Elliptic Problems in Nonsmooth Domains || 3. Second-Order Elliptic Boundary Value Problems in Convex Domains

3

Secondworder ellipticboundary value problems inconvex domains

3.1 A priori estimates and the curvature of the boundary

One of our basic tools throughout Chapter 2 has been the a prioriinequality (2,3,3,7) proved in Theorem 2.3.3.6. In the present chapter wepropose an alternative proof of this inequality in the particular case whenp =2 and when the function u E H2(f) under consideration fulfils thehomogeneous boundary condition

yBu=O.

We shall mainly consider boundary value problems for the Laplaceoperator (in order to avoid some extra technical difficulties). However, weshall allow some nonlinear boundary conditions.

The main idea of the forthcoming alternative proof is to bypass the useof local coordinates. These were used in Section 2.3 to reduce theproblem to the case when the boundary I , of the domgin D is flat. Herewe shall perform straightforward integration by parts to prove the in-equality

1U112,2,n^ C(f2 ) LLAuI1o,2,n (3,1,1)

for all u e H2(Q) such that yBu =0 on T. The constant C(Q) takes intoprecise account the curvature of T.

This inequality has various applications, all of which are along thefollowing lines. We shall consider very rough domains fl, such as generalconvex domains or domains whose boundary has turning points. We shallapproximate these domains by sequences of domains with a C 2 boundaryfor which the constant in inequality (3,1,1) can easily be controlled.Taking the limit will prove smoothness results for the solution of aboundary value problem in f, although D is far from having a C 2

boundary.

132

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3.1 CURVATURE OF THE BOUNDARY 133

3.1.1 An identity based on integration by parts

In this section we shall consider a bounded open subset D of or with aC 2 boundary F together with its second fundamental quadratic form,denoted by 99. Let us recall briefly an elementary definition of 3L For thatpurpose, let P be any point on T. It is possible to find n —1 curves of classC 2 in a neighbourhood of P, passing through P, and being orthogonalthere. Let us denote by '9, , ... , 'g n _ 1 those curves, by z 1 , ... , zn _ 1 theunit tangent vectors to , ... ,'g respectively, and by s 1 , ... , sn _ 1 thearc lengths along 1 , ... , <Pn _, respectively. We can assume that{T1, ... , Tn _ 1} has the direct orientation at P.

Then, at P, 93 1 is the bilinear form (we shall often drop the subscript P.and write Ok instead of P f whenever this does not lead to any misunder-standing)

av— • Tkj lik

i.k asi

where and iq are the tangent vectors to T at P, whose components areand { -r 1, ... , -r1n _ 1}, respectively, in the basis {r 1 , ... , Tn --1}•

In other words, we have

where aIa denotes differentiation in the direction of g. (Actually, wecould also extend the definition of 93 to sets D with a C" 1 boundary withjust a little more extra work. All the subsequent results hold for domainswith a C' -1 boundary instead of C 2 .)

Another point of view is this. Let us consider a point P of F and(according to Definition 1.2.1.1) related new coordinates {y,, ... , y,} withorigin at P as follows: there exists a hypercube

V={(y1,..., Yn)1 —ai <yi <a1, 1_jEn}

and a function cp of class C 2 in V', where

V ={(y 1 ,...,Yn - 1 ) — a;<y;<a;, 1<j_n-1}

such that

l co (y') j an/2 for every y' E f''

Q(1 V={Y=(Y', Yn)E V 1 Yn<^P(Y')}

FnV={y=(Y', Yn)E V1 Yn =^P(Y')}.

Let us assume further that V p(0) = 0. This means that the new coordi-nates have been chosen in such a way that the hyperplane y n = 0 isD

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134 SECOND-ORDER PROBLEMS IN CONVEX DOMAINS

tangent to r at P. Then, it is easily checked that the form 99 isn-1 a2^

(0)k1;,;,k =1 aYk ay;

where {^1, ... ,n _ 1} and {111, ... , 71n -1} are the components of t and inrespectively in the directions of { Y 1, ... , Yn _ 1}. In particular, when d2 isconvex, the function —p is also convex, and consequently the form !$ isnonpositive.

Let us observe, finally, that in the general case of a domain fl with aC 2 boundary the form .B is uniformly bounded on T. In other words,there exists K such that

^n)I K Iti nfor all P E T, where g and 11 are tangent vectors to T at P. Indeed, v is aC' vector field on T. This is why domains with a C 2 boundary (or evenC 1 ' 1 ) are said to have a boundary with bounded curvature.

Now we introduce some more notation. Let v be any vector field on T;we shall denote by v„ the component of v in the direction of v, while weshall denote by VT the projection of v on the tangent hyperplane to T. Inother words

v„ = v • v and VT=V—VVV.

In the same way, we shall denote by VT the projection of the gradientoperator on the tangent hyperplane:

au0TU =VU — v.

av

We can now state the following.

Theorem 3.1.1.1 Let Dl be a bounded open subset of ll with a C 2

boundary, and let v E H'(,f2)n. Then, we have

J n av• avIdiv v12 dx — E' dx

j,,= t JQ ax; axj

_ —2((yv)T; VT[YV • v]) { ((yv)T; (YV)T)r

+ (tr 93)[(yv) • v]2} dc. (3,1,1,1)

Here tr 9B is the trace of the bilinear form 93, i.e.,

n-' avtr24=—^ •z;

=1 ast

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Proof First, we apply repeatedly the Green formula of Theorem 1.5.3.1which holds as soon as I' is Lipschitz. We thus get, for v E C 2(f2)n,

J navi avidiv vit dx = ' dx

n i,j=1 n axi axj

n a2Uj_ — v, ndx + v; ay, vt dar= 1 axi ax1 i,j = 1 r axj

= n ave avj dx + n v • av't v dert

i,j=1 n axj axi i,j=l r 3x1

n— 1 v^ a,Uj v; dar.i.j=1 r ax^


't av^ avI(v) _ (div v( 2 dx — ` ' dx

S2 i, j =1 n axj axi

= V. div v do* — {(v • V)v} • v da - . (3,1,1,2)r r

We shall now transform the integrand on the boundary. This can bedone locally. Let us consider any point P on I' and choose an openneighbourhood W of P in T small enough to allow the existence of (n — 1)families of C 2 curves on W with these properties: a curve of each familypasses through every piont of W and the unit tangent vectors to thesecurves form an orthonormal system (which we assume to have the directorientation) at every point of W. The lengths s,, ... , s n _, along eachfamily of curves, respectively, are a possible system of coordinates in W.We denote by z,, ... , zn _, the unit tangent vectors to each family ofcurves, respectively.

With this notation, we haven-1

V = VT+VV V, VT = 1 VkZk,k=1

where vj = v • T,. We also have for any cp E C 1 (n):

Vtp =V p+ v, V p = 'rav i=1 ast

and consequently

n-1 av avdivv= 1 • r1 + • v.

j=1 asj0v

Consider now the first integrand on T in the right-hand side ofDow

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(3,1,1,2). We haveR1 av av

divv= 1 •T+ v;= 1 as;avn-1 ravk a?k n-1 ovy av

_ L 1 k+vk '?j+ L v+vv Ti•j , k = 1 LOSJ OSj J j=1 LOSJ OSj

n-1 ávk 0kl FOv„ áv+?k + Vk • v + v + vv• v

k=1 av OvJ L Ov av

and thusn-1 av n-1 a? n-1 av

kV. div v = v„ti=E ' + 1 Vk • ?i + vv ' Ti

1 OSj 1,k=1 as; ;=1 asj

n -1 aTk a vv+ vk• v + (3,1,1,3)

k =1 av av

since (av/av) • v = 0.

We then consider the second integrand on T in the right-hand side of(3,1,1,2). We have

n-1 av av(v•V)v=y v; +v

asj av

i1 avk + 0 k\

+ n-1 av,,v+ v av— vj ?k vk L vj vj,k =1 OSj asj 1 = 1 as; OS;

n -1 jóVk a?k avv av+ vv ?k + vk

+ vvv

+ v„ ,

k=1 av av av av

and thusn-1 n-1 av n-1

kav

{(V • V)V} ' V = 1 vjvk k V+ i Uj v + 1 vvvk • V + vv vj,k=1 asj j=1 OSj k=1 av av

(3,1,1,4)

because (av/as; ) • v = 0.Subtracting identities (3,1,1,3) and (3,1,1,4), we finally obtain

n-1 av n-1 a?vv div v — {(v • V)v} • v = v v 2., , +v,, E vk k • ?j

;=1 asj j,k=1 asj

►i-1 0 0k n-1 avv+vv L —L vjuk •v- - vj

j=1 asi ;.k=1 asj j=1 asj

(3,1,1,5)

On the other hand, using s 1 ,. . . , s_ as coordinates in W, we knowDow

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that Ok is defined by

n --1 C^1^ n - 1c^?

q) = —Tk^jÍk = + 1 v k ^j1kj,k = 1 0Sj j.k = 1 (9Sj

and consequently we haven-1 ov

tr =--^T;j=1 ast

It follows that (3,1,1,5) may be rewritten as:n-1

(3Vn-1

(J,Tkv r, div v — {(v • V)v} • v = v r, J + v

= 1 asj ;,k = l (Sj

n-1 ave— (tr )v — (vT ; VT) — Y V1 . (3,1,1,6)

j=1 asj

Let us finally calculate► i-1 a(V,V-1 ) n1 a(V

Vk?k)d I v T (v,, V,) = Z • T _ k • T.

