Let , , then is a Banach space. Denote , , .
For the reader’s convenience, we present some necessary definitions from fractional calculus theory and lemmas. They can be found in the recent literature; see [14–17].
Definition 2.1 The Riemann-Liouville fractional integral of order of a function is given by
provided the right-hand side is pointwise defined on .
Definition 2.2 The Riemann-Liouville fractional derivative of order of a continuous function is given by
where , denotes the integer part of the number α, provided that the right-hand side is pointwise defined on .
From the definition of the Riemann-Liouville derivative, we can obtain the statement.
Lemma 2.1 ()
Let. If we assume, then the fractional differential equation
has, , , as unique solutions, where N is the smallest integer greater than or equal to α.
Lemma 2.2 ()
Assume thatwith a fractional derivative of orderthat belongs to.
for some, , where N is the smallest integer greater than or equal to α.
In the following, we present Green’s function of the fractional differential equation boundary value problem.
Lemma 2.3 Given, the problem
where, , , , is equivalent to
Here, , is called the Green function of BVP (2). Obviously, is continuous on.
Proof We may apply Lemma 2.2 to reduce (2) to an equivalent integral equation
for some . Consequently, the general solution of (2) is
By , one gets that . On the other hand, combining with
Therefore, the unique solution of the problem (2) is
For , one has
For , one has
The proof is complete. □
Lemma 2.4 The functiondefined by (3) satisfies
(a1) , ;
(a2) , ;
(a3) , ;
(a4) , andis not decreasing on;
(a5) , ,
where, , .
Proof For , ,
For , ,
From above, (a1), (a2), (a3), (a5) are complete. Clearly, (a4) is true. The proof is complete. □
Throughout this article, we adopt the following conditions.
(H1) and there exist , , such that
(H2) There exists such that uniformly for ;
(H3) There exists such that
Obviously, Q is a cone in a Banach space E and is an ordering Banach space.
where is defined as that in (H1). It follows from Lemma 2.4 and (H3) that
So, and it satisfies
For any , , . Consequently, by (6) and Lemma 2.4, we have
For any , denote
We define an operator A as follows:
Lemma 2.5 Suppose that ()-() hold. Thenis completely continuous.
Proof For any , it is clear that . By (H1), we get
By (10) and Lemma 2.4, we have
which together with (H3) means that operator A defined by (9) is well defined.
Now, we show that .
For any , by (H1) we have by (9) and Lemma 2.4 that
which means that
It follows from (12) and Lemma 2.4 that
Thus, A maps Q into Q.
Finally, we prove that A maps Q into Q is completely continuous.
Let be any bounded set. Then there exists a constant such that for any . Notice that , for any , , by (H3) and (11), we have
Therefore, is uniformly bounded.
On the other hand, since is continuous on , it is uniformly continuous on as well. Thus, for fixed and for any , there exists a constant such that for any and ,
Therefore, for any , we get by (10) and (13)
which implies that the operator A is equicontinuous. Thus, the Ascoli-Arzela theorem guarantees that is a relatively compact set.
Let , (). Then is bounded. Let , by (10), we get
By (9), we have
It follows from (14), (15), (H1), (H3), and the Lebesgue dominated convergence theorem that A is continuous. Thus, we have proved the continuity of the operator A. This completes the complete continuity of A. □
To prove the main result, we need the following well-known fixed point theorem.
Lemma 2.6 (Fixed point theorem of cone expansion and compression of norm type )
Letandbe two bounded open sets in a Banach space E such thatandbe a completely continuous operator, where θ denotes the zero element of E and P a cone of E. Suppose that one of the two conditions holds:
, ; , ;
, ; , .
Then A has a fixed point in.