# Some New Coupled Coincidence Point and Coupled Fixed Point Results in Partially Ordered Metric-Like Spaces and an Application

Authors: Min Liang; Chuanxi Zhu; Zhaoqi Wu; Chunfang Chen
Journal: Journal of Function Spaces
 (2018)

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## Abstract

Some new coupled coincidence point and coupled fixed point theorems are established in partially ordered metric-like spaces, which generalize many results in corresponding literatures. An example is given to support our main results. As an application, we discuss the existence of the solutions for a class of nonlinear integral equations.


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## Body

## 1. Introduction and Preliminaries

As we all know, the fixed point theory is one of the most important tools in the field of nonlinear analysis. In particular, the coupled fixed point theorems in partially ordered metric-like spaces are very valuable for discussing the existence and uniqueness of solutions of any nonlinear problem in fields of mathematics and physics.In order to solve the more complex nonlinear analysis problems, the concept of a metric space has been extended in many aspects. In 1992, the notion of a partial metric space was introduced by Matthews [1]. Considering d(x,x) may be greater than zero, he tried to extend the concept of a metric space. Since then, many authors proved various fixed point theorems in partial metric spaces (see [2–9]). In the interest of studying the existence of solutions of ordinary differential equations, the mixed monotone mapping was established. Afterward, Lakshmikantham andĆirić [10] substituted the mixed g-monotone mapping for the mixed monotone mapping and proved the coupled common fixed point theorems. On the other hand, Hitzler and Seda [11] first studied the dislocated space in 2000, and then Amini-Harandi [12] named it as a metric-like space. Subsequently, some authors discussed the fixed point and coincidence point results in generalized metric spaces. And several applications to operator equations and integral equations are given in a line of research (see [13–19]). Recently, B. Hazarika et al. [20] generalized some previous results and gave some new common fixed point theorems in metric-like spaces.In this paper, inspired by the above literatures, we propose some new coupled coincidence point and coupled fixed point theorems in partially ordered metric-like spaces, which extend the theorems of B. Hazarika et al. [20]. As an application, we discuss the existence of solutions for a system of nonlinear integral equations to illustrate our main results.First, we review some concepts which are going to be used later.Throughout this paper, letR=(-∞,+∞), R+=(0,+∞), and R0+=[0,+∞). Let N be the set of all natural numbers and N+ be the set of all positive integer numbers.Definition 1 (see [12]).
LetX be a nonempty set. A function σ:X×X→R0+ is said to be a dislocated (metric-like) metric on X if, for all x,y,z∈X, the following conditions hold:
(σ1) σx,y=0⇒x=y.
(σ2) σx,y=σy,x.
(σ3) σx,y≤σx,z+σz,y.Definition 2 (see [8]).
Let(X,σ) be a metric-like space.
(a) A sequence {xn} in X is said to be a Cauchy sequence if limn,m→∞σ(xn,xm) exists and is finite.
(b) (X,σ) is said to be complete if every Cauchy sequence {xn} in X converges with respect to τσ to a point x∈X such that (1)limn→∞σxn,x=σx,x=limn,m→∞σxn,xm.Definition 3 (see [6]).
Letψ,ϕ:[0,∞)→[0,∞) be two functions. A pair of functions (ψ,ϕ) is said to belong to the class Ψ if they satisfy the following conditions:
(i) Foru,v∈[0,+∞) if ψ(u)≤ϕ(v), then u≤v.
(ii) Forun,vn∈[0,+∞) with limn→∞un=limn→∞vn=w, if ψ(un)≤ϕ(vn) for all n, then w=0.Definition 4 (see [7]).
LetF be the class of all functions F:[0,∞)3→[0,∞) satisfying the following conditions:
(F1) max{a,b}≤F(a,b,c) for all a,b,c≥0.
(F2) F(a,0,0)=a for all a≥0.
(F3) F is continuous.Definition 5 (see [9]).
Let(X,⪯) be a partially ordered set and T:X×X→X. Then the map T is said to have mixed monotone property if T(x,y) is monotone nondecreasing in x and is monotone nonincreasing in y; that is, (2)x1⪯x2impliesTx1,y⪯Tx2,yfor  ally∈X,y1⪯y2impliesTx,y2⪯Tx,y1forallx∈X.Definition 6 (see [10]).
Let(X,⪯) be a partially ordered set and T:X×X→X. Then the map T is said to have mixed g-monotone property if T(x,y) is monotone g-nondecreasing in x and is monotone g-nonincreasing in y; that is, (3)gx1⪯gx2impliesTx1,y⪯Tx2,yfor  ally∈X,gy1⪯gy2impliesTx,y2⪯Tx,y1for  allx∈X.Definition 7 (see [9]).
An element(x,y)∈X×X is called a coupled fixed point for the mapping T:X×X→X if T(x,y)=x and T(y,x)=y.Definition 8 (see [10]).
LetX be a nonempty set. We say that the mappings T:X×X→X and g:X→X are commutative if gT(x,y)=T(gx,gy), for all x,y∈X.Definition 9 (see [10]).
An element(x,y)∈X×X is called a coupled coincidence point of the mapping T:X×X→X and g:X→X if T(x,y)=gx and T(y,x)=gy.


## 2. Main Results

In this section, we put forward a new coupled coincidence point theorem in a partially ordered metric-like space, and ifg=I, which is a self-mapping, we get a new coupled fixed point theorem. Then, we establish a common coupled fixed point theorem in a partially ordered metric-like space and prove the uniqueness of the coupled fixed point.Theorem 10.
