#### Abstract

We provide sufficient conditions for the existence of a unique common fixed point for a pair of mappings , where is a nonempty set endowed with a certain metric. Moreover, a numerical algorithm is presented in order to approximate such solution. Our approach is different to the usual used methods in the literature.

#### 1. Introduction and Problem Formulation

Let be a complete metric space and be two given operators. In this paper, we are interested on the problem:We provide sufficient conditions for the existence of one and only one solution to (1). Moreover, we present a numerical algorithm in order to approximate such solution. Our approach is different to the existing methods in the literature.

System (1) arises in the study of different problems from nonlinear analysis. For example, when we deal with the solvability of a system of integral equations, such problem can be formulated as a common fixed point problem for a pair of self-mappings , where and are two operators that depend on the considered problem. For some examples in this direction, we refer to  and references therein.

The most used techniques for the solvability of problem (1) are based on a compatibility condition introduced by Jungck . Such techniques are interesting and can be useful for the solvability of certain problems (see  and references therein). However, two major difficulties arise in the use of such approach. At first, the compatibility condition is not always satisfied, and in some cases it is not easy to check such condition. Moreover, the numerical approximation of the common fixed point is constructed via the axiom of choice using certain inclusions, which makes its numerical implementation difficult.

In this paper, problem (1) is investigated under the following assumptions.

Assumption (A1). We suppose that is equipped with a partial order . Recall that is a partial order on if it satisfies the following conditions:(i), for every .(ii) and imply that , for every .(iii) and imply that , for every .

Assumption (A2). The operator is level closed from the left; that is, the set is nonempty and closed.

In order to make the lecture for the reader easy, let us give an example.

Example 1. Let be the set of real valued and continuous functions on . We consider the metric on defined by We endow with the partial order given by Next, define the operator by Clearly, is well-defined. Now, consider the set that is, Let be a sequence that converges to some (with respect to ); that is, Since the uniform convergence implies the point-wise convergence, for all , we have Moreover, for all ,Therefore,which proves that is level closed from the left.

Remark 2. Note that the fact that is level closed from the left does not imply that is closed. Several counterexamples can be obtained. We invite the reader to check this fact by himself.

Assumption (A3). For every , we have

In order to fix our next assumption, we need to introduce the following class of mappings. We denote by the set of functions satisfying the conditions: () is nondecreasing. ()For all , we have Here, is the th iterate of . Any function is said to be a (c)-comparison function.

We have the following properties of (c)-comparison functions.

Lemma 3 (see ). Let . Then (i), for all ,(ii),(iii) is continuous at ,(iv) is nondecreasing and continuous at .

Our next assumption is the following.

Assumption (A4). There exists a function such that, for every , we have

Now, we are ready to state and prove our main result.

#### 2. A Common Fixed Point Result and Approximations

Our main result is given by the following theorem.

Theorem 4. Suppose that Assumptions (A1)–(A4) are satisfied. Then (i)for any , the Picard sequence converges to some , which is a solution to (1),(ii) is the unique solution to (1),(iii)the following estimateshold, where

Proof. Let be an arbitrary element of ; that is, Such an element exists from Assumption (A2). From Assumption (A3), we have where . Again, from Assumption (A3), we have where . Now, let us consider the Picard sequence defined by Proceeding as above, by induction we getTherefore, by Assumption (A4), we have Again, by Assumption (A4), we have As a consequence, we haveFrom (26), since is a nondecreasing function, for every , we haveSuppose that In this case, from (23), we have Since is a partial order, this proves that is a solution to (1). Now, we may suppose that . Let From (27), we haveUsing the triangle inequality and (31), for all , we haveOn the other hand, since , we havewhich implies that is a Cauchy sequence in . Then there is some such thatOn the other hand, from (23), we have Since is level closed from the left (from Assumption (A2)), passing to the limit as and using (34), we obtain that is,Now, using (23), (37), and Assumption (A4), we obtain that is, Passing to the limit as , using (34), the continuity of at , and the fact that (see Lemma 3), we get that is,Next, using (37), (41), and Assumption (A3), we obtain that is,Since is a partial order, inequalities (37) and (43) yieldFurther, (41) and (44) yield that is a solution to problem (1). Therefore, (i) is proved.
Suppose now that is another solution to (1) with . Using Assumption (A4) and the result (i) in Lemma 3, we obtain which is a contradiction. Therefore, is the unique solution to (1), which proves (ii).
Passing to the limit as in (32), we obtain estimate (16). In order to obtain estimate (17), observe that, by (26), we inductively obtain and hence, similar to the derivation of (32), we obtain Now, passing to the limit as , (17) follows.
The proof is complete.

Observe that Theorem 4 holds true if we replace Assumption (A2) by the following.

Assumption (A2). The operator is level closed from the right; that is, the set is nonempty and closed.

As a consequence, we have the following result.

Theorem 5. Suppose that Assumptions (A1) and (A2)′–(A4) are satisfied. Then (i)for any , the Picard sequence converges to some , which is a solution to (1), (ii) is the unique solution to (1), (iii)the following estimateshold.

Taking (the identity operator), we obtain immediately from Theorem 4 (or from Theorem 5) the following fixed point result.

Corollary 6. Let be a complete metric space. Let be a given mapping. Suppose that there exists some such that Then  (i)for any , the Picard sequence converges to some , which is a fixed point of , (ii) is the unique fixed point of , (iii)the following estimates hold.

Remark 7. Observe that all the obtained results hold true if we replace the partial order by any binary relation which is antisymmetric; that is, satisfies

We end the paper with the following illustrative example.

Example 8. Let and be the metric on defined by Then is a complete metric space. Let be the binary relation on defined by Consider the partial order on defined by Let us define the pair of mappings by Observe that, in this case, we have which is nonempty and closed set. Therefore, the operator is level closed from the left, and Assumption (A2) is satisfied. Moreover, we have In order to check the validity of Assumption (A3), let be such that ; that is, . If , then and . Then . If , then and . Then . Now, let be such that ; that is, . In this case, we have and . Then . Therefore, Assumption (A3) is satisfied. Now, let be such that and ; that is, and . For , we have for every . For , we have for every . Therefore, Assumption (A4) is satisfied. Now, applying Theorem 4, we deduce that problem (1) has a unique solution . Clearly, in our case, we have .

Remark 9. Note that Theorem 4 (or Theorem 5) provides us just the existence and uniqueness of a common fixed point of the operators . However, the uniqueness of the fixed points of is not satisfied in general. As we observe in Example 8, the operator has infinitely many fixed points.

#### Competing Interests

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

#### Acknowledgments

The authors extend their appreciation to Distinguished Scientist Fellowship Program (DSFP) at King Saud University (Saudi Arabia).