#### Abstract

The object of this paper is to establish the existence and uniqueness of coupled fixed points under a (, )-contractive condition for mixed monotone operators in the setup of partially ordered metric spaces. Presented work generalizes the recent results of Berinde (2011, 2012) and weakens the contractive conditions involved in the well-known results of Bhaskar and Lakshmikantham (2006), and Luong and Thuan (2011). The effectiveness of our work is validated with the help of a suitable example. As an application, we give a result of existence and uniqueness for the solutions of a class of nonlinear integral equations.

#### 1. Introduction and Preliminaries

Fixed point theory is an important tool for studying the phenomenon of nonlinear analysis and is a bridge bond between pure and applied mathematics. The theory has its wide applications in engineering, computer science, physical and life sciences, economics, and other fields. Banach [1] introduced the well-known classical and valuable theorem in nonlinear analysis, which is named after him, known as the Banach contraction principle. This celebrated principle has been extended and improved by various authors in many ways over the years (see for instance [2–17]). Nowadays, fixed point theory has been receiving much attention in partially ordered metric spaces, that is, metric spaces endowed with a partial ordering. Ran and Reurings [17] were the first to establish the results in this direction. The results were then extended by Nieto and Rodríguez-López [10] for nondecreasing mappings. Works noted in [18–24] are some examples in this direction.

The work of Bhaskar and Lakshmikantham [25] is worth mentioning, as they introduced the new notion of fixed points for the mappings having domain the product space , which they called coupled fixed points, and thereby proved some coupled fixed point theorems for mappings satisfying the mixed monotone property in partially ordered metric spaces. As an application, they discussed the existence and uniqueness of a solution for a periodic boundary value problem.

*Definition 1 (see [25]). *Let be a partially ordered set. The mapping is said to have the mixed monotone property if is monotone nondecreasing in and monotone nonincreasing in ; that is, for any , ,

*Definition 2 (see [25]). *An element is called a coupled fixed point of the mapping if and .

Bhaskar and Lakshmikantham [25] gave the following result.

Theorem 3 (see [25]). *Let be a partially ordered set and suppose there exists a metric on such that is a complete metric space. Let be a continuous mapping having the mixed monotone property on . Assume that there exists a with
**
for all , , , with and .**If there exist two elements , with and , then there exist such that and .*

It has also been shown in [25, Theorem 2.2] that the continuity assumption of in Theorem 3 can be replaced by an alternative condition imposed on the convergent nondecreasing and nonincreasing sequences in the space .

*Assumption 4. * has the property that(i)if a nondecreasing sequence converges to , then for all ,(ii)if a nonincreasing sequence converges to , then for all .

These results were then extended and generalized by several authors in the last six years (see [26–40]). Luong and Thuan [26, Theorem 2.1] extended the results of Bhaskar and Lakshmikantham [25] under the following contractive condition:
with and , where , are the functions such that is continuous and nondecreasing; if and only if ;, for all , ,and the function satisfies

Berinde [27, Theorem 3] in an alternative way generalized the results of Bhaskar and Lakshmikantham [25] under a weaker contraction as follows:
with and , where .

Berinde [28, Thoerem 2] further generalized and complemented the results noted in [25] and extended the work presented in [27] by considerably weakening the involved contractive condition as
with and , where , : are the functions such that is continuous and (strictly) increasing; for all ;, for all ,and the function satisfies .

#### 2. Coupled Fixed Points

In order to proceed with developing of our work and obtain our results, we need to consider the following.

Let denote all functions which satisfy the following: is lower semicontinuous and (strictly) increasing; for all ; for all , .We note that if and only if for .

Also, for , denote all functions which satisfy the following: if ; if for *.*

Now we are ready to prove our first main result as follows.

Theorem 5. *Let be a partially ordered set and there is a metric on such that is a complete metric space. Suppose is a mapping having mixed monotone property on . Assume there exist and such that
**
for all , , , with and (or and ).**Suppose either (a) is continuous or (b) satisfies Assumption 4.**If there exist two elements , with
**
or
**
then there exist , such that
**
that is, has a coupled fixed point in .*

*Proof. *Without loss of generality, assume that there exist two elements , such that (8) holds; that is, and . Let and . Then and . Similarly, let and . Since has the mixed monotone property, then we have and . Continuing in the same way, we can easily construct two sequences and in such that , and
If for some , we have , then and ; that is, has a coupled fixed point. So from now onwards, we suppose , for all ; that is, we suppose that either or .

Now applying inequality (7) with and , for all , we get
then for all , we have
where .

Clearly, for all . Then for any , we obtain
By monotonicity of , together with (14), we can easily see that is a nonnegative decreasing sequence in . So we have for some . If , then using the properties of , we get
which is a contradiction. Hence and so we have
We now prove that and are Cauchy sequences. Suppose, to the contrary, that at least one of the sequences , is not a Cauchy sequence. Then there exists for which we can find subsequences , of and , of with such that
Further, corresponding to , we can choose in such a way that it is the smallest integer with and satisfies (17). Then,
By (17), (18), and triangle inequality, we have
Letting and using (16) in the last inequality, we have
Again, by triangle inequality
By monotonicity of and property , we have
Since , , and .

Then by (7), we have
By (22) and (23), we have
Since is lower semicontinuous by taking limit as , we get
a contradiction. Thus and are Cauchy sequences in complete metric space and hence and for some , .

Now suppose that assumption (a) holds. Then
which shows that is a coupled fixed point of .

Suppose now assumption (b) holds. Since is a nondecreasing sequence that converges to , we have that for all . Similarly, for all .

