Variational inequality in hilbert spaces and their applications

Table Of Contents

Acknowledgment i Certication ii Approval iii Dedication v Abstract vi Introduction viii

Chapter ONE

1 1 Linear Functional Analysis 1 1.1 Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Function spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.3 Sobolev spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2 Variational Inequalities in RN 20 2.1 Basic Theorems and Denition about Fixed point . . . . . . . . . . . 20 2.2 First Theorem about variational inequalities . . . . . . . . . . . . . . 21 2.3 Some problems leading to variational inequality . . . . . . . . . . . . 24 3 Variational Inequality in Hilbert Spaces 30 3.1 Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 3.2 Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 4 CONCLUSION 38

Project Abstract

The study of variational inequalities frequently deals with a mapping F from a vector space X or a convex subset of X into its dual X0 . Let H be a real Hilbert space and a(u; v) be a real bilinear form on H. Assume that the linear and continuous mapping A H ô€€€! H0 determines a bilinear form via the pairing a(u; v) = hAu; vi. Given K H and f 2 H0 . Then, Variational inequality(VI) is the problem of nding u 2 K such that a(u; v ô€€€ u) hf; v ô€€€ ui, for all v 2 K. In this work, we outline some results in theory of variational inequalities. Their relationships with other problems of Nonlinear Analysis and some applications are also discussed Keywords Sobolev spaces, Variational inequalities,Hilbert Spaces, Elliptic variational inequalities

Project Overview

Linear Functional Analysis The aim of this Chapter is to recall basic results from functional analysis and Distribution theory. The chapter is divided into three sections. The rst section introduces Hilbert spaces and some basic properties of Hilbert spaces. The second section introduces basic concept of Distribution theory and the last section deals with basic results about Sobolev spaces that are of important in the remaining chapters. 1.1 Hilbert Spaces Let us recall some denitions, theorems, and elementary properties on Hilbert spaces. Denition 1.1.1 Let E be a linear space over K, (K = R or C). An inner product on E is a function h:; :i : E E ô€€€! K such that the following are satised, for x; y; z 2 E; ; 2 K: (i) hx; xi 0 and hx; xi = 0 if and only if x = 0 (ii) hx; yi = hx; yi (iii) hx + y; zi = hx; zi + hy; zi. The pair (E; h:; :i) is called an inner product space. Remark 1.1.2 A complete inner product space is called a Hilbert space. Examples 1 Euclidean space: The space RN is a Hilbert space with the inner product dened by hx; yi = PN i=1 xiyi where, x = (x1; x2; :::; xN) and y = (y1; y2; :::; yN): We obtain that 1 kxk = p hx; xi = ( PN i=1 x2i ) 1 2 . 