EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
2025, Vol. 18, Issue 4, Article Number 7033
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

The Ordered Implicit Relations: Fixed Point Problems
and Applications

Anam Arif1, Muhammad Nazam2, Hijaz Ahmad3,4,5,9,∗, Ibrahim Alraddadi3,∗,
Changjin Xu6, Waleed Mohammed Abdelfattah7,8

1 Department of Mathematics, Government College University, Lahore, Pakistan
2 Department of Mathematics, Allama Iqbal Open University, H-8, Islamabad, Pakistan
3 Department of Mathematics, Faculty of Science, Islamic University of Madinah, Madinah,
Saudi Arabia
4 Operational Research Center in Healthcare, Near East University, TRNC Mersin 10,
Nicosia, 99138, Turkey
5 Department of Mathematics, College of Science, Korea University, 145 Anam-ro,
Seongbuk-gu, Seoul 02841, South Korea
6 Guizhou Key Laboratory of Economics System Simulation, Guizhou University of Finance
and Economics, Guiyang, P.R. China
7 College of Engineering, University of Business and Technology, Jeddah 23435,
Saudi Arabia
8 Department of Engineering Mathematics and Physics, Faculty of Engineering,
Zagazig University, P.O. 44519, Egypt
9 Jadara University Research Center, Jadara University, Irbid, Jordan

Abstract. We introduce an ordered implicit relation and obtain fixed point theorems in rectan-
gular cone b-metric space, as an extension of the results on cone metric, rectangular cone metric
and cone b-metric space. We provide some examples as an explanation of established outcomes.
Our theorems universalize many fixed point outcomes in literature([1], [2]). A homotopy result as
an application of main theorem is given, which further applied to the human aging process. The
obtained fixed point results is an extension of [3], as the present article deals with nonlinear con-
tractions. The convergence of the sequence generated by Urysohn integral operator is also shown
by using fixed point technique.

2020 Mathematics Subject Classifications: 47H09, 47H10, 54H25

Key Words and Phrases: Fixed point, implicit relation, cone rectangular b-metric space, con-
traction

∗Corresponding author.
∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v18i4.7033

Email addresses: amanarif970@gmail.com (A. Arif),
muhammad.nazam@aiou.edu.pk (M. Nazam), hijazahmad@korea.ac.kr (H. Ahmad),
ialraddadi@iu.edu.sa (I. Alraddadi), xcj403@126.com (C. Xu),
w.abdelfattah@ubt.edu.sa (W. M. Abdelfattah)

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



A. Arif et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7033 2 of 29

1. Introduction

Ran [4] applied the concept of partial order to obtain a new generalization of BCP [5].
The appearance of this idea set a new direction in metric fixed point theory, and many
authors contributed with remarkable research articles ( see [6, 7] ). Popa [1] introduced the
concept of an implicit relation to present generalization of BCP. Beg et al. [8, 9], Berinde
et al. [10, 11] and Sedghi et al. [12] contributed with several new fixed point results of self-
operators satisfying implicit relations. Altun and Simsek applied the concept of implicit
relation ( see[13] ), and obtained a generalization of [4, 14]. Fixed point theory deals with
metric structures and their generalizations. Recently, Jachymski [15] generalized the BCP
subject to a graphic metric space, Nadler [16] used Hausdorff metric space to manifest
the fixed point results for multivalued mappings. Rashem at al. [17] presented some
fixed point outcomes in a modular like metric structure with graph. Hammad at al. [18],
contributed some tripled fixed point techniques in partially ordered metric spaces( see
[19, 20] ). Huang and Zhang [21] proposed the idea of cone metric, and hence proved the
BCP and Kannan fixed point theorem in this setting. They used the idea of normal cones
in their work (see [21]), later on, Rezapour et al. [22] improved these results. Motivated
by b-metric space [23], Hussain and Shah [24] inaugurated the theory of cone b-metric
structure. Huang and Xu [25] presented some fixed point results in the cone b-metric
space. Recently, Anam et al. investigated some necessary conditions for the convergence
of a sequence generated by implicit contraction in cone b-metric space [26]. For more
information on this direction, we suggest reading of ([23, 27–35] ). George [2] defined the
rectangular cone b-metric space that universalizes the idea of [36], and established some
well-known consequences in this setting.

• The fixed point theorem and the variational iteration method (VIM) are related
being fundamental tools for resolving a variety of mathematical issues, including
differential and integral equations.

(a) Iterative procedures are the foundation of both VIM and the fixed point the-
orem. Similar to finding a fixed point, the correction process in VIM can be
thought of as a mapping that seeks to move successive approximations closer
to the genuine solution.

(b) It is feasible to understand the convergence of the VIM iterations as the discov-
ery of a fixed point in the space of potential solutions. In particular, if VIM’s
iterative process is successful, it indicates that, much like when a fixed point is
found, the series of approximations converges to a point (the solution) where
the correction functional no longer changes.

(c) In some VIM analysis, by proving that the iterative process is, in fact, a contrac-
tion mapping, the fixed point theorem (more precisely, Banach’s Fixed Point
Theorem) can be used to establish the existence and uniqueness of the solution
to the problem.



A. Arif et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7033 3 of 29

We observe that the implicit relation defined by Popa, can be generalized to vector spaces
([37, 38]). In this research paper, motivated by Beg et al. [8, 9], Berinde et al. [10, 11] and
Sedghi [12], we define an ordered implicit relation in a rectangular cone b-metric space.
We contribute with a fixed point problem subject to monotone mappings, satisfying an
implicit contraction. We aslo solve a homotopy problem and show existence of solution
to a Urysohn Integral Equation as applications of the obtained fixed point theorems. The
obtained fixed point theorems are independent of the observations presented by Ercan [3].
Ercan worked on linear contractions only, while in this paper, we considered nonlinear
contractions. So, the obtained results are real generalizations and could not be followed
from known ones in literature.

This paper has been organized in eight sections. Section 1, contains the introduction
of the topic and motivation for this research work. Section 2, consists of related basic
notions and results. Section 3, shed light on the notion of ordered implicit relation and its
associated properties. Section 4, explains the methodology for the existence of fixed points
of the self mapping satisfying an implicit contractive condition under ordered implicit
relation. Section 5, contains the examples that explain the hypotheses of the fixed point
theorems stated in Section 4. It also contains the corollaries of the main theorem. Section
6, consists of a homotopy result as an application of the main theorem and a result about
human aging process that uses homotopy result. Section 7, shed light on the process of
application of main theorem to show the existence of solution to Urysohn integral equation.
Section 8, contains the conclusion of the research work done in this paper.

2. Preliminaries

This section consists of some basic notions and related axioms of cone, cone metric
space, cone b-metric space, rectangular cone metric space and rectangular cone b-metric
space. Throughout in this article, we will take E as a real Banach space.

Definition 1. [21] Let C ⊆ E, then C is said to be a cone if it admits the following axioms:

(1) C is non empty and closed set, and C ̸= {0};

(2) px+ qy ∈ C, ∀ x, y ∈ C where p, q ∈ R and a, b ≥ 0;

(3) C ∩ (−C) = {0}.

For C ⊂ E, the partial order ⪯ in C is taken as:

x ⪯ y ⇔ y − x ∈ Cfor allx, y ∈ E .

The expression x≪ y shows that y − x ∈ C◦(the interior of C).

The cone C ⊆ E is called normal, if there exists K > 0 such that

0 ⪯ x ⪯ y ⇒ ∥x∥ ≤ K∥y∥.

Let ℜ denotes the partial order in an ordinary set X and ⪯ be a partial order in the cone
C ⊆ E . For X ⊆ E , ℜ and ⪯ assumed to be same.



A. Arif et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7033 4 of 29

Definition 2. [21] Let X ̸= ∅ and dc : X ×X 7→ E satisfies the following axioms:

(d1) dc(ℓ, r) ⪰ 0, ∀ ℓ, r ∈ X and dc(ℓ, r) = 0 ⇔ ℓ = r;

(d2) dc(ℓ, r) = dc(r, ℓ);

(d3) dc(ℓ, r1) ⪯ dc(ℓ, r) + dc(r, r1), ∀ ℓ, r, r1 ∈ X.

Then dc is called cone metric and the pair (X,dc) represents a cone metric space.

Example 1. [36] Consider X = R, E = R2 with cone C = {(ℓ, y) ∈ E : ℓ, y ≥ 0}. Take
dc : X ×X → E, in a way

dc(ℓ, y) = (| ℓ− y |, δ | ℓ− y |),

here δ ≥ 0 (scalar). Subsequently dc represents a cone metric, and (X, dc) a cone metric
space.

Proposition 3. [36] Let C be a cone with cone metric space (X, dc). Then for x, ℓ, ϱ ∈ E,
we have

(1) If x ⪯ αx and α ∈ [0, 1), then x = 0.

(2) If 0 ⪯ x≪ ℓ for each 0 ≪ ℓ, then x = 0.

(3) If x ⪯ ℓ and ℓ≪ ϱ, then x≪ ϱ.

Definition 4. [34] Let db : X ×X 7→ E fulfills the following axioms:

(db1) 0 ⪯ db(p, y) along with db(p, y) = 0 ⇔ p = y;

(db2) db(p, y) = db(y, p);

(db3) db(p, r) ⪯ α[db(p, x) + db(x, r)] for some α ≥ 1, ∀ p, x, r ∈ X.

