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 [1–5]. Recently, the existence of solutions (or positive solutions) of nonlinear fractional differential equation has been investigated in many papers (see [6–28] and references cited therein). However, in terms of the eigenvalue problem of fractional differential equation, there are only a few results [29–33].
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).