EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
2025, Vol. 18, Issue 1, Article Number 5794
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Fixed Point Theorems in Controlled Rectangular
Modular Metric Spaces with Solution of Fractional

Differential Equations

Umar Ishtiaq1, Naeem Saleem2,3,∗, Muhammad Farhan4, Maggie Aphane3,
Mohammad S. R. Chowdhury5

1 Office of Research, Innovation and Commercialization, University of Management and
Technology, Lahore 54770, Pakistan
2 Department of Mathematics, University of Management and Technology, Lahore, Pakistan
3 Department of Mathematics and Applied Mathematics, Sefako Makgatho Health Sciences
University, Ga-Rankuwa, Pretoria, Medunsa-0204, South Africa
4 Department of Mathematics, Numl University Multan Campus, Multan, Pakistan
5 Department of Mathematics and Statistics, The University of Lahore, Lahore, Pakistan

Abstract. In this paper, we establish the notion of controlled rectangular modular metric space as
a generalization of modular b−metric space and rectangular b−metric space. We used contraction
mappings to find the existence and uniqueness of a fixed point in the framework of controlled
rectangular modular metric space. We give several non-trivial examples and show the validity of
contraction mappings via graphs. At the end, we utilize our main result to solve a non-linear
fractional differential equation.

2020 Mathematics Subject Classifications: 47H10, 54H25

Key Words and Phrases: Controlled metric space, modular metric space, fixed point; existence
and uniqueness, non-linear fractional differential equations

1. Introduction and Preliminaries

The Banach fixed-point theorem [10] ensures the existence and uniqueness of fixed
points of particular self-maps of metric spaces (MSs) and provides a constructive approach
to identify those fixed points. Picard’s method [20] of consecutive approximations might
be viewed as an abstract formulation of this method. In 1922 Banach established the
following famous result.

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v18i1.5794

Email addresses: umarishtiaq000@gmail.com (U. Ishtiaq),
naeem.saleem2@gmail.com (N. Saleem), maggie.aphane@smu.ac.za (M. Aphane),
muhammadfarhan@numl.edu.pk (M. Farhan),
msrchowdhury@hotmail.com showkat.rahim@math.uol.edu.pk (M S. R. Chowdhury)

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



U. Ishtiaq et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5794 2 of 20

Definition 1. [10] Suppose (ℵ,∆) be a MS. Then a mapping T : ℵ → ℵ is known as
contraction mapping on ℵ if there exists q ∈ [0, 1) such that

∆(Tϑ, Ty) ≤ q∆(ϑ, y)

for all ϑ, y ∈ ℵ.

Theorem 1. [10] Let (ℵ,∆) be a complete MS and T : ℵ → ℵ be a contraction mapping.
Then T has a unique fixed point ϑ

∗
in ℵ.

Many authors established various kinds of contraction inequalities in an attempt to
generalize the famous Banach contraction principle by using different generalizations of
Chistyakov [13] established the notion of modular MS and proved some new results.

Definition 2. Let ℵ be a non-empty set and the function ∆ξ : (0,+∞)×ℵ×ℵ → [0,+∞],
which satisfies the following axioms for all µ, κ, ϑ ∈ ℵ :

(M1) ∆ξ(µ, κ) = 0 for all ξ > 0 if and only if µ = κ;

(M2) ∆ξ(µ, κ) = ∆ξ(κ, µ) for all ξ > 0;

(M3) ∆ξ+ρ(µ, κ) ≤ ∆ξ(µ, ϑ) + ∆ρ(ϑ, κ) for all ξ, ρ > 0.

Then ∆ξ be known as a modular metric on ℵ.

Recently, Abdou [3, 4] proved various interesting fixed point results in the sense of
modular MS. In 2018, Mlaiki et al. [21] established the notion of controlled type MS as
follows:

Definition 3. Consider a non-empty set ℵ and the function α : ℵ × ℵ → [0,+∞). Then
a function ∆ : ℵ × ℵ → [0,+∞) is said to be controlled MS if the following axioms holds
for all µ, κ, ϑ ∈ ℵ :

(C1) ∆(µ, κ) = 0 if and only if µ = κ;

(C2) ∆(µ, κ) = ∆(κ, µ);

(C3) ∆(µ, κ) ≤ α(µ, ϑ)∆(µ, ϑ) + α(ϑ, κ)∆(y, κ).

Then the pair (∆,ℵ) is called a controlled MS.

In addition, Mlaiki et al. [21] generalized the Banach fixed point theorem.
In 2000, Branciari [11] coined the concept of rectangular (generalized) MSs as follows:

Definition 4. Consider a non-empty set ℵ and the mapping ∆ : ℵ×ℵ → [0,+∞) satisfies
the following conditions:

(R1) ∆(µ, κ) = 0 if and only if µ = κ;



U. Ishtiaq et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5794 3 of 20

(R2) ∆(µ, κ) = ∆(κ, µ), for all µ, κ ∈ ℵ;

(R3) ∆(µ, κ) ≤ ∆(µ, ϑ) + ∆(ϑ, y) + ∆(y, κ);

for all µ, κ ∈ ℵ and all distinct points ϑ, y ∈ ℵ. Then the pair (∆,ℵ) is called a
rectangular MS.

In 2021, Alamgir et al. [5] coined the concept of controlled rectangular MS and proved
some fixed point results for contraction mappings.

