# Rational Type Compatible Single-Valued Mappings via Unique Common Fixed Point Findings in Complex-Valued b-Metric Spaces with an Application

Authors: Shahid Mehmood; Saif Ur Rehman; Naeem Jan; Mabrook Al-Rakhami; Abdu Gumaei
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we establish some new generalized rational type common fixed point results for compatible three self-mappings in complex-valued b-metric space, in which a one self-map is continuous. In support of our results, we present some illustrative examples to verify the validity of our main work. Moreover, we present the application of two Urysohn integral type equations (UITEs) for the existence of a common solution to support our work. The UITEs arev1p=∫k1k2Q1p,r,v1rdr+ℏ1p and v2p=∫k1k2Q2p,r,v2rdr+ℏ2p, where p∈k1,k2,v1,v2,ℏ1,ℏ2∈V, where V=Ck1,k2,ℝn is the set of all real-valued continuous functions defined on k1,k2 and Q1,Q2:k1,k2×k1,k2×ℝn⟶ℝn.


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## Body

## 1. Introduction

The theory of fixed point is one of the most interesting area of research in Mathematics. Initially, the concept of this theory was given by Banach [1] and he proved “a Banach contraction theorem for fixed point” which is stated as “a single-valued contractive type mapping in a complete metric space has a unique fixed point.” After the publication of this research article, many authors have contributed their ideas to the theory of fixed point in the context of metric spaces and proved different contractive type fixed point results for single-valued and multivalued mappings with different types of applications. Some fixed soft points and a-fixed soft points results can be found in [2, 3].The concept of b-metric space was first introduced by Bakhtin [4], while Czerwik [5] proved some fixed point results for nonlinear set-valued contractive type mappings in b-metric spaces. Later on, Akkouchi [6] established some common fixed point theorems (CFP-theorems) for single-valued mappings under an implicit relation in b-metric spaces. In [7], Aghajani et al. proved CFP-results under the generalized weak contraction in partially ordered b-metric spaces. Further, Aydi et al. [8, 9] presented some FP-theorems and CFP-theorems for set-valued quasi-contraction and weak ϕ-contraction, respectively, in b-metric spaces. In 2013, Roshan et al. [10] established some generalized contractive type CFP-theorems in b-metric spaces and they proved that the b-metric function used in the theorems and results are not necessarily continuous. Some more FP-results in b-metric space can be found in [11–19]; the references are therein.In 2011, Azam et al. [20] introduced the notion of complex-valued metric space and proved some CFP-theorems for a pair of self-mappings. Though complex-valued metric space forms a special class of cone metric space, so far this concept is proposed to define rational type expressions that are not significant in cone metric spaces, and therefore, some results of the analysis cannot be generalized to cone metric spaces. Properly the notion of complex-valued metric space was introduced by Rouzkard and Imdad [21] which generalized the expression of Azam et al. [20] and proved some CFP-theorems. Some more FP-results in the context of complex-valued metric spaces can be found in [22–24].In 2013, Rao et al. [25] introduced the notion of complex-valued b-metric space which generalized the notion of complex-valued metric spaces given by Azam et al. [20] in 2011. They presented some CFP-results for generalized contraction conditions in complex-valued b-metric space. Later on, Mukheimer [26] extended and improved the results of [20, 25] and established some unique CFP-theorems in complex-valued metric spaces with illustrative examples.In this paper, we establish some new generalized rational type CFP-theorems for compatible three self-mappings on complex-valued b-metric spaces in which one is a continuous self-map. Our results extend and modify many results given in the literature. This paper is organized as follows: Section2 consists of preliminary concepts. In Section 3, we present some generalized unique CFP-theorems for compatible three self-mappings in complex-valued b-metric spaces with some illustrative examples to verify the validity of our work. In Section 4, we present an application of the two UITEs for the existence of a common solution to support our main result, while in Section 5, we discuss the conclusion.


## 2. Preliminaries

Considerℂ represents a set of 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:(C1) Rezi=Rezii and Imzi=Imzii,(C2) Rezi<Rezii and Imzi=Imzii,(C3) Rezi=Rezii and Imzi<Imzii,(C4) Rezi<Rezii and Imzi<Imzii,In particular, we can writezi⪇zii if zi≠zii and one of (C2), (C3), and (C4) is satisfied.Remark 1 (see [26]).
The properties given below hold and can be verified:
(R1) if a1,a2∈ℝ and a1≤a2⇒a1y≤a2y∀y∈ℂ,
(R2) 0≤zi≤zii⇒zi<zii,
(R3) zi≤zii and zii<ziii⇒zi<ziii.Definition 2 (see [5]).
LetV be a nonempty set and let b≥1 be a given real number. A function δ:V×V⟶0,∞ is said to be a b-metric on V if it holds the following conditions:
(bm1) δv1,v2=0⇔v1=v2,
(bm2) δv1,v2=δv2,v1,
(bm3) δv1,v2≤bδv1,v3+δv3,v2,
for allv1,v2,v3∈V. The pair V,δ is called a b-metric space, where b is a coefficient of V,δ.Definition 3 (see [25]).
LetV be a nonempty set and let b≥1 be a given real number. A function δ:V×V⟶ℂ is said to be a complex-valued b-metric on V if it holds the following conditions:
(Cbm1) δv1,v2≥0 and δv1,v2=0 if and only if v1=v2,
(Cbm2) δv1,v2=δv2,v1,
(Cbm3) δv1,v2≤bδv1,v3+δv3,v2,
for allv1,v2,v3∈V. The pair V,δ is called a complex-valued b-metric space, where b is a coefficient of V,δ.Example 1.
LetV=0,∞. The mapping δ:V×V⟶ℂ is defined by
(1)δv1,v2=717v1−v22+i717v1−v22,∀v1,v2∈V.
ThenV,δ is a complex-valued b-metric space with b=2.Definition 4 (see [25, 26]).
LetV,δ is a complex-valued b-metric space and vn be a sequence in V and v∈V. Consider the following:
(1)
If there isN1∈ℕ for every c1∈ℂ and 0<c1 such that for all n>N1, δvn,v<c1, then vn is called convergent, vn converges to v, and v is a limit point of vn. Mathematically, it can be written as limn⟶∞vn=v or vn⟶v as n⟶∞.(2)
If there isN1∈ℕ for every c1∈ℂ and 0<c1 such that for all n>N1, δvn,vn+m<c1, where m∈ℕ, then vn is said to be Cauchy sequence.(3)
If every Cauchy sequence is convergent, thenV,δ is said to be complete complex-valued b-metric space.Lemma 5 (see [25, 26]).
