• Research Article
• Open Access

Iterative Solutions of Singular Boundary Value Problems of Third-Order Differential Equation

Boundary Value Problems20112011:483057

https://doi.org/10.1155/2011/483057

• Accepted: 6 March 2011
• Published:

Abstract

By using the cone theory and the Banach contraction mapping principle, the existence and uniqueness results are established for singular third-order boundary value problems. The theorems obtained are very general and complement previous known results.

Keywords

• Unique Solution
• Ordinary Differential Equation
• Functional Equation
• Natural Number
• Electromagnetic Wave

1. Introduction

Third-order differential equations arise in a variety of different areas of applied mathematics and physics, such as the deflection of a curved beam having a constant or varying cross section, three-layer beam, electromagnetic waves, or gravity-driven flows [1]. Recently, third-order boundary value problems have been studied extensively in the literature (see, e.g., [213], and their references). In this paper, we consider the following third-order boundary value problem:
(1.1)

where , .

Three-point boundary value problems (BVPs for short) have been also widely studied because of both practical and theoretical aspects. There have been many papers investigating the solutions of three-point BVPs, see [25, 10, 12] and references therein. Recently, the existence of solutions of third-order three-point BVP (1.1) has been studied in [2, 3]. Guo et al. [2] show the existence of positive solutions for BVP (1.1) when and is separable by using cone expansion-compression fixed point theorem. In [3], the singular third-order three-point BVP (1.1) is considered under some conditions concerning the first eigenvalues corresponding to the relevant linear operators, where , is separable and is not necessary to be nonnegative, and the existence results of nontrivial solutions and positive solutions are given by means of the topological degree theory. Motivated by the above works, we consider the singular third-order three-point BVP (1.1). Here, we give the unique solution of BVP (1.1) under the conditions that and is mixed nonmonotone in and does not need to be separable by using the cone theory and the Banach contraction mapping principle.

2. Preliminaries

Let , . By [2, Lemma  2.1], we have that is a solution of (1.1) if and only if
(2.1)
where
(2.2)

It is shown in [2] that is the Green's function to , , and .

Let
(2.3)

It is easy to see that .

Lemma 2.1 (Guo [14, 15]).

is generating if and only if there exists a constant such that every element can be represented in the form , where and

3. Singular Third-Order Boundary Value Problem

This section discusses singular third-order boundary value problem (1.1).

Let . Obviously, is a normal solid cone of Banach space ; by [16, Lemma  2.1.2], we have that is a generating cone in .

Theorem 3.1.

Suppose that , and there exist two positive linear bounded operators and with such that for any , , , , we have
(3.1)
and there exists , such that
(3.2)
Then (1.1) has a unique solution in . And moreover, for any , the iterative sequence
(3.3)

converges to .

Remark 3.2.

Recently, in the study of BVP (1.1), almost all the papers have supposed that the Green's function is nonnegative. However, the scope of is not limited to in Theorem 3.1, so, we do not need to suppose that is nonnegative.

Remark 3.3.

The function in Theorem 3.1 is not monotone or convex; the conclusions and the proof used in this paper are different from the known papers in essence.

Proof.

It is easy to see that, for any , can be divided into finite partitioned monotone and bounded function on , and then by (3.2), we have
(3.4)
For any , let , , then , by (3.1), we have
(3.5)
Hence
(3.6)
Following the former inequality, we can easily have
(3.7)
thus
(3.8)
Similarly, by and being converged, we have that
(3.9)
Define the operator by
(3.10)
Then is the solution of BVP (1.1) if and only if . Let
(3.11)
By (3.1) and (3.10), for any , , , we have
(3.12)
(3.13)
so we can choose an , which satisfies , and so there exists a positive integer such that
(3.14)
Since is a generating cone in , from Lemma 2.1, there exists such that every element can be represented in
(3.15)
This implies
(3.16)
Let
(3.17)
By (3.16), we know that is well defined for any . It is easy to verify that is a norm in . By (3.15)–(3.17), we get
(3.18)
On the other hand, for any which satisfies , we have . Thus , where denotes the normal constant of . Since is arbitrary, we have
(3.19)

It follows from (3.18) and (3.19) that the norms and are equivalent.

Now, for any and which satisfies , let
(3.20)

then ,   ,   ,   ,   and .

It follows from (3.12) that
(3.21)
(3.22)
(3.23)
Subtracting (3.22) from (3.21) + (3.23), we obtain
(3.24)
Let ; then we have
(3.25)
As and are both positive linear bounded operators, so, is a positive linear bounded operator, and therefore . Hence, by mathematical induction, it is easy to know that for natural number in (3.14), we have
(3.26)
Since , we see that
(3.27)
which implies by virtue of the arbitrariness of that
(3.28)

By , we have . Thus the Banach contraction mapping principle implies that has a unique fixed point in , and so has a unique fixed point in ; by the definition of has a unique fixed point in , that is, is the unique solution of (1.1). And, for any , let ; we have . By the equivalence of and again, we get . This completes the proof.

Example 3.4.

In this paper, the results apply to a very wide range of functions, we are following only one example to illustrate.

Consider the following singular third-order boundary value problem:
(3.29)
where and there exists , such that for any ,   ,   , we have
(3.30)
Applying Theorem 3.1, we can find that (3.29) has a unique solution provided . And moreover, for any , the iterative sequence
(3.31)

converges to .

To see that, we put
(3.32)

Then (3.1) is satisfied for any ,   ,   ,  and .

In fact, if , then
(3.33)
If , then
(3.34)
Similarly,
(3.35)
Next, for any , by (3.30) and (3.32), we get
(3.36)
Then, from (3.32) and (3.36), we have
(3.37)
so it is easy to know by induction, for any , we get
(3.38)
thus
(3.39)
so
(3.40)
then we get
(3.41)
Let ; then
(3.42)

Thus all conditions in Theorem 3.1 are satisfied.

Declarations

Acknowledgment

The author is grateful to the referees for valuable suggestions and comments.

Authors’ Affiliations

(1)
Department of Elementary Education, Heze University, Heze, Shandong, 274000, China

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