Definition 5. Consider a nonempty set ℵ and the function α : ℵ × ℵ → [0,+∞). Then
a function ∆ : ℵ × ℵ → [0,+∞) is said to be a controlled rectangular MS if the following
axioms hold:

(CR1) ∆(µ, κ) = 0 for all ξ > 0 if and only if µ = κ;

(CR2) ∆(µ, κ) = ∆(κ, µ) for all ξ > 0;

(CR3) ∆(µ, κ) ≤ α(µ, ϑ)∆(µ, ϑ) + α(ϑ, y)∆(ϑ, y) + α(y, κ)∆(y, κ),

for all µ, κ ∈ ℵ and all distinct points ϑ, y ∈ ℵ. Then the pair (∆,ℵ) is called a
controlled rectangular MS.

We refer [1, 2, 9, 15, 24–27, 29] for more detail. Aydi et. al. [6] proved a fixed point
theorem for set-valued quasi-contractions in b−metric spaces. Karapinar et. al. [16–18]
proved several interesting fixed point theorems under nonlinear contractive conditions in
partially ordered metric spaces. Souayah and Mrad [28] proved some fixed point results
for contraction mappings in the context of controlled partial metric type spaces. Debnath
and Sen [14] proved various fixed point results of interpolative Ćirić-Reich–Rus-Type con-
tractions in b−metric spaces. Roy et. al. [23] provided an extended -metric-type space and
related fixed point theorems with an application to nonlinear integral equations. Rossafi
and Kari [22] gave some fixed point results in the sense of controlled rectangular metric
spaces. Budhia et. al. [12] provided some new fixed point results in rectangular met-
ric spaces with an application to fractional-order functional differential equations. Aydi
et. al. [7] presented a common Jungck type fixed point result in extended rectangular
b−metric spaces. Aydi et. al. [8] gave fixed-discs in rectangular metric spaces. Kari et.
al. [19] established contraction mapping on complete rectangular metric spaces.

In this manuscript, we introduce the notions of rectangular modular metric space
(RMMS) and controlled rectangular modular metric space (CRMMS). We generalize and
prove the well-known Banach fixed point theorem in the sense of CRMMS. We also give
some non-trivial examples to ensure the validity of provided fixed point results. At the
end, we use a fixed point technique to ensure the existence and uniqueness of non-linear
fractional differential equations. From now to onward, we use ∆ξ(µ, κ) = ∆(µ, κ, ξ), for
all µ, κ ∈ ℵ which denotes the map ∆ξ : ℵ × ℵ × (0,+∞) → [0,+∞), where ξ ∈ (0,+∞).



U. Ishtiaq et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5794 4 of 20

2. Main Results

In this section, we will introduce the concepts of RMMS and CRMMS and develop
some fixed-point results.

Definition 6. Consider a non-empty set ℵ and the mapping ∆ξ : ℵ × ℵ × (0,+∞) →
[0,+∞), which satisfies the following axioms:

(RM1) ∆ξ(µ, κ) = 0 if and only if µ = κ;

(RM2) ∆ξ(µ, κ) = ∆ξ(κ, µ);

(RM3) ∆ξ(µ, κ) ≤ ∆ξ(µ, ϑ) + ∆ξ(ϑ, y) + ∆ξ(y, κ);

for all ξ > 0, µ, κ ∈ ℵ and all distinct points ϑ, y ∈ ℵ. Then the pair (∆ξ,ℵ, ξ) is called
RMMS.

Definition 7. Consider a non-empty set ℵ and α : ℵ × ℵ → [0,+∞). Then a function
∆ξ : ℵ× ℵ× (0,+∞) → [0,+∞) is said to be controlled rectangular modular metric if for
all ξ > 0, µ, κ ∈ ℵ and distinct ϑ, y ∈ ℵ the following axioms are satisfied:

(CM1) ∆ξ(µ, κ) = 0 if and only if µ = κ;

(CM2) ∆ξ(µ, κ) = ∆ξ(κ, µ) for all µ, κ ∈ ℵ;

(CM3) ∆ξ(µ, κ) ≤ α(µ, ϑ)∆ξ(µ, ϑ) + α(ϑ, y)∆ξ(ϑ, y) + α(y, κ)∆ξ(y, κ).

Then the pair (∆ξ,ℵ) is called CRMMS.

Example 1. Let ℵ = N define ∆ξ : ℵ × ℵ × (0,+∞) → [0,+∞), by

∆(µ, κ, ξ) =


0, if µ = κ;
4η
ξ , if µ, κ ∈ {1, 2 and µ ̸= κ;

η
ξ , if µ or κ /∈ {1, 2, · · · , 10} and µ ̸= κ;

where η > 0 is a constant. Then (∆ξ,ℵ) is a CRMMS with controlled function

α = (µ, κ) =

{
1 if µ ̸= κ;
1 + µ+ κ if µ = κ;

but not RMMS. Let µ = 1 , κ = 2, ϑ = 3 and y = 4, then from triangular inequality (RM3),
we have

∆(µ, κ, ξ) ≤ ∆(µ, ϑ, ξ) + ∆(ϑ, y, ξ) + ∆(y, κ, ξ),



U. Ishtiaq et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5794 5 of 20

∆(1, 2, ξ) ≤ ∆(1, 3, ξ) + ∆(3, 4, ξ) + ∆(4, 2, ξ).

That is,

4η

ξ
≤ η

ξ
+
η

ξ
+
η

ξ

≤ 3η

ξ

After simplification, we have
4 ≤ 3,
which is contradiction. Hence, CRMMS need not to be RMMS. Also observe that it is not
modular b−MS and rectangular b−MS.

