# Contraction Mappings in Intuitionistic Fuzzy Rectangular Extended B-Metric Spaces

Authors: Doha Kattan; Amjad Owaidh Alzanbaqi; Sahidul Islam
Journal: Mathematical Problems in Engineering
 (2022)

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## Abstract

In this study, we present the notion of intuitionistic fuzzy rectangular extended b-metric spaces as a generalization of intuitionistic fuzzy metric spaces and intuitionistic fuzzy rectangular b-metric spaces. Some well-known fixed-point results in metric fixed-point theory are generalized in the sense of intuitionistic fuzzy rectangular extended b-metric spaces. Several nontrivial examples and an application to nonlinear fractional differential equations are also imparted in this work to examine the validity of given results.


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## Body

## 1. Introduction

In our daily life to impart knowledge and information, there is the great deal of imprecision and vagueness or fuzziness. Such statements as “Umar is tall” and “Gulzar is intelligent” are simple examples. In the above examples, terms such as “tall” and “intelligent” are fuzzy in the sense that they cannot be sharply defined. Here was a need to represent terms such as “tall” and “intelligent” in sharp contrast. Zadeh [1] introduced “fuzzy concept” in 1965 to represent imprecise terms (like in above examples) in sharp contrast.Two concepts fuzzy sets (FSs) are introduced in [1] and metric spaces are presented in [2] generalized under one notion named fuzzy metric spaces (FMSs) in which membership function is used. The concept of FMSs was generalized by many authors in different ways. We recall the notion of FMSs introduced by Kramosil and Michalak [3] in 1975 and modified by George and Veeramani [4, 5] in 1994. Garbiec [6] gave the fuzzy interpretation of Banach contraction principle in FMSs.Intuitionistic fuzzy sets (IFSs) introduced by Atanassov [7], that is, generalization of FSs in which degrees of membership and nonmembership are involved. Park [8] presented intuitionistic fuzzy metric spaces (IFMSs) by using the combined idea of IFSs and metric spaces, in which, he used George and Veeramani’s [4, 5] approach of applying continuous triangular norms (CTNs) and continuous triangular conorm (CTCNs) to the FMSs.Mehmood et al. [9] presented the concept of fuzzy rectangular metric spaces and fuzzy rectangular b-metric spaces (FRBMSs) and proved Banach contraction principle in the sense of FRBMSs. Hezarjaribi [10] originated the idea of orthogonal FMSs and proved various fixed-point results. Authors, in [11–14], are proved several fixed-point results for contraction mappings in FMSs. Park and Jeong [15] proved fixed-point results for fuzzy mappings. Later, Konwar [16] presented intuitionistic fuzzy b-metric space. Authors, in [17, 18], proved several fixed-point results for contraction mappings in the sense of IFMS. Authors, in [19, 20], did nice work on fractional differential equations. Javed et al. [21] proved several fixed-point results in the context of fuzzy b-metric-like spaces. Uddin et al. [22] proved various fixed-point theorems for contraction mappings in the sense of orthogonal controlled fuzzy metric spaces. Many mathematicians have applied different new models of fuzzy sets to several algebraic structures [23–27]. Recently, Aftab et al. [28] presented the notion of pentagonal controlled FMSs and proved several fixed-point results. Main objectives of manuscript areTo present the notion of intuitionistic fuzzy rectangular extended b-metric spacesTo prove certain fixed-point theoremsTo enhance existing literature of fuzzy metric spaces and fuzzy fixed-point theoryWe will present the notions of intuitionistic fuzzy rectangular extended b-metric spaces (IFREBMSs) by using the concepts of FRBMSs in [13] and IFMSs in [14]. Some well-known fixed-point results in metric fixed-point theory are generalized in the sense of IFREBMSs. Several nontrivial examples and an application to nonlinear fractional differential equations are also imparted in this work to examine the validity of given results.


## 2. Preliminaries

This section is devoted for some necessary notions that are helpful to understand main section.Definition 1.
(see [8]). A binary operation ∗:0,1×0,1⟶0,1 is a CTN if it satisfies the following conditions:(i)
∗ is associative and commutative(ii)
∗ is continuous(iii)
a∗1=a, for all a∈0,1(iv)
a∗b≤c∗d whenever a≤c and b≤d, for all a,b,c,d∈0,1Example 1.
(see [8]). a∗b=ab,a∗b=mina,b are CTN.Definition 2.
(see [8]). A binary operation ○:0,1×0,1⟶0,1 is called a CTCN if it meets the below assertions:(T1)
a○b=b○a,for alla,b∈0,1(T2)
○ is continuous(T3)
a○0=0,for alla∈0,1(T4)
a○b○c=a○b○c,for alla,b,c∈0,1(T5)
Ifa≤c and b≤d with a,b,c,d∈0,1, then a○b≤c○dExample 2.
(see [8]). a○b=maxa,b is CTCN.Definition 3.
(see [4]). An ordered triple V,M,∗ is a FMS if V is nonempty.
∗ is a CTN, M is a FS on V×V×0,+∞, satisfying the below assertions, for all ϰ,ζ,z∈V and σ,τ>0:(F1)
Mϰ,ζ,σ>0(F2)
Mϰ,ζ,σ=1 if and only if ϰ=ζ(F3)
Mϰ,ζ,σ=Mζ,ϰ,σ(F4)
Mϰ,ζ,σ∗Mζ,z,τ≤Mϰ,z,σ+τ(F5)
Mϰ,ζ,.:0,++∞⟶0,1 is continuousDefinition 4.
(see [9]). An ordered triple V,M,∗ is a FRBMS if V is nonempty, ∗ is a CTN, M is a FS on V×V×0,+∞ and fulfills the following assertions, for all ϰ,ζ,z∈V and σ,τ>0:(S1)
Mϰ,ζ,0=0(S2)
Mϰ,ζ,σ=1 if and only if ϰ=ζ(S3)
Mϰ,ζ,σ=Mζ,ϰ,σ(S4)
Mϰ,ζ,σ∗Mζ,u,τ∗Mu,z,w≤Mϰ,z,bσ+τ+w(S5)
Mϰ,ζ,.:0,++∞⟶0,1 is left continuous and limσ⟶+∞Mϰ,ζ,σ=1Definition 5.
(see [8]). Take V≠∅. Let ∗ be a CTN, ○ be a CTCN, and F,V be FSs on V×V×0,+∞. If V,F,V,∗,○ verifies the following assertions for all ϰ,ζ∈V,τ,σ>0,(F1)
Fϰ,ζ,σ+Vϰ,ζ,σ≤1(F2)
Fϰ,ζ,σ>0(F3)
Fϰ,ζ,σ=1⇔ϰ=ζ(F4)
Fϰ,ζ,σ=Fζ,ϰ,σ(F5)
Fϰ,z,σ+τ≥Fϰ,ζ,σ∗Fζ,z,τ(F6)
Fϰ,ζ,⋅:0,+∞⟶0,1 is continuous and limσ⟶+∞Fϰ,ζ,σ=1for allσ>0(F7)
Vϰ,ζ,σ>0(F8)
Vϰ,ζ,σ=0⇔ϰ=ζ(F9)
Vϰ,ζ,σ=Vζ,ϰ,σ(F10)
Vϰ,z,σ+τ≤Vϰ,ζ,σ○Vζ,z,τ(F11)
Vϰ,ζ,⋅:0,++∞⟶0,1 is continuous and limσ⟶+∞Vϰ,ζ,σ=0for allσ>0
Then,V,F,V,∗,○ is an IFMS.


