# An Approach of Integral Equations in Complex-Valuedb-Metric Space Using Commuting Self-Maps

Authors: Rashad A. R. Bantan; Saif Ur Rehman; Shahid Mehmood; Waleed Almutiry; Amani Abdullah Alahmadi; Mohammed Elgarhy
Journal: Journal of Function Spaces
 (2022)

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## Abstract

This paper is aimed at establishing some unique common fixed point theorems in complex-valuedb-metric space under the rational type contraction conditions for three self-mappings in which the one self-map is continuous. A continuous self-map is commutable with the other two self-mappings. Our results are verified by some suitable examples. Ultimately, our results have been utilized to prove the existing solution to the two Urysohn integral type equations. This application illustrates how complex-valued b-metric space can be used in other types of integral operators.


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## Body

## 1. Introduction

In 1922, Banach [1] proved a fixed point theorem (FP-theorem), which is stated as the following: “a single-valued contractive type mapping on a complete metric space has a unique fixed point.” After the publication of the Banach FP-theorem, many researchers have contributed their ideas to the theory of FP. Chandok [2, 3], Jungck and Rhoades [4], and Al-Shami and Abo-Tabl [5, 6] proved different contractive types of FPs and a-fixed soft point results in the context of metric spaces.Bakhtin [7] introduced the idea of b-metric space, while Czerwik [8] proved some fixed point results for nonlinear set-valued contractive type mappings in b-metric spaces. Suzuki in [9] proved basic inequality and some FP-theorems. Jain and Kaur [10] presented a new class of functions to define new contractive maps and established FP-results for these maps. They also extended some results in the framework of b-metric-like spaces. They presented examples and established the application of their main results. They also presented some open problems. Petrusel et al. [11] considered coupled FP-problems for single-valued operators satisfying contraction in said space. They discussed uniqueness, data dependence, and shadowing-property of coupled FP-problem and also established an application for main results. Ameer et al. [12], Boriceanu [13, 14], Bota et al. [15], Czerwik et al. [16, 17], Hussain and Shah [18], Karapinar et al. [19], and Samreen et al. [20] established different contractive type FP and common FP (CFP) results in the context of b-metric spaces.In 2011, the concept of complex-valued metric space was given by Azam et al. [21], and they proved some CFP-theorems for self-mappings. The notion of said space was proposed by Rouzkard and Imdad [22] which generalizes the results of Azam et al. [21] and established some CFP-results. Abbas et al. [23] presented some generalized CFP-results by using cocyclic mappings in complex-valued metric space. They provided examples to indicate the authenticity of his expressions. Sarwar and Zada [24] used the ideas of E.A and CLR properties and proved FP-results for six self-mappings. They showed the existence of their results by establishing some examples. Abbas et al. [25], Nashine et al. [26], Mohanta and Maitra [27], Sintunavarat and Kumam [28], and Verma and Pathak [29] proved some results in the context of complex-valued metric space.In 2013, Rao et al. [30] introduced the notion of complex-valued b-metric space. Mukheimer et al. [31] established CFP-results on said space by extending and generalizing the results of [30, 31]. In [32], Chantakun et al. extend the work of Dubey et al. [33] by introducing sufficient conditions to prove some CFP-results in complex-valued b-metric space. Yadav et al. [34] used compatible and weakly compatible maps to find CFP-results. They proved the validity of the results by providing some examples and establishing an application. Berrah et al. [35], Hasana [36], Mehmood et al. [37], and Mukheimer [38] established some FP and CFP theorems in complex-valued b-metric spaces.In this paper, we provide some extended and effective CFP-results for commuting three self-maps on complex-valuedb-metric spaces. To verify the validity of our work, we present some illustrative examples in the main section. Further, our results have been utilized to prove the existing solution to the two Urysohn integral type equations. This application is also illustrative of how complex-valued b-metric space can be used in other integral type operators. This paper is organized as follows: In Section 2, we present the preliminary concepts. In Section 3, we establish some extended and modified CFP-results for commuting self-maps in complex-valued b-metric space under the generalized rational type conditions. We also provide authentic examples to indicate the effectiveness of these results. In Section 4, we present an application of the two UITEs to support our main work. Finally, in Section 5, we discuss the conclusion.


## 2. Preliminaries

Letℂ be the set of all complex numbers and zi,zii∈ℂ. Define ≤ as zi≤zii, iff Rezi≤Rezii and Imzi≤Imzii, where Re denotes the real part and Im denotes the imaginary part of a complex number. Accordingly, zi≤zii, if any one of the following conditions holds:C1Rezi=Rezii and Imzi=ImziiC2Rezi<Rezii and Imzi=ImziiC3Rezi=Rezii and Imzi<ImziiC4Rezi<Rezii and Imzi<ImziiKnow thatzi⪇zii if zi≠zii and one of (C2), (C3), and (C4) is satisfied.Remark 1 (see [31]).
We can easily check the following:(i)
Ifa1,a2∈ℝ and a1≤a2⇒a1y≤a2y, ∀y∈ℂ(ii)
0≤zi⪇zii⇒zi<zii(iii)
zi≤zii and zii<ziii⇒zi<ziiiDefinition 2 (see [8]).
LetΩ be a nonempty set and b≥1 a given real number. A mapping ð:Ω×Ω⟶0,∞is called a b-metric on Ω if the following conditions are satisfied:
bm1ðρ1,ρ2=0 if and only if ρ1=ρ2
bm2ðρ1,ρ2=ðρ2,ρ1
bm3ðρ1,ρ2≤bðρ1,ρ3+ðρ3,ρ2,
for allρ1,ρ2,ρ3∈Ω. Then, Ω,ð is called a b-metric space.Definition 3 (see [30]).
LetΩ be a nonempty set and b≥1 a given real number. A mapping ð:Ω×Ω⟶ℂis called a complex-valued b-metric on Ω if the following conditions are satisfied:
Cbm1ðρ1,ρ2≥0 and ðρ1,ρ2=0 if and only if ρ1=ρ2
Cbm2ðρ1,ρ2=ðρ2,ρ1
Cbm3ðρ1,ρ2≤bðρ1,ρ3+ðρ3,ρ2,
for allρ1,ρ2,ρ3∈Ω. Then, Ω,ð is called a complex-valued b-metric space.Example 4.
LetΩ=R+. Define the mapping ð:Ω×Ω⟶ℂbyðρ1,ρ2=7/17ρ1−ρ22+i7/17ρ1−ρ22, for all ρ1,ρ2∈Ω.
Then,Ω,ð is a complex-valued b-metric space with b=2.Definition 5 (see [30, 31]).
LetΩ,ð be a complex-valued b-metric space and ρn a sequence in Ω and ρ∈Ω. Then,
(1)
ρn is said to converge to ρ if for every 0<c∗∈ℂ there exists N∗∈ℕ such that ðρn,ρ<c∗, ∀n>N∗. We denote this by limn⟶∞ρn=ρ or ρn⟶ρ as n⟶∞(2)
if for every0<c∗∈ℂ there exists N∗∈ℕ such that ðρn,ρn+m<c∗ for all n>N∗, m∈ℕ, then ρn is called a Cauchy sequence(3)
if every Cauchy sequence is convergent, thenΩ,ð is called a complete complex-valued b-metric spaceLemma 6 (see [30, 31]).
LetΩ,ð be a complex-valued b-metric space and let ρn be a sequence in Ω. Then, ρn converges to ρ iff ðρn,ρ⟶0 as n⟶∞.Lemma 7 (see [30, 31]).
LetΩ,ð be a complex-valued b-metric space and let ρn be a sequence in Ω. Then, ρn is a Cauchy sequence iff ðρn,ρn+m⟶0 as n⟶∞.Definition 8 (see [39]).
LetΩ,ð be a complex-valued b-metric space. The self-mappings f1 and f2 are said to be commuting if f1f2ρ=f2f1ρ for all ρ∈Ω.


