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`Journal of Function SpacesVolume 2016, Article ID 8567679, 5 pageshttp://dx.doi.org/10.1155/2016/8567679`
Research Article

## On Positive Periodic Solutions to Nonlinear Fifth-Order Differential Equations with Six Parameters

1College of Mechanical Engineering, Guizhou Institute of Technology, Guiyang 550003, China
2College of Science, Hohai University, Nanjing 210098, China

Received 6 November 2015; Revised 4 February 2016; Accepted 15 February 2016

Copyright © 2016 Yunhai Wang and Fanglei Wang. 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 study the existence and multiplicity of positive periodic solutions to the nonlinear differential equation: ,  ,  ,  , where ,  , is a 1-periodic function. The proof is based on the Krasnoselskii fixed point theorem.

#### 1. Introduction

Fifth-order boundary value problems (BVPs) are known to arise in the mathematical modeling of viscoelastic flow and other branches of mathematical, physical, and engineering sciences; see [14]. So the existence, multiplicity, and nonexistence of solutions to the fifth-order BVPs are noticed by many authors. For example, Agarwal and Odda [5, 6] gave some theorems which list conditions for the existence and uniqueness of solutions to the fifth-order BVPs by the topological methods; Gamel and Lv [7, 8] focused their attention on the study of numerical solution to the fifth-order BVPs.

On the other hand, all kinds of topological methods, such as the method of upper and lower solutions, degree theory, and some fixed point theorems in cones, have been widely applied to study the singular and regular periodic boundary value problems; see [919].

Inspired by the above references, we establish the existence and multiplicity results of  the nonlinear differential equation in this paper:where ,  , and is a 1-periodic function.

Both of regular and singular cases are considered. The proof is based on the Krasnoselskii fixed point theorem in a cone.

Lemma 1 (see [20]). Let be a Banach space, and let be a cone in . Assume and   are open subsets of with ,  , and let be a completely continuous operator such that either(i),  ,  and  ,   or(ii),  ,  and  ,  .Then, has a fixed point in .

This paper is organized as follows: in Section 2, some preliminaries are given; in Section 3, we give the main results; in Section 4, we give an example to illustrate our main results.

#### 2. Preliminaries

If , , , , satisfy the assumption(H1),  ,  ,  ,  , ,then (1) can be rewritten as Furthermore, we have

From Lemma 3 in [12], the fourth-order linear problem has a unique continuous solution and ,  . Let For given constants and , the minimum and maximum can be computed explicitly. Let denote the Banach space with the maximum norm .

Lemma 2 (see [12]). Assume that (H1) holds. For , the linear boundary value problem has a unique solution , which is given by expression where

Lemma 3 (see [13]). For any and , the linear boundary value problem has a unique solution , which can be expressed as where satisfying the estimates

From Lemmas 2 and 3, it follows that is the solution of (1) if and only if satisfies We define an operator on by

Let be a cone in defined as where . For , let . Note that .

Lemma 4. Assume that (H1) holds. In addition, we suppose the following conditions hold:(H2) and .(H3) is continuous.(H4),   for .Then, and is completely continuous.

Proof. For any , on one hand, we have On the other hand, we have Thus, Therefore, . Using the Ascoli-Arzela theorem (Chapter 2 [21]), it is easy to show that is compact and continuous.

From Lemma 4 and [18], it follows that

Lemma 5. Assume that (H1) and (H2) hold. In addition, we suppose the following conditions hold:(H5).(H6) is continuous.Then, and is completely continuous.

#### 3. Main Results

For convenience, we give the notations

Theorem 6 (regular case). Assume that (H1)–(H4) hold. (a)If and , then, for all , (1) has a positive solution.(b)If and , then, for all , (1) has a positive solution.(c)If or , then there exists a such that, for all , (1) has a positive solution.(d)If or , then there exists a such that, for all , (1) has a positive solution.(e)If , then there exists a such that, for all , (1) has two positive solutions.(f)If , then there exists a such that, for all , (1) has two positive solutions.

Proof. (a) Since , there exists a such that , for any , where satisfies . Then, for any , we have Since , there exists a such that , for any , where satisfies . Take . Then, for any , we have and By the above inequalities and Lemma 1, has a fixed point in .
(b) Since , there exists a such that , for any , where satisfies . Then, for any , we have Since , there exists a such that , for any , where satisfies .
Take . Then, for any , we have and By the above inequalities and Lemma 1, has a fixed point in .
(c) For any , let . Then, there exists a such that for any ,  .
If , from the proof of (a), it follows that there exists a such that If , from the proof of (b), it follows that there exists a such that By Lemma 1, has a fixed point in or .
(d) For any , let Then, there exists a such that for any ,  .
From the part proof of (a) and (b), the results of (d) follow.(e)From the proof of (c), it is easy to obtain the result.(f)From the proof of (d), it is easy to obtain the result.

Theorem 7 (singular case). Assume that (H1), (H2), (H5), and (H6) hold. (A)There exists a such that (1) has a positive solution for .(B)If , then, for any , (1) has a positive solution.(C)If , then (1) has two positive solutions for all sufficiently small .

Proof. (A) On one hand, from the proof of (d) in Theorem 6, it is clear to see that there exists a such that . On the other hand, since , there exists a with such that , for , where satisfies . For any , we have . Therefore, by Lemma 1, has a fixed point in .
Finally, from (A) and (b), result (B) follows. From (A) and (d), result (C) follows.

#### 4. Examples

Example 1. Consider the fourth-order nonlinear problems whereFirst, via some simple computations, it is easy to see that (H1) and (H2) hold. Second, if , then (H3) and (H4) obviously hold. Now we illustrate the cases of Theorem 6 via the discussions of and  . (a)If and  , then, for all , (28) has a positive solution.(b)If and  , then, for all , (28) has a positive solution.(c)If or , then there exists a such that, for all , (28) has a positive solution.(d)If or , then there exists a such that, for all , (28) has a positive solution.(e)If and  , then there exists a such that, for all , (28) has two positive solutions.(f)If and , then there exists a such that, for all , (28) has two positive solutions.Finally, if , , then (H5) and (H6) hold. Now we illustrate the cases of Theorem 7 via the discussions of and  . (A)If , then there exists a such that (28) has a solution for .(B)If , , then, for any , (28) has a positive solution.(C)If , , then (28) has two positive solutions for all sufficient small .

#### Disclosure

The first author is supported by Doctoral Scientific Research Foundation of Guizhou Province, The Joint Science and Technology FundQKH. The second author is supported by by NSF of China (no. 11501165), NSF of Jiangsu Province (no. BK20130825), and the Fundamental Research Funds for the Central Universities.

#### Conflict of Interests

The authors declare that they have no competing interests.

#### Authors’ Contribution

Both authors contributed to each part of this work equally and read and approved the final paper.

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