= 1 as; j.k = 1 as;

Thus, we havert-1 Óv

v aVk aTkdiVT (v„ VT) _ L 1 vk Tk + v„ Tk + V,Vk • Ti

j,k =1 (iS1 9sj asj

'^ -' ave n -1 a ,U; i1-1 ^Tk_ V;+VV +V„ Vk 'T j . (3,1,1,7)1=1 ast = 1 asj j,k = 1 asj

Then, from (3,1,1,6) and (3,1,1,7), we deduce

v„ div v—{(v • V)v} • v

=d1VT (vvVT)—(tr )vv - 19(VT; VT)-2 v;i=1 asi

= divT (v„vT ) — (tr PJ3 )v —19(vT ; vT ) — 2vT • VTV,. (3,1,1,8)

This expression of the integrand on T in (3,1,1,2) no longer involves theparticular coordinates in W. Varying W, it is consequently trueeverywhere on T. Thus, we have

J 2dx

av e av;Idiv v^ dx —

(2 i,; =1 ax; alx,

= _2J VT • VTv„ dif — {(tr )v? + (v T , VT)}dom. (3,1,1,9)i1 j

Indeed, the integral of div T (VV VT) is zero since the vector field V v VT iseverywhere tangent to I'.Dow

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We have proved (3,1,1,9) assuming that v E C 2(.2)". However, since theboundary of D is of class C 2, the space C 2(f) is dense in H' (f) . From(3,1,1,9) we deduce (3,1,1,1) by approximating v E H'(f2)" by a sequencevm, m = 1, 2,... of elements of C 2(f2)" for which (3,1,1,9) holds. •

Theorem 3.1.1.1 is thus completely proved. Most of the previous proofcan be carried out under weaker assumptions on f2. We shall say that abounded open subset of ff " with a Lipschitz boundary T has a piecewiseC 2 boundary if T = To U Tl , where

(a) T'0 has zero measure (for the surface measure du).(b) T1 is open in T, and each point x E T 1 has the property of Definition

1.2.1.1 with a function cp of class C 2 .

Theorem 3.1.1.2 Let f2 be a bounded open subset of R" with a Lipschitzboundary T. Assume, in addition, that T is piecewise C 2. Then for allv E H2(f2) n we have

Jn av av

ldiv v) 2 dx — ' dx = {divT (vv vT) — 2vT - VTvv } do-n i,j = 1 12 axj axi r,

— {(tr ) U + 14(vT ; VT)} do'.

(3,1,1,10)

Proof This is quite similar to the proof of Theorem 3.1.1.1. Indeed,(3,1,1,2) holds since T is Lipschitz. Then identity (3,1,1,8) holds at anypoint of I,1 . Integrating (3,1,1,8), we obtain (3,1,1,10) since To hasmeasure zero. Finally, we extend (3,1,1,10) from V E C 2(f )" to VE

H2(f)" only, since it is now impossible to give a meaning to the bracketof VT and VTv when v E H'(Q)" and I' is only Lipschitz globally. n

3.1.2 A priori inequalities for the Laplace operator revisited

We now take advantage of the results of Section 3.1.1 to prove inequality(3,1,1). Such an inequality has been proved by Caccioppoli (1950-51) andLadyzhenskaia and Ural'ceva (1968) in the case when f has a C1,1

boundary. These latter authors call it `the second fundamental a prioriestimate'. All of them make use of local coordinates in order to flattenthe boundary. The proof given below follows Grisvard and Iooss (1975);it allows better control of the constant C(f). It also allows one toconsider some nonlinear boundary conditions. A slightly different point ofview is developed in Lewis (1978) for two-dimensional domains.

For the sake of clarity, we shail first consider the particular case whenthe operator A is the Laplace operator %, or the modified LaplaceDow

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operator A — A with A > 0. The first inequality concerns a Dirichletboundary condition.

Theorem 3.1.2.1 Let D be a convex, bounded open subset of R" with a C 2

boundary T. Then there exists a constant C(fl), which depends only on thediameter of D, such that

1142,2,n = C(f) IIAUIIO,2,n (3,1,2,1)

for all u E H2((2) n H 1 (a).

Proof We first apply identity (3,1,1,1) to v = Vu, observing that, sinceyu =0 on T, we also have (Yv) T = -YVTU =0 on T. Thus, we have

J na2 u 2^ dx Y dx= — (tr 3)(yyv•v)2 dcr.

fl i,j=i 17 axt a; t

Due to the convexity of t, we have tr 9B = 0 and consequently'2a2 u

dx _1 1ul2 dx. (3,1,2,2)i,j =1 n axi axj ,fl

So far, we have estimated the second derivatives of u. The estimate forthe first derivatives is well known to be obtainable by the straightforwardintegration by parts which follows. We have

n au 2dx = — du • u dx ^ (

i=1 f2 ax

where the norm is the norm of L2(12). On the other hand, the Poincaréinequality implies that

2 =K1 2 au 2IIUJ1 ( ) dxIaxJ

where K(f) depends only on the diameter of D (see Theorem 1.4.3.4). Itfollows that

n auf dxK(f2)2 IIuII2 (3,1,2,3)i=1 n IaX1 I

and that

I K(Q)2 IJAUJI . (3,1,2,4)

Adding up inequalities (3,1,2,2) to (3,1,2,4), we obtain inequality(3,1,2,1) with C(S) 2 1 + K(Q)2 + K(fl)4 .

Remark 3.1.2.2 In the case when we assume 12 to be only a boundedDow

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open subset of R" with a C2 boundary without the assumption of convex-ity, we again obtain inequality (3,1,2,1). This is achieved with the aid ofidentity (3,1,1,1), inequality (1,5,1,2) and by using an upper bound fortr 92. Consequently, the constant C(Q) depends not only on the diameterof (l but also on an upper bound for tr 99. In other words, the constantC(fl) depends on the curvature of F. This is nothing but an alternativeproof of the corresponding inequality in Section 2.3.

Let us now consider a Neumann boundary condition and even somenonlinear boundary conditions closely related to the 'third boundaryproblem'. Here we consider a real-valued, nondecreasing function J3defined on the real line. In addition we assume 3(0) = 0, and we assume 0to be uniformly Lipschitz continuous. We now deal with the followingboundary problem for a function u E H2(f):

— Du +Au = f in D,

au (3,1,2,5)—y = (3(yu) on T.

av

The corresponding estimate is the following:

Theorem 3.1.2.3 Let f be a convex, bounded open subset of R" with a C 2

boundary P, and let 13 be a uniformly Lipschitz, nondecreasing functionsuch that (3(0) =0. Then we have

11 lil11u^^2,2,f ^ 2 + — + 411— ááU +AuII0,2.n (3,1,2,6)A A

for all u E H2(f) such that —y au/av = (3(-yu) on T and all A > 0.

The particular case when the function Q is identically zero is just aNeumann problem. Obviously the interest of inequality (3,1,2,6) is thatthe constant depends neither on f? nor on 0. This will allow us to extendwidely the possible ,fis and the possible (3s, in the following sections.

Proof We again apply identity (3,1,1,1) to v = Vu. The boundary condi-tion now means that —(-yv) - v = 13(^yu) on T. Thus we get

n21jU1 2 dx— 1

a2 u dx

f2 i,j=I n axi ax1

= +2(VT(yu); VTf(yu))

— l,9 ((yv)T; (yv)T) + (tr 9)[(yv) - v]2} dif.

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Due to the convexity of 12, 99 is nonpositive. On the other hand,yu E H3'2(r), and since 13 is uniformly Lipschitz, we also have (3(yu) EH 1 (T ). This allows one to rewrite the bracket as an integral. Conse-quently, we have

Jn a2u 2

láU12 dx — dx ; +2 J VT (^yu) - O T(3(yu) do~.f2 i,; = 1 0J Lxi ax; r

The integrand on r is

2VT('iu)VTI3 (yu) = 20 '(yu) IVT(yu)1 2 ;

this is a nonnegative function since 13 is nondecreasing. We conclude thatn a2u 2

dx d ( dx . (3,1,2,7)i,; =1ff2 l axj ax1 {2

The estimate of the remaining terms in the H2(Q) norm of u isobtained, as usual, by integrating (—au + Au)u. Indeed, we have

— u+Au u d = nauL1 2 Jdx+A dx— ^ u u dcr.

f2 i=1 f2 l axi f2 f,- a v

Consequently, we have

,1 J dx+^ j1l2 ij—u + Au Iluil — J R(yu)yud.n

Since we assume 13 to be nondecreasing and 3(0) = 0, it follows that

(3(yyu)yyu'0 a.e.

on F. Then, we have

a J J laxj

IdX-11—vu+ûllllulln

and consequently

lu 11 2:-s^1

11—áu + Aull

(3,1,2,8)

and

au 21dx =— ij_au + Âu1h 2 . (3,1,2,9)i=1 ƒ£2 l axi /t

The conclusion follows from inequalities (3,1,2,7) to (3,1,2,9). n

Remark 3.1.2.4 Again we can drop the convexity assumption on f andlet D be any bounded open subset of R' with a C 2 boundary. Then, weDow

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deduce from (1,5,1,2) and (3,1,1,1) the inequality

IIUI12,2.n = C(A; D) (L —au + AuIIo,2,n (3,1,2,10)

for every u E H2(Q) such that --y a ul a v = (3 (-yu) and every )A > 0. Here theconstant C(A, f) depends only on A and on the curvature of L (or moreprecisely, on an upper bound for 9B ). It is important to observe thatC(A, (2) does not depend on (3.

Remark 3.1.2.5 A priori bounds in H 2 for solutions of the Laplaceoperator under oblique boundary conditions are also proved in Subsec-tion 3.2.4 in the particular case n = 2.

3.1.3 A priori inequalities for more general operators

The purpose of this subsection is to extend the resuits of the previoussubsection to the more general operators A that we introduced inChapter 2. Here we shall no longer consider nonlinear boundary condi-tions. This is to avoid some very cumbersome calculations which can befound in Grisvard and Iooss (1975).

Accordingly, we consider an operator A defined by

Au = Di (a1 ,;D;u)i,j=1

with ai ,; = a; , i E C°"(1). We assume again that —A is strongly elliptic; i.e.there exists a >0 such that

n

a (x)^j^; — a 1 1 2 (3,1,3,1)

for all x E and EWe only consider here Dirichlet and Neumann boundary conditions.

Theorem 3.1.3.1 Let fl be a convex, bounded open subset of R with a C 2

boundary. Then there exists a constant C(fl; A), which depends only on thediameter of (2 and on the Lipschitz norms of the coefficientsicients a 1, 1 i, j < n,such that

11u112,2,.^ = C(Q; A) 1lAu110,2,n (3,1,3,2)

for all u E H2(fl) fl H 1 ((2).

Proof We could again use identity (3,1,1,1) with

v=,W Vu

where .sol is the matrix of the a^,1. However, the less natural method ofDow

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proof that we shall follow here is simpler. Namely, we shall deduceinequality (3,1,3,2) directly from inequality (3,1,2,1) through the sameperturbation procedure that we already used in Section 2.3.3. The mainstep is the following:

Lemma 3.1.3.2 Each point y E ,fl has a neighbourhood V, such that

142,2,fi = Cy{llAull0,2,n +IIUII1,2,s} (3,1,3,3)0

for all u E H2(f2) n HP(Q) whose support is contained in V. Furthermore,the constant Cy depends only on the diameter of f2 and on the Lipschitznorm of the a. The neighbourhood Vy depends only on the Lipschitz normof the a^, 1 .

