# Fixed Point Theorems for Generalizedαs-ψ-Contractions with Applications

Authors: Zhenhua Ma; Muhammad Nazam; Sami Ullah Khan; Xiangling Li
Journal: Journal of Function Spaces
 (2018)

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## Abstract

We study the sufficient conditions for the existence of a unique common fixed point of generalizedαs-ψ-Geraghty contractions in an αs-complete partial b-metric space. We give an example in support of our findings. Our work generalizes many existing results in the literature. As an application of our findings we demonstrate the existence of the solution of the system of elliptic boundary value problems.


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## Body

## 1. Introduction and Preliminaries

The Banach contraction mapping principle is very important in modern mathematics. For decades, several authors have studied existence of fixed points by contraction mappings, such as fuzzy mappings and others, and also get some important results, for details we can see [1–5, 5–12]. Now the theory of fixed point has been applied in many applied mathematics [13, 14] besides integral equations and differential equations [15]. For decades, people have done a lot of research on this issue and got a lot of important results [16–20].As is well known, the existence of the solution of boundary value problems is an important of differential equations. In this paper we study the sufficient conditions for the existence of a unique common fixed point of generalizedαs-ψ-Geraghty contractions in an αs-complete partial b-metric space. As an application of our findings we demonstrate the existence of the solution of the system of elliptic boundary value problems.We first give some conceptions of this paper. In 1973, Geraghty studied a generalization of Banach contraction principle. In 2013, Cho introduced the notion ofα-Geraghty contractive type mappings and established some unique fixed point theorems for such mappings in complete metric spaces. Popescu defined the concept of triangular α-orbital admissible mappings and proved the unique fixed point theorems for the mentioned mappings which are generalized α-Geraghty contraction type mappings. On the other hand, Karapinar proved the existence of a unique fixed point theorem for a triangular α-admissible mapping which is a generalized α-ψ-Geraghty contraction type mapping. Shukla [21] introduced the concept of partial b-metric space and established some fixed point theorems. We have Figure 1 where arrows stand for inclusions. The inverse inclusions do not hold.Figure 1In this paper, we introduce the notion of generalizedαs-ψ-Geraghty contraction type mappings and develop some new common fixed point theorems for such mappings in an αs-complete partial b-metric space. An example and an application are given to support the theory.We denote the set of natural numbers, rational numbers,-∞,+∞, 0,+∞, and 0,+∞ by N, Q, R, R+, and R0+, respectively.First we recall some definitions and properties of a partialb-metric space.Shukla generalized the notion ofb-metric, as follows.Definition 1 (see [21]).
LetX be a nonempty set and s≥1 be a real number. A mapping pb:X×X→R0+ is said to be a partial b-metric if it satisfies following axioms, for all x,y,z∈X, (pb1)
x=y if and only if pb(x,y)=pb(x,x)=pb(y,y)(pb2)
pb(x,x)≤pb(x,y)(pb3)
pb(x,y)=pb(y,x);(pb4)
pb(x,y)≤spb(x,z)+pb(z,y)-pb(z,z).
The triplet(X,pb,s) is called a partial b-metric space.Remark 1.
The self-distancepb(x,x), referred to the size or weight of x, is a feature used to describe the amount of information contained in x.Remark 2.
Obviously, every partial metric space is a partialb-metric space with coefficient s=1 and every b-metric space is a partial b-metric space with zero self-distance. However, the converse of this fact needs not to hold.Example 3.
LetX=R+ and k>1; the mapping pb:X×X→R+ defined by(1)pbx,y=x∨yk+x-ykfor  allx,y∈Xis a partial b-metric on X with s=2k. For x=y, pb(x,x)=xk≠0, so pb is not a b-metric on X.Letx,y,z∈X such that x>z>y. Then following inequality always holds:(2)x-yk>x-zk+z-yk.Since pb(x,y)=xk+(x-y)k and pb(x,z)+pb(z,y)-pb(z,z)=xk+(x-z)k+(z-y)k,(3)pbx,y>pbx,z+pbz,y-pbz,z.This shows that pb is not a partial metric on X.Example 4 (see [21]).
LetX be a nonempty set and p be a partial metric defined on X. The mapping pb:X×X→R+ defined by(4)pbx,y=px,yqfor  allx,y∈Xandq>1defines a partial b-metric.Definition 5.