2 Space L2( ). L2( ) := ff : ô€€€! R : f is measurablbe and Zf2dx < 1g where is an open set in RN, is a Hilbert space with the inner product dened by hf; gi = Zf(x)g(x)dx; and kfk = ( Zjf(x)j2dx) 1 2 : Proposition 1.1.3 [2] (Cauchy-Schwart’s Inequality) Let E be an inner prod- uct space. For arbitary x; y 2 E we have jhx; yij2 hx; xihy; yi Proof. Let x; y 2 E be arbitrary. Take z 2 C with jzj = 1 and let t 2 R. Then, 0 htzx + y; tzx + yi = htzx; tzxi + htzx; yi + hy; tzxi + hy; yi = t2zzhx; xi + tzhx; yi + tzhy; xi + hy; yi = t2jzjhx; xi + tzhx; yi + tzhx; yi + hy; yi = t2hx; xi + 2tRe(zhx; yi) + hy; yi t2hx; xi + 2tjzhx; yij + hy; yi = t2hx; xi + 2tjzjjhx; yij + hy; yi = t2hx; xi + 2tjhx; yij + hy; yi: (1.1.1) t2hx; xi + 2tjhx; yij + hx; xi is a quadratic function with variable t 2 R. Since, t2hx; xi + 2tjhx; yij + hy; yi 0, for arbitrary t 2 R, Hence, jhx; yij2 hx; xihy; yi, for all x; y 2 E. Theorem 1.1.4 [2] Let h:; :i be an inner product on E, then the mapping x 7ô€€€! kxk = p hx; xi is a norm on E. Proof. Let x; y 2 E be arbitrary. From the denition of the inner product, we have 2 hx; xi = 0 , x = 0, hence kxk2 = hx; xi = 0 , x = 0. Also, hx; xi = jjhx; xi. And kxk = p hx; xi = p jj2hx; xi = jj p hx; xi = kxk: Let x; y 2 E. Then, kx + yk = hx + y; x + yi = hx; xi + hx; yi + hy; xi + hy; yi = kxk2 + 2Re(hx; yi) + kyk2 kxk2 + 2jhx; yij + kyk2 kxk2 + 2 p (hx; xihy; yi) + kyk2 = kxk2 + 2(kxk + kyk) + kyk2 = (kxk + kyk)2: Denition 1.1.5 Let E be a linear space and F E is said to be convex if for each x; y 2 F and 2 [0; 1] we have x + (1 ô€€€ )y 2 E. Proposition 1.1.6 [4] (Parallelogram Law) Let E be an inner product space, then for x; y 2 E kx + yk2 + kx ô€€€ yk2 = 2(kxk2 + kyk2). Theorem 1.1.7 [4] Let H denote a real Hilbert space and let K be a closed, convex subset of H. Then for each x 2 H there exists unique y 2 K such that kx ô€€€ yk = inffkx ô€€€ k : 2 Kg: (1.1.2) Proof. Let k 2 K be a minimizing sequence such that lim k!1 kk ô€€€ xk = d = inf2Kk ô€€€ xk. Since H is a Hilbert Space, then by the Parallelogram Law, we have kx + yk2 + kx ô€€€ yk2 = 2(kxk2 + kyk2), for all x; y 2 H. 3 Thus, kk ô€€€ pk2 + kk + pk2 = 2(kkk2 + kpk2); for k; p 2 K. By convexity of K, we have 1 2k + (1 ô€€€ 1 2 )p = 1 2 (k + p) 2 K. But d = inf2Kkx ô€€€ k kx ô€€€ k, for all 2 K. Take = 1 2 (k + p). Then, d kx ô€€€ 1 2 (k + p)k ) d2 kx ô€€€ 1 2 (k + p)k2 ) ô€€€d2 ô€€€kx ô€€€ 1 2 (k + p)k2: Now, using the Parallelogram Law and setting y = xô€€€k 2 H and x = xô€€€p 2 H, we obtain that 0 kk ô€€€ pk2 = 2kx ô€€€ kk2 + 2kx ô€€€ pk2 ô€€€ k2x ô€€€ (k + p)k2 = 2kx ô€€€ kk2 + 2kx ô€€€ pk2 ô€€€ 4kx ô€€€ 1 2 (k + p)k2 2kx ô€€€ kk2 + 2kx ô€€€ pk2 ô€€€ 4d2: But lim k!1 kx ô€€€ kk = d. Then, 2kx ô€€€ kk2 + 2kx ô€€€ pk2 ô€€€ 4d2 ô€€€! 0 as n ô€€€! 1. Consequently, kk ô€€€ pk2 ô€€€! 0 as n ô€€€! 