Then db is called a cone b-metric, and (X, db) represents a cone b-metric space. It is
observed that every dc is a db metric space, but converse may not be true.

Example 2. [36] Let E = R2, C = {(ℓ, r) ∈ E : ℓ, r ≥ 0} ⊂ R2, and X = {1, 2, 3, 4}.
Define

db(ℓ, r) =

{
(|ℓ− r|−1, |ℓ− r|−1) if ℓ ̸= r
0 if ℓ = r.

Here (X, db) represents a cone b-metric space for s = 6
5 , and since db(1, 2) ≻ db(1, 4) +

db(4, 2), db is not a cone metric.

Definition 5. [34] Let dr : X×X 7→ E, ∀ f1, f2, f3, f4 ∈ X, satisfies the following axioms:

(dR1) 0 ⪯ dr(f1, f2) and dr(f2, f1) = 0 if and only if f1 = f2;

(dR2) dr(f1, f2) = dr(f2, f1);



A. Arif et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7033 5 of 29

(dR3) dr(f1, f4) ⪯ dr(f1, f2) + dr(f2, f3) + dr(f3, f4) for all distinct f3, f4 ∈ X \ {f1, f2}.

Then dr is known as rectangular cone metric and the pair (X, dr) represents a rectangular
cone metric space.

Remark 6. A cone metric space always represents a rectangular cone metric space, but
converse may not hold.

Definition 7. [34] Let R : X ×X 7→ E, ∀ ℓ, y, x, υ ∈ X, satisfies the following axioms:

(dRb1) 0 ⪯ R(ℓ, y) and R(ℓ, y) = 0 ⇔ ℓ = y;

(dRb2) R(ℓ, y) = R(y, ℓ);

(dRb3) R(ℓ, υ) ⪯ s[R(ℓ, y)+R(y, x)+R(x, υ)] for all distinct y, x ∈ X \ {ℓ, υ}, where s ≥ 1.

Then R stands for rectangular cone b-metric, and (X,R) represents a rectangular cone
b-metric space (RCBMS).

Example 3. [36] Consider X = [0, 2] and E = CR2(X). For x = (ℓ, y) and ℘ = (u, v) in
E, we define x.℘ = (ℓu, yv) and ∥x∥ = max(∥ℓ∥, ∥y∥), where ∥℘∥ = supx∈X | ℘(x) |. Then
E is a Banach algebra with unit element e = (1, 1) and zero element ω = (0, 0). Suppose
that C = {(ℓ, y) ∈ E : ℓ(x), y(x) ≥ 0, x ∈ X} ⊂ CR2(X), clearly C represents a cone in E.
Define R : X ×X → E, such that

R(ℓ, y)(x) =


(0, 0) if ℓ = y
(a+ bx, c+ dx2) if ℓ, y ∈ [0, 12)

and a,b,c,d are some fixed real numbers.

(a+bx
n3 , c+dx2

n3 ) if ℓ = 1
n(n ≥ 2) ∈ [0, 12) and y ∈ {1, 2}

(|ℓ− y|2(a+ bx), |ℓ− y|2(c+ dx2)) if ℓ, y ∈ [0, 12) otherwise.

Here (X,R) represents rectangular cone b-metric space with s = 2 over E, but is not cone
b-metric space.

Remark 8. A cone b-metric space always represents a rectangular cone b-metric space,
converse may not hold, as illustrated in the above mentioned examples.

Definition 9. [36] Let E be a real Banach space, (X,R) is taken as a rectangular cone
b-metric space and ϵ ∈ E with 0 ≪ ϵ(arbitrary).

(1) Any sequence {ℓn} is referred as a Cauchy sequence (CS), whenever we get K ∈ N
with R(ℓn, ℓm) ≪ c ∀ n,m > K.

(2) Any sequence {ℓn} is referred to as convergent if we have an K ∈ N with R(ℓn, ℓ) ≪ ϵ
∀ n ≥ K and ℓ ∈ X.

(3) Any (RCBMS) referred as complete, whenever every (CS) has a convergent point in
X.



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The main theorem of Simsek [13] paper is as follows:

Theorem 10. [13] Let (Y, d,⪯) be a partially ordered metric space, and L : Y → Y
satisfies the following inequality:

T (d(Lx,Ly), d(x, y), d(x, Lx), d(y, Ly), d(x, Ly), d(y, Lx)) ≤ 0, (1)

∀ x, y ∈ Y with x ⪯ y, where T : [0,∞)6 → (−∞,∞). Then L admits a fixed point in Y
provided L is continuous or (Y, d,⪯) is as a regular space.

A number of contractive conditions were obtained by using (1), for instance, define
T : [0,∞)6 → (−∞,∞) such that

T (f1, f2, f3, f4, f5, f6) = f1 − ψ

(
max

{
f2, f3, f4,

1

2
(f5 + f6)

})
,

which leads to main theorems proved in [14]. Equivalently by choosing

T (f1, f2, f3, f4, f5, f6) = f1 − kf2; k ∈ [0, 1),

in (1), we get results proved in [4]. So, we get different contractive conditions for different
definitions of T : [0,∞)6 → (−∞,∞).

Note: Throughout in this article we will take (E , ∥.∥) as a real Banach space and
B(E , E) taken as collection of all bounded linear mappings.

3. Ordered Implicit Relations

Motivated by [8–11, 39, 40], we construct an ordered implicit relation as follows:

Definition 11. Assume (E , ∥.∥) is taken as a real Banach space, along with B(E , E) (the
collection of all bounded linear operators T ), T : E → E where ∥T ∥1 <

1
s for s ≥ 1, we

take ∥.∥1 as the usual norm. The mapping J : E6 → E will define an ordered implicit
relation, if J taken as continuous on E6 and also it satisfy the given conditions:

(J1) r1 ⪯ υ1, r5 ⪯ υ5 and r6 ⪯ υ6
⇒ J (υ1, r2, r3, r4, υ5, υ6) ⪯ J (r1, r2, r3, r4, r5, r6).

(J2) if J (r1, r2, r2, r1, α[r1 + r2 + r3], r1) ⪯ 0E
or
if J (r1, r2, r1, r2, r1, α[r1 + r2 + r3]) ⪯ 0E , then ∃ T ∈ B(E , E) so that r1 ⪯ 1

αT (r2)
and r3 ⪯ 1

αT (r1) (∀ r1, r2 ∈ E) for some R ∋ α ≥ 1.

(J3) J (αr,0E ,0E , r, αr,0E) ≻ 0E whenever ∥r∥ > 0 and α ≥ 1.

Let F = {J : E6 → E |J owns J1,J2,J3}.



A. Arif et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7033 7 of 29

Example 4. Consider the partial order ⪯ in a cone C. For all ri ∈ E( for all i = 1 to 6)
and for some α < 1, define J : E6 → E such that

J (r1, r2, r3, r4, r5, r6) = r5 − α{r1 + r2} − r1.

Then the operator J ∈ F . Indeed
(J1). Let r1 ⪯ γ1, r5 ⪯ γ5 and r6 ⪯ γ6, then γ1 − r1 ∈ C, γ5 − r5 ∈ C and γ6 − r6 ∈ C.
Now we show that J (r1, r2, r3, r4, r5, r6)− J (γ1, r2, r3, r4, γ5, γ6) ∈ C. Consider,

J (r1, r2, r3, r4, r5, r6)− J (γ1, r2, r3, r4, γ5, γ6)

= r5 − α{r1 + r2} − r1 − (γ5 − α{γ1 + r2}+ r1)

= −(γ5 − r5) + α(γ1 − r1) ∈ C.

Thus, J (γ1, r2, r3, r4, γ5, γ6) ⪯ J (r1, r2, r3, r4, r5, r6).
(J2). Let r1, r2, r3 ∈ E be such that 0E ⪯ r1, 0E ⪯ r2, 0E ⪯ r3. If J (r1, r2, r1, r2, s[r1 +
r2 + r3], r1) ⪯ 0E then by definition of J , we have −s[r1 + r2 + r3] + α(r1 + r2) + r1 ∈ C.
So we get two equations,

αr2 − (s− α− 1)r1 ∈ C, (2)

(α+ 1− s)r1 − sr3 ∈ C. (3)

For (2) if r1 = 0E , then αr2 ∈ C. So, we get a T : E → E such that T (r2) = αr2 (α < 1 is
fixed) and ∥ T ∥= α < 1. If r1 ̸= 0E , then, (2) implies r1 ⪯ α

(s−α−1)r2. So ∃, T : E → E
such that T (r2) = yr2 (y = α

(s−α−1) is a scalar) such that r1 ⪯ T (r2), for α < 1.

For (3) if r3 = 0E , then (α + 1− s)r1 ∈ C. So, we have T : E → E taken as T (r1) = yr1
(where y = (α + 1 − s) is a scalar with α < 1 ≤ s implies α − s < 0 and α − s + 1 < 1)

such that r3 ⪯ T (r1). Now if r3 ̸= 0E , then, r3 ⪯ (1+α−s)
s r1. So, we get T : E → E taken

as T (r2) = yr2 (y = (1+α−s)
s is fixed) with r3 ⪯ (r1), for α < 1.

(J3). Take r ∈ E and ∥r∥ > 0 with, 0E ⪯ J (sr,0E ,0E , r, sr,0E) then (s − αs − 1)r ∈ C,
which is true for ∥r∥ > 0.