Let(X,pb,s) be a partial b-metric space. The mapping dpb:X×X→R0+ defined by(5)dpbx,y=2pbx,y-pbx,x-pby,yfor  allx,y∈Xdefines a metric on X, called induced metric.In partialb-metric space (X,pb,s), we immediately have a natural definition for the open balls:(6)Bpbx;ϵ=y∈X∣pbx,y<pbx,x+ϵfor  allx∈X.Remark 6.
The open balls in a partialb-metric space (X,pb,s) may not be open set.The following example justifies Remark6.Example 7.
LetX={a,b,c} and define pb as follows: pb(a,a)=pb(c,c)=1, pb(b,b)=1/2, pb(a,b)=pb(b,a)=3, pb(a,c)=p(c,a)=3/2, pb(b,c)=pb(c,b)=1. Then p is a partial b-metric, c∈Bpb(a;1) but, for any r>0, Bpb(c;r) does not lie in Bpb(a;1). This implies that Bpb(a;1) is not an open set in (X,pb,s).Definition 8.
Let(X,pb,s) be a partial b-metric space. (1)
A sequence{xn}n∈N in (X,pb,s) is called a Cauchy sequence if limn,m→∞pb(xn,xm) exists and is finite.(2)
A partialb-metric space (X,pb,s) is said to be complete if every Cauchy sequence {xn}n∈N in X converges with respect to topology induced by its convergence, to a point υ∈X such that (7)pbx,x=limn,m→∞⁡pbxn,xm.Lemma 9 (see [21]).
Let(X,pb,s) be a partial b-metric space. Then (1)
every Cauchy sequence in(X,dpb) is also a Cauchy sequence in (X,pb,s) and vice versa;(2)
a partialb-metric (X,pb,s) is complete if and only if the metric space (X,dpb) is complete;(3)
a sequence{xn}n∈N in X converges to a point υ∈X if and only if (8)limn→∞⁡pbυ,xn=pbυ,υ=limn→∞⁡pbxn,xm.Remark 10.
We know that in a metric space limit of a convergent sequence is always unique but in a partialb-metric space the limit of a convergent sequence may not be unique. Indeed, if X=R+, let σ>0 be any constant. Define pb:X×X→R+ by pb(x,y)=x∨y+σ for all x,y∈X, then (X,pb,s) is a partial b-metric space with arbitrary coefficient s≥1. Define the sequence xn in X by xn=ρ for all n∈N. One can note that if y≥ρ then pb(xn,y)=y+σ=pb(y,y); thus limn→∞⁡pb(xn,y)=pb(y,y) for all y≥ρ. Hence, the limit of a convergent sequence is not unique.Definition 11 (see [22]).
LetT:X→X and α:X×X→R0+ be two mappings. We say that T is α-admissible if α(x,y)≥1 implies α(Tx,Ty)≥1.Definition 12 (see [22]).
LetT:X→X and α:X×X→R0+ be two mappings. Then T is said to be triangular α-admissible if T satisfies the following conditions:(T1)
T is α-admissible.(T2)
α(x,u)≥1 and α(u,y)≥1 imply α(x,y)≥1.Definition 13 (see [23]).
LetT:X→X and α:X×X→R0+ be two mappings. Then T is said to be α-orbital admissible if(T3)
α(x,Tx)≥1 implies α(Tx,T2x)≥1.Definition 14 (see [23]).
LetT:X→X and α:X×X→R0+ be two mappings. Then T is said to be triangular α-orbital admissible if T is α-orbital admissible and (T4)
α(x,y)≥1 and α(y,Ty)≥1 imply α(x,Ty)≥1.LetΩ denote the class of all mappings β:R0+→[0,1/s) such that, for any bounded sequence tn of positive reals, βtn→1/s implies tn→0.We letΨ denote the class of the functions ψ:[0,∞)→R0+ satisfying the following conditions:(1)
ψ is nondecreasing.(2)
ψ is continuous.(3)
ψt=0 if and only if t=0.


## 2. Main Results

Throughout this paper we letX=X,pb,s be a partial b-metric space, αs:X×X→R0+ be a mapping, and(9)Mx,y=max⁡pbx,y,pbx,Sx,pby,Ty,pbx,Ty+pby,Sx2s.Definition 15.
The spaceX,pb,s is said to be αs-complete if every Cauchy sequence xn in X satisfying αs(xn,xn+1)≥s2 for all n∈N converges in X.Remark 16.
