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 ?! 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 ? u) hf; v ? 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

 

 

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 ?! 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 :
?! 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-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 7?! 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 ? )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 ? 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 ? yk = inffkx ? k : 2 Kg: (1.1.2)
Proof.
Let k 2 K be a minimizing sequence such that
lim
k!1
kk ? xk = d = inf2Kk ? xk.
Since H is a Hilbert Space, then by the Parallelogram Law, we have
kx + yk2 + kx ? yk2 = 2(kxk2 + kyk2), for all x; y 2 H.
3
Thus,
kk ? pk2 + kk + pk2 = 2(kkk2 + kpk2); for k; p 2 K.
By convexity of K, we have
1
2k + (1 ? 1
2 )p = 1
2 (k + p) 2 K.
But d = inf2Kkx ? k kx ? k, for all 2 K.
Take = 1
2 (k + p). Then,
d kx ?
1
2
(k + p)k
) d2 kx ?
1
2
(k + p)k2
) ?d2 ?kx ?
1
2
(k + p)k2:
Now, using the Parallelogram Law and setting y = x?k 2 H and x = x?p 2 H,
we obtain that
0 kk ? pk2
= 2kx ? kk2 + 2kx ? pk2 ? k2x ? (k + p)k2
= 2kx ? kk2 + 2kx ? pk2 ? 4kx ?
1
2
(k + p)k2
2kx ? kk2 + 2kx ? pk2 ? 4d2:
But lim
k!1
kx ? kk = d. Then,
2kx ? kk2 + 2kx ? pk2 ? 4d2 ?! 0 as n ?! 1.
Consequently,
kk ? pk2 ?! 0 as n ?! 1
) kk ? pk ?! 0 as n ?! 1
Hence, (k)k0 is Cauchy sequence in K. Since H is a Hilbert space, then there
exists ^x 2 H such that
k ?! ^x.
But K is a closed subset of H and k 2 K, thus ^x 2 K.
Therefore,
kx ? ^xk = lim
k!1
kx ? kk = d.
For uniqueness: Let ^x; ^y 2 K such that
kx ? ^xk = inf2Kkx ? k
and
kx ? ^yk = inf2Kkx ? k.
4
By the Parallelogram law and convexity of K, we obtain
0 k^x ? ^yk
= 2kx ? ^xk2 + 2kx ? ^yk2 ? 4kx ?
1
2
(^x ? ^y)k2
2d2 + 2d2 ? 4d2 = 0:
Then k^x ? ^yk = 0 , ^x = ^y. Hence, there is a unique y 2 K such that
kx ? yk = inf2Kkx ? 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 ? PKx0k kx ? 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; ? yi hx; ? yi, for all 2 K,
and
hy0 ; ? y0i hx0 ; ? y0i , for all 2 K.
Setting = y0 and = y in the rst and second inequality respectively we obtain
hy; y0 ? yi hx; y0 ? yi and hy0 ; y ? y0i hx0 ; y ? y0i.
Adding we obtain that,
hy; y0 ? yi + hy0 ; y ? y0i hx; y0 ? yi + hx0 ; y ? y0i,
hence
hy; y0 ? yi ? hy0 ; y0 ? yi hx; y0 ? yi ? hx0 ; y0 ? yi,
and
hy ? y0 ; y0 ? yi hx ? x0 ; y0 ? yi.
Consequently,
?hy ? y0 ; y ? y0i hx ? x0 ; y0 ? yi.
Then,
hy ? y0 ; y ? y0i hx ? x0 ; y ? y0i
5
ky ? y
0
k2 = hy ? y
0
; y ? y
0
i
hx ? x
0
; y ? y
0
i
jhx ? x
0
; y ? y
0
ij
kx ? x
0
kky ? y
0
k;
and thus
ky ? y0k kx ? x0k.
Therefore,
kPKx ? PKx0k kx ? 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; ? yi hx; ? yi, for all 2 K:
Proof.
Let x 2 H and y = PKx. Since K is convex, then
t + (1 ? t)y = y + t( ? y) 2 K, for all 2 K, 0 t 1.
Set (t) = kx ? (y + t( ? y))k2, 0 t 1.
(t) = kx ? y ? t( ? y)k2
= kx ? yk2 ? 2tRehx ? y; ? yi + t2k ? yk2
= kx ? yk2 ? 2thx ? y; ? yi + t2k ? yk2:
Then,
0(t) = ?2hx ? y; ? yi + 2tk ? yk2,
thus, 0(0) = ?2hx ? y; ? yi. Therefore, the function attains its minimum at
t = 0. Thus,

0
(0) 0 , hx ? y; ? yi 0; 2 K
, hx; ? yi ? hy; ? y > 0
, hy; ? yi hx; ? yi; 2 K:
Let y 2 K. Then,
hy; ? yi hx; ? yi, 2 K.
Thus,
hy; ? yi ? hx; ? yi 0
and
hy ? x; ? yi 0
6
0 hy ? x; ? yi
= hy ? x; ( ? x) + (x ? y)i
?kx ? yk2 + hy ? x; ? xi:
Therefore,
ky ? xk2 hy ? x; ? xi
jhy ? x; ? xij
ky ? xkk ? xk:
Thus,
ky ? xk k ? xk.
Hence for each x 2 H there exists y 2 K such that
ky ? xk = inf2Kk ? 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 ? ~yk kx ? 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 :
?! 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 :
?! 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; ) = (?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 :
?! R; measurable :
Z

