# Fixed Point Results for an Almost Generalizedα-Admissible Z-Contraction in the Setting of Partially Ordered b-Metric Spaces

Authors: Solomon Gebregiorgis Teweldemedhin; Kidane Koyas Tola
Journal: Abstract and Applied Analysis
 (2021)

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## Abstract

In this paper, we introduce an almost generalizedα-admissible Z-contraction with the help of a simulation function and study fixed point results in the setting of partially ordered b-metric spaces. The presented results generalize and unify several related fixed point results in the existing literature. Finally, we verify our results by using two examples. Moreover, one of our fixed point results is applied to guarantee the existence of a solution of an integral equation.


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## Body

## 1. Introduction

Metric fixed point theory is a vivid topic, which furnishes useful methods and notions for dealing with various problems. In particular, we refer to the existence of solutions of mathematical problems reducible to equivalent fixed point problems. Thus, we recall that Banach contraction principle [1] is at the foundation of this theory. Due to its usefulness, Banach contraction principle has been extended and generalized in various spaces using different conditions either by modifying the basic contractive condition or by generalizing the ambient spaces or both. For some extensions of Banach contraction principle in metric spaces, see References [2–12].The existence of fixed point in partially ordered sets has been considered by Turinici in ordered metrizable uniform spaces [11]. The applications of fixed point results in partially ordered metric spaces were studied by Ran and Reurings [13] to solve matrix equations and by Nieto and Rodríguez-López [14] to obtain solutions of certain partial differential equations with periodic boundary conditions. Many researchers have focused on different contractive conditions in complete metric spaces endowed with a partial order and obtained many fixed point results in such spaces. For further works in this direction, see References [3, 15–19]. The concept of metric spaces has been generalized in many directions. The notion of a b-metric space was introduced by Bakhtin in [20] and later extensively used by Czerwik in [21, 22]. Since then, several papers have been published on the fixed point theory of various classes of single-valued and multivalued operators in (ordered) b-metric spaces. For further works in this direction, see References [2, 4, 23–27].Khojasteh et al. [28] introduced the notion of Z-contraction and studied existence and uniqueness of fixed points for Z-contraction type operators. This class of Z-contractions unifies large types of nonlinear contractions existing in the literature. Afterwards, Karapinar [25] originated the concept of α-admissible Z-contraction. For more works in this line of research, see References [3, 4, 9, 12, 25]. Recently, Melliani et al. [9] introduced a new concept of α-admissible almost type Z-contraction and proved the existence of fixed points for admissible almost type Z-contractions in a complete metric space.Inspired and motivated by the works of [9, 25], the purpose of this paper is to introduce a new class of mappings, namely, an almost generalized α-admissible Z-contraction, and prove the existence and uniqueness of fixed points for such mappings in the setting of partially ordered b-metric spaces.


