ABSTRACT

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:

(1.1)

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

(2.1)

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

satisfy

where

, 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

(3.1)

where

(3.2)

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

and

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,.

3).

4).

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

(a)

(b)

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

(3.3)

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

(3.4)

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

Theorem 3.1. Suppose that there exists such that

(3.5)

for any and, and satisfies

(3.6)

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

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

then

(3.7)

Let. The above inequality is

(3.8)

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

(3.9)

Similarly, from (3.5), one has

(3.10)

Let, , one has

(3.11)

Let. It is clear that, and now let

(3.12)

where is chosen such that

For any, we define

(3.13)

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

and from (3.9) we have

(3.14)

Then from (3.4) and (3.13) we have

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

(3.15)

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.

Let

Thus and for any , ,

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

Funding

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

References