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Abstract and Applied Analysis
Volume 2013 (2013), Article ID 704160, 7 pages
http://dx.doi.org/10.1155/2013/704160
Research Article

Existence of Solutions for a Periodic Boundary Value Problem via Generalized Weakly Contractions

1Department of Mathematics, Faculty of Science, Arak University, Arak 38156-8-8349, Iran
2Department of Mathematics, Atilim University, İncek, 06836 Ankara, Turkey
3Université de Sousse, Institut Supérieur d'Informatique et des Technologies de Communication de Hammam Sousse, Route GP1-4011 H., Sousse, Tunisia
4Department of Mathematics, Jubail College of Education, Dammam University 31961, Saudi Arabia

Received 21 December 2012; Accepted 19 February 2013

Academic Editor: Abdul Latif

Copyright © 2013 Sirous Moradi et al. 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 discuss the existence of solutions for a periodic boundary value problem and for some polynomials. For this purpose, we present some fixed point theorems for weakly and generalized weakly contractive mappings in the setting of partially ordered complete metric spaces.

1. Introduction

Existence of solutions for a periodic boundary value problem by using upper and lower solution methods has attracted the attention of many authors (see, e.g., [15]).

We consider a special case of the following boundary value problem: where , is a continuous map and is constant.

Obviously, if , then the problem (1) becomes the following periodic boundary value problem:

Definition 1. A lower solution for (1) is a function such that

Let stand for the class of functions , which satisfy the following conditions: (i)is nondecreasing, (ii), for each , (iii).

Very recently, Amini-Harandi and Emami [1] proved the following existence theorem, which extended the main theorem of Harjani and Sadarangani [2].

Theorem 2. Consider problem (2), with being continuous. Suppose that there exists such that for with , where . Then, the existence of a lower solution for (2) provides the existence of a unique solution of (2).

In this paper, we solve (2) by extending a fixed point theorem in the context of partially ordered metric space. Our results improve/extend/generalize some results in the literature, in particular, the results of Amini-Harandi and Emami [1] and Harjani and Sadarangani [2]. Finally, in the last section, we prove the existence of a solution for some polynomials, as applications.

2. Preliminaries

In this section, we state a necessary background on the topic of fixed point theory, one of the core subjects of nonlinear analysis, for the sake of completeness of the paper. Fixed point theory has a wide potential application not only in the branches of mathematics, but also in several disciplines such as economics, computer science, and biology (see, e.g., [6, 7]). The most beautiful and elementary result in this direction is the Banach contraction mapping principle [8]. After this substantial result of Banach, several authors have extended this principle in many different ways (see, e.g., [17, 931]). In particular, the authors have introduced new type of contractions and researched the existence and uniqueness of the fixed point in various spaces. One of the important contraction types, a -contraction, was introduced by Boyd and Wong [14]. In 1997, Alber and Duerre-Delabriere [10] defined the concept of a weak--contraction which is a generalization of the -contraction. A self-mapping on a metric space is said to be weak--contractive if there exists a map with and for all such that for all .

Later, Zhang and Song [31] introduced the notion of a generalized weak--contraction which is a natural extension of the weak--contraction. A self-mapping on a metric space is said to be generalized weak--contractive if there exists a map with and for all such that for all , where For more details on weak -contractions, we refer to, for example, [20, 21, 28].

On the other hand, the existence and uniqueness of a fixed point in the context of partially ordered metric spaces were first investigated in 1986 by Turinici [30]. After this pivotal paper, a number of results were reported in this direction with applications to matrix equations, ordinary differential equations, and integral equations (see, e.g., [1, 2, 4, 5, 7, 9, 1113, 1519, 22, 2527]).

Recently, the main theorem of Geraghty [16, Theorem 2.1] is reproved by Amini-Harandi and Emami [1] in the context of partially ordered metric space. On the other hand, the main theorem of Amini-Harandi and Emami [1, Theorem 2.1] extends the theorem of Harjani and Sadarangani [2]. The authors in [1, 2] also proved the existence and uniqueness of a solution for a periodic boundary value problem.

Before stating the main theorem in [1], we recall the following class of functions introduced by Geraghty [16]. Let denote the set of all functions such that

Theorem 3. Let be a partially ordered set and suppose that there exists a metric in such that is a complete metric space. Let be a nondecreasing mapping such that there exists an element with . Suppose that there exists such that Assume that either (a) is continuous or (b)for every nondecreasing sequence if , then for all .
Moreover, if for each there exists which is comparable to and , then has a unique fixed point.

Let denote the set of fixed points of .

We give the following classes of functions. Let denote the set of all mappings verifying that It is clear that if , we have that

3. Some Auxiliary Fixed Point Theorems

In the following theorem, we prove the existence and uniqueness of a fixed point for generalized weak--contractive mappings in partially ordered complete metric spaces.

Theorem 4. Let be a partially ordered set and suppose that there exists a metric in such that is a complete metric space. Let be a nondecreasing mapping such that there exists an element with . Suppose that there exists such that for each with (i.e., a generalized weak--contraction).
Suppose also that either (a) is continuous or (b)for every nondecreasing sequence if , then for all .
Then has a fixed point. Moreover, if every is comparable, then the fixed point of is unique.

