Phuong Trinh Vi Phan_Dao Ham Rieng

7/31/2019 Phuong Trinh Vi Phan_Dao Ham Rieng

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Seminar

Phng trnhvi phno hm ring

H Ni, 2004


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i Phng trnh o hm ring

Mc lc

Li ni u iv

Ti liu tham kho v

Bi 1 Phng php b in p hn v b i ton Dirichlet i vi phng trnh elliptic

cp hai 1

1.1 Cch t bi ton bin phn . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Bin phn v gradient ca mt phim hm . . . . . . . . . . . . . . . . 2

1.3 iu kin cn ca cc tr . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.4 Ton t xc nh dng . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.4.1 Khng gian nng lng ca ton t xc nh dng . . . . . . . 5

1.5 Cc tiu phim hm nng lng . . . . . . . . . . . . . . . . . . . . . . 5

1.6 Nghim suy rng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.7 Bi ton Dirichlet i vi phng trnh elliptic cp 2 t lin hp . . . . 8

1.7.1 Biu thc elliptic v biu thc elliptic cp hai t lin hp . . . . 8

1.7.2 Bi ton Dirichlet i vi phng trnh elliptic cp 2 t lin hp 9

Bi 2 nh l Lax-Milgr am v bi ton bin elliptic 17

2.1 nh l Lax-Milgram . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.2 Bi ton bin Dirichlet i vi phng trnh elliptic tuyn tnh cp hai . 21

2.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.2.2 iu kin ca tnh coercitive. . . . . . . . . . . . . . . . . 23

2.2.3 Bi ton bin Dirichlet trong min b chn . . . . . . . . . . . . 25

Bi 3 Khng gian Sobolev Wm,p() 283.1 Mt s k hiu v khi nim . . . . . . . . . . . . . . . . . . . . . . . . 28

3.2 i ngu ca khng gian Sobolev Wm,p() . . . . . . . . . . . . . . . . 30

3.3 Xp x khng gian Sobolev Wm,p() bng cc hm trn trn . . . . . 36

3.3.1 Xp x bi cc hm Cm() . . . . . . . . . . . . . . . . . . . . . 36

3.3.2 Xp x bi cc hm Cm() . . . . . . . . . . . . . . . . . . . . . 38

3.3.3 Xp x bi cc hm C0 () . . . . . . . . . . . . . . . . . . . . 41

Bi 4 Bt ng thc Gar ding v bi ton Dirichlet i vi phng trnh elliptic

i


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ii Mc lc

cp cao 48

4.1 Mt s k hiu v kt qu cn thit . . . . . . . . . . . . . . . . . . . . 48

4.2 Bt ng thc Garding . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

4.2.1 Bi ton Dirichlet . . . . . . . . . . . . . . . . . . . . . . . . . . 55

Bi 5 L thuyt Fredholm-Riesz-Schauder v bi ton Dirichlet thu n nht vi

phng tr nh elliptic cp cao 57

5.1 L thuyt Fredholm-Riesz-Schauder . . . . . . . . . . . . . . . . . . . . 57

5.2 p dng l thuyt Fredholm - Schauder vo bi ton Dirichlet thunnht i vi phng trnh elliptic cp cao . . . . . . . . . . . . . . . . . 64

Bi 6 Ph ca ton t Elliptic 68

6.1 Bi ton Dirichlet i vi phng trnh Laplace . . . . . . . . . . . . . . 68

6.1.1 Gii thiu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 686.1.2 Ton t ca bi ton Dirichlet . . . . . . . . . . . . . . . . . . . 68

6.1.3 S tn ti v tnh trn ca nghim ca bi ton Dirichlet . . . . 69

6.2 Nguyn l cc i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

6.3 Ph ca ton t elliptic tuyn tnh cp 2 t lin hp . . . . . . . . . . . 79

6.4 Mt s v d ng dng l thuyt ph ca ton t elliptic . . . . . . . . 85

6.4.1 Bi ton bin i vi phng trnh truyn nhit . . . . . . . . . 85

6.4.2 Bi ton bin i vi phng trnh truyn sng . . . . . . . . . 87

Bi 7 Hm iu ho di v bi ton Dir ichlet 887.1 Gii thiu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

7.2 Cc hm iu ho v iu ho di . . . . . . . . . . . . . . . . . . . . 89

7.2.1 Biu din tch phn ca hm iu ho . . . . . . . . . . . . . . 89

7.2.2 Cng thc tch phn biu din hm iu ho . . . . . . . . . . . 91

7.2.3 Hm iu ha di v tnh cht . . . . . . . . . . . . . . . . . . 93

7.3 Min chnh quy v bi ton Dirichlet . . . . . . . . . . . . . . . . . . . 100

Bi 8 Na nhm v bi ton bin i vi phng tr nh par abolic 106

8.1 Gii thiu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1068.2 Mt s k hiu v kin thc b sung . . . . . . . . . . . . . . . . . . . . 107

8.2.1 Na nhm G(s) . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

8.3 Khng gian L2(a,b,X) . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

8.4 nh l tn ti duy nht nghim ca bi ton bin i vi phng trnhparabolic tru tng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

8.4.1 t bi ton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

8.4.2 Ton t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

8.4.3 Na nhm lin hp G(s) . . . . . . . . . . . . . . . . . . . . . 116


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iii Mc lc

8.4.4 nh l ng cu . . . . . . . . . . . . . . . . . . . . . . . . . . 117

8.4.5 S tn ti v duy nht nghim ca bi ton bin i vi phngtrnh Parabolic . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

Bi 9 Phng php im bt ng v p dng vo bi ton bin ca ph ng

tr nh o hm r ing phi tuyn 122

9.1 Gii thiu mt s nh l im bt ng c bn . . . . . . . . . . . . . 122

9.1.1 Nguyn l im bt ng Brouwer . . . . . . . . . . . . . . . . 122

9.1.2 Nguyn l nh x co Banach . . . . . . . . . . . . . . . . . . . . 122

9.1.3 nh l im bt ng Schauder . . . . . . . . . . . . . . . . . . 123

9.1.4 nh l im bt ng Leray-Schauder-Schaefer . . . . . . . . . 126

9.1.5 nh l im bt ng Tikhonov . . . . . . . . . . . . . . . . . . 127

9.2 S tn ti nghim ca bi ton Dirichlet i vi mt lp phng trnhelliptic cp 2 phi tuyn. . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

9.3 ng dng nh l Leray- Schaefer gii bi ton gi tr bin i viphng trnh HR ta tuyn tnh. . . . . . . . . . . . . . . . . . . . . . 130

Bi 10 Dng song tuyn tnh v m r ng Fr iedr ichs 134

10.1 Dng song tuyn tnh v dng ton phng ca ton t . . . . . . . . . 134

10.2 Ton t lin kt vi dng ton phng. M rng Friedrichs . . . . . . . 137


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iv Phng trnh o hm ring

Li ni u

Phng trnh vi phn o hm ring l mt nhm chuyn mn c truyn thngthuc b mn Gii tch, Khoa Ton-C-Tin hc, i hc Tng hp H Ni. Sau nhiubin c, vic sinh hot chuyn mn v PTHR b gin on lm nh hng nhiun cng tc o to v nghin cu khoa hc.

Bt u t nm 2003, chng ti c gng xy dng li seminar PTHR vi s thamgia ca thy tr thuc b mn Gii tch, Khoa Ton-C-Tin hc, trng i hc Khoa

hc T nhin v nhiu hc vin cao hc.Trc ht chng ti bt u vi mt s bi ging v bo co nhm cung cp nhng

kin thc c bn nht v l thuyt PTHR m trong chng trnh o to i hccng nh Cao hc cha c hc, ri dn dn nh hng mt s nghin cu, cp nhtnhng thng tin khoa hc, nhng hng nghin cu mi nht v PTHR ngy naytrong nc cng nh trn th gii.

Bng s n lc v nhit tnh ca cc thnh vin m c bit l nhng gig vinmi c b sung cho b mn, chng ti ra c mt tuyn tp ghi li nhng biging v cc bo co trnh by seminar trong sut nm qua nh mt ti liu hc

tp v nghin cu. V l do v thi gian, tp cc bo co ch

a

c bin tp mt cchk lng nn khng trnh khi nhng sai st. Hy vng nhng sai st s c honthin trong cc ln sau.

Chng ti rt mong s gip v tham gia ca cc cn b v ging vin trong bmn cng nh trong ton khoa.

Qu kh ngt ngo, nhng tm li hng v ca n qu l kh khn, i hi thigian, ngh lc v c s nhit tnh. Nhng chng ti tin rng chng ti s lm c.

H Ni, thng 09 nm 2004


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v Phng trnh o hm ring

Ti liu tham kho

[1] R. Adams, Sobolev spaces, Academic Press, New York, San Francisco, London,1975.

[2] L. Schwartz, Thories des Distributions, Hermann, 1978.

[3] S. Stuckless, Brouwers fixed point theorem: Method of proof and generalizations,B. Sc., Memorial University of Newfoundland, 1999.

[4] Hng Tn, Nguyn Th Thanh H, Cc nh l im bt ng, NXB HSP2003.

[5] Trn c Vn, Phng trnh vi phn o hm ring, Tp 2, NXB HQGHN 2001.

[6] E. Zeidler, Nonlinear Functional Analysis and its Applications, I: Fixed-Point The-orems. Springer-Verlag New York Berlin Heidelberg, 1992.

[7] E. Zeidler, Introduction to applied functional analysis and mathematical physics,Chapter 1. Springer-Verlag, 1993.

[8] M. Taylor, Partial Differential Equations, Vol I. Basic theory, Springer-Verlag,1997.

[9] M. Taylor, Partial Differential Equations, Vol II. Qualitative Studies of LinearEquations, Springer-Verlag, 1997.

[10] M. Taylor, Partial Differential Equations, Vol III. Nonlinear Equations, Springer-Verlag, 1997.

[11] M. Strunwe, Variational methods, 2nd Edition, Springer-Verlag, 2000.

[12]

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1 Phng trnh o hm ring

Bi 1Phng php bin phn v bi ton Dirichlet

i vi phng trnh elliptic cp hai

1.1 Cch t bi ton bin phn

Gi sX l mt khng gian Hilbert thc, F(u) l phim hm thc trn X, D(F) Xl min xc nh ca phim hm F(u). Bi ton t ra l: Tm u0 D(F) sao cho

F(u0) = inf uD(F)

F(u), (1.1)

hay

F(u0) = supuD(F)

F(u). (1.2)

Nu X l khng gian hu hn chiu th bi ton t ra tr thnh bi ton cc tr hmnhiu bin, v vy ta lun gi thit rng X l khng gian v hn chiu.

nh ngha 1.1. Phim hm F(u) t cc tiu tuyt i ti u0 nu

F(u0) F(u), u D(F),

Phim hm F(u) t cc tiu a phng ti u0 nu

> 0 : F(u

0)

F(u),

u

D(F)

B(u0, ),

trong B(u0, ) = {u X : u u0 < }.

Gi s u X l im c nh, M l tp tuyn tnh ca X. Ta nh ngha

M = {u X : u = u + , M}.

Ta chng minh c M tr mt trong X khi v ch khi M tr mt trong X. M cgi l a tp tuyn tnh trong X.

Xt phim hm F tho mn cc iu kin sau


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2 Bi 1. Phng php bin phn

iu kin 1 Min xc nh D(F) l a tp tuyn tnh tr mt trong X, c ngha lD(F) tr mt trong X v

D(F) =

{u

X : u = u + ,

M

X

}.

iu kin 2 Ly 1, 2, . . . , n l n phn t c nh trong M. Khi F(u + c11 +c22 + cnn) = F(c1, c2, . . . , cn) l mt hm kh vi theo c = (c1, c2, . . . , cn) cpk 1.

1.2 Bin phn v gradient ca mt phim hm

Gi thit phim hm F(u) tho mn cc iu kin 1,2, u0 D(F), M, R.Khi u0 + D(F). Tht vy, do u0 D(F) nn c dng u0 = u + 0, 0 M.Vy

u0 + = u + 0 + = u

+ (0 + ).

V 0 + M nn u = u + (0 + ) D(F). Theo iu kin 2, F(u0 + ) lphim hm kh vi theo .

nh ngha 1.2. i lng

F(u0, ) =dF(u0 + )

d =0c gi l bin phn (hay bin phn cp 1) ca phim hm F(u) ti im u0 D(F).Nhn xt.

(i) Vi u0 c nh, F(u0, ) l mt phim hm i vi .

(ii) F(u0, ) l phim hm thun nht i vi , tc l F(u0, t) = tF(u0, ). Tht

vy,

F(u0, t) = lim

0

F(u0 + t) F(u0)

= t lim0

F(u0 + ) F(u0)

= tF(u0, ).

(iii) Ni chung F(u0, ) khng phi l mt phim hm tuyn tnh.

iu kin 3 Gi thit ti im u0 bt k, F(u0, ) l phim hm tuyn tnh i vi .

(iv) Vi u0 bt k, ni chung F(u0, ) khng phi l mt phim hm gii ni theo .


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Ta k hiu

N = {u D(F) : F(u0, ) l phim hm tuyn tnh gii ni theo },

gi thit N = . Vi mi u N, tn ti phim hm tuyn tnh gii ni gu trn M saocho

F(u, ) = gu(), M.V D(F) tr mt trong X nn M tr mt trong X, gu xc nh trn M c th m rngthnh phim hm gu X. Nh vy tn ti ton t

P :N Xu gu = P u X.

nh ngha 1.3. Ton t P c gi l gradient ca phim hm F(u):

gu = P u = gradientF(u) = grad F(u).

Khi X l khng gian Hilbert, X = X, ta xem grad F(u) N = D(grad F). Do vi u N D(F),

F(u, ) = gu() = grad F(u) = (grad F(u), ), M.

