Abstract
We discuss the existence of solutions, under the Pettis integrability assumption, for a class of boundary value problems for fractional differential inclusions involving nonlinear nonseparated boundary conditions. Our analysis relies on the MΓΆnch fixed point theorem combined with the technique of measures of weak noncompactness.
1. Introduction
This paper is mainly concerned with the existence results for the following fractional differential inclusion with non-separated boundary conditions: where is a real number, is the Caputo fractional derivative. is a multivalued map, is a Banach space with the norm , and is the family of all nonempty subsets of .
Recently, fractional differential equations have found numerous applications in various fields of physics and engineering [1, 2]. It should be noted that most of the books and papers on fractional calculus are devoted to the solvability of initial value problems for differential equations of fractional order. In contrast, the theory of boundary value problems for nonlinear fractional differential equations has received attention quite recently and many aspects of this theory need to be explored. For more details and examples, see [3β18] and the references therein.
To investigate the existence of solutions of the problem above, we use MΓΆnchβs fixed point theorem combined with the technique of measures of weak noncompactness, which is an important method for seeking solutions of differential equations. This technique was mainly initiated in the monograph of BanaΕ and Goebel [19] and subsequently developed and used in many papers; see, for example, BanaΕ and Sadarangani [20], Guo et al. [21], KrzyΕka and Kubiaczyk [22], Lakshmikantham and Leela [23], MΓΆnchβs [24], OβRegan [25, 26], Szufla [27, 28], and the references therein.
In 2007, Ouahab [29] investigated the existence of solutions for -fractional differential inclusions by means of selection theorem together with a fixed point theorem. Very recently, Chang and Nieto [30] established some new existence results for fractional differential inclusions due to fixed point theorem of multivalued maps. Problem (1.1) was discussed for single valued case in the paper [31]; some existence results for single- and multivalued cases for an extension of (1.1) to non-separated integral boundary conditions were obtained in the article [32] and [33]. About other results on fractional differential inclusions, we refer the reader to [34]. As far as we know, there are very few results devoted to weak solutions of nonlinear fractional differential inclusions. Motivated by the above mentioned papers, the purpose of this paper is to establish the existence results for the boundary value problem (1.1) by virtue of the MΓΆnch fixed point theorem combined with the technique of measures of weak noncompactness.
The remainder of this paper is organized as follows. In Section 2, we present some basic definitions and notations about fractional calculus and multivalued maps. In Section 3, we give main results for fractional differential inclusions. In the last section, an example is given to illustrate our main result.
2. Preliminaries and Lemmas
In this section, we introduce notation, definitions, and preliminary facts that will be used in the remainder of this paper. Let be a real Banach space with norm and dual space , and let denote the space with its weak topology. Here, let be the Banach space of all continuous functions from to with the norm and let denote the Banach space of functions that are the Lebesgue integrable with norm We let to be the Banach space of bounded measurable functions equipped with the norm Also, will denote the space of functions that are absolutely continuous and whose first derivative, , is absolutely continuous.
Let be a Banach space, and let , , , and . A multivalued map is convex (closed) valued if is convex (closed) for all . We say that is bounded on bounded sets if is bounded in for all (i.e., . The mapping is called upper semicontinuous (u.s.c.) on if for each , the set is a nonempty closed subset of and if for each open set of containing , there exists an open neighborhood of such that . We say that is completely continuous if is relatively compact for every . If the multivalued map is completely continuous with nonempty compact values, then is u.s.c. if and only if has a closed graph (i.e., imply ). The mapping has a fixed point if there is such that . The set of fixed points of the multivalued operator will be denoted by . A multivalued map is said to be measurable if for every , the function is measurable. For more details on multivalued maps, see the books of Aubin and Cellina [35], Aubin and Frankowska [36], Deimling [37], Hu and Papageorgiou [38], Kisielewicz [39], and Covitz and Nadler [40].
Moreover, for a given set of functions , let us denote by , , and .
For any , let be the set of selections of defined by
Definition 2.1. A function is said to be weakly sequentially continuous if takes each weakly convergent sequence in to a weakly convergent sequence in (i.e., for any in with in then in for each ).
Definition 2.2. A function has a weakly sequentially closed graph if for any sequence , for with in for each and in for each , then .
Definition 2.3 (see [41]). The function is said to be the Pettis integrable on if and only if there is an element corresponding to each such that for all , where the integral on the right is supposed to exist in the sense of Lebesgue. By definition, .
Let be the space of all -valued Pettis integrable functions in the interval .
Lemma 2.4 (see [41]). If is Pettisβ integrable and is a measurable and essentially bounded real-valued function, then is Pettisβ integrable.
