# Related Fixed Point Theorems in Partially Orderedb-Metric Spaces and Applications to Integral Equations

Authors: Youssef Errai; El Miloudi Marhrani; Mohamed Aamri
Journal: Abstract and Applied Analysis
 (2021)

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## Abstract

In this research paper, we have set some related fixed point results for generalized weakly contractive mappings defined in partially ordered completeb-metric spaces. Our results are an extension of previous authors who have already worked on fixed point theory in b-metric spaces. We state some examples and one sample of the application of the obtained results in integral equations, which support our results.


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## Body

## 1. Introduction

The concept ofb-metric space has been dealt with by distinctly different authors since it first appeared and became largely used. Bakhtin in [1] was the first who introduced this concept, and later on, Czerwik in [2, 3] invested it in the convergence of measurable functions. Accordingly, several interesting results about the existence of fixed points have been obtained concerning both single-valued and multivalued operators in b-metric spaces (see, e.g., [4, 5]). In the same way, Hussain and Shah [6] conducted a research and then came up with results from KKM mappings in cone b-metric spaces. Likewise, Roshan et al. [7] used the notion of almost generalized contractive mappings in ordered complete b-metric spaces and set some fixed and common fixed point results. Regarding partially ordered metric spaces, Ran and Reurings (see [8]) had first set up their assumptions before Nieto and Rodríguez-López used them (see [9, 10]). Afterwards, many authors presented several interesting and significant results in such spaces (see [4, 7, 11–19]).The purpose of this research paper is to prove some fixed point theorems for generalized contractive conditions for four mappings in completeb-metric spaces.Our work goes through the following steps. First, we have demonstrated a twob-metric space theorem with four mappings. Second, we have stated a relevant example which backs up the mentioned theorem. Third, we have given three corollaries related to the theorem. Besides, an example with only two related corollaries is provided for the second theorem. To sum up, we conclude the manuscript by an application to solve a system of integral equations.Throughout this paper,ℝ and ℝ+ denote the sets of all real numbers and nonnegative real numbers, respectively.Consistent with [3, 4], the following definitions and outcomes are going to be vital and required in the ending.Definition 1 (see [3]).
Given a nonempty setX. A function d:X×X⟶ℝ+ is called b-metric if there is a real number s∈1,∞ such that for all x,y,z∈X, the following conditions hold:
(i)
dx,y=0 if and only if x=y;(ii)
dx,y=dy,x;(iii)
dx,z≤sdx,y+dy,z.
The pairX,d is called a b-metric space.Definition 2 (see [20]).
LetX,d be a b-metric space. Then, a sequence xn in X is
(a)
convergent if there existsx∈X such that dxn,x⟶0 as n⟶∞. In this case, we write limn⟶∞xn=x.(b)
Cauchy sequence ifdxn,xm⟶0 as n,m⟶∞.Definition 3.
LetX be a nonempty set. Then, X,d,δ,≼ is called a partially ordered two b-metric space if d and δ are b-metrics on the partially ordered set X,≼.A subsetY of a partially ordered set X is said to be well ordered if every two elements of Y are comparable.Definition 4 (see [4]).
LetX,≼ be a partially ordered set. A mapping f on X is called dominating (resp. dominated) if x≼fx (resp. fx≼x) for each x in X.Definition 5.
Letxn be a sequence in a b-metric space X,d, g,h:X⟶X, and x∈X.
(i)
x is said to be a coincidence point of pair g,h if gx=hx(ii)
g,h is said to be compatible if dghxn,hgxn⟶0 as n⟶∞(iii)
g,h is said to be weakly compatible if ghx=hgx, where gx=hxRemark 6 (see [20]).
In ab-metric space X,d, the following assertions hold:
(R1)
A convergent sequence has a unique limit(R2)
Each convergent sequence is a Cauchy sequence(R3)
In general, ab-metric is not continuous(R4)
In general, ab-metric does not induce a topology on X.The fact in (R3) requires the following Lemma concerningb-convergent sequences to prove our findings:Lemma 7 (see [4]).
LetX,d be a b-metric space with s∈1,∞ and suppose that xn and yn are convergent to x,y, respectively. Then, we have
(1)1s2dx,y≤liminfn⟶∞dxn,yn≤limsupn⟶∞dxn,yn≤s2dx,y.
In particular, ifx=y, then we have limn⟶∞dxn,yn=0.
Moreover, for eachz∈X, we have
(2)1sdx,z≤liminfn⟶∞dxn,z≤limsupn⟶∞dxn,z≤sdx,z.


## 2. Main Results

Throughout this paper, letf,g,P, and T be four self-maps on X, and d and δ are two b-metrics with constant s and r, respectively. Set
(3)Mdx,y=maxdPx,Ty,dfx,Px,dgy,Ty,dPx,gy+dfx,Ty2s,Mδx,y=maxδPx,Ty,δfx,Px,δgy,Ty,δPx,gy+δfx,Ty2r,Ndx,y=maxdx,y,dfx,x,dgy,y,dx,gy+dfx,y2s,Nδx,y=maxδx,y,δfx,x,δgy,y,δx,gy+δfx,y2r,Qdx,y=a1dPx,Ty+a2dfx,Px+a3dgy,Ty+a4dPx,gy+a5dfx,Ty,Rdx,y=a1dx,y+a2dfx,x+a3dgy,y+a4dx,gy+a5dfx,y,with
(4)0<a3,a4,0≤a1,a2,a5,0<a2+a5s≤1,a1+a2+a3+2a4s≤1,a1+a4+a5≤1.Let,(5)Ψ=ψ:0,+∞⟶0,+∞iscontinuousmonotone,nondecreasingandψt=0⇔t=0,Φ=φ:0,+∞⟶0,+∞islowersemicontinuousandφt=0⇔t=0.Finally, we consider the following assumptions:Assumption 8.