IfX is a complete partial b-metric space, then X is also an αs-complete partial b-metric space but the converse is not true. The following example explains this fact.Example 17.
LetX=(0,∞) and the partial b-metric pb:X×X→[0,∞) be defined by pb(x,y)=(x∨y)2, for all x,y∈X. Define α2:X×X→[0,∞) by(10)α2x,y=4ex-yifx,y∈1,3;0otherwise.It is easy to see that(X,pb,2) is not a complete partial b-metric space, but (X,pb,2) is an α2-complete partial b-metric space. Indeed, if {xn} is a Cauchy sequence in X such that α2(xn,xn+1)≥4, for all n∈N, then xn∈[1,3], for all n∈N. Since [1,3] is a closed subset of R, we see that ([1,3],pb,2) is a complete partial b-metric space and then there exists x∈[1,3] such that xn→x as n→∞.Definition 18.
Let(X,⪯) be a partially ordered set. Two mappings S,T:X→X are said to be weakly increasing mappings, if S(x)⪯TS(x) and T(y)⪯ST(y) hold for all x,y∈X.Example 19.
LetX=R+. Define S,T:X→X by (11)Sx=x1/2ifx∈0,1x2ifx∈1,∞andTx=xifx∈0,12xifx∈1,∞Then, S,T are weakly increasing mappings.Definition 20.
The self-mappingsS,T:X→X are said to be αs-orbital admissible if the following condition holds.
α s ( x , S x ) ≥ s 2 and αs(x,Tx)≥s2 imply αs(Sx,TSx)≥s2 and αs(Tx,STx)≥s2.We note that Definitions13 and 18 are particular cases of Definition 20 (set S=T and define αs(x,y)≥s2 whenever x⪯y or y⪯x, respectively, in Definition 20).Definition 21.
LetS,T:X→X be two mappings. The pair S,T is said to be triangular αs-orbital admissible, if(i)
the self-mappingsS,T are αs-orbital admissible,(ii)
αs(x,y)≥s2, αs(y,Sy)≥s2 and αs(y,Ty)≥s2 imply αs(x,Sy)≥s2 and αs(x,Ty)≥s2.Example 22.
LetM=R0+ and pb(r1,r2)=(r1∨r2)2 for all r1,r2∈M be a partial b-metric with s=2:(12)Sr=rifr∈0,1;1ifr∈1,∞,Tr=r1/3ifr∈0,1;1ifr∈1,∞.Define αs:M×M→R0+ by(13)αsr1,r2=4+r2-r1ifr1,r2∈0,1;0ifr1,r2∈1,∞.Then it is easy to show that the mappings S,T satisfy conditions (i) and (ii) in Definition 21.Lemma 23.
LetS,T:X→X be two mappings such that the pair (S,T) is triangular αs-orbital admissible. Assume that there exists x0∈X such that αs(x0,Sx0)≥s2. Define a sequence xn in X by x2i+1=S(x2i) and x2i+2=T(x2i+1), where i=0,1,2,…. Then for n,m∈N∪{0} with m>n, we have αsxn,xm≥s2.Proof.
Sinceαs(x0,Sx0)=αs(x0,x1)≥s2 and S,T are αs-orbital admissible self-mappings,(14)αsx0,Sx0≥s2impliesαsSx0,TSx0=αsx1,Tx1=αsx1,x2≥s2αsx1,Tx1≥s2impliesαsTx1,STx1=αsx2,Sx2=αsx2,x3≥s2αsx2,Sx2≥s2implies(15)αsSx2,TSx2=αsx3,Tx3=αsx3,x4≥s2.Applying the above argument repeatedly, we obtain αsxn,xm≥s2for  alln,m∈N∪{0} with m=n+1. Since S,T are triangular αs-orbital admissible mappings, αsxn,xm≥s2for  alln,m∈N∪{0}withm>n.Definition 24.
We say the self-mappingS:X→X is an αs-pb-continuous mapping if whenever xn is a sequence in X with αs(xn,xn+1)≥s2 for all n∈N and x∈X such that limn→∞⁡pb(xn,x)=0, then limn→∞⁡pb(S(xn),S(x))=0.Now, we introduce the concept of generalizedαs-ψ-Geraghty contractions as follows.Definition 25.