1.3 iu kin cn ca cc tr

Gi sF(u) l phim hm xc nh trn X tho mn ba iu kin phn trc. Githit phim hm F(u) t cc tiu a phng ti im u0 D(F), tc l F(u) F(u0),u : u u0 < . Khi vi mi M, R sao cho || b th

F(u0 + ) F(u0),

c ngha l F(u0 + ) t cc tiu ti = 0. Do

dF(u0 + )

d =0 = F(u0, ) = 0.Vy nu phim hm F(u) t cc tiu ti u0 D(F) th

F(u0, ) = 0 M.

Nh vy u0 N = D(grad F). Do

F(u0, ) = gu0() = (grad F(u0), ) = 0 M.

V M tr mt trong X nn ta suy ra grad F(u0) = 0. Ta c nh l sau y ni ln

iu kin cn ca cc tr


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nh l 1.1. Nu phim hm F(u) tho mn cc iu kin 1,2,3 v t cc tiu aphng ti u0 D(F) th u0 D(grad F) v

grad F(u0) = 0 (phng trnh Euler.)

Ch . iu kin ny trng vi iu kin cn ca cc tr hm nhiu bin.

Chng minh. Gi s F(x) = F(x1, x2, . . . , xn) t cc tr a phng ti

x0 = (x01, x02, . . . , x

0n).

Khi

F(x0, x) =dF(x0 + x

d =0 = ddF(x01 + x1, x02 + x2, . . . , x0n + xn)= Fx1(x

0)x1 + Fxn(x0)xn= grad F(x0)x = 0 (x = (x1, . . . , xn).

T , grad F(x0) = 0, suy ra

F(x0)

xi= 0, (i = 1, 2, . . . , n).

1.4 Ton t xc nh dng

nh ngha 1.4. X l khng gian Hilbert, A : X X l ton t tuyn tnh. Ton tA c gi l ton t i xng nu

1. Min xc nh D(A) tr mt X,

2. u, v D(A), (Au,v) = (u,Av).

T iu kin 2, (Au,v) l phim hm i xng v (Au,u) l mt dng ton phngi xng.

nh ngha 1.5. Ton t i xng A c gi l ton t dng nu

(Au,u) 0, u D(A),(Au,u) = 0 u = 0.

Ton t dng A c gi l xc nh dng nu tn ti R sao cho

(Au,u) 2

u2

, u D(A).


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1.4.1 Khng gian nng lng ca ton t xc nh dng

Gi s X l khng gian Hilbert, A l ton t xc nh dng trong X. Vi miu, v

D(A) ta xc nh phim hm song tuyn tnh

a(u, v) = (Au,v)

tho mn cc tnh cht

(i) a(u, v) = (Au,v) = (u,Av) = a(v, u), u, v D(A),(ii) a(u1 + u2, v) = a(u1, v) + a(u2, v),

(iii) a(u, u) 0, u D(A), du bng xy ra khi v ch khi u = 0.Vi cc tnh cht trn, a(u, v) l mt tch v hng trong D(A). t

|||u||| = (a(u, u))1/2 = (Au,u)1/2

Khi |||||| l mt chun trong D(A).K hiu HA l b sung ca D(A) theo chun ||||||, ch rng v A l ton t xc

nh dng nn(Au,u) 2u2, u D(A).

T

u 1|||u|||, u D(A).

Vy, nu {un} D(A) hi t trong HA th cng hi t trong X, do HA X. Khnggian HA l khng gian Hilbert, c gi l khng gian nng lng ca ton t xcnh dng A.

1.5 Cc tiu phim hm nng lng

Cho X l khng gian Hilbert, A : X X l ton t xc nh dng, f X. Xtphng trnh

Au = f. (1.3)

Cng vi phng trnh (1.3) ta xt phim hm

F(u) = (Au,u) 2(u, f). (1.4)Phim hm F(u) c gi l phim hm nng lng ca ton t xc nh dng A,min xc nh ca A v F l trng nhau.

nh l 1.2. iu kin cn v phn tu0 D(A) lm phim hm F(u) t cctiu l u0 l nghim ca phng trnh Au0 = f. ng thi u0 l phn t duy nht.


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Chng minh.

iu kin cn Gi s u0 D(A) lm cc tiu phim hm F(u), u0 D(grad F). Khi grad F(u0) = 0. Vi u D(F), R, X, ta c

F(u, ) =d

dF(u + )

=0

.

Ch rng

F(u + ) = A(u + ,u + ) 2(u + ,f)= (Au,u) + (Au,) + (u,A) + 2(A,) 2(u, f) 2(, f)= (Au,u)

2(u, f) + 2(Au,)

2(, f) + 2(A,)

= F(u) + 2(Au f, ) + 2(A,)

Suy ra F(u, ) = 2(Au f, ). V Au X, f X, X nn (Au f, ) latch v hng, do vi mi u D(A), (Au f, ) l phim hm tuyn tnhgii ni i vi X, tc l u D(grad F). Vy D(grad F) = D(A) = D(F).T suy ra grad F(u) = 2(Au f) m rng thnh phim hm trn X v ta c

F(u0, ) = 2(Au0 f, = 0, X, Au0 f = 0 Au0 = f.

iu kin Gi s u0 D(A) l nghim ca phng trnh Au = f. Ly u D(A),u = u0. t = u u0 X. Ta c

F(u) = F(u0 + ) = F(u0) + (A,)

V A xc nh dng nn (A,) > 0, t suy ra F(u) > F(u0), vy u0 l phnt cc tiu ca F.

Chng minh duy nht Gi s tn ti u1 D(A) sao cho F(u) > F(u1) vi miu D(A). Khi ta c F(u0) > F(u1) > F(u0), v l.

1.6 Nghim suy r ng

Cho X l khng gian Hilbert, A l ton t xc nh dng, f X. Xt

F(u) = (Au,u) 2(u, f) (1.5)


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l phim hm nng lng ca ton t A. Vi u D(A) ta c (Au,u) = |||u|||2. Khi phim hm nng lng c th vit di dng

F(u) =

|||u

|||2

2(u, f). (1.6)

ng thc ny chng t rng phim hm F(u) c th m rng thnh phim hm trnHA. Vy vn t ra l ta i tm cc tiu phim hm nng lng F(u) vi u HA.nh l 1.3. Trong khng gian nng lng HA, tn ti mt v ch mt phn t u0 lmphim hm F(u) t cc tiu.

Chng minh. Xt F(u) = |||u|||2 2(u, f), f X, u HA X. Ta c

|(u, f) uf f

|||u|||.

iu c ngha l |(u, f) M|||u|||, u HA, tc l F(u) l phim hm tuyn tnhb chn trong HA. Theo nh l Riesz-Frchet, tn ti u0 HA sao cho

(f, u) = a(u, u0), u HA.

T

F(u) =|||

u|||

2

2a(u, u0)

= a(u, u) 2a(u, u0) + a(u0, u0) a(u0, u0)= a(u u0, u u0) a(u0, u0)= |||u u0|||2 |||u0|||2

T , F(u0) = |||u0|||2 v F(u) > F(u0) vi mi u HA. Nh vy u0 l phn tlm cc tiu phim hm nng lng F(u). R rng u0 l nghim duy nht (theo nhl 1.2, u0 l nghim duy nht thuc HA.)

nh ngha 1.6. Phn t u0 lm cc tiu phim hm nng lng

F(u) = |||u|||2 2(u, f)

c gi l nghim suy rng ca phng trnh Au = f. Nu u0 D(A) th u0 lnghim c in ca phng trnh

Ch . nh l trn vn ng nu F(u) = |||u|||2 2(u), trong (u) l phim hmtuyn tnh gii ni trong HA.


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1.7 Bi ton Dir ichlet i vi phng trnh elliptic cp 2 t lin hp

1.7.1 Biu thc elliptic v biu thc elliptic cp hai t lin hp

Gi s Rn l tp m vi bin trn . Ta xt biu thc vi phn cp hai tuyntnh trong :

Au =n

i,j=1

aij(x)2u

xixj+

nj=1

aj(x)u

xj+ c(x)u. (1.1)

trong aij, aj, c l nhng hm xc nh trong , tho mn

aij = aji, c, cj , aji C1().Ta ni A l biu thc vi phn elliptic (hoc ton t (vi phn) elliptic) trong nu:

ni,j=1

aijij ||2 Rn, = 0, > 0.

Gi s u, v C0 (). Xt biu thc:

(Au,v) =

Au.vdx

=

ni,j=1

aij2u

xixjvdx +

nj=1

aju

xjvdx +

c(x)uvdx

Ch rng:

aij2u

xixjv =

xj

aijv

u

xi

xj(aijv)

u

xi.

p dng cng thc Green v ch v

= 0, ta c:

aij2u

xixjvdx =

xj(aijv)

u

xidx +

aijvu

xicos(xiv)dx

=

xj(aijv)

u

xidx

=

u2(a

ijv)

xjxi dx

Tng t:

aju

xjvdx =

u

xj(ajv)dx.

Kt hp li ta c:

(Au,v) =

n

i,j=1

2(aijv)

xixj

nj=1

xj(ajv) + cv

udx

= (u, Av)


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trong

Av =n

i,j=12(aijv)

xixj

n

j=1

xj(ajv) + cv (1.2)

c gi l biu thc vi phn cp hai lin hp ca Au. Mt cch tng qut, A cgi l biu thc vi phn lin hp (hoc ton t lin hp) ca A nu:

(Au,v) = (u, Av) u, v C0 (). (1.3)

Ch rng biu thc Au c th vit di dng:

Au =

n

i,j=1

xi aij

u

xj+

n

j=1 aj aij

xi u

xj+ cv.

T , ta thy nu bj = aj aijxi = 0 (j = 1, 2, . . . , n) th

(Au,v) = (u,Av) u, v C0 ().

tc l A = A. Khi , ta ni A l biu thc vi phn t lin hp (ton t t lin hp).

Tng qut, ta c nh ngha: Nu

(Au,v) = (u,Av) u, v C0 (). (1.4)

th A l biu thc vi phn t lin hp.Vy Au l biu thc vi phn t lin hp khi v ch khi Au c dng

Au =n

i,j=1

xi

aij

u

xj

+ cu, aij = aji. (1.5)

1.7.2 Bi ton Dir ichlet i vi phng trnh elliptic cp 2 t lin hp

Cho tp m, b chn Rn vi bin trn . Xt bi ton Dirichlet:

Au = n

i,j=1

xj

aij

u

xi

+ c(x)u = f trong , (1.6)

u

= 0, (1.7)

trong A l ton t elliptic t lin hp trong , c(x) 0, aij = aji L().


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1.7.2.1 Dng song tuyn tnh

Gi s u C2() v v(x) C0 (). p dng cng thc Green, ta c:

(Au,v) = n

i,j=1

xj

aij uxi

vdx +

cuvdx

=

n

i,j=1

aiju

xi

v

xj+ cuv

dx

Vi mi u C2(), v C0 (), ta t

a(u, v) =

n

i,j=1aij

u

xi

v

xj+ cuv

dx.

Nhn xt rng nu u C2() l nghim ca bi ton Dirichlet (1.6)-(1.7), trong f C() th:

(Au,v) = a(u, v) = (f, v) v C0 ().B 1.4. Vi mi u, v C0 (), a(u, v) l mt dng song tuyn tnh i xng xcnh dng.

Chng minh b .

a) T nh ngha suy ra a(u, v) l dng song tuyn tnh trong C0 ().

b) Ta ch ra rng a(u, v) l dng song tuyn tnh i xng trong C0 (). Tht vy,

a(u, v) = (Au,v) =n

i,j=1

aiju

xi

v

xjdx +

cuvdx.

V aij = aji nn t ta suy ra u, v C0 ()

(Au,v) = (u,Av) hay l a(u, v) = a(v, u).

c) a(u, v) l dng song tuyn tnh xc nh dng.

Trc ht, ta chng minh bt ng thc Friedrichs: Cho gii ni Rn, bin trn. Khi vi mi u C1(), u| = 0, tn ti 1 > 0 sao cho:

u2(x)dx 1ni=1

u

xi

2dx.


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Chng minh bt ng thc Friedrichs. = {x : 0 xi ai}, . tu = 0 x .

u(x) = u(x1, x) = u(x1, x

)

u(0, x) = x1

0

u(t, x)

tdt

u2(x) =

x10

u(t, x)

tdt

2x10

dt

x10

u(t, x)

t

2dt

a10

u

x1

2dx1.

u2dx a10

dx1

a20

dx2 . . .

an0

u2(x)dxn

a1

0

. . .

an

0

dx

a1

a1

0 u

x12

dx

a21

u

x12

dx.

T ,

u2(x)dx 1

nk=1

u

xk

2dx, 1 = a

21.

p dng tnh elliptic v bt ng thc Friedrichs,

a(u, u) = (Au,u) =

n

i,j=1

aiju

xi

u

xj+ cu2

dx

nj=1

uxk2

dx +

cu2dx

1 .

Ch rng nu c(x) 0 x , chn 0 < 0 < ,

a(u, u)

nj=1

u

xk

2dx

( 0)

nj=1

u

xk

2dx +

01

|u|2dx

20

k = 1n uxk

2 + |u|2 dxa(u, u) 20

(|u|2 + |u|2)dx = 20u2H10 ().

Ch rng theo cch chng minh trn, khi c(x) 0 th dng a(u, v) vn xcnh dng.

Vy ta c kt lun ca b .


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1.7.2.2 Ton t ca bi ton Dirichlet

nh l Lax-Milgram Cho [u, v] l phim hm song tuyn tnh trong khng gian HilbertX nhn gi tr thc tho mn cc iu kin

i) |[u, v]| cuv, u, v X.ii) u2 k[u, v] u X, v X.

