# Generalization Contractive Mappings on Rectangularb-Metric Space

Authors: Jayshree Patil; Basel Hardan; Ahmed A. Hamoud; Amol Bachhav; Homan Emadifar; Hatira Günerhan
Journal: Advances in Mathematical Physics
 (2022)

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## Abstract

In this paper, we introduce new coincidence fixed point theorems for generalizedϕ,ψ-contractive mappings fulfilling kind of an admissibility provision in a Hausdorff b-rectangular metric space with the support of C-functions. We applied our results to establish the existence of a solution for some integralitions. Finally, an example is presented to clarify our theorem.


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## Body

## 1. Introduction

One of the main results in the development area of fixed point theorems is the Banach contraction principle [1]. It has been evolutionized and generalized in several directions. From those directions, we faced some new kinds of metric spaces in literature as the one established by Branciari [2] and gave the principle of a rectangular metric space in which the replacement of the triangle inequality with a weaker hypothesis called quadrilateral inequality and an analogue of the Banach contraction principle is shown. Then next, fixed point results in these spaces were studied by many authors. For more informations on fixed point theorems, in rectangular metric space, see [3–10]. In fact, a contraction principle in rectangulat b-metric space and its properties appeared by George [11]. Many definitions of various mathematical concepts and terms in rectangular b-metric space can be found in [12–19]. Separation of the Hausdorff space from rectangular b-metric space is not useful for our theorem, as Hausdorff space plays an important role in Theorem 16 and its corollaries. On another hand, Samet in [20] introduced the principle of (ϕ,ψ)-contractive mapping. Newly, two separate evaluations of α-admissible mapping were introduced in which the researcher Ansari in [21] used the notion of C-functions, whereas Budhia et al. in [22] used a rectangular metric space. By ideas from [21, 22], we prove several coincide fixed point results in rectangular b-metric space. That should be considered as development of [23], which are applied to find the existence and uniqueness of a solution for many problems in different mathematical branches. Moreover, one of the most attractive research subjects in fixed point theorem is the investigation of the existence and uniqueness of coincidence points of various operators in the setting of metric spaces (see [24–29]).


## 2. Mathematical Preliminaries

We recall some basic notions and needful results on the work in the literature.Definition 1 (see [2]).
SupposeX is a nonempty set. A function d:X×X⟶0,∞ is a rectangular metric (RM) on X if, for all s1,s2∈X and all distinct points r1,r2∈X such that r1,r2∉s1,s2, the following terms hold
(i)
ds1,s2=0, if and only if s1=s2(ii)
ds1,s2=ds2,s1(iii)
ds1,r1≤ds1,r2+dr2,s2+ds2,r1 (rectangular inequality)Then,X,d is called a rectangular-metric space.Definition 2 (see [11, 30]).
LetX be a nonempty set and the mapping d:X×X⟶0,∞ satisfies
(i)
ds1,s2=0, if and only if s1=s2 for all s1,s2∈X(ii)
ds1,s2=ds2,s1 for all s1,s2∈X(iii)
there exists a real numberb≥1 such that
(1)ds1,r1≤bds1,r2+dr2,s2+ds2,r1,b‐rectangularinequality
For alls1,s2∈X and all r1≠r2∉s1,s2.
Then,d is called a bRM on X and X,d is called a rectangular b-metric space with coefficient b.Remark 3 (see [31]).
The type of rectangularb-metric space is greater than the type of metric space, where a b-metric space is a metric space when b=1.Example 1 (see [31]).
LetX=ℝ and d:X×X⟶ℝ+, such that ds1,s2=s1−s23. Then, X,d is a bMS, with b=3.Remark 4 (see [11]).
Every metric space is a rectangular metric space, and every rectangular metric space is a rectangularb-metric space, with b=1. However, the opposite of implying above is not valid.Example 2 (see [11]).
