# Project Materials

## Existence and Uniqueness of Solutions of Integral Equations of Hammerstein Type

ABSTRACT

Let X be a real Banach space, X its conjugate dual space. Let A be a monotone angle-bounded continuous linear mapping of X into X with constant of angle-boundedness c 0. Let N be a hemicontinuous (possibly nonlinear) mapping of X into X such that for a given constant k 0;
hv1 􀀀 v2; Nv1 􀀀 Nv2i 􀀀kkv1 􀀀 v2k2
X
for all v1 and v2 in X. Suppose finally that there exists a constant R with

Dedication iii
Preface iv
Acknowledgement vi
Abstract vii
1 General Introduction
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Definition and examples of some basic terms . . . . . . . . . . . 2
1.3 Hammerstein Equations . . . . . . . . . . . . . . . . . . . . . . 10
2 Existence and Uniqueness s Using Factorization of Operators
2.1 Existence and uniqueness theorem . . . . . . . . . . . . . . . . . 13
2.2 of Minty [5] . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.3 of of theorem (2.1.1) . . . . . . . . . . . . . . . . . . . . . . 17
3 Existence and Uniqueness s Using Methods 20
3.1 Gateaux derivative and gradient . . . . . . . . . . . . . . . . . . 20
3.2 Maxima and minima of functions . . . . . . . . . . . . . . . . . 22
3.3 Fundamental theorems of optimization . . . . . . . . . . . . . . 23
3.4 Extension of Vain-berg's result to real
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
Bibliography 30

CHAPTER ONE

General Introduction

1.1 Introduction

The contribution of this falls within the general area of nonlinear functional . Within this area, our attention is focused on the topic: “Existence and Uniqueness of Solutions of Nonlinear Hammerstein Integral Equations” in . We study theorems that establish existence and uniqueness of solutions of these equations using factorization of operators and variational methods.

Several ical problems in the theory of differential equations lead to s. In many cases, these equations can be dealt with in a more satisfactory manner using the integral form than directly with differential
equations.

Interest in Hammerstein equations stem mainly from the fact that several problems that arise in differential equations, for instance, elliptic boundary value problems whose linear parts possess 's function can, as a rule be transformed into a nonlinear integral equation of Hammerstein type.

Elliptic boundary value problems are a of problems which do not involve time variable but only depend on the space variables. That is, they are of problems which are typically associated with steady state behavior. An example is a Laplace's equation: r2u = 0 e.g @2u
@x2 +
@2u
@y2 = 0 in 2D :

Consequently, solvability of such differential equations is equivalent to the solvability of the corresponding Hammerstein equation.

1.2 Definition and examples of some basic terms

In this section, definitions of basic terms used are given. Throughout this chapter, X denotes a real Banach space and X denotes its corresponding dual. We shall denote by the pairing hx; xi or x(x) the value of the functional x 2 X at x 2 X: The norm in X is denoted by k:k, while the norm in X is denoted by k:k. If there is no danger of confusion, we omit the asterisk and denote both norms in X and X by the symbol k:jj. We shall use the symbol ! to indicate strong and * to indicate weak convergence.

We shall also use w! to indicate the weak-star convergence.

The first term we de is monotone map. The concept of monotonicity pertains to nonlinear functional , and its use in the theory of functional equations (ordinary differential equations, s, integro differential equations, delay equations) is probably the most powerful method in obtaining existence theorems.

Definition 1.2.1 (Monotone Operator): A map A : D(A) X ! 2X is said to be monotone if 8 x; y 2 D(A); x 2 Ax; y 2 Ay, we have hx 􀀀 y; x 􀀀 yi 0:

From the definition above, a single-valued map A : D(A) X ! X is monotone if
hAx 􀀀 Ay; x 􀀀 yi 0; 8 x; y 2 D(A):

Remark 1.2.1 For a linear map A, the above definition reduces to hAu; ui 0 8 u 2 D(A):

The following are some examples of monotone operators.

Example 1.2.1 Every nondecreasing function on R is monotone.

of.

