This paper deals with the integral boundary value problems of fractional differential equations at resonance. By Mawhin’s coincidence degree theory, we present some new results on the existence of solutions for a class of differential equations of fractional order with integral boundary conditions at resonance. An example is also included to illustrate the main results.

1. Introduction

In this paper, we are concerned with the following integral boundary value problem for nonlinear fractional differential equation: where , is the standard Riemann-Liouville differentiation, , and satisfies Carathéodory conditions; is right continuous on and left continuous at ; denotes the Riemann-Stieltjes integrals of with respect to . Our problem are at resonance, in the sense that, under the integral boundary conditions, we study the linear equation , , which has nontrivial solutions.

Recently, fractional differential equations have received considerable attentions not only because of a generalization of ordinary differential equations but also because they have played a significant role in science, engineering, economy, and other fields; see, for example, [13].

When , problem (1) is nonresonant. In [4], the authors studied the existence of positive solutions for the nonresonant case by Krasnosel’skii’s fixed point theorem. In [5], the author investigated the uniqueness of solutions for the nonresonant case by use of the -positive operator under a Lipschitz condition on .

In present, many papers are devoted to the integral boundary value problem for fractional differential equation under nonresonance conditions; see [420]. On the other hand, there are some papers studying integral boundary value problem for differential equation under resonant conditions; we refer the reader to [2129].

Motivated by the above results, in this paper, we consider the existence of solutions for the resonance integral boundary value problem (1) under nonlinear growth restriction of . Our method is based upon the coincidence degree theorem of Mawhin.

Now, we recall the essentials of the coincidence degree theory. Let and be real Banach spaces, and let be a Fredholm operator of index zero. If and are continuous projectors such that , , , and , then the inverse operator of exists and is denoted 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.

The abstract equation is shown to be solvable in view of Theorem IV.13 [30].

Theorem 1 (see [30]). Let be a Fredholm operator of index zero and let be -compact on . Assume 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 .

Throughout this paper, we always suppose that, ; satisfies the Carathéodory conditions; that is, is measurable for each fixed , is continuous for a.e. , and for each , there exists such that for all , , and and for a.e. .

2. Preliminaries and Lemmas

In this section, first we provide recall some necessary basic definitions and lemmas of the fractional calculus theory, which will be used in this paper. For more details, we refer to books [13] for details.

Definition 2 (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 3 (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 4 (see [1]). Assume that . If , , , then

Lemma 5 (see [1]). Assume that , . Then one has where .

Lemma 6 (see [4]). Let and , then the unique solution ofis given by where is Green’s function given by

Lemma 7 (see [4]). The function defined by (8) satisfies and .

We use the classical Banach space with the norm and with the norm . We also use the Banach space with the norm .

Define and as follows: where .

Then integral boundary value problems (1) can be rewritten as follows:

Lemma 8. The operator is a Fredholm operator of index zero.

Proof. Firstly, we show that By Lemma 5, means that . It follows from that . That is, .
Now we prove In fact, if and , then by Lemma 5, By the boundary condition, we obtain , The above equalities imply that On the other hand, if satisfies , let By a simple computation, we can obtain that and ; that is, .
Clearly, and is closed. It follows from that where . In fact, for each , we have which shows that . This together with implies that . Note that and thus . Therefore, is a Fredholm operator of index zero. The proof is completed.

Next, define the projections by and byClearly, and .

The generalized inverse operator of , can be defined by

In fact, if , then

For , , we have Then by Lemma 6, we obtain whenever .

Using (21) and (22), we write

By a standard method, we obtain the following lemma.

Lemma 9. is completely continuous.

Lemma 10. For , Moreover,

Proof. It is easy to see that Then by Lemma 7, we obtain It follows that The proof is completed.

3. Main Results

In this section, we will use Theorem 1 to prove the existence of solutions to IBVP (1). To obtain our main theorem, we need the following conditions:There exist functions such that, for all , There exists a constant such that either for each or for each , There exists a constant such that if for all , then .There exists a constant such that if for all , then .

Theorem 11. Suppose ()–() hold. Then IBVP (1) has at least one solution in , provided

Proof. Set Take . Since , so , , and hence Thus, from (), there exists such that Noticing thatwe obtainObserve that for all . Then by Lemma 10, Using (39) and (41), we have so that that is, for all , Applying (), we have Therefore, is bounded.
Let Then for , for some . So, . By (), we have . Therefore, is bounded.
We define the isomorphism by If (32) holds, then let For , If , then . Otherwise, if , in view of (), one has which contradicts . Thus is bounded.
If (33) holds, then define the set where is as above. Similar to the above argument, we can show that is bounded too.
Next, we will prove that all the assumptions of Theorem 1 are satisfied. Let be given any bounded open subset of such that . By Lemma 9, is compact; thus is -compact on . Clearly, assumptions (i) and (ii) of Theorem 1 are fulfilled.
At last, we will prove that (iii) of Theorem 1 is satisfied. Let . According to above argument, we know Thus, by the homotopy property of degree Then by Theorem 1, has at least one solution in , so that IBVP (1) has a solution. The proof is completed.

Theorem 12. Suppose ()–() and () hold. Then IBVP (1) has at least one solution in , provided

Proof. As in the proof of Theorem 11, , implies from (), there exist such thatNoticing that we obtainHence, we have In view of , . From this together with Lemma 5, for , we have Thus, by (58), Considering (57), (58), and (60), and applying , we get Therefore, is bounded. The rest of the proof repeats that of Theorem 11.

Example 13. Consider the IBVP Let and , then and ; thus () is satisfied. Let so that , , and , which verifies and (34).
Taking , we have , Hence holds. Finally, taking , when , then we obtain ; that is, condition () is satisfied. It follows from Theorem 11 that IBVP (62) has at least one solution.

Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.


The project was supported by the National Natural Science Foundation of China (11371221, 11571207, and 51774197).