1. Introduction

Existence of fixed point in partially ordered sets has been considered recently in [1–17]. Tarski's theorem is used in [9] to show the existence of solutions for fuzzy equations and in [11] to prove existence theorems for fuzzy differential equations. In [2, 6, 7, 10, 13] some applications to ordinary differential equations and to matrix equations are presented. In [3–5, 17] some fixed point theorems are proved for a mixed monotone mapping in a metric space endowed with partial order and the authors apply their results to problems of existence and uniqueness of solutions for some boundary value problems.

In the context of ordered metric spaces, the usual contraction is weakened but at the expense that the operator is monotone. The main tool in the proof of the results in this context combines the ideas in the contraction principle with those in the monotone iterative technique [18].

Let denote the class of the class of the functions which satisfies the condition

(1.1)

In [19] the following generalization of Banach's contraction principle appears.

Theorem 1.1.

Let be a complete metric space and let be a mapping satisfying

(1.2)

where . Then has a unique fixed point and converges to for each .

Recently, in [2] the authors prove a version of Theorem 1.1 in the context of ordered complete metric spaces. More precisely, they prove the following result.

Theorem 1.2.

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

(1.3)

where . Assume that either is continuous or satisfies the following condition:

(1.4)

Besides, suppose that for each there exists which is comparable to and . If there exists with , then has a unique fixed point.

The purpose of this paper is to generalize Theorem 1.2 with the help of the altering functions.

We recall the definition of such functions.

Definition 1.3.

An altering function is a function which satisfies the following.

(a) is continuous and nondecreasing.

(b) if and only if .

Altering functions have been used in metric fixed point theory in recent papers [20–22].

In [7] the authors use these functions and they prove some fixed point theorems in ordered metric spaces.

2. Fixed Point Theorems

Definition 2.1.

If is a partially ordered set and , we say that is monotone nondecreasing if for ,

(2.1)

This definition coincides with the notion of a nondecreasing function in the case and represents the usual total order in .

In the sequel, we prove the main result of the paper.

Theorem 2.2.

Let be a partially ordered set and suppose that there exists a metric in such that is a complete metric space. Let be a continuous and nondecreasing mapping such that

(2.2)

where is an altering function and .

If there exist with , then has a fixed point.

Proof.

If , then the proof is finished. Suppose that . Since and is a nondecreasing mapping, we obtain by induction that

(2.3)

Put . Taking into account that and since for each then, by (2.2), we get

(2.4)

Using the fact that is nondecreasing, we have

(2.5)

If there exists such that , then and is a fixed point and the proof is finished. In another case, suppose that for all . Then, taking into account (2.5), the sequence is decreasing and bounded below, so

(2.6)

Assume that .

Then, from (2.4), we have

(2.7)

Letting in the last inequality and by the fact that is an altering function, we get

(2.8)

and, consequently, Since this implies that = and this contradicts our assumption that Hence,

(2.9)

In what follows, we will show that is a Cauchy sequence.

Suppose that is not a Cauchy sequence. Then, there exists for which we can find subsequences and of with such that

(2.10)

Further, corresponding to , we can choose in such a way that it is the smallest integer with and satisfying (2.10), then

(2.11)

Using (2.10), (2.11), and the triangular inequality, we have

(2.12)

Letting and using (2.9), we get

(2.13)

Again, the triangular inequality gives us

(2.14)

Letting in the above two inequalities and using (2.9) and (2.13), we have

(2.15)

As and , by (2.2), we obtain

(2.16)

Taking into account (2.13) and (2.15) and the fact that is continuous and letting in (2.16), we get

(2.17)

As is an altering function, , the last inequality gives us

(2.18)

Since , this means that

(2.19)

This fact and (2.15) give us which is a contradiction.

This shows that is a Cauchy sequence.

Since is a complete metric space, there exists such that Moreover, the continuity of implies that

(2.20)

and this proves that is a fixed point.

In what follows, we prove that Theorem 2.2 is still valid for not necessarily continuous, assuming the following hypothesis in (which appears in [10, Theorem ]):

(2.21)

Theorem 2.3.

Let be a partially ordered set and suppose that there exists a metric in such that is a complete metric space. Assume that satisfies (2.21). Let be a nondecreasing mapping such that

(2.22)

where is an altering function and . If there exists with , then has a fixed point.

Proof.

Following the proof of Theorem 2.2, we only have to check that . As is a nondecreasing sequence in and then, by (2.21), we have for all , and, consequently,

(2.23)

Letting and using the continuity of , we have

(2.24)

or, equivalently,

(2.25)

As is an altering function, this gives us and, thus,

Now, we present an example where it can be appreciated that the hypotheses in Theorems 2.2 and 2.3 do not guarantee uniqueness of the fixed point. This example appears in [10].

Let and consider the usual order

(2.26)

is a partially ordered set whose different elements are not comparable. Besides, is a complete metric space considering as the Euclidean distance. The identity map is trivially continuous and nondecreasing and condition (2.2) of Theorem 2.2 is satisfied since elements in are only comparable to themselves. Moreover, and has two fixed points in .