1 ) kk ô€€€ pk ô€€€! 0 as n ô€€€! 1 Hence, (k)k0 is Cauchy sequence in K. Since H is a Hilbert space, then there exists ^x 2 H such that k ô€€€! ^x. But K is a closed subset of H and k 2 K, thus ^x 2 K. Therefore, kx ô€€€ ^xk = lim k!1 kx ô€€€ kk = d. For uniqueness: Let ^x; ^y 2 K such that kx ô€€€ ^xk = inf2Kkx ô€€€ k and kx ô€€€ ^yk = inf2Kkx ô€€€ k. 4 By the Parallelogram law and convexity of K, we obtain 0 k^x ô€€€ ^yk = 2kx ô€€€ ^xk2 + 2kx ô€€€ ^yk2 ô€€€ 4kx ô€€€ 1 2 (^x ô€€€ ^y)k2 2d2 + 2d2 ô€€€ 4d2 = 0: Then k^x ô€€€ ^yk = 0 , ^x = ^y. Hence, there is a unique y 2 K such that kx ô€€€ yk = inf2Kkx ô€€€ k. Remark 1.1.8 The point y 2 H satisfying (1.1.2) is called a projection of x on K and y = PKx. Corollary 1.1.9 [2] Let K be a closed, convex subset of a Hilbert space H. Then the operator PK is nonexpansive, that is kPKx ô€€€ PKx0k kx ô€€€ x0k, for all x; x0 2 H. Proof. Let x; x0 2 H such that y = PKx, y0 = PKx0 . Then, for y; y0 2 K we have hy; ô€€€ yi hx; ô€€€ yi, for all 2 K, and hy0 ; ô€€€ y0i hx0 ; ô€€€ y0i , for all 2 K. Setting = y0 and = y in the rst and second inequality respectively we obtain hy; y0 ô€€€ yi hx; y0 ô€€€ yi and hy0 ; y ô€€€ y0i hx0 ; y ô€€€ y0i. Adding we obtain that, hy; y0 ô€€€ yi + hy0 ; y ô€€€ y0i hx; y0 ô€€€ yi + hx0 ; y ô€€€ y0i, hence hy; y0 ô€€€ yi ô€€€ hy0 ; y0 ô€€€ yi hx; y0 ô€€€ yi ô€€€ hx0 ; y0 ô€€€ yi, and hy ô€€€ y0 ; y0 ô€€€ yi hx ô€€€ x0 ; y0 ô€€€ yi. Consequently, ô€€€hy ô€€€ y0 ; y ô€€€ y0i hx ô€€€ x0 ; y0 ô€€€ yi. Then, hy ô€€€ y0 ; y ô€€€ y0i hx ô€€€ x0 ; y ô€€€ y0i 5 ky ô€€€ y 0 k2 = hy ô€€€ y 0 ; y ô€€€ y 0 i hx ô€€€ x 0 ; y ô€€€ y 0 i jhx ô€€€ x 0 ; y ô€€€ y 0 ij kx ô€€€ x 0 kky ô€€€ y 0 k; and thus ky ô€€€ y0k kx ô€€€ x0k. Therefore, kPKx ô€€€ PKx0k kx ô€€€ x0k, for all x; x0 2 H: Theorem 1.1.10 [8] Let K be a closed convex subset of a real Hilbert space H. Then y = PKx, the projection of x on K, if and only if y 2 K such that hy; ô€€€ yi hx; ô€€€ yi, for all 2 K: Proof. Let x 2 H and y = PKx. Since K is convex, then t + (1 ô€€€ t)y = y + t( ô€€€ y) 2 K, for all 2 K, 0 t 1. Set (t) = kx ô€€€ (y + t( ô€€€ y))k2, 0 t 1. (t) = kx ô€€€ y ô€€€ t( ô€€€ y)k2 = kx ô€€€ yk2 ô€€€ 2tRehx ô€€€ y; ô€€€ yi + t2k ô€€€ yk2 = kx ô€€€ yk2 ô€€€ 2thx ô€€€ y; ô€€€ yi + t2k ô€€€ yk2: Then, 0(t) = ô€€€2hx ô€€€ y; ô€€€ yi + 2tk ô€€€ yk2, thus, 0(0) = ô€€€2hx ô€€€ y; ô€€€ yi. Therefore, the function attains its minimum at t = 0. Thus,0 (0) 0 , hx ô€€€ y; ô€€€ yi 0; 2 K , hx; ô€€€ yi ô€€€ hy; ô€€€ y > 0 , hy; ô€€€ yi hx; ô€€€ yi; 2 K: Let y 2 K. Then, hy; ô€€€ yi hx; ô€€€ yi, 2 K. Thus, hy; ô€€€ yi ô€€€ hx; ô€€€ yi 0 and hy ô€€€ x; ô€€€ yi 0 6 0 hy ô€€€ x; ô€€€ yi = hy ô€€€ x; ( ô€€€ x) + (x ô€€€ y)i ô€€€kx ô€€€ yk2 + hy ô€€€ x; ô€€€ xi: Therefore, ky ô€€€ xk2 hy ô€€€ x; ô€€€ xi jhy ô€€€ x; ô€€€ xij ky ô€€€ xkk ô€€€ xk: Thus, ky ô€€€ xk k ô€€€ xk. Hence for each x 2 H there exists y 2 K such that ky ô€€€ xk = inf2Kk ô€€€ xk. Corollary 1.1.11 [2] Let H be a real Hilbert space and K a closed subspace of H. Then, for arbitrary vector x 2 H, there exist a unique vector ~y 2 K such that kx ô€€€ ~yk kx ô€€€ yk, for all y 2 K. Theorem 1.1.12 [2] (Reiesz Theorem) Let H be a Hilbert space. Then H0 = H, where H0 denote the dual of H. Theorem 1.1.13 [4] (Riesz Representation Theorem) Let H be a Hilbert space and let f be a bounded linear functional on H. Then, there exists a unique vector of x0 2 H such that f(x) = hx; x0i, for each x 2 H. Moreover, kfk = kx0k. 1.2 Function spaces We recall some denitions of function spaces used in this thesis Denition 1.2.1 An open connected set RN is called a domain. By , we denote the closure of ; @ is the boundary and o is the interior of . x = (x1; x2; :::; xN) 2 RN and = (1; 2; :::; N) 2 N is a multi-index. jj = 1 + 2 + ::: + N Du := @jju @x1 1 @x2 2 :::@xN N ru = (@1u; @2u; :::; @Nu) jruj = ( XN j=1 [email protected]) 1 2 7 Denition 1.2.2 Let f : ô€€€! R be continuous. We dene support of f by supp(f) = fx 2 : f(x) 6= 0g. The function is said to be of compact support on if the support is a compact set contained inside . The space of test functions in is denoted by D( ) and dened D( ) := ff : ô€€€! R;C1 : support(f ) is compactg = ff 2 C1( ) : supp(f) is compactg Denition 1.2.3 Let ( n)n1 be a sequence in D( ) and 2 D( ). Then, n ! in D( ) if (i) there exists a compact set K such that, supp( n) K, for all n 1. (ii) D n ! D uniformly on K as n ! 1 and for all 2 Nn. Denition 1.2.4 A distribution on is any continuous linear mapping T : D( ) ! R. The set of all distribution on is denoted by D0( ). Means that if n ! 0 in D( ) , then (T; n) ! 0 in R. Example. The map : D(R) ! R dened by h; i = ( ) = (0) is a distribution. It is usually called Dirac distribution. To see this, we have that is linear, since for 1; 2 2 D( ) and 2 R, ( 1 + 2) = h; 1 + 2i = ( 1 + 2)(0) = 1(0) + 2(0) = h; 1i + h; 2i = h; 1i + h; 2i = ( 1) + ( 2): Hence, is linear. Let f ngn1 D(R) such that n ! 0 as n ! 1 on D(R). But n ! 0 on D(R) implies that there exists a compact set K R such that supp( n) K and for all j 2 N, (j) n ! 0 uniformly on R. Thus, 0 j n(0)j supx2Kj n(x)j! 0 as n ! 1. And then h; ni = n(0) ! 0 as n ! 1. Therefore, is continuous and hence is a distribution. Denition 1.2.5 A funtion f : ! R is said to be locally integrable if for any compact set, K , we have that Z K jf(x)jdx < 1: 8 The collection of all locally integrable funtionals is denoted by L1l oc( ). For any f 2 L1l oc( ), f gives a distribution Tf dened by (Tf ; ) = Zf(x) (x)dx; for all 2 D( ): Remark 1.2.6 L1l oc( ) D0( ). Theorem 1.2.7 [6] Let T 2 D0( ) be a distribution on an open set in RN, and , a multi index. Then, for all 2 Nn, n 1 (DT; ) = (ô€€€1)jj(T;D ), D 2 D0( ), for all 2 D( ). Denition 1.2.8 By Ck( ), we denote the space of k times dierentiable (real valued) functions on . Denition 1.2.9 [2] By Ck;( ), 0 < < 1, we indicate the functions k times continuously dierentiable in whose derivative of order k are continuous , 0 < < 1. 