Let(X,⪯) be a partially ordered set and (X,σ) be a complete metric-like space. Let T:X×X→X and g:X→X be two mappings such that the following conditions are satisfied:
(i)T(X×X)⊆g(X).
(ii)g(X) is closed.
(iii)T has the mixed g-monotone property.
(iv) There existx0,y0∈X such that gx0⪯T(x0,y0) and T(y0,x0)⪯gy0.
(v)(4)ψFσTx,y,Tu,v,φTx,y,φTu,v≤ϕMx,u,y,v,where(5)Mx,u,y,v=max⁡Fσgx,gu,φgx,φgu,Fσgy,gv,φgy,φgv,Fσgx,Tx,y,φgx,φTx,y,Fσgy,Ty,x,φgy,φTy,x,with gx⪯gu and gv⪯gy, or gu⪯gx and gy⪯gv for all x,y,u,v∈X,F∈F,(ψ,ϕ)∈Ψ, ψ is nondecreasing, and φ:X→[0,+∞) is a lower semicontinuous function.
ThenT and g have a coupled coincidence point.Proof.
Letx0,y0∈X such that gx0⪯T(x0,y0) and T(y0,x0)⪯gy0. Since T(X×X)⊆g(X), we can find x1,y1∈X such that gx1=T(x0,y0) and gy1=T(y0,x0). Similarly, there exist x2,y2∈X such that gx2=T(x1,y1) and gy2=T(y1,x1). Repeating the above process, we can construct two sequences {xn},{yn} such that gxn+1=T(xn,yn) and gyn+1=T(yn,xn). Since T has the mixed g-monotone property, we get gx0⪯gx1⪯gx2⪯⋯⪯gxn⪯⋯ and gy0⪰gy1⪰gy2⪰⋯⪰gyn⪰⋯. If for some n∈N and we have (gxn+1,gyn+1)=(gxn,gyn), then T(xn,yn)=gxn+1=gxn,T(yn,xn)=gyn+1=gyn. This means T and g have a coupled coincidence point.
First, we show that ifT and g have a coupled coincidence point, i.e., T(x,y)=gx,T(y,x)=gy, then σ(gx,gx)=σ(T(x,y),T(x,y))=σ(gy,gy)=σ(T(y,x),T(y,x))=0. Indeed, we have gx⪯gx, and supposing σ(gx,gx)>0 and using (4), we obtain(6)ψFσgx,gx,φgx,φgx=ψFσTx,y,Tx,y,φTx,y,φTx,y≤ϕMx,x,y,y=ϕmaxFσgx,gx,φgx,φgx,Fσgy,gy,φgy,φgy,Fσgx,Tx,y,φgx,φTx,y,Fσgy,Ty,x,φgy,φTy,x=ϕmaxFσgx,gx,φgx,φgx,Fσgy,gy,φgy,φgy.Similarly, since gy⪯gy, we get(7)ψFσgy,gy,φgy,φgy=ψFσTy,x,Ty,x,φTy,x,φTy,x≤ϕMy,y,x,x=ϕmax⁡Fσgy,gy,φgy,φgy,Fσgx,gx,φgx,φgx,Fσgy,Ty,x,φgy,φTy,x,Fσgx,Tx,y,φgx,φTx,y=ϕmax⁡Fσgx,gx,φgx,φgx,Fσgy,gy,φgy,φgy.Noting that ψ is nondecreasing and using (6) and (7), we have (8)ψmaxFσgx,gx,φgx,φgx,Fσgy,gy,φgy,φgy=maxψFσgx,gx,φgx,φgx,ψFσgy,gy,φgy,φgy≤ϕmaxFσgx,gx,φgx,φgx,Fσgy,gy,φgy,φgy.By Definition 3, we obtain that if ψ(w)≤ϕ(w), then w=0. On the other hand, through the definition of F, we can reach the following conclusions: (9)maxFσgx,gx,φgx,φgx,Fσgy,gy,φgy,φgy=0,maxσgx,gx,σgy,gy,φgx,φgy≤maxFσgx,gx,φgx,φgx,Fσgy,gy,φgy,φgy=0.This is a contradiction. Hence, if there exist x,y∈X such that T(x,y)=gx,T(y,x)=gy, we have(10)σgx,gx=σgy,gy=φgx=φgy=0.
