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Mathematical Problems in Engineering
Volume 2013, Article ID 157849, 9 pages
http://dx.doi.org/10.1155/2013/157849
Research Article

Existence of Positive Solution for a Third-Order BVP with Advanced Arguments and Stieltjes Integral Boundary Conditions

Department of Applied Mathematics, Lanzhou University of Technology, Lanzhou, Gansu 730050, China

Received 20 June 2013; Revised 1 August 2013; Accepted 1 August 2013

Academic Editor: Fazal M. Mahomed

Copyright © 2013 Jian-Ping Sun 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

A class of third-order boundary value problems with advanced arguments and Stieltjes integral boundary conditions is discussed. Some existence criteria of at least one positive solution are established. The main tool used is the Guo-Krasnoselskii fixed point theorem.

1. Introduction

Third-order differential equations arise in a variety of different areas of applied mathematics and physics, for example, in the deflection of a curved beam having a constant or varying cross-section, a three-layer beam, electromagnetic waves or gravity-driven flows, and so on [1].

Recently, third-order boundary value problems (BVPs for short) with integral boundary conditions, which cover third-order multipoint BVPs as special cases, have attracted much attention from many authors; see [26] and the references therein. In particular, in 2012, by using a fixed point theorem due to Avery and Peterson [7], Jankowski [4] established the existence of at least three nonnegative solutions to the following BVP: where denoted a linear functional on given by involving a Stieltjes integral with a suitable function of bounded variation. The measure could be a signed one. The situation with a signed measure was first discussed in [8, 9] for second-order differential equations; it was also discussed in [10, 11] for second-order impulsive differential equations. For some other related results, one can refer to [1214].

Among the boundary conditions in (1), only is related to a Stieltjes integral. In this paper, we are concerned with the following third-order BVP with advanced arguments and Stieltjes integral boundary conditions: Throughout this paper, we always assume that is continuous and for , , , is a suitable function of bounded variation, and is defined as in (2). It is important to indicate that it is not assumed that is positive to all positive . Some existence criteria of at least one positive solution to the BVP (3) are obtained by using the following well-known Guo-Krasnoselskii fixed point theorem [15, 16].

Theorem 1. Let be a Banach space, and let be a cone in . Assume that and are bounded open subsets of such that , , and let be a completely continuous operator such that either for and for or for and for . Then, has a fixed point in .

2. Preliminaries

Let . Then, .

Lemma 2. For any , the BVP has the unique solution where

Proof. By integrating the differential equation in (4) three times from to and using the boundary condition , we get So, In view of (7), (8), and the boundary conditions and , we have So, Substituting (10) into (9), we get

Lemma 3 (see [4]). Consider that , .

Throughout, we assume that the following conditions are fulfilled:(C1) , (C2)

For convenience, we denote Obviously, . In the remainder of this paper, we always assume that .

Let be equipped with the maximum norm. Then, is a Banach space. Define where Then, is a cone in .

Now, we define operators and on by where

Lemma 4. Consider that .

Proof. Let . Then, it is easy to verify that which shows that is concave down on . In view of we have So, , .
Now, we prove that . To do it, we consider two cases.
Case  1. Let . Then , and there exists such that . Moreover, So, which together with implies that that is,
Case  2. Let and . Note that in this case .
If , then So, which together with implies that that is,
If , then So, which together with (23) and (28) implies that that is,
It follows from (25), (30), and (34) that
Finally, we need to show that . In view of we have
This shows that . Similarly, we can prove that .

Lemma 5. The operators and have the same fixed points in .

Proof. Suppose that is a fixed point of . Then, which shows that So, which indicates that is a fixed point of .
Suppose that is a fixed point of . Then, which shows that So, which indicates that is a fixed point of .

Lemma 6. is completely continuous.

Proof. First, by Lemma 4, we know that .
Next, we show that is compact. Let be a bounded set. Then, there exists such that for any . Since is a function of bounded variation, there exists such that for any partition . Let Then, for any , which shows that is uniformly bounded.
On the other hand, for any , since is uniformly continuous on , there exists such that for any with , Let . Then, for any , with , we have which shows that is equicontinuous. It follows from the Arzela-Ascoli theorem that is relatively compact. Thus, we have shown that is a compact operator.
Finally, we prove that is continuous. Suppose that , and . Then, there exists such that and . For any , since is uniformly continuous on , there exists such that for any with , At the same time, since , there exists positive integer such that for any , It follows from (48) and (49) that for any , which indicates that is continuous. Therefore, is completely continuous. Similarly, we can prove that is also completely continuous.

3. Main Results

For convenience, we define

Theorem 7. If , then the BVP (3) has at least one positive solution.

Proof. Since , there exists such that By the definition of , we may choose so that Let . Then, for any , which together with (52) and (53) implies that This shows that
On the other hand, since , there exists such that By the definition of , we may choose so that Let and . Then, for any , which together with (57) and (58) implies that This indicates that
Therefore, it follows from (56), (61), and Theorem 1 that the operator has one fixed point , which is a positive solution of the BVP (3).

Theorem 8. If , then the BVP (3) has at least one positive solution.

Proof. Since , there exists such that By the definition of , we may choose so that Let . Then, for any , which together with (62) and (63) implies that This shows that
On the other hand, since , there exists so that By the definition of , we may choose such that which implies that where Now, we choose , and let . Then, for any , which together with (69) implies that This indicates that
Therefore, it follows from (66), (73), and Theorem 1 that the operator has one fixed point , which is a positive solution of the BVP (3).

Example 9. Consider the following BVP:
In view of , we have At the same time, since and , a simple calculation shows that So, If we let , , then it is easy to compute that which together with (77) implies that Therefore, it follows from Theorem 7 that the BVP (74) has at least one positive solution.

Acknowledgment

This paper is supported by the Natural Science Foundation of Gansu Province of China (1208RJZA240).

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