LetV,δ be a complex-valued b-metric space and let vn be a sequence in V. Then, vn converges to viffδvn,v⟶0 as n⟶∞.Lemma 6 (see [25, 26]).
LetV,δ be a complex-valued b-metric space and let vn be a sequence in V. Then, vn is a cauchy sequence iffδvn,vn+m⟶0 as n⟶∞.
To prove the main result, we will use following lemma.Lemma 7 (see [10]).
LetV,δ be a complex-valued b-metric space. Consider vn and wn be two sequences such that limn⟶∞δvn,wn=0, whenever vn is a sequence in V such that limn⟶∞vn=z for some z∈V; then, limn⟶∞wn=z.Proof.
Given that(2)limn⟶∞δvn,wn=0,limn⟶∞vn=z.
By triangular property ofV,δ,
(3)δwn,z≤bδwn,vn+δvn,z⇒δwn,z≤bδwn,vn+δvn,z.
Now by applyinglimn⟶∞ and using (2), we have
(4)limn⟶∞δwn,z≤limn⟶∞bδwn,vn+limn⟶∞bδvn,z=0+0.
Hence, we proved thatlimn⟶∞wn=z.Definition 8 (see [27]).
LetV,δ be a complex-valued b-metric space. A pair J,K is said to be compatible ifflimn⟶∞δJKvn,KJvn=0, whenever vn is a sequence in V such that
(5)limn⟶∞Jvn=limn⟶∞Kvn=zforsomez∈V.


## 3. Main Result

Theorem 9.
LetV,δ be a complete complex-valued b-metric space and let J,K,f:V⟶V be three self-mappings satisfying the following:
(6)δJv1,Kv2≤η1δfv1,fv2+η2δfv1,Jv1δfv2,Kv21+δfv1,fv2+η3δfv1,Jv1δfv1,Kv2+δfv2,Kv2δfv2,Jv1δfv1,Kv2+δfv2,Jv1+η4maxδfv1,Jv1,δfv2,Kv2,δfv1,Kv2,δfv2,Jv1,for all v1,v2∈V, and η1,η2,η3,η4∈0,1 with η1+η2+η3+2bη4<1, where b≥1. If f is continuous and f,J, f,K are compatible, then f,J and K have a unique common fixed point in V.Proof.
Fixv0∈V, and we define some sequences in V such that
(7)ℏ2n=fv2n+1=Jv2n,ℏ2n+1=fv2n+2=Kv2n+1,∀n≥0.
Now by using (6),
(8)δℏ2n,ℏ2n+1=δJv2n,Kv2n+1≤η1δfv2n,fv2n+1+η2δfv2n,Jv2nδfv2n+1,Kv2n+11+δfv2n,fv2n+1+η3δfv2n,Jv2nδfv2n,Kv2n+1+δfv2n+1,Kv2n+1δfv2n+1,Jv2nδfv2n,Kv2n+1+δfv2n+1,Jv2n+η4maxδfv2n,Jv2n,δfv2n+1,Kv2n+1,δfv2n,Kv2n+1,δfv2n+1,Jv2n=η1δℏ2n−1,ℏ2n+η2δℏ2n−1,ℏ2nδℏ2n,ℏ2n+11+δℏ2n−1,ℏ2n+η3δℏ2n−1,ℏ2nδℏ2n−1,ℏ2n+1+δℏ2n,ℏ2n+1δℏ2n,ℏ2nδℏ2n−1,ℏ2n+1+δℏ2n,ℏ2n+η4maxδℏ2n−1,ℏ2n,δℏ2n,ℏ2n+1,δℏ2n−1,ℏ2n+1,δℏ2n,ℏ2n.
This implies that(9)δℏ2n,ℏ2n+1≤η1δℏ2n−1,ℏ2n+η2δℏ2n−1,ℏ2nδℏ2n,ℏ2n+11+δℏ2n−1,ℏ2n+η3δℏ2n−1,ℏ2nδℏ2n−1,ℏ2n+1+δℏ2n,ℏ2n+1δℏ2n,ℏ2nδℏ2n−1,ℏ2n+1+δℏ2n,ℏ2n+η4maxδℏ2n−1,ℏ2n,δℏ2n,ℏ2n+1,δℏ2n−1,ℏ2n+1,δℏ2n,ℏ2n.
After simplification, we get that(10)δℏ2n,ℏ2n+1≤η1+η3δℏ2n−1,ℏ2n+η2δℏ2n,ℏ2n+1+η4maxδℏ2n−1,ℏ2n,δℏ2n,ℏ2n+1,δℏ2n−1,ℏ2n+1.
Now there are three possibilities:(i)
Ifδℏ2n−1,ℏ2n is a maximum term in ∣δℏ2n−1,ℏ2n∣,∣δℏ2n,ℏ2n+1∣,∣δℏ2n−1,ℏ2n+1∣, then after simplification, (10) can be written as follows:(11)δℏ2n,ℏ2n+1≤a1δℏ2n−1,ℏ2n,wherea1=η1+η3+η41−η2<1.(ii)
Ifδℏ2n,ℏ2n+1 is a maximum term in δℏ2n−1,ℏ2n,δℏ2n,ℏ2n+1,δℏ2n−1,ℏ2n+1, then after simplification, (10) can be written as follows:(12)δℏ2n,ℏ2n+1≤a2δℏ2n−1,ℏ2n,wherea2=η1+η31−η2−η4<1.(iii)
Ifδℏ2n−1,ℏ2n+1 is a maximum term in δℏ2n−1,ℏ2n,δℏ2n,ℏ2n+1,δℏ2n−1,ℏ2n+1, and by using the triangular property of complex-valued b-metric space, then after simplification, (10) can be written as follows:(13)δℏ2n,ℏ2n+1≤a3δℏ2n−1,ℏ2n,wherea3=η1+η3+η4b1−η2−η4b<1.
Leta≔maxa1,a2,a3<1; then, from (11), (12), and (13), for all n≥0, we have
(14)δℏ2n,ℏ2n+1≤aδℏ2n−1,ℏ2n.
Similarly,(15)δℏ2n−1,ℏ2n≤aδℏ2n−2,ℏ2n−1.
Now from (15) and (14), and by induction, we have that,
(16)δℏ2n,ℏ2n+1≤aδℏ2n−1,ℏ2n≤a2δℏ2n−2,ℏ2n−1≤⋯≤a2nδℏ0,ℏ1.