Now, suppose that(gxn+1,gyn+1)≠(gxn,gyn) for all n∈N, which means that either gxn≠T(xn,yn) or gyn≠T(yn,xn). Since gxn-1⪯gxn and gyn⪯gyn-1, we obtain(11)ψFσgxn,gxn+1,φgxn,φgxn+1=ψFσTxn-1,yn-1,Txn,yn,φTxn-1,yn-1,φTxn,yn≤ϕMxn-1,xn,yn-1,yn,where (12)Mxn-1,xn,yn-1,yn=maxFσgxn-1,gxn,φgxn-1,φgxn,Fσgyn-1,gyn,φgyn-1,φgyn,Fσgxn-1,Txn-1,yn-1,φgxn-1,φTxn-1,yn-1,Fσgyn-1,Tyn-1,xn-1,φgyn-1,φTyn-1,xn-1=maxFσgxn-1,gxn,φgxn-1,φgxn,Fσgyn-1,gyn,φgyn-1,φgyn. Similarly, we have(13)ψFσgyn,gyn+1,φgyn,φgyn+1=ψFσTyn-1,xn-1,Tyn,xn,φTyn-1,xn-1,φTyn,xn≤ϕMyn-1,yn,xn-1,xn,where (14)Myn-1,yn,xn-1,xn=maxFσgyn-1,gyn,φgyn-1,φgyn,Fσgxn-1,gxn,φgxn-1,φgxn,Fσgyn-1,Tyn-1,xn-1,φgyn-1,φTyn-1,xn-1,Fσgxn-1,Txn-1,yn-1,φgxn-1,φTxn-1,yn-1=maxFσgxn-1,gxn,φgxn-1,φgxn,Fσgyn-1,gyn,φgyn-1,φgyn. Noting that ψ is nondecreasing, using (11) and (13) we get (15)ψmaxFσgxn,gxn+1,φgxn,φgxn+1,Fσgyn,gyn+1,φgyn,φgyn+1=maxψFσgxn,gxn+1,φgxn,φgxn+1,ψFσgyn,gyn+1,φgyn,φgyn+1≤ϕmaxFσgxn-1,gxn,φgxn-1,φgxn,Fσgyn-1,gyn,φgyn-1,φgyn. Using condition (i) of Definition 3, we get (16)maxFσgxn,gxn+1,φgxn,φgxn+1,Fσgyn,gyn+1,φgyn,φgyn+1≤maxFσgxn-1,gxn,φgxn-1,φgxn,Fσgyn-1,gyn,φgyn-1,φgyn, which implies that {max{F(σ(gxn,gxn+1),φ(gxn),φ(gxn+1)), F(σ(gyn,gyn+1),φ(gyn),φ(gyn+1))}} is a decreasing nonnegative sequence. Therefore, there exists r≥0, such that (17)limn→∞maxFσgxn,gxn+1,φgxn,φgxn+1,Fσgyn,gyn+1,φgyn,φgyn+1=r. Using condition (ii) of Definition 3, we get r=0. Then (18)limn→∞maxσgxn,gxn+1,σgyn,gyn+1,φgxn,φgyn≤limn→∞maxFσgxn,gxn+1,φgxn,φgxn+1,Fσgyn,gyn+1,φgyn,φgyn+1=0. Thus(19)limn→∞σgxn,gxn+1=limn→+∞σgyn,gyn+1=limn→∞φgxn=limn→∞φgyn=0.
Next, we show thatlimn,m→∞σ(gxn,gxm)=0 and limn,m→∞σ(gyn,gym)=0. Assume that this is not true; that is, (20)limn,m→∞σgxn,gxm≠0orlimn,m→∞σgyn,gym≠0.Then there exist ϵ>0, k1∈N+, and two subsequences {gxm(k)},{gxn(k)} of {gxn} with m(k)>n(k)>k1, such that(21)maxσgxnk,gxmk,σgynk,gymk≥ϵ,maxσgxnk,gxmk-1,σgynk,gymk-1<ϵ.With the definitions of metric-like spaces, we obtain (22)σgxnk,gxmk≤σgxnk,gxmk-1+σgxmk-1,gxmk<ϵ+σgxmk-1,gxmk,σgynk,gymk≤σgynk,gymk-1+σgymk-1,gymk<ϵ+σgymk-1,gymk,which yields that (23)limk→∞σgxnk,gxmk≤limk→∞σgxnk,gxmk-1≤ϵ. Similarly, we have (24)limk→∞σgynk,gymk≤limk→∞σgynk,gymk-1≤ϵ. Thus, using (19) and (21), we obtain(25)ϵ≤maxlimk→∞σgxnk,gxmk,limk→∞σgynk,gymk≤maxlimk→∞σgxnk,gxmk-1,limk→∞σgynk,gymk-1≤ϵ.Since (26)σgxnk,gxmk≤σgxnk,gxnk-1+σgxnk-1,gxmk-1+σgxmk-1,gxmk≤σgxnk,gxnk-1+σgxnk-1,gxnk+σgxnk,gxmk-1+σgxmk-1,gxmk<σgxnk,gxnk-1+σgxnk-1,gxnk+ϵ+σgxmk-1,gxmk,taking the limit as k→+∞, we get (27)limk→∞σgxnk,gxmk≤limk→∞σgxnk-1,gxmk-1≤ϵ.Similarly, we have (28)limk→∞σgynk,gymk≤limk→∞σgynk-1,gymk-1≤ϵ.Then it follows from the above inequalities that(29)ϵ≤maxlimk→∞σgxnk,gxmk,limk→∞σgynk,gymk≤maxlimk→∞σgxnk-1,gxmk-1,limk→∞σgynk-1,gymk-1≤ϵ.Now, using (4), we have (30)ψFσgxnk,gxmk,φgxnk,φgxmk=ψFσTxnk-1,ynk-1,Txmk-1,ymk-1,φTxnk-1,ynk-1,φTxmk-1,ymk-1≤ϕMxnk-1,xmk-1,ynk-1,ymk-1, where (31)Mxnk-1,xmk-1,ynk-1,ymk-1=maxFσgxnk-1,gxmk-1,φgxnk-1,φgxmk-1,Fσgynk-1,gymk-1,φgynk-1,φgymk-1,Fσgxnk-1,Txnk-1,ynk-1,φgxnk-1,φTxnk-1,ynk-1,Fσgynk-1,Tynk-1,xnk-1,φgynk-1,φTynk-1,xnk-1. Since (32)limk→∞maxFσgxnk-1,Txnk-1,ynk-1,φgxnk-1,φTxnk-1,ynk-1,Fσgynk-1,Tynk-1,xnk-1,φgynk-1,φTynk-1,xnk-1=0, there exists k2∈N+ with k2>k1, such that (33)maxFσgxnk-1,gxmk-1,φgxnk-1,φgxmk-1,Fσgynk-1,gymk-1,φgynk-1,φgymk-1≥maxFσgxnk-1,Txnk-1,ynk-1,φgxnk-1,φTxnk-1,ynk-1,Fσgynk-1,Tynk-1,xnk-1,φgynk-1,φTynk-1,xnk-1, for all m(k)>n(k)>k2. Hence, we get (34)Mxnk-1,xmk-1,ynk-1,ymk-1=maxFσgxnk-1,gxmk-1,φgxnk-1,φgxmk-1,Fσgynk-1,gymk-1,φgynk-1,φgymk-1, for all m(k)>n(k)>k2. Similarly, there exists k3∈N+ with k3>k1, such that (35)ψFσgynk,gymk,φgynk,φgymk=ψFσTynk-1,xnk-1,Tymk-1,xmk-1,φTynk-1,xnk-1,φTymk-1,xmk-1≤ϕMynk-1,ymk-1,xnk-1,xmk-1=ϕmaxFσgxnk-1,gxmk-1,φgxnk-1,φgxmk-1,Fσgynk-1,gymk-1,φgynk-1,φgymk-1,for all m(k)>n(k)>k3. Letting k′=max{k2,k3} and noting that ψ is nondecreasing, we obtain (36)ψmaxFσgxnk,gxmk,φgxnk,φgxmk,Fσgynk,gymk,φgynk,φgymk=maxψFσgxnk,gxmk,φgxnk,φgxmk,ψFσgynk,gymk,φgynk,φgymk≤ϕmaxFσgxnk-1,gxmk-1,φgxnk-1,φgxmk-1,Fσgynk-1,gymk-1,φgynk-1,φgymk-1, for all m(k)>n(k)>k′. Taking the limit as k→+∞ and using (25) and (29), we have (37)limk→∞maxFσgxnk,gxmk,φgxnk,φgxmk,Fσgynk,gymk,φgynk,φgymk=limk→∞maxσgxnk,gxmk,σgynk,gymk=ϵ.limk→∞maxFσgxnk-1,gxmk-1,φgxnk-1,φgxmk-1,Fσgynk-1,gymk-1,φgynk-1,φgymk-1=limk→∞maxσgxnk-1,gxmk-1,σgynk-1,gymk-1=ϵ. According to the properties of Ψ, we get ϵ=0, which is a contradiction. So, {gxn} and {gyn} are σ-Cauchy sequences in (X,σ), which is a complete metric-like space. Since T(X×X)⊆g(X) and g(X) is closed, there exist x,y∈X, such that limn→∞gxn=gx,limn→∞gyn=gy. Correspondingly, we have(38)σgx,gx=limn→∞σgxn,gx=0,σgy,gy=limn→∞σgyn,gy=0.Since φ is a lower semicontinuous function, then 0≤φ(gx)≤liminfn→∞φ(gxn)=0. We have φ(gx)=0. By same arguments, we can derive that φ(gy)=0.
Finally, we claim thatT and g have a coupled coincidence point. It follows from (4) that (39)ψFσTxn,yn,Tx,y,φTxn,yn,φTx,y≤ϕMxn,x,yn,y=ϕmaxFσgxn,gx,φgxn,φgx,Fσgyn,gy,φgyn,φgy,Fσgxn,Txn,yn,φgxn,φTxn,yn,Fσgyn,Tyn,xn,φgyn,φTyn,xn=ϕmaxFσgxn,gx,φgxn,φgx,Fσgyn,gy,φgyn,φgy,Fσgxn,gxn+1,φgxn,φgxn+1,Fσgyn,gyn+1,φgyn,φgyn+1 Similarly, we have (40)ψFσTyn,xn,Ty,x,φTyn,xn,φTy,x≤ϕMyn,y,xn,x=ϕmaxFσgyn,gy,φgyn,φgy,Fσgxn,gx,φgxn,φgx,Fσgyn,Tyn,xn,φgyn,φTyn,xn,Fσgxn,Txn,yn,φgxn,φTxn,yn=ϕmaxFσgyn,gy,φgyn,φgy,Fσgxn,gx,φgxn,φgx,Fσgyn,gyn+1,φgyn,φgyn+1,Fσgxn,gxn+1,φgxn,φgxn+1Letting n→∞ and using (38), we obtain (41)limn→∞Mxn,x,yn,y=limn→∞maxFσgxn,gx,φgxn,φgx,Fσgyn,gy,φgyn,φgy,Fσgxn,gxn+1,φgxn,φgxn+1,Fσgyn,gyn+1,φgyn,φgyn+1=0,limn→∞Myn,y,xn,x=limn→∞maxFσgyn,gy,φgyn,φgy,Fσgxn,gx,φgxn,φgx,Fσgyn,gyn+1,φgyn,φgyn+1,Fσgxn,gxn+1,φgxn,φgxn+1=0. Therefore, we get (42)limn→∞FσTxn,yn,Tx,y,φTxn,yn,φTx,y=0,limn→∞FσTyn,xn,Ty,x,φTyn,xn,φTy,x=0. We can take advantage of the properties of F to get (43)limn→∞σTxn,yn,Tx,y=limn→∞σTyn,xn,Ty,x=limn→∞φTxn,yn=limn→∞φTyn,xn=0. By triangle inequality in metric-like space, we have (44)σgx,Tx,y≤σgx,gxn+1+σgxn+1,Tx,y=σgx,gxn+1+σTxn,yn,Tx,y.Letting n→+∞, we get σ(gx,T(x,y))=0. Similarly, we have σ(gy,T(y,x))=0, that is, gx=T(x,y),gy=T(y,x). Therefore, T and g have a coupled coincidence point.Theorem 11.
