#### Abstract

By employing known Guo-Krasnoselskii fixed point theorem, we investigate the eigenvalue interval for the existence and nonexistence of at least one positive solution of nonlinear fractional differential equation with integral boundary conditions.

#### 1. Introduction

Fractional calculus has been receiving more and more attention in view of its extensive applications in the mathematical modelling coming from physical and other applied sciences; see books [15]. Recently, the existence of solutions (or positive solutions) of nonlinear fractional differential equation has been investigated in many papers (see [628] and references cited therein). However, in terms of the eigenvalue problem of fractional differential equation, there are only a few results [2933].

To the best of author’s knowledge, no paper has considered the eigenvalue problem of the following nonlinear fractional differential equation with integral boundary conditions: where , is the Caputo fractional derivative, and is a continuous function.

Our proof is based upon the properties of the Green function and Guo-Krasnoselskii’s fixed point theorem given in [34]. Our purpose here is to give the eigenvalue interval for nonlinear fractional differential equation with integral boundary conditions. Moreover, according to the range of the eigenvalue , we establish some sufficient conditions for the existence and nonexistence of at least one positive solution of the problem (1).

#### 2. Preliminaries

For the convenience of the readers, we first present some background materials.

Definition 1. For a function , the Caputo derivative of fractional order is defined as where denotes the integer part of the real number .

Definition 2. The Riemann-Liouville fractional integral of order for a function is defined as provided that such integral exists.

Lemma 3. Let ; then for some ,  , .

Lemma 4 (see [34]). Let be a Banach space, and let be a cone. Assume that , are open subsets of with , , and let be a completely continuous operator such that(i), , and , , or(ii), , and , .
Then has a fixed point in .

Lemma 5. Let , , , and . Assume ; then the unique solution of the problem is given by the expression where

Proof. It is well known that the equation can be reduced to an equivalent integral equation: for some   .
By the conditions and , we can get that and Hence, we have
Put ; then, from (10), we deduce that which implies that
Replacing this value in (10), we obtain the following expression for function : This completes the proof.

Lemma 6. Let be the Green function, which is given by the expression (7). For , the following property holds:

The proof is similar to that of Lemma 2.4 in [7], so we omit it.

Consider the Banach space with general norm Define the cone .

Suppose is a solution of (1). It is clear from Lemma 5 that

Define the operator as follows:

Lemma 7. is completely continuous.

Proof. Since , it is obvious that . So we have Therefore, . The other proof is similar to that in [7], so we omit it.

#### 3. Main Result

For convenience, we list the denotation:

Next, we will establish some sufficient conditions for the existence and nonexistence of positive solution for problem (1).

Theorem 8. Let be a constant. Then for each problem (1) has at least one positive solution.

Proof. First, for any , from (20) we have
On the one hand, by the definition of , there exists such that, for any , we have Choose . For , we have
On the other hand, by the definition of , there exists such that, for any , we have Take . For , we have According to (23),  (25), and Lemma 4, has at least one fixed point with , which is a positive solution of (1).

Remark 9. If and , then we can get Theorem 8 implies that, for , problem (1) has at least one positive solution.

Theorem 10. Let be a constant. Then for each problem (1) has at least one positive solution.

Proof. First, it follows from (27) that, for any ,
By the definition of , there exists such that, for any , we have Choose . For , we have . Similar to the proof in Theorem 8, it holds from (28) and (29) that
Note . There exists , such that We consider the problem on two cases. (I) Suppose is bounded. There exists , such that , . Choose . Let . For , we have
(II) Suppose is unbounded. There exists such that
Let . For , we have Combining (I) and (II), take ; here, . Then for , we have
Hence, (30) and (42) together with Lemma 4 imply that has at least one fixed point with , which is a positive solution of (1).

Theorem 11. Assume and . Problem (1) has no positive solution provided where is a constant defined in (38).

Proof. Since and , together with the definitions of and , there exist positive constants , , , and satisfying such that Take
It follows that for any . Suppose that is a positive solution of (1). That is, In sequence, which is a contradiction. Hence, (1) has no positive solution.

Theorem 12. Assume and . Problem (1) has no positive solution provided where is a constant defined in (43).

Proof. Since and , together with the definitions of and , there exist positive constants , , , and satisfying such that Take It follows that for any . Suppose that is a positive solution of (1). That is, In sequence, which is a contradiction. Hence, (1) has no positive solution.

Example 13. Consider the fractional differential equation In this example, take Obviously, we have
Since and , through a computation, we can get
Choose ; we have Theorem 8 implies that, for , , the problem (46) has at least one positive solution.

Remark 14. In particular, if we take in Example 13, then and . Remark 9 implies that problem (46) has at least one positive solution for .

#### Conflict of Interests

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

#### Acknowledgments

This work is supported by the NNSF of China (no. 61373174) and the Natural Science Foundation for Young Scientists of Shanxi Province, China (no. 2012021002-3).