(uixminus ui

xRs

)x

st dz = 0

and summing over i j and

int ∣∣uminus uRs

∣∣22x t dx+int

upminus2 2u22xt dz

le cint (upminus2 + uqminus2

)2u uminus uRsst dz

+int

uminus uRs22x sdx

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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ORDER REPRINTS

640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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ORDER REPRINTS

642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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ORDER REPRINTS

644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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Request PermissionOrder Reprints

Reprints of this article can also be ordered at

httpwwwdekkercomservletproductDOI101081PDE120037327

Request Permission or Order Reprints Instantly

Interested in copying and sharing this article In most cases US Copyright Law requires that you get permission from the articlersquos rightsholder before using copyrighted content

All information and materials found in this article including but not limited to text trademarks patents logos graphics and images (the Materials) are the copyrighted works and other forms of intellectual property of Marcel Dekker Inc or its licensors All rights not expressly granted are reserved

Get permission to lawfully reproduce and distribute the Materials or order reprints quickly and painlessly Simply click on the Request Permission Order Reprints link below and follow the instructions Visit the US Copyright Office for information on Fair Use limitations of US copyright law Please refer to The Association of American Publishersrsquo (AAP) website for guidelines on Fair Use in the Classroom

The Materials are for your personal use only and cannot be reformatted reposted resold or distributed by electronic means or otherwise without permission from Marcel Dekker Inc Marcel Dekker Inc grants you the limited right to display the Materials only on your personal computer or personal wireless device and to copy and download single copies of such Materials provided that any copyright trademark or other notice appearing on such Materials is also retained by displayed copied or downloaded as part of the Materials and is not removed or obscured and provided you do not edit modify alter or enhance the Materials Please refer to our Website User Agreement for more details

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 637

for some c independent of Integrating with respect to s on minusR we haveint

minusR

ds suptisinR

intBR

ux tminus uRs22 dx

le c2minuspR2intSR

upminus22u uminus uRs st dz

+intSR

uminus uRs22 dx ds = I + II (525)

First we assume that p ge 2 We estimate II from Houmllder inequality such that

II =intSR

uminus uRs22 dx ds

le SRpminus2p

[ intSR

uminus uRsp2 dx ds

] 2p

le c(2minuspRn+2

) pminus2p

[ intSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds] 2

p

(526)

from Houmllder inequality Using GagliardondashNirenberg inequality (see Theorem 22 ofChapter II in Ladyzhenskaya et al 1968) and Houmllder inequality we have

intSR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx ds

leintR

intBR

(upminus22u2

) nn+1

dxRn

n+1

[intBR

∣∣∣u pminus22 uminus(u pminus2

2 u)Rs

∣∣∣22dx]1

n+1

dt

le cp

n+1R2nn+1

intSR

(upminus22u2

) nn+1

dz (527)

We define SR def= x t isin SR u2 ge 1minus 2 Now from the assumption

(57) we get SRSR le cSR for some c and we obtainintSRSR

(upminus22u2) nn+1 dz

le ∣∣SR SR ∣∣ 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1 2Rn

nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (528)

On the other hand from Houmllder inequality we get

intSR

(upminus22u2

) nn+1

dz le (2minuspRn+2

) 1n+1

( intSR

upminus22u2 dz) n

n+1

(529)

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638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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ORDER REPRINTS

640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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ORDER REPRINTS

642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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ORDER REPRINTS

644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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ity] A

t 02

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t 200

7

ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

Dow

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7

Request PermissionOrder Reprints

Reprints of this article can also be ordered at

httpwwwdekkercomservletproductDOI101081PDE120037327

Request Permission or Order Reprints Instantly

Interested in copying and sharing this article In most cases US Copyright Law requires that you get permission from the articlersquos rightsholder before using copyrighted content

All information and materials found in this article including but not limited to text trademarks patents logos graphics and images (the Materials) are the copyrighted works and other forms of intellectual property of Marcel Dekker Inc or its licensors All rights not expressly granted are reserved

