# Fixed Point Theorems for Mizoguchi-Takahashi Type Contraction in Bicomplex-Valued Metric Spaces and Applications

Authors: Fuli He; Z. Mostefaoui; M. Abdalla
Journal: Journal of Function Spaces
 (2020)

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## Abstract

The main aim of this paper is to study and establish some new fixed point theorems for contractive maps that satisfied Mizoguchi-Takahashi’s condition in the setting of bicomplex-valued metric spaces. These new results improve and generalize the Banach contraction principle and some well-known results in the literature. Finally, as applications of our results, we give the existence and uniqueness of the solution of a nonlinear integral equation.


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## Body

## 1. Introduction

Fixed point theorems concern mapsf of a set X into itself that, under certain conditions, admit a fixed point, that is, a point x∈X such that fx=x. The knowledge of the existence of fixed points has relevant applications in many branches of analysis and topology. In 2011, Azam et al. [1] introduced the so-called complex-valued metric spaces and proved the existence of fixed points under some contraction conditions. Very recently, Choi et al. [2] introduced the notion of bicomplex-valued metric space which is a generalization of the complex-valued metric space and established sufficient conditions for the existence of common fixed points of a pair of mappings satisfying a contractive condition.The following new fixed point theorem is one of the main results of this paper. It can be considered a bicomplex-valued metric version of the Banach contraction principle and will generalize and improve some well-known results in the literature. Finally, we strengthen our results by giving an application to find an analytical solution for a nonlinear integral equation. The results thus derived in this study are generally extended and generalized in character. Moreover, the results presented here may be very useful in the study of engineering problems like boundary value problems (see, e. g., [3–7]) in future works.


## 2. Preliminaries

In this section, we recall some definitions and terminologies, used to prove the main results.


### 2.1. Bicomplex Numbers

The set of bicomplex numbers denoted byBℂ is the first setting in an infinite sequence of multicomplex sets which are generalizations of the set of complex numbers ℂ. Here, we recall the set of bicomplex numbers Bℂ (see, for example, [8–10]):
(1)Bℂ=w=x0+i1x1+i2x2+i1i2x3;xk∈ℝ,k=0,1,2,3.Since each elementw in Bℂ can be written as w=x0+i1x1+i2x2+i1x3 or w=z1+i2z2;z1,z2∈ℂ, we can also express Bℂ as
(2)Bℂ≔w=z1+i2z2,z1,z2∈ℂ,where z1=x0+i1x1, z2=x2+i1x3, and i1,i2 are independent imaginary units such that i12=−1=i22. The product of i1 and i2 defines a hyperbolic unit j such that j2=1. The product of units is commutative and is defined as
(3)i1i2=j,i1j=−i2,i2j=−i1.With the addition and multiplication of two bicomplex numbers defined in the obvious way, the setBℂ makes up a commutative ring. Three important subsets of Bℂ can be specified as
(4)ℂik≔x+yik;x;y∈ℝ,k=1,2,D≔x+yj;x,y∈ℝ.Each of the setℂik is isomorphic to the field of complex numbers, while D is the set of the so-called hyperbolic numbers.


### 2.2. Conjugation and Moduli

Three kinds of conjugationw†1,w†2, and w†3 can be defined on bicomplex numbers. With the bicomplex number w specified as in (2) and the bar − denoting complex conjugation in ℂ, we define
(5)w†1≔z¯1+z¯2i2,w†2≔z1−z2i2,w†3≔z¯1−z¯2i2.It is easy to check that each conjugation has the following properties:(6)u+v†k=u†k+v†k,u†k†k=u,u.v†k=u†k.v†k.Here,u,v∈Bℂ and k=1,2,3.With each kind of conjugation, one can define a specific bicomplex modulus as(7)wi12≔w.w†2=z12+z22∈ℂi1w122≔w.w†1=z12−z22+2Rez1z¯2i2∈ℂi2wj2≔w.w†3=z12+z22−2Imz1z¯2j∈DIt can be shown thatu.vs2=us2.vs2, where s=i1,i2 or j.A norm of a bicomplex numberw=z1+i2z2 denoted by w is defined by
(8)w=z1+i2z2=z12+z221/2,which, upon choosing w=x0+i1x1+i2x2+i1i2x3,xk∈ℝ,k=0,1,2,3, gives
(9)w=x02+x12+x22+x321/2.For any two bicomplex numbersu,v∈Bℂ, one can easily verify that 0≺i2u≺i2v⟹u≤v,u+v≤u+v,αu=αu, where α is a nonnegative real number. Further, for any two bicomplex numbers, u,v∈Bℂ,uv≤2uv holds.Next, we recall some necessary definitions and lemmas.First, define a partial order≺¯ on Bℂ as follows:Letu=u1+i2u2∈Bℂ and v=v1+i2v2∈Bℂ. Define a partial order relation ≺¯i2 on Bℂ as follows:u≺¯i2v if and only if u1≺¯i1v1 and u2≺¯i1v2. We find that u≺¯i2v if any one of the following properties holds:
(10)a1u1=v1u2=v2a2u1≺i1v1u2=v2a3u1=v1u2≺i1v2a4u1≺i1v1u2≺i1v2where ≺¯i1 is a partial order relation on ℂ. So that
(11)b1Reu1=Rev1,Imu1=Imv1,Reu2=Rev2,Imu2=Imv2b2Reu1<Rev1,Imu1<Imv1,Reu2=Rev2,Imu2=Imv2b3Reu1=Rev1,Imu1=Imv1,Reu2<Rev2,Imu2<Imv2b4Reu1<Rev1,Imu1<Imv1,Reu2<Rev2,Imu2<Imv2We writeu≺∼i2v if u≺¯i2v and u≠v, i.e., one of b2, b3, and b4 is satisfied and we write u≺i2v if only b4 is satisfied.Given the definition of the complex metric space introduced in [1], we extend the definition of bicomplex analysis as follows.Definition 1.
LetXbe a nonempty set. A functionaldBℂ:X×X⟶Bℂis called a bicomplex-valued metric onXif forx,y,z∈Xthe following conditions are satisfied:(12)m10≺¯i2dBℂx,ym2dBℂx,y=0ifandonlyifx=ym3dBℂx,y=dBℂy,xm4dBℂx,y≺¯i2dBℂx,z+dBℂz,yThen,X,dBℂ is called a bicomplex-valued metric space.Some known examples of bicomplex-valued metric, which show a bicomplex-valued metric space, are the following.Example 2.
The set of real numbers together with the functional(13)dBℂ1x,y=1+i1+i2+i1i2x−y,dBℂ2x,y=i1i2x−y,for all x,y∈ℝ where ⋅ is the usual real modulus. One can easily check that X,dBℂ1, X,dBℂ2 are bicomplex-valued metric spaces.Definition 3.
Letxnbe a sequence in a bicomplex-valued metric spaceX,dBℂ. The sequencexnis said to converge tox∈Xif and only if for any0≺i2ε∈Bℂ, there existsN∈ℕdepending onεsuch thatdBℂxn,x≺i2εfor alln>N. It is denoted byxn⟶xasn⟶+∞orlimn→+∞xn=x.Definition 4.
A sequencexnin a bicomplex-valued metric spaceX,dBℂis said to be a Cauchy sequence if and only if for any0≺i2ε∈Bℂ, there existsN∈ℕdepending onεsuch thatdBℂxn,xm≺i2εfor alln,m>N.Definition 5.
A bicomplex-valued metric spaceX,dBℂis said to be complete if and only if every Cauchy sequence inXconverges inX.Lemma 6 (see [2]).
LetX,dBℂbe a bicomplex-valued metric space andxnbe a sequence inX. Then,xnconverges tox∈Xif and only ifdBℂxn,x⟶0asn⟶+∞.Lemma 7 (see [2]).
LetX,dBℂbe a bicomplex-valued metric space andxnbe a sequence inXsuch thatlimn→+∞xn=x. Then, for anya∈X, limn→+∞dBℂxn,a=dBℂx,a.


## 2.1. Bicomplex Numbers

The set of bicomplex numbers denoted byBℂ is the first setting in an infinite sequence of multicomplex sets which are generalizations of the set of complex numbers ℂ. Here, we recall the set of bicomplex numbers Bℂ (see, for example, [8–10]):
(1)Bℂ=w=x0+i1x1+i2x2+i1i2x3;xk∈ℝ,k=0,1,2,3.Since each elementw in Bℂ can be written as w=x0+i1x1+i2x2+i1x3 or w=z1+i2z2;z1,z2∈ℂ, we can also express Bℂ as
(2)Bℂ≔w=z1+i2z2,z1,z2∈ℂ,where z1=x0+i1x1, z2=x2+i1x3, and i1,i2 are independent imaginary units such that i12=−1=i22. The product of i1 and i2 defines a hyperbolic unit j such that j2=1. The product of units is commutative and is defined as
(3)i1i2=j,i1j=−i2,i2j=−i1.With the addition and multiplication of two bicomplex numbers defined in the obvious way, the setBℂ makes up a commutative ring. Three important subsets of Bℂ can be specified as
(4)ℂik≔x+yik;x;y∈ℝ,k=1,2,D≔x+yj;x,y∈ℝ.Each of the setℂik is isomorphic to the field of complex numbers, while D is the set of the so-called hyperbolic numbers.


## 2.2. Conjugation and Moduli

Three kinds of conjugationw†1,w†2, and w†3 can be defined on bicomplex numbers. With the bicomplex number w specified as in (2) and the bar − denoting complex conjugation in ℂ, we define
(5)w†1≔z¯1+z¯2i2,w†2≔z1−z2i2,w†3≔z¯1−z¯2i2.It is easy to check that each conjugation has the following properties:(6)u+v†k=u†k+v†k,u†k†k=u,u.v†k=u†k.v†k.Here,u,v∈Bℂ and k=1,2,3.With each kind of conjugation, one can define a specific bicomplex modulus as(7)wi12≔w.w†2=z12+z22∈ℂi1w122≔w.w†1=z12−z22+2Rez1z¯2i2∈ℂi2wj2≔w.w†3=z12+z22−2Imz1z¯2j∈DIt can be shown thatu.vs2=us2.vs2, where s=i1,i2 or j.A norm of a bicomplex numberw=z1+i2z2 denoted by w is defined by
(8)w=z1+i2z2=z12+z221/2,which, upon choosing w=x0+i1x1+i2x2+i1i2x3,xk∈ℝ,k=0,1,2,3, gives
(9)w=x02+x12+x22+x321/2.For any two bicomplex numbersu,v∈Bℂ, one can easily verify that 0≺i2u≺i2v⟹u≤v,u+v≤u+v,αu=αu, where α is a nonnegative real number. Further, for any two bicomplex numbers, u,v∈Bℂ,uv≤2uv holds.Next, we recall some necessary definitions and lemmas.First, define a partial order≺¯ on Bℂ as follows:Letu=u1+i2u2∈Bℂ and v=v1+i2v2∈Bℂ. Define a partial order relation ≺¯i2 on Bℂ as follows:u≺¯i2v if and only if u1≺¯i1v1 and u2≺¯i1v2. We find that u≺¯i2v if any one of the following properties holds:
(10)a1u1=v1u2=v2a2u1≺i1v1u2=v2a3u1=v1u2≺i1v2a4u1≺i1v1u2≺i1v2where ≺¯i1 is a partial order relation on ℂ. So that
(11)b1Reu1=Rev1,Imu1=Imv1,Reu2=Rev2,Imu2=Imv2b2Reu1<Rev1,Imu1<Imv1,Reu2=Rev2,Imu2=Imv2b3Reu1=Rev1,Imu1=Imv1,Reu2<Rev2,Imu2<Imv2b4Reu1<Rev1,Imu1<Imv1,Reu2<Rev2,Imu2<Imv2We writeu≺∼i2v if u≺¯i2v and u≠v, i.e., one of b2, b3, and b4 is satisfied and we write u≺i2v if only b4 is satisfied.Given the definition of the complex metric space introduced in [1], we extend the definition of bicomplex analysis as follows.Definition 1.
LetXbe a nonempty set. A functionaldBℂ:X×X⟶Bℂis called a bicomplex-valued metric onXif forx,y,z∈Xthe following conditions are satisfied:(12)m10≺¯i2dBℂx,ym2dBℂx,y=0ifandonlyifx=ym3dBℂx,y=dBℂy,xm4dBℂx,y≺¯i2dBℂx,z+dBℂz,yThen,X,dBℂ is called a bicomplex-valued metric space.Some known examples of bicomplex-valued metric, which show a bicomplex-valued metric space, are the following.Example 2.
The set of real numbers together with the functional(13)dBℂ1x,y=1+i1+i2+i1i2x−y,dBℂ2x,y=i1i2x−y,for all x,y∈ℝ where ⋅ is the usual real modulus. One can easily check that X,dBℂ1, X,dBℂ2 are bicomplex-valued metric spaces.Definition 3.
Letxnbe a sequence in a bicomplex-valued metric spaceX,dBℂ. The sequencexnis said to converge tox∈Xif and only if for any0≺i2ε∈Bℂ, there existsN∈ℕdepending onεsuch thatdBℂxn,x≺i2εfor alln>N. It is denoted byxn⟶xasn⟶+∞orlimn→+∞xn=x.Definition 4.
A sequencexnin a bicomplex-valued metric spaceX,dBℂis said to be a Cauchy sequence if and only if for any0≺i2ε∈Bℂ, there existsN∈ℕdepending onεsuch thatdBℂxn,xm≺i2εfor alln,m>N.Definition 5.
A bicomplex-valued metric spaceX,dBℂis said to be complete if and only if every Cauchy sequence inXconverges inX.Lemma 6 (see [2]).
LetX,dBℂbe a bicomplex-valued metric space andxnbe a sequence inX. Then,xnconverges tox∈Xif and only ifdBℂxn,x⟶0asn⟶+∞.Lemma 7 (see [2]).
LetX,dBℂbe a bicomplex-valued metric space andxnbe a sequence inXsuch thatlimn→+∞xn=x. Then, for anya∈X, limn→+∞dBℂxn,a=dBℂx,a.


## 3. Statement of Results

Definition 8 (see [11, 12]).
A functionψ:0,+∞⟶0,1is said to be a MT-function if it satisfies Mizoguchi-Takahashi’s condition, i.e.,limsups→t+ψs<1for allt∈0,+∞.Clearly, ifψ:0,+∞⟶0,1 is a nondecreasing function, then ψ is a MT-function.Notice thatψ:0,+∞⟶0,1 is a MT-function if and only if for each t∈0,+∞ there exist rt∈0,1 and εt>0 such that ψs≤rt for all s∈t,t+εt, for more detail, (see [4], Remark 2.5 (iii)).An example which is not an MT-function is given hereunder. Letψ:0,+∞⟶0,1 be defined by
(14)ψt=sintt,ift∈0,π2,0,otherwise.Sincelimsups→0+ψs=1, ψ is not a MT-function.Recently, Du [13] first proved the following characterizations of MT-functions which are quite useful for proving our main results.Theorem 9.
[13] Letψ:0,+∞⟶0,1be a function. Then, the following statements are equivalent.(a)
ψ is a MT-function(b)
For eacht∈0,+∞, there exist rt1∈0,1 and εt1>0 such that ψs≤rt1 for all s∈t,t+εt1(c)
For eacht∈0,+∞, there exist rt2∈0,1 and εt2>0 such that ψs≤rt2 for all s∈t,t+εt2(d)
For eacht∈0,+∞, there exist rt3∈0,1 and εt3>0 such that ψs≤rt3 for all s∈t,t+εt3(e)
For eacht∈0,+∞, there exist rt4∈0,1 and εt4>0 such that ψs≤rt4 for all s∈t,t+εt4(f)
For any nonincreasing sequencexnn∈ℕ in 0,+∞, we have 0≤supn∈ℕψxn<1(g)
ψ is a function of contractive factor, that is, for any strictly decreasing sequence xnn∈ℕ in 0,+∞, we have 0≤supn∈ℕψxn<1Theorem 10.
LetX,dBℂbe a complete bicomplex-valued metric space and letf:X⟶Xbe a mapping onX. Suppose that there exists a MT-functionψ:0,+∞⟶0,1such that(15)dBℂfx,fy≺¯i2ψdBℂx,ydBℂx,y,forallx,y∈X.
Then,f has a unique fixed point on X.Notice thatf need not be continuous.Proof. Letx0∈X be given. Define the sequence xnn∈ℕ by
(16)xn=fnx0=fxn−1,foreachn∈ℕ.For eachn∈ℕ, by (15), we have
(17)dBℂxn,xn+1=dBℂfxn−1,fxn≺¯i2ψdBℂxn−1,xndBℂxn−1,xn,which implies
(18)dBℂxn,xn+1≤ψdBℂxn−1,xndBℂxn−1,xn.Letδn=dBℂxn−1,xn for n∈ℕ . Then, by (18), we have
(19)δn+1≤ψδnδn<δn,∀n∈ℕ.So, we know thatδnn∈ℕ is a strictly decreasing sequence in 0,+∞. Applying (g) of Theorem 9, we obtain
(20)0≤supψδn<1.That is,(21)0≤supψdBℂxn−1,xn<1.Let(22)k=supψdBℂxn−1,xn.Then,k∈0,1. For each n∈ℕ, by (18) again, we have
(23)dBℂxn,xn+1≤ψdBℂxn−1,xndBℂxn−1,xn≤kdBℂxn−1,xn≤k2dBℂxn−2,xn−1≤⋯≤kndBℂx0,x1.For anym,n∈ℕ with m>n, we get
(24)dBℂxn,xm≤dBℂxn,xn+1+dBℂxn+1,xn+2+⋯+dBℂxm−1,xm≤kn+kn+1+⋯+km−1dBℂx0,x1<kn1−kdBℂx0,x1.Sincek∈0,1, limn→+∞kn/1−kdBℂx0,x1=0. Hence, by the last inequality, we obtain
(25)∥dBℂxn,xm∥⟶0,asm,n⟶+∞,that is, dBℂxn,xm⟶0 as n⟶+∞. Therefore, xn is a Cauchy sequence in the complete bicomplex-valued metric space. Then, there exists u∈X such that
(26)limn→+∞dBℂxn,u=0.Next, we prove thatfu=u. Assume that fu≠u. For each n∈ℕ, by (15), we have
(27)dBℂxn+1,fu=dBℂfxn,fu≺¯i2ψdBℂxn,udBℂxn,u,which deduces
(28)∥dBℂxn+1,fu∥≤ψ∥dBℂxn,u∥∥dBℂxn,u∥<∥dBℂxn,u.Sincexn⟶u as n⟶+∞, from Lemma 6 and Lemma 7 and by taking the limit from both sides of (29), we get
(29)∥dBℂu,fu∥≤∥dBℂu,u∥=0.Since0≺dBℂu,fu, then
(30)0<∥dBℂu,fu∥≤0,which is a contradiction. Therefore, fu=u, that is, u is a fixed point of f.Finally, we want to show the uniqueness of the fixed point off. Let v be a fixed point of f and v≠u. By (15), we obtain
(31)dBℂfv,fu=dBℂv,u≺¯i2ψdBℂv,udBℂv,u,which implies
(32)∥dBℂv,u∥≤ψ∥dBℂv,u∥∥dBℂv,u∥,then
(33)1−ψ∥dBℂv,u∥∥dBℂv,u∥≤0.Sinceψ∥dBℂv,u∥∈0,1, we have
(34)∥dBℂv,u∥≤0,which deduces
(35)∥dBℂv,u∥=0,and hence dBℂv,u=0 which is a contradiction. Therefore v=u, that is, f has a unique fixed point.Example 11.
LetX=ℝanddBℂ:X×X⟶Bℂbe defined bydBℂx,y=i1i2x−y,x,y∈X, where⋅is the usual real modulus. One can easily check thatX,dBℂis a complete bicomplex-valued metric onXand∥dBℂx,y∥=∣x−y∣for allx,y∈X.
Definef:X⟶X and ψ:0,+∞⟶0,1 by fx=1/12xforx∈X,
(36)ψt=56ift=014ift>0
Then,ψ is a MT-function.For allx,y∈X we have
(37)dBℂfx,fy=i1i2fx−fy=12i1i2x−yifx≠y0ifx=yψdBℂx,ydBℂx,y=ψx−ydBℂx,y=14i1i2x≠yifx≠y0ifx=ythen, f satisfies (15).So all the hypotheses of Theorem10 are fulfilled. It is therefore possible to apply Theorem 10 to get the fact that f has a unique fixed point on X (Here, 0 is the unique fixed point of f).The following fixed point theorem established in metric space is immediate from Theorem10.Theorem 12.
LetX,dbe a complete metric space and letf:X⟶Xbe a mapping onX. Suppose that there exists a MT-functionψ:0,+∞⟶0,1such that(38)dfx,fy≤ψdx,ydx,y,forallx,y∈X.
Then,f has a unique fixed point on X.Since any nondecreasing function or any nonincreasing functionψ:0,+∞⟶0,1 is an MT-function, by applying Theorem 10, we have the following results.Corollary 13.
LetX,dBℂbe a complete bicomplex-valued metric space and letf:X⟶Xbe a mapping onX. Suppose that there exists a nondecreasing (or nonincreasing) functionψ:0,+∞⟶0,1such that(39)dBℂfx,fy≺¯i2ψdBℂx,ydBℂx,y,forallx,y∈X,
Then,f has a unique fixed point on X.Corollary 14.
LetX,dbe a complete metric space and letf:X⟶Xbe a mapping onX. Suppose that there exists a nondecreasing (nonincreasing) functionψ:0,+∞⟶0,1such that(40)dfx,fy≤ψdx,ydx,y,forallx,y∈X.
Then,f has a unique fixed point on X.Corollary 15 (bicomplex-valued metric version of Banach contraction principle).
LetX,dBℂbe a bicomplex-valued complete metric space and letf:X⟶Xbe a contraction mapping onX, that is,(41)dBℂfx,fy≺¯i2kdBℂx,y,for all x,y∈X, where k∈0,1. Then, f has a unique fixed point on X.The aim of the next work is to generalize our results by finding a common fixed point fork functions fi,i=1,2,⋯,k defined on a bicomplex-valued complete metric space.Theorem 16.
LetX,dBℂbe a complete bicomplex-valued metric space andfi:X⟶Xbekmappings onX. Suppose that there exists a MT-functionψ:0,+∞⟶0,1such that(42)dBℂfix,fi+1y≺¯i2ψdBℂx,ydBℂx,y,for all x,y∈X and i=1,2,⋯,k, then, there exists a unique common fixed point u∈X such that u=f1u=f2u=⋯=fkuProof. For any x0∈X Putting x1=fi+1x0 and x2=fix1, then let x3=fi+1x2 and x4=fix3. Inductively, choose a sequence xnn∈ℕ in X so that x2n+2=fix2n+1 and x2n+1=fi+1x2n for all n≥0.It follows from the property of the functionψ that if n is odd and for xn+1≠xn, then
(43)dBℂxn+1,xn=dBℂfixn,fi+1xn−1≺¯i2ψdBℂxn,xn−1dBℂxn,xn−1.Similarly, ifn is even, we also obtain that
(44)dBℂxn+1,xn=dBℂfi+1xn,fixn−1≺¯i2ψdBℂxn−1,xndBℂxn−1,xn,which implies that, for all n∈ℕ(45)dBℂxn+1,xn<dBℂxn,xn−1Arguing like in the proof of Theorem16, we prove that xn is a Cauchy sequence in the complete bicomplex-valued metric space. Then, there exists u∈X such that
(46)limn→+∞dBℂxn,u=0.Next, we prove thatfiu=fi+1u=u,i=1,2,⋯,k. Assume that fiu≠u. For each n∈ℕ, by (42), we have
(47)dBℂx2n+1,fiu=dBℂfi+1x2n,fiu≺¯i2ψdBℂu,x2ndBℂu,x2n,then
(48)∥dBℂx2n+1,fiu∥≤ψ∥dBℂu,x2n∥∥dBℂu,x2n∥<∥dBℂu,x2n∥,then letting n⟶+∞ and by Lemma 6 and Lemma 7, we have
(49)∥dBℂu,fiu∥≤∥dBℂu,u∥=0.We obtain a contradiction. Hence,fiu=u. As
(50)dBℂu,fi+1u=dBℂfiu,fi+1u≺¯i2ψdBℂu,udBℂu,u,then
(51)∥dBℂu,fi+1u∥<∥dBℂu,u∥=0.Therefore,∥dBℂu,fi+1u∥=0, i.e., u=fi+1u=fiu. If there exists another point v∈X such that v=fiv=fi+1v, then using an argument similar to the above, we get
(52)dBℂu,v=dBℂfiu,fi+1v≺¯i2ψdBℂu,vdBℂu,v,i.e.,
(53)∥dBℂu,v∥<∥dBℂu,v.We obtain a contradiction, henceu=v.By the same way, if there exists a unique pointw∈X such that w=fi+1w=fi+2w. We will show that u=w. Suppose that u≠w, we have
(54)dBℂu,w=dBℂfiu,fi+1w≺¯i2ψdBℂu,wdBℂu,w,then
(55)∥dBℂu,w∥<∥dBℂu,w∥,which is a contradiction. Then, u=w so u=fiu=fi+1u=fi+2u.The proof is completed.


## 4. Application

In this section, we give existence theorems for the Volterra-type integral equation. For this purpose, letX=Ca,b,ℝ be the space of all continuous realvalued functions on a,b. Note that X is a complete bicomplex-valued metric space by considering dBℂx,y=i1i2maxt∈a,bxt−yt∞, then dBℂx,y=maxt∈a,bxt−yt∞.Consider the Volterra-type integral equation as(56)xt=∫abKt,s,xsds+gt,where xt is the unknown solution, t∈a,b⊂ℝ and x,g∈X.The kernelK of the integral equation is defined on a,b×a,b×ℝ.Theorem 17.
Assume that the following conditions are satisfied:
(i)
K:a,b×a,b×ℝ⟶ℝ is continuous function(ii)
there existsλ∈0,1 such that(57)∣Kt,s,y1−Kt,s,y2∣≤λb−a∣y1−y2∣for all t,s∈a,b and y1,y2∈ℝ.Then, the nonlinear Volterra Equation (56) has a unique solution.Proof. Define a mapping f:X⟶X as follows:
(58)fxt=∫abKt,s,xsds+gt.Letx,y∈X. By the simple calculation and applying assumption (i2), for each t∈a,b, we can write
(59)∣fxt−fyt∣≤∫ab∣Kt,s,xs−Kt,s,ys∣ds≤λb−a∫ab∣xs−ys∣ds.Then,(60)dBℂfx,fy=i1i2maxt∈a,bfxt−fyt∞≺¯λi1i2maxt∈a,bxt−yt∞=λdBℂx,y.It follows thatf satisfies all the conditions of Theorem 10 where wt=λ for all t∈a,b. Hence, f has a unique fixed point. This yields that there exists a unique solution of the nonlinear Volterra integral equation.



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*Source: 4070324-2020-06-04.xml*