Letxn and yn be two sequences such that
(i)
xn is nonincreasing(ii)
foralln,yn≼xnandyn→u⇒u≼xn,foralln.Assumption 9.
Letf,g,P, and T be four self-maps on X such that either condition
(a)
g,T is weakly compatible, f,P is compatible and f or P is continuous, or(b)
f,P is weakly compatible, g,T is compatible and g or T is continuous.Our main result is the following theorem:Theorem 10.
We consider four self-mappingsf,g,P, and T in ordered complete two b-metric space X,d,δ,≼ that fulfill the following conditions:
(i)
f,g is dominated and P,T is dominating(ii)
fX⊆TX and gX⊆PX(iii)
ψ∈Ψ, φ∈Φ and for every two comparable elements x,y∈X, we have(6)ψs4dfx,gy≤ψMδx,y−φMδx,y,ψr4δfx,gy≤ψMdx,y−φMdx,y.(iv)
Assumptions8 and 9 are satisfied
Then,f,g,P, and T have a unique common fixed point in X.Proof.
Letx0 be an arbitrary point in X. By condition (ii), we can define inductively two sequences xn and yn in X as follows:
(7)y2n+1=fx2n=Tx2n+1,y2n+2=gx2n+1=Px2n+2,n=0,1,2,⋯.
We have(8)x2n+1≼Tx2n+1=fx2n≼x2nandx2n≼Px2n=gx2n−1≼x2n−1.
Thus,xn+1≼xn for all n≥0, and
(9)Mδx2n,x2n+1=maxδy2n,y2n+1,δy2n+1,y2n+2,Mdx2n,x2n+1=maxdy2n,y2n+1,dy2n+1,y2n+2.Step 1.
Ify2k+1=y2k for some k, then ynn is constant. Indeed, we have
(10)ψδy2k+1,y2k+2≤ψMdx2k,x2k+1−φMdx2k,x2k+1,ψdy2k+1,y2k+2≤ψMδx2k,x2k+1−φMδx2k,x2k+1.
which implies(11)δy2k+1,y2k+2≤Mdx2k,x2k+1=dy2k+1,y2k+2,dy2k+1,y2k+2≤Mδx2k,x2k+1=δy2k+1,y2k+2.If we insert in (10), we obtain φdy2k+1,y2k+2=0, and then yn is a constant sequence. Its value is a common fixed point of f,g,P, and T.In the following, we can assume thatdy2n,y2n+1>0 for each n.Step 2.
The sequencesdy2n+1,y2n and δy2n+1,y2n are monotone nonincreasing. Indeed, since x2n and x2n+1 are comparable, we obtain
(12)ψdy2n+1,y2n+2≤ψs4dy2n+1,y2n+2=ψs4dfx2n,gx2n+1≤ψMδx2n,x2n+1−φMδx2n,x2n+1≤ψMδx2n,x2n+1.Consequently, if we permuted and δ, we obtain
(13)dy2n+1,y2n+2≤Mδx2n,x2n+1,δy2n+1,y2n+2≤Mdx2n,x2n+1.Ifδy2n+1,y2n+2>δy2n,y2n+1 for some n, we obtain
(14)dy2n+1,y2n+2≤Mδx2n,x2n+1=δy2n+1,y2n+2.We havedy2n+1,y2n+2≤dy2n+1,y2n. Otherwise, we obtain Mdx2n,x2n+1=dy2n+2,y2n+1, and consequently,
(15)ψdy2n+1,y2n+2≤ψδy2n+1,y2n+2≤ψdy2n+2,y2n+1−φdy2n+2,y2n+1,which gives φdy2n+2,y2n+1=0, which contradicts our hypothesis. Consequently,
(16)δy2n+1,y2n+2≤Mδx2n,x2n+1=δy2n+1,y2n,dy2n+1,y2n+2≤Mdx2n,x2n+1=dy2n+1,y2n.Taking account of (6), we obtain
(17)ψδy2n+1,y2n+2≤ψdy2n+1,y2n−φdy2n+1,y2n,which gives
(18)δy2n+1,y2n+2≤dy2n+1,y2n.On the other hand, the inequality(19)ψdy2n+1,y2n≤ψMδx2n,x2n−1−φMδx2n,x2n−1,gives
(20)dy2n+1,y2n≤Mδx2n,x2n−1=δy2n+1,y2n<δy2n+1,y2n+2,which contradicts (18). Thus, dy2n+1,y2n+2 and δy2n+1,y2n+2 are monotone nonincreasing sequences. It follows that there exists a nonnegative real number c such that
(21)limndy2n+1,y2n+2=limnδy2n+1,y2n+2=c,Using (10), we deduce that ψc≤ψc−φc, and consequently, c=0.Step 3.