## 2. Preliminaries

In this section, we present some notions, definitions, and theorems used in the sequel.Throughout this paper, we shall useℝ and ℝ+ to represent the set of real numbers and the set of nonnegative real numbers, respectively.Definition 1 (see [21]).
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=0ifandonlyifx=y(ii)
dx,y=dy,x(iii)
dx,z≤sdx,y+d(y,z)
The tripletX,d,s is called a b-metric space.Definition 2 (see [29]).
LetX,d be a b-metric space. Then, a sequence xn in X is said to be
(a)
b-convergent if there existsx∈X such that dxn,x⟶0 as n⟶∞. In this case, we write limn⟶∞xn=x(b)
b-Cauchy sequence ifdxn,xm⟶0 as n,m⟶∞(c)
b-complete if every b-Cauchy sequence inX is b-convergent.Remark 3 (see [29]).
In a b-metric spaceX,d, the following assertions hold:
(1)
(R1) A convergent sequence has a unique limit.(2)
(R2) Each convergent sequence is a Cauchy sequence.(3)
(R3) In general, a b-metric is not continuous.(4)
(R4) In general, a b-metric does not induce a topology onXDefinition 4.
A partially ordered set (poset) is a systemX,≼, where X is nonempty set and ≼ is a binary relation of X satisfying
(i)
x≼x(reflexivity).(ii)
ifx≼yandy≼x, then x=y (antisymmetry).(iii)
ifx≼yandy≼z, thenx≼z (transitivity) for all x,y,z∈XDefinition 5.
LetX be a nonempty set. Then, X,d,≼ is called partially ordered b-metric spaces if
(i)
X,d is a b-metric space and(ii)
X,≼ is a partially ordered set.
Now, we give an example to show that a b-metric is not necessarily metric.Example 1 (see [23]).
LetX,d be a metric space and ρx,y=dx,yp, where p>1 is a real number. Then, ρ is a b-metric with s=2p−1. However, if X,d is a metric space, then ρx,y is not necessarily a metric space. For example, if X=ℝ and dx,y=x−y, then ρx,y=d(x,ys is a b-metric on ℝ with s=2 but it is not a metric on ℝ.Definition 6 (see [18]).
LetX,≼ be a partially ordered set and T:X⟶X is a self-mapping; we say T is monotone nondecreasing with respect to ≼ if for x,y∈X,(1)x≼y⟹Tx≼Ty.Definition 7.
Let (X,≼) be a partially ordered set and x,y∈X; then, x and y are said to be comparable elements of X if
(2)x≼yory≼x.Theorem 8 (see [30]).
LetX,d be a complete metric space and T:X⟶X be a map satisfying
(3)dTx,Ty≤αdy,Ty1+dx,Tx1+dx,y+βdx,y,for all x,y∈X where α,β≥0withα+β<1. Then, T has a unique fixed point.Theorem 9 (see [18]).
LetX,d,≼ be a complete partially ordered metric space. Let T:X⟶X be a continuous and nondecreasing map satisfying
(4)dTx,Ty≤αdy,Ty1+dx,Tx1+dx,y+βdx,y,for all x,y∈X with x≼y and α,β≥0withα+β<1. If there exists x0∈X such that x0≼Tx0, then T has a fixed point.Definition 10 (see [5]).
LetX,d be a metric space. A map T:X⟶X is called an almost contraction or δ,L contraction if there exist constants δ∈0,1 and L≥0 such that
(5)dTx,Ty≤δdx,y+Ldy,Tx,for all x,y∈X.Definition 11 (see [8]).
A functionϕ:ℝ+⟶ℝ+ is called an altering distance function if
(i)
ϕis nondecreasing and continuous;(ii)
ϕt=0if and only ift=0Definition 12 (see [28]).
A simulation function is a mapζ:ℝ+×ℝ+⟶ℝ that satisfies the following conditions:
ζ1ζ0,0=0
ζ2ζt,s<s−tforeacht,s>0
ζ3Ifsn,tnaresequencesin0,∞suchthat(6)limn⟶∞sn=limn⟶∞tn=l∈ℝ+,then limn⟶∞supζtn,sn<0.
The collection of all simulation functions is denoted byZ.Example 2 (see [28]).
Letζi:0,+∞×0,+∞⟶ℝ where i=1,2,3 is defined as follows:
(i)
ζ1t,s=λs−t for all s,t∈0,+∞ where λ∈0,1(ii)
ζ2t,s=φs−t for all s,t∈0,+∞ where φ:0,+∞⟶0,+∞ is an upper semicontinuous function such that φt=0 if and only if t=0 and φt<t for all t>0(iii)
ζ3t,s=φs−θt for all s,t∈0,+∞ where φ,θ:0,+∞⟶0,+∞ are continuous functions such that φt=θt=0 if and only if t=0 and φt<t≤θt for all t>0Definition 13 (see [22]).
LetX,d be a metric space, T:X⟶X be a map, and ζ∈Ζ. Then, T is called Ζ-contraction with respect to ζ if the following condition is satisfied:
(7)ζdTx,Ty,dx,y≥0,for all x,y∈X.Definition 14 (see [10]).
LetX be a nonempty set. Let T:X⟶X and α:X×X⟶ℝ+ be the maps. Then, T is called α-admissible if αx,y≥1 implies αTx,Ty≥1, for each x,y∈X.Definition 15 (see [31]).
LetX be a nonempty set. Let T:X⟶X and α:X×X⟶ℝ+ be the maps. Then, T is said to be α-orbital admissible mapping if αx,Tx≥1 implies αTx,T2x≥1, for each x∈X.Definition 16 (see [31]).
LetX be a nonempty set. Let T:X⟶X and α:X×X⟶ℝ+ be the maps. Then, T is said to be a triangular α-orbital admissible if
(i)
T is α-orbital admissible;(ii)
αx,y≥1 and αy,Ty≥1 implies αx,Ty≥1for each x,y∈X.Definition 17 (see [25]).
LetX be a nonempty set and T:X⟶X be a self-map defined on a metric space X,d. If there exist ζ∈Z and α:X×X⟶ℝ+ such that
(8)ζαx,ydTx,Ty,dx,y≥0,for all x,y∈X, then T is called an α-admissible Z-contraction with respect to ζ.Theorem 18 (see [25]).
LetX,d be a complete metric space and T be an α-admissible Z-contraction with respect to ζ. Suppose that
(i)
T is triangular α-orbital admissible(ii)
There existsx0∈X such that αx0,Tx0≥1(iii)
T is continuous
Then, there existsu∈X such that u=Tu.