The self-mappingsS,T defined on (X,pb,s) are called generalized αs-ψ-Geraghty contractions with respect to pb, if there exist β∈Ω, ψ∈Ψ, and L≥0 such that(16)ψαsx,ypbSx,Ty≤βψMx,y.ψMx,y+Lψmin⁡pbx,Sx,pby,Sx,for x,y∈X satisfying αsx,y≥s2.The main result of this section is given by the following:Theorem 26.
LetX,pb,s be an αs-complete partial b-metric space. Let S,T:X→X be generalized αs-ψ-Geraghty contractions satisfying the following conditions: (i)
there existsx0∈X such that αsx0,Sx0≥s2;(ii)
the mappingsS,T are triangular αs-orbital admissible;(iii)
(a)
the mappingsS,T are αs-pb-continuous(b)
xn is a sequence in X such that αsxn,xn+1≥s2 for all n∈N and xn→x∗∈X as n→∞, then there exists a subsequence xnk of xn such that αsxnk,x∗≥s2 for all k∈N.
ThenS and T have a common fixed point in X. In addition, if y∗ is also a common fixed point of the pair (S,T) such that αs(x∗,y∗)≥s2, then x∗=y∗.Proof.
Firstly we prove that the self-mappingsS,T have at most one common fixed point. Suppose that υ and ω are two different common fixed points of S and T. Then S(υ)=υ≠ω=T(ω). It follows that pb(S(υ),T(ω))=pb(υ,ω)>0, pb(υ,υ)=0 and pb(ω,ω)=0. Since αs(υ,ω)≥s2, contractive condition (16) implies (17)ψpbυ,ω=ψpbSυ,Tω≤ψαsυ,ωpbSυ,Tω≤βψMυ,ω·ψMυ,ω+Lψmin⁡pbυ,Sυ,pbω,Sυ<ψMυ,ω=ψpbυ,ω.which is a contradiction. Hence, the pair (S,T) has at most one common fixed point.
(a). By assumption (i) and Lemma 23, we have (18)αsxn,xn+1≥s2,for  alln∈N.For i∈N, we have (19)0<ψpbx2i+1,x2i+2≤ψαsx2i,x2i+1pbSx2i,Tx2i+1≤βψMx2i,x2i+1.ψMx2i,x2i+1+Lψmin⁡pbx2i,Sx2i,pbx2i+1,Sx2i,where (20)Mx2i,x2i+1=max⁡pbx2i,x2i+1,pbx2i,Sx2i,pbx2i+1,Tx2i+1,pbx2i,Tx2i+1+pbx2i+1,Sx2i2s=max⁡pbx2i,x2i+1,pbx2i,x2i+1,pbx2i+1,x2i+2,pbx2i,x2i+2+pbx2i+1,x2i+12s≤max⁡pbx2i,x2i+1,pbx2i,x2i+1,pbx2i+1,x2i+2,pbx2i,x2i+1+pbx2i+1,x2i+22s=max⁡pbx2i,x2i+1,pbx2i+1,x2i+2.If max⁡pbx2i,x2i+1,pbx2i+1,x2i+2=pbx2i+1,x2i+2, then by (29) we have(21)ψpbx2i+1,x2i+2≤βψpbx2i+1,x2i+2.ψpbx2i+1,x2i+2<ψpbx2i+1,x2i+2,which is a contradiction. Thus we conclude that (22)max⁡pbx2i,x2i+1,pbx2i+1,x2i+2=pbx2i,x2i+1.By (29), we get that ψpbx2i+1,x2i+2<ψpbx2i,x2i+1. Since ψ is nondecreasing, we have (23)pbx2i+1,x2i+2<pbx2i,x2i+1.This implies that (24)pbxn+1,xn+2<pbxn,xn+1,for  alln∈N.Hence, we deduce that the sequence pbxn,xn+1 is nonincreasing. Therefore, there exists r≥0 such that limn→∞⁡pbxn,xn+1=r. Now, we shall prove that r=0. Suppose that r>0. By (16), we have (25)ψpbxn+1,xn+2≤ψαsxn,xn+1pbSxn,Txn+1≤βψMxn,xn+1.ψMxn,xn+1+Lψmin⁡pbxn,Sxn,pbxn+1,Sxn,which implies (26)ψpbxn+1,xn+2≤βψpbxn,xn+1.