Khi , mi phim hm tuyn tnh gii ni F(u) trn X u tn ti f X saocho:

F(u) = [u, f] u X.Khng gian H10 () Trong khng gian C

0 (), ta a vo chun

uH10() = |u|2dx1/2

u C

0 ().

vi

u =

x1,

u

x2, . . . ,

u

xn

l vector gradient v tch v hng

(u, v)1 =

u vdx, u, v C0 ().

Ch rng trong C0 (), ta c th xc nh mt chun tng ng (nh btng thc Friedrichs):

uH10() =

(|u|2 + |u|2)dx1/2

, u C0 ().

v tch v hng:

(u, v)1 =

(u v + u v)dx, u C0 ().

nh ngha 1.7. Khng gian H10 () c coi l b sung ca C0 () theo

chun H10 (). Php nhng H10 () L2(), l tr mt v compact.

Mt cch khc, ta c th nh ngha khng gian H10 () gm cc hm u H

1

()trit tiu trn bin cng vi cc o hm suy rng theo ngha vt:

H10 () =

u H1()|u = 0, u

xi= 0 trn theo ngha vt

.

Ta xc nh ton t A : H10 () L2() sao cho min xc nhD(A) = {u H10 () : Au L2()}

(Au,v) = a(u, v) u D(A), v C0 ().

Ch rng:


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i) V C0 () D(A) H10 () nn D(A) tr mt trong H10 (), do tr mttrong L2(). Do ,

(Au,v) = a(u, v)

u

D(A),

v

H10 ().

ii) Vi u C2() l nghim ca bi ton (1.6)- (1.7), ta c:

a(u, v) =

n

i,j=1

aiju

xi

u

xj+ cuv

dx v C0 ()

=

ni,j=1

xj

aij

u

xi

+ cu

vdx

= (Au,v) = (f, v), v C0 (),

trong

Au = n

i,j=1

xj

aij

u

xi

+ cu

Ton t A c xy dng nh trn c gi l ton t ca bi ton (1.6)- (1.7).

nh ngha 1.8. Vi f L2(), hm u0 H10 () c gi l nghim suy rngca bi ton (1.6)- (1.7) nu

a(u0, v) = (f, v) v

C0 ()

hay

(Au0, v) = (f, v) v C0 ()(Au0 f, v) = 0 v C0 ()

Nh vy, vic tn ti nghim suy rng ca bi ton Dirichlet (1.6)- (1.7) cquy v vic tn ti nghim ca phng trnh ton t Au = f trong L2().

1.7.2.3 p dng php tnh bin phn gii bi ton Dirichlet i vi

phng trnh vi phn tuyn tnh cp hai t lin hp

Gi s l tp m b chn Rn c bin trn. Ta xt bi ton Dirichlet

Au = n

i,j=1

xj

aij

u

xj

+ c(x)u = f trong , (1.8)

u = 0 trn , (1.9)


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trong A : H10 () L2() l ton t elliptic, t lin hp trong , aij = aji, c(x) 0thuc C(), f L2(), xc nh dng song tuyn tnh tng ng

(Au,v) = a(u, v)

v

C0 ()

c min xc nh D(A) tr mt trong H10 ().

Theo chng minh ca b 1.4, ta suy ra: dng song tuyn tnh

a(u, v) =

n

i,j=1

aiju

xi

u

xj+ cuv

dx

xc nh mt tch v hng trong D(A) v

|||u

|||= a(u, u) u D(A)

l chun lin kt vi tch v hng a(u, v) trong D(A).

Hn na:

i) A l ton t i xng

(Au,v) = a(u, v) = a(v, u) = (u,Av), u, v D(A).

ii) A l ton t xc nh dng

(Au,u) 20u2H10() u D(A).

K hiu HA l b sung ca D(A) theo chun ||||||. Ta gi HA l khng gian nnglng ca ton t A.

nh l 1.5. Khng gian nng lng HA bao gm cc hm u(x) c tnh cht:

1. u(x) L2() c cc o hm suy rng uxi

L2().2. Tn ti dy {uk} D(A) sao cho:

limk

uk u = 0, limk

ukxj uxj = 0

Chng minh. Nu u(x) HA, c hai kh nng

1. Nu u D(A) th u L2(), uxj

L2(). iu khng nh lun ng.


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2. Nu u HA, u / D(A). Khi , tn ti dy {uk} D(A) H10 () sao choliml,k

|||uk ul||| = 0

tc l

liml,k

ni,j=1

aij(uk ul)

xi

(uk ul)xj

dx +

c(uk ul)2dx = 0

T suy ra: limk

uk u = 0 u L2() v

liml,k

ni,j=1

aij(uk ul)

xi

(uk ul)xj

dx = 0

Mt khc, ta c:

ni,j=1

aij(uk ul)

xi

(uk ul)xj

dx

nj=1

ukxj

ulxj

2dx

T , ta suy ra:ukxj

k=1

l dy c bn trong L2() nn tn ti vj L2() sao cho:

limk

ukxj vj = 0.

Theo tnh cht ca o hm suy rng, ta suy ra: vj =ukxj

v

limk

ukxj uxj = 0.Vy HA() H10 ()

Phim hm nng lng ca ton t xc nh dng A l phim hm

F(u) = a(u, u) 2(u, f), u HA.T chng minh nh l 1.2, ta suy ra:

i) Phn t u0

D(A) lm cc tiu phim hm F(u) khi v ch khi u0 l nghim

ca phng trnh Au0 = f.

ii) Trong khng gian nng lng HA, tn ti duy nht mt phn t u0 lm cc tiuphim hm nng lng F(u). Phn t u0 l nghim suy rng ca phng trnhAu0 = f, tc l

(Au0 f, v) = 0, v C0 (),hay

(Au0, v) = (f, v), v C0 (),a(u0, v) = (f, v),

v

C

0

().


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Vy u0 l nghim suy rng ca bi ton Dirichlet. T cc chng minh trn, ta cnh l:

nh l 1.6. Bi ton Dirichlet (1.6)- (1.7) c duy nht nghim suy rng u0

HA

H10 () vi mi f L2().Ch . Vi mi f L2(), v HA L2(), ta c:

|(f, v)| fv f|||v|||A.

Nh vy, (f, v) l phim hm tuyn tnh lin tc trong HA. Theo nh l Riesz, tn ti duy

nht u0 HA H10 () sao cho

(f, v) = a(u0, v) v HA.

T suy ra

(f, v) = a(u0, v) v C0 ().Theo nh ngha, u0 l nghim suy rng ca bi ton Dirichlet (1.6)- (1.7).


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Bi 2nh l Lax-Milgram v bi ton bin elliptic

2.1 nh l Lax-Milgram

nh l 2.1. Gi s X l khng gian Hilbert thc, a(u, v) l phim hm song tuyntnh thc trn X. Gi thit rng a(u, v) tho mn cc iu kin:

(i) Tn ti c > 0 sao cho |a(u, v)| cuv u, v X,(ii) Tn ti > 0 sao cho a(u, u) u2 u X.

Khi , mi phim hm tuyn tnh lin tc F(u) trn X u tn ti f X sao cho

F(u) = a(u, f) u X.

Ch . nh l Lax- Milgram suy ra t nh l Riesz-Frchet.

Chng minh. Ly u X c nh. Khi , u(v) = a(u, v) l phim hm tuyn tnh trnX. Theo (i), ta c:

|u(v)| = |a(u, v)| uv v X.iu chng t u(v) l phim hm tuyn tnh lin tc trn X. Theo nh l Riesz-Frechet, tn ti mt phn t, k hiu l Au X sao cho:

u(v) = (Au,v)

v

X.

Nh vy, a(u, v) = (Au,v) v X, v ta c mt ton t

A : X Xu Au

R rng A l ton t tuyn tnh. Theo gi thit (ii), ta c:

Au2 = (Au,Au)

a(u,Au)

c

u

Au

v

X.


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18 Bi 2. nh l Lax-Milgram v bi ton bin elliptic

T , ta c: Au cu u X. Bt ng thc ny chng t rng A : X X lton t lin tc. Hn na, vi u1, u2 X m

Au1

= Au2

u1

= u2 (2.1)

Mt khc, vi mi u X,

u2 a(u, u) = 1

(Au,u) cAuu.

T ,

u cAuu X.

Do , vi u1, u2

X m

u1 = u2 Au1 = Au2. (2.2)

T (2.1)-(2.2), suy ra: A : X X l nh x 1-1. K hiu A(X) = {u X : Au X}.Ta chng minh A(X) l ng trong X. Tht vy, gi s{Auj} l dy hi t n v X.V {Auj} l dy Cauchy trong X, ta c

limj,k+

Auj Auk = 0.

T (2.1), ta c uj uk cAuj Auk. iu ny chng t {uj} l dy Cauchy

trong X, cho nn tn ti u X sao cho: limj+ uj = u trong X.Do A l nh x lin tc nn Au = v A(X), tc l A(X) ng trong X.By gi, ta chng minh A(X) = X. Gi sA(X) X, A(X) ng. Ta ly u X

m u A(X), trc giao vi A(X), tc l

(u,Au) = a(u, u) = 0

V u2 1

a(u, u) = 0 nn u = 0, tc l A(X) = X. Vy A : X X l song nh.Gi s F(u) l phim hm tuyn tnh lin tc trn X. Theo nh l Riesz-Frecht,

tn ti duy nht g

X sao cho

F(u) = (g, u).

Khi , tn ti f X sao cho g = Af. Do ,

F(u) = (g, u) = (Au,u) = a(f, u), u X.

nh l c chng minh.

Ch .

- ng cu A : X X c xy dng trong nh l Lax-Milgram sao cho:

(Au,v) = a(u, v) u, v X


25/144


c gi l ton t lin kt vi dng song tuyn tnh a(u, v) trn khng gian Hilbert X,

hay ngc li, a(u, v) c gi l dng song tuyn tnh lin kt vi ton t A.

- Khng gian Banach X c gi l khng gian Hilbertian hay l khng gian Hilbert

ho cnu c th xy dng mt tch v hng sinh ra mt chun tng ng.

Xt V l khng gian Hilbert phc v tch v hng (u, v), u, v V, tho mn iukin

(u, v) = (v, u), u, v V.V l khng gian i ngu ca V, (khng gian cc phim hm tuyn tnh lin tc

trong V).

Theo inh l Riesz, vi mi L V, tn ti duy nht u V sao cho L(v) =(u, v)V, u, v V.V d 1. l tp m trong Rn, H1

0

() l b sung ca C0

() trong H1(). t V =

H10 (), V = H1() l i ngu ca H10 (), ng nht vi khng gian cc hm suy rng

f D():|(f, )| cH1 C0 ().

Gi sa(u, v), u, v V l dng song tuyn tnh lin tc trong V. Vi mi u V,ta c th xc nh mt phim hm tuyn tnh L trn V theo cng thc:

L(v) = a(u, v) v V.

R rng L l phim hm tuyn tnh lin tc trn V:

L(v) |a(u, v)| uv v V.

Theo nh l Riesz, tn ti duy nht Au V sao cho:

L(v) = (Au, v)V = a(u, v) v V,

trong A : V V l ton t lin kt vi dng song tuyn tnh a(u, v).nh ngha 2.1. Dng song tuyn tnh lin tc a(u, v) c gi l c tnh cht coerci-tive nu tn ti hng s c > 0 sao cho

|a(u, u)| cu2, u V. (2.3)

nh l 2.2 (nh l Lax - Milgram). Nu a(, ) l dng song tuyn tnh lin tc ctnh cht coercitive th ton t A lin kt vi dng song tuyn tnh a(u, v) l mt ngcu tV ln V.

Chng minh. Chng minh hon ton tng t nh trng hp a(u, v) thc. Ta c


26/144


27/144


28/144


Gi s a(u, v) l dng song tuyn tnh lin tc:

a(u, v) =

n

i,j=1aij

u

xi

v

xj+

n

i=1bi

u

xiv +

n

i=1biu

v

xi+ cuv

vdx (2.3)

u, v C0 ().

K hiu V = H10 (), H = L2(). Khi , V H v php nhng l lin tc v tr

mt, ng thi dng song tuyn tnh xc nh trn V c th m rng vi u, v H10 ().Hn na, vi u H10 () = V, phim hm a(u, v) = l(v), v V l phim hm tuyntnh lin tc trn V = H10 (). Do , ta xc nh ton t:

A : V V = H10 = H1()sao cho (Au, v)L2() = a(u, v) v V. Khi , A l ton t tuyn tnh lin tctrn V:

Au2 = (Au, Au) = a(u,Au) uAu u H Au u u V

nh l 2.1. Tn ti mt hng s0 R sao cho vi 0 th ton tA + I l mtng cu t H10 () ln H

1().

Chng minh. Theo gi thit (2.1) v tnh elliptic, ta c:

Re

ni,j=1

aiju

xi

u

xjdx

|u|2dxu =

u

x1, . . . ,

u

xn

l vector gradient

Mt khc, cc h s ca ton t A xc nh bi (2.1) l b chn trong nn ta c:

bi uxi udx , biu uxidx c uxiL2 uL2v cu2dx u cu2T , ta suy ra tn ti cc hng s c1, c2 sao cho:

Re a(u, u) u2L2() c1uL2()uL2() c2u2L2().


29/144


Ch bt ng thc:

|ab| a2 + ( 14

)b2, > 0

Ta c: uL2uL2 u2 + ( 14)u2.t =

2c1, ta nhn c bt ng thc

Re a(u, u) u2L2() c1

2c1u2 + ( 1

4)u2

c2u2L2()

2u2L2()

2c21

+ c2

u2L2().

Vy tn ti hng s c > 0 sao cho:

Re a(u, u) 2u2L2() cu2L2()

hay l

Re a(u, u) + cu2L2()

2u2H10 (), u V (2.4)

p dng h qu ca nh l Lax - Milgram, ta suy ra ton t A + I l ng cu tV = H10 () ln V

vi > c.