Next, we show thatℏn is a Cauchy sequence. Let m,n∈ℕ and m>n, then we have
(17)δℏn,ℏm≤bδℏn,ℏn+1+bδℏn+1,ℏm≤bδℏn,ℏn+1+b2δℏn+1,ℏn+2+⋯+bm−nδℏm−1,ℏm≤banδℏ0,ℏ1+b2an+1δℏ0,ℏ1+⋯+bm−nam−1δℏ0,ℏ1≤ban+b2an+1+⋯+bm−nam−1δℏ0,ℏ1=ban+b2an+1+⋯+bm−nam−1δℏ0,ℏ1=ban1+ba+b2a2⋯+bm−n+1am−n+1δℏ0,ℏ1=ban∑t=0m−n+1btatδℏ0,ℏ1≤ban∑t=0∞btatδℏ0,ℏ1=ban1−baδℏ0,ℏ1⟶0,asn⟶∞.
Hence,ℏn is a Cauchy sequence. Since V is a complete complex-valued b-metric space, there exists p∈V such that ℏn⟶p, as n⟶∞ or limn⟶∞ℏn=p, and from (7), we have
(18)limn⟶∞fv2n+1=p,limn⟶∞Jv2n=p,limn⟶∞Kv2n+1=p.
Sincef is a continuous self-map on V, therefore
(19)limn⟶∞ffv2n+1=fp,limn⟶∞fJv2n=fp,limn⟶∞fKv2n+1=fp.
As a pairf,J is compatible, so for some sequence v2n in V and by the definition of compatibility, we have that
(20)limn⟶∞δJfv2n,fJv2n=0.
Now from (19), (20), and by using Lemma 7, we have
(21)limn⟶∞Jfv2n=fp.
Next, we have to show thatfp=p, so by putting v1=fv2n and v2=v2n+1, in (6),
(22)δJfv2n,Kv2n+1≤η1δffv2n,fv2n+1+η2δffv2n,Jfv2nδfv2n+1,Kv2n+11+δffv2n,fv2n+1+η3δffv2n,Jfv2nδffv2n,Kv2n+1+δfv2n+1,Kv2n+1δfv2n+1,Jfv2nδffv2n,Kv2n+1+δfv2n+1,Jfv2n+η4maxδffv2n,Jfv2n,δfv2n+1,Kv2n+1,δffv2n,Kv2n+1,δfv2n+1,Jfv2n.
This implies that(23)δJfv2n,Kv2n+1≤η1δffv2n,fv2n+1+η2δffv2n,Jfv2nδfv2n+1,Kv2n+11+δffv2n,fv2n+1+η3δffv2n,Jfv2nδffv2n,Kv2n+1+δfv2n+1,Kv2n+1δfv2n+1,Jfv2nδffv2n,Kv2n+1+δfv2n+1,Jfv2n+η4maxδffv2n,Jfv2n,δfv2n+1,Kv2n+1,δffv2n,Kv2n+1,δfv2n+1,Jfv2n.
Now applyinglimn⟶∞ on both sides and from (18), (19), and (21), we get that
(24)δfp,p≤η1δfp,p+η2δfp,fpδp,p1+δfp,p+η3δfp,fpδfp,p+δp,pδp,fpδfp,p+δp,fp+η4maxδfp,fp,δp,p,δfp,p,δp,fp.
After simplification, we get that(25)δfp,p≤η1+η4δfp,p⇒1−η1−η4δfp,p≤0.
Since1−η1−η4≠0⇒δfp,p=0, hence we get that
(26)fp=p.
Next, we have to show thatJp=p, and by the view of (6),
(27)δJp,fv2n+2=δJp,Kv2n+1≤η1δfp,fv2n+1+η2δfp,Jpδfv2n+1,Kv2n+11+δfp,fv2n+1+η3δfp,Jpδfp,Kv2n+1+δfv2n+1,Kv2n+1δfv2n+1,Jpδfp,Kv2n+1+δfv2n+1,Jp+η4maxδfp,Jp,δfv2n+1,Kv2n+1,δfp,Kv2n+1,δfv2n+1,Jp.
This implies that(28)δJp,fv2n+2≤η1δfp,fv2n+1+η2δfp,Jpδfv2n+1,Kv2n+11+δfp,fv2n+1+η3δfp,Jpδfp,Kv2n+1+δfv2n+1,Kv2n+1δfv2n+1,Jpδfp,Kv2n+1+δfv2n+1,Jp+η4maxδfp,Jp,δfv2n+1,Kv2n+1,δfp,Kv2n+1,δfv2n+1,Jp.
Now again applyinglimn⟶∞ on both sides and by using (18) and (26), we have that
(29)δJp,p≤η1δfp,p+η2δfp,Jpδp,p1+δfp,p+η3δfp,Jpδfp,p+δp,pδp,Jpδfp,p+δp,Jp+η4maxδfp,Jp,δp,p,δfp,p,δp,Jp=η4δp,Jp.
This implies that1−η4δJp,p≤0. Since 1−η4≠0⇒δJp,p=0, hence
(30)Jp=p.
Now, we have to show thatKp=p, and by using (6),
(31)δfv2n+1,Kp=δJv2n,Kp≤η1δfv2n,fp+η2δfv2n,Jv2nδfp,Kp1+δfv2n,fp+η3δfv2n,Jv2nδfv2n,Kp+δfp,Kpδfp,Jv2nδfv2n,Kp+δfp,Jv2n+η4maxδfv2n,Jv2n,δfp,Kp,δfv2n,Kp,δfp,Jv2n.
This implies that(32)δfv2n+1,Kp≤η1δfv2n,fp+η2δfv2n,Jv2nδfp,Kp1+δfv2n,fp+η3δfv2n,Jv2nδfv2n,Kp+δfp,Kpδfp,Jv2nδfv2n,Kp+δfp,Jv2n+η4maxδfv2n,Jv2n,δfp,Kp,δfv2n,Kp,δfp,Jv2n.
Applyinglimn⟶∞ on both sides and by using (18) and (26), we have that
(33)δp,Kp≤η1δp,fp+η2δp,pδfp,Kp1+δp,fp+η3δp,pδp,Kp+δfp,Kpδfp,pδp,Kp+δfp,p+η4maxδp,p,δfp,Kp,δp,Kp,δfp,p=η4δp,Kp.
This implies that1−η4δp,Kp≤0. Since 1−η4≠0⇒δp,Kp=0, hence
(34)Kp=p.
Now from (26), (30), and (34), we get that p is a common fixed point of f,J and K, i.e.,
(35)fp=Jp=Kp=p.
Uniqueness: assume that p∗∈V is an other common fixed point of f,J and K along with p, i.e.,
(36)fp=Jp=Kp=p,fp∗=Jp∗=Kp∗=p∗.