yn is a Cauchy sequence.Assume that there existε,τ>0 for which we can find subsequences y2mk and y2nk of y2n such that 2nk>2mk>k for all k, we have
(22)dy2mk,y2nk≥ε,dy2mk,y2nk−2<ε,δy2mk,y2nk≥τ,δy2mk,y2nk−2<τ.Using the triangle inequality inb-metric spaces, we obtain
(23)ε≤dy2mk,y2nk≤sdy2nk−2,y2mk+s2dy2nk−1,y2nk−2+s2dy2nk,y2nk−1<εs+s2dy2nk−1,y2nk−2+s2dy2nk,y2nk−1,which leads to
(24)ε≤limsupk⟶∞dy2nk,y2mk≤εs,and since
(25)ε≤dy2nk,y2mk≤sdy2nk,y2nk−1+sdy2nk−1,y2mk≤sdy2nk,y2nk−1+s2dy2nk−1,y2nk+s2dy2nk,y2mk,we obtain
(26)εs≤limsupk⟶∞dy2nk−1,y2mk≤s2ε.By the same arguments, we obtain(27)εs2≤limsupk⟶∞dy2mk+1,y2nk−1≤εs3,and consequently
(28)εs≤limsupk⟶∞dy2mk+1,y2nk.Similarly, we obtain(29)ε≤liminfk⟶∞dy2nk,y2mk≤εs,(30)εs≤liminfk⟶∞dy2nk−1,y2mk≤s2ε,(31)εs2≤liminfk⟶∞dy2mk+1,y2nk−1≤εs3.Moreover, we have(32)Mdx2mk,x2nk−1=maxdy2mk,y2nk−1,dy2mk+1,y2mk,dy2nk,y2nk−1,dy2mk,y2nk+dy2mk+1,y2nk−12s.Using (24) and (26)–(28), we obtain
(33)εs≤limsupk⟶∞Mdx2mk,x2nk−1≤εs3.By the same arguments, we obtain(34)τr≤limsupkMδx2mk,x2nk−1≤τr3.Using (31), we get
(35)ε2s+ε2s3=minεs,ε+ε/s22s≤maxliminfk⟶∞dy2mk,y2nk−1,liminfk⟶∞dy2nk,y2mk+liminfk⟶∞dy2mk+1,y2nk−12s≤maxεs2,εs+εs32s=εs2.Then,(36)ε2s+ε2s3≤limk⟶∞infMdx2mk,x2nk−1≤εs2.Similarly, we obtain(37)τ2r+τ2r3≤limk⟶∞infMδx2mk,x2nk−1≤τr2.Consequently,(38)ψεs3≤ψs4limk⟶∞supdy2mk+1,y2nk≤ψlimk⟶∞supMδx2mk,x2nk−1−limk⟶∞infφMδx2mk,x2nk−1≤ψτr3−φlimk⟶∞infMδx2mk,x2nk−1,ψτr3≤ψr4limk⟶∞supδy2mk+1,y2nk≤ψlimk⟶∞supMdx2mk,x2nk−1−limk⟶∞infφMdx2mk,x2nk−1≤ψεs3−φlimk⟶∞infMdx2mk,x2nk−1,which implies that(39)ψεs3≤ψτr3≤ψεs3−φlimk⟶∞infMdx2mk,x2nk−1,ψτr3≤ψεs3≤ψτr3−φlimk⟶∞infMδx2mk,x2nk−1.It follows that(40)φlimk⟶∞infMδx2mk,x2nk−1=0.We have also(41)φlimk⟶∞infMdx2mk,x2nk−1=0.Then,(42)limk⟶∞infMdx2mk,x2nk−1=limk⟶∞infMδx2mk,x2nk−1=0,which contradicts with (36) and (37). It follows that y2n is a Cauchy sequence with b-metric d; it is also a Cauchy sequence in X,δ. And then there exists y∈X such that
(43)limn⟶∞fx2n=limn⟶∞Tx2n+1=limn⟶∞gx2n+1=limn⟶∞Px2n+2=y.Step 4.
