# Some Fixed Point Theorems for Boyd and Wong Type Contraction Mapping in Ordered Partial Metric Spaces with an Application

Authors: Santosh Kumar; Faustine Nziku
Journal: Journal of Function Spaces
 (2022)

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## Abstract

This paper establishes and proves a fixed point theorem for Boyd and Wong type contraction in ordered partial metric spaces. In doing so, we have extended several existing results into ordered complete partial metric spaces. An illustrative example is given to demonstrate the validity of our results. Finally, the existence of the solution of nonlinear integral equation is discussed as an application of the main result.


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## Body

## 1. Introduction and Preliminaries

Banach’s fixed point theorem [1] has been extensively studied to solve the problems in nonlinear analysis since many years. This theorem provides the existence and uniqueness of the solution. It states that, if M,ϱ is a complete metric space and T:M⟶M is a self-contractive mapping, then T has a unique fixed point u∈M. Due to its usefulness and applications, this theorem has been massively investigated and generalized by different researchers. In 1969, Boyd and Wong [2] gave an important generalization of the Banach fixed point theorem by the application of control function in the Banach contraction condition. Boyd and Wong [2] took into account the condition as follows:(1)ϱTu,Tv≤χϱu,v,∀u,v∈M,whereby M,ϱ is a complete metric space, and a mapping χ:0,∞⟶0,∞ is upper semicontinuous from the right on 0,∞ such that χt<t, ∀t>0. Consequently, T has a unique fixed point z∈M and ϱTnu,z⟶0 as n⟶∞, ∀u∈M.Imdad and Kumar [3] extended the existing results by relaxing “continuity” and lightening the “commutativity” requirement besides increasing the number of involved maps from “two” to “four.” Several other researchers extended these results in different directions. Some of them are [4–6] and the references therein. A beautiful survey of the fixed point theory was given by Kumar [7]. Naziku and Kumar [8, 9] proved results using the Boyd and Wong type contractive condition. Recently, Kumar [10] established and proved a fixed point theorem for Boyd and Wong type contraction for a pair of maps in complete metric spaces. This theorem gives conditions for a pair of mappings that possess a fixed point but not continuous at the fixed point and can be applied for both continuous and discontinuous mappings.In the last few decades, fixed point results in a partially ordered set have been revealed as a very important area of interest to many researchers. In particular, the existence of the fixed point in partially ordered sets has been massively considered in [11–16] and others as they appear in the literature.In literature [17], Matthews introduced the study of partial metric spaces as the important subject in the approach of formalizing the meaning of programming languages by formulating mathematical objects called “denotations.” Partial metric was introduced to ensure that partial order semantics should have a metric-based tools for program verification in which the notion of size of data object in a domain is used in quantifying how data object is well defined in the domain.Definition 1 (see [18]).
LetM be a nonempty set. A function p:M×M⟶0,∞ is called a partial metric on M if it satisfies the followings:(PM0):0≤pu,u≤pu,v (non-negativity and small self distance).(PM1):pu,v=pu,u=pv,v⇒u=v (indistancy implies equality).(PM2):pu,v=pv,u (symmetric).(PM3):pu,v+pz,z≤pu,z+pz,v (triangularity), for all u,v,z∈M.
M,p is called a partial metric space.Note thatpu,v=0 implies u=v (by PM0 through PM2); the converse is always not true. Therefore, a metric space is a partial metric space with all self distances zero.Several researchers generalized the results of metric fixed point theory using partial metric space setting in different directions. Some of them are [19–22] and the references therein.Now, we will recall some definitions and lemmas which will be utilized in the proof of main results of this paper.Definition 2 (see [18]).
Letun be a sequence in a partial metric space M,p; then,(i)
A sequenceun⟶u∈M if and only if pu,u=limn⟶∞pu,un=limn⟶∞pun,un.(ii)
A sequenceun is called a Cauchy sequence if there exists ϵ>0 such that for all n,m>N, we have pun,um<ϵ for some integers N≥0; that is limn,m⟶+∞pun,um exists and it is finite.(iii)
A partial metric spaceM,p is complete if every Cauchy sequence un converges to a point u∈M such that pu,u=limn,m⟶+∞pun,um.Definition 3 (see [18]).
A contraction on a partial metric spaceM is a function f:M⟶M such that there exist a constant 0≤k<1 for all u,u∈M satisfies that(2)pfu,fv≤k×pu,v.Definition 4 (see [17]).
LetM be a nonempty set. Partial ordering is a relation ≪⊆M2 such that(PO 1) for allu∈M, u≪u (reflexive).(PO 2) for allu,v∈M, u≪v, and v≪u⇒u=v (antisymmetric).(PO 3) for allu,v,z∈M, u≪v, and v≪z⇒u≪z (transitivity).Definition 5 (see [17]).
For each partial metric spacep:M×M⟶0,∞, ≪p⊆M2 is a binary relation such that for all u,v∈M, u≪pv⇔pu,u=pu,v.
Note that for each partial metric spacep, ≪p is a partial ordering.Definition 6 (see [23]).
LetM,ϱ be a metric space. Let a mapping T:M⟶M to be injective (one to one) and continuous (ICS) mapping with the property that if Tun is convergent, then the sequence un is also convergent for all sequences un∈M.Definition 7 (see [23]).
LetΦ be the set of functions ϕ:0,∞⟶0,∞ satisfying,(i)
ϕt<t for all t>0.(ii)
ϕ is an upper semicontinuous from right; that is, for any sequence tn∈0,∞ such that tn⟶t as n⟶∞ as tn>t, we have limsupn⟶∞ϕtn≤ϕt.
Aydi and Karapinar [24] generalized results of Harjani et al. [11] and Luong and Thun [13] by using an ICS mapping and involved Boyd and Wong type contractive condition and provided the following theorem:Theorem 1 (see [23]).
LetM,≪ be a partially ordered set. Suppose there exists a metric ϱ such that M,ϱ is a complete metric space. Let f,T:M⟶M be a mapping such that T is an ICS mapping and f is a nondecreasing mapping satisfying,
ϱTfu,Tfv≤ϕNu,v for all u,v∈M with u≤v where ϕ∈Φ and(3)Nu,v=maxϱTu,TfuϱTv,TfvϱTu,Tv,ϱTu,Tv.Also assume that either(i)
f is continuous, or(ii)
If the sequenceun is nondecreasing in M such that un⟶u, then u=supun.If there exists a pointu0∈M such that u0≤fu0, then f has a unique fixed point.In the next section, the letterℕ will be used to refer to the set of all positive integer numbers.