Suppose,X=ℕ and d:X×X⟶X as
(2)ds1,s2=0,ifs1=s2,4η,ifs1,s2∈1,2,s1≠s2,η,ifs1ors2∈1,2,s1≠s2.
Considerη∈0,∞. Hence, X,d is a rectangular b-metric space, with b=4/3, but X,d is not rectangular metric space, take
(3)d1,2=4η>3η=d1,3+d3,4+d4,2.In 1914, German mathematician Felix Hausdorff [32] defined a new distance idea called Hausdorff metric, denoted by HX,Y, as
(4)HX,Y=maxδ¯X,Y,δ¯Y,X,where
(5)δ¯X,Y=maxδs,Y:s∈X.The Hausdorff distance is the largest one of all the distances measured from one set to another.Definition 5 (see [33]).
A Hausdorff spaceH is a topological space where for any two distinct points s1,s2∈H, there exist neighbourhoods of each are disjoint from each other.Definition 6 (see [11]).
LetX,d be a rectangular b-metric space, ni be a sequence in X, and n∈X. Then,
(i)
a sequencesi is said to be convergent in X,d and converges to s, if for all ε>0 there exists i0∈ℕ such that dsi,s<ε, for all i>i0 and this truth is acted via limi⟶∞si=sorsi⟶sasi⟶∞(ii)
a sequencesi is said to be Cauchy sequence in X,d if for all ε>0 there exists i0∈ℕ such that dsi,si+η<ε, for all i>i0,η>0 or equivalently, if limi⟶∞dsi,si+η=0, for all η>0(iii)
X,d is said to be a complete rectangular b-metric space if every Cauchy sequence in X converges to some s∈XThe next lemmas are helpful in providing principle outcomes.Lemma 7 (see [34]).
LetX,d be a rectangular b-metric space with b≥1 and let si be a Cauchy sequence in X such that si≠sj when it was i≠j. Then, si be able convergence at most one point.Lemma 8 (see [34]).
LetX,d be a rectangular b-metric space with b≥1.
(i)
Suppose that the sequencessi,ri∈X where si⟶s,ri⟶rasi⟶∞, such that si≠s,ri≠r, for all i∈ℕ. Thus, we have(6)1bds,r≤limi⟶∞infdsi,ri≤limi⟶∞supdsi,ri≤bds,r(ii)
Supposes,m∈X and si are a Cauchy sequence in X where si≠sj, for all i,j∈ℕ,i≠j. where si⟶s,sj⟶masi,j⟶∞,s≠m. Thus, we have(7)1bds,m≤limi⟶∞infdsi,m≤limi⟶∞supdsi,m≤bds,m,1bdm,s≤limi⟶∞infdsj,s≤limi⟶∞supdsj,s≤bdm,sDefinition 9 (see [35, 36]).
LetF:X⟶X be a self-mapping on a metric space X,d and let α:X×X⟶0,∞ be a function. F is called a α-admissible function if
(8)αFs1,Fs2≥1wheneverαs1,s2≥1,∀s1,s2∈X.Definition 10 (see [35, 36]).
LetF be a self-mapping on a metric space X,d. A map F is called a (ϕ,ψ)-contractive mapping if there exist two functions ϕ:X×X⟶0,∞ and ψ:0,+∞⟶0,+∞ such that
(9)ϕs1,s2dFs1,Fs2≤ψs1,s2∀s1,s2∈X,where ψ is a nondecreasing functions such that ∑i=1∞ψit<+∞,∀t>0, where ψi is the ith iteration of ψ.
For more informations forα-admissible and (ϕ,ψ)-contractive mappings, see [35–37].Definition 11 (see [38]).
LetT be a self-mapping on a metric space X,d and let α,η:X×X⟶0,∞ are two mappings. A function T is called α-admissible with regard to η if
(10)αTs1,Ts2≥ηTs1,Ts2whereαs1,s2≥ηs1,s2,∀s1,s2∈X.Observe that, ifαs1,s2=1,∀s1,s2∈X. Thus, this definition led to Definition 9. Likewise, if we pick αs1,s2=1, then we state that T is a η-subadmissible functions.Definition 12 ([21]).