## 3. Main Result

In this section, we present our main findings.Definition 19.
LetX,d,≼ be a partially ordered b-metric space with parameter s≥1. Let T:X⟶X and α:X×X⟶ℝ+ be the maps. Assume that there exist a simulation function ζ and a constant L≥0 such that
(9)ζαx,ydTx,Ty,Mx,y+L∙mx,y≥0,for all x,y∈X with x≼y where
(10)Mx,y=maxdy,Ty1+dx,Tx1+dx,y,dy,Tx1+dx,Tys21+dx,y,dx,ys,(11)mx,y=mindx,Tydy,Tx1+dx,y,dx,Txdy,Ty1+dx,y.
Then,T is called an almost generalized α-admissible Z-contraction with respect to ζ.Theorem 20.
LetX,d,≼ be a complete partially ordered b-metric space and T be an almost generalized α-admissible Z-contraction with respect to ζ. Suppose that
(i)
T is nondecreasing and continuous(ii)
There existsx0∈X such that x0≼Tx0 and αx0,Tx0≥1(iii)
T is triangular α-orbital admissible
Then,T has a fixed point.Proof.
Due to condition (ii), there existsx0∈X such that x0≼Tx0 and αx0,Tx0≥1.
Define an iterative sequencexn in X as follows:
(12)xn+1=Txn,for every n≥0.If there exists some nonnegative integern0 such thatxn0=xn0+1=Txn0, then xn0 is a fixed point of T.Now, we shall assume thatxn≠xn+1 for all n≥0. So, we have dxn+1,xn≠0 for all integers n≥0. Due to (ii) and T is α-admissible, we have αx0,x1=αx0,Tx0≥1 which in turn implies that αTx0,Tx1=αx1,x2≥1.Inductively, we obtain that(13)αxn,xn+1≥1.SinceT is nondecreasing map and x0≼Tx0, we have the following:
(14)x0≼Tx0=x1≼Tx1=x2≼Tx2=x3≼⋯≼Txn−1=xn≼Txn=xn+1≼⋯.Hence,xn≼xn+1 for n≥0.Using Equation (9), putting x=xn−1 and y=xn for n≥1, we get the following:
(15)ζαxn−1,xndTxn−1,Txn,Mxn−1,xn+L∙mxn−1,xn≥0,where
(16)Mxn−1,xn=maxdxn,xn+11+dxn−1,xn1+dxn−1,xn,dxn,xn1+dxn−1,xn+1s21+dxn−1,xn,dxn−1,xns=maxdxn,xn+1,dxn−1,xns,mxn−1,xn=mindxn−1,xn+1dxn,xn1+dxn−1,xn,dxn−1,xndxn,xn+11+dxn−1,xn=min0,dxn−1,xndxn,xn+11+dxn−1,xn=0.IfMxn−1,xn=maxdxn,xn+1,dxn−1,xn/s=dxn,xn+1, then Equation (15) becomes the following:
(17)ζαxn−1,xndxn,xn+1,dxn,xn+1≥0.It follows thatαxn−1,xndxn,xn+1<dxn,xn+1.Using Equation (13), we have the following:
(18)dxn,xn+1≤αxn−1,xndxn,xn+1<dxn,xn+1,which is a contradiction.Hence,maxdxn,xn+1,dxn−1,xn/s=dxn−1,xn/sNow Equation (15) becomes ζαxn−1,xndxn,xn+1,dxn−1,xn/s≥0.It follows that(19)αxn−1,xndxn,xn+1<dxn−1,xns.Furthermore, using Equation (15) and since s≥1, we have the following:
(20)dxn,xn+1≤αxn−1,xndxn,xn+1<dxn−1,xns<dxn−1,xn.It follows thatdxn,xn+1<dxn−1,xn.Hence, the sequencedxn−1,xn is nonincreasing and bounded below. Accordingly, there exists r≥0 such that
(21)limn⟶∞dxn−1,xn=r.Assumingr>0 and using Equation (20), we get the following:
(22)limn⟶∞αxn−1,xndxn,xn+1=r>0.Lettingtn=αxn−1,xndxn,xn+1, sn=dxn−1,xn, and using ζ3, we obtain the following:
(23)0≤limn⟶∞supζtn,sn<0,which is a contradiction.Thus, we have the following:(24)limn⟶∞dxn,xn+1=0.Now, we need to show thatxn is a Cauchy sequence in X.Supposexnis not a Cauchy sequence in X, then there exists an ε>0such that