2.2.2 iu kin ca tnh coercitive

Ch rng vi gi thit bi, bi C1() nn bi, bi v cc o hm ca chng b chntrong . Do , vi f H1(), b H1() v k(bif) = (kbi)f + bi(kf). Ta chngminh b sau y:

B 2.2. Gi sf, g H1() v mt trong chng thuc H10 (). Ta c:

(kf)gdx =

(kg)fdx.

Chng minh. Nu f H10 (), xp x f bi dy hm j C0 () theo chun trongH1() v p dng tch phn tng phn

(kj)gdx =

jgdx

(kg)jdx

=

(kg)jdx.

Cho j , ta nhn c ng thc cn chng minh.


30/144


p dng b , ta c:

bi(iu)udx =

(ibi)uudx

biuudx

Do ,

Re

bi(iu)udx = 12

(ibi)|u|2dx.

Tng t ta cng c:

Re

bi(iu)dx = 1

2

(ibi)|u|2dx.

p dng tnh elliptic, vi u V = H10 (),

Re a(u, u) = Re

ni,j=1

aijuxi

uxj

+ Re

ni=1

biuxi

u +ni=1

biuuxi

dx ++ Re

c|u|2dx

u2L2 +

c 1

2

ni=1

bixi

12

ni=1

bixi

|u|2dx.

K hiu B = (b1, b2, . . . , bn), B = (b1, b2, . . . , b

n). Ta nhn c c lng:

Re a(u, u) u2L2() + c 12 div(B + B ) |u|2 dx, u H10 ()Vy nu tn ti > 0 sao cho:

c 12

div(B + B )

x , (2.5)

th dng song tuyn tnh a(u, v) l coercitive

Re a(u, u)

c0

u

2H10()

u

H10 (),

c0 = min(, ).

nh l 2.3. Gi thit tn ti hng s > 0 sao choc 1

2div (B + B)

x .

Khi , ton t A lin kt vi dng song tuyn tnh a(u, v) l mt ng cu t H10 ()ln H1().


31/144


p dng: Gi s l min khng b chn vi bin trn. Ta xt bi ton Dirichleti vi phng trnh elliptic cp 2:

Au =

n

i,j=1 xj aij uxi +n

i=1 bi uxi n

i=1 xi (biu) + cu = f trong (2.6)u

= 0, u(x) 0 khi |x| + (2.7)nh ngha 2.2. Vi f L2(), hm u0 H10 () l nghim suy rng ca bi ton(2.6) - (2.7) nu:

a(u0, v) = (f, v)L2() v C0 ().T nh l 2.3, ta suy ra:

nh l 2.4. Gi thit: tn ti hng s > 0 sao cho

c(x) 1

2div (B + B)

x .Khi , vi mi f L2(), tn ti duy nht nghim suy rng u0 H10 () ca bi ton(2.6) - (2.7).

Vi iu kin ca nh l, ton t A lin kt vi dng song tuyn tnh a(u, v) lng cu tH10 () ln H

1().

Gi s f L2(), f H1(), tn ti u0 H10 () sao cho: Au0 = f trongL2(), tc l

(Au0, v) = (f, v) v C0 ()

Do , a(u0, v) = (f, v) v C0 ().iu c ngha l: u0 l nghim suy rng ca bi ton (2.6) - (2.7).

2.2.3 Bi ton bin Dir ichlet trong min b chn

Gi s l tp m b chn. Trong cho ton t Au xc nh bi (2.1) v dngsong tuyn tnh a(u, v) xc nh bi cng thc (2.2).

Trc ht, ta ch vi l min b chn, nh x nhng H10 () vo L2() (cng

chnh l nh x nhng H10 () vo H1()) l ton t compact.Ta c nh l sau y.

nh l 2.5. Gi s l tp m b chn trong Rn. Ton t vi phn tuyn tnh ellipticcp 2 c xc nh theo cng thc (2.1) v dng song tuyn tnh a(u, v) xc nh bi(2.2). Khi ,

a) Ton tA lin kt vi dng song tuyn tnh a(u, v) l mt ng cu tH10 () lnH1()) nu

c

1

2

div(B + B )

0

x

.


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b) Trong trng hp tng qut, ton tA l ton t Fredholm vi ch s 0 tH10 ()ln H1().

Chng minh. Theo gi thit,

c 12

div(B + B ) 0 x .

T bt ng thc c chng minh trn y:

Re a(u, u) u2L2 +

c 1

2div(B + B )

|u|2 dx

ta suy ra

Re a(u, u) u2L2 u H10 ().p dng bt ng thc Poincareq:

u2L2 ku2L2 u H10(), k > 0.

( l tp m b chn), ta nhn c c lng

Re a(u, u) u2L2

2u2L2 +

k

2u2L2

u2H10() u H10 ().

T suy ra a(u, u) l coercitive.

Theo nh l Lax - Milgram, A l ng cu tH10 () ln H1().

Trong trng hp tng qut: A + I l ng cu t H10 () ln H1() vi

ln, cn I l ton t compact t H10 () vo H1(). Do , A = (A + I) I l

tng ca mt ng cu vi mt ton t compact, cho nn A l ton t Fredholm vich s 0. nh l c chng minh.

nh l 2.6. Gi s m Rn, ton t A v dng song tuyn tnh a(u, v) xc nhbi (2.1) v (2.1). Gi thit rng a(u, v) l coercitivetrn H10 (). Khi , vi

f H1(), tn ti mt v ch mt u H1() sao choAu = f

u w H10 () (u w) = 0

Chng minh. t v = u w, Av = Au Aw Av = f Aw H1().Khi , tn ti duy nht v H10 () sao cho:

Av = f

Aw


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hay

A(v + w) = f, u = v + w H1()

tho mn iu kin ca dnh l: u w = v H10 ().nh l 2.7. Gi s l tp m b chn vi bin trn, A l ton t xc nhbi (2.1). Khi :

a) nh x u (Au,u) l mt ton t Fr edholm vi ch s 0 tH1() ln H1()H

12 ().

b) Vi gi thit

c(x)

1

2div (B + B )

0

x

,

nh x u (Au,u) l mt ng cu tH1() ln H1() H12 (), c nghal vi f H1() v h H12 (), tn ti duy nht u H1() sao cho

Au = f trong

u = h trn

Chng minh. a) Ton t (A, ) l tng ca ton t ((A + I), ) vi ln l mt

ton t ng cu vi ton t compact (I, 0). Vy nn(A, ) = ((A + I), ) + (I, 0)

l ton t Fredholm vi ch s 0.

b) Vi iu kin b) th a(u, v) l coercitive. Do , p dng nh l 2.6, vi H1() sao cho = h, tn ti duy nht u H1() sao cho: Au = f vu w H10 () hay u = h. nh l c chng minh.


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Bi 3Khng gian Sobolev Wm,p()

3.1 Mt s k hiu v khi nim

Trong phn ny, l mt tp m trong Rn, m l s nguyn khng m, p l s thctho mn 1 p . Khi , ta nh ngha cc phim hm sau:

um,p = 0||m

Dupp1/p

, 1 p < ,

um,p = max0||mDu, p = .

Cc khng gian Hm,p(), Wm,p(), Wm,p0 () c xc nh nh sau:

Hm,p() = bao y ca {u Cm() : um,p < } theo chun m,p,

W

m,p

() = {u Lp

() : D

u Lp

(), 0 || m} vi chun m,p,Wm,p0 () = bao ng ca C

0 () trongW

m,p().

Nhn xt. (i) W0,p() = Lp(), 1 p .(ii) Khi 1 p < , C0 () tr mt trong Lp() nn

W0,p0 () = Lp(), 1 p < .

Ch , khi p = ni chung W0,0 () = L(). Chng hn, khi = R, d cf(x) = 1

L(R). Nhng,

f 1, C0 (R).

(iii) Wm,p0 () Wm,p(), 1 p .(iv) Wm

,p() Wm,p() Lp() = W0,p(), 0 m m, 1 p .(v) Khi Rn l tp m b chn, t php nhng Lp() Lq(), 1 q p, ta c

php nhng Wm,p() Wm,q(), 1 q p.

nh l 3.1. Wm,p

() l khng gian Banach.


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29 Bi3. Khng gian Sobolev Wm,p()

Chng minh. Ly {uk}k=1 l mt dy Cauchy trong Wm,p(). Ta phi chng minh:

u Wm,p

() ukm,p

u.Do {uk}k=1 l dy Cauchy trong Wm,p(), nn t nh ngha c {Duk}k=1 l dyCauchy trong Lp(), vi 0 || m. M Lp() l khng gian nn

u Lp() : Duk p u, 0 || m. (3.1)

Nu c Du = u(u = u0) theo ngha suy rng, ngha l

Du u, = 0 C0 () (3.2)

th

Du = u Lp(),Duk

p u, 0 || m,

hay

u Wm,p(),Duk

m,p u.

Nh vy ta ch cn phi chng minh (3.2). Vi mi k, 0 || m, C0 (), c|Du u, | | Du Duk, | + |Duk u, |

m p dng bt ng thc Holder

|Du Duk, | = |u uk, D | u ukpDp,|Du u, | Duk upp,

v C0 () : Dp C, 0 || m, nn

|Du u, | C(u ukp + Duk up).

Do , t (3.1) cho k tin ra v cng

Du u, = 0, C0 (), 0 || m.

Ch C0 () {u Cm() : um,p < } Wm,p(),

ta d c nh l sau


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nh l 3.2. Wm,p0 () Hm,p() Wm,p().K hiu

LpN =

p

j=1 Lp() (N = 0||m 1),vi chun c xc nh nh sau, u = (u1, . . . , uN) LpN,

u; LpN =

Nj=1

ujpp1/p

, 1 p < ,

u; LN = max1jNuj.D thy, nh x

P : Wm,p() LpNP u = (Du)0||m

l mt ng c n cu. Do , Wm,p() ng c ng cu vi khng gian con ngW = ImP ca LpN.M khng gian LpN l khng gian kh li khi 1 p < , l khng gian phn x, liu khi 1 < p < . c bit khi p = 2, khng gian L2N l khng gian Hilbert kh livi tch trong(hay tch v hng)

(u, v)m =

0jN

uj(x)vj(x)dx

nn ta c nh l sau.nh l 3.3. (i) Wm,p() l khng gian kh li khi 1 p < .

(ii) Wm,p() l khng gian phn x, li u khi 1 < p < .(iii) Khi p = 2, khng gian Wm() = Wm,2() l khng gian Hilbert kh li vi tch

trong

(u, v)m =

0||m

Du(x)Dv(x)dx.

3.2 i ngu ca khng gian Sobolev Wm,p()

B 3.4. (LpN) = LpN, 1 p < ,

ngha l

L (LpN)!v Lp

N

L(u) =

Nj=1

uj , vj =Nj=1

uj(x)vj(x)dx, (3.3)

L; (Lp

N

)

=

v; Lp

N. (3.4)


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Chng minh. Vi w Lp() k hiu w(j) = (0, . . . , w

j

, . . . , 0).

t Ljw = L(w(j)). T gi thit L (Lp

N)

, c Lj (Lp

())

.Khi , c duy nht mt hm vj Lp() sao cho Lj(w) = w, vj.t v = (v1, . . . , vN), c v LpN v vi mi u = (u1, . . . , uN) LpN

L(u) = L(Nj=1

uj(j)) =Nj=1

L(uj(j))

=

Nj=1

Lj(uj) =

Nj=1

uj , vj.

Nh

vy, c duy nht mt hm v Lp

N c (3.3).Ta ch cn phi chng minh (3.4). Theo bt dng thc Hoder, vi u LpN

|L(u)| = |Nj=1

uj , vj| Nj=1

ujpvjp

Nj=1

ujpp1/p N

j=1

vjpp1/p

u; LpN v; Lp

N

nnL; (LpN) v; Lp

N. (3.5)Khi 1 < p < , t

uj(x) =

|vj(x)|p2vj(x), nu vj(x) = 00, nu vj(x) = 0,

c

uj, vj = |vj(x)|p2vj(x)vj(x)dx = |vj(x)|pdx = vjpp,ujpp =

|vj(x)|(p1)pdx = vjp

p,

nn

u; LpN v; Lp

N = Nj=1

ujpp1/p N

j=1

vjpp1/p

=

N

j=1vjpp

=N

j=1uj, vj = L(u).


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Do , t (3.5), c (3.4) khi 1 < p < .Khi p = 1, p = , c

k

{1, . . . , N

}:

vk

= max

1jN

vj

=

v; Lp

N

,

> 0, A , (A) > 0 : |vk(x)| vk , x A.

Nu vk = 0 th v = 0. Do , t (3.3), L = 0 v L; (LpN) = 0 = v; Lp

N.Nu vk = 0, ly 0 < < vk. t

uj(x) =

vj(x)/|vj(x)|, nu x A0, nu x / A,

c

L(u(k)) = u, vk = A

(vj(x)/|vj(x)|)vk(x)dx

=

A

|vk(x)|dx (vk )(A),

u(k); L1N =A

vj(x)/|vj(x)|dx = (A),nn vi mi 0 < < vk : L(u(k)) (v; LN )u(k); L1N.Do , t (3.5), c (3.4) khi p = 1.

nh l 3.5. Vi 1 p < , cL (Wm,p()) v LpN

L(u) =

0||m

Du, v, u Wm,p(), (3.6)

L; (Wm,p()) = infv; LpN, (3.7)

trong , infimum ly, v t c, trn tt c cc hm v LpN tho mn (3.6).

Chng minh. tL(P u) = Lu,u Wm,p().