## 3. Main Result

In this section, we prove some CFP theorems in complex-valuedg-metric space under the generalized rational type contraction conditions for three self-mappings in which one is continuous. We present some examples for the validation of our work.Theorem 9.
LetΩ,ð be a complete complex-valued b-metric space and f1,f2,f:Ω⟶Ω be three self-maps satisfying
(1)ðf1ρ1,f2ρ2≤κ1ðfρ1,fρ2+κ2ðfρ1,f2ρ2·ðfρ2,f1ρ11/2ðfρ1,f2ρ2+ðfρ2,f1ρ1+κ3minðfρ1,f1ρ1,ðfρ2,f2ρ2,ðfρ1,f1ρ1·ðfρ2,f2ρ21+ðfρ1,fρ2,ðfρ1,f1ρ1·ðfρ1,f2ρ2ðfρ1,f2ρ2+ðfρ2,f1ρ1,ðfρ2,f2ρ2·ðfρ2,f1ρ1ðfρ1,f2ρ2+ðfρ2,f1ρ1,for all ρ1,ρ2∈Ω, κ1,κ2,κ3∈0,1 such that κ1+κ2<1 and b≥1. If f is continuous and f,f1, f,f2 are commutable pairs, then f,f1, and f2 have a unique CFP in Ω.Proof.
Fixρ0∈Ω, and define a sequence ρn sequences in Ω such that
(2)Γ2n=fρ2n+1=f1ρ2n,Γ2n+1=fρ2n+2=f2ρ2n+1,∀n≥0.Now, by using (1),
(3)ðΓ2n,Γ2n+1=ðf1ρ2n,f2ρ2n+1≤κ1ðfρ2n,fρ2n+1+κ2ðfρ2n,f2ρ2n+1·ðfρ2n+1,f1ρ2n1/2ðfρ2n,f2ρ2n+1+ðfρ2n+1,f1ρ2n+κ3minðfρ2n,f1ρ2n,ðfρ2n+1,f2ρ2n+1,ðfρ2n,f1ρ2n·ðfρ2n+1,f2ρ2n+11+ðfρ2n,fρ2n+1,ðfρ2n,f1ρ2n·ðfρ2n,f2ρ2n+1ðfρ2n,f2ρ2n+1+ðfρ2n+1,f1ρ2n,ðfρ2n+1,f2ρ2n+1·ðfρ2n+1,f1ρ2nðfρ2n,f2ρ2n+1+ðfρ2n+1,f1ρ2n=κ1ðΓ2n−1,Γ2n+κ2ðΓ2n−1,Γ2n+1·ðΓ2n,Γ2n1/2ðΓ2n−1,Γ2n+1+ðΓ2n,Γ2n+κ3minðΓ2n−1,Γ2n,ðΓ2n,Γ2n+1,ðΓ2n−1,Γ2n·ðΓ2n,Γ2n+11+ðΓ2n−1,Γ2n,ðΓ2n−1,Γ2n·ðΓ2n−1,Γ2n+1ðΓ2n−1,Γ2n+1+ðΓ2n,Γ2n,ðΓ2n,Γ2n+1·ðΓ2n,Γ2nðΓ2n−1,Γ2n+1+ðΓ2n,Γ2n.This implies that(4)ðΓ2n,Γ2n+1≤κ1ðΓ2n−1,Γ2n+κ2ðΓ2n−1,Γ2n+1·ðΓ2n,Γ2n1/2ðΓ2n−1,Γ2n+1+ðΓ2n,Γ2n+κ3minðΓ2n−1,Γ2n,ðΓ2n,Γ2n+1,ðΓ2n−1,Γ2n·ðΓ2n,Γ2n+11+ðΓ2n−1,Γ2n,ðΓ2n−1,Γ2n·ðΓ2n−1,Γ2n+1ðΓ2n−1,Γ2n+1+ðΓ2n,Γ2n,ðΓ2n,Γ2n+1·ðΓ2n,Γ2nðΓ2n−1,Γ2n+1+ðΓ2n,Γ2n.After simplification, we get that(5)ðΓ2n,Γ2n+1≤κ1ðΓ2n−1,Γ2n.Again, by using (1) and (2),
(6)ðΓ2n−1,Γ2n=ðf2ρ2n−1,f1ρ2n=ðf1ρ2n,f2ρ2n−1≤κ1ðfρ2n,fρ2n−1+κ2ðfρ2n,f2ρ2n−1·ðfρ2n−1,f1ρ2n1/2ðfρ2n,f2ρ2n−1+ðfρ2n−1,f1ρ2n+κ3minðfρ2n,f1ρ2n,ðfρ2n−1,f2ρ2n−1,ðfρ2n,f1ρ2n·ðfρ2n−1,f2ρ2n−11+ðfρ2n,fρ2n−1,ðfρ2n,f1ρ2n·ðfρ2n,f2ρ2n−1ðfρ2n,f2ρ2n−1+ðfρ2n−1,f1ρ2n,ðfρ2n−1,f2ρ2n−1·ðfρ2n−1,f1ρ2nðfρ2n,f2ρ2n−1+ðfρ2n−1,f1ρ2n=κ1ðΓ2n−1,Γ2n−2+κ2ðΓ2n−1,Γ2n−1·ðΓ2n−2,Γ2n1/2ðΓ2n−1,Γ2n−1+ðΓ2n−2,Γ2n+κ3minðΓ2n−1,Γ2n,ðΓ2n−2,Γ2n−1,ðΓ2n−1,Γ2n·ðΓ2n−2,Γ2n−11+ðΓ2n−1,Γ2n−2,ðΓ2n−1,Γ2n·ðΓ2n−1,Γ2n−1ðΓ2n−1,Γ2n−1+ðΓ2n−2,Γ2n,ðΓ2n−2,Γ2n−1·ðΓ2n−2,Γ2nðΓ2n−1,Γ2n−1+ðΓ2n−2,Γ2n.This implies that(7)ðΓ2n−1,Γ2n≤κ1ðΓ2n−1,Γ2n−2+κ2ðΓ2n−1,Γ2n−1·ðΓ2n−2,Γ2n1/2ðΓ2n−1,Γ2n−1+ðΓ2n−2,Γ2n+κ3minðΓ2n−1,Γ2n,ðΓ2n−2,Γ2n−1,ðΓ2n−1,Γ2n·ðΓ2n−2,Γ2n−11+ðΓ2n−1,Γ2n−2,ðΓ2n−1,Γ2n·ðΓ2n−1,Γ2n−1ðΓ2n−1,Γ2n−1+ðΓ2n−2,Γ2n,ðΓ2n−2,Γ2n−1·ðΓ2n−2,Γ2nðΓ2n−1,Γ2n−1+ðΓ2n−2,Γ2n.After simplification, we get that(8)ðΓ2n−1,Γ2n≤κ1ðΓ2n−2,Γ2n−1.Now, from (8) and (5) and by induction, we have that
(9)ðΓ2n,Γ2n+1≤κ1ðΓ2n−1,Γ2n≤κ12ðΓ2n−2,Γ2n−1≤⋯≤κ12nðΓ0,Γ1.So, form,n∈ℕ with m>n,
(10)ðΓn,Γm≤bðΓn,Γn+1+bðΓn+1,Γm≤bðΓn,Γn+1+b2ðΓn+1,Γn+2+⋯+bm−nðΓm−1,Γm≤bκ1nðΓ0,Γ1+b2κ1n+1ðΓ0,Γ1+⋯+bm−nκ1m−1ðΓ0,Γ1≤bκ1n+b2κ1n+1+⋯+bm−nκ1m−1ðΓ0,Γ1=bκ1n+b2κ1n+1+⋯+bm−nκ1m−1ðΓ0,Γ1=bκ1n1+bκ1+b2κ12⋯+bm−n+1κ1m−n+1ðΓ0,Γ1=bκ1n∑t=0m−n+1btκ1tðΓ0,Γ1≤bκ1n∑t=0∞btκ1tðΓ0,Γ1=bκ1n1−bκ1ðΓ0,Γ1⟶0,asn⟶∞.Therefore, the sequenceΓn is Cauchy. Since Ω is complete, there exists s∈Ω such that Γn⟶s, as n⟶∞, or limn⟶∞Γn=s, and from (2), we have
(11)limn⟶∞fρ2n+1=s,limn⟶∞f1ρ2n=s,limn⟶∞f2ρ2n+1=s.Asf is continuous, so
(12)limn⟶∞ffρ2n+1=fs,limn⟶∞ff1ρ2n=fs,limn⟶∞ff2ρ2n+1=fs.Since,f,f1 and f,f2 are commutable pairs, therefore, from (12), we have that
(13)limn⟶∞f1fρ2n=fs,limn⟶∞f2fρ2n+1=fs.Now, we have to show thatfs=s, so by putting ρ1=fρ2n and ρ2=ρ2n+1, in (1),
(14)ðf1fρ2n,f2ρ2n+1≤κ1ðffρ2n,fρ2n+1+κ2ðffρ2n,f2ρ2n+1·ðfρ2n+1,f1fρ2n1/2ðffρ2n,f2ρ2n+1+ðfρ2n+1,f1fρ2n+κ3minðffρ2n,f1fρ2n,ðfρ2n+1,f2ρ2n+1,ðffρ2n,f1fρ2n·ðfρ2n+1,f2ρ2n+11+ðffρ2n,fρ2n+1,ðffρ2n,f1fρ2n·ðffρ2n,f2ρ2n+1ðffρ2n,f2ρ2n+1+ðfρ2n+1,f1fρ2n,ðfρ2n+1,f2ρ2n+1·ðfρ2n+1,f1fρ2nðffρ2n,f2ρ2n+1+ðfρ2n+1,f1fρ2n.This implies that(15)ðf1fρ2n,f2ρ2n+1≤κ1ðffρ2n,fρ2n+1+κ2ðffρ2n,f2ρ2n+1·ðfρ2n+1,f1fρ2n1/2ðffρ2n,f2ρ2n+1+ðfρ2n+1,f1fρ2n+κ3minðffρ2n,f1fρ2n,ðfρ2n+1,f2ρ2n+1,ðffρ2n,f1fρ2n·ðfρ2n+1,f2ρ2n+11+ðffρ2n,fρ2n+1,ðffρ2n,f1fρ2n·ðffρ2n,f2ρ2n+1ðffρ2n,f2ρ2n+1+ðfρ2n+1,f1fρ2n,ðfρ2n+1,f2ρ2n+1·ðfρ2n+1,f1fρ2nðffρ2n,f2ρ2n+1+ðfρ2n+1,f1fρ2n.Takinglimn⟶∞ and using (11), (12), and (13), we get that
(16)ðfs,s≤κ1ðfs,s+κ2ðfs,s·ðs,fs1/2ðfs,s+ðs,fs+κ3minðfs,fs,ðs,s,ðfs,fs·ðs,s1+ðfs,s,ðfs,fs·ðfs,sðfs,s+ðs,fs,∣ðs,s∣·∣ðs,fs∣∣ðfs,s∣+∣ðs,fs∣.After simplification, we get that(17)ðfs,s≤κ1+κ2ðfs,s⇒1−κ1−κ2ðfs,s≤0.Since1−κ1−κ2≠0⇒∣ðfs,s∣=0, hence, we get that