y is a common fixed point of f,g,P, and T.SinceP is continuous, we have
(44)limn⟶∞P2x2n+2=limn⟶∞Pfx2n=Py.Moreover, the pairf,P is compatible, then limn⟶∞dfPx2n,Pfx2n=0. The triangle inequality implies
(45)limn⟶∞fPx2n=Py.SincePx2n+2=gx2n+1≤x2n+1, we obtain by (6),
(46)ψs4dfPx2n+2,gxn+1≤ψMδPx2n+2,x2n+1−φMδPx2n+2,x2n+1,ψr4δfPx2n+2,gxn+1≤ψMdPx2n+2,x2n+1−φMdPx2n+2,x2n+1.Using Lemma (7), we obtain
(47)1s2dPy,y≤limn⟶∞supMdPx2n+2,x2n+1≤s2dPy,y,1r2δPy,y≤limn⟶∞supMδPx2n+2,x2n+1≤r2δPy,y.Consequently,(48)ψs2dPy,y≤ψr2δPy,y−φ1r2δPy,y,ψr2δPy,y≤ψs2dPy,y−φ1s2dPy,y,and then
(49)ψs2dPy,y≤ψs2dPy,y−φ1s2dPy,y,ψr2δPy,y≤ψr2δPy,y−φ1r2δPy,y,which givesφ1/s2dPy,y=0 or equivalently Py=y.We havegx2n+1≼x2n+1 and gx2n+1⟶y, then y≼x2n+1, which gives
(50)ψs4dfy,gx2n+1≤ψMδy,x2n+1−φMδy,x2n+1,ψr4δfy,gx2n+1≤ψMdy,x2n+1−φMdy,x2n+1,with
(51)limnMdy,x2n+1=dfy,y,limnMδy,x2n+1=δfy,y.Ifn⟶∞ in (50) and using Lemma (7), we obtain
(52)ψdfy,y≤ψs41sdfy,y≤ψδfy,y−φδfy,y,ψδfy,y≤ψr41rδfy,y≤ψdfy,y−φdfy,y,which leads to φdfy,y=φδfy,y=0, and consequently, fy=y.On the other hand, we havefX⊆TX, then there exists a point v∈X such that fy=Tv.Moreover,v≼Tv=fy≼y gives
(53)ψdTv,gv=ψdfy,gv≤ψMδy,v−φMδy,v,ψδTv,gv=ψδfy,gv≤ψMdy,v−φMdy,v,with Mdy,v=dgv,Tv and Mδy,v=δgv,Tv. Consequently,
(54)ψdTv,gv≤ψδgv,Tv−φδgv,Tv,ψδTv,gv≤ψdgv,Tv−φdgv,Tv,which implies gv=Tv. And since g,T is weakly compatible, we obtain
(55)gy=gfy=gTv=Tgv=Tfy=Ty.fx2n≼x2n and fx2n⟶y, we obtain y≼x2n and then
(56)ψs4dfx2n,gy≤ψMδx2n,y−φMδx2n,y,ψr4δfx2n,gy≤ψMdx2n,y−φMdx2n,y.Ifn⟶+∞, and using Lemma (7), we obtain
(57)1sdy,gy≤limninfMdx2n,y≤limnsupMdx2n,y≤sdy,gy,1sδy,gy≤limninfMδx2n,y≤limnsupMδx2n,y≤sdy,gy.Ifn⟶+∞ in (56), and using Lemma (7), we obtain
(58)ψsdy,gy≤ψs41sdy,gy≤ψrδy,gy−φ1rδy,gy,ψrδy,gy≤ψr41rδy,gy≤ψsdy,gy−φ1sdy,gy,which implies dy,gy=δy,gy=0 and then y=gy.We conclude thatfy=gy=Py=Ty=y.Iff is continuous, the proof is the same. The same goes for condition (b) of Assumption 9.Step 5.
Suppose thatu and v are two common fixed points of f,g,P, and T but du,v>0 with u and v are comparable. We have
(59)ψdu,v=ψdfu,gv≤ψs4dfu,gv≤ψMδu,v−φMδu,v,ψδu,v=ψδfu,gv≤ψr4δfu,gv≤ψMdu,v−φMdu,v,with Mδu,v=δu,vandMdu,v=du,v.So from (59), we have
(60)ψdu,v≤ψδu,v−φδu,v,ψδu,v≤ψdu,v−φdu,v.Hence,ψdu,v≤ψdu,v−φdu,v. Therefore, u=v.Example 11.
LetX=0,1; we define on X the b-metrics dx,y=x−y2 with s=2 and δx,y=x−y3 with r=4. We endow X with the partial order ≼ given by
(61)y≼x⇔x≤yforallx,y∈X.
And we definef,g,P, and T on X by
(62)fx=0,ifx≤18,164,ifx∈18,1,gx=0forallx∈X,Tx=0,ifx=0,x,ifx∈0,18,1,ifx∈18,1,Px=0,ifx=0,18,ifx∈0,18,1,ifx∈18,1.
Obviously, conditions (i) and (ii) of Theorem10 are satisfied. Moreover, f,P is weakly compatible, g,T is compatible and g is continuous.The control functionsψ,φ:0,∞⟶0,∞ are defined as ψt=9/8t and φt=1/8t for all t∈0,∞.To prove thatf,g,P, and T satisfy (6), for this, we consider the following cases:
(i)
Ifx∈0,1/8 and y∈0,1, then dfx,gy=0=δfx,gy and (6) are satisfied.(ii)
Ifx∈1/8,1 and y=0, then(63)ψs4dfx,gy=ψ16dfx,gy=ψ1256=92048<1=δPx,Ty≤Mδx,y=ψMδx,y−φMδx,y,ψr4δfx,gy=ψ256δfx,gy=ψ11024=98192<1=dPx,Ty≤Mdx,y=ψMdx,y−φMdx,y.(iii)
Ifx∈1/8,1 and y∈0,1/8, then(64)ψs4dfx,gy=ψ16dfx,gy=ψ1256=92048<343512≤δPx,Ty≤Mδx,y=ψMδx,y−φMδx,y,ψr4δfx,gy=ψ256δfx,gy=ψ11024=98192<4964≤dPx,Ty≤Mdx,y=ψMdx,y−φMdx,y.(iv)
Ifx∈1/8,1 and y∈1/8,1,(65)ψs4dfx,gy=ψ16dfx,gy=ψ1256=92048<1=δgy,Ty≤Mδx,y=ψMδx,y−φMδx,y,ψr4δfx,gy=ψ256δfx,gy=ψ11024=98192<1=dgy,Ty≤Mdx,y=ψMdx,y−φMdx,y.Thus, all conditions of Theorem (10) are satisfied. Moreover, 0 is the unique common fixed point of f,g,P, and T.If we takeP and T as the identity maps on X in Theorem (10), we conclude the following corollary.Corollary 12.