Get permission to lawfully reproduce and distribute the Materials or order reprints quickly and painlessly Simply click on the Request Permission Order Reprints link below and follow the instructions Visit the US Copyright Office for information on Fair Use limitations of US copyright law Please refer to The Association of American Publishersrsquo (AAP) website for guidelines on Fair Use in the Classroom

The Materials are for your personal use only and cannot be reformatted reposted resold or distributed by electronic means or otherwise without permission from Marcel Dekker Inc Marcel Dekker Inc grants you the limited right to display the Materials only on your personal computer or personal wireless device and to copy and download single copies of such Materials provided that any copyright trademark or other notice appearing on such Materials is also retained by displayed copied or downloaded as part of the Materials and is not removed or obscured and provided you do not edit modify alter or enhance the Materials Please refer to our Website User Agreement for more details

Dow

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ORDER REPRINTS

638 Bae and Choe

To estimate the last term in (529) we apply uix

(u2 minus 1minus 22)+

132 as a testfunction to (31) where 13 is a standard cutoff function with support containingSR From a usual calculation we haveint

S1+Rupminus22u2(u2minus1minus22

)+132dz+

intS1+R

upminus2∣∣(u2)∣∣2132dz

le cintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

+cintS1+R

[(u2minus1minus22)+]213t13dz (530)

for some c independent of From Youngrsquos inequality we also haveintS1+R

uqminus2∣∣(u2)∣∣(u2minus1minus22

)+1313dz

le 15

intS1+R

upminus2∣∣(u2)∣∣2132dz+c

pminus2

R2

intS1+R

[(u2minus1minus22)+]2

dz

(531)

for some c independent of Since for all z isin S1+R 2 le u2minus 1minus 22 le 22 we have from (530) and (531)int

SRupminus22u2132 dz le c2Rn (532)

Combining (528) (529) and (532) we getintSR

(upminus22u2) nn+1 dz le c

1n+1

2n+2minuspn+1 R

n2+n+2n+1 + c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (533)

for some c independent of Substituting (533) in (527) we see that

intSR

∣∣∣u pminus22 u minus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le cp

n+1R2nn+1

1n+1

2n+2minuspn+1 R

n2+n+2n+1 = c

1n+1 2Rn+2 (534)

for some c independent of Therefore employing (534) in (526)we estimate

II =intSR

uminus uRs22 dx ds le c2minusppminus2

p Rn+2pminus2

p[

1n+1 2Rn+2

] 2p

= c2

pn+1 4minuspRn+2 (535)

Finally we estimate I Clearly we have

I le c2minus p2 R

intSR

u pminus22 2udz

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Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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t 200

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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Regularity for Certain Nonlinear Parabolic Systems 639

Thus following the same argument for the case II we haveintSR

u pminus22 2udz le

intSRSR

(upminus22u2) 12 dz

+intSR

(upminus22u2) 12 dz le c

12 4minuspRn+2 (536)

Therefore combining (535) and (536) we conclude the proof when p ge 2Now we assume that p isin 0 1 We estimate II in (525) Applying Gagliardondash

Nirenberg inequality and Houmllderrsquos inequality we obtain

II le c2minuspintSR

uminus uRsp2 dx ds

le c2minuspintSR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣22 dx ds

le c2minuspintR

intBR

(upminus22u2

) nn+1

dx

timesRn

n+1

[ intBR

∣∣∣u pminus22 uminus (u pminus2

2 u)Rs

∣∣∣2 dx]1

n+1

dt

le c2minusp+ pn+1R

2nn+1

intSR

(upminus22u2

) nn+1

dz (537)

for some c independent of Following the same calculation for the case p ge 2we estimateint