Proof As in Lemma 2.3.3.3 we freeze the coefficients of A at y. Thus,we set li ,j = a^, j (y). This defines a strictly negative symmetric matrix L, andconsequently there exists a nonsingular matrix R such that — RLR is theidentity matrix (R is the inverse of the square root of —L.) If we set

v(x) = u(R -'x), g(x) = f(R -1 x),

then the equation

11,jD1Dj u = fi,j=1

is equivalent to

--av = g.

We can apply inequality (3,1,2,1) to v.Precisely, our assumptions on u imply that

v E H2(Rf) n H 1(Rfl).

On the other hand, Rf is convex and has a C' • 1 boundary. Consequently,we have

1 = C(Rfl) llgLIo,2,rzn-

Going back to the original variables, we also have

n

11u^^2,2,n K(R, D) Y lt,jDjDju (3,1,3,4)i;=1 0212,,.

where K is a continuous function of the matrix R and of the diameter off.

We compare the right-hand side in (3,1,3,4) with the norm of Au inDow

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L2(fJ). Actually, we haven n

li,;DiD; u — Au = Di (li ,j — aij )D,ui,j=1 i,;-1

n n

_ „r ( li,j — a j )DD; u — (D a ;)(L;u) •i,j=1 i,j=1

It follows that

11 li ,;DiDju —Aui,j=1

n 2 max 1ai; (y) — ai.;(x)I ((UI12,2,n+ M I141,2,n (3,1,3,5)XEV,

where M is a common bound for the Lipschitz norms of all the a ij .From (3,1,3,4) and (3,1,3,5), it follows that H42,2,n:^^- K(R, 1l)

1lAu110,2,n+ Mn2 max lx — y i i1U112,2.n + 11U111,2,nY

The inequality (3,1,3,3) follows by choosing the neighbourhood V ysmall enough to ensure that

11x Yl^ Mn2122K(R, )

for all x E V.The proof of Lemma 3.1.3.2 is complete. The claim in Theorem 3.1.3.1

follows easily with the aid of a partition of the unity on 11 n

We turn now to the Neumann problem.

Theorem 3.1.3.3 Let Q be a convex, bounded open subset of ll with aC 2 boundary. Then there exists a constant C(A, A), which depends only onA, a t and the C° ' 1 norm of the aij, such that

11U112,2,&:_^_: C(A, A) ( Au(Io,2,Q (3,1,3,6)

for all u E H2(Q) such that —y aulavA = 0 on 1, and all A >0.

Proof We first apply identity (3,1,1,1) to

v= vu

where is the matrix of the a i; . We observe that

Au=divv in f

t We recall that a is the ellipticity constant which occurs in (3,1,3,1).Dow

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and that

aU n aUy -- L vàz;y = (yv) • v on Í:

a vA i ,^ =1 dxx

Accordingly, we have

J n1

av av ' dx = — 9B ((yv)T; (yv) T) df 0

f2 i,j = I ax; ax

(3,1,3,7)

since D is assumed to be convex.We then use the following lemma, whose proof is postponed to the

completion of the proof of Theorem 3.1.3.3 (cf. also Lemma 7.1, p. 152in Ladyzhenskaia and Ural'ceva (1968)).

Lemma 3.1.3.4 The following inequality holds for all u E H2(fl):

a2 u 2 n a2u a2 u« 2 ------- _ E aika;,t (3,1,3358)i,; = , l axj ax1 i ,;, k , l = ,,

ax; axk ax^ axj

a. e. in il

From (3,1,3,8) it follows that

2 a2u 12 +2 n a2u aa« c.,k

i,; =1I 9xax; ^^ =1 ax, ax i,j,k,(=1 ax; aXk aX.i axI 1a.e. in (2. Integrating, we have

2 r1 a2u 2 n avi av;a dx dx

i,; =i n axi ax; i,;=1 n ax; a

4 2 n aU n+ 2 n M

a2 U^^ dx,

.n ^ _ ^ ^,- 1 i,; = 1 It3xi ax;

where M is a common bound for the C°" norms of all the ai;. This,together with inequality (3,1,3,7), implies:

n 2

a2 1 92u dx

i,; = 1 l axn ax;

2 + 4 2 n au n a2UÀu ` dx 2n M dx

n 0 i= ÎaxI 1 axi ax;

Ja 2 n a 22u 2 n8M4 n au

1Au ^ 2 dx + dx + 2 2dx.2 i,;=1 a ax; n=^ aXiD

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Thus we get2 92u 2 2 n2M4 n au 2

a dx ; 2 IAu^ dx + 4 2 dx.i,;=1 1 ;)Xjax; n ;=^ n a

(3,1,3,9)

The estimate of the remaining terms in the norm of u in H2(,J) is theclassical one. Indeed, we have

n óU áU(Au+ Â u)udx= aL; dx + A IUJ2 dx

=i n axj ax; n

and consequentlyn

Â JuIIZ+a ^ liuji.

It follows that

11u11:!^-:: I

(3,1,3,10)

and thatn 1

IIIAu + Aull2 . (3,1,3,11)i=1 aA

Adding inequalities (3,1,3,9) to (3,1,3,11), we obtain (3,1,3,6). Indeed,we have

n n

IIUJ12,2,n =IIU11 2 + 1 1 I►,j_1

22 n 2M4 22 + IIAu +aull + 2 IlAujl + 4 IIAu + Aull( 1

ax a aA a 4

1 4nâ

2M4 82 + + 5 + IIAu +pull

2 . n

A aA

Proof of Lemma 3.1.3.4 By density, it is enough to prove inequality(3,1,3,8) for u E C 2(L). At a particular fixed point x, let A 1 ,. . . , An be theeigenvalues of the matrix whose entries are the ;. Also, let y 1 , ... , yn bea new system of orthogonal axes which diagonalize this matrix. Theinequality (3,1,3,8) is equivalent to

a2 u 2 n a2 u 2a = Y A jA;

i,;=1 1 03 yi ay; i,;=1 layi ay;

This is evident since all the eigenvalues are —a.Dow

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3.2 PROBLEMS IN CONVEX DOMAINS 147

3.2 Boundary value problems in convex domains

For lome of the boundary value problems introduced at the beginning ofChapter 2, we now have two kinds of results. First an existence anduniqueness result for a solution in H2(fl) provided ,f2 is bounded and hasa C' - ' boundary (see Section 2.4 mainly). On the other hand, we proved(in Section 3.1) a priori bounds for solutions in H 2(D), where theconstants depend very weakly on f provided it is convex and has a C 2

boundary. In most of the inequalities the constants do not depend on thecurvature of T, i.e. on the fact that r is C''. This will allow us to takelimits with respect to fl, i.e. to let D vary among convex domains. Thuswe shall extend our previous results to general bounded convex domains.

The first result of this kind is due to Kadlec (1964) and concerns theDirichlet problem. The extension of this result to other boundary condi-tions has been achieved in Grisvard and Iooss (1975).

3.2.1 Linear boundary conditions

The possibility of approximating a general convex domain by domainswith C 2 boundaries foliows easily from the results in Eggleston (1958).

Lemma 3.2.1.1 Let (2 be a convex, bounded and open subset of Ir. Thenfor every e > 0, there exist two convex open subsets ,fl, and (2 2 in Ir suchthat

(a) D, C D C D2(b) Di has a C 2 boundary fl, j = 1, 2.(c) d (F , 1 '2) = E,

where d(F,, T2) denotes the distance from T, to T2 .

This lemma allows us to approximate a given f either from the insideor from the outside by a domain with a smoother boundary. The insideapproximation is more convenient for studying the Dirichlet boundarycondition while the outside approximation is more suitable for dealingwith boundary conditions of the Neumann type. By the way, we recallthat we already proved in Section 1.2 that a bounded convex open subsetof II " always has a Lipschitz boundary.

In the following results A denotes the same operator as in 3.1.3,fulfilling the assumption (3,1,3,1).

Theorem 3.2.1.2 Let (2 be a convex, bounded and open subset of Ir.Then for each f E L2(fl), there exists a unique u E H2(f), the solution of

Au=f inD(3' 2 1)

Vyu = 0 on T.' 1'

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Proof We choose a sequence D , m = 1, 2, ... of convex open subsets ofR'1 with C 2 boundaries F,. such that flm c f and d (Tm, T) tends to zeroas m --> +00. We consider the solution um E H2(S2m ) of the Dirichletproblem in ,klm , i.e.

Ç AUm =f in Q,

=on T(3,2,1,2)

.Ym um ,

where y1z denotes the trace operator on Tm , m = 1, 2, .... Such a solutionU. exists by Theorem 2.2.2.3.

It follows from Theorem 1.5.1.5 that um E H' ((lm ) ; in other words, wehave um E H'(R"). Then from Theorem 3.1.3.1 we know that there existsa constant C such that

IC. (3,2,1,3)

This implies that u,n is a bounded sequence in H 1 (P),), and in additionthat

vm,i,j = (DiDjum Y, m =1, 2, .. .

are bounded sequences in L2(R") for 1; i, j = n. Consequently there existU E H 1 (Rn) and V, 1 E L2(ffr) and a suitable increasing sequence of integersmk, k = 1, 2,... such that

U U weakly in H 1 ([}^n ), k —+00

v,n , ; ,1 --^ V in L2([Jr ), k — 00 •

First, we shall check that the restriction u of U to (2 is solution of theDirichlet problem in ,f2. Indeed we have u E H'(fl). In addition, all the u,.,,have their support in £2; it follows that U also has its support in ƒ2, i.e.U = ü. By Definition 1.3.2.5, this means that u E H' (f2) and finallyCorollary 1.5.1.5 implies that yu =0 on T (here the Corollary is appliedwith k =0, which is possible owing to Corollary 1.2.2.3). Finally, letcp E 2 (fl) ; then there exists k (9 ). such that the support of cp is containedin DMk for all k k (cp). Thus for k k(Q) we have

n

fcp dx = Aumkcp dx = – ajjDjumkDjcp dxi j ° 1 Lk

n

_ – a^jDj u,nkD1 p dx.i,j=1 n

Taking the limit in k, we obtain

Jn

f q dx = – Y, a.jD^ uDjcp dx.S2 i,j=1 ff2D

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This identity is valid for all cp E g(Q); it means that

Au =f in f

in the sense of distributions.So far we proved the existence of u E H'(Q), the solution of (3,2,1,1).