(18)fs=s.Next, we have to show thatf1s=s, by the view of (1),
(19)ðf1s,fρ2n+2=ðf1s,f2ρ2n+1≤κ1ðfs,fρ2n+1+κ2ðfs,f2ρ2n+1·ðfρ2n+1,f1s1/2ðfs,f2ρ2n+1+ðfρ2n+1,f1s+κ3minðfs,f1s,ðfρ2n+1,f2ρ2n+1,ðfs,f1s·ðfρ2n+1,f2ρ2n+11+ðfs,fρ2n+1,ðfs,f1s·ðfs,f2ρ2n+1ðfs,f2ρ2n+1+ðfρ2n+1,f1s,ðfρ2n+1,f2ρ2n+1·ðfρ2n+1,f1sðfs,f2ρ2n+1+ðfρ2n+1,f1s.This implies that(20)ðf1s,fρ2n+2≤κ1ðfs,fρ2n+1+κ2ðfs,f2ρ2n+1·ðfρ2n+1,f1s1/2ðfs,f2ρ2n+1+ðfρ2n+1,f1s+κ3minðfs,f1s,ðfρ2n+1,f2ρ2n+1,ðfs,f1s·ðfρ2n+1,f2ρ2n+11+ðfs,fρ2n+1,ðfs,f1s·ðfs,f2ρ2n+1ðfs,f2ρ2n+1+ðfρ2n+1,f1s,ðfρ2n+1,f2ρ2n+1·ðfρ2n+1,f1sðfs,f2ρ2n+1+ðfρ2n+1,f1s.Now, again applyinglimn⟶∞ on both sides and by using (11) and (18), we have that
(21)ðf1s,s≤κ1ðs,s+κ2ðs,s·ðs,f1s1/2ðs,s+ðs,f1s+κ3min∣ðs,f1s∣,∣ðs,s∣,∣ðs,f1s∣·∣ðs,s∣∣1+ðs,s∣,∣ðs,f1s∣·∣ðs,s∣∣ðs,s∣+∣ðs,f1s∣,∣ðs,s∣·∣ðs,f1s∣∣ðs,s∣+∣ðs,f1s∣.This implies that∣ðf1s,s∣≤0. Hence,
(22)f1s=s.Now, we have to show thatf2s=s, by using (1),
(23)ðfρ2n+1,f2s=ðf1ρ2n,f2s≤κ1ðfρ2n,fs+κ2ðfρ2n,f2s·ðfs,f1ρ2n1/2ðfρ2n,f2s+ðfs,f1ρ2n+κ3minðfρ2n,f1ρ2n,ðfs,f2s,ðfρ2n,f1ρ2n·ðfs,f2s1+ðfρ2n,fs,ðfρ2n,f1ρ2n·ðfρ2n,f2sðfρ2n,f2s+ðfs,f1ρ2n,ðfs,f2s·ðfs,f1ρ2nðfρ2n,f2s+ðfs,f1ρ2n.This implies that(24)ðfρ2n+1,f2s≤κ1ðfρ2n,fs+κ2ðfρ2n,f2s·ðfs,f1ρ2n1/2ðfρ2n,f2s+ðfs,f1ρ2n+κ3minðfρ2n,f1ρ2n,ðfs,f2s,ðfρ2n,f1ρ2n·ðfs,f2s1+ðfρ2n,fs,ðfρ2n,f1ρ2n·ðfρ2n,f2sðfρ2n,f2s+ðfs,f1ρ2n,ðfs,f2s·ðfs,f1ρ2nðfρ2n,f2s+ðfs,f1ρ2n.Takinglimn⟶∞ and using (11) and (18), we get
(25)ðs,f2s≤κ1ðs,s+κ2ðs,f2s·ðs,s1/2ðs,f2s+ðs,s+κ3minðs,s,ðs,f2s,ðs,s·ðs,f2s1+ðs,s,ðs,s·ðs,f2sðs,f2s+ðs,s,∣ðs,f2s∣·∣ðs,s∣∣ðs,f2s∣+∣ðs,s∣.This implies that∣ðs,f2s∣≤0. Hence,
(26)f2s=s.Thus, from (18), (22), and (26), we find that s is a CFP of f,f1, and f2, i.e.,
(27)fs=f1s=f2s=s.Uniqueness: suppose thats∗∈Ω is another CFP of f,f1, and f2 such that
(28)fs=f1s=f2s=s,fs∗=f1s∗=f2s∗=s∗.Then, from (1), we have that
(29)ðs,s∗=ðf1s,f2s∗≤κ1ðfs,fs∗+κ2ðfs,f2s∗·ðfs∗,f1s1/2ðfs,f2s∗+ðfs∗,f1s+κ3minðfs,f1s,ðfs∗,f2s∗,ðfs,f1s·ðfs∗,f2s∗1+ðfs,fs∗,ðfs,f1s·ðfs,f2s∗ðfs,f2s∗+ðfs∗,f1s,ðfs∗,f2s∗·ðfs∗,f1sðfs,f2s∗+ðfs∗,f1s=κ1+κ2ðs,s∗.This implies that∣ðs,s∗∣≤κ1+κ2∣ðs,s∗∣⇒1−κ1−κ2∣ðs,s∗∣≤0. Since 1−κ1−κ2≠0⇒∣ðs,s∗∣=0⇒s=s∗. Hence, prove that f,f1, and f2 have a unique CFP in Ω.If we putκ3=0 in Theorem 9, we get the following corollary.Corollary 10.
LetΩ,ð be a complete complex-valued b-metric space and f1,f2,f:Ω⟶Ω be three self-maps satisfying
(30)ðf1ρ1,f2ρ2≤κ1ðfρ1,fρ2+κ2ðfρ1,f2ρ2·ðfρ2,f1ρ11/2ðfρ1,f2ρ2+ðfρ2,f1ρ1,
for allρ1,ρ2∈Ω, κ1,κ2∈0,1 such that κ1+κ2<1 and b≥1. If f is continuous and f,f1, f,f2 are commutable pairs, then f,f1, and f2 have a unique CFP in Ω.If we putκ2=0 in Theorem 9, we can get the following corollary.Corollary 11.
LetΩ,ð be a complete complex-valued b-metric space and f1,f2,f:Ω⟶Ω be three self-maps satisfying:
(31)ðf1ρ1,f2ρ2≤κ1ðfρ1,fρ2+κ3minðfρ1,f1ρ1,ðfρ2,f2ρ2,ðfρ1,f1ρ1·ðfρ2,f2ρ21+ðfρ1,fρ2,ðfρ1,f1ρ1·ðfρ1,f2ρ2ðfρ1,f2ρ2+ðfρ2,f1ρ1,ðfρ2,f2ρ2·ðfρ2,f1ρ1ðfρ1,f2ρ2+ðfρ2,f1ρ1,
for allρ1,ρ2∈Ω, κ1,κ3∈0,1 and b≥1. If f is continuous and f,f1, f,f2 are commutable pairs, then f,f1, and f2 have a unique CFP in Ω.Theorem 12.
LetΩ,ð be a complete complex-valued b-metric space and f1,f2,f:Ω⟶Ω be three self-maps satisfying
(32)ðf1ρ1,f2ρ2≤κ1ðfρ1,fρ2+κ2ðfρ1,f2ρ2·ðfρ2,f1ρ11/2ðfρ1,f2ρ2+ðfρ2,f1ρ1+κ3maxðfρ1,f1ρ1,ðfρ2,f2ρ2,ðfρ1,f1ρ1·ðfρ2,f2ρ21+ðfρ1,fρ2,ðfρ1,f1ρ1·ðfρ1,f2ρ2ðfρ1,f2ρ2+ðfρ2,f1ρ1,ðfρ2,f2ρ2·ðfρ2,f1ρ1ðfρ1,f2ρ2+ðfρ2,f1ρ1,for all ρ1,ρ2∈Ω, κ1,κ2,κ3∈0,1 such that κ1+κ2<1, κ1+κ3<1 and b≥1. If f is continuous and f,f1, f,f2 are commutable pairs, then f,f1, and f2 have a unique CFP in Ω.Proof.
Fixρ0∈Ω, and define a sequence ρn sequences in Ω such that
(33)Γ2n=fρ2n+1=f1ρ2n,Γ2n+1=fρ2n+2=f2ρ2n+1,∀n≥0.Now, by using (32),
(34)ðΓ2n,Γ2n+1=ðf1ρ2n,f2ρ2n+1≤κ1ðfρ2n,fρ2n+1+κ2ðfρ2n,f2ρ2n+1·ðfρ2n+1,f1ρ2n1/2ðfρ2n,f2ρ2n+1+ðfρ2n+1,f1ρ2n+κ3maxðfρ2n,f1ρ2n,ðfρ2n+1,f2ρ2n+1,ðfρ2n,f1ρ2n·ðfρ2n+1,f2ρ2n+11+ðfρ2n,fρ2n+1,ðfρ2n,f1ρ2n·ðfρ2n,f2ρ2n+1ðfρ2n,f2ρ2n+1+ðfρ2n+1,f1ρ2n,ðfρ2n+1,f2ρ2n+1·ðfρ2n+1,f1ρ2nðfρ2n,f2ρ2n+1+ðfρ2n+1,f1ρ2n=κ1ðΓ2n−1,Γ2n+κ2ðΓ2n−1,Γ2n+1·ðΓ2n,Γ2n1/2ðΓ2n−1,Γ2n+1+ðΓ2n,Γ2n+κ3maxðΓ2n−1,Γ2n,ðΓ2n,Γ2n+1,ðΓ2n−1,Γ2n·ðΓ2n,Γ2n+11+ðΓ2n−1,Γ2n,ðΓ2n−1,Γ2n·ðΓ2n−1,Γ2n+1ðΓ2n−1,Γ2n+1+ðΓ2n,Γ2n,ðΓ2n,Γ2n+1·ðΓ2n,Γ2nðΓ2n−1,Γ2n+1+ðΓ2n,Γ2n.This implies that,(35)ðΓ2n,Γ2n+1≤κ1ðΓ2n−1,Γ2n+κ2ðΓ2n−1,Γ2n+1·ðΓ2n,Γ2n1/2ðΓ2n−1,Γ2n+1+ðΓ2n,Γ2n+κ3maxðΓ2n−1,Γ2n,ðΓ2n,Γ2n+1,ðΓ2n−1,Γ2n·ðΓ2n,Γ2n+11+ðΓ2n−1,Γ2n,ðΓ2n−1,Γ2n·ðΓ2n−1,Γ2n+1ðΓ2n−1,Γ2n+1+ðΓ2n,Γ2n,ðΓ2n,Γ2n+1·ðΓ2n,Γ2nðΓ2n−1,Γ2n+1+ðΓ2n,Γ2n.After simplification, we get that(36)ðΓ2n,Γ2n+1≤κ1ðΓ2n−1,Γ2n+κ3max∣ðΓ2n−1,Γ2n∣,∣ðΓ2n,Γ2n+1∣.Now, there are two possibilities:(i)