## 3. Main Section

This section is devoted to present the notion of IFREBMS and to prove fixed-point results.Definition 6.
LetV be a nonempty set. A five tuple V,M,N,∗,○ is called an IFREBMS if ∗ is a CTN, ○ is a CTCN, and M,N are FSs on V×V×0,+∞,f:V×V⟶1,+∞ fulfilling the following assertions for all ϰ,ζ,z∈V and σ,τ,w>0:(1)
Mϰ,ζ,σ+Nϰ,ζ,σ≤1(2)
Mϰ,ζ,0=0(3)
Mϰ,ζ,σ=1 if and only if ϰ=ζ(4)
Mϰ,ζ,σ=Mζ,ϰ,σ(5)
Mϰ,z,fϰ,zσ+τ+w≥Mϰ,ζ,σ∗Mζ,u,τ∗Mu,z,w for all distinct ζ,u∈V\ϰ,z(6)
Mϰ,ζ,.:0,+∞⟶0,1 is continuous and limσ⟶+∞Mϰ,ζ,σ=1(7)
Nϰ,ζ,0=1(8)
Nϰ,ζ,σ=0 if and only if ϰ=ζ(9)
Nϰ,ζ,σ=Nζ,ϰ,σ(10)
Nϰ,z,fϰ,zσ+τ+w≤Nϰ,ζ,σ○Nζ,u,τ○Nu,z,w, for all distinct ζ,u∈V\ϰ,z(11)
Nϰ,ζ,.:0,+∞⟶0,1 is continuous and limσ⟶+∞Nϰ,ζ,σ=0Example 3.
LetV,d be a rectangular extended b-metric space, define f:V×V⟶1,+∞ by fϰ,ζ=1+ϰ+ζ, and define M,N:V×V×0,+∞⟶0,1 by(1)Mϰ,ζ,σ=σσ+dϰ,ζ,Nϰ,ζ,σ=dϰ,ζσ+dϰ,ζ,for allϰ,ζ∈V,σ>0,with CTN a∗b=mina,b and CTCN a○b=maxa,b. Then, V,M,N,∗,○ is an IFREBMS.Proof.
Properties 1–4, 6–9, and 11 are easy to examine. Here, we prove 5 and 11.
(5)Mϰ,z,fϰ,zσ+τ+w≥Mϰ,ζ,σ∗Mζ,u,τ∗Mu,z,w, for all distinct ζ,u∈V\ϰ,z
Without loss of any generality, we consider that(2)Mϰ,ζ,σ≤Mζ,u,τ,(3)Mϰ,ζ,σ≤Mu,z,w,which implies that(4)σσ+dϰ,ζ≤ττ+dζ,u,(5)σσ+dϰ,ζ≤ww+du,z.
We obtain(6)σdζ,u≤τdϰ,ζ,σdu,z≤wdϰ,ζ.
This implies(7)τ+wdϰ,ζ≥σdζ,u+du,z.
Now, observe that(8)Mϰ,z,fϰ,zσ+τ+w≥Mϰ,ζ,σ,⟺fϰ,zσ+τ+wfϰ,zσ+τ+w+dϰ,z≥σσ+dϰζ⟺fϰ,zσ+τ+wfϰ,zσ+τ+w+fϰ,zdϰ,ζ+dζ,u+du,z≥σσ+dϰ,ζ⟺σ+τ+wσ+τ+w+dϰ,ζ+dζ,u+du,z≥σσ+dϰ,ζ⟺τ+wdϰ,ζ≥σdζ,u+du,z,which is the same as (1). Hence, it follows that(9)Mϰ,z,fϰ,zσ+τ+w≥Mϰ,ζ,σ∗Mζ,u,τ∗Mu,z,w.
(10)Nϰ,z,fϰ,zσ+τ+w≤Nϰ,ζ,σ○Nζ,u,τ○Nu,z,w, for all distinct ζ,u∈V\ϰ,z.
Recall that(10)dϰ,z=dϰ,zmaxdϰ,ζdϰ,ζ,dζ,udζ,u,du,zdu,z.
Therefore,(11)dϰ,z≤σ+τ+w+dϰ,zmaxdϰ,ζdϰ,ζ,dζ,udζ,u,du,zdu,z.
Definef:V×V⟶1,+∞ by fϰ,ζ=1+ϰ+ζ. Then,(12)dϰ,z≤fϰ,zσ+τ+w+dϰ,zmaxdϰ,ζdϰ,ζ,dζ,udζ,u,du,zdu,z.
Also, observe the fact that(13)dϰ,z≤fϰ,zσ+τ+w+dϰ,zmaxdϰ,ζσ+dϰ,ζ,dζ,uτ+dζ,u,du,zw+du,z.
This implies(14)dϰ,zfϰ,zσ+τ+w+dϰ,z≤maxdϰ,ζσ+dϰ,ζ,dζ,uτ+dζ,u,du,zw+du,z.
Then,(15)Nϰ,z,fϰ,zσ+τ+w≤maxNϰ,ζ,σ,Nζ,u,τ,Nu,z,w.
We defined continuous t-conorm bya○b=maxa,b. Hence,(16)Nϰ,z,fϰ,zσ+τ+w≤Nϰ,ζ,σ○Nζ,u,τ○Nu,z,w.
Therefore,V,M,N,∗,○ is an IFREBMS.Remark 1.
In the above example, if we takef=1, then it also satisfied the properties of intuitionistic fuzzy metric spaces.Example 4.
LetV=0,+∞. Define f:V×V⟶1,+∞ by fϰ,ζ=1+ϰ2+ζ3 and define M,N:V×V×0,+∞⟶0,1 by(17)Mϰ,ζ,σ=1,ifϰ=ζ,σσ+maxϰ,ζp,otherwise,(18)Nϰ,ζ,σ=0,ifϰ=ζ,maxϰ,ζpσ+maxϰ,ζp,otherwise,for allϰ,ζ∈V,σ>0.
CTN and CTCN are, respectively, defined bya∗b=a⋅b,a○b=maxa,b,p≥1; then, it is easy to see that V,M,N,∗,○ is an IFREBMS on the lines of the above example.Example 5.
LetV=0,+∞. Define f:V×V⟶1,+∞ by fϰ,ζ=1+ϰ/ζ, and define M,N:V×V×0,+∞⟶0,1 by(19)Mϰ,ζ,σ=1,ifϰ=ζ,σσ+ϰ+ζp,otherwise,(20)Nϰ,ζ,σ=0,ifϰ=ζ,ϰ+ζpσ+ϰ+ζp,otherwise.for allϰ,ζ∈V,σ>0.
CTN and CTCN are, respectively, defined bya∗b=a⋅b,a○b=maxa,b,p≥1; then, it is easy to see that V,M,N,∗,○ is an IFREBMS.Example 6.
LetV=0,+∞. Define f:V×V⟶1,+∞ by fϰ,ζ=1ifϰ=ζ1+ϰ+ζotherwise, and define M,N:V×V×0,+∞⟶0,1 by(21)Mϰ,ζ,σ=σσ+ϰ−ζp,Nϰ,ζ,σ=ϰ−ζpσ+ϰ−ζp,for allϰ,ζ∈V,σ>0.
CTN and CTCN are, respectively, defined bya∗b=a⋅b,a○b=maxa,b,p≥1; then, it is easy to see that V,M,N,∗,○ is an IFREBMS.Example 7.
LetV,d be a rectangular extended b-metric space. Define f:V×V⟶1,+∞ by(22)fϰ,ζ=1,ifϰ=ζ,1+maxϰ,ζ,otherwise.
DefineM,N:V×V×0,+∞⟶0,1 by(23)Mϰ,ζ,σ=0,ifσ=0,e−dϰ,ζ/σ,otherwise,(24)Nϰ,ζ,σ=1,ifσ=0,1−e−dϰ,ζ/σ,otherwise,for allϰ,ζ∈V,σ>0.
