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Abstract and Applied Analysis
Volume 2015, Article ID 263748, 11 pages
http://dx.doi.org/10.1155/2015/263748
Research Article

Positive Solutions for Class of State Dependent Boundary Value Problems with Fractional Order Differential Operators

School of Mathematics and Physics, University of South China, Hengyang 421001, China

Received 13 April 2014; Revised 4 August 2014; Accepted 4 August 2014

Academic Editor: Ali H. Bhrawy

Copyright © 2015 Dongyuan Liu 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

We consider the following state dependent boundary-value problem , ; , where is the standard Riemann-Liouville fractional derivative of order , , , , the function is defined as , and the function is defined as , are continuous on and , . Using Banach contraction mapping principle and Leray-Schauder continuation principle, we obtain some sufficient conditions for the existence and uniqueness of the positive solutions for the above fractional order differential equations, which extend some references.

1. Introduction

Fractional order differential equations has useful applications in many fields, such as physics, mechanics, chemistry, engineering, biology, and so on. There has been a significant development in fractional differential equations (e.g., [19]). In the previous papers, some authors investigated fractional order partial differential equations [1015]. For example, Wu [15] used the wavelet operational method for solving fractional partial differential equations numerically. Since it is one of the important fields to be concerned with the boundary value problems for fractional order differential equations, some authors considered the existence of positive solutions for fractional differential equations or systems with boundary value conditions [1625] and the stability [26].

As early as 1994, Delbosco [27] investigated the nonlinear Dirichlet-type problem where is order Riemann-Liouville derivative. The author had proved that if is a Lipschitz function, then the problem has at least one solution in a certain subspace of in which the fractional derivative has a Hölder property.

Later, using some fixed point theorems, Bai and Lü [20] obtained the existence of positive solutions of the following equation with boundary value conditions

More recently, Bai [9] also considered the following boundary value problem By constructing a Green’s function, and using contraction map principle, the author obtained some existence conditions of positive solutions for (3).

Motivated by the above references, we consider a state dependent boundary value problem with fractional order differential operators where is the standard Riemann-Liouville fractional derivative of order , , , , , ; the function is defined as , and ; the function is defined as and , are continuous on and .

By using Banach contraction mapping principle and Leray-Schauder continuation principle, we obtain some sufficient conditions for the existence and uniqueness of the positive solutions for boundary value problem (4). Furthermore, we give an example to illustrate our results.

2. Preliminary

In this section, we introduce some definitions and preliminary facts which are used in this paper.

Definition 1 (see [8, 16]). The fractional integral of order with the lower limit for a function is defined as provided the right-side is point-wise defined on , where is the Gamma function.

Definition 2 (see [8, 16]). Riemann-Liouville derivative of order with the lower limit for a function can be written as

Definition 3 (see [8, 16]). Caputo’s derivative of order with the lower limit for a function can be written as

It is well known that if , then , . Furthermore, if and , then for , we have which is with the semigroup property for and .

We also need to introduce some Lemmas as follows, which will be used in the proof of our main theorems.

Lemma 4 (see [19, 20]). Let ; then the fractional equation has solutions

Lemma 5 (see [19, 20]). Let ; then for some , , .

Lemma 6 (see [28], the Banach contraction mapping theorem). Let be a contraction mapping of a complete metric space . Then has one and only one fixed point.

Lemma 7 (see [2830], the Leray-Schauder continuation principle). Let be a Banach space with being closed and convex. Assume that is a relatively open subset of with and is completely continuous. Then either(a)has a fixed point in , or(b)there exists and with .

Throughout this paper, we assume that and we satisfy the following(H):(i) is Lebesgue measurable with respect to on ;     (ii) is continuous with respect to on .

3. Main Results

For convenience, we rewrite (4) as follows:

Integrating both sides of (13) of order with respect to , it follows that From (14) and (15), we have

Combining (14) with (17), we obtain According to (15) and (18), it follows that Let According to (19)–(21), it follows that which means that if satisfies (13)-(14), then it satisfies (22). It is easy to show that if satisfies (22), then it also satisfies (13)-(14). Thus, the boundary value problem (13)-(14) is actually equivalent to integral equation (22). Therefore, we have the following.

Lemma 8. Problem (13)-(14) is equivalent to (22).

Lemma 9. For any , are continuous and , .

Proof. It is obvious that are continuous on . We first prove that on . Let We first show that , , . We rewrite as follows: Let Since , then Differentiating both sides of (25) with respect to , it follows that which means that is nondecreasing with respect to on . Thus, for any , , therefore, .
Using the similar method, we can prove that , and it is obvious that , . Hence, combining (20) and (23), we obtain that .
Now, we prove that . Denote Let Since , then Differentiating both sides of (29) with respect to , it follows that which means that is nonincreasing with respect to on . Thus, for any , , therefore, .
Using the similar method, we can prove that , and the case that , is obvious. Combining (21) and (28) and using the above argument, we obtain that . The proof is complete.

Lemma 10. For any , are nondecreasing functions with respect to ; that is, for any , .

Proof. According to the proof of Lemma 9, we notice that At the same time, for , we have The proof is complete.

Now, we present our main results.

Theorem 11. Assume that (H) holds. Suppose that there are two functions such that If then the problem (13)-(14) has a unique positive solution.

Proof. Set with the maximum norm It is easy to show that is a complete metric space. We denote a operator as follows: From Lemma 9 and the conditions , , , it follows that maps into itself and we only need to prove the contraction. In fact, according (H) and (34), for any , we have which means that By the Banach contraction mapping principle (Lemma 6), we obtain that has a unique fixed point which is a positive solution of (13)-(14). The proof is complete.

Remark 12. If , then problem (13)-(14) is problem (3).

Theorem 13. Assume that (H) holds. Suppose that there exists four nonnegative real-valued functions such that If then the problem (13)-(14) has at least one positive solution.

Proof. We also consider the operator defined in (38). We divide the proof into four steps.
Step 1.   is continuous.
Let be a sequence in such that as . Noticing that are continuous with respect to , then for each , we have thus, which implies that Step 2. maps bounded sets into bounded sets in .
Indeed, it suffices to show that for any , there exists a positive constant such that for , we have . From (38)–(43), we have which means Step 3. maps bounded sets into equicontinuous sets in .
For any , and for each , we have Because are continuous on , it is uniformly continuous on , which means that for any , there exists , when , Thus which means that is equicontinuous.
Step 4. A priori bounds.
Let
Assume that there exists and such that and we claim that . In fact, which implies that which means that That is, there is no , such that for .
From Step 1 to Step 3, it follows that is completely continuous. Along with Step 4 and Lemma 7, it follows that has at least a fixed point in . The proof is complete.

4. Examples

Example 1. Consider where It is easy to show that Let then thus, which satisfies (28), (29). Because thus which satisfies Theorem 11. Thus (56) has a unique positive solution on .

Conflict of Interests

The authors declare that they have no competing interests.

Acknowledgments

The work was supported by NSF of Hunan Province (no. 13JJ3074), the Project of Department of Education of Hunan Province (13A088 and 12C0361), the Scientific Research Foundation of Hengyang City (nos. J1, 2012KJ2, and 2012KJ3) and the construct program in USC.

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