We investigate the existence of solutions for the following multipoint boundary value problem of a fractional order differential inclusion , where is the standard Riemann-Liouville fractional derivative, , satisfies ,??and?? is a set-valued map. Several results are obtained by using suitable fixed point theorems when the right hand side has convex or nonconvex values.

1. Introduction

In this paper, we will consider the existence of solutions for the following multipoint boundary value problem of a fractional order differential inclusion where is the standard Riemann-Liouville fractional derivative, , , satisfies , and is a set-valued map.

The present paper is motivated by a recent paper of Liang and Zhang [1], where it is considered problem (1) with single valued, and several existence results are provided.

Fractional differential equations have been of great interest recently. This is because of both the intensive development of the theory of fractional calculus itself and the applications of such constructions in various scientific fields such as physics, mechanics, chemistry, and engineering. For details, see [24] and the references therein.

The existence of solutions of initial value problems for fractional order differential equations has been studied in the literature [517] and the references therein. The study of fractional differential inclusions was initiated by El-Sayed and Ibrahim [18]. Also, recently, several qualitative results for fractional differential inclusions were obtained in [1923] and the references therein.

The aim here is to establish existence results for problem (1) when the right hand side is convex as well as nonconvex valued. In the first result (Theorem 21), we consider the case when the right hand side has convex values and prove an existence result via nonlinear alternative for Kakutani maps. In the second result (Theorem 25), we will use the fixed point theorem for contraction multivalued maps according to Covitz and Nadler. The paper is organized as follows.

In Section 2 we recall some preliminary facts that we need in the sequel, and in Section 3 we prove our main results. Finally, in Section 4, an example is given to demonstrate the application of one of our main results.

2. Preliminaries

In this section, we present some notations and preliminary lemmas that will be used in the proof of the main result.

Let be a metric space with the corresponding norm and let . We denote by the -algebra of all Lebesgue measurable subsets of , by the family of all nonempty subsets of , and by the family of all Borel subsets of . If then denotes the characteristic function of . For any subset , we denote by the closure of .

Recall that the Pompeiu-Hausdorff distance of the closed subsets is defined by the following: where . Define

Also, we denote by the Banach space of all continuous functions endowed with the norm and by the Banach space of all (Bochner) integrable functions endowed with the norm .

Let and be two metric spaces. If is a set-valued map, then a point is called a fixed point for if . is said to be bounded on bounded sets if is a bounded subset of for all bounded sets in . is said to be compact if is relatively compact for any bounded sets in . is said to be totally compact if is a compact subset of . is said to be upper semicontinuous if for any open set , the set is open in . is called completely continuous if it is upper semicontinuous and, for every bounded subset , is relatively compact. It is well known that a compact set-valued map with nonempty compact values is upper semicontinuous if and only if has a closed graph.

We define the graph of to be the set and recall a useful result regarding connection between closed graphs and upper semicontinuity.

Lemma 1 (see [24, Proposition 1.2]). If is upper semicontinuous, then is a closed subset of , that is, for every sequence and , if when , , , and , then . Conversely, if is completely continuous and has a closed graph, then it is upper semicontinuous.

For the convenience of the reader, we present here the following nonlinear alternative of the Leray-Schauder type and its consequences.

Theorem 2 (nonlinear alternative for Kakutani maps [25]). Let be a Banach space, a closed convex subset of , an open subset of , and . Suppose that is an upper semicontinuous compact map; here denotes the family of nonempty, compact convex subsets of . Then, either(i)has a fixed point in or(ii)there is a and with .

Definition 3. The multifunction is said to be lower semicontinuous if for any closed subset , the subset is closed.

If is a set-valued map with compact values and , we define Then, is of a lower semicontinuous type if is a lower semicontinuous with closed and decomposable values.

Theorem 4 (see [26]). Let be a separable metric space and be a lower semicontinuous set-valued map with closed decomposable values. Then has a continuous selection (i.e., there exists a continuous mapping such that for all ).

Definition 5. Consider the following.(i)A set-valued map with nonempty compact convex values is said to be measurable if for any the function is measurable.(ii)A set-valued map is said to be Carathéodory if is measurable for all and is upper semicontinuous for almost all .(iii) is said to be -Carathéodory if for any there exists such that .