## 2. Main Results

We now present an extension of Definition6 in partial metric spaces.Definition 8.
LetM,p be a partial metric space. Let a mapping T:M⟶M to be injective (one to one) and continuous (ICS) mapping with the property that if Tun is convergent, then un is also convergent for all sequences un∈M.
Corresponding to Theorem1, we state and prove our main results and then provide an illustrative example to demonstrate our results.Theorem 2.
LetM,≤ be a partially ordered set (Poset). Let p be a partial metric such that M,p is a complete partial metric space. Also, let T:M⟶M be an ICS mapping, and f:M⟶M be a nondecreasing mapping satisfying(4)pTfu,Tfv≤ϕλu,v,with u≤v for all distinct u,v∈m, where ϕ∈Φ and(5)λu,v=maxpTu,TfupTv,TfvpTu,Tv,pTu,Tv.Furthermore, we assume that either(i)
a mappingf is continuous, or(ii)
If a sequenceun∈M is a nondecreasing sequence such that un⟶u as n⟶∞, then u=supnun.Therefore, if there existsu0∈M such that u0≤fu0, then f has a unique fixed point u0 with pu0,u0=0.Proof. Let a pointu0∈M such that u1=fu0. We define a sequence un∈M as(6)un=fun−1,for all integers n≥1. Since f is a nondecreasing mapping and M,≤ is a Poset, then we can have(7)u0≤u1=fu0≤u2=fu1≤⋯≤un−1≤un=fun−1≤⋯,for all integers n≥1. By induction, we can have that(8)u0≤u1≤u2≤⋯≤un−1≤un≤un+1≤⋯.If we suppose that there existsn∈ℕ such that un=un+1=fun, then f has a fixed point un, which ends the proof.Now suppose thatun≠un+1 for all integers n∈ℕ, then (8) becomes(9)u0<u1<u2<⋯<un−1<un<un+1<⋯.From (4), we can have that(10)pTun,Tun+1=pTfun−1,Tfun≤ϕλun−1,un,where(11)λun−1,un=maxpTun−1,Tfun−1pTun,TfunpTun−1,Tun,pTun−1,Tun=maxpTun,Tun+1,pTun−1,Tun.Suppose thatλun−1,un=pTun,Tun+1 for some integers n≥1, then (10) becomes,(12)pTun,Tun+1≤ϕpTun,Tun+1.From Definition7 (i), we see that (12) becomes(13)pTun,Tun+1<pTun,Tun+1,which is a contradiction. Hence, λun−1,un=pTun−1,Tun for all integers n≥1; hence (10) becomes(14)pTun,Tun+1=pTfun−1,Tfun≤ϕλun−1,un=ϕpTun−1,Tun<pTun−1,Tun.Therefore, from the above equation, we can observe that the sequencepTun−1,Tun is a decreasing sequence, and it is bounded below.Letcn=pTun−1,Tun, for all integers n≥1. Therefore, there exists a real number c≥0 such that limn⟶∞cn=c.We claim thatc=0. In contrary, we suppose that c>0, then by the semicontinuity property of ϕ and considering (10) above, we can have that(15)0<c=limsupn⟶∞pTun,Tun+1=limsupn⟶∞pTfun−1,Tfun≤limsupn⟶∞ϕλun−1,un=limsupn⟶∞ϕpTun−1,Tun=ϕc<c,which is a contradiction. Hence,(16)limn⟶∞pTun−1,Tun=c=0.Now, we need to prove that the sequenceTun∈X is a Cauchy sequence. For the sake of contradiction, we suppose that ∃ϵ>0 and the sequence of integers nr,mr≥r for some r≥0 such that(17)pTumr,Tunr≥ϵ.Furthermore, suppose thatnr is chosen as the smallest integer such that (17) above holds so that we can have(18)pTumr,Tunr−1<ϵ.Thus, by triangle inequality, we obtain(19)ϵ≤pTumr,Tunr≤pTumr,Tunr−1+pTunr−1,Tunr−pTunr−1,Tunr−1≤pTumr,Tunr−1+pTunr−1,Tunr.Asr⟶∞ in (19) above and considering (16), we obtain(20)ϵ≤limr⟶∞pTumr,Tunr<ϵ.Thus,(21)limr⟶∞pTumr,Tunr=ϵ.Similarly, we have(22)ϵ≤pTumr−1,Tunr−1≤pTumr−1,Tumr+pTumr,Tunr−1−pTumr,Tumr≤pTumr−1,Tumr+pTumr,Tunr−1.Also, asr⟶∞ in (22) and again considering (26), we obtain(23)ϵ≤limr⟶∞pTumr−1,Tunr−1<ϵ.Thus,(24)limr⟶∞pTumr−1,Tunr−1=ϵ.Now from (1) and for all positive integers nr>mr, we obtain(25)pTumr,Tunr=pTfumr−1,Tfunr−1≤ϕλumr−1,unr−1,where(26)λumr−1,unr−1=maxpTumr−1,Tfumr−1pTunr−1,Tfunr−1pTumr−1,Tunr−1,pTumr−1,Tunr−1=maxpTumr−1,TumrpTunr−1,TunrpTumr−1,Tunr−1,pTumr−1,Tunr−1.Asr⟶∞ in (25) and in (26) and by considering (16), (21), and (24), then (25) becomes(27)ϵ≤ϕmax0,ϵ=ϕϵ<ϵ,which is a contradiction. Hence, the sequence Tun∈M is a Cauchy sequence.SinceM,p is a complete partial metric space; therefore, there exists a point u0∈M such that a sequence Tun converges to a point u0.Given thatT is an ICS mapping, and a sequence Tun converges, then there exists a point u∈M such that(28)u=limn⟶∞un.Furthermore, sinceT is also continuous, then(29)Tu=limn⟶∞Tun=u0.Now, we need to prove thatu is a fixed point for a mapping f.(i)
We suppose that the first assumption of Theorem2 holds; that is, f is a continuous mapping. Therefore,(30)u=limn⟶∞un=limn⟶∞fun−1=flimn⟶∞un−1=fu.Hence,u is a fixed point for a mapping f.(ii)
Now, we suppose that the second assumption of Theorem2 holds. Given that a sequence un is a nondecreasing sequence such that un⟶u as n⟶∞ and supnun=u, then for all integers n≥0, we have that un≤u.Sincef is a nondecreasing mapping, consequently we obtain that fun≤fu, And respectively we have(31)un≤un+1≤⋯≤fu,for all integers n≥0.Sinceun⟶u and supnun=u as n⟶∞; hence,(32)u≤fu.Now, we construct a new sequencevn∈M such that vn⟶v as n⟶∞ and supnvn=v which is defined as follows:(33)v0=u,vn=fvn−1,∀n∈ℕ.Since the mappingT is continuous, then(34)limn⟶∞Tvn=Tv.Consequently,v0≤fv0 since u≤fu and u=v0. Similar to the above discussion, we can conclude that a sequence Tvn is a Cauchy sequence.Sincesupnvn=v, then vn≤v. Therefore, from (32), we obtain(35)u=v0≤fu=fv0≤v2≤v3≤⋯≤vn≤vn+1≤⋯≤v,for all integers n≥0. If we suppose u=v, then(36)u=u0≤fu=fu0≤u2≤⋯≤un≤un+1≤⋯≤u,and hence, u≤fu≤u, which ends the proof that is fu=u.Otherwise, suppose thatu≠v, then Tu≠Tv since T is an injective map. Therefore, pTu,Tv>0.From (1), we have(37)pTun+1,Tvn+1=pTfun,Tfvn≤ϕλun,vn,where(38)λun,vn=maxpTun,TfunpTvn,TfvnpTun,Tvn,pTun,Tvn=maxpTun,Tun+1pTvn,Tvn+1pTun,Tvn,pTun,Tvn.Considering (16), (29), and (34) and by letting n⟶∞ in (37) and (38), then (37) becomes(39)pTu,Tv≤ϕpTu,Tv<pTu,Tv,which is a contradiction; thus u=v, and therefore we have u≤fu≤u; hence u=fu.Therefore,u is a unique fixed point of the mapping f.Remark 1.
If we letϕt=kt for all t∈0,∞ and k∈0,1 in Theorem 2, we get the following corollary:Corollary 1.