(25)dxmk,xnk≥ε,where mkandnk are two sequences of positive integerswith nk>mk>k for all positive integersk. Moreover, mk is chosen as the smallest integer satisfying Equation (25).Thus, we have the following:(26)dxmk,xnk−1<ε.By applying the triangle inequality and using Equations (25) and (26), we get the following:
(27)ε≤dxmk,xnk≤sdxmk,xnk−1+sdxnk−1,xnk<sε+sdxnk−1,xnk.Now, taking the upper limit ask⟶∞ in Equation (27) and using Equation (24), we get the following:
(28)ε≤lim¯dxmk,xnk<sε.Similarly, applying the triangle inequality twice, we get the following:(29)ε≤dxmk,xnk≤sdxmk,xnk−1+sdxnk−1,xnk≤s2dxmk,xmk−1+s2dxmk−1,xnk−1+sdxnk−1,xnk.Furthermore,(30)dxmk−1,xnk−1≤sdxmk−1,xmk+sdxmk,xnk−1≤sdxmk−1,xmk+sε.Taking the upper limit ask⟶∞ in Equations (29) and (30) and combining, we get the following:
(31)εs2≤lim¯dxmk−1,xnk−1<sε.Again,(32)ε≤dxmk,xnk≤sdxmk,xnk−1+sdxnk−1,xnk.Furthermore,(33)dxmk,xnk−1≤sdxmk,xmk−1+sdxmk−1,xnk−1.Taking the upper limit ask⟶∞ in Equations (32) and (33), using Equation (31), and combining, we get the following:
(34)εs≤lim¯dxmk,xnk−1<s2ε.Similarly, we can show that(35)εs≤lim¯dxmk−1,xnk<s2ε.Now, for simplicity, we denote the following expressions as follows:(36)ak=dxnk−1,xnk1+dxmk−1,xmk1+dxmk−1,xnk−1,bk=dxnk−1,xmk1+dxmk−1,xnks21+dxmk−1,xnk−1,ck=dxmk−1,xnk−1s,dk=dxmk−1,xnkdxnk−1,xmk1+dxmk−1,xnk−1,ek=dxmk−1,xmkdxnk−1,xnk1+dxmk−1,xnk−1,which give Mxmk−1,xnk−1=maxak,bk,ck and mxmk−1,xnk−1=mindk,ek.Using Equations (24), (34), and (35), and the property of lim sup, we can see that
(37)lim¯mxmk−1,xnk−1=minlim¯dk,lim¯ek=minlim¯dk,0=0.SinceT is triangular α-orbital admissible, we have the following:
(38)αxmk−1,xnk−1≥1.Using Equation (38), we have the following:
(39)dxmk,xnk≤αxmk−1,xnk−1dxmk,xnk.Using Equation (9) with x=xmk−1 and y=xnk−1, we obtain the following:
(40)ζαxmk−1,xnk−1dxmk,xnk,Mxmk−1,xnk−1+L∙mxmk−1,xnk−1≥0.It follows that(41)αxmk−1,xnk−1dxmk,xnk<Mxmk−1,xnk−1+L∙mxmk−1,xnk−1.Combining Equations (39) and (41), we get the following:
(42)dxmk,xnk≤αxmk−1,xnk−1dxmk,xnk<Mxmk−1,xnk−1+L∙mxmk−1,xnk−1.IfMxmk−1,xnk−1=bk, then after rearranging, collecting like terms, and applying the triangle inequality, Equation (42) becomes the following:
(43)s21+dxmk−1,xnk−1dxmk,xnk<dxnk−1,xmk1+dxmk−1,xnk+s2L∙mindxmk−1,xnkdxnk−1,xmk,dxmk−1,xmkdxnk−1,xnk≤sdxnk−1,xmk−1+sdxmk−1,xmk1+sdxmk−1,xmk+sdxmk,xnk+s2L∙dxmk−1,xmkdxnk−1,xnk=sdxnk−1,xmk−1+s2dxnk−1,xmk−1dxmk−1,xmks2dxnk−1,xmk−1dxmk,xnk+sdxmk−1,xmk1+sdxmk−1,xmk+sdxmk,xnk+s2Ldxmk−1,xmkdxnk−1,xnk.Taking the upper limit ask⟶∞ in Equation (43) and using Equation (31), we get the following:
(44)lim¯dxmk,xnk≤lim¯dxmk−1,xnk−1<ε,which is a contradiction due to Equation (9).Hence,(45)Mxmk−1,xnk−1=maxak,bk,ck=ck.Using Equations (32) and (45), we get the following:
(46)lim¯ck=1slim¯dxmk−1,xnk−1<sεs=ε.Using Equations (45) and (46) together with Equations (24), (34), and (35) and applying the property of limsup, it follows that
(47)lim¯Mxmk−1,xnk−1=maxlim¯ak,lim¯bk,lim¯ck<ε.Taking the lower limit in Equation (39) as k⟶∞ and using Equation (25), we get the following:
(48)ε≤lim¯dxmk,xnk≤lim¯αxmk−1,xnk−1dxmk,xnk.Taking the upper limit in Equation (40) and using Equations (37), (47), and (48), we get the following:
(49)0≤lim¯ζαxmk−1,xnk−1dxmk,xnk,Mxmk−1,xnk−1+L∙mxmk−1,xnk−1<lim¯Mxmk−1,xnk−1+L∙mxmk−1,xnk−1−αxmk−1,xnk−1dxmk,xnk≤lim¯Mxmk−1,xnk−1+L∙lim¯mxmk−1,xnk−1−lim¯αxmk−1,xnk−1dxmk,xnk<ε−ε=0,which is a contradiction.Hence,xnis a Cauchy sequence in a complete b-metric spaceX. So, there exists u∈X such that
(50)limn⟶∞xn=u.SinceT is continuous, we obtain the following:
(51)u=limn⟶∞xn+1=limn⟶∞Txn=Tlimn⟶∞xn=Tu.Thus,u is a fixed point of T.Theorem 21.
LetX,d,≼ be a complete partially ordered b-metric space and T is an almost generalized α-admissible Z-contraction with respect to ζ. Suppose that
(i)
T is nondecreasing(ii)
There existsx0∈X such that x0≼Tx0 and αx0,Tx0≥1(iii)
T is triangular α-orbital admissible(iv)
There exists a nondecreasing sequencexn in X such that xn⟶x with xn≼x and αxn,x≥1
Then,T has a fixed point.Proof.
Following the proof of Theorem20, we know that the sequence xn defined by xn+1=Txn for all n≥0 converges to some u∈X.
Hence, by (iii), we have the following:(52)αxn,u≥1,for all n.By (i), there existsx0∈X with x0≼Tx0 and since T is a nondecreasing map, we have the following:
(53)x0≼Tx0=x1≼Tx1=x2≼Tx2=x3≼⋯≼Txn−1=xn≼Txn=xn+1≼⋯.Thus,xn is a nondecreasing sequence that converges to u.It follows thatxn≼u,foralln∈N.Now, we show thatu=Tu.Applying Equation (9) with x=xn and y=u, we get the following:
(54)0≤ζαxn,udxn+1,Tu,Mxn,u+L∙mxn,u,where
(55)Mx,y=Mxn,u=maxdu,Tu1+dxn,xn+11+dxn,u,du,xn+11+dxn,Tus21+dxn,u,dxn,us,mx,y=mxn,u=mindxn,xn+1du,Tu1+dxn,u,dxn,Tudu,xn+11+dxn,u.From Equation (54), it follows that
(56)αxn,udxn+1,Tu<Mxn,u+L∙mxn,u.Using Equations (52) and (56), we have the following:
(57)dxn+1,Tu≤αxn,udxn+1,Tu<Mxn,u+L∙mxn,u.Sincelimn⟶∞dxn+1,Tu=limn⟶∞Mxn,u+L∙mxn,u=du,Tu, we have the following:
(58)limn⟶∞αxn,udxn+1,Tu=du,Tu.Ifdu,Tu=0, then u=Tu, then we are done. Assume du,Tu≠0.Lettingqn=αxn,udxn+1,Tu and pn=Mxn,u and using ζ3, we obtain the following:
(59)0≤limn⟶∞supζqn,pn<0,which is a contradiction. Thus, we havedu,Tu=0; that is, u=Tu.Theorem 22.
In addition to the hypotheses of Theorem20 or Theorem 21, suppose that for everyx,y∈X, there exists u∈X such that u≼xandu≼y and αu,Tu≥1. Then, T has a unique fixed point.Proof.
Referring to Theorem20 or Theorem 21, the sets of fixed points of T are nonempty. Now, we shall show the uniqueness of fixed point. To prove the uniqueness of the fixed point, assume that there exist z1,z2∈X such that z1=Tz1 and z2=Tz2 with z1≠z2.
Assume that there existsu0∈X such that u0≼z1 and u0≼z2, then as in the proof of Theorem 20, we define the sequence such that
(60)un+1=Tun=Tn+1u0,for all n≥0.Using the hypotheses of Theorem22, Equation (60), and proceeding inductively, we get the following:
(61)αun,un+1≥1.Due to the monotone property ofT, we have the following:
(62)Tnu0=un≼z1=Tnz1,Tnu0=un≼z2=Tnz2.Sincez1,z2,un∈X for all n≥0, then z1=um for some positive integer m, and hence, z1=Tz1=Tun=un+1 for all n≥m. It follows that un⟶z1 as n⟶∞.Using Equation (61) and the fact that un⟶z1 as n⟶∞, we have αun,z1≥1 for all n≥0.Now, suppose thatz1≠un for all n≥0, so un≼z1 for all n≥0, then dun,z1≠0 for all n≥0.Applying Equation (9) with x=un and y=z1, we get the following:
(63)0≤ζαun,z1dTun,Tz1,Mun,z1+L∙mun,z1<Mun,z1+L∙mun,z1−αun,z1dTun,Tz1.It follows that(64)αun,z1dTun,Tz1<Mun,z1+L∙mun,z1.Here,(65)Mun,z1=maxdz1,Tz11+dun,Tun1+dun,z1,dz1,Tun1+dun,Tz1s21+dun,z1,dun,z1s=maxdz1,z11+dun,un+11+dun,z1,dz1,un+1s2,dun,z1s=maxdz1,un+1s2,dun,z1s.Moreover,(66)mun,z1=mindz1,Tz1dun,Tun1+dun,z2,dz1,Tundun,Tz11+dun,z2=mindz1,z1dun,un+11+dun,z2,dz1,un+1dun,z11+dun,z2=max0,dz1,un+1dun,z11+dun,z2=0.Ifmaxdz1,un+1/s2,dun,z1/s=dz1,un+1/s2, then substituting the corresponding values of Mun,z1 and mun,z1 in Equation (64) and using the fact that αun,z1≥1, it becomes the following:
(67)αun,z1dun+1,z1<dz1,un+1s2+L∙0.It follows that(68)dun+1,z1≤αun,z1dun+1,z1<dz1,un+1s2,which is a contradiction. Hence, maxdz1,un+1/s2,dun,z1/s=dun,z1/s.Again, substituting the corresponding values ofMun,z1 and mun,z1 in Equation (64) and using the fact that αun,z1≥1, it becomes the following:
(69)dun+1,z1≤αun,z1dun+1,z1<dun,z1s+L∙0≤dun,z1.Hence, the sequencedun,z1 is nonincreasing and bounded below. Accordingly, there exists r≥0 such that limn⟶∞dun,z1=r. Assume r>0.Using Equation (69), we get the following:
(70)limn⟶∞αun,z1dun+1,z1=r.Lettingbn=αun,z1dun+1,z1 and an=dun,z1, and using ζ3, we obtain the following:
(71)0≤limn⟶∞supζbn,an<0,which is a contradiction.Thus, we havelimn⟶∞dun,z1=0; that is, un⟶z1 asn⟶∞.Similarlyun⟶z2 asn⟶∞.Due to the uniqueness of the limit, it implies thatz1=z2. Thus, T has a unique fixed point.Now, we give corollaries of our main theorem, Theorem20.If we takeαx,y=1 for all x,y∈X in Theorem 20, then we have the following result.Corollary 23.
LetX,d,≼ be a complete partially ordered b-metric space with parameter s≥1. Let T:X⟶X be a map. Suppose there exists a simulation function ζ such that
(72)ζdTx,Ty,Mx,y+L∙mx,y≥0,for all x,y∈X with x≼y where Mx,y and mx,y are the same as in Theorem 20.
Also, assume that the following conditions hold;(i)
T is nondecreasing and continuous(ii)
There existsx0∈X such that x0≼Tx0
Then,T has a fixed point.
Similarly, we can deduce the following results.Corollary 24.
LetX,d,≼ be a complete partially ordered b-metric space with parameter s≥1. Let T:X⟶X and α:X×X⟶0,∞ be the maps. Suppose there exists a simulation function ζ such that
(73)ζαx,ydTx,Ty,Mx,y≥0,for all x,y∈X with x≼y where Mx,y is defined in Theorem 20.
Also, assume that the following conditions hold;(i)
T is nondecreasing and continuous(ii)
There existsx0∈X such that x0≼Tx0 and αx0,Tx0≥1(iii)
T is triangular α-orbital admissible
Then,T has a fixed point.Corollary 25.
LetX,d,≼ be a complete partially ordered b-metric space with parameter s≥1. Let T:X⟶X be a map. Suppose there exists a simulation function ζ such that
