Received 19 June 2016; accepted 14 August 2016; published 17 August 2016
The aim of this paper is to investigate the existence of positive solutions to the semipositone fractional differential equation
where, is the standard Riemann-Liouville fractional derivative of order which is defined as follows:
where denotes the Euler gamma function and denotes the integer part of number, provided that the right side is pointwise defined on, see  . Here, by a positive solution to the problem (1), we mean a function, which is positive in, and satisfies (1).
Fractional differential equations have gained much importance and attention due to the fact that they have been proved to be valuable tools in the modelling of many phenomena in engineering and sciences such as physics, mechanics, economics and biology. In recent years, there exist a great deal of researches on the existence and/or uniqueness of solutions (or positive solutions) to boundary value problems for fractional-order differential equations. Sun  studied the existence of positive solutions for the following boundary value pro- blems:
where, is continuous and on. But paper  did not give the results of the existence of positive solution when the nonlinearity can take negative value, i.e. semipositone problems.
The purpose of the present paper is to apply the method of varying translation together with the fixed point theorems in cone to discuss (1) without nonnegativity imposed on the nonlinearity. Meanwhile, we also allow the nonlinearity to have many finite singularities on.
2. Preliminaries and Lemmas
In this section, we present several lemmas that are useful to the proof of our main results. For the forthcoming analysis, we need the following assumptions:
(H1) is continuous. For any, there exist constants
(H2) with and,
will be defined in the following text.
In  , the authors obtained the Green function associated with the problem (1). More precisely, the authors proved the following lemma.
Lemma 2.1  . For any, the unique solution of the boundary value problem
is given by
Lemma 2.2  . The Green function defined by (4) satisfies the inequality
Remark 2.1. A simple computation shows that there exists a constant such that
Remark 2.2  . If satisfies (H1), then for any is increasing on
and for any,
Lemma 2.3  . Let X be a real Banach space, be a bounded open subset of X with and is a completely continuous operator, where P is a cone in X.
(i) Suppose that then.
(ii) Suppose that then.
Consider the Banach space with the usual supremum norm and define the
cone. Let, then is the unique solution
to (2) for. Now we first consider the singular nonlinear boundary value problem
where We have the following Lemma.
Lemma 2.4. If the singular nonlinear boundary value problem (2) has a positive solution such that for any. Then boundary value problem (1) has a positive solution.
Proof. In fact, if u is a positive solution to (6) such that for any. Let, then. Since is the unique solution to (2) for for any, we
have, which implies that. So
. Consequently is positive solution to (1). This complete the proof of Lemma 2.4.
For any, define an operator
Since for any fixed, we can choose such that. Note that
so by (H1), we have
Consequently, for any, we have
Therefore, the operator T is well defined and
Lemma 2.5. Assume that (H1), (H2) hold. Then is a completely continuous operator.
Proof. For any, in view of (2) we conclude that
Whence, it follows from (8) that which implies
Next we show that is continuous. Suppose, and Then, there exists a constant such that. Since for any
, by Remark 2.2, we have
Thus, we have
and. It follows from the Lebesgue control convergence theorem that
which implies is continuous.
In what follows, we need to prove that is relatively compact.
Let be any bounded set. Then there exists a constant such that for any. Similarly as (9), for any we have
Therefore is uniformly bounded.
Now we show that is equicontinuous on. For any, by (9), (11) and the Lebesgue control convergence theorem, and noticing the continuity of, we have
Thus, is equicontinuous on [0,1]. The Arezlà-Ascoli Theorem guarantees that is relatively compact set. Therefore is completely continuous operator.
Lemma 2.6. Let then.
Proof. Assume that there exists such that Then and Thus we have
This contradiction shows that.
Lemma 2.7. There exists a constant such that, where
Proof. Choose constants and N such that
From Remark (2.2), there exists, such that
Let Obviously, Now we show that In
fact, otherwise, there exists such that By (2), for any we have
Consequently, That is This
contradiction shows that.
3. Main Results
Theorem 3.1. Suppose that (H1), (H2) hold. Then, the boundary value problems (1) has at least one positive solution, and exists a constant such that
Proof of Theorem 3.1. Applying Lemma 2.6 and Lemma 2.7 and the definition of the fixed point index, we have Thus T has a fixed point in with. Since, we have
Let It follows from Lemma (2.4) that is a positive solution to boundary value problem (1), and there exists a constant such that
We thank the Editor and the referee for their comments. This research was supported financially by the National Natural Science Foundation of China (11471187, 11571197), the Natural Science Foundation of Shandong Province of China (ZR2014AL004) and the Project of Shandong Province Higher Educational Science and Technology Program (J14LI08), the Project of Scientific and Technological of Qufu Normal University (XKJ201303).