LetM,≤,M,p,T,andf be the same as in Theorem 2 such that(40)pTfu,Tfv≤kλu,v,with u≤v for all distinct u,v∈M, and λu,v is defined as in Theorem 2.Also assume either(i)
f is continuous, or(ii)
If a sequenceun∈M is a nondecreasing sequence such that un⟶u as n⟶∞, then u=supnun.Therefore, if there exists a pointu0∈M such that u0≤fu0, then f has a unique fixed point.Remark 2.
If we takek=a+b in Corollary 1 with a,b∈0,1 such that a+b<1, we get the following corollary:Corollary 2.
LetM,≤,M,p,T,andf be the same as in Theorem 2 such that(41)pTfu,Tfv≤apTu,TfupTv,TfvpTu,Tv+bpTu,Tv,for all u,v∈M with u≤v.Assume either(i)
f is continuous, or(ii)
If a sequenceun∈M is a nondecreasing sequence such that un⟶u as n⟶∞, then u=supnun.Therefore, if there exists a pointu0∈M such that u0≤fu0, then f has a unique fixed point.Remark 3.
If we take the mappingTu=u in Corollary 2, we get an extension to the work of Harjani et al. [11] in partial metric spaces which runs as follows:Corollary 3.
LetM,≤,M,p,andf be the same as in Theorem 2 such that(42)pfu,fv≤apu,fupv,fvpu,v+bpu,v,for all u,v∈M with u≤v.Assume either(i)
f is continuous, or(ii)
If a sequenceun∈M is a nondecreasing sequence such that un⟶u as n⟶∞, then u=supnun.Therefore, if there exists a pointu0∈M such that u0≤fu0, then f has a unique fixed point.Remark 4.
If we setϕt=kt and set T to be the identity mapping in Theorem 2, then (4), which is(43)pTfu,Tfv≤ϕλu,v,becomes(44)pfu,fv≤kλu,v,which leads to the extension of the work of Ran and Reurings [15] and Nieto and Rodriguez-Lopez [14] in partial metric spaces which runs as follows:Corollary 4.
LetM,≤,M,p,andf be the same as in Theorem 2 such that(45)pfu,fv≤kλu,v,for all u,v∈M with u≤v.Assume either(i)
f is continuous, or(ii)
If a sequenceun∈M is a nondecreasing sequence such that un⟶u as n⟶∞, then u=supnun.Therefore, if there exists a pointu0∈M such that u0≤fu0, then f has a unique fixed point.The following theorem results are from the additional assumption made on the assumptions to the hypotheses of Theorem2.Theorem 3.
LetM be a partially ordered set, and p be a partial metric on M such that M,p is a complete partial metric space. Also, let T:M⟶M be an ICS mapping, and f:M⟶M be a nondecreasing mapping satisfying(46)pTfu,Tfv≤ϕλu,v,with u≤v for all distinct u,v∈M, where ϕ∈Φ, and(47)λu,v=maxpTu,TfupTv,TfvpTu,Tv,pTu,Tv.Furthermore, assume that either(i)
A mappingf is continuous, or(ii)
If a sequenceun∈M is a nondecreasing sequence such that un⟶u as n⟶∞, then u=supnun(iii)
For everyu,v∈M, there exists z∈M, that is comparable to u and vTherefore, if there existsu0∈M such that u0≤fu0, then f has a unique fixed point u0.Proof. In contrary, suppose that a mappingf has two distinct fixed points, say u and v.From assumption (iii) of Theorem3, there exists z∈M which is comparable to u and v. Without the lose of generality, we choose z≤u.Now, we construct a sequencezn as follows:zn=fzn−1 for all n∈ℕ such that z=z0.Sincef is a nondecreasing mapping, then z≤u implies