ψpbxn,xn+1.Hence (27)ψpbxn+1,xn+2ψpbxn,xn+1≤βψpbxn,xn+1<1.This implies that limn→∞⁡βψpbxn,xn+1=1. Since β∈Ω, we have(28)limn→∞⁡ψpbxn,xn+1=0,which yields(29)r=limn→∞⁡pbxn,xn+1=0,a contradiction. Now, we claim that {xn} is a Cauchy sequence in (X,pb,s). Suppose, on thw contrary, that {xn} is not a Cauchy sequence; that is, limn,m→∞pb(xn,xm)≠0. Then there exists ϵ>0 for which we can find two subsequences {xmk} and {xnk} of {xn} such that nk is the smallest index for which nk>mk>k, (30)pbxmk,xnk≥ϵ.This means that (31)pbxmk,xnk-1<ϵ.By the triangle inequality, we have (32)ϵ≤pbxmk,xnk≤spbxmk,xnk-1+pbxnk-1,xnk-pbxnk-1,xnk-1≤spbxmk,xnk-1+pbxnk-1,xnk.Thus,(33)ϵs≤pbxmk,xnk<ϵ+pbxnk-1,xnkfor all k∈N. In the view of (33) and (29), we have(34)ϵ≤limk→∞⁡pbxmk,xnk<sϵ.Again by triangle inequality, we have (35)pbxmk,xnk≤spbxmk,xmk+1+pbxmk+1,xnk-pbxmk+1,xmk+1≤spbxmk,xmk+1+pbxmk+1,xnk≤spbxmk,xmk+1+s2pbxmk+1,xnk+1+s2pbxnk+1,xnk-pbxnk+1,xnk+1≤spbxmk,xmk+1+s2pbxmk+1,xnk+1+s2pbxnk+1,xnkand (36)pbxmk+1,xnk+1≤spbxmk+1,xmk+pbxmk,xnk+1-pbxmk,xmk≤spbxmk+1,xmk+spbxmk,xnk+1≤spbxmk+1,xmk+s2pbxmk,xnk+s2pbxnk,xnk+1-pbxnk,xnk≤spbxmk+1,xmk+s2pbxmk,xnk+s2pbxnk,xnk+1.By (29) and (34), we deduce that(37)ϵs2≤limk→+∞⁡pxmk+1,xnk+1<s3ϵ.Also by application of triangle inequality, it follows that(38)ϵs≤limk→∞⁡pbxnk+1,xmk≤s2ϵ.By Lemma 23, since αs(xnk+1,xmk)≥s2, we have(39)εs=max⁡εs,sε4≤limk→∞⁡sup⁡Mxnk+1,xmk≤max⁡s2ε,s2ε4=s2ε.Similarly, we can show that(40)εs=max⁡εs,sε4≤limk→∞⁡inf⁡Mxnk+1,xmk≤max⁡s2ε,s2ε4=s2ε.Thus, concluding above arguments we have (41)ψs2ε≤ψαsxnk+1,xmklimk→∞⁡sup⁡pbxnk+2,xmk+1≤βψlimk→∞⁡sup⁡Mxnk+1,xmkψlimk→∞⁡sup⁡Mxnk+1,xmk+0≤βψs2εψs2ε<ψs2ε,which is a contradiction. Thus xn is a Cauchy sequence in (X,pb,s). Since (X,pb,s) is an αs-complete partial b-metric space, by Lemma 9(2), (X,dpb) is an αs-complete b-metric space. Therefore, the sequence {xn} converges to some x∗∈(X,dpb). By Lemma 9(3), there exists x∗∈X such that limn→∞⁡dpbxn,x∗=0 if and only if(42)limn→∞⁡pbx∗,xn=pbx∗,x∗=limn,m→∞⁡pbxn,xm.Since dpb(x,y)=2pb(x,y)-p(x,x)-pb(y,y), thus, considering (29) and axiom (pb2) with limn→∞⁡dpbxn,x∗=0, we conclude that(43)limn→∞⁡pbxn,xm=0.Combining (42) and (43), we have(44)limn→∞⁡pbx∗,xn=pbx∗,x∗=limn,m→∞⁡pbxn,xm=0.Now limn→∞pb(x∗,xn)=0 implies that limi→∞⁡pbx2i+1,x∗=0 and limi→∞⁡pbx2i+2,x∗=0. As S and T are αs-pb-continuous mappings, we limi→∞⁡pbSx2i+1,Sx∗=0. Thus (45)pbx∗,Sx∗=limi→∞⁡pbx2i+2,Sx∗=limi→∞⁡pbSx2i+1,Sx∗=0,and so x∗=Sx∗, and, similarly, x∗=Tx∗. Therefore S and T have a common fixed point x∗∈X.