Let f : R ! R be a nondecreasing function. Then for arbitrary x; y 2 R, both
(f(x) 􀀀 f(y)) and (x 􀀀 y) have the same sign. Thus we see that
hf(x) 􀀀 f(y); x 􀀀 yi = (f(x) 􀀀 f(y))(x 􀀀 y) 0 8 x; y 2 R. Hence, f is monotone.

Example 1.2.2 Let h : R2 ! R2 be ded as h(x; y) = (2x; 5y); 8 (x; y) 2 R2. Then h is montone.

of.

For arbitrary (x1; y1); (x2; y2) 2 R2; we have hh(x1; y1) 􀀀 h(x2; y2); (x1; y1) 􀀀 (x2; y2)i = 2(x1 􀀀 x2)2 + 5(y1 􀀀 y2)2 0:

Thus, h is monotone.

Example 1.2.3 Let H be a real Hilbert space, I is the identity map of H and

T : H ! H be a non-expansive map (i:e kTx 􀀀 Tyk kx 􀀀 yk 8 x; y 2 H).

Then the operator I 􀀀 T is monotone.

of.
Let x; y 2 H; then
h(I 􀀀 T)x 􀀀 (I 􀀀 T)y; x 􀀀 yi = h(x 􀀀 y) 􀀀 (Tx 􀀀 Ty); x 􀀀 yi
= kx 􀀀 yk2 􀀀 hTx 􀀀 Ty; x 􀀀 yi
kx 􀀀 yk2 􀀀 kTx 􀀀 Tyk:kx 􀀀 yk
kx 􀀀 yk2 􀀀 kx 􀀀 yk2 = 0 (T is nonexpansive).
Thus we have that I 􀀀 T is monotone on H.
Example 1.2.4 Let A = (1 0
0 0) and x = (xy
). Consider the function
g : R2 ! R2 ded by g(x) = Ax: Then g is monotone.
of.
Since g is linear, by remark (1.2.1) it suffices to show that hg(x); xi 0. For
arbitrary x = (xy
) 2 R2; we have Ax = (1 0
0 0)(xy
) = (x0
).
Thus hg(x); xi = hAx; xi = x2 + 0 = x2 0. Hence g is monotone.
Example 1.2.5 Let X be a real Banach space. The duality map J : X ! 2X
ded by
Jx := fx 2 X : hx; xi = kxk:kxk; kxk = kxk; x 2 Xg
is monotone.
of.
Let x; y 2 X and x 2 Jx; y 2 Jy. Then
hx 􀀀 y; x 􀀀 yi = hx 􀀀 y; xi 􀀀 hx 􀀀 y; yi
= hx; xi 􀀀 hy; xi 􀀀 hx; yi + hy; yi
= kxk2 + kyk2 􀀀 hy; xi 􀀀 hx; yi
kxk2 + kyk2 􀀀 kyk:kxk 􀀀 kxk:kyk
= kxk2 + kyk2 􀀀 2kxk:kyk
= (kxk 􀀀 kyk)2 0:
Thus, J is monotone.
Example 1.2.6 Let f : X ! R[f+1g be convex and proper. The subdifferential
of f; @f : X ! 2X ded as
@f(x) =

