Abstract

We initiate the concept of a new generalized -contraction satisfying some contractive conditions involving four maps on a partial metric space. We set up an example to elucidate our main result. An application is derived where a system of elliptic boundary value equations has a common solution.

1. Introduction

The Banach Contraction Principle (BCP) has many useful applications in various directions like differential, integral, functional, and matrix equations (both linear and nonlinear). This contraction principle is being generalized extensively in different distance spaces. One of its best generalizations is the -contraction presented by Wardowski [1]. Aydi el al. [2] (see also [3]) modified the concept of -contractions via -admissibility. Many authors have formed different generalizations of Wardowski [1] result (see [46]). Recently, in [7] authors provided a fixed point result for a iri type -contraction. Berinde [8] and Berinde and Vetro [9] have established some common fixed point results for mappings satisfying compatible conditions (resp., implicit contractions) in a complete metric space.

Here, we discuss a new generalized -contraction based on four self-mappings related to elliptic boundary value problem; in particular, we collect a common fixed point result for four self-mappings to show the existence of a common solution for operators satisfying an elliptic boundary value problem. It is remarked that the notion of the -contraction in partial metric spaces is more general with respect to the metric space.

Definition 1 (see [10]). Let be a nonempty set. If the function satisfies the following properties:(p1) (p2)(p3)(p4), for all , then is called a partial metric on

Example 2. Let . Define the classical partial metric as . Note that is not a metric on . Indeed for each .

Matthews [10] explored the following aspects of a partial metric on .(1)The function defined byfor all , defines a metric on (called the induced metric by )(2)Open balls: ; (3)Each partial metric on generates a topology on . The base of topology consists of the set (4)A sequence in converges to iff (5)A sequence in is said Cauchy if exists and is finite(6) is said to be complete if every Cauchy sequence in converges, with respect to , to so thatAfter the result of Matthews [10], numerous authors obtained various applicable fixed point results in partial metric spaces. Kumam et al. [11] showed the existence of fixed points for weak -contraction mappings in partial metric spaces. Radenovi et al. [12] investigated the existence of fixed points via simulation functions. Similarly various authors have established (common) fixed point theorems under the notion of a partial metric; see, for example, [1319].

Lemma 3 (see [10]). Given a sequence in a partial metric space , the following are provided.(1) is Cauchy in iff it is Cauchy in (2) is complete iff the metric space is complete(3) converges to in iff

Definition 4 (see [20]). Given , the following is provided. If for some , then is said to be a coincidence point of and .
Such maps and are called weakly compatible if they commute at their coincidence points; i.e., if , then .

Example 5. Let . Consider asHere, are weakly compatible and the coincidence point is .

Definition 6. The pair of mappings is compatible in a metric space iffWardowski [1] considered functions verifying the following:(F1) is strictly increasing(F2)For each positive sequence , we have (F3)There exists such that The family of the above functions satisfying (F1) – (F3) is denoted by .

Example 7. The following functions are in :(a)(b)(c)(d)

Definition 8 (see [1]). A self-map on the metric space is an -contraction if there are and so that

Definition 9. Let be a partial metric space. The mapping is an -contraction if there exist and such thatFollowing example indicates that an -contraction is more general than an -contraction.

Example 10. Consider for all . Here, for all . Define by and byNote that, for all with or , we haveSo is an -contraction, but is neither continuous, nor an -contraction. Indeed, for and , and we haveIt is a contradiction for each .

Lemma 11. Let be a metric space. If there are two sequences and so thatthen .

Proof. By triangle inequality, we haveTaking , the result follows.

2. Fixed Point Theorems

First, we introduce a new generalized -contraction involving four self-mappings . By imposing the conditions and , we generate a Cauchy sequence whose terms satisfy new generalized -contractions. We then (under some assumptions) prove that this sequence converges to a point such that .

Note that, in view of Example 10, the results in this article are independent of remarks given in [21]. Consider

Definition 12. Let be self-mappings defined on a partial metric space . These maps form a new generalized -contraction if there are and such that, for all ,The following theorem is our main result.

Theorem 13. Let be four self-mappings defined on a complete partial metric space . Assume that the mappings form a new generalized -contraction with and . Assume that either or holds:(i) is a compatible pair, or is continuous on , and is a weakly compatible pair(ii) is a compatible pair, or is continuous on , and is a weakly compatible pair Then have a unique common fixed point.