We consider two self-mappingsf and g in ordered complete two b-metric space X,d,δ,≼ that fulfill the following conditions:
(i)
f,g is dominated(ii)
ψ∈Ψ, φ∈Φ and for every two comparable elements x,y∈X, we have(66)ψs4dfx,gy≤ψNδx,y−φNδx,y,ψr4δfx,gy≤ψNdx,y−φNdx,y.(iii)
Assumption8 is satisfied
Thenf and g have a unique common fixed point in X.
If we takeψt=t for t∈0,∞ in Corollary (12), we have the following corollary.Corollary 13.
We consider two self-mappingsf and g in ordered complete two b-metric space X,d,δ,≼ that fulfill the following conditions:
(i)
f,g is dominated(ii)
φ∈Φ and for every two comparable elements x,y∈X, we have(67)s4dfx,gy≤Nδx,y−φNδx,y,r4δfx,gy≤Ndx,y−φNdx,y.(iii)
Assumption8 is satisfied
Then,f and g have a unique common fixed point in X.
If we takeφt=1−kt for k∈0,1 in Corollary (13), we obtain the following result.Corollary 14.
We consider two self-mappingsf and g in ordered complete two b-metric space X,d,δ,≼ that fulfill the following conditions:
(i)
f,g is dominated(ii)
k∈0,1 and for every two comparable elements x,y∈X, we have(68)s4dfx,gy≤kNδx,y,r4δfx,gy≤kNdx,y.(iii)
Assumption8 is satisfied
Then,f and g have a unique common fixed point in X.Theorem 15.
We consider two self-mappingsf,g,P, and T in ordered complete two b-metric space X,d,δ,≼ that fulfill the following conditions:
(i)
f,g is dominated and P,T is dominating(ii)
fX⊆TX and gX⊆PX(iii)
ψ∈Ψ, φ∈Φ and for every two comparable elements x,y∈X, we have(69)ψs4dfx,gy≤ψQdx,y−φQdx,y.(iv)
Assumptions8 and 9 are satisfied
Then,f,g,P, and T have a unique common fixed point in X.Proof.
If we follow similar arguments to those given in Theorem (10), we have the following steps.Step 1.
Ify2k+1=y2k for some k, then ynn is constant.So, from (69), we have
(70)ψdy2k+1,y2k+2≤ψs4dy2k+1,y2k+2=ψs4dfx2k,gx2k+1≤ψQdx2k,x2k+1−φQdx2k,x2k+1,where
(71)Qdx2k,x2k+1=a1dPx2k,Tx2k+1+a2dfx2k,Px2k+a3dgx2k+1,Tx2k+1+a4dPx2k,gx2k+1+a5dfx2k,Tx2k+1=a1dy2k,y2k+1+a2dy2k+1,y2k+a3dy2k+2,y2k+1+a4dy2k,y2k+2+a5dy2k+1,y2k+1=a3dy2k+2,y2k+1+a4dy2k,y2k+2≤a3+a4sdy2k+1,y2k+2.So, from (70) and (71), we obtain
(72)ψdy2k+1,y2k+2≤ψa3+a4sdy2k+1,y2k+2−φQdx2k,x2k+1≤ψdy2k+1,y2k+2−φQdx2k,x2k+1,which gives φQdx2k,x2k+1=0, and so, Qdx2k,x2k+1=0, since a3,a4>0 which further implies that y2k=y2k+1=y2k+2. Hence, yn is a constant sequence and y2k is a common fixed point of f,g,P, and T.In the following, we can assume thatdy2n,y2n+1>0 for each n.Step 2.