CTN and CTCN are, respectively, defined bya∗b=a⋅b,a○b=maxa,b; then, it is easy to see that V,M,N,∗,○ is an IFREBMS.Remark 2.
Examples5–7 are also satisfied if a∗b=mina,b,a○b=maxa,b.
Now, following Gabriec, we present the notions of a convergent sequence and Cauchy sequence in an IFREBMS.Definition 7.
LetV,M,N,∗,○ be an IFREBMS, and assume ϰn is a sequence in V. Then,ϰn is named to be a convergent sequence if there exists ϰ∈V such that(25)limn⟶+∞Mϰn,ϰ,σ=1,for allσ>0,(26)limn⟶+∞Nϰn,ϰ,σ=0,for allσ>0,ϰn is named to be Cauchy sequence if(27)limn⟶+∞Mϰn,ϰn+q,σ=1,for allσ>0,(28)limn⟶+∞Nϰn,ϰn+q,σ=0,for  allσ>0.If every Cauchy sequence is convergent inV, then V,M,N,∗,○ is said to be a complete IFREBMS.Example 8.
LetV=0,+∞. Define f:V×V⟶1,+∞ by fϰ,ζ=1ifϰ=ζ1+ϰ+ζotherwise, and define M,N:V×V×0,+∞⟶0,1 by(29)Mϰ,ζ,σ=σσ+ϰ−ζp,Nϰ,ζ,σ=ϰ−ζpσ+ϰ−ζp,for allϰ,ζ∈V,σ>0.
CTN and CTCN are, respectively, defined bya∗b=a⋅b,a○b=maxa,b,p≥1; then, V,M,N,∗,○ is an IFREBMS.
Letϰn=1/n, for all n∈1,2,3,…,, be a sequence in V; then, ϰn converges to 0. Now,(30)Mϰ,ζ,σ=σσ+ϰ−ζp,limn⟶+∞Nϰn,0,σ=limn⟶+∞1/npσ+1/np=0.
Hence,ϰn is a convergent sequence.Example 9.
Consider an IFREBMS in the above example, and consider a sequenceϰn=1/n, for all n∈1,2,3,…,. Now, for all q∈1,2,3,…,, we deduce(31)limn⟶+∞Mϰn,ϰn+q,σ=limn⟶+∞σσ+1/n+1p+⋯+1/n+qp=1,limn⟶+∞Nϰn,ϰn+q,σ=limn⟶+∞1/n+1p+⋯+1/n+qpσ+1/n+1p+⋯+1/n+qp=0.
Hence,ϰn is a Cauchy sequence.Theorem 1.
(Banach contraction theorem in an IFREBMS). LetV,M,N,∗,○ be a complete IFREBMS such that(32)limσ⟶+∞Mϰ,ζ,σ=1,limσ⟶+∞Nϰ,ζ,σ=0,for allϰ,ζ∈V.
LetΨ:V⟶V be a mapping satisfying(33)MΨϰ,Ψζ,ησ≥Mϰ,ζ,σ,NΨϰ,Ψζ,ησ≤Nϰ,ζ,σ,for all ϰ,ζ∈V,η∈0,1. Then, Ψ has a unique fixed point, u∈V.Proof.
Fix an arbitrary pointϰ0∈V and for n=0,1,2,…, start an iterative process ϰn+1=Ψϰn. Successively applying inequality (33), we get, for all n,σ>0,(34)Mϰn,ϰn+1,σ≥Mϰ0,ϰ1,σηn,Nϰn,ϰn+1,σ≤Nϰ0,ϰ1,σηn.
SinceV,M,N,∗,○ is a complete IFREBMS for the sequence ϰn, writing σ=σ/3+σ/3+σ/3 and using the rectangular inequality given in (5) and (10) on Mϰn,ϰn+p,σandNϰn,ϰn+p,σ in the following cases.Case 1.
Ifp is odd, then p=2m+1, where m∈1,2,3,…; we have(35)Mϰn,ϰn+2m+1,σ≥Mϰn,ϰn+1,σ3fϰn,ϰn+2m+1∗Mϰn+1,ϰn+2,σ3fϰn,ϰn+2m+1∗Mϰn+2,ϰn+2m+1,σ3fϰn,ϰn+2m+1≥Mϰn,ϰn+1,σ3fϰn,ϰn+2m+1∗Mϰn+1,ϰn+2,σ3fϰn,ϰn+2m+1∗Mϰn+2,ϰn+3,σ32fϰn,ϰn+2m+1fϰn+2,ϰn+3∗Mϰn+3,ϰn+4,σ32fϰn,ϰn+2m+1fϰn+2,ϰn+3∗Mϰn+4,ϰn+2m+1,σ32fϰn,ϰn+2m+1fϰn+2,ϰn+3,≥Mϰn,ϰn+1,σ3fϰn,ϰn+2m+1∗Mϰn+1,ϰn+2,σ3fϰn,ϰn+2m+1∗Mϰn+2,ϰn+3,σ32fϰn,ϰn+2m+1fϰn+2,ϰn+3∗Mϰn+3,ϰn+4,σ32fϰn,ϰn+2m+1fϰn+2,ϰn+3∗Mϰn+4,ϰn+5,σ33fϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5∗Mϰn+5,ϰn+6,σ33fϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5,(36)Mϰn+2m,ϰn+2m+1,σ3mfϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5⋯fϰn+2m,ϰn+2m+1.(37)Nϰn,ϰn+2m+1,σ≤Nϰn,ϰn+1,σ3fϰn,ϰn+2m+1○Nϰn+1,ϰn+2,σ3fϰn,ϰn+2m+1○Nϰn+2,ϰn+2m+1,σ3fϰn,ϰn+2m+1≤Nϰn,ϰn+1,σ3fϰn,ϰn+2m+1○Nϰn+1,ϰn+2,σ3fϰn,ϰn+2m+1○Nϰn+2,ϰn+3,σ32fϰn,ϰn+2m+1fϰn+2,ϰn+3○Nϰn+3,ϰn+4,σ32fϰn,ϰn+2m+1fϰn+2,ϰn+3○Nϰn+4,ϰn+2m+1,σ32fϰn,ϰn+2m+1fϰn+2,ϰn+3≤Nϰn,ϰn+1,σ3fϰn,ϰn+2m+1○Nϰn+1,ϰn+2,σ3fϰn,ϰn+2m+1○Nϰn+2,ϰn+3,σ32fϰn,ϰn+2m+1fϰn+2,ϰn+3○Nϰn+3,ϰn+4,σ32fϰn,ϰn+2m+1fϰn+2,ϰn+3○Nϰn+4,ϰn+5,σ33fϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5○Nϰn+5,ϰn+6,σ33fϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5,(38)Nϰn+2m,ϰn+2m+1,σ3mfϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5⋯fϰn+2m,ϰn+2m+1.
