EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
2025, Vol. 18, Issue 4, Article Number 6251
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Solving Fractional Differential Equations and Integral
Equations via Neutrosophic Bipolar Metric Space

Rajagopalan Ramaswamy

Department of Mathematics, College of Science and Humanities in Alkharj, Prince Sattam
Bin Abdulaziz University, Alkharj 11942, Saudi Arabia

Abstract. The theory of metric spaces forms the basis of metric fixed point theory, which has
varied applications in areas such as engineering, economics, medicine, and even space science
(e.g., satellite launch). In many generalizations of metric and metric-like spaces, fuzzy metric
spaces, intuitionistic fuzzy sets, and neutrosophic sets have evolved. While both metric and bipolar
metric are distance functions, the classical metric considers a single set, whereas the bipolar metric
considers the distance between two potentially different sets. It is also well known that fractional
calculus has broad applications. In this work, we introduce neutrosophic bipolar metric spaces and
establish fixed point theorems in these spaces. Our main results generalize several proven results in
the literature. The derived results are supported with non-trivial illustrations. Three applications
are presented to supplement the theoretical findings.

2020 Mathematics Subject Classifications: 47H10, 54H25

Key Words and Phrases: Fixed point, neutrosophic metric space, neutrosophic bipolar metric
space, integral equation

1. Introduction

The foundation of metric fixed point theory lies on the concept of metric spaces and
the Banach contraction principle [1] . An axiomatic grasp of metric space draws thousands
of scholars to spaciousness. Metric spaces have seen various changes in the past. Here, we
notify that the beauty, attraction, and expansion of the concept of metric spaces. Fractal
or the Hausdorff derivative of mathematical analysis, which is a non neutonion derivative,
deals with fractals defined in fractal geometry. It has vast applications. To know about
certain fundamentals of fractal calculus and its application, one can refer to [2–4].

The notion of fuzzy set (FS) was introduced by L. Zadeh [5] in 1965, where each
element had a degree of membership (t). The intuitionistic fuzzy set (IFS) on a universe
X was introduced by K. Atanassov [6] in 1986 as a generalization of FS, where besides

DOI: https://doi.org/10.29020/nybg.ejpam.v18i4.6251

Email address: r.gopalan@psau.edu.sa (R. Ramaswamy)

https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0)



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 2 of 40

the degree of membership µA(x) ∈ [0, 1] of each element x ∈ X to a set A ,there was
considered a degree of non-membership νA(x) ∈ [0, 1],

∀x ∈ X,µA(x) + νA(x) ≤ 1 (1)

The neutrosophic set (NS) was introduced by F. Smarandache [7] degree of indeterminacy
as independent component. In the study on analytical and practical applicability of fuzzy
sets and their generalisations, mathematicians reported many results including analysing
the impact of fuzzy ideal extension, evaluating solutions of generalised fuzzy differential
equations, intutionistic fuzzy linear system of equations, etc., to name a few., see [8–11].

In contemporary examinations of the set-theoretical and logical underpinnings of math-
ematics, the word “fuzzy” appears to be prevalent. The primary rationale for this unex-
pected development, our judgment, is straightforward. Because the information we collect
from our surroundings, we employ the idea of resulting from our observations or measure-
ments are all hazy and erroneous, the world around us is full of ambiguity. So, every usual
illustration is an approximation or idealization of truth of the real world or a part of it.
Fuzzy sets (orderings, languages, etc.) and other ideas allow us to deal with analyze the
mentioned in a purely mathematical and formal manner of uncertainty.

Many mathematical structures have moved within the concept of the fuzzy set. Schweizer
et al. [12] pioneered the conceit of continuous criteria. Kramosil et al. [13] initiated fuzzy
metric spaces (shortly, FMS). They used continued norms to apply the idea of fuzziness
to standard concepts of probabilistic, statistical extensions of metric spaces and compared
the results to these obtained from other. In [14], Garbiec established the Banach con-
traction concept in FMS. Rehmam et al. [15] discovered numerous α − ϕ contraction in
fuzzy cone using the integral type, mostly considered membership functions in FMS. Park
[16] developed an intuitionistic FMS for dealing without membership and nonmembership
functions. Konwar [17] introduced an intuitionistic fuzzy b-metric space (shortly, FbMS)
and proved many fixed-point theorems.

Mutlu et al. [18], initiated the concept of bipolar metric spaces (shortly, BMS) and
established fixed point theorems. Many researchers have recently produced a slew of fixed
point outcomes in the constructions of BMS using various extension of these spaces using
different contractions [19–30].

In 2019, Kiricsci et al. [31] introduced the concept of neutrosophic metric spaces
(NMS), which deals with membership, non-membership, and naturalness. Again in 2020,
Simsek et al. [32] established various fixed point results in the setting of NMS. Later in
2020, Sowndarrajan et al. [33] showed several fixed point discoveries in the context of
neutrosophic metric spaces.

In the recent past, applications of fixed point thoerems to fractal calculus is a mat-
ter of interest. Many mathematicians have applied the results of fixed point theorems



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 3 of 40

to fractional calculus. Baleanu et al. [34] studied the existence and uniqueness of a so-
lution to non-linear fractional differential equation. More recently, in 2021, Zhu et al.
[35] applied some fixed-point theorems to discuss the existence of solutions for fractional
m-point boundary value problems. More recently, Chandok et al. [36] presented appli-
cation to fractional calculus via orthogonal contractions. In the recent past, Mani et al.
[37] improved fixed point results and to find analytical solution of integral equation using
Neutrosophic triple controlled metric spaces.

Inspired by the reported results in the setting of Bipolar as well as the Neutrospohic
metric spaces, in the present work, the notion of NBMS and some related topological
concepts are introduced and fixed point results have been established in the setting of this
space and the derived results are supplemented with non-trivial examples. This work also
presents three types of applications of the derived fixed point results: (a) to find analytical
and closed form solutions of an Integral Equation, (b) to find the voltage in an electrical
circuit and finally (c) to find the analytical solution of a fractional differential equation.
The results proven in this manuscript are extensions or generalizations of the result proven
in the past.

The rest of the paper is organized as follows: Some definitions and theories are reviewed
in Section 2.In Section 3, the proposed neutrosophic bipolar metric space and associated
concepts are defined and discussed. Furthermore, the main fixed point result is presented
in this section supported with non trivial examples to supplement the derived results.
In Section 4, an application of the derived fixed point result to find the solution of the
Fredholm integral equation is given. This is followed by the finding an analytical solution
for the voltage in an electric circuit in Section-5 along with the closed form of BVP and
finally, an application to find the analytical solution of the fractional differential equation
is also presented.

2. Preliminaries

We commence this section, with certain abbreviations and some symbols used in the
manuscript:

Table 1: List of Acronyms

Acronym Full Form

FMS Fuzzy Metric Space

NBMS Neutrosophic Bipolar Metric Space

NMS Neutrosophic Metric Space

NB Neutrosophic Bipolar



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Table 2: List of symbols

Symbol Meaning

ct−||.|| Continuous triangle norm

ct−co−||.|| Continuous triangle co-norm

Π,Ψ,Ξ,Ω Functions

𭟋, S Non empty sets

⋇ Continuous triangle norm

♢ Continuous triangle co-norm

u̇, ż, ṫ, ṡ, ẇ Positive Reals

The following definitions are required in the sequel.

Definition 1. ([16]) A binary operation ⋇ : [0, 1]2 → [0, 1] is said to be a continuous
triangle norm (shortly, ct−||.||) if:

i. i⋇ ♭ = ♭⋇ i, (∀)i, ♭ ∈ [0, 1];

ii. ⋇ is continuous;

iii. i⋇ 1 = i, (∀)i ∈ [0, 1];

iv. (i⋇ ♭)⋇ ℏ = i⋇ (♭⋇ ℏ), ∀ i, ♭, ℏ ∈ [0, 1];

v. If i ≤ ℏ and ♭ ≤ j, with i, ♭, ℏ, j ∈ [0, 1], then i⋇ ♭ ≤ ℏ ⋇ j.

Definition 2. ([16]) A binary operation ♢ : [0, 1]2 → [0, 1] is said to be a continuous
triangle co-norm (shortly, ct−co−||.||) if:

i. i♢♭ = ♭♢i, ∀ i, ♭ ∈ [0, 1];

ii. ♢ is continuous;

iii. i♢0 = 0, ∀ i ∈ [0, 1];

iv. (i♢♭)♢ℏ = i♢(♭♢ℏ), ∀ i, ♭, ℏ ∈ [0, 1];

v. If i ≤ ℏ and ♭ ≤ j, with i, ♭, ℏ, j ∈ [0, 1], then i♢♭ ≤ ℏ♢j.

Definition 3. ([17]) Take 𭟋 ̸= ∅. Let ⋇ be a ct−||.||, ♢ be a ct−co−||.||, b ≥ 1 and Π,Ψ
be defined on fuzzy sets on 𭟋 × 𭟋 × (0,+∞). If (𭟋,Π,Ψ,⋇,♢) fulfills all ς,ϖ ∈ 𭟋 and
u̇, ż > 0:

i. Π(ς,ϖ, ż) + Ψ(ς,ϖ, ż) ≤ 1;

ii. Π(ς,ϖ, ż) > 0;

iii. Π(ς,ϖ, ż) = 1 ⇔ ς = ϖ;



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 5 of 40

iv. Π(ς,ϖ, ż) = Π(ϖ, ς, ż);

v. Π(ς,⋏, b(ż+ u̇)) ≥ Π(ς,ϖ, ż)⋇Π(ϖ,⋏, u̇);

vi. Π(ς,ϖ, ·) is a non-decreasing function of R+ and limż→+∞Π(ς,ϖ, ż) = 1;

vii. Ψ(ς,ϖ, ż) > 0;

viii. Ψ(ς,ϖ, ż) = 0 ⇔ ς = ϖ;

ix. Ψ(ς,ϖ, ż) = Ψ(ϖ, ς, ż);

x. Ψ(ς,⋏, b(ż+ u̇)) ≤ Ψ(ς,ϖ, ż)♢Ψ(ϖ,⋏, u̇);

xi. Ψ(ς,ϖ, ·) is a function of non-increasing R+ and limż→+∞Ψ(ς,ϖ, ż) = 0,

Then, (𭟋,Π,Ψ,⋇,♢) is an intuitionistic FbMS.

Definition 4. ([31]) Let 𭟋 ̸= ∅,⋇ and ♢ are a ct−||.|| and ct−co−||.||. Here Π,Ψ,Ω defined
on the neutrosophic sets 𭟋 × 𭟋 × (0,+∞) is said to be a neutrosophic metric on 𭟋, if
∀ ς,ϖ,⋏ ∈ 𭟋, the following axioms are fulfilled:

i. Π(ς,ϖ, ż) + Ψ(ς,ϖ, ż) + Ω(ς,ϖ, ż) ≤ 3;

ii. Π(ς,ϖ, ż) > 0;

iii. Π(ς,ϖ, ż) = 1 ∀ ż > 0 ⇔ ς = ϖ;

iv. Π(ς,ϖ, ż) = Π(ϖ, ς, ż);

v. Π(ς,⋏, ż+ u̇) ≥ Π(ς,ϖ, ż)⋇Π(ϖ,⋏, u̇);

vi. Π(ς,ϖ, ·) : (0,+∞) → [0, 1] is continuous and limż→+∞Π(ς,ϖ, ż) = 1;

vii. Ψ(ς,ϖ, ż) < 1;

viii. Ψ(ς,ϖ, ż) = 0 ∀ ż > 0 ⇔ ς = ϖ;

ix. Ψ(ς,ϖ, ż) = Ψ(ϖ, ς, ż);

x. Ψ(ς,⋏, ż+ u̇) ≤ Ψ(ς,ϖ, ż)♢Ψ(ϖ,⋏, u̇);

xi. Ψ(ς,ϖ, ·) : (0,+∞) → [0, 1] is continuous and limż→+∞Ψ(ς,ϖ, ż) = 0;

xii. Ω(ς,ϖ, ż) < 1;

xiii. Ω(ς,ϖ, ż) = 0 ∀ ż > 0 ⇔ ς = ϖ;

xiv. Ω(ς,ϖ, ż) = Ω(ϖ, ς, ż);

xv. Ω(ς,⋏, ż+ u̇) ≤ Ω(ς,ϖ, ż)♢Ω(ϖ,⋏, u̇);



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 6 of 40

xvi. Ω(ς,ϖ, ·) : (0,+∞) → [0, 1] is continuous and limż→+∞Ω(ς,ϖ, ż) = 0;

xvii. If ż ≤ 0, then Π(ς,ϖ, ż) = 0,Ψ(ς,ϖ, ż) = 0;

Then, (𭟋,Π,Ψ,Ω,⋇,♢) is said to be a neutrosophic metric space.