Finally, the following results are easily deduced from the theoretical limit set properties.

Lemma 6 (see [27, Lemma ]). Let be a sequence of subsets where is a compact subset of a separable Banach space . Then, where refers to the closure of the convex hull of .

Lemma 7 (see [27, Lemma ]). Let and be two metric spaces. If is an upper semicontinuous, then, for each ,

Definition 8. Let be a Banach space. A sequence is said to be semicompact if(a)it is integrably bounded; that is, there exists such that (b)the image sequence is relatively compact in for a.e. .

The following important result follows from the Dunford-Pettis theorem (see [28, Proposition ]).

Lemma 9. Every semicompact sequence is weakly compact in .

When the nonlinearity takes convex values, Mazur’s Lemma, 1933, may be useful.

Lemma 10 (see [29, Theorem 21.4]). Let be a normed space and a sequence weakly converging to a limit . Then, there exists a sequence of convex combinations with for and which converges strongly to .

Lemma 11 (see [30]). Let be defined as before and . Then is relatively compact in if the following conditions hold:(a) is uniformly bounded in ;(b)the functions from are equicontinuous on any compact interval of ;(c)the functions from are equiconvergent; that is, for any given , there exists a such that , for any , .

Definition 12 (see [6]). The Riemann-Liouville fractional integral operator of order , of function , is defined as where is the Euler gamma function.

Definition 13 (see [31]). The Riemann-Liouville fractional derivative of order , , is defined as where the function has absolutely continuous derivatives up to order .

Lemma 14 (see [31]). The equality , holds for .

Lemma 15 (see [31]). Let and . Then, the differential equation has a unique solution , , , where .

Lemma 16 (see [31]). Let . Then, the following equality holds for , : , , where .

By we denote the space of continuous real-valued functions whose first derivative exists and it is absolutely continuous on . In this paper, we will use the following space to the study (1) which is denoted by

From [32], we know that is a Banach space equipped with the norm

In what follows, , , and . Next, we need the following technical result proved in [1].

Lemma 17 (see [1]). For any , the problem has a unique solution that where

Note that , (e.g., Lemma??3.2 in [1]) and from the definition of , we have the following (e.g., Remark??3.1 in [1]): where Also, one can get where

Lemma 18. The function defined by (16) satisfies

By calculation, it is easy to prove that Lemma 18 holds. So, we omit its proof here.

3. Main Results

Now we are able to present the existence results for problem (1).

3.1. The Upper Semicontinuous Case

To obtain the complete continuity of existence solutions operator, the following lemma is still needed.

Lemma 19 (see [32]). Let . If?? is equicontinuous on any compact interval of and equiconvergent at infinity, then is relatively compact on .

Definition 20. is called equiconvergent at infinity if and only if for all , there exists such that for all , it holds

Theorem 21. The Carathéodory multivalued map has nonempty, compact, convex values and satisfies the following.?? There exists a continuous nondecreasing function and , such that??and??each??. There exists a constant such that
Then, problem (1) has at least one solution.