SRSR

(upminus22u2) nn+1 dz le SRSR 1

n+1

( intSR

upminus22u2 dz) n

n+1

le c(2minuspRn+2

) 1n+1

(2Rn

) nn+1

le c1

n+1 2n+2minusp

n+1 Rn2+n+2

n+1 (538)

for some c independent of andintSR

(upminus22u2) nn+1 dz le c

nn+1

2n+2minuspn+1 R

n2+n+2n+1 (539)

for some c independent of Thus combining (537) (538) and (539) we get

II le c1

n+1 4minuspRn+2 (540)

for some c By the same argument in the case p ge 2 we also estimate I in (525) asfollows

I le c12 4minuspRn+2 (541)

for some c Therefore combining (525) (540) and (541) we conclude theproof

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ORDER REPRINTS

640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

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Request PermissionOrder Reprints

Reprints of this article can also be ordered at

httpwwwdekkercomservletproductDOI101081PDE120037327

Request Permission or Order Reprints Instantly

Interested in copying and sharing this article In most cases US Copyright Law requires that you get permission from the articlersquos rightsholder before using copyrighted content

All information and materials found in this article including but not limited to text trademarks patents logos graphics and images (the Materials) are the copyrighted works and other forms of intellectual property of Marcel Dekker Inc or its licensors All rights not expressly granted are reserved

Get permission to lawfully reproduce and distribute the Materials or order reprints quickly and painlessly Simply click on the Request Permission Order Reprints link below and follow the instructions Visit the US Copyright Office for information on Fair Use limitations of US copyright law Please refer to The Association of American Publishersrsquo (AAP) website for guidelines on Fair Use in the Classroom

The Materials are for your personal use only and cannot be reformatted reposted resold or distributed by electronic means or otherwise without permission from Marcel Dekker Inc Marcel Dekker Inc grants you the limited right to display the Materials only on your personal computer or personal wireless device and to copy and download single copies of such Materials provided that any copyright trademark or other notice appearing on such Materials is also retained by displayed copied or downloaded as part of the Materials and is not removed or obscured and provided you do not edit modify alter or enhance the Materials Please refer to our Website User Agreement for more details

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640 Bae and Choe

Using the estimate (524) we can show the smallness of mean oscillation of uIndeed a similar argument has been used in proving Poincareacute inequality for u inSec 5

Lemma 511 For each given there exists a small such that if

∣∣x isin S1+R uz2 le 1minus 2∣∣ le S1+R

then we have

ndashintS R

2

∣∣vminus u R2

∣∣2 dz le 2∣∣u R

2

∣∣ ge 12

Consequently we have for each r le R2

ndashintSr

vminus ur 2 dz le c( r

R

)

ndashintSR

vminus uR2 dz

for some gt 0 and c

Proof We observe thatintS R

2

∣∣vminus u R2

∣∣2 dz le intSR

vminus uRs22 dz

le cpminus2

R2

intminus

R

dsintSR

vminus uRs22 dz

le c suptisinR

intminus

R

dsintSR

vminus uRs22x tdx

le c1

n+1 4minuspRn+2

and dividing SR2 = c2minuspRn+2 we prove that ndash

intS R

2

vminus u R22 dz le 2 for

sufficiently small Finally we have that

∣∣u R2

∣∣ ge[ndashintS R

2

u2dz] 1

2

minus[ndashintS R

2

∣∣vminus u R2

∣∣2 dz] 1

2

ge[ndashintS R

2

u2dz] 1

2

minus 12

Hence for small we obtain(see also Lemma 51 in DiBenedetto 1986)

∣∣u R2

∣∣ ge 12

and considering Lemma 59 we complete the proof

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Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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ORDER REPRINTS

644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

Dow

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ded

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[200

7 A

jou

Uni

vers

ity] A

t 02

49

31 A

ugus

t 200

7

ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

Dow

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[200

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31 A

ugus

t 200

7

Request PermissionOrder Reprints

Reprints of this article can also be ordered at

httpwwwdekkercomservletproductDOI101081PDE120037327

Request Permission or Order Reprints Instantly

Interested in copying and sharing this article In most cases US Copyright Law requires that you get permission from the articlersquos rightsholder before using copyrighted content

