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In this paper we prove the existence of mild solutions of a general class of nonlinear evolution integrodifferential equation in Banach spaces. Based on the resolvent operator and the Schaefer fixed point theorem, a sufficient condition for the existence of general integrodifferential evolution equations is established.

Pazy [

Lin and Liu [

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.

Consider the nonlinear delay integrodifferential evolution equation with nonlocal condition of the form

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 [

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

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

where is defined as

Then froms (3) and (4) we have

Denoting the right hand side of the above inequality as. Then and

.

This implies

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

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.