IfðΓ2n−1,Γ2n is a maximum term in ∣ðΓ2n−1,Γ2n∣,∣ðΓ2n,Γ2n+1∣, then after simplification, (36) can be written as(37)ðΓ2n,Γ2n+1≤g1ðΓ2n−1,Γ2n,whereg1=κ1+κ3<1(ii)
IfðΓ2n,Γ2n+1 is a maximum term in ðΓ2n−1,Γ2n∣,∣ðΓ2n,Γ2n+1∣, then after simplification, (36) can be written as(38)ðΓ2n,Γ2n+1≤g2ðΓ2n−1,Γ2n,whereg2=κ11−κ3<1Letg≔maxg1,g2<1, then from (37) and (38), for all n≥0, we have
(39)ðΓ2n,Γ2n+1≤gðΓ2n−1,Γ2n.Similarly,(40)ðΓ2n−1,Γ2n≤gðΓ2n−2,Γ2n−1.Now, from (40) and (39) and by induction, we have that
(41)ðΓ2n,Γ2n+1≤gðΓ2n−1,Γ2n≤g2ðΓ2n−2,Γ2n−1≤⋯≤g2nðΓ0,Γ1.Now, form,n∈ℕ with m>n,
(42)ðΓn,Γm≤bðΓn,Γn+1+bðΓn+1,Γm≤bðΓn,Γn+1+b2ðΓn+1,Γn+2+⋯+bm−nðΓm−1,Γm≤bgnðΓ0,Γ1+b2gn+1ðΓ0,Γ1+⋯+bm−ngm−1ðΓ0,Γ1≤bgn+b2gn+1+⋯+bm−ngm−1ðΓ0,Γ1=bgn+b2gn+1+⋯+bm−ngm−1ðΓ0,Γ1=bgn1+bg+b2g2⋯+bm−n+1gm−n+1ðΓ0,Γ1=bgn∑t=0m−n+1btgtðΓ0,Γ1≤bgn∑t=0∞btgtðΓ0,Γ1=bgn1−bgðΓ0,Γ1⟶0,asn⟶∞.Therefore, sequenceΓn is Cauchy. Since Ω is complete, there exists s∈Ω such that Γn⟶s, as n⟶∞, or limn⟶∞Γn=s, and from (33), we have
(43)limn⟶∞fρ2n+1=s,limn⟶∞f1ρ2n=s,limn⟶∞f2ρ2n+1=s.Asf is continuous, so
(44)limn⟶∞ffρ2n+1=fs,limn⟶∞ff1ρ2n=fs,limn⟶∞ff2ρ2n+1=fs.Since,f,f1 and f,f2 are commutable pairs, therefore, from (44), we have that
(45)limn⟶∞f1fρ2n=fs,limn⟶∞f2fρ2n+1=fs.Now, we have to show thatfs=s, so by putting ρ1=fρ2n and ρ2=ρ2n+1, in (32):
(46)ðf1fρ2n,f2ρ2n+1≤κ1ðffρ2n,fρ2n+1+κ2ðffρ2n,f2ρ2n+1·ðfρ2n+1,f1fρ2n1/2ðffρ2n,f2ρ2n+1+ðfρ2n+1,f1fρ2n+κ3maxðffρ2n,f1fρ2n,ðfρ2n+1,f2ρ2n+1,ðffρ2n,f1fρ2n·ðfρ2n+1,f2ρ2n+11+ðffρ2n,fρ2n+1,ðffρ2n,f1fρ2n·ðffρ2n,f2ρ2n+1ðffρ2n,f2ρ2n+1+ðfρ2n+1,f1fρ2n,ðfρ2n+1,f2ρ2n+1·ðfρ2n+1,f1fρ2nðffρ2n,f2ρ2n+1+ðfρ2n+1,f1fρ2n.This implies that(47)ðf1fρ2n,f2ρ2n+1≤κ1ðffρ2n,fρ2n+1+κ2ðffρ2n,f2ρ2n+1·ðfρ2n+1,f1fρ2n1/2ðffρ2n,f2ρ2n+1+ðfρ2n+1,f1fρ2n+κ3maxðffρ2n,f1fρ2n,ðfρ2n+1,f2ρ2n+1,ðffρ2n,f1fρ2n·ðfρ2n+1,f2ρ2n+11+ðffρ2n,fρ2n+1,ðffρ2n,f1fρ2n·ðffρ2n,f2ρ2n+1ðffρ2n,f2ρ2n+1+ðfρ2n+1,f1fρ2n,ðfρ2n+1,f2ρ2n+1·ðfρ2n+1,f1fρ2nðffρ2n,f2ρ2n+1+ðfρ2n+1,f1fρ2n.Takinglimn⟶∞ and using (43), (44), and (45), we get that
(48)ðfs,s≤κ1ðfs,s+κ2ðfs,s·ðs,fs1/2ðfs,s+ðs,fs+κ3maxðfs,fs,ðs,s,ðfs,fs·ðs,s1+ðfs,s,ðfs,fs·ðfs,sðfs,s+ðs,fs,ðs,s·ðs,fsðfs,s+ðs,fs.After simplification, we get that(49)ðfs,s≤κ1+κ2ðfs,s⇒1−κ1−κ2ðfs,s≤0.Since1−κ1−κ2≠0⇒∣ðfs,s∣=0; hence, we get that
(50)fs=s.Next, we have to show thatf1s=s, by the view of (32),
(51)ðf1s,fρ2n+2=ðf1s,f2ρ2n+1≤κ1ðfs,fρ2n+1+κ2ðfs,f2ρ2n+1·ðfρ2n+1,f1s1/2ðfs,f2ρ2n+1+ðfρ2n+1,f1s+κ3maxðfs,f1s,ðfρ2n+1,f2ρ2n+1,ðfs,f1s·ðfρ2n+1,f2ρ2n+11+ðfs,fρ2n+1,ðfs,f1s·ðfs,f2ρ2n+1ðfs,f2ρ2n+1+ðfρ2n+1,f1s,ðfρ2n+1,f2ρ2n+1·ðfρ2n+1,f1sðfs,f2ρ2n+1+ðfρ2n+1,f1s.This implies that(52)ðf1s,fρ2n+2≤κ1ðfs,fρ2n+1+κ2ðfs,f2ρ2n+1·ðfρ2n+1,f1s1/2ðfs,f2ρ2n+1+ðfρ2n+1,f1s+κ3maxðfs,f1s,ðfρ2n+1,f2ρ2n+1,ðfs,f1s·ðfρ2n+1,f2ρ2n+11+ðfs,fρ2n+1,ðfs,f1s·ðfs,f2ρ2n+1ðfs,f2ρ2n+1+ðfρ2n+1,f1s,ðfρ2n+1,f2ρ2n+1·ðfρ2n+1,f1sðfs,f2ρ2n+1+ðfρ2n+1,f1s.Now, again applyinglimn⟶∞ on both sides and by using (43) and (50), we have that
(53)∣ðf1s,s∣≤κ1∣ðs,s∣+κ2ðs,s·ðs,f1s1/2ðs,s+ðs,f1s+κ3maxðs,f1s,ðs,s,ðs,f1s·ðs,s1+ðs,s,ðs,f1s·ðs,sðs,s+ðs,f1s,ðs,s·ðs,f1sðs,s+ðs,f1s=κ3ðs,f1s.This implies that1−κ3∣ðf1s,s∣≤0. Since 1−κ3≠0⇒∣ðf1s,s∣=0. Hence,
(54)f1s=s.Now, we have to show thatf2s=s, by using (32),
(55)ðfρ2n+1,f2s=ðf1ρ2n,f2s≤κ1ðfρ2n,fs+κ2ðfρ2n,f2s·ðfs,f1ρ2n1/2ðfρ2n,f2s+ðfs,f1ρ2n+κ3maxðfρ2n,f1ρ2n,ðfs,f2s,ðfρ2n,f1ρ2n·ðfs,f2s1+ðfρ2n,fs,ðfρ2n,f1ρ2n·ðfρ2n,f2sðfρ2n,f2s+ðfs,f1ρ2n,ðfs,f2s·ðfs,f1ρ2nðfρ2n,f2s+ðfs,f1ρ2n.This implies that(56)ðfρ2n+1,f2s≤κ1ðfρ2n,fs+κ2ðfρ2n,f2s·ðfs,f1ρ2n1/2ðfρ2n,f2s+ðfs,f1ρ2n+κ3maxðfρ2n,f1ρ2n,ðfs,f2s,ðfρ2n,f1ρ2n·ðfs,f2s1+ðfρ2n,fs,ðfρ2n,f1ρ2n·ðfρ2n,f2sðfρ2n,f2s+ðfs,f1ρ2n,ðfs,f2s·ðfs,f1ρ2nðfρ2n,f2s+ðfs,f1ρ2n.Takinglimn⟶∞ and using (43) and (50), we get
(57)∣ðs,f2s∣≤κ1∣ðs,s∣+κ2ðs,f2s·ðs,s1/2ðs,f2s+ðs,s+κ3maxðs,s,ðs,f2s,ðs,s·ðs,f2s1+ðs,s,ðs,s·ðs,f2sðs,f2s+ðs,s,ðs,f2s·ðs,sðs,f2s+ðs,s=κ3ðs,f2s.This implies that1−κ3∣ðs,f2s∣≤0. Since 1−κ3≠0⇒∣ðs,f2s∣=0. Hence,
(58)f2s=s.Thus, from (50), (54), and (58), we find that s is a CFP of f,f1, and f2, i.e.,
(59)fs=f1s=f2s=s.Uniqueness: suppose thats∗∈Ω is another CFP of f,f1, and f2 such that
(60)fs=f1s=f2s=s,fs∗=f1s∗=f2s∗=s∗.Then, from (32), we have that