Define the partial order in(X×X,⪯) by (x,y)⪯(x∗,y∗)⇔x⪯x∗,y∗⪯y. Add to the hypotheses of Theorem 10 the following conditions:
(i)T and g commute at their coincidence points.
(ii) For every(x,y) and (x∗,y∗) in X×X, there exists (u,v)∈X×X such that (T(u,v),T(v,u)) is comparable to (T(x∗,y∗),T(y∗,x∗)) and (T(x,y),T(y,x)).
ThenT and g have a unique common coupled fixed point.Proof.
First, by Theorem10, we know that the set of coupled coincidence points is nonempty. Let (x,y) and (x∗,y∗) be coupled coincidence points, i.e., gx=T(x,y),gy=T(y,x) and gx∗=T(x∗,y∗),gy∗=T(y∗,x∗). Then we need to verify(45)gx=gx∗,gy=gy∗.
From (ii), there exists(u,v)∈X×X such that (T(u,v),T(v,u)) is comparable to (T(x∗,y∗), T(y∗,x∗)), and (T(x,y),T(y,x)). Letting u0=u,v0=v, we can choose u1,v1∈X such that gu1=T(u0,v0),gv1=T(v0,u0). Using this method, we can choose u2,v2∈X such that gu2=T(u1,v1),gv2=T(v1,u1). Repeating the procedures, we have {gun} and {gvn} such that gun=T(un-1,vn-1),gvn=T(vn-1,un-1). Similarly, letting x0=x,y0=y,x0∗=x∗,y0∗=y∗, we get {gxn}, {gyn} and {gxn∗}, {gyn∗}. Since gx=T(x,y),gy=T(y,x), we get (gx0,gy0)=(gx,gy)=(T(x0,y0),T(y0,x0))=(gx1,gy1). And we have (gu1,gv1)=(T(u,v),T(v,u)), so (gx,gy)⪯(gu1,gv1), that is, gx⪯gu1,gv1⪯gy. By condition (iii) of Theorem 10, we have gx⪯gun,gvn⪯gy. Using (4), we obtain (46)ψFσgx,gun+1,φgx,φgun+1=ψFσTx,y,Tun,vn,φTx,y,φTun,vn≤ϕMx,un,y,vn, where (47)Mx,un,y,vn=maxFσgx,gun,φgx,φgun,Fσgy,gvn,φgy,φgvn,Fσgx,Tx,y,φgx,φTx,y,Fσgy,Ty,x,φgy,φTy,x=maxFσgx,gun,φgx,φgun,Fσgy,gvn,φgy,φgvn. Similarly, (48)ψFσgy,gvn+1,φgy,φgvn+1=ψFσTy,x,Tvn,un,φTy,x,φTvn,un≤ϕMy,vn,x,un=maxFσgx,gun,φgx,φgun,Fσgy,gvn,φgy,φgvn. Noting that ψ is nondecreasing, we have (49)ψmaxFσgx,gun+1,φgx,φgun+1,Fσgy,gvn+1,φgy,φgvn+1≤ϕmaxFσgx,gun,φgx,φgun,Fσgy,gvn,φgy,φgvn, so {max{F(σ(gx,gun+1),φ(gx),φ(gun+1)),F(σ(gy,gvn+1), φ(gy),φ(gvn+1))}} is a nonnegative decreasing sequence, and (50)limn→∞maxFσgx,gun+1,φgx,φgun+1,Fσgy,gvn+1,φgy,φgvn+1=0, which implies that (51)limn→∞σgx,gun=φgx=0,limn→∞σgy,gvn=φgy=0. Through the same process, we can prove that (52)limn→∞σgx∗,gun=φgx∗=0,limn→∞σgy∗,gvn=φgy∗=0. Using triangle inequality, we have (53)σgx,gx∗≤σgx,gun+σgx∗,gun,σgy,gy∗≤σgy,gvn+σgy∗,gvn. Letting n→+∞, we get σ(gx,gx∗)=0,σ(gy,gy∗)=0, which implies that gx=gx∗,gy=gy∗.
Denotegx=z,gy=w. We now show that (z,w) is a coupled common fixed point. Since T and g commute at their coincidence points, then we have (54)gz=ggx=gTx,y=Tgx,gy=Tz,w,gw=ggy=gTy,x=Tgy,gx=Tw,z.Hence, (z,w) is a coupled coincidence point of T and g. Using (45), we have gz=gx,gw=gy, and, therefore, z=gz=T(z,w),w=gw=T(w,z), which means (z,w) is a coupled common fixed point.
Finally, we prove the existence and uniqueness of coupled common fixed point. If(t,s) is also a coupled common fixed point, i.e., t=gt=T(t,s),s=gs=T(s,t), then (t,s) is also a coupled coincidence point. We have t=gt=gz=z,s=gs=gw=w. Therefore, T and g have a unique common coupled fixed point.Corollary 12.
Define the partial order in(X×X,⪯) by (x,y)⪯(x∗,y∗)⇔x⪯x∗,y∗⪯y. Add to the hypotheses of Theorem 10 the following conditions:
(i)T and g commute at their coincidence points.
(ii)g(X) is a totally ordered subset of X.
ThenT and g have a unique common coupled fixed point.Proof.
It is easy to see that condition (ii) of Theorem11 can be naturally established.Theorem 13.
Under the hypotheses of Theorem11, add the following condition: suppose that gy0⪯gx0. Then T and g have a unique common coupled fixed point of the form (x,x).Proof.