Example 2. Let ℵ = N define ∆ξ : ℵ × ℵ × (0,+∞) → [0,+∞) by

∆(µ, κ, ξ) =


0, if µ = κ;

10ηξ, if µ, κ ∈ {1, 2, · · · , 10} and µ ̸= κ;
2ηξ
3 , if µ or κ /∈ {1, 2, · · · , 10} and µ ̸= κ;

where η > 0 is a constant. Then (∆ξ,ℵ) is a CRMMS with control function

α = (µ, κ) =

{
3 if µ ̸= κ;
2(µ+ κ) if µ = κ;

but not RMMS.
Let µ = 5, κ = 8, ϑ = 11 and y = 12, then from triangular inequality (RM3), we have

∆(µ, κ, ξ) ≤ ∆(µ, ϑ, ξ) + ∆(ϑ, y, ξ) + ∆(y, κ, ξ),

∆(5, 8, ξ) ≤ ∆(5, 11, ξ) + ∆(11, 12, ξ) + ∆(12, 8, ξ).

That is,

(10η)ξ ≤ (2η)
ξ

3
+ (2η)

ξ

3
+ (2η)

ξ

3
.

That is,
10 ≤ 6,

which is a contradiction. Hence, CRMMS does not need to be RMMS. Also, observe
that it is not modular b−MS and rectangular b−MS.

Remark 1.

a) Every rectangular MS and control MS is a CRMMS.



U. Ishtiaq et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5794 6 of 20

b) Every CRMMS need not be RMMS, but the converse is true, as shown in the above
examples.

In terms of CRMMS, the concepts of convergence, Cauchy, and completeness can be
easily generalized.

Definition 8. Assume (∆ξ,ℵ) be a CRMMS. Then

a) a sequence (ϑn) in ℵ is said to be convergent to ϑ ∈ ℵ if

lim
n−→+∞

∆ξ(ϑn, ϑ) = 0

b) a sequence (ϑn) in ℵ is called Cauchy sequence if

lim
n,m−→+∞

∆ξ(ϑn, ϑm) = 0

c) the pair (∆ξ,ℵ) is said to be a complete CRMMS if every Cauchy sequence in ℵ
converges in ℵ.

Definition 9. Assume (∆ξ,ℵ) be CRMMS, let ϑ ∈ ℵ and r > 0. Then,

i) the open ball denoted and defined by

B(ϑ, r) = {ϑ0 ∈ ℵ,∆ξ(ϑ, ϑ0) < r},

ii) the mapping g : ℵ → ℵ is said to be continuous at ϑ ∈ ℵ if for every γ > 0 and δ > 0
such that g(B(ϑ, δ)) ⊆ B(g(ϑ, γ)). If g is continuous at ϑ, then for any sequence
(ϑn) converges to ϑ, we have

lim
n−→+∞

g(ϑn) = g(ϑ).

Lemma 1. Assume(∆ξ,ℵ) be a CRMMS and (ϑn) be a Cauchy sequence in ℵ and (ϑn) ̸=
(ϑm) whenever n ̸= m. If

lim
n,m−→+∞

∆ξ(ϑn, ϑm) < +∞

for all (ϑn), (ϑm) ∈ ℵ, then (ϑn) has a unique fixed point.



U. Ishtiaq et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5794 7 of 20

Proof. Assume s, t are two fixed point of the sequence (ϑn) in ℵ. Then

lim
n−→+∞

g(ϑn) = s.

and
lim

n−→+∞
g(ϑn) = t.

Here (ϑn) is a Cauchy sequence. Then from the triangular inequality (CM3) of definition
(7), we have

∆ξ(µ, κ) ≤ α(µ, ϑn)∆ξ(µ, ϑn)+α(ϑn, ϑm)∆ξ(ϑn, ϑm)+α(ϑm, κ)∆ξ(ϑm, κ) → 0 as n, r → +∞.
(2.1)

This implies that
∆ξ(µ, κ) = 0.

Hence (ϑn) has a unique fixed point in ℵ.

Definition 10. Suppose (∆ξ,ℵ) be a CRMMS. Then the mapping

i) g : ℵ → ℵ defined by

Ω(ϑ, n) = {ϑ, gϑ, g2ϑ, · · · gnϑ},
Ω(ϑ,+∞) = {ϑ, gϑ, g2ϑ, · · · gnϑ · · · },

where ϑ ∈ ℵ and n ∈ N. Here, Ω(ϑ,+∞) is known as orbit of g.

ii) g : ℵ → ℵ is known as g−orbitally continuous, if

lim
k−→+∞

gnkϑ = ϑ

implies
lim

k−→+∞
g (gnkϑ) = gϑ

for ϑ ∈ ℵ.

Theorem 2. Suppose g : ℵ → ℵ is a mapping in a CRMMS (∆ξ,ℵ). Assume that the
following conditions hold:

a) for all µ, κ ∈ ℵ,
∆ξ(gµ, gκ) ≤ λ∆ξ(µ, κ),

b) sup
q≥1

lim
i−→+∞

α(µi, µq)
(
α(µi,µ(i+1))

α(µ(i−1),µi)

)
λ < 1 for any µi ∈ ℵ,

c) (∆ξ,ℵ) is g is orbitally complete,

d) g is orbitally continuous,



U. Ishtiaq et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5794 8 of 20

e) For each µ ∈ ℵ lim
n→+∞

α(µn, µ) and lim
n→+∞

α(µ, µn) exist and finite.

Then g has a unique fixed point.