AC-function ϑ:0,∞×0,∞⟶ℝ is a continuous functions such that
(i)
ϑs1,s2≤s1(ii)
ϑs1,s2=s1⇒s1=0ors2=0
For alls1,s2∈0,∞.Definition 13 (see [39]).
A nondecreasing continuous mapϕ:0,∞⟶0,∞ is called an altering distance mapping whenever ϕβ=0⇔β=0.Definition 14 (see [29]).
LetX,d be a rectangular metric space and let α,η in Definition 10. X is said to be α-orderly with respect to η if for si∈X with αsi,si+1≥ηsi,si+1, for all i≥N where si⟶sasi⟶∞; therefore, αsi,s≥ηsi,s, for all i≥N.Remark 15.
In altering the type of distance mapping, we denoted it by symbolΦ.In the next section, we present a new coincide FPT by the generalization (ϕ,ψ)-contractive mappings on rectangular b-metric space fulfilling α-admissibility by the concept of C-functions.A lot of authors used altering functions to prove the existence and uniqueness of the fixed point; see [3, 22, 34]. We will use generalizing some of them to prove our results.


## 3. Main Results

We will introduce new results of coincide fixed point in RMS. Let us start with the following.Theorem 16.
LetX,d be a Hausdorff rectangular b-metric space, with b≥1. Assume that T,F:X⟶X be an α-admissible function with respect to η such that TX⊂FX. Assume that FX,d is a complete rectangular metric space. Let Θ∈C-functions and ϕ,ψ∈Φ such that, for all s1,s2∈X,
(11)αFs1,Fs2≥ηFs1,Fs2⇒ϕ1bdTs1,Ts2≤ΘϕϑFs1,Fs2,ψϑFs1,Fs2,where
(12)ϑFs1,Fs2=sup1bdFs1,Fs2,1bdFs1,Ts1,1bdFs2,Ts2,dFs1,Ts1dFs2,Ts2b+bdFs1,Fs2,dFs1,Ts1dFs2,Ts2b+bdTs1,Ts2.
Let that(a)
αFs1,Fs2≥ηFs1,Fs2andαFs2,Fs3≥ηFs2,Fs3⇒αFs1,Fs3≥ηFs1,Fs3, for all s1,s2,s3∈X(b)
eitherT is continuous or X is α-orderly with respect to η
Then, there existw∈X such that Tkw=Fkw, for some k∈ℕ; i.e., w is a periodic point. If all periodic point w satisfies
(13)αFw,Tw≥ηFw,Tw,
we can decide thatT and F have a fixed point. The fixed point is unique if for all z1,z2∈ΘT=w∈X:Fw=Tw=w, such that
(14)αFz1,Fz2≥ηFz1,Fz2.Proof.
We shall prove the existence of coincide point ofT and F. Assume that s0∈X be an arbitrary point such that
(15)αFs0,Ts0≥ηFs0,Ts0.Consider the iteration(16)Tsi=Fsi+1=vi,such thatvi≠vi+1, for all i∈ℕ. Since T satisfied Definition 11 and by (16) using (15), we have
(17)αvi,vi+1≥ηvi,vi+1,∀i∈ℕ.In the beginning, we will show thatdvi,vi+1⟶0,asi⟶∞; i.e., dvi,vi+1 is nonincreasing.By (11), we get
(18)ϕ1bdvi,vi+1=ϕdTsi,Tsi+1≤ΘϕϑFsi,Fsi+1,ψϑFsi,Fsi+1,where
(19)ϑFsi,Fsi+1=sup1bdFsi,Fsi+1,1bdFsi,Tsi,1bdFsi+1,Tsi+1,dFsi,TsidFsi+1,Tsi+1b+bdFsi,Fsi+1,dFsi,TsidFsi+1,Tsi+1b+bdTsi,Tsi+1=sup1bdFsi,Fsi+1,1bdFsi,Fsi+1,1bdFsi+1,Fsi+2,dFsi,Fsi+1dFsi+1,Fsi+2b+bdFsi,Fsi+1,dFsi,Fsi+1dFsi+1,Fsi+2b+bdFsi+1,Fsi+2=sup1bdvi−1,vi,1bdvi−1,vi,1bdvi,vi+1,dvi−1,vidvi,vi+1b+bdvi−1,vi,dvi−1,vidvi,vi+1b+bdvi,vi+1=sup1bdvi−1,vi,1bdvi,vi+1.We have two cases.Case 1.