Through the proof of Theorem10, we construct the sequences {xn},{yn}∈X such that gxn+1=T(xn,yn), gyn+1=T(yn,xn) and gx0⪯gx1⪯gx2⪯⋯⪯gxn⪯⋯ and gy0⪰gy1⪰gy2⪰⋯⪰gyn⪰⋯. And g(X) is closed; then there exist x,y∈X, such that limn→∞gxn=gx,limn→∞gyn=gy. In Theorem 11, we prove that (x,y) is the unique coupled common fixed point of T and g, i.e., x=gx=T(x,y),y=gy=T(y,x).
Next, we shall showgx=T(x,y)=gy=T(y,x). Owing to the supplementary condition, we have gy⪯gyn⪯⋯⪯gy0⪯gx0⪯gx1⪯⋯⪯gxn⪯gx. From (4), we obtain(55)ψFσgx,gy,φgx,φgy≤ϕMx,y,y,x=ϕmaxFσgx,gy,φgx,φgy,Fσgy,gx,φgy,φgx.By (10), we get φ(gx)=φ(gy)=0, and thus we have (56)Fσgx,gy,φgx,φgy=σgx,gy=Fσgy,gx,φgy,φgx.So (55) can be converted into (57)ψσgx,gy≤ϕσgx,gy, which implies that gx=gy. Then T and g have a unique common coupled fixed point of the form (x,x).Example 14.
LetX=[0,+∞) and σ:X×X→X be defined by (58)σx,y=2x,ifx=y,maxx,y,otherwise.Then (X,σ) is a complete metric-like space, rather than a partial metric space. Indeed, if x<y<2x, we obtain σ(x,x)>σ(x,y). Let T:X×X→X be defined as(59)Tx,y=x-y6,x≥y,0,x<y.Define function g:X→X by gx=x. Let ψ,φ:X→X be defined by (60)ψt=ln⁡1+t,ϕt=ln⁡1+t2. Let φ:X→X be defined by φ(t)=t and F:[0,∞)3→[0,∞) be the function defined by F(a,b,c)=a+b+c.It is clear thatT has the mixed g-monotone property. Easily, we can know g(X) is closed and T(X×X)⊆g(X). Letting x0=y0=0, then x0≤T(x0,y0),T(y0,x0)≤y0 is true. And if ψ(u)=ln(1+u)≤ϕ(v)=ln(1+v/2), we have u≤v/2; i.e., u≤v is established, so ψ,φ∈Ψ. And gT(x,y)=T(gx,gy), and then g and T are commutative. Next, we prove that (4) holds for all x,y,u,v∈X with x≤u,v≤y. We can distinguish three cases.Case 1 (v≤y≤x≤u).
IfT(x,y)=T(u,v), we get v=y and x=u.(61)ψFσTx,y,Tu,v,φTx,y,φTu,v=ln1+2·x-y6+x-y6+u-v6=ln1+x-y2+u-v6,ϕMx,u,y,v=ln1+12×max4x,4y,2x+x-y6,2y=ln1+2u.If T(x,y)≠T(u,v), we get v≠y or x≠u.(62)ψFσTx,y,Tu,v,φTx,y,φTu,v=ln1+maxTx,y,Tu,v+Tx,y+Tu,v=ln1+u-v3+x-y6,Mx,u,y,v=max4u,2y+v,2x+x-y6,2y,ifv≠yandx=u,max2u+x,4y,2x+x-y6,2y,ifx≠uandv=y,max2u+x,2y+v,2x+x-y6,2y,ifv≠yandx≠u. Then we have (63)ϕMx,u,y,v≥ln1+u+x2.Hence, ψ(F(σ(T(x,y),T(u,v)),φ(T(x,y)),φ(T(u,v))))≤ϕ(M(x,u,y,v)) is proved.Case 2 (x≤u,x<y,v≤y,u≥v).
IfT(x,y)=T(u,v), we have T(x,y)=0 and u=v.(64)ψFσTx,y,Tu,v,φTx,y,φTu,v=0IfT(x,y)≠T(u,v), we have T(x,y)=0 and u≠v.(65)ψFσTx,y,Tu,v,φTx,y,φTu,v=ln1+u-v3,Mx,u,y,v=max4u,2y+v,2x,2y+y-x6,ifv≠yandx=u,max4u,4y,2x,2y+y-x6,ifx=uandv=y,max2u+x,4y,2x,2y+y-x6,ifv=yandx≠u,max2u+x,2y+v,2x,2y+y-x6,ifv≠yandx≠u. Then we have (66)ϕMx,u,y,v≥ln1+u+x2.Hence, ψ(F(σ(T(x,y),T(u,v)),φ(T(x,y)),φ(T(u,v))))≤ϕ(M(x,u,y,v)) is tenable.Case 3 (x≤u,x<y,v≤y,v<u).
(67) ψ F σ T x , y , T u , v , φ T x , y , φ T u , v = 0 ≤ ϕ M x , u , y , v . Hence, (4) is established. And (4) still holds with u≤x,y≤v.Note that all conditions of Theorems10–13 are satisfied. Hence, T and g have a unique common coupled fixed point, which is (0,0).


## 3. Consequences of the Main Result

By choosing suitable mappingsψ,ϕ,g, and F, one can deduce subsequent corollaries.Lettingg=I (the identity mapping on X) in Theorem 10, we get the following corollary.Corollary 15.
Let(X,⪯) be a partially ordered set and (X,σ) be a complete metric-like space. Let T:X×X→X be a mapping, satisfying the following conditions:
(i)T has the mixed monotone property.
(ii) There existx0,y0∈X such that x0⪯T(x0,y0),T(y0,x0)⪯y0 and y0⪯x0.