1.3 Sobolev spaces Denition 1.3.1 Let be open set in RN. Let p 2 R with 1 p < +1. LP ( ) := ff : ô€€€! R; measurable : Zjfjpd < +1g where is a measure on and kfkp := ( Zjfjpd) 1 p L1( ) := ff : ô€€€! R : f is essentially boundedg, i.e f 2 L1( ) , there exists c > 0 such that jf(x) c a.e on and kfk1 = inffc > 0: jf(x)j c a.e on g. Theorem 1.3.2 [6] LP ( )is a Banach space for 1 p 1. Proposition 1.3.3 [6] (Holder’s Inequality) Let f 2 Lp( ); g 2 Lq( ) such that 1 p + 1 q = 1. Then fg 2 L1( ). Moreover, Zjfgj ( Zjfjp) 1 p ( Zjgjq) 1 q = kfkpkkgkq: (1.3.1) Denition 1.3.4 The space Hm( ) is called Sobolev space of order m and it is dened as Hm( ) := ff 2 L2( ) : Df 2 L2( ); jj mg, endowed with the inner product < f; g >Hm( )=< f; g >L2( ) +jm < Df;Dg >L2( ) : (1.3.2) 9 and kfkHm( ) = (kfkL2( ) + jjmjDfj2) 1 2 , for all f; g 2 Hm( ). Theorem 1.3.5 [6] The spaces Hm( ) , m 0 endowed with the inner product (1.1.1) are Hilbert spaces. Proof. Let (fn)n1 be a Cauchy sequence in Hm( ). Let > 0 be given. Then, there exists no 2 N such that kfn ô€€€ fmkHm( ) for all n;m no. Thus kfn ô€€€ fmk2 L2( ) + P jjmkDfn ô€€€ Dfmk2 < 2, for all n;m no which implies kfn ô€€€ fmk2 L2( ) < 2 and P jjmkDfm ô€€€ Dfmk2 < 2, for all n;m no. then kfn ô€€€ fmk < and P jjmkDfn ô€€€ Dfmk 2, for all n;m no. Thus, we obtain that (fn)n1 is a Cauchy sequence in L2( ) and (D(fn))n is a Cauchy sequence in L2( ). Since L2( ) is complete, then there exist f; fi 2 L2( ) such that fn ô€€€! f in L2( ) as n ô€€€! 1 and Dfn ô€€€! fi in L2( ) as n ô€€€! 1. Since L2( ) D0( ) we obtain that fn ô€€€! f in D0( ) and Dfn ô€€€! fi. But Df ô€€€! Df in D0( ) as n ô€€€! 1. By uniqueness of limit we obtain that Df = fi in D0( ). Thus, fn ô€€€! f in L2( ) and Dfn ô€€€! Df in L2( ), jj m. Thus f 2 Hm( ) with kfn ô€€€ fkL2( ) ô€€€! 0 and P jjmkDfn ô€€€ DfkL2( ) ô€€€! 0 an n ô€€€! 1. Hence, f 2 Hm( ) and kfn ô€€€ fkHm( ) ô€€€! 0 as n ô€€€! 1. Therefore, Hm( ) is a Hilbert space. For m = 1 we obtain H1( ) := ff 2 L2( ) : @f @xi 2 L2( ); i = 1; 2; :::;Ng 10 and on H1( ) we have the following inner product hf; giH1( ) = hf; giL2( ) + XN i=1 h @f @xi ; @g @xi iL2( ) = Zfg + XN i=1 Z@f @xi @g @xi and kfkH1( ) = q hf; giH1( ); for all f 2 H1( ) = vuut kfk2 L2( ) + XN i=1 k @f @xi kL2( ): Denition 1.3.6 We dene H1 0( ) := D( ) jH1( ). H1 0( ) is a Hilbert space with the norm k:kH1( ) and we dene the norm on H1 0( ) by kukH1 0 ( ) := sZjruj2: Theorem 1.3.7 [6] (Poincare’s inequality) Suppose is bounded. Then, there exists c > 0 such that Zu2dx c2 