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`Discrete Dynamics in Nature and SocietyVolume 2013 (2013), Article ID 943961, 9 pageshttp://dx.doi.org/10.1155/2013/943961`
Research Article

## Solvability of Nonlocal Fractional Boundary Value Problems

Department of Mathematics, Yanbian University, Yanji 133002, China

Received 28 January 2013; Revised 30 March 2013; Accepted 11 April 2013

Copyright © 2013 Zhongmin Huang and Chengmin Hou. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

This paper is devoted to introduce a new approach to investigate the existence of solutions for a three-point boundary value problem of fractional difference equations as fllows: , , and = We present an existence result at resonance case. The proof relies on coincidence degree theory.

#### 1. Introduction

In this paper, we consider a discrete fractional boundary value problem, for , of the form subject to the conjugate boundary conditions where is a continuous function, , and , and satisfies both and .

The three-point boundary value problem (1), (2) happens to be at resonance in the sense that associated linear homogeneous boundary value problem has nontrivial solution .

The research into the boundary value problems for differential equation and fractional differential equation have been always very active subjects. Rich results has been obtained due to the various powerful devices such as coincidence degree theory and cone theory. For details, see [17] and the references therein. Discrete fractional calculus has generated interest in recent years. There are many literatures dealing with the discrete fractional difference equation subject to various boundary value conditions or initial value conditions. We refer to [818] and references therein. However, we note that these results were usually obtained by analytic techniques and various fixed point theorems. For example, in [1316], authors investigated the existence to some boundary value problems by fixed point theorems on a cone. In [17], we given the existence of multiple solutions for a fractional difference boundary value problem with parameter by establishing the corresponding variational framework and using the mountain pass theorem, linking theorem, and Clark theorem in critical point theory. As we know, the coincidence degree theory has played an important role in dealing with the existence and multiple solutions for differential equations, which include the boundary value problems. To the best of our knowledge, it has not be used in discrete fractional boundary value problems. The aim of this paper is to establish the existence conditions for boundary value problem (1), (2). The proof relies on the coincidence degree theory.

Now, we will briefly recall some notations and an abstract existence result.

Let be real Banach space, let be a Fredholm map of index zero, and let , be continuous projectors such that , , and . It follows that is invertible. We denote the inverse of the map by . If is an open bounded subset of such that the map will be called L-compact on , if is bounded and is compact.

We need the following known result for the sequel.

Theorem 1 (Mawhin continuation theorem, see [2]). Let be a Fredholm operator of index zero, and let be on . Assume that the following conditions are satisfied:(i) for every ;(ii) for every ;(iii), where is a projection as above with , and is any isomorphism.

Then the equation has at least one solution in .

#### 2. Preliminaries

We first collect some basic lemmas for manipulating discrete fractional operators.

First, for any real number , we let . We define , for any and for which the right-hand side is defined. We also appeal to the convention that if is a pole of the Gamma function and is not a pole, then .

Definition 2 (see [13]). The fractional sum of defined on , for , is defined to be where . We also define the fractional difference, where and with , to be where .

Lemma 3 (see [13]). Let and be any numbers for which and are defined. Then

Lemma 4 (see [13]). Let . Then for some , with .

Lemma 5 (see [11]). For and all , for which the following is defined, we find that

Lemma 6 (see [10]). Let be a positive integer, and let . Then .

Lemma 7 (see [17]). A real symmetric matrix is positive definite if there exists a real nonsingular matrix such that , where is the transpose.

We define the Banach space with the norm and Banach space with the norm . For , since we can easily see that .

Define to be the linear operator from to with

We define as Then the boundary value problem (1), (2) can be written by

Lemma 8. is a Fredholm operator of index zero.

Proof. By Lemma 4 and the condition , we have .
Let , and then and where
In fact, if , then there exists such that .
In view of Lemma 4, we have
By , we get . By Lemmas 5 and 6, we get then
By and , we have
Therefore, that is,
Next, we will show that , for all .
Obviously for . Suppose that , then
Since , we may imply that for , then .
Let , consider continuous linear mapping defined by where .
Note that
That is, the map is idempotent. In fact, is a continuous linear projector. Note that implies . Conversely, if , then . Therefore, .
Take in the form , so that and . Thus, . Let , and assume that . Then, since , we have , which is a contradiction. Hence, , thus .
Now, , and so is a Fredholm operator of index zero. The proof is accomplished.

Define a mapping by we see that

Next, we will show that .

In fact, for , there exists such that and , and then

That is, , and hence . Next, if , then there exists a such that ; therefore . Since and , we see that .

Note that the projectors and are exact. Define by

If , then

Also, if , then

We can easily see that ; then and ; in view of , we have , then , thus hence, .

Lemma 9. is completely continuous.

Proof. Since is continuous and is a finite sum for , is completely continuous.

#### 3. Existence Results

Observe by Lemma 6 that for ,

We let and then

That is, , where .

One has

By Lemma 7, is a positive definite matrix. Let and denote, respectively, the minimum and the maximum eigenvalues of .

Since , we may easyly see that furthermore,

Theorem 10. Let be a continuous function in the second and the third variables. Assume that andthere exist nonnegative constants   and constants which satisfy such that for all , there exists such that for , if for any , then there exists such that for any , if , then either or else Then the boundary value problem (1), (2) has at least one solution in .

Proof. Set . Then for , thus , and hence
From , there exists such that .
Next, for , we have , and hence then we get then from (41), then we can get
Furthermore, since and for any , holds, and we may see that can be expressed by a linear combination with . Therefore, is bounded. Thus, is bounded.
Let . For ,, since , then , thus
From Lemma 5 and in view of , we see that , then from we get ; that is, . Hence, is bounded. Next, according to , for any , if , then either
or
If (53) holds, set where is the linear isomorphism given by . Since , then
If , then . Otherwise, if , in view of (53), one has which is a contradiction. Thus is bounded.
If (54) holds, then the set where is mentioned as above. By the analogous argument, we can show that is bounded too.
Now, we will prove that all the conditions of Theorem 1 are satisfied. Set to be bounded open set of such that . By Lemma 9, is compact; hence is on ; then by the above argument, we have (i) for every ,(ii) for every ,then the conditions of Theorem 1 are hold.
At last we will prove that the condition (iii) of Theorem 1 is satisfied. In fact, let for all , thus, by the homotopy property of degree as follows:
Then by Theorem 1, has at least one solution in ; therefore the boundary value problem (1), (2) has at least one solution in .

Remark 11. We may obtain similarly result when the third variable of is instead by fractional difference of with the order which satisfies

Example 12. Consider the boundary value problem where .

It is clearly that is continuous. , then and ; therefore, the boundary value problem (1), (2) is at resonance case. By , we have

By calculating, . Choosing , we can get , furthermore, for all ,

Therefore, the condition of Theorem 10 holds. Moreover, we can choose ; then the condition of Theorem 10 hold too. In fact, by the expression of , it is clear that if , then . Therefore, we can see

Thus, for every ,, so holds, and if , then for , or

Furthermore, or so (53) or (54) holds; that is, of Theorem 10 holds. Then, all the assumptions of Theorem 10 hold. Thus, with Theorem 10, the boundary value problem (60) has at least one solution in .

#### Acknowledgment

This project was supported by the National Natural Science Foundation of China (11161049).

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