# Fixed Point Results for Dualistic Contractions with an Application

Authors: Muhammad Nazam; Aiman Mukheimer; Hassen Aydi; Muhammad Arshad; Raheel Riaz
Journal: Discrete Dynamics in Nature and Society
 (2020)

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## 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.


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## Body

## 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 setA is a function η:A×A→0,∞ so that
η
1
σ
=
ς
⟺
η
σ
,
σ
=
η
σ
,
ς
=
η
ς
,
ς;
η
2
η
σ
,
σ
≤
η
σ
,
ς;
η
3
η
σ
,
ς
=
η
ς
,
σ;
η
4
η
σ
,
ς
≤
η
σ
,
ξ
+
η
ξ
,
ς
−
η
ξ
,
ξ
,
for allσ,ς,ξ∈A.
The pairA,η is called a partial metric space. O’Neill [22] did one significant change to the definition of the partial metric η by extending its range from 0,∞ to −∞,∞. The partial metric η with extended range is called a dualistic partial metric (DPM), denoted by η∗.Definition 2 ([22]).
LetA be a non empty set. If a function η∗:A×A→R is such that
η
1
∗
σ
=
ς
⟺
η
∗
σ
,
σ
=
η
∗
ς
,
ς
=
η
∗
σ
,
ς;
η
2
∗
η
∗
σ
,
σ
≤
η
∗
σ
,
ς;
η
3
∗
η
∗
σ
,
ς
=
η
∗
ς
,
σ;
η
4
∗
η
∗
σ
,
ξ
+
η
∗
ς
,
ς
≤
η
∗
σ
,
ς
+
η
∗
ς
,
ξ,
for allσ,ς,ξ∈A, then η∗ is called a dualistic partial metric and the pair A,η∗ is known as a dualistic partial metric space.
IfA,η∗ is a dualistic partial metric space, then dη∗:A×A→0,∞ defined by(1)dη∗σ,ς=η∗σ,ς−η∗σ,σ,forallσ,ς∈A is called a quasi metric on A such that τη∗=τdη∗. Moreover, if dη∗ is a dualistic quasi metric on X, then dη∗sσ,ς=maxdη∗σ,ς,dη∗ς,σ is a metric on A (an induced metric).Remark 1.
Unlike partial metric case, note that ifη∗ is a dualistic partial metric, then η∗σ,ς=0 may not imply σ=ς. The self-distance η∗σ,σ is a feature utilized to describe the amount of information contained in σ. The restriction of η∗ to 0,∞ is a partial metric.Example 1.
We defineη∗:R×R→R as η∗σ,ς=max{σ,ς}. It is easy to check that η∗ satisfies η1∗−η4∗ and hence η∗ is a dualistic partial metric on R. Mention that η∗ is not a partial metric on R because that η∗−m,−m=−m for each m>0.Example 2.
Letη be a partial metric defined on a non empty set A. The function η∗:A×A→R defined by(2)η∗σ,ς=ησ,ς−ησ,σ−ης,ςforallσ,ς∈A
verifiesη1∗−η4∗ and so it defines a dualistic partial metric on A. Note that η∗σ,ς may have negative values.
In the following, a new example of a dualistic partial metric is stated.Example 3.
TakeA=R and b>0. We define η∗:A×A→R as(3)η∗σ,ς=σ−ςifσ≠ς,−bifσ=ς.
The axiomsη1∗, η2∗, and η3∗ can be proved immediately. We just prove axiom η4∗ in details, for all σ,ς,ν∈A.Case 1.
Ifσ≠ς=ν, then ησ,ν≤ησ,ς+ης,ν−ης,ς implies σ−ν=σ−ς.Case 2.
Ifσ=ς≠ν, then ησ,ν≤ησ,ς+ης,ν−ης,ς implies σ−ν=ς−ν.Case 3.
Ifσ=ς=ν, then ησ,ν≤ησ,ς+ης,ν−ης,ς implies −b=−b.Case 4.