(74)ζdTx,Ty,Mx,y≥0,for all x,y∈X with x≼y where Mx,y is defined in Theorem 20.
Also, assume that the following conditions hold;(i)
T is nondecreasing and continuous(ii)
There existsx0∈X such that x0≼Tx0
Then,T has a fixed point.
Now, we provide an example in support of Theorem20.Example 3.
LetX=0,1,2,3,4 and let d:X×X⟶ℝ+ be defined by the following:
(75)dx,y=x−y2,for all x,y∈X.
Hence,d is a complete b-metric space with parameter s=2.
Define a partial order onX as follows:
(76)≼:0,0,0,3,0,4,1,1,2,2,2,4,3,3,3,4,4,4.Then, X,≼ is a partially ordered set.
We define the mapsT:X⟶X and α:X×X⟶ℝ+as follows:
(77)T0=T1=2,T2=3,T3=T4=4,αx,y=1,ifx,y∈A,0,ifx,y∈X×X/A,where
(78)A=0,3,0,4,2,4,3,4,4,4.
Clearly,T is a continuous, nondecreasing, and triangular α-orbital admissible map.
Moreover, we choosex0=3∈X; then, 3=x0≼Tx0=T3=4 and αx0,Tx0=α3,4=1≥1.
Now, we verify Equation (9) by choosing L=10 and ζt,s=s−ϕs−t for all t,s∈0,∞ where ϕs=s/2. For simplicity, let t=αx,ydTx,Ty and s=Mx,y+Lmx,y.
The cases where we havex=y are trivial. So, we consider only the following three cases.
Case (i): Whenx=0 and y=3.(79)t=αx,ydTx,Ty=α0,3dT0,T3=α0,3d2,4=4,s=Mx,y+Lmx,y=M0,3+L∙m0,3=92+10∙410=172.
Hence,ζt,s=ζ4,17/2=17/2−ϕ17/2−4=17/2−17/4−4=1/4≥0.
Case (ii): Whenx=0 and y=4.(80)t=αx,ydTx,Ty=α0,4d2,4=4,s=Mx,y+Lmx,y=M0,4+L∙m0,4=8+20∙0=8.
Hence,ζt,s=s−ϕs−t⇒ζ4,8=8−ϕ8−4=8−4−4=0≥0.
Case (iii): Whenx=2 and y=4.(81)t=αx,ydTx,Ty=α2,4d3,4=1,s=Mx,y+Lmx,y=M2,4+L∙m2,4=2+20∙0=2.
Hence,ζt,s=ζ1,2=2−ϕ2−1=0≥0.
Case (iv): Whenx=3 and y=4.(82)t=αx,ydTx,Ty=α3,4d4,4=0,s=Mx,y+Lmx,y=M3,4+L∙m3,4=12+20∙0=12.
Hence,ζt,s=ζ0,1/2=1/2−ϕ1/2−0=1/4≥0.
From cases (i)-(iv) considered above,T satisfies Equation (9) and hence all the hypotheses of Theorem 20. Thus, T has a fixed point which is x=4.
The following is an example in support of Corollary24.Example 4.
LetX=0,4 and let d:X×X⟶ℝ+is the same as in Example 3..
Define a partial order onX as follows:
(83)≼:x,y:x,y∈0,2,x=y∪x,y∈2,4,x≤y.Then, X,≼ is a partially ordered set.
We define the following mappings:(84)Tx=x+2,if0≤x<1,12x+5,if1≤x<3,4,if3≤x≤4,αx,y=1,if2≤x≤4andy=4,0,otherwise.
Clearly,T is a continuous, nondecreasing, and triangular α-orbital admissible mapping.
We choosex0=3∈X and then 3=x0≼Tx0=T3=4 and αx0,Tx0=α3,4=1≥1.
Now, we verify Equation (9) by choosing the simulation function:
(85)ζt,s=s−ϕs−t,for all t,s∈0,∞ where ϕs=1/2∗s.Case 1.
2≤x<3andy=4.
For simplicity, lett=αx,ydTx,Ty and s=Mx,y+Lmx,y.
In this case,Tx∈7/2,4 and Ty=4. Moreover,
(86)t=αx,ydTx,Ty=αx,4d12x+5,4=14x−32,Mx,y=Mx,4=116x−32,12x−42,mx,y=mx,4=0.
Hence,(87)ζt,s=ζ14x−32,12x−42=12x−42−14x−42−14x−32=7−2x4≥0.Case 2.
3≤x≤4andy=4.
In this case,Tx=4 and Ty=4. Moreover,
(88)t=αx,ydTx,Ty=αx,4d4,4=0,Mx,y=Mx,4=12x−42,mx,y=mx,4=0.
Hence,(89)ζt,s=ζ0,12x−42=12x−42−14x−42−0=14x−42≥0.Therefore,T satisfies Equation (73) and hence all the assertions of Corollary 24. Thus, T has a unique fixed point, namely, x=4.