thatz1=fz0=fz≤fz1≤fz2≤⋯≤fu=u. Inductively, we obtain zn≤u.If we suppose that there existsn0∈ℕ such that u=zn0, then zn=fzn−1=fu=u, for all n≥n0−1.This implies thatlimn⟶∞zn=u. Similarly, limn⟶∞zn=v, which completes the proof.Now, if we suppose thatu≠zn, for all integers n≥0, then pTu,Tzn>0 since the mapping T is one to one.From (46), we have(48)pTu,Tzn=pTfu,Tfzn−1≤ϕλu,zn−1,where(49)λu,zn−1=maxpTu,TfupTzn−1,Tfzn−1pTu,Tzn−1,pTu,Tzn−1=maxpTu,TupTzn−1,TznpTu,Tzn−1,pTu,Tzn−1.Analogously to Theorem2, we have(50)λu,zn−1=pTu,Tzn−1.Thus, (48) becomes(51)pTu,Tzn=pTfu,Tfzn−1≤ϕpTu,Tzn−1<pTu,Tzn−1.We observe that the sequencepTu,Tzn−1 is the decreasing sequence which is bounded below. Therefore, there exists a constant c≥0 such that limn⟶∞pTu,Tzn−1=c. We claim that c=0. In contrary, suppose that c>0.From (51), we can have that(52)limsupn⟶∞pTu,Tzn≤limsupn⟶∞ϕpTu,Tzn−1<limsupn⟶∞pTu,Tzn−1,which leads to(53)0<c≤ϕc<c,which is a contradiction since ϕ is an upper semicontinuous mapping. Hence, c=0. Therefore, we conclude that(54)limn⟶∞pTu,Tzn=0.Similarly,(55)limn⟶∞pTv,Tzn=0.Sinceu and v are distinct fixed points of f and T is one to one mapping, then(56)0<pTu,Tv=pTfu,Tfv≤ϕλu,v<λu,v,where(57)λu,v=maxpTu,TfupTv,TfvpTu,Tv,pTu,Tv=maxpTu,TupTv,TvpTu,Tv,pTu,Tv=pTu,Tv.Therefore, from (56), we have(58)pTu,Tv<λu,v=pTu,Tv,which is a contradiction. Hence, u=v, which ends the proof.Example 1.
LetM=0,∞ be a set equipped with a partial metric pu,v=maxu,v for all u,v∈M. Let the order ≪p be defined by u≪pv⇔u=v, for all u,v.
It is easy to check thatM,≪p is a partially ordered set (Poset) by proving PO1 through PO3 from Definition 4.
Tu=1/2eu, for all u∈M, is defined. Also,(59)fu=12u2,u∈0,1,u−12,u∈1,∞.
is defined. It is easy to see that a mappingT is an ICS mapping as it is injective and continuous. Also, we observe that a function fu is continuous and nondecreasing f.
Now, we show that inequality (4) holds.
From left hand side of (4), for all u,v∈0,1, we have(60)pTfu,Tfv=maxTfu,Tfv=maxT12u2,T12v2=max12eu2/2,12ev2/2≤max12eu2,12ev2≤maxeu2,ev2.
From the right hand side of (4), we have(61)λu,v=maxpTu,TfupTv,TfvpTu,Tv,pTu,Tv,where(62)pTu,Tfu=maxTu,Tfu=maxeu2,Tu22=maxeu2,eu2/22=eu2.
Similarly,(63)pTv,Tfv=ev2,and pTu,Tv=maxTu,Tv=maxeu/2,ev/2.Case 1.
Suppose,maxeu/2,ev/2=eu/2.
Then,λu,v=eu/2. Hence, pTfu,Tfv≤eu/2=λu,v.Case 2.
Suppose,maxeu/2,ev/2=ev/2.
Then,λu,v=ev/2. Hence, pTfu,Tfv≤ev/2=λu,v.
From left hand side of (4), for all u,v∈1,∞, we have(64)pTfu,Tfv=maxTfu,Tfv=maxTu−12,Tv−12=max12eu−1/2,12ev−1/2≤max12eu,12ev.
From the right hand side of (4), we have(65)λu,v=maxpTu,TfupTv,TfvpTu,Tv,pTu,Tv,where(66)pTu,Tfu=maxTu,Tfu=maxeu2,Tu−12=maxeu2,eu−1/22=eu2.
Similarly,(67)pTv,Tfv=ev2,and pTu,Tv=maxTu,Tv=maxeu/2,ev/2.
In the similar way, we find thatpTfu,Tfv≤eu/2=λu,v.
Now, if we defineϕt=1/2t, for all t=λu,v∈0,∞, we obtain that pTfu,Tfv≤ϕλu,v. Therefore, (4) holds, and u0=0 is the fixed point of a mapping f.