Proof. Suppose µ0 be any point in ℵ. We describe the iterative sequence (µn) over µ0

as follows
g(µ0) = µ1, g(µ1) = µ2, g(µ2) = µ3 · · · gn(µ0) = µn,

then, we obtain

∆ξ(µ1, µ2) = ∆ξ(g(µ0), g
2(µ0)) ≤ λ∆ξ(µ0, g(µ0)) = λ∆ξ(µ0, µ1)

Recursively, we get

∆ξ(µn, µ(n+1)) ≤ ∆ξ(g
n(µ0), g

(n+1)(µ0)) ≤ λ∆ξ(g
(n−1)(µ0), g

n(µ0))

...

≤ λn∆ξ(µ0, µ1)

Taking lim
n→+∞

∆ξ(µn, µ(n+1)) = 0. Similarly, lim
n→+∞

∆ξ(µ(n+1), µ(n+2)) = 0. Now, we exam-

ine that (µn) is a Cauchy sequence. Here, we make the following cases:
Case 1:
Suppose p be an odd number, then p = 2n+ 1 and r ≥ 1, we have

∆ξ(µn, µ(n+2r+1)) ≤ ξ

3
α(µn, µ(n+1))∆ξ(µn, µ(n+1)) +

ξ

3
α(µ(n+1), µ(n+2))∆ξ(µ(n+1), µ(n+2))

+
ξ

3
α(µ(n+2), µ(n+2r+1))∆ξ(µ(n+2), µ(n+2r+1)),

=
ξ

3
[λnα(µn, µ(n+1)) + λ(n+1)α(µ(n+1), µ(n+2))]∆ξ(µ0, µ1)

+
ξ

3
α(µ(n+2), µ(n+2r+1))∆ξ(µ(n+2), µ(n+2r+1)),

≤ ξ

3
[λnα(µn, µ(n+1)) + λ(n+1)α(µ(n+1), µ(n+2))]∆ξ(µ0, µ1)

+
ξ

3
α(µ(n+2), µ(n+2r+1))[

ξ

3
α(µ(n+2), µ(n+3))∆ξ(µ(n+2), µ(n+3))

+
ξ

3
α(µ(n+3), µ(n+4))∆ξ(µ(n+3), µ(n+4)) +

ξ

3
α(µ(n+4), µ(n+2r+1))∆ξ(µ(n+4), µ(n+2r+1))],

=
ξ

3
[λnα(µn, µ(n+1)) + λ(n+1)α(µ(n+1), µ(n+2))]∆ξ(µ0, µ1)

+
ξ2

32
[α(µ(n+2), µ(n+2r+1))α(µ(n+2), µ(n+3))λ

(n+2)

+α(µ(n+2), µ(n+2r+1))α(µ(n+3), µ(n+4))λ
(n+3)]∆ξ(µ0, µ1)

+
ξ2

32
α(µ(n+2), µ(n+2r+1))α(µ(n+4), µ(n+2r+1))∆ξ(µ(n+4), µ(n+2r+1)),



U. Ishtiaq et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5794 9 of 20

≤ ξ

3
[λnα(µn, µ(n+1)) + λ(n+1)α(µ(n+1), µ(n+2))]∆ξ(µ0, µ1)

+
ξ2

32
[α(µ(n+2), µ(n+2r+1))α(µ(n+2), µ(n+3))λ

(n+2)

+α(µ(n+2), µ(n+2r+1))α(µ(n+3), µ(n+4))λ
(n+3)]∆ξ(µ0, µ1)

+
ξ2

32
α(µ(n+2), µ(n+2r+1))α(µ(n+4), µ(n+2r+1))

[
ξ

3
α(µ(n+4), µ(n+5))∆ξ(µ(n+4), µ(n+5)) +

ξ

3
α(µ(n+5), µ(n+6))∆ξ(µ(n+5), µ(n+6))

+
ξ

3
α(µ(n+6), µ(n+2r+1))∆ξ(µ(n+6), µ(n+2r+1))],

≤ ξ

3
[λnα(µn, µ(n+1)) + λ(n+1)α(µ(n+1), µ(n+2))]∆ξ(µ0, µ1)

+
ξ2

32
[α(µ(n+2), µ(n+2r+1))α(µ(n+2), µ(n+3))λ

(n+2)

+α(µ(n+2), µ(n+2r+1))α(µ(n+3), µ(n+4))λ
(n+3)]∆ξ(µ0, µ1)

+
ξ2

32
α(µ(n+2), µ(n+2r+1))α(µ(n+4), µ(n+2r+1))

[
ξ

3
λ(n+4)α(µ(n+4), µ(n+5)) +

ξ

3
λ

(n+5)
α(µ(n+5), µ(n+6))]∆ξ(µ o, µ1)

+
ξ

3
α(µ(n+6), µ(n+2r+1))∆ξ(µ(n+6), µ(n+2r+1)),

≤ ξ

3
[λnα(µn, µ(n+1)) + λ(n+1)α(µ(n+1), µ(n+2))]∆ξ(µ0, µ1)

+
ξ2

32
[λ(n+2)α(µ(n+2), µ(n+2r+1))α(µ(n+2), µ(n+3))

+λ(n+3)α(µ(n+2), µ(n+2r+1))α(µ(n+3), µ(n+4))]∆ξ(µ0, µ1)

+
ξ3

33
[λ(n+4)α(µ(n+2), µ(n+2r+1))α(µ(n+4), µ(n+2r+1))α(µ(n+4), µ(n+5))

+λ(n+5)α(µ(n+2), µ(n+2r+1))α(µ(n+4), µ(n+2r+1))α(µ(n+5), µ(n+6))]

∆ξ(µ0, µ1)

...