Then, from (6), we have that
(37)δp,p∗=δJp,Kp∗≤η1δfp,fp∗+η2δfp,Jpδfp∗,Kp∗1+δfp,fp∗+η3δfp,Jpδfp,Kp∗+δfp∗,Kp∗δfp∗,Jpδfp,Kp∗+δfp∗,Jp+η4maxδfp,Jp,δfp∗,Kp∗,δfp,Kp∗,δfp∗,Jp=η1+η4δp,p∗.
This implies thatδp,p∗≤η1+η4δp,p∗⇒1−η1−η4δp,p∗≤0, since 1−η1−η4≠0⇒δp,p∗=0⇒p=p∗. Hence, we proved that f,J and K have a unique common fixed point in V.Remark 10.
(i)
If we putη1=λ,η2=μ, and η3=η4=0 in Theorem 9, we get the results of [26] Theorem 15.(ii)
If we putη2=a and η1=η3=η4=0 in Theorem 9, we get the results of [26] Theorem 19.Example 2.
LetV,δ be a complex-valued b-metric space, where V=0,1 and δ:V×V⟶ℂ with δv1,v2=4v1−v22/9+i4v1−v22/9, for all v1,v2∈V. Now to find the value of b, we have that
(38)δv1,v2=4v1−v229+i4v1−v229=4v1−v3+v3−v229+i4v1−v3+v3−v229≤4v1−v329+4v3−v229+492v1−v3v3−v2+i4v1−v329+4v3−v229+492v1−v3v3−v2≤4v1−v329+4v3−v229+4v1−v329+4v3−v229+i4v1−v329+4v3−v229+4v1−v329+4v3−v229=24v1−v329+4v3−v229+i24v1−v329+4v3−v229=24v1−v329+i4v1−v329+4v3−v229+i4v3−v229=2δv1,v3+δv3,v2.
That is,δv1,v2≤bδv1,v3+δv3,v2 with b=2. Now we define J,K,f:V⟶V as
(39)Jv1=Kv1=2v135andfv1=v15.
Notice that(40)δfv1,fv2,δfv1,Jv1δfv2,Kv2∣1+δfv1,fv2∣,δfv1,Jv1δfv1,Kv2+δfv2,Kv2δfv2,Jv1δfv1,Kv2+δfv2,Jv1,maxδfv1,Jv1,δfv2,Kv2,δfv1,Kv2,δfv2,Jv1≥0.
In all regards, it is enough to show thatδJv1,Kv2≤η1δfv1,fv2, for all v1,v2∈0,1 and η1,η2,η3,η4∈0,1, with η1+η2+η3+2η4b<1.
(41)δJv1,Kv2=4Jv1−Kv229+i4Jv1−Kv229=42v1/35−2v2/3529+i42v1/35−2v2/3529=2724v1/5−v2/529+i4v1/5−v2/529=4494v1/5−v2/529+i4v1/5−v2/529.
And(42)δfv1,fv2=4fv1−fv229+i4fv1−fv229=4v1/5−v2/529+i4v1/5−v2/529.
Forv1,v2∈0,1, we discuss different cases with η1=2/5,η2=1/5,η3=1/10,η4=1/20, and b=2. Hence,
(43)η1+η2+η3+2η4b=25+15+110+21202<1.Case 1.
Letv1=0,v2=0; then, from (41) and (42), directly we get that δJv1,Kv2≤η1δfv1,fv2. Hence, (6) is satisfied with η1=2/5,η2=1/5,η3=1/10,η4=1/20, and b=2.Case 2.
Letv1=1,v2=0; then, from (41) and (42), we find δJv1,Kv2≤η1δfv1,fv2 is satisfied with η1=2/5, i.e,
(44)44941/5−0/529+i41/5−0/529≤η141/5−0/529+i41/5−0/5290.00141+i≤0.00711+i.
Hence, (6) is satisfied with η1=2/5,η2=1/5,η3=1/10,η4=1/20, and b=2.Case 3.
Letv1=1/2,v2=1/4; then from (41) and (42), we find δJv1,Kv2≤η1δfv1,fv2 is satisfied with η1=2/5, i.e,
(45)44943600+i43600≤η143600+i436000.0000901+i≤0.000441+i.
Hence, (6) is satisfied with η1=2/5,η2=1/5,η3=1/10,η4=1/20, and b=2.Case 4.
Letv1=1,v2=1; then, from (41) and (42), directly we get that δJv1,Kv2≤η1δfv1,fv2. Hence, (6) is satisfied with η1=2/5,η2=1/5,η3=1/10,η4=1/20, and b=2. Thus, all the conditions of Theorem 9 are satisfied with noticing that the point 0∈V, which remains fixed under mappings f,J and K, is indeed unique.Corollary 11.
LetV,δ be a complete complex-valued b-metric space and let J,K,f:V⟶V be three self-mappings satisfying the following:
(46)δJv1,Kv2≤η1δfv1,fv2+η2δfv1,Jv1δfv2,Kv21+δfv1,fv2+η3δfv1,Jv1δfv1,Kv2+δfv2,Kv2δfv2,Jv1δfv1,Kv2+δfv2,Jv1+η4δfv1,Jv1+δfv2,Kv2,for all v1,v2∈V and η1,η2,η3,η4∈0,1 such that η1+η2+η3+2η4<1. If f is a continuous self-mapping and f,J, f,K are compatible, then f,J and K have a unique common fixed point in V.Corollary 12.
LetV,δ be a complete complex-valued b-metric space and let J,K,f:V⟶V be three self-mappings satisfying the following:
(47)δJv1,Kv2≤η1δfv1,fv2+η2δfv1,Jv1δfv2,Kv21+δfv1,fv2+η3δfv1,Jv1δfv1,Kv2+δfv2,Kv2δfv2,Jv1δfv1,Kv2+δfv2,Jv1+η4δfv1,Kv2+δfv2,Jv1,for all v1,v2∈V and η1,η2,η3,η4∈0,1 such that η1+η2+η3+2η4b<1, where b≥1. If f is continuous and f,J, f,K are compatible, then f,J and K have a unique common fixed point in V.Theorem 13.
LetV,δ be a complete complex-valued b-metric space and let J,K,f:V⟶V be three self-mappings satisfying:
(48)δJv1,Kv2≤η1δfv1,fv2+η2δfv1,Jv1δfv2,Kv21+δfv1,fv2+η3δfv1,Jv1δfv1,Kv2+δfv2,Kv2δfv2,Jv1δfv1,Kv2+δfv2,Jv1+η4δfv1,Jv1+δfv2,Kv2+δfv1,Kv2+δfv2,Jv1,for all v1,v2∈V, η1,η2,η3∈0,1, and η4∈0,1/4, such that η1+η2+η3+2η4+2η4b<1, where b≥1. If f is a continuous self-mapping and f,J, f,K are compatible, then f,J and K have a unique common fixed point in V.Proof.
