Abstract
A class of nonlinear fractional order differential equation
is investigated in this paper, where
1 Introduction
Consider the following boundary value problem
where
In the last few decades, many authors have investigated fractional differential equations which have been applied in many fields such as physics, mechanics, chemistry, engineering etc. (see references [1,6,10,2123]). Especially, many works have been devoted to the study of initial value problems and bounded value problems for fractional order differential equations [12,13,15,24].
Recently, the existence of positive solutions of fractional differential equations has attracted many authors’ attention [25,8,9,12,14,1720,25,26]. Using some fixed point theorems, they obtained some nice existence conditions for positive solutions.
In more recent works, Jiang and Yuan [7] considered the following boundary value problem of fractional differential equations
where
Further, Li, Luo, and Zhou [4] investigated the following fractional order threepoint boundary value problems
where
In this paper, we discuss the boundary value problem (1.1)(1.2). Using some properties
of the Green function
The paper is arranged as follows: In Section 2, we introduce some definitions for fractional order differential equations and give our main results for the boundary value problem (1.1)(1.2). We give some lemmas for our results in Section 3. In Section 4, we prove our main result; and finally, we give an example to illustrate our results.
2 Main results
In this section, we introduce some definitions and preliminary facts which are used in this paper.
The fractional integral of order α with the lower limit
provided that the integral on the righthand side is pointwise defined on
RiemannLiouville derivative of order α with the lower limit
where n is a positive integer.
We call the function
We also need to introduce some lemmas as follows, which will be used in the proof of our main theorems.
Lemma 2.1 ([26])
Assume that
Lemma 2.2 ([26])
Assume that
for some
Lemma 2.3 ([16])
Suppose thatXbe a Banach space,
(i) Thas a fixed point in
(ii) there exist
Throughout this paper, we assume that
(H)
where
for any
We have our main results:
Theorem 2.1Suppose that (H) holds. If
then the boundary value problem (1.1)(1.2) has at least one solution, where
3 Some lemmas
Let
then Ω is a Banach space.
We first give some lemmas as follows:
Lemma 3.1Problem (1.1)(1.2) is equivalent to the following integral equation
Proof The sufficiency is obvious, we only need to prove the necessity.
Suppose that
According to (1.2) and (3.4), we have
Combining (3.4) and (3.5), we obtain
According to (3.3), it is easy to show that (3.2) holds. The proof is completed. □
Lemma 3.2For any
Proof The continuity of
Let
we only need to show that
for
Let
then
The new Green’s function
Lemma 3.3
Furthermore,
Lemma 3.4For any
Let
Then,
Let
Then,
We can divide our proof into the following two steps:
First, we replace
It suffices to show that for any given real number μ, (3.13) has a solution
Second, we show that there exists a μ such that the solution
In this section, we will prove the first step. For convenience sake, we define an operator T on the set Ω as follows:
Lemma 3.5Suppose that
Proof It is easy to show that the operator T maps Ω into itself. We divide the proof into the following three steps.
Step 1.
In fact, suppose that
which follows from (3.14)(3.15) that
Thus, the operator T is continuous in Ω.
Step 2. T maps bounded set in Ω into bounded set.
Suppose that
This gives that the operator T maps bounded set into bounded set in Ω.
Step 3. T is equicontinuous in Ω.
It suffices to show that for any
(i)
(ii)
(iii)
We only prove the case (i), the rest two cases are similar. Since B is bounded, then there exists a
According to Step 1Step 3, the operator T is completely continuous in Ω. The proof is completed. □
Further, we have
Lemma 3.6Suppose that
Proof We only need to show that the operator T is priori bounded. Let
Define a set
To show the existence of a fixed point of T by Lemma 2.3, we need to verify that the second possibility in Lemma 2.3 cannot happen.
In fact, assume that there exists
and
Here we have the use of the inequality
It is obvious that (3.18) contradicts our assumption that
4 The proof of the main results
Now, we prove Theorem 2.1 by Lemma 3.43.5 and the intermediate value theorem.
Proof of Theorem 2.1 It is obvious that the righthand side of (3.14) is continuously dependent on the
parameter μ, so we need to find a μ such that
For any given real number μ, we rewrite (3.13) as follows:
From (4.1), it suffices to show that there exists a μ such that
It is obvious that
Now, we show that
as
for all
for all
by (4.2), (4.5)(4.6), we have
which contradicts our assumption.
Now, for large
Then,
Further, we divide the set
It is easy to know that
From (H) again, the function
Let
From the definitions of
and it follows that
for
from which it follows that
which implies that
This contradicts (4.9).
Now, we have proved that
Notice that
5 Examples
Example 5.1 Consider the following boundary value problem
where
and
It is easy to show that
and
Thus, the conditions of Theorem 2.1 are satisfied. Therefore, the problem (5.1) has at least a nontrivial solution.
Competing interests
The authors declare that they have no competing interests.
Author’s contributions
Each of the authors, ZO and GL contributed to each part of this study equally and read and approved the final version of the mnanuscript.
Acknowledgements
Supported partially by China Postdoctoral Science Foundation under Grant No.20110491280 and the Subject Lead Foundation of University of South China No. 2007XQD13.
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