Proof. First, we prove the existence of a fixed point of . Since the self-mapping is nondecreasing and , we get that
Define , . Then, expression (13) is equivalent to Assume that for each . Otherwise, the proof is completed. From (12), we derive that where If for some , then from (15) and (16), we have This is a contradiction. Hence, for all . So by (15) and (16), we have for all , Thus, we conclude that the nonnegative sequence is decreasing. Therefore, there exists such that . By using (18), we find that Taking in (19), we get . Since , we obtain that ; that is, .
We prove that the iterative sequence is Cauchy. Take , then . From (12), we obtain that and thus, where Hence, by (21), This shows that ; that is, is Cauchy. Since is a complete metric space, then there exists such that . Now, we prove that is a fixed point of .
If (a) holds, that is, if is continuous, then Suppose that (b) holds. By using (12), we derive that where So . Taking in (25), we get . Since , we conclude that . So and hence .
Now, we show that this fixed point of the self-mapping is unique. If for each , and are comparable, then the fixed point is unique. Let be two fixed points of . Then and from (12), we conclude that . Thus, and hence, . This completes the proof.

The following consequence of Theorem 4 plays a crucial role in the proof of our main result, Theorem 9.

Theorem 5. Let be a partially ordered set and suppose that there exists a metric in such that is a complete metric space. Let be a nondecreasing mapping such that there exists an element with . Suppose that there exists such that for each with (i.e., weak--contraction). Suppose also that either (a) is continuous or (b)for every nondecreasing sequence if , then for all .
Then has a fixed point. Moreover, if for each there exists which is comparable to and , then the fixed point of is unique.

Remark 6. In Theorem 4, if the condition “every is comparable” is replaced by the condition “for each there exists which is comparable to and ,” then we cannot conclude that the fixed point is unique. The following example illustrates our claim.

Example 7. Let be endowed with the relation given as follows: and for each . Obviously, is a partially ordered set. Also, we may endow with the following metric: and for each . Define by , , , and . Obviously, the mapping is nondecreasing and for all with , where . Also , but and .

Remark 8. If , then . But if , then we can not conclude that the function belongs to . Consider, for example, which illustrates our claim. As a result, Theorem 5 is a proper extension of Theorem 3.

4. Applications

4.1. Solving a Boundary Value Problem

In this paragraph, we prove the existence of a solution of the problem (1).

Theorem 9. Consider problem (1) with being continuous. Suppose that there exists such that for with where and is nondecreasing. Then the existence of a lower solution for (1) provides the existence of a unique solution for (1).

Proof. Define . Then, problem (1) becomes as follows Suppose . So and hence problem (34) can be rewritten as where is defined by and . Obviously, is continuous. Also the lower solution of (34) is replaced by the lower solution of (35). Now we prove that the problem (35) has a unique solution. Obviously, if and , then for every , and hence from (33), Inequality (36) implies that if , Problem (35) is equivalent to the following integral equation: where Let be the set of continuous functions defined on . Consider given by Note that if is a fixed point of , then is a solution of (35). Now, we check that hypotheses of Theorem 5 are satisfied.
Take . The space can be equipped with a partial order given by Also, can be equipped with the following metric: We have that is complete. For every and for every , we have and by hypothesis, Therefore, and since for , hence for all with .
Also, for all with , we find (using the fact that is nondecreasing) Finally, let be a lower solution for (35). We can show that by a method similar to that in [1, 2]. Also, is totally ordered. Hence, due to Theorem 5, has a unique fixed point. Therefore, problem (35) has a unique solution . Thus, is the unique solution of (34) and this completes the proof.

Remark 10. If the mapping satisfies the condition (33), then for with and for , Hence, for all and all , we have Therefore, by using Banach contraction principle, for every , the problem has a unique solution . So there exists a unique such that is a solution of (1).
Now let be a mapping such that for all and all , for some . We know that for every , problem (49) has a unique solution .

Question 1. It is natural to ask whether there is an where is a solution of problem ((2), i.e., ?

The following example shows that the above question is not true.

Example 11. Let be defined by . Obviously, (50) holds for . Let be a solution for problem (2). From , we conclude that for all . Hence, is monotone nondecreasing. Using , we conclude that . Since and , then for all and this is a contradiction. So, problem (2) has no solution.

Example 12. Let be defined by and let be defined by . Take . One can show that inequality (33) holds. Suppose that is defined by . Obviously, is a lower solution of problem (2). Hence, problem (2) has a unique solution, which is where .

4.2. Solving Some Polynomials

In this paragraph, we prove the existence and uniqueness of a solution of some polynomials.

Theorem 13. Let be such that and . Then, has a unique solution on .

Proof. Suppose that is defined by If , then . So is nondecreasing. Also for with , we derive that Suppose that is defined by , for . Since , then is monotone nondecreasing. Hence, if , then . So, . Therefore, from (54), we get where . Also . Thus, using Theorem 5, the mapping has a unique fixed point . Moreover, the sequence converges to this fixed point. Note that here the space is taken to be , which is equipped with the usual Euclidian metric and the usual partial order.
On the other hand, there exists a unique such that . So, from , we have and therefore we find Also the sequence converges to and this completes the proof.

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