(61)ðs,s∗=ðf1s,f2s∗≤κ1ðfs,fs∗+κ2ðfs,f2s∗·ðfs∗,f1s1/2ðfs,f2s∗+ðfs∗,f1s+κ3maxðfs,f1s,ðfs∗,f2s∗,ðfs,f1s·ðfs∗,f2s∗1+ðfs,fs∗,ðfs,f1s·ðfs,f2s∗ðfs,f2s∗+ðfs∗,f1s,ðfs∗,f2s∗·ðfs∗,f1sðfs,f2s∗+ðfs∗,f1s=κ1+κ2ðs,s∗.This implies that∣ðs,s∗∣≤κ1+κ2∣ðs,s∗∣⇒1−κ1−κ2∣ðs,s∗∣≤0. Since 1−κ1−κ2≠0⇒∣ðs,s∗∣=0⇒s=s∗, hence proving that f,f1, and f2 have a unique CFP in Ω.If we putκ2=0 in Theorem 12, we can get the following corollary.Corollary 13.
Let (Ω,ð) be a complete complex-valued b-metric space and f1, f2, f:Ω⟶Ω be three self-maps satisfying
(62)ðf1ρ1,f2ρ2≤κ1ðfρ1,fρ2+κ3maxðfρ1,f1ρ1,ðfρ2,f2ρ2,ðfρ1,f1ρ1·ðfρ2,f2ρ21+ðfρ1,fρ2,ðfρ1,f1ρ1·ðfρ1,f2ρ2ðfρ1,f2ρ2+ðfρ2,f1ρ1,ðfρ1,f1ρ1·ðfρ1,f2ρ2ðfρ1,f2ρ2+ðfρ2,f1ρ1,for all ρ1,ρ2∈Ω, κ1,κ2,κ3∈0,1 such that κ1+κ2<1, κ1+κ3<1 and κ1/1−κ3<1, where b≥1. If f is continuous and f,f1, f,f2 are commutable pairs, then f,f1, and f2 have a unique common fixed point in Ω.Corollary 14.
LetΩ,ð be a complete complex-valued b-metric space and f1,f2,f:Ω⟶Ω be three self-maps satisfying
(63)ðf1ρ1,f2ρ2≤κ1ðfρ1,fρ2+κ3maxðfρ1,f1ρ1,ðfρ2,f2ρ2,ðfρ1,f1ρ1·ðfρ2,f2ρ21+ðfρ1,fρ2,ðfρ1,f1ρ1·ðfρ1,f2ρ2ðfρ1,f2ρ2+ðfρ2,f1ρ1,ðfρ2,f2ρ2·ðfρ2,f1ρ1ðfρ1,f2ρ2+ðfρ2,f1ρ1,for all ρ1,ρ2∈Ω, κ1,κ2,κ3∈0,1 such that κ1+κ2<1, κ1+κ3<1 and κ1/1−κ3<1, where b≥1. If f is continuous and f,f1, f,f2 are commutable pairs, then f,f1, and f2 have a unique common fixed point in Ω.Example 15.
LetΩ,ð be a complex-valued b-metric space, where Ω=0,1 and ð:Ω×Ω⟶ℂ with ðρ1,ρ2=4ρ1−ρ22/9+i4ρ1−ρ22/9, for all ρ1,ρ2∈Ω. Now, we find b,
(64)ðρ1,ρ2=4ρ1−ρ229+i4ρ1−ρ229≤4ρ1−ρ3+ρ3−ρ229+i4ρ1−ρ3+ρ3−ρ229≤4ρ1−ρ329+4ρ3−ρ229+492ρ1−ρ3ρ3−ρ2+i4ρ1−ρ329+4ρ3−ρ229+492ρ1−ρ3ρ3−ρ2≤4ρ1−ρ329+4ρ3−ρ229+4ρ1−ρ329+4ρ3−ρ229+i4ρ1−ρ329+4ρ3−ρ229+4ρ1−ρ329+4ρ3−ρ229=24ρ1−ρ329+4ρ3−ρ229+2i4ρ1−ρ329+4ρ3−ρ229=24ρ1−ρ329+i4ρ1−ρ329+4ρ3−ρ229+i4ρ3−ρ229=2ðρ1,ρ3+ðρ3,ρ2.That isðρ1,ρ2≤bðρ1,ρ3+ðρ3,ρ2, where b=2. Now, define f1,f2,f:Ω⟶Ω as
(65)f1ρ1=f2ρ1=3ρ120,fρ1=ρ14forρ1∈Ω.Notice that(66)∣ðfρ1,fρ2∣,∣ðfρ1,f2ρ2∣·∣ðfρ2,f1ρ1∣1/2∣ðfρ1,f2ρ2∣+∣ðfρ2,f1ρ1∣,maxðfρ1,f1ρ1,ðfρ2,f2ρ2,ðfρ1,f1ρ1·ðfρ2,f2ρ21+ðfρ1,fρ2,ðfρ1,f1ρ1·ðfρ1,f2ρ2ðfρ1,f2ρ2+ðfρ2,f1ρ1,∣ðfρ2,f2ρ2∣·∣ðfρ2,f1ρ1∣∣ðfρ1,f2ρ2∣+∣ðfρ2,f1ρ1∣≥0.In all regards, it is enough to show thatðf1ρ1,f2ρ2≤κ1ðfρ1,fρ2, for all ρ1,ρ2∈0,1 and κ1,κ2,κ3∈0,1, such that κ1+κ2<1 and κ1+κ3<1, where b≥1, we have
(67)ðf1ρ1,f2ρ2=4f1ρ1−f2ρ229+i4f1ρ1−f2ρ229=43ρ1/20−3ρ2/2029+i43ρ1/20−3ρ2/2029=3524ρ1/4−ρ2/429+i4ρ1/4−ρ2/429=9254ρ1/4−ρ2/429+i4ρ1/4−ρ2/429,(68)ðfρ1,fρ2=4fρ1−fρ229+i4fρ1−fρ229=4ρ1/4−ρ2/429+i4ρ1/4−ρ2/429.Forρ1,ρ2∈0,1, we discuss different cases with κ1=2/5,κ2=1/5,κ3=1/10, and b=2. Hence,
(69)κ1+κ2=25+15=35<1,κ1+κ3=25+110=12<1.Case 1.
Letρ1=0,ρ2=0, then from (67) and (68), directly, we get that ðf1ρ1,f2ρ2≤κ1ðfρ1,fρ2. Hence, (32) is satisfied with κ1=2/5,κ2=1/5,κ3=1/10, and b=2.Case 2.
Letρ1=1,ρ2=0, then from (67) and (68), we find ðf1ρ1,f2ρ2≤κ1ðfρ1,fρ2, satisfied with κ1=2/5, i.e.,
(70)92541/4−0/429+i41/4−0/429≤κ141/4−0/429+i41/4−0/4290.00991+i≤0.01101+i.
Thus, (32) is true for κ1=2/5,κ2=1/5,κ3=1/10, and b=2.Case 3.
Letρ1=1/2,ρ2=1/4; then, from (67) and (68), we find ðf1ρ1,f2ρ2≤κ1ðfρ1,fρ2 is satisfied with κ1=2/5, i.e.,
(71)92541/8−1/1629+i41/8−1/1629≤κ141/8−1/1629+i41/8−1/16290.000611+i≤0.000681+i.
Thus, (32) is true for κ1=2/5,κ2=1/5,κ3=1/10, and b=2.Case 4.
Letρ1=1/2,ρ2=1; then, from (67) and (68), we find ðf1ρ1,f2ρ2≤κ1ðfρ1,fρ2 is satisfied with κ1=2/5, i.e.,
(72)92541/8−1/429+i41/8−1/429≤κ141/8−1/429+i41/8−1/4290.00241+i≤0.00271+i.Hence, (32) is satisfied with κ1=2/5,κ2=1/5,κ3=1/10, and b=2. The pairs of self-mappings f,f1 and f,f2 are commutable; that is,
(73)f1fρ1=ff1ρ1=3ρ180,f2fρ1=ff2ρ1=3ρ180,∀ρ1∈Ω.Thus, all the conditions of Theorem12 are satisfied with noticing that the point 0∈Ω, which remains fixed under mappings f,f1, and f2, is indeed unique.Theorem 16.
LetΩ,ð be a complete complex-valued b-metric space and f1,f2,f:Ω⟶Ω be three self-maps satisfying
(74)ðf1ρ1,f2ρ2≤κ1ðfρ1,fρ2+κ2ðfρ1,f2ρ2·ðfρ2,f1ρ11/2ðfρ1,f2ρ2+ðfρ2,f1ρ1+κ3ðfρ1,f1ρ1+ðfρ2,f2ρ2+ðfρ1,f1ρ1·ðfρ2,f2ρ21+ðfρ1,fρ2+ðfρ1,f1ρ1·ðfρ1,f2ρ2ðfρ1,f2ρ2+ðfρ2,f1ρ1+ðfρ2,f2ρ2·ðfρ2,f1ρ1ðfρ1,f2ρ2+ðfρ2,f1ρ1,
for allρ1,ρ2∈Ω, κ1,κ2,κ3∈0,1, such that κ1+κ2<1, κ1+4κ3<1 and b≥1. If f is a continuous self-map and f,f1, f,f2 are commutable pairs, then f,f1, and f2 have a unique CFP in Ω.Proof.
Fixρ0∈Ω, and define a sequence ρn sequences in Ω such that
(75)Γ2n=fρ2n+1=f1ρ2n,Γ2n+1=fρ2n+2=f2ρ2n+1,∀n≥0.Now, by the view of (74) and (75),