IfϑFsi,Fsi+1=1/bdvi−1,vi for some i∈ℕ. Thus, inequality (18) will become
(20)ϕ1bdvi,vi+1≤Θϕ1bdvi−1,vi,ψ1bdvi−1,vi,≤ϕ1bdvi−1,vi
Sinceϕ∈Φandϕ is nondecreasing function,
(21)1bdvi,vi+1≤1bdvi−1,viwas obtained; dvi,vi+1 is a nonincreasing sequence which v1≥0 and satisfies
(22)limi⟶∞dvi,vi+1=v1,limi⟶∞ϑvi−1,vi=v1.
Also, sinceϕ,ψ are continuous functions,
(23)limi⟶∞ϕdvi,vi+1≤limi⟶∞Θϕϑvi−1,vi,ψϑvi−1,vi=Θlimi⟶∞ϕϑvi−1,vi,limi⟶∞ψϑvi−1,vi.
Thus,(24)ϕv1≤Θϕv1,ψv1≤ϕv1.
By Definition13, we get v1=0; hence,
(25)limi⟶∞dvi,vi+1=0.Case 2.
IfϑFsi,Fsi+1=1/bdvi,vi+1 for some i∈ℕ. Thus, inequality (18) becomes
(26)ϕ1bdvi,vi+1≤Θϕ1bdvi+1,vi,ψ1bdvi+1,vi≤ϕ1bdvi+1,vi.
By Definition13 we get either ϕ1/bdvi+1,vi=0 or ψ1/bdvi+1,vi=0 and then 1/bdvi+1,vi=0, but this is a contradiction with vi+1≠vi.In the next step, we will show that1/bdvi,vi+2⟶0,asi⟶∞.By (11), we have
(27)ϕ1bdvi,vi+2=ϕ1bdTsi,Tsi+2≤ΘϕϑFsi,Fsi+2,ψϑFsi,Fsi+2≤ϕϑFsi,Fsi+2.Well, it could be(28)ϕ1bdvi,vi+2≤ϕϑFsi,Fsi+2sinceϕ is altering distance. We find that
(29)1bdvi,vi+2≤ϑFsi,Fsi+2=sup1bdFsi,Fsi+2,1bdFsi,Tsi,1bdFsi+2,Tsi+2,dFsi,TsidFsi+2,Tsi+2b+bdFsi,Fsi+2,dFsi,TsidFsi+2,Tsi+2b+bdTsi,Tsi+2=sup1bdvi−1,vi+1,1bdvi−1,vi,1bdvi+1,vi+2,dvi−1,vidvi+1,vi+2b+bdvi−1,vi+1,dvi−1,vidvi+1,vi+2b+bdvi,vi+2≤sup1bdvi−1,vi+1,1bdvi−1,vi,1bdvi+1,vi+2,dvi−1,vidvi+1,vi+2,dvi−1,vidvi+1,vi+2≤sup1bdvi−1,vi+1+dvi−1,vi+dvi+1,vi+2,dvi−1,vidvi+1,vi+2=supdvi,vi+2,dvi−1,vidvi+1,vi+2.We obtainϑFsi,Fsi+2=dvi,vi+2, when i⟶∞. Therefore, by (27)
(30)ϕlimi⟶∞dvi,vi+2≤Θϕlimi⟶∞ϑFsi,Fsi+2,ψlimi⟶∞ϑFsi,Fsi+2≤ϕlimi⟶∞ϑFsi,Fsi+2=limi⟶∞ϕdvi,vi+2.Consequently, by Definition13, we get
(31)limi⟶∞dvi,vi+2⟶0,asi⟶∞.The next lemma is useful for the rest and its proof is classical. We omit it.Lemma 17.