fx 2 X : hy 􀀀 x; xi f(y) 􀀀 f(x); y 2 Xg ; if f(x) 6= 1
;; if f(x) = 1;
is monotone.
3
of.
Let x; y 2 X; x 2 @f(x) and y 2 @f(y).
x 2 @f(x) ) hy 􀀀 x; xi f(y) 􀀀 f(x) 8 y 2 X: (1.2.1)
y 2 @f(y) ) hx 􀀀 y; yi f(x) 􀀀 f(y) 8 x 2 X
) 􀀀hy 􀀀 x; yi f(x) 􀀀 f(y) 8 x 2 X: (1.2.2)
Adding inequalities (1.2.1) and (1.2.2), we have
hy 􀀀 x; xi 􀀀 hy 􀀀 x; yi 0:
This implies that hy 􀀀 x; x 􀀀 yi 0, i.e hx 􀀀 y; x 􀀀 yi 0.
Definition 1.2.2 (Hemicontinuity): A mapping A : D(A) X ! X is
said to be hemicontinuous if it is continuous from each line segment of X to
the weak topology of X. That is, 8 u 2 D(A); 8 v 2 X and (tn)n1 R+
such that tn ! 0+ and u + tnv 2 D(A) for n sufficiently large, we have
A(u + tnv) * A(u).
position 1.2.1 Let X denote a Banach space and X its corresponding
dual. Let A : D(A) X ! X be a continuous mapping . Then A is
hemicontinuous.
of
Let u 2 D(A); v 2 X, (tn)n1 be a sequence of positive numbers such that
tn ! 0+ as n ! 1 and (u + tnv) 2 D for n large enough. We observe that
(u + tnv) ! u as n ! 1 because tn ! 0+ as n ! 1. By the continuity
of A, we have A(u + tnv) ! A(u) as n ! 1. Since strong convergence
implies weak convergence we have A(u + tnv) * A(u) as n ! 1: Hence A is
hemicontinuous.
Remark 1.2.2 The converse of proposition (1.2.1) is false.
Consider the function f : R2 ! R2 ded by
f(x; y) =
(
( x2+xy2
x2+y4 ; x); if (x; y) 6= (0; 0)
(1; 0); if (x; y) = (0; 0):
Clearly, f is not continuous at (0; 0). For,
f(x; 0) = ( x2
x2 ; x) = (1; x) for all x 6= 0: This implies lim
x!0
f(x; 0) = (1; 0).
f(0; y) = (0; 0); 8y 6= 0. This implies lim
y!0
f(0; y) = (0; 0). Thus, the
limit does not exist at (0; 0). Hence, f is not continuous at (0,0).
4
However, f is hemicontinuous. Indeed, let u = (0; 0); v = (v1; v2) and
ftngn1 be arbitrary such that tn ! 0+ as n ! 1. Then,
f(u + tnv) = f(tnv1; tnv2)) =

v2
1+tnv1v2
2
v2
1+t2
nv4
2
; tnv1

! (1; 0); as n ! 1: Therefore,
lim
n!1
f(u + tnv) = (1; 0) = f(0; 0). Thus, f(u + tnv) ! f(u) as tn ! 0+.
Hence, f is hemicontinuous on R2 since strong and weak convergence are the
same on R2.
Definition 1.2.3 (Coercivity): An operator A : X ! X is said to be
coercive if for any x 2 X; hx;Axi
kxk ! 1 as kxk ! 1:
Example 1.2.7 Let H be a real Hilbert space and f : H ! H be ded by
f(x) = 1
2u. Then, f is coercive.
of.
Let x 2 H be arbitrary. Then,
hf(x); xi
kxk
=
1
2 hx; xi
kxk
=
1
2kxk2
kxk
=
1
2
kxk ! +1 as kxk ! 1:
Hence f is coercive.
Definition 1.2.4 (Symmetry): Let A : X ! X be a bounded linear mapping.
A is said be symmetric if for all u and v in X, we have hAu; vi = hAv; ui :
Example 1.2.8 Let A : l2(R) ! l2(R) be a map ded by Au = 1
2u. Then
A is symmetric.
of.
For arbitrary u; v 2 l2;
hAu; vi =

1
2
u; v

=
1
2
h(u1; u2; :::); (v1; v2; :::)i =
1
2
X1
i=1
uivi
=
1
2
X1
i=1
viui =
1
2
h(v1; v2; :::); (u1; u2; :::)i
=

1
2
v; u

= hu; Avi :
Hence A is symmetric.

Definition 1.2.5 (Skew-symmetry): Let A : X ! X be a bounded linear mapping. A is said be skew-symmetric if for all u and v in X, we have hAu; vi = 􀀀hAv; ui :

Definition 1.2.6 (Angle-boundedness): Let A : X ! X be a bounded monotone linear mapping . A is said be angle-bounded with constant c 0 if for all u, v in X, j hAu; vi􀀀hAv; ui j 2c fhAu; uig
1
2 fhAv; vig
1
2 . (This is well ded since hAu; ui 0 and hAv; vi 0 by the linearity and monotonicity of A).

Example 1.2.9 A symmetric map. It follows that every symmetric mapping A of X into X is angle-bounded with constant of angle-boundedness c = 0:

Definition 1.2.7 (Adjoint Operators): Let X and Y be normed linear spaces and A 2 B(X; Y ): The adjoint of A, denoted by A, is the operator A : Y ! X ded by hAy; xi = hy; Axi for all y 2 Y and all
x 2 X.

