Existence Results on General Integrodifferential Evolution Equations in Banach Space ()
1. Introduction
Pazy [1] has discussed the existence and uniqueness of mild, strong and classical solutions of semilinear evolution equations by using semigroup theory. The nonlocal Cauchy problem for the same equation has been studied by Byszewskii [2,3]. Balachandran and Chandrasekaran [4] investigated the nonlocal Cauchy problem for semilinear integrodifferential equation with deviating argument. Balachandran and Park [5] have discussed about the existence of solutions and controllability of nonlinear integrodifferential systems in Banach spaces. Grimmer [6] obtained the representation of solutions of integrodifferential equations by using resolvent operators in a Banach space. Liu [7] discussed the Cauchy problem for integrodifferential evolution equations in abstract spaces and also in [8] he discussed nonautonomous integrodifferential equations.
Lin and Liu [9] studied the nonlocal Cauchy problem for semilinear integrodifferential equations by using resolvent operators. Liu and Ezzinbi [10] investigated nonautonomous integrodifferential equations with nonlocal conditions. Byszewskii and Acka [11] studied the classical solution of nonlinear functional differential equation with time varying delays. There are several papers appeared on the existence of differential and integrodifferential equations in Banach spaces [12,13]. The purpose of this paper is to prove the existence of mild solutions for time varying delay integrodifferential evolution equations with the help of Schaefer’s fixed point theorem. The results generalize the results of [14].
The paper is organized as follows: In Section 2, we give the necessary definition and gave a description of the idea of the proof of the main results formulated and proved in Section 3. Moreover in Section 3, we prove the existence of solution of general integrodifferential evolution equation with nonlocal condition.
2. Preliminaries
Consider the nonlinear delay integrodifferential evolution equation with nonlocal condition of the form
(1)
(2)
where A(t) and B(t,s) are closed linear operators on a Banach space X with dense domain D(A) which is independent of t, 
and 
are given functions. Here 
.
We shall make the following conditions:
A(t) generates a strongly continuous semigroup of evolution operators.
Suppose Y is a Banach space formed from D(A) with the graph norm. A(t) and B(t,s) are closed operators it follows that A(t) and B(t,s) are in the set of bounded linear operators from Y to X, B(Y,X), for 
and 
, respectively. A(t) and B(t,s) are continuous on
, respectively, into B(Y,X).
Definition 2.1. A resolvent operator for (1) and (2) is a bounded operator valued function 
, the space of bounded linear operators on X, having the following properties.
(i) R(t,s) is strongly continuous in s and t. R(t,t)=I, the identity operator on X. 
and 
are constants.
(ii) 
is strongly continuous in s and t on Y.
(iii) For 
is continuously differentiable in s and t, and for
,
with 
and 
are strongly continuous on
. Here R(t,s) can be extracted from the evolution operator of the generator A(t). The resolvent operator is similar to the evolution operator for nonautonomous differential equations in Banach spaces.
Definition 2.2. A continuous function x(t) is said to be a mild solution of the nonlocal Cauchy problems (1) and (2), if
is satisfied.
Schaefer’s Theorem [15]. Let E be a normed linear space. Let 
be a completely continuous operator, that is, it is continuous and the image of any bounded set is contained in a compact set and let
Then either 
is unbounded or F has a fixed point.
Assume that the following conditions hold:
There exists a resolvent operator R(t,s) which is compact and continuous in the uniform operator topology for
. Further, there exists a constant 
such that
The function 
is continuous and there exists a constant 
such that 
for any
.
For each
, the function
is continuous and for each
the function
is strongly measurable.
There exists an integrable function 
such that
for any 
where 
is a continuous nondecreasing function.
Ther exists an integrable function 
such that
where 
is a continuous nondecreasing function.
The function 
is completely continuous and there exists a constant 
such that
and is equicontinuous in (J,X)
The function 
is completely continuous and there exists a constant 
such that
and is equicontinuous in (J,X)
There are function 
such that
The function
where 
3. Existence of Mild Solutions
The main result is as follows.
Theorem 3.1. If the assumptions 
are satisfied then the problems (1) and (2) has a mild solution on J.
Proof: Consider the Banach space Z = C(J,X). We establish the existence of a mild solution of the problems (1) and (2) by applying the Schaefer’s fixed point theorem.
First we obtain a priori bounds for the operator equation
(3)
where 
is defined as
(4)
Then froms (3) and (4) we have
Denoting the right hand side of the above inequality as
. Then 
and
.
This implies
(5)
where 
Inequality (5) implies that there is a constant K such that 
and hence we have
where K depends only on T and on the functions
.
We shall now prove that the operator 
is a completely continuous operator. Let 
for some
. We first show that 
maps 
into an equicontinuous family.
Let 
and
. Then if 
The right hand side is independent of 
and tends to zero as
, since f is completely continuous and by 
for 
is continuous in the uniform operator topology. Thus 
maps 
into an equicontinuous family of functions.
It is easy to see that 
is uniformly bounded. Nextwe show 
is compact. Since we have shown 
is equicontinuous collection, by the Arzela-Ascoli theorem it suffices to show that 
maps 
into a precompact set in X.
Let 
be fixed and let 
be a real number satisfying
. For
, we define
(6)
Since R(t,s) is a compact operator, the set 
is precompact in X for every
. Moreover, for every 
we have
Therefore there are precompact sets arbitrarily close to the set
.
Hence, the set 
is precompact in X.
It remains to show that 
is continuous. Let 
with 
in Z. Then there is an integer q such that 
for all n and
, so 
and
. By
,
for each 
and since
and
we have by dominated convergence theorem
Thus 
is continuous. This completes the proof that 
is completely continuous.
Finally the set 
is bounded, as we proved in the first step. Consequently, by Schaefer’s theorem, the operator 
has a fixed point in Z. This means that any fixed point of 
is a mild solution of (1) and (2) on J satisfying
.
NOTES