We first offer some definitions and fundamental facts of fractional calculus theory, which can be found in [5–9].
Definition 2.1 (see [7, 8], [[6], pp.36-37])
The Riemann-Liouville fractional derivative of order of a continuous function is given by
where , denotes the integer part of number α, provided that the right-hand side is pointwise defined on .
Definition 2.2 (see [[6], Definition 2.1])
The Riemann-Liouville fractional integral of order of a function is given by
provided that the right-hand side is pointwise defined on .
From the definition of the Riemann-Liouville derivative, we can obtain the following statement.
Lemma 2.1 (see [11])
Let . If we assume , then the fractional differential equation has a unique solution
where N is the smallest integer greater than or equal to α.
Lemma 2.2 (see [11])
Assume that with a fractional derivative of order that belongs to . Then
where N is the smallest integer greater than or equal to α.
In what follows, we need to consider the following fractional integral boundary value problem:
(2.1)
then we present Green’s function for (2.1), and study the properties of Green’s function. In our paper, we always assume that the following two conditions are satisfied:
(H0) .
(H1) is bounded from below, i.e., there is a positive constant M such that , .
Lemma 2.3 Let (H0), (H1) hold. Then problem (2.1) is equivalent to
where
(2.2)
and
(2.3)
Proof By Lemmas 2.1 and 2.2, we can reduce the equation of problem (2.1) to an equivalent integral equation
(2.4)
where () are fixed constants. By , there is . Thus,
(2.5)
Differentiating (2.5), we have
(2.6)
By (2.6) and , we have . Then
(2.7)
From , we arrive at
and thus
Therefore, we obtain by (2.7)
(2.8)
where is defined by (2.3). From (2.8), we have
(2.9)
and by (H0) we find
(2.10)
Combining (2.8) and (2.10), we see
(2.11)
where is determined by (2.2). This completes the proof. □
Lemma 2.4 (see [[10], Lemma 3.2])
For any , let , , . Then the following two inequalities are satisfied:
-
(i)
,
-
(ii)
.
Proof (i) For , we have , then
(2.12)
On the other hand, for , since , we have
Consequently,
Moreover, for , note that , , and , then we find
On the other hand, for , we have
Therefore, we get
-
(ii)
If , since , we have and
For , we have , then by (2.12) we get
This completes the proof. □
Lemma 2.5 Let and . Then the following inequality holds:
(2.13)
Proof By (i) of Lemma 2.4, we have
and we easily obtain (2.13), as claimed. This completes the proof. □
Let
Then is a real Banach space and P is a cone on E.
The norm on is defined by , . Note that is a real Banach space under the above norm, and is a positive cone on .
By Lemma 2.3, we can obtain that system (1.1) is equivalent to the system of nonlinear Hammerstein integral equations
(2.14)
where is defined by (2.2).
Lemma 2.6 (i) If is a positive solution of (2.1), then is a positive solution of the following differential equation:
(2.15)
where
the function , is continuous,
(2.16)
-
(ii)
If is a solution of (2.15) and , , then is a positive solution of (2.1).
Proof If is a positive solution of (2.1), then we obtain
By a simple computation, we easily get , and
i.e., satisfies (2.15). Therefore, (i) holds, as claimed. Similarly, it is easy to prove that (ii) is also satisfied. This completes the proof. □
By Lemma 2.3, we obtain that (2.15) is equivalent to the integral equation
(2.17)
where is determined by (2.2). Clearly, the continuity and nonnegativity of G and F imply that is a completely continuous operator.
Lemma 2.7 Put . Then , where μ and T are defined by Lemma 2.4 and (2.17), respectively.
Proof By (i) of Lemma 2.4, we easily find
and thus
This completes the proof. □
In this paper, we assume that () satisfy the following condition:
(H2) and there is a positive constant M such that , .
By (H2) and Lemma 2.6, (2.14) is turned into the following integral equation:
(2.18)
where
the function , and is denoted by (2.16). By Lemma 2.6, we know if is a solution of (2.18) and , , then (, ) is a positive solution of (1.1).
Define the operator A as follows:
(2.19)
where
It is obvious that , are completely continuous operators. Clearly, is a positive solution of (1.1) if and only if is a fixed point of A and , .
The following two lemmas play some important roles in our proofs involving fixed point index.
Lemma 2.8 ([19])
Let be a bounded open set, and let be a completely continuous operator. If there exists such that for all and , then .
Lemma 2.9 ([19])
Let be a bounded open set with . Suppose that is a completely continuous operator. If for all and , then .