Proof. Let and consider as in (25). It is obvious that the existence of solutions to problem (1) is reduced to the existence of the solutions of the integral inclusion where is defined by (16) and (17). Consider the set-valued map, is defined by
We show that satisfies the hypotheses of Theorem 2.
Claim 1. We show that is convex for any . If , then, there exist such that for any one has Let . Then, for any , we have The values of are convex; thus, is a convex set and hence .
Claim 2. We show that is bounded on bounded sets of . Let be any bounded subset of . Then, there exists such that for all . If , then there exists such that . One may write the following for any : On the other hand, and therefore for all ; that is, is bounded.
Claim 3. We show that maps bounded the sets into equi-continuous sets. Let be any bounded subset of as before and for some . Then, there exists such that . So, for any and , without loss of generality, we may assume that and one can get the following:
On the other hand, we get Similar to (34), we have From (34) and (35), we have Similar to (36), one can get Therefore, is an equicontinuous set in .
Claim 4. We show that is equiconvergent at . Let for some . Then, there exists such that . So, we have the following: and, similarly, one has Therefore, is equiconvergent at infinity.
Therefore, with Lemma 11, Lemma 19 and Claims 2–4, we conclude that is completely continuous.
Claim 5. is upper semicontinuous. To this end, it is sufficient to show that has a closed graph. Let such that and , as . Then, there exists such that . We will prove that means that there exists such that, for a.e. , we have . Then, we need to show that .
Condition (H1) implies that . Then, is integrably bounded in . Since has compact values, we deduce that is semicompact. By Lemma 9, there exists a subsequence, still denoted as , which converges weakly to some limit . Moreover, the mapping defined by is a continuous linear operator. Then, it remains continuous if these spaces are endowed with their weak topologies [29, 33]. Moreover, for a.e. , converges to . Then, we have It remains to prove that , a.e. . Mazur’s Lemma (see Lemma 10) yields the existence of , such that and the sequence of convex combinations converges strongly to in . Using Lemma 6, we obtain that However, the fact that the multivalued is upper semicontinuous and has compact values, together with Lemma 7, implies that This along with (42) yields that . Finally, has closed, convex values; hence, , a.e. . Thus, , proving that has a closed graph. Finally, with Lemma 1 and the compactness of , we conclude that is upper semicontinuous.
Claim 6. A priori bounds on solutions. Let be a solution of (1). Then, there exists with such that . In view of (H1), and using the computations in Claim 2 above, for each , we obtain Consequently,
In view of (H2), there exists such that . Let us consider the following:
Note that the operator is upper semicontinuous and completely continuous. From the choice of , there is no such that for some . Consequently, by the nonlinear alternative of the Leray-Schauder type (Theorem 2), we deduce that has a fixed point which is a solution of the problem (1). This completes the proof.

3.2. The Lipschitz Case

Now we prove the existence of solutions for the problem (1) with a nonconvex-valued right hand side by applying a fixed point theorem for multivalued maps according to Covitz and Nadler [34].

Definition 22. A multivalued operator is called the following:(a)-Lipschitz if and only if there exists such that for each ;(b)a contraction if and only if it is -Lipschitz with .

Lemma 23 (Covitz-Nadler, [34]). Let be a complete metric space. If is a contraction, then .

Definition 24. A measurable multivalued function is said to be integrably bounded if there exists a function such that for all , for a.e. .

Theorem 25. Assume that the following condition holds: ?? such that is measurable for each ; There exist which are not identical zero on any closed subinterval of , and such that for almost all , for all , and with for almost all .
Then, the boundary value problem (1) has at least one solution on if

Proof. We transform problem (1) into a fixed point problem. Consider the set-valued map defined at the beginning of the proof of Theorem 21. It is clear that the fixed point of are solutions of the problem (1).
Note that since the set-valued map is measurable with the measurable selection theorem (e.g., Theorem in [35]) it admits a measurable selection . Moreover, since is integrably bounded, . Therefore, .
We will prove that fulfills the assumptions of Covitz-Nadler contraction principle (Lemma 23).
First, we note that since , for any .
Second, we prove that is closed for any . Let such that in . Then and there exists such that Since has compact values, we may pass onto a subsequence (if necessary) to obtain that converges to in . In particular, , and for any we have that is, and is closed.
Next we show that is a contraction on . Let and . Then there exist such that Consider the set-valued map By (H5), we have hence has nonempty closed values. Moreover, since is measurable (e.g., Proposition in [35]), there exists which is a measurable selection of . It follows that and for any , Define and one can get Similarly, we have Therefore, From an analogous reasoning by interchanging the roles of and , it follows that Since is a contraction, it follows by the Lemma 23 that admits a fixed point which is a solution to problem (1).

4. Application

Consider the fractional boundary value problem, Here , , and , and is a multivalued map given by For , we have Thus, with , .

Also, by direct calculation, we can obtain that and . Further, by using the following condition: we find that . Clearly, all the conditions of Theorem 21 are satisfied. So, there exists at least one solution of problem (1) on .


The authors thank the referees for their careful reading of this paper and useful suggestions. This paper was funded by King Abdulaziz University, under Grant no. (130-1-1433/HiCi). The authors, therefore, acknowledge technical and financial support of KAU.