+
ξm

3m
[α(µ(n+2), µ(n+2r+1))α(µ(n+4), µ(n+2r+1)) . . .

α(µ(n+2r−2), µ(n+2r−1))λ
(n+2r−2)

+α(µ(n+2), µ(n+2r−1)) · · ·α(µ(n+2r−1), µ(n+2r))λ
(n+2r−1)

+α(µ(n+2), µ(n+2r+1)) . . . α(µ(n+2r), µ(n+2r+1))λ
(n+2r)]∆ξ(µ0, µ1).

From the above inequality, we get

lim
n→+∞

∆ξ(µn, µ(n+2r+1)) = 0.



U. Ishtiaq et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5794 10 of 20

Case: 2
Let p be an even number then p = 2n and r ≥ 1, then

∆ξ(µn, µ(n+2r)) ≤ ξ

3
[α(µn, µ(n+1))∆ξ(µn, µ(n+1)) +

ξ

3
α(µ(n+1), µ(n+2))∆ξ(µ(n+1), µ(n+2))

+
ξ

3
α(µ(n+2), µ(n+2r))∆ξ(µ(n+2), µ(n+2r))]

=
ξ

3
[λnα(µn, µ(n+1)) + λ(n+1)α(µ(n+1), µ(n+2))]∆ξ(µ0, µ1)

+
ξ

3
α(µ(n+2), µ(n+2r))∆ξ(µ(n+2), µ(n+2r)),

≤ ξ

3
[λnα(µn, µ(n+1)) + λ(n+1)α(µ(n+1), µ(n+2))]∆ξ(µ0, µ1)

+
ξ

3
α(µ(n+2), µ(n+2r))[

ξ

3
α(µ(n+2), µ(n+3))∆ξ(µ(n+2), µ(n+3))

+
ξ

3
α(µ(n+3), µ(n+4))∆ξ(µ(n+3), µ(n+4)) +

ξ

3
α(µ(n+4), µ(n+2r))∆ξ(µ(n+4), µ(n+2r))],

≤ ξ

3
[λnα(µn, µ(n+1)) + λ(n+1)α(µ(n+1), µ(n+2))]∆ξ(µ0, µ1)

+
ξ2

32
[λ(n+2)α(µ(n+2), µ(n+2r))α(µ(n+2), µ(n+3))

+λ(n+3)α(µ(n+2), µ(n+2r))α(µ(n+3), µ(n+4))]∆ξ(µ0, µ1)

+
ξ2

32
α(µ(n+2), µ(n+2r))α(µ(n+4), µ(n+2r))∆ξ(µ(n+4), µ(n+2r)),

≤ ξ

3
[λnα(µn, µ(n+1)) + λ(n+1)α(µ(n+1), µ(n+2))]∆ξ(µ0, µ1)

+
ξ2

32
[λ(n+2)α(µ(n+2), µ(n+2r))α(µ(n+2), µ(n+3))

+λ(n+3)α(µ(n+2), µ(n+2r))α(µ(n+3), µ(n+4))]∆ξ(µ0, µ1)

...

+
ξ(m−1)

3(m−1)
[α(µ(n+2), µ(n+2r)) · · ·α(µ(n+2−4), µ(n+2r−3))λ

(n+2r−4)

+α(µ(n+2), µ(n+2r)) · · ·α(µ(n+2−3), µ(n+2r−2))λ
(n+2r−3)

+α(µ(n+2), µ(n+2r)) · · ·α(µ(n+2r−2), µ(n+2r))λ
(n+2r−2)]∆ξ(µ0, µ1).

From the above inequality, we get lim
n→+∞

∆ξ(µn, µ(n+2r+1)) = 0.

Hence, both cases show that {µn} is a Cauchy sequence. As ℵ is g−orbitally complete,
so there exist µ ∈ ℵ such that

lim
n→+∞

µn = µ.

Now we show that µ is a fixed point of g. As ℵ is g−orbitally continuous, we get

∆ξ(µ, gµ) ≤
ξ

3
α(µ, µn)∆ξ(µ, µn)+

ξ

3
α(µ, µ(n+1))∆ξ(µ, µ(n+1))+

ξ

3
α(µ(n+1), gµ)∆ξ(µ(n+1), gµ).



U. Ishtiaq et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5794 11 of 20

Since for each µ ∈ ℵ, lim
n→+∞

α(µn, µ) and lim
n→+∞

α(µ, µn) exist and finite, so by taking limit

and utilizing
lim

n→+∞
∆ξ(µn, µ(n+1)) = 0.

We get
lim

n−→+∞
∆ξ(µ, gµ) = 0.

That is,
gµ = µ.

Hence µ is a fixed point of g. In view of Lemma (2), µ is unique fixed point of g.

Corollary 1. Suppose g : ℵ → ℵ be a mapping on a complete CRMMS (∆ξ,ℵ). Assume
that the following conditions hold:

a) For all µ, κ ∈ ℵ we have

∆ξ(gµ, gκ) ≤ λ∆ξ(µ, κ), λ ∈ [0, 1) ,

b) sup(q≥1) lim
n−→+∞

α(µi, µq)
(
α(µ(i+1),µ(i+2))

α(µ(i−1),µi)

)
λ < 1, for any µi ∈ ℵ,

c) g is continuous.

Then g has a unique fixed point.