Zjruj2dx; for all u 2 H1 0( ): Proof. Since is bounded. Then Ni =1[ai; bi]. We proceed this way, we rst prove it in D( ). Let ‘ 2 D( ) such that ‘(t; x) = ‘(t; x2; x3; :::; xN). But R t a1 @ @s'(s; x)ds = ‘(t; x) ô€€€ ‘(a1; x) = ‘(t; x): Then using Cauchy Schwartz inequality, we obtain that ‘2(t; x) = ( Z t a1 @ @s ‘(s; x)ds)2 (t ô€€€ a1) Z t a1 ( @ @s ‘(s; x))2ds: 11 Integrating, we obtain that Z‘2(t; x)dtdx Z((t ô€€€ a1) Z t a1 ( @ @s ‘(s; x))2)ds)dtdxZZ b1 a1 (t ô€€€ a1)( @ @s ‘(s; x))2)dsdtdx = Z b1 a1 (t ô€€€ a1)dt Z( @ @s ‘(s; x))2dsdx = 1 2 (b1 ô€€€ a1)2 Z( @ @s ‘(s; x))2dsdx1 2 (b1 ô€€€ a1)2 Zjr’j2: Choose c2 = 1 2 (b1 ô€€€ a1)2 > 0. Then Z‘2 c2 Zjr’j2; for all ‘ 2 D( ): (1.3.3) Now, let u 2 H1 0( ) = D( ) jH1( ). Then, there exist (‘p)p1 D( ) such that k’p ô€€€ uk ! 0 as p ! 1. We have Z‘2 p c2 Zjr’pj2: (1.3.4) Zj’p ô€€€ uj2 + Zjr(‘p ô€€€ u)j2 ! 0; as n ! 1: And thus, Z‘2 p ! Zu2and Zjr’pj2 ! Zjruj2: Letting p ! 1 in equation (1.3.4), we obtain that Zu2 c2 Zjruj2; for all u 2 H1 0( ): Theorem 1.3.8 [6] (Poincare-Wirtinger’s inequality) Suppose is smooth and connected, then for any u 2 H1( ) there exists c > 0 such that Zju ô€€€ ^uj2 c2 Zjruj2; for all u 2 H1( ); where ^u = 1 mes( ) Zu: 12 Corollary 1.3.9 The norm k:kH1 0 ( ) is equivalent to k:kH1( ). Proof of Corollary. Let u 2 H1 0 ( ). Then, kuk2 H1( ) = kuk2 L2( ) + Zjruj2Zjruj2 = kuk2 H1 0 ( ): Thus, kukH1 o ( ) kukH1( ): (1.3.5) Now using Poincare’s inequality we obtain that kuk2 H1( ) = Zu2dx + Zjruj2dx c2 Zjruj2dx + Zjruj2 = (c2 + 1) Zjruj2 = (c2 + 1)kuk2 H1 0 ( ); which implies kukH1( ) (c2 + 1)kukH1 0 ( ) and thus kukH1( ) kukH1 0 ( ); = 1 (c2 + 1) : (1.3.6) Therefore, from equation (1.3.5) and (1.3.6) we obtain kukH1( ) kukH1 0 ( ) kukH1( ). Therefore, the two norms are equivalent on H1 0 ( ). Theorem 1.3.10 [6] Let be smooth in RN;N 2 and D( ) = fu j : u 2 D(RN)g. D( ) = H1( ). Then, : D( ) ! L2(@ ) is continuous with the H1( ) norm. Hence, is exten- sible by continuity over H1( ). i.e : H1( ) ! L2(@ ) is continuous u ! @u = u [email protected] . Moreover, there exists > 0 such that Z @u2d 2kukH1( ); for all u 2 H1( ): 13 Application Let be smooth and connected in RN. Dene ^ V = fu 2 H1( ) : Z @ud = 0g: Then ^ V is closed in H1( ). To see this, let (un)n1 be a sequence in ^ V such that un ! u in H1( ). Since (un)n1 ^ V , then Z @und = 0: Thus, we obtain that Z @jun ô€€€ uj2d 2kun ô€€€ ukH1( ): But un ! u in H1( ), thus kun ô€€€ ukH1( ) ! 0 as n ! 1. Which implies Z @jun ô€€€ uj2d ! 0 as n ! 1: But Z @jun ô€€€ ujd ( Z @jun ô€€€ uj2d) 1 2 (mes(@ )) 1 2 : Hence, since mes(@ ) > 0 we have Z @jun ô€€€ ujd ! 