Using (34) in the above inequalities, we deduce(39)Mϰn,ϰn+2m+1,σ≥Mϰ0,ϰ1,σ3ηnfϰn,ϰn+2m+1∗Mϰ0,ϰ1,σ3ηn+1fϰn,ϰn+2m+1,∗Mϰ0,ϰ1,σ32ηn+2fϰn,ϰn+2m+1fϰn+2,ϰn+3∗Mϰ0,ϰ1,σ32ηn+3fϰn,ϰn+2m+1fϰn+2,ϰn+3∗Mϰ0,ϰ1,σ33ηn+4fϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5∗Mϰ0,ϰ1,σ33ηn+5fϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5,(40)Mϰ0,ϰ1,σ3mηn+2mfϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5⋯fϰn+2m,ϰn+2m+1≥Mϰ0,ϰ1,σ3ηnfϰn,ϰn+2m+1∗Mϰ0,ϰ1,σ3ηηnfϰn,ϰn+2m+1∗Mϰ0,ϰ1,σ3η2ηnfϰn,ϰn+2m+1fϰn+2,ϰn+3∗Mϰ0,ϰ1,σ3η2ηn+1fϰn,ϰn+2m+1fϰn+2,ϰn+3∗Mϰ0,ϰ1,σ3η3ηn+1fϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5∗Mϰ0,ϰ1,σ3η3ηn+2fϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5,(41)Mϰ0,ϰ1,σ3ηmηn+mfϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5⋯fϰn+2m,ϰn+2m+1,(42)Nϰn,ϰn+2m+1,σ≤Nϰ0,ϰ1,σ3ηnfϰn,ϰn+2m+1○Nϰ0,ϰ1,σ3ηn+1fϰn,ϰn+2m+1,○Nϰ0,ϰ1,σ32ηn+2fϰn,ϰn+2m+1fϰn+2,ϰn+3,○Nϰ0,ϰ1,σ32ηn+3fϰn,ϰn+2m+1fϰn+2,ϰn+3,○Nϰ0,ϰ1,σ33ηn+4fϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5,○Nϰ0,ϰ1,σ33ηn+5fϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5,(43)Nϰ0,ϰ1,σ3mηn+2mfϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5⋯fϰn+2m,ϰn+2m+1≤Nϰ0,ϰ1,σ3ηnfϰn,ϰn+2m+1○Nϰ0,ϰ1,σ3ηηnfϰn,ϰn+2m+1○Nϰ0,ϰ1,σ3η2ηnfϰn,ϰn+2m+1fϰn+2,ϰn+3○Nϰ0,ϰ1,σ3η2ηn+1fϰn,ϰn+2m+1fϰn+2,ϰn+3○Nϰ0,ϰ1,σ3η3ηn+1fϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5○Nϰ0,ϰ1,σ3η3ηn+2fϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5,(44)Nϰ0,ϰ1,σ3ηmηn+mfϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5⋯fϰn+2m,ϰn+2m+1.Case 2.
Ifp is even, then p=2m,m∈1,2,3,⋯; we have(45)Mϰn,ϰn+2m,σ≥Mϰn,ϰn+1,σ3fϰn,ϰn+2m∗Mϰn+1,ϰn+2,σ3fϰn,ϰn+2m∗Mϰn+2,ϰn+2m,σ3fϰn,ϰn+2m≥Mϰn,ϰn+1,σ3fϰn,ϰn+2m∗Mϰn+1,ϰn+2,σ3fϰn,ϰn+2m∗Mϰn+2,ϰn+3,σ32fϰn,ϰn+2mfϰn+2,ϰn+3∗Mϰn+3,ϰn+4,σ32fϰn,ϰn+2mfϰn+2,ϰn+3∗Mϰn+4,ϰn+2m,σ32fϰn,ϰn+2mfϰn+2,ϰn+3≥Mϰn,ϰn+1,σ3fϰn,ϰn+2m∗Mϰn+1,ϰn+2,σ3fϰn,ϰn+2m∗Mϰn+2,ϰn+3,σ32fϰn,ϰn+2mfϰn+2,ϰn+3∗Mϰn+3,ϰn+4,σ32fϰn,ϰn+2mfϰn+2,ϰn+3∗Mϰn+4,ϰn+5,σ33fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5∗Mϰn+5,ϰn+6,σ33fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5,(46)Mϰn+2m−2,ϰn+2m,σ3m−1fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5⋯fϰn+2m−2,ϰn+2m,(47)Nϰn,ϰn+2m,σ≤Nϰn,ϰn+1,σ3fϰn,ϰn+2m,○Nϰn+1,ϰn+2,σ3fϰn,ϰn+2m○Nϰn+2,ϰn+2m,σ3fϰn,ϰn+2m≤Nϰn,ϰn+1,σ3fϰn,ϰn+2m○Nϰn+1,ϰn+2,σ3fϰn,ϰn+2m,○Nϰn+2,ϰn+3,σ32fϰn,ϰn+2mfϰn+2,ϰn+3○Nϰn+3,ϰn+4,σ32fϰn,ϰn+2mfϰn+2,ϰn+3○Nϰn+4,ϰn+2m,σ32fϰn,ϰn+2mfϰn+2,ϰn+3≤Nϰn,ϰn+1,σ3fϰn,ϰn+2m○Nϰn+1,ϰn+2,σ3fϰn,ϰn+2m○Nϰn+2,ϰn+3,σ32fϰn,ϰn+2mfϰn+2,ϰn+3○Nϰn+3,ϰn+4,σ32fϰn,ϰn+2mfϰn+2,ϰn+3○Nϰn+4,ϰn+5,σ33fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5○Nϰn+5,ϰn+6,σ33fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5,(48)Nϰn+2m−2,ϰn+2m,σ3mfϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5⋯fϰn+2m−2,ϰn+2m.
Using (34) in the above inequalities, we deduce(49)Mϰn,ϰn+2m,σ≥Mϰ0,ϰ1,σ3ηnfϰn,ϰn+2m∗Mϰ0,ϰ1,σ3ηn+1fϰn,ϰn+2m,∗Mϰ0,ϰ1,σ32ηn+2fϰn,ϰn+2mfϰn+2,ϰn+3,∗Mϰ0,ϰ1,σ32ηn+3fϰn,ϰn+2mfϰn+2,ϰn+3∗Mϰ0,ϰ1,σ33ηn+4fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5,∗Mϰ0,ϰ1,σ33ηn+5fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5,(50)Mϰ0,ϰ1,σ3m−1ηn+2m−2fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5⋯fϰn+2m−2,ϰn+2m≥Mϰ0,ϰ1,σ3ηnfϰn,ϰn+2m∗Mϰ0,ϰ1,σ3ηηnfϰn,ϰn+2m∗Mϰ0,ϰ1,σ3η2ηnfϰn,ϰn+2mfϰn+2,ϰn+3∗Mϰ0,ϰ1,σ3η2ηn+1fϰn,ϰn+2mfϰn+2,ϰn+3∗Mϰ0,ϰ1,σ3η3ηn+1fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5∗Mϰ0,ϰ1,σ3η3ηn+2fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5,(51)Mϰ0,ϰ1,σ3ηm−1ηn+