(76)ðΓ2n,Γ2n+1=ðf1ρ2n,f2ρ2n+1≤κ1ðfρ2n,fρ2n+1+κ2ðfρ2n,f2ρ2n+1·ðfρ2n+1,f1ρ2n1/2ðfρ2n,f2ρ2n+1+ðfρ2n+1,f1ρ2n+κ3ðfρ2n,f1ρ2n+ðfρ2n+1,f2ρ2n+1+ðfρ2n,f1ρ2n·ðfρ2n+1,f2ρ2n+11+ðfρ2n,fρ2n+1+ðfρ2n,f1ρ2n·ðfρ2n,f2ρ2n+1ðfρ2n,f2ρ2n+1+ðfρ2n+1,f1ρ2n+ðfρ2n+1,f2ρ2n+1·ðfρ2n+1,f1ρ2nðfρ2n,f2ρ2n+1+ðfρ2n+1,f1ρ2n=κ1ðΓ2n−1,Γ2n+κ2ðΓ2n−1,Γ2n+1·ðΓ2n,Γ2n1/2ðΓ2n−1,Γ2n+1+ðΓ2n,Γ2n+κ3ðΓ2n−1,Γ2n+ðΓ2n,Γ2n+1+ðΓ2n−1,Γ2n·ðΓ2n,Γ2n+11+ðΓ2n−1,Γ2n+ðΓ2n−1,Γ2n·ðΓ2n−1,Γ2n+1ðΓ2n−1,Γ2n+1+ðΓ2n,Γ2n+ðΓ2n,Γ2n+1·ðΓ2n,Γ2nðΓ2n−1,Γ2n+1+ðΓ2n,Γ2n.This implies that(77)ðΓ2n,Γ2n+1≤κ1ðΓ2n−1,Γ2n+κ2ðΓ2n−1,Γ2n+1·ðΓ2n,Γ2n1/2ðΓ2n−1,Γ2n+1+ðΓ2n,Γ2n+κ3ðΓ2n−1,Γ2n+ðΓ2n,Γ2n+1+ðΓ2n−1,Γ2n·ðΓ2n,Γ2n+11+ðΓ2n−1,Γ2n+ðΓ2n−1,Γ2n·ðΓ2n−1,Γ2n+1ðΓ2n−1,Γ2n+1+ðΓ2n,Γ2n+ðΓ2n,Γ2n+1·ðΓ2n,Γ2nðΓ2n−1,Γ2n+1+ðΓ2n,Γ2n.After simplification, we get that(78)ðΓ2n,Γ2n+1≤gðΓ2n−1,Γ2n,whereg=κ1+2κ31−2κ3<1.Again, by the view of (74) and (75),
(79)ðΓ2n−1,Γ2n=ðf2ρ2n−1,f1ρ2n=ðf1ρ2n,f2ρ2n−1≤κ1ðfρ2n,fρ2n−1+κ2ðfρ2n,f2ρ2n−1·ðfρ2n−1,f1ρ2n1/2ðfρ2n,f2ρ2n−1+ðfρ2n−1,f1ρ2n+κ3ðfρ2n,f1ρ2n+ðfρ2n−1,f2ρ2n−1+ðfρ2n,f1ρ2n·ðfρ2n−1,f2ρ2n−11+ðfρ2n,fρ2n−1+ðfρ2n,f1ρ2n·ðfρ2n,f2ρ2n−1ðfρ2n,f2ρ2n−1+ðfρ2n−1,f1ρ2n+ðfρ2n−1,f2ρ2n−1·ðfρ2n−1,f1ρ2nðfρ2n,f2ρ2n−1+ðfρ2n−1,f1ρ2n=κ1ðΓ2n−1,Γ2n−2+κ2ðΓ2n−1,Γ2n−1·ðΓ2n−2,Γ2n1/2ðΓ2n−1,Γ2n−1+ðΓ2n−2,Γ2n+κ3ðΓ2n−1,Γ2n+ðΓ2n−2,Γ2n−1+ðΓ2n−1,Γ2n·ðΓ2n−2,Γ2n−11+ðΓ2n−1,Γ2n−2+ðΓ2n−1,Γ2n·ðΓ2n−1,Γ2n−1ðΓ2n−1,Γ2n−1+ðΓ2n−2,Γ2n+ðΓ2n−2,Γ2n−1·ðΓ2n−2,Γ2nðΓ2n−1,Γ2n−1+ðΓ2n−2,Γ2n.This implies that(80)ðΓ2n−1,Γ2n≤κ1ðΓ2n−1,Γ2n−2+κ2ðΓ2n−1,Γ2n−1·ðΓ2n−2,Γ2n1/2ðΓ2n−1,Γ2n−1+ðΓ2n−2,Γ2n+κ3ðΓ2n−1,Γ2n+ðΓ2n−2,Γ2n−1+ðΓ2n−1,Γ2n·ðΓ2n−2,Γ2n−11+ðΓ2n−1,Γ2n−2+ðΓ2n−1,Γ2n·ðΓ2n−1,Γ2n−1ðΓ2n−1,Γ2n−1+ðΓ2n−2,Γ2n+ðΓ2n−2,Γ2n−1·ðΓ2n−2,Γ2nðΓ2n−1,Γ2n−1+ðΓ2n−2,Γ2n.After simplification, we get that(81)ðΓ2n−1,Γ2n≤gðΓ2n−2,Γ2n−1,sinceg=κ1+2κ31−2κ3<1.Now, from (81) and (78) and by induction, we have
(82)∣ðΓ2n,Γ2n+1∣≤g∣ðΓ2n−1,Γ2n∣≤g2∣ðΓ2n−2,Γ2n−1∣≤⋯≤g2nðΓ0,Γ1.So, form,n∈ℕ with m>n,
(83)ðΓn,Γm≤bðΓn,Γn+1+bðΓn+1,Γm≤bðΓn,Γn+1+b2ðΓn+1,Γn+2+⋯+bm−nðΓm−1,Γm≤bgnðΓ0,Γ1+b2gn+1ðΓ0,Γ1+⋯+bm−ngm−1ðΓ0,Γ1≤bgn+b2gn+1+⋯+bm−ngm−1ðΓ0,Γ1=bgn1+bg+b2g2⋯+bm−n+1gm−n+1ðΓ0,Γ1=bgn∑t=0m−n+1btgtðΓ0,Γ1≤bgn∑t=0∞btgtðΓ0,Γ1=bgn1−bgðΓ0,Γ1⟶0,asn⟶∞.Therefore, sequenceΓn is Cauchy. Since Ω is complete, there exists s∈Ω such that Γn⟶s, as n⟶∞, and from (75), we have that
(84)limn⟶∞fρ2n+1=s,limn⟶∞f1ρ2n=s,limn⟶∞f2ρ2n+1=s.Asf is continuous, so
(85)limn⟶∞ffρ2n+1=fs,limn⟶∞ff1ρ2n=fs,limn⟶∞ff2ρ2n+1=fs.Since,f,f1 and f,f2 are commutable pairs, therefore, from (85), we have that
(86)limn⟶∞f1fρ2n=fs,limn⟶∞f2fρ2n+1=fs.Now, we provefs=s. So, for this, we put ρ1=fρ2n and ρ2=ρ2n+1 in (74),
(87)ðf1fρ2n,f2ρ2n+1≤κ1ðffρ2n,fρ2n+1+κ2ðffρ2n,f2ρ2n+1·ðfρ2n+1,f1fρ2n1/2ðffρ2n,f2ρ2n+1+ðfρ2n+1,f1fρ2n+κ3ðffρ2n,f1fρ2n+ðfρ2n+1,f2ρ2n+1+ðffρ2n,f1fρ2n·ðfρ2n+1,f2ρ2n+11+ðffρ2n,fρ2n+1+ðffρ2n,f1fρ2n·ðffρ2n,f2ρ2n+1ðffρ2n,f2ρ2n+1+ðfρ2n+1,f1fρ2n+ðfρ2n+1,f2ρ2n+1·ðfρ2n+1,f1fρ2nðffρ2n,f2ρ2n+1+ðfρ2n+1,f1fρ2n.This implies that(88)ðf1fρ2n,f2ρ2n+1≤κ1ðffρ2n,fρ2n+1+κ2ðffρ2n,f2ρ2n+1·ðfρ2n+1,f1fρ2n1/2ðffρ2n,f2ρ2n+1+ðfρ2n+1,f1fρ2n+κ3ðffρ2n,f1fρ2n+ðfρ2n+1,f2ρ2n+1+ðffρ2n,f1fρ2n·ðfρ2n+1,f2ρ2n+11+ðffρ2n,fρ2n+1+ðffρ2n,f1fρ2n·ðffρ2n,f2ρ2n+1ðffρ2n,f2ρ2n+1+ðfρ2n+1,f1fρ2n+ðfρ2n+1,f2ρ2n+1·ðfρ2n+1,f1fρ2nðffρ2n,f2ρ2n+1+ðfρ2n+1,f1fρ2n.Takinglimn⟶∞ and using (84), (85), and (86), we get that
(89)ðfs,s≤κ1ðfs,s+κ2ðfs,s·ðs,fs1/2ðfs,s+ðs,fs+κ3ðfs,fs+ðs,s+∣ðfs,fs∣·∣ðs,s∣∣1+ðfs,s∣+∣ðfs,fs∣·∣ðfs,s∣∣ðfs,s∣+∣ðs,fs∣+∣ðs,s∣·∣ðs,fs∣∣ðfs,s∣+∣ðs,fs∣=κ1+κ2ðfs,s.This implies that1−κ1−κ2∣ðfs,s∣≤0. Since, 1−κ1−κ2≠0⇒∣ðfs,s∣=0. Hence,
(90)fs=s.Next, we have to show thatf1s=s, by using (74),
(91)ðf1s,fρ2n+2=ðf1s,f2ρ2n+1≤κ1ðfs,fρ2n+1+κ2ðfs,f2ρ2n+1·ðfρ2n+1,f1s1/2ðfs,f2ρ2n+1+ðfρ2n+1,f1s+κ3ðfs,f1s+ðfρ2n+1,f2ρ2n+1+ðfs,f1s·ðfρ2n+1,f2ρ2n+11+ðfs,fρ2n+1+ðfs,f1s·ðfs,f2ρ2n+1ðfs,f2ρ2n+1+ðfρ2n+1,f1s+ðfρ2n+1,f2ρ2n+1·ðfρ2n+1,f1sðfs,f2ρ2n+1+ðfρ2n+1,f1s.This implies that(92)ðf1s,fρ2n+2≤κ1ðfs,fρ2n+1+κ2ðfs,f2ρ2n+1·ðfρ2n+1,f1s1/2ðfs,f2ρ2n+1+ðfρ2n+1,f1s+κ3ðfs,f1s+ðfρ2n+1,f2ρ2n+1+ðfs,f1s·ðfρ2n+1,f2ρ2n+11+ðfs,fρ2n+1+ðfs,f1s·ðfs,f2ρ2n+1ðfs,f2ρ2n+1+ðfρ2n+1,f1s+ðfρ2n+1,f2ρ2n+1·ðfρ2n+1,f1sðfs,f2ρ2n+1+ðfρ2n+1,f1s.Takinglimn⟶∞ and using (84) and (90), we get
(93)ðf1s,s≤κ1ðs,s+κ2ðs,s·ðs,f1s1/2ðs,s+ðs,f1s+κ3ðs,f1s+ðs,s+∣ðs,f1s∣·∣ðs,s∣∣1+ðs,s∣+∣ðs,f1s∣·∣ðs,s∣∣ðs,s∣+∣ðs,f1s∣+ðs,s·ðs,f1sðs,s+ðs,f1s.Thus, we get that∣ðf1s,s∣≤κ3∣ðs,f1s∣⇒1−κ3∣ðf1s,s∣≤0. Since 1−κ3≠0, therefore, ∣ðf1s,s∣=0. Hence,
(94)f1s=s.Now, we have to show thatf2s=s, by using (74),