0 in L1( ): And we obtan that Z @und ! Z @ud: But Z @und = 0: Then by uniqueness of limit we obtain that Z @ud = 0: 14 Hence u 2 ^ V and therefore ^ V is closed. Theorem 1.3.11 [6] (Rellich Theorem) If is smooth, then H1( ) ,! L2( ) is compact. Moreover, if (un)n1 is a bounded sequence in H1( ), then there exists a subsequence (unk)k1 of (un)n1 such that (unk)k1 converges in L2( ). Application Let be smooth and connected in RN. Dene V = fu 2 H1( ) : Zu = 0g: Then, Poincare’s inequality is true on V . Thus, there exists c > 0 such that Zu2 Zjruj2; for all u 2 V: We proceed by contradiction. Suppose for all n 2 N there exists (un) 2 V such that Zu2 n > ( p n)2 Zjrunj2; thus Zu2 n > n Zjrunj2: But Zu2 n + Zjrunj2 Zu2 n > n Zjrunj2: Hence, kunkH1( ) > n Zjrunj2: (1.3.7) Let vn = un kunkH1( ) . Then kvnkH1( ) = 1, for all n 1. Since (vn)n1 is bounded in H1( ). Then by Rellich Theorem, there exists a subsequence (vnk)k1 of (vn)n1 and f 2 L2( ) such that vnk ! f in L2( ). Multiplying equation (1.3.7) by 1 kunk2 H1( ) , we obtain that n 1 kunk2 H1( ) Zjrunj2 < 1; for all n 1: 15 Which implies that Zjrvnj2 < 1 n ; for all n 1: Thus Zjrvnj2 ! 0 asn ! 1: And hence Zjrvnk j2 ! 0 as k ! 1: Then, for all i = 1; 2; :::;N @vnk @xi ! 0 in L2( ). But vnk ! f in L2( ) as k ! 1 and since L2( ) D0( ), we obtain that vnk ! f in D0( ) and @vnk @xi ! 0 in D0( ). And by convergence in D0( ), we have that @vnk @xi ! @f @xi ! in D0( ). Thus, by uniqueness of limits @f @xi = 0, for all i = 1; 2; :::;N. And therefore f is constant. Thus f = ~c and by the above argument, we have that vnk ! ~c in H1( ). But V is closed and vnk 2 V , then ~c 2 V . It implies that Z~cdx = 0 and thus ~c = 0. Hence, vnk ! 0 in H1( ) as k ! 1. But kvnkkH1( ) = 1, a contracdiction. Therefore, the claim is true. Denition 1.3.12 More generally, we dene for every 1 p < 1 and for m 0, the Sobolev spaces Wm;p( ) = ff 2 Lp( ) : Df 2 Lp( ); jj mg endowed with the following norm kfkWm;p( ) = kfkp Lp( ) + (jjmkDfkp LP ( )) 1 p . We dene Wm;q 0 = D( ) jWm;q( ) . Thus, Wm;q 0 is the closure of D( ) with respect to the norm k:kWm;q( ). When q = 2, we write Hm( ) = Wm;2( ) and Hm 0 ( ) = Wm;2 0 ( ). For m = 0 we have that W0;q( ) = Lq( ). Theorem 1.3.13 [2] Suppose is smooth, then Wm;q 0 ( ) := ff 2 Wm;q( ) : f = Df = ::: = :::Dmô€€€1f = 0 on @ g. 16 For p = 2, we obtain that Wm;2 0 ( ) := ff 2 Wm;2( ) : f = Df = ::: = Dmô€€€1f = 0 on @ g. Theorem 1.3.14 [6] Wm;p( ) is Banach space. Proof. Let (fn)n1 be a Cauchy in Wm;q( ). Let > 0 be given, then there exists n0 2 N such that kfn ô€€€ fkkWm;q( ) < , for all n; k n0. Then, (kfn ô€€€ fkkq Lq( ) + P jjmkDfn ô€€€ Dfkkq Lq( )) 1 q < , for all n; k n0. And kfn ô€€€ fkkq Lq( ) + P jjmkDfn ô€€€ Dfkkq Lq( ) < q, for all n; k n0. Consequently, kfn ô€€€ fkkq Lq( ) < q and P jjmkDfn ô€€€ Dfkkq Lq( ) < q, for all n; k n0, thus kfn ô€€€ fkkLq( ) < and kDfn ô€€€ DfkkLq( ) < , for all n; k n0. Hence, (fn)n1 and (Dfn)n are Cauchy sequences in Lq( ) and since Lq( ) is complete, then there exists f; fi 2 Lq( ) such that fn ô€€€! f in Lq( ) as n ô€€€! 