m−1fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5⋯fϰn+2m−2,ϰn+2m,(52)Nϰn,ϰn+2m,σ≤Nϰ0,ϰ1,σ3ηnfϰn,ϰn+2m○Nϰ0,ϰ1,σ3ηn+1fϰn,ϰn+2m○Nϰ0,ϰ1,σ32ηn+2fϰn,ϰn+2mfϰn+2,ϰn+3○Nϰ0,ϰ1,σ32ηn+3fϰn,ϰn+2mfϰn+2,ϰn+3○Nϰ0,ϰ1,σ33ηn+4fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5○Nϰ0,ϰ1,σ33ηn+5fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5,(53)Nϰ0,ϰ1,σ3mηn+2m−2fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5⋯fϰn+2m−2,ϰn+2m≤Nϰ0,ϰ1,σ3ηnfϰn,ϰn+2m○Nϰ0,1,σ3ηηnfϰn,ϰn+2m○Nϰ0,ϰ1,σ3η2ηnfϰn,ϰn+2mfϰn+2,ϰn+3○Nϰ0,ϰ1,σ3η2ηn+1fϰn,ϰn+2mfϰn+2,ϰn+3○Nϰ0,ϰ1,σ3η3ηn+1fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5○Nϰ0,ϰ1,σ3η3ηn+2fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5,(54)Nϰ0,ϰ1,σ3ηm−1ηn+m−1fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5⋯fϰn+2m−2,ϰn+2m.
Therefore, fromlimσ⟶+∞Mϰ,ζ,σ=1,limσ⟶+∞Nϰ,ζ,σ=0 Case (1), Case (2), and together with (32), it follows that, for all p∈1,2,3,…, we have(55)limn⟶+∞Mϰn,ϰn+p,σ=1,limn⟶+∞Nϰn,ϰn+p,σ=0.
Hence,ϰn is a Cauchy sequence. Since, V,M,N,∗,○ is a complete IFREBMS, so there exists u∈V such that(56)limn⟶+∞Mϰn,u,σ=1,limn⟶+∞Nϰn,u,σ=0,for allσ>0,q≥1.
Now, we examine thatu is a fixed point of Ψ:(57)Mu,Ψu,σ≥Mu,ϰn,σ3fu,Ψu∗Mϰn,ϰn+1,σ3fu,Ψu∗Mϰn+1,Ψu,σ3fu,Ψu≥Mu,ϰn,σ3fu,Ψu∗MΨϰn−1,Ψϰn,σ3fu,Ψu∗MΨϰn,Ψu,σ3fu,Ψu≥Mu,ϰn,σ3fu,Ψu∗Mϰn−1,ϰn,σ3ηfu,Ψu∗Mϰn,u,σ3ηbfu,Ψu⟶1∗1∗1=1asn⟶+∞,(58)Nu,Ψu,σ≤Nu,ϰn,σ3fu,Ψu○Nϰn,ϰn+1,σ3fu,Ψu○Nϰn+1,Ψu,σ3fu,Ψu≤Nu,ϰn,σ3fu,Ψu○NΨϰn−1,Ψϰn,σ3fu,Ψu○NΨϰn,Ψu,σ3fu,Ψu≤Nu,ϰn,σ3fu,Ψu○Nϰn−1,ϰn,σ3ηfu,Ψu○Nϰn,u,σ3ηbfu,Ψu⟶0○0○0=0asn⟶+∞,which shows that u is a fixed point of Ψ. Now, we show the uniqueness.
Assumev is another fixed point of Ψ, for some v∈V; then,(59)Mv,u,σ=MΨv,Ψu,σ≥Mv,u,ση=MΨv,Ψu,ση≥Mv,u,ση2≥⋯≥Mv,u,σηn⟶1asn⟶+∞,and(60)Nv,u,σ=NΨv,Ψu,σ≤Nv,u,ση=NΨv,Ψu,ση≤Nv,u,ση2≤⋯≤Nv,u,σηn⟶0asn⟶+∞.
Thus,u=v. Hence, the fixed point is unique.Remark 3.
In Theorem1, if we take f=1, then it has also a unique fixed point in the context of intuitionistic fuzzy metric spaces.Definition 8.
LetV,M,N,∗,○ be a complete IFREBMS. A mapping Ψ:V⟶V is an intuitionistic fuzzy rectangular contraction, such that(61)1MΨϰ,Ψζ,σ−1≤η1Mϰ,ζ,σ−1,MΨϰ,Ψζ,σ≤ηNϰ,ζ,σ,for all ϰ,ζ∈V,η∈0,1,andσ>0.Theorem 2.
LetV,M,N,∗,○ be a complete IFREBMS, such that(62)limσ⟶+∞Mϰ,ζ,σ=1,limσ⟶+∞Mϰ,ζ,σ=1,for allϰ,ζ∈V.
LetΨ:V⟶V be an intuitionistic fuzzy rectangular contraction. Then, Ψ has a unique fixed point u∈V.Proof.
LetV,M,N,∗,○ is a complete IFREBMS. For arbitrary ϰ0∈V, define a sequence ϰn in V by ϰ1=Ψϰ0, ϰ2=Ψ2ϰ0=Ψϰ1,…,ϰn=Ψnϰ0=Ψϰn−1, for all. n∈ℕ.
Ifϰn=ϰn−1, for some n∈ℕ, then ϰn is a fixed point of Ψ. We assume that ϰn≠ϰn−1, for all n∈ℕ. For σ>0 and n∈ℕ, we get from (61) the following equation:(63)1Mϰn,ϰn+1,σ−1=1MΨϰn−1,Ψϰn,σ−1≤η1Mϰn−1,ϰn,σ−1.
We have(64)1Mϰn,ϰn+1,σ≤ηMϰn−1,ϰn,σ+1−η,∀σ>0,=ηMΨϰn−2,Ψϰn−1,σ+1−η≤η2Mϰn−2,ϰn−1,σ+η1−η+1−η.
Continuing this way, we obtain(65)1Mϰn,ϰn+1,σ≤ηnMϰ0,ϰ1,σ+ηn−11−η+ηn−21−η+…+η1−η+1−η≤ηnMϰ0,ϰ1,σ+ηn−1+ηn−2+…+11−η≤ηnMϰ0,ϰ1,σ+1−ηn.
We have(66)1ηn/Mϰ0,ϰ1,σ+1−ηn≤Mϰn,ϰn+1,σ,∀σ>0,n∈ℕ⋬(67)Nϰn,ϰn+1,σ=NΨϰn−1,Ψϰn,σ≤ηNϰn−1,ϰn,σ=ηNΨϰn−2,Ψϰn−1,σ,≤η2Nϰn−2,ϰn−1,σ≤⋯≤ηnϰ0,ϰ1,σ.
SinceV,M,N,∗,○ is a complete IFREBMS for the sequence ϰn, write σ=σ/3+σ/3+σ/3 and use the rectangular inequalities given in (5) and (10) on Mϰn,ϰn+p,σandNϰn,ϰn+p,σ in the following cases.Case 3.
