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In this paper, we study the existence of mild solutions for fractional semilinear integro-differential equations in an arbitrary Banach space associated with operators generating compact semigroup on the Banach space. The arguments are based on the Schauder fixed point theorem.

The purpose of the present paper is to present an alternative approach to the existence of solution of fractional semilinear integro-differential equations in an arbitrary Banach space of the form

where and generates an evolution system, satisfying:

• , where denotes the Banach space of bounded linear operators from into

• (is the identity operator in)• for

• the mapping is strongly continuous in and is a given function.

Differential equations of fractional order have recently proved to be valuable tools in the modeling of many phenomena in various fields of science and engineering. This equations also serve as an tool for the description of hereditary properties of various materials and processes. For details, see [1-5]. The most important problem examined up to now is that concerning the existence of solutions of considered equations. In order to solve (1), many different methods have been applied in the literature. Most of these methods use the notion of a measure of noncompactness in Banach spaces, see [6-10]. Such a method can be to apply in this work. The method we are going to use is to reduce the existence of mild solutions of fractional semilinear integro-differential equations of type (0.1) to searching a fixed points of a suitable map on the space tempered by an arbitrary positive real continuous function defined on. In order to prove the existence of fixed points, we shall rely on the Schauder theorem. Moreover, an application to fractional differential equations is provided to illustrate the results of this work.

In what follows, will represent a Banach space with norm. Denote by the space of continuous functions. Now, let us assume thet is a given function defined and continuous on the interval with real positive values. Denote by

the Banach space consisting of all functions defined and continuous on with values in the Banach space such that

The space is furnished with the following standard norm

Let us recall two facts:

• The convergence in is the uniform convergence in the compact intervals, i.e. converge to in if and only if is uniformly convergent to on compact subsets of

• A subset is relatively compact if and only if the restrictions to of all functions from X form an equicontinuous set for each and is relatively compact in for each ,where, See [

Definition 1 A nonempty subset is said to be bounded if the there is a function such that for each and.

Namely, denote by the space of real functions defined and Lebesgue integrable on and equiped with the standard norm. For and for a fixed number we define the Riemann-Liouville fractional integral of order of the function by putting

It may be shown that the fractional integral operator transforms the space into itself and has some other properties (see [6-8], for example). More generally, we can consider the operator on the function space consisting of real functions being locally integrable over.

The following result is well known, one can see Michalski [

Lemma 1 For all and.

Our consideration is based on following Schauder fixed point theorem.

Theorem 1 [

The following hypotheses well be needed in the sequel.

• (A) is a bounded linear operator on for each and generates a uniformly continuous evolution system such that

•

• (C_{f}) (i) satisfies the Caratheodory type conditions, i.e. is measurable for and continuous for a.e., (ii) there exists a continuous positive function such that

for a.e. and all.

• (C_{u}) (i) is continuous on

• (ii) being continuous such that

where is continuous and increasing function with

• (iii) For all positive function there exist such that

.

where

Definition 2 [

Our main result is given by the following theorem.

Theorem 2 If the Banach space is separable.

Assume that the hypotheses and are satisfied. Then for each, the problem (0.1) has at least one mild solution in, for

Proof. Consider the operator defined by the formula

for and. Let

where

and

The estimate (0.3) guarantee the convergence of the integral. In the other hand, observe that if is nondecreasing function, then the function is also nondecreasing on.Therefore, the function is will defined and nondecreasing on Next, put

Obviously, the function is continuous, positive and decreasing. In the space let us consider the set

Clearly is closed convex of. Next, let Applying assumptions (1) and (2) we have

From the estimate (7), we deduce that transforms into itself. In what follows we show that is continuous. To do this, let us fix and take arbitrary sequence such that converge to in. Further, let us fix. Applying the properties of and we get

Then, keeping in mind that, we obtain, that there exists so big that

Next, for, denote the operator defined by

For, we have,

Next, by the Lebesgue dominated convergence theorem and (0.8) we derive that for suitable large we have this fact proves that is continuous on.

Next, from (**) we see that to prove the compactness of, we should prove that is equicontinuous on and is relatively compact for each and. For any and we get,

Thus,

Observe that for any there exists such that for all,

and, such that

Then, by the monotonicity of and for all , we get

where. Keeping in mind the continuity of, the right-hand side of the above inequality tends to zero as.

• If, then we have

• If, note that implies that and. According to the above results, we have

converging to 0 as.

So for, is equicontinuous. Meanwhile, is relatively compact because that is uniformly bounded. Thus is completely continuous on. By Schauder fixed point theorem, we deduce that has a fixed point in.

The last result in this article is to prove the existence of solutions to (0.1) but with the following conditions.

• satisfies the Caratheodory type conditions, i.e. is measurable for and continuous for a.e.• for a.e. and all.

Theorem 3 If the Banach space is separable. Assume that the hypotheses and are satisfied. Then for each, the problem (0.1) has at least one mild solution in, for

Proof. Define the operator by:

for Kipping in mind the result of lemma (0.2), we get, for.

Put and

Next, define the set

Then, we have that is a self-mapping of. We omit the proof of continuity and is relatively compact, because are similar to that in Theorem 2.

In this section, we illustrate the main result contained in Theorem 2 by the following quadratic fractional differential equation

for Let be a complete probability measure space. Let the space of

-measurable maps with

Consider the operator

defined by

Put.

Clearly is densely defined in and is the infinitesimal generator of a strongly continuous semigroup

in. Observe that the above equation is a special case of Equation (1) if we put and

Bay using the Jensen’s inequality it is not difficult to see that

To check conditions and it is enough to take

Let be a positive function defined on.

Thus, on the basis of Theorem 2, we conclude that Equation (4.1) has at least one mild solution in the space

.

I thank the referee for their invaluable advices, comments and suggestions.