Abstract
In this paper, by introducing a convergence comparison property of a self-mapping, we establish some new fixed point theorems for Bianchini type, Reich type, and Dass-Gupta type dualistic contractions defined on a dualistic partial metric space. Our work generalizes and extends some well known fixed point results in the literature. We also provide examples which show the usefulness of these dualistic contractions. As an application of our findings, we demonstrate the existence of the solution of an elliptic boundary value problem.
1. Introduction
The recent study in the metric fixed point theory is due to the Banach Contraction Principle, which has been modified, improved, and generalized in many ways in metric spaces (see for example, [1–17]).
This Contraction principle has also been studied in partial metric spaces (PMS) introduced by Matthews [18]. The PMS generalizes the metric space where the self-distance may be not equal to zero. The topological concepts, like convergence, Cauchy sequence, continuity, and completeness in this class can be found in [18–21] and references there in.
O’Neill [22] initiated the notion of a dualistic partial metric space. This class generalizes the notion of partial metric spaces. In [22], a relationship between a dualistic partial metric and a quasi metric has been obtained. O’Neill [22] also studied various topological properties of a dualistic partial metric space, while fixed point theory on dualistic partial metric spaces was presented by Oltra and Valero [23], who proved a Banach fixed point theorem and gave convergence properties of sequences on complete dualistic partial metric spaces. Later, Nazam et al. [24] ensured fixed point results for rational type contraction mappings in this setting (see also the related paper [25]).
In this paper, motivated by [26, 27, 12], we establish some new fixed point theorems in dualistic partial metric spaces, generalizing fixed point results of Bianchini [26], Reich [12], and Dass and Gupta [27]. We also provide examples and an application to show significance of the obtained results involving dualistic contractive conditions.
2. Preliminaries
We recall some mathematical basics and definitions.
Definition 1 ([18]). A partial metric on a non empty set is a function so that
;;;for all .
The pair is called a partial metric space. O’Neill [22] did one significant change to the definition of the partial metric by extending its range from to . The partial metric with extended range is called a dualistic partial metric (DPM), denoted by .
Definition 2 ([22]). Let be a non empty set. If a function is such that
;;;,for all , then is called a dualistic partial metric and the pair is known as a dualistic partial metric space.
If is a dualistic partial metric space, then defined by is called a quasi metric on such that . Moreover, if is a dualistic quasi metric on , then is a metric on (an induced metric).
Remark 1. Unlike partial metric case, note that if is a dualistic partial metric, then may not imply . The self-distance is a feature utilized to describe the amount of information contained in . The restriction of to is a partial metric.
Example 1. We define as . It is easy to check that satisfies and hence is a dualistic partial metric on . Mention that is not a partial metric on because that for each .
Example 2. Let be a partial metric defined on a non empty set . The function defined byverifies and so it defines a dualistic partial metric on . Note that may have negative values.
In the following, a new example of a dualistic partial metric is stated.
Example 3. Take and . We define asThe axioms , , and can be proved immediately. We just prove axiom in details, for all .
Case 1. If , then implies .
Case 2. If , then implies .
Case 3. If , then implies .
Case 4. If , then implies .
Thus, the axiom holds in all cases. Hence is a dualistic partial metric space.
O’Neill [22] established that each dualistic partial metric on generates a topology on having a base, the family of -balls where
Definition 3 ([22]). Let be a dualistic partial metric space, then
(1)A sequence in converges to a point if and only if (2)A sequence in is called a Cauchy sequence if exists and is finite.(3) is said to be complete if every Cauchy sequence in converges, with respect to , to a point such that .The following lemma will be helpful in the sequel.
Lemma 1 ([22, 22]). (1) A dualistic partial metric space is complete iff the metric space is complete.
(2) A sequence in converges to a point , with respect to iff
3. Main Results
We begin with the following useful property.
Definition 4. Let be a self-mapping on a dualistic partial metric space . If there is a convergent sequence in with , such thatthen is said to have the convergence comparison property [in short, (CCP)].
Example 4. Let . Define by Clearly, is a dualistic partial metric space. Consider . We have . That is, in . Define for all . For such , observe that , so has the convergence comparison property.
The following theorem corresponds to the unique fixed point result of Bianchini type dualistic contraction.
Theorem 1. Let be a self-mapping on a complete dualistic partial metric space satisfying (CCP). If there is so thatfor all , then possesses a unique fixed point.
Proof. We generate a Picard iterative sequence with an initial point such that for all . If there exists such that , then is a fixed point of , so the proof is completed. From now on, assume that for all , then by the contractive condition (8), we haveThus,Ifthen (10) leads to a contradiction. As a result, we haveArguing like above, we haveThus, inequality (12) entailsProceeding further in a similar way, we getNow considerthe inequality (16) impliesWe deduce from (1) thatand using inequalities (16) and (18), we obtainNow, for , we haveAs , , thus, is a Cauchy sequence in . Since is a complete dualistic partial metric space, by Lemma 1 (1), is a complete metric space. Thus, there exists such that as , that is and by Lemma 1 (2), we know thatSince, , by (1) and (18), we haveThis shows that is a Cauchy sequence converging to . We shall show that is a fixed point of . By (8), we haveas , which implies that . Since has (CCP), we getOn the other hand, by axiom , we have which implies thatThe inequalities (26) and (27) imply that . Thus,By using axiom , we have . This shows that is a fixed point of . To prove the uniqueness, suppose that is another fixed point of , then and . By (8), we obtainThis implies that and using axiom , we have , which proves the uniqueness of .
Corollary 1. Let be a complete partial metric space and let be a mapping satisfying for all and . Then, has a unique fixed point.