Definition 2.5 (see [42]). Let be a Banach space, the set of all bounded subsets of , and the unit ball in . The De Blasi measure of weak noncompactness is the map defined by
Lemma 2.6 (see [42]). The De Blasi measure of noncompactness satisfies the following properties:(a);(b) is relatively weakly compact;(c);(d), where denotes the weak closure of ;(e);(f);(g);(h).
The following result follows directly from the Hahn-Banach theorem.
Lemma 2.7. Let be a normed space with . Then there exists with and .
For completeness, we recall the definitions of the Pettis-integral and the Caputo derivative of fractional order.
Definition 2.8 (see [25]). Let be a function. The fractional Pettis integral of the function of order is defined by where the sign ββ denotes the Pettis integral and is the gamma function.
Definition 2.9 (see [3]). For a function , the Caputo fractional-order derivative of is defined by where and denotes the integer part of .
Lemma 2.10 (see [43]). Let be a Banach space with a nonempty, bounded, closed, convex, equicontinuous subset of . Suppose has a weakly sequentially closed graph. If the implication holds for every subset of , then the operator inclusion has a solution in .
3. Main Results
Let us start by defining what we mean by a solution of problem (1.1).
Definition 3.1. A function is said to be a solution of (1.1), if there exists a function with for a.e. , such that
and satisfies conditions .
To prove the main results, we need the following assumptions:(H1) has weakly sequentially closed graph;(H2)for each continuous , there exists a scalarly measurable function with a.e. on and is Pettis integrable on ;(H3)there exist and a continuous nondecreasing function such that
(H4)for each bounded set , and each , the following inequality holds:
(H5)there exists a constant such that
where and are defined by (3.9).
Theorem 3.2. Let be a Banach space. Assume that hypotheses (H1)β(H5) are satisfied. If then the problem (1.1) has at least one solution on .
Proof. Let be a given function; it is obvious that the boundary value problem [18]
has a unique solution
where is defined by the formula
From the expression of and , it is obvious that is continuous on and is continuous on . Denote by
We transform the problem (1.1) into fixed point problem by considering the multivalued operator defined by
and refer to [31] for defining the operator . Clearly, the fixed points of are solutions of Problem (1.1). We first show that (3.10) makes sense. To see this, let ; by (H2) there exists a Pettisβ integrable function such that for a.e. . Since , then is Pettis integrable and thus is well defined.
Let , and consider the set
clearly, the subset is a closed, convex, bounded, and equicontinuous subset of . We shall show that satisfies the assumptions of Lemma 2.10. The proof will be given in four steps.
Stepββ1. We will show that the operator is convex for each .
Indeed, if and belong to , then there exists Pettisβ integrable functions , such that, for all , we have
Let . Then, for each , we have
Since has convex values, and we have .
Stepββ2. We will show that the operator maps into .
To see this, take . Then there exists with and there exists a Pettis integrable function with for a.e. . Without loss of generality, we assume for all . Then, there exists with and . Hence, for each fixed , we have
Therefore, by (H5), we have
Next suppose and , with so that . Then, there exists such that . Hence,
this means that .
Stepββ3. We will show that the operator has a weakly sequentially closed graph.
Let be a sequence in with in for each , in for each , and for . We will show that . By the relation , we mean that there exists such that
We must show that there exists such that, for each ,
Since has compact values, there exists a subsequence such that
Since has a weakly sequentially closed graph, . The Lebesgue dominated convergence theorem for the Pettis integral then implies that for each ,
that is, in . Repeating this for each shows .
Stepββ4. The implication (2.9) holds. Now let be a subset of such that . Clearly, for all . Hence, , is bounded in .
Since function is continuous on , the set is compact, so . By assumption (H4) and the properties of the measure , we have for each
which gives
This means that
By (3.5) it follows that ; that is, for each , and then is relatively weakly compact in . In view of Lemma 2.10, we deduce that has a fixed point which is obviously a solution of Problem (1.1). This completes the proof.
In the sequel we present an example which illustrates Theorem 3.2.
4. An Example
Example 4.1. We consider the following partial hyperbolic fractional differential inclusion of the form
Set , , , then . So .
Let
with the norm
Set
For each and , we have
Hence conditions , , and hold with , and . For any bounded set , we have
Hence (H4) is satisfied. From (3.8), we have
So, we get
A simple computation gives
We shall check that condition (3.5) is satisfied. Indeed
which is satisfied for some , and (H5) is satisfied for . Then by Theorem 3.2, the problem (4.1) has at least one solution on for values of satisfying (4.10).
Acknowledgments
The first authorβs work was supported by NNSF of China (11161027), NNSF of China (10901075), and the Key Project of Chinese Ministry of Education (210226). The authors are grateful to the referees for their comments according to which the paper has been revised.