All information and materials found in this article including but not limited to text trademarks patents logos graphics and images (the Materials) are the copyrighted works and other forms of intellectual property of Marcel Dekker Inc or its licensors All rights not expressly granted are reserved

Get permission to lawfully reproduce and distribute the Materials or order reprints quickly and painlessly Simply click on the Request Permission Order Reprints link below and follow the instructions Visit the US Copyright Office for information on Fair Use limitations of US copyright law Please refer to The Association of American Publishersrsquo (AAP) website for guidelines on Fair Use in the Classroom

The Materials are for your personal use only and cannot be reformatted reposted resold or distributed by electronic means or otherwise without permission from Marcel Dekker Inc Marcel Dekker Inc grants you the limited right to display the Materials only on your personal computer or personal wireless device and to copy and download single copies of such Materials provided that any copyright trademark or other notice appearing on such Materials is also retained by displayed copied or downloaded as part of the Materials and is not removed or obscured and provided you do not edit modify alter or enhance the Materials Please refer to our Website User Agreement for more details

Dow

nloa

ded

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[200

7 A

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Uni

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ugus

t 200

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 641

Theorem 512 Suppose SR sub T and r le R2 Then there is a positive constant gt 0

such that

ndashintSr

uminus uSr 2 dz le c1

( r

R

)

ndashintSR

uminus uSR 2 dz+ c2

( r

R

)

(542)

where c2 depends only on p q and n

Proof Let R0 isin 0 1 be fixed Define Irdef= ndash

intSr

uminus uSr 2 dz and rdef=

supSru

First we prove (542) when p ge 2 We define Ri by Ri+1 = Ri1+ for small

and i = 0 1 2 and hence we see that Ri = 1 1+iminus1R

1+i

0 Note that if

Ri ge Rp

i then

SRi+1sub SRi+1

Ri sub SRi

Consequently if assumption (57) is true for

R = Ri and = Ri ge Rp

i

then we have from Lemma 511

IRi+1 le(Ri+1

Ri

)

IRi+ Ri

for some If Ri le Rp

i then we also have

IRi+1 le(Ri+1

Ri

)

IRi+ Ri (543)

for some Now if assumption (57) is false for

R = Ri and = Ri ge Rp

i

then from Lemma 54 we have 2Ri+1 le 2Ri for some isin 0 1 On the other

hand if Ri le Rp

i we have

2Ri+1 le 2Ri+ Ri (544)

Let Rk0be the switching radius ie assumption (57) holds for SRk0

Rk0minus1

for the first time Define i0def= s0 + t0 where s0 and t0 are decided later If k0 le s0

then from (543) we see that

IRi0 le

(Ri0

Ri0minus1

)

IRi0minus1+ Ri0minus1

le(

Ri0

Ri0minus1

)(Ri0minus1

Ri0minus2

)

IRi0minus2+ Ri0minus1 +

(Ri0

Ri0minus1

)

Ri0minus2

le(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

le(Ri0

Rk0

) SR0

SRk0 IR0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

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ORDER REPRINTS

642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

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ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

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ORDER REPRINTS

644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

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7 A

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ity] A

t 02

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ugus

t 200

7

ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

Dow

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ded

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vers

ity] A

t 02

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31 A

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t 200

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Request PermissionOrder Reprints

Reprints of this article can also be ordered at

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Get permission to lawfully reproduce and distribute the Materials or order reprints quickly and painlessly Simply click on the Request Permission Order Reprints link below and follow the instructions Visit the US Copyright Office for information on Fair Use limitations of US copyright law Please refer to The Association of American Publishersrsquo (AAP) website for guidelines on Fair Use in the Classroom