Ifσ≠ς≠ν, then ησ,ν≤ησ,ς+ης,ν−ης,ς implies σ−ν≤σ−ς+ς−ν+b.Thus, the axiomη4∗ holds in all cases. Hence A,η∗ is a dualistic partial metric space.O’Neill [22] established that each dualistic partial metric η∗ on A generates a T0 topology τη∗ on A having a base, the family of η∗-balls Bη∗σ,ϵ:σ∈A,ϵ>0 where(4)Bη∗σ,ϵ=ς∈A:η∗σ,ς<ϵ+η∗σ,σ.Definition 3 ([22]).
LetA,η∗ be a dualistic partial metric space, then
(1)
A sequenceσnn∈N in A,η∗ converges to a point σ∈A if and only if η∗σ,σ=limn→∞η∗σ,σn.
(2)
A sequenceσnn∈N in A,η∗ is called a Cauchy sequence if limn,m→∞η∗σn,σm exists and is finite.
(3)
A
,
η
∗ is said to be complete if every Cauchy sequence σnn∈N in A converges, with respect to τη∗, to a point σ∈A such that η∗σ,σ=limn,m→∞η∗σn,σm.
The following lemma will be helpful in the sequel.Lemma 1 ([22, 22]).
(1) A dualistic partial metric spaceA,η∗ is complete iff the metric space A,dη∗s is complete.
(2) A sequenceσnn∈N in A converges to a point σ∈A, with respect to τdη∗s iff(5)limn→∞η∗σ,σn=η∗σ,σ=limn,m→∞η∗σn,σm.


## 3. Main Results

We begin with the following useful property.Definition 4.
LetT be a self-mapping on a dualistic partial metric space A,η∗. If there is a convergent sequence σn in A with σn→σ, such that(6)η∗σ,σ≤η∗Tσ,Tσ,
thenT is said to have the convergence comparison property [in short, (CCP)].Example 4.
LetA=Q. Define η∨∗:A×A→R by (7)η∨∗σ,ς=maxσ,ςforallσ,ς∈A.
Clearly,A,η∨∗ is a dualistic partial metric space. Consider σn=1/n−1,n≥1n∈N⊂A. We have limn→∞η∨∗σn,−1=η∨∗−1,−1. That is, σn→σ=−1 in A,η∨∗. Define Tu=eu for all u∈A. For such σ=−1, observe that η∨∗σ,σ=−1≤e−1=η∨∗Tσ,Tσ, so T has the convergence comparison property.
The following theorem corresponds to the unique fixed point result of Bianchini type dualistic contraction.Theorem 1.
LetT be a self-mapping on a complete dualistic partial metric space A,η∗ satisfying (CCP). If there is k∈0,1 so that(8)η∗Tσ,Tς≤kmaxη∗σ,Tσ,η∗ς,Tς,
for allσ,ς∈A, then T possesses a unique fixed point.Proof.
We generate a Picard iterative sequenceσnn∈N with an initial point σ0∈A such that σn=Tσn−1 for all n∈N. If there exists n0∈N such that σn0=σn0+1=Tσn0, then σn0 is a fixed point of T, so the proof is completed. From now on, assume that σn≠σn+1 for all n∈N, then by the contractive condition (8), we have(9)η∗σn,σn+1=η∗Tσn−1,Tσn≤kmaxη∗σn−1,Tσn−1,η∗σn,Tσn=kmaxη∗σn−1,σn,η∗σn,σn+1.
Thus,(10)η∗σn,σn+1≤kmaxη∗σn−1,σn,η∗σn,σn+1.
If(11)maxη∗σn−1,σn,η∗σn,σn+1=η∗σn,σn+1,
then (10) leads to a contradiction. As a result, we have(12)η∗σn,σn+1≤kη∗σn−1,σn,(13)η∗σn−1,σn=η∗Tσn−2,Tσn−1≤kmaxη∗σn−2,Tσn−2,η∗σn−1,Tσn−1=kmaxη∗σn−2,σn−1,η∗σn−1,σn.
Arguing like above, we have(14)maxη∗σn−2,σn−1,η∗σn−1,σn=η∗σn−2,σn−1.
Thus, inequality (12) entails(15)η∗σn,σn+1≤k2η∗σn−2,σn−1.
Proceeding further in a similar way, we get(16)η∗σn,σn+1≤knη∗σ0,σ1.