Fixv0∈V, and we define some sequences in V such that
(49)ℏ2n=fv2n+1=Jv2n,ℏ2n+1=fv2n+2=Kv2n+1,foralln≥0.
Now by view of (48) and (49),
(50)δℏ2n,ℏ2n+1=δJv2n,Kv2n+1≤η1δfv2n,fv2n+1+η2δfv2n,Jv2nδfv2n+1,Kv2n+11+δfv2n,fv2n+1+η3δfv2n,Jv2nδfv2n,Kv2n+1+δfv2n+1,Kv2n+1δfv2n+1,Jv2nδfv2n,Kv2n+1+δfv2n+1,Jv2n+η4δfv2n,Jv2n+δfv2n+1,Kv2n+1+δfv2n,Kv2n+1+δfv2n+1,Jv2n=η1δℏ2n−1,ℏ2n+η2δℏ2n−1,ℏ2nδℏ2n,ℏ2n+11+δℏ2n−1,ℏ2n+η3δℏ2n−1,ℏ2nδℏ2n−1,ℏ2n+1+δℏ2n,ℏ2n+1δℏ2n,ℏ2nδℏ2n−1,ℏ2n+1+δℏ2n,ℏ2n+η4δℏ2n−1,ℏ2n+δℏ2n,ℏ2n+1+δℏ2n−1,ℏ2n+1+δℏ2n,ℏ2n.
This implies that(51)δℏ2n,ℏ2n+1≤η1δℏ2n−1,ℏ2n+η2δℏ2n−1,ℏ2nδℏ2n,ℏ2n+11+δℏ2n−1,ℏ2n+η3δℏ2n−1,ℏ2nδℏ2n−1,ℏ2n+1+δℏ2n,ℏ2n+1δℏ2n,ℏ2nδℏ2n−1,ℏ2n+1+δℏ2n,ℏ2n+η4δℏ2n−1,ℏ2n+δℏ2n,ℏ2n+1+δℏ2n−1,ℏ2n+1+δℏ2n,ℏ2n.
Now by using triangular inequality ofV,δ and after simplification, we get that
(52)δℏ2n,ℏ2n+1≤η1+η3+η4+η4b1−η2−η4−η4bδℏ2n−1,ℏ2n.
Again by view of (48) and (49),
(53)δℏ2n−1,ℏ2n=δKv2n−1,Jv2n=δJv2n,Kv2n−1≤η1δfv2n,fv2n−1+η2δfv2n,Jv2nδfv2n−1,Kv2n−11+δfv2n,fv2n−1+η3δfv2n,Jv2nδfv2n,Kv2n−1+δfv2n−1,Kv2n−1δfv2n−1,Jv2nδfv2n,Kv2n−1+δfv2n−1,Jv2n+η4δfv2n,Jv2n+δfv2n−1,Kv2n−1+δfv2n,Kv2n−1+δfv2n−1,Jv2n=η1δℏ2n−1,ℏ2n−2+η2δℏ2n−1,ℏ2nδℏ2n−2,ℏ2n−11+δℏ2n−1,ℏ2n−2+η3δℏ2n−1,ℏ2nδℏ2n−1,ℏ2n−1+δℏ2n−2,ℏ2n−1δℏ2n−2,ℏ2nδℏ2n−1,ℏ2n−1+δℏ2n−2,ℏ2n+η4δℏ2n−1,ℏ2n+δℏ2n−2,ℏ2n−1+δℏ2n−1,ℏ2n−1+δℏ2n−2,ℏ2n.
This implies that(54)δℏ2n−1,ℏ2n≤η1δℏ2n−1,ℏ2n−2+η2δℏ2n−1,ℏ2nδℏ2n−2,ℏ2n−11+δℏ2n−1,ℏ2n−2+η3δℏ2n−1,ℏ2nδℏ2n−1,ℏ2n−1+δℏ2n−2,ℏ2n−1δℏ2n−2,ℏ2nδℏ2n−1,ℏ2n−1+δℏ2n−2,ℏ2n+η4δℏ2n−1,ℏ2n+δℏ2n−2,ℏ2n−1+δℏ2n−1,ℏ2n−1+δℏ2n−2,ℏ2n.
By using triangular inequality ofV,δ and after simplification we get that
(55)δℏ2n−1,ℏ2n≤η1+η3+η4+η4b1−η2−η4−η4bδℏ2n−2,ℏ2n−1.
Now from (55) and (52) and by induction, we have
(56)δℏ2n,ℏ2n+1≤qδℏ2n−1,ℏ2n≤q2δℏ2n−2,ℏ2n−1≤⋯≤q2nδℏ0,ℏ1,where q=η1+η3+η4+η4b/1−η2−η4−η4b<1. Next, we have to show that ℏn is a Cauchy sequence. Let m,n∈ℕ and m>n. Then, we have
(57)δℏn,ℏm≤bδℏn,ℏn+1+bδℏn+1,ℏm≤bδℏn,ℏn+1+b2δℏn+1,ℏn+2+⋯+bm−nδℏm−1,ℏm≤bqnδℏ0,ℏ1+b2qn+1δℏ0,ℏ1+⋯+bm−nqm−1δℏ0,ℏ1≤bqn+b2qn+1+⋯+bm−nqm−1δℏ0,ℏ1=bqn1+bq+b2q2⋯+bm−n+1qm−n+1δℏ0,ℏ1=bqn∑t=0m−n+1btqtδℏ0,ℏ1≤bqn∑t=0∞btqtδℏ0,ℏ1=bqn1−bqδℏ0,ℏ1⟶0,asn⟶∞.
Hence,ℏn is a Cauchy sequence. Since V is complete, there exists some p∈V, such that ℏn⟶p, as n⟶∞, and from (49), we have that
(58)limn⟶∞fv2n+1=p,limn⟶∞Jv2n=p,limn⟶∞Kv2n+1=p.
f is a continuous self-mapping on V, so that
(59)limn⟶∞ffv2n+1=fp,limn⟶∞fJv2n=fp,limn⟶∞fKv2n+1=fp.
Sincef,J is compatible and for some sequence v2n in V, we have that
(60)limn⟶∞δJfv2n,fJv2n=0.
From (59), (60), and by using Lemma 7, we get that
(61)limn⟶∞Jfv2n=fp.
Now, we have to show thatfp=p. So, by putting v1=fv2n and v2=v2n+1 in (48),
(62)δJfv2n,Kv2n+1≤η1δffv2n,fv2n+1+η2δffv2n,Jfv2nδfv2n+1,Kv2n+11+δffv2n,fv2n+1+η3δffv2n,Jfv2nδffv2n,Kv2n+1+δfv2n+1,Kv2n+1δfv2n+1,Jfv2nδffv2n,Kv2n+1+δfv2n+1,Jfv2n+η4δffv2n,Jfv2n+δfv2n+1,Kv2n+1+δffv2n,Kv2n+1+δfv2n+1,Jfv2n.
This implies that(63)δJfv2n,Kv2n+1≤η1δffv2n,fv2n+1+η2δffv2n,Jfv2nδfv2n+1,Kv2n+11+δffv2n,fv2n+1+η3δffv2n,Jfv2nδffv2n,Kv2n+1+δfv2n+1,Kv2n+1δfv2n+1,Jfv2nδffv2n,Kv2n+1+δfv2n+1,Jfv2n+η4δffv2n,Jfv2n+δfv2n+1,Kv2n+1+δffv2n,Kv2n+1+δfv2n+1,Jfv2n.
Applyinglimn⟶∞ on both sides and using (58), (59), and (61), we get that
(64)δfp,p≤η1δfp,p+η2δfp,fpδp,p1+δfp,p+η3δfp,fpδfp,p+δp,pδp,fpδfp,p+δp,fp+η4δfp,fp+δp,p+δfp,p+δp,fp=η1+2η4δfp,p.
This implies that1−η1−2η4δfp,p≤0. Since 1−η1−2η4≠0⇒δfp,p=0, hence,
(65)fp=p.
Next, we have to show thatJp=p, and by using (48),
(66)δJp,fv2n+2=δJp,Kv2n+1≤η1δfp,fv2n+1+η2δfp,Jpδfv2n+1,Kv2n+11+δfp,fv2n+1+η3δfp,Jpδfp,Kv2n+1+δfv2n+1,Kv2n+1δfv2n+1,Jpδfp,Kv2n+1+δfv2n+1,Jp+η4δfp,Jp+δfv2n+1,Kv2n+1+δfp,Kv2n+1+δfv2n+1,Jp.
This implies that(67)δJp,fv2n+2≤η1δfp,fv2n+1+η2δfp,Jpδfv2n+1,Kv2n+11+δfp,fv2n+1+η3δfp,Jpδfp,Kv2n+1+δfv2n+1,Kv2n+1δfv2n+1,Jpδfp,Kv2n+1+δfv2n+1,Jp+η4δfp,Jp+δfv2n+1,Kv2n+1+δfp,Kv2n+1+δfv2n+1,Jp.
Applyinglimn⟶∞ on both sides and using (58) and (65), we have
(68)δJp,p≤η1δfp,p+η2δfp,Jpδp,p1+δfp,p+η3δfp,Jpδfp,p+δp,pδp,Jpδfp,p+δp,Jp+η4δfp,Jp+δp,p+δfp,p+δp,Jp=η1δp,p+η2δp,Jpδp,p1+δp,p+η3δp,Jpδp,p+δp,pδp,Jpδp,p+δp,Jp+η4δp,Jp+δp,p+δp,p+δp,Jp.
Thus, we get thatδJp,p≤2η4δp,Jp⇒1−2η4δJp,p≤0. Since 1−2η4≠0, as η1+η2+η3+2η4+2η4b<1, therefore δJp,p=0. Hence,
(69)Jp=p.
Now, we have to show thatKp=p, and by using (48),
(70)δfv2n+1,Kp=δJv2n,Kp≤η1δfv2n,fp+η2δfv2n,Jv2nδfp,Kp1+δfv2n,fp+η3δfv2n,Jv2nδfv2n,Kp+δfp,Kpδfp,Jv2nδfv2n,Kp+δfp,Jv2n+η4δfv2n,Jv2n+δfp,Kp+δfv2n,Kp+δfp,Jv2n.
This implies that(71)δfv2n+1,Kp≤η1δfv2n,fp+η2δfv2n,Jv2nδfp,Kp1+δfv2n,fp+η3δfv2n,Jv2nδfv2n,Kp+δfp,Kpδfp,Jv2nδfv2n,Kp+δfp,Jv2n+η4δfv2n,Jv2n+δfp,Kp+δfv2n,Kp+δfp,Jv2n.
Applyinglimn⟶∞ on both sides and by view of (58) and (65), we have that
(72)δp,Kp≤η1δp,fp+η2δp,pδfp,Kp1+δp,fp+η3δp,pδp,Kp+δfp,Kpδfp,pδp,Kp+δfp,p+η4δp,p+δfp,Kp+δp,Kp+δfp,p=2η4δp,Kp.
This implies that1−2η4δp,Kp≤0. Since 1−2η4≠0, therefore δp,Kp=0. Hence,
(73)Kp=p.
Thus, from (65), (69), and (73), we get that p is a common fixed point of f,J and K, i.e.,
(74)fp=Jp=Kp=p.
Uniqueness: we contrary suppose that p∗ is another common fixed point of f,J and K such that fp∗=Jp∗=Kp∗=p∗. Now, by (48),
(75)δp,p∗=δJp,Kp∗≤η1δfp,fp∗+η2δfp,Jpδfp∗,Kp∗1+δfp,fp∗+η3δfp,Jpδfp,Kp∗+δfp∗,Kp∗δfp∗,Jpδfp,Kp∗+δfp∗,Jp+η4δfp,Jp+δfp∗,Kp∗+δfp,Kp∗+δfp∗,Jp=η1+2η4δp,p∗.
This implies thatδp,p∗≤η1+2η4δp,p∗⇒1−η1−2η4δp,p∗≤0.Since 1−η1−η4≠0, therefore δp,p∗≤0⟶p=p∗. Hence, we proved that f,J and K have a unique common fixed point in V.Example 3.
LetV=0,∞ and δ:V×V⟶ℂ defined by δv1,v2=3v1−v22/13+i3v1−v22/13 for all v1,v2∈V is a b-metric on V and V,δ is a complex-valued b-metric space. Now, first, we show that V is a b-metric with b=2, so that
(76)δv1,v2=3v1−v2213+i3v1−v2213≤3v1−v3+v3−v2213+i3v1−v3+v3−v2213≤3v1−v3213+3v3−v2213+3132v1−v3v3−v2+i3v1−v3213+3v3−v2213+3132v1−v3v3−v2≤3v1−v3213+3v3−v2213+3v1−v3213+3v3−v2213+i3v1−v3213+3v3−v2213+3v1−v3213+3v3−v2213=23v1−v3213+3v3−v2213+i23v1−v3213+3v3−v2213=23v1−v3213+i3v1−v3213+3v3−v2213+i3v3−v2213=2δv1,v3+δv3,v2.