jfjpd < +1g
where is a measure on
and
kfkp := (
Z

jfjpd)
1
p
L1(
) := ff :
?! 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,
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 ? fmkHm(
) for all n;m no.
Thus
kfn ? fmk2
L2(
) +
P
jjmkDfn ? Dfmk2 < 2, for all n;m no
which implies
kfn ? fmk2
L2(
) < 2 and
P
jjmkDfm ? Dfmk2 < 2, for all n;m no.
then
kfn ? fmk < and
P
jjmkDfn ? 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 ?! f in L2(
) as n ?! 1 and Dfn ?! fi in L2(
) as n ?! 1.
Since L2(
) D0(
) we obtain that
fn ?! f in D0(
) and Dfn ?! fi.
But Df ?! Df in D0(
) as n ?! 1. By uniqueness of limit we obtain that
Df = fi in D0(
). Thus,
fn ?! f in L2(
) and Dfn ?! Df in L2(
), jj m.
Thus f 2 Hm(
) with
kfn ? fkL2(
) ?! 0 and
P
jjmkDfn ? DfkL2(
) ?! 0 an n ?! 1.
Hence,
f 2 Hm(
) and kfn ? fkHm(
) ?! 0 as n ?! 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] (Poincare’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 ‘ 2 D(
) such that ‘(t; x) = ‘(t; x2; x3; :::; xN).
But
R t
a1
@
@s'(s; x)ds = ‘(t; x) ? ‘(a1; x) = ‘(t; x): Then using Cauchy Schwartz
inequality, we obtain that
‘2(t; x) = (
Z t
a1
@
@s
‘(s; x)ds)2
(t ? a1)
Z t
a1
(
@
@s
‘(s; x))2ds:
11
Integrating, we obtain that
Z

‘2(t; x)dtdx
Z

((t ? a1)
Z t
a1
(
@
@s
‘(s; x))2)ds)dtdx

Z

Z b1
a1
(t ? a1)(
@
@s
‘(s; x))2)dsdtdx
=
Z b1
a1
(t ? a1)dt
Z

(
@
@s
‘(s; x))2dsdx
=
1
2
(b1 ? a1)2
Z

(
@
@s
‘(s; x))2dsdx

1
2
(b1 ? a1)2
Z

jr’j2:
Choose c2 = 1
2 (b1 ? a1)2 > 0. Then
Z

‘2 c2
Z

jr’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 ? uk ! 0 as p ! 1.
We have Z

‘2
p c2
Z

jr’pj2: (1.3.4)
Z

j’p ? uj2 +
Z

jr(‘p ? u)j2 ! 0; as n ! 1:
And thus,
Z

‘2
p !
Z

u2and
Z

jr’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-Wirtinger’s inequality) Suppose
is smooth
and connected, then for any u 2 H1(
) there exists c > 0 such that
Z

ju ? ^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 Poincare’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 [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 ? uj2d 2kun ? ukH1(
):
But un ! u in H1(
), thus kun ? ukH1(
) ! 0 as n ! 1. Which implies
Z
@

jun ? uj2d ! 0 as n ! 1:
But
Z
@

jun ? ujd (
Z
@

jun ? uj2d)
1
2 (mes(@
))
1
2 :
Hence, since mes(@
) > 0 we have
Z
@

jun ? 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, Poincare’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 = ::: = :::Dm?1f = 0 on @
g.
16
For p = 2, we obtain that
Wm;2
0 (
) := ff 2 Wm;2(
) : f = Df = ::: = Dm?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 ? fkkWm;q(
) < , for all n; k n0.
Then,
(kfn ? fkkq
Lq(
) +
P
jjmkDfn ? Dfkkq
Lq(
))
1
q < , for all n; k n0.
And
kfn ? fkkq
Lq(
) +
P
jjmkDfn ? Dfkkq
Lq(
) < q, for all n; k n0.
Consequently,
kfn ? fkkq
Lq(
) < q and
P
jjmkDfn ? Dfkkq
Lq(
) < q, for all n; k n0,
thus
kfn ? fkkLq(
) < and kDfn ? 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 ?! f in Lq(
) as n ?! 1 and Dfn ?! fi in Lq(
) as n ?! 1.
But Lq(
) D0(
), we obtain that
fn ?! f in D0(
) as n ?! 1 and Dfn ?! Df as n ?! 1 in D0(
)
By uniqueness of limit we obtain that Df = fi in D0(
). Thus
fn ?! f as n ?! 1 in Lq(
) and Dfn ?! Df as n ?! 1 in Lq(
),
jj m. Hence
kfn ? fkkLq(
) ?! 0 as n ?! 1 and
P
jjmkDfn ? DfkLq(
) ?! 0 as n ?! 1
which implies that kfn ? fkWm;q(
) ?! 0 as n ?! 1. Thus,
f 2 Wm;q(
) and kfn ? fkWm;q(
) ?! 0 as n ?! 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,
Z

rurv = ?
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 @
= ?0 [ ?1. Dene
H = fu 2 H1(
) : u j?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 [email protected]
! ^c in L2(@
).
Thus
vnk j?0! ^c in L2(?0).
18
Hence Z
?0
jvnk j2d ! (^c)2mes(?0):
But Z
?0
jvnk j2d ! 0:
Therefore, (^c)2mes(?0) = 0. Since mes(?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 ?! 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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