Abstract
This paper is concerned with fractional differential inclusions with threepoint fractional integral boundary conditions. We consider the fractional differential inclusions under both convexity and nonconvexity conditions on the multivalued term. Some new existence results are obtained by using standard fixed point theorems. Two examples are given to illustrate the main results.
MSC: 34A60, 26A33, 34B15.
Keywords:
fractional differential inclusions; boundary value problems; existence results; multivalued maps1 Introduction
Fractional differential equations have recently gained much importance and attention due to the fact that they have been proved to be valuable tools in the modeling of many physical phenomena [13]. For some recent developments on the existence results of fractional differential equations, we can refer, for instance, to [417] and the references therein.
Differential inclusions arise in the mathematical modeling of certain problems in economics, optimal control, etc. and are widely studied by many authors, see [18,19] and the references therein. For some recent works on differential inclusions of fractional order, we refer the reader to the references [4,5,2029].
Motivated by the above papers, in this article, we study a new class of fractional boundary value problems, i.e., the following fractional differential inclusions with threepoint fractional integral boundary conditions:
where denotes the Caputo fractional derivative of order p, the RiemannLiouville fractional integral of order q, is a multifunction and a, b, c are real constants with .
We remark that when , and third variable of the function F in (1) vanishes, problem (1) reduces to a threepoint fractional integral boundary value problem (see [17] with a given continuous function).
The rest of this paper is organized as follows. In Section 2 we present the notations, definitions and give some preliminary results that we need in the sequel, Section 3 is dedicated to the existence results of problem (1), in the final Section 4, two examples are given to illustrate the main results.
2 Preliminaries
In this section, we introduce notations, definitions and preliminary facts that will be used in the remainder of this paper.
Let be a normed space. We use the notations: , , , , and so on.
Let , the PompeiuHausdorff distance of A, B is defined as
A multivalued map is convex (closed) valued if is convex (closed) for all . F is said to be completely continuous if is relatively compact for every . F is called upper semicontinuous on X if, for every , the set is a nonempty closed subset of X, and for every open set O of X containing , there exists an open neighborhood U of x such that . Equivalently, F is upper semicontinuous if the set is open for any open set O of X. F is called lower semicontinuous if the set is open for each open set O in X. If a multivalued map F is completely continuous with nonempty compact values, then F is upper semicontinuous if and only if F has a closed graph, i.e., if and , then implies [30].
A multivalued map is said to be measurable if, for every , the function is a measurable function.
Definition 2.1 A multivalued map is called
(1) γLipschitz if there exists such that
(2) a contraction if it is γLipschitz with .
Definition 2.2 A multivalued map is said to be Carathéodory if:
(2) is upper semicontinuous for a.e. .
Further, a Carathéodory function F is said to be Carathéodory if
(3) for each , there exists such that
The following lemmas will be used in the sequel.
Lemma 2.1 (see [31])
LetXbe a Banach space. Letbe anCarathéodory multivalued map andPbe a linear continuous map fromto, then the operator
is a closed graph operator in.
Here the set of selections
Definition 2.3 ([32])
The RiemannLiouville fractional integral of order q for a function f is defined as
provided the integral exists.
Definition 2.4 ([32])
For at least ntimes differentiable function f, the Caputo derivative of order q is defined as
where denotes the integer part of the real number q.
Lemma 2.2 ([16])
Let, then the differential equation
Lemma 2.3For any, the unique solution of the threepoint boundary value problem
is given by
Proof For and some constants , the general solution of the equation can be written as
From , it follows that . Using the integral boundary conditions of (2), we obtain
Therefore, we have
Substituting the values of , , we obtain the result. This completes the proof. □
Let us define what we mean by a solution of problem (1).
Definition 2.5 A function is a solution of problem (1) if it satisfies the boundary conditions in (1) and there exists a function such that a.e. on and
Let be the space of all continuous functions defined on . Define the space endowed with the norm . Obviously, is a Banach space.
Theorem 2.1 (Nonlinear alternative of LeraySchauder type)
LetXbe a Banach space, Cbe a closed convex subset ofX, Ube an open subset ofCwith. Suppose thatis an upper semicontinuous compact map. Then either (1) Fhas a fixed point in, or (2) there areandsuch that.
Theorem 2.2 (Covitz and Nadler)
Letbe a complete metric space. Ifis a contraction, thenFhas a fixed point.
3 Existence results
In this section, three existence results of problem (1) are presented. The first one concerns the convex valued case, and the others are related to the nonconvex valued case.
Now let us begin with the convex valued case.
Theorem 3.1Suppose that the following (H1), (H2) and (H3) are satisfied.
(H1) is a Carathéodory multivalued map.
(H2) There existandcontinuous, nondecreasing such that
(H3) There exists a constantsuch that
where
Then boundary value problem (1) has at least one solution on.