That is,δv1,v2≤bδv1,v3+δv3,v2, with b=2.
DefineJ,K,f:V⟶V by Jv1=Kv1=ln1+v1/3+v1, and fv1=e3v1−1.
Notice that(77)δfv1,fv2,δfv1,Jv1δfv2,Kv21+δfv1,fv2,δfv1,Jv1δfv1,Kv2+δfv2,Kv2δfv2,Jv1δfv1,Kv2+δfv2,Jv1δfv1,Jv1+δfv2,Kv2+δfv1,Kv2+δfv2,Jv1≥0,in all regards. It is enough to show that δJv1,Kv2≤η1δfv1,fv2, for all v1,v2∈0,∞ and η1,η2,η3∈0,1, η4∈0,1/4 such that η1+η2+η3+2η4+2η4b<1, where b≥1, and we have
(78)δJv1,Kv2=3Jv1−Kv2213+i3Jv1−Kv2213=3ln1+v1/3+v1−ln1+v2/3+v2213+i3ln1+v1/3+v1−ln1+v2/3+v2213≤3v1/3+v1−v2/3+v2213+i3v1/3+v1−v2/3+v2213≤33v1−3v2/9213+i33v1−3v2/9213=19233v1−3v2213+i33v1−3v2213≤1813e3v1−e3v2213+i3e3v1−e3v2213.
And(79)δfv1,fv2=3fv1−fv2213+i3fv1−fv2213=3e3v1−1−e3v2−1213+i3e3v1−1−e3v2−1213=3e3v1−e3v2213+i3e3v1−e3v2213.
Forv1,v2∈0,∞, we discuss different cases with η1=1/5,η2=1/4,η3=1/10,η4=1/20, and b=2. Notice that η1+η2+η3+2η4b=1/5+1/4+1/10+21/202<1.Case 1.
Letv1=0,v2=0. Then, from (78) and (79), directly we get that δJv1,Kv2≤η1δfv1,fv2. Hence, (48) is satisfied with η1=1/5,η2=1/4,η3=1/10,η4=1/20, and b=2.Case 2.
Letv1=0,v2=1; then, from (78) and (79), we find δJv1,Kv2≤η1δfv1,fv2 is satisfied with η1=1/5, as
(80)18131−e3213+i31−e3213≤η131−e3213+i31−e3213.
By usingη1=1/5 and after simplifying, we get that
(81)1813−19.0855213+i3−19.0855213≤153−19.0855213+i3−19.08552131.041+i≤16.811+i.
Hence, (48) is satisfied with η1=1/5,η2=1/4,η3=1/10,η4=1/20, and b=2.Case 3.
Letv1=1/2,v2=1/4; then, from (78) and (79), we find δJv1,Kv2≤η1δfv1,fv2, is true for η1=1/5, as
(82)1813e3/2−e3/4213+i3e3/2−e3/4213≤η13e3/2−e3/4213+i3e3/2−e3/4213.
By usingη1=1/5 and after simplifying, we get that
(83)18132.3646213+i32.3646213≤1532.3646213+i32.36462130.021+i≤0.261+i.
Hence, (48) is satisfied with η1=1/5,η2=1/4,η3=1/10,η4=1/20, and b=2.Case 4.
Letv1=1/2,v2=1; then, from (78) and (79), we get that δJv1,Kv2≤η1δfv1,fv2 is true for η1=1/5, as
(84)1813e3/2−e3213+i3e3/2−e3213≤η13e3/2−e3213+i3e3/2−e3213.
By usingη1=1/5 and after simplifying, we get that
(85)1813−15.6038213+i3−15.6038213≤153−15.6038213+i3−15.60382130.691+i≤11.241+i.
Hence, (48) is satisfied with η1=1/5,η2=1/4,η3=1/10,η4=1/20, and b=2.Case 5.
Letv1=1,v2=5; then, from (78) and (79), we find δJv1,Kv2≤η1δfv1,fv2 is true for η1=1/5, as
(86)1813e3−e15213+i3e3−e15213≤η13e3−e15213+i3e3−e15213.
By usingη1=1/5 and after simplifying, we get that
(87)1813−3268997.28213+i3−3268997.28213≤153−3268997.28213+i3−3268997.282130.304×10111+i≤4.932×10111+i.
Hence, (48) is satisfied with η1=1/5,η2=1/4,η3=1/10,η4=1/20, and b=2.Thus, all conditions of Theorem13 are satisfied with noticing that the point 0∈V remains fixed under mappings f,J and K and is indeed unique.Corollary 14.
LetV,δ be a complete complex valued b-metric space and let J,K,f:V⟶V be three self-mappings satisfying the following:
(88)δJv1,Kv2≤η1δfv1,fv2+η2δfv1,Jv1δfv1,Kv2+δfv2,Kv2δfv2,Jv11+δfv1,fv2+η3δfv1,Jv1+δfv2,Kv2+δfv1,Kv2+δfv2,Jv1,for all v1,v2∈V and η1,η2∈0,1, η3∈0,1/4, such that η1+2η2b+2η3+2η3b<1, where b≥1. If f is continuous and f,J, f,K are compatible, then f,J and K have a unique common fixed point in V.


## 4. Applications

In this section, we present an integral type application to support our main work. For this purpose, we use the two UITEs to get the existing result of a common solution to verify the validity of our work. LetV=Ck1,k2,ℝn be the set of all real-valued continuous functions defined on k1,k2. In the following, we apply Theorem 9 to get the existing result of a common solution by using the two UITEs. Now we are in the position to present a theorem based on the two UITEs to get the existing result of a common solution to support our work.Theorem 15 (see [23]).
LetV=Ck1,k2,ℝn, where k1,k2⊆ℝ and δ:V×V⟶ℂ is defined as
(89)δv1,v2=v1p−v2p21+k12eicotk1,for all v1,v2∈V and p∈k1,k2. Consider the UITEs are
(90)v1p=∫k1k2Q1p,r,v1rdr+ℏ1p,v2p=∫k1k2Q2p,r,v2rdr+ℏ2p,where r∈k1,k2. Let Q1,Q2:k1,k2×k1,k2×ℝn⟶ℝn are such that Dv1,Ev2∈V for every v1,v2∈V, and we have that
(91)Dv1p=∫k1k2Q1p,r,v1rdr,Ev2p=∫k1k2Q2p,r,v2rdr.
If there existsμ∈0,1 such that for all v1,v2∈V,
(92)Dv1p−Ev2p+ℏ1p−ℏ2p21+k12eicotk1≤μMv1,v2,where
(93)Mv1,v2=maxA1v1,v2p,A2v1,v2p,A3v1,v2p,A4v1,v2pwith
(94)A1v1,v2p=v1p−v2p21+k12eicotk1,(95)A2v1,v2p=Dv1p+ℏ1p−v1p2Ev2p+ℏ2p−v2p21+k12eicotk121+v1p−v2p21+k12eicotk1,(96)A3v1,v2p=Dv1p+ℏ1p−v1p2Ev2p+ℏ2p−v1p2+Ev2p+ℏ2p−v2p2Dv1p+ℏ2p−v2p2Ev2p+ℏ2p−v1p2+Dv1p+ℏ1p−v2p2×1+k12eicotk1,(97)A4v1,v2p=maxa1v1,v2p,a2v1,v2p,a3v1,v2p,a4v1,v2p,where
(98)a1v1,v2p=Dv1p+ℏ1p−v1p21+k12eicotk1,a2v1,v2p=Ev2p+ℏ2p−v2p21+k12eicotk1,a3v1,v2p=Ev2p+ℏ2p−v1p21+k12eicotk1,a4v1,v2p=Dv1p+ℏ1p−v2p21+k12eicotk1.
Then, the two UITEs, i.e., (90), have a unique common solution.Proof.
DefineJ,K,f:V⟶V as
(99)Jv1=Jv1p=Dv1p+ℏ1p=Dv1+ℏ1,fv1=fv1p=v1p=v1,Kv2=Kv2p=Ev2p+ℏ2p=Ev2+ℏ2,fv2=fv2p=v2p=v2.
Then, we have the following four cases:(1)
IfA1v1,v2p is the maximum term in A1v1,v2p,A2v1,v2p,A3v1,v2p,A4v1,v2p, then from (92), (93), and (99), we have that
(100)δJv1,Kv2≤μv1−v221+k12eicotk1,for all v1,v2∈V. Hence, the mappings J,K and f satisfy all the conditions of Theorem 9 with μ=η1 and η2=η3=η4=0 in (6). Then, the given two UITEs, i.e., (90), have a unique common solution in V.(2)
IfA2v1,v2p is the maximum term in A1v1,v2p,A2v1,v2p,A3v1,v2p,A4v1,v2p, then from (92), (93), and (99), we have that
(101)δJv1,Kv2≤μDv1+ℏ1−v12Ev2+ℏ2−v221+k12eicotk121+v1−v221+k12eicotk1,for all v1,v2∈V. Hence, the mappings J,K and f satisfy all the conditions of Theorem 9 with μ=η2 and η1=η3=η4=0 in (6). Then, the given two UITEs, i.e., (90), have a unique common solution in V.(3)
IfA3v1,v2p is the maximum term in A1v1,v2p,A2v1,v2p,A3v1,v2p,A4v1,v2p, then from (92), (93), and (99), we have that
(102)δJv1,Kv2≤μDv1+ℏ1−v12Ev2+ℏ2−v12+Ev2+ℏ2−v22Dv1+ℏ1−v221+k12eicotk1Ev2+ℏ2−v12+Dv1+ℏ1−v22,for all v1,v2∈V. Hence, the mappings J,K and f satisfy all the conditions of Theorem 9 with μ=η3 and η1=η2=η4=0 in (6). Then, the given two UITEs, i.e., (90), have a unique common solution in V.(4)
IfA4v1,v2p is the maximum term in A1v1,v2p,A2v1,v2p,A3v1,v2p,A4v1,v2p, then from (93), we have that
(103)Mv1,v2=A4v1,v2p,
Then, there are furthermore four subcases arise:(i)
Ifa1v1,v2p is the maximum term in a1v1,v2p,a2v1,v2p,a3v1,v2p,a4v1,v2p. Then, from (92), (97), (99), and (103), we have that(104)δJv1,Kv2≤μDv1+ℏ1−v121+k12eicotk1,for all v1,v2∈V. Hence, the mappings J,K and f satisfy all the conditions of Theorem 9 with μ=η4 and η1=η2=η3=0 in (6). Then, the given two UITEs, i.e., (90), have a unique common solution in V.
(ii)
Ifa2v1,v2p is the maximum term in a1v1,v2p,a2v1,v2p,a3v1,v2p,a4v1,v2p. Then, from (92), (97), (99), and (103), we have that(105)δJv1,Kv2≤μEv2+ℏ2−v221+k12eicotk1,for all v1,v2∈V. Hence, the mappings J,K and f satisfy all the conditions of Theorem 9 with μ=η4 and η1=η2=η3=0 in (6). Then, the given two UITEs, i.e., (90), have a unique common solution in V.
(iii)
Ifa3v1,v2p is the maximum term in a1v1,v2p,a2v1,v2p,a3v1,v2p,a4v1,v2p. Then, from (92), (97), (99), and (103), we have that(106)δJv1,Kv2≤μEv2+ℏ2−v121+k12eicotk1,for all v1,v2∈V. Hence, the mappings J,K and f satisfy all the conditions of Theorem 9 with μ=η4 and η1=η2=η3=0 in (6). Then, the given two UITEs, i.e., (90), have a unique common solution in V.
(iv)
Ifa4v1,v2p is the maximum term in a1v1,v2p,a2v1,v2p,a3v1,v2p,a4v1,v2p. Then, from (92), (97), (99), and (103), we have that(107)δJv1,Kv2≤μDv1+ℏ1−v221+k12eicotk1,for all v1,v2∈V. Hence, the mappings J,K and f satisfy all the conditions of Theorem 9 with μ=η4 and η1=η2=η3=0 in (6). Then, the given two UITEs, i.e., (90), have a unique common solution in V.


## 5. Conclusions

We proved some unique CFP-theorems in complex-valued b-metric spaces under the more generalized rational type contraction conditions for compatible three self-mappings in which a one self-map is continuous. Our results extend and improved many results given in the literature (e.g., see [26, 23]). In the support of our work, we presented some illustrative examples for three self-mappings in complex-valued b-metric spaces. Moreover, we presented an application of the two UITEs to get the existing result of a common solution to support our main work. In this direction, many results can be contributed to the complex-valued b-metric spaces by using different contractive type single-valued mappings with different types of applications.



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*Source: 9938959-2021-05-26.xml*