Now consider(17)η∗σn,σn=η∗Tσn−1,Tσn−1≤kmaxη∗σn−1,Tσn−1,η∗σn−1,Tσn−1=kmaxη∗σn−1,σn,η∗σn−1,σn=kη∗σn−1,σn,
the inequality (16) implies(18)η∗σn,σn≤knη∗σ0,σ1.
We deduce from (1) that(19)dη∗σn,σn+1+η∗σn,σn≤η∗σn,σn+1,
and using inequalities (16) and (18), we obtain(20)dη∗σn,σn+1≤2knη∗σ0,σ1.
Now, form>n, we have(21)dη∗σn,σm≤dη∗σn,σn+1+dη∗σn+1,σn+2+⋯+dη∗σm−1,σm≤2knη∗σ0,σ1+2kn+1η∗σ0,σ1+⋯+2km−1η∗σ0,σ1≤2kn+kn+1+⋯+km−1η∗σ0,σ1≤2kn1−km−n1−kη∗σ0,σ1(22)dη∗σm,σn≤dη∗σm,σm−1+dη∗σm−1,σm−2+⋯+dη∗σn+1,σn≤2km−1η∗σ0,σ1+2km−2η∗σ0,σ1+⋯+2knη∗σ0,σ1≤2kn+kn+1+⋯+km−1η∗σ0,σ1≤2kn1−km−n1−kη∗σ0,σ1.
Asm,n→∞, dη∗sσn,σm=maxdη∗σn,σm,dη∗σm,σn→0, thus, σn is a Cauchy sequence in A,dη∗s. Since A,η∗ is a complete dualistic partial metric space, by Lemma 1 (1), A,dη∗sis a complete metric space. Thus, there exists σ∗∈A,dη∗s such that σn→σ∗ as n→∞, that is limn→∞dη∗σn,σ∗=0 and by Lemma 1 (2), we know that(23)η∗σ∗,σ∗=limn→∞η∗σn,σ∗=limn,m→∞η∗σn,σm.
Since,limm,n→∞dη∗σm,σn=0, by (1) and (18), we have(24)η∗σ∗,σ∗=limn→∞η∗σn,σ∗=limn,m→∞η∗σn,σm=0.
This shows thatσn is a Cauchy sequence converging to σ∗∈A,η∗. We shall show that σ∗ is a fixed point of T. By (8), we have(25)η∗σn,Tσ∗=η∗Tσn−1,Tσ∗≤kmaxη∗σn−1,Tσn−1,η∗σ∗,Tσ∗,
asn→∞, η∗σ∗,Tσ∗≤kη∗σ∗,Tσ∗ which implies that η∗σ∗,Tσ∗=0. Since T has (CCP), we get(26)0=η∗σ∗,σ∗≤η∗Tσ∗,Tσ∗.
On the other hand, by axiomη4∗, we have η∗σ∗,σ∗≤η∗σ∗,Tσ∗+η∗Tσ∗,σ∗−η∗Tσ∗,Tσ∗ which implies that(27)η∗Tσ∗,Tσ∗≤0.
The inequalities (26) and (27) imply that η∗Tσ∗,Tσ∗=0. Thus,(28)η∗σ∗,Tσ∗=η∗Tσ∗,Tσ∗=η∗σ∗,σ∗.
By using axiomη1∗, we have σ∗=Tσ∗. This shows that σ∗ is a fixed point of T. To prove the uniqueness, suppose that y∗ is another fixed point of T, then Ty∗=y∗ and η∗y∗,y∗=0. By (8), we obtain(29)η∗σ∗,y∗=η∗Tσ∗,Ty∗≤kmaxη∗σ∗,Tσ∗,η∗y∗,Ty∗=0.
This implies thatη∗σ∗,y∗=η∗σ∗,σ∗=η∗y∗,y∗=0 and using axiom η1∗, we have σ∗=y∗, which proves the uniqueness of σ∗.Corollary 1.
LetA,η be a complete partial metric space and let T:A→A be a mapping satisfying (30)ηTσ,Tς≤kmaxησ,Tσ,ης,Tς,
for allσ,ς∈A and 0≤k<1. Then, T has a unique fixed point.Proof.