The sequencedy2n,y2n+1are monotone nonincreasing.Sincex2n and x2n+1 are comparable, from (69), we have
(73)ψdy2n+1,y2n+2≤ψs4dy2n+1,y2n+2=ψs4dfx2n,gx2n+1≤ψQdx2n,x2n+1−φQdx2n,x2n+1≤ψQdx2n,x2n+1.Hence,(74)dy2n+1,y2n+2≤Qdx2n,x2n+1,where
(75)Qdx2n,x2n+1=a1dPx2n,Tx2n+1+a2dfx2n,Px2n+a3dgx2n+1,Tx2n+1+a4dPx2n,gx2n+1+a5dfx2n,Tx2n+1=a1dy2n,y2n+1+a2dy2n+1,y2n+a3dy2n+2,y2n+1+a4dy2n,y2n+2+a5dy2n+1,y2n+1=a1+a2dy2n,y2n+1+a3dy2n+2,y2n+1+a4dy2n,y2n+2≤a1+a2dy2n,y2n+1+a3dy2n+2,y2n+1+a4sdy2n,y2n+1+dy2n+1,y2n+2≤a1+a2+a4sdy2n,y2n+1+a3+a4sdy2n+1,y2n+2≤a1+a2+a3+2a4smaxdy2n,y2n+1,dy2n+1,y2n+2≤maxdy2n,y2n+1,dy2n+1,y2n+2.If for somen, dy2n+2,y2n+1≥dy2n+1,y2n>0, then (74) gives that Qdx2n,x2n+1=dy2n+2,y2n+1, and from (69), we have
(76)ψdy2n+1,y2n+2≤ψs4dy2n+2,y2n+1≤ψQdx2n,x2n+1−φQdx2n,x2n+1=ψdy2n+2,y2n+1−φdy2n+2,y2n+1.Andφdy2n+2,y2n+1=0 or equivalently dy2n+2,y2n+1=0, a contradiction.Then,y2n+2,y2n+1≤Qdx2n,x2n+1=dy2n+1,y2n. By the same arguments, we have
(77)dy2n+3,y2n+2≤Qdx2n+2,x2n+1=dy2n+2,y2n+1.Therefore,dyn,yn+1 is a nonincreasing sequence, and so, there exists c≥0 so that
(78)limn⟶+∞dyn,yn+1=limn⟶+∞Qdxn,xn+1=c.Now, we demonstrate thatc=0. Suppose that c>0. Since we have
(79)ψdy2n+1,y2n+2≤ψs4dy2n+2,y2n+1≤ψQdx2n,x2n+1−φQdx2n,x2n+1.Then, passing the upper limit asn⟶+∞, it implies that
(80)ψc≤ψc−liminfn⟶+∞φQdx2n,x2n+1=ψc−φliminfn⟶+∞Qdx2n,x2n+1=ψc−φc,which shows a contradiction. Thus, c=0. Hence,
(81)limn⟶+∞dyn,yn+1=0.Step 3.
yn is a Cauchy sequence.Assume that there existsε>0 for which we can find subsequences y2mk and y2nk of y2n such that nk is the smallest index for which 2nk>2mk>k, dy2mk,y2nk≥ε, and dy2mk−2,y2nk<ε.Following the same argument as in Theorem (10), we have
(82)ε≤limsupk⟶∞dy2mk,y2nk≤εs,(83)εs≤limsupk⟶∞dy2mk−1,y2nk≤εs2,(84)εs2≤limsupk⟶∞dy2nk+1,y2mk−1≤εs3,(85)εs≤limsupk⟶∞dy2nk+1,y2mk.As(86)Qdx2nk,x2mk−1=a1dPx2nk,Tx2mk−1+a2dfx2nk,Px2nk+a3dgx2mk−1,Tx2mk−1+a4dPx2nk,gx2mk−1+a5dfx2nk,Tx2mk−1=a1dy2nk,y2mk−1+a2dy2nk+1,y2nk+a3dy2mk,y2mk−1+a4dy2nk,y2mk+a5dy2nk+1,y2mk−1.So, from (86), (81), (82), (83), and (84), we obtain
(87)a1εs+a4ε+a5εs2≤limsupk⟶∞Qdx2nk,x2mk−1≤a1εs2+a4εs+a5εs3,which implies(88)εs2a1s+a4s2+a5≤limsupk⟶∞Qdx2nk,x2mk−1≤a1+a4+a5εs3,which implies(89)εs2a1+a4+a5≤limsupk⟶∞Qdx2nk,x2mk−1≤εs3.Similarly, we have(90)εs2a1+a4+a5≤liminfk⟶∞Qdx2nk,x2mk−1≤εs3.As(91)ψs4dy2nk+1,y2mk=ψs4dfx2nk,gx2mk−1≤ψQdx2nk,x2mk−1−φQdx2nk,x2mk−1.Taking the upper limit ask⟶∞ and using (85) and (89), we obtain
(92)ψεs3≤ψs4limsupk⟶∞dy2nk+1,y2mk≤ψlimsupk⟶∞Qdx2nk,x2mk−1−liminfk⟶∞φQdx2nk,x2mk−1≤ψεs3−φliminfk⟶∞Qdx2nk,x2mk−1,which implies that(93)φliminfk⟶∞Qdx2nk,x2mk−1=0.So,liminfk⟶∞Qdx2nk,x2mk−1=0, which contradicts (90). It follows that y2n is a Cauchy sequence in X. Since X is complete, there exists y∈X so that
(94)limn⟶∞fx2n=limn⟶∞Tx2n+1=limn⟶∞gx2n+1=limn⟶∞Px2n+2=y.Step 4.