Definition 5. [18] Let 𭟋 and S be non-void sets and ϱ : 𭟋×S → [0,+∞) be a function,
such that

i. ϱ(ς,ϖ) = 0 iff ς = ϖ, ∀ (ς,ϖ) ∈ 𭟋× S

ii. ϱ(ς,ϖ) = ϱ(ς,ϖ), ∀ (ς,ϖ) ∈ 𭟋 ∩ S

iii. ϱ(ς,ϖ) ≤ ϱ(ς, γ) + ϱ(ς1, γ) + ϱ(ς1, ϖ), ∀ ς, ς1 ∈ 𭟋 and γ,ϖ ∈ S.

We say that the pair (𭟋,S, ϱ) is a bipolar metric space.

3. Main Results

In this section, we define NBMS and prove few fixed-point theorems.

Definition 6. Let 𭟋 ̸= ∅, S ̸= ∅ be two sets and ⋇ be a ct−||.||, ♢ be a ct−co−||.||. Then,
Π,Ψ,Ξ defined on neutrosophic sets 𭟋×S×(0,+∞) is called a neutrosophic bipolar metric
on 𭟋× S, if ∀ ς, x ∈ 𭟋, ϖ,⋏ ∈ S and ż, ŝ, ŵ > 0, the following axioms are fulfilled:

i. Π(ς,ϖ, ż) + Ψ(ς,ϖ, ż) + Ξ(ς,ϖ, ż) ≤ 3;

ii. Π(ς,ϖ, ż) > 0;

iii. Π(ς,ϖ, ż) = 1 ∀ ż > 0 ⇔ ς = ϖ;

iv. Π(ς,ϖ, ż) = Π(ϖ, ς, ż);

v. Π(ς,⋏, ż+ u̇+ ṫ) ≥ Π
(
ς,ϖ, ż

)
⋇Π

(
x, ϖ, u̇

)
⋇Π

(
x,⋏, ṫ

)
;

vi. Π(ς,ϖ, ·) : (0,+∞) → [0, 1] is continuous and lim
ż→+∞

Π(ς,ϖ, ż) = 1;

vii. Ψ(ς,ϖ, ż) < 1;

viii. Ψ(ς,ϖ, ż) = 0 ∀ ż > 0 ⇔ ς = ϖ;

ix. Ψ(ς,ϖ, ż) = Ψ(ϖ, ς, ż);

x. Ψ(ς,⋏, ż+ u̇+ ṫ) ≤ Ψ
(
ς,ϖ, ż

)
♢Ψ

(
x, ϖ, u̇

)
♢Ψ

(
x,⋏, ṫ

)
;

xi. Ψ(ς,ϖ, ·) : (0,+∞) → [0, 1] is continuous and lim
ż→+∞

Ψ(ς,ϖ, ż) = 0;

xii. Ξ(ς,ϖ, ż) < 1;

xiii. Ξ(ς,ϖ, ż) = 0 ∀ ż > 0 ⇔ ς = ϖ;



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 7 of 40

xiv. Ξ(ς,ϖ, ż) = Ξ(ϖ, ς, ż);

xv. Ξ(ς,⋏, ż+ u̇+ ṫ) ≤ Ξ
(
ς,ϖ, ż

)
♢Ξ

(
x, ϖ, u̇

)
♢Ξ

(
x,⋏, ṫ

)
;

xvi. Ξ(ς,ϖ, ·) : (0,+∞) → [0, 1] is continuous and lim
ż→+∞

Ξ(ς,ϖ, ż) = 0;

xvii. If ż ≤ 0, then Π(ς,ϖ, ż) = 0,Ψ(ς,ϖ, ż) = 1 and Ξ(ς,ϖ, ż) = 1.

Then, (𭟋,S,Π,Ψ,Ξ,⋇,♢) is said to be a NBMS.

An illustrative example of NBMS is presented below:

Example 1. Let 𭟋 = {1, 3, 5, 7}, S = {1, 2, 6, 4}. Define Π,Ψ,Ξ: 𭟋×S×(0,+∞) → [0, 1]
as

Π(ς,ϖ, ż) =

{
1, if ς = ϖ

ż
ż+max{ς,ϖ} , if otherwise,

Ψ(ς,ϖ, ż) =

{
0, if ς = ϖ
max{ς,ϖ}

ż+max{ς,ϖ} , if otherwise,

and

Ξ(ς,ϖ, ż) =

{
0, if ς = ϖ
max{ς,ϖ}

ż , if otherwise.

Let ς = 1, ϖ = 2, x = 3 and ⋏ = 4. Then from, (v), (x) and (xv) and obviously others.

Π(1, 4, ż+ u̇+ ṫ) =
ż+ u̇+ ṫ

ż+ u̇+ ṫ+max{1, 4}
=

ż+ u̇+ ṫ

ż+ u̇+ ṫ+ 4
.

Further,

Π
(
1, 2, ż

)
=

ż
˙̇z+max{1, 2}

=
ż

ż+ 2
=

ż

ż+ 2
,

Π
(
2, 3, u̇

)
=

u̇

u̇+max{2, 3}
=

u̇

u̇+ 3
=

u̇

u̇+ 3

and

Π
(
3, 4, ṫ

)
=

ṫ

ṫ+max{3, 4}
=

ṫ

ṫ+ 4
=

ṫ

ṫ+ 4
.



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 8 of 40

That is,

ż+ u̇+ ṫ

ż+ u̇+ ṫ+ 3
≥ ż

ż+ 2
· u̇

u̇+ 3
.
ṫ

ṫ+ 4
.

Since each of ż, u̇, ṫ > 0, the above inequality holds.

Π(ς,⋏, ż+ u̇+ ṫ) ≥ Π
(
ς,ϖ, ż

)
⋇Π

(
ϖ, x, u̇

)
⋇Π

(
x,⋏, ṫ

)
.

Now,

Ψ(1, 4, ż+ u̇+ ṫ) =
max{1, 4}

ż+ u̇+ ṫ+max{1, 4}
=

4

ż+ u̇+ ṫ+ 4
.

On the other hand,

Ψ
(
1, 2, ż

)
=

max{1, 2}
ż+max{1, 2}

=
2

ż+ 2
=

2

ż+ 2
,

Ψ
(
2, 3, u̇

)
=

max{2, 3}
u̇+max{2, 3}

=
3

u̇+ 3
=

3

u̇+ 3

and

Ψ
(
3, 4, ṫ

)
=

max{3, 4}
ṫ+max{3, 4}

=
4

ṫ+ 4
=

4

ṫ+ 4
.

That is,

4

ż+ u̇+ ṫ+ 4
≤ max

{
2

ż+ 2
,

3

u̇+ 3
,

4

ṫ+ 4

}
.

Here again, ż, u̇, ṫ > 0 the above inequality is true and so,

Ψ(ς,⋏, ż+ u̇+ ṫ) ≤ Ψ
(
ς,ϖ, ż

)
♢Ψ

(
x,⋏, u̇

)
♢Ψ

(
x,⋏, ṫ

)
.

Finally,

Ξ(1, 3, ż+ u̇+ ṫ) =
max{1, 3}
ż+ u̇+ ṫ

=
3

ż+ u̇+ ṫ
.

On the other hand,

Ξ
(
1, 2, ż

)
=

max{1, 2}
ż

=
2

ż
,

Ξ
(
2, 3, u̇

)
=

max{2, 3}
u̇

=
3

u̇
=

3

u̇



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 9 of 40

and

Ξ
(
3, 4, ṫ

)
=

max{3, 4}
ṫ

=
4

ṫ
=

4

ṫ
.

That is,

3

ż+ u̇+ ṫ
≤ max

{
2

ż
,
3

u̇
,
4

ṫ

}
.

The above inequality holds as ż, u̇ > 0. Thus,,

Ξ(ς,⋏, ż+ u̇+ ṫ) ≤ Ξ
(
ς,ϖ, ż

)
♢Ξ

(
ϖ, x, u̇

)
♢Ξ

(
x,⋏, u̇

)
.

Thus, (𭟋,Π,Ψ,Ξ,⋇,♢) is a NBMS with i ⋇ ♭ = i♭ and i♢♭ = max{i, ♭} by ct−||.|| and
ct−co−||.||, respectively.

Remark 1. It is to be noted that every NBMS is a neutrosophic metric space NMS, but
the converse is not always true because, NMS is a particular case of NBMS where F = S.

Definition 7. Let p : 𭟋1 ∪S1 → 𭟋2 ∪S2 be a mapping, where (𭟋1,S1) and (𭟋2,S2) pairs
of sets.

i. If p(𭟋1) ⊆ 𭟋2 and p(S1) ⊆ S2, then p is said to be a covariant map, or a map from
(𭟋1,S1,Π1,Ψ1,Ξ1,⋇,♢) to (𭟋2,S2,Π2,Ψ2,Ξ2,⋇,♢) and this is written as,
p : (𭟋1,S1,Π1,Ψ1,Ξ1,⋇,♢) ⇒ (𭟋2,S2,Π2,Ψ2,Ξ2,⋇,♢).

ii. If p(𭟋1) ⊆ S2 and p(S1) ⊆ 𭟋2, then p is said to be a contravariant map from
(𭟋1,S1,Π1,Ψ1,Ξ1,⋇,♢) to (𭟋2,S2,Π2,Ψ2,Ξ2,⋇,♢) and this is denoted as:
p : (𭟋1,S1,Π1,Ψ1,Ξ1,⋇,♢) ⇆ (𭟋2,S2,Π2,Ψ2,Ξ2,⋇,♢).

3.1. Some topological properties of neutrosophic bipolar metric space

Definition 8. Let (𭟋,S,Π,Ψ,Ξ,⋇,♢) is a NBMS, and define a right open ball B(ς, r, ż)
with center ς ∈ 𭟋, radius r, r ∈ (0, 1), ż > 0 as follows:

B(ς, r, ż) = {ϖ ∈ S : Π(ς,ϖ, ż) > 1− r,Ψ(ς,ϖ, ż) < r,Ξ(ς,ϖ, ż) < r}.

Definition 9. Let (𭟋,S,Π,Ψ,Ξ,⋇,♢) is a NBMS, and define a left open ball B(ϖ, r, ż)
with center ϖ ∈ S, radius r, r ∈ (0, 1), ż > 0 as follows:

B(ϖ, r, ż) = {ς ∈ 𭟋 : Π(ς,ϖ, ż) > 1− r,Ψ(ς,ϖ, ż) < r,Ξ(ς,ϖ, ż) < r}.

Definition 10. Let (𭟋,S,Π,Ψ,Ξ,⋇,♢) is a NBMS. A subset P of S is said to be right
open set if for every ϖ ∈ P, there exists r such that

ϖ ∈ B(ς, r, ż) ⊆ P.



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Definition 11. Let (𭟋,S,Π,Ψ,Ξ,⋇,♢) is a NBMS. A subset K of 𭟋 is said to be left
open set if for every ς ∈ K, there exists r such that

ς ∈ B(ϖ, r, ż) ⊆ K.

Definition 12. Let (𭟋,S,Π,Ψ,Ξ,⋇,♢) is a NBMS. Let G ⊆ 𭟋 and H ⊆ S. Then G ×H
is called an open set if G is left open set and H is right open set.

Definition 13. Let (𭟋,S,Π,Ψ,Ξ,⋇,♢) is a NBMS. Let G ⊆ 𭟋 and H ⊆ S. We say that
a subset G ×H of 𭟋× S is closed if (𭟋− G)× (S −H) is open.