(95)ðfρ2n+1,f2s=ðf1ρ2n,f2s≤κ1ðfρ2n,fs+κ2ðfρ2n,f2s·ðfs,f1ρ2n1/2ðfρ2n,f2s+ðfs,f1ρ2n+κ3ðfρ2n,f1ρ2n+ðfs,f2s+ðfρ2n,f1ρ2n·ðfs,f2s1+ðfρ2n,fs+ðfρ2n,f1ρ2n·ðfρ2n,f2sðfρ2n,f2s+ðfs,f1ρ2n+ðfs,f2s·ðfs,f1ρ2nðfρ2n,f2s+ðfs,f1ρ2n.This implies that(96)ðfρ2n+1,f2s≤κ1ðfρ2n,fs+κ2ðfρ2n,f2s·ðfs,f1ρ2n1/2ðfρ2n,f2s+ðfs,f1ρ2n+κ3ðfρ2n,f1ρ2n+ðfs,f2s+ðfρ2n,f1ρ2n·ðfs,f2s1+ðfρ2n,fs+ðfρ2n,f1ρ2n·ðfρ2n,f2sðfρ2n,f2s+ðfs,f1ρ2n+ðfs,f2s·ðfs,f1ρ2nðfρ2n,f2s+ðfs,f1ρ2n.Takinglimn⟶∞ and using (84) and (90), we get
(97)ðs,f2s≤κ1ðs,s+κ2ðs,f2s·ðs,s1/2ðs,f2s+ðs,s+κ3ðs,s+ðs,f2s+∣ðs,s∣·∣ðs,f2s∣∣1+ðs,s∣+∣ðs,s∣·∣ðs,f2s∣∣ðs,f2s∣+∣ðs,s∣+∣ðs,f2s∣·∣ðs,s∣∣ðs,f2s∣+∣ðs,s∣.So, we get that∣ðs,f2s∣≤κ3∣ðs,f2s∣⇒1−κ3∣ðs,f2s∣≤0. Since 1−κ3≠0, therefore, ∣ðs,f2s∣=0. Hence,
(98)f2s=s.Thus, from (90), (94), and (98), we find that s is a CFP of f,f1, and f2, i.e.,
(99)fs=f1s=f2s=s.Uniqueness: suppose thats∗∈Ω is another CFP of f,f1, and f2 such that
(100)fs=f1s=f2s=s,fs∗=f1s∗=f2s∗=s∗.Then, from (74), we have that
(101)ðs,s∗=ðf1s,f2s∗≤κ1ðfs,fs∗+κ2ðfs,f2s∗·ðfs∗,f1s1/2ðfs,f2s∗+ðfs∗,f1s+κ3ðfs,f1s+ðfs∗,f2s∗+ðfs,f1s·ðfs∗,f2s∗1+ðfs,fs∗+ðfs,f1s·ðfs,f2s∗ðfs,f2s∗+ðfs∗,f1s+ðfs∗,f2s∗·ðfs∗,f1sðfs,f2s∗+ðfs∗,f1s=κ1+κ2ðs,s∗.This implies that∣ðs,s∗∣≤κ1+κ2∣ðs,s∗∣⇒1−κ1−κ2∣ðs,s∗∣≤0. Since 1−κ1−κ2≠0, therefore, ∣ðs,s∗∣=0⇒s=s∗, hence proving that f,f1, and f2 have a unique CFP in Ω.Corollary 17.
LetΩ,ð be a complete complex-valued b-metric space and f1, f2, f:Ω⟶Ω be three self-maps satisfying
(102)ðf1ρ1,f2ρ2≤κ1ðfρ1,fρ2+κ2ðfρ1,f2ρ2⋅ðfρ2,f1ρ11/2ðfρ1,f2ρ2+ðfρ2,f1ρ1+κ3ðfρ1,f1ρ1⋅ðfρ2,f2ρ21+ðfρ1,fρ2+ðfρ1,f1ρ1⋅ðfρ1,f2ρ2ðfρ1,f2ρ2+ðfρ2,f1ρ1+ðfρ2,f2ρ2⋅ðfρ2,f1ρ1ðfρ1,f2ρ2+ðfρ2,f1ρ1,
for allρ1, ρ2∈Ω, κ1, κ2, κ3∈0,1, such that κ1+κ2<1 and κ1+κ3/1−κ3<1, with b≥1. If f is a continuous self-mapping and (f,f1), (f,f2) are commutable pairs, then f, f1, and f2 have a unique common fixed point in Ω.Example 18.
LetΩ=0,∞ and ð:Ω×Ω⟶ℂ be defined as ðρ1,ρ2=3ρ1−ρ22/13+i3ρ1−ρ22/13 for all ρ1,ρ2∈Ω. Then, Ω,ð is a complex-valued b-metric space. Now, we find b:
(103)ðρ1,ρ2=3ρ1−ρ2213+i3ρ1−ρ2213≤3ρ1−ρ3+ρ3−ρ2213+i3ρ1−ρ3+ρ3−ρ2213≤3ρ1−ρ3213+3ρ3−ρ2213+3132ρ1−ρ3ρ3−ρ2+i3ρ1−ρ3213+3ρ3−ρ2213+3132ρ1−ρ3ρ3−ρ2≤3ρ1−ρ3213+3ρ3−ρ2213+3ρ1−ρ3213+3ρ3−ρ2213+i3ρ1−ρ3213+3ρ3−ρ2213+3ρ1−ρ3213+3ρ3−ρ2213=23ρ1−ρ3213+3ρ3−ρ2213+i23ρ1−ρ3213+3ρ3−ρ2213=23ρ1−ρ3213+i3ρ1−ρ3213+3ρ3−ρ2213+i3ρ3−ρ2213=2ðρ1,ρ3+ðρ3,ρ2.That isðρ1,ρ2≤bðρ1,ρ3+ðρ3,ρ2, where b=2. Now, we define f1,f2,f:Ω⟶Ω by
(104)f1ρ1=f2ρ1=ln1+ρ14+ρ1,fρ1=e4ρ1−1,forallρ1∈Ω.Notice that(105)∣ðfρ1,fρ2∣,∣ðfρ1,f2ρ2∣·∣ðfρ2,f1ρ1∣1/2∣ðfρ1,f2ρ2∣+∣ðfρ2,f1ρ1∣,ðfρ1,f1ρ1+ðfρ2,f2ρ2+∣ðfρ1,f1ρ1∣·∣ðfρ2,f2ρ2∣∣1+ðfρ1,fρ2∣+∣ðfρ1,f1ρ1∣·∣ðfρ1,f2ρ2∣∣ðfρ1,f2ρ2∣+∣ðfρ2,f1ρ1∣+∣ðfρ2,f2ρ2∣·∣ðfρ2,f1ρ1∣∣ðfρ1,f2ρ2∣+∣ðfρ2,f1ρ1∣≥0.In all regards, it is enough to show thatðf1ρ1,f2ρ2≤κ1ðfρ1,fρ2, for all ρ1,ρ2∈0,∞ and κ1,κ2,κ3∈0,1, such that κ1+κ2<1 and κ1+4κ3<1, where b≥1, we have
(106)ðf1ρ1,f2ρ2=3f1ρ1−f2ρ2213+i3f1ρ1−f2ρ2213=3ln1+ρ1/4+ρ1−ln1+ρ2/4+ρ2213+i3ln1+ρ1/4+ρ1−ln1+ρ2/4+ρ2213≤3ρ1/4+ρ1−ρ2/4+ρ2213+i3ρ1/4+ρ1−ρ2/4+ρ2213≤34ρ1−4ρ2/16213+i34ρ1−4ρ2/16213=116234ρ1−4ρ2213+i34ρ1−4ρ2213≤12563e4ρ1−e4ρ2213+i3e4ρ1−e4ρ2213,(107)ðfρ1,fρ2=3fρ1−fρ2213+i3fρ1−fρ2213=3e4ρ1−1−e4ρ2−1213+i3e4ρ1−1−e4ρ2−1213=3e4ρ1−e4ρ2213+i3e4ρ1−e4ρ2213.Forρ1,ρ2∈0,∞, we discuss different cases with κ1=1/5,κ2=1/4,andκ3=1/10, where b=2. Hence,
(108)κ1+κ2=15+14=920<1,κ1+4κ3=15+25=35<1.Case 1.
Letρ1=0,ρ2=0. Then, from (106) and (107), directly, we get that ðf1ρ1,f2ρ2≤κ1ðfρ1,fρ2. Hence, (74) is satisfied with κ1=1/5,κ2=1/4,κ3=1/10, and b=2.Case 2.
Letρ1=0,ρ2=1, then from (106) and (107), we find ðf1ρ1,f2ρ2≤κ1ðfρ1,fρ2 is satisfied with κ1=1/5, as
(109)12563e0−e4213+i3e0−e4213≤κ13e0−e4213+i3e0−e4213.By usingκ1=1/5 and after simplifying, we get that
(110)12563−53.5981213+i3−53.5981213≤153−53.5981213+i3−53.59812132.58961+i≤132.58871+i.Thus, (74) is true for κ1=1/5,κ2=1/4,κ3=1/10, and b=2.Case 3.
Letρ1=1/2,ρ2=1/4, then from (106) and (107), we find ðf1ρ1,f2ρ2≤κ1ðfρ1,fρ2 is true for κ1=1/5, as
(111)12563e2−e1213+i3e2−e1213≤κ13e2−e1213+i3e2−e1213.By usingκ1=1/5 and after simplifying, we get that
(112)125634.6708213+i34.6708213≤1534.6708213+i34.67082130.01961+i≤1.00691+i.Thus, (74) is true for κ1=1/5,κ2=1/4,κ3=1/10, and b=2.Case 4.
Letρ1=1/2,ρ2=1, then from (106) and (107), we get that ðf1ρ1,f2ρ2≤κ1ðfρ1,fρ2 is true for κ1=1/5, as
(113)12563e2−e4213+i3e2−e4213≤κ13e2−e4213+i3e2−e4213.By usingκ1=1/5 and after simplifying, we get that
(114)12563−51.8799213+i3−51.8799213≤153−51.8799213+i3−51.87992132.42621+i≤124.22411+i.Thus, (74) is true for κ1=1/5,κ2=1/4,κ3=1/10, and b=2.So, all conditions of Theorem16 are satisfied to get a unique CFP, that is “0” of the mappings f,f1, and f2.