## 3. The Existence Solution of Nonlinear Integral Equations

In this section, we studied the existence of solutions for nonlinear integral equations, as an application to the fixed point theorems proved in the previous section.The following integral equations were inspired by [25, 26]. An unknown function x is considered. Now, an application of Theorem 3 is presented as a study of the existence and uniqueness of solution to nonlinear integral equations.(68)ut=ht+∫abGt,s,Ks,usds,where K:a,b×ℝ⟶ℝ and G:a,b2⟶0,∞ are given continuous functions.LetX be the set Ca,b of real continuous function on a,b and let p:X×X⟶ℝ+ be given by(69)pu,v=maxa≤t≤bu,v.It is easy to see thatp is a partial metric and that X,p is a complete partial metric space.Next, we prove a theorem to establish the existence of a common fixed point for a pair of self mappings:Theorem 4.
Let us consider the integral equation (68) as above. Also, suppose that it satisfies the following conditions:(i)
fort,s∈a,b and u,v∈X, there exists a nondecreasing function ϕ∈Φ such that the following inequality holds:(70)Kt,s,us−Kt,s,vs≤ϕλu,v,where(71)λu,v=maxpTu,TfupTv,TfvpTu,Tv,pTu,Tv.(ii)
(72)maxa≤t≤b∫abGt,sds≤1b−a.Then, the integral equation (68) has a unique common solution u0∈X.Proof. LetT,f:X⟶X be a mapping defined by(73)Tfut=ht+∫abGt,sKt,s,usds.This implies that f∈T and f∈X possess a fixed point u0∈Tf. To prove the existence of the fixed point of Tf, we prove that Tf is a contraction. On contrary, we assume that Tfu≠Tfv, for all u,v∈a,b. Using conditions (i) and (ii) of Theorem 4, we have(74)pTfu,Tfv=maxa≤t≤bTfut,Tfvt,≤maxa≤t≤bht+∫abGt,sKt,s,us,ht+∫abGt,sKt,s,vsds,≤maxa≤t≤b∫abGt,sdsht+Kt,s,us,ht+Kt,s,vs,≤∫abGt,sdsmaxa≤t≤bht+Kt,s,us,ht+Kt,s,vs,≤1b−amaxa≤t≤bht+Kt,s,us,ht+Kt,s,vs,≤1b−amaxpTu,TfupTv,TfvpTu,Tv,pTu,Tv,≤1b−aϕλu,v.For b = 1, a = 0, we have(75)pTfu,Tfv≤ϕλu,v,which is a contradiction. Hence, u is a common fixed of T and f, also a solution to integral equation (74). Thus, Theorem 3 is satisfied.


## 4. Conclusion

The main contribution of this paper to fixed point theory is Definition8, Theorems 2, and 3. Here, the results have proved for Boyd and Wong type contraction in ordered partial metric spaces. Several existing results in the literature are generalized and extended into ordered complete partial metric spaces. Suitable examples are given to demonstrate the validity of the results. Finally, the existence of the solution of nonlinear integral equation is discussed as an application of the main result.



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