Theorem 1. Every right open ball B(ς, r, ż) is a right open set.
Proof. Take B(ς, r, ż) be a right open ball. Choose ϖ ∈ B(ς, r, ż). Therefore, Π(ς,ϖ, ż) >

1 − r,Ψ(ς,ϖ, ż) < r,Ξ(ς,ϖ, ż) < r. There exists ż0 ∈ (0, ż) such that Π(ς,ϖ, ż0) >
1−r,Ψ(ς,ϖ, ż0) < r,Ξ(ς,ϖ, ż0) < r because of Π(ς,ϖ, ż) > 1−r. If we take r0 = Π(ς,ϖ, ż0),
then for r0 > 1−r, θ ∈ (0, 1) will exist such that r0 > 1−θ > 1−r. Give r0 and θ such that
r0 > 1−θ. Then, r1, r2, r3 ∈ (0, 1) will exist such that r0⋇ r1 > 1−θ, (1− r0)⋄ (1− r2) ≤ θ
and (1 − r0) ⋄ (1 − r3) ≤ θ. Choose r4 = max{r1, r2, r3}. Consider the right open ball
B(ϖ, 1 − r4, ż − ż0). We will show that B(ϖ, 1 − r4, ż − ż0) ⊂ B(ς, r, ż). If we take
θ1 ∈ B(ϖ, 1−r4, ż− ż0), then Π(ϖ, θ1, ż− ż0) > r4,Ψ(ϖ, θ1, ż− ż0) < r4,Ξ(ϖ, θ1, ż− ż0) < r4.
Then,

Π(ς, θ1, ż) ≥ Π(ς,ϖ, ż0)⋇Π(ϖ, θ1, ż− ż0) ≥ r0 ⋇ r4 ≥ r0 ⋇ r1 ≥ 1− θ > 1− r,

Ψ(ς, θ1, ż) ≤ Ψ(ς,ϖ, ż0)♢Ψ(ϖ, θ1, ż− ż0) ≤ (1− r0)♢(1− r4) ≤ (1− r0)♢(1− r2) ≤ θ < r,

Ξ(ς, θ1, ż) ≤ Ξ(ς,ϖ, ż0)♢Ξ(ϖ, θ1, ż− ż0) ≤ (1− r0)♢(1− r4) ≤ (1− r0)♢(1− r2) ≤ θ < r.

Therefore θ1 ∈ B(ς, r, ż).

Similarly, we can prove the following theorems.

Theorem 2. Every left open ball B(ϖ, r, ż) is a left open set.

Theorem 3. Every open ball is an open set.

Remark 2. We can say that
τp = {G ⊂ 𭟋 : there exist ż > 0 and r ∈ (0, 1) such that B(ς, r, ż) ⊆ G for each ς ∈
G}×{H ⊂ S : there exist ż > 0 and r ∈ (0, 1) such that B(ϖ, r, ż) ⊆ G for each ϖ ∈
H} is a product topology on 𭟋×S. In that case every NBMS p on 𭟋×S produces a product
topology τp on 𭟋× S which has a base the family of open sets.

Theorem 4. Every NBMS is Hausdorff.
Proof. Let (𭟋,S,Π,Ψ,Ξ,⋇,♢) is a NBMS. Choose ς and ϖ as two distinct points

in 𭟋 and S. Hence, 0 < Π(ς,ϖ, ż) < 1, 0 < Ψ(ς,ϖ, ż) < 1, 0 < Ξ(ς,ϖ, ż) < 1. Take



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 11 of 40

r1 = Π(ς,ϖ, ż), r2 = Ψ(ς,ϖ, ż), r3 = Ξ(ς,ϖ, ż) and r = max{r1, 1 − r2, 1 − r3}. If we
take r0 ∈ (r, 1), then there exist r4, r5, r6 such that r4 ⋇ r4 ≥ r0, (1 − r5)♢(1 − r5) ≤
1− r0,(1− r6)♢(1− r6) ≤ 1− r0. Let r7 = max{r4, r5, r6}. If we consider the right open ball
B(ς, r7, ż2) and left open ball B(ϖ, r7, ż2), then clearly B(ς, r7, ż2)∩B(ϖ, r7, ż2) = ∅. Suppose

that θ1 ∈ B(ς, r7, ż2) ∩ B(ϖ, r7, ż2), then

r1 = Π(ς,ϖ, ż) ≥ Π(ς, θ1,
ż

2
)⋇Π(θ1, ϖ,

ż

2
) ≥ r7 ⋇ r7 ≥ r4 ⋇ r4 ≥ r0 > r1,

r2 = Ψ(ς,ϖ, ż) ≤ Ψ(ς, θ1,
ż

2
)♢Ψ(θ1, ϖ,

ż

2
) ≤ (1− r7)♢(1− r7) ≤ (1− r5)♢(1− r5)

≤ 1− r0 < r2,

r3 = Ξ(ς,ϖ, ż) ≤ Ξ(ς, θ1,
ż

2
)♢Ξ(θ1, ϖ

ż

2
) ≤ (1− r7)♢(1− r7) ≤ (1− r6)♢(1− r6)

≤ 1− r0 < r3,

which is a contradiction.

Definition 14. Let (𭟋,S,Π,Ψ,Ξ,⋇,♢) is a NBMS.

i. A point ς ∈ 𭟋 ∪ S is said to be a left point if ς ∈ 𭟋, a right point if ς ∈ S and a
central point if both hold.

ii. A sequence {ςµ} ⊂ 𭟋 is said to be a left sequence and a sequence {βn} ⊂ S is said
to be a right sequence.

iii. A sequence {ςµ} ⊂ 𭟋∪ S is said to converge to a point ς if and only if {ςµ} is a left
sequence, ς is a right point and

lim
µ→+∞

Π(ςµ, ς, ż) = 1, lim
µ→+∞

Ψ(ςµ, ς, ż) = 0, lim
µ→+∞

Ξ(ςµ, ς, ż) = 0 ∀ ż > 0

or {ςµ} is a right sequence, ς is a left point and

lim
µ→+∞

Π(ς, ςµ, ż) = 1, lim
µ→+∞

Ψ(ς, ςµ, ż) = 0, lim
µ→+∞

Ξ(ς, ςµ, ż) = 0 ∀ ż > 0.

iv. A sequence {(ςµ, βµ)} ⊂ 𭟋× S is said to be a bisequence. If the sequences {ςµ} and
{βµ} both converge then the bisequence {(ςµ, βµ)} is said to be convergent in 𭟋×S.

v. If {ςµ} and {βµ} both converge to a point β ∈ 𭟋 ∩ S then the bisequence {(ςµ, βµ)}
is said to be biconvergent. A sequence {(ςµ, βµ)} is a Cauchy bisequence if

lim
µ,m→+∞

Π(ςµ, βm, ż) = 1, lim
µ,m→+∞

Ψ(ςµ, βm, ż) = 0, lim
µ,m→+∞

Ξ(ςµ, βm, ż) = 0,

∀ ż > 0.

vi. A NBMS is said to be complete if every Cauchy bisequence is convergent.

Now we establish our main results.



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 12 of 40

3.2. Main Results

Lemma 1. Let {ςµ} be a Cauchy sequence in NBMS (𭟋,S,Π,Ψ,Ξ,⋇,♢) such that ςµ ̸= ςm
for every m, µ(̸= m) ∈ N. Then, at most, the sequence {ςµ} converge to one limit point.

Proof. Conversely, suppose that ςµ → ς ∈ S and ςµ → ϖ ∈ 𭟋 ∩ S, for ς ̸= ϖ.
Then, limµ→+∞Π(ςµ, ς, ż) = 1, limµ→+∞Ψ(ςµ, ς, ż) = 0, limµ→+∞ Ξ(ςµ, ς, ż) = 0, and
limµ→+∞Π(ςµ, ϖ, ż) = 1, limµ→+∞Ψ(ςµ, ϖ, ż) = 0, limµ→+∞ Ξ(ςµ, ϖ, ż) = 0, ∀ ż > 0.
Suppose

Π(ς,ϖ, ż) ≥ Π

(
ς, ςµ,

ż

3

)
⋇Π

(
ςµ, ςµ+1,

ż

3

)
⋇Π

(
ςµ+1, ϖ,

ż

3

)
→ 1⋇ 1⋇ 1, as µ→ +∞,

Ψ(ς,ϖ, ż) ≤ Ψ

(
ς, ςµ,

ż

3

)
♢Ψ

(
ςµ, ςµ+1,

ż

3

)
♢Ψ

(
ςµ+1, ϖ,

ż

3

)
→ 0♢0♢0, as µ→ +∞,

Ξ(ς,ϖ, ż) ≤ Ξ

(
ς, ςµ,

ż

3

)
♢Ξ

(
ςµ, ςµ+1,

ż

3

)
♢Ξ

(
ςµ+1, ϖ,

ż

3

)
→ 0♢0♢0, as µ→ +∞.

That is Π(ς,ϖ, ż) ≥ 1 ⋇ 1 ⋇ 1 = 1,Ψ(ς,ϖ, ż) ≤ 0♢0♢0 = 0 and Ξ(ς,ϖ, ż) ≤ 0♢0♢0 = 0.
Hence ς = ϖ, i.e., the sequence converges to unique limit point at most.

Lemma 2. Let (𭟋,S,Π,Ψ,Ξ,⋇,♢) be a NBMS. If ζ ∈ (0, 1) and for some ς,ϖ ∈ 𭟋, ż > 0,

Π(ς,ϖ, ż) ≥ Π

(
ς,ϖ,

ż

ζ

)
,Ψ(ς,ϖ, ż) ≤ Ψ

(
ς,ϖ,

ż

ζ

)
,Ξ(ς,ϖ, ż) ≤ Ξ

(
ς,ϖ,

ż

ζ

)
(2)

then ς = ϖ.
Proof. (2) implies that

Π(ς,ϖ, ż) ≥ Π

(
ς,ϖ,

ż

ζµ

)
,Ψ(ς,ϖ, ż) ≤ Ψ

(
ς,ϖ,

ż

ζµ

)
,Ξ(ς,ϖ, ż)

≤ Ξ

(
ς,ϖ,

ż

ζµ

)
, µ ∈ N, ż > 0.

Now

Π(ς,ϖ, ż) ≥ lim
µ→+∞

Π

(
ς,ϖ,

ż

ζµ

)
= 1,

Ψ(ς,ϖ, ż) ≤ lim
µ→+∞

Ψ

(
ς,ϖ,

ż

ζµ

)
= 0,

Ξ(ς,ϖ, ż) ≤ lim
µ→+∞

Ξ

(
ς,ϖ,

ż

ζµ

)
= 0, ż > 0.

Also, by definition of iii, viii, xiii, that is, ς = ϖ.



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 13 of 40

Theorem 5. Suppose (𭟋,S,Π,Ψ,Ξ,⋇,♢) is a complete NBMS with 0 < ζ < 1. Let
p : 𭟋 ∪ S → 𭟋 ∪ S be a mapping satisfying

i. p(𭟋) ⊆ 𭟋 and p(S) ⊆ S;

ii.

Π(pς, pϖ, ζ ż) ≥ Π(ς,ϖ, ż),

Ψ(pς, pϖ, ζ ż) ≤ Ψ(ς,ϖ, ż) and Ξ(pς, pϖ, ζ ż) ≤ Ξ(ς,ϖ, ż) (3)

∀ ς ∈ 𭟋 , ϖ ∈ S and ż > 0.

Then p has a unique fixed point.
Proof. Let ς0 ∈ 𭟋 and ϖ0 ∈ S and assume that p(ςµ) = ςµ+1 and p(ϖµ) = ϖµ+1 ∀

µ ∈ N ∪ {0}. Then we get (ςµ, ϖµ) as a bisequence on NBMS (𭟋,S,Π,Ψ,Ξ,⋇,♢). Now,
we have

Π(ς1, ϖ1, ż) = Π(pς0, pϖ0, ż) ≥ Π(ς0, ϖ0,
ż

ζ
),

Ψ(ς1, ϖ1, ż) = Ψ(pς0, pϖ0, ż) ≤ Ψ(ς0, ϖ0,
ż

ζ
)

and

Ξ(ς1, ϖ1, ż) = Ξ(pς0, pϖ0, ż) ≤ Ξ(ς0, ϖ0,
ż

ζ
),

∀ ż > 0 and µ ∈ N. By simple induction, we get

Π(ςµ, ϖµ, ż) = Π(pςµ−1, pϖµ−1, ż) ≥ Π(ςµ−1, ϖµ−1,
ż

ζ
) ≥ Π

(
ςµ−2, ϖµ−2,

ż

ζ2

)
≥ Π

(
ςµ−3, ϖµ−3,

ż

ζ3

)
≥ · · · ≥ Π

(
ς0, ϖ0,

ż

ζµ

)
,

Ψ(ςµ, ϖµ, ż) = Ψ(pςµ−1, pϖµ−1, ż) ≤ Ψ(ςµ−1, ϖµ−1,
ż

ζ
) ≤ Ψ

(
ςµ−2, ϖµ−2,

ż

ζ2

)
≤ Ψ

(
ςµ−3, ϖµ−3,

ż

ζ3

)
≤ · · · ≤ Ψ

(
ς0, ϖ0,

ż

ζµ

)
.

and

Ξ(ςµ, ϖµ, ż) = Ξ(pςµ−1, pϖµ−1, ż) ≤ Ξ(ςµ−1, ϖµ−1,
ż

ζ
) ≤ Ξ

(
ςµ−2, ϖµ−2,

ż

ζ2

)
≤ Ξ

(
ςµ−3, ϖµ−3,

ż

ζ3

)
≤ · · · ≤ Ξ

(
ς0, ϖ0,

ż

ζµ

)
.