(iii)ψ(F(σ(T(x,y),T(u,v)),φ(T(x,y)),φ(T(u,v))))≤ϕ(M(x,u,y,v)), where (68)Mx,u,y,v=maxFσx,u,φx,φu,Fσy,v,φy,φv,Fσx,Tx,y,φx,φTx,y,Fσy,Ty,x,φy,φTy,x, with x⪯u and v⪯y, or u⪯x and y⪯v, for all x,y,u,v∈X,F∈F,(ψ,ϕ)∈Ψ, ψ is nondecreasing, and φ:X→[0,+∞) is a lower semicontinuous function.
ThenT has a coupled fixed point.SettingF(a,b,c)=a+b+c and φ(t)=0,t≥0 in Theorem 10, we get another corollary.Corollary 16.
Let(X,⪯) be a partially ordered set and (X,σ) be a complete metric-like space. Let T:X×X→X and g:X→X be two mappings such that the following conditions are satisfied:
(i)T(X×X)⊆g(X).
(ii)g(X) is closed.
(iii)T has the mixed g-monotone property.
(iv) There existx0,y0∈X such that gx0⪯T(x0,y0),T(y0,x0)⪯gy0.
(v)(69)ψσTx,y,Tu,v≤ϕmaxσgx,gu,σgy,gv,σgx,Tx,y,σgy,Ty,x,with x⪯u and v⪯y, or u⪯x and y⪯v, for all x,y,u,v∈X,(ψ,ϕ)∈Ψ, and ψ is nondecreasing.
ThenT and g have a coupled coincidence point.Corollary 17.
Let(X,⪯) be a partially ordered set and (X,σ) be a complete metric-like space. Let T:X×X→X and g:X→X be two mappings, such that the following conditions are satisfied:
(i)T(X×X)⊆g(X).
(ii)g(X) is closed.
(iii)T has the mixed g-monotone property.
(iv) There existx0,y0∈X such that gx0⪯T(x0,y0),T(y0,x0)⪯gy0.
(v)(70)σTx,y,Tu,v≤aσgx,gu+bσgy,gv+cσgx,Tx,y+dσgy,Ty,x,with x⪯u and v⪯y, or u⪯x and y⪯v, for all x,y,u,v∈X,a,b,c,d∈R+, a+b+c+d∈[0,1).
ThenT and g have a coupled coincidence point.Proof.
Denoteψ(t)=t,ϕ(t)<t, we can note that (71)aσgx,gu+bσgy,gv+cσgx,Tx,y+dσgy,Ty,x≤a+b+c+dmaxσgx,gu,σgy,gv,σgx,Tx,y,σgy,Ty,x. Hence, the conditions of Corollary 16 can be satisfied.Since metric spaces are metric-like spaces, corresponding theorems in metric spaces can be derived. On the other hand, according to the definition ofΨ, if (ψ,ϕ)∈Ψ, we obtain ψ-ϕ∈Ψ. Now we have the following result in metric spaces.Theorem 18.
Let(X,⪯) be a partially ordered set and (X,d) be a complete metric space. Let T:X×X→X and g:X→X be two mappings such that the following conditions are satisfied:
(i)T(X×X)⊆g(X).
(ii)g(X) is closed.
(iii)T has the mixed g-monotone property.
(iv) There existx0,y0∈X such that gx0⪯T(x0,y0) and T(y0,x0)⪯gy0.
(v)ψ(F(d(T(x,y),T(u,v)),φ(T(x,y)),φ(T(u,v))))≤ψ(M(x,u,y,v))-ϕ(M(x,u,y,v)), where (72)Mx,u,y,v=maxFdgx,gu,φgx,φgu,Fdgy,gv,φgy,φgv,Fdgx,Tx,y,φgx,φTx,y,Fdgy,Ty,x,φgy,φTy,x, with x⪯u and v⪯y, or u⪯x and y⪯v, for all x,y,u,v∈X,F∈F,(ψ,ϕ)∈Ψ, ψ is nondecreasing, and φ:X→[0,+∞) is a lower semicontinuous function.
ThenT and g have a coupled coincidence point.


## 4. An Application

In this section, we study the existence of a unique solution to a system of nonlinear integral equations by applying a coupled fixed point result in Section3. Consider the following system of equations:(73)xt=ht+∫01k1t,sf1s,xsds∫01k2t,sf2s,ysdsyt=ht+∫01k1t,sf1s,ysds∫01k2t,sf2s,xsdsNow, we discuss the existence of the unique solution.Theorem 19.
LetX=C(I) be the set of all continuous functions defined on I=[0,1] with a dislocated metric given by σ(x,y)=supt∈Ixt-yt. Assume that T:X×X→X has the mixed monotone property, and T is defined by(74)Tx,yt=ht+∫01k1t,sf1s,xsds∫01k2t,sf2s,ysdsTy,xt=ht+∫01k1t,sf1s,ysds∫01k2t,sf2s,xsdswhich satisfy the following conditions:
(i)h:[0,1]→R is continuous.
(ii)ki:[0,1]×[0,1]→R+ and fi:[0,1]×R→R+(i=1,2) are continuous. There exists a constant Di∈R+ such that ∫01ki(t,s)ds≤Di. And ∫01ki(t,s)fi(s,x(s))ds≤K, where K≥0 is a constant real number.
