Open Access
Issue
Wuhan Univ. J. Nat. Sci.
Volume 29, Number 1, February 2024
Page(s) 13 - 20
DOI https://doi.org/10.1051/wujns/2024291013
Published online 15 March 2024

© Wuhan University 2023

Licence Creative CommonsThis is an Open Access article distributed under the terms of the Creative Commons Attribution License (https://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

0 Introduction

Fixed point theory provides essential tools for solving the existence of unique solutions to many problems in mathematics and applications. The Banach contraction mapping principle, one of the fundamental and most widely applied fixed point theorems, has been generalized. These expansions generally proceed along two lines: one is extending the domain of mappings, and another is considering a more general contractive condition on mappings; the fixed points or common fixed points of mappings satisfying certain contraction conditions on a specific space have received much research[1-6]. Bashirov et al introduced the notion of multiplicative metric spaces and studied some fundamental theorems of multiplicative calculus[7]. O¨zavsar and Cevikel studied some topological properties of multiplicative metric spaces and proved an analogous result to the Banach contraction principle in multiplicative metric spaces[8]. Since then, some fixed-point and common fixed-point results have been obtained in multiplicative metric spaces[9-13].

This paper shows some common fixed-point results for two mappings satisfying specific multiplicative contraction conditions with exponents of fraction expression in multiplicative metric spaces.

1 Preliminaries

Definition 1   [ 7 ] Let X be a nonempty set. If mapping d:X×XR+ satisfies the following conditions:

(1) d(x,y)1 for all x,yX, and d(x,y)=1 if and only if x=y;

(2) d(x,y)=d(y,x) for all x,yX;

(3) d(x,z)d(x,y)d(y,z) for all x,y,zX(multiplicative triangle inequality).

We say d is a multiplicative metric on X, and (X,d) a multiplicative metric space.

Definition 2   [ 8 ] Let (X,d) be a multiplicative metric space, {xn}X,xX, {xn} is said to be multiplicative convergent to x, if for arbitrary ε>1, there exists a natural number N such that d(xn,x)<ε for all n>N, denoted by xnx(n).

Definition 3   [ 8 ] Let (X,d) be a multiplicative metric space, {xn}X, {xn} is called a multiplicative Cauchy sequence, if for arbitrary ε>1, there exists a natural number N such that d(xn,xm)<ε for all n,m>N.

We say that (X,d) is complete if every multiplicative Cauchy sequence in (X,d) is multiplicative convergent to xX.

Proposition 1   [ 8 ] The uniqueness of the limit holds for a convergent sequence in a multiplicative metric space.

Definition 4   [ 7 ] Multiplicative absolute value function ||*: R+R+ is defined as:

| x | * = { x , x 1 1 x , x < 1

Remark 1   Multiplicative absolute value function ||*: R+R+ satisfies :

1) |x|*1; 2) |xx0|*<ax0a<x<ax,x0,xR+.

Proposition 2   [ 8 ] ( R + , | | * ) is a complete multiplicative metric space.

Proposition 3   [ 8 ] Let (X,d) be a multiplicative metric space, {xn}X, xX, {xn} multiplicative converges to x if and only if d(xn,x)1(n).

{ x n } is a multiplicative Cauchy sequence if and only if d(xn,xm)1(n,m).

Definition 5   [ 7 ] The multiplicative derivative of a function f: RR is defined by limh0(f(x+h)f(x))1h. Denoted it by f*(x).

If f(x) is a positive function and its derivative at x exists, then f*(x)=e(lnf(x))'.

Definition 6   [ 7 ] Let f: [a,b]R be a positive bounded function, P={x0,x1,,xn} be a partition of [a,b], and ξi[xi-1,xi]. The function f is said to be integral in the multiplicative sense if there exists a number P having the properties: for every ε>0, there exists a partition Pε of [a,b] such that |i=1nf(ξi)(xi-xi-1)-P|<ε for every refinement P of Pε independently on selection of the numbers ξi[xi-1,xi](i=1,2,,n). P is called the multiplicative integral of f on [a,b], we denote it with abf(x)dx.

It is easily seen that if f is positive and Riemann integral on [a,b], then abf(x)dx=eablnf(x)dx.