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 14 of 40

We obtain

Π(ςµ, ϖµ, ż) ≥ Π

(
ς0, ϖ0,

ż

ζµ

)
,

Ψ(ςµ, ϖµ, ż) ≤ Ψ

(
ς0, ϖ0,

ż

ζµ

)
, Ξ(ςµ, ϖµ, ż) ≤ Ξ

(
ς0, ϖ0,

ż

ζµ

)
(4)

and

Π(ςµ+1, ϖµ, ż) ≥ Π

(
ς1, ϖ0,

ż

ζµ

)
,

Ψ(ςµ+1, ϖµ, ż) ≤ Ψ

(
ς1, ϖ0,

ż

ζµ

)
, Ξ(ςµ+1, ϖµ, ż) ≤ Ξ

(
ς1, ϖ0,

ż

ζµ

)
. (5)

Letting µ < m, for µ,m ∈ N. Then,

Π(ςµ, ϖm, ż) ≥ Π(ςµ, ϖµ,
ż

3
)⋇Π(ςµ+1, ϖµ,

ż

3
)⋇Π(ςµ+1, ϖm,

ż

3
)

...

≥ Π(ςµ, ϖµ,
ż

3
)⋇Π(ςµ+1, ϖµ,

ż

3
)⋇ · · ·⋇Π(ςm−1, ϖm−1,

ż

3m−1
)

⋇Π(ςm, ϖm−1,
ż

3m−1
)⋇Π(ςm, ϖm,

ż

3m−1
),

Ψ(ςµ, ϖm, ż) ≤ Ψ(ςµ, ϖµ,
ż

3
)♢Ψ(ςµ+1, ϖµ,

ż

3
)♢Ψ(ςµ+1, ϖm,

ż

3
)

...

≤ Ψ(ςµ, ϖµ,
ż

3
)♢Ψ(ςµ+1, ϖµ,

ż

3
)♢ · · ·♢Ψ(ςm−1, ϖm−1,

ż

3m−1
)

♢Ψ(ςm, ϖm−1,
ż

3m−1
)♢Ψ(ςm, ϖm,

ż

3m−1
),

and

Ξ(ςµ, ϖm, ż) ≤ Ξ(ςµ, ϖµ,
ż

3
)♢Ξ(ςµ+1, ϖµ,

ż

3
)♢Ξ(ςµ+1, ϖm,

ż

3
)

...

≤ Ξ(ςµ, ϖµ,
ż

3
)♢Ξ(ςµ+1, ϖµ,

ż

3
)♢ · · ·♢Ξ(ςm−1, ϖm−1,

ż

3m−1
)

♢Ξ(ςm, ϖm−1,
ż

3m−1
)♢Ξ(ςm, ϖm,

ż

3m−1
).

Therefore,

Π(ςµ, ϖm, ż) ≥ Π(ςµ, ϖµ,
ż

3
)⋇Π(ςµ+1, ϖµ,

ż

3
)⋇ · · ·⋇Π(ςm−1, ϖm−1,

ż

3m−1
)



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 15 of 40

⋇Π(ςm, ϖm−1,
ż

3m−1
)⋇Π(ςm, ϖm,

ż

3m−1
)

≥ Π(ς0, ϖ0,
ż

3ζµ
)⋇Π(ς1, ϖ0,

ż

3ζµ
)⋇ · · ·⋇Π(ς0, ϖ0,

ż

3m−1ζm−1
)

⋇Π(ς1, ϖ0,
ż

3m−1ζm−1
)⋇Π(ς0, ϖ0,

ż

3m−1ζm
),

Ψ(ςµ, ϖm, ż) ≤ Ψ(ςµ, ϖµ,
ż

3
)♢Ψ(ςµ+1, ϖµ,

ż

3
)♢ · · ·♢Ψ(ςm−1, ϖm−1,

ż

3m−1
)

♢Ψ(ςm, ϖm−1,
ż

3m−1
)♢Ψ(ςm, ϖm,

ż

3m−1
)

≤ Ψ(ς0, ϖ0,
ż

3ζµ
)♢Ψ(ς1, ϖ0,

ż

3ζµ
)♢ · · ·♢Ψ(ς0, ϖ0,

ż

3m−1ζm−1
)

♢Ψ(ς1, ϖ0,
ż

3m−1ζm−1
)♢Ψ(ς0, ϖ0,

ż

3m−1ζm
),

and

Ξ(ςµ, ϖm, ż) ≤ Ξ(ςµ, ϖµ,
ż

3
)♢Ξ(ςµ+1, ϖµ,

ż

3
)♢ · · ·♢Ξ(ςm−1, ϖm−1,

ż

3m−1
)

♢Ξ(ςm, ϖm−1,
ż

3m−1
)♢Ξ(ςm, ϖm,

ż

3m−1
)

≤ Ξ(ς0, ϖ0,
ż

3ζµ
)♢Ξ(ς1, ϖ0,

ż

3ζµ
)♢ · · ·♢Ξ(ς0, ϖ0,

ż

3m−1ζm−1
)

♢Ξ(ς1, ϖ0,
ż

3m−1ζm−1
)♢Ξ(ς0, ϖ0,

ż

3m−1ζm
).

Which implies that,

Π(ςµ, ϖm, ż) ≥ Π(ς0, ϖ0,
ż

3ζµ
)⋇Π(ς1, ϖ0,

ż

3ζµ
)⋇ · · ·⋇Π(ς0, ϖ0,

ż

3m−1ζm−1
)

⋇Π(ς1, ϖ0,
ż

3m−1ζm−1
)⋇Π(ς0, ϖ0,

ż

3m−1ζm
),

Ψ(ςµ, ϖm, ż) ≤ Ψ(ς0, ϖ0,
ż

3ζµ
)♢Ψ(ς1, ϖ0,

ż

3ζµ
)♢ · · ·♢Ψ(ς0, ϖ0,

ż

3m−1ζm−1
)

♢Ψ(ς1, ϖ0,
ż

3m−1ζm−1
)♢Ψ(ς0, ϖ0,

ż

3m−1ζm
),

and

Ξ(ςµ, ϖm, ż) ≤ Ξ(ς0, ϖ0,
ż

3ζµ
)♢Ξ(ς1, ϖ0,

ż

3ζµ
)♢ · · ·♢Ξ(ς0, ϖ0,

ż

3m−1ζm−1
)

♢Ξ(ς1, ϖ0,
ż

3m−1ζm−1
)♢Ξ(ς0, ϖ0,

ż

3m−1ζm
).



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 16 of 40

As µ,m → +∞, we deduce

lim
µ,m→+∞

Π(ςµ, ϖm, ż) = 1⋇ 1⋇ · · ·⋇ 1 = 1,

lim
µ,m→+∞

Ψ(ςµ, ϖm, ż) = 0♢0♢ · · ·♢0 = 0

and

lim
µ,m→+∞

Ξ(ςµ, ϖm, ż) = 0♢0♢ · · ·♢0 = 0.

Which implies that bisequence (ςµ, ϖµ) is a Cauchy bisequence.
Since (𭟋,S,Π,Ψ,Ξ,⋇,♢) is a complete NBMS. Then, {ςµ} → ς and {ϖµ} → ς, where
ς ∈ 𭟋 ∩ S. Using v, x and xv, we get

Π(ς, pς, ż) ≥ Π

(
ς, ςµ+1,

ż

3

)
⋇Π

(
ςµ+1, ςµ+1,

ż

3

)
⋇Π

(
ςµ+1, pς,

ż

3

)
= Π

(
ς, ςµ+1,

ż

3

)
⋇Π

(
pςµ, pςµ,

ż

3

)
⋇Π

(
pςµ, pς,

ż

3

)
→ 1⋇ 1⋇ 1 = 1 as µ→ +∞,

Ψ(ς, pς, ż) ≤ Ψ

(
ς, ςµ+1,

ż

3

)
♢Ψ

(
ςµ+1, ςµ+1,

ż

3

)
♢Ψ

(
ςµ+1, pς,

ż

3

)
= Ψ

(
ς, ςµ+1,

ż

3

)
♢Ψ

(
pςµ+1, pςµ+1,

ż

3

)
♢Ψ

(
pςµ+1, pς,

ż

3

)
→ 0♢0♢0 = 0 as µ→ +∞

and

Ξ(ς, pς, ż) ≤ Ξ

(
ς, ςµ+1,

ż

3

)
♢Ξ

(
ςµ+1, ςµ+1,

ż

3

)
♢Ξ

(
ςµ+1, pς,

ż

3

)
= Ξ

(
ς, ςµ+1,

ż

3

)
♢Ξ

(
pςµ+1, pςµ+1,

ż

3

)
♢Ξ

(
pςµ+1, pς,

ż

3

)
→ 0♢0♢0 = 0 as µ→ +∞.

Hence, pς = ς. Let pη = η for any η ∈ 𭟋 ∩ S, then

1 ≥ Π(η, ς, ż) = Π(pη, pς, ż) ≥ Π

(
η, ς,

ż

ζ

)
= Π

(
pη, pς,

ż

ζ

)
≥ Π

(
η, ς,

ż

ζ2

)
≥ · · · ≥ Π

(
η, ς,

ż

ζµ

)
→ 1 as µ→ +∞,

0 ≤ Ψ(η, ς, ż) = Ψ(pη, pς, ż) ≤ Ψ

(
η, ς,

ż

ζ

)
= Ψ

(
pη, pς,

ż

ζ

)



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 17 of 40

≤ Ψ

(
η, ς,

ż

ζ2

)
≤ · · · ≤ Ψ

(
η, ς,

ż

ζµ

)
→ 0 as µ→ +∞,

and

0 ≤ Ξ(η, ς, ż) = Ξ(pη, pς, ż) ≤ Ξ

(
η, ς,

ż

ζ

)
= Ξ

(
pη, pς,

ż

ζ

)
≤ Ξ

(
η, ς,

ż

ζ2

)
≤ · · · ≤ Ξ

(
η, ς,

ż

ζµ

)
→ 0 as µ→ +∞,

since iii, viii and xiii, we get ς = η.

Theorem 6. Suppose (𭟋,S,Π,Ψ,Ξ,⋇,♢) is a complete NBMS with 0 < ζ < 1. Let
p : 𭟋 ∪ S → 𭟋 ∪ S be a mapping satisfying

i. p(𭟋) ⊆ S and p(S) ⊆ 𭟋;

ii.

Π(pς, pϖ, ζ ż) ≥ Π(ϖ, ς, ż),

Ψ(pς, pϖ, ζ ż) ≤ Ψ(ϖ, ς, ż) and Ξ(pς, pϖ, ζ ż) ≤ Ξ(ϖ, ς, ż) (6)

∀ ς ∈ 𭟋 , ϖ ∈ S and ż > 0.

Then p has a unique fixed point.
Proof. Let ς0 ∈ 𭟋 and ϖ0 ∈ S and assume that p(ςµ) = ϖµ and p(ϖµ) = ςµ+1 ∀

µ ∈ N ∪ {0}. Then we get (ςµ, ϖµ) as a bisequence on NBMS (𭟋,S,Π,Ψ,Ξ,⋇,♢). Now,
we have

Π(ς1, ϖ0, ż) = Π(pϖ0, pς0, ż) ≥ Π(ς0, ϖ0,
ż

ζ
),

Ψ(ς1, ϖ0, ż) = Ψ(pϖ0, pς0, ż) ≤ Ψ(ς0, ϖ0,
ż

ζ
)

and

Ξ(ς1, ϖ0, ż) = Ξ(pϖ0, pς0, ż) ≤ Ξ(ς0, ϖ0,
ż

ζ
),

∀ ż > 0 and µ ∈ N. By simple induction, we get

Π(ςµ, ϖµ, ż) = Π(pϖµ−1, pςµ, ż) ≥ Π

(
ς0, ϖ0,

ż

ζ2µ

)
,

Ψ(ςµ, ϖµ, ż) = Ψ(pϖµ−1, pςµ, ż) ≤ Ψ

(
ς0, ϖ0,

ż

ζ2µ

)
,



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 18 of 40

Ξ(ςµ, ϖµ, ż) = Ξ(pϖµ−1, pςµ, ż) ≤ Ξ

(
ς0, ϖ0,

ż

ζ2µ

)
and

Π(ςµ+1, ϖµ, ż) = Π(pϖµ, pςµ, ż) ≥ Π

(
ς0, ϖ0,

ż

ζ2µ+1

)
,

Ψ(ςµ+1, ϖµ, ż) = Ψ(pϖµ, pςµ, ż) ≤ Ψ

(
ς0, ϖ0,

ż

ζ2µ+1

)
,

Ξ(ςµ+1, ϖµ, ż) = Ξ(pϖµ, pςµ, ż) ≤ Ξ

(
ς0, ϖ0,

ż

ζ2µ+1

)
.