Proof Consider the multivalued operator defined as
with
Clearly, by Lemma 2.3, we know that the fixed points of N are solutions of problem (1). From (H1) and (H2), we have, for each , that the set is nonempty [31]. Next we will show that N satisfies the assumptions of the nonlinear alternative of LeraySchauder type. The proof is given in the following five steps.
Step 1: is convex valued. Since F is convex valued, we know that is convex and therefore it is obvious that for each , is convex.
Step 2: Nmaps bounded sets into bounded sets in. Let
be a bounded subset of . We need to prove that there exists a constant such that for each , one has for each . Let and , then there exists such that
By simple calculations, we have
Similarly, we can obtain
Therefore, we have
Hence, we obtain
Step 3: Nmaps bounded sets into equicontinuous sets in. Let be as in Step 2 and . Then, for each and , there exists such that for . Since
and
we deduce that
Step 4: Nhas a closed graph. Let , and , we need to show that . Since , there exists such that for . We must prove that there exists such that for .
Now, let us consider the continuous linear operator
and denote
By the definition of P, we have
It follows from Lemma 2.1 that is a closed graph operator. Since , we have
for some . This implies that .
Step 5: A priori bounds for solutions. Let for some . Then there exists such that for . By a similar discussion as in Step 2, we have
Thus
By the assumption of (H3), there exists M such that . Let us set
As a consequence of Steps 14, together with the ArzelaAscoli theorem, we can obtain that is an upper semicontinuous and completely continuous map. From the choice of U, there is no such that for some . Hence, by Theorem 2.1, we deduce that N has a fixed point which is a solution of problem (1). This is the end of the proof. □
Next we study the case when F is not necessarily convex valued.
Let A be a subset of . A is measurable if A belongs to the σalgebra generated by all sets of the form , where J is Lebesgue measurable in and D is a Borel set of ℝ. A subset A of is decomposable if for all and Lebesgue measurable, then , where χ stands for the characteristic function.
Theorem 3.2Let (H2) and (H3) hold and assume:
(H4) is such that: (1) ismeasurable; (2) the mapis lower semicontinuous for a.e. .
Then problem (1) has at least one solution on.
Proof From (H2), (H4) and Lemma 4.4 of [27], the map
is lower semicontinuous and has nonempty closed and decomposable values. Then, from a selection theorem due to Bressan and Colombo [33], there exists a continuous function such that for all , a.e. . Now consider the problem
with the boundary conditions in (2). Note that if is a solution of problem (7), then x is a solution to problem (1).
Problem (7) is then reformulated as a fixed point problem for the operator defined by
It can easily be shown that is continuous and completely continuous and satisfies all conditions of the LeraySchauder nonlinear alternative for singlevalued maps [34]. By a discussion similar to the one in Theorem 3.1, Theorem 3.2 follows. □
Theorem 3.3We assume that:
(H5) is such that: (1) the mapis measurable for all; (2) there existssuch that for a.e. and all,
and
then problem (1) has at least one solution on.
Proof From (H5), for each , the multivalued map is measurable and closed valued. Hence it has measurable selection (Theorem 2.2.1 [30]) and the set is nonempty. Let N be defined in (5). We will show that N satisfies the requirements of Theorem 2.2.
Step 1: For each , . Let be such that in . Then and there exists such that
By (H5), the sequence is integrable bounded. Since F has compact values, we may pass to a subsequence if necessary to get that converges to v in . Thus and for each ,
This implies that and is closed.
Step 2: There exists such that
Let and , then there exists such that
From (H5), we know that
Hence, for a.e. , there exists such that
Consider the multivalued map given by
Since , are measurable, Theorem III.41 in [35] implies that V is measurable. It follows from (H5) that the map is measurable. Hence, by (9) and Proposition 2.1.43 in [30], the multivalued map with nonempty closed values is measurable. Therefore, we can find and
and
we obtain that
Define
By using an analogous relation obtained by interchanging the roles of x and y, we get
Therefore from condition (8), Theorem 2.2 implies that N has a fixed point which is a solution of problem (1). This completes the proof. □
4 Examples
In this section, we give two examples to illustrate the results.
Example 1 Consider the following threepoint fractional integral boundary value problem:
where , , , , , , and is a multivalued map given by
In the context of this problem, we have
It is clear that F is convex compact valued and is of Carathéodory type. Let and , , we get that for , ,
As for condition (4), since (see O, Q in (H3)) is a constant, we can choose M large enough so that
Thus, by the conclusion of Theorem 3.1, boundary value problem (10) has at least one solution on .
Example 2 Consider the following threepoint fractional integral boundary value problem:
From the data given above, we have
Hence it follows from Theorem 3.3 that problem (11) has at least one solution on .
Competing interests
The author declares that he has no competing interests.
Author’s contributions
The author carried out the proofs of the theorems and approved the final manuscript.
Acknowledgements
The author would like to express his thanks to the referees for their helpful suggestions.
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