Since the restriction of a dualistic partial metricη∗ to 0,∞, which is η∗0,∞=η, is a partial metric, so arguments follow the same lines as in the proof of Theorem 1.Corollary 2 ([26]).
LetA,d be a complete metric space and let T:A→A be a mapping satisfying(31)dTσ,Tς≤kmaxdσ,Tσ,dς,Tς,
for allσ,ς∈A and 0≤k<1. Then, T has a unique fixed point.Proof.
Setησ,σ=0 for all σ∈A, in Corollary 1.Example 5.
LetA=−∞,0. Define the mapping η∗:A×A→R by(32)η∗σ,ς=σ−ςifσ≠ς,σ∨ςifσ=ς,
then,A,η∗ is a complete dualistic partial metric space. Define a mapping T:A→A by(33)Tσ=−1ifσ∈−∞,−4,0ifσ∈−4,0.Note that the mappingT has the convergence comparison property (CCP). Indeed, consider σn=−1/n−2n=1,2,…. Here σn is a sequence in A. Clearly, we have that σn→σ=−2 as n→∞ in A,η∗. We have η∗−2,−2=−2≤0=η∗T−2,T−2. Choose k=3/4. On the other hand, the contractive condition (8) is also satisfied. For this, let σ,ς∈A. Here, we need the following cases:Case 1.
(a)
σ
≠
ς and σ,ς∈−4,0. We have
(34)
η
∗
T
σ
,
T
ς
=
0
.
(b)
σ
≠
ς and σ,ς∈−∞,−4. We have
(35)
η
∗
T
σ
,
T
ς
=
1
≤
3
k
≤
k
max
η
∗
σ
,
T
σ
,
η
∗
ς
,
T
ς
.
(c)
σ
≠
ς and σ∈−∞,−4, ς∈−4,0 and vice versa. In this case,
(36)
η
∗
T
σ
,
T
ς
=
1
≤
3
k
≤
k
max
η
∗
σ
,
T
σ
,
η
∗
ς
,
T
ς
.Case 2.
(a)
σ
=
ς
∈
−
4
,
0. We have
(37)
η
∗
T
σ
,
T
ς
=
0
.
(b)
σ
=
ς
∈
−
∞
,
−
4. We have
(38)
η
∗
T
σ
,
T
ς
=
1
≤
3
k
≤
k
max
η
∗
σ
,
T
σ
,
η
∗
ς
,
T
ς
.Hence,T verifies all conditions of Theorem 1. Note that σ=0 is the unique fixed point of T.Example 6.
LetA=−∞,0. Define the mapping η∗:A×A→R by
(39)
η
∗
σ
,
ς
=
σ
+
ς
if
σ
≠
ς
,
σ
∨
ς
if
σ
=
ς
,
thenA,η∗ is a complete dualistic partial metric space. Define a mapping T:A→A as in Example 5. It is obvious that T verifies all conditions of Theorem 1. But, classical Bianchini contraction (contraction without the absolute value function) is not applicable. Indeed, for σ=−3 and ς=−2, we have
(40)
η
∗
T
σ
,
T
ς
=
η
∗
0
,
0
=
0
>
−
2
k
=
k
max
η
∗
σ
,
T
σ
,
η
∗
ς
,
T
ς
for eachk∈0,1.
In what follows, we investigate a unique fixed point of Reich type dualistic contraction mappings.Theorem 2.
LetA,η∗ be a complete dualistic partial metric space and let T:A→A be a mapping satisfying (CCP). If there exist a,b,c≥0 with a+b+c<1 such that (41)η∗Tσ,Tς≤aη∗σ,y+bη∗σ,Tσ+cη∗ς,Tς
for allσ,ς∈A, then, T has a fixed point.Proof.
Letσnn∈N be a sequence with an initial point σ0∈X such that σn=Tσn−1 for all n∈N. If there exists n0∈N such that σn0=σn0+1=Tσn0, then σn0 is a fixed point of T. On the other hand, if σn≠σn+1 for all n∈N, then by contractive condition (41), we have(42)η∗σn,σn+1=η∗Tσn−1,Tσn≤aη∗σn−1,σn+bη∗σn−1,Tσn−1+cη∗σn,Tσn=aη∗σn−1,σn+bη∗σn−1,σn+cη∗σn,σn+1,(43)η∗σn,σn+1≤a+b1−cη∗σn−1,σn.