Ifp is odd, then p=2m+1, where m∈1,2,3,…; we have(68)Mϰn,ϰn+2m+1,σ≥Mϰn,ϰn+1,σ3fϰn,ϰn+2m+1∗Mϰn+1,ϰn+2,σ3fϰn,ϰn+2m+1∗Mϰn+2,ϰn+2m+1,σ3fϰn,ϰn+2m+1≥Mϰn,ϰn+1,σ3fϰn,ϰn+2m+1∗Mϰn+1,ϰn+2,σ3fϰn,ϰn+2m+1∗Mϰn+2,ϰn+3,σ32fϰn,ϰn+2m+1fϰn+2,ϰn+3∗Mϰn+3,ϰn+4,σ32fϰn,ϰn+2m+1fϰn+2,ϰn+3∗Mϰn+4,ϰn+2m+1,σ32fϰn,ϰn+2m+1fϰn+2,ϰn+3≥Mϰn,ϰn+1,σ3fϰn,ϰn+2m+1∗Mϰn+1,ϰn+2,σ3fϰn,ϰn+2m+1∗Mϰn+2,ϰn+3,σ32fϰn,ϰn+2m+1fϰn+2,ϰn+3∗Mϰn+3,ϰn+4,σ32fϰn,ϰn+2m+1fϰn+2,ϰn+3∗Mϰn+4,ϰn+5,σ33fϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5∗Mϰn+5,ϰn+6,σ33fϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5,(69)Mϰn+2m,ϰn+2m+1,σ3mfϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5⋯fϰn+2m,ϰn+2m+1,(70)Nϰn,ϰn+2m+1,σ≤Nϰn,ϰn+1,σ3fϰn,ϰn+2m+1,○Nϰn+1,ϰn+2,σ3fϰn,ϰn+2m+1○Nϰn+2,ϰn+2m+1,σ3fϰn,ϰn+2m+1,≤Nϰn,ϰn+1,σ3fϰn,ϰn+2m+1○Nϰn+1,ϰn+2,σ3fϰn,ϰn+2m+1,○Nϰn+2,ϰn+3,σ32fϰn,ϰn+2m+1fϰn+2,ϰn+3,○Nϰn+3,ϰn+4,σ32fϰn,ϰn+2m+1fϰn+2,ϰn+3,○Nϰn+4,ϰn+2m+1,σ32fϰn,ϰn+2m+1fϰn+2,ϰn+3,≤Nϰn,ϰn+1,σ3fϰn,ϰn+2m+1○Nϰn+1,ϰn+2,σ3fϰn,ϰn+2m+1,○Nϰn+2,ϰn+3,σ32fϰn,ϰn+2m+1fϰn+2,ϰn+3,○Nϰn+3,ϰn+4,σ32fϰn,ϰn+2m+1fϰn+2,ϰn+3,○Nϰn+4,ϰn+5,σ33fϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5,○Nϰn+5,ϰn+6,σ33fϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5,(71)Nϰn+2m,ϰn+2m+1,σ3mfϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5⋯fϰn+2m,ϰn+2m+1.
By using (66) and (67) in the above inequalities, we have(72)Mϰn,ϰn+2m+1,σ≥1ηn/Mϰ0,ϰ1,σ/3fϰn,ϰn+2m+1+1−ηn∗1ηn+1/Mϰ0,ϰ1,σ/3fϰn,ϰn+2m+1+1−ηn+1∗1ηn+2/Mϰ0,ϰ1,σ/32fϰn,ϰn+2m+1fϰ2,ϰn+3+1−ηn+2∗1ηn+3/Mϰ0,ϰ1,σ/32fϰn,ϰn+2m+1fϰn+2,ϰn+3+1−ηn+3,(73)1ηn+2m/Mϰ0,ϰ1,σ/3mfϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5⋯fϰn+2m,ϰn+2m+1+1−ηn+2m≥1ηn/Mϰ0,ϰ1,σ/3fϰn,ϰn+2m+1+1−ηn∗1ηηn/Mϰ0,ϰ1,σ/3fϰn,ϰn+2m+1+1−ηηn∗1η2ηn/Mϰ0,ϰ1,σ/32fϰn,ϰn+2m+1fϰ2,ϰn+3+1−η2ηn∗1η2ηn+1/Mϰ0,ϰ1,σ/32fϰn,ϰn+2m+1fϰn+2,ϰn+3+1−η2ηn+1,(74)1ηmηn+m/Mϰ0,ϰ1,σ/3mfϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5⋯fϰn+2m,ϰn+2m+1+1−ηmηn+m,(75)Nϰn,ϰn+2m+1,σ≤ηnNϰ0,ϰ1,σ3fϰn,ϰn+2m+1○ηn+1Nϰ0,ϰ1,σ3fϰn,ϰn+2m+1○ηn+2Nϰ0,ϰ1,σ32fϰn,ϰn+2m+1fϰn+2,ϰn+3○ηn+3Nϰ0,ϰ1,σ32fϰn,ϰn+2m+1fϰn+2,ϰn+3○ηn+4Nϰ0,ϰ1,σ33fϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5○ηn+5Nϰ0,ϰ1,σ33fϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5,(76)ηn+2mNϰ0,ϰ1,σ3mfϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5⋯fϰn+2m,ϰn+2m+1≤ηnNϰ0,ϰ1,σ3fϰn,ϰn+2m+1○ηηnNϰ0,ϰ1,σ3fϰn,ϰn+2m+1○η2ηnNϰ0,ϰ1,σ32fϰn,ϰn+2m+1fϰn+2,ϰn+3○η2ηn+1Nϰ0,ϰ1,σ32fϰn,ϰn+2m+1fϰn+2,ϰn+3○η3ηn+1Nϰ0,ϰ1,σ33fϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5○η3ηn+2Nϰ0,ϰ1,σ33fϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5,(77)ηmηn+mNϰ0,ϰ1,σ3mfϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5⋯fϰn+2m,ϰn+2m+1.Case 4.
Ifp is even, then p=2m,m∈1,2,3,⋯; we have(78)Mϰn,ϰn+2m,σ≥Mϰn,ϰn+1,σ3fϰn,ϰn+2m∗Mϰn+1,ϰn+2,σ3fϰn,ϰn+2m∗Mϰn+2,ϰn+2m,σ3fϰn,ϰn+2m≥Mϰn,ϰn+1,σ3fϰn,ϰn+2m∗Mϰn+1,ϰn+2,σ3fϰn,ϰn+2m∗Mϰn+2,ϰn+3,σ32fϰn,ϰn+2mfϰn+2,ϰn+3∗Mϰn+3,ϰn+4,σ32fϰn,ϰn+2mfϰn+2,ϰn+3∗Mϰn+4,ϰn+2m,σ32fϰn,ϰn+2mfϰn+2,ϰn+3≥Mϰn,ϰn+1,σ3fϰn,ϰn+2∗Mϰn+1,ϰn+2,σ3fϰn,ϰn+2m∗Mϰn+2,ϰn+3,σ32fϰn,ϰn+2mfϰn+2,ϰn+3∗Mϰn+3,ϰn+4,σ32fϰn,ϰn+2mfϰn+2,ϰn+3∗Mϰn+4,ϰn+5,σ33fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5∗Mϰn+5,ϰn+6,σ33fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5,(79)Mϰn+2m−2,ϰn+2m,σ3m−1fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5⋯fϰn+2m−2,ϰn+2m,(80)Nϰn,ϰn+2m,σ≤Nϰn,ϰn+1,σ3fϰn,ϰn+2m○Nϰn+1,ϰn+2,σ3fϰn,ϰn+2m○Nϰn+2,ϰn+2m,σ3fϰn,ϰn+2m≤Nϰn,ϰn+1,σ3fϰn,ϰn+2m○Nϰn+1,ϰn+2,σ3fϰn,ϰn+2m○Nϰn+2,ϰn+3,σ32fϰn,ϰn+2mfϰn+2,ϰn+3○Nϰn+3,ϰn+4,σ32fϰn,ϰn+2mfϰn+2,ϰn+3,○Nϰn+4,ϰn+2m,σ32fϰn,ϰn+2mfϰn+2,ϰn+3,≤Nϰn,ϰn+1,σ3fϰn,ϰn+2m○Nϰn+1,ϰn+2,σ3fϰn,ϰn+2m○Nϰn+3,ϰn+4,σ32fϰn,ϰn+2mfϰn+2,ϰn+3○Nϰn+4,ϰn+5,σ33fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5○Nϰn+5,ϰn+6,σ33fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5,(81)Nϰn+2m−2,ϰn+2m,σ3mfϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5⋯fϰn+2m−2,ϰn+2m.
