Abstract and Applied Analysis

Volume 2013 (2013), Article ID 324848, 9 pages

http://dx.doi.org/10.1155/2013/324848

Research Article

Existence Results for a Coupled System of Nonlinear Fourth-Order Differential Equations

Bessem Samet

Department of Mathematics, College of Science, King Saud University, P.O. Box 2455, Riyadh 11451, Saudi Arabia

Received 10 November 2013; Accepted 3 December 2013

Academic Editor: Mohamed Boussairi Jleli

Copyright © 2013 Bessem Samet. 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.

Sufficient conditions are obtained for the existence of solutions to a coupled system of nonlinear fourth-order differential equations.

1. Introduction

The purpose of this paper is to study the existence of solutions to the following coupled system of nonlinear fourth-order differential equations:

where and are given continuous functions.

In recent years, boundary value problems for fourth-order ordinary differential equations have been extensively studied. Gupta [1] considered the boundary value problem

which describes the unstable static equilibrium of an elastic beam which is supported by sliding clamps at both ends, and . Many authors (see [2–8] and references therein) studied the existence of solutions to the problem

which describes the deformation of the equilibrium state of an elastic beam with its two ends being simply supported. For other results on the study of fourth-order ordinary differential equations, we refer to [9–13] and references therein. On the other hand, to our knowledge, there are only few papers concerning existence results for systems of fourth-order ordinary differential equations. For example, in [14], via the fixed point index theory, the authors studied the existence of solutions to the system of bending elastic beam equations

In this work, we provide four sufficient conditions for the existence of solutions to the problem (1). Some illustrative examples are presented.

The paper is organized as follows. In Section 2, we present the main results. In Section 3, we give some illustrative examples. In Section 4, we present some basic lemmas that will be used for the proofs. Finally, in Section 5, we prove our results.

2. Main Results

We obtained the following existence results.

Theorem 1. Suppose the following conditions hold:

(i)there exist such that, for all and , one has

(ii)there exist such that, for all , one has

(iii), where is the matrix given by

and denotes the spectral radius of . Then

(I)equation (1) has a unique solution ,(II)for any , the sequences defined by

converge, respectively, to and ; that is,

Theorem 2. Suppose the following conditions hold:

(i)there exists a nonnegative function such that, for all and , one has

(ii)there exists a nonnegative function such that, for all and , one has

(iii)for all , one has

Then (1) has at least one solution .

Theorem 3. Suppose the following conditions hold:

(i)for all and , one has

(ii)for all and , one has

(iii)for all , one has

Then (1) has at least one solution .

Theorem 4. Suppose the following conditions hold:

(i)for all and , one has

(ii)for all and , one has

(iii)for all , one has

(iv).

Then (1) has at least one solution .

3. Illustrative Examples

Example 1. Consider the initial value problem

In this case, for all and , we have

Conditions (i) and (ii) of Theorem 1 are satisfied with

Moreover, the matrix

has the eigenvalues and . From Theorem 1, the considered problem has a unique solution .

Example 2. Consider the initial value problem

In this case, for all and , we have

Conditions (i) and (ii) of Theorem 2 are satisfied with

From Theorem 2, the considered problem has at least one solution .

Example 3. Consider the initial value problem

In this case, for all and , we have

Conditions (i) and (ii) of Theorem 3 are satisfied with

From Theorem 3, the considered problem has at least one solution .

Example 4. Consider the initial value problem

In this case, for all and , we have

Conditions (i), (ii), and (iii) of Theorem 4 are satisfied with

From Theorem 4, the considered problem has at least one solution .

4. Preliminaries

In order to obtain an integral formulation of the problem (1), we need the following lemma. At first, we denote by the set of continuous functions from into .

Lemma 5. Let and . The problem

has a unique solution

where

Proof. We omit the proof as it employs standard arguments.

From Lemma 5, problem (1) is equivalent to the following system of integral equations:

Let . Define the mappings by

for every . Let be the mapping defined by

Observe that is a solution to (35) if and only if is a fixed point of .

Lemma 6. One has

Proof. For , we have

Similarly, for , we have

Then, for all , we have

Now, for , we have

Similarly, for , we have

We endow the product set with the norm

where

Lemma 7. The mapping is continuous with respect to the norm .