Example 5. Similarly, the operator J : E6 → E defined by

(i). J ∗
1 (r1, r2, r3, r4, r5, r6) = r1 + r5 − α(r2 + r4);α <

1
2 .

(ii). J ∗
2 (r1, r2, r3, r4, r5, r6) = (1− β)r5 − β(r3 + r4)− r3;β ∈ (−∞, 14).

(iii). J ∗
3 (r1, r2, r3, r4, r5, r6) = γr1 − r2; γ > 1.

(iv). J ∗
4 (r1, r2, r3, r4, r5, r6) = r1 + r5 − ϱ{r2 + r4}; ϱ < 3

2 .

Then J ∗
i ∈ F for each i = 1, 2, 3, 4.

We will apply this implicit relation along with some additional conditions to construct
a sequence, and to solve the following problem fixed-point problem:
“find z∗ ∈ (X,R) along f(z∗) = z∗”, when T ∈ B(E , E), I : E → E considered as



A. Arif et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7033 8 of 29

an identity operator, J ∈ F and 𭟋 : X → X admits (4), for each pair of comparable
elements r, x ∈ X when s ≥ 1.

(I − T )2(I + T )(R(r,𭟋(r))) ⪯ sR(r, x) further implies

F (R(𭟋(r),𭟋(x)),R(r, x),R(r,𭟋(r)),R(x,𭟋(x)),R(r,𭟋(x)),R(x,𭟋(r))) ⪯ 0E . (4)

Remark 12. For H ∈ B(E , E), I + H + H2 + · · · + Hn + · · · converges for ∥H∥1 < 1,
otherwise diverge. In addition if ∥H∥1 < 1, then we get a y > 0, so that ∥H∥1 < y < 1
along with ∥Hn∥1 ≤ yn < 1.

4. Results on Ordered Implicit Contractions

Popa[1] considered a self-mapping satisfying an implicit contractive conditions and estab-
lished well-known fixed point results. Ran et al. [4] used monotone functions to gener-
alize the famous Banach fixed point theorem in partially ordered metric space. Haung
et al.[21] introduced cone metric space, Hussain and Shah [24] introduced cone b-metric
space, Huang and Xu [25] practiced with cone b-metric structure to establish new fixed
point results. Azam et al. [34] introduced the concept of rectangular cone metric space
and generalized BCP. Anam et al.[26] investigated necessary convergence axioms for an
implicit contraction in a cone b-metric space. Motivated by [34, 36], George [2] defined
the rectangular cone b-metric space, that generalizes the rectangular b-metric space and
established some well-known theorems in this setting. In this section, we utilize the ideas
of Popa[1] and George [2] to develop the notion of ordered implicit relation, and prove
some fixed point results in rectangular cone b-metric spaces.

Theorem 13. Let (W,R) be a complete rectangular cone b-metric space along with a cone
C ⊂ E and ℏ : W → W . Let T ∈ B(E , E) such that ∥T ∥1 <

1
s (s ≥ 1), I : E → E and

J ∈ F such that, for each pair of comparable elements f, x ∈ W and for some s ≥ 1, the
following holds

(I − T )2(I + T )(R(f, ℏ(f))) ⪯ sR(f, x) implies

J
(
R(ℏ(f), ℏ(x)),R(f, x),R(f, ℏ(f)),R(x, ℏ(x)),R(f, ℏ2(x)),R(x, ℏ2(f))

)
⪯ 0E , (5)

and

(1) there exists f0 ∈W satisfying f0ℜℏ(f0);

(2) for every pair f, x ∈W , fℜx ⇒ ℏ(f)ℜℏ(x);

(3) for any sequence {fn} such that fnℜfn+1 and fn → z∗, we have fnℜz∗ ∀ n ∈ N and
R(z∗, ℏ(z∗)) ⪯ R(z∗, ℏ2(z∗)).

Then, there exists z∗ ∈W such that z∗ = ℏ(z∗).



A. Arif et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7033 9 of 29

Proof. Let f0 ∈ W satisfying (1), define {fn} by ℏ(fn−1) = fn, then f0ℜf1. By
assumption (2), we get f1ℜf2, f2ℜf3, · · · , fn−1ℜfn. In (5), put f = f0 and x = f1,

(I − T )2(I + T )(R(f0, ℏ(f0))) = (I − T )2(I + T )(R(f0, f1)) ⪯ sR(f0, f1) implies

J
(
R(ℏ(f0), ℏ(f1)),R(f0, f1),R(f0, ℏ(f0)),R(f1, ℏ(f1)),R(f0, ℏ2(f1)),R(f1, ℏ2(f0))

)
⪯ 0E ,

that is,

J (R(f1, f2),R(f0, f1),R(f0, f1),R(f1, f2),R(f0, f3),R(f1, f2)) ⪯ 0E . (6)

By (dR3), we have

R(f0, f3) ⪯ s[R(f0, f1) + R(f1, f2) + R(f2, f3)].

By (J1), and rewriting (6) we get:

J (R(f1, f2),R(f0, f1),R(f0, f1),R(f1, f2), s[R(f0, f1) + R(f1, f2) + R(f2, f3)],R(f1, f2)) ⪯ 0E .

By (J2), there exists T ∈ B(E , E) along with ∥T ∥1 < 1, s ≥ 1, so that

R(f1, f2) ⪯
1

s
T (R(f0, f1)) ⪯ T (R(f0, f1)) and R(f2, f3) ⪯

1

s
T (R(f1, f2)) ⪯ T (R(f1, f2)).

Now put f = f1 and x = f2 in (5) to have

(I − T )2(I + T )(R(f1, ℏ(f1))) = (I − T )2(I + T )(R(f1, f2)) ⪯ sR(f1, f2) implies

J (R(ℏ(f1), ℏ(f2)),R(f1, f2),R(f1, ℏ(f1)),R(f2, ℏ(f2)),R(f1, ℏ(f3)),R(f2, ℏ(f2))) ⪯ 0E ,

that is,

J (R(f2, f3),R(f1, f2),R(f1, f2),R(f2, f3),R(f1, f4),R(f2, f3)) ⪯ 0E .

By (dRb3), we get

R(f1, f4) ⪯ s[R(f1, f2) + R(f2, f3) + R(f3, f4)],

(J1) implies

J (R(f2, f3),R(f1, f2),R(f1, f2),R(f2, f3), s[R(f1, f2) + R(f2, f3) + R(f3, f4)],R(f2, f3)) ⪯ 0E .

Now (J2) guarantees the existence of T ∈ B(E , E) along with ∥T ∥1 < 1 so that

R(f3, f4) ⪯
1

s
T (R(f2, f3)) ⪯

1

s2
T 2(R(f2, f3)) ⪯

1

s3
T 3(R(f0, f1)) ⪯ T 3(R(f0, f1)).

Continuing, we construct a sequence {fn} so that fnℜfn+1 with fn+1 = ℏ(fn), hence for

(I − T )2(I + T )(R(fn−1, ℏ(fn−1))) = (I − T )2(I + T )(R(fn−1, fn)) ⪯ sR(fn−1, fn),



A. Arif et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7033 10 of 29

we get

R(fn, fn+1) ⪯
1

s
T (R(fn−1, fn)) ⪯

1

s2
T 2(R(fn−2, fn−1)) ⪯ · · · ⪯ 1

sn
T n(R(f0, f1)) ⪯ T n(R(f0, f1)).

Assume two situations for R(fn, fn+p), for if p is odd, that is p = 2m + 1 and R∗
n =

R(fn, fn+1)

R(fn, fn+2m+1) ⪯ s[R∗
n + R∗

n+1 + R(fn+2, fn+2m+1)]

⪯ s[R∗
n + R∗

n+1] + s2[R∗
n+2 + R∗

n+3 + R(fn+3, fn+2m+1)]

⪯ s[R∗
n + R∗

n+1] + s2[R∗
n+2 + R∗

n+3] + · · ·+ smR(fn+2m, fn+2m+1)

⪯ s[T nR∗
0 + T n+1R∗

1] + s2[T n+2R∗
0 + T n+3R∗

0] + · · ·+ smT n+2mR∗
0

≺ sT n(I + sT 2 + s2T 4 + · · · )R∗
0 + sT n+1(I + sT 2 + (sT 2)2 + · · · )R∗

0

= sT n(I − sT 2)−1R∗
0 + sT n+1(I − sT 2)−1R∗

0

= sT n(I − sT 2)−1R∗
0(I + T ).( By Remark 12)

For if p is even, that is p = 2m and R∗
n = R(fn, fn+1).

R(fn, fn+2m) ⪯ s[R∗
n + R∗

n+1 + R(fn+2, fn+2m)]

⪯ s[R∗
n + R∗

n+1] + s2[R∗
n+2 + R∗

n+3 + R(fn+3, fn+2m+1)]

⪯ s[R∗
n + R∗

n+1] + s2[R∗
n+2 + R∗

n+3] + · · ·+ sm−1[R∗
2m−4 + R∗

2m−3]

+ sm−1R(fn+2m−2, fn+2m)

⪯ s[T nR∗
0 + T n+1R∗

1] + s2[T n+2R∗
0 + T n+3R∗

0] + · · ·+ sm−1T n+2m−2R∗
0

≺ (sT n + sT n+1)(I + sT 2 + s2T 4 + · · · )R∗
0 + sm−1T n+2m−2R∗

0

= (sT n + sT n+1)(I − sT 2)−1R∗
0 + sm−1T 2m−2T nR∗

0

=
1

s
(sT n + sT n+1)(

1

s
− T 2)−1R∗

0 + sm−1T 2m−2T nR∗
0.

