Abstract

We present a fixed point theorem for generalized contraction in partially ordered complete metric spaces. As an application, we give an existence and uniqueness for the solution of a periodic boundary value problem.

1. Introduction

The contraction mapping theorem and the abstract monotone iterative technique are well known and are applicable to a variety of situations. Recently, there has been a trend to weaken the requirement on the contraction by considering metric spaces endowed with partial order (see [17]). It is of interest to determine if it is still possible to establish the existence of a unique fixed point assuming that the operator considered is monotone in such a setting. Such a fixed point theorem is useful, for example, in establishing the existence of a unique solution to periodic boundary value problems, besides many others.

That approach was initiated by Ran and Reurings in [8], where some applications to matrix equations were studied. This fixed point theorem was refined and extended in [7, 9] and applied to the periodic boundary value problem in the monotone case. In this paper, we consider a special case of the following periodic boundary value problem where and is continuous function.

Definition 1.1. A lower solution for (1.1) is a function such that

Very recently, Harjani and Sadarangani [4] proved the following existence theorem.

Theorem 1.2. Consider problem (1.1) with is continuous and suppose that there exists such that for with where can be written by with continuous increasing positive in , and . Then the existence of a lower solution of (1.1) provides the existence of a unique solution of (1.1).

In Section 2, we prove a new fixed point theorem in partially ordered complete metric spaces. In Section 3, existence of a unique solution for problem (1.1) is obtained under suitable conditions.

2. Fixed Point Theorem

Let denote the class of those functions which satisfies the condition We prove the main theorem of the paper as follows.

Theorem 2.1. Let be a partially order metric space that there exists a metric in such that is a complete metric space. Let be an increasing mapping such that there exists with . Suppose that there exists such that for all comparable . If or Besides, if then have a unique fixed point.

Proof. We first show that has a fixed point. Since and is increasing function, we obtain by induction that Put , . Since for each then by (2.2) And so the sequence is nonincreasing and bounded below. Thus there exists such that . Since and then there exist and such that for all . We can take such that for all with . Then since for all with we have and hence is a Cauchy sequence. Since is complete, converges to some point . To prove that is a fixed point of , if is a continuous, then hence . If case (2.4) holds then Since then we get . To prove the uniqueness of the fixed point, let be another fixed point of . From (2.5) there exists which is comparable to and . Monotonicity implies that is comparable to and for . Moreover, Consequently, the sequence is nonnegative and decreasing and so . Similarly we can show that the sequence is nonnegative and decreasing and so . Now similarly the above method we can choose in and such that for all with . Finally for all with . Therefor if in (2.14) taking yields .

3. Application to Ordinary Differential Equation

In this section we present an example where Theorem 2.1 can be applied. This example is inspired in [2, 4, 7].

Definition 3.1. Let denote the class of those functions which satisfies the following condition:(i) is increasing,(ii)for each , ,(iii).

For example, , where , , and are in .

Theorem 3.2. Consider problem (1.1) with is continuous and suppose that there exists such that for with where . Then the existence of a lower solution of (1.1) provides the existence of a unique solution of (1.1).

Proof. Problem (1.1) is equivalent to the integral equation where Define by Note that if is a fixed point of then is a solution of (1.1). Now, we check that hypotheses in Theorem 2.1 are satisfied. Indeed, is a partially ordered set if we define the following order relation in : The mapping is increasing since, by hypotheses, for which implies for , using that for , that Beside, for As the function is increasing and then we obtain Then for Finally, let be a lower solution of (1.1), and we will show that .
Indeed, Multiplying by we get and this gives us which implies that and so From this equality and (3.13) we obtain and, consequently, Hence Finally, Theorem 2.1 gives that has a unique fixed point.

Example 3.3. Let be a defined Let be continuous and suppose that there exists such that for with Then be Theorem 2.1, the existence of a lower solution for (1.1) provides the existence of a unique solution of (1.1).
The example discussed above cannot be the result of Harjani and Sadarangani noted Theorem  1.2, because is not increasing.

Acknowledgment

The third author of this work was partially supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (Grant no.: 2010-0010243).