y is a common fixed point of f,g,P, and T.SinceP is continuous, we have
(95)limn⟶∞P2x2n+2=Py,limn⟶∞Pfx2n=Py.Using the triangle inequality inb-metric space, we get
(96)dfPx2n,Py≤sdfPx2n,Pfx2n+dPfx2n,Py.Since the pairf,P is compatible, limn⟶∞dfPx2n,Pfx2n=0. So, passing the upper limit when n⟶∞ from the above inequality and Lemma (7), we have
(97)limsupn⟶∞dfPx2n,Py≤slimsupn⟶∞dfPx2n,Pfx2n+limsupn⟶∞dPfx2n,Py=0.Hence,(98)limn⟶∞fPx2n=Py.AsPx2n+2=gx2n+1≼x2n+1 so, from (69), we have
(99)ψs4dfPx2n+2,gx2n+1≤ψQdPx2n+2,x2n+1−φQdPx2n+2,x2n+1,where(100)QdPx2n+2,x2n+1=a1dP2x2n+2,Tx2n+1+a2dfPx2n+2,P2x2n+2+a3dgx2n+1,Tx2n+1+a4dP2x2n+2,gx2n+1+a5dfPx2n+2,Tx2n+1.Now, by using Lemma (7), we get
(101)limsupn⟶∞QdPx2n+2,x2n+1≤a1s2dPy,y+a4s2dPy,y+a5s2dPy,y≤a1+a4+a5s2dPy,y≤s2dPy,y.Similarly,(102)1s2a1+a4+a5dPy,y≤liminfn⟶∞QdPx2n+2,x2n+1.Hence, by taking the upper limit in (99) and using Lemma (7), we obtain
(103)ψs2dPy,y≤ψs2dPy,y−φ1s2a1+a4+a5dPy,y,which givesφ1/s2a1+a4+a5dPy,y=0 or equivalently Py=y. Now, since gx2n+1≼x2n+1 and gx2n+1⟶y as n⟶∞, then y≼x2n+1, and from (69), we have
(104)ψs4dfy,gx2n+1≤ψQdy,x2n+1−φQdy,x2n+1,where(105)Qdy,x2n+1=a1dPy,Tx2n+1+a2dfy,Py+a3dgx2n+1,Tx2n+1+a4dPy,gx2n+1+a5dfy,Tx2n+1.By using Lemma (7), we get
(106)limsupn⟶∞Qdy,x2n+1≤a1sdPy,y+a2dfy,Py+a3s2dy,y+a4sdPy,y+a5sdfy,y≤a2+a5sdfy,y≤dfy,y≤sdfy,y.Similarly,(107)a2+a5sdfy,y≤liminfn⟶∞Qdy,x2n+1.Taking the upper limit asn⟶∞ in (104) and using Lemma (7), we have
(108)ψs3dfy,y=ψs41sdfy,y≤ψsdfy,y−liminfn⟶∞φQdy,x2n+1≤ψs3dfy,y−φliminfn⟶∞Qdy,x2n+1≤ψs3dfy,y−ψa2+a5sdfy,y,which impliesψa2+a5/sdfy,y=0. So, fy=y.On the other hand, we havefX⊆TX, so there exists a point v∈X such that fy=Tv.Suppose thatgv≠Tv. Since v≼Tv=fy≼y, from (69), we have
(109)ψdTv,gv=ψdfy,gv≤ψQdy,v−φQdy,v,where(110)Qdy,v=a1dPy,Tv+a2dfy,Py+a3dgv,Tv+a4dPy,gv+a5dfy,Tv=a3dgv,Tv+a4dPy,gv=a3+a4dgv,Tv≤dgv,Tv.So, from (109), we have
(111)ψdTv,gv≤ψdgv,Tv−φQdy,v.This implies thatφQdy,v=0, so dgv,Tv=0 which shows a contradiction; therefore, gv=Tv. Since the pair g,T is weakly compatible, gy=gfy=gTv=Tgv=Tfy=Ty, and y is the coincidence point of g and T.Sincefx2n≼x2n and fx2n⟶y as n⟶∞, it implies that y≼x2n, and from (69), we obtain
(112)ψs4dfx2n,gy≤ψQdx2n,y−φQdx2n,y,where(113)Qdx2n,y=a1dPx2n,Ty+a2dfx2n,Px2n+a3dgy,Ty+a4dPx2n,gy+a5dfx2n,Ty.By using Lemma (7), we have
(114)1sa1+a4+a5dy,gy≤liminfn⟶∞Qdx2n,y≤limsupn⟶∞Qdx2n,y≤sa1+a4+a5dy,gy≤sdy,gy.Passing the upper limit asn⟶∞, and using (114), we have
(115)ψs3dy,gy=ψs41sdy,gy≤ψsdy,gy−φ1sa1+a4+a5dy,gy≤ψs3dy,gy−φ1sa1+a4+a5dy,gy,which implies thaty=gy; therefore, fy=gy=Py=Ty=y.Iff is continuous, the proof is the same. The same goes for condition (b) of Assumption 9.Step 5.
Suppose thatu and v are two common fixed points of f,g,P, and T, but du,v>0 with u and v are comparable. By assumption, we can apply (69) to obtain
(116)ψdu,v=ψdfu,gv≤ψs4dfu,gv≤ψQdu,v−φQdu,v,
where(117)Qdu,v=a1dPu,Tv+a2dfu,Pu+a3dgv,Tv+a4dPu,gv+a5dfu,Tv=a1du,v+a4du,v+a5du,v=a1+a4+a5du,v≤du,v.
Hence,(118)ψdu,v≤ψdu,v−φQdu,v,implies φQdu,v=0, which is a contradiction. Therefore, u=v. The converse is clear.Example 16.