LetX,d be a rectangular b-metric space with b≥1 and let vi be a sequence in X such that
(32)limi⟶∞vi,vi+1=limi⟶∞vi,vi+2=0,where vi≠mj, for all i≠j. If vi is not a b-Cauchy sequence, then there exist ε>0 and two subsequences vik,vjk⊂vi, where ik>jk>k,k∈ℕ. Also,
(33)dvik,vjk≥bε,dvik,vjk−1≤bε,
such that for the next sequences(34)dvik,vjk,dvik−1,vjk,dvik,vjk−1,dvik+1,vjk+1,
it satisfies(35)ε≤limi⟶∞infdvik,vjk≤limi⟶∞supdvik,vjk≤bε,ε≤limi⟶∞infdvik−1,vjk≤limi⟶∞supdvik−1,vjk≤b3ε,ε≤limi⟶∞infdvik,vjk≤limi⟶∞supdvik,vjk≤b3ε,ε≤limi⟶∞infdvik−1,vjk−1≤limi⟶∞supdvik−1,vjk−1≤b5ε.Now, replaces1 by xik and s2 by xjk in (11). Then,
(36)ϕ1bdvjk,vik=ϕ1bdTsjk,Tsik≤ΘϕϑFsjk,Fsik,ψϑFsjk,Fsik,where
(37)ϑFsjk,Fsik=sup1bdFsjk,Fsik,1bdFsjk,Tsjk,1bdFsik,Tsik,dFsjk,TsjkdFsik,Tsikb+bdFsjk,Fsik,dFsjk,TsjkdFsik,Tsikb+bdTsjk,Tsik=sup1bdvjk−1,vik−1,1bdvjk−1,vjk,1bdvik−1,vik,dvjk−1,vjkdvik−1,vikb+bdvjk−1,vik−1,dvjk−1,vjkdvik−1,vikb+bdvjk,vjk=supb4ε,0,0,0,0.Takingk⟶∞ in (36), we obtain
(38)0≤ϕb4ε≤Θϕb4ε,ψb4ε≤ϕb4ε.This implies thatϕb4ε=0 or ψb4ε=0; thus, ε=0, but this is a contradiction with the fact ε>0. Hence, vi is a Cauchy sequence in a rectangular b-metric space. Since FX,d is complete in a rectangular b-metric space, then there exist s∈FX such that vi⟶sasi⟶∞.For the case thatT is continuous and by relation (16) we have
(39)limi⟶∞Tvi=limi⟶∞Fvi+1⟶Ts.X is Hausdorff; then Ts=Fs. Hence, T and F have a common fixed point. On the other case, assume that X is α-orderly with respect to η, then from αFvi,Fw≥ηFvi,Fs, for all i∈ℕ we have
(40)ϕ1bdTvi,Ts≤ΘϕϑFvi,Fs,ψϑFvi,Fs,where
(41)ϑFvi,Fs=sup1bdFvi,Fs,1bdFvi,Tvi,1bdFs,Ts,dFvi,TvidFs,Tsb+bdFvi,Fs,dFvi,TvidFs,Tsb+bdTvi,Ts=sup1bdFs,Fs,1bdFs,Ts,1bdFs,Ts.Sincevi⟶swherei⟶∞.