Letk=a+b/1−c, then 0≤k<1 and repeating arguments given above, we have(44)η∗σn,σn+1≤knη∗σ0,σ1.
Now, consider the self-distance(45)η∗σn,σn=η∗Tσn−1,Tσn−1≤aη∗σn−1,σn−1+bη∗σn−1,Tσn−1+cη∗σn−1,Tσn−1=aη∗σn−1,σn−1+bη∗σn−1,σn+cη∗σn−1,σn=aη∗σn−1,σn−1+b+cη∗σn−1,σn.
Due to inequality (44), we have(46)η∗σn,σn≤aη∗σn−1,σn−1+b+ckn−1η∗σ0,σ1,(47)η∗σn−1,σn−1=η∗Tσn−2,Tσn−2≤aη∗σn−2,σn−2+bη∗σn−2,Tσn−2+cη∗σn−2,Tσn−2≤aη∗σn−2,σn−2+b+cη∗σn−2,σn−1≤aη∗σn−2,σn−2+b+ckn−2η∗σ0,σ1.
The inequality (46) implies that(48)η∗σn,σn≤a2η∗σn−2,σn−2+ab+ckn−2η∗σ0,σ1+b+ckn−1η∗σ0,σ1≤a3η∗σn−3,σn−3+a2kn−3+akn−2+kn−1b+cη∗σ0,σ1.
Continuing further, we get(49)η∗σn,σn≤anη∗σ0,σ0+an−1+an−2k+an−3k2+⋯+akn−2+kn−1b+cη∗σ0,σ1,(50)η∗σn,σn≤anη∗σ0,σ0+kn−ank−ab+cη∗σ0,σ1.
The equation (1) implies that(51)dη∗σn,σn+1≤η∗σn,σn+1−η∗σn,σn≤η∗σn,σn+1+η∗σn,σn≤knη∗σ0,σ1+anη∗σ0,σ0+kn−ank−ab+cη∗σ0,σ1≤kn+b+ckn−ank−aη∗σ0,σ1+anη∗σ0,σ0≤kn+kn−ank−aη∗σ0,σ1+anη∗σ0,σ0=knk−a+kn−ank−aη∗σ0,σ1+anη∗σ0,σ0=kn+kn−1+akn−2+⋯+an−1η∗σ0,σ1+anη∗σ0,σ0.
Letρn=kn+kn−1+akn−2+⋯+an−1. Then,(52)dη∗σn,σn+1≤ρnη∗σ0,σ1+anη∗σ0,σ0.
Now, form>n, we have(53)dη∗σn,σm≤dη∗σn,σn+1+dη∗σn+1,σn+2+...+dη∗σm−1,σm≤ρnη∗σ0,σ1+anη∗σ0,σ0+ρn+1η∗σ0,σ1+an+1η∗σ0,σ0+⋯+ρm−1η∗σ0,σ1+am−1η∗σ0,σ0=ρn+ρn+1+⋯+ρn+1η∗σ0,σ1+an+an+1+⋯+am−1η∗σ0,σ0≤ρn+ρn+1+⋯+ρn+1+⋯η∗σ0,σ1+an+an+1+⋯+am−1+⋯η∗σ0,σ0=ρn1−ρη∗σ0,σ1+an1−aη∗σ0,σ0.
We conclude thatlimm,n→∞dη∗sσn,σm=0. This shows that σn is a Cauchy sequence in A,dη∗s. Since A,η∗ is a complete dualistic partial metric space, by Lemma 1 (1), A,dη∗sis a complete metric space. Thus, there exists j∗∈A,dη∗s such that σn→j∗ as n→∞, that is limn→∞dη∗σn,j∗=0 and by Lemma 1 (2), we know that(54)η∗j∗,j∗=limn→∞η∗σn,j∗=limn,m→∞η∗σn,σm.
Sincelimm,n→∞dη∗σm,σn=0, by (1) and (50), we have(55)η∗j∗,j∗=limn→∞η∗σn,j∗=limn,m→∞η∗σn,σm=0.