We consider the data of the previous example and takea1=3/32,a2=a5=3/64,a3=5/16, and a4=7/64. This permits only to verify the condition (69) of Theorem (15). We shall distinguish two cases:
(i)
Ifx∈0,1/8 and y∈0,1, then dfx,gy=0 and (69) is satisfied(ii)
Ifx∈1/8,1 and y∈0,1, then(119)ψs4dfx,gy=ψ16dfx,gy=ψ128=9211<92111323128=32626−12212=a2dfx,Px≤Qdx,y=ψQdx,y−φQdx,y.Thus, (69) is satisfied for all x,y∈X. Therefore, all conditions of Theorem (15) are satisfied. Moreover, 0 is the unique common fixed point of f,g,P, and T.If we takeP and T as the identity maps on X in Theorem (15), we conclude the following corollary.Corollary 17.
We consider two self-mappingsf and g in ordered complete two b-metric space X,d,δ,≼ that fulfill the following conditions:
(i)
f,g is dominated(ii)
ψ∈Ψ, φ∈Φ and for every two comparable elements x,y∈X, we have(120)ψs4dfx,gy≤ψRdx,y−φRdx,y.(iii)
Assumption8 is satisfied
Then,f and g have a unique common fixed point in X.If we takeψt=t for t∈0,∞ in Corollary (17), we have the following corollary.Corollary 18.
We consider two self-mappingsf and g in ordered complete two b-metric space X,d,δ,≼ that fulfill the following conditions:
(i)
f,g is dominated(ii)
φ∈Φ and for every two comparable elements x,y∈X, we have(121)s4dfx,gy≤Rdx,y−φRdx,y.(iii)
Assumption8 is satisfied
Then,f and g have a unique common fixed point in X.


## 3. Applications

Consider the following system of integral equations:(122)xt=∫0TKt,u,yudu+L1t,forallt∈0,T,yt=∫0TKt,u,xudu+L2t,whereT>0,K:0,T×0,T×ℝ⟶ℝ and L1,L2:0,T⟶ℝ are continuous functions.LetX=C0,T×C0,T, where C0,T is the set of real continuous on 0,T.We endowX with the two b-metric:
(123)dx,y,x1,y1=maxmaxt∈0,Tx1t−xtp,maxt∈0,Ty1t−ytp,δx,y,x1,y1=maxmaxt∈0,Tx1t−xtq,maxt∈0,Ty1t−ytq,for all x,y,x1,y1∈X. Clearly, X,d,δ is a complete two b-metric space with s=2p−1,r=2q−1, and p,q∈1,∞ with 1/p+1/q=1. We endow X with the partial order ≼ given by
(124)x,y≼x1,y1⇔xt≤x1t,yt≤y1tforallt∈0,T.Clearly, the spaceX,d,δ,≼ is an ordered complete two b-metric space.Consider the mappingF:X⟶X defined as follows: Fx,y=F1x,y,F2x,y with
(125)F1x,yt=∫0TKt,u,yudu+L1t,forallt∈0,T.F2x,yt=∫0TKt,u,xudu+L2t,Suppose that the following hypotheses hold:(i)
there exists a continuous functionε:0,T×0,T⟶ℝ+, for all t,u∈0,T; i=p,q and for all comparable x,y∈C0,T, we have(126)Kt,u,xu−Kt,u,yui≤εt,uxu−yui/i−1.(ii)
maxi∈p,q16Ti−1maxt∈I∫0Tεt,udu<1.Theorem 19.
Under the assumptions (i) and (ii), the system integral equations (122) have a unique solution in the set C0,T.Proof.
By the condition (i), we have for allt∈0,T and for all comparable x,y∈X,
(127)F1x,yt−F1x1,y1tp≤∫0T∣Kt,u,yu−Kt,u,y1u∣dup≤∫0T1q1/q∫0T∣Kt,u,yu−Kt,u,yupdu1/pp≤Tp/q∫0T∣Kt,u,yu−Kt,u,y1u∣pdu≤Tp−1∫0Tεt,uyu−y1up/p−1du≤Tp−1∫0Tεt,umaxu∈0,Tyu−y1uqdu≤Tp−1∫0Tεt,uduδx,y,x1,y1.
By a similar calculus, we obtain(128)F2x,yt−F2x1,y1tp≤Tp−1∫0Tεt,uduδx,y,x1,y1.
It follows that(129)dFx,y,Fx1,y1≤Tp−1maxt∈0,T∫0Tεt,uduNδx,y,x1,y1.
Similarly,(130)δFx,y,Fx1,y1≤Tq−1maxt∈0,T∫0Tεt,uduNdx,y,x1,y1,
which implies that(131)s4dfx,y,gx1,y1≤kNδx,y,x1,y1,r4δfx,y,gx1,y1≤kNdx,y,x1,y1,with
(132)k=maxi∈p,q16Ti−1maxt∈I∫0Tεt,udu<1.
Thus, all the hypotheses of Corollary (14) (with F=f=g) are satisfied, and hence, the mapping F has a unique fixed point in X, which is a solution of the system of nonlinear integral equations (122).


## 4. Conclusion

In this manuscript, we have obtained interesting results on the coupled fixed point theorems that give a generalization of well-known results in the field. We have shown that these results are very useful to solve the system of integral equations.



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