Advances in Pure Mathematics
Vol.4 No.8(2014), Article ID:49141,7 pages DOI:10.4236/apm.2014.48053

Existence Theory for Single Positive Solution to Fourth-Order Boundary Value Problems

Ying He

School of Mathematics and Statistics, Northeast Petroleum University, Daqing, China


Copyright © 2014 by author and Scientific Research Publishing Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY).

Received 25 June 2014; revised 25 July 2014; accepted 5 August 2014


By fixed point theorem of a mixed monotone operator, we study boundary value problems to nonlinear singular fourth-order differential equations, and provide sufficient conditions for the existence and uniqueness of positive solution. The nonlinear term in the differential equation may be singular.

Keywords:Mixed Monotone Operator, Singular, Existence, Uniqueness

1. Introduction

Fourth-order differential equations play an important role in various fields of science and engineering. With the help of boundary value conditions, we can describe the natural phenomena and mathematical model more accurately. Therefore, the fourth-order differential equations have received much attention and the theory and application have been greatly developed (see [1] -[4] and their references). Most of the results told us that the equations had at least single and multiple positive solutions. In papers [1] -[3] , the authors obtained some newest results for the singular fourth-order boundary value problems. But there is no result on the uniqueness of solution in them.

In this paper, we consider the following singular fourth-order boundary value problem:


Throughout this paper, we always suppose that

Moreover, may be singular at, , or.

Equation (1.1) is often referred to as the deformation for an elastic beam under a variety of boundary conditions. A brief discussion of the physical interpretation under some boundary conditions associated with the linear beam equation can be found in Zill and Cullen [5] . In this article, we consider the existence and uniqueness of positive solutions for fourth-order singular boundary value problems by using mixed monotone method.

2. Preliminary

Let P be a normal cone of a Banach space E, and with,. Define

Now we give a definition(see [7] ).

Definition 2.1. Assume. A is said to be mixed monotone if is nondecreasing in x and nonincreasing in y, i.e. if implies for any, and implies for any. is said to be a fixed point of A if.

Theorem 2.1. Suppose that is a mixed monotone operator and a constant, , such that


Then A has a unique fixed point. Moreover, for any



, r is a constant from.

Theorem 2.2. (See [7] ): Suppose that is a mixed monotone operator and a constant such that (2.1) holds. If is a unique solution of equation

in, then,. If, then implies, , and

3. Uniqueness Positive Solution of Problem (1.1)

This section discusses the problem

Throughout this section, we assume that




Let and,. We denote the Green’s functions for the following boundary value problems


by and, respectively. It is well known that and can be written by

where and

Lemma 3.1. Suppose that holds, then the Green’s function, possesses the following properties:

1) is increasing and,.

2) is decreasing and,.



5) is a positive constant. Moreover,.

6) is continuous and symmetrical over Q.

7) has continuously partial derivative over,.

8) For each fixed, satisfies for,. Moreover, for.

9) has discontinuous point of the first kind at and

Following from Lemma 3.1, it is easy to see that



Suppose that x is a positive solution of (1.1). Then


By using (3.3) and (a), we see that for every positive solution x one has

where. Let

Thus by (3.3) one has

by (a) one has


Let. Obviously, P is a normal cone of Banach space.

Theorem 3.1. Suppose that there exists such that


for any and, and satisfies


Then (1.1) has a unique positive solution. And moreover, implies,. If, then

Proof. Since (3.5) holds, let, one has



Let. The above inequality is


From (3.5), (3.7) and (3.8), one has


Similarly, from (3.5), one has


Let, , one has


Let. It is clear that, and now let


where is chosen such that

For any, we define


First we show that. Let, from (3.10) and (3.11) we have

and from (3.9) we have


Then from (3.4) and (3.13) we have

On the other hand, for any, from (3.9) and (3.10), we have


Thus, from (3.15), we have

So, is well defined and

Next, for any, one has

So the conditions of Theorems 2.1 and 2.2 hold. Therefore there exists a unique such that

. It is easy to check that is a unique positive solution of (1.1) for given. MoreoverTheorem 2.2 means that if then, , and if, then

This completes the proof.

Example. Consider the following singular fourth-order boundary value problem:

where, satisfies.


Thus and for any , ,

Now Theorem 3.1 guarantees that the above equation has a positive solution.


Project was supported by Heilongjiang Province Education Department Natural Science Research Item, China (12541076).


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