This shows thatσn is a Cauchy sequence converging to j∗∈A,η∗. We show that j∗ is a fixed point of T. By (41), we have(56)η∗σn,Tj∗=η∗Tσn−1,Tj∗≤aη∗σn−1,j∗+bη∗σn−1,Tσn−1+cη∗j∗,Tj∗.
Applying limit asn→∞ and using equation (55), we have η∗j∗,Tj∗=0. Since T has (CCP), we get(57)0=η∗j∗,j∗≤η∗Tj∗,Tj∗.
By axiomη2∗, we have(58)η∗Tj∗,Tj∗≤η∗j∗,Tj∗=0.
The inequalities (57) and (58) imply that η∗Tj∗,Tj∗=0. Thus,(59)η∗j∗,Tj∗=η∗Tj∗,Tj∗=η∗j∗,j∗.
By using axiomη1∗, we have j∗=Tj∗. This shows that j∗ is a fixed point of T.Example 7.
LetA=0,−1,−2,−3,−0.3,−0.1 and η∗:A×A→R be as in Example 5. Then A,η∗ is a complete dualistic partial metric space. Define the mapping T:A→A by (60)Tσ=0ifσ∈0,−1,−0.3,−0.1,−0.3ifσ=−2,−0.1ifσ=−3.
The mappingT satisfies (CCP) for any convergent sequence in A. Indeed, for a convergent sequence σn in A such that σn→σ, σ∈A due to completeness of A,η∗. Thus for every such σ we have η∗σ,σ≤η∗Tσ,Tσ. Observe that for each case (as shown in Table 1) there exist some a,b,c≥0 with a+b+c<1 (however, assuming different values in each case) for which the contractive condition (41) is satisfied.Table 1
Evaluation of contractive condition.
σ
,
ς
η
∗
T
σ
,
T
ς
a
η
∗
σ
,
y
+
b
η
∗
σ
,
T
σ
+
c
η
∗
ς
,
T
ς
(0,0)
0
0
(0,−1)
0
a
+
c
(0,−2)
0.3
2
a
+
1.7
c
(0,−3)
0.1
3
a
+
2.9
c
(0,−0.3)
0
0.3
a
+
c
(0,−0.1)
0
0.1
a
+
c
(−1,−2)
0.3
a
+
b
+
2
c
(−1,−3)
0.1
2
a
+
b
+
2.9
c
(−1,−0.1)
0
0.9
a
+
b
+
0.1
c
(−1,−0.3)
0
0.7
a
+
b
+
0.3
c
(−1,−1)
0
a
+
b
+
c
(−2,−2)
0.3
2
a
+
1.7
b
+
c
(−2,−3)
0.2
a
+
1.7
b
+
2.9
c
(−2,−0.3)
0.3
1.7
a
+
b
+
0.3
c
(−2,−0.1)
0.3
1.9
a
+
1.7
b
+
0.1
c
(−3,−3)
0.1
3
a
+
2.9
b
+
c
(−3,−0.1)
0.1
2.9
a
+
b
+
0.1
c
(−3,−0.3)
0.1
2.7
a
+
2.9
b
+
0.3
c
(−0.3,−0.3)
0
0.3
a
+
b
+
c
(−0.1,−0.1)
0
0.1
a
+
b
+
cExample 8.
LetA=0,−1,−2,−3,−0.3,−0.1 and η∗:A×A→R be as in Example 6. Then A,η∗ is a complete dualistic partial metric space. Define the mapping T:A→A as in Example 7. It is obvious that (41) holds for all σ,ς∈A, while, the classical Reich fixed point theorem is not applicable. Indeed, for σ=ς=−1, we have (61)η∗Tσ,Tς=0>−a−b−c=aη∗σ,ς+bη∗σ,Tσ+cη∨∗ς,Tς
for alla,b,c∈0,1.Corollary 3.
LetA,η be a complete partial metric space and let T:A→A be a mapping satisfying(62)ηTσ,Tς≤aησ,ς+bησ,Tσ+cης,Tς
for allσ,ς∈A and a,b,c≥0 such that a+b+c<1. Then, T has a fixed point.Proof.