Proof. (i) Let . As , there exists such that . Since , we can choose such that . In general and are chosen in such that and . Define a sequence in as and for each . We prove that is Cauchy. Assume that ; then by (16), we getfor each , whereFor if , thenwhich is a contradiction to . Therefore,for each . Similarly,for each Hence, by (20) and (21), we havefor all andAgain, we getwhich implies . By , we haveBy the property of an -contraction, there exists such thatThe inequality (24) impliesUsing (25) and (26) and taking in (27), we getBy (28), there is , such that for each ; that is,Using (29), we get, for ,Since the series converges, we get ; i.e., is Cauchy in . By Lemma 3(1), it is also Cauchy in , which is a complete metric space. Hence, there is so that . Moreover, Lemma 3(3) impliesSince , we deduce from (31) thatWe also have the following:The map is continuous, soThe pair is compatible on ; thenNow put and in (16) and suppose on contrary that to obtainwhereWe claim that . Sincewe haveLetting , we obtain from (39) thatThe upper limit in (36) yields thata contradiction. Hence, , so . Now, assuming, on the contrary, , we have by (16)whereTaking the upper limit in (42), we havea contradiction. Hence, ; i.e., . Since , there is such that . Assume on contrary that and so, by (16), we obtainwhereThus, by (45), we havea contradiction. Hence, . By and we have . As and are weakly compatible so . Thus is a coincidence point of g and . If , from (16) we havewhereTaking the upper limit in (48), we havea contradiction, so . Hence , so is a common fixed point of the four mappings . Similarly, is a common fixed point of in the case that (ii) holds. If is another common fixed point of , then, by (16), we havewhereThus, from (51),which is a contradiction. Thus .

Theorem 13 is supported by the following.

Example 14. Take the partial metric on . Consider as ,Note that , , is continuous on , is a weakly compatible pair, and is a pair of compatible mappings. Choose as , for each . Let be such that and . Take ThenThus, the contractive condition (16) holds for all . Hence, all the hypotheses of Theorem 13 are satisfied (for case ) and have a unique common fixed point .

Following results can be derived as consequences of Theorem 13.

Corollary 15. Let be self-mappings defined on a complete partial metric space such that and . If there are and such that, for all ,where, for ,where . If either condition (i) or (ii) in the statement of Theorem 13 holds, then have a unique common fixed point.

Proof. Recall thatApplying , we see that the inequality (56) reduces to the inequality (16).

Corollary 16. Let be self-mappings defined on a complete partial metric space such that , . Assume there exist and such that, for all and ,where . If either condition (i) or (ii) in the statement of Theorem 13 holds, then have a unique common fixed point.

Proof. Using and , the inequality (59) reduces to the inequality (16). Thus, this proof follows the pattern of the proof of Theorem 13.

Corollary 17. Let be a self-mapping on a complete partial metric space . Suppose there are and such that, for all ,whereIf either or is continuous, then has a unique fixed point.

Proof. Set, in Theorem 13, and (identity mapping).

Corollary 18 (see [7, Theorem 2.2]). Let be a self-mapping on a complete metric space . Suppose there exist and such that, for all ,whereIf either or is continuous, then has a unique fixed point.

Proof. Set , for all , in Corollary 17.

Remark 19. In Example 10, is not an -contraction. Hence we cannot apply Corollary 18, so condition (61) is more general than [7] and hence that of Wardowski (see [1]) is not applicable.

Corollary 20. Let be a self-mapping on a complete partial metric space . If there are and such that, for all ,whereprovided such that . If either or is continuous, then has a unique fixed point.

Proof. Set and in Corollary 15.

Corollary 21 (see [22, Theorem 3.1]). Let be a self-mapping on a complete metric space . Assume there are and such thatfor all , whereprovided and . If or is continuous, then has a unique fixed point.

Proof. Set in Corollary 20 for all .

Remark 22. We have seen in Example 10 that is not an -contraction, so we cannot apply Corollary 21. Hence condition (65) is more general than the contractive condition provided by Cosentino et al. [22].

Remark 23. The reader interested in fixed points of multivalued contraction mappings in a partial metric space and their applications may be referred to Khan et al. [23].

3. Applications

Denote by the set of all continuous functions defined on . Define the partial metric byNote that is complete.

We consider a system of elliptic boundary value equations given bywhere are continuous mappings. The Green function associated to (70) is defined byNote that the system (70) has a solution if and only if the operators , defined byhave a common fixed point.

Theorem 24. Let and let be as in (69). If(A1)there are , , and continuous mappings such that for each and such that(A2), for all ;(A3) there is a sequence in such that and , wheneverthen the system (70) has at least one solution (say) .

Proof. We note that is a solution of (70) iff is a solution of the integral equations given by (72) and (73).
Let . By assumption , we haveSince for all , we have . Taking the supremum over in the resulting inequality and using assumption (A2), we get for all . verifies all the hypotheses of Corollary 16(a) and so its conclusion ensures that the problem (70) has a solution.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest regarding the publication of this paper.

Authors’ Contributions

All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.

Acknowledgments

The third and fourth authors would like to acknowledge the grant: UKM Grant DIP-2017-011 and FRGS/1/2017/STG06/UKM/01/1, for financial support.