In what follows, we give a sufficient condition for the uniqueness of the fixed point in Theorems 2.2 and 2.3. This condition appears in [16] and says that

(2.27)

In [10] it is proved that condition (2.27) is equivalent to

(2.28)

Theorem 2.4.

Adding condition (2.28) to the hypotheses of Theorem 2.2 (resp., Theorem 2.3), we obtain uniqueness of the fixed point of .

Proof.

Suppose that there exist which are fixed points of and . We distinguish two cases.

Case 1.

If and are comparable, then and are comparable for Using the contractive condition appearing in Theorem 2.2 (or Theorem 2.3) and the fact that , we get

(2.29)

which is a contradiction.

Case 2.

Using condition (2.28), there exists comparable to and . Monotonicity of implies that is comparable to and to , for Moreover, as , we get

(2.30)

Since is nondecreasing the above inequality gives us

(2.31)

Thus,

Assume that .

Taking into account that is an altering function and letting in (2.30), we obtain

(2.32)

and this implies that .

Since then we get

(2.33)

and, consequently, , which is a contradiction.

Hence, .

Analogously, it can be proved that

(2.34)

Finally, as

(2.35)

and taking limit, we obtain .

This finishes the proof.

Remark 2.5.

Under the assumptions of Theorem 2.4, it can be proved that for every , , where is the fixed point (i.e., the operator is Picard).

In fact, for and comparable to then using the same argument that is in Case 1 of Theorem 2.4 can prove that and, hence, .

If is not comparable to , we take that is comparable to and . Using a similar argument that is in Case 2 of Theorem 2.4, we obtain

(2.36)

Finally,

(2.37)

and taking limit as , we obtain or, equivalently, = .

Remark 2.6.

Notice that if is totally ordered, condition (2.28) is obviously satisfied.

Remark 2.7.

Considering the identity mapping in Theorem 2.4, we obtain Theorem 1.2, being the main result of [2].

3. Application to Ordinary Differential Equations

In this section we present an example where our results can be applied.

This example is inspired by [10].

We study the existence of solution for the following first-order periodic problem

(3.1)

where and is a continuous function.

Previously, we considered the space () of continuous functions defined on . Obviously, this space with the metric given by

(3.2)

is a complete metric space. can also be equipped with a partial order given by

(3.3)

Clearly, satisfies condition (2.28), since for the function .

Moreover, in [10] it is proved that with the above-mentioned metric satisfies condition (2.21).

Now, let denote the class of functions satisfying the following.

(i) is nondecreasing.

(ii), .

(iii),

where is the class of functions defined in Section 1.

Examples of such functions are , with , , and .

Recall now the following definition

Definition 3.1.

A lower solution for (3.1) is a function such that

(3.4)

Now, we present the following theorem about the existence of solution for problem (3.1) in presence of a lower solution.

Theorem 3.2.

Consider problem (3.1) with continuous and suppose that there exist with

(3.5)

such that for with

(3.6)

where . Then the existence of a lower solution for (3.1) provides the existence of a unique solution of (3.1).

Proof.

Problem (3.1) can be written as

(3.7)

This problem is equivalent to the integral equation

(3.8)

where is the Green function given by

(3.9)

Define by

(3.10)

Notice that if is a fixed point of , then is a solution of (3.1).

In the sequel, we check that hypotheses in Theorem 2.4 are satisfied.

The mapping is nondecreasing since, by hypothesis, for

(3.11)

and this implies, taking into account that for , that

(3.12)

Besides, for , we have

(3.13)

Using the Cauchy-Schwarz inequality in the last integral, we get

(3.14)

The first integral gives us

(3.15)

As is nondecreasing, the second integral in (3.14) can be estimated by

(3.16)

Taking into account (3.14), (3.15), and (3.16), from (3.13) we get

(3.17)

Since , the last inequality gives us

(3.18)

or, equivalently,

(3.19)

This implies that

(3.20)

Putting , which is an altering function, and because , we have

(3.21)

This proves that the operator satisfies condition (2.2) of Theorem 2.2.

Finally, letting be a lower solution for (3.1), we claim that

In fact,

(3.22)

Multiplying by ,

(3.23)

and this gives us

(3.24)

As , the last inequality implies that

(3.25)

and so

(3.26)

This and (3.24) give us

(3.27)

and, consequently,

(3.28)

Finally, Theorem 2.4 gives that has a unique fixed point.

Remark 3.3.

Notice that if , then . In fact, as , then is nondecreasing and, consequently, is also nondecreasing.

Moreover, as , then , and, thus, .

Finally, as , and as , then it is easily seen that .

Example 3.4.

Consider given by

(3.29)

It is easily seen that . Taking into account Remark 3.3, .

Now, we consider problem (3.1) with continuous and suppose that there exist with

(3.30)

such that for with

(3.31)

where is the function above mentioned.

This example can be treated by our Theorem 3.2 but it cannot be covered by the results of [6] because is not increasing.