Here ∥T ∥1 <
1
s , which further implies that limn→∞ T n = 0. Hence, limn→∞R(fn, fn+p) =

0E and {fn} is a Cauchy sequence in W . Now (W,R) is a complete rectangular cone b-
metric space, so fn → z∗ for some z∗ ∈W if n being very large. So

R(fn, z
∗) ≪ ϵ for all n ≥ N2 ∈ N and 0 ≪ ϵ.

Let
(I − T )2(I + T )(R(fn, ℏ(fn))) ≻ sR(fn, z

∗) and

(I − T )2(I + T )(R(fn+2, ℏ(fn+2))) ≻ sR(fn+2, z
∗) for some n ∈ N and s ≥ 1.

Using (5) and (dRb3), implies

R(fn, ℏ(fn)) ⪯ s[R(fn, z
∗) + R(z∗, fn+2) + R(fn+2, fn+1)]

≺ s

s
(I − T )2(I + T )(R(fn, ℏ(fn))) +

s

s
(I − T )2(I + T )R(fn+2, ℏ(fn+2)))



A. Arif et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7033 11 of 29

+ s
1

s
T (R(fn, ℏ(fn)))

≺ (I − T )2(I + T )(R(fn, ℏ(fn))) + (I − T )2(I + T )T 2R(fn, ℏ(fn)))
+ T (R(fn, ℏ(fn)))

(I − T )R(fn, ℏ(fn)) ⪯ (I − T )2(I + T )(I + T 2)R(fn, ℏ(fn))). Thus

T 4(R(fn, ℏ(fn))) ≺ 0E ,

that steers to contradiction. Thus, for every n ≥ 1 and for some s ≥ 1

(I − T )2(I + T )(R(fn, ℏ(fn))) ⪯ sR(fn, z
∗).

This implies that (due to (5))

J
(

R(ℏ(fn), ℏ(z∗)),R(fn, z∗),R(fn, ℏ(fn)),R(z∗, ℏ(z∗)),
R(fn−1, ℏ2(z∗)),R(ℏ2(fn), ℏ(fn))

)
⪯ 0E . (7)

Now, we claim that ∥R(z∗, ℏ(z∗))∥ = 0, on contrary suppose that ∥R(z∗, ℏ(z∗))∥ > 0, as a
result we get

R(ℏ(fn), ℏ(z∗)) ⪯ s[R(ℏ(fn), ℏ(fn+1)) + R(ℏ(fn+1), z
∗)) + R(z∗, ℏ(z∗))]

limn→∞R(ℏ(fn), ℏ(z∗)) ⪯ sR(z∗, ℏ(z∗))

R(fn−1, ℏ2(z∗)) ⪯ s[R(fn−1, fn) + R(fn, z
∗) + R(z∗, ℏ2(z∗))]

limn→∞R(fn−1, ℏ2(z∗)) ⪯ sR(z∗, ℏ2(z∗))

Taking n→ ∞, and in view of assumption (3) and (7) , we get

J
(
sR(z∗, ℏ(z∗)),0E ,0E ,R(z∗, ℏ(z∗)), sR(z∗, ℏ2(z∗)),0E

)
⪯ 0E .

By (J1) and condition (3), we have

J (sR(z∗, ℏ(z∗)),0E ,0E ,R(z∗, ℏ(z∗)), sR(z∗, ℏ(z∗)),0E) ⪯ 0E .

This contradicts (J3). Thus, ∥R(z∗, ℏ(z∗))∥ = 0. Hence, R(z∗, ℏ(z∗)) = 0E and by (dRb1),
we get z∗ = ℏ(z∗).

In next theorem, we present a result for non-increasing self-operators.

Theorem 14. Let (W,R) be a complete rectangular cone b-metric space along with a cone
C ⊂ E and let ℏ : W → W . Consider T ∈ B(E , E) with ∥T ∥1 <

1
s (s ≥ 1), I : E → E and

J ∈ F such that, for each pair of comparable elements f, x ∈ W with s ≥ 1, following
holds

(I − T )2(I + T )(R(f, ℏ(f))) ⪯ sR(f, x) implies

J
(
R(ℏ(f), ℏ(x)),R(f, x),R(f, ℏ(f)),R(x, ℏ(x)),R(f, ℏ2(x)),R(x, ℏ2(f))

)
⪯ 0E (8)

and



A. Arif et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7033 12 of 29

(1) there exists f0 ∈W satisfying ℏ(f0)ℜf0;

(2) for every pair of f, x ∈W , fℜx ⇒ ℏ(x)ℜℏ(f);

(3) for a sequence {fn} satisfying fnℜfn+1 and fn → z∗, we get fnℜz∗ ∀ n ∈ N. More-
over, R(z∗, ℏ(z∗)) ⪯ R(z∗, ℏ2(z∗)).

Then, ℏ admits a fixed point z∗ ∈W .

Proof. Let f0 ∈W satisfies assumtion (1). Construct the sequence {fn} such that fn =
ℏ(fn−1) ∀ n ∈ N . Since f1 = ℏ(f0)ℜf0 and by using condition (2), f1 = ℏ(f0)ℜℏ(f1) = f2,
and assumption (2) further implies fn ⪯ fn−1. Letting f = f1 and x = f0 in (8), we get

(I − T )2(I + T )(R(ℏ(f0), f0) = (I − T )2(I + T )(R(f1, f0)) ⪯ sR(f1, f0) implies

J
(
R(ℏ(f1), ℏ(f0)),R(f1, f0),R(f1, ℏ(f1)),R(f0, ℏ(f0)),R(f1, ℏ2(f0)),R(f0, ℏ2(f1))

)
⪯ 0E

⇒ J (R(f1, f2),R(f0, f1),R(f1, f2),R(f0, f1),R(f1, f2),R(f0, f3)) ⪯ 0E .

By (dR3), we have

R(f0, f3) ⪯ s[R(f0, f1) + R(f1, f2) + R(f2, f3)]

and then using J1, we obtain

J (R(f1, f2),R(f0, f1),R(f1, f2),R(f0, f1),R(f1, f2), s[R(f0, f1)+R(f1, f2)+R(f2, f3)]) ⪯ 0E .

By (J2), there exists T ∈ B(E , E) with ∥T ∥1 < 1 and

R(f1, f2) ⪯
1

s
T (R(f0, f1)) ⪯ T (R(f0, f1)) and R(f2, f3) ⪯

1

s
T (R(f1, f2)) ⪯ T (R(f1, f2)).

By (2), ℏ(f0)ℜℏ(f1), letting f = f1 and x = f2 in (8) we get

(I − T )2(I + T )(R(f1, ℏ(f1)) = (I − T )2(I + T )(R(f1, f2)) ⪯ sR(f1, f2) implies

J
(
R(ℏ(f1), ℏ(f2)),R(f1, f2),R(f1, ℏ(f1)),R(f2, ℏ(f2)),R(f1, ℏ2(f2)),R(f2, ℏ2(f1))

)
⪯ 0E

⇒ J (R(f2, f3),R(f1, f2),R(f1, f2),R(f2, f3),R(f1, f4),R(f2, f3)) ⪯ 0E .

By (dR3), (J1) and (J2), we get

R(f3, f4) ⪯
1

s
T (R(f2, f3)) ⪯

1

s2
T 2(R(f1, f2)) ⪯

1

s3
T 3(R(f0, f1)) ⪯ T 3(R(f0, f1)).

Following this pattern, we get a sequence {fn} that satisfies

R(fn, fn+1) ⪯
1

s
T (R(fn−1, fn)) ⪯

1

s2
T 2(R(fn−2, fn−1)) ⪯ · · · ⪯ 1

sn
T n(R(f0, f1)).

Following the pattern for the proof of Theorem 13, there exists a convergence point z∗

satisfying z∗ = ℏ(z∗).

The next result summarizes the results of Theorem 13 and Theorem 14.



A. Arif et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7033 13 of 29

Theorem 15. Let (W,R) be a complete rectangular cone b-metric space with C ⊂ E as a
cone and ℏ :W →W . If T ∈ B(E , E) with ∥T ∥1 <

1
s (s ≥ 1), I : E → E and J ∈ F . Also

let, for each pair of comparable elements ϱ, x ∈W and for some s ≥ 1, the following holds

(I − T )2(I + T )(R(f, ℏ(f))) ⪯ sR(f, x) implies

J
(
R(ℏ(f), ℏ(x)),R(f, x),R(f, ℏ(f)),R(x, ℏ(x)),R(f, ℏ2(x)),R(k, ℏ2(f))

)
⪯ 0E , (9)

and

(1) ∃ f0 ∈W with f0ℜℏ(f0) or ℏ(f0)ℜf0;

(2) for a sequence {fn} satisfying fnℜfn+1 and fn → z∗, we get fnℜz∗ ∀ n ∈ N and let
R(z∗, ℏ(z∗)) ⪯ R(z∗, ℏ2(z∗)).

Then, ℏ admits a fixed point z∗ ∈W .

