Abstract and Applied Analysis

Volume 2014, Article ID 314083, 8 pages

http://dx.doi.org/10.1155/2014/314083

## The Existence of Solutions for Four-Point Coupled Boundary Value Problems of Fractional Differential Equations at Resonance

^{1}Department of Statistics and Finance, Shandong University of Science and Technology, Qingdao 266590, China^{2}School of Mathematical Sciences, Qufu Normal University, Qufu, Shandong 273165, China^{3}Department of Mathematics, Shandong University of Science and Technology, Qingdao, Shandong 266590, China

Received 18 December 2013; Accepted 14 February 2014; Published 23 March 2014

Academic Editor: Xinguang Zhang

Copyright © 2014 Yumei Zou et al. 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

A four-point coupled boundary value problem of fractional differential equations is studied. Based on Mawhin’s coincidence degree theory, some existence theorems are obtained in the case of resonance.

#### 1. Introduction

In this paper, we are concerned with the following four-point coupled boundary value problem for nonlinear fractional differential equation. Consider where , and are the standard Riemann-Liouville differentiation and integration, , , , and

The subject of fractional calculus has gained considerable popularity and importance because of its intensive development of the theory of fractional calculus itself and its varied applications in many fields of science and engineering. As a result, the subject of fractional differential equations has attracted much attention; see [1–11] for a good overview.

At the same time, we notice that coupled boundary value problems, which arise in the study of reaction-diffusion equations and Sturm-Liouville problems, have wide applications in various fields of sciences and engineering, for example, the heat equation [12–14] and mathematical biology [15, 16]. In [17], Asif and Khan used the Guo-Krasnosel’skii fixed-point theorem to show the existence of positive solutions to the nonlinear differential system with coupled four-point boundary value conditions where , , and are two given continuous functions.

In [18], the authors considered the existence of positive solutions of four-point coupled boundary value problem for systems of the nonlinear semipositone fractional differential equation where is a parameter, satisfy , , is a real number and , and is Riemann-Liouville’s fractional derivative.

Recently, Cui and Sun [19] showed the existence of positive solutions of singular superlinear coupled integral boundary value problems for differential systems where are bounded linear functionals on given by with being functions of bounded variation with positive measures.

A key assumption in the above papers is that the case studied is not at resonance; that is, the associated fractional (or ordinary) linear differential operators are invertible. In this paper, instead, we are interested in the resonance case due to the critical condition (2). Boundary value problems at resonance have been studied by several authors including the most recent works [20–31]. In this paper, we establish new results on the existence of a solution for the couple boundary value problems at resonance. Our method is based on the coincidence degree theorem of Mawhin [32, 33].

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

Let be real Banach spaces and let be a Fredholm operator of index zero. and are continuous projectors such that It follows that is invertible. We denote the inverse of the mapping by (generalized inverse operator of ). If is an open bounded subset of such that , the mapping will be called -compact on if is bounded and is compact.

Theorem 1 (see [32, 33]). *Let be a Fredholm operator of index zero and let be -compact on . Assume that the following conditions are satisfied.*(i)* for every .*(ii)* for every .*(iii)*, where is a projector as above with .**Then the equation has at least one solution in .*

For convenience, let us set the following notations:

#### 2. Preliminaries and Lemmas

In this section, first we provide recall of some basic definitions and lemmas of the fractional calculus, which will be used in this paper. For more details, we refer to books [1, 2, 4].

*Definition 2 (see [1, 4]). *The Riemann-Liouville fractional integral of order of a function is given by
provided that the right-hand side is pointwise defined on .

*Definition 3 (see [1, 4]). *The Riemann-Liouville fractional derivative of order of a continuous function is given by
where , provided that the right-hand side is pointwise defined on .

We use the classical Banach space with the norm and with the norm . We also use the space defined by and the Banach space () with the norm .

Lemma 4 (see [1]). *Let , . Assume that with a fractional integration of order that belongs to . Then the equality
**
holds almost everywhere on .*

In the following lemma, we use the unified notation of both for fractional integrals and fractional derivatives assuming that for .