By using and (66) (BA) in the above inequalities, we have(82)Mϰn,ϰn+2m,σ≥1ηn/Mϰ0,ϰ1,σ/3fϰn,ϰn+2m+1−ηn∗1ηn+1/Mϰ0,ϰ1,σ/3fϰn,ϰn+2m+1−ηn+1∗1ηn+2/Mϰ0,ϰ1,σ/32fϰn,ϰn+2mfϰ2,ϰn+3+1−ηn+2∗1ηn+3/Mϰ0,ϰ1,σ/32fϰn,ϰn+2mfϰn+2,ϰn+3+1−ηn+3,(83)1ηn+2m−2/Mϰ0,ϰ1,σ/3m−1fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5⋯fϰn+2m−2,ϰn+2m+1−ηn+2m−2≥1ηn/Mϰ0,ϰ1,σ/3fϰn,ϰn+2m+1−ηn∗1ηηn/Mϰ0,ϰ1,σ/3fϰn,ϰn+2m+1−ηηn∗1η2ηn/Mϰ0,ϰ1,σ/32fϰn,ϰn+2mfϰ2,ϰn+3+1−η2ηn∗1η2ηn+1/Mϰ0,ϰ1,σ/32fϰn,ϰn+2mfϰn+2,ϰn+3+1−η2ηn+1.(84)1ηm−1ηn+m−1/Mϰ0,ϰ1,σ/3m−1fϰn,ϰn+2m+1fϰn+2,ϰn+3fϰn+4,ϰn+5⋯fϰn+2m,ϰn+2m+1+1−ηm−1ηn+m−1,(85)Nϰn,ϰn+2m,σ≤ηnNϰ0,ϰ1,σ3fϰn,ϰn+2m○ηn+1Nϰ0,ϰ1,σ3fϰn,ϰn+2m,○ηn+2Nϰ0,ϰ1,σ32fϰn,ϰn+2mfϰn+2,ϰn+3,○ηn+3Nϰ0,ϰ1,σ32fϰn,ϰn+2mfϰn+2,ϰn+3,○ηn+4Nϰ0,ϰ1,σ33fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5,○ηn+5Nϰ0,ϰ1,σ33fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5,(86)ηn+2m−2Nϰ0,ϰ1,σ3mfϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5⋯fϰn+2m−2,ϰn+2m≤ηnNϰ0,ϰ1,σ3fϰn,ϰn+2m○ηηnNϰ0,1,σ3fϰn,ϰn+2m○η2ηnNϰ0,ϰ1,σ32fϰn,ϰn+2mfϰn+2,ϰn+3○η2ηn+1Nϰ0,ϰ1,σ32fϰn,ϰn+2mfϰn+2,ϰn+3○η3ηn+1Nϰ0,ϰ1,σ3η3ηn+1fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5○η3ηn+2Nϰ0,ϰ1,σ33fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5,(87)ηm−1ηn+m−1Nϰ0,ϰ1,σ3m−1fϰn,ϰn+2mfϰn+2,ϰn+3fϰn+4,ϰn+5⋯fϰn+2m−2,ϰn+2m.
Therefore, fromlimσ⟶+∞Mϰ,ζ,σ=1,limσ⟶+∞Nϰ,ζ,σ=0, Case 3, Case 4, and together with (32), it follows that, for all p∈1,2,3,⋯s, we have(88)limn⟶+∞Mϰn,ϰn+p,σ=1,limn⟶+∞Nϰn,ϰn+p,σ=0.
Hence,ϰn is a Cauchy sequence. Since V,M,N,∗,○ is a complete IFREBMS, so there exists u∈V such that(89)limn⟶+∞Mϰn,u,σ=1,limn⟶+∞Nϰn,u,σ=0,for allσ>0,q≥1.
Now, we prove thatu is a fixed point for Ψ. For this, we obtain from (61) that(90)1MΨϰn,Ψ,σ−1≤η1Mϰn,u,σ−1=ηMϰn,u,σ−η,1η/Mϰn,u,σ+1−η≤MΨϰn,Ψu,σ,(91)1MΨϰn,Ψ,σ−1≤η1Mϰn,u,σ−1=ηMϰn,u,σ−η,1η/Mϰn−1,ϰn,σ+1−η≤MΨϰn−1,Ψϰn,σ.
Using the above inequalities, we deduce(92)Mu,Ψu,σ≥Mu,ϰn,σ3fu,Ψu∗Mϰn,ϰn+1,σ3fu,Ψu∗Mϰn+1,Ψu,σ3fu,Ψu≥Mu,ϰn,σ3fu,Ψu∗MΨϰn−1,Ψϰn,σ3fu,Ψu∗MΨϰn,Ψu,σ3fu,Ψu≥Mu,ϰn,σ3fu,Ψu∗1η/Mϰn−1,ϰn,σ/3fu,Ψu+1−η∗1η/Mϰn,u,σ/3fu,Ψu+1−η,⟶1∗1∗1=1asn⟶+∞,(93)Nu,Ψu,σ≤Nu,ϰn,σ3fu,Ψu○Nϰn,ϰn+1,σ3fu,Ψu○Nϰn+1,Ψu,σ3fu,Ψu≤Nu,ϰn,σ3fu,Ψu○NΨϰn−1,Ψϰn,σ3fu,Ψu○NΨϰn,Ψu,σ3fu,Ψu≤Nu,ϰn,σ3fu,Ψu○ηNϰn−1,ϰn,σ3fu,Ψu○ηNϰn,u,σ3fu,Ψu,⟶0○0○0=0asn⟶+∞.
Hence,Ψu=u; therefore, u is a fixed point of Ψ.
Now, we investigate the uniqueness of fixed pointu of Ψ. Let v be another fixed point of Ψ, such that Mu,v,t<1, for some σ>0, and following from (61),(94)1Mu,v,σ−1=1MΨu,Ψv,σ−1,≤η1Mu,v,σ−1<1Mu,v,σ−1,a contradiction. And(95)Nu,v,σ=NΨu,Ψv,σ≤ηNu,v,σ,a contradiction. Therefore, we must have Mu,v,σ=1andNu,v,σ=0, for all σ>0, and hence, u=vExample 10.
LetV=0,1. Define f:V×V⟶1,+∞ by fϰ,ζ=1ifϰ=ζ1+ϰ+ζotherwise, and define M,N:V×V×0,+∞⟶0,1 by(96)Mϰ,ζ,σ=σσ+ϰ−ζp,(97)Nϰ,ζ,σ=ϰ−ζpσ+ϰ−ζp,for allϰ,ζ∈V,σ>0.
CTN and CTCN are, respectively, defined bya∗b=a⋅b,a○b=maxa,b,p≥1; then, V,M,N,∗,○ is a complete IFREBMS.
DefineΨ:V⟶V by Ψϰ=ηpϰ. Then,(98)MΨϰ,Ψζ,ησ=Mηpϰ,ηpζ,ησ=ησησ+ηpϰ−ηpζp,=σσ+ϰ−ζp=Mϰ,ζ,σ,(99)NΨϰ,Ψζ,ησ=Nηpϰ,ηpζ,ησ=ηpϰ−ηpζpησ+ηpϰ−ηpζp,=ϰ−ζpσ+ϰ−ζp=Nϰ,ζ,σ.
Also, contraction conditions of Theorem2,(100)1MΨϰ,Ψζ,σ−1≤η1Mϰ,ζ,σ−1,MΨϰ,Ψζ,σ≤ηNϰ,ζ,σ,are satisfied. Hence, all conditions of Theorems 1 and 2 are fulfilled and 0 is a unique fixed point.