Proof. Let f0 ∈ W and define the sequence {fn} by fn = ℏ(fn−1) ∀ n ∈ N. By
assumption (1), we get f0ℜℏ(f0) = f1. Also f2ℜf1(as f is order reversing). By (9), we
obtain

(I − T )2(I + T )(R(f0, ℏ(f0))) = (I − T )2(I + T )(R(f0, f1)) ⪯ sR(f0, f1) implies

J
(

R(ℏ(f0), ℏ(f1)),R(f0, f1),R(f0, ℏ(f0)),
R(f1, ℏ(f1)),R(f0, ℏ(f2)),R(f1, ℏ(f1))

)
⪯ 0E ,

that is,

J
(

R(f1, f2),R(f0, f1),R(f0, f1),R(f1, f2),
R(f0, f3),R(f1, f2)

)
⪯ 0E . (10)

By (dR3) and (J1),

J (R(f1, f2),R(f0, f1),R(f0, f1),R(f1, f2), s[R(f0, f1)+R(f1, f2)+R(f2, f3)],R(f2, f1) ⪯ 0E .

By (J2), ∃ T ∈ B(E , E) so that ∥T ∥1 < 1 with

R(f1, f2) ⪯
1

s
T (R(f0, f1)) ⪯ T (R(f0, f1)),

and

R(f2, f3) ⪯
1

s
T (R(f1, f2)) ⪯ T (R(f1, f2)).

using assumption (2), we get f2ℜf1(f being order preserving), hence using (9)

(I − T )2(I + T )R(ℏ(f1), f1) = (I − T )2(I + T )R(f2, f1) ⪯ sR(f2, f1) implies

J
(

R(ℏ(f2), ℏ(f1)),R(f2, f1),R(f2, ℏ(f2)),R(f1, ℏ(f1),
R(f2, ℏ2(f1)),R(f1, ℏ2(f2))

)
⪯ 0E .



A. Arif et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7033 14 of 29

By (dR3), (J1) and (J2), we get

R(f3, f4) ⪯
1

s
T (R(f2, f3)) ⪯

1

s2
T 2(R(f1, f2)) ⪯

1

s3
T 3(R(f0, f1)) ⪯ T 3(R(f0, f1)).

This pattern enable us to construct a sequence {fn} so that

R(fn, fn+1) ⪯
1

s
T (R(fn−1, fn)) ⪯

1

s2
T 2(R(fn−2, fn−1)) ⪯ · · · ⪯ 1

sn
T n(R(f0, f1)).

By following same pattern as for the proof of Theorem 13, we will obtain z∗ = ℏ(z∗).

Remark 16. We get different ordered contractive conditions for different definitions of
J . Also, Theorem 15 generalizes the results in [8–11, 37, 38].

Remark 17. In a non-normal cone, by taking an“ upper bound or lower bound” for every
pair ϱ, ω ∈W , one may find a unique fixed point in Theorem 15.

5. Examples and consequences

In this section, we give some examples for the explanation of hypotheses of main
theorem.

Example 6. Let E = (R3, ∥ · ∥), then it is a real Banach space. Define a cone C ∈ E
such that C = {(ϑ, x, ν) ∈ R3 : ϑ, x, ν ≥ 0} and ∥ϑ∥ = max {(|ϑ1|, |ϑ2|, |ϑ3|)}. Consider
X = {(0, 0, 0) ,

(
1
4 , 0, 0

)
,
(
1
4 ,

1
4 , 0

)
,
(
1
4 ,

1
4 ,

1
4

)
} ⊂ E with ℏ : X → X defined as

ℏ (0, 0, 0) = ℏ
(
1

4
, 0, 0

)
= (0, 0, 0) , ℏ

(
1

4
,
1

4
, 0

)
=

(
1

4
, 0, 0

)
, ℏ

(
1

4
,
1

4
,
1

4

)
=

(
1

4
,
1

4
, 0

)
.

Define R : X ×X → E,

R(ϑ1, ϑ2) =


(0, 0, 0) if ϑ1 = ϑ2(
3
4 ,

3
4 ,

3
4

)
if ϑ1, ϑ2 ∈ {(0, 0, 0) ,

(
1
4 ,

1
4 , 0

)
}(

1
4 ,

1
4 ,

1
4

)
if ϑ1, ϑ2 ∈ {

(
1
4 ,

1
4 ,

1
4

)
,
(
1
4 ,

1
4 , 0

)
}(

0, 14 ,
1
4

)
otherwise.

Define T : E → E by

T (ϑ) =
ϑ

2
, clearly ∥T ∥1 < 1.

(
3

4
,
3

4
,
3

4

)
= R

(
(0, 0, 0) ,

(
1

4
,
1

4
, 0

))
⪰ R

(
(0, 0, 0) ,

(
1

4
, 0, 0

))
+ R

((
1

4
, 0, 0

)
,

(
1

4
,
1

4
,
1

4

))
+ R

((
1

4
,
1

4
,
1

4

)
,

(
1

4
,
1

4
, 0

))
=

(
0,

1

4
,
1

4

)
+

(
0,

1

4
,
1

4

)
+

(
1

4
,
1

4
,
1

4

)



A. Arif et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7033 15 of 29

=

(
1

4
,
3

4
,
3

4

)
.

Note that R is RCBM but not RBM, for s = 3. Consider ϑ = (0, 0, 0) and ω =
(
1
4 , 0, 0

)
.

Then ℏ(ϑ) = (0, 0, 0) = ℏ(ω) = ℏ2(ω)

R(ϑ, ℏ(ϑ)) = R(ϑ, ℏ2(ω)) = (0, 0, 0) , R(ϑ, ω) =

(
0,

1

4
,
1

4

)
,

(I − T )2(I + T )R(ϑ, ℏ(ϑ)) = (0, 0, 0).

Take ϑ =
(
1
4 , 0, 0

)
and ω =

(
1
4 ,

1
4 , 0

)
. Then ℏ(ϑ) = (0, 0, 0) = ℏ2(ω) and ℏ(ω) =

(
1
4 , 0, 0

)
.

R(ϑ, ℏ(ϑ)) = R(ϑ, ℏ2(ω)) = R(ϑ, ω) =

(
0,

1

4
,
1

4

)
,

(I − T )2(I + T )R(ϑ, ℏ(ϑ)) =

(
0,

3

32
,
3

32

)
.

For ϑ =
(
1
4 ,

1
4 , 0

)
and ω =

(
1
4 ,

1
4 ,

1
4

)
. Then ℏ(ϑ) = (14 , 0, 0) = ℏ2(ω) and ℏ(ω) =

(
1
4 ,

1
4 , 0

)
.

R(ℏ(ϑ), ℏ(ω)) = R(ϑ, ℏ(ϑ)) = R(ϑ, ℏ2(ω)) =
(
0,

1

4
,
1

4

)
, R(ϑ, ω) =

(
1

4
,
1

4
,
1

4

)
,

(I − T )2(I + T )R(ϑ, ℏ(ϑ)) =

(
0,

3

32
,
3

32

)
.

Take ϑ = (0, 0, 0) and ω =
(
1
4 ,

1
4 , 0

)
. Then ℏ(ϑ) = (0, 0, 0) = ℏ2(ω) and ℏ(ω) =

(
1
4 , 0, 0

)
.

R(ℏ(ϑ), ℏ(ω)) = R(ϑ, ω) =

(
0,

1

4
,
1

4

)
and R(ϑ, ℏ2(ω)) = (0, 0, 0),

(I − T )2(I + T )R(ϑ, ℏ(ϑ)) = (0, 0, 0) .

Take ϑ = (0, 0, 0) and ω =
(
1
4 ,

1
4 ,

1
4

)
. Then ℏ(ϑ) = (0, 0, 0) = ℏ2(ω) and ℏ(ω) =

(
1
4 ,

1
4 , 0

)
.

R(ℏ(ϑ), ℏ(ω)) = R(ϑ, ω) =

(
0,

1

4
,
1

4

)
and R(ϑ, ℏ2(ω)) = (0, 0, 0),

(I − T )2(I + T )R(ϑ, ℏ(ϑ)) = (0, 0, 0) .

Take ϑ =
(
1
4 , 0, 0

)
and ω =

(
1
4 ,

1
4 ,

1
4

)
. Then ℏ(ϑ) = (0, 0, 0) = ℏ2(ω) and ℏ(ω) =

(
1
4 ,

1
4 , 0

)
.

R(ℏ(ϑ), ℏ(ω)) = R(ϑ, ω) =

(
3

4
,
3

4
,
3

4

)
and R(ϑ, ℏ2(ω)) = (0, 0, 0),



A. Arif et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7033 16 of 29

(I − T )2(I + T )R(ϑ, ℏ(ϑ)) =
(
0,

3

32
,
3

32

)
.

Define for α ∈ [1,∞), and J ∈ F

J
(

R(ℏ(ϑ), ℏ(ω)),R(ϑ, ω),R(ϑ, ℏ(ϑ))
,R(ω, ℏ(ω)),R(ϑ, ℏ2(ω)),R(ω, ℏ2(ϑ))

)
= R(ϑ, ℏ2(ω))− [αR(ℏ(ϑ), ℏ(ω)) + R(ϑ, ω) ].

Thus, (I − T )2(I + T )R(ϑ, ℏ(ϑ)) ⪯ R(ϑ, ω) implies

J
(

R(ℏ(ϑ), ℏ(ω)),R(ϑ, ω),R(ϑ, ℏ(ϑ)),R(ω, ℏ(ω)),
R(ϑ, ℏ2(ω)),R(ω, ℏ2(ϑ))

)
⪯ 0.