Let X be a nonempty set, recall that mappings f, g: XX are weakly compatible if, for every xX, fgx=ggx holds whenever fx=gx. If f and g are weakly compatible and have an unique point of coincidence w=fx=gx, then w is the unique common fixed point of f and g.

2 Common Fixed Points

Theorem 1   Let (X,d) be a multiplicative metric space, mappings f, g: XX satisfy: for all x,yX,

d ( f x , f y ) d ( g x , g y ) λ 1 d ( f x , g x ) λ 2 d ( f y , g y ) λ 3 d ( f x , g y ) λ 4 d ( f y , g x ) λ 5 (1)

holds, where λi(0,1)(i=1,2,3,4,5), λ1+λ4+λ5<1,λ3+λ5<1,λ1+λ2+λ3+2λ4<1.

If g(X) is a complete subspace of X, f(X)g(X) and f and g are weakly compatible, then f and g have an unique common fixed point.

Proof   Let x0 is an arbitrary point in X, since f(X)g(X), there exists x1,x2X such that fx0=gx1,fx1=gx2. Continuing this process, we can obtain a sequence {xn}X such that fxn=gxn+1(n=0,1,2,). From (1), we have

d ( g x n + 1 , g x n ) = d ( f x n , f x n - 1 ) d ( g x n , g x n - 1 ) λ 1 d ( f x n , g x n ) λ 2 d ( f x n - 1 , g x n - 1 ) λ 3 d ( f x n , g x n - 1 ) λ 4 d ( f x n - 1 , g x n ) λ 5 = d ( g x n , g x n - 1 ) λ 1 d ( g x n + 1 , g x n ) λ 2 d ( g x n , g x n - 1 ) λ 3 d ( g x n + 1 , g x n - 1 ) λ 4 d ( g x n , g x n ) λ 5 d ( g x n , g x n - 1 ) λ 1 + λ 3 d ( g x n + 1 , g x n ) λ 2 d ( g x n + 1 , g x n ) λ 4 d ( g x n , g x n - 1 ) λ 4 = d ( g x n + 1 , g x n ) λ 2 + λ 4 d ( g x n , g x n - 1 ) λ 1 + λ 3 + λ 4

which implies that d(gxn+1,gxn)1-λ2-λ4d(gxn,gxn-1)λ1+λ3+λ4, and so d(gxn+1,gxn)d(gxn,gxn-1)λ1+λ3+λ41-λ2-λ4.

Let q=λ1+λ3+λ41-λ2-λ4, then

d ( g x n + 1 , g x n ) d ( g x n , g x n - 1 ) q d ( g x n - 1 , g x n - 2 ) q 2 d ( g x 1 , g x 0 ) q n (2)

for all n=1,2,. Since λ1+λ2+λ3+2λ4<1, we get q<1.

For arbitrary natural numbers n,m,n>m, from the multiplicative inequality and (2), we have

d ( g x n , g x m ) d ( g x n , g x n - 1 ) d ( g x n - 1 , g x n - 2 ) d ( g x m + 1 , g x m ) d ( g x 1 , g x 0 ) q n - 1 d ( g x 1 , g x 0 ) q n - 2 d ( g x 1 , g x 0 ) q m = d ( g x 1 , g x 0 ) q n - 1 + q n - 2 + + q m = d ( g x 1 , g x 0 ) q m ( 1 - q n - m - 1 ) 1 - q < d ( g x 1 , g x 0 ) q m 1 - q

For every ε>1, by taking N=[ln(1-q)lnεln(d(gx1,gx0))lnq], we have|d(gxn,gxm)1|*<ε for n>N, that is, {gxn} is a multiplicative Cauchy sequence in g(X).

Since g(X) is a complete, there exists pX such that gxngp. From (1) and the multiplicative inequality,

d ( g x n , f p ) = d ( f x n - 1 , f p ) d ( g x n - 1 , g p ) λ 1 d ( f x n - 1 , g x n - 1 ) λ 2 d ( f p , g p ) λ 3 d ( f x n - 1 , g p ) λ 4 d ( f p , g x n - 1 ) λ 5 d ( g x n - 1 , g p ) λ 1 d ( g x n , g x n - 1 ) λ 2 d ( f p , g x n ) λ 3 d ( g x n , g p ) λ 3 d ( g x n , g p ) λ 4 d ( f p , g x n ) λ 5 d ( g x n , g x n - 1 ) λ 5

We get that d(gxn,fp)1-λ3-λ5d(gxn-1,gp)λ1d(gxn,gp)λ3+λ4d(gxn,gxn-1)λ2+λ5, that is

d ( g x n , f p ) d ( g x n - 1 , g p ) λ 1 1 - λ 3 - λ 5 d ( g x n , g p ) λ 3 + λ 4 1 - λ 3 - λ 5 d ( g x n , g x n - 1 ) λ 2 + λ 5 1 - λ 3 - λ 5

Since gxngp, for every ε>1, there exists a natural number N such that

| d ( g x n - 1 , g p ) λ 1 1 - λ 3 - λ 5 1 | * < ε 3 , | d ( g x n , g p ) λ 3 + λ 4 1 - λ 3 - λ 5 1 | * < ε 3 , | d ( g x n , g x n - 1 ) λ 2 + λ 5 1 - λ 3 - λ 5 1 | * < ε 3

for n>N. Therefore, we obtain that

| d ( g x n , f p ) 1 | * = d ( g x n , f p ) d ( g x n - 1 , g p ) λ 1 1 - λ 3 - λ 5 d ( g x n , g p ) λ 3 + λ 4 1 - λ 3 - λ 5 d ( g x n , g x n - 1 ) λ 2 + λ 5 1 - λ 3 - λ 5

= | d ( g x n - 1 , g p ) λ 1 1 - λ 3 - λ 5 1 | * | d ( g x n , g p ) λ 3 + λ 4 1 - λ 3 - λ 5 1 | * | d ( g x n , g x n - 1 ) λ 2 + λ 5 1 - λ 3 - λ 5 1 | * < ε 3 ε 3 ε 3 = ε

for n>N, which implies that gxnfp. From the uniqueness of the limit, fp=gp.

If there exists another point p*X such that fp*=gp*. From (1), we get

d ( g p , g p * ) = d ( f p , f p * ) d ( g p , g p * ) λ 1 d ( f p , g p ) λ 2 d ( f p * , g p * ) λ 3 d ( f p , g p * ) λ 4 d ( f p * , g p ) λ 5 = d ( g p , g p * ) λ 1 d ( g p , g p * ) λ 4 d ( g p * , g p ) λ 5 = d ( g p , g p * ) λ 1 + λ 4 + λ 5 .

Since λ1+λ4+λ5<1, the above inequality implies that d(gp,gp*)=1, so gp=gp*. Since f and g are weakly compatible, fp(=gp) is a unique common fixed point of f and g.

Corollary 1   Let (X,d) be a multiplicative metric space, mappings f, g: XX satisfy the followings: for arbitrary x,yX, inequality (1) holds, where λi(i=1,2,3,4,5) are non-negative real numbers and λ1+λ2+λ3+λ4+λ51,λ3<λ2,λ4<λ5. If g(X) is a complete subspace of X, f(X)g(X) and f and g are weakly compatible. Then f and g have a unique common fixed point.

Proof   Since λ1+λ2+λ3+λ4+λ51,λ3<λ2,λ4<λ5 , we get

λ 1 + λ 4 + λ 5 < λ 1 + λ 2 + λ 3 + λ 4 + λ 5 1 ,   λ 3 + λ 5 < λ 1 + λ 2 + λ 3 + λ 4 + λ 5 1 ,   λ 1 + λ 2 + λ 3 + 2 λ 4 < λ 1 + λ 2 + λ 3 + λ 4 + λ 5 1

The conditions in Theorem 1 are satisfied. The conclusion is true from Theorem 1.

Corollary 2   Let (X,d) be a multiplicative metric space, mappings f, g: XX satisfy the following: for arbitrary x,yX,

d ( f x , f y ) d ( g x , g y ) α ( d ( f x , g x ) d ( f y , g y ) ) β ( d ( f x , g y ) d ( f y , g x ) ) γ

holds, where α,β,γ are non-negative real numbers and α+2β+2γ<1. If g(X) is a complete subspace of X, f(X)g(X) and f and g are weakly compatible. Then f and g have a unique common fixed point.

Proof   Let λ1=α,λ2=λ3=β,λ4=λ5=γ, it is easy to see that the conclusion is true from Theorem 1.

Corollary 3   [ 6 ] Let (X,d) be a multiplicative metric space, mappings f: XX satisfy: for arbitrary x,yX,

d ( f x , f y ) d ( x , y ) λ

holds, where 0<λ<1. Then f has a unique fixed point.

Proof   By taking g=IX and λ1=λ,λ2=λ3=λ4=λ5=0 in Theorem 1, we see that f has a unique fixed point.

Theorem 2   Let (X,d) be a multiplicative metric space, mappings f, g: XX satisfying followings: for arbitrary x,yX,

d ( f x , f y ) m a x { d ( g x , g y ) α , ( d ( f x , g x ) d ( f y , g y ) ) β d ( g x , g y ) , ( d ( f y , g x ) d ( f x , g y ) ) γ d ( f x , g y ) d ( f x , f y ) d ( g x , g y ) d ( f y , g y ) } (3)

holds, where α, β, γ are non-negative real numbers and,2max{α,β,γ}<1. If g(X) is a complete subspace of X, f(X)g(X) , f and g are weakly compatible, then f and g have a unique common fixed point.

Proof   As we do in the proof of Theorem 1, we can obtain a sequence {xn}X such that fxn=gxn+1(n=0,1,2,). From (3), we have

d ( g x n + 1 , g x n ) = d ( f x n , f x n - 1 ) m a x { d ( g x n , g x n - 1 ) α , ( d ( f x n , g x n ) d ( f x n - 1 , g x n - 1 ) ) β d ( g x n , g x n - 1 ) , ( d ( f x n - 1 , g x n ) d ( f x n , g x n - 1 ) ) γ d ( f x n , g x n - 1 ) d ( f x n , f x n - 1 ) d ( g x n , g x n - 1 ) d ( f x n - 1 , g x n - 1 ) } .

Since

( d ( f x n , g x n ) d ( f x n - 1 , g x n - 1 ) ) β d ( g x n , g x n - 1 ) = ( d ( g x n + 1 , g x n ) d ( g x n , g x n - 1 ) ) β d ( g x n , g x n - 1 ) ( d ( g x n + 1 , g x n ) d ( g x n , g x n - 1 ) ) β ( d ( g x n + 1 , g x n ) d ( g x n , g x n - 1 ) ) m a x { α , β , γ } ,

( d ( f x n - 1 , g x n ) d ( f x n , g x n - 1 ) ) γ d ( f x n , g x n - 1 ) d ( f x n , f x n - 1 ) d ( g x n , g x n - 1 ) d ( f x n - 1 , g x n - 1 ) ( d ( g x n , g x n ) d ( g x n + 1 , g x n - 1 ) ) γ d ( g x n + 1 , g x n ) d ( g x n , g x n - 1 ) d ( g x n + 1 , g x n ) d ( g x n , g x n - 1 ) d ( g x n , g x n - 1 ) = ( d ( g x n , g x n ) d ( g x n + 1 , g x n - 1 ) ) γ d ( g x n , g x n - 1 ) ( d ( g x n + 1 , g x n ) d ( g x n , g x n - 1 ) ) m a x { α , β , γ } ,

and

d ( g x n , g x n - 1 ) α ( d ( g x n + 1 , g x n ) d ( g x n , g x n - 1 ) ) α ( d ( g x n + 1 , g x n ) d ( g x n , g x n - 1 ) ) m a x { α , β , γ }

We get

d ( g x n + 1 , g x n ) ( d ( g x n + 1 , g x n ) d ( g x n , g x n - 1 ) ) m a x { α , β , γ }

and so

d ( g x n + 1 , g x n ) 1 - m a x { α , β , γ } d ( g x n , g x n - 1 ) m a x { α , β , γ }

Let λ=max{α,β,γ}1-max{α,β,γ}, then 0<λ<1 and

d ( g x n + 1 , g x n ) d ( g x n , g x n - 1 ) λ (4)

From (4), we get

d ( g x n + 1 , g x n ) d ( g x 1 , g x 0 ) λ n (5)

For natural numbers n,m,n>m, from the multiplicative triangle inequality and (5), we get

d ( g x n , g x m ) d ( g x n , g x n - 1 ) d ( g x n - 1 , g x n - 2 ) d ( g x m + 1 , g x m ) d ( g x 1 , g x 0 ) λ n - 1 + λ n - 2 + + λ m d ( g x 1 , g x 0 ) λ m ( 1 - λ n - m - 1 ) 1 - λ d ( g x 1 , g x 0 ) λ m 1 - λ .