(b). From (a) we know that the sequence xn in X defined by x2i+1=Sx2i and x2i+2=Tx2i+1, where i=0,1,2,… with αsxn,xn+1≥s2, for all n∈N converges to x∗∈X. There exists a subsequence xnk of xn such that αsxnk,x∗≥s2 for all k. Therefore, (46)ψpbx2nk+1,Tx∗=ψpbSx2nk,Tx∗≤ψαsx2nk,x∗pbSx2nk,Tx∗≤βψMx2nk,x∗.ψMx2nk,x∗+Lψminpbx2nk,Sx2nk,pbx∗,Sx2nk,which implies(47)ψpbx2nk+1,Tx∗≤βψMx2nk,x∗.ψMx2nk,x∗+Lψmin⁡pbx2nk,Sx2nk,pbx∗,Sx2nk,where(48)Mx2nk,x∗=max⁡pbx2nk,x∗,pbx2nk,Sx2nk,pbx∗,Tx∗,pbx2nk,Sx∗+pbx∗,Tx2nk2s≤max⁡pbx2nk,x∗,pbx2nk,x2nk+1,pbx∗,Tx∗,pbx2nk,Tx∗+pbx∗,Sx2nk2s.Since (49)limk→∞⁡sup⁡pbx2nk,Tx∗+pbx∗,Sx2nk2s≤pbx∗,Tx∗+pbx∗,x∗2s,then by letting k→∞ we have limk→∞⁡Mx2nk,x∗=pbx∗,Tx∗. Suppose that pbx∗,Tx∗>0. By (47), we have(50)ψpbx2nk+1,Tx∗ψMx2nk,x∗≤βψMx2nk,x∗<1.Letting k→∞ in above inequality, we obtain that (51)limk→∞⁡βψMx2nk,x∗=1.So limk→∞⁡Mx2nk,x∗=0. Hence pbx∗,Tx∗=0, and due to (pb1) and (pb2) we obtain so x∗=Tx∗. Similarly we can show that x∗=Sx∗. Thus S and T have a common fixed point x∗∈X.Remark 27.
We note that Theorem26 is more general than the results established in [24–26].Example 28.
LetX=0,1. Define a function pb:X×X→[0,+∞) by pb(x,y)=(x∨y)2+x-y2. Clearly, (X,pb,s) is a complete partial b-metric space with the constant s=4. Let β be a function on [0,∞] defined by βt=1/4+t for all t≥0. Then β∈Ω. Also, ψ be a function on [0,+∞) defined by ψt=t/2. Then ψ∈Ψ. Define the mappings S,T:X→X by (52)Tx=2245x,ifx∈0,121,ifx∈12,1andSx=0∀x∈X.Also, we define the function αs:X×X→0,∞ by (53)αsx,y=s2,if0≤x,y≤120,otherwise.If xn is a Cauchy sequence such that αsxn,xn+1≥s2 for all n∈N, then xn⊆0,1/2. Since 0,1/2,pb is a complete partial b-metric space, then the sequence xn converges in 0,1/2⊆X. Thus X,pb is an αs-complete partial b-metric space. Let αsx,Sx≥s2 and αsx,Tx≥s2, and thus x∈0,1/2 and Sx, Tx∈0,1/2 and so αsSx,TSx≥s2 and αsTx,STx≥s2. Thus, S,T is αs-orbital admissible. Let x,y∈X be such that αsx,y≥2, αsy,Sy≥1 and αsy,Ty≥s2. Then we have x,y,Sy,Ty∈0,1/2, which implies that αsx,Sy≥s2 and αsx,Ty≥s2. Therefore S,T is triangular αs-orbital admissible. Let xn be a Cauchy sequence such that xn→x as n→∞ and αsxn,xn+1≥s2 for all n∈N. Then xn⊆0,1/2 for all n∈N. So limn→∞⁡Txn=limn→∞⁡2/245xn=2/245x=Tx. Hence T is αs-continuous. Similarly, we can show that S is αs-continuous. Let x0=1/4. Then (54)αs14,S14=αs14,0≥s2.Let x,y∈X such that αsx,y≥s2. Then x,y∈0,1/2 and hence (55)16245≤ψαsx,ypbSx,Ty≤βψMx,y.ψMx,y+Lψmin⁡pbx,Sx,pby,Sx≤19,with L≥0. Thus all conditions of Theorem 26 are satisfied. Hence S and T have a common fixed point x=0.