The Materials are for your personal use only and cannot be reformatted reposted resold or distributed by electronic means or otherwise without permission from Marcel Dekker Inc Marcel Dekker Inc grants you the limited right to display the Materials only on your personal computer or personal wireless device and to copy and download single copies of such Materials provided that any copyright trademark or other notice appearing on such Materials is also retained by displayed copied or downloaded as part of the Materials and is not removed or obscured and provided you do not edit modify alter or enhance the Materials Please refer to our Website User Agreement for more details

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642 Bae and Choe

So if is sufficiently small and t0 gtgt s0 then

(Ri0

Rk0

) SR0

SRk0 le

1+i0minus1

1+k0minus1

Rn+2+1+i0

0

Rn+2+1+k0

0

le 12

Since Rk+1 = Rk1+

Rk

Rk+1

= 1R

k

le 1R

i0

for all k le i0 we have

Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]le R

i0i0 minus k0

1

Ri0

= i0 minus k0 1+i0minus1+R

1minusi0

le 12

for sufficiently large i0 Hence if k0 le s0 we have

IRi0 le 1

2IR0+

12 (545)

Now suppose k0 gt s0 then from (544) we see that

IRi0 le 2Rs0

le s0minus12R1+ s0minus2R1 + s0minus3R

2 + middot middot middot + Rs0minus1

le s0minus12R1+14 (546)

if s0 is sufficiently large and is small enough On the other hand by weak Harnackinequality (39) in Theorem 32 we have

2R1 = 2R1+0 le c ndash

intSR0

u2 dz+ c (547)

with c depending only on n p and Thus using (547) in (546) we have

2Ri0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ cs0minus1uR0

2 + cs0minus1 + 14

If uR0 le 2Rs0

then

2Rs0 le cs0minus1 ndash

intSR0

uminus uR02 dz+ 4cs0minus12Rs0

+ cs0minus1 + 14

1minus 4cs0minus12Rs0 le cs0minus1IR0+ cs0minus1 + 1

4

Dow

nloa

ded

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[200

7 A

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Uni

vers

ity] A

t 02

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t 200

7

ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

Dow

nloa

ded

By

[200

7 A

jou

Uni

vers

ity] A

t 02

49

31 A

ugus

t 200

7

ORDER REPRINTS

644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

Dow

nloa

ded

By

[200

7 A

jou

Uni

vers

ity] A

t 02

49

31 A

ugus

t 200

7

ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

Dow

nloa

ded

By

[200

7 A

jou

Uni

vers

ity] A

t 02

49

31 A

ugus

t 200

7

Request PermissionOrder Reprints

Reprints of this article can also be ordered at

httpwwwdekkercomservletproductDOI101081PDE120037327

Request Permission or Order Reprints Instantly

Interested in copying and sharing this article In most cases US Copyright Law requires that you get permission from the articlersquos rightsholder before using copyrighted content

All information and materials found in this article including but not limited to text trademarks patents logos graphics and images (the Materials) are the copyrighted works and other forms of intellectual property of Marcel Dekker Inc or its licensors All rights not expressly granted are reserved

Get permission to lawfully reproduce and distribute the Materials or order reprints quickly and painlessly Simply click on the Request Permission Order Reprints link below and follow the instructions Visit the US Copyright Office for information on Fair Use limitations of US copyright law Please refer to The Association of American Publishersrsquo (AAP) website for guidelines on Fair Use in the Classroom

The Materials are for your personal use only and cannot be reformatted reposted resold or distributed by electronic means or otherwise without permission from Marcel Dekker Inc Marcel Dekker Inc grants you the limited right to display the Materials only on your personal computer or personal wireless device and to copy and download single copies of such Materials provided that any copyright trademark or other notice appearing on such Materials is also retained by displayed copied or downloaded as part of the Materials and is not removed or obscured and provided you do not edit modify alter or enhance the Materials Please refer to our Website User Agreement for more details

Dow

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ded

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[200

7 A

jou

Uni

vers

ity] A

t 02

49

31 A

ugus

t 200

7

ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 643

Therefore if s0 is sufficiently large we have

IRi0 le 2Rs0

le 12IR0+

12 (548)