So all assumptions of Theorem 15 are justified and ℏ(0, 0, 0) = (0, 0, 0).

Example 7. Let E = (R, ∥ · ∥), then it is a real Banach space and C = {ϑ ∈ R : ϑ ≥ 0}
is a cone in E. Let X = {1, 2, 3, 4} and define ℏ : X → X by ℏ(1) = ℏ(2) = 1 and
ℏ(4) = ℏ(3) = 2. Define R by

R(ϑ1, ϑ2) =


0 if ϑ1 = ϑ2
et

3 if ϑ1, ϑ2 ∈ {1, 2}
et

2 if (ϑ1, ϑ2) ∈ {3, 4}
et otherwise .

Note that

et = R(2, 4) ≥ R(2, 1) + R(1, 3) + R(3, 4) =
5et

6
.

For s = 6
5 , R represents RCBM but not RCM. Define T : E → E by

T (ϑ) =
ϑ

3
, clearly ∥T ∥1 < 1.

Take ϑ = 1 and ω = 2, then

R(ω, ℏ(ω)) = R(ϑ, ω) = R(ω, ℏ2(ϑ)) =
et

3
, R(ϑ, ℏ(ϑ)) = R(ℏ(ϑ), ℏ(ω)) = 0.

Take ϑ = 2 and ω = 3, then

R(ϑ, ω) = R(ϑ, ℏ(ϑ)) = R(ω, ℏ2(ϑ)) = R(ω, ℏ(ω)) =
et

3
.

As
(I + T )(I − T )2R(ϑ, ℏ(ϑ)) = (I − T )(I − T 2)R(ϑ, ℏ(ϑ)).

We get

(I − T )(I − T 2)R(ϑ, ℏ(ϑ)) =
17et

81
.



A. Arif et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7033 17 of 29

Take ϑ = 3 and ω = 4, then

R(ω, ℏ(ω)) = R(ω, ℏ2(ϑ)) = et, R(ϑ, ℏ(ϑ)) =
et

3
, R(ϑ, ω) =

et

2
.

(I − T )(I − T 2)R(ϑ, ℏ(ϑ)) =
17et

81
.

For J ∈ F , define

J
(

R(ℏ(ϑ), ℏ(ω)),R(ϑ, ω),R(ϑ, ℏ(ϑ))
,R(ω, ℏ(ω)),R(ϑ, ℏ2(ω)),R(ω, ℏ2(ϑ))

)
= R(ω, ℏ2(ϑ))− [R(ϑ, ℏ(ϑ)) + R(ω, ℏ(ω))] .

Thus, (I − T )2(I + T )R(ϑ, ℏ(ϑ)) ⪯ R(ϑ, ω) implies

J
(

R(ℏ(ϑ), ℏ(ω)),R(ϑ, ω),R(ϑ, ℏ(ϑ)),R(ω, ℏ(ω)),
R(ϑ, ℏ2(ω)),R(ω, ℏ2(ϑ))

)
⪯ 0.

Note that ℏ(1) = 1.

5.1. Consequences of the main theorem.

In this subsection, we state some corollaries that can be deduced directly from the
main theorem.

Corollary 18. Let (X,R) be a complete rectangular cone b-metric space along with self
mapping ℏ : X → X and a cone C. Let T ∈ B(E , E) and I : E → E an identity operator,
s ≥ 1. If there exists J ∈ F satisfying, for each pair of comparable elements ϑ, ω ∈ X,

(I − T )2(I + T )(R(ϑ, ℏ(ϑ))) ⪯ sR(ϑ, ω) implies

R(ℏ(ϑ), ℏ(ω)) ⪯ 1

s
T (R(ϑ, ω)), (11)

and

(1) ∃ ϑ ∈ X so that ϑ0ℜℏ(ϑ0) or ℏ(ϑ0)ℜϑ0;

(2) ∀ ϑ, ω ∈ X, ϑℜω ⇒ ℏ(ϑ)ℜℏ(ω) or ℏ(ω)ℜℏ(ϑ);

(3) for a sequence {ϑn} such that ϑnℜϑn+1 and ϑn → z∗, we get ϑnℜz∗ ∀ n ∈ N and let
R(z∗, ℏ(z∗)) ⪯ R(z∗, ℏ2(z∗)).

Then there exists z∗ ∈ X so that z∗ = ℏ(z∗).

Proof. Define J as in Example 5 (iv), and T : E → E by T (ϑ) = βϑ ∀ ϑ ∈ E ,
0 ≤ β < 1. Clearly ∥T ∥1 < 1, we get T ∈ B(E , E), and the proof is obvious as an
application of Theorem 15.



A. Arif et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7033 18 of 29

Corollary 19. Let (X,R) be a complete rectangular cone b-metric space, ℏ : X → X, C
be a cone, T : E → E such that ∥T ∥1 <

1
s , I : E → E ba an identity operator and s ≥ 1. If

there exists J ∈ F , for each pair of comparable elements ϑ, ω ∈ X, satisfying

(I − T )2(I + T )(R(ϑ, ℏ(ϑ))) ⪯ sR(ϑ, ω) implies

R(ℏ(ϑ), ℏ(ω)) ⪯ T
s
(R(ϑ, ω))

and

(1) there exists ϑ0 ∈ X satisfying ϑ0ℜℏ(ϑ0) or ℏ(ϑ0)ℜϑ0;

(2) ∀ ϑ, ω ∈ X, ϑℜω ⇒ ℏ(ϑ)ℜℏ(ω) or ℏ(ω)ℜℏ(ϑ);

(3) for a sequence {ϑn} such that ϑnℜϑn+1 and ϑn → z∗, we get ϑnℜz∗ for all n ∈ N
and let R(z∗, ℏ(z∗)) ⪯ R(z∗, ℏ2(z∗)).

Then there exists z∗ ∈ X satisfying z∗ = ℏ(z∗).

Proof. Define T : E → E by T (ϑ) = ϑ ∀ ϑ ∈ E . Then the proof is obvious from
Corollary 18 .

Example 8. Let E = (R, ∥ · ∥), then it is real Banach space. The set C = {ϑ ∈ R : ϑ ≥ 0}
is a cone. Let W = {0, 1, 2, 3}, define ℏ :W →W by ℏ(0) = ℏ(2) = 0, ℏ(1) = 2, ℏ(3) = 1
and R :W ×W → E by

R(ϑ1, ϑ2) =


0 if ϑ1 = ϑ2
3 if ϑ1, ϑ2 ∈ {1, 2}
10 if ϑ1, ϑ2 ∈ {0, 1}
22 if ϑ1, ϑ2 ∈ {2, 3}
0.5 otherwise .

22 = R (2, 3) ≥ R (2, 0) + R (0, 1) + R (1, 3) = 0.5 + 10 + 0.5 = 11.

We notice that for s = 2, R is a rectangular cone b-metric space, but not cone rectangular
metric space. Define T : E → E by

T (ϑ) =
ϑ

2
, clearly ∥T ∥1 < 1.

For if ϑ = 0, ω = 1, then

R(ϑ, ω) = 10, R(ℏ(ϑ), ℏ(ω)) = 0.5, R(ϑ, ℏ(ϑ)) = 0.

T R(ϑ, ω) =
10

2
= 5.

For if ϑ = 1, ω = 2, then

R(ϑ, ω) = R(ϑ, ℏ(ϑ)) = 3, R(ℏ(ϑ), ℏ(ω)) = 1.125.



A. Arif et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7033 19 of 29

T R(ϑ, ω) =
3

2
= 1.5, (I − T )2(I + T )R(ϑ, ℏ(ϑ)) = 0.25.

For if ϑ = 2, ω = 3, then

R(ϑ, ω) = 22, R(ℏ(ϑ), ℏ(ω)) = 10, R(ϑ, ℏ(ϑ)) = 0.5.

T R(ϑ, ω) =
0.5

2
= 0.25, (I − T )2(I + T )R(ϑ, ℏ(ϑ)) = 0.1875.

Hence for all ϑ, ω ∈W ,

(I − T )2(I + T )R(ϑ, ℏ(ϑ)) ⪯ sR(ϑ, ω)

implies
R(ℏ(ϑ), ℏ(ω)) ⪯ T R(ϑ, ω).

Corollary 18 leads us to have a point 0 satisfying ℏ(0) = 0.

Example 9. Let E = C1
R[1, 2], and ∥ϑ∥ = ∥ϑ∥∞ + ∥ϑ́∥∞, C = {ϑ(s) ∈ E : ϑ(s) > 0, s ∈

[1, 2]}. For q ≥ 1, let ϑ = x and ω = x2k. By definition ∥ϑ∥ = 2 and ∥ω∥ = 2k + 1 also
ϑ ⪯ ω, with q∥ϑ∥ ≤ ∥ω∥. So C represents a non normal cone. Define T : E → E by

(T ϑ)(t) = 1

2

∫ t

0
ϑ(t)ds.

(T (ϑ+ ω))(t) =
1

2

∫ t

0
(ϑ+ ω)(t)ds =

1

2

∫ t

0
ϑ(t)ds+

1

2

∫ t

0
ω(t)ds.