Since the restriction of a dualistic partial metricη∗ to 0,∞, η∗0,∞=η, is a partial metric, so arguments follow the same lines as in the proof of Theorem 2.Corollary 4 ([12]).
LetA,d be a complete metric space and let T:A→A be a mapping satisfying(63)dTσ,Tς≤adσ,ς+bdσ,Tσ+cdς,Tς
for allσ,ς∈A and a,b,c≥0 such that a+b+c<1. Then, T has a fixed point.Proof.
Setησ,σ=0 for all σ∈A, in Corollary 3.The following theorem implies the uniqueness of the fixed point of a new Rational type dualistic contraction.Theorem 3.
LetA,η∗ be a complete dualistic partial metric space and let T:A→A be a mapping satisfying (CCP). If there exist a,b,c≥0 with a+b+c<1 such that(64)η∗Tσ,Tς≤aη∗σ,Tσ·η∗ς,Tςη∗σ,y+bη∗ς,Tς+cη∗σ,y,
for allσ,ς∈A. Then, T has a fixed point.Proof.
Let{σn}n∈N be a sequence with an initial point σ0∈A such that σn=Tσn−1 for all n∈N. If there exists n0∈N such that σn0=σn0+1=Tσn0, then σn0 is a fixed point of T. On the other hand, if σn≠σn+1 for all n∈N, then from contractive condition (64), we have(65)η∗σn,σn+1=η∗Tσn−1,Tσn≤aη∗σn−1,Tσn−1⋅η∗σn,Tσnη∗σn−1,σn+bη∗σn,Tσn+cη∗σn−1,σn=aη∗σn−1,σn⋅η∗σn,σn+1η∗σn−1,σn+bη∗σn,σn+1+cη∗σn−1,σn=a+bη∗σn,σn+1+cη∗σn−1,σnη∗σn,σn+1≤c1−a−bη∗σn−1,σn.
Letλ=c/1−a−b<1, so(66)η∗σn,σn+1≤λη∗σn−1,σn≤λ2η∗σn−2,σn−1⋯λnη∗σ0,σ1.
Now,(67)η∗σn,σn=η∗Tσn−1,Tσn−1≤aη∗σn−1,Tσn−1⋅η∗σn−1,Tσn−1η∗σn−1,σn−1+bη∗σn−1,Tσn−1+cη∗σn−1,σn−1=aη∗σn−1,σn⋅η∗σn−1,σnη∗σn−1,σn−1+bη∗σn−1,σn+cη∗σn−1,σn−1.
Asη∗σn−1,σn/η∗σn−1,σn−1≥1, we get(68)η∗σn,σn≤aη∗σn−1,σn+bη∗σn−1,σn+cη∗σn−1,σn−1=a+bη∗σn−1,σn+cη∗σn−1,σn−1=a+bλn−1η∗σ0,σ1+cη∗σn−1,σn−1.
Arguing like above, we have(69)η∗σn−1,σn−1≤a+bλn−2η∗σ0,σ1+cη∗σn−2,σn−2.
The inequality (68) leads to(70)η∗σn,σn≤a+bλn−1+cλn−2η∗σ0,σ1+c2η∗σn−2,σn−2.
Continuing in the same way, we get(71)η∗σn,σn≤a+bλn−1+cλn−2+c2λn−3+⋯+cn−1η∗σ0,σ1+cnη∗σ0,σ0,(72)η∗σn,σn≤a+bcn−1−λn−1c−λη∗σ0,σ1+cnη∗σ0,σ0.
The equation (1) implies(73)dη∗σn,σn+1≤η∗σn,σn+1+η∗σn,σn≤λnη∗σ0,σ1+a+bλn−1−cn−1λ−cη∗σ0,σ1+cnη∗σ0,σ0≤λn+λn−2+cλn−3+⋯+cn−2η∗σ0,σ1+cnη∗σ0,σ0.
Setλn+λn−2+cλn−3+⋯+cn−2=γn. Then, (74)dη∗σn,σn+1≤γnη∗σ0,σ1+cnη∗σ0,σ0.