(42)limi⟶∞ϑFvi,Fs=dFs,Ts.Then,dFs,Ts=0 and Fs=Ts. Hence, T and F have a periodic coincide fixed point.Now, we will prove thatT and F have a coincide fixed point. Assume that v1∈X is the coincide of T and F such that v1=Tks=Fks. When k=1, then v1, is a coincide of T and F. We shall show v2=Tk−1s=Fk−1s is the coincide of T and F in case k>1. Assume that Tk−1s≠Tks and Fk−1s≠Fks, for all k>1, such that αFs,Ts≥ηFs,Ts for a periodic point s. Therefore, from (11) and (12),
(43)ϕ1bdTk−1s,Tks≤ΘϕϑFTk−2s,FTk−1s,ψϑFTk−2s,FTk−1s≤ΘϕϑFFk−2s,FFk−1s,ψϑFFk−2s,FFk−1s=ΘϕϑFk−1s,Fks,ψϑFk−1s,Fks,where
(44)ϑFk−1s,Fks=sup1bdFk−1s,Fks,1bdFk−1s,TFk−2s,1bdFks,TFi−1s,dFk−1s,TFk−2sdFks,TFk−1sb+bdFk−1s,,Fks,dFk−1s,TFk−2sdFis,TFk−1sb+bdTFk−1s,TFk−2s=sup1bdFk−1s,Fks,1bdFk−1s,Tk−1s,1bdFks,Tis,dFk−1s,FksdFks,Tksb+bdFk−1s,,Fks,dFk−1s,FksdFks,Tksb+bdTks,Tk−1s=1bdFk−1s,Fks.TakeϑFk−1s,Fks=1/bdFk−1s,Fks. Then,
(45)ϕ1bdTk−1s,Tks≤Θϕ1bdFk−1s,Fks,ψ1bdFk−1s,Fks≤Θϕ1bdFk−1s,Fks.Eitherϕ1/bdFi−1s,Fis=0orψ1/bdFi−1s,Fis=0, i.e., dFi−1s,Fis=0; this leads to Fi−1s=Fis which is a contradiction. Hence, our assumption that v2=Ti−1s is a coincide fixed point of T and F is not true. Accordingly, T and F have a coincide fixed point.To make sure the uniqueness of the coincide fixed point, let us assume thatv1,v2∈X such that w1≠w2 are two coincide fixed points of T and F. By the inequality αFv1,Fv2≥ηFv1,Fv2 and (11) and (12), we get
(46)ϕ1bdFv1,Fv2=ϕ1bdTv1,Tv2≤ΘϕϑFv1,Fv2,ψϑFv1,Fv2,where
(47)ϑFv1,Fv2=supdFv1,Fv2,dFv1,Tv1,dFv2,Tv2,dFv1,Tv1dFv2,Tv2b+bdFv1,Fv2,dFv1,Fv2dTv1,Tv2b+bdTv1,Tv2=sup1bdv1,v2,1bdv1,v1,1bdv2,v2,dv1,v1dv2,v2b+bdv1,v2,dv1,v2dv1,v2b+bdv1,v2.Then,ϑFv1,Fv2=1/bdv1,v2. Applying it in (46), we get
(48)ϕ1bdFv1,Fv2=ϕ1bdTv1,Tv2≤Θϕ1bdv1,v2,ψ1bdv1,v2≤ϕ1bdv1,v2.Thus, eitherϕ1/bdv1,v2=0 or ψ1/bdv1,v2=0, which implies that dv1,v2=0. Hence, v1=v2. This proves the uniqueness coincide fixed point of T and F on X.Corollary 18.
LetX,d be a complete Hausdorff rectangular b-metric space, with b≥1. Let T,F:X×X be a self-mapping such that TX⊂FX satisfies
(49)dTs1,Ts2≤ηsup1bdFs1,Fs2,1bdFs1,Ts1,1bdFs2,Ts2,dFs1,Ts1dFs2,Ts2b+bdFs1,Fs2,dFs1,Ts1dFs2,Ts2b+bdTs1,Ts2
For alls1,s2∈X, such that 0≤η<1. Then, T and F have a unique coincide fixed point in X.Corollary 19.