(T nϑ)(t) ⪯ T n

2nn!
∥ϑ∥

1
2∞ ⪯ 2n

2nn!
∥ϑ∥

1
2 for each t ∈ [1, 2] and n ≥ 1.

So

∥(T nϑ)(t)∥∞ ⪯ 1

n!
∥ϑ∥

1
2 .

∥(T nϑ)′(t)∥∞ ⪯ 1√
(n− 1)!

∥ϑ∥
1
4 for n ≥ 2.

∥(T nϑ)(t)∥ = ∥(T nϑ)(t)∥∞ + ∥(T nϑ)′(t)∥∞ ⪯ 1

n!
∥ϑ∥

1
2 +

1√
(n− 1)!

∥ϑ∥
1
4 for n ≥ 2,

∥(T nϑ)(t)∥ = 0 when n ≥ n1 for n1 ∈ N.

So T ∈ B(E , E). Take X = { et

3 ,
et

2 , e
t, 2et} and ℏ : X → X, such that ℏ( et3 ) = ℏ( et2 ) =

et

3 ,

ℏ(et) = 2et and ℏ(2et) = et

2 . Define R : X ×X → E by

R(ϑ1, ϑ2) =



0 if ϑ1 = ϑ2
10et if (ϑ1, ϑ2) = ( e

t

3 , e
t),

et

3 if (ϑ1, ϑ2) = ( e
t

2 , 2e
t),

6et if (ϑ1, ϑ2) ∈ {( et2 , e
t), (et, 2et)}

3et

2 otherwise .



A. Arif et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7033 20 of 29

10et = R

(
et

3
, et

)
≥ R

(
et

3
,
et

2

)
+ R

(
et

2
, 2et

)
+ R

(
2et, et

)
=

3et

2
+
et

3
+ 6et =

29et

6
.

One can easily check that (db3) holds for s = 3, so R represents rectangular cone b-metric
space, but not rectangular cone metric space. For if ϑ = et

3 and ω = et

2 , then

R(ϑ, ℏ(ϑ)) = R(ℏ(ϑ), ℏ(ω)) = 0, R(ϑ, ω) =
3et

2

(I − T )2(I + T )R(ϑ, ℏ(ϑ)) = 0.

For if ϑ = et

2 and ω = et, then

R(ϑ, ω) = 6et, R(ℏ(ϑ), ℏ(ω)) = R(ϑ, ℏ(ϑ)) =
3et

2
,

T
s
R(ϑ, ω)(t) = et.

(I − T )2(I + T )R(ϑ, ℏ(ϑ)) =
15et

4
.

For if ϑ = et and ω = 2et, then

R(ϑ, ω) = 6et, R(ℏ(ϑ), ℏ(ω)) =
3et

2
, R(ϑ, ℏ(ϑ)) = et

T
s
R(ϑ, ω)(t) = et.

(I − T )2(I + T )R(ϑ, ℏ(ϑ)) =
3et

8
.

Now if t ∈ [1, 2], and s = 3, then

(I − T )2(I + T )R(ϑ, ℏ(ϑ)) ⪯ sR(ϑ, ω)

implies

R(ℏ(ϑ), ℏ(ω)) ⪯ 1

s
T R(ϑ, ω).

By Corollary 19, we get ℏ( et3 ) =
et

3 .

6. A homotopy result

In this section, we apply Corollary 19 to obtain the following homotopy result.

Theorem 20. Let (E , ∥.∥) be a real Banach space and C ⊂ E be a cone. Let (X,R) be a
complete rectangular cone b-metric space. Assume an open set V ⊂ X. Let T ∈ B(E , E)
with ∥T ∥1 < 1 and T (C) ⊂ C and g : V× [0, 1] → X be a monotonic mapping, and satisfies
the axioms of Corollary 19 in the first variable and

(1) ϑ ̸= g(ϑ, ℘) for each ϑ ∈ ∂V (∂V stands for the boundary of V in X);

(2) we have M ≥ 0 with
∥R(g(ϑ, ℘), g(ϑ, σ))∥ ≤M |℘− σ|,

for some ϑ ∈ V and σ, ℘ ∈ [0, 1];



A. Arif et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7033 21 of 29

(3) for any ϑ ∈ V there is ω ∈ X so that ∥R(ϑ, ω)∥ ≤ r, then ϑℜω, r is the radius of V.

If g(·, 0) admits a fixed point in V, then g(·, 1) also admits a fixed point in V.

Proof. Let
C = {℘ ∈ [0, 1] | ϑ = g(ϑ, ℘); for ϑ ∈ V}.

Define the partial order ⪯ in E by p ⪯ h ⇔ ∥p∥ ≤ ∥h∥ ∀ p,h ∈ E . It is obvious that 0 ∈ C,
because g(., 0) possesses a fixed point in V. Hence C ̸= ϕ. Defining R(ϑ, g(ϑ, ℘)) = R(ϑ, ω),
(I − T )2(I + T )(R(ϑ, g(ϑ, ℘))) ⪯ sR(ϑ, ω) ∀ ϑℜω, then by using Corollary 19, we get

R(g(ϑ, ℘), g(ω, ℘)) ⪯ 1

s
T (R(ϑ, ω)).

Firstly, we need to show that C is closed in [0, 1]. Take {℘n}∞n=1 ⊆ C along with
limn→∞ ℘n → ℘ ∈ [0, 1]. To show C is closed, we need to show that ℘ ∈ C. As ℘n ∈ C ∀
n ∈ N, so we have ϑn ∈ V so that ϑn = g(ϑn, ℘n). Because, g(ϑ, ·) is monotone, we get
ϑmℜϑn ∀ n,m ∈ N. Also for s ≥ 1,

(I − T )2(I + T )(R(ϑn, g(ϑm, ℘m))) = (I − T )2(I + T )(R(ϑn, ϑm)) ⪯ sR(ϑn, ϑm),

we have

R(g(ϑn, ℘m), g(ϑm, ℘m)) ⪯ 1

s
T (R(ϑnϑm)),

and

R(ϑn, ϑm+1) = R(g(ϑn, ℘n), g(ϑm+1, ℘m+1))

⪯ s[R(g(ϑn, ℘n), g(ϑn, ℘m)) + R(g(ϑn, ℘m), g(ϑn, ℘m+1))]

+ sR(g(ϑn, ℘m+1), g(ϑm+1, ℘m+1))

∥R(ϑn, ϑm)∥ ≤ s

[
M |℘n − ℘m|+M |℘m − ℘m+1|+

∥∥∥∥1sT (R(ϑn, ϑm))

∥∥∥∥]
∥R(ϑn, ϑm)∥ ≤ sM

1− ∥T ∥
[|℘n − ℘m|+ |℘m − ℘m+1|].

Since {ϑn}∞n=1 is a Cauchy sequence in [0, 1], so

lim
n,m→∞

R(ϑn, ϑm) = 0E .

This shows that {ϑn} represents a Cauchy sequence in X. Since X is a complete RCBM
space, hence limn→∞R(ϑn, ϑ) ≪ c for some ϑ ∈ U . As a result ϑnℜϑ ∀ n ∈ N. By (dRb3)

R(ϑ, g(ϑ, ℘)) ⪯ s[R(ϑ, ϑn) + R(ϑn, g(ϑn, ℘)) + R(g(ϑn, ℘), g(ϑ, ℘))]

⪯ s[R(ϑ, ϑn) + R(g(ϑn, ℘n), g(ϑn, ℘)) + R(g(ϑn, ℘), g(ϑ, ℘))]

∥R(ϑ, g(ϑ, ℘))∥ ≤ sR(ϑ, ϑn) + s

[
M |℘n − ℘|+

∥∥∥∥1sT (R(ϑn, ϑ))

∥∥∥∥] .



A. Arif et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7033 22 of 29

Hence, R(ϑ, g(ϑ, ℘)) = 0E . So ℘ ∈ C, hence C is closed in [0, 1]. Now we show that C is
open in [0, 1]. Take ℘2 ∈ C with g(℘2, ϑ2) = ϑ2 for ϑ2 ∈ V. Now V is open, so we can find
some r > 0 with C(ϑ2, r) ⊆ V. Assume

l = R(ϑ2, ∂V) = inf{R(ϑ2, x) : x ∈ ∂V}.

We get r = l > 0. Given ϵ > 0 taking ϵ < (1−∥T ∥)l
2sM . Take ℘ ∈ (℘1 − ϵ, ℘1 + ϵ) with

℘1 ∈ (℘2 − ϵ, ℘2 + ϵ). Subsequently

ϑ ∈ C(ϑ2, r) = {ϑ ∈ X : ∥R(ϑ, ϑ2)∥ ≤ r}, as ϑℜϑ2.

Consider

R(g(ϑ, ℘), ϑ2) = R(g(ϑ, ℘), g(ϑ2, ℘2)

⪯ s[R(g(ϑ, ℘), g(ϑ, ℘1) + R(g(ϑ, ℘1), g(ϑ, ℘2) + R(g(ϑ, ℘2), g(ϑ2, ℘2))]

∥R(g(ϑ, ℘), ϑ2)∥ ≤ sM |℘1 − ℘|+ sM |℘2 − ℘1|+ ∥T (R(ϑ2, ϑ))∥
≤ sMϵ+ sMϵ+ ∥T ∥ l = 2sMϵ+ ∥T ∥ l < l.