Now, form>n, we have(75)dη∗σn,σm≤dη∗σn,σn+1+dη∗σn+1,σn+2+...+dη∗σm−1,σm≤γnη∗σ0,σ1+cnη∗σ0,σ0+γn+1η∗σ0,σ1+cn+1η∗σ0,σ0+⋯+γm−1η∗σ0,σ1+cm−1η∗σ0,σ0≤γn+γn+1+⋯+γn+1η∗σ0,σ1+cn+cn+1+⋯+cm−1η∗σ0,σ1.
We deduce thatlimn,m→∞dη∗σn,σm=0 and similarly, limn,m→∞dη∗σm,σn=0 which implies that limn,m→∞dη∗sσn,σm=0. The remaining part of this proof is similar to the proof of Theorem 2.Corollary 5.
LetA,η be a complete partial metric space and let T:A→A be a mapping satisfying(76)ηTσ,Tς≤aησ,Tσ·ης,Tςησ,y+bης,Tς+cησ,y,
for allσ,ς∈A and a,b,c are non negative numbers such that a+b+c<1. Then, T has a fixed point.Proof.
Since the restriction of a dualistic partial metricη∗ to 0,∞, η∗0,∞=η is a partial metric, the result is obvious.Corollary 6.
LetA,d be a complete metric space and let T:A→A be a mapping satisfying(77)dTσ,Tς≤adσ,Tσ·dς,Tςdσ,ς+bdς,Tς+cdσ,ς
for allσ,ς∈A and a,b,c are non negative numbers such that a+b+c<1. Then, T has a fixed point.Proof.
Take in Corollary5, ησ,σ=0 for all σ∈A.


## 4. Application

In this section, we present an application of Theorem1 to ensure the existence of the solution of the boundary value problem given by(78)−d2σdt2=ft,σt,t∈0,1,σ0=σ1=0,wheref:0,1×R→R is a continuous mapping. The Green function associated to the boundary value problem (78) is defined by(79)Vt,τ=t1−τ,0≤t≤τ≤1,τ1−t,0≤τ≤t≤1.LetC0,1 be the space of all continuous mappings defined on 0,1. Let A=C0,1,R. Define the mapping η∗:A×A→R by(80)η∗σ,ς=σ−ς∞+L=supt∈0,1σt−ςt+L,L∈R.It is known thatA,η∗ is a complete dualistic partial metric space. Define T:A→A by(81)Tσt=∫01Vt,τfτ,στdτfor allt∈0,1. Note that the problem (78) has a solution iff the operator T has a fixed point.Theorem 4.
LetA=C0,1,R. Define the mapping T:A→A by(82)Tσt=∫01Vt,τfτ,στdτ,
wheref:0,1×A→R is a continuous mapping and σt is such that η∗σt,σt≤η∗Tσt,Tσt. Assume that(i)
The mappingf:0,1×A→R satisfies
(83)
f
t
,
σ
−
f
t
,
ς
≤
8
ln
e
k
ω
ρ
for allt∈0,1, σ,ς∈A, ω=maxη∗σ,Tσ,η∗ς,Tς and ρ>0;
Then, boundary value problem (78) has a solution.Proof.
We note thatqt∈C20,1,R (say) is a solution of (78) if and only if qt∈A is a solution of the integral equation (82). The solution of (82) is given by the fixed point of T, i.e., qt=Tqt.
Letσ,ς∈A and t∈0,1, by assumption i, we get(84)Tσt−Tςt=∫01Vt,τfτ,στ−fτ,ςτdτ≤∫01Vt,τfτ,στ−fτ,ςτdτ≤8∫01Vt,τlnekωρdτ≤8∫01Vt,τlnekωρdτ=8lnekωρsupt∈0,1∫01Vt,τdτ.
Since∫01Vt,τdτ=−t2/2+t/2 for all t∈0,1, we have supt∈0,1∫01Vt,τdτ=1/8, which implies that(85)η∗Tσ,Tς=supt∈0,1Tσt−Tςt+L≤kω,whereL=lnρ=kmaxη∗σ,Tσ,η∗ς,Tς.
Hence, application of Theorem1 ensures that T has at least one fixed point qt∈A, that is, Tqt=qt which is a solution of (82).



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*Source: 6428671-2020-01-22.xml*