LetX,d be a complete Hausdorff rectangular b-metric space, with b≥1. Let T,F:X×X be a self-mappings such that TX⊂FX satisfies
(50)dTs1,Ts2≤η1bdFs1,Fs2+dFs1,Ts1+dFs2,Ts2+dFs1,Ts1dFs2,Ts2b+bdFs1,Fs2+dFs1,Ts1dFs2,Ts2b+bdTs1,Ts2.
For alls1,s2∈X such that 0≤η<1. Then, T and F have a unique coincide fixed point in X.Corollary 20.
LetX,d be a complete bRMS Hausdorff, with b≥1. Let T,F:X×X be the self-mappings such that TX⊂FX satisfies
(51)dTs1,Ts2≤ΘϑFs1,Fs2,ψϑFs1,Fs2,where
(52)ϑFs1,Fs2≤η1bdFs1,Fs2,1bdFs1,Ts1,1bdFs2,Ts2,dFs1,Ts1dFs2,Ts2b+bdFs1,Fs2,dFs1,Ts1dFs2,Ts2b+bdTs1,Ts2.
For alls1,s2∈X such that 0≤η<1. Then, T and F have a unique coincide fixed point in X.Theorem 21.
LetX,d be a complete Haudorff rectangular b-metric space, with b≥1. Let T,F:X⟶X be the self-mappings such that TX⊂FX satisfies
(53)dTs1,Ts2≤ΘϕϑFs1,Fs2,ψϑFs1,Fs2.For all s1,s2∈Xandϕ,ψ∈Φ, such that
(54)ϑFs1,Fs2≤sup1bdFs1,Fs2,1bdFs2,Ts2b+bdFs1,Ts1b+bdFs1,Fs2,dFs1,Ts1dFs2,Ts2b+bdFs1,Fs2,dFs1,Ts1dFs2,Ts2b+bdTs1,Ts2.
Then,T and F have a unique coincide fixed point.Proof.
Assume thats0∈X be an arbitrary point. Since TX⊂FX, consider the sequence si⊂X such that Fsi=Tsi−1 for all i≥1. Let that Fsi≠Fsi+1=Tsi for all i≥0. We have from (11)
(55)ϕ1bdFsi,Fsi+1=ϕ1bdTsi−1,Tsi≤ΘϕϑFsi−1,Fsi,ψϑFsi−1,Fsi,where
(56)ϑFsi−1,Fsi≤sup1bdFsi−1,Fsi,1bdFsi,Tsib+bdFsi−1,Tsi−1b+bdFsi−1,Fsi,dFsi−1,Tsi−1dFsi,Tsib+bdFsi−1,Fsi,dFsi−1,Tsi−1dFsi,Tsib+bdTsi−1,Tsi=sup1bdFsi−1,Fsi,1bdFsi,Fsi+1b+bdFsi−1,Fsib+bdFsi−1,Fsi,dFsi−1,FsidFsi,Fsi+1b+bdFsi−1,Fsi,dFsi−1,FsidFsi,Fsi+1b+bdFsi,Fsi+1=sup1bdFsi−1,Fsi,1bdFsi,Fsi+1.
The remainder of the proof is identical as the proof of Theorem16.


## 4. Applications

Definition 22.
LetΓ be the class of functions τ:0,∞⟶0,∞ such the following are satisfying
(i)
τ is Lebesgue integral function for all compact subset of 0,∞(ii)
∫0ετxdx>0 for each ε>0Theorem 23.
LetX,d be a complete Hausdorff rectangular b-metric space, with b≥1. Let T,F:X⟶X be a self-mappings. Assume that FX,d is a complete rectangular b-metric space and that the next condition holds
(57)∫01/bdTs1,Ts2τsds≤∫0ϑFs1,Fs2τsds−∫0ϑFs1,Fs2χsds.
