Related Fixed Point Theorems in Partially Ordered <span class="nowrap"><svg xmlns:xlink="http://www.w3.org/1999/xlink" xmlns="http://www.w3.org/2000/svg" style="vertical-align:-0.2723999pt" id="M1" height="12.533pt" version="1.1" viewBox="-0.0657574 -12.2606 8.20973 12.533" width="8.20973pt"><g transform="matrix(.017,0,0,-0.017,0,0)"><path id="g113-99" d="M452 333C452 398 425 448 375 448C329 448 235 404 140 292H138L229 683C233 701 234 712 225 712C208 712 149 686 66 677L64 652H97C135 652 140 651 129 598L38 167C23 96 29 57 44 33C60 7 98 -12 132 -12C169 -12 232 3 290 41C383 102 452 223 452 333ZM365 316C365 199 308 87 239 49C221 39 200 35 183 35C137 35 99 70 112 163C116 191 123 215 129 233C190 316 290 392 330 392C348 392 365 372 365 316Z"/></g></svg>-</span>Metric Spaces and Applications to Integral Equations

Errai, Youssef; Marhrani, El Miloudi; Aamri, Mohamed

doi:https://doi.org/10.1155/2021/6672724

Abstract and Applied Analysis

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Research Article | Open Access

Volume 2021 | Article ID 6672724 | https://doi.org/10.1155/2021/6672724

Related Fixed Point Theorems in Partially Ordered -Metric Spaces and Applications to Integral Equations

Youssef Errai,¹El Miloudi Marhrani,¹and Mohamed Aamri¹

Academic Editor: Paul Eloe

Received28 Nov 2020

Accepted23 Feb 2021

Published16 Mar 2021

Abstract

In this research paper, we have set some related fixed point results for generalized weakly contractive mappings defined in partially ordered complete -metric spaces. Our results are an extension of previous authors who have already worked on fixed point theory in -metric spaces. We state some examples and one sample of the application of the obtained results in integral equations, which support our results.

1. Introduction

The concept of -metric space has been dealt with by distinctly different authors since it first appeared and became largely used. Bakhtin in [1] was the first who introduced this concept, and later on, Czerwik in [2, 3] invested it in the convergence of measurable functions. Accordingly, several interesting results about the existence of fixed points have been obtained concerning both single-valued and multivalued operators in -metric spaces (see, e.g., [4, 5]). In the same way, Hussain and Shah [6] conducted a research and then came up with results from KKM mappings in cone -metric spaces. Likewise, Roshan et al. [7] used the notion of almost generalized contractive mappings in ordered complete -metric spaces and set some fixed and common fixed point results. Regarding partially ordered metric spaces, Ran and Reurings (see [8]) had first set up their assumptions before Nieto and Rodríguez-López used them (see [9, 10]). Afterwards, many authors presented several interesting and significant results in such spaces (see [4, 7, 11–19]).

The purpose of this research paper is to prove some fixed point theorems for generalized contractive conditions for four mappings in complete -metric spaces.

Our work goes through the following steps. First, we have demonstrated a two -metric space theorem with four mappings. Second, we have stated a relevant example which backs up the mentioned theorem. Third, we have given three corollaries related to the theorem. Besides, an example with only two related corollaries is provided for the second theorem. To sum up, we conclude the manuscript by an application to solve a system of integral equations.

Throughout this paper, and denote the sets of all real numbers and nonnegative real numbers, respectively.

Consistent with [3, 4], the following definitions and outcomes are going to be vital and required in the ending.

Definition 1 (see [3]). Given a nonempty set . A function is called -metric if there is a real number such that for all , the following conditions hold: (i) if and only if (ii)(iii)The pair is called a -metric space.

Definition 2 (see [20]). Let be a -metric space. Then, a sequence in is (a)convergent if there exists such that as . In this case, we write .(b)Cauchy sequence if as .

Definition 3. Let be a nonempty set. Then, is called a partially ordered two -metric space if and are -metrics on the partially ordered set .

A subset of a partially ordered set is said to be well ordered if every two elements of are comparable.

Definition 4 (see [4]). Let be a partially ordered set. A mapping on is called dominating (resp. dominated) if (resp. ) for each in .

Definition 5. Let be a sequence in a -metric space , , and . (i) is said to be a coincidence point of pair if (ii) is said to be compatible if as (iii) is said to be weakly compatible if , where

Remark 6 (see [20]). In a -metric space , the following assertions hold: (R1)A convergent sequence has a unique limit(R2)Each convergent sequence is a Cauchy sequence(R3)In general, a -metric is not continuous(R4)In general, a -metric does not induce a topology on .