Example 3. Suppose ℵ = R and a mapping ∆ξ : ℵ × ℵ × (0,+∞) −→ ℵ define by

∆(µ, κ, ξ) =
|µ− κ|

ξ + |µ− κ|

Then (∆ξ,ℵ) is a complete CRMMS with controlled function

α = (µ, κ) =

{
1 if µ ̸= κ;
1 + µ+ κ if µ = κ;

but not RMMS. Define a mapping g : ℵ −→ ℵ by

g(µ) =
µ

5
+ 7

Now, we examine the contraction condition. Let 3
4 ≤ λ < 1, then

∆(gµ, gκ, ξ) =
|gµ− gκ|

ξ + |gµ− gκ|
=

|µ5 − κ
5 |

ξ + |µ5 − κ
5 |

=
|µ− κ|

5ξ + |µ− κ|
≤ λ

|µ− κ|
ξ + |µ− κ|

= λ∆(µ, κ, ξ).

Observe that all circumstances of Corollary 1 are fulfilled and 35
4 is a unique fixed point of

g. See Figures 1, 2 and 3 for more details.



U. Ishtiaq et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5794 12 of 20

Figure 1: First view of contraction mapping ∆(gµ, gκ, ξ) ≤ λ∆(µ, κ, ξ) when ξ = 1 and 3
4
≤ λ < 1.

Table 1: The matrix of values of λ∆(µ, κ, ξ)

0 0.4500 0.6000 0.6750 0.7200 0.7500 0.7714 0.7875 0.8000 0.8100 0.8181
0.4500 0 0.4500 0.6000 0.6750 0.7200 0.7500 0.7714 0.7875 0.8000 0.8100
0.6000 0.4500 0 0.4500 0.6000 0.6750 0.7200 0.7500 0.7714 0.7875 0.8000
0.6750 0.6000 0.4500 0 0.4500 0.6000 0.6750 0.7200 0.7500 0.7714 0.7875
0.7200 0.6750 0.6000 0.4500 0 0.4500 0.6000 0.6750 0.7200 0.7500 0.7714
0.7500 0.7200 0.6750 0.6000 0.4500 0 0.4500 0.6000 0.6750 0.7200 0.7500
0.7714 0.7500 0.7200 0.6750 0.6000 0.4500 0 0.4500 0.6000 0.6750 0.7200
0.7875 0.7714 0.7500 0.7200 0.6750 0.6000 0.4500 0 0.4500 0.6000 0.6750
0.8000 0.7875 0.7714 0.7500 0.7200 0.6750 0.6000 0.4500 0 0.4500 0.6000
0.8100 0.8000 0.7875 0.7714 0.7500 0.7200 0.6750 0.6000 0.4500 0 0.4500
0.8181 0.8100 0.8000 0.7875 0.7714 0.7500 0.7200 0.6750 0.6000 0.4500 0

Theorem 3. Suppose g : ℵ → ℵ be a mapping on a CRMMS (∆ξ,ℵ). Assume that the
following conditions hold:

a) For all µ, κ ∈ ℵ,

∆ξ(gµ, gκ) ≤ λ[∆ξ(µ, gµ) + ∆ξ(κ, gκ)], λ ∈
[
0,

1

2

)
(A)

b) sup(q≥1) lim
i−→+∞

α(µi, µq)
(
α(µ(i),µ(i+1))

α(µ(i−1),µi)

)
λ < 1, for any µi ∈ ℵ, where λ ̸= 1

α(µ1,µ2)
for

each µ1, µ2 ∈ ℵ,

c) For each µ ∈ ℵ lim
n−→+∞

α(µn, µ(n+1)) ≤ 1, lim
n−→+∞

α(µ, µn) and lim
n−→+∞

α(µn, µ) exist

and finite.



U. Ishtiaq et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5794 13 of 20

Figure 2: Second view of contraction mapping ∆(gµ, gκ, ξ) ≤ λ∆(µ, κ, ξ) when ξ = 1 and 3
4
≤ λ < 1.

Table 2: The matrix of values of ∆(gµ, gκ, ξ)

0 0.1666 0.2857 0.3750 0.4444 0.5000 0.5454 0.5833 0.6153 0.6428 0.6666
0.1666 0 0.1666 0.2857 0.3750 0.4444 0.5000 0.5454 0.5833 0.6153 0.6428
0.2857 0.1666 0 0.1666 0.2857 0.3750 0.4444 0.5000 0.5454 0.5833 0.6153
0.3750 0.2857 0.1666 0 0.1666 0.2857 0.3750 0.4444 0.5000 0.5454 0.5833
0.4444 0.3750 0.2857 0.1666 0 0.1666 0.2857 0.3750 0.4444 0.5000 0.5454
0.5000 0.4444 0.3750 0.2857 0.1666 0 0.1666 0.2857 0.3750 0.4444 0.5000
0.5454 0.5000 0.4444 0.3750 0.2857 0.1666 0 0.1666 0.2857 0.3750 0.4444
0.5833 0.5454 0.5000 0.4444 0.3750 0.2857 0.1666 0 0.1666 0.2857 0.3750
0.6153 0.5833 0.5454 0.5000 0.4444 0.3750 0.2857 0.1666 0 0.1666 0.2857
0.6428 0.6153 0.5833 0.5454 0.5000 0.4444 0.3750 0.2857 0.1666 0 0.1666
0.6666 0.6428 0.6153 0.5833 0.5454 0.5000 0.4444 0.3750 0.2857 0.1666 0

Then g has a unique fixed point in ℵ.

Proof. Suppose µ0 be any point in ℵ. We describe the iterative sequence (µn) over

µ0, g(µ0) = µ1, g(µ1) = µ2, g(µ2) = µ3 · · · gn(µ0) = µn, then from (A), we obtain

∆ξ(µ1, µ2) = ∆ξ(gµ0, gµ1),

∆ξ(µ1, µ2) ≤ λ[∆ξ(µ0, gµ0) + ∆ξ(µ1, gµ1)]

∆ξ(µ1, µ2) = λ[∆ξ(µ0, gµ0) + ∆ξ(µ1, µ2)]

∆ξ(µ1, µ2)− λ∆ξ(µ1, µ2) = λ∆ξ(µ0, µ1),

∆ξ(µ1, µ2) ≤ λ

1− λ
∆ξ(µ0, µ1).



U. Ishtiaq et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5794 14 of 20

Figure 3: Graph of g(µ) = µ. It is easy to see that shows that 35
4

is a unique fixed point.

Let, λ
(1−λ) = α < 1, as λ ≤ 1

2 . Then by continuously applying (A), we obtain

∆ξ(µn, µ(n+1)) ≤ αn∆ξ(µ0, µ1).

Taking limit on both sides, we get

lim
n−→+∞

∆ξ(µn, µ(n+1)) = 0. (B)

Now again from (A), we have

∆ξ(µn, µ(n+2)) = ∆ξ(gµ(n−1), gµ(n+1)),

∆ξ(µ1, µ2) ≤ λ[∆ξ(µ(n−1), gµ(n−1)) + ∆ξ(µ(n+1), gµ(n+1))],

≤ λ[∆ξ(µ(n−1), µn) + ∆ξ(µ(n+1), µ(n+2))]

Again, applying limit on the both sides, we get

lim
n−→+∞

∆ξ(µn, µ(n+1)) = 0. (C)

Now we prove that the sequence {µn} is a Cauchy sequence. Using equations (B) and (C),
repeat the method as in Theorem (0.2.6), we examine that {µn} is a Cauchy sequence.
As ℵ is complete, so there exist µ ∈ ℵ such that

lim
n−→+∞

∆ξ(µn, µ) = 0. (D)

Now we examine that µ is a fixed point of g

∆ξ(µ, gµ) ≤ ξ

3
[α(µ, µn)∆ξ(µ, µn) + α(µn, µ(n+1))∆ξ(µn, µ(n+1))



U. Ishtiaq et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5794 15 of 20

+α(µ(n+1), gµ)∆ξ(µ(n+1), gµ)]

≤ ξ

3
[α(µ, µn)∆ξ(µ, µn) + α(µn, µ(n+1))∆ξ(µn, µ(n+1))

+α(µ(n+1), gµ)∆ξ(gµn, gµ)],

≤ ξ

3
[α(µ, µn)∆ξ(µ, µn) + α(µn, µ(n+1))∆ξ(µn, µ(n+1))

+α(µ(n+1), gµ)λ(∆ξ(µn, gµn) + ∆ξ(µ, gµ))]

≤ ξ

3
[α(µ, µn)∆ξ(µ, µn)] +

ξ

3
[α(µn, µ(n+1))∆ξ(µn, µ(n+1))]

+
ξ

3
α(µ(n+1), gµ)λ∆ξ(µn, gµn)

+
ξ

3
λα(µ(n+1), gµ)∆ξ(µ, gµ),

∆ξ(µ, gµ)−
ξ

3
λα(µ(n+1), gµ)∆ξ(µ, gµ) ≤ ξ

3
[α(µ, µn)∆ξ(µ, µn)]

+
ξ

3
[α(µn, µ(n+1))∆ξ(µn, µ(n+1))]

+
ξ

3
α(µ(n+1), gµ)λ∆ξ(µn, gµn),

∆ξ(µ, gµ)(1−
ξ

3
λα(µ(n+1), gµ)) ≤ ξ

3
[α(µ, µn)∆ξ(µ, µn)] +

ξ

3
[α(µn, µ(n+1))∆ξ(µn, µ(n+1))]

+
ξ

3
α(µ(n+1), gµ)λ∆ξ(µn, gµn),

∆ξ(µ, gµ) ≤
( ξ3 [α(µ, µn)∆ξ(µ, µn)])

(1− ξ
3λα(µ(n+1), gµ))

E (1)

+
( ξ3 [α(µn, µ(n+1)) + λα(µ(n+1), gµ)])

(1− ξ
3λα(µ(n+1), gµ))

for each µ ∈ ℵ, lim
n−→+∞

∆ξ(µn, µ(n+1)) ≤ 1, lim
n−→+∞

∆ξ(µn, µ),and lim
n−→+∞

∆ξ(µ, µn) exist

and finite. Therefore, by taking lim
n−→+∞

in (E) and using (B) and (C), we obtain

∆ξ(µ, gµ) = 0. (F)

This shows that
µ = gµ.

For uniqueness, consider µ ̸= κ with κ be another fixed point of g then from (A), we obtain

∆ξ(µ, κ) = ∆ξ(gµ, gκ),

≤ λ[∆ξ(µ, gµ) + ∆ξ(κ, gκ)]

≤ λ[∆ξ(µ, µ) + ∆ξ(κ, κ)]

We get, ∆ξ(µ, κ) = 0, where ∆ξ(µ, µ) = 0 and ∆ξ(κ, κ) = 0. Hence ∆ξ(µ, κ) = 0, this
implies that µ = κ. Hence µ is a unique fixed point of g.



U. Ishtiaq et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5794 16 of 20

3. Application fractional calculus

In this section, we use Corollary 1 to find the existence and uniqueness of a solution
of nonlinear fractional differential equations given by

Dα
c µ(l) = f(l, µ(l)) (l ∈ (0, 1), α ∈ (1, 2]),

with boundary conditions µ(0) = 0, µ′(0) = Iµ(l)l ∈ (0, 1),where Dα
c means Caputo

fractional derivative of order α, defined by

Dα
c f(l) =

1

(Γ(n− α))

l∫
0

(l − ω̄)(n−α−1)fn(ω̄)∆ω (n− 1 < α < n, n = [α] + 1)

and f : [0, 1]× R −→ R+ is a continuous function. We assume ℵ = C([0, 1],R) from[0, 1]
into R with supremum |µ| = sup

l∈[0,1]
|µ(l)|.The Riemann-Liouville fractional integral of order

α is given by

Iαf(l) =
1

Γ(α)

l∫
0

(l − ω)(α−1)f(ω)∆ω. (α > 0)

Initially, we give reasonable form of a nonlinear fractional differential equation and then
inquest the existence of a solution. Now, we assume the fractional differential equation
given by

Dα
c µ(l) = f(l, µ(l)) (l ∈ (0, 1), α ∈ (1, 2]), (3.1)

with the boundary conditions

µ(0) = 0, µ′(0) = Iµ(l)(l ∈ (0, 1)),

where

i f : [0, 1]×R −→ R+ is a continuous function,

ii µ(l) : [0, 1] −→ R is continuous,

meet the below conditions

|f(l, µ)− f(l, κ)| ≤ L|µ− κ|,

for all l ∈ [0, 1], L is a constant with LΠ < 1, where

Π =
1

Γ(α+ 1)
+

2κ(α+1)Γ(α)

(2− κ2)Γ(α+ 1)
.



U. Ishtiaq et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5794 17 of 20

Then the equation (3.1) has a unique solution.
Proof. We define a CRMMS by

∆(µ, κ, ξ) =
|µ(l)− κ(l)|

(ξ + |µ(l)− κ(l)|)

for all µ, κ ∈ ℵ, we consider |µ− κ| = sup
(l∈[0,1])

|µ(l)− κ(l)|. We define a mapping ψ : ℵ → ℵ

by

µ(l) =
1

Γ(α)

l∫
0

(l−ω)(α−1)f(ω, µ(ω))∆ω+
2l

(2− κ2)Γ(α)

κ∫
0

 ω∫
0

(ω −m)(α−1)f(m,µ(m))∆m)

∆ω

(3.2)
for all l ∈ [0, 1]. An equation (3.1) has a solution, for a function µ ∈ ℵ iff µ(l) = ψµ(l) for
all l ∈ [0, 1]. For all l ∈ [0, 1], we have

∆(ψµ, ψκ, ξ) =
|ψµ(l)− ψκ(l)|

(ξ + |ψµ(l)− ψκ(l)|)
. (3.3)

Now,

|ψµ(l)− ψκ(l)| =
1

Γ(α)

l∫
0

(l − ω)(α−1)f(ω, µ(ω))∆ω

+
2l

(2− κ2)Γ(α)

κ∫
0

 ω∫
0

(ω −m)(α−1)f(m,µ(m))∆m)

∆ω

− 1

Γ(α)

l∫
0

(l − ω)(α−1)f(ω, κ(ω))∆ω

+
2l

(2− κ2)Γ(α)

κ∫
0

 ω∫
0

(ω −m)(α−1)f(m,κ(m))∆m)

∆ω

≤ 1

Γ(α)

l∫
0

(l − ω)(α−1) |f(ω, µ(ω))− f(ω, κ(ω))|∆ω

+
2l

(2− κ2)Γ(α)

κ∫
0

 ω∫
0

(ω −m)(α−1) |f(ω, µ(m))− f(ω, κ(m))|∆m


≤ L |µ− κ|

Γ(α)

l∫
0

(l − ω)(α−1)∆ω +
2L |µ− κ|

Γ(α)

κ∫
0

 ω∫
0

(ω −m)(α−1)∆m

∆ω



U. Ishtiaq et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5794 18 of 20

≤ L |µ− κ|
Γ(α+ 1)

+
2κα+1L |µ− κ|Γ(α)
(2− κ2) Γ(α+ 2)

≤ L |µ− κ|
(

1

Γ(α+ 1)
+

2κα+1Γ(α)

(2− κ2) Γ(α+ 2)

)
= LΠ|µ− κ|.

From the fact LΠ < 1, 34 ≤ λ < 1, and (3.3), we get

∆(ψµ, ψκ, ξ) =
|ψµ(l)− ψκ(l)|

(ξ + |ψµ(l)− ψκ(l)|)

≤ LΠ|µ− κ|
(ξ + LΠ|µ− κ|)

≤ λ|µ− κ|
(ξ + |µ− κ|)

= λ∆(µ, κ, ξ).

Observe that all conditions of Corollary 1 are fulfilled. This implies that µ(l) is a unique
fixed point of ψ.

Open Problem 3.1 Introduce a new notion to combine the structure of fuzzy sets and
CRMMS and then prove Theorem (2) and Theorem (3) in the context of fuzzy CRMMS.

4. Conclusion

In this manuscript, we established the notions of rectangular modular metric space
and controlled rectangular modular metric space and proved several fixed point results.
Also, we provide some non-trivial examples with some graphical views and an application
to fractional calculus. These results are in more generalized form in the existing literature.
This work can easily be extended in the combined structure of fuzzy sets and controlled
rectangular modular metric space.

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