Then
So
and hence
which imply, by the monotonicity of and condition (7),
Since is lower semicontinuous, by taking the limit as in the above inequality, we get
Hence, we get and . Therefore, has a coupled fixed point .

*Remark 6. *In Theorem 5, substituting for implies the main result of Berinde [28, Theorem 2]. Note that the function satisfies all the conditions of our result. It is easy to verify this as in Theorem 2 [28] is a continuous function and
for all such that . On the other hand,
if for .

*Remark 7. *In Theorem 5, let be . Then it is easy to see that substituting for , where , yields the main result of Berinde [27, Theorem 3].

The following example shows that Theorem 5 is more general than Theorem 2.1 [25] (stated as Theorem 3 in the present paper) and Theorem 2.1 in [26], since the contractive condition (7) is more general than the contractive conditions (2) and (3).

*Example 8. *Let . Then is a partially ordered set with the natural ordering of real numbers. Let be defined by
Then is a complete metric space.

Define by , .

Then is continuous, has mixed monotone property, and satisfies condition (7) but does not satisfy either condition (2) or condition (3). Indeed, assume there exists , such that (2) holds. Then, we must have
by which, for , we get
which for implies , a contradiction, since . Hence does not satisfy contractive condition (2).

Further, the contractive condition (3) is also not satisfied. Assume, to the contrary, that there exist , satisfying and , respectively, such that (3) holds. Then, we must have
for all and . On taking , and letting , we obtain
But by we have and hence, we deduce that, for all , , that is, , which contradicts . This shows that does not satisfy (3).

Now, we prove that (7) holds. Indeed, for and , we have

and by summing up the last two inequalities we get exactly (7) with , . Also, , are the two points in such that and .

So by Theorem 5 we obtain that has a coupled fixed point but none of Theorem 2.1 in [25] and Theorem 2.1 in [26] can be applied to in this example.

Next, we prove the uniqueness of coupled fixed point for our main result by defining a partial ordering on . If is a partially ordered set, then we endow the product with the following partial order: Then, we say , .

We say that and are comparable if or .

Theorem 9. *Let all the conditions of Theorem 5 be fulfilled and for every , in , there exists a in that is comparable to and . Then has a unique coupled fixed point.*

*Proof. *From Theorem 5, the set of coupled fixed points of is nonempty. Assume that and are two coupled fixed points of . We will show that and .

By assumption, there exists that is comparable to and . We define the sequences and as follows:
Since is comparable to , we may assume . Following the proof of Theorem 5 we obtain inductively
and therefore, by (7), we get
Thus, we have , where . Inspired with the proof of Theorem 5 we can conclude that converges to for some . If , then we have
which shows a contradiction. Thus ; that is, , which implies
Similarly, we obtain that , and hence, by the uniqueness of limit and .

Theorem 10. *Let all the conditions of Theorem 5 be fulfilled and suppose that , are comparable. Then has a unique fixed point; that is, there exist such that .*

*Proof. *By Theorem 5, without loss of generality, we assume that
Since and are comparable, we have or . We assume the second case. Then, by the mixed monotone property of , we have
and, hence, by making use of induction, we can get
Also, we have
then, by the continuity of the distance function , we can obtain
Further, on setting and in (7), we have
Taking the limit as in the last inequality, we get
If , then by we get
a contradiction. So and hence . Similarly, uniqueness of can be easily established.

#### 3. Application

As an application of the results proved in Section 2, we study the existence of unique solution of the following integral equation: Let denote the class of functions which satisfies the following conditions: for .

Clearly, is nonempty, as for given , is in , where , .

We assume that the functions , , , and fulfill the following conditions.

*Assumption 11. *(i) , , for all , .

(ii) There exist the positive numbers , , and , such that for all and , , with , the following conditions hold:

(iii) There exists some satisfying (57), such that
where

*Definition 12. *An element with is called a coupled lower and upper solution of the integral equation (54) if and

for all .

Theorem 13. *Consider the integral equation (54) with , for and . Suppose that there exists a coupled lower-upper solution for (54) and the Assumption 11 is satisfied. Then the integral equation (54) has a unique solution in .*

*Proof. *Consider the natural order relation on ; that is, for ,
It is well known that is a complete metric space with respect to the sup metric
It is easy to verify that condition (b) in Theorem 5 holds on the complete metric space . Also, is a partially ordered set under the following order relation in :
For any , , and , for each , are in and are the upper and lower bounds of , respectively. Therefore, for every , , there exists a that is comparable to and .

Now we define the mapping by
First, we show that has the mixed monotone property. To do this, let , and ; that is, for all . Then using Assumption 11, for any and all we obtain
which implies that . Similarly, if , and , then for any . Let be as given in Assumption 11. Then for , , , such that and we get
Since the function is nondecreasing and and , we have
hence by (68), we obtain
Similarly, we can obtain
By summing up (70) and (71), multiplying by and dividing by 2, and then taking supremum with respect to we get, by using (60)-(61),
Now, since is nondecreasing, we have
and so
by the definition of . Thus we get
which is just the contractive condition (7) for , where . Now, let be a coupled upper-lower solution of (54). Then we have
for all . Finally, Theorems 5 and 9 give that has a unique coupled fixed point . Since , then the hypotheses of Theorem 10 is satisfied and, therefore, there exists a unique such that for all ; that is, the integral equation (54) has a unique solution.

*Remark 14. *Trivial choices of the functions are , , , and , . Note that here we will assume that and .

#### 4. Conclusion

We conclude that the obtained results improve, generalize, and enrich various recent coupled fixed point theorems in the framework of partially ordered metric spaces in a way that is essentially more natural, by considering the more general (symmetric) contractive condition (7). The theoretical results are accompanied by an applied example and an application to the nonlinear integral equation.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.