## 4. An Application to an Integral Equation

In this section, we give an application of our result to an integral equation.Consider the following integral equation:(90)ur=vr+θ∫abHr,zfz,uzdz,r∈I=a,b,where v:I⟶ℝ,H:I×I⟶ℝ,andf:I×ℝ⟶ℝ are given continuous functions and θ is a constant such that θ≥0.LetX be the set of continuous real functions defined on a,b. Define the b-metric by the following:
(91)du,v=supr∈Iur−vrs,for all x,y∈X. Consider s>1. Then, X,d,≼ is a complete partially ordered b-metric space with the usual order.Now, define a mappingT:X⟶X by the following:
(92)Tur≔vr+θ∫abHr,zfz,uzdz,r∈I.We will use the following assumptions to prove the existence of a solution of Equation (90):
(a)
θ≤1/s(b)
supr∈I∫abHr,zdz≤1/b−a(c)
For allu,v∈ℝ, fz,u−fz,v≤u−v(d)
There exists a functionζ:ℝ+×ℝ+⟶ℝ such that for r∈I and for all a,b∈X with ζb,a≥0The existence of a solution to Equation (90) is equivalent to the existence of a fixed point of T. Now, we prove the following result.Theorem 26.
Under the assumptions (a)-(d), Equation (90) has a unique solution in X.Proof.
(93)dTu1,Tu2=supr∈ITu1r−Tu2rs=supr∈Ivr+θ∫abHr,zfz,u1zdz−vr+θ∫abHr,zfz,u2zdzs=supr∈Iθ∫abHr,zfz,u1z−fz,u2zdzs=θssupr∈I∫abHr,zfz,u1z−fz,u2zdzs≤θssupr∈I∫abHr,zdzs∫abfz,u1z−fz,u2zdzs≤θs1b−assupr∈I∫abdu1,u21/sdzs=θsdu1,u2≤1sdu1,u2≤Mu1,u2+L∙Nu1,u2,where M and L are defined as in Theorem 20. Hence, T satisfies all the conditions of Corollary 23. Therefore, T has a fixed point; that is, Equation (90) has a solution in X.


## 5. Conclusion

In this paper, we introduce a new class of maps, namely, an almost generalizedα-admissible Z-contraction, and establish fixed point theorems. Moreover, we prove the existence and uniqueness of fixed points in the setting of partially ordered b-metric spaces with the help of simulation function. Our results unify several related results in the existing literature. The given results not only unify several existing results but also extend and improve them. Finally, we verify the established theorems by some examples and provide an application of our result to an integral equation.



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