The uniqueness of u is a classical result by the energy method (see forinstance Necas (1967)). To complete the proof we have to check that thesecond derivatives of u are square integrable. We again let cp belong to.9(D). Then for k k (q ) , we have

uffkDiD;cp dx = umkDjD;cp dx = DiDiuynk cP dx =

dx.

Taking the limit in k, we get

uDiD1c dx = Vij<p dx.n n

In other words, the distributional derivative DiDu is the restriction of VV.i

to , T; this is a square integrable function for all i, j = 1, ... , n. n

Theorem 3.2.1.3 Let ( be a convex, bounded and open subset of D.Then for each f E L(f2) and for each A >0 there exists a unique u E H2(,^)which is the solution of

a^;DûD;v dx + uv dx = fv dx (3,2,1,4)i,i =1 sl st st

for all v E H'(,fl).

Identity (3,2,1,4) is the weak form of the Neumann problem for theequation

Au + Au = f in D. (3,2,1,5)

As we saw in the proof of Theorem 2.2.2.5, identity (3,2,1,4) is equival-ent to equation (3,2,1,5) together with the boundary condition

n

vry(ai;D;u) = 0 a.e. on T. (3,2,1,6)

This makes sense since T is Lipschitz and a i;D; u E H'(Q).

Proof This time, we choose a sequence ,klm, m = 1, 2,... , of boundedconvex open subsets of Rn with C 2 boundaries I'm such that fl C ,klm andd (Tm , T) tends to zero as m --* oc. We consider the solution um E H2(Dm)Dow

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of the Neumann problem in flm , i.e.

(3.2.1.7)

Obviously f E L2(flrn ) and Urn exists by Corollary 2.2.2.6.From Theorem 3.1.3.3, we know that there exists a constant C such

that

(3.2.1.8)

Consequently, restricting the Urn to {}, we obtain a bounded sequence inH 2(fl ) and then a weakly convergent subsequence. In other words, thereexists an increasing sequence of integers mk and a function U E H 2

( {}, )

such that

weakly in H 2(!J ), as k ~ 00.

(3,2,1,9)

We now complete the proof by checking that u is a solution of(3,2,1,4). Indeed let v E H 1

( {}, ) . Since T is Lipschitz, we can applyTheorem 1.4.3.1 to find a V E Hl(~n) such that V'n = v. It is clear thatVlo

mE H1(!Jm ) and from (3,2,1,7) we deduce that

-. ~ i ai.jDiUm.DjV dx +Ai Um. V dx = i fv dx.I.J. - 1 Omk n"'k n

We shall now consider the limit of (3,2,1,9) when k ~OO. We have first

J u.n. V dx - i uvdx= J u.n. V dx +J(u.n. -u)vdxllmk n nmk\n 0

and consequently

If. um• V dx -1 uv dxlnUl/( n

~1Iu.nJo.2.n",. (J v2dX) 1/2 + lIum k - ullo.2.nllVllo.2.n.fl"'k\fl

The right-hand side of this inequality converges to zero due to (3,2,1,8)and the compactness of the injection of H 2

( {}, ) in L 2 ( {} ) (see Theorem1.4.3.2). In the same way, we prove that

J a· o« D· V dx ~ J a· ouo» dxq I mk 1 1.1 I IOm

kn

owing to the compactness of the injection of H 2( {}, ) in H 1

( {}, ) . Summingup we obtain identity (3,2,1,4) as the limit of (3,2,1,9) when k~ 00. IIDow

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Remark 3.2.1.4 One can prove results similar to those of Theorems3.2.1.2 and 3.2.1.3 when f is a plane bounded domain with Lipschitz andpiecewise C2 boundary whose angles are all convex.

3,2.2 Nonlinear boundary conditions (review)

In the next subsection we shall take advantage of inequality (3,1,2,6),which concerns the nonlinear boundary condition.

au _— "av —

0(yu) on F

where f3 is a uniformly Lipschitz continuous and nondecreasing functionsuch that f3(0) = 0. We shall take limits with respect to D and 0. In thissubsection we review some known results about monotone operators.Here we follow Brezis (1971).

Let H be a Hilbert space and A a mapping from H into the family of allsubsets of H. In other words, A is a (possibly multivalued) mapping from

D(A)={xEH; Ax 0}

into H. A is said to be monotone if

(Y — Y2; x, — x2) r0,

Vx 1 , x2 E D(A) and y, E Ax,, y 2 E Ax2 . Then A is said to be maximalmonotone if it is maximal in the sense of inclusions of graphs; i.e., itadmits no proper monotone extension. For each A >0 we define aninverse for the multivalued mapping (AA + I) as follows:

(AA +[) - 'y = {x E D(A) I ^ (y —x) E Ax}.

It turns out that (AA + I) -- ' is univalued and is a contraction in Hprovided A is monotone. It was shown by Minty (1962) that A ismaximal if and only if (AA +1) is onto for A >0, or equivalently(AA + I) - ' is defined everywhere.

In what follows we shall only consider monotone operators which are insome sense the gradient of a convex function. More precisely, let cp be aconvex lower semicontinuous function from H into ]—oo+oc]. We assumethat cp is proper, i.e. that cp +oe. Let

D(p)={xEHJ p(x)<+oc}.

For x E D(cp) the set

acp(x)={yEHI p(z) — ^P(x)=(Y; z —x), VzED(cp)}

is called the subdifferential of cp at x. It was shown by Minty (1964) thatthe operator x H a p(x) is maximal monotone.Dow

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Following Moreau (1965), such a convex function can be approximatedby smooth convex functions cp,, defined for A >0, by

x 1 x—z 2 + ^/^ 2^Á( =min 3) ZEH I I `f' (

( 1221f )2A

It turns out that cp> is convex and Frechet differentiable. Thus

a p (x) _ {fr(x)}for every x E H. In addition cp,, (x) is a decreasing function of A and

^p (x) ---> p(x)

for every x E H as k ---> 0. Finally

' x = x-- Aa +I^'x .(3 ,2 ,2 ,2)

This is simply the so-called Yosida approximation of A = acp, which ismonotone and Lipschitz continuous with Lipschitz constant 1/A.

Not all maximal monotone operators are subdifferentials of convexfunctions; however, in the particular case when H is just the real line El,this does hold. Here are two typical examples of subdifferentials ofconvex functions in R. First, if we assume that

Ç

0^ x#0

then it is easy to check that

r x0l x=0.

This is a maximal monotone operator in ff. . The Moreau approximation ofcp is

x2

and accordingly the Yosida approximation of acp is

çoá(x)= •

On the other hand, let

+w x<0

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then it is easy to check that

Q^ x<0c^cp(x) _ [-00, 0] x = 0

{O} x>0

x 22^ , x;U

px(x)x^0

xx _^^ x`0

^x( )l o.) xO.

Turning back to the general case, an important existence result is thefollowing:

Lemma 3.2.2.1 Let cp be a convex, lower semi -continuous and properfunction on H. Assume that cp is coercive, i.e. that

cp(x) - 0 when lixli--->+oo. (3,2,2,3)

Then cp has a minimum in H. The minimum is unique when cp is strictlyconvex.

Accordingly, if x o is such a minimum, we have

0e app(x0)•

This is an existence result for the subdifferential of cp.

Proof of Lemma 3.2.2.1 We denote by m the g.l.b. of cp. There exists asequence xn, n = 1, 2,... of elements of H such that

cp (x) —+ m

when n - +oe. Since cp is proper, we have m < +oo and condition (3,2,2,3)implies that the sequence x, 1, n = 1, 2,... is bounded in H.

Consequently, by possibly replacing the original sequence by a suitablesubsequence, we can assume that x,,, n = 1, 2,... is weakly convergent tosome limit x e H. By the very definition of m we have

cp(x); m.

On the other hand, since cp is lower semi-continuous (for either the strongDow

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topology or the weak topology on H), we have

cp(x) =lim sup cp(x) o M.n-POC

Summing up we have proved that cp (x) = m and x is the desiredminimum.

The uniqueness of the minimum, when cp is strictly convex isobvious. n

We shall use this existence result as follows. We consider a maximaloperator (3 in ff and the corresponding convex function j on R such that

a = ai.

Then we build a new convex function on L 2(Q) by setting

2 1VV1 2 dx+j(-yv)do-

^P(v)=n r

1 (3,2,2,4)if vEH (D) and i(yv)eL 1 (F)

+00 otherwise

We are looking for solutions of the following boundary value problemwhere c > 0, f E L2(I) and D is, say, a bounded open subset of ion with aLipschitz boundary T:

f—iu +cu=f in D

— Vu - v E (u) a.e. on I' (3,2,2,5)

The function cp allows a weak formulation for problem (3,2,2,5). Indeedwe have this lemma.

Lemma 3.2.2.2 Let u e H2(Q) be a solution of (3,2,2,5), then we have

cp(v)+^lIvll2 — j fvdx,Q(u)+^llull2— ƒni

fudx (3,2,2,6)2 n 2

for all v e L2(Q).

Proof The boundary condition in (3,2,2,5) implies that for v E H 1 (f )such that j(-yv) E L 1 (T) we have

j(yv) — j( -yu) % — (y Vu) •v(yv — yu)

and consequently that

{j(-yv) — J— (v - y Vu)( -yv --yu) do-.

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Then by the Green formula, we have

f(v—u)dx= (—Au+cu)(v—u)dxn .^

_— (v•-yVu)(^yv—yu)dur

+J Vu .V(v—u)dx+c J u(v — u) dx^ n

and consequently

f(v—u)dx_ {j(yv)—j(yu)}du+ Vu •V(v—u)dxn n .o

+cJ u(v — u)dx.n

Finally, observing that

2u(v—u)_v 2 —u 2

and that

2Vu . V(v— u)Vv 2- 1Vuj 2

we conclude that

2J0IVVI

2 dx+c 1V1 2 dx+ j(yv)do^— fvdx 2 rz r o

J cIVul 2 dx+ IuI 2 dx+ J i(u)d_ J fu dx.

n 2 n r siThis is exactly inequality (3,2,2,6) when v E H 1 ([2) and j (yv) e L 1 (T ). Inthe other cases, we have cp (v) = +oo and inequality (3,2,2,6) isobvious. •

In other words, we have reduced the problem of solving (3,2,2,5) tothat of minimizing the function

v : cp (v)+ C J — fv dx. (3,2,2,7)2 n

This is easily achieved, owing to the following lemma.