The fact in (R3) requires the following Lemma concerning -convergent sequences to prove our findings:

Lemma 7 (see [4]). Let be a -metric space with and suppose that and are convergent to , respectively. Then, we have In particular, if , then we have .
Moreover, for each we have

2. Main Results

Throughout this paper, let , and be four self-maps on , and and are two -metrics with constant and , respectively. Set with

Let,

Finally, we consider the following assumptions:

Assumption 8. Let and be two sequences such that (i) is nonincreasing(ii).

Assumption 9. Let , and be four self-maps on such that either condition (a) is weakly compatible, is compatible and or is continuous, or(b) is weakly compatible, is compatible and or is continuous.

Our main result is the following theorem:

Theorem 10. We consider four self-mappings , and in ordered complete two -metric space that fulfill the following conditions: (i) is dominated and is dominating(ii) and (iii), and for every two comparable elements , we have(iv)Assumptions 8 and 9 are satisfiedThen, , and have a unique common fixed point in .

Proof. Let be an arbitrary point in . By condition (ii), we can define inductively two sequences and in as follows: We have Thus, for all , and

Step 1. If for some , then is constant. Indeed, we have which implies

If we insert in (10), we obtain and then is a constant sequence. Its value is a common fixed point of , and

In the following, we can assume that for each .

Step 2. The sequences and are monotone nonincreasing. Indeed, since and are comparable, we obtain

Consequently, if we permute and we obtain

If for some we obtain

We have Otherwise, we obtain and consequently, which gives , which contradicts our hypothesis. Consequently,

Taking account of (6), we obtain which gives

On the other hand, the inequality gives which contradicts (18). Thus, and are monotone nonincreasing sequences. It follows that there exists a nonnegative real number such that

Using (10), we deduce that and consequently,

Step 3. is a Cauchy sequence.

Assume that there exist for which we can find subsequences and of such that for all , we have

Using the triangle inequality in -metric spaces, we obtain which leads to and since we obtain

By the same arguments, we obtain and consequently

Similarly, we obtain

Moreover, we have

Using (24) and (26)–(28), we obtain

By the same arguments, we obtain

Using (31), we get

Then,

Similarly, we obtain

Consequently,

which implies that

It follows that

We have also

Then, which contradicts with (36) and (37). It follows that is a Cauchy sequence with -metric ; it is also a Cauchy sequence in And then there exists such that

Step 4. is a common fixed point of , and .

Since is continuous, we have

Moreover, the pair is compatible, then The triangle inequality implies

Since we obtain by (6),

Using Lemma (7), we obtain

Consequently, and then

which gives or equivalently .

We have and then , which gives with

If in (50) and using Lemma (7), we obtain which leads to and consequently, .

On the other hand, we have then there exists a point such that

Moreover, gives with and Consequently, which implies . And since is weakly compatible, we obtain

and we obtain and then

If and using Lemma (7), we obtain

If in (56), and using Lemma (7), we obtain which implies and then .

We conclude that .

If is continuous, the proof is the same. The same goes for condition (b) of Assumption 9.

Step 5. Suppose that and are two common fixed points of , and but with and are comparable. We have with

So from (59), we have

Hence, Therefore, .

Example 11. Let we define on the -metrics with and with . We endow with the partial order given by And we define , and on by Obviously, conditions (i) and (ii) of Theorem 10 are satisfied. Moreover, is weakly compatible, is compatible and is continuous.

The control functions are defined as and for all .

To prove that , and satisfy (6), for this, we consider the following cases: (i)If and , then and (6) are satisfied.(ii)If and , then(iii)If and , then(iv)If and ,

Thus, all conditions of Theorem (10) are satisfied. Moreover, is the unique common fixed point of , and .

If we take and as the identity maps on in Theorem (10), we conclude the following corollary.

Corollary 12. We consider two self-mappings and in ordered complete two -metric space that fulfill the following conditions: (i) is dominated(ii), and for every two comparable elements , we have(iii)Assumption 8 is satisfiedThen and have a unique common fixed point in .
If we take for in Corollary (12), we have the following corollary.

Corollary 13. We consider two self-mappings and in ordered complete two -metric space that fulfill the following conditions: (i) is dominated(ii) and for every two comparable elements , we have(iii)Assumption 8 is satisfiedThen, and have a unique common fixed point in .
If we take for in Corollary (13), we obtain the following result.

Corollary 14. We consider two self-mappings and in ordered complete two -metric space that fulfill the following conditions: (i) is dominated(ii) and for every two comparable elements , we have(iii)Assumption 8 is satisfiedThen, and have a unique common fixed point in .

