International Journal of Differential Equations

Volume 2014 (2014), Article ID 632434, 7 pages

http://dx.doi.org/10.1155/2014/632434

Research Article

Existence of Solutions for Two-Point Boundary Value Problem of Fractional Differential Equations at Resonance

Lei Hu,1,2 Shuqin Zhang,² and Ailing Shi³

¹Department of Mathematics, China University of Mining and Technology, Beijing 100083, China

²School of Science, Shandong Jiaotong University, Jinan, Shandong 250357, China

³School of Science, Beijing University of Civil Engineering and Architecture, Beijing 100044, China

Received 19 February 2014; Accepted 21 July 2014; Published 5 August 2014

Academic Editor: Patricia J. Y. Wong

Copyright © 2014 Lei Hu 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.

We establish the existence results for two-point boundary value problem of fractional differential equations at resonance by means of the coincidence degree theory. Furthermore, a result on the uniqueness of solution is obtained. We give an example to demonstrate our results.

1. Introduction

Fractional differential equations have been studied extensively. It is caused both by the intensive development of the theory of fractional calculus itself and by the applications such as physics, chemistry, phenomena arising in engineering, economy, and science; see, for example, [1–5].

Recently, more and more authors have paid their attentions to the boundary value problems of fractional differential equations; see [6–21]. Moreover, there have been many works related to the existence of solutions for boundary value problems at resonance; see [12–21]. It is considerable that there are many papers that have dealt with the solutions of multipoint boundary value problems of fractional differential equations at resonance (see, e.g., [12, 16]).

In [12], Bai and Zhang considered a three-point boundary value problem of fractional differential equations with nonlinear growth given by

where , , , , is Riemann-Liouville fractional derivative, and are given functions.

In [13], Hu et al. have studied a two-point boundary value problem for fractional differential equation at resonance

where , is Caputo fractional derivative, and satisfies Carathéodory conditions.

As far as we know, there are few works on the existence of two-point boundary value problems of the fractional differential equations at resonance. Motivated by the works above, we discuss the existence and uniqueness of solutions for the following nonlinear fractional differential equation:

where , , is Riemann-Liouville fractional derivative, and is continuous function.

More precisely, we use the coincidence degree theorem due to Mawhin [22]. The rest of this paper is organized as follows. In Section 2, we give some necessary notations, definitions, and lemmas. In Section 3, we study the existence of solutions of (3) by the coincidence degree theory. Finally, an example is given to illustrate our results in Section 4.

The two-point boundary value problem (3) happens to be at resonance in the sense that the associated linear homogeneous boundary value problem

has as a nontrivial solution.

2. Preliminaries

In this section, we present the necessary definitions and lemmas from fractional calculus theory. These definitions and properties can be found in the literature. For more details see [1–3].

Definition 1 (see [1]). The Riemann-Liouville fractional integral of order of a function is given by

provided that the right-hand side is pointwise defined on .

Definition 2 (see [1]). 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 .

Lemma 3 (see [1]). Let , ; then

where , .

Lemma 4 (see [1]). If , and . If the fractional derivatives and exist, then

Lemma 5 (see [1]). The relation

is valid in following cases , , and .

Now let us recall some notations about the coincidence degree continuation theorem.

Let , be real Banach spaces, 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 this map by . If is an open bounded subset of , the map will be called -compact on if is bounded and is compact.

Theorem 6. Let be a Fredholm operator of index zero and be -compact on . Suppose that the following conditions are satisfied:

(1) for each ;(2) for each ;(3), where is a continuous projection as above with and is any isomorphism.

Then the equation has at least one solution in .

3. Main Results

In this section, we will prove the existence results for (3).

We use the Banach space with the norm . For , , we define a linear space

By means of the functional analysis theory, we can prove that is a Banach space with the norm .

Define to be the linear operator from to with and

We define by

Then the problem (3) can be written by .

Lemma 7. The mapping is a Fredholm operator of index zero.

