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Existence of the solutions for a class of nonlinear fractional order three-point boundary value problems with resonance
Boundary Value Problems volume 2012, Article number: 68 (2012)
A class of nonlinear fractional order differential equation
is investigated in this paper, where is the standard Riemann-Liouville fractional derivative of order , , . Using intermediate value theorem, we obtain a sufficient condition for the existence of the solutions for the above fractional order differential equations.
Consider the following boundary value problem
where is the standard Riemann-Liouville fractional derivative of order , and .
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, 21–23]). 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 [2–5, 8, 9, 12, 14, 17–20, 25, 26]. Using some fixed point theorems, they obtained some nice existence conditions for positive solutions.
In more recent works, Jiang and Yuan  considered the following boundary value problem of fractional differential equations
where is the standard Riemann-Liouville fractional derivative of order and is continuous. Using some properties of the Green function , they obtain some new sufficient conditions for the existence of positive solutions for the above problem.
Further, Li, Luo, and Zhou  investigated the following fractional order three-point boundary value problems
where is the standard Riemann-Liouville fractional derivative of order , and is continuous.
In this paper, we discuss the boundary value problem (1.1)-(1.2). Using some properties of the Green function and intermediate value theorem, we establish some sufficient conditions for the existence of the positive solutions of the problem (1.1)-(1.2).
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 for a function f is defined as
provided that the integral on the right-hand side is point-wise defined on , where Γ is the Gamma function.
Riemann-Liouville derivative of order α with the lower limit for a function can be written as
where n is a positive integer.
We call the function a solution of (1.1)-(1.2) if with a fractional derivative of order α belongs to and satisfies Equation (1.1) and the boundary condition (1.2).
We also need to introduce some lemmas as follows, which will be used in the proof of our main theorems.
Lemma 2.1 ()
Assume thatwith a fractional derivative of orderbelongs to. Then, the fractional equation
Lemma 2.2 ()
Assume thatwith a fractional derivative of orderbelongs to. Then
for some, , .
Lemma 2.3 ()
Suppose that X be a Banach space, is closed and convex. Assume that U is a relatively open subset of C with, andis a completely continuous operator. Then, either
T has a fixed point in , or
there exist and with .
Throughout this paper, we assume that satisfies the following:
(H), and there exist two positive functionsandsuch that
We have our main results:
Theorem 2.1 Suppose that (H) holds. If
then the boundary value problem (1.1)-(1.2) has at least one solution, where
3 Some lemmas
Let , equipped the norm
then Ω is a Banach space.
We first give some lemmas as follows:
Lemma 3.1 Problem (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 is a solution of the problem (1.1)-(1.2). Integrating both sides of (1.1) of α order from 0 to t with respect to t, it follows 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 3.2 For any, is continuous, andfor any.
Proof The continuity of for is obvious.
we only need to show that for , the rest of the proof is similar or obvious. From the definition of , we have
for . The proof is completed. □
The new Green’s function has the following properties:
Lemma 3.3is continuous for, and
Lemma 3.4 For any, is nonincreasing with respect to. Especially, for any, for, andfor. That is, where
Then, , we have from Lemma 3.1, (3.6) and (3.9) that the integral Equation (3.2) can be rewritten as follows:
Then, and (3.10) is equivalent to the following
We can divide our proof into the following two steps:
First, we replace by any real number μ, then (3.12) can be rewritten as
It suffices to show that for any given real number μ, (3.13) has a solution , which implies that Equation (1.1) has a solution which satisfies the first boundary value condition .
Second, we show that there exists a μ such that the solution of (3.13) satisfies , which implies that the solution of (1.1) also satisfies the boundary value condition .
In this section, we will prove the first step. For convenience sake, we define an operator T on the set Ω as follows:
Lemma 3.5 Suppose that, and (2.4) hold, then the operator T is completely continuous in Ω.
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. is continuous with respect to .
In fact, suppose that is a sequence in Ω, and converges to . Since is continuous with respect to , and it is obvious that is uniformly continuous with respect to from Lemma 3.3, then for any , there exists an integer N, when ,
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 is a bounded set with for any . Then, we have from (2.4) and (3.14) 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 and any , as . We consider the following three cases:
We only prove the case (i), the rest two cases are similar. Since B is bounded, then there exists a such that . According to (3.14), we have
According to Step 1-Step 3, the operator T is completely continuous in Ω. The proof is completed. □
Further, we have
Lemma 3.6 Suppose that, and (2.4) and (2.6) holds, then, for any real number μ, the integral Equation (3.13) has at least one solution.
Proof We only need to show that the operator T is priori bounded. Let
Define a set as follows
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 with and such that . It follows that
Here we have the use of the inequality
It is obvious that (3.18) contradicts our assumption that . Therefore, by Lemma 2.3, it follows that T has a fixed point . Hence, the integral Equation (3.14) has at least a solution . The proof is completed. □
4 The proof of the main results
Now, we prove Theorem 2.1 by Lemma 3.4-3.5 and the intermediate value theorem.
Proof of Theorem 2.1 It is obvious that the right-hand side of (3.14) is continuously dependent on the parameter μ, so we need to find a μ such that , which implies 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 is continuously dependent on the parameter μ. In order to prove that there exists a such that , we only need to show that , and .
Now, we show that . On the contrary, we assume that . Then, there exists a sequence , such that , which implies that the sequence is bounded from above. Notice that the function is continuous with respect to and . We first claim that it is impossible to have
as is large enough. Indeed, assume that (4.3) is true. Then, by (4.1), we have
for all . Thus, we get
for all . Since we have assumed in (H) that
by (4.2), (4.5)-(4.6), we have
which contradicts our assumption.
Now, for large , we define
Then, is not empty.
Further, we divide the set into two sets and as follows:
It is easy to know that , and , and we have from (H) that is not empty.
From (H) again, the function is bounded below by a constant for and . Thus, there exists a constant M (<0), independent of t and , such that
From the definitions of and , we have
and it follows that as (since if is bounded below by a constant as , then (4.7) holds). Therefore, we can choose large enough so that
for . From (H), (4.1), (4.8)-(4.9), and the definitions of and , for any , we have
from which it follows that
which implies that
This contradicts (4.9).
Now, we have proved that . By a similar method, we can prove that . The detail is omitted.
Notice that is continuous with respect to . It follows from the intermediate value theorem  that there exists a such that , that is , which satisfies the second boundary value condition of (1.2). The proof is completed. □
Example 5.1 Consider the following boundary value problem
It is easy to show that
Thus, the conditions of Theorem 2.1 are satisfied. Therefore, the problem (5.1) has at least a nontrivial solution.
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.
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Supported partially by China Postdoctoral Science Foundation under Grant No.20110491280 and the Subject Lead Foundation of University of South China No. 2007XQD13.
The authors declare that they have no competing interests.