(iii)ϕ∈Ψ is nondecreasing and ϕ(t)<t. There exists a constant Li∈[0,1) such that for all x,y∈X, (75)fit,x-fit,y≤Liϕx-yi=1,2.(iv)2K·max{L1D1,L2D2}<1.Ifx⪯u,v⪯y, or u⪯x,y⪯v, and there exist x0,y0∈X such that x0⪯T(x0,y0) and T(y0,x0)⪯y0, then T has a coupled fixed point; i.e., (83) has a unique solution in C(I).Proof.
Defineσ(x,y)=supt∈Ixt-yt. Letting x⪯u,v⪯y, or u⪯x,y⪯v, and x,y,u,v∈X, we have (76)Tx,yt-Tu,vt=∫01k1t,sf1s,xsds∫01k2t,sf2s,ysds-∫01k1t,sf1s,usds∫01k2t,sf2s,vsds≤∫01k1t,sf1s,xsds∫01k2t,sf2s,ys-f2s,vsds+∫01k1t,sf1s,xs-f1s,usds∫01k2t,sf2s,vsds≤K·∫01k2t,sL2ϕσys,vsds+K·∫01k1t,sL1ϕσus,xsds≤K·∫01k2t,sL2ϕσy,vds+K·∫01k1t,sL1ϕσu,xds≤2K·maxL1D1,L2D2·ϕmaxσy,v,σu,x≤ϕmaxσy,v,σu,x,σTx,y,u,σTy,x,v
Hence, we getσ(T(x,y),T(u,v))≤ϕ(max{σ(y,v),σ(u,x),σ(T(x,y),u),σ(T(y,x),v)}). On the other hand, by the assumptions of the theorem, the other conditions of Corollary 16 are satisfied with gt=t. Then T has a coupled fixed point; i.e., (73) has a unique solution in C(I).Example 20.
Consider the following system of nonlinear integral equations:(77)xt=lnt+1+∫01set·s231+s4·xs1+xsds∫01s1+t2·131+s21+ys2+ys2dsyt=lnt+1+∫01set·s231+s4·ys1+ysds∫01s1+t2·131+s21+xs2+xs2ds,where t∈[0,1]. Then we can define the functions by (78)ht=lnt+1k1t,s=set,k2t,s=s1+t2f1s,xs=s231+s4·xs1+xs,f2s,xs=131+s2·ys2+ys2 It is clear that h and fi,ki (i=1,2) are continuous functions, and fi(t,x)≥0. Supposing ϕt=(1/3)t, we obtain ϕ∈Ψ with ϕ(t)<t and ϕ is nondecreasing. Since (79)f1t,xt-f1t,ut=t231+t4·xt1+xt-ut1+ut≤16x-u=12ϕx-u,f2t,yt-f2t,vt=131+t2·1+yt2+yt2-1+vt2+vt2≤16y-v=12ϕy-v,we can get L1=L2=1/2. And we have (80)∫01k1t,sf1s,xsds≤∫01set·s231+s4ds=112ln⁡2·et≤112ln⁡2·e,∫01k2t,sf2s,xsds≤∫01s1+t2·131+s2ds≤16ln⁡2. So, letting K=1/12ln⁡2·e, we have ∫01ki(t,s)fi(s,x(s))ds≤K.(81)∫01k1t,sds≤12et≤12e=D1,∫01k2t,sds≤121+t2≤14=D2. Through calculation, we can see 2K·max⁡{L1D1,L2D2}<1. Hence, all conditions of Theorem 19 are satisfied; then the integral equations have a unique solution in C(I), where I=[0,1].Remark 21.
Notice that the givenσ, which is defined by σ(x,y)=supt∈Ixt-yt, is a metric. Next, we consider an application in a metric-like space. Let X=C(I) be the set of all continuous functions with a dislocated metric given by (82)σx,y=supt∈Ixt+yt,∀x,y∈X. Clearly, (X,σ) is a complete metric-like space but not a metric space.Theorem 22.
LetX=C(I) be the set of all continuous functions defined on I=[0,1] with a dislocated metric given by σ(x,y)=supt∈I(xt+yt). Assume that T:X×X→X has the mixed monotone property, and T is defined by(83)Tx,yt=∫01k1t,sf1s,xsds∫01k2t,sf2s,ysdsTy,xt=∫01k1t,sf1s,ysds∫01k2t,sf2s,xsdswhich satisfy the following conditions:
(i)ki:[0,1]×[0,1]→R+ and fi:[0,1]×R→R+(i=1,2) are continuous. There exists a constant Di∈R+ such that ∫01ki(t,s)ds≤Di. And ∫01ki(t,s)fi(s,x(s))ds≤K, where K≥0 is a constant real number.
(ii)ϕ∈Ψ is nondecreasing and ϕ(t)<t. There exists a constant Li∈[0,1) such that, for all x,y∈X, (84)fit,x+fit,y≤Liϕx+yi=1,2.
(iii)2K·max{L1D1,L2D2}<1.
Ifx⪯u,v⪯y, or u⪯x,y⪯v, and there exist x0,y0∈X such that x0⪯T(x0,y0) and T(y0,x0)⪯y0, then T has a coupled fixed point; i.e., (83) has a unique solution in C(I).Proof.
Denoteσ(x,y)=supt∈I(xt+yt). Let x⪯u,v⪯y, or u⪯x,y⪯v, and x,y,u,v∈X, and we have (85)
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Hence, we getσ(T(x,y),T(u,v))≤ϕ(max{σ(y,v),σ(u,x),σ(T(x,y),u),σ(T(y,x),v)}). Then, all the conditions of Corollary 16 are satisfied with gt=t. Then T has a coupled fixed point; i.e., (83) has a unique solution in C(I).



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*Source: 1378379-2018-08-01.xml*