1 and Dfn ô€€€! fi in Lq( ) as n ô€€€! 1. But Lq( ) D0( ), we obtain that fn ô€€€! f in D0( ) as n ô€€€! 1 and Dfn ô€€€! Df as n ô€€€! 1 in D0( ) By uniqueness of limit we obtain that Df = fi in D0( ). Thus fn ô€€€! f as n ô€€€! 1 in Lq( ) and Dfn ô€€€! Df as n ô€€€! 1 in Lq( ), jj m. Hence kfn ô€€€ fkkLq( ) ô€€€! 0 as n ô€€€! 1 and P jjmkDfn ô€€€ DfkLq( ) ô€€€! 0 as n ô€€€! 1 which implies that kfn ô€€€ fkWm;q( ) ô€€€! 0 as n ô€€€! 1. Thus, f 2 Wm;q( ) and kfn ô€€€ fkWm;q( ) ô€€€! 0 as n ô€€€! 1 in Wm;q( ). Therefore, Wm;q( ) is a Banach space. Theorem 1.3.15 [6] (Green’s Formula) Let be smooth in Rn, u 2 H2( ) and v 2 H1( ). Then, Zrurv = ô€€€ Zuv + Z @( ) @u @n vd; n 2 17 where @u @n denotes the normal derivatives dened by @u @n = ru:~n and ~n denote the normal vector. Denition 1.3.16 The bilinear form a : H H ! R is coercive on H if there exists > 0 such that a(v; v) kvk2, for all v 2 H Example Let be smooth and connected with @ = ô€€€0 [ ô€€€1. Dene H = fu 2 H1( ) : u jô€€€0= 0g. Then, the bilinear form a(u; v) = Zrurv is coercive on H. To see this, we proceed by contradiction. suppose it is not coercive then for all n 1 there exists (un)n 2 H such that a(un; un) < 1 nkunk2 H1( ). Thus, Zjrunj2 < 1 n kunk2 H1( ): (1.3.8) Let vn = un kunkH1( ) . Then, kvnkH1( ) = 1 Multiplying equation (1.3.8) by 1 kunk , we obtain that Zjrvnj2 < 1 n : Which implies that Zjrvnj2 ! 0 in L2( ): But jvnj = 1, hence (vn)n1 is bounded and by Rellich theorem there exists a subsequence (vnk)k1 (vn)n1 such that (vnk) ! g in L2( ). Thus, (vnk) ! g in D0( ) and @vnk @xi ! @g @xi in D0( ). By uniquness of limit in D0( ). Thus, @g @xi = 0. Since is connected we have that g = ^c, a constant. Thus, vnk ! ^c in H1( ). By Trace theorem we obtain that vnk [email protected] ! ^c in L2(@ ). Thus vnk jô€€€0! ^c in L2(ô€€€0). 18 Hence Z ô€€€0 jvnk j2d ! (^c)2mes(ô€€€0): But Z ô€€€0 jvnk j2d ! 0: Therefore, (^c)2mes(ô€€€0) = 0. Since mes(ô€€€0) > 0, then ^c = 0. And hence vnk ! 0 in H1( ). But kvnkkH1( ) = 1, a contradiction. Therefore the bilinear form is coercive on H. Denition 1.3.17 A bilinear form a : H H ô€€€! R is said to be continuous if there exists a constant c > 0 such that ja(u; v)j ckukkvk, for u; v 2 H Example The bilinear form a : H1( ) H1( ) ! R dened by a(u; v) = Zrurv + Z @(x)u(x)v(x)d is continuous; 2 L1(@ ): To see this we apply Cauchy schwartz inequality. Let u; v 2 H1( ), thus ja(u; v)j = j Zrurv + Z @(x)u(x)v(x)djZjrurvj + Z @j(x)u(x)v(x)jd ( Zjruj2) 1 2 ( Zjrvj2) 1 2 + jj Z @ju(x)v(x)jd kukL2( )kvkL2( ) + kk1kukL2(@ )kvkL2(@ ) kukH1( )kvkH1( ) + 2kk1kukH1( )kvkH1( ) = (1 + 2kk1)kukH1( )kvkH1( ): Take c = (1 + 2kk1), then ja(u; v)j ckukH1( )kvkH1( ). Therefore, a is continuous. 19

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