Thus for each ℘ ∈ (℘2 − ϵ, ℘2 + ϵ), g(·, ℘) : C(ϑ, r) → C(ϑ, r) has a fixed point in V as an
implementation of Corollary 19. So ℘ ∈ C, for every ℘ ∈ (℘2 − ϵ, ℘2 + ϵ) and hence C is
open in [0, 1]. Using connectedness, C = [0, 1]. So g(·, 1) has a fixed point in V.

6.1. Application of homotopy to human aging process

This segment use homotopy to explain the procedure of aging of human body. We
assume aging process by taking appropriate values for the time parameters t of the ho-
motopy η(t, α). Here t and α control the procedure of aging. Human body will be con-
sidered as one year old, if we have a homotopy η(t, α) : ℏ(α) → g(α) for t ∈ [0, 1] along
η(0, α) = ℏ(α) and η(1, α) = g(α). Human body will be considered as n years old, if we
have η(t, α) : ℏ(α) → w(α) for t ∈ [1, n], n > l with η(l, α) = ℏ(α) and η(n, α) = w(α).
Now the true age of human body is the least upper bound η(n, α) = y(α), where t ∈ [l, n].
Topologically the toddler is the same as a fully grown person, because the toddler con-
stantly becomes an adult. We establish an algebraic method to relate homotopy with the
procedure of aging of human body. The cylinder X = S × I is to be considered topo-
logically equivalent to a compact connected human body, when S is taken as circle and
I = [0, β]. The infant is topologically taken as X = S × I. The family of continuous
functions η(t, α) on the interval I = [0, β] is called homotopy. Consider X as human body,
then homotopy is an increasing sequence of the function η(t, α). We use homotopy to
relate topologically a toddler to topologically an adult. Consider X as a human body,
assume α ∈ X define growth of the body and t ∈ I define age of the body. Because we
are not sure about the life duration of human body, consider t = ∞ as the final age of
the body, where t ∈ [θ,∞] represents the age interval from t = θ to t = ∞. The aging
procedure is the sequence of the functions η(t, α) so that, ∀ t ∈ [0,∞] η(0, α) = ℏ(α) and
η(∞, α) = y(α).



A. Arif et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7033 23 of 29

Theorem 21. Consider X = S × I as a cylinder. Take a homotopy η(t, α), if we have a
fixed point for η(0, α), then η(∞, α) also has fixed point.

Proof. Because of compactness and connectedness, human body is topologically equiv-
alent to a cylinder X = S × I. Since constant changing in the shape of cylinder has many
invariant points, which leads to proof.

7. The existence of a solution to Urysohn Integral Equation (UIE)

In this section, we will get a unique solution for an UIE as an application of Theorem 13:

ℓ(℘) = c(℘) +

∫
IR

K1(℘, s,ϖ(s))ds. (12)

Above mentioned equation is a summarization of Volterra Integral Equation (VIE) and
Fredholm Integral Equation (FIE). This integral equation is dependent on the range of
integration (IR). UIE will converted into VIE by fixing a in IR = (a, x) and UIE will
become FIE by fixing a, b in IR = (a, b). Many authors have found a unique solution to
UIE (see [41–43]). Here we get a unique solution to UIE by using a fixed point way, which
further helps to find a convergence point to many mathematical structures.

Take IR as the set with finite measure along J 2
IR =

{
ϖ |

∫
IR |ϖ(s)|2ds <∞

}
. Take the

norm ∥.∥ : J 2
IR → [0,∞) such that

∥ϖ∥2 =

√∫
IR

|ϖ(s)|2ds, for all ϖ, r ∈ J 2
IR.

Equivalently we define the norm in the following manner:

∥ϖ∥2,y =

√
sup{e−y

∫
IR α(s)ds

∫
IR

|ϖ(s)|2ds}, for all ϖ ∈ J 2
IR, y > 1.

Then E = (J 2
IR, ∥.∥2,y) represents a Banach space. Consider a cone D = {ϖ ∈ J 2

IR :
ϖ(s) > 0 for almost every s}. The rectangular cone b-metric Ry is taken as Ry(ϖ, r) =
ϖ ∥ϖ − r∥22,y ∀ ϖ, r ∈ D. Take ⪯ as a partial order on D, so that

a ⪯ υ ⇔ a(s)υ(s) ≥ υ(s), ∀a, υ ∈ D.

Subsequently (E ,⪯,Ry) represents a complete RCBM space. Assume

(D1) The kernel K1 : IR× IR× R → R holding Carathéodory axioms along with

|K1(c, s,ϖ(s))| ≤ w(℘, s) + e(℘, s) |ϖ(s)| ; w, e ∈ J 2(IR× IR), e(℘, s) > 0.

D2) Take c : IR → [1,∞) as a continuous and bounded function over IR.



A. Arif et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7033 24 of 29

(D3) ∃ a scalar C > 0 so that

sup
℘∈IR

∫
IR

|K1(℘, s)| ds ≤ C.

(D4) For every ϖ0 ∈ J 2
IR, ∃ ϖ1 = R(ϖ0) so that ϖ1 ⪯ ϖ0 or ϖ0 ⪯ ϖ1.

(D4’) ϖn−1 ⪯ ϖn and ϖn → p holds for any sequence {ϖn}, implies ϖn ⪯ p, for each
natural number n.

(D5) We get a non-negative integrable and measurable operator q : IR× IR → R over IR,
so that

α(℘) :=

∫
IR
q2(℘, s) ds ≤ 1

y
, where y ≥ 1

and

|K1(℘, s,ϖ(s))−K1(℘, s, r(s))| ≤ q(℘, s)|ϖ(s)− r(s)|

for all ℘, s ∈ IR and ϖ, r ∈ E with ϖ ⪯ r.

Theorem 22. Assume c and K1 satisfying all axioms (D1)-(D5), then we get a unique
solution for UIE.

Proof.
Take a function R : E → E , along with the mentioned symbols, so that

(Rϖ) (℘) = c(℘) +

∫
IR
K1(℘, s,ϖ(s))ds.

R is taken as ⪯-preserving:
Consider ϖ, r ∈ E such that ϖ ⪯ r, then ϖ(s)r(s) ≥ r(s). Now, for almost every ℘ ∈ IR,

(Rϖ) (℘) = c(℘) +

∫
IR
K1(℘, s,ϖ(s))ds ≥ 1,

implies that (Rϖ) (℘) (Rr) (℘) ≥ (Rr) (℘). So, (Rϖ) ⊥ (Rr).
Self-operator:
Using (D1) and (D3) we get R : D → D as continuous and compact function (see [41,
Lemma 3]).

Using (D4), assures the existence of ϖ1 = R(ϖ0) so that ϖ1 ⪯ ϖ0 or ϖ0 ⪯ ϖ1,
for every ϖ0 ∈ D and R is ⪯-preserving, so we get ϖn = Rn(ϖ0) with ϖn ⪯ ϖn+1 or
ϖn+1 ⪯ ϖn ∀ n ≥ 0.

Using (D5) and Holder inequality will lead us to the contractive condition of Theorem
13.

ϖ |(Rϖ)(℘)− (Rr)(℘)|2 = ϖ

∣∣∣∣∣∣
∫
IR

K1(℘, s,ϖ(s)) ds−
∫
IR

K1(℘, s, r(s)) ds

∣∣∣∣∣∣
2



A. Arif et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7033 25 of 29

⪯ ϖ

∫
IR

|K1(℘, s,ϖ(s))−K1(℘, s, r(s))| ds

2

⪯ ϖ

∫
IR

q(℘, s)|ϖ(s)− r(s)| ds

2

⪯ ϖ

∫
IR

q2(℘, s) ds ·
∫
IR

|ϖ(s)− r(s)|2 ds

= ϖα(℘)

∫
IR

|ϖ(s)− r(s)|2 ds.

By integrating with respect to ℘, we get

ϖ

∫
IR

|(Rϖ)(℘)− (Rr)(℘)|2 d℘ ⪯ ϖ

∫
IR

α(℘)∫
IR

|ϖ(s)− r(s)|2 ds

 d℘

= ϖ

∫
IR

α(℘)ey ∫IR α(s)ds · e−y
∫
IR α(s)ds

∫
IR

|ϖ(s)− r(s)|2 ds

 d℘

⪯ ϖ ∥ϖ − r∥22,y
∫
IR

α(℘)ey
∫
IR α(s)ds d℘

⪯ ϖ
1

y
∥ϖ − r∥22,y e

y
∫
IR α(s)ds.

Hence, we get

ϖe−y
∫
IR α(s)ds

∫
IR

|(Rϖ)(℘)− (Rr)(℘)|2 d℘ ⪯ ϖ
1

y
∥ϖ − r∥22,y .

Which further implies that

ϖ ∥Rϖ −Rr∥22,y ⪯ ϖ
1

y
∥ϖ − r∥22,y .

So,

cy(Rϖ,Rr) ⪯
1

y
cy(ϖ, r) implies

ycy(Rϖ,Rr) ⪯ cy(ϖ, r)

Define J : E6 → E by

J (p1, p2, p3, p4, p5, p6) = kp1 − p2; k > 1,



A. Arif et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 7033 26 of 29

we have

J (c(Rϖ,Rr), c(ϖ, r), c(ϖ,Rϖ), c(r,Rr), c(ϖ,Rr), c(r,Rϖ)) ⪯ 0E .

By using 13, we get a unique fixed point for R, which implies UIE (12) has a unique
solution.

8. Conