# Exciting Fixed Point Results on a Novel Space with Supportive Applications

Authors: Hasanen A. Hammad; Hassen Aydi; Yaé Ulrich Gaba
Journal: Journal of Function Spaces
 (2021)

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## Abstract

The goal of this paper is to present a new space, a complex valued controlled rectangularb-metric space (for short, υℂ-metric space). Some examples and topological properties of υℂ-metric spaces are given. Also, some related common fixed point results are discussed. Our results generalize a lot of works in this direction. Moreover, we apply the theoretical results to find a unique solution of a complex valued Atangana-Baleanu fractional integral operator and a system of complex linear equations. Finally, a numerical example to find the current that passes through the RLC circuit is illustrated.


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## Body

## 1. Introduction

In 2011, the concept of a complex valued metric space has been initiated by Azam et al. [1]. Later, Al-Mezel et al. [2] emphasized that this space is just a particular case of cone metric spaces [3, 4]. Several fixed point results have been investigated on complex valued metric spaces; see [5–9]. The notion of a b-metric space, as a generalization of metric spaces, was introduced by Bakhtin [10] and Czerwik [11]. Combining the above concepts, Rao et al. [12] initiated the notion of complex valued b-metric spaces. For related results on this setting, see [13, 14]. Going in the same direction, recently, Ullah et al. [15] presented complex valued extended b-metric spaces to extend the idea of extended b-metric spaces [16].On the other hand, in the last few years, the fixed point theory became an essential tool to resolve several problems in nonlinear analysis, including fractional calculus. The contribution of our paper is to solve a complex valued Atangana-Baleanu fractional integral operator and a system of complex linear equations. The idea is to introduce the concept of a new space, the so-called complex valued controlled rectangularb-metric space (for short, υℂ-metric space). We also provide a numerical example illustrating the fact that the current passes through the RLC circuit.


## 2. Statements aboutυℂ-Metric Spaces

Letℂ be the set of complex numbers and η1,η2∈ℂ. Define the below partial order ≾ on ℂ as
(1)η1≾η2iffReη1≤Reη2andImη1≤Imη2.Hence,η1≾η2 if one of the below statements holds:ℷ1Reη1=Reη2andImη1<Imη2ℷ2Reη1<Reη2andImη1=Imη2ℷ3Reη1<Reη2andImη1<Imη2ℷ4Reη1=Reη2andImη1=Imη2Also, we can writeη1⋦η2 if η1≠η2 and one of ℷ1, ℷ2, and ℷ3 holds. By following the same pattern, we write η1≺η2 if only ℷ3 is fulfilled. This implies thatℏ10≾η1⋦η2⇒η1<η2ℏ2η1≾η2andη2≺η3⇒η1≺η3ℏ30≾η1≾η2⇒η1<η2ℏ4ifl,m∈ℝ,0≤l≤mandη1≾η2,thenlη1≾mη2This important part is devoted to present new definitions and examples of aυℂ-metric space.Definition 1.
A mappingυℂ:∇×∇→ℂis called aυℂ-metric on a nonempty set∇, if it satisfies the following hypotheses, for alll,m∈∇,
υℂ1υℂl,m=0⇔l=m
υℂ2υℂl,m=υℂm,l
υℂ3 there is a function ρ:∇×∇→1,∞ such that υℂl,m≾ρl,mυℂl,i+υℂi,j+υℂj,m (the so-called controlled rectangular inequality), for all distinct points i,j∈∇\l,mThen,∇,υℂ is called a υℂ-metric space.Example 1.
Assume that∇,ωis a complex valued extended rectangularb-metric with coefficients≥1andτ:0,∞→0,∞is a strictly increasing continuous function withμ≤τμ, for allμ>0andτ0=0. Suppose thatυℂm,l=τωm,lfor alll,m∈∇. For any two distinct pointsi,j∈∇\l,m, one can obtain that(2)υℂm,l=τωm,l≾τsωl,i+sωi,j+sωj,m≾τsτωl,i+sτωi,j+sτωi,j=ρsυℂl,i+sυℂi,j+sυℂj,m.Thus, the pair∇,υℂ is a υℂ-metric space with ρμ=τsμ.Example 2.
Let∇=0,∞. Defineυℂ:∇×∇→ℂ, ρ:∇→0,∞byυℂm,l=m−il2, andρl,m=l+m+3ℓ−1, ℓ>1, respectively. The conditionsυℂ1andυℂ2are clear and the conditionυℂ3follows immediately by the Jensen inequality(3)υℂm,l=m−il2≾3ℓ−1m−ip2+p−iq2+q−il2≾l+m+3ℓ−1m−ip2+p−iq2+q−il2=ρυℂm,p+υℂp,q+υℂq,l,for all l,m,p,q∈∇. This implies that ∇,υℂ is a υℂ-metric space with ρl,m=l+m+3ℓ−1. Note that in general υℂ is not continuous.Example 3.
Let∇=0,1. Defineυℂ:∇×∇→ℂ, ρ:∇×∇→1,∞byυℂm,l=4/3m−l2+im−l2, andρl,m=l+m+4, respectively. The conditionsυℂ1andυℂ2are trivial. We only show the conditionυℂ3. For alll,m,p,q,r,s∈∇, we have(4)υℂm,l=43m−l2+im−l2=43m−p+p−q+q−l2+im−p+p−q+q−l2≾43m−p2+p−q2+q−l2+2m−pp−q+2m−pq−l+2p−qq−l+im−p2+p−q2+q−l2+2m−pp−q+2m−pq−l+2p−qq−l≾43m−p2+p−q2+q−l2+m−p2+p−q2+m−p2+q−l2+p−q2+q−l2+im−p2+p−q2+q−l2+m−p2+p−q2+m−p2+q−l2+p−q2+q−l2=4m−p2+p−q2+q−l2+im−p2+p−q2+q−l2≾l+m+443m−p2+im−p2+43p−q2+ip−q2+43q−l2+iq−l2=ρl,mυℂm,p+υℂp,q+υℂq,l.Thus,∇,υℂ is a υℂ-metric space with ρl,m=l+m+4.Example 4.
Let∇=Y1∪Y2, whereY1=1/n:n∈ℕandY2=0,2, and choose a functionρl,m=l+m+2for alll,m∈∇. Define a functionυℂ:∇×∇→ℂby(5)vcl,m=0,ifm=l,i,ifl≠m,l,m∈ϒ1orl,m∈ϒ2,im,ifl∈ϒ2andm∈ϒ1,l,ifl∈ϒ1andm∈ϒ2.It is clear that the pair∇,υℂ is a υℂ-metric space with ρl,m=l+m+2. If we take the sequence 1/nn∈ℕ, it is easy to prove that this sequence converges to both 0 and 2, so it is not a Cauchy sequence. Moreover, υℂ is not continuous, since
(6)limn→∞1n=0,limn→∞υ1n,13=i≠i3=υ0,13.This means that there exists a sequence in∇,υℂ which is not convergent to a unique point in ∇. Hence, ∇,υℂ is not metrizable.Example 5.
Assume that∇=1,2,3,4. Defineυℂ:∇×∇→ℂby(7)υℂ1,3=υℂ3,1=1−i,υℂ1,4=υℂ4,1=1+i,υℂ1,2=υℂ2,1=4+4i,υℂ2,3=υℂ3,2=i,υℂ2,4=υℂ4,2=2+2i,υℂ1,1=υℂ2,2=υℂ3,3=υℂ4,4=0.υℂ3,4=υℂ4,3=1+2i,Defineρ:∇×∇→1,∞ by
(8)ρ1,3=ρ3,1=1,ρ1,4=ρ4,1=32,ρ1,2=ρ2,1=2,ρ2,3=ρ3,2=43,ρ2,4=ρ4,2=54,ρ3,4=ρ4,3=87.Now, we examine the conditionυℂ3 in the following cases:
(i)
Whenl=m, then the result follows immediately(ii)
Whenl=1,m=2 and i=3,j=4, then(9)υℂl,m=υℂ1,2=4+4i≾8+6i=24+3i=ρl,mυℂl,i+υℂi,j+υℂj,m.(iii)
Whenl=1,m=3 and i=2,j=4, then(10)υℂl,m=υℂ1,3=1−i≾7+8i=ρl,mυℂl,i+υℂi,j+υℂj,m.(iv)
Whenl=1,m=4 and i=2,j=3, then(11)υℂl,m=υℂ1,4=1+i≾152+212i=325+7i=ρl,mυℂl,i+υℂi,j+υℂj,m.(v)
Whenl=2,m=3 and i=1,j=4, then(12)υℂl,m=υℂ2,3=i≾8+283i=436+7i=ρl,mυℂl,i+υℂi,j+υℂj,m.(vi)
Whenl=2,m=4 and i=1,j=3, then(13)υℂl,m=υℂ2,4=2+2i≾152+254i=546+5i=ρl,mυℂl,i+υℂi,j+υℂj,m.(vii)
Whenl=3,m=4 and i=1,j=2, then(14)υℂl,m=υℂ3,4=1+2i≾8+407i=877+5i=ρl,mυℂl,i+υℂi,j+υℂj,m.Note that(15)υℂl,m=υℂ1,2=4+4i≻4+3i=ρl,mυℂl,i+υℂi,j+υℂj,m.Thus, according to the above cases,υℂ is not a complex valued rectangular b-metric for the function ρ, but υℂ is a υℂ-metric (conditions υℂ1 and υℂ2 are trivial).Example 6.
Let∇=ℂnand defineυℂl,m=4/3max1≤t≤nlt−mt2, wherel,m∈∇, then∇,υℂis a completeυℂ-metric space with the functionρl,m=4+l+m.Verification: initially, we show that∇,υℂ is a υℂ-metric space by discussing the following conditions. Assume that l,m∈∇ and i,j∈∇\l,m, where i≠j.υℂ1 Since lt−mt≽0, one writes 4/3max1≤t≤nlt−mt≽0. So that υℂl,m±0.⇒ Let υℂl,m=0, then 4/3max1≤t≤nlt−mt=0⇒lt=mt for 1≤t≤n⇒l=m.⇐ If l=m, then for each 1≤t≤n, lt−mt=0⇒4/3max1≤t≤nlt−mt=0⇒υℂl,m=0.υℂ2υℂl,m=4/3max1≤t≤nlt−mt=4/3max1≤t≤nmt−lt=υℂm,l.υℂ3 Since for all 1≤t≤n, one gets
(16)43lt−mt2=43lt−it+it−jt+jt−mt2≾4lt−it2+it−jt2+jt−mt2≾4+m+l43lt−it2+43it−jt2+43jt−mt2.Taking the maximum on two sides for all1≤t≤n, we have
(17)43max1≤t≤nlt−mt2≾4+m+l43max1≤t≤nlt−it2+43max1≤t≤nit−jt2+43max1≤t≤njt−mt2.This leads toυℂl,m≾ρl,mυℂl,i+υℂi,j+υℂj,m. From the above condition, we conclude that ∇,υℂ is a υℂ-metric space.Finally, we show the completeness property of∇,υℂ. It means that for every Cauchy sequence in ∇,υℂ is convergent. Let mn be a Cauchy sequence in ∇,υℂ, then for every ε≻0, there exists n∘∈ℕ so that
(18)υℂmn,mj≺ε,∀n,j>n∘.That is, for1≤t≤n, it satisfies
(19)υℂmn,mj=43max1≤t≤nmtn−mtj2≺ϵ.Hence, for everyt, lt1,lt2,⋯ is a Cauchy sequence in ℂ. Since ℂ is complete, ltn→lt for n→∞. Now, define m=m1,m2,⋯∈ℂn so that mn→m. Passing j→∞ in (19), we have 4/3max1≤t≤nmtn−mt2≺ϵ, for 1≤t≤n, i.e., υℂmn,m≺ϵ. This implies that mn→m, so ∇,υℂ is complete.Now, we introduce topological properties of aυℂ-metric space.Definition 2.
Assume that∇,υℂis aυℂ-metric space,mnis a sequence in∇andm∈∇. For everyc∈ℂwith0≺c, there existsN∈ℕsuch that(i)
for alln≥ℕ, υℂmn,m≺c, then mn is called convergent, mn converges to m, or we say that m is a limit point of mn and write limn→∞mn=m or mn→n→∞m(ii)
for alln≥ℕ, υℂmn,mn+j≺c, j∈ℕ, then mn is called a Cauchy sequence(iii)
if every Cauchy sequence in∇ is convergent, then the pair ∇,υC is called a complete υℂ-metric spaceLemma 1.
Assume that∇,υℂis aυℂ-metric space andmnis a sequence in∇, thenmnis a Cauchy sequence iffυℂmn,mj→n,j→∞0, wheren,j∈ℕ.Proof.
⇒ Let mn be a Cauchy sequence in ∇, then for every ϵ≻0, ∃n∘∈ℕ so that
(20)υℂmn,mj≺ϵ,∀n,j>n∘.Sinceε∈ℂ, there are α,β∈ℝ such that ϵ=α+iβ. Consider δ=ϵ=α2+β2, and take the absolute value in two sides of (20), we get υℂmn,mj≺δ. From definition of the limit, we get υℂmn,mj→0 as n,j→∞.⇐ Let υℂmn,mj→0 as n,j→∞. It follows that υℂmn,mj≺δ, for every δ>0. Since υℂmn,mj∈ℂ, there exists ϵ∈ℂ such that δ=ϵ. It holds that υℂmn,mj≺ϵ, for every ϵ≻0. Thus, there is n∘∈ℕ so that for n,j>n∘, mn is a complex valued Cauchy sequence.Lemma 2.
Assume that∇,υℂis aυℂ-metric space andmnis a sequence in∇,thenmis the limit point ofmniffυℂmn,m→n→∞0.Proof.
Passingj→∞ in the proof of Lemma 1, we get the result.Remark 1.
If we takeυℂl,m=s≥1, for allm,l∈∇, then anυℂ-metric space reduces to a complex valued rectangularb-metric space.The below lemma is very important in the equal.Lemma 3.
Let∇,υℂbe aυℂ-metric space andmnbe a Cauchy sequence in∇such thatmn≠mj, whenevern≠j. Then,mnconverges at most to one point.Proof.
Suppose that the sequencemn has two limit points m,m∗∈∇, that is, limn→∞υℂmn,m=0=limn→∞υℂmn,m∗. Since mn is Cauchy, from υℂ3, for mn≠mj, whenever n≠j, we can write
(21)υℂm,m∗≾ρm,m∗υℂm,mn+υℂmn,mj+υℂmj,m∗→0asn,j→∞.It follows thatυℂm,m∗=0, i.e., m=m∗. Thus, mn converges at most to one point.The following diagram shows the latest generalizations of complex valued metric spaces. However, the inverse implications need not be true.(22)c.v.m.s→c.v.b‐m.s→extendedc.v.b‐m.s↓↓↓c.v.rectangularm.s→c.v.rectangularb‐m.s→c.v.controlledrectangularb‐m.swhere c.v.m.s. refers to a complex valued metric space.


## 3. Exciting Fixed Point Results onυℂ-Metric Spaces

This section is aimed at introducing two fixed point theorems in the context ofυℂ-metric spaces.Theorem 1.
Let∇,υℂbe a completeυℂ-metric space and let the mappingsΘ,Ξ:∇→∇satisfy(23)κμυℂΘl,Ξm≾ζυℂl,m+ϖυℂl,Θlυℂm,Ξm1+υℂl,m,∀l,m∈∇,where 0≺υℂl,m, κ,μ≥1 and ζ,ϖ are nonnegative real numbers with ζ+ϖ<1.Assume also(24)limn→∞ρmn,mn+1<1D,whereD=ζκμ−ϖ<1,and for all m∈∇,limn→∞ρmn,m exists and is finite. Then, Θ and Ξ have a common fixed point (cfp).Proof.
Letm0 be an arbitrary point such that the sequence mn=Ξnm∘ in ∇ satisfies the conditions of theorem and iterates below as follows:
(25)m2n+1=Θm2nandm2n+2=Ξm2n+1,n≥0.
Assume that(26)κμυℂm2n+1,m2n+2=κμυℂΘm2n,Ξm2n+1ζυℂm2n,m2n+1≾ϖυℂm2n,Θm2nυℂm2n+1,Ξm2n+11+υℂm2n,m2n+1=ζυℂm2n,m2n+1+ϖυℂm2n,m2n+1υℂm2n+1,m2n+21+υℂm2n,m2n+1ζυℂ≾ϖυℂm2n+1,m2n+2.This implies that
(27)υℂm2n+1,m2n+2≾ζκμ−ϖυℂm2n,m2n+1.Similarly,(28)υℂm2n+2,m2n+3≾ζκμ−ϖυℂm2n+1,m2n+2.Now, withζ/κμ−ϖ=D, we can write
(29)υℂmn,mn+1≾Dυℂmn−1,mn≾D2υℂmn−2,mn−1⋮≾Dnυℂm0,m1.Now, we want to show thatmn is a Cauchy sequence. So, we distinguish two cases.Case 1.
Letj be odd, that is, j=2k+1 for any k≥1. Now, by υℂ3 for any n∈ℕ, we get
(30)υℂmn,mn+j=υℂmn,mn+2k+1≾ρmn,mn+2k+1υℂmn,mn+1+υℂmn+1,mn+2+υℂmn+2,mn+2k+1≾ρmn,mn+2k+1υℂmn,mn+1+υℂmn+1,mn+2+ρmn,mn+2k+1υℂmn+2,mn+2k+1≾ρmn,mn+2k+1Dnυℂm0,m1+Dn+1υℂm0,m1+ρmn,mn+2k+1υℂmn+2,mn+2k+1+ρmn,mn+2k+1Dn+Dn+1υℂm0,m1+ρmn,mn+2k+1υℂmn+2,mn+2k+1=ρmn,mn+2k+1Dn+Dn+1υℂm0,m1≾ρmn,mn+2k+1ρmn+2,mn+2k+1Dn+2+Dn+3υℂm0,m1+.⋯+ρmn,mn+2k+1⋯ρmn+2k−2,mn+2k+1Dn+2k−2+Dn+2k−1υℂm0,m1+ρmn,mn+2k+1⋯ρmn+2k−2,mn+2k+1Dn+2kυℂm0,m1=Dn1+Dυℂm0,m1∑i=0k−1D2i∏t=0iρmn+2t,mn+2k+1+Dn+2k∏t=0k−1ρmn+2t,mn+2k+1υℂm0,m1.This implies that(31)∑i=0k−1D2i∏t=0iρmn+2t,mn+2k+1≤∑i=0k−1D2i∏t=0iρm2t,mn+2k+1.As, in view of (24), we get limn,k→∞υℂmn,mkD<1; therefore, by the ratio test, we conclude the convergence of the series ∑i=0∞D2i∏t=0iρm2t,mn+2k+1 for all k∈ℕ. Consider
(32)R=∑i=0∞D2i∏t=0iρm2t,mn+2k+1,Rn=∑i=0nD2i∏t=0iρm2t,mn+2k+1.It follows by (30) that
(33)υℂmn,mn+2k+1≾Dn1+Dυℂm0,m1Rk−1−Rn−1+Dn+2k∏t=0k−1ρmn+2t,mn+2k+1υℂm0,m1.Passingn→∞ in (33), we can write
(34)limn→∞υℂmn,mn+2k+1=0.Case 2.
Letj be even, that is, j=2k for any k≥1. Then,
(35)υℂmn,mn+j=υℂmn,mn+2k≾ρmn,mn+2kυℂmn,mn+1+υℂmn+1,mn+2+υℂmn+2,mn+2k≾ρmn,mn+2kυℂmn,mn+1+υℂmn+1,mn+2+ρmn,mn+2kυℂmn+2,mn+2k≾ρmn,mn+2kDnυℂm0,m1+Dn+1υℂm0,m1+ρmn,mn+2kυℂmn+2,mn+2k=ρmn,mn+2kDn+Dn+1υℂm0,m1+ρmn,mn+2kυℂmn+2,mn+2k+1≾ρmn,mn+2kDn+Dn+1υℂm0,m1+ρmn,mn+2kρmn+2,mn+2kDn+2+Dn+3υℂm0,m1+.⋯+ρmn,mn+2k⋯ρmn+2k−2,mn+2kDn+2k−2+Dn+2k−1υℂm0,m1+ρmn,mn+2k⋯ρmn+2k−2,mn+2kDn+2kυℂm0,m1=Dn1+Dυℂm0,m1∑i=0k−1D2i∏t=0iρmn+2t,mn+2k+Dn+2k−2∏t=0k−1ρmn+2t,mn+2kυℂm0,m2.Similar to Case1, we conclude that
(36)υℂmn,mn+2k≾Dn1+Dυℂm0,m1Rk−1−Rn−1Dn+2k−2∏t=0k−1ρmn+2t,mn+2kυℂm0,m2.Applying the limit on (36) as n→∞, we have
(37)limn→∞υℂmn,mn+2k=0.It follows from (34) and (37) that limn→∞υℂmn,mn+j=0, j∈ℕ, which illustrates that the sequence mn is Cauchy in ∇. The completeness of ∇,υℂ leads that there exists an element ξ∈∇ such that mn is complex valued convergent to ξ, i.e.,
(38)limn→∞υℂmn,ξ=0.By using controlled rectangular inequalityυℂ3, we can write
(39)υℂξ,mn+2≾ρξ,mn+2υℂξ,mn+υℂmn,mn+1+υℂmn+1,mn+2.Passing the limit asn→∞ in (39) and using (38), we have
(40)limn→∞υℂmn+2,ξ=0=limn→∞υℂmn+1,ξ.Now, let us demonstrate thatξ=Θξ, that is, ξ is a fixed point of Θ; otherwise, 0≺υℂΘξ,ξ, and for n∈ℕ, by (23), we have
(41)κμυℂΘξ,ξ≾κμρΘξ,ξυℂΘξ,mn+2+υℂmn+2,mn+1+υℂmn+1,ξ=κμρΘξ,ξυℂmn+2,mn+1+υℂmn+1,ξ+κμρΘξ,ξυℂΘξ,Ξmn+1≾κμρΘξ,ξυℂmn+2,mn+1+υℂmn+1,ξ+κμρΘξ,ξζυℂξ,mn+1+ϖυℂξ,Θξυℂmn+1,Ξmn+11+υℂξ,m≾κμρΘξ,ξυℂmn+2,mn+1+υℂmn+1,ξ+κμρΘξ,ξζυℂξ,mn+1+ϖυℂξ,Θξυℂmn+1,mn+21+υℂξ,mn+1.Taking the limit in (41) as n→∞ and applying (38) and (40), we conclude that υℂΘξ,ξ=0, a contradiction. Hence, Θξ=ξ. Similarly, we can prove ξ=Ξξ. Observe that, in view of Lemma 3, the sequence mn converges uniquely at point ξ∈∇.To proveξ is the unique cfp of Θ and Ξ, suppose that η is a distinct cfp of Θ and Ξ, then 0≺υℂξ,η. So from (23), we get
(42)κμυℂξ,η=κμυℂΘξ,Ξη≾ζυℂξ,η+ϖυℂξ,Θξυℂη,Ξη1+υℂξ,η=ζυℂξ,η+ϖυℂξ,ξυℂη,η1+υℂξ,η=ζυℂξ,η⇒κμ−ζζυℂξ,η¶0.This implies thatυℂξ,η=0, a contradiction. Thus, ξ=η. Hence, Θ and Ξ have a unique cfp. This finishes the proof.The following theorem is more general than Theorem1.Theorem 2.
Let∇,υℂbe a completeυℂ-metric space and let the mappingsΘ,Ξ:∇→∇satisfy(43)κμυℂΘl,Ξm≾αmaxυℂl,m,υℂl,Θlυℂm,Ξm1+υℂl,m,υℂl,Ξmωm,Θl1+υℂl,m,for all l,m∈∇,κ,μ≥1 and ζ∈0,1, where 0≺υℂl,m. If limn→∞ρmn,mn+1<1/A, where A=α/κμ<1 and for all m∈∇,limn→∞ρmn,m exists and is finite. Then, Θ and Ξ have a unique cfp.Proof.
Letm∘∈∇ be an arbitrary point. Define a sequence mn as follows:
(44)m2n+1=Θm2nandm2n+2=Ξm2n+1,n≥0.Consider(45)κμυℂm2n+1,m2n+2=κμυℂΘm2n,Ξm2n+1≾αmaxυℂm2n,m2n+1,υℂm2n,Θm2nυℂm2n+1,Ξm2n+11+υℂm2n,m2n+1,υℂm2n,Ξm2n+1υℂm2n+1,Θm2n1+υℂm2n,m2n+1≾αmaxυℂm2n,m2n+1,υℂm2n,m2n+1υℂm2n+1,m2n+21+υℂm2n,m2n+1,υℂm2n,m2n+2υℂm2n+1,m2n+11+υℂm2n,m2n+1≾αmaxυℂm2n,m2n+1,υℂm2n+1,m2n+2.Ifυℂm2n,m2n+1¶υℂm2n+1,m2n+2, we can write
(46)κμυℂm2n+1,m2n+2≾αυℂm2n+1,m2n+2,⇒κμ−αυℂm2n+1,m2n+20.This is a contradiction. Hence,(47)κμυℂm2n+1,m2n+2≾αυℂm2n,m2n+1,or
(48)υℂm2n+1,m2n+2≾ακμυℂm2n,m2n+1=Aυℂm2n,m2n+1.Similarly, we can obtain(49)υℂm2n+2,m2n+3≾Aυℂm2n+1,m2n+2.In general, we have(50)υℂmn,mn+1≾Aυℂmn−1,mn≾A2υℂmn−2,mn−1⋮≾A2υℂm0,m1.Using the same scenario as Cases1 and 2 of Theorem 1, we conclude that the sequence mn is Cauchy in ∇. Since ∇,υℂ is complete, there is an element σ∈∇ such that
(51)limn→∞υℂmn,σ=0.By Lemma3, a sequence mn converges uniquely at point σ∈∇. By the condition υℂ3, we get
(52)υℂσ,mn+2≾ρσ,mn+2υℂσ,mn+υℂmn,mn+1+υℂmn+1,mn+2.Passing the limit asn→∞ in (52) and using (51), we have
(53)limn→∞υℂmn+2,σ=0=limn→∞υℂmn+1,σ.Now, we show thatξ is a fixed point of Θ; otherwise, 0≺υℂΘξ,ξ, and for n∈ℕ, by (43), we can get
(54)κμυℂΘσ,σ≾κμρΘσ,συℂΘσ,mn+2+υℂmn+2,mn+1+υℂmn+1,σ=κμρΘσ,συℂmn+2,mn+1+υℂmn+1,ξ+κμρΘσ,συℂΘσ,Ξmn+1≾κμρΘσ,συℂmn+2,mn+1+υℂmn+1,σ+ακμρΘσ,σmaxυℂσ,mn+1,υℂσ,Θσυℂmn+1,Ξmn+11+υℂσ,mn+1,υℂσ,Ξmn+1ωmn+1,Θσ1+υℂσ,mn+1≾κμρΘσ,συℂmn+2,mn+1+υℂmn+1,σ+ακμρΘσ,σmaxζυℂξ,mn+1+ϖυℂσ,Θσυℂmn+1,mn+21+υℂξ,mn+1,υℂσ,mn+2ωmn+1,Θσ1+υℂσ,mn+1.After entering the limit asn→∞ in (54) and using (51) and (53), we conclude that υℂΘσ,σ=0, a contradiction. Hence, Θσ=σ. Similarly, we can prove σ=Ξσ.The uniqueness comes easy from the condition (51), and it is similar to Theorem 1. Therefore, the mappings Θ and Ξ have a unique cfp in ∇. This ends the proof.


## 4. Examples and Some Sequences on the Main Theorems

This part is devoted to obtain some sequences of Theorems1 and 2.If we considerΘ=Ξ in Theorem 1, we have the below result.Corollary 1.
Let∇,υℂbe a completeυℂ-metric space and let the mappingΘ:∇→∇satisfy(55)κμυℂΘl,Θm≾ζυℂl,m+ϖυℂl,Θlυℂm,Θm1+υℂl,m,∀l,m∈∇,where 0≺υℂl,m,κ,μ≥1 and ζ,ϖ are nonnegative real numbers with ζ+ϖ<1.Assume also(56)limn→∞ρmn,mn+1<1D,where D=ζ/κμ−ϖ<1 and limn→∞ρmn,m exists and is finite for all m∈∇. Then, Θ has a unique fixed point.Assume thatκ=μ=1 and ϖ=0 in Corollary 1, we get the important below result.Corollary 2.
Let∇,υℂbe a completeυℂ−metric space and let the mappingΘ:∇→∇satisfy(57)υℂΘl,Θm≾ζυℂl,m,∀l,m∈∇,where ζ is a non-negative real number with ζ<1. Assume also
(58)limn→∞ρmn,mn+1<1D,where D=ζ/κμ−ϖ<1 and limn→∞ρmn,m exists and is finite for all m∈∇. Then, Θ has a unique fixed point.Remark 2.
If we takeΘ=Ξin Theorem2, we have the following immediate consequences.Corollary 3.
Let∇,υℂbe a completeυℂ-metric space and let the mappingΘ:∇→∇satisfy(59)κμυℂΘl,Θm≾αmaxυℂl,m,υℂl,Θlυℂm,Θm1+υℂl,m,υℂl,Θmωm,Θl1+υℂl,m,for all l,m∈∇,κ,μ≥1 and ζ∈0,1, where 0≺υℂl,m. Assume that limn→∞ρmn,mn+1<1/A, where A=α/κμ<1 and for all m∈∇,limn→∞ρmn,m exists and is finite. Then, Θ has a unique fixed point.Corollary 4.
Let∇,υℂbe a completeυℂ-metric space and letΘ:∇→∇. Suppose that there existsn∈ℕsuch that(60)κμυℂΘnl,Θnm≾αmaxυℂl,m,υℂl,Θnlυℂm,Θnm1+υℂl,m,υℂl,Θnmυℂm,Θnl1+υℂl,m,for all l,m∈∇,κ,μ≥1 and ζ∈0,1, where 0≺υℂl,m. Assume that limn→∞ρmn,mn+1<1/A, where A=α/κμ<1 and for all m∈∇,limn→∞ρmn,m exists and is finite. Then, Θ possesses a unique fixed point.Proof.
By Corollary3, we have κ∈∇ such that
(61)Θnκ=κ.So(62)κμυℂΘκ,κ=κμυℂΘΘnκ,Θnκ=κμυℂΘnΘκ,Θnκ≾αmaxυℂΘκ,κ,υℂΘκ,ΘnΘκυℂκ,Θnκ1+υℂΘκ,κ,υℂΘκ,Θnκυℂκ,ΘnΘκ1+υℂΘκ,κ≾αmaxυℂΘκ,κ,υℂΘκ,ΘΘnκυℂκ,Θnκ1+υℂΘκ,κ,υℂΘκ,Θnκυℂκ,ΘΘnκ1+υℂΘκ,κ=αυℂΘκ,κ.Therefore, the result follows.The below examples support Theorems1 and 2.Example 7.
Assume that all requirements of Example3hold. It is clear that∇,υℂis a completeυℂ-metric space withρl,m=l+m+4. Define the mappingsΘ,Ξ:∇→∇byΘl=1/2landm=1/2m,m,l∈∇. We Only prove the contractive condition ((23)) of Theorem1withκ=2andμ=1. Indeed, for alll,m∈∇,(63)κμυℂΘl,Ξm=2υℂl2,m2=83l2−m22+il2−m22=23l−m2+il−m2≾23υℂl,m34+16υℂl,m≾ζυℂl,m+ϖυℂl,Θlυℂm,Ξm1+υℂl,m.Hence, condition (23) of Theorem 1 is satisfied with ζ=3/4 and ϖ=1/6 such that 3/4+1/6=11/12<1 and the point 0∈∇ is the unique cfp of Θ and Ξ.Example 8.
Suppose that the statements of Example2are fulfilled. DefineΘ,Ξ:∇→∇byΘl=l/4andΞm=m/4for alll,m∈∇.It is clear that∇,υℂis a completeυℂ-metric space withρl,m=l+m+3ℓ−1,ℓ>1.Now, we will go to fulfill condition ((43)) forκ=2andμ=1. We have(64)κμυℂΘl,Ξm=2υℂl4,m3=2l4−im42≾2142l−im2=18υℂl,m≾18maxυℂl,m,υℂl,Θlυℂm,Ξm1+υℂl,m,υℂl,Ξmωm,Θl1+υℂl,m.Hence, all required assumptions of Theorem2 are fulfilled with α=1/8<1 and 0 is the unique cfp of Θ and Ξ.Remark 3.
(i) Theorems1and2extend some previous works in the different spaces. Some of them are as follows.In Theorem1(i)
whenϖ=0,κμ=s>1 and Θ=Ξ, we have the work of Ege [17] in complex valued rectangular b-metric spaces, where ζ∈0,1/s(ii)
whenκ=μ=1 and Θ=Ξ, we have the exciting results of Azam [1] in complex valued metric spaces(iii)
if we reduce a completeυℂ-metric space to a complete complex valued b-metric space with sζ+ϖ<1, we get directly the work of Mukheimer [13](iv)
whenκ=μ=1 and ζ,ϖ:ℂ+→0,1 are two mappings in a complete complex valued metric space, we have the results of Sintunavarat et al. [5]In Theorem2(i)
if we reduce a completeυℂ-metric space to a complete complex valued metric space with α∈0,1 and κ=μ=1, we obtain the results of Hammad [6](ii)
the conclusion of Theorems1 and 2 is still valid by using one of the following contractive conditions:(⋆1)
(65)κμυℂΘl,Θm≾ζυℂm,Θm+ϖυℂ2l,m1+υℂl,m,for all l,m∈∇,κ,μ≥1 and ζ,ϖ are nonnegative real numbers with ζ+ϖ<1,Θ=Ξ and 0≺υℂl,m.(⋆2)
(66)κμυℂΘl,Θm≾ζυℂl,m+ϖυℂl,Θlυℂm,Θmυℂl,m+υℂl,Θm+υℂm,Θl,for all l,m∈∇,κ,μ≥1 and ζ,ϖ are nonnegative real numbers with ζ+ϖ<1,Θ=Ξ and υℂl,m+υℂl,Θm+υℂm,Θl≠0.(⋆3)
(67)κμ∫0υℂΘl,Ξmγtdt≾ζ∫0υℂl,mγtdt+ϖ∫0υℂl,Θlυℂm,Ξm/1+υℂl,mγtdt,for all l,m∈∇,κ,μ≥1 and ζ,ϖ are nonnegative real numbers with ζ+ϖ<1, and 0≺υℂl,m, where γ:0,∞→0,∞ is a Lebesgue integrable mapping on a compact subset of 0,∞ and for all ϵ>0,∫0ϵγtdt>0. This direction involves fixed point results of integral type.(⋆4)
(68)κμeυℂΘl,Ξm≾αemaxυℂl,m,υℂl,Θlυℂm,Ξm1+υℂl,m,υℂl,Ξmωm,Θl1+υℂl,m,for all l,m∈∇,κ,μ≥1 and ζ∈0,1, where 0≺υℂl,m, where e:0,∞→0,∞ is a Lebesgue integrable mapping on a compact subset of 0,∞ and for all ϵ>0,eϵ>0. This direction concerns fixed point results of exponential type.


## 5. Solving a Complex Valued Atangana-Baleanu Fractional Integral Operator

Let∇ be the set of all complex analytic functions and υℂ:∇×∇→ℂ be defined as
(69)υℂmr,lr=43mr−lr2+imr−lr2,for all mr,lr∈∇. Also, define ρ:∇×∇→1,∞ by ρlr,mr=lr+mr+4. Then, ∇,υℂ is a complete υℂ-metric space. We will use the form cABYrυmr to express the complex valued Atangana-Baleanu integral (AB-fractional integral), where c is a fixed complex number, m is a complex function which is analytic on an open star-domain ℝ centered at c,υ∈ℂ and r∈ℝ\c. It is defined by
(70)cABYrυmr=1−υβυmr+υβυcRLYrυmr,where cRLYrυmr refers to the Reimann-Lioville fractional integral equation for a complex analytical function m.Another form of the complex valued Atangana-Baleanu integral is presented by Fernandez [18] as follows:
(71)cABYrυmr=12πiβυ∫H1−υu−r+υΓ1−υu−r1−υmudr,where u,r∈ℝ\c;υ∈ℂ/ℕ and H refers to the Hankel contour [19].Define the operatorΘAb:∇→∇ by
(72)ΘAbmr=12πiβυ∫H1−υu−r+υΓ1−υu−r1−υmudr.Now, we will seek for the existence and uniqueness of a fixed point for the operator (72), which is considered as the unique solution for the operator (70) under the following condition:
(73)1π2β2υυ−1r−ulnr−ur−u−Γ1−υu−rυ2≤1.Consider(74)υℂΘAbmr,ΘAbmr=43ΘAbmr−ΘAblr2+iΘAbmr−ΘAblr2=4312πiβυ∫H1−υu−r+υΓ1−υu−r1−υmudr−12πiβυ∫H1−υu−r+υΓ1−υu−r1−υludr2+i12πiβυ∫H1−υu−r+υΓ1−υu−r1−υmudr−12πiβυ∫H1−υu−r+υΓ1−υu−r1−υludr2=4312πiβυ∫H1−υu−r+υΓ1−υu−r1−υdrmu−lu2+i12πiβυ∫H1−υu−r+υΓ1−υu−r1−υdrmu−lu2≾131πiβυ2∫H1−υu−r+υΓ1−υu−r1−υdr2mu−lu2+imu−lu2≾131π2β2υυ−1r−ulnr−ur−u−Γ1−υu−rυ2×43mu−lu2+imu−lu2≾13υℂmr,lr=ζυℂmr,lr.Thus, the conditions of Corollary2 are satisfied. Hence, ΘAB has a unique fixed point, which concludes that the fractional integral (70) has a unique solution.


## 6. Solving a System of Complex Linear Equations

In this part, we shall try to find a unique solution to a system of complex linear equations and apply it to solve the RLC circuit. Consider the below system:(75)ϑ11m1+ϑ12m2+⋯+ϑ1nmn=ℓ1ϑ21m1+ϑ22m2+⋯+ϑ2nmn=ℓ2⋮ϑn1m1+ϑn2m2+⋯+ϑnnmn=ℓn,in n linear equations and n variables, where m∈∇=ℂn and ϑts is the complex coefficients matrix. Assume that ∇=ℂn. Define υℂ:∇×∇→ℂ,ρ:∇→1,∞ by υℂl,m=4/3max1≤t≤nlt−mt2 and ρl,m=l+m+4, respectively. According to Example 6, the pair ∇,υℂ is a complete υℂ-metric space.The following theorem is the main result of this section.Theorem 3.
Suppose that∇,υℂis a completeυℂ-metric space, then the system (75) has only one solution if∑snϑts<0.46fort=1,2,⋯,n.Proof.
Define the linear mappingΘ:∇→∇ by Θm=Qm+L, where m=m1,m2,⋯,mn,L=ℓ1,ℓ2,⋯,ℓn∈ℂn and
(76)Q=ϑ11ϑ12⋯ϑ1nϑ21ϑ22⋯ϑ2n⋮⋮⋮⋮ϑ31ϑ32⋯ϑ3n.
If we can prove thatΘ is a contraction mapping, the proof is over. Consider κ=3 and μ=1, then we have
(77)κμυℂΘl,Θm=3×43max1≤t≤n∑snϑtsls+Lj−∑snϑtsms+Lj2=4max1≤t≤n∑snϑtsls−ms2≾4max1≤t≤n∑snϑts2ls−ms2≾4max1≤t≤n∑snϑts2×max1≤s≤nls−ms2≾0.8464max1≤s≤nls−ms2≾0.8464×43max1≤s≤nls−ms2=αυℂl,Θ≾αmaxυℂl,m,υℂl,Θlυℂm,Θm1+υℂl,m,υℂl,Θmωm,Θl1+υℂl,m,where α=0.8464<1, then Θ is a contraction mapping. Thus, by Corollary 3, the system (75) has a unique solution.To illustrate the usefulness of Theorem3, we give the following example.Example 9.
Consider the RLC circuit containing three as follows.
The symbolsR,C,V, and I refer to the resistance, capacity, voltage, and current, respectively. Also, let λ denote the imaginary number with current. Let us take the values V1=0.7+1.2λ and V2=1.65+0.3λ and ω=1. It is obvious that the RLC circuit forms a complex valued linear equation system that has one solution.
Verification: from Kirchhoff voltage law “the sum of all voltage in a closed loop is zero”; in this example, we observe it in three loops (Figure1). From our simple physical information, we know that V=IR=IΦC=IΦL, where ΦC=1/λωC and ΦL=λωL.(i)
From loop 1, we haveΦC1=1/λωC1=1/λ×1×9=1/9λ×λ/λ=−0.111λ and(78)−V1+R1I1+ΦC1I1−ΦC1I2=0,⇒−0.7+1.2λ+0.04I1−0.11λI1+0.111λI2=0,⇒0.04−0.111λI1+0.111λI2=0.7+1.2λ.(ii)
From loop 2, we haveΦC2=1/λωC2=1/7λ=−0.142λ,ΦL=λωL=0.14λ and(79)ΦC1I2−ΦC1I1+R2I2+ΦC2I2+ΦLI2−ΦLI3=0,⇒−0.11λI2+0.11λI1+0.1I2−0.142λI2+0.14λI2−0.14λI3≈0,⇒0.11λI1+0.1−0.112λI2−0.14λI3=0.(iii)
From loop 3, we haveΦC3=1/λωC3=1/24λ=−0.042λ, and ∑V=0; hence,(80)ΦLI3−ΦLI2+ΦC3I3+R3I3−V2=0,⇒0.14λI3−0.14λI2−0.042λI3+0.015I3−1.65+0.3λ=0,⇒−0.14λI2+0.015+0.098λI3=1.65+0.3λ.Figure 1
RLC circuit.So we have the below complex linear equation system:(81)0.04−0.111λI1+0.111λI2=0.7+1.2λ,0.11λI1+0.1−0.112λI2−0.14λI3=0,−0.14λI2+0.015+0.098λI3=1.65+0.3λ.By testing the sum of the coefficientsIpq, where p=1,2,3 and q=1,2,3, we have
(i)
forp=1,∑q=13I1q=0.042+0.1112+0.11≈0.223<0.46(ii)
forp=2,∑q=13I2q=0.111+0.12+0.1122+0.14≈0.401<0.46(iii)
forp=3,∑q=13I3q=0.14+0.0152+0.0982≈0.239<0.46Hence, according to Theorem3, the system (81) has a unique solution in the form of I1=−0.9001+19.8529λ, I2=2.7565+13.2222λ, and I3=6.5320+3.0520λ.


## 7. Conclusion

In this paper, we ensured the existence of a unique common fixed point for several contraction type self-mappings in the class of complex valued controlled rectangularb-metric spaces. Our presented results complement and generalize several significant known fixed point theorems in literature. As applications, we resolve a complex valued Atangana-Baleanu fractional integral operator and a system of complex linear equations, and we presented a related numerical example. We think our paper will be the source of motivation for researchers to improve and extend our results for new applications.



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