This implies that d(gxn,gxm)1 (m). Hence {gxn} is a multiplicative Cauchy sequence in g(X). By the completeness of g(X), there exists pX such that gxngp(n). From (3), we have

d ( g x n , f p ) = d ( f x n - 1 , f p )

m a x { d ( g x n - 1 , g p ) α , ( d ( f x n - 1 , g x n - 1 ) d ( f p , g p ) ) β d ( g x n - 1 , g p ) , ( d ( f p , g x n - 1 ) d ( f x n - 1 , g p ) ) γ d ( f x n - 1 , g p ) d ( f x n - 1 , f p ) d ( g x n - 1 , g p ) d ( f p , g p ) } m a x { d ( g x n - 1 , g p ) α , ( d ( g x n , g x n - 1 ) d ( f p , g p ) ) β d ( g x n - 1 , g p ) , ( d ( f p , g x n - 1 ) d ( g x n , g p ) ) γ d ( g x n , g p ) d ( g x n , f p ) d ( g x n - 1 , g p ) d ( f p , g p ) } .

Since

d ( g x n - 1 g p ) α d ( g x n - 1 g p ) m a x { α , β , γ } ( d ( g x n , g p ) d ( g x n - 1 , g p ) d ( f p , g p ) ) m a x { α , β , γ } ( d ( g x n , g p ) ) 2 m a x { α , β , γ } ( d ( g x n - 1 , g p ) ) m a x { α , β , γ } ( d ( g x n , f p ) ) m a x { α , β , γ } ,

( d ( g x n , g x n - 1 ) d ( f p , g p ) ) β d ( g x n - 1 , g p ) ( d ( g x n , g p ) d ( g x n - 1 , g p ) d ( f p , g p ) ) m a x { α , β , γ } d ( g x n - 1 , g p ) ( d ( g x n , g p ) d ( g x n - 1 , g p ) d ( f p , g p ) ) m a x { α , β , γ } ( d ( g x n , g p ) ) 2 m a x { α , β , γ } ( d ( g x n - 1 , g p ) ) m a x { α , β , γ } ( d ( g x n , f p ) ) m a x { α , β , γ } ,

and

( d ( f p , g x n - 1 ) d ( g x n , g p ) ) γ d ( g x n , g p ) d ( g x n , f p ) d ( g x n - 1 , g p ) d ( f p , g p ) ( d ( f p , g p ) d ( g x n - 1 , g p ) d ( g x n , g p ) ) m a x { α , β , γ } d ( g x n , g p ) d ( g x n , f p ) d ( g x n - 1 , g p ) d ( f p , g p )

( d ( f p , g p ) d ( g x n - 1 , g p ) d ( g x n , g p ) ) m a x { α , β , γ } d ( g x n , f p ) d ( f p , g p ) d ( g x n , f p ) d ( g x n - 1 , g p ) d ( f p , g p )

                       ( d ( f p , g p ) d ( g x n - 1 , g p ) d ( g x n , g p ) ) m a x { α , β , γ }                        ( d ( g x n , g p ) ) 2 m a x { α , β , γ } ( d ( g x n - 1 , g p ) ) m a x { α , β , γ } ( d ( g x n , f p ) ) m a x { α , β , γ } ,

we get

d ( g x n , f p ) ( d ( g x n , g p ) ) 2 m a x { α , β , γ } ( d ( g x n - 1 , g p ) ) m a x { α , β , γ } ( d ( g x n , f p ) ) m a x { α , β , γ }

That is,

d ( g x n , f p ) ( d ( g x n , g p ) ) 2 m a x { α , β , γ } 1 - m a x { α , β , γ } ( d ( g x n - 1 , g p ) ) m a x { α , β , γ } 1 - m a x { α , β , γ }

Since gxngp(n), for every ε>1, there exists a natural number N, such that d(gxn,fp)<ε for n>N. Hence, gxnfp(n). From the uniqueness of the limit, fp=gp.