Letting µ < m, for µ,m ∈ N. Then,

Π(ςµ, ϖm, ż) ≥ Π(ςµ, ϖµ,
ż

3
)⋇Π(ςµ+1, ϖµ,

ż

3
)⋇Π(ςµ+1, ϖm,

ż

3
)

...

≥ Π(ςµ, ϖµ,
ż

3
)⋇Π(ςµ+1, ϖµ,

ż

3
)⋇ · · ·⋇Π(ςm−1, ϖm−1,

ż

3m−1
)

⋇Π(ςm, ϖm−1,
ż

3m−1
)⋇Π(ςm, ϖm,

ż

3m−1
),

Ψ(ςµ, ϖm, ż) ≤ Ψ(ςµ, ϖµ,
ż

3
)♢Ψ(ςµ+1, ϖµ,

ż

3
)♢Ψ(ςµ+1, ϖm,

ż

3
)

...

≤ Ψ(ςµ, ϖµ,
ż

3
)♢Ψ(ςµ+1, ϖµ,

ż

3
)♢ · · ·♢Ψ(ςm−1, ϖm−1,

ż

3m−1
)

♢Ψ(ςm, ϖm−1,
ż

3m−1
)♢Ψ(ςm, ϖm,

ż

3m−1
)

and

Ξ(ςµ, ϖm, ż) ≤ Ξ(ςµ, ϖµ,
ż

3
)♢Ξ(ςµ+1, ϖµ,

ż

3
)♢Ξ(ςµ+1, ϖm,

ż

3
)

...

≤ Ξ(ςµ, ϖµ,
ż

3
)♢Ξ(ςµ+1, ϖµ,

ż

3
)♢ · · ·♢Ξ(ςm−1, ϖm−1,

ż

3m−1
)

♢Ξ(ςm, ϖm−1,
ż

3m−1
)♢Ξ(ςm, ϖm,

ż

3m−1
).

Therefore,

Π(ςµ, ϖm, ż) ≥ Π(ςµ, ϖµ,
ż

3
)⋇Π(ςµ+1, ϖµ,

ż

3
)⋇ · · ·⋇Π(ςm−1, ϖm−1,

ż

3m−1
)



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 19 of 40

⋇Π(ςm, ϖm−1,
ż

3m−1
)⋇Π(ςm, ϖm,

ż

3m−1
)

≥ Π(ς0, ϖ0,
ż

3ζ2µ
)⋇Π(ς0, ϖ0,

ż

3ζ2µ+1
)⋇ · · ·⋇Π(ς0, ϖ0,

ż

3m−1ζ2m−2
)

⋇Π(ς0, ϖ0,
ż

3m−1ζ2m−1
)⋇Π(ς0, ϖ0,

ż

3m−1ζ2m
),

Ψ(ςµ, ϖm, ż) ≤ Ψ(ςµ, ϖµ,
ż

3
)♢Ψ(ςµ+1, ϖµ,

ż

3
)♢ · · ·♢Ψ(ςm−1, ϖm−1,

ż

3m−1
)

♢Ψ(ςm, ϖm−1,
ż

3m−1
)♢Ψ(ςm, ϖm,

ż

3m−1
)

≤ Ψ(ς0, ϖ0,
ż

3ζ2µ
)♢Ψ(ς0, ϖ0,

ż

3ζ2µ+1
)♢ · · ·♢Ψ(ς0, ϖ0,

ż

3m−1ζ2m−2
)

♢Ψ(ς0, ϖ0,
ż

3m−1ζ2m−1
)♢Ψ(ς0, ϖ0,

ż

3m−1ζ2m
),

and

Ξ(ςµ, ϖm, ż) ≤ Ξ(ςµ, ϖµ,
ż

3
)♢Ξ(ςµ+1, ϖµ,

ż

3
)♢ · · ·♢Ξ(ςm−1, ϖm−1,

ż

3m−1
)

♢Ξ(ςm, ϖm−1,
ż

3m−1
)♢Ξ(ςm, ϖm,

ż

3m−1
)

≤ Ξ(ς0, ϖ0,
ż

3ζ2µ
)♢Ξ(ς0, ϖ0,

ż

3ζ2µ+1
)♢ · · ·♢Ξ(ς0, ϖ0,

ż

3m−1ζ2m−2
)

♢Ξ(ς0, ϖ0,
ż

3m−1ζ2m−1
)♢Ξ(ς0, ϖ0,

ż

3m−1ζ2m
).

Which implies that,

Π(ςµ, ϖm, ż) ≥ Π(ς0, ϖ0,
ż

3ζ2µ
)⋇Π(ς0, ϖ0,

ż

3ζ2µ+1
)⋇ · · ·⋇Π(ς0, ϖ0,

ż

3m−1ζ2m−2
)

⋇Π(ς0, ϖ0,
ż

3m−1ζ2m−1
)⋇Π(ς0, ϖ0,

ż

3m−1ζ2m
),

Ψ(ςµ, ϖm, ż) ≤ Ψ(ς0, ϖ0,
ż

3ζ2µ
)♢Ψ(ς0, ϖ0,

ż

3ζ2µ+1
)♢ · · ·♢Ψ(ς0, ϖ0,

ż

3m−1ζ2m−2
)

♢Ψ(ς0, ϖ0,
ż

3m−1ζ2m−1
)♢Ψ(ς0, ϖ0,

ż

3m−1ζ2m
),

and

Ξ(ςµ, ϖm, ż) ≤ Ξ(ς0, ϖ0,
ż

3ζ2µ
)♢Ξ(ς0, ϖ0,

ż

3ζ2µ+1
)♢ · · ·♢Ξ(ς0, ϖ0,

ż

3m−1ζ2m−2
)

♢Ξ(ς0, ϖ0,
ż

3m−1ζ2m−1
)♢Ξ(ς0, ϖ0,

ż

3m−1ζ2m
).



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 20 of 40

As µ,m → +∞, we deduce

lim
µ,m→+∞

Π(ςµ, ϖm, ż) = 1⋇ 1⋇ · · ·⋇ 1 = 1,

lim
µ,m→+∞

Ψ(ςµ, ϖm, ż) = 0♢0♢ · · ·♢0 = 0

and

lim
µ,m→+∞

Ξ(ςµ, ϖm, ż) = 0♢0♢ · · ·♢0 = 0.

Which implies that bisequence (ςµ, ϖµ) is a Cauchy bisequence.
Since (𭟋,S,Π,Ψ,Ξ,⋇,♢) is a complete NBMS. Then, {ςµ} → ς and {ϖµ} → ς, where
ς ∈ 𭟋 ∩ S. Using v, x and xv, we get

Π(ς, pς, ż) ≥ Π

(
ς, ςµ+1,

ż

3

)
⋇Π

(
ςµ+1, ςµ+1,

ż

3

)
⋇Π

(
ςµ+1, pς,

ż

3

)
= Π

(
ς, ςµ+1,

ż

3

)
⋇Π

(
pςµ, pςµ,

ż

3

)
⋇Π

(
pςµ, pς,

ż

3

)
→ 1⋇ 1⋇ 1 = 1 as µ→ +∞,

Ψ(ς, pς, ż) ≤ Ψ

(
ς, ςµ+1,

ż

3

)
♢Ψ

(
ςµ+1, ςµ+1,

ż

3

)
♢Ψ

(
ςµ+1, pς,

ż

3

)
= Ψ

(
ς, ςµ+1,

ż

3

)
♢Ψ

(
pςµ+1, pςµ+1,

ż

3

)
♢Ψ

(
pςµ+1, pς,

ż

3

)
→ 0♢0♢0 = 0 as µ→ +∞

and

Ξ(ς, pς, ż) ≤ Ξ

(
ς, ςµ+1,

ż

3

)
♢Ξ

(
ςµ+1, ςµ+1,

ż

3

)
♢Ξ

(
ςµ+1, pς,

ż

3

)
= Ξ

(
ς, ςµ+1,

ż

3

)
♢Ξ

(
pςµ+1, pςµ+1,

ż

3

)
♢Ξ

(
pςµ+1, pς,

ż

3

)
→ 0♢0♢0 = 0 as µ→ +∞.

Hence, pς = ς. Let pη = η for some η ∈ 𭟋 ∩ S, then

1 ≥ Π(η, ς, ż) = Π(pς, pη, ż) ≥ Π

(
η, ς,

ż

ζ

)
= Π

(
pς, pη,

ż

ζ

)
≥ Π

(
η, ς,

ż

ζ2

)
≥ · · · ≥ Π

(
η, ς,

ż

ζµ

)
→ 1 as µ→ +∞,

0 ≤ Ψ(η, ς, ż) = Ψ(pς, pη, ż) ≤ Ψ

(
η, ς,

ż

ζ

)
= Ψ

(
pς, pη,

ż

ζ

)



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 21 of 40

≤ Ψ

(
η, ς,

ż

ζ2

)
≤ · · · ≤ Ψ

(
η, ς,

ż

ζµ

)
→ 0 as µ→ +∞,

and

0 ≤ Ξ(η, ς, ż) = Ξ(pς, pη, ż) ≤ Ξ

(
η, ς,

ż

ζ

)
= Ξ

(
pς, pη,

ż

ζ

)
≤ Ξ

(
η, ς,

ż

ζ2

)
≤ · · · ≤ Ξ

(
η, ς,

ż

ζµ

)
→ 0 as µ→ +∞,

since iii, viii and xiii, we get ς = η.

Definition 15. Let (𭟋,S,Π,Ψ,Ξ,⋇,♢) be a NBMS. A map p : 𭟋 ∪ S → 𭟋 ∪ S is an
NB(neutrosophic bipolar)-contraction if we can find 0 < ζ < 1, satisfying

1

Π(pς, pϖ, ż)
− 1 ≤ ζ

[
1

Π(ς,ϖ, ż)
− 1

]
(7)

Ψ(pς, pϖ, ż) ≤ ζΨ(ς,ϖ, ż), (8)

and

Ξ(pς, pϖ, ż) ≤ ζΞ(ς,ϖ, ż), (9)

∀ ς ∈ 𭟋 , ϖ ∈ S and ż > 0.

Now, we present the following theorem for NB(neutrosophic bipolar)-contraction.

Theorem 7. Let (𭟋,S,Π,Ψ,Ξ,⋇,♢) be a complete NBMS. Let p : 𭟋 ∪ S → 𭟋 ∪ S be a
mapping satisfying

i. p(𭟋) ⊆ 𭟋 and p(S) ⊆ S;

ii. p is NB-contraction, ∀ ς ∈ 𭟋 , ϖ ∈ S and ż > 0.

Then, p has a unique fixed point.
Proof. Let ς0 ∈ 𭟋 and ϖ0 ∈ S and assume that p(ςµ) = ςµ+1 and

p(ϖµ) = ϖµ+1 ∀ µ ∈ N ∪ {0}. Then we get (ςµ, ϖµ)as a bisequence on NBMS
(𭟋,S,Π,Ψ,Ξ,⋇,♢). By using (7), (8) and (9) ∀ ż > 0, we deduce

1

Π(ςµ, ϖµ, ż)
− 1 =

1

Π(pςµ−1, pϖµ−1, ż)
− 1 ≤ ζ

[
1

Π(ςµ−1, ϖµ−1, ż)

]
=

ζ

Π(ςµ−1, ϖµ−1, ż)
− ζ

⇒ 1

Π(ςµ, ϖµ, ż)
≤ ζ

Π(ςµ−1, ϖµ−1, ż)
+ (1− ζ)

≤ ζ2

Π(ςµ−2, ϖµ−2, ż)
+ ζ(1− ζ) + (1− ζ).