Do L (Wm,p()),v W = P(Wm,p()) l ng cu ng c vi Wm,p()nn

L W, L; W = L; (Wm,p()). (3.8)Theo nh l Hahn-Banach

L

(Lp

N

) : L|W = L

,

L; (Lp

N

)

=

L; W

. (3.9)


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Theo B 4, c duy nht mt hm v LpN sao cho

L(u) =

0||mu, v, u LpN, (3.10)

L; (LpN) = v; LpN. (3.11)

T (3.8), (3.9), (3.10) v (3.11), c mt hm v LpN sao cho vi mi u Wm,p()

L(u) = L(P u) = L(P u) =

0||m

Du, v, u LpN,

L; (Wm,p()) = v; LpN. (3.12)

Nh vy, ta ch cn phi chng minh nu v LpN tho mn (3.6) th

L; (Wm,p()) v; Lp

N.Tht vy, p dng bt dng thc Holder, vi mi u Wm,p()

|L(u)| =

0||m

u, v

0||m

Dup vp

um,pvm,p

nn L; (Wm,p()) v; LpN.Ch . Do vi L

W c th c nhiu thc trin L

(Lp

N) tho mn (3.9), nn vi mi

L Wm,p() c th c nhiu hm v LpN tho mn (3.6), (3.12).Tuy nhin, khi 1 < p < , th vi mi L Wm,p() c duy nht mt hm v LpN thomn (3.6), (3.12). Tng ng ny l mt ng cu ng c t(Wm,p()) vo Lp

N.

Tht vy, vi 1 < p < , LpN, Lp

N=(LpN) l cc khng gian li u, v khng gian conng W ca LpN cng li u. Khi , vi mi L

W c duy nht mt hm u Wsao cho

u; W = 1, L(u) = L; W.Nu c L, L (LpN) l cc thc trin bo ton chun ca L th

L(u) = L(u) = L; W = L; (LpN),L(u) = L(u) = L; W = L; (LpN),

nn

L + L2

; (LpN) L(u) + L(u)

2=

1

2(L; (LpN) + L; (LpN)).

iu ny ch xy ra khi L = L, v (LpN) l li u. Do , vi L W c duy nht mt

thc trin L (LpN) tho mn (3.9).Nh vy, c th coi (Wm,p()) l khng gian con ng ca khng gian Lp

N, khi 1 < p 1/k}, 0 = 1 = ,Uk = k+1 (k1)c.

C Uk v {Uk}k=1 l mt ph m ca nn t ch ta c th chn mt h{k}k=1 tho mn

(i) k C

0 (Rn

), 0 k(x) 1, x Rn

, k = 1, 2, . . . ,


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(ii) suppk Uk, k = 1, 2, . . . ,(iii) K ch c hu hn k sao cho k|K = 0,

(iv) k=1 k(x) = 1, x .Ly > 0. Vi k(chn sau), 0 < k < 1(k+1)(k+2) c

Jk (ku) Vk = k+2 (k2)c,supp(ku) Uk Vk ,

nn theo B 3.9 ta c th chn k sao cho

Jk (ku) kum,p, = Jk (ku) kum,p,Vk 0 nh

u 1,p

|u(x) (x)|pdx1/p

> , C1().


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iu ny l do khi i qua {x = 0, 0 < y < 1} hm u c gin on, cn hm C1()th lin tc. Hay do nm c v hai pha i vi mt phn ca bin {x = 0, 0 < y < 1}.Nh vy c th xp x khng gian Wm,p() bi Cm(), th t nht ch nm vmt pha i vi mt phn ca bin. C th l ta cn n khi nim sau.Tp m Rn c gi l c tnh cht segmentnu vi mi x c mt tpm Ux cha x, v mt vecto yx khc 0 sao cho

nu z Ux th z+ tyx , 0 < t < 1.

Nhn xt. Tp m c tnh cht segmentth phi c bin (n 1)chiu v khng thnm c hai pha i vi bt k phn no ca bin .

nh l sau khng nh tnh cht segmentl iu kin C0 (Rn) tr mt

trong Wm,p(), v do Cm() tr mt trong Wm,p().

nh l 3.11. Nu c tnh cht segmentth tp cc hn ch xung ca cchm thuc khng gian C0 (R

n) l tr mt trong Wm,p(), vi 1 p < .

Chng minh. Vi mi u Wm,p() t

Ku = {x : u(x) = 0},

u(x) = u(x), nu x ,0, nu x c.

Trc ht ta chng minh tp cc hm u Wm,p() c Ku b chn l tr mt trongWm,p().Ly f C0 (Rn) c nh tho mn

(i) f(x) = 1, |x| 1,(ii) f(x) = 0, |x| 2,

(iii) f(x) 0, 1 < |x| < 2.

Khi , c M > 0 sao cho

|Df(x)| M, x Rn, 0 || m.

Vi 1 > > 0 t f(x) = f(x) c

(i) f(x) = 1, |x| 1/,(ii) f(x) = 0,

|x|

2/,


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(iii) |Df(x)| M|| M, x Rn, 0 || m.Khi , theo cng thc Leibnitz

Df(x)u(x)

Du(x)Df(x) M

|Du(x)|nn fu Wm,p(). t = {x : |x| > } c

fu um,p, = fu um,p, um,p, + fum,p, Cum,p,

m lim0+

um,p, = 0, v u Wm,p(), nn

lim0+ fu um,p, = 0,Kfu {|x| < /2} l tp b chn.

By gi ta chng minh tp cc hn ch xung ca cc hm thuc C0 (Rn) tr mt

trong tp cc hm u Wm,p() c Ku b chn. Ly u Wm,p() c Ku b chn. tF = K (xUx)c l tp compact trong , trong Ux l cc tp m c xc nht tnh cht segmentca . Khi , c mt tp m U0 m

F U0 .

C K l tp compact mK U0 (xUx)

nn c hu hn cc im x1, . . . , xk sao cho

K U0 (kj=1Uxj ).

Ta chn U0 = U0 l tp m, compact tng i.Li c V1 = K

U0 (kj=2Uxj)

cl tp compact trong Ux1. Nn c mt tp m U1

sao choV1

U1 Ux1 .C th ta xy dng c mt h cc tp m {Uj}kj=0 sao choU0 = U0, Uj Uxj , 1 j k, K kj=0 Uj .Khi ta xy c mt h {j}kj=0 tho mn

(i) j C0 (Rn), 0 j(x) 1, x Rn, j = 0, 1, . . . , k ,(ii) suppj Uj, j = 0, 1, . . . , k ,

(iii) A ch c hu hn k sao cho j|A = 0,


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(iv)kj=0 j(x) = 1, x K.

t uj = ju, j = 0, 1, . . . , k . Ly > 0. Nu vi mi j, ta u chn c hm

C0 (R

n) sao cho

uj jm,p, < /(k + 1)th hm =

kj=0 j tho mn

u m,p, =kj=0

uj jm,p, .

C suppu0 U0 nn, theo B 3.8 v 3.9, vi > 0 nh hm cn tm l0 = J u0 C0 (Rn). Ta cn phi tm hm j vi 1 j k. C uj Wm,p(Rn\j)vi j = Uj . Ly yj l vecto c xc nh t tnh cht segmenti vi imxj v ln cn Uxj . t tj = j tyj vi t c chn sao cho

0 < t < min{1, d(Uj ,Rn\Uxj)/|yj |}.C tj Uxj v tj = (theo tnh cht segment). t utj(x) = uj(x + tyj). C

utj Wm,p(Rn\tj), v Dutj Duj trong Lp() khi t 0+, 0 || m.

Nn utj uj trong Wm,p() khi t 0+.Li c Uxj Rn\tj nn theo B 3.9 c

J utj utj trong Wm,p( Uxj ) khi 0+.

Do vi t > 0, > 0 nh th hm cn tm l j = J utj.T nh l trn, vi ch Rn tho mn tnh cht segment, ta c H qu sau.

H qu 3.12. Wm,p0 (Rn) = Wm,p(Rn).

3.3.3 Xp x bi cc hm C0

()

Trong mc ny, ta lun gi s rng 1 < p < , p l s lin hp ca p. Mt tp ngF trong Rn c gi l cha gi ca mt hm suy rng T D(), k hiu suppT F,nu T() = 0 vi mi C0 (Rn), |F = 0. Ta ni tp ng F l (m, p)polar nuch c duy nht mt hm suy rng T Wm,p(Rn) c gi trong F l T = 0.Ch . Nu F c o dng th n khng th l (m, p)polar. Thc vy, nu F c o dng th n cha mt tp con compact K c o dng. Khi , hm c trng 1Kca tp K thuc Lp

(Rn), v do thuc Wm,p

(), c supp1K F nhng 1K = 0.Nu mp > n th ch c duy nht tp rng l (m, p)polar. Tht vy, nu mp > n ta


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c php nhng lin tc Wm,p(Rn) C0 (Rn), theo ngha vi mi u Wm,p() c mtu0 C0 (Rn) m u(x) = u0(x)a.e, v

u0(x)

C

u

m,p,

x

Rn,

trong , hng s C khng ph thuc vo x v u. Nh vy, vi mi x Rn hm Diracx() = (x) thuc vo (Wm,p(Rn)) = (W

m,p0 (R

n))=Wm,p(Rn), c suppx = {x}.C php nhng Wm+1,p(Rn) Wm,p(Rn) nn Wm,p(Rn) Wm1,p(Rn), do mttp l (m + 1, p)polar th n cng la (m, p)polar.

Xt nh x thc trin u u c xc nh nh sau

u(x) =

u(x), nu x ,0, nu x

c.

B sau ch ra rng nh x thc trin ny l php nhng ng c t Wm,p0 () voWm,p(Rn).

B 3.13. Cho u Wm,p0 (). Khi , vi 0 || m c Du = Du theo nghasuy rng trong Rn. Do , u Wm,p(Rn).

Chng minh. T nh ngha, c mt dy {k}k=1 trong C0 () m k hi t n utrong Wm,p0 (), khi k

. Vi 0

|

| m,

C0 (R

n) cRn

u(x)D(x)dx =

u(x)D(x)dx

= limk

k(x)D(x)dx

= limk

(1)||

D(x)(x)dx

= (1)||

Du(x)(x)dx

= (1)|| Rn Du(x)(x)dx

nn, Du = Du theo ngha suy rng trong Rn. Do , um,p,Rn = um,p,.nh l sau cho thy khi no th nh x thc trin ny l ng c ln.

nh l 3.14. C0 () tr mt trong Wm,p(Rn) khi v ch khi c l (m, p)polar.


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Chng minh. Trc ht ta chng minh iu kin cn, ngha l gi s C0 () tr mttrong Wm,p(Rn), T Wm,p(Rn), suppT c, ta phi chng minh T = 0. Tht vy, lyu Wm,p(). T gi thit, tn ti mt dy {k}k=1 trong C0 () m k

m,p u, k .

Khi , do suppT c

nn

T(u) = limk

T(k) = 0

hay T = 0.

By gi ta i chng minh iu kin . T h qu ca nh l Hahn-Banach, chng minh C0 () tr mt trong W

m,p() ta ch cn chng minh nu T Wm,p

(), T|C0 () = 0 th T = 0. y chnh l nh ngha ca c l (m, p)polar chng minh nh l chnh ca mc ny ta cn mt s b sau.

B 3.15. Cho B = (a1, b1) (a2, b2) (an, bn) l mt hp ch nht m trongRn, v C0 (B). Nu

B

(x)dx = 0 th ta c phn tch (x) =nj=1 j(x) vi

j(x) C0 (B) sao cho bjaj

j(x)dxj = 0 (3.14)

vi mi xi R, i = j, i = 1, 2, . . . , n c nh.

Chng minh. Vi mi 1 j m chn uj C0 (aj, bj) sao cho bjaj uj(t)dt = 1.t

Bj = (aj, bj) (aj+1, bj+1) (an, bn),

j(xj, . . . , xn) =

b1a1

dt1

b2a2

dt2 bj1aj1

(t1, . . . , tj1, xj, . . . , xn)dtj1,

j(x) = u1(x1) . . . uj1(xj1)j(xj, . . . , xn).

C j C0 (B) v Bj

j(xj, . . . , xn)dxj . . . d xn =B

(x)dx = 0.

t 1 = 2, j = j j+1(2 j n 1), n = n. C j C0 (B), 1 j nv =

nj=1 j.


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Li c

b1

a1

1(x)dx1 =

=b1a1

(x)dx1 2(x2, . . . , xn)b1a1

u1(x1)dx1 = 0bjaj

j(x)dxj =

= u1(x1) . . . uj1(xj1)

bj

aj

j(xj, . . . , xn)dxj j+1(xj+1, . . . , xn)bjaj

uj(xj)dxj

= 0

bn

an

n(x)dxn = u1(x1) . . . un1(xn1)bn

an

(xn)dxn = 0

B 3.16. Nu T D(B), DjT = 0 j = 1, . . . , n , th tn ti hng s k sao cho

T() = k

B

(x)dx, C0 (B).

Chng minh. Trc ht ch rng, nu B (x)dx = 0 th theo b 3.15 trc ta cphn tch = nj=1 j vi j C0 (B), tho mn (3.14) v j = Djj vi j C0 (B)c xc nh bi

j(x) =

xjaj

j(x1, . . . , tj, . . . , xn)dtj.

Khi , T() =nj=1 T(j) =

nj=1 T(Djj) =

nj=1 DjT(j) = 0.

Nu T = 0 hin nhin ta c iu phi chng minh.Nu T = 0 th c 0 C0 (B) sao cho T(0) = k1 = 0. T ch trn, c

B

0(x)dx =

k2 = 0 v T(0) = kB

0(x)dx vi k = k1/k2. Vi mi C0 (B) t K() =

B(x)dx. C

B(x) K

k20(x)dx = 0.

T ch , c

T() =K

k2T(0) = k

B

(x)dx.

B 3.17. Nu u L1loc(),

u(x)(x)dx = 0, C0 () th u = 0a.e trong .


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Chng minh. Nu C0(), th vi > 0 nh c J thuc C0 () v J hit u n trong khi 0+. Do , t gi thit c

u(x)(x)dx = 0 vi mi

C0().Ly K

v > 0. T gi thit c

k=0

K{k|u(x)| 0 sao cho vi bt k tp con Aca K c (A) < th

A

|u(x)|dx < /2.Ta p dng nh l Lusin cho hm o c 1Ksignu trong tp K c o (K) < ,c C0() vi supp K v |(x)| 1, x K sao cho

{

x

: (x)= 1K(x)signu(x)

}=

{x

K : (x)

= 1K(x)signu(x)

}< .

Khi K

|u(x)|dx =

u(x)1K(x)signu(x)dx

=

u()(x)dx +

u(x)(1K(x)signu(x) )dx

2{x:(x)=1K(x)signu(x)}

|u(x)|dx .

Nh

vy u(x) = 0a.e trn K, v do trn .T hai B 3.16 v 3.17 trn ta d c

H qu 3.18. Nu u L1loc(B), Dju = 0, j = 1, . . . , n , th tn ti mt hng s k saocho u(x) = k a.e trong B.

nh l 3.19. (1) Nu Wm,p() = Wm,p0 () th c l (m, p)polar.

(2) Nu c l (1, p)polar v (m, p)polar th Wm,p() = Wm,p0 ().

Chng minh. (1) Gi s c Wm,p() = Wm,p0 (). Trc ht, ta chng minh (c) = 0.V nu khng ta lun chn c mt hp ch nht mB Rn sao cho (B ) > 0 v(B c) > 0. Ly u l hn ch xung ca mt hm v C0 (Rn) m v(x) = 1, x B . C u Wm,p() = Wm,p0 (). Theo B 3.13, c u Wm,p() v Dj u = Djutheo ngha suy rng trong Rn, vi 1 j m. M Dju|B = 0 nn Dj u|B = 0. Do ,theo H qu trn, u l hng hu khp trn B. iu ny tri vi gi thit u|B = 1, vu|Bc = 0. Nh vy c c o 0.Ly v Wm,p(Rn) v u = v| c u Wm,p() = Wm,p0 (). T B 3.13 cu Wm,p(Rn) c th xp x bng cc hm thuc C0 (). C Dv = Du trong , vimi 0 || m, v do D

v = D

u hu khp trong Rn

(v c

c o 0), vi mi


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0 || m. Nh vy, v c th xp x bi cc hm thuc C0 (), nn theo nh l3.14 c c l (m, p)polar.(2)Gi s c l (1, p)polar v (m, p)polar. Ly u Wm,p(), ta phi chng minhu

Wm,p

0(). V u

Lp() nn D

ju

W1,p(Rn) theo ngha

Dj u, v = limk

Dj u, k = limk

u, Djk

vi k C0 (Rn), k v trong W1,p(Rn) khi k . Li c Dju Lp(Rn) H1,p(Rn) nn Dj u Dju W1,p(Rn). M (Dj u Dju)| = 0 v c l (1, p)polarnn Dj u = Dju theo ngha suy rng trong Rn. Bng phng php quy np, ta cDu = Du theo ngha suy rng trong Rn, vi 0 || m. Do , u Wm,p(Rn).Theo nh l 3.14, do c l (m, p)polar, hn chu ca u xung thuc Wm,p0 ().Nu (m, p)polar suy ra (1, p)polar, th t nh l trn, (m, p)polar l iu kincn v W

m,p0 () = Wm,p(). Ta s i tm xem khi no iu xy ra. Trc

ht ta cn mt s B sau.

B 3.20. F Rn l (m, p)polar khi v ch khi F K l (m, p)polar vi mitp compact K Rn.

Chng minh. R rng l nu F l (m, p)polar th F K l (m, p)polar vi mitp compact K Rn. Do ta ch cn phi chng minh iu kin . Gi sT Wm,p(Rn) c suppT F. Theo nh l 3.5 tn ti v LpN(Rn) sao cho

T(u) = 0||m

Du, v, u Wm,p(Rn).

Ly c nh mt hm f C0 (Rn) tho mn(i) f(x) = 1, |x| 1,(ii) f(x) = 0, |x| 2,(iii) f(x) 0, 1 < |x| < 2.

Vi mi > 0 t f(x) = f(x). C Df(x) = ||Df(x) hi t u theo x

Rn

n 0 khi 0+, 1 || m. Do fT Wm,p

(Rn) v vi mi C0 (Rn) c|T() fT()| = |T() T(f)|

= 0||m

|Rn

vD

(x)(1 f(x))

dx

= 0||m

Rn

v(x)D(x)D(1 f(x))dx

0||m Rn|w(x)D(x)|dx m,pw; LpN


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trong

w(x) =

||m

vD

(1 f(x))

= v(x)(1 f(x)) ||m,=

vD

f(x),

m supp(1f) {|x| 1/} v Df(x) hi t u theo x Rn n 0 khi 0+, 1 || m nn lim

0+wp = 0. Do fT T trong Wm,p(Rn) khi 0+. Li c

supp(fT) {|x| 2/} l tp compact nn, theo gi thit (fT)F{|x|2/}

= 0, vF{|x| 2/} l (m, p)polar, c fT = 0 v do T = 0 hay F l (m, p)polar.

B 3.21. Nu p q v F Rn l (m, p)polar th F cng l (m, q)polar.

Chng minh. Ly K Rn l tp compact. Theo B 3.20 ta ch cn chng minhKF l (m, q)polar. Ly G Rn l tp m, b chn cha K. D c php nhng lintc Wm,p0 (G) Wm,q0 (G) nn Wm,q(G) Wm,p(G). Vi bt k T Wm,q(Rn)c suppT K G th T Wm,q(G) v do T Wm,p(G). M K F l(m, p)polar nn T = 0 hay K F l (m, q)polar.nh l 3.22. Vi

p 2th

W

m,p

0 () = W

m,p

()khi v ch khi

c l(m, p

)polar.

Chng minh. Do p 2 nn p p. Theo b 3.21, nu c l (m, p)polar th c l(m, p)polar v do cng l (1, p)polar. Nh vy theo nh l 3.19 ta c iu phichng minh.

Ch . T nh l nhng Sobolev, vi (m 1)p < n c php nhng lin tc

Wm,p(Rn) W1,q(Rn), q = npn

(m

1)p

nn W1,q

(Rn) Wm,p(Rn). Nu p 2n/(n + m 1) th q p, nn theo b 3.21,nu c l (m, p)polar th cng l (1, p)polar.Nu (m 1)p n th ch khi = Rn th c mi l (m, p)polar v do hin nhin l(1, p)polar.Nh vy nu p min{n/(m 1), 2n/(n + m 1)} th c l (1, p)polar khi c l(m, p)polar.


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Bi 4Bt ng thc Garding v bi ton Dirichlet i

vi phng trnh elliptic cp cao

4.1 Mt s k hiu v kt qu cn thit

Gi s l tp m, b chn trong Rn vi bin trn. Trong , cho ton t viphn cp r:

A(x, D) =||r

a(x)D,

trong a(x) C(), = (1, 2, . . . , n) l a ch s. Vi cc k hiu quenthuc:

|| = 1 + 2 + + n, Dj = i xj

, D = D1 . . . Dn ,

ta nh ngha ton tA0(x, D) =

||=r

a(x)D,

l phn chnh ca ton t vi phn A(x, D).

nh ngha 4.1. Ton t vi phn A(x, D) c gi l elliptic u trong nu tn tihng s dng 0 > 0 sao cho

Re(A0(x, )) = Re||=r

a(x) 0||r, x , Cn. (4.1)

Nu n > 2 th ton t elliptic A(x, D) c cp chn r = 2m. Vi n 2, iu nykhng ng, chng hn c ton t elliptic vi a thc c trng 1 + i2 nhng cpkhng chn. T by gi tr i, ta ch xt ton t elliptic c cp r = 2m:

A(x, D) =

||2m

a(x)D, a(x) C(). (4.2)

Ton t vi phn A(x, D) trong c gi l lin hp hnh thc ca ton t vi phnA(x, D) nu

(A(x, D)u, v) = (u, A(x, D)v)

u, v

C

0

(), (4.3)


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49 Bi4. Bt ng thc Garding v bi ton Dirichlet...

trong (, ) k hiu tch v hng trong L2().Nu A(x, D) l ton t elliptic u trong th A(x, D) cng l ton t elliptic u

trong .

K hiu H0

() = L2

() l khng gian cc hm o

c, bnh ph

ng kh tchtrong , tc l vi u L2() th

uL2() =

|u(x)|2dx1/2

< +.

Khi ta c: H0() l khng gian Hilbert vi tch v hng

(u, v)0 =

u(x)v(x)dx, u(x), v(x) H0().

Vi m nguyn dng, m

1, ta k hiu Hm() l khng gian cc hm u(x)

L2(),

c o hm suy rng n cp m thuc L2(), tc l

Hm() = {u : Du L2(), : || m}.Hm() l khng gian Hilbert vi tch v hng

(u, v)m =||m

(Du, Dv)L2(), u(x), v(x) Hm()

v chun:

um = ||m

|D|2dx1/2

, (u Hm()). (4.4)

Ta c: Hm1 Hm() L2() = H0() nu m1 > m v php nhng l tr mt vcompact.

Hm0 () l b sung ca C0 () theo chun (4.4):

um =||m

|D|2dx1/2 , u C0 ().

Khi Hm0 () l khng gian con ng trong Hm(), c tnh cht:

(i) Hm0 () =

u : u Hm(),

ju

j= 0, 0 j m 1

.

(ii) Hm0 () l khng gian cc hm u m c th thc trin khng ra ngoi thnhmt hm Hm(Rn), tc l nu u Hm0 (),

u =

u(x), x ,0, x / .

th u Hm

(Rn

).


56/144


(iii) nh l v vt: Gi s l min m b chn trong Rn c bin trn. Khi, nh x

u

ju

j, j = 0, 1, . . . , m 1

(trong u

l o hm theo php tuyn trong ca bin) tC() vo C()

c m rng thnh ton t tuyn tnh lin tc t Hm() vom1j=0

Hmj+1/2().

Hn na, l nh x ln, ng thi tn ti nh x tuyn tnh lin tc

g = {gj}m1j=1 Rg = um1j=0

Hmj+1/2() Hm()

sao cho j

jRg =

ju

j= gj , 0 j m 1.

K hiu Hm() l i ngu ca Hm0 (): Hm() = (Hm0 ())

.

V C0 () tr mt trong Hm0 () nn H

m() l khng gian cc hm suy rngtrong sao cho |(u, )| cm, C0 ().Ch rng nu u L2(), u xc nh mt phim hm tuyn tnh lin tc trnC0 (), do u Hm(). Bng cch ng nht L2() vi mt khng gian conca Hm(), ta c:

Hm0 () L2() Hm() (4.5)

trong php nhng Hm() L2() l lin tc v tr mt.

4.2 Bt ng thc Garding

Cho ton t elliptic n cp 2m trong

A(x, D)u = ||2ma(x)Du,trong a(x) C(). Khi , A(x, D)u c th vit di dng

A(x, D)u =

||m,||m

D(aDu), a C() (4.6)

Ta xc nh a(u, v) l dng song tuyn tnh

a(u, v) =

||m,||maD

u(x)Dv(x)dx (4.7)


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vi mi u(x), v(x) C0 ().Khi , a(u, v) c th m rng thnh dng song tuyn tnh lin tc trn Hm0 ().

Tht vy, ta c: vi mi u(x), v(x) C0 () th

|a(u, v)| ||m,||m

|Du(x)Dv(x)|dx

c ||m,|

|Du(x)|2dx1/2

||m,|

|Du(x)|2dx1/2

cuHm0 ()vHm0 ().

T suy ra a(u, v) c th m rng thnh dng song tuyn tnh lin tc trn Hm0 ().

Hn na, vi mi u, v

C0 (), p dng cng thc Green, ta c

a(u, v) = (A(x, D), v),

theo nh l Lax-Milgram, tn ti duy nht ton t lin kt ca dng song tuyn tnha(u, v):

A : Hm0 () Hm()sao cho:

a(u, v) = (Au,v) u, v Hm0 ().

iu ny c ngha l ton t vi phn

A(x, D) : C0 () C0 ()

c th m rng thnh ton t tuyn tnh lin tc A : Hm0 () Hm() xc nh bicng thc (4.10).

nh l 4.1 (Bt ng thc Garding). Gi s

A(x, D)u =

||m,||mD(aD

u)

l ton t elliptic u trong . Khi , tn ti cc hng sc1 > 0, c2 > 0 sao cho:

Re a(u, u) c1u2Hm0 () c2u2L2() u Hm0 (). (4.8)

Chng minh. Bt ng thc Garding i vi ton t ellitic cp hai c chng minhtrc y. Ta ch cn chng minh (4.8) vi u C0 ().

chng minh bt ng thc Garding (4.8), ta phi tin hnh theo tng bc sau

y:


58/144


Bc 1: Gi s A(x, D) l ton t elliptic thun nht cp 2m vi h s hng s:

A(x, D) = A(D) =

||=m,||=maD

+,

trong a l hng s.

Khi , ta c

a(u, u) =

||=m,||=m

aDuDudx, u, v C0 ().

Thc trin u(x) bng 0 bn ngoi , ta c th xem u C0 (). Khi ,

a(u, u) = ||=m,||=mRn aDuDudx

=

||=m,||=m

(2)nRn

a+|u()|2d.

V A(D) l ton t elliptic u, nn ta c:

Re

||=m,||=m

a++ 0||2m.

Do :

Re a(u, u) 0(2)n

Rn ||2m

|u()|2

d.

Ta bin i

||2m|u()|2 = (1 + ||2m)|u()|2 |u()|2

1 + ||2m

(1 + ||2)m (1 + ||2)m|u()|2 |u()|2

T , ta nhn cRe a(u, u) cu2m cu2Hm1.

Bc 2: Gi s u C0 (), supp u B(x0, ) l hnh cu tm x0, bn knh .Theo gi thit, a(x) C0 () nn a(x) lin tc u trong . Do , tn ti hmkhng ph thuc x0 v dn v 0 khi 0 sao cho vi mi ,

|a(x) a(x0)| () x B(x0, ).

K hiu a0(u, v) l dng song tuyn tnh nhn c bng cch thay cc hm a(x) bia(x0). Gi sA(x, D) l ton t elliptic thun nht cp 2m:

A(x, D) = ||m,||ma(x)D+u.


59/144


Khi ,

a(u, u) = a0(u, u) +

||m,||m

(a(x) a(x0))DuDudx.

Theo chng minh Bc 1, ta nhn c:

Re a(u, u) cu2Hm cu2Hm1 ()u2Hm.Chn b sao cho () < c

2. Khi , vi n C0 (), supp u B(x0, ), ta c

Re a(u, u) c2u2Hm cu2Hm1.

Bc 3: Gi s{Bj}j l ph hu hn ca gm cc hnh cu Bj bn knh v {j}l phn hoch n v ng vi ph {Bj} sao choj

2j(x) = 1, x , supp j Bj.Ta c

a(u, u) =j

||=m, ||=m

a(x)(jDu) (jDu)dx + a1(u, u).

Trong a1(u, u) gm nhng s hng ng vi , m t nht | hoc || m 1.Khi

|a1(u, u)| cuHm1 uHm.Khai trin D(ju) v D(ju) theo cng thc Leibniz, sau vit a(u, u) di dng

a(u, u) = j

||=m, ||=m

a(x)(jDu) (jDu)dx + a2(u, u),

vi a2(u, u) gm cc s hng cha Du v Du m | hoc || m 1. Khi |a2(u, u)| cuHm1 uHm.

p dng chng minh bc 2 ta c

Rea(u, u) c2

j

ju2Hm cuHm1uHm.

Ch rng j

ju2Hm =j

||=m

|D(ju)|2dx + a3(u)

=j

||=m

j|D(u)|2dx + a4(u)

=j

||=m

|D(u)|2dx + a4(u)

=

u2

Hm + a4(u).


60/144


Tnh ton tng t nh trn ta c

Rea(u, u) c2u2Hm cuHm1u2Hm, u C0 (). (4.9)

B 4.2 (Tnh li logarithm ca chun Sobolev). Vi s1 s s2, gi s [0, 1]sao cho s = s1 + (1 )s2. Vi mi u Hs2(Rn) ta cuHs uHs1 u1Hs2 . (4.10)

Hn na nu s1 < s2, vi mi > 0 tn ti C > 0 sao cho

uHs uHs2 + CuHs1 . (4.11)

Chng minh. Tht vy, ta c

uHs = (1 + ||2)s1|u()|2(1 + ||2)(1)s2|u()|2(1) d. (4.12)p dng bt ng thc Holder:

fg1

f

g

1, f, g 0, (4.13)

ta c

u2Hs u2Hs1u1Hs2 . (4.14) ta c bt ng thc cn chng minh.

chng minh phn hai ca b , ta ch rng

ab1 = (a)

/(1)b1 maxa,/(1)b

a + /(1)b.T ta suy ra

uHs uHs1 + /(1)uHs2 = uHs1 + CuHs2 . (4.15)

p dng b vi s1 = 0, s2 = m, s = m

1, = /4c ta nhn c

cuHm1uHm c4u2Hm + C1uHmuL2. (4.16)

Cui cng p dng bt ng thc Cauchy vi = c/8c1, a = uHm, b = uL2 ta cVy ta nhn c bt ng thc Garding vi c1 = c/8:

Rea(u, u) u2Hm 3c

8u2Hm c2u2L2

c8u2Hm c2u2L2 = c1u2Hm c2u2L2.

Ta c iu phi chng minh.


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4.2.1 Bi ton Dir ichlet

Gi s A(x, D) l ton t elliptic cp 2m

A(x, D) = ||m,||m

D

(aD

u)

trong min b chn Rn vi bin trn, a C(). Khi , ta c bt ngthc Garding:

Re a(u, u) + c2u2L2() c1u2Hm0 (), u Hm0 ().

p dng h qu ca nh l Lax-Milgram, ta c nh l sau y:

nh l 4.3. Gi s > c2. Khi , vi mi f(x)

L2(), tn ti duy nht nghim

ca bi ton

(A + )u = f

u Hm0 ()

trong nghim u Hm0 () ca bi ton c hiu theo ngha suy rng:

(u, Av + v) = (f, v), v C0 ().

vi A l ton t vi phn lin hp hnh thc ca A.

Chng minh. K hiua1(u, v) = a(u, v) + (u, v)

trong a(u, v) l dng song tuyn tnh trong Hm0 () xc nh bi (4.7). Theo btng thc Garding, ta c:

Re a1(u, u) = Re a(u, u) + u2 c1u2Hm0 (), u Hm0 ().

Mt khc, ta li c:

|a1(u, v)| cuHm0 ()vHm0 (), u Hm0 ().

Gi sf(x) L2(). Khi , vi mi v Hm0 (), ta c

|(f, v)| f0v0 cvHm0 ().

iu c ngha (f, v) l phim hm tuyn tnh b chn trong Hm0 (). Theo nh lLax-Milgram, tn ti phn t u0 Hm0 () sao cho

a1(u0, v) = (f, v), v Hm

0 ().


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Ch rng vi mi v C0 () tha1(u0, v) = a(u0, v) + (u0, v) = (Au0, v) + (u0, v)

= (u0, Av + v).

Do , ta c(u0, A

v + v) = (f, v), v C0 ().Gi s nu u0 l mt nghim khc ca bi ton

(u0, Av + v) = (f, v), v C0 ().

t = u0 u0. Khi , a1(, ) = 0 v Hm0 (). T , a1(, ) = 0 nn = 0.Vy nghim u0 ca bi ton Dirichlet

(A + )u0 = f, u0 Hm0 ()l duy nht.

By gi, gi thit f L2(), hj Hmj1/2(), (j = 1, 2, . . . , m 1). Khi ,tn ti duy nht h Hm() sao cho:

jh

j= jh = hj, (j = 1, 2, . . . , m 1).

t f = f (A + )h Hm(), > c2. Theo h qu ca nh l Lax-Milgram,

tn ti duy nht nghim v ca bi ton:(A + )v = f

v Hm0 ().Khi , u = v + h l nghim duy nht ca bi ton:

(A + )u = f

u Hm()ju

j

= ju = jv + jh = hj, (j = 1, 2, . . . , m

1).

Vy ta c nh l:

nh l 4.4. Gi s > c2. Khi , vi mi f L2(), hj Hmj1/2(), (j =1, 2, . . . , m 1), tn ti duy nht nghim suy rng ca bi ton:

(A + )u = f trong

u Hm()ju

j= hj trn , (j = 1, 2, . . . , m 1).


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Bi 5L thuyt Fredholm-Riesz-Schauder v bi ton

Dirichlet thun nht vi phng trnh elliptic

cp cao

5.1 L thuyt Fr edholm-Riesz-Schauder

Gi s H l mt khng gian Hilbert thc. Theo truyn thng, ta k hiu (, ) v l tch v hng v chun trong khng gian H. Trong phn sau y ta s s dnghai kt qu quan trng c chng minh trong cc gio trnh gii tch hm

Mnh 5.1. Mi khng gian con hu hn chiu ca khng gian Hilbert H u ng.

Chng minh. Gi s H1 l khng gian con hu hn chiu ca H, x1, x2, . . . , xn l cs trc chun trong H1. Khi , mi u H1 u c biu din

u = 1x1 + 2x2 + + nxn,

trong j = (u, xj), j = 1, 2, . . . , n. Nu uk = k1x1 + k2x2 + + knxn l dy hi

t trong H1 n u, th tc lng

|kj lj| = |(uk ul, xj)| uk ulxj k,l+

0

ta suy ra dy {kj }+k=1, (j = 1, 2, . . . , n) hi t trong R, tc l

limk+

kj = 0j , j = 1, 2, . . . , n.

t u0 = 01x1 + 02x2 + + 0nxn H1. Khi uk u0 trong H. T tnh duy nht

ca gii hn ta suy ra u = u0 H. Vy mi dy {uk} H1 hi t n u th u H1.Do H1 l khng gian con ng ca H.

Mnh 5.2. Khng gian con H1 trong H hu hn chiu khi v ch khi mi dy b

chn trong H1 u cha mt dy con hi t.


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58 Bi 5. L thuyt Fredholm-Riesz-Schauder v bi ton Dirichlet...

Chng minh. Mnh 5.2 c suy t nh l: Khng gian compact a phng thhu hn chiu trong l thuyt gii tch hm.

Gi s T l ton t tuyn tnh hon ton lin tc trong H, c ngha l nu {uk} ldy b chn trong H th dy {T uk} cha dy con hi t. K hiu L = I T, trong I l ton t n v trong H, v N = ker L = {u H : Lu = 0}.Mnh 5.3. Khng gian con N ca khng gian H l hu hn chiu.

Chng minh. R rng N l khng gian con ca H. Gi s{uk} l dy b chn trongN, ta c Luk = 0, k = 1, 2, . . . . Khi T uk = uk. V T l ton t hon ton lin tc

nn trch c mt dy con hi t {T ukl}+l=1 . V T ukl = ukl nn {ukl} l dy con hi

t ca dy b chn {uk} trong N. T mnh 5.2 ta suy ra N hu hn chiu.K hiu M = {u H : (u, N) = 0}. R rng M l khng gian con ng trong H.

Mnh 5.4. Tn ti hng s c0 > 0 sao cho

T u c0u, u H.

Chng minh. Gi s ngc li khng tn ti hng s c0

nh vy. Khi trong H tnti dy {uk} sao cho uk = 1, T uk +. Nhng iu khng th v T l tont hon ton lin tc.

Mnh 5.5. Tn ti hng s c1 > 0 sao cho

c11 u Lu c1u, u M.

Chng minh. Bt ng thc bn phi c suy trc tip t mnh 5.4. Gi s bt

ng thc bn tri khng tho mn. Khi tn ti dy {uk} M sao cho uk = 1v Luk 0 khi k +. V T l ton t hon ton lin tc v {uk} l dy b chnnn {T uk} cha mt dy con hi t {T ukl}.

Mt khc ta c ukl T ukl = Lukl , hay ukl = T ukl + Lukl . V dy {Lukl} hi t v0 v dy {T ukl} hi t, nn dy {ukl} hi t. Gi s

limk+

ukl = u trong H.

Theo mnh 5.4,

T(ukl

u)

c0

ukl

u

.


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Cho l +, ta c T ukl T u trong H. Vy t ng thc ukl = T ukl + Lukl , chol + ta s c u = T u, hay Lu = 0. iu ny c ngha l u N. V N v Mtrc giao nn

(u, v) = 0

v

M.

Mt khc M l khng gian con ng v {ukl} l dy con hi t nn liml+

ukl = u M.T suy ra u = 0. iu ny l v l v ukl = 1 v lim

l+ukl = u = 0. Mnh

c chng minh xong.

Vi mi v c nh, t mnh 5.4 ta c

|(v , T u)| vT u cu u H.V vy (v , T u) l phim hm tuyn tnh b chn trong H. Theo nh l Frchet-Riesz,

tn ti phn t g H sao cho(v , T u) = (g, u), u H.

t g = Tv, trong T : v Tv = g H, l ton t tuyn tnh trong H, tho mniu kin

(v , T u) = (Tv, u), u, v H.T c gi l ton t lin hp ca T. Hn na, T = T.

Mnh 5.6. Ton t T l ton t hon ton lin tc.

Chng minh. Gi s{uk} l mt dy b chn trong H, tc l uk C, k = 1, 2, . . . .Khi

Tuk2 = (Tuk, Tuk) = (uk, T Tuk) ukT Tuk c0cTuk

T ta c Tuk c0c, tc l dy {Tuk} l dy b chn. V T l ton t hon tonlin tc nn tn ti dy con hi t {T Tukl}. Do

T(ukl ukr )2 = (T(ukl ukr ), T(ukl ukr))= (ukl ukr , T T(ukl ukr)) 2cT T(ukl ukr) 0 khi l, r +.

Vy {Tukl} l dy hi t. T suy ra T l ton t hon ton lin tc.nh l 5.7. Phng trnh

v Tv = f (5.1)

c nghim khi v ch khi f M.


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Chng minh. Nu phng trnh (5.1) c nghim th vi mi w N,

(f, w) = (v T

v, w) = (v, w) (T

v, w)= (v, w T w) = (v,Lw) = 0,

t suy ra f M. Ngc li gi thit f M. Theo mnh 5.5, Lu v u lcc chun tng ng trong M. Do , vi u M, ta c

(u, f)| uf cu,

c ngha l (u, f) l phim hm tuyn tnh b chn trong M. V M l khng giancon ng trong H nn M l khng gian Hilbert. V vy theo nh l Frchet - Riesz,

tn ti phn t w M sao cho (u, f) = (Lu, Lw), u M. Ta s chng minh rng(u, f) = (Lu, Lw) trong H. Tht vy, trc ht ch rng khng gian con N = ker Lhu hn chiu nn N l ng trong H. V vy vi mi u H, ta c u = u + u, trong u M, u N. T , ch rng Lu = 0 v (u, f) = 0 ta s c

((Lu, Lw) = (Lu, Lw) = (u, f) = (u, f) + (u, f) = (u, f).

Vy,(u, f) = (Lu, Lw), u H.

Gi sv = Lw. Khi (Lu,v) = (u, f),

u

H. T (u, v

Tv) = (u, f),

u

H.

V vyv Tv = f.

K hiu L = I T, N, M l cc khng gian i ngu ca N v M. Khi tanhn c

H qu 5.8. Phng trnh Lu = f c nghim khi v ch khi f M.