Lemma 5 (see [1]). *Assume that ; then *(i)*let . If and exist, then
*(ii)*if , then
* *satisfies at any point on for and ;*(iii)*let . Then holds on ;*(iv)*note that, for , , one has
*

*Remark 6. *If and satisfies and , then . In fact, with Lemma 4, one has
Combined with , there is . So

*Lemma 7 (see [34]). is a sequentially compact set if and only if is uniformly bounded and equicontinuous. Here to be uniformly bounded means that there exists such that for every
and to be equicontinuous means that for all, and for all , , , and , the following holds:
*

*We also use the following two Banach spaces with the norm
and with the norm
*

*Let the linear operator with
be defined by
where and are defined by
*

*Let the nonlinear operator be defined by
where are defined by
Then four-point coupled boundary value problems (1) can be written as
*

*Lemma 8. Let be the linear operator defined as above. If (2) holds, then
*

*Proof. *Let and let . Then by Lemma 5, we have , , , and . So
For every , if , then
Considering that , and , we can obtain that and . It yields the following:

Let ; then there is such that ; that is, , and , . By Lemma 4,
and by the couple boundary conditions, we have
It yields the following:
On the other hand, suppose that satisfy (36). Let and , and then , , and
Therefore, (30) holds.

*Lemma 9. If (2) holds, then is a Fredholm operator of index zero and . Furthermore, the linear operator can be defined by
Also
*

*Proof. *Define operator as follows:
where is defined by
It is easy to see that ; that is, is a continuous linear projector. Furthermore, . For , set . Then and . It follows from and that . So we have
Now, , and so is a Fredholm operator of index 0.

Let be continuous linear operator defined by
Obviously, is a linear projector and
It is easy to know that .

Define by
Since
then

In fact, if , then
By Lemma 4, for ,
( since and since ). Hence,
The proof is complete.

*3. Main Results*

*3. Main Results**In this section, we will use Theorem 1 to prove the existence of solutions to BVP (1). To obtain our main theorem, we use the following assumptions.(H1)There exist functions such that for all ,
(H2)There exists a constant such that, for , if for all , then or .(H3)There exists a constant such that either, for each ,
or, for each ,
*

*Theorem 10. Suppose (2) and ()–() hold. Then (1) has at least one solution in Y, provided that
*

*Proof. *Set
Take . Since , so and ; hence,
Thus, from (H2), there exist such that
Noticing that
so
Thus
For all , . Considering Lemma 9, we get . Together with (39), we have
From (60) and (61), we have
From (62), we discuss various cases.*Case 1* (). From (H1), we have
which yield
Thus, is bounded.*Case 2* (). From (H1), we have
which yield
Thus, is bounded. Let
For and , so , , . Noticing that , then we get , and thus and . From (H2), we get , and thus is bounded.

We define the isomorphism by
If the first part of (H3) is satisfied, then let
For ,
If , then . Otherwise, if , in view of (H3) and , one has
which contradict . Thus is bounded.

If the second part of (H3) holds, then define the set
and here is as above. Similar to the above argument, we can show that is bounded too.

In the following, we will prove that all conditions of Theorem 1 are satisfied. Let be a bounded open subset of such that . By standard arguments, we can prove that is compact, and thus is -compact on . Then by the above argument we have (i), for every ,(ii) for .Finally, we will prove that (iii) of Theorem 1 is satisfied. Let . According to the above argument, we know that
Thus, by the homotopy property of degree,
Then by Theorem 1, has at least one solution in so that BVP (1) has a solution in . The proof is complete.

*Conflict of Interests*

*Conflict of Interests*

*The authors declare that there is no conflict of interests regarding the publication of this paper.*

*Acknowledgments*

*Acknowledgments*

*The authors are very grateful to the anonymous referees for many valuable comments and suggestions which helped to improve the presentation of the paper. The project was supported by the National Natural Science Foundation of China (11371221, 61304074), the Specialized Research Foundation for the Doctoral Program of Higher Education of China (20123705110001), the Program for Scientific Research Innovation Team in Colleges and Universities of Shandong Province, the Postdoctoral Science Foundation of Shandong Province (201303074), and Foundation of SDUST.*

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