Lemma 3.2.2.3 For c >0 and f E L2([2) the function (3,2,2,7) is convex,lower semicontinuous, proper and coercive on L2(,3), provided 13(0) D 0.

Proof All but the coerciveness is obvious. To prove (3,2,2,3), we ob-serve that j (x) : 0 everywhere. Indeed, the condition that a j (0) = 13(0)30Dow

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means that the graph of j is contained in the upper half -plane. Then itfollows that

+c v 2 — dx^'-IVV 2 dx+^ 1 V1 2 dx- 1ƒ1 2 dx.^(v) II II fv 2 I 2 n 4 nc n

This lower bound obviously tends to +oe when livil--->oo. n

3.2.3 Nonlinear boundary conditions (continued)

Let us first state the result which is the purpose of this subsection.

Theorem 3.2.3.1 Let ( be a bounded convex open subset of R". Let 13 be amaximal monotone operator on EI such that 13(0) 30. Then for eachf E L(f2) and for each c >0, there exists a unique u E H2(f) which is thesolution of (3,2,2,5).

Before proving this theorem, let us take a look at some examples. Letus assume first that

[ +00 x0

l(x)0 x=0.

Then obviously we have

02 ] if v E H l (,^l)

^P(v)= in+ w otherwise.

Then, surprisingly enough, problem (3,2,2,5) is just a Dirichlet problem.Indeed, the boundary condition means that a.e. on T,

(yu, —v ' yV u)

is a point of R 2 , which actually lies on the vertical axis. In other words,yu =0 a.e. on T. With this special choice of j. Theorem 3.2.3.1 is just aparticular case of Theorem 3.2.1.2.

Let us assume now that

+0° x<0

Then we have

2 IVV 12 dx if v E H 1 (Q) and yv >— 0 a.e. on Tcp(v)=1

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The boundary condition in (3,2,2,5) means that a.e. on T

(yu,—v•yVu)EG

where G is the graph of (3 = aj, i.e.

G={(x, Y)ER2 I x =0, y=0, x• y =0}.

Accordingly we have

-yu . 0, v • -y Vu 0, (-yu) (v • y V u) = 0 (3,29391)

a..e. on F and this is the famous Signorini boundary condition.Finally let us observe that if we assume that

x 2

J(x)=b 2

where b 0, then

J ,fa 1 1yvl 2 da- if vcH I (,l)^P(v)= 2 r

t.. +oo otherwise.

Accordingly, we have (3(x) = j'(x) = bx and the boundary condition in(3,2,2,5) is just

—v • (-y Vu) = byu a.e. on r.

In particular, when b = 0, this is a Neumann boundary condition and wehave a particular case of Theorem 3.2.1.3.

Before proving Theorem 3.2.3.1, we need some preliminary results onthe approximation of ,l.

Lemma 3.2.3.2 Let D be a convex, bounded and open subset of Di" andlet [2,, , m = 1, 2,... be a sequence of convex, bounded and open subsets ofR" such that £2 C A, ,gym has a C 2 boundary Em and

d(Tm, F)- +0 when m--°o.

Then, for large enough m, there exists a finite number of open subsets Vk ,k -= 1, 2, ... , K in Rn with the following properties:

(a) For each k there exist new coordinates { y i, ... , y } in which Vk is thehypercube

{(y , ... , Yn) (—ai <yi <ai 1 j s n}

(b) For each k there exist Lipschitz functions cp k and cp m defined in

Vk = {(Yi, ... , Yn-1) 1 — a; < Y; < a, 1=1= n —1}Dow

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and such that

kj^pk(Zk)j„ Im(Z

k ){ an for every Zk E V,t2

V = {y k (Z k ^ Yn) I yk < pk (Z k )I

Vk = {Y k = (Z k , Y k ) I yn < ^p k (Z k )}

,rrÍ Vk = y k = (Zk, k )1

Yk = ^ k (Z k )}r j k k k k k1 m ^ÍVk={y =(Z ,Yn)yn — ^m(Z )}

K K

(c) -r C U Vk, rm C U Vk•k=1 k=1

In addition _cp k and —cpm are convex functions, cp , is of class C 2 forall large enough m and

(d) cp m converges uniformly to cp k on Vk and there exists L such that

1V^Pk(Zk)j. J L for every Z k E Vk, 1k _ K.(3,2,3,2)

Finally, Vq ,, —* VQ k a.e. in V.

It follows from Corollary 1.2.2.3 that D and all the f,n have Lipschitzboundaries. Accordingly, properties (a) and (b) just refer to the corres-ponding properties in Definition 1.2.1.1. We actually just have to checkthat property (d) holds.

Proof of property (d) in Lemma 3.2.3.2 Since the distance from T to 1„1

converges to zero when m --> oc, it foliows that the distance from thegraph of cp k to the graph of cp m converges to zero. This means that cp mconverges uniformly to cp k .

Inequality (3,2,3,2) follows from the geometry. Indeed let us consider afixed k and a fixed m. Let y k = (z k , cp n(z k )) be a point on T,,, fl V k , with

—a; +e<y; <a; —r, 1jn-1

for a given e > 0. The setk

Qs= Z k , — n Z k € Vk

2

is included in ,gym . So is the line segment from y k to any point of 8S. Theslope of such a line has a modulus less than or equal to ak/e. This impliesthat

Iv^m(Zk)I an/£.

We conclude by replacing all the a, 1 j = n —1 by a, — e with anDow

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Figure 3.1

e >0 small enough to preserve the condition I, U I< Vk. Since

d (I'm , F) —*0 as m ---* +00, the condition Tm c Uk =1 Vk is also preservedfor in large enough. We can define L as follows:

KL= max an/E.

k=1

Let us now complete the proof by looking at the convergence of Vcp ,to V p k . Since cp k is Lipschitz continuous, it has a gradient a.e. in V. Letus consider such a point z E Vk such that V p k (z) exists. The tangenthyperplane at (z, cp ,,(z )) to the graph of cp, is above the graph of cp k ,since cp k and cp, are concave functions and cpn, cp k . In other words, wehave

n(Z)+vcpm(Z) —Z)opk^ )

for all E V. Since 4k has a gradient at z in the usual sense, it followsthat

pk,(Z)+O^Pk,(z) ' ( Z) k(z)+V(pk(z) ' ( — Z) — C(I zI)Dow

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for all V. Then for each j = 1, 2, ... , n —1, we have

(p m(Z)+ tDjpm(Z) k(Z) tDD (p k (Z) +0 (Itl)

for itl small enough. If we denote by a ; and /3 ; the limits

a; = lim inf Djcp m(z ), j3; = lim sup D;cp m(z ),m—° rn —^ c

we easily see that

a; i D;cp k (z)+®(1), p;D;cpk(z)+C(1).

This shows that

D;pm(z) — D;çc k (z)

when m —+00 and completes the proof. •

Proof of Theorem 3.2.3.1 We shall approximate D by a sequence ofconvex open subsets Dm, m = 1, 2,... of ff as in Lemma 3.2.3.2. Weshall also approximate j by its Moreau approximation j, or equivalently 13by its Yosida approximation (3,, = j. Thus we start from u m E L2(D ),which minimizes the functional

2 1VV12 dx + c IUJ2 dx + j„(ymv)doem — fvdx

^Ym,1 (v) =lf v E H 1 ,, G( m)

+00 otherwise.

We observe that since j', is uniformly Lipschitz, its primitive j,, does notgrow faster than a quadratic function. Accordingly, when v E H'(L m ), wehave j,(ymv) E L1 (rm)•

There are four main steps in the proof. .

1st step We check that u, E H2(Q,,, ).2nd step We prove that 11UI,mI12.2,0. remains bounded uniformly in rn

and A.3rd step We take the limit in m.4th step We take the limit in A.

In the first step we use the fact that u, is the solution of

AUX,m + CUA,nj =1 1n f2m

ó UX , m

— Ym av (x (ymux,m) onm

Indeed, we first observe that since gym.,, (Up) < +oe, u,,, m must belong toDow

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H' ffl,n ). Thus it is the minimum for qjm.,, on H'((2»,). It is easily checkedthat gym,,, is Frechet differentiable on H' (Qm ) and consequently we have

m,^,(u^,m) = 0.

In other words we have

Vu,,m • Vv dx + cJ uv dx + J 3^ (Ymu^,m)Yv dam

frn ^m rm

— J fv dx = 0 (3,2,3,3)

0

for all v E H 1 (Qm ). Making use of (3,2,3,3) with only v E H 1 (,fZm ), wereadily see that

-- Llux,r, + Cux,r = f in fl,» .

Then applying the Green formula (1,5,3,10), we rewrite (3,2,3,3) asfollows:

au).'mv ( ) v dQYm V Ym = (^ Ymu , m ma v

m rm

for all -ymv E H"2(Fm ). This implies the boundary condition on u m , i.e.-- Ym a u,,,m/a vm = (3A ('ym u,m) in the sense of H-' ^2(Fm ) .

Let us now consider (3,, ( ,yu m ) as the Neumann data for u,. Sinceu) ,,1, E H' (Ilm ), we have ym u t , m E H"2(I'm ). Then, taking advantage of thefact that (3,, is uniformly Lipschitz continuous, we conclude that

l3 (YmL¼m) E H"2(rm ).

Now it follows from Corollary 2.2.2.6 that u,,, m E H2(fm ) since T,,, is C 2 .The second step is just an application of Theorem 3.1.2.3. This theorem

can be used here because ,kl m is convex and has a C2 boundary, while (3,,is uniformly Lipschitz, non-decreasing and fulfils the condition

/3(0)=0.

Indeed, we have —3 ).(0) _ (1/A)(A(3 + I) '0 and (A(3 + I) - '0 = 0 since0 E (AR + I)(0). Thus, we have

1 1\1/2

II'4x,»,I2,2J2,„ = + — + 4)0110,2,6 (3,2,3,4)(C2 c

u,m, m = 1, 2,... is consequently a bounded sequence in H 2(d2). Bypossibly consiclering a suitable subsequence we can therefore assume thatthere exists u,, e H2(f) such that

weakly in H2(D) when m -- +oo.Dow

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In the third step we take the limit in m in the equation which expressesthat &ym,,, has a minimum at u,m . Actually we have

c2 IV Ux,m I dx + 1Um l2 dx + jJ^ (ym UA,m) dam — u^, J dx

^m2 ^mrm

C1VV12 dx+ 2IVI2 dx+ I,,(-ymV)da-m-- Vf dx

1`m 2 „m r,n 1L

(3,2,3,5)

for all V E H' (ff ") . It is easy to check that

2 ly V dx+^

V dx —2 1VV12dx+c (V dx.