Finally we assume that

uR0 gt 2Rs0

Then we have

IRi0 le

(Ri0

Rk0

)

IRk0+ R

i0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)] (549)

IRk0 le k0minuss02Rk0

+ (s0minus2R

s0+1 + s0minus3Rs0+2 + middot middot middot + R

s0

) (550)

Since we are assuming uR0 ge 2Rs0

we have

uzminus uR0 ge uR0

minus uz ge Rs0

for all z isin SRs0and

2Rs0 le ndash

intSRs0

uminus uR02 dz le SR0

SRs0

ndashintSR0

uminus uR02 dz (551)

Combining (549) (550) and (551) we have

IRi0 le

(Ri0

Rk0

)

k0minuss0SR0

SRs0

IR0+(Ri0

Rk0

)(s0minus2R

s0+1 + middot middot middot + Rs0

)

+Ri0

[(Rk0

Rk0+1

)

+ middot middot middot +(Ri0minus1

Ri0

)]

Hence if t0 13 s0 then we have

IRi0 le 1

2IR0+

12

(552)

Therefore combining (545) (548) and (552) and iterating them we conclude

Ir le( r

R

)

IR+( r

R

)

for some small This prove (542) when p ge 2Now we assume p isin 1 2 We define Ri by Ri = Ri

1+ for small Therefore if Ri ge R

2i then we note that

SRi+1sub SRi

2Ri sub SRi

The remaining proof follows exactly the same way as the case p ge 2

Dow

nloa

ded

By

[200

7 A

jou

Uni

vers

ity] A

t 02

49

31 A

ugus

t 200

7

ORDER REPRINTS

644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

Dow

nloa

ded

By

[200

7 A

jou

Uni

vers

ity] A

t 02

49

31 A

ugus

t 200

7

ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

Dow

nloa

ded

By

[200

7 A

jou

Uni

vers

ity] A

t 02

49

31 A

ugus

t 200

7

Request PermissionOrder Reprints

Reprints of this article can also be ordered at

httpwwwdekkercomservletproductDOI101081PDE120037327

Request Permission or Order Reprints Instantly

Interested in copying and sharing this article In most cases US Copyright Law requires that you get permission from the articlersquos rightsholder before using copyrighted content

All information and materials found in this article including but not limited to text trademarks patents logos graphics and images (the Materials) are the copyrighted works and other forms of intellectual property of Marcel Dekker Inc or its licensors All rights not expressly granted are reserved

Get permission to lawfully reproduce and distribute the Materials or order reprints quickly and painlessly Simply click on the Request Permission Order Reprints link below and follow the instructions Visit the US Copyright Office for information on Fair Use limitations of US copyright law Please refer to The Association of American Publishersrsquo (AAP) website for guidelines on Fair Use in the Classroom

The Materials are for your personal use only and cannot be reformatted reposted resold or distributed by electronic means or otherwise without permission from Marcel Dekker Inc Marcel Dekker Inc grants you the limited right to display the Materials only on your personal computer or personal wireless device and to copy and download single copies of such Materials provided that any copyright trademark or other notice appearing on such Materials is also retained by displayed copied or downloaded as part of the Materials and is not removed or obscured and provided you do not edit modify alter or enhance the Materials Please refer to our Website User Agreement for more details

Dow

nloa

ded

By

[200

7 A

jou

Uni

vers

ity] A

t 02

49

31 A

ugus

t 200

7

ORDER REPRINTS

644 Bae and Choe

ACKNOWLEDGMENTS

The authors thank the referees The first author was supported by KoreaResearch Foundation Grant (KRF-2001-015-DP0020)

REFERENCES

Campanato S (1966) Equazioni paraboliche del secondo ordine e spazi 2 Ann Mat Pura Appl 73(4)55ndash102