## 3. Consequences

Corollary 29.
LetX,pb,s be an αs-complete partial b-metric space. Assume that (i)
there existβ∈Ω and L≥0 such that, for all x,y∈X with αsx,y≥s2 the self-mappings S,T satisfy the following inequality: (56)αsx,ypbSx,Ty≤βMx,y.Mx,y+Lminpbx,Sx,pby,Sx;(ii)
S,T are triangular αs-orbital admissible mappings;(iii)
there existsx0∈X such that αsx0,Sx0≥s2(iv)
(a)
S and T are αs-continuous mappings;(b)
xn is a sequence in X with αsxn,xn+1≥s2 for all n∈N such that xn→x∗∈X as n→∞, then there exists a subsequence xnk of xn such that αsxnk,x∗≥s2 for all k∈N.
ThenS and T have a common fixed point x∗∈X. In addition, if y∗ is also a common fixed point of the pair (S,T) such that αs(x∗,y∗)≥s2, then x∗=y∗.Proof.
Defineψ:R0+→R0+ by ψt=t for all t∈R0+.Corollary 30.
LetY=Y,d,s be an αs-complete b-metric space. Let S,T:Y→Y be a generalized αs-ψ-Geraghty contractions with respect to d satisfying the following conditions: (i)
There existsy0∈Y such that αsy0,Sy0≥s2.(ii)
The mappingsS,T are triangular αs-orbital admissible.(iii)
(a)
The mappingsS,T are αs-d-continuous.(b)
yn is a sequence in Y such that αsyn,yn+1≥s2 for all n∈N and yn→y∗∈Y as n→∞, then there exists a subsequence ynk of yn such that αsynk,y∗≥s2 for all k∈N.
ThenS and T have a common fixed point in Y. In addition, if y∗ is also a common fixed point of the pair (S,T) such that αs(x∗,y∗)≥s2, then x∗=y∗.Proof.
Setpb(x,x)=0 for all x∈X in Theorem 26.Corollary 31.
LetX,⪯ be a partially ordered set and X,⪯,pb,s be an ordered complete partial b-metric space. Assume that the weakly increasing mappings S,T:X→X satisfy the following conditions: (i)
there existβ∈Ωψ∈Ψ and L≥0 such that (57)ψs2pbSx,Ty≤βψMx,y.ψMx,y+Lψminpbx,Sx,pby,Sx,for all comparablex,y∈X (i.e.x⪯y or y⪯x);(ii)
there existsx0∈X such that x0⪯Sx0(iii)
(a)
eitherS or T is continuous;(b)
xn is a nondecreasing sequence such that xn→x∗∈X as n→∞, then there exists a subsequence xnk of xn such that xnk⪯x∗ for all k∈N.
ThenS and T have a common fixed point x∗∈X. In addition, if y∗ is also a common fixed point of the pair (S,T) such that x∗⪯y∗, then x∗=y∗.Proof.
Define the relation⪯ on X by (58)αsx,y=s2,ifx⪯yory⪯x0,otherwise.Proof follows from the proof of Theorem 26.Definition 32.
The self-mappingsT defined on (X,pb,s) is called a generalized αs-ψ-Geraghty contraction if there exist β∈Ω, ψ∈Ψ, and L≥0 such that (59)ψαsx,ypbTx,Ty≤βψCx,y.ψCx,y+Lψmin⁡pbx,Tx,pby,Tx,for x,y∈X satisfying αsx,y≥s2, where(60)Cx,y=max⁡pbx,y,pbx,Tx,pby,Ty,pbx,Ty+pby,Tx2s.Corollary 33.
LetX,pb,s be an αs-complete partial b-metric space. Let T:X→X be a generalized αs-ψ-Geraghty contraction satisfying the following conditions: (i)
there existsx0∈X such that αsx0,Tx0≥s2;(ii)
the mappingT is triangular αs-orbital admissible;(iii)
(a)
the mappingT is αs-pb-continuous;(b)
xn is a sequence in X such that αsxn,xn+1≥s2 for all n∈N and xn→x∗∈X as n→∞, then there exists a subsequence xnk of xn such that αsxnk,x∗≥s2 for all k∈N.
