#### Abstract

We discuss the existence of solution for a multipoint boundary value problem of fractional differential equation. An existence result is obtained with the use of the coincidence degree theory.

#### 1. Introduction

In this paper, we study the multipoint boundary value problem where , satisfying the CarathΓ©odory conditions, . and are the standard Riemann-Liouville derivative and integral, respectively. We assume, in addition, that where is the Gamma function. Due to condition (1.3), the fractional differential operator in (1.1), (1.2) is not invertible.

Fractional differential equation can describe many phenomena in various fields of science and engineering. Many methods have been introduced for solving fractional differential equations, such as the popular Laplace transform method, the iteration method. For details, see [1, 2] and the references therein.

Recently, there are some papers dealing with the solvability of nonlinear boundary value problems of fractional differential equation, by use of techniques of nonlinear analysis (fixed-point theorems, Leray-Schauder theory, etc.), see, for example, [3β6]. But there are few papers that consider the fractional-order boundary problems at resonance. Very recently [7], Y. H. Zhang and Z. B. Bai considered the existence of solutions for the fractional ordinary differential equation subject to the following boundary value conditions: where is a natural number, is a real number, is continuous, and , and are given constants such that . and are the standard Riemann-Liouville derivative and integral, respectively. By the conditions, the kernel of the linear operator is one dimensional.

Motivated by the above work and recent studies on fractional differential equations [8β18], in this paper, we consider the existence of solutions for multipoint boundary value problem (1.1), (1.2) at resonance. Note that under condition (1.3), the kernel of the linear operator in (1.1), (1.2) is two dimensional. Our method is based upon the coincidence degree theory of Mawhin [18].

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

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 the map by . If is an open-bounded subset of such that , the map will be called -compact on if is bounded and is compact.

The theorem that we used is Theoremββ2.4 of [18].

Theorem 1.1. *Let be a Fredholm operator of index zero and be -compact 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 . *

The rest of this paper is organized as follows. In Section 2, we give some notation and Lemmas. In Section 3, we establish an existence theorem of a solution for the problem (1.1), (1.2).

#### 2. Background Materials and Preliminaries

For the convenience of the reader, we present here some necessary basic knowledge and definitions for fractional calculus theory, and these definitions can be found in the recent literature [1, 2].

*Definition 2.1. *The fractional integral of order of a function is given by
provided the right side is pointwise defined on . And we let for every continuous .

*Definition 2.2. *The fractional derivative of order of a function is given by
where , provided the right side is pointwise defined on .

Lemma 2.3 (see [3]). *Assume that with a fractional derivative of order that belongs to , then
**
for some , where is the smallest integer greater than or equal to .*

We use the classical space with the norm . Given and , one can define a linear space where and . By means of the linear function analysis theory, one can prove that with the norm is a Banach space.

Lemma 2.4 (see [7]). * is a sequentially compact set if and only if is uniformly bounded and equicontinuous. Here, uniformly bounded means that there exists such that for every ,
**
and equicontinuous means that such that
*

Let with the norm . , where , with the norm , and is a Banach space.

*Definition 2.5. *By a solution of the boundary value problem (1.1), (1.2), we understand a function such that is absolutely continuous on and satisfies (1.1), (1.2).

*Definition 2.6. *We say that the map satisfies the CarathΓ©odory conditions with respect to if the following conditions are satisfied:(i)for each , the mapping is Lebesgue measurable,(ii)for almost every , the mapping is continuous on ,(iii)for each , there exists such that, for a.e., and every , we have .

Define to be the linear operator from to with We define by setting Then boundary value problem (1.1), (1.2) can be written as

Lemma 2.7. *Let condition (1.3) and (1.4) hold, then is a Fredholm map of index zero.*

*Proof. *It is clear that .

Let and
then and, if
hold. Then satisfies the boundary conditions (1.2), that is, , and we have
Let , then for , we have
which, due to the boundary value condition (1.2), implies that satisfies (2.11). In fact, from , we have , from , we have
and from , we have
Hence,
Therefore,
Consider the continuous linear mapping and defined by
Using the above definitions, we construct the following auxiliary maps :
Since the condition (1.4) holds, the mapping defined by
is well defined.

Recall (1.4) and note that
and similarly we can derive that
So, for , it follows from the four relations above that
that is, the map is idempotent. In fact, is a continuous linear projector.

Note that implies . Conversely, if , then we must have ; since the condition (1.4) holds, this can only be the case if , that is, . In fact, .

Take in the form , so that and . Thus, . Let and assume that is not identically zero on , then, since , from (2.11) and the condition (1.4), we derive , which is a contradiction. Hence, ; thus, .

Now, , and so is a Fredholm operator of index zero.

Let be defined by Note that is a continuous linear projector and It is clear that .

Note that the projectors and are exact. Define by Hence, we have then , and thus In fact, if , then . Also, if , then from boundary value condition (1.2) and the fact that , we have . Thus, Using (2.19), we write With arguments similar to those of [7], we obtain the following Lemma.

Lemma 2.8. * is completely continuous.*

#### 3. The Main Results

Assume that the following conditions on the function are satisfied:

(H1) there exists a constant , such that for satisfying for all , we have

(H2) there exist functions and a constant such that for all and a.e., , one of the following inequalities is satisfied:

(H3) there exists a constant such that for every satisfying , then either or else

*Remark 3.1. * and from (H3) stand for the images of under the maps and , respectively.

Theorem 3.2. * If (H1)β(H3) hold, then boundary value problem (1.1)-(1.2) has at least one solution provided that
**
where .*

*Proof. *Set
then for ; thus, , and hence for all . By the definition of , we have . It follows from (H1) that there exists such that . Now,
so
Now by (3.8), we have
Note that for , then, by (2.28) and (2.30),
Using (3.9) and (3.10), we obtain
where is a constant. This is for all ,
If the first condition of (H2) is satisfied, then we have
and consequently,
Note that and , so there exists such that for all . The inequalities (3.14) and (3.15) show that there exist such that for all . Therefore, for all , that is, is bounded given the first condition of (H2). If the other conditions of (H2) hold, by using an argument similar to the above, we can prove that is also bounded.

Let
For , and ; thus, . By (H3), , that is, is bounded.

We define the isomorphism by

If the first part of (H3) is satisfied, let
For every ,
If , then , and if , then by (H3),
which, in either case, obtain a contradiction. If the other part of (H3) is satisfied, then we take
and, again, obtain a contradiction. Thus, in either case,
for all , that is, is bounded.

In the following, we will prove that all the conditions of Theorem 1.1 are satisfied. Set to be a bounded open set of such that . by Lemma 2.8, the operator is compact; thus, is -compact on , then by the above argument, we have(i), for every ,(ii), for every .Finally, we will prove that (iii) of Theorem 1.1 is satisfied. Let , where is the identity operator in the Banach space . According to the above argument, we know that
and thus, by the homotopy property of degree,
then by Theorem 1.1, has at least one solution in , so boundary problem (1.1), (1.2) has at least one solution in the space . The proof is finished.

#### Acknowledgments

This work is supported by the NSFC (no. 11061030, no. 11026060), the Fundamental Research Funds for the Gansu Universities, the nwnu-kjcxgc-03-69, nwnu-kjcxgc-03-61.