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. and Cevikel studied some topological properties of multiplicative metric spaces and proved an analogous result to the Banach contraction principle in multiplicative metric spaces. 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   Let be a nonempty set. If mapping satisfies the following conditions:

(1) for all , and if and only if ;

(2) for all ;

(3) for all (multiplicative triangle inequality).

We say is a multiplicative metric on , and a multiplicative metric space.

Definition 2   Let be a multiplicative metric space, , is said to be multiplicative convergent to , if for arbitrary , there exists a natural number such that for all , denoted by .

Definition 3   Let be a multiplicative metric space, , is called a multiplicative Cauchy sequence, if for arbitrary , there exists a natural number such that for all .

We say that is complete if every multiplicative Cauchy sequence in is multiplicative convergent to .

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

Definition 4   Multiplicative absolute value function is defined as:

Remark 1   Multiplicative absolute value function satisfies :

1) ; 2) .

Proposition 2   is a complete multiplicative metric space.

Proposition 3   Let be a multiplicative metric space, , multiplicative converges to if and only if .

is a multiplicative Cauchy sequence if and only if.

Definition 5   The multiplicative derivative of a function is defined by . Denoted it by .

If is a positive function and its derivative at exists, then .

Definition 6   Let be a positive bounded function, be a partition of , and . The function is said to be integral in the multiplicative sense if there exists a number having the properties: for every there exists a partition of such that for every refinement of independently on selection of the numbers . is called the multiplicative integral of on , we denote it with .

It is easily seen that if is positive and Riemann integral on , then .

Let be a nonempty set, recall that mappings are weakly compatible if, for every , holds whenever . If and are weakly compatible and have an unique point of coincidence , then is the unique common fixed point of and .

2 Common Fixed Points

Theorem 1   Let be a multiplicative metric space, mappings satisfy: for all ,

(1)

holds, where , .

If is a complete subspace of , and and are weakly compatible, then and have an unique common fixed point.

Proof   Let is an arbitrary point in , since , there exists such that . Continuing this process, we can obtain a sequence such that . From (1), we have

which implies that , and so .

Let , then

(2)

for all . Since , we get .

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

For every , by taking , we have for , that is, is a multiplicative Cauchy sequence in .

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

We get that , that is

Since , for every , there exists a natural number such that

for . Therefore, we obtain that

for , which implies that . From the uniqueness of the limit, .

If there exists another point such that . From (1), we get

Since , the above inequality implies that , so . Since and are weakly compatible, is a unique common fixed point of and .

Corollary 1   Let be a multiplicative metric space, mappings satisfy the followings: for arbitrary , inequality (1) holds, where are non-negative real numbers and . If is a complete subspace of , and and are weakly compatible. Then and have a unique common fixed point.

Proof   Since , we get

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

Corollary 2   Let be a multiplicative metric space, mappings satisfy the following: for arbitrary ,

holds, where are non-negative real numbers and . If is a complete subspace of , and and are weakly compatible. Then and have a unique common fixed point.

Proof   Let , it is easy to see that the conclusion is true from Theorem 1.

Corollary 3   Let be a multiplicative metric space, mappings satisfy: for arbitrary ,

holds, where . Then has a unique fixed point.

Proof   By taking and in Theorem 1, we see that has a unique fixed point.

Theorem 2   Let be a multiplicative metric space, mappings satisfying followings: for arbitrary ,

(3)

holds, where are non-negative real numbers and,. If is a complete subspace of , , and are weakly compatible, then and have a unique common fixed point.

Proof   As we do in the proof of Theorem 1, we can obtain a sequence such that . From (3), we have

Since

and

We get

and so

Let , then and

(4)

From (4), we get

(5)

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

This implies that . Hence is a multiplicative Cauchy sequence in . By the completeness of , there exists such that . From (3), we have

Since

and

we get

That is,

Since , for every , there exists a natural number , such that for . Hence, . From the uniqueness of the limit, .

If there exists another point such that , from (3), we get

Since , it is a contradiction. Because and are weakly compatible, and have a unique common fixed point.

3 Applications

Let be a usual metric space, we define as follows: for any ,

(6)

It is easy to see that is a multiplicative metric space. Conversely, if is a multiplicative metric space, by defining : for any ,

(7)

We have a metric space .

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

1) is a complete metric space if and only if is a complete multiplicative metric space;

2) is an open set in if and only if is a multiplicative open set in ;

3) is a closed set in if and only if is a multiplicative closed set in .

Proof   1) Let be a complete metric space and a multiplicative Cauchy sequence . For every , there exists a natural number such that for . So we have for . This implies that is a Cauchy sequence in . Since is complete, there exists . For any , there exists a natural number such that for . Hence, for , and so multiplicative converges to in , is a complete multiplicative metric space.

Similarly, the inverse state is true.

2) Let is an open set in . For every , there exists such that ( is an open ball of radius with center in ). Let ( is an open ball of radius with center in ), then . From (7), , that is . This implies , Hence, is a multiplicative open set in .

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 has the same topological properties as multiplicative metric space if and satisfies equality (6) or (7).

Consider multiplicative initial value problem:

(8)

where is the multiplicative derivative of , is defined on some subset , and , .

We say that satisfy multiplicative Lipschitz type condition concerning second coordinate on , if for any , holds, where is a constant.

Theorem 4   Let be a positive continuous function on a rectangular region . If satisfies the multiplicative Lipschitz type condition concerning the second coordinate on , then multiplicative initial value problem (6) has a unique solution on , where , .

Proof   Let . is defined as for , since is a complete metric space. is a complete multiplicative metric space, where . We define as :.

Let . It is easy to see that is a closed set in , is complete in , and so it is complete in . Since , we have .

That is, . Hence , which implies that . Then, .

For , since , we have :

We get:

That is

Thus, we obtain that

We conclude that

where . By Corollary 3, has a unique fixed point in , denoted it by , that is . Thus, , and . Hence, is a solution of (6).

If is another solution of (6), since

we have (a constant), then . From , we get , which implies that is also a fixed point of . Therefore, , we conclude that the solution of (6) is unique.

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