∀s1,s2∈Xandτ,χ∈Γ, such that T and F satisfy inequality (11), where
(58)ϑFs1,Fs2≤sup1bdFs1,Fs2,1bdFs1,Ts1,1bdFs2,Ts2,dFs1,Ts1dFs2,Ts2b+bdFs1,Fs2,dFs1,Ts1dFs2,Ts2b+bdTs1,Ts2.
Then,T and F have a unique coincide fixed point.Proof.
Let thatϕs=∫0sτudu and ψs=∫0sχudu. Then, ϕandψ∈Φ. Hence, by Theorem 16, T and F have a unique coincide fixed point.Theorem 24.
LetX,d be a complete bRMS Hausdorff with b≥1. Let T,F:X⟶X be a self-mappings, such that
(59)∫01/bdTs1,Ts2τrdr≤η∫0ϑFs1,Fs2τrdr.
For alls1,s2∈Xandτ∈Γ and 0≤η<1, such that T and F satisfy inequality (11), where
(60)ϑFs1,Fs2≤sup1bdFs1,Fs2,1bdFs1,Ts1,b+bdFs1,Ts1b+bdFs1,Fs2dFs2,Ts2,dFs1,Ts1dFs2,Ts2b+bdFs1,Fs2,dFs1,Ts1dFs2,Ts2b+bdTs1,Ts2.
Then,T and F have a unique coincide fixed point.Proof.
Supposegs=τs−ητs. Hence, by Theorem 23, T and F have a unique fixed point.Theorem 25.
LetX,d be a complete Hausdorff rectangular b-metric space, with b≥1. Assume that T,F:X⟶X be a self-mappings. Let
(61)∫01/bdTs1,Ts2τsds≤∫0ϑFs1,Fs2τsds−∫0ϑFs1,Fs2χsds.
For alls1,s2∈Xandτ,χ∈Γ, such that T and F satisfy inequality (11), where
(62)ϑFs1,Fs2≤sup1bdFs1,Fs2+1bdFs1,Ts1+1bdFs2,Ts2+dFs1,Ts1dFs2,Ts2b+bdFs1,Fs2+dFs1,Ts1dFs2,Ts2b+bdTs1,Ts2.
ThenT and F have a unique coincide fixed point.Proof.
Supposeϕs=∫0sζudu and ψs=∫0sfudu. Then, ϕandψ∈Φ. Hence, by Theorem 21, T and F have a unique coincide fixed point.Example 3.
SupposeX=0,1 and define T,F:X⟶X such that
(63)Ts1=s1+12,s1∈0,12,12,s1∈12,1.
Considerα,η:X×X⟶0,∞ since αFs1,Fs2=3 and ηFs1,Fs2=2, for all s1,s2∈X. Assume d:X×X⟶0,1 be rectangular b-metric space, with b=3, where
(64)1bdFs1,Fs2=15,s1,s2∈0,12,120,s1,s2∈12,0,19s1∈0,12,s2∈0,12.It is very well maybe the next step is to:(i)
dTs1,Ts2=1/20 and ϑFs1,Fs2=1/5, if s1,s2∈0,1/2(ii)
dTs1,Ts2=0 and ϑFs1,Fs2=1/20, if s1,s2∈1/2,1(iii)
dFs1,Fs2=1/20 and ϑFs1,Fs2=1/9, if s1∈0,1/2,s2∈0,1Consider the functionΘ:0,∞×0,∞⟶ℝ and ϕ,ψ:0,∞⟶0,∞ defined as Θs1,s2=s1−s2,ϕs2=4s2/5 and ψs2=s2/3. Then, Theorem 16 has been fulfilled. Hence, v1=1/2 is a unique coincide fixed point of T and F.



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*Source: 7291001-2022-05-14.xml*