Chen Y Z DiBenedetto E (1988) On the local behavior of solutions of singularparabolic equations Arch Rational Mech Anal 103319ndash345

Chen YZ DiBenedetto E (1989) Boundary estimates for solutions of nonlineardegenerate parabolic systems J Reine Angew Math 395(4)102ndash131

Choe H J (1991) Houmllder regularity for the gradient of solutions of certain singularparabolic equations Comm Part Diff Equations 16(11)1709ndash1732

Choe H J (1992a) Interior behavior of minimizers for certain functional withnonstandard growth Nonlinear Anal TMA 19(10)933ndash945

Choe H J (1992b) Houmllder continuity for solutions of certain degenerate parabolicsystems Nonlinear Anal TMA 18(3)235ndash243

Choe H J Lewis J (1991) On the obstacle problem for quasilinear ellipticequations of p-Laplacian type SIAM J Math Anal 22(3)623ndash638

Da Prato G (1965) Spazi Lp e lora proprietagrave Ann Mat Pura Appl69383ndash392

DiBenedetto E (1986) On the local behavior of solutions of degenerate parabolicequations with measurable coefficients Annali Sc Norm Sup Pisa 13(4)487ndash535

DiBenedetto E Friedman A (1984) Regularity of solutions of nonlineardegenerate parabolic system J Reine Angew Math 34983ndash128

DiBenedetto E Friedman A (1985) Houmllder estimates for nonlinear degenerateparabolic system J Reine Angew Math 3571ndash22

Fusco N Sbordone C (1990) Higher integrability of the gradient of minimizersof functionals with nonstandard growth conditions Comm Pure Appl Math43(5)673ndash683

Ladyzhenskaya O A Solonnikov V A Uralrsquotzeva N N (1968) Linear andQuasilinear Equations of Parabolic Type Trans Math Mono 23 ProvidenceRI Amer Math Soc

Lewis J (1983) Regularity of derivatives of solutions to certain degenerate ellipticequations Indiana Univ Math J 32849ndash858

Lieberman G M (1991) The natural generalization of the natural conditionsof Ladyzhenskaya and Uralrsquotzeva for elliptic equations Comm PDE 16311ndash361

Liberman G M (1993) Boundary and initial regularity for solutions of degenerateparabolic equations Nonlinear Anal TMA 20(5)551ndash569

Marcellini P (1989) Regularity for minimizers of integrals of the calculusof variations with nonstandard growth conditions Arch Rat Mech Anal105267ndash284

Dow

nloa

ded

By

[200

7 A

jou

Uni

vers

ity] A

t 02

49

31 A

ugus

t 200

7

ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

Dow

nloa

ded

By

[200

7 A

jou

Uni

vers

ity] A

t 02

49

31 A

ugus

t 200

7

Request PermissionOrder Reprints

Reprints of this article can also be ordered at

httpwwwdekkercomservletproductDOI101081PDE120037327

Request Permission or Order Reprints Instantly

Interested in copying and sharing this article In most cases US Copyright Law requires that you get permission from the articlersquos rightsholder before using copyrighted content

All information and materials found in this article including but not limited to text trademarks patents logos graphics and images (the Materials) are the copyrighted works and other forms of intellectual property of Marcel Dekker Inc or its licensors All rights not expressly granted are reserved

Get permission to lawfully reproduce and distribute the Materials or order reprints quickly and painlessly Simply click on the Request Permission Order Reprints link below and follow the instructions Visit the US Copyright Office for information on Fair Use limitations of US copyright law Please refer to The Association of American Publishersrsquo (AAP) website for guidelines on Fair Use in the Classroom

The Materials are for your personal use only and cannot be reformatted reposted resold or distributed by electronic means or otherwise without permission from Marcel Dekker Inc Marcel Dekker Inc grants you the limited right to display the Materials only on your personal computer or personal wireless device and to copy and download single copies of such Materials provided that any copyright trademark or other notice appearing on such Materials is also retained by displayed copied or downloaded as part of the Materials and is not removed or obscured and provided you do not edit modify alter or enhance the Materials Please refer to our Website User Agreement for more details

