## 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

**Â **

## 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

## CHAPTER ONE

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

Z

f2dx < 1g

where

is an open set in RN, is a Hilbert space with the inner product dened by

hf; gi =

Z

f(x)g(x)dx;

and

kfk = (

Z

jf(x)j2dx)

1

2 :

Proposition 1.1.3 [2] (Cauchy-Schwartâ€™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

j@juj2)

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 ; ) =

Z

f(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 :

Z

jfjpd < +1g

where is a measure on

and

kfkp := (

Z

jfjpd)

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] (Holderâ€™s Inequality) Let f 2 Lp(

); g 2 Lq(

) such that

1

p + 1

q = 1. Then fg 2 L1(

). Moreover,

Z

jfgj (

Z

jfjp)

1

p (

Z

jgjq)

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(

)

=

Z

fg +

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 (

) :=

sZ

jruj2:

Theorem 1.3.7 [6] (Poincareâ€™s inequality) Suppose

is bounded. Then, there

exists c > 0 such that

Z

u2dx c2

Z

jruj2dx; 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 â€˜ 2 D(

) such that â€˜(t; x) = â€˜(t; x2; x3; :::; xN).

But

R t

a1

@

@s'(s; x)ds = â€˜(t; x) ô€€€ â€˜(a1; x) = â€˜(t; x): Then using Cauchy Schwartz

inequality, we obtain that

â€˜2(t; x) = (

Z t

a1

@

@s

â€˜(s; x)ds)2

(t ô€€€ a1)

Z t

a1

(

@

@s

â€˜(s; x))2ds:

11

Integrating, we obtain that

Z

â€˜2(t; x)dtdx

Z

((t ô€€€ a1)

Z t

a1

(

@

@s

â€˜(s; x))2)ds)dtdx

Z

Z b1

a1

(t ô€€€ a1)(

@

@s

â€˜(s; x))2)dsdtdx

=

Z b1

a1

(t ô€€€ a1)dt

Z

(

@

@s

â€˜(s; x))2dsdx

=

1

2

(b1 ô€€€ a1)2

Z

(

@

@s

â€˜(s; x))2dsdx

1

2

(b1 ô€€€ a1)2

Z

jrâ€™j2:

Choose c2 = 1

2 (b1 ô€€€ a1)2 > 0. Then

Z

â€˜2 c2

Z

jrâ€™j2; for all â€˜ 2 D(

): (1.3.3)

Now, let u 2 H1

0(

) = D(

) jH1(

). Then, there exist (â€˜p)p1 D(

) such that

kâ€™p ô€€€ uk ! 0 as p ! 1.

We have Z

â€˜2

p c2

Z

jrâ€™pj2: (1.3.4)

Z

jâ€™p ô€€€ uj2 +

Z

jr(â€˜p ô€€€ u)j2 ! 0; as n ! 1:

And thus,

Z

â€˜2

p !

Z

u2and

Z

jrâ€™pj2 !

Z

jruj2:

Letting p ! 1 in equation (1.3.4), we obtain that

Z

u2 c2

Z

jruj2; for all u 2 H1

0(

):

Theorem 1.3.8 [6] (Poincare-Wirtingerâ€™s inequality) Suppose

is smooth

and connected, then for any u 2 H1(

) there exists c > 0 such that

Z

ju ô€€€ ^uj2 c2

Z

jruj2; for all u 2 H1(

);

where

^u =

1

mes(

)

Z

u:

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(

) +

Z

jruj2

Z

jruj2

= kuk2

H1

0 (

):

Thus,

kukH1

o (

) kukH1(

): (1.3.5)

Now using Poincareâ€™s inequality we obtain that

kuk2

H1(

) =

Z

u2dx +

Z

jruj2dx

c2

Z

jruj2dx +

Z

jruj2

= (c2 + 1)

Z

jruj2

= (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 j@

.

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(

) :

Z

u = 0g:

Then, Poincareâ€™s inequality is true on V . Thus, there exists c > 0 such that

Z

u2

Z

jruj2; for all u 2 V:

We proceed by contradiction. Suppose for all n 2 N there exists (un) 2 V such that

Z

u2

n > (

p

n)2

Z

jrunj2;

thus

Z

u2

n > n

Z

jrunj2:

But

Z

u2

n +

Z

jrunj2

Z

u2

n > n

Z

jrunj2:

Hence,

kunkH1(

) > n

Z

jrunj2: (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(

)

Z

jrunj2 < 1; for all n 1:

15

Which implies that

Z

jrvnj2 <

1

n

; for all n 1:

Thus Z

jrvnj2 ! 0 asn ! 1:

And hence Z

jrvnk 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] (Greenâ€™s Formula)

Let

be smooth in Rn, u 2 H2(

) and v 2 H1(

). Then,

Z

rurv = ô€€€

Z

uv +

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) =

Z

rurv

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, Z

jrunj2 <

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

Z

jrvnj2 <

1

n

:

Which implies that Z

jrvnj2 ! 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 j@

! ^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) =

Z

rurv +

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

Z

rurv +

Z

@

(x)u(x)v(x)dj

Z

jrurvj +

Z

@

j(x)u(x)v(x)jd

(

Z

jruj2)

1

2 (

Z

jrvj2)

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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