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 22 of 40

In this manner, we obtain

1

Π(ςµ, ϖµ, ż)
≤ ζµ

Π(ς0, ϖ0, ż)
+ ζµ−1(1− ζ)

+ ζµ−2(1− ζ) + · · ·+ ζ(1− ζ) + (1− ζ)

≤ ζµ

Π(ς0, ϖ0, ż)
+ (ζµ−1 + ζµ−2 + · · ·+ 1)(1− ζ)

≤ ζµ

Π(ς0, ϖ0, ż)
+ (1− ζµ)

We obtain

1
ζµ

Π(ς0,ϖ0,ż)
+ (1− ζµ)

≤ Π(ςµ, ϖµ, ż) (10)

Ψ(ςµ, ϖµ, ż) = Ψ(pςµ−1, pϖµ−1, ż) ≤ ζΨ(ςµ−1, ϖµ−1, ż) = Ψ(pςµ−2, pϖµ−2, ż)

≤ ζ2Ψ(ςµ−2, ςϖ−2, ż) ≤ · · · ≤ ζµΨ(ς0, ϖ0, ż), (11)

Ξ(ςµ, ϖµ, ż) = Ξ(pςµ−1, pϖµ−1, ż) ≤ ζΞ(ςµ−1, ϖµ−1, ż) = Ξ(pςµ−2, pϖµ−2, ż)

≤ ζ2Ξ(ςµ−2, ϖµ−2, ż) ≤ · · · ≤ ζµΞ(ς0, ϖ0, ż) (12)

and

1
ζµ

Π(ς1,ϖ0,ż)
+ (1− ζµ)

≤ Π(ςµ+1, ϖµ, ż) (13)

Ψ(ςµ+1, ϖµ, ż) = Ψ(pςµ, pϖµ−1, ż) ≤ ζΨ(ςµ, ϖµ−1, ż) = Ψ(pςµ−1, pϖµ−2, ż)

≤ ζ2Ψ(ςµ−1, ςϖ−2, ż) ≤ · · · ≤ ζµΨ(ς1, ϖ0, ż), (14)

Ξ(ςµ+1, ϖµ, ż) = Ξ(pςµ, pϖµ−1, ż) ≤ ζΞ(ςµ, ϖµ−1, ż) = Ξ(pςµ−1, pϖµ−2, ż)

≤ ζ2Ξ(ςµ−1, ϖµ−2, ż) ≤ · · · ≤ ζµΞ(ς1, ϖ0, ż). (15)

Letting µ < m, for µ,m ∈ N. Then,

Π(ςµ, ϖm, ż) ≥ Π(ςµ, ϖµ,
ż

3
)⋇Π(ςµ+1, ϖµ,

ż

3
)⋇Π(ςµ+1, ϖm,

ż

3
)

...

≥ Π(ςµ, ϖµ,
ż

3
)⋇Π(ςµ+1, ϖµ,

ż

3
)⋇ · · ·⋇Π(ςm−1, ϖm−1,

ż

3m−1
)

⋇Π(ςm, ϖm−1,
ż

3m−1
)⋇Π(ςm, ϖm,

ż

3m−1
),



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 23 of 40

Ψ(ςµ, ϖm, ż) ≤ Ψ(ςµ, ϖµ,
ż

3
)♢Ψ(ςµ+1, ϖµ,

ż

3
)♢Ψ(ςµ+1, ϖm,

ż

3
)

...

≤ Ψ(ςµ, ϖµ,
ż

3
)♢Ψ(ςµ+1, ϖµ,

ż

3
)♢ · · ·♢Ψ(ςm−1, ϖm−1,

ż

3m−1
)

♢Ψ(ςm, ϖm−1,
ż

3m−1
)♢Ψ(ςm, ϖm,

ż

3m−1
),

and

Ξ(ςµ, ϖm, ż) ≤ Ξ(ςµ, ϖµ,
ż

3
)♢Ξ(ςµ+1, ϖµ,

ż

3
)♢Ξ(ςµ+1, ϖm,

ż

3
)

...

≤ Ξ(ςµ, ϖµ,
ż

3
)♢Ξ(ςµ+1, ϖµ,

ż

3
)♢ · · ·♢Ξ(ςm−1, ϖm−1,

ż

3m−1
)

♢Ξ(ςm, ϖm−1,
ż

3m−1
)♢Ξ(ςm, ϖm,

ż

3m−1
).

Therefore,

Π(ςµ, ϖm, ż) ≥ Π(ςµ, ϖµ,
ż

3
)⋇Π(ςµ+1, ϖµ,

ż

3
)⋇ · · ·⋇Π(ςm−1, ϖm−1,

ż

3m−1
)

⋇Π(ςm, ϖm−1,
ż

3m−1
)⋇Π(ςm, ϖm,

ż

3m−1
)

≥ 1
ζµ

Π(ς0,ϖ0,
ż
3
)
+ (1− ζµ)

⋇
1

ζµ

Π(ς1,ϖ0,
ż
3
)
+ (1− ζµ)

⋇ · · ·

⋇
1

ζm−1

Π(ς0,ϖ0,
ż

3m−1 )
+ (1− ζm−1)

⋇
1

ζm−1

Π(ς1,ϖ0,
ż

3m−1 )
+ (1− ζm−1)

⋇
1

ζm

Π(ς0,ϖ0,
ż

3m−1 )
+ (1− ζm)

,

Ψ(ςµ, ϖm, ż) ≤ Ψ(ςµ, ϖµ,
ż

3
)♢Ψ(ςµ+1, ϖµ,

ż

3
)♢ · · ·♢Ψ(ςm−1, ϖm−1,

ż

3m−1
)

♢Ψ(ςm, ϖm−1,
ż

3m−1
)♢Ψ(ςm, ϖm,

ż

3m−1
)

≤ ζµΨ(ς0, ϖ0,
ż

3
)♢ζµΨ(ς1, ϖ0,

ż

3m−1
)♢ · · ·♢ζm−1Ψ(ς0, ϖ0,

ż

3m−1
)

♢ζm−1Ψ(ς1, ϖ0,
ż

3m−1
)♢ζmΨ(ς0, ϖ0,

ż

3m−1
),

and

Ξ(ςµ, ϖm, ż) ≤ Ξ(ςµ, ϖµ,
ż

3
)♢Ξ(ςµ+1, ϖµ,

ż

3
)♢ · · ·♢Ξ(ςm−1, ϖm−1,

ż

3m−1
)



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 24 of 40

♢Ξ(ςm, ϖm−1,
ż

3m−1
)♢Ξ(ςm, ϖm,

ż

3m−1
)

≤ ζµΞ(ς0, ϖ0,
ż

3
)♢ζµΞ(ς1, ϖ0,

ż

3m−1
)♢ · · ·♢ζm−1Ξ(ς0, ϖ0,

ż

3m−1
)

♢ζm−1Ξ(ς1, ϖ0,
ż

3m−1
)♢ζmΞ(ς0, ϖ0,

ż

3m−1
).

Which implies that,

Π(ςµ, ϖm, ż) ≥
1

ζµ

Π(ς0,ϖ0,
ż
3
)
+ (1− ζµ)

⋇
1

ζµ

Π(ς1,ϖ0,
ż
3
)
+ (1− ζµ)

⋇ · · ·

⋇
1

ζm−1

Π(ς0,ϖ0,
ż

3m−1 )
+ (1− ζm−1)

⋇
1

ζm−1

Π(ς1,ϖ0,
ż

3m−1 )
+ (1− ζm−1)

⋇
1

ζm

Π(ς0,ϖ0,
ż

3m−1 )
+ (1− ζm)

,

Ψ(ςµ, ϖm, ż) ≤ ζµΨ(ς0, ϖ0,
ż

3
)♢ζµΨ(ς1, ϖ0,

ż

3m−1
)♢ · · ·♢ζm−1Ψ(ς0, ϖ0,

ż

3m−1
)

♢ζm−1Ψ(ς1, ϖ0,
ż

3m−1
)♢ζmΨ(ς0, ϖ0,

ż

3m−1
),

and

Ξ(ςµ, ϖm, ż) ≤ ζµΞ(ς0, ϖ0,
ż

3
)♢ζµΞ(ς1, ϖ0,

ż

3m−1
)♢ · · ·♢ζm−1Ξ(ς0, ϖ0,

ż

3m−1
)

♢ζm−1Ξ(ς1, ϖ0,
ż

3m−1
)♢ζmΞ(ς0, ϖ0,

ż

3m−1
).

As µ,m → +∞, we deduce

lim
µ,m→+∞

Π(ςµ, ϖm, ż) = 1⋇ 1⋇ · · ·⋇ 1 = 1,

lim
µ,m→+∞

Ψ(ςµ, ϖm, ż) = 0♢0♢ · · ·♢0 = 0

and

lim
µ,m→+∞

Ξ(ςµ, ϖm, ż) = 0♢0♢ · · ·♢0 = 0.

Which implies that bisequence (ςµ, ϖµ) is a Cauchy bisequence.
Since (𭟋,S,Π,Ψ,Ξ,⋇,♢) is a complete NBMS. Then, {ςµ} → ς and {ϖµ} → ς, where
ς ∈ 𭟋 ∩ S. Using v, x and xv, we get

Π(ς, pς, ż) ≥ Π

(
ς, ςµ+1,

ż

3

)
⋇Π

(
ςµ+1, ςµ+1,

ż

3

)
⋇Π

(
ςµ+1, pς,

ż

3

)



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 25 of 40

= Π

(
ς, ςµ+1,

ż

3

)
⋇Π

(
pςµ, pςµ,

ż

3

)
⋇Π

(
pςµ, pς,

ż

3

)
≥ Π

(
ς, ςµ+1,

ż

3

)
⋇

1
ζµ+1

Π(ς0,ϖ0,
ż
3
)
+ (1− ζµ+1)

⋇Π

(
pςµ, pς,

ż

3

)
→ 1⋇ 1⋇ 1 = 1 as µ→ +∞,

Ψ(ς, pς, ż) ≤ Ψ

(
ς, ςµ+1,

ż

3

)
♢Ψ

(
ςµ+1, ςµ+1,

ż

3

)
♢Ψ

(
ςµ+1, pς,

ż

3

)
= Ψ

(
ς, ςµ+1,

ż

3

)
♢Ψ

(
pςµ+1, pςµ+1,

ż

3

)
♢Ψ

(
pςµ+1, pς,

ż

3

)
≤ Ψ

(
ς, ςµ+1,

ż

3

)
♢ζµ+1Ψ(ς0, ϖ0,

ż

3
)♢Ψ

(
pςµ+1, pς,

ż

3

)
→ 0♢0♢0 = 0 as µ→ +∞

and

Ξ(ς, pς, ż) ≤ Ξ

(
ς, ςµ+1,

ż

3

)
♢Ξ

(
ςµ+1, ςµ+1,

ż

3

)
♢Ξ

(
ςµ+1, pς,

ż

3

)
= Ξ

(
ς, ςµ+1,

ż

3

)
♢Ξ

(
pςµ+1, pςµ+1,

ż

3

)
♢Ξ

(
pςµ+1, pς,

ż

3

)
≤ Ξ

(
ς, ςµ+1,

ż

3

)
♢ζµ+1Ψ(ς0, ϖ0,

ż

3
)♢Ξ

(
pςµ+1, pς,

ż

3

)
→ 0♢0♢0 = 0 as µ→ +∞.

Hence, pς = ς. Let pη = η for some η ∈ 𭟋, then

1

Π(ς, η, ż)
− 1 =

1

Π(pς, pη, ż)
− 1

≤ ζ

[
1

Π(ς, η, ż)
− 1

]
<

1

Π(ς, η, ż)
− 1,

which is a contradiction.

Ψ(ς, η, ż) = Ψ(pς, pη, ż) ≤ ζΨ(ς, η, ż) < Ψ(ς, η, ż),

which is a contradiction and

Ξ(ς, η, ż) = Ξ(pς, pη, ż) ≤ ζΞ(ς, η, ż) < Ξ(ς, η, ż),

which is a contradiction. Therefore, we get Π(ς, η, ż) = 1,Ψ(ς, η, ż) = 0 and Ξ(ς, η, ż) = 0,
hence, ς = η.



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 26 of 40

Example 2. Let 𭟋 = [0, 1] and S = {0}∪N−{1}. Define Π,Ψ,Ξ: 𭟋×S×(0,+∞) → [0, 1]
as

Π(ς,ϖ, ż) =
ż

ż+ |ς −ϖ|
,

Ψ(ς,ϖ, ż) =
|ς −ϖ|

ż+ |ς −ϖ|
,

Ψ(ς,ϖ, ż) =
|ς −ϖ|

ż
.