## 4. Applications

Here, we provide an application to support our main result. To do this, we take a couple of UITEs to obtain the existing result of a common solution to check the effectiveness of our result. Let the setΩ=Ck1,k2,ℝ contain real-valued continuous functions defined on k1,k2. In the following, we use Theorem 9 to obtain the existing result of a common solution. This enables us to establish a theorem based on UITEs to attain the existing result of a common solution.Theorem 19 (see [28]).
LetΩ=Ck1,k2,ℝ, where k1,k2⊆ℝ and ð:Ω×Ω⟶ℂ is defined as
(115)ðρ1,ρ2=ρ1y−ρ2y21+k12eicotk1
for allρ1,ρ2∈Ω and y∈k1,k2. Consider that the UITEs are
(116)ρ1y=∫k1k2Q1y,r,ρ1rdr+Γ1y,ρ2y=∫k1k2Q2y,r,ρ2rdr+Γ2y,where r∈k1,k2. Let Q1,Q2:k1,k2×k1,k2×ℝ⟶ℝ be such that Dρ1,Eρ2∈Ω for every ρ1,ρ2∈Ω, we have that
(117)Dρ1y=∫k1k2Q1y,r,ρ1rdr,Eρ2y=∫k1k2Q2y,r,ρ2rdr.If there existsμ∈0,1 such that, for all ρ1,ρ2∈Ω,
(118)Dρ1y−Eρ2y+Γ1y−Γ2y21+k12eicotk1≤μMρ1,ρ2,where
(119)Mρ1,ρ2=maxA1ρ1,ρ2y,A2ρ1,ρ2y,A3ρ1,ρ2y,with
(120)A1ρ1,ρ2y=ρ1y−ρ2y21+k12eicotk1,A2ρ1,ρ2y=Eρ2y+Γ2y−ρ1y2Dρ1y+Γ1y−ρ2y21+k12eicotk11/2Eρ2y+Γ2y−ρ1y2+Dρ1y+Γ1y−ρ2y2,(121)A3ρ1,ρ2y=mina1ρ1,ρ2y,a2ρ1,ρ2y,a3ρ1,ρ2y,a4ρ1,ρ2y,a5ρ1,ρ2y,where
(122)a1ρ1,ρ2y=Dρ1y+Γ1y−ρ1y21+k12eicotk1,a2ρ1,ρ2y=Eρ2y+Γ2y−ρ2y21+k12eicotk1,a3ρ1,ρ2y=Dρ1y+Γ1y−ρ1y2Eρ2y+Γ2y−ρ2y21+k12eicotk121+ρ1y−ρ2y21+k12eicotk1,a4ρ1,ρ2y=Dρ1y+Γ1y−ρ1y2Eρ2y+Γ2y−ρ1y21+k12eicotk1Eρ2y+Γ2y−ρ1y2+Dρ1y+Γ1y−ρ2y2,a5ρ1,ρ2y=Eρ2y+Γ2y−ρ2y2Dρ1y+Γ1y−ρ2y21+k12eicotk1Eρ2y+Γ2y−ρ1y2+Dρ1y+Γ1y−ρ2y2.Then, the two UITEs, i.e., (41), have a unique common solution.Proof.
Definef1,f2,f:Ω⟶Ω as
(123)f1ρ1=f1ρ1y=Dρ1y+Γ1y=Dρ1+Γ1,fρ1=fρ1y=ρ1y=ρ1,f2ρ2=f2ρ2y=Eρ2y+Γ2y=Eρ2+Γ2,fρ2=fρ2y=ρ2y=ρ2.Then, we have the following three cases:(1)
IfA1ρ1,ρ2y is the maximum term in A1ρ1,ρ2y,A2ρ1,ρ2y,A3ρ1,ρ2y, then from (118), (119), and (123), we have that(124)ðf1ρ1,f2ρ2≤μρ1−ρ221+k12eicotk1,for allρ1,ρ2∈Ω. Thus, f1,f2, and f satisfy all conditions of Theorem 9 with μ=κ1 and κ2=κ3=0 in (1). Then, two UITEs, i.e., (116), have a unique common solution in Ω.
(2)
IfA2ρ1,ρ2y is the maximum term in A1ρ1,ρ2y,A2ρ1,ρ2y,A3ρ1,ρ2y, then from (118), (119), and (123), we have that(125)ðf1ρ1,f2ρ2≤μEρ2+Γ2−ρ12Dρ1+Γ1−ρ221+k12eicotk11/2Eρ2+Γ2−ρ12+Dρ1+Γ1−ρ22,for allρ1,ρ2∈Ω. Thus, f1,f2, and f satisfy all conditions of Theorem 9 with μ=κ2 and κ1=κ3=0 in (1). Then, two UITEs, i.e., (116), have a unique common solution in Ω.
(3)
IfA3ρ1,ρ2y is the maximum term in A1ρ1,ρ2y,A2ρ1,ρ2y,A3ρ1,ρ2y, then from (119), we have that(126)Mρ1,ρ2=A3ρ1,ρ2y.Then, there are furthermore five subcases arising:(i)
Ifa1ρ1,ρ2y is the minimum term in a1ρ1,ρ2y,a2ρ1,ρ2y,a3ρ1,ρ2y,a4ρ1,ρ2y,a5ρ1,ρ2y. Then from (118), (121), (123), and (126), we have that(127)ðf1ρ1,f2ρ2≤μDρ1+Γ1−ρ121+k12eicotk1,for allρ1,ρ2∈Ω(ii)
Ifa2ρ1,ρ2yis the minimum term in a1ρ1,ρ2y,a2ρ1,ρ2y,a3ρ1,ρ2y,a4ρ1,ρ2y,a5ρ1,ρ2y. Then from (118), (121), (123), and (126), we have that(128)ðf1ρ1,f2ρ2≤μEρ2+Γ2−ρ221+k12eicotk1,for allρ1,ρ2∈Ω(iii)
Ifa3ρ1,ρ2y is the minimum term in a1ρ1,ρ2y,a2ρ1,ρ2y,a3ρ1,ρ2y,a4ρ1,ρ2y,a5ρ1,ρ2y. Then from (118), (121), (123), and (126), we have that(129)ðf1ρ1,f2ρ2≤μDρ1+Γ1−ρ12Eρ2+Γ2−ρ221+k12eicotk121+ρ1−ρ221+k12eicotk1,for allρ1,ρ2∈Ω(iv)
Ifa4ρ1,ρ2y is the minimum term in a1ρ1,ρ2y,a2ρ1,ρ2y,a3ρ1,ρ2y,a4ρ1,ρ2y,a5ρ1,ρ2y. Then from (118), (121), (123), and (126), we have that(130)ðf1ρ1,f2ρ2≤μDρ1+Γ1−ρ12Eρ2+Γ2−ρ121+k12eicotk1Eρ2+Γ2−ρ12+Dρ1+Γ1−ρ22,for allρ1,ρ2∈Ω(v)
Ifa5ρ1,ρ2y is the minimum term in a1ρ1,ρ2y,a2ρ1,ρ2y,a3ρ1,ρ2y,a4ρ1,ρ2y,a5ρ1,ρ2y. Then from (118), (121), (123), and (126), we have that(131)ðf1ρ1,f2ρ2≤μEρ2+Γ2−ρ22Dρ1+Γ1−ρ221+k12eicotk1Eρ2+Γ2−ρ12+Dρ1+Γ1−ρ22,for allρ1,ρ2∈Ω. Thus, the subcases of Case 3 (Case (i-v)) for the mappings f1,f2, and f satisfy all the conditions of Theorem 9 with μ=κ3 and κ1=κ2=0 in (1). Then, two UITEs, i.e., (116), have a unique common solution in Ω.


## 5. Conclusions

We have established some unique CFP-results in complex-valuedb-metric space by using rational contraction conditions for three self-mappings in which one self-map is continuous and commutable with the other two self-mappings. In our main work, we have generalized the results (e.g., see [28, 37, 38]). To show the authenticity of our results, we have given some useful examples in the main section. We have also provided an application for our main result to indicate its utility. In this direction, many results can be contributed to the said space by applying different contractions with different types of integral operators.



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*Source: 5862251-2022-01-07.xml*