Proof. It is clear that

Let , so there exists a function which satisfies . By (11) and Lemma 3, we have

By , we can obtain . Hence

Then, we have

Taking into account , we obtain

On the other hand, suppose satisfy . Let , we can easily prove .

Thus, we conclude that

Consider the linear operators defined by

Take ; then

We can see .

For in the type , obviously, and . That is to say, . If , we have ; then . As a result , and we get .

Note that . Then is a Fredholm mapping of index zero.

We can define the operators , where

For ,

So we have .

Note that

Since , it is easy to say that and . So we have . If , then . We can derive from . Then

For ,

where .

We define by .

For , we have

For , we have . And for , the coefficients in the expressions

are all equal to zero. Thus, we obtain

This shows that . Again for each ,

where .

Lemma 8. Assume is an open bounded subset such that ; then map is -compact on

Proof. By the continuity of , we can get that and are bounded. So, in view of the Arzela-Ascoli theorem, we need only to prove that is equicontinuous. From the continuity of , there exists a constant , such that , for all , .

For , , we have

Furthermore, we have

where . Since and are uniformly continuous on , we can get that is compact. The proof is completed.

To obtain our main results, we need the following conditions.

(H₁)There exist functions , , such that for all , ,

(H₂)There exists a constant such that for every , if for all , then

(H₃)There exists a constant such that, for each , satisfying . We have either at least one of the following:

 or

(H₄), where , .

Lemma 9. is bounded.

Proof. For , and . By (12), ; that is,

By the integral mean value theorem, there exits a constant such that

Form (H₂), we can get .

Again for , and . From (29), we have

Now by Lemma 4

That is,

From (25) and (38), we have

Furthermore, it follows from (40) and (H₁) that

By the definition and (H₄), it is easy to see that and are bounded. So, is bounded.

Lemma 10. is bounded.

Proof. Let , so we have , . For ,

By the integral mean value theorem, there exits a constant such that

From (H₂), it follows that . Hence, is bounded.

Lemma 11. is bounded.

Proof. Let , so we have , . If , then . If , we have .

If and , then

It follows that

Then we get

which, together with (H₃), implies . Here, is bounded.

Remark 12. If the other parts of (H₃) hold, then the set is bounded.

Theorem 13. Suppose (H₁)–(H₄) hold; then the problem (3) has at least one solution in .

Proof. Let be a bounded open set of , such that . It follows from Lemma 8, is -compact on . By Lemmas 9, 10, and 11, we get the following:

(1), for every ;(2) for every ;(3)let , where is the identical operator. Via the homotopy property of degree, we obtain that

Applying Theorem 6, we conclude that has at least one solution in .

Under the stronger conditions imposed on , we can prove the uniqueness of solutions to the (3) studied above.

Theorem 14. Suppose the conditions (H₁) in the theorem are replaced by the following conditions.

(H₁)′There exist positive constants , , such that, for all , , one has

(H₁)′′There exist constants , , such that for all , , one has

Then, the BVP (3) has a unique solution, provided that

Proof. Let , , and ; then the condition (H₁) is satisfied. According to Theorem 13, BVP (3) has at least one solution. Suppose , are two solutions of (3); then

Note that , so satisfy the equation

According to , we have

By the integral mean value theorem, there exists , such that

By (H₁)′′, we have

We can have

Thus, we can obtain

According to (25), (38), and (58), we have

From the definition of and the assumption (51), we have , so that .

4. Example

Let us consider the following boundary value problems:

Corresponding to the problem (3), we have that and

Moreover,

We can get that the condition (H₁) holds; that is, , , and . Taking , , we can calculate that (H₂)–(H₄) hold.

Hence, by Theorem 13, we obtain that (60) has at least one solution.

Conflict of Interests

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

Authors’ Contribution

All authors typed, read, and approved the final paper.

Acknowledgments

Research was supported by the National Natural Science Foundation of China (11371364) and 2013 Science and Technology Research Project of Beijing Municipal Education Commission (KM201310016001).

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