Then obviously we have

lim inf ( 1 2 dx = lim ( dx = ly u 2 dxm —COO ^m 1t1--^^ S L

!2

lim inf J IU,ml 2 dx lim

I 1x ,m L 2 dx =I u,,1 2 dxm-o fZ

sZ

since f,,, 2 f2. Also, we have

J fU.„. dx —> J fu,, dx.

n ^

Thus the only difficult point in taking the limit in (3,2,3,5), is to provethat

IIt (mV) do-m jJ^ (yV) dff (3,2,3,6)rmr

lim inf j,, (y nUX,m) dam >— j,, ( -yu,,) du. (3,2,3,7)m- I „, ƒri

For this purpose, we fix a partition of unity on T and T m corresponding tothe covering Vk , 1k K, introduced in Lemma 3.2.3.2, i.e. we consider8k E 2(ll ), 1 k K, such that ek has its support in Vk and

K

1= Ok (x)k=1

for all x E T and all x E T„t (for m large enough). We have to prove that

ek1A (-ym V) do m —^ OkIX ( -YV) do-. (3,2,3,8)rm rD

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We drop the index k and set q = oj,, (V). It follows that

r ^m71 dffm = (Yml)(Z, ^m (Z))[ 1 + Ivgm (2)1

21 112 dZ.

Clearly we have

(ym'i)(Z, Qm(Z))[ 1 + Ivpm(Z)1 2] 112 - (y^)(Z, p(Z))[ 1 + vp(Z)12]

112

a.e. in V'. In addition, we have

1(ymi)(Z cp,,, (Z))1 [1 + 1VQ m (Z)I2] 1/2

^ L 1 + L2^ 1/2 I(,yi)(Z ^(Z))I + [g. (L) ^(Z)j 1/2

a„/2 1/2%C lDnq (Z, y)`2 dy

Thus we have a fixed square integrable bound, and applying Lebesgue'sdominated convergence theorem, we conclude that (3,2,3,8) holds.

To prove (3,2,3,7) we introduce U,, E H 2(llr) such that U u.Then, we have

J jj (ymu^,m) dan, — jA (yu,) daI,rn

.

lj, (ym Ux,m) 1 x (ym vx )I damr,„

+ IJS(%nUA) dfm 1>.(yuA) da.-5It follows from (3,2,3,6) that

jj(yrUU) dam 1A(Yu„)da.r+m

r.

Then we observe that for x, y E G we have

lx (X) — Jx (Y) = RA (Y)(x — Y)

and thus

l^) (^mu^,m^ ^^ (y. UA )} dcrm Px (-ym v^ )f -ymu.,m -y. U } dam.

(3,2,3,9)

We shall show that the right-hand side of this inequality converges toZero. Indeed, we have

U 2 do- 1U 2 d^ K O U 2 + U 2 dxI PA (-m a^ l m 2 l ^m .1 ►n 2 [ I .1 I .1rm 1'm ^mD

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with a constant K which does not depend on m, due to Lemma 3.2.3.3below and Theorem 1.5.1.10.

Lemma 3.2.3.3 Under the assumptions of Lemma 3.2.3.2, there exists aLipschitz vector field defined on ion and a constant b >0 such that

vm > b on Tm for all m.

We postpone the proof of this lemma until the proof of Theorem3.2.3.1 is completed.

This shows that ^^ ^3, (y U,, )jô2r remains bounded when m — oo. Thenlet us set

'IÍ m = 0k (UX , m — Uk) •

Use local coordinates and drop k. We have

le{imux, ► n — Ym UU}I 2 dfm =

m(Z, ^m(Z))I 2 [ 1 -i- IV^m(2)1211/2 dZ.

We write

w„,(z)

im(z, pm(z)) = Tlm(Z, ^P(Z Dnlm(Z, y) dyw(z)

and consequently

(Tlm (Z, <Pm (z))l Iri,(z , p(z))I

p(z) 1/2

+ ['Pm (z) Q(z)]U2 (jIDn7,m (Z.> y)I2 dy)

cp(z)

Thus we have1/2

1 pm(Z))I2 [1 + I dZ

1/2

[1 + L2]114 I q(Z))I2 [1 + ly^p(Z)I2]112 dZƒv o

+[1+L2]1/4 max [p(Z) <P(Z)] 1/2zFV

cpm (z) 1/2

x jj J Dn7lm (Z,y) 2 dZdy JV' w(z)

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In other words

1/2

6y , (uJk,, — UA )1 2 d Qm

1/2

[1 + .,2]4 lOY (U.m — [J )I2 dcrr

+ max LOm(Z) <p(z)]112 110 (U>^

^m —

This shows that

^m (ua, m v ^) 1 2 d ►ri 0

when m --> oc since u, — U,, in H 1 (t2) and'IUA,,n lI l , 2 ,Q,„ remains bounded.Summing up we have proved that

J — ym U} 0in

and remembering (3,3,2,9), this implies (3,2,3,7).Taking the infimum limit in m, of inequality (3,2,3,5), we eventually

obtain

g(u) = 0X(v) (3,2,3,10)

for all v E H 1 (Sl ), where 4„ is defined by

,,(v) = IVVf 2 dx + (vI2 dx + J j,,(Yv) do-- vf dx.

n 2 s^ r s1

In addition, taking the limit in (3,2,3,4) we also have

1 1 `/2lUx 112.2,0 = (—C + — + 4 IIf IIo,2,si• (3,2, 3,11 )

c

We can now perform the last step of the proof. Due to (3,2,3,11) wecan find a sequence A ;, j = 1, 2,... such that

Ai --* 0, j--O0

and there exists u E H2(,2) such that

u^ --> u, j --* x

weakly in H2(fl) and consequently strongly in H2-, (f) for r >0. Inaddition, by the Lebesgue theorem on subsequences, we can also assume

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that

yuxj --> yu l

v • y V U> --> v • -y V u, j ---v oc

a.e. on T.On the other hand, as in the first step of the proof, inequality (3,2,3,10)

implies that

—4 u,, + cu,, = f in Sl(3,2,3,12)

1. —v • y Vu _ 0. (yu) a.e. on T.

It is easy to take the limit in this equation. We thus obviously obtain

—du + cu = f in D.

To take the limit in the boundary condition, we use the following trick.

Lemma 3.2.3.4 Let (3 be a maximal monotone operator in R and let (3 ), beits Yosida approxirnation. Let x,n, ym, Am be three sequences of real numberssuch that

xm — ym—

Ani — 0

ym = !3Jx m (xm )

Then

YE0(x)•

We consequently obtain

—v•yVuE(('yu)

a.e. on T. Summing up, u E H2(fl) is the solution of problem (3,2,2,5).The uniqueness of u follows from Lemma 3.2.2.2 since the functionalwhich is minimized there is strictly convex (see also Lemma 3.2.2.3). Thiscompletes the proof of Theorem 3.2.3.1. n

Proof of Lemma 3.2.3.3 We fix a point x0 E .fl and a ball B of radiusp >0 and centre at x o such that B c 12. Then we fix a function 0 E 2(R")such that 0 =1 on all S2m for large enough m. Then we can define µ by

µ(x) = 0 (x)(x — x4 ).

It is clear that the angle of g with vm is less than or equal to

Irp-- arcsin2IX—x0LD

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at x E T,,,. Consequently we have

1^'vm =^µ^ Plx — X

Thus a 8 >0 obviously exists and on the other hand, it is easy to checkthat g is Lipschitz continuous. n

Proof of Lemma 3.2.3.4 Using the definition of (3, it is easy to checkthat

Y,,, E 13(x, — Ani Yrn )

Since y1, — y and x,,, — A,n y r„ --> x, it follows that

y E (3 (x)

since the graph of (3 is closed (by maximality). n

3.2.4 Oblique boundary conditions

Here, for the sake of simplicity, we shall restrict our purpose to boundaryvalue problems in a plane domain f2. Thus let us assume that fl is abounded convex open subset of D2; its boundary is a closed Lipschitzcurve I' along which the arc-length s is well defined. We assume, inaddition, that c is a given Lipschitz function in f2. We shall solve thefollowing problem: for a given f E L2(f2) and a > 0, find u E H2(,l) whichis a solution of

— 4u + Au = f in f

-- au =c u on T(3,2,4,1)

av as

The main result below is due to Moussaoui (1974).

Theorem 3.2.4.1 Let fl be a bounded convex open subset of 1 2 and letc E C"''(Q). Then there exists A, ) such that for each A > A,) and for eachf L2(f2) there exists a unique u E H2(Q) which is a solution of

aVu - Vv dx + A uv dx = fv dx —( c— -yu; yv (3,2,4,2)

as

for all vEH 1 ffl).

Of course, identity (3,2,4,2) is a weak form for problem (3,2,4,1).Indeed, applying the Green formula of Theorem 1.5.3.1, it is easy tocheck that for u E H2(f2), (3,2,4,1) and (3,2,4,2) are equivalent.Dow

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Exactly as we did in the proof of Theorem 3.2.3.1, we shall approxi-mate L by a decreasing sequence of smoother convex domains ,lm ,m = 1, 2,... (see Lemma 3.2.3.2). In each of these domains, we shallsolve a problem similar to (3,2,4,1). Then it will be possible to take thelimit in m with the help of an a priori estimate. Let us first prove thisestimate.

Theorem 3.2.4.2 Let ( be a bounded convex open subset of R2 with a C 2boundary. Let c E C ° ° 1 (fl) and µ E C°' 1 (D)2 be such that p- v _>- 6>0everywhere on r; then there exists k 0 and k such that

11U112,2,n:t—::^ k II—au + Aullo,z,fi (3,2,4,3)

for all u E H2(1Z), such that -y au/av = c a(yu)/as on F. Moreover, k and Ao

depend only on 6, the Lipschitz norm of and maxr l

Note that the existence of g follows from Lemma 1.5.1.9.