## 4. Application to Nonlinear Fractional Differential Equation

In this section, we use Theorem1 to find the existence and uniqueness of a solution in nonlinear fractional differential equations (see [19]) given by(101)Dcαϰl=fl,ϰll∈0,1,α∈1,2.With boundary conditions,(102)ϰ0=0,ϰ'0=Iϰll∈0,1.where Dcα means Caputo fractional derivative of order α, defined by(103)Dcαfl=1Γn−α∫0ll−gn−α−1fngdgn−1<α<n,n=α+1,and f:0,1×ℝ⟶ℝ+ is a continuous function. We assume V=C0,1,ℝ from 0,1 into ℝ with supremum ϰ=Supθ∈0,1ϰl.The Riemann–Liouville fractional integral of orderα (see [20]) is given by(104)Iαfl=1Γα∫0ll−gα−1fgdgα>0.Firstly, we give reasonable form of a nonlinear fractional differential equation and then inquest the existence of a solution. Now, we assume below fractional differential equations:(105)Dcαϰl=fl,ϰll∈0,1,α∈1,2.With the boundary conditions,(106)ϰ0=0,ϰ′0=Iϰll∈0,1,where(i
))f:0,1×ℝ⟶ℝ+ is a continuous function(ii)
ϰl:0,1⟶ℝ is continuousMeet the below conditions:(107)fl,ϰ−fl,ζ≤ЛLϰ−ζ.For alll∈0,1 and L is a constant with LЛ<1,(108)Л=1Γα+1+2ζα+1Γα2−ζ2Γα+1.Then, equation (105)has a unique solution.Proof.
We take(109)Mϰ,ζ,σ=σσ+ϰ−ζp,(110)Nϰ,ζ,σ=ϰ−ζpσ+ϰ−ζp,for allϰ,ζ∈Vσ>0.
CTN and CTCN are, respectively, defined bya∗b=a⋅b,a○b=maxa,b. For all ϰ,ζ∈V, we consider ϰ−ζ=Supl∈0,1ϰl−ζl. Then, V,M,N,∗,○ is a complete IFREBMS. We define a mapping Ψ:V⟶V by(111)Ψϰl=1Γα∫0ll−gα−1fg,ϰgdg+2l2−ζ2Γα∫0ζ∫0gg−mα−1fm,ϰmdmdg,for all l∈0,1. (105) has a solution a function ϰ∈V iff ϰl=Ψϰl for all l∈0,1.
For alll∈0,1, we obtain(112)Mϰl,ζl,σ=σσ+ϰl−ζlp,Nϰl,ζl,σ=ϰl−ζlpσ+ϰl−ζlp,(113)Ψϰl−Ψζl=1Γα∫0ll−gα−1fg,ϰgdg+2l2−ζ2Γα∫0ζ∫0gg−mα−1fm,ϰmdmdg−1Γα∫0ll−gα−1fg,ζgdg+2l2−ζ2Γα∫0ζ∫0gg−mα−1fm,ζmdmdg,Ψϰl−Ψζl=1Γα∫0ll−gα−1fg,ϰgdg+2l2−ζ2Γα∫0ζ∫0gg−mα−1fm,ϰmdmdg−1Γα∫0ll−gα−1fg,ζgdg−2l2−ζ2Γα∫0ζ∫0gg−mα−1fm,ζmdmdg,≤1Γα∫0ll−gα−1fg,ϰg−fg,ζgdg,+2l2−ζ2Γα∫0ζ∫0gg−mα−1fm,ϰm−fm,ζmdmdg,≤Lϰ−ζΓα∫0ll−gα−1dg+2Lϰ−ζΓα∫0ζ∫0gg−mα−1dmdg,≤Lϰ−ζΓα+1+2ζα+1Lϰ−ζΓα2−ζ2Γα+2,≤Lϰ−ζ1Γα+1+2ζα+1Γα2−ζ2Γα+2=LЛϰ−ζ.
From the factLЛ<1 and (112), we obtain(114)MΨϰl,Ψζl,ησ=ησησ+Ψϰl−Ψζlp≥ησησ+LЛϰl−ζlp≥σσ+ϰl−ζlp=Mϰl,ζl,σand(115)NΨϰl,Ψζl,ησ=Ψϰl−Ψζlpησ+Ψϰl−Ψζlp≤LЛϰl−ζlpησ+LЛϰl−ζlp≤ϰl−ζlpσ+ϰl−ζlp=Nϰl,ζl,σ.
Hence, all circumstances of Theorem1 are fulfilled, which imply that Ψ has a unique fixed point.


## 5. Conclusion

The aim of this study is to present the notions of intuitionistic fuzzy rectangular extended b-metric spaces and prove several contraction theorems in this space, nontrivial examples, and an application to nonlinear fractional differential equations is also given to support our results. Due to diverse range applications of metric fixed-point theory in mathematics, science, and economics, it researched widely. Different types of fixed-point results for single and multivalued mappings can be proved in the sense of the above defined notion in this study. Also, presented notion can be extend in different mathematical structures, i.e., intuitionistic fuzzy rectangular controlled metric spaces, intuitionistic fuzzy rectangular double controlled metric spaces, and neutrosophic rectangular extended b-metric spaces.



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*Source: 1814291-2022-04-23.xml*