Then, (𭟋,S,Π,Ψ,Ξ,⋇,♢) is a complete NBMS with ct−||.|| ė ⋇ ȧ = ėȧ and ct−co−||.||
ė♢ȧ = max{ė, ȧ}.
Define p : (𭟋,S,Π,Ψ,Ξ,⋇,♢) ⇒ (𭟋,S,Π,Ψ,Ξ,⋇,♢) by

p(ς) =

{
1−3−ς

5 , if ς ∈ [0, 1],

0, if ς ∈ N− {1},

∀ ς ∈ 𭟋 ∪ S and take ζ ∈ [12 , 1), then

Π(pς, pϖ, ζ ż) = Π

(
1− 3−ς

5
,
1− 3−ϖ

5
, ζ ż

)
=

ζ ż

ζ ż+

∣∣∣∣1−3−ς

5 − 1−3−ϖ

5

∣∣∣∣ =
ζ ż

ζ ż+ |3−ς−3−ϖ|
5

≥ ζ ż

ζ ż+ |ς−ϖ|
5

=
5ζ ż

5ζ ż+ |ς −ϖ|
≥ ż

ż+ |ς −ϖ|
= Π(ς,ϖ, ż),

Ψ(pς, pϖ, ζ ż) = Ψ

(
1− 3−ς

5
,
1− 3−ϖ

5
, ζ ż

)

=

∣∣∣∣1−3−ς

5 − 1−3−ϖ

5

∣∣∣∣
ζ ż+

∣∣∣∣1−3−ς

5 − 1−3−ϖ

5

∣∣∣∣ =
|3−ς−3−ϖ|

5

ζ ż+ |3−ς−3−ϖ|
5

=
|3−ς − 3−ϖ|

5ζ ż+ |3−ς − 3−ϖ|
≤ |ς −ϖ|

5ζ ż+ |ς −ϖ|
≤ |ς −ϖ|

ż+ |ς −ϖ|
= Ψ(ς,ϖ, ż)

and

Ξ(pς, pϖ, ζ ż) = Ξ

(
1− 3−ς

5
,
1− 3−ϖ

5
, ζ ż

)

=

∣∣∣∣1−3−ς

5 − 1−3−ϖ

5

∣∣∣∣
ζ ż

=
|3−ς−3−ϖ|

5

ζ ż



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 27 of 40

=
|3−ς − 3−ϖ|

5ζ ż
≤ |ς −ϖ|

5ζ ż
≤ |ς −ϖ|

ż
= Ξ(ς,ϖ, ż).

Therefore, all the hypothesis of Theorem 5 are satisfied, and 0 is the only fixed point for
p.

Example 3. Let 𭟋 = {Uµ(R) : Uµ(R) is an upper triangular matrices over R} and S =
{Lµ(R) : Lµ(R) is an upper triangular matrices over R}.
Define Π,Ψ,Ξ: 𭟋× S × (0,+∞) → [0, 1] as

Π(R,Q, ż) = ż

ż+
∑µ

i,j=1 |rij − qij|
,

Ψ(R,Q, ż) =
∑µ

i,j=1 |rij − qij|
ż+

∑µ
i,j=1 |rij − qij|

,

Ψ(R,Q, ż) =
∑µ

i,j=1 |rij − qij|
ż

,

for all R = (rij)µ×µ ∈ 𭟋 and Q = (qij)µ×µ ∈ S, Then, (𭟋,S,Π,Ψ,Ξ,⋇,♢) is a complete
NBMS with ct−||.|| ė⋇ ȧ = ėȧ and ct−co−||.|| ė♢ȧ = max{ė, ȧ}.
Define p : (𭟋,S,Π,Ψ,Ξ,⋇,♢) ⇒ (𭟋,S,Π,Ψ,Ξ,⋇,♢) by

p((rij)µ×µ) =

(
rij
5

)
µ×µ

,

∀ (rij)µ×µ ∈ 𭟋 ∪ S and take ζ ∈ [12 , 1), then

Π(pR, pQ, ζ ż) = Π

((
rij
5

)
µ×µ

,

(
qij
5

)
µ×µ

, ζ ż

)
=

ζ ż

ζ ż+ 1
5

∑µ
i,j=1 |rij − qij|

≥ ζ ż

ζ ż+
∑µ

i,j=1 |rij − qij|

≥ ż

ż+
∑µ

i,j=1 |rij − qij|
= Π(R,Q, ż),

Ψ(pR, pQ, ζ ż) = Π

((
rij
5

)
µ×µ

,

(
qij
5

)
µ×µ

, ζ ż

)
=

1
5

∑µ
i,j=1 |rij − qij|

ζ ż+ 1
5

∑µ
i,j=1 |rij − qij|

=

∑µ
i,j=1 |rij − qij|

5ζ ż+
∑µ

i,j=1 |rij − qij|
≤

∑µ
i,j=1 |rij − qij|

ż+
∑µ

i,j=1 |rij − qij|
= Ψ(R,Q, ż)



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 28 of 40

and

Ξ(pR, pQ, ζ ż) = Ξ

((
rij
5

)
µ×µ

,

(
qij
5

)
µ×µ

, ζ ż

)
=

1
5

∑µ
i,j=1 |rij − qij|

ζ ż

≤
∑µ

i,j=1 |rij − qij|
ζ ż

≤
∑µ

i,j=1 |rij − qij|
ż

= Ξ(R,Q, ż).

Therefore, all the hypothesis of Theorem 5 are satisfied, and Oµ×µ is the unique fixed point
for p, where Oµ×µ is the null matrix of order µ.

4. Application 1

Consider the set of all continuous functions 𭟋 = C([c, a], [0,+∞)) defined on [c, a] with
values in the interval [0,+∞) and S = C([c, a], (−∞, 0]) defined on [c, a] with values in the
interval (−∞, 0].
Suppose the integral equation:

ς(l) = ∧(l) + δ

∫ a

c
℧(l, ε)ς(l)dε for l, ε ∈ [c, a] (16)

where δ > 0,℧ : C([c, a] × R) → R+, ∧(ε) is a fuzzy function of ε : ε ∈ [c, a]. Define Π, Ψ
and Ξ by

Π(ς(l), ϖ(l), ż) = sup
l∈[c,a]

ż

ż+ |ς(l)−ϖ(l)|
∀ ς,ϖ ∈ 𭟋 and ż > 0,

Ψ(ς(o), ϖ(o), ż) = 1− sup
o∈[c,a]

ż

ż+ |ς(o)−ϖ(o)|
∀ ς,ϖ ∈ 𭟋 and ż > 0,

and

Ξ(ς(o), ϖ(o), ż) = sup
o∈[c,a]

|ς(o)−ϖ(o)|
ż

∀ ς,ϖ ∈ 𭟋 and ż > 0,

with ct−||.|| and ct−co−||.|| define by i⋇ ♭ = i♭ and i♢♭ = max{i, ♭}. Then (𭟋,Π,Ψ,Ξ,⋇,♢)
is a complete NBMS. Consider |℧(o, ε)ς(o)− ℧(o, ε)ϖ(o)| ≤ |ς(o)−ϖ(o)| for ς ∈ 𭟋, ϖ ∈
S, ζ ∈ (0, 1) and ∀o, ε ∈ [c, a].

Also, let ℧(o, ε)(δ
∫ a
c dε) ≤ ζ < 1. Then, the integral Equation (16) has a unique

solution.
Proof. Define p : (𭟋,S,Π,Ψ,Ξ,⋇,♢) ⇒ (𭟋,S,Π,Ψ,Ξ,⋇,♢) by

pς(o) = ∧(o) + δ

∫ a

c
℧(o, ε)ς(o)dε ∀ o, ε ∈ [c, a].



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 29 of 40

Now, ∀ ς,ϖ ∈ 𭟋 ∪ S, we deduce

Π(pς(o),pϖ(o), ζ ż) = sup
o∈[c,a]

ζ ż

ζ ż+ |pς(o)− pϖ(o)|

= sup
o∈[c,a]

ζ ż

ζ ż+ | ∧ (o) + δ
∫ a
c ℧(o, ε)ς(o)dε− ∧(o)− δ

∫ a
c ℧(o, ε)ς(o)dε|

= sup
o∈[c,a]

ζ ż

ζ ż+ |δ
∫ a
c ℧(o, ε)ς(o)dε− δ

∫ a
c ℧(o, ε)ς(o)dε|

= sup
o∈[c,a]

ζ ż

ζ ż+ |℧(o, ε)ς(o)− ℧(o, ε)ϖ(o)|(δ
∫ a
c dε)

≥ sup
o∈[c,a]

ż

ż+ |ς(o)−ϖ(o)|

≥ Π(ς(o), ϖ(o), ż),

Ψ(pς(o), pϖ(o), ζ ż) = 1− sup
o∈[c,a]

ζ ż

ζ ż+ |pς(o)− pϖ(o)|

= 1− sup
o∈[c,a]

ζ ż

ζ ż+ | ∧ (o) + δ
∫ a
c ℧(o, ε)ς(o)dε− ∧(o)− δ

∫ a
c ℧(o, ε)ς(o)dε|

= 1− sup
o∈[c,a]

ζ ż

ζ ż+ |δ
∫ a
c ℧(o, ε)ς(o)dε− δ

∫ a
c ℧(o, ε)ς(o)dε|

= 1− sup
o∈[c,a]

ζ ż

ζ ż+ |℧(o, ε)ς(o)− ℧(o, ε)ϖ(o)|(δ
∫ a
c dε)

≤ 1− sup
o∈[c,a]

ż

ż+ |ς(o)−ϖ(o)|

≤ Ψ(ς(o), ϖ(o), ż),

and

Ξ(pς(o), pϖ(o),ζ ż) = sup
o∈[c,a]

|pς(o)− pϖ(o)|
ζ ż

= sup
o∈[c,a]

| ∧ (o) + δ
∫ a
c ℧(o, ε)ς(o)dε− ∧(o)− δ

∫ a
c ℧(o, ε)ς(o)dε|

ζ ż

= sup
o∈[c,a]

|δ
∫ a
c ℧(o, ε)ς(o)dε− δ

∫ a
c ℧(o, ε)ς(o)dε|

ζ ż

= sup
o∈[c,a]

|℧(o, ε)ς(o)− ℧(o, ε)ϖ(o)|(δ
∫ a
c dε)

ζ ż

≤ sup
o∈[c,a]

|ς(o)−ϖ(o)|
ż



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 30 of 40

≤ Ξ(ς(o), ϖ(o), ż).

Therefore, all the hypothesis of Theorem 5 are satisfied and p has a unique fixed point
and the integral equation (16) has a unique solution.

Example 4. Consider the the non-linear integral equation.

ς(o) = | cos o|+ 1

9

∫ 1

0
ες(ε)dε, ∀ ε ∈ [0, 1]

Then it has a solution in 𭟋.
Proof. Let p : (𭟋,S,Π,Ψ,Ξ,⋇,♢) ⇒ (𭟋,S,Π,Ψ,Ξ,⋇,♢) be defined by

pς(o) = | cos o|+ 1

9

∫ 1

0
ες(ε)dε,

and set ℧(o, ε)ς(o) = 1
9ες(ε) and ℧(o, ε)ϖ(o) = 1

9εϖ(ε), where ς,ϖ ∈ 𭟋 ∪ S, and ∀ o, ε ∈
[0, 1]. Then we have

|℧(o, ε)ς(o)− ℧(o, ε)ϖ(o)| = |1
9
ες(ε)− 1

9
εϖ(ε)|

=
ε

9
|ς(ε)−ϖ(ε)| ≤ |ς(ε)−ϖ(ε)|.

Furthermore, we have 1
9

∫ 1
0 εdε =

1
9

( (1)2
2 − (0)2

2

)
= 1

18 = ζ < 1, with δ = 1
9 . Hence, all the

conditions of the application are easy to verify and the integral equation (16) has a unique
solution 𭟋 ∪ S.

Using Mathematica Software, near to the unique solution for the integral equation of
Example 4 is found to be

a(τ) = | cos τ |+ 0.0444392,

and the graph of the solution is shown in Figure 1.

5. Application 2

Consider the definition of the intensity of a series electric circuit I = dϖ
dl , where ϖ

denote the electric charge and l-the time, let us recall the following usually formulas

• V = IR;

• V = ϖ
C



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 31 of 40

Figure 1: Solution of Example 4.1.

• V = L dI
dl ,

here

i. R (Ohms) is a resistor,

ii. C (Faradays) is a capacitor,

iii. L (Henries) is an inductor,

iv. V (Volts) is an voltage and

v. E (Volts) is an electromotive force.

Because there is only one current flowing in a series circuit, mathcalI has the same value
throughout the circuit. Kirchhoff’s Voltage Law is the second of his fundamental laws that
can be used to analyse circuits. His voltage law states that the algebraic sum of all voltages
around any closed loop in a circuit is equal to zero for a closed loop series path. The



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 32 of 40

algebraic sum of all the voltages around any closed loop in a circuit equals zero, according
to Kirchhoff’s Voltage Law.

The main idea behind Kirchhoff’s Voltage Law is that as you move around a closed
loop/circuit, you will end up back where you started. As a result, you return to the same
initial potential without any voltage losses around the loop. As a result, any voltage drop
around the loop must be equal to any voltage source encountered along the way. The math-
ematical expression for this consequence of Kirchhoff’s Voltage Law is: the sum of voltage
rises across any loop is equal to the sum of voltage drops across that loop. Then we have
the following relation:

IR+
ϖ

C
+ LdI

dl
= V(l).

The voltage equation can be expressed as in the second-order differential equations with
parameters as follows.

