BOUNDARY VALUE PROBLEMS FOR FRACTIONAL DIFFERENTIAL EQUATION IN SPECIAL BANACH SPACE

This paper studies the existence of solutions of boundary value problem for fractional differential equations on the half-line in a special Banach space. The main result is based on Mönch fixed point theorem combining with a suitable measure of non-compactness, an example is given to illustrate our approach.


Introduction
Fractional differential equations play a very important role in describing some real world problems, for example, in the description of hereditary properties of various materials and processes. They are also widely applied in the mathematical modeling of processes in physics, chemistry, aerodynamics, electro-dynamics of complex medium, polymer rheology, etc. Consequently, the fractional calculus and its applications in various fields of science and engineering have received much attention and have developed very rapidly (cf. [18,20,23] for instance).
Very recently, many research papers have appeared concerning the fractional differential equations in Banach spaces, some of them investigated the existence results of solutions on finite intervals by classical tools from functional analysis; see, for example References. [1, 3, 6-8, 19, 21, 22].
In [4]. A. Arara and M. Benchohra studied the following problem    c D α 0 + y(t) = f (t, y(t)), t ∈ J = (0, +∞), 1 < α ≤ 2, where c D α 0 + is the Caputo fractional derivative of order α, f : J × R → R is a is a continuous function and y 0 ∈ R. The main approach is based on Schauder's fixed point theorem. my results is to generalize the previous work.
This paper is organized in the following way. In Section 2, we give some general results and preliminaries and in Section 3, we present existence results for the problem (1.1)-(1.3), using the Mönch's fixed point theorem combined with the technique of measure of noncompactness. Finally an illustrative example will be presented in Section 4.

Preliminary results
In this section, we introduce some notation and technical results which are used throughout this paper.
Let I ⊂ J be a compact interval and denote by C(I, E) the Banach space of continuous functions y : I → E with the usual norm y ∞ = sup{ y(t) , t ∈ I}.
L 1 (J, E) the space of E-valued Bochner integrable functions on J with the norm We consider the space of functions C α ([0, ∞), E) = {y ∈ C(J, E) : lim t→0 + t 2−α y(t) exists and finite}.
A norm in this space is given by We begin with some definitions from the theory of fractional calculus. Let α > 0, n = α + 1 (the least integer greater than or equal to α) and h ∈ C(J, E).
(2) The Riemann-Liouville fractional derivative of the function h of order α is defined by for all t > 0. Where Γ is the gamma function.
For the existence of solutions for the problem (1.1)-(1.3), we need the following auxiliary lemmas.

Remark 2.2.
If h is suitabe function (see for instance [18,20,23] ), we have the composition relations We note that γ, γ C and γ Xα the Kuratowski noncompactness measure of bounded sets in the spaces E, C(I, E) and X α , respectively. As for the definition of the Kuratowski noncompactness measure, we refer to references [5,17]. The following properties of the Kuratowski measure of noncompactness and Mönch fixed point theorem are needed for our discussion.
is bounded and equicontinuous, then γ(H(t)) is continuous on I and Theorem 2.1. [2,24] Let D be a bounded, closed and convex subset of a Banach space E such that 0 ∈ D , and let N be a continuous mapping of D into itself. If the implication holds for every subset V of D, then N has a fixed point.

Main result
We will need to introduce the following hypotheses which are assumed here after.
(H 1 ) There exists a nonnegative functions a, b ∈ C(J, R + ) such that Lemma 3.1. Let 1 < α < 2 and let h : J → E be continuous. A function y is a solution of the fractional integral equation if and only if y is solution of the problem Proof. Assume that y satisfies the problem (3.2)-(3.4). We may apply Lemma 2.2 to reduce equation (3.2) to an equivalent integral equation for some c 1 , c 2 ∈ R. Applying I 2−α 0 + to both side of (3.5), we have From Remark 2.1, we then get As t −→ 0, we obtain .
Applying D α−1 0 + to both side of (3.5), we have From Remark 2.1 and Remark 2.2, we then get Thus, we have Conversely. The proof is simple.
Consider the operator N : The following several lemmas present some properties of the operator N , which are necessary for the proof of our main result. Proof. For y ∈ X α ([0, ∞), E), it is easy to deduce from (H 1 ), and that N y ∈ X α (J, E). Furthermore, (H 1 ) guarantees that Hence, N : X α (J, E) → X α (J, E) is bounded. Next we prove that N is continuous. Let {y n } ∞ n=1 ⊂ X α (J, E) and y ∈ X α (J, E) such that y n → y as n → ∞. Then, {y n } ∞ n=1 is a bounded set of X α (J, E), i.e. there exists M > 0 such that y n α ≤ M for n > 1. We also have by taking limit that y α ≤ M . In view of condition (H 1 ), for any ε > 0, there exists L > 0 such that and there exists N ∈ N such that, for all n ≥ N , we have f (s, y n (s)) − (s, y(s)) < Γ(α) 3L ε.
Therefore, for all t ∈ J and n > N , we can obtain from If t ≤ L and n > N , we can obtain from f (s, y n (s)) − f (s, y(s)) ds The case when t > L and n > N is treated similarly. Thus we conclude that y n − y α → 0 as n → ∞, namely, N is continuous and the conclusion of the lemma follows. (ii) For given ε > 0, there exists a constant N 1 > 0 such that In view of condition (H 1 ) and the boundedness of B, there exists M > 0 such that let [a, b] ⊂ J be a compact interval and t 1 , t 2 ∈ [a, b] with t 1 < t 2 . Then where a * = max t∈[a,b] a(t) and b * = max t∈[a,b] b(t). As t 2 → t 1 the right-hand side of the above inequality tends to zero. Then Next we verify assertion (ii). Let ε > 0, we heve It is sufficient to prove that Relation (3.6) yields that there exits N 0 > 0 such that On the other hand, since lim t→∞ Now taking t 1 , t 2 ≥ N 1 , from (3.7), (3.8) we can arrive at f (s, y(s)) ds < ε.
Therefore, we complete the proof of lemma 3.3.   Then we can derive that N : B → B. Indeed, for any y ∈ B, by condition (H 1 ) we get Hence, from (H 4 ) we have N y α ≤ R, and we conclude that N : B → B. Clearly B is a bounded, convex and closed subset of X α ([0, ∞), E), together with Lemma 3.2 we know that N : B → B is continuous.
Finally we need to prove the following implication and t ∈ J, we choose ξ > 0 and n > 0 such that ξ < t < n.
For each y ∈ V , we consider Then from (H 1 ), we obtain that 1+t α → 0 as ξ → 0 and n → ∞, t ∈ J. Where H d denotes the Hausdorff metric in space E. By the prorerty of noncompactness mearure we get From (3.9), we know that Consequently, by condition (H 3 ). We get γ Xα (N (V )) = 0; that is γ Xα (V ) = 0. From the theorem 2.1, we conclude that N has a fixed point y ∈ B which is a solution of problem (1.1)-(1.3).

Example
We consider the following problem. With the aid of simple computation we find that  Since ∞ 0 e −10t dt = 0.1 < Γ( 3 2 ), we conclude that condition (H 3 ) is satisfied. Therefore, Theorem 3.1 ensures that problem (4.1)-(4.3) has a solution.