Chng minh. Ta ch rng T l ton t hon ton lin tc v T = T, do h quc suy trc tip t nh l 5.7

Ta nh ngha ton t Ln(n = 1, 2, . . . ) bng quy nap nh sau;

L1 = L, Ln+1 = L Ln,Nn = ker L

n = {u H : Lnu = 0}.

Mnh 5.9. Ton tLn c th biu din di dng Ln = I Tn, trong Tn l tont hon ton lin tc.


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Ln = (I T)n =n=0

Cn(1)T

= I (nT C2nT2 + + Tn) = I Tn,trong Tn = nT C2nT2 + + (1)nTn l ton t hon ton lin tc.H qu 5.10. Mi khng gian con Nn trong H u l hu hn chiu.

Chng minh. Suy trc tip t mnh 5.3, 5.9

Mnh 5.11. Nn Nn+1.

Chng minh. T nh ngha ca khng gian Nn ta suy ngay ra iu phi chng minh.

Mnh 5.12. Tn ti s nguyn dng k sao cho Nn = Nn+1 vi n < k v Nn = Nkvi n > k.

Chng minh. Trc ht ta chng minh rng: Nu Nn = Nn+1 th Nn+2 = Nn. Gi su Nn+2. Khi Ln+1Lu = Ln+2u = 0. V vy Lu Nn+1 = Nn. Nhng iu c ngha l LnLu = 0, tc l u Nn+1 = Nn. Vy Nn+2 Nn+1 = Nn. Mtkhc theo mnh 5.11, Nn = Nn+1 = Nn+2. Vy Nn+2 = Nn. Ta gi thit ngcli rng mnh 5.12 khng ng, tc l Nn = Nn+1 vi mi n. Theo nh l vphp chiu trong H, tn ti dy {un} Nn, un = 1 sao cho (un, Nn) = 0. Ta cT(un um) = un (Lun + T um). Nu n > m th

Ln(Lun + T um) = Ln+1un + T L

num = 0.

T suy ra Lun + T um Nn vT(un um)2 = un2 + Lun + T um2 1.

V vy dy {T un} khng cha dy con hi t, iu ny mu thun v T l ton t honton lin tc v {un} l dy b chn. Mnh c chng minh.K hiu T l min gi tr ca ton t L, cn R l min gi tr ca L, tc l

R = {v H : u H,Lu = v},R =

{v

H :

u

H, Lu = v

}.


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Mnh 5.13. Nu R = H th N = {0}.

Chng minh. Gi thit R = H v tn ti u0 = 0 sao cho Lu0 = 0. V R = H, ta gi su1, u2, . . . , un ln lt l nghim ca phng trnh

Lu1 = u0, Lu2 = u1, . . . , Lun = un1

nhng khi Lnun = u0 = 0 v Ln+1un = Lu0 = 0. Nh vy un / Nn v un Nn+1.iu ny c ngha l Nn+1 = Nn vi mi n. iu ny mu thun vi mnh 5.12.nh l 5.14. R = H khi v ch khi N = {0}.

Chng minh. Theo mnh 5.13, nu R = H th N = {0}. Ta ch cn chng minhrng nu N = {0} th R = H. Tht vy, nu N = {0} th M = H v R = H v theonh l 5.7, vi mi f H , tn ti nghim v ca phng trnh Lv = (I T)v = f.

Mt khc, theo mnh 5.6, T l ton t hon ton lin tc, min gi tr caL = I T trng vi H nn N = {0}. iu c ngha l M = H. Theo hqu ca nh l 5.7 th vi mi f H, tn ti u H sao cho Lu = f, chng t rngR = H. iu ny kt thc chng minh.

nh l 5.15. Cc khng gian con ker L = N v ker L = N c s chiu hu hn v

bng nhau, tc ldim N = dim N < +.

Chng minh. Theo mnh 5.3 v 5.6 ta suy ra N v N l hu hn chiu. Gi sdim N = n, dim N = m. Gi thit m > n. Ta s chng minh rng iu ny l khngth xy ra.

K hiu u1, u2, . . . , un l c s trc chun trong N, v1, v2, . . . , vm l c s trcchun trong N, ta c

(uj, ui) = ij, (vj, vi) = ij.t

Su = T u ni=1

(ui, u)vi, S : H H,

W u = u Su = u T u ni=1

(ui, u)vi.

Khi S l ton t hon ton lin tc. Tht vy, ta c ton t S T nh x mi dy bchn trong H thnh dy b chn trong khng gian con N

hu hn chiu. Theo mnh


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5.2 th dy b chn trong N cha dy con hi t. T suy ra: nu {uk}k=1 l dyb chn trong H th dy {Suk}k=1 = {T uk

ni=1

(ui, uk)vi}k=1 cha dy con hi t.Nu W u = 0 th

0 = (vj, W u) = (vj, Lu) +ni=1

(ui, u)(vi, vj)

= (Lvj, u) + (uj, u) = (uj, u),

(j = 1, 2, . . . , n).

V vy ta c

0 = W u = Lu +ni=1

(ui, u)vi = Lu.

iu ny c ngha l ker W ker L. Nh vy nu u ker W th (u, uj) = 0, (i =1, 2, . . . , n, tc l uN = ker L, nhng u ker L nn ta suy ra u = 0, v phng trnhW u = 0 c nghim duy nht u = 0. Theo nh l 5.14, phng trnh W u0 = un+1 cnghim u0, nhng

(vn+1, W u0) = (vn+1, Lu0) +ni=1

(ui, u0)(vn+1, vi) = 0,

hay l (vn+1, vn+1) = 0. iu ny l v l v vn+1 = 1. Trng hp m < n ta kho

st t

ng t bng cch xt ton t S

v = T

v +

mj=1(vj, v)uj. Vy m = n.nh l 5.16. Phng trnh u T u = 0 c nghim khng tm thng ch i vi mttp hp m c s thc khng c im gii hn hu hn.

Chng minh. Ta s chng minh rng khng th tn ti mt dy b chn {n}nN vin = m, n = m, n, n = 1, 2, . . . , sao cho phng trnh

u

nT u = 0

c nghim u = 0. Ta s chng minh bng phn chng. Gi s un = 0 l nghimphng trnh

un nT un = 0, (n = 1, 2, . . . ).Khi vi mi n, cc phn t u1, u2, . . . , un l c lp tuyn tnh. Tht vy, ta schng minh iu ny bng quy np. Gi s u1, u2, . . . , uk1 l c lp tuyn tnh, v

c1u1 + c2u2 + + ck1uk1 + ckuk =k

j=1cjuj. (5.2)


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Ta c

0 = k

k

j=1cjT uj =

k

j=1kcj

juj. (5.3)

Tr hai v ca hai ng thc trn ta c

k1j=1

1 k

j

cjuj = 0. (5.4)

V {uk} c lp tuyn tnh, nn t ta suy ra cj = 0 vi j = k. Do t (5.2) ta cckuk = 0. V uk = 0 nn ck = 0, v do u1, u2, . . . , un c lp tuyn tnh.

K hiu En l khng gian sinh bi h vect c lp tuyn tnh u1, u2, . . . , uk. Theonh l v php chiu: Tn ti cc phn t vn En sao cho vn = 1 v (vn, En1) = 0.Ngoi ra nu

En th

nT

En1. Tht vy, vi

En th

=nj=1

juj,

v

nT =nj=1

juj nnj=1

jujj

=

n1j=1

1 n

j

juj.

Do ta c nT En1.Nu {n} l dy b chn th {nvn} l dy b chn trong H, do dy {T(nvn}

cha mt dy con hi t. Tuy nhin vi n > m,T(nvn mvm) = vn (vn nT vn + mT vm),

vi ch rng vn nT vn v T vm = vmm thuc En1 ta cvn nT vn + mT vm En1,

t suy ra(vn, vn nT vn + mT vm) = 0.

Do , T(nvn mvm)2 = vn2 + vn nT vn+ mT vm2 1. T ta c khngnh rng dy

{T(nvn)

}khng th cha mt dy con hi t.

Vy khng th tn ti dy cc gi tr phn bit {n}n=1, n = m, n, m.

5.2 p dng l thuyt Fr edholm - Schauder vo bi ton Dir ichlet thun

nht i vi phng trnh elliptic cp cao

Gi s l tp m, b chn trong Rn vi bin trn .

A(x, D) = ||2ma(x)D


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l ton t elliptic cp 2m, a(x) C().Bi ton Dirichlet. Cho f(x) L2(). Tm nghim ca bi ton

A(x, D)u = f(x) trong ,juj

= 0 trong (j = 0, 1, . . . , m 1). (D)

trong

l o hm theo php tuyn trong ca .

Theo nh l 5.7, vi 0 ln, f(x) L2(), tn ti duy nht nghim u0 Hm0 ()ca bi ton:

(A + 0)u = f(x) trong ,

u u0 Hm0 ()

sao cho a1(u0, v) = (u0, Av + 0v) = (f, v), v C0 ().T c lng

|a1(u0, u0)| Re a1(u0, u0) + Re a(u0, u0) + 0u02 c1u02Hm0 (),

ta cc1u02Hm0 () |a1(u0, u0)| = |(f, u0)| f0u00.

Do ,

u0

2

Hm

0 () c

f0

.

t u0 = (A + 0)1f, ta c

(A + 0)1f2Hm0 () cf0.

Nh vy, (A + 0)1 : L2() Hm0 () l nh x lin tc. V Hm0 () L2() lnh x nhng compact. Do c th coi (A + 0)1 l ton t hon ton lin tctrong L2(). iu ny cho php ta p dng l thuyt Fredholm - Schauder cho bi tonDirichlet i vi cc phng trnh

Au = f, (5.5)Av = f, (5.6)

Au = 0, (5.7)

Av = 0. (5.8)

K hiuT = (A + 0)

1 : L2() L2()l ton t hon ton lin tc. Tng t,

T

= (A

+ 0)1

: L2

() L2

()


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cng l ton t hon ton lin tc.

Hn na, ta ch rng

Au = 0

0u = (A + 0)u

0(A + 0)

1u = u.

Nh vy, phng trnh Au = 0 c vit di dng tng ng

u 0T u = 0. (5.7 )

Tng t, phng trnh Av = 0 c vit di dng tng ng

v 0Tv = 0. (5.8 )

Cc phng trnh (5.5) v (5.6) c vit di dng tng ng

u 0T u = T f, (5.5 )v 0Tv = Tf, (5.6 )

trong T v T l cc ton t hon ton lin tc trong L2(). Ngoi ra, ta ch Tv T l cc ton t lin hp trong L2(). Tht vy, ta c (5.7 )

(Au,v) = (u, Av) u, v C0 ().

Do , vi u, v C0 () th

(T u , v) = (T u, (A + 0)Tv)

= (Tu , ATv + 0Tv) = (Tu , ATv) + (Tu , 0T

v)

= (AT u + 0T u , T v) = (u, Tv).

iu chng t T l ton t lin hp ca T. p dng l thuyt Fredholm - Schauder,ta c nh l sau y

nh l 5.17. Cc phng trnh thun nht

Au = 0 v A

v = 0

c cng mt s hu hn nghim c lp tuyn tnh.

Chng minh. Tht vy, cc phng trnh (5.7) v (5.8) tng ng vi cc phngtrnh (5.7 ) v (5.8 ), trong T l ton t compact trong L2() v T l ton t linhp ca n. Theo l thuyt Fredholm-Schauder, cc phng trnh (5.7 ) v (5.8 ) ccng mt s hu hn nghim c lp tuyn tnh. T , ta suy ra kt qu cn chngminh.


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nh l 5.18. Phng trnh Au + u = 0 c nghim khng tm thng ch i vi mttp m c cc gi tr khng c im gii hn hu hn.


Au + u = 0 (A + 0)u + ( 0)u = 0 u + ( 0)T u = 0

Phng trnh Au + u = 0 tng ng vi phng trnh

u + T u = 0, = 0trong T l ton t hon ton lin tc trong L2().

Theo nh l Fredholm, phng trnh u + T u = 0 c nghim khng tm thngi vi mt tp hp m c cc gi tr , khng c im gii hn hu hn. T ,ta suy ra kt qu cn chng minh.

nh l 5.19. Phng trnh Au = f gii c khi v ch khi (f, v) = 0 vi mi nghimv ca phng trnh Av = 0.

Chng minh. Vi f L2(), phng trnh Au = f c vit di dng tng ng lphng trnh (5.5 ), trong T = (A + 0)1 l ton t hon ton lin tc trong L2().

Theo nh l Fredholm, phng trnh (5.7 ) gii c khi v ch khi (T f , v) = 0 vimi nghim v ca phng trnh v 0Tv = 0. Do :

(f, v) = 0(f, Tv) = 0(T f , v) = 0.

T , ta suy ra: phng trnh (5.5 ) gii c khi v ch khi (f, v) = 0 vi mi nghimv ca phng trnh (5.8 ).

V Av = 0 khi v ch khiv

0T

v = 0.

Do , phng trnh (1) khi v ch khi (f, v) = 0 vi mi nghim v ca phng trnhAv = 0.

H qu 5.20. Phng trnh Au = f gii c vi mi f L2() khi v chi khi phngtrnh Av = 0 chi c nghim tm thng.


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Bi 6Ph ca ton t Elliptic

6.1 Bi ton Dir ichlet i vi phng trnh Laplace

6.1.1 Gii thiuGi s l tp m b chn vi bin trong Rn, C0 () l khng gian cc hm

kh vi v hn vi gi compact trong .

Trong C0 () ta ch mt bt ng thc quan trng l bt ng thc Poincar:

u2L2() ku2L2(), k > 0, u C0 ()

Trong u l vect gradient: u :=ux1

, ux2

, . . . , uxn

. Nh c bt ng thc

Poincar, trong C0 () ta c th xc nh hai chun tng ng:

u =

|u|2dx1/2 , u C0 ()v

u =

|u|2 + |u|2dx1/2

, u C0 ().

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Phuong Trinh Vi Phan_Dao Ham Rieng