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Journal of Function Spaces
Volume 2017, Article ID 3187492, 9 pages
https://doi.org/10.1155/2017/3187492
Research Article

Positive Solutions of Fractional Differential Equations with -Laplacian

1College of Mathematics and Finance, Xiangnan University, Chenzhou 423000, China
2College of Mathematics and System Science, Shandong University of Science and Technology, Qingdao 266590, China

Correspondence should be addressed to Sujing Sun; moc.361@jssdk

Received 21 June 2017; Revised 5 October 2017; Accepted 23 October 2017; Published 14 November 2017

Academic Editor: Lishan Liu

Copyright © 2017 Yuansheng Tian et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

The multiplicity of positive solution for a new class of four-point boundary value problem of fractional differential equations with -Laplacian operator is investigated. By the use of the Leggett-Williams fixed-point theorem, the multiplicity results of positive solution are obtained. An example is given to illustrate the main results.

1. Introduction

In recent years, boundary value problems of nonlinear fractional differential equations have been studied extensively (see [123] and the references therein). By the use of some fixed-point theorem, the existence results of positive solutions are obtained for singular factional problem [2, 4, 24, 25], impulsive fractional problem [3, 20], nonlocal problem [1, 12, 14, 16, 22, 23], numerical solution problem [17, 18], initial value problem [19, 20], Dirichlet value problem [6, 9, 15], iterative solution problem [5], and so on. Cui [7] considered the following boundary values problems: where is a real number and is the standard Riemann-Liouville differentiation. Under the assumption that is a Lipschitz continuous function, by the use of -positive operator, they studied the uniqueness results for the fractional differential equation.

On the other hand, because of the wide mathematical and physical background, the existence of positive solutions for nonlinear integer-order boundary values problems with -Laplacian operator has received wide attention (see [8, 12, 13, 21, 2429]). For example, Su et al. [28] considered the following four-point boundary values problems with -Laplacian operator: where . Liu et al. [13], Dong et al. [8], and Zhang et al. [25] studied -Laplacian boundary value problems with fractional derivative. By using the fixed-point index theory, they obtained the existence of positive solutions.

In this paper, we investigate the multiplicity of positive solution for a new class of four-point boundary value problem of fractional differential equations with -Laplacian operator: where and and is the standard Riemann-Liouville differentiation and . By using the Leggett-Williams fixed-point theorem on a cone, the multiplicity results of positive solution are obtained.

2. Preliminaries

Lemma 1 (see [10]). Assume that with a fractional derivative of order that belongs to . Then where is the smallest integer greater than or equal to .

Lemma 2. If and , then the problem has a unique solution where

Proof. Suppose is a solution of the problem (5). By Lemma 1, there is for some . Taking into account the fact that , we have and Thus, By , (10), it holds that So, Applying Lemma 1, we can reduce (13) to an equivalent integral equation for some . With the condition , there is . Consequently, By (15), one has So, By , (17), we have So, the unique solution of problem (5) is The proof is completed.

Lemma 3. Suppose that . The functions and satisfy the following:(1) for ,(2) for ,(3)There exists a positive function such that

Proof.
The Proof of the Statement (1). From definitions, it is clear that . By , we have If , let It is obvious that for . Hence, we have By using the analogous argument, it holds that for other situations. Hence, for .
If , let It is obvious that for . Hence, we have Similarly, it holds that for or or . Hence, for .
The Proof of the Statement (2). Let It is easy to check that and are increasing with respect to on . We will show that and are decreasing with respect to on . For , let then we have For , let We have If there exists such that , then Therefore, we have In fact, taking into account the fact that and , one has So, we have , which implies that . Hence is decreasing with respect to on .
For , let ; then we have which implies that is decreasing with respect to on .
We can conclude that is increasing with respect to for and is decreasing with respect to for . Hence, for .
By using the analogous method, we can conclude that is increasing with respect to for and is decreasing with respect to for . Hence, for .
The Proof of the Statement (3). Taking into account the definition of , there is where . It is easy to see that is continuous on and for all . Let where Then The proof is completed.

Lemma 4 (see [11]). Let be a cone in a real space be a nonnegative continuous concave functional on a cone such that , for all , and Suppose is completely continuous and there exist constants such that() and for ;() for ;() for with . Then has at least three fixed points satisfying and with .

3. Main Result

Let be a Banach space with . Define the cone by Let the nonnegative continuous concave functional on the be defined by

Lemma 5. Let be an operator defined by Then is completely continuous.

Proof. is continuous in view of nonnegativity and continuity of and . Furthermore it is easy to see that by the Arzela-Ascoli theorem and Lebesgue dominated convergence theorem is completely continuous.

For convenience, we introduce the following notations:

Theorem 6. Suppose and there exist constants such that() , for ;() , for ;() , for . Then problem (3) possesses at least three positive solutions , and with

Proof. In order to apply Lemma 4, we divide the proof into four steps.
Step  1. If , then . Assumption implies for . Consequently, Hence .
Step  2. We claim that the condition of Lemma 4 is satisfied. Let . we can easily see that ; consequently, . Hence, if , then for . From assumption , we have So Then, we have This implies that condition of Lemma 4 is satisfied.
Step  3. We now prove the condition of Lemma 4 is satisfied. If , then . Assumption implies for . Thus Hence, the condition of Lemma 4 is also satisfied.
Step  4. Finally, we prove that the condition of Lemma 4 is satisfied. If , by Step  2, we have . Hence, the condition of Lemma 4 is also satisfied.
By Lemma 4, problem (3) has at least three positive solutions , and with The proof is complete.

Example 7. Consider the following boundary value problem: where Choose It is easy to check that and ; by simple computation, we have Let ; one can check that the function satisfies() , for ;() , for ;() , for .That is to say that all the conditions of Theorem 6 hold. Thus Theorem 6 implies the problem (50) has at least three positive solutions