Proof We apply again identity (3,1,1,1) to v = Vu. The two-dimensionalversion of this identity and the convexity of L, imply that i

J 2 avi aviIdiv v1

2 dx - dx ?-2 (yv)T d(-yv),.

ax; ax^ r


2 a2u 2a au{dx dx a —2 —(u)d( _,y y

i ,; = 1axi ax; ras0V

Due to the boundary condition, we have

a au _ a a a 2-2 -- (-yu ) d y

2 f,-as (^yu) d cas yu

_ las yu dc.r as av ir

Denoting by M an upper bound for laclasj on T, we have2 a2 u 2 a 2

dx I dx + M -- 'yu ds. (3,2,4,4)^,; = 1 axi ax; as

Besides this, we have as usual

-au+Au u dx = Vu 2 dx+Adx- au u( ) I I y y^ r

dx. a v

It follows from the boundary condition that

- au uds= c(5-s

u uds=-'i ac u 2 dsr av r ras

For any vector field a, a„ and a T are the normal and tangential components of a on T.

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and consequently

J 1VUJ2 dx+,i Jn n

= J^ (—du+Âu)u dx+2 Jr as lds

^ll—du+AulI Iluli + MJ i _YU 12 do-r

11_áU+aullll ull+MK./E ^ dx+M/K j û^2 dxS2 v E fl

owing to inequality (1,5,1,2) of Lemma 1.5.1.10. Choosing r smallenough (e.g. such that Mk '/r = 2 and A 0 (MKRJr) + 1, we finally obtain

2 IJVUI12 + llull2 = j + Au112 . (3,2,4,5)

On the other hand, we haveaasyly Vul

consequently it follows from inequality (3,2,4,4) that2 1 a2 u 2

dxEJ 1jU12 dx + M k Dul 2 ds^, i = 1 ff2 Iax ax; st r

2 MKJ

dx + MK ^/E

a2 u 2 dx + llVull

2

a Iax ax;

by a new application of Theorem 1.5.1.10. Then we have^ a2u 2

dx = 2 1jU12 dx + L jlVujj2 (3,2,4,6)^,; =1 axj ax;

where L = MK/-/s, MK -✓r = 2.Combining inequalities (3,2,4,5) and (3,2,4,6), we plainly obtain the

desired result. n

Remark 3.2.4.3 The same proof can be worked out, without theconvexity assumption on Q; of course in that case, the constants k and A o

will also depend on the curvature of F.

Remark 3.2.4.4 We can combine the two kinds of computations thatwe did in the proof of Theorems 3.1.2.3 and 3.2.4.2 to deal with the morecomplicated boundary condition

au a—y av —g as

yu + a(yu)

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assuming that !3 is a nondecreasing Lipschitz function. Unfortunately, weshall not be able to take advantage of such a result in what follows.

We now turn to the

Proof of Theorem 3.2.4.1 We use a sequence of plane convex domainsm = 1, 2, ... , as in Lemma 3.2.3.2. We extend c to the whole plane

in such a f ashion that

C E C° . 1 (l2).

Consequently c is defined on Tm . Then, in each D m, we consider u, EH2(Qm ), a solution of

r um +Âum = i in Qm

auma (3,2,4,7)— Ym = C Ymurn on

avm ás,

The existence of such a function um follows from Theorem 2.4.2.7 forA>0. We shall now use inequality (3,2,4,3) to show that l remainsbounded when m --> oo.

Here, we take advantage of Lemma 3.2.3.3 again. Obviouslymax ,, j remains bounded uniformly in m. This implies the exis-tence of constants k and Iv o , both independent of m, such that

`1 um 11Z,Z,n„, k 11f 11o,2,n„, (3,2,4,8)

provided A ; h o .Now we proceed as in the proof of Theorem 3.2.3.1. We first observe

that (3,2,4,7) implies this:

` u •VVdx+A u Vdx= dx— c u Vdsm m ^ a S sim m ^m m„, n,^ r„, m

(3,2,4,9)

for all V e- H 1 ((Im ).On the other hand by inequality (3,2,4,8), there exists a subsequence of

the sequence um, m = 1, 2, ... which is weakly convergent in H 2(f) tosome u e H2(Q). Let us again denote this subsequence u m, m = 1, 2, .. .for the sake of avoiding further complications in the notation. We shallshow that u is the desired solution of (3,2,4,2); this will be achieved bytaking the limit in identity (3,2,4,9).

First let V E C 1 (R2) be fixed. We shall set v = Vin. Exactly as in theproof of Theorem 3.2.1.3 we show that

O um - V V dx --^ Vu - V v dx (3,2,4,10)^^n ,`D

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and that

.L um V dx uv dx. (3,2,4,11)

To complete the proof, we shall show that

c a u V ds = c( V u) V dsa S ym m ^rn m ym m T ^m mr„ m r„,

---^ c V U) v ds = c a u; v. (3,2,4,12)(^ T^ aS ^ yr

This requires much more care. First we fix a partition of unity (Ok ,1 k _ K) on I, and Fm corresponding to the covering Vk, 1k = K inLemma 3.2.3.2. The limit in (3,2,4,12) will follow by adding these limits:

O C(ym Vurn)-rV ds,,, — J Vu)Tu ds. (3,2,4,13)

r,„ r

We shall use the local coordinates of Lemma 3.2.3.2 in each Vk . Let usfirst introduce some auxiliary notation:

W,n = VUm , 'n = ekCV.

From now on, we shall drop the subscript k to simplify the notation. Inaddition, we consider a function U E H2(12) such that UJn = u (seeTheorem 1.4.3.1) we shall consider separately

J -n L -irn (Wm — W)]T„, ds#n

and

i [yrn W]T dsm — -1 [y WIT dsr,,, r

where W=VU.According to the notation in Lemma 3.2.3.2, we have (after dropping

the k) :

-i I-Y►n (Wm — W)]T ►►► dSm a I ym (Wm — W) I dsm

r,,, r,,, n v

= af I%m(Wm — W)(Z, rn(Z))IV'

X J( 1 +pm(z)2)J dzDow

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where a does not depend on m. Owing to (3,2,3,2) we have

n [-Ym (wm — W)]T,n dsm

a v/(1 + L2) I-y(w. — W)(z, Q(z)){ dzv '

+

-ym(wm — W)(z, m(Z)) ^(wm — W)(z, (z))I dz }

V'

b I — Vu)1(0,2,r+ a J(1 + L2)(v„,(z)x J J Dy(Wm_W)(Z , Y)IdYIdZ

V' p(z)

b ( — Vu)II0,2,r+ max I pm () — p(Z)I112 b

(lum — UIt2,2,s2„.ZEV

This implies that

J Í [ J m (wm - W)]Tm dsm -> 0 (3,2,4,14)

since um u weakly in H2(,2) and cpm -^ cp uniformly in V.Next we have

J (YmW)TSrn = 7gim - iym W dsm

= [^ÍTm - ymWI(Z , m(Z)) v [ 1 + m(z)2] dzV'

= (Z, QPm (z))[YmD1 U(z, QPm (Z))V'

+cPm(z)YmD2U(Z, 9Pm(z))idz. (3,2,4,15)

Obviously q (z, (p m (z)) converges uniformly to -q (z, cp (z)) in V'. On theother hand, we have

p(z)

-vmW(Z, (pm(Z)) yW(Z, p(Z))y E^D2W(Z, y)j dyco(z)

Ipm (Z) ^(Z)I1/2

a,12 1/2x jj D2W(z, Y)I2 dY

-a,,/2

and consequently 'y„1 W(z, (Pm (z )) converges to -yW(z, cp (z )) almosteverywhere in V'. Summing up this shows that the integrand in (3,2,4,15)has a limit almost everywhere in V'.Dow

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In addition, we have

I-yr?l W(Z

^ ^p^^. (Z))I ffty1V(z, cp(Z))I + kpm(Z) — 9(Z)I l/2

u,/2 1/2

X 1D2W(Z, y)1 2 dy—a„/2

Thus the integrand in (3,2,4,15) is bounded by a fixed integrable f unctionon V'. One of Lebesgue's theorems implies, therefore, that

J ,7 (ym w)-„„, dsm 'n (yW)T ds. (3,2,4,16)r ,„

Now (3,2,4,13) follows from (3,2,4,14) and (3,2,4,16). Accordingly, wehave shown that (3,2,4,2) hoids for all v e C 1 (,Q). This identity is ex-tended to all v E H' (Q) by density. Finally, the uniqueness of u followsfrom (3,2,4,2) by substituting u for v and assuming A large enough. n

Remark 3.2.4.5 - We can make the lower bound A (, more precise in thestatement of Theorem 3.2.4.1. Indeed, let us set v = u in (3,2,4,2); thuswe obtain

1 IV U 12 dx +,1 j lul 2 dx = J fu dx +i Jr ás lyu l 2 ds.

Then let K be the best possible constant in inequality (1,5,1,2) and set

a^M = max as

r

it follows that

I J (dx^ j fudz+KM

n n r^

x {✓E J dx+Ê Jn (dx}.

Choosing ./£ = 2/KM, we obtain

((

KJ\4)2} jul2dxJ fudx

accordingly we have

- "2) J llull 11fl

and this shows that (KM/2)2 is a possible value for Â.Dow

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Remark 3.2.4.6 In all the previous results the convexity of L can bereplaced by the weaker assumptions that there exists a sequence ,flm ,m = 1, 2,... of bounded open subsets of R", with C 2 boundaries Vm suchthat d (Fm, T) --> 0 as u --> +ce, the sequence of the corresponding secondfundamental forms 9B m , m = 1, 2, ... is uniformly bounded from aboveindependently of m, and Qm f to solve the Dirichlet problem orklm f to solve the other boundary value problems. This assumption isobviously f ulfilled when Q is a bounded open subset of R 2 whoseboundary is a curvilinear polygon of class C l ', provided all the angles arestrictly convex.

3.3 Boundary value problems in domains with turning points

As we observed in Remark 3.2.4.6, the good domains L for the regular-ity in H2(f) are those which are piecewise C 2 with convex corners. Thisis an upper bound on the measure of the possible angles and this leadsnaturally to the question whether turning points (i.e. angles with measurezero) allow the solution of an elliptic boundary value problem to belongto H2(L ). The answer is yes for several boundary conditions as it is shownin Khelif (1978).