Ld2ϖ

dl2
+Rdϖ

dl
+
ϖ

C
= V(l) = h(l, ς(l)),

with the initial conditions, ϖ(0) = 0, ϖ
′
(0) = 0, (17)

where C = 4L
R2 and τ = R

2L - the non dimensional time for Physics. The following Green
function associated with equation 17 is

Q(l, s) =

{
−si−τ(s−l), if0 ≤ s ≤ l ≤ 1;

−li−τ(s−l), if0 ≤ l ≤ s ≤ 1.

In equation 17 can be expressed as in the integral equation with the above condition is

ς(l) =

∫ l

0
Q(l, s)h(s, ς(s))ds, for all l ∈ [0, 1] (18)

and h(s, ·) : [0, 1]× R → R is a monotone non-decreasing mapping ∀ s ∈ [0, 1].
Consider the set of all continuous functions 𭟋 = (C[0, 1], [0,+∞)) defined on [0, 1] with
values in [0,+∞) and S = (C[0, 1], (−∞, 0]) defined on [0, 1] with values in (−∞, 0].
Define Π, Ψ and Ξ by

Π(ς(o), ϖ(o), ż) = sup
o∈[c,a]

ż

ż+ |ς(o)−ϖ(o)|
∀ ς,ϖ ∈ 𭟋 and ż > 0,

Ψ(ς(o), ϖ(o), ż) = 1− sup
o∈[c,a]

ż

ż+ |ς(o)−ϖ(o)|
∀ ς,ϖ ∈ 𭟋 and ż > 0,

and

Ξ(ς(o), ϖ(o), ż) = sup
o∈[c,a]

|ς(o)−ϖ(o)|
ż

∀ ς,ϖ ∈ 𭟋 and ż > 0,



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 33 of 40

Figure 2: Series RLC circuit.

with ct−||.|| and ct−co−||.|| define by i⋇ ♭ = i♭ and i♢♭ = max{i, ♭}. Then (𭟋,Π,Ψ,Ξ,⋇,♢)
is a complete NBMS.

Theorem 8. Let p : (𭟋,S,Π,Ψ,Ξ,⋇,♢) ⇒ (𭟋,S,Π,Ψ,Ξ,⋇,♢) be a map such that the
following axioms hold:

i. Q : [0, 1]2 → [0,∞) is a continuous function;

ii. h(s, ·) : [0, 1] × R → R is a monotone nondecreasing mapping ∀ s ∈ [0, 1] satisfying
(ς,ϖ) ∈ (𭟋,S),

|h(l, ς)− h(l, ϖ) ≤ |ς(l)−ϖ(l)|.

iii.
∫ l
0 Q(l, s)ds ≤ ζ < 1

Then the voltage differential equation (17) has a unique solution.

Proof. Define p : (𭟋,S,Π,Ψ,Ξ,⋇,♢) ⇒ (𭟋,S,Π,Ψ,Ξ,⋇,♢) by

pς(l) =

∫ l

0
Q(l, s)h(s, ς(s))ds, where l ∈ [0, 1]



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 34 of 40

Now, ∀ ς,ϖ ∈ 𭟋 ∪ S, we deduce

Π(pς(l), pϖ(l), ζ ż) = sup
l∈[0,1]

ζ ż

ζ ż+ |pς(l)− pϖ(l)|

= sup
l∈[0,1]

ζ ż

ζ ż+ |
∫ l
0 Q(l, s)h(s, ς(s))ds−

∫ l
0 Q(l, s)h(s, ϖ(s))ds|

= sup
l∈[0,1]

ζ ż

ζ ż+
∫ l
0 Q(l, s)|h(s, ς(s))− h(s, ϖ(s))|ds

= sup
l∈[0,1]

ζ ż

ζ ż+ |h(s, ς(s))− h(s, ϖ(s))|

≥ sup
l∈[0,1]

ż

ż+ |ς(l)−ϖ(l)|

≥ Π(ς(l), ϖ(l), ż),

Ψ(pς(l), pϖ(l), ζ ż) = 1− sup
l∈[0,1]

ζ ż

ζ ż+ |pς(l)− pϖ(l)|

= 1− sup
l∈[0,1]

ζ ż

ζ ż+ |
∫ l
0 Q(l, s)h(s, ς(s))ds−

∫ l
0 Q(l, s)h(s, ϖ(s))ds|

= 1− sup
l∈[0,1]

ζ ż

ζ ż+
∫ l
0 Q(l, s)|h(s, ς(s))− h(s, ϖ(s))|ds

≤ 1− sup
l∈[0,1]

ż

ż+ |ς(l)−ϖ(l)|

≤ Ψ(ς(l), ϖ(l), ż),

and

Ξ(pς(l), pϖ(l), ζ ż) = sup
l∈[0,1]

|pς(l)− pϖ(l)|
ζ ż

= sup
l∈[0,1]

|
∫ l
0 Q(l, s)h(s, ς(s))ds−

∫ l
0 Q(l, s)h(s, ϖ(s))ds|

ζ ż

= sup
l∈[0,1]

∫ l
0 Q(l, s)|h(s, ς(s))− h(s, ϖ(s))|ds

ζ ż

≤ sup
l∈[0,1]

|ς(l)−ϖ(l)|
ż

≤ Ξ(ς(l), ϖ(l), ż).

It can be seen that all conditions of Theorem (5) are satisfied and p has a unique fixed
point and the differential voltage equation (17) has a unique solution.



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 35 of 40

Now assume that R = 10, C = 1 and L = 1 whereas the voltage source is given by
V(t) = 5sin(t). Thus, the nearer form of the unique solution for circuit IVP is found
using Mathematica Software and expressed as

q(t) =
1

24
e−5t

(
5
√
6 sinh

(
2
√
6t
)
+ 12 cosh

(
2
√
6t
))

− cos(t)

2
,

and the graph of the solution is shown in Figure 3.

Figure 3: Solution of (5.1).



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 36 of 40

6. Application to Fractional Differential Equation

Consider the definition of Caputo derivative of a continuous function
p̄ : [0,+∞) → R order β̄ > 0 (See[2, 3]):

CDβ̄(p̄(l)) : =
1

Γ(µ− β̄)

∫ l

0
(l− s)µ−β̄−1p̄(µ)(s)ds(µ− 1 < β̄ < µ, µ = [β̄] + 1),

where Γ is a gamma function and [β̄] denotes the integer part of the real number β̄ > 0.
Additionally, we provide an application of the Theorem 5 for proving the existence solution
of the nonlinear fractional differential equation

CDβ̄(ς(l)) + h(l, ς(l)) = 0 (0 ≤ l ≤ 1, β̄ < 1) (19)

with ς(0) = 0 = ς(1) and h : [0, 1] × R → R is a continuous function (See[4, 34–36]). The
Green function related with (19) is

Q(l, s) =

{
(l(1− s))α−1 − (l− s)α−1, if 0 ≤ s ≤ l ≤ 1
(l(1−s))α−1

Γ(α) , if 0 ≤ l ≤ s ≤ 1.

Obviously ς∗ ∈ 𭟋 is a solution of (19) if and only if ς∗ ∈ 𭟋 is a solution of the equation

ς(l) =

∫ l

0
Q(l, s)h(s, ς(s))ds ∀ l ∈ [0, 1].

Let the set of all continuous functions 𭟋 = (C[0, 1], [0,+∞)) defined on [0, 1] with values
in [0,+∞) and S = (C[0, 1], (−∞, 0]) defined on [0, 1] with values in (−∞, 0].

Define Π, Ψ and Ξ by

Π(ς(o), ϖ(o), ż) = sup
o∈[c,a]

ż

ż+ |ς(o)−ϖ(o)|
∀ ς,ϖ ∈ 𭟋 and ż > 0,

Ψ(ς(o), ϖ(o), ż) = 1− sup
o∈[c,a]

ż

ż+ |ς(o)−ϖ(o)|
∀ ς,ϖ ∈ 𭟋 and ż > 0,

and

Ξ(ς(o), ϖ(o), ż) = sup
o∈[c,a]

|ς(o)−ϖ(o)|
ż

∀ ς,ϖ ∈ 𭟋 and ż > 0,

with ct−||.|| and ct−co−||.|| define by i⋇ ♭ = i♭ and i♢♭ = max{i, ♭}. Then (𭟋,Π,Ψ,Ξ,⋇,♢)
is a complete NBMS.

Theorem 9. Let p : (𭟋,S,Π,Ψ,Ξ,⋇,♢) ⇒ (𭟋,S,Π,Ψ,Ξ,⋇,♢) be a map such that the
following axioms hold:

i. Q : [0, 1]2 → [0,∞) is a continuous function;



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 37 of 40

ii. h(s, ·) : [0, 1] × R → R is a monotone non decreasing function ∀ s ∈ [0, 1] such that
(ς,ϖ) ∈ (𭟋,S), we have

|h(l, ς)− h(l, ϖ) ≤ |ς(l)−ϖ(l)|;

iii. supl∈[0,1]
∫ l
0 Q(l, s) ≤ ζ < 1.

Then the fractional differential equation (19) has a unique solution.

Proof. Define p : (𭟋,S,Π,Ψ,Ξ,⋇,♢) ⇒ (𭟋,S,Π,Ψ,Ξ,⋇,♢) by

pς(l) =

∫ l

0
Q(l, s)h(s, ς(s))ds, where l ∈ [0, 1]

Now, ∀ ς,ϖ ∈ 𭟋 ∪ S, we deduce

Π(pς(l), pϖ(l),ζ ż) = sup
l∈[0,1]

ζ ż

ζ ż+ |pς(l)− pϖ(l)|

= sup
l∈[0,1]

ζ ż

ζ ż+ |
∫ l
0 Q(l, s)h(s, ς(s))ds−

∫ l
0 Q(l, s)h(s, ϖ(s))ds|

= sup
l∈[0,1]

ζ ż

ζ ż+
∫ l
0 Q(l, s)|h(s, ς(s))− h(s, ϖ(s))|ds

= sup
l∈[0,1]

ζ ż

ζ ż+ |h(s, ς(s))− h(s, ϖ(s))|

≥ sup
l∈[0,1]

ż

ż+ |ς(l)−ϖ(l)|

≥ Π(ς(l), ϖ(l), ż),

Ψ(pς(l),pϖ(l), ζ ż) = 1− sup
l∈[0,1]

ζ ż

ζ ż+ |pς(l)− pϖ(l)|

= 1− sup
l∈[0,1]

ζ ż

ζ ż+ |
∫ l
0 Q(l, s)h(s, ς(s))ds−

∫ l
0 Q(l, s)h(s, ϖ(s))ds|

= 1− sup
l∈[0,1]

ζ ż

ζ ż+
∫ l
0 Q(l, s)|h(s, ς(s))− h(s, ϖ(s))|ds

≤ 1− sup
l∈[0,1]

ż

ż+ |ς(l)−ϖ(l)|

≤ Ψ(ς(l), ϖ(l), ż),

and

Ξ(pς(l), pϖ(l), ζ ż) = sup
l∈[0,1]

|pς(l)− pϖ(l)|
ζ ż



R. Ramaswamy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6251 38 of 40

= sup
l∈[0,1]

|
∫ l
0 Q(l, s)h(s, ς(s))ds−

∫ l
0 Q(l, s)h(s, ϖ(s))ds|

ζ ż

= sup
l∈[0,1]

∫ l
0 Q(l, s)|h(s, ς(s))− h(s, ϖ(s))|ds

ζ ż

≤ sup
l∈[0,1]

|ς(l)−ϖ(l)|
ż

≤ Ξ(ς(l), ϖ(l), ż).

It can be seen that all conditions of Theorem 5 are satisfied and p has a unique fixed point
and the fractional differential equation (19) has a unique solution.

7. Conclusion

In this paper, the notion of neutrosophic bipolar metric space has been introduced and
fixed point results in NBMS have been established. Some of the topological properties of
the NBMS have also been presented in the manuscript. It can be seen that an analogue
of the Banach Fixed Point theorem has been established supplemented with suitable non
trivial examples. The results have been applied to find solution to integral equation,
voltage differential equation and fractional differential equation. Simulation has also been
presented for the analytical results using Mathematica Software. Since the space NBMS
generalises neutrosophic metric space NMS and its seeds, the results established vide the
contractions considered in this manuscript will not be satisfied in the setting of NMS or
general metric spaces. It will also be an open question to establish fixed point results using
different types of contractions, such as Kannan Type, Ciric Type, Reich Type, Meir-keeler
type, to name a few in the setting of neutrosophic bipolar metric spaces and also finding
applications in other fields such as neural networking, stochastic process etc.

Acknowledgements

The author extend his appreciation to Prince Sattam Bin Adbulaziz University for
funding this research work through the Project Number PSAU/2025/01/33096.

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