#### 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.(H_{1})There exist functions , , such that for all , ,
(H_{2})There exists a constant such that for every , if for all , then
(H_{3})There exists a constant such that, for each , satisfying . We have either at least one of the following:
or
(H_{4}), 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 (H_{2}), 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 (H_{1}) that
By the definition and (H_{4}), 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 (H_{2}), 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 (H_{3}), implies . Here, is bounded.

*Remark 12. *If the other parts of (H_{3}) hold, then the set is bounded.

Theorem 13. *Suppose (H _{1})–(H_{4}) 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 (H _{1}) in the theorem are replaced by the following conditions.*(H

_{1})′

*There exist positive constants , , such that, for all , , one has*(H

_{1})′′

*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 (H_{1}) 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 (H_{1})′′, 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 (H_{1}) holds; that is, , , and . Taking , , we can calculate that (H_{2})–(H_{4}) 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).