We note that A is well-ded. Indeed, 8 y 2 Y ; x1; x2 2 X and 2 R,
we have
hAy; x1 + x2i = hy;A(x1 + x2)i = hy; Ax1i + hy; Ax2i
= hy; Ax1i + hy; Ax1i
which shows that Ay is linear.
For boundedness, given y 2 Y and x 2 X;
j hAy; xi j = j hy; Axi j
kyk:kAxk since y 2 Y .
kyk:kAk:kxk since A 2 B(X; Y ).
Therefore, for all y 2 Y ,
j hAy; xi j Kykxk 8 x 2 X; where Ky = kyk:kAk 0:
Hence, for all y 2 Y ;Ay 2 X:
1.2.1 Let A : X ! Y be a bounded linear maps with adjoint A.
Then,
(a) A 2 B(Y ;X);
(b) kAk = kAk.
of.
(a) Let y; z 2 Y and 2 R. We show that
A (y + z) = Ay + Az;
6
i.e
8 x 2 X; hA (y + z) ; xi = hAy; xi + hAz; xi :
Let x 2 X: Then
hA (y + z) ; xi = hy + z; Axi = hy; Axi + hz; Axi
= hAy; xi + hAz; xi :
So, A is linear.
Furthermore, for any y 2 Y and x 2 X,
j hAy; xi j = j hy; Axi j kyk:kAk:kxk; since A 2 B(X; Y ) :
Thus, kAyk = sup
kxk=1
j hAy; xi j kAk:kyk: Therefore, kAyk
Kkyk; where K = kAk 0: Hence A 2 B(Y ;X).
(b)
kAk = sup
kxk=1
kAxk = sup
kxk=1

sup
kyk=1
hy; Axi
!
= sup
kxk=1

sup
kyk=1
hAy; xi
!
= sup
kyk=1

sup
kxk=1
hAy; xi
!
= sup
kyk=1
kAyk = kAk:

Definition 1.2.8 (Weak Topology): Let (X; !) denote a Banach space endowed with the weak topology. For an arbitrary sequence fxngn1 X and x 2 X, we say that fxng converges weakly to x if f(xn) ! f(x) for each
f 2 X. We denote this by xn * x:

Definition 1.2.9 (Weak Star Topology): Let (X; !) denote a Banach space endowed with the weak star topology. For an arbitrary sequence ffngn1 X and f 2 X we say that ffng converges to f in weak-star topology, denoted fn!
􀀀! f, if fn(x) ! f(x) for each x 2 X.
position 1.2.2 Let fxng be a sequence and x a point in X. Then the following hold.
(a) xn ! x ) xn * x;
(b) xn * x ) fxng is bounded and kxk lim inf kxnk;
(c) xn * x (in X), fn ! f (in X) ) fn(xn) ! f(x) (in R).
Definition 1.2.10 (Reflexive Space): Let X be a Banach space and let
J : X ! X be the canonical injection from X into X, that is hJ(x); fi =
hf; xi ; 8 x 2 X; f 2 X. Then X is said to be reflexive if J is surjective, i.e
J(X) = X:
Definition 1.2.11 (Uniformly convex ): A Banach space X is called uniformly convex if for any 2 (0; 2], there exists a = () > 0
such that if x; y 2 X, with kxk 1; kyk 1 and kx 􀀀 yk , then
k1
2 (x + y)k 1 􀀀 .
Hilbert spaces, Lp and lp spaces, 1 algebra): A collection M of subsets of a nonempty
set
is called a 􀀀algebra if
(a) ;
2M,
(b) A2 M ! Ac 2 M,
(c) [1 n=1An 2 M whenever An 2 M 8 n.
Definition 1.2.14 (Measurable Space): If M is a algebra of, then the pair (;M) is referred to as a measurable space.
Definition 1.2.15 (Measure): A measure on (; M) is a function
: M! [0; 1] such that
(a) (A) 0 for all A 2M;
(b) () = 0;
(c) if Ai 2M are pairwise disjoint, then ([1i
Ai) =
P1
i=1 (Ai).

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