Proof. Let be an arbitrary point. Let be a nonnegative real number such that . To show the continuity of at the point , let us consider a sequence converging to ; that is,

Without restriction to the generality, we can suppose that , for every ( is the set of natural numbers). Since is continuous, we can define

For all , we have

Using Lemma 6, for all , we have

which implies that

From the continuity of , we have

The continuity of implies that, for all ,

On the other hand, for all , we have

Then, by the dominated convergence theorem, we have

Consequently, we have

Similarly, we can show that

We deduce that

which yields the desired result.

Lemma 8. Let be the subset of defined by

for some . Under condition (iii) of Theorem 2 (or conditions (iii)-(iv) of Theorem 4), is an equicontinuous set.

Proof. Suppose that (iii) of Theorem 2 holds. Since and are continuous, we can define

Let . Since is continuous in , there exists such that

whenever . Let . For , for , we have

Then, the equicontinuity of is proved. If conditions (iii)-(iv) of Theorem 4 hold, using the same technique with , we get the desired result.

Now, we are ready to prove our main results.

5. The Proofs

5.1. Proof of Theorem 1

The proof is based on Perov fixed point theorem [15]. At first, let us recall some concepts.

Let be a positive integer. We endow the set with the partial order defined by

We denote by the zero vector of .

Let be a nonempty set. A mapping is called a vector valued metric on if the following properties are satisfied:

(d1), for all ;(d2), for all ;(d3), for all ;(d4), for all .

A set equipped with a vector valued metric is called a generalized metric space. We will denote such a space by . For the generalized metric spaces, the notions of convergent sequence, Cauchy sequence, completeness, open subset, and closed subset are similar to those for usual metric spaces.

We denote by the set of all matrices with positive elements. For , we denote by the spectral radius of , that is, the supremum among the absolute values of the elements in its spectrum.

Lemma 9. Let be a complete generalized metric space and the mapping with the property that there exists a matrix such that

If , then has a unique fixed point . Moreover, for any , the iterative sequence defined by converges to .

Proof. We endow with the vector valued metric defined by

It is easy to show that is a complete generalized metric space. Let . For any , we have

Using Lemma 6, for all , we have

Using conditions (i) and (ii), we get that

for all . This implies that

Using similar calculations, we obtain that

It follows from (68) and (69) that

where

Since , the desired result follows from Lemma 9.

5.2. Proof of Theorem 2

The proof is based on Schauder fixed point theorem [16].

Lemma 10. Let be a closed bounded convex subset of a normed space . If is a continuous compact map such that , then has a fixed point.

Proof. Let be the set defined by

where

We will prove that

Let ; that is, and . For all , we have

Using Lemma 6 and conditions (i) and (iii), we obtain that

Thus, we have

Similarly, for all , we have

Using Lemma 6 and conditions (ii) and (iii), we obtain that

Thus, we have

Then, (74) follows from (77) and (80).

Now, using Lemmas 7 and 8, Arzelá-Ascoli theorem, and Lemma 10, we obtain the desired result.

5.3. Proof of Theorem 3

The proof is based on Schaefer fixed point theorem [17].

Lemma 11. Let be a Banach space. Assume that is a continuous compact operator and the set

is bounded. Then has a fixed point in .

Proof. The continuity and the compactness of the mapping follow from Lemmas 7 and 8. We have just to show that the set

is bounded with respect to the norm . Then, the result will be obtained by applying Lemma 11.

Let such that and for some . For all , using (i), (iii), and Lemma 6, we obtain that

which implies that

Similarly, using (ii), (iii), and Lemma 6, for all , we have

which implies that

Consequently, is a bounded set.

5.4. Proof of Theorem 4

The following fixed point theorem due to Agarwal et al. [18] will be used in the proof.

Lemma 12. Let be a Banach space, let be a bounded closed convex subset of , and let such that for every . If is a contraction and is continuous and compact, then the equation has a solution .

Proof. We define the operators as follows: for all ,

where, for every ,

Clearly, for every , we have

Let . We define the set as follows:

Let . For every , we have

Similarly, for every , we have

Thus, we proved that, for every , we have

Now, we will prove that is a contraction. Let . For every , we have

Similarly, for every , we have

Since , then is a contraction (with respect to the norm ). Finally, the continuity and the compactness of the operator follow by using the same arguments of the proof of Theorem 2. Applying Lemma 12, we obtain the existence of such that is a fixed point of .

Acknowledgment

The author would like to extend his sincere appreciation to the Deanship of Scientific Research at King Saud University for the funding of this research through the Research Group Project no. RGP-VPP-237.

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