We shall consider here only the simplest problem, namely the Dirichletproblem for the Laplace equation in a plane domain D with a boundary Twhich is C 2 everywhere except in one point, which is a turning point. Tobe more precise we assume that this turning point is at 0 and that thereexists p >0 such that, denoting by V the disc with centre at zero andradius p, we have

V (1 Q ={(x, y) E V; x > 0, (P (x) < y < <P2(x)}

where cp, and P 2 are a pair of C 2 functions such that

I (0) = 'P2(0) = 0

<Pí(0)_^P2(0)= 0

(P (x)<<PZ(x), 0<x<p.

Thus, near the origin, the boundary of L is a pair of C 2 curves whichmeet at zero and which are tangent there to the positive half x-axis.

We are going to prove the following result of Ibuki (1974) applying themethod of Khelif (1978) which is simpler and more general.

0

Theorem 3.3.1 Given f E L 2([2) there exists a unique u e H2(fi) n H 1 (,J)

such that

áu = fDow

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3.3 DOMAINS WITH TURNING POINTS 175

Figure 3.2

in D provided

(IV 1(x)I + Itp2(x)I)[p2(x) p1(x)] 2lim sup

x—'O [p2(x) — p1(x))2

(One can observe that those conditions are fulfilled in the following exam-ple: cp 1 (x) = 0 and Q 2(x) = x", where a is any real number >1.)

Exactly as in the previous section the method consists in approximating!Z by a sequence ,f^ m , m = 1, 2, ... , of 'better' domains. For this purposewe consider a decreasing sequence a„ 1 , m = 1, 2,... of positive realnumbers and we set

,klm = n (1{(x, y); x > aan}.

Clearly, we have

x

U m,m=1

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Consequently there exists a unique

um E H2(Qm ) (1 H 1 (Qm )

which is a solution of

Lt urn = f.

in fl,. We are going to show that the sequence

I IUvn 112,2,fi,.

is bounded as m -+ +00.

Lemma 3.3.2 Under the assumptions of Theorem 3.3.1 there exists aconstant K such that

Hum 112,2,f2,,. K lIf IIo,2,n (3, 3,1 )

for every m and every f E L2 (,Q).

Proof Integration by parts of (Aum )um and the Poincaré inequality(Theorem 1.4.3.4) yield a constant C, such that

11Um111.2,n,n Ci lifil0,2,0- (3,3,2)

Then we apply Theorem 3.1.1.2 in order to bound the second deriva-tives of um . For this purpose, we assume that m is fixed and we set

v=DX um , w=Drum . (3,3,3)

The functions v and w only belong to H' (,2 m ) and we approximate themby functions belonging to H2(2m ).

We observe that v and w fulfil the following boundary conditions:

'iw(am, y)=0, p1(a)<y<p2(a)

Âyv + µyw =0 on T fl{(x, y) 1 x > am },

where A and µ are the components of the unit tangent vector to T.Accordingly, there exists a couple of sequences of functions V k , Wk ,k = 1, 2, ... , such that

Vk –^V, Wk—w

in H' (Q11 ) as k --> +x and such that

vk, Wk E H2 (Qm )

ywk(am, y) =0 9 p1(am )<y<QP2(am)

[ Jk ^YVk + WYWk =0 on Ffl{(x, y) I x>a i }.

(We skip the proof of this density result due to its similarity to Lemma4.3.1.3.)Dow

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Applying identity (3,1,1,10) to the vector function {v k , Wk }, we obtain

( +D,,Wkl 2 dx dy = [IDxvki 2 +IDy wkl 2 + 2DXWkDyvk] dx dy^m^m

— (tr 2A) I VI'YVk + V2-YWk 12 do-.rm

Then, taking the limit in k, we have

I + Dwf 2 dx dy = [JDx vj 2 + IDw 12 + 2DX wDv] dx dy^m ^rn

- (tr ) 1 v1 yv + v2-YW 1 2 difrm

and consequently, using (3,3,3), we have

J 1f1 2 dx dy = [IDXum12+IDyum12+2 I DXDY Ul 12 ] dx dyY rni ^nt

áUm 2— (tr 9) y av do.. (3,3,4)

r,n

Let us now consider the boundary integral in (3,3,4). The secondfundamental form vanishes on the segment

{(am, Y), ^P 1(x) < Y < <P2(x)}

which is curvature free. On the other hand the form is bounded oneach C 2 curve; in addition 9B is bounded by Icp1(x)I at (x, p; (x)). Thus weshall consider differently the points of the boundary Tm according to theirdistance to the origin. For this purpose let 6>0 be small enough toensure that the points (b, cp l (b)) and (b, P2(6)) lie in V. Assuming that inis large enough to ensure a m <6, there exists a constant M2 such that

2 2(ûm áum(tr ^) y v da _ M2l -Iydif

r, ► ^ n, cnv

au,, 2+ (x, i (x)) I(P ï(x)I dx

an, Ó v

s aum 2

+ y av (x, 2(x)) j^p'4x)1 dx. (3,3,5)ani

Since

n,=nn{(x,y):x>8}

has a Lipschitz boundary, Theorem 1.5.1.10 implies the existence of aDow

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constant K8 such that

au„, 2

Jrn{(x,y),x-S} y dâv

KSte1/2 [umI 2 +IDy I2+2 IDXDVUMI21 dx dy

+ e L2 I dx dy (3,3,6)

for every e >0.Next on the graph of cp ; (j = 1, 2) we have

Yum (x, ppi (x )) = 0 (3,3,7)

and consequently

y(Dxum)(x, ij (x)) + j(x)y(Dyu )(x, cj (x)) = 0

for a.e. x. On the other hand we have

aura _ — ^P2YDxum + yDyumy av ✓(1 + cp2

at (x, p2(x)). It follows that

au,,, = ^/ 2 + 1 'YLx um + í iDy umy (3v (^P2 ) , _ ,

<P 1 (P 2

and there exists a constant Ns such that

aum N 1 D u + D1(—'s , Iy x m ^1y yumiay ,o ' )^2 t

for am <x S (and N8 --> 1 as S --> 0).The boundary integral corresponding to the graph of (P 2 in (3,3,5) is

bounded by

s Icp2(x)I dxI = N8 I-YDxum (x, ço2(x)) + i (x)Dyum (x, Q2(x))1 2 la.

1p ; (x) — p(x)I 2

(3,3,8)Since we have

,iDxum(x, ^P1(x))+^Pí (x) -iD,u.(x, ^P1(x))=0,

by (3,3,7), we can write

yDDU. (x, (p2(x)) + i(x)'Dyum (x, (P2(x))<o 2(x)

=Dy {Dx um +cpí(x)Dy um }dy.qp (x )D

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Hence

I ,yD. u,. (x, (P2(x)) + 1(x)yDyum (x, p2(x))I, f,(x ) 1/2

[p2(x) — pl(x)]112 iDyDxurn I2 d yw1(x)

cp `(x ► 1/2

+ (x)I [P2(x) p1(x)]'/2 ID2Um I2 dywi(x) I

Finally this implies that

1/2 C 1/2 0(P2"Wi [p2(x) — 1(x)])112

I _ Ns max()<x=8 I^ 1(X) — ^2(x)I

s p,(x) 1/2

X I (2 dx dy

S cp,(x) 1/2

+ max4 11(X), (x) 1 ID,2,um ( 2 dx dy (3,3,9)(1<x

t(x)

We have a similar inequality for the boundary integral corresponding to'Pi' Summing up, there exists a constant M2 such that

2

(tr ^) c^ um

diy a y

M2K4£1/2 [IDzum 1 2 +JDyum(2 + 2 II^xDyuIn1 21dx dyn

+ E A 1/2 J„1 ( 2 dx dy

[Ip 1(x)I + Ip2(x)IJ[p2(x) — pl(x)]+ N max 20<x-8 1£D,(x1—[D-, (x1I

x (1 + ,t) PvDx U rn I 2 dx dy

+ 1+ 1 max ' x 2 + x 2 I D2u 2 dx d(^<xx[I^ 1( )i I^2( )I ] y rn l yƒn. \ fi8

(3,3,10)

for every n >O.Now we can make precise the choice of 6. We choose b small enough

so that

A = ax( ^ [iep' (x)i + 1^0 5(x)][gP2(x) — (Pl(x)]m^ x2 2.

[p 2(X) -

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Then we choose -q in such a way that again

A(1+'ri)<2.

Then we can replace b by a smaller one to ensure that for this value of r^we have

NSA(l+rl)<2

and

NA1+ 1 max [kp(x )1 2 +' p(x ) 2 < 1'.s IIJO<X -s

This is clearly possible since cp i(0) = cp(0) =0 and NS --^ 1 as S --> 0.Finally we choose E small enough so that

M2Kse 112 < 1.

Then the inequality (3,3,10) may be rewritten merely as

`ti ✓J)ly avm

I dQl

a [ Dxum(2 +(DyUm( 2 + 2 IDXDy Un12] dx dy

+13J J 1 2 dx dy (3,3,11)

where a <1 and neither a nor 13 depend on m.Let us go back to the identity (3,3,4). Together with (3,3,11) it yields

[1DXu ►nI2 +I I2] dx dyom

j-- [J 11ƒ 12 dx dy + 13 (V um 1 2 dx dy . (3,3,12)– a nm nm

This last inequality combined with (3,3,2) implies (3,3,1).

Proof of Theorem 3.3.1 This is very similar to the proof of Theorem3.2.1.2. By the same technique we find an increasing sequence of integersmk , k = 1, 2, ... and a function u such that

umk U

weakly in H 1 (fJ),

DiD;umk —D^D;uDow

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weakly in L2((2) such that

au=f

in fl.Obviously u belongs to H 2(L2) and we just have to check that u

belongs to H'((2). Here (2 has no Lipschitz boundary and H'((2) must beunderstood as the closure of 2(D) in H'(fl) (according to Definition1.3.2.2). Indeed u„„ belongs to the closure of g((2mk ) for the norm ofH'(Q,nk ). Therefore umk belongs to the closure of 2(Q) for the norm ofH' (Q) since ,fl lk is a subset of (2. By taking the limit in k it is clear that ubelongs to the closure of 2((2) in H'(Q). n

Remark 3.3.3 In the work of Khelif (1978) conditions on cp, and cP 2 aregiven for the smoothness of the solution of the Laplace equation underother boundary conditions. For instance the conditions corresponding tothe Neumann problem are

l im I^P2W — ^^(x)I c1.lim sup 2

x-Ô [(X) - 1(X)1

Mixed boundary conditions are also considered.

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Elliptic Problems in Nonsmooth Domains || 3. Second-Order Elliptic Boundary Value Problems in Convex Domains