ThenT has a fixed point in X. In addition, if y∗ is also a common fixed point of the pair (S,T) such that αs(x∗,y∗)≥s2, then x∗=y∗.Proof.
SetS=T in Theorem 26.We extend Definition25 for all x,y∈X as followsDefinition 34.
The self-mappingsS,T defined on (X,pb,s) are called generalized ψ-Geraghty contractions, if there exist β∈Ω, ψ∈Ψ, and L≥0 such that (61)ψpbSx,Ty≤βψMx,y.ψMx,y+Lψmin⁡pbx,Sx,pby,Sx,for x,y∈X.Theorem 35.
LetS,T:X→X be two pb-continuous generalized ψ-Geraghty contractions defined on a complete partial b-metric space (X,pb,s); then S and T have a common fixed point.Proof.
The arguments follow as the same lines in proof of Theorem26.


## 4. Application

In this section, we present an application on existence of a solution of a pair of elliptic boundary value problems. LetCI be the space of all continuous function defined on I=0,1. Consider the following pair of differential equations:(62)-d2xdt2=ft,xt,t∈0,1x0=x1=0.and-d2ydt2=Kt,yt,t∈0,1y0=y1=0,where f,K:I×C(I)→R are continuous functions. The Green function associated with (62) is defined by (63)Gt,s=t1-τ,0≤t≤τ≤1τ1-t,0≤τ≤t≤1.Define the function pb:CI×CI→[0,∞) by (64)pbx,y=supt∈I⁡xt-yt2+k,for  allx,y∈CIandk>0.It is known that (CI,pb) is a complete partial b-metric space with constant s=4. Now, define the operators S,T:CI→CI defined by(65)Sxt=∫01Gt,τfτ,xτdτandTxt=∫01Gt,τKτ,yτdτ,for all t∈I. Remark that (62) has a solution if and only if operators S and T have a common fixed point.Theorem 36.
Assume that there exist continuous functionsf,K:I×C(I)→R such that, for all x,y∈C(I) and ρ∈R, we have (66)ft,x-Kt,y2≤64ln⁡Mxt,yt+1ρfor  allt∈I,where M(x(t),y(t)) is defined by (9) such that M(x(t),y(t))>ρ>0Proof.
It is well known thatx∗∈C2I is a solution of (62) if and only if x∗∈CI is a common solution of the integral equations given by (65). Define the mappings S,T:CI→CI by (65). Hence the solution of (62) is equivalent to find a common fixed point x∗∈CI of T and S. Let x,y∈CI. By (i), we get (67)Sxt-Tyt2=∫01Gt,τfτ,xτ-Kτ,yτdτ2≤∫01Gt,τfτ,xτ-Kτ,yτdτ2≤8∫01Gt,τln⁡Mxτ,yτ+1ρdτ2≤8∫01Gt,τln⁡Mxτ,yτ+1ρdτ2=82ln⁡Mxτ,yτ+1ρsupt∈I⁡∫01Gt,τdτ2.Since ∫01‍G(t,τ)dτ=-t2/2+t/2 for all t∈I, then we have supt∈I⁡∫01‍G(t,τ)dτ2=1/82, which implies that(68)Sxt-Tyt2≤ln⁡Mxτ,yτ+1ρ.It can easily be proved that M(x,y)=supτ∈IM(x(τ),y(τ)). Thus, (69)ln⁡pbSx,Ty+1≤ln⁡ln⁡Mx,y+1+1=ln⁡ln⁡Mx,y+1+1ln⁡Mx,y+1ln⁡Mx,y+1.Define the functions ψ:0,∞→0,∞ and β:0,∞→0,1/s by(70)ψx=ln⁡x+1(71)andβx=ψxx,ifx≥100,otherwise.Note that ψ:0,∞→0,∞ is continuous, nondecreasing, positive in 0,∞, ψ0=0, and ψx<x.
Henceβ∈Ω and(72)ψpbSx,Ty≤βψMx,y.ψMx,yfor all x,y∈C(I). Therefore all assumptions of Theorem 35 are satisfied with L=0. Hence S and T have a common fixed point x∗∈CI; that is, Sx∗=Tx∗=x∗ which is a solution of (62).



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