Proof. Since the restriction of a dualistic partial metric to , which is , is a partial metric, so arguments follow the same lines as in the proof of Theorem 1.
Corollary 2 ([26]). Let be a complete metric space and let be a mapping satisfyingfor all and . Then, has a unique fixed point.
Proof. Set for all , in Corollary 1.
Example 5. Let . Define the mapping bythen, is a complete dualistic partial metric space. Define a mapping by
Note that the mapping has the convergence comparison property (CCP). Indeed, consider . Here is a sequence in . Clearly, we have that as in . We have . Choose . On the other hand, the contractive condition (8) is also satisfied. For this, let . Here, we need the following cases:
Case 1. (a) and . We have(b) and . We have(c) and , and vice versa. In this case,
Case 2. (a). We have(b). We have
Hence, verifies all conditions of Theorem 1. Note that is the unique fixed point of .
Example 6. Let . Define the mapping by
then is a complete dualistic partial metric space. Define a mapping as in Example 5. It is obvious that verifies all conditions of Theorem 1. But, classical Bianchini contraction (contraction without the absolute value function) is not applicable. Indeed, for and , we have
for each .
In what follows, we investigate a unique fixed point of Reich type dualistic contraction mappings.
Theorem 2. Let be a complete dualistic partial metric space and let be a mapping satisfying (CCP). If there exist with such that for all , then, has a fixed point.
Proof. Let be a sequence with an initial point such that for all . If there exists such that , then is a fixed point of . On the other hand, if for all , then by contractive condition (41), we haveLet , then and repeating arguments given above, we haveNow, consider the self-distanceDue to inequality (44), we haveThe inequality (46) implies thatContinuing further, we getThe equation (1) implies thatLet . Then,Now, for , we haveWe conclude that . This shows that is a Cauchy sequence in . Since is a complete dualistic partial metric space, by Lemma 1 (1), is a complete metric space. Thus, there exists such that as , that is and by Lemma 1 (2), we know thatSince , by (1) and (50), we haveThis shows that is a Cauchy sequence converging to . We show that is a fixed point of . By (41), we haveApplying limit as and using equation (55), we have . Since has (CCP), we getBy axiom , we haveThe inequalities (57) and (58) imply that . Thus,By using axiom , we have . This shows that is a fixed point of .
Example 7. Let and be as in Example 5. Then is a complete dualistic partial metric space. Define the mapping by The mapping satisfies (CCP) for any convergent sequence in . Indeed, for a convergent sequence in such that , due to completeness of . Thus for every such we have . Observe that for each case (as shown in Table 1) there exist some with (however, assuming different values in each case) for which the contractive condition (41) is satisfied.
Example 8. Let and be as in Example 6. Then is a complete dualistic partial metric space. Define the mapping as in Example 7. It is obvious that (41) holds for all , while, the classical Reich fixed point theorem is not applicable. Indeed, for , we have for all .
Corollary 3. Let be a complete partial metric space and let be a mapping satisfyingfor all and such that . Then, has a fixed point.
Proof. Since the restriction of a dualistic partial metric to , , is a partial metric, so arguments follow the same lines as in the proof of Theorem 2.
Corollary 4 ([12]). Let be a complete metric space and let be a mapping satisfyingfor all and such that . Then, has a fixed point.
Proof. Set for all , in Corollary 3.
The following theorem implies the uniqueness of the fixed point of a new Rational type dualistic contraction.
Theorem 3. Let be a complete dualistic partial metric space and let be a mapping satisfying (CCP). If there exist with such thatfor all . Then, has a fixed point.
Proof. Let be a sequence with an initial point such that for all . If there exists such that , then is a fixed point of . On the other hand, if for all , then from contractive condition (64), we haveLet , soNow,As , we getArguing like above, we haveThe inequality (68) leads toContinuing in the same way, we getThe equation (1) impliesSet . Then, Now, for , we haveWe deduce that and similarly, which implies that . The remaining part of this proof is similar to the proof of Theorem 2.
Corollary 5. Let be a complete partial metric space and let be a mapping satisfyingfor all and are non negative numbers such that . Then, has a fixed point.
Proof. Since the restriction of a dualistic partial metric to , is a partial metric, the result is obvious.
Corollary 6. Let be a complete metric space and let be a mapping satisfyingfor all and are non negative numbers such that . Then, has a fixed point.
Proof. Take in Corollary 5, for all .
4. Application
In this section, we present an application of Theorem 1 to ensure the existence of the solution of the boundary value problem given by
where is a continuous mapping. The Green function associated to the boundary value problem (78) is defined by
Let be the space of all continuous mappings defined on . Let . Define the mapping by
It is known that is a complete dualistic partial metric space. Define by
for all Note that the problem (78) has a solution iff the operator has a fixed point.
Theorem 4. Let . Define the mapping bywhere is a continuous mapping and is such that . Assume that(i) The mapping satisfiesfor all , , and ;
Then, boundary value problem (78) has a solution.
Proof. We note that (say) is a solution of (78) if and only if is a solution of the integral equation (82). The solution of (82) is given by the fixed point of , i.e., .
Let and by assumption , we getSince for all we have which implies thatHence, application of Theorem 1 ensures that has at least one fixed point , that is, which is a solution of (82).
Data Availability
The data used to support the findings of this study are available from the corresponding author upon request.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Authors’ Contributions
All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.
Acknowledgments
The third author would like to thank Prince Sultan University for funding this work through research group Nonlinear Analysis Methods in Applied Mathematics (NAMAM) group number RG-DES-2017-01-17.