Theorem 15. We consider two self-mappings , and T in ordered complete two -metric space that fulfill the following conditions: (i) is dominated and is dominating(ii) and (iii), and for every two comparable elements , we have(iv)Assumptions 8 and 9 are satisfiedThen, , and have a unique common fixed point in .

Proof. If we follow similar arguments to those given in Theorem (10), we have the following steps.

Step 1. If for some , then is constant.

So, from (69), we have where

So, from (70) and (71), we obtain which gives , and so, , since which further implies that . Hence, is a constant sequence and is a common fixed point of , and .

In the following, we can assume that for each .

Step 2. The sequence are monotone nonincreasing.

Since and are comparable, from (69), we have

Hence, where

If for some , , then (74) gives that , and from (69), we have

And or equivalently , a contradiction.

Then, . By the same arguments, we have

Therefore, is a nonincreasing sequence, and so, there exists so that

Now, we demonstrate that . Suppose that . Since we have

Then, passing the upper limit as , it implies that which shows a contradiction. Thus, . Hence,

Step 3. is a Cauchy sequence.

Assume that there exists for which we can find subsequences and of such that is the smallest index for which , , and .

Following the same argument as in Theorem (10), we have

As

So, from (86), (81), (82), (83), and (84), we obtain

which implies

Similarly, we have

As

Taking the upper limit as and using (85) and (89), we obtain

which implies that

So, , which contradicts (90). It follows that is a Cauchy sequence in . Since is complete, there exists so that

Step 4. is a common fixed point of , and .

Since is continuous, we have

Using the triangle inequality in -metric space, we get

Since the pair is compatible, . So, passing the upper limit when from the above inequality and Lemma (7), we have

Hence,

As so, from (69), we have

where

Now, by using Lemma (7), we get

Similarly,

Hence, by taking the upper limit in (99) and using Lemma (7), we obtain

which gives or equivalently . Now, since and as , then , and from (69), we have

where

By using Lemma (7), we get

Similarly,

Taking the upper limit as in (104) and using Lemma (7), we have

which implies . So, .

On the other hand, we have , so there exists a point such that .

Suppose that . Since , from (69), we have

where

So, from (109), we have

This implies that , so which shows a contradiction; therefore, . Since the pair is weakly compatible, , and is the coincidence point of and .

Since and as , it implies that , and from (69), we obtain

where

By using Lemma (7), we have

Passing the upper limit as , and using (114), we have

which implies that ; therefore, .

If is continuous, the proof is the same. The same goes for condition (b) of Assumption 9.

Step 5. Suppose that and are two common fixed points of , and , but with and are comparable. By assumption, we can apply (69) to obtain where Hence, implies , which is a contradiction. Therefore, . The converse is clear.

Example 16. We consider the data of the previous example and take , and . This permits only to verify the condition (69) of Theorem (15). We shall distinguish two cases: (i)If and , then and (69) is satisfied(ii)If and , then

Thus, (69) is satisfied for all . Therefore, all conditions of Theorem (15) are satisfied. Moreover, is the unique common fixed point of , and .

If we take and as the identity maps on in Theorem (15), we conclude the following corollary.

Corollary 17. We consider two self-mappings and in ordered complete two -metric space that fulfill the following conditions: (i) is dominated(ii), and for every two comparable elements , we have(iii)Assumption 8 is satisfiedThen, and have a unique common fixed point in .

If we take for in Corollary (17), we have the following corollary.

Corollary 18. We consider two self-mappings and in ordered complete two -metric space that fulfill the following conditions: (i) is dominated(ii) and for every two comparable elements , we have(iii)Assumption 8 is satisfiedThen, and have a unique common fixed point in .

3. Applications

Consider the following system of integral equations:

where and are continuous functions.

Let , where is the set of real continuous on .

We endow with the two -metric: for all . Clearly, is a complete two -metric space with , and with . We endow with the partial order given by

Clearly, the space is an ordered complete two -metric space.

Consider the mapping defined as follows: with

Suppose that the following hypotheses hold: (i)there exists a continuous function , for all ; and for all comparable , we have(ii)

Theorem 19. Under the assumptions (i) and (ii), the system integral equations (122) have a unique solution in the set .

Proof. By the condition (i), we have for all and for all comparable , By a similar calculus, we obtain It follows that Similarly, which implies that with Thus, all the hypotheses of Corollary (14) (with ) are satisfied, and hence, the mapping has a unique fixed point in , which is a solution of the system of nonlinear integral equations (122).

4. Conclusion

In this manuscript, we have obtained interesting results on the coupled fixed point theorems that give a generalization of well-known results in the field. We have shown that these results are very useful to solve the system of integral equations.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

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Copyright

Copyright © 2021 Youssef Errai 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.

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