Dow

nloa

ded

By

[200

7 A

jou

Uni

vers

ity] A

t 02

49

31 A

ugus

t 200

7

ORDER REPRINTS

Regularity for Certain Nonlinear Parabolic Systems 645

Tolksdorff P (1983) Everywhere regularity for some quasi-linear systems with lackof ellipticity Ann Mat Pura Appl 134(4)241ndash266

Uhlenbeck K (1977) Regularity for a class of nonlinear elliptic systems Acta Math138219ndash240

Uraltseva N N (1968) Degenerate quasilinear elliptic systems Zap Naucn SemLeningrad Otdel Mat Inst Steklov (LOMI) 7184ndash222

Wiegner M (1986) On C-regularity of the gradient of solutions of degenerateparabolic systems Ann Mat Pura Appl 145(4)385ndash405

Received August 2002Revised December 2003

Dow

nloa

ded

By

[200

7 A

jou

Uni

vers

ity] A

t 02

49

31 A

ugus

t 200

7

Request PermissionOrder Reprints

Reprints of this article can also be ordered at

httpwwwdekkercomservletproductDOI101081PDE120037327

Request Permission or Order Reprints Instantly

Interested in copying and sharing this article In most cases US Copyright Law requires that you get permission from the articlersquos rightsholder before using copyrighted content

All information and materials found in this article including but not limited to text trademarks patents logos graphics and images (the Materials) are the copyrighted works and other forms of intellectual property of Marcel Dekker Inc or its licensors All rights not expressly granted are reserved

Get permission to lawfully reproduce and distribute the Materials or order reprints quickly and painlessly Simply click on the Request Permission Order Reprints link below and follow the instructions Visit the US Copyright Office for information on Fair Use limitations of US copyright law Please refer to The Association of American Publishersrsquo (AAP) website for guidelines on Fair Use in the Classroom

The Materials are for your personal use only and cannot be reformatted reposted resold or distributed by electronic means or otherwise without permission from Marcel Dekker Inc Marcel Dekker Inc grants you the limited right to display the Materials only on your personal computer or personal wireless device and to copy and download single copies of such Materials provided that any copyright trademark or other notice appearing on such Materials is also retained by displayed copied or downloaded as part of the Materials and is not removed or obscured and provided you do not edit modify alter or enhance the Materials Please refer to our Website User Agreement for more details

Dow

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7 A

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vers

ity] A

t 02

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ugus

t 200

7

Request PermissionOrder Reprints

Reprints of this article can also be ordered at

httpwwwdekkercomservletproductDOI101081PDE120037327

Request Permission or Order Reprints Instantly

Interested in copying and sharing this article In most cases US Copyright Law requires that you get permission from the articlersquos rightsholder before using copyrighted content

All information and materials found in this article including but not limited to text trademarks patents logos graphics and images (the Materials) are the copyrighted works and other forms of intellectual property of Marcel Dekker Inc or its licensors All rights not expressly granted are reserved

Get permission to lawfully reproduce and distribute the Materials or order reprints quickly and painlessly Simply click on the Request Permission Order Reprints link below and follow the instructions Visit the US Copyright Office for information on Fair Use limitations of US copyright law Please refer to The Association of American Publishersrsquo (AAP) website for guidelines on Fair Use in the Classroom

The Materials are for your personal use only and cannot be reformatted reposted resold or distributed by electronic means or otherwise without permission from Marcel Dekker Inc Marcel Dekker Inc grants you the limited right to display the Materials only on your personal computer or personal wireless device and to copy and download single copies of such Materials provided that any copyright trademark or other notice appearing on such Materials is also retained by displayed copied or downloaded as part of the Materials and is not removed or obscured and provided you do not edit modify alter or enhance the Materials Please refer to our Website User Agreement for more details