If there exists another point qX such that fq=gq, from (3), we get

d ( f p , f q ) m a x { d ( g p , g q ) α , ( d ( f p , g p ) d ( f q , g q ) ) β d ( g p , g q ) , ( d ( f q , g p ) d ( f p , g q ) ) γ d ( f p , g q ) d ( f p , f q ) d ( g p , g q ) d ( f q , g q ) } = m a x { d ( f p , f q ) α , ( d ( f p , f p ) d ( f q , f q ) ) β d ( f p , f q ) , ( d ( f p , f q ) ) 2 γ d ( f p , f q ) d ( f q , f q ) } = m a x { d ( f p , f q ) α , ( d ( f p , f q ) ) 2 γ d ( f p , f q ) d ( f q , f q ) } d ( f p , f q ) 2 m a x { α , β , γ } .

Since 2max{α,β,γ}<1, it is a contradiction. Because f and g are weakly compatible, f and g have a unique common fixed point.

3 Applications

Let (X,ρ) be a usual metric space, we define d: X×X[1,+) as follows: for any x, yX,

d ( x , y ) = e ρ ( x , y ) (6)

It is easy to see that (X,d) is a multiplicative metric space. Conversely, if (X,d) is a multiplicative metric space, by defining ρ: X×XR+: for any x,yX,

ρ ( x , y ) = l n d ( x , y ) (7)

We have a metric space (X,ρ).

Theorem 3   Let X be a nonempty set, ρ and d as (6) or (7), and AX.

1) (X,ρ) is a complete metric space if and only if (X,d) is a complete multiplicative metric space;

2) A is an open set in (X,ρ) if and only if A is a multiplicative open set in (X,d);

3) A is a closed set in (X,ρ) if and only if A is a multiplicative closed set in (X,d).

Proof   1) Let (X,ρ) be a complete metric space and {xn} a multiplicative Cauchy sequence (X,d). For every ε>0, there exists a natural number N such that d(xn,xm)<eε for n,m>N. So we have ρ(xn,xm)<ε for n,m>N. This implies that {xn} is a Cauchy sequence in (X,ρ). Since (X,ρ) is complete, there exists x0X. For any ε>1, there exists a natural number N such that ρ(xn,x0)<lnε for n>N. Hence, d(xn,x0)<ε for n>N, and so {xn} multiplicative converges to x0 in (X,d), (X,d) is a complete multiplicative metric space.

Similarly, the inverse state is true.

2) Let A is an open set in (X,ρ). For every xA, there exists ε>0 such that Bρ(x,ε)A (Bρ(x,ε) is an open ball of radius ε>0 with center x in (X,ρ)). Let yBd(x,eε) (Bd(x,eε) is an open ball of radius eε>1 with center x in (X,d)), then d(x,y)<eε. From (7), ρ(x,y)=lnd(x,y)<ε, that is yBρ(x,ε). This implies Bd(x,eε)Bρ(x,ε)A, Hence, A is a multiplicative open set in (X,d).

Similarly, the inverse state is true.

3) From 2), we see that the conclusion in 3) is true.

Remark 2   Theorem 3 shows that metric space (X,ρ) has the same topological properties as multiplicative metric space (X,d) if ρ and d satisfies equality (6) or (7).

Consider multiplicative initial value problem:

{ y * = f ( x , y ) y ( x 0 ) = y 0 (8)

where y* is the multiplicative derivative of y, f(x,y) is defined on some subset GR×R+, and f(x,y)>0, (x0,y0)G.

We say that f(x,y) satisfy multiplicative Lipschitz type condition concerning second coordinate on G, if for any (x,y),(x,z)G, |f(x,y)f(x,z)|*L|y-z| holds, where L>1 is a constant[7].

Theorem 4   Let f(x,y) be a positive continuous function on a rectangular region D={(x,y)||x-x0|a,|yy0|*b}(a>0,b>1). If f(x,y) satisfies the multiplicative Lipschitz type condition concerning the second coordinate on D, then multiplicative initial value problem (6) has a unique solution on [x0-δ,x0+δ], where δ<min{a,lnblnM,1y0blnL}, M=max(x,y)Df(x,y).

Proof   Let X=C[x0-δ,x0+δ].ρ: X×XR is defined as ρ(y1(x),y2(x))=maxx[x0-δ,x0+δ]|y1(x)-y2(x)| for y1(x),y2(x)X, since (X,ρ) is a complete metric space. (X,d) is a complete multiplicative metric space, where d(y1(x),y2(x))=eρ(y1(x),y2(x)). We define T: XX as :Ty(x)=y0x0xf(x,y(x))dx.

Let C˜={y(x)|y(x)X,|y(x)y0|*Mδ}. It is easy to see that C˜ is a closed set in (X,ρ), C˜ is complete in (X,ρ), and so it is complete in (X,d). Since f(x,y)M, we have -δlnMx0xlnf(x,y(x))dxδlnM.

That is, 1Mδ=e-δlnMex0xlnf(x,y(x))dxeδlnM=Mδ. Hence 1MδTy(x)y0Mδ, which implies that |Ty(x)y0|*Mδ. Then, Ty(x)C˜.

For y1(x),y2(x)C˜, since |f(x,y1(x))f(x,y2(x))|*L|y1(x)-y2(x)|, we have :

- | y 1 ( x ) - y 2 ( x ) | l n L l n f ( x , y 1 ( x ) ) f ( x , y 2 ( x ) ) | y 1 ( x ) - y 2 ( x ) | l n L

We get:

- l n L | x 0 x | y 1 ( x ) - y 2 ( x ) | d x | x 0 x l n f ( x , y 1 ( x ) ) f ( x , y 2 ( x ) ) d x l n L | x 0 x | y 1 ( x ) - y 2 ( x ) | d x |

That is

| x 0 x l n f ( x , y 1 ( x ) ) f ( x , y 2 ( x ) ) d x | l n L | x 0 x | y 1 ( x ) - y 2 ( x ) | d x | δ ρ ( y 1 ( x ) , y 2 ( x ) ) l n L

Thus, we obtain that

| e x 0 x l n f ( x , y 1 ( x ) ) d x - e x 0 x l n f ( x , y 2 ( x ) ) d x | = | e x 0 x l n f ( x , y 2 ( x ) ) d x | | e x 0 x l n f ( x , y 1 ( x ) ) d x - x 0 x l n f ( x , y 2 ( x ) ) d x - 1 |

e | x - x 0 | l n M | e x 0 x l n f ( x , y 1 ( x ) ) f ( x , y 2 ( x ) ) d x - 1 | M δ | x 0 x l n f ( x , y 1 ( x ) ) f ( x , y 2 ( x ) ) d x | < δ M δ ρ ( y 1 ( x ) , y 2 ( x ) ) l n L .

We conclude that

d ( T y 1 ( x ) , T y 2 ( x ) ) = e m a x x [ x 0 - δ , x 0 + δ ] { | T y 1 ( x ) - T y 2 ( x ) | } = e m a x x [ x 0 - δ , x 0 + δ ] { | y 0 x 0 x f ( x , y 1 ( x ) ) d x - y 0 x 0 x f ( x , y 2 ( x ) ) d x | }

= e m a x x [ x 0 - δ , x 0 + δ ] { | y 0 e x 0 x l n f ( x , y 1 ( x ) ) d x - y 0 e x 0 x l n f ( x , y 2 ( x ) ) d x | } < e y 0 δ M δ l n L ρ ( y 1 ( x ) , y 2 ( x ) ) = ( e ρ ( y 1 ( x ) , y 2 ( x ) ) ) y 0 δ M δ l n L = d ( y 1 ( x ) , y 2 ( x ) ) y 0 δ M δ l n L = d ( y 1 ( x ) , y 2 ( x ) ) λ ,

where λ=y0δMδlnL<1. By Corollary 3, T has a unique fixed point in C˜, denoted it by y˜(x), that is y˜(x)=y0x0xf(x,y˜(x))dx. Thus, y˜*(x)=f(x,y˜(x)), and y˜(x0)=y0. Hence, y˜(x) is a solution of (6).

If y˜1(x) is another solution of (6), since

( y ˜ 1 ( x ) x 0 x f ( x , y ˜ 1 ( x ) ) d x ) * = y ˜ 1 * ( x ) ( x 0 x f ( x , y ˜ 1 ( x ) ) d x ) * = 1

we have y˜1(x)x0xf(x,y˜1(x))dx=C (a constant), then y˜1(x)=Cx0xf(x,y˜1(x))dx. From y˜1(x0)=y0, we get C=y0, which implies that y˜1(x) is also a fixed point of T. Therefore, y˜(x)=y˜1(x), we conclude that the solution of (6) is unique.

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