Existence solutions for boundary value problem of nonlinear fractional q-difference equations
© Zhou and Liu; licensee Springer 2013
Received: 10 December 2012
Accepted: 4 April 2013
Published: 19 April 2013
In this paper, we discuss the existence of weak solutions for a nonlinear boundary value problem of fractional q-difference equations in Banach space. Our analysis relies on the Mönch’s fixed-point theorem combined with the technique of measures of weak noncompactness.
Keywordsboundary value problem fractional q-difference equations Caputo fractional derivative weak solutions
Fractional differential calculus is a discipline to which many researchers are dedicating their time, perhaps because of its demonstrated applications in various fields of science and engineering . Many researchers studied the existence of solutions to fractional boundary value problems, for example, [2–10].
The fractional q-difference calculus had its origin in the works by Al-Salam  and Agarwal . More recently, maybe due to the explosion in research within the fractional differential calculus setting, new developments in this theory of fractional q-difference calculus were made, for example, q-analogues of the integral and differential fractional operators properties such as Mittage-Leffler function , just to mention some.
where and is the fractional q-derivative of the Caputo type.
where , , , and is a fixed constant, and is a fixed real number.
where , , and is a fixed constant, and is a real number.
where and is the fractional q-derivative of the Caputo type. is a given function satisfying some assumptions that will be specified later, and E is a Banach space with norm .
To investigate the existence of solutions of the problem above, we use Mönch’s 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 Banaś and Goebel , and subsequently developed and used in many papers; see, for example, Banaś et al. , Guo et al. , Krzyska and Kubiaczyk , Lakshmikantham and Leela , Mönch , O’Regan [30, 31], Szufla [32, 33] and the references therein. As far as we know, there are very few results devoted to weak solutions of nonlinear fractional differential equations [34–38]. 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 Mönch’s fixed-point theorem combined with the technique of measures of weak noncompactness.
The remainder of this article is organized as follows. In Section 2, we provide some basic definitions, preliminaries facts and various lemmas, which are needed later. In Section 3, we give main results of the problem (1.1). In the end, we also give an example for the illustration of the theories established in this paper.
2 Preliminaries and lemmas
In this section, we present some basic notations, definitions and preliminary results, which will be used throughout this paper.
and satisfies .
Remark 2.1 We note that if and , then .
Let and denote the Banach space of real-valued Lebesgue integrable functions on the interval J, denote the Banach space of real-valued essentially bounded and measurable functions defined over J with the norm .
Let E be a real reflexive Banach space with norm and dual , and let denote the space E with its weak topology. Here, is the Banach space of continuous functions with the usual supremum norm .
Moreover, for a given set V of functions , let us denote by , and .
Definition 2.1 A function is said to be weakly sequentially continuous if h takes each weakly convergent sequence in E to a weakly convergent sequence in E (i.e. for any in E with in then in for each ).
Definition 2.2 
The function is said to be Pettis integrable on J 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 E-valued Pettis integrable functions in the interval J.
Lemma 2.1 
If is Pettis integrable and is a measurable and an essentially bounded real-valued function, then is Pettis integrable.
Definition 2.3 
Lemma 2.2 
is relatively weakly compact;
, where denotes the weak closure of S;
The following result follows directly from the Hahn-Banach theorem.
Lemma 2.3 Let E be a normed space with . Then there exists with and .
Definition 2.4 
Definition 2.5 
where is the smallest integer greater than or equal to α.
Definition 2.6 
where is the smallest integer greater than or equal to α.
Lemma 2.4 
Lemma 2.5 
holds for every subset V of D, then A has a fixed point.
3 Main results
Let us start by defining what we mean by a solution of the problem (1.1).
Definition 3.1 A function is said to be a solution of the problem (1.1) if u satisfies the equation on J, and satisfy the conditions , .
For the existence results on the problem (1.1), we need the following auxiliary lemmas.
Lemma 3.1 
Lemma 3.2 
We derive the corresponding Green’s function for boundary value problem (1.1), which will play major role in our next analysis.
Here, is called the Green’s function of boundary value problem (3.1).
which completes the proof. □
To prove the main results, we need the following assumptions:
(H1) For each , the function is weakly sequentially continuous;
(H2) For each , the function is Pettis integrable on J;
then the problem (1.1) has at least one solution on J.
where is the Green’s function defined by (3.3). It is well known the fixed points of the operator are solutions of the problem (1.1).
First notice that, for , we have (assumption (H2)). Since, , then is Pettis integrable for all by Lemma 2.1, and so the operator is well defined.
Clearly, the subset D is closed, convex and equicontinuous. We shall show that satisfies the assumptions of Lemma 2.5. The proof will be given in three steps.
Step 1: We will show that the operator maps D into itself.
this means that .
Step 2: We will show that the operator is weakly sequentially continuous.
Let be a sequence in D and let in for each . Fix . Since f satisfies assumptions (H1), we have converge weakly uniformly to . Hence, the Lebesgue dominated convergence theorem for Pettis integrals implies converges weakly uniformly to in . Repeating this for each shows . Then is weakly sequentially continuous.
thus, V is relatively weakly compact in E. In view of Lemma 2.5, we deduce that has a fixed point, which is obviously a solution of the problem (1.1). This completes the proof. □
Remark 3.2 In Theorem 3.1, we presented an existence result for weak solutions of the problem (1.1) in the case where the Banach space E is reflexive. However, in the nonreflexive case, conditions (H1)-(H3) are not sufficient for the application of Lemma 2.5; the difficulty is with condition (2.1).
Theorem 3.2 Let E be a Banach space, and assume assumptions (H1), (H2), (H3), (H4) are satisfied. If (3.9) holds, then the problem (1.1) has at least one solution on J.
Theorem 3.3 Let E be a Banach space, and assume assumptions (H1), (H2), (H3)′, (H4), (H5) are satisfied. If (3.9) holds, then the problem (1.1) has at least one solution on J.
Proof Assume that the operator is defined by the formula (3.10). It is well known the fixed points of the operator are solutions of the problem (1.1).
First notice that, for , we have (assumption (H2)). Since, , then for all is Pettis integrable (Lemma 2.1), and thus, the operator makes sense.
clearly, the subset is closed, convex and equicontinuous. We shall show that satisfies the assumptions of Lemma 2.5. The proof will be given in three steps.
Step 1: We will show that the operator maps into itself.
this means that .
Step 2: We will show that the operator is weakly sequentially continuous.
Let be a sequence in and let in for each . Fix . Since f satisfies assumptions (H1), we have , converging weakly uniformly to . Hence, the Lebesgue dominated convergence theorem for Pettis integral implies converging weakly uniformly to in . We do it for each so . Then is weakly sequentially continuous.
By (3.9), it follows that , that is for each , and then is relatively weakly compact in E. In view of Lemma 2.5, we deduce that has a fixed point which is obviously a solution of the problem (1.1). This completes the proof. □
Dedicated to Professor Hari M Srivastava.
This research was supported by the National Natural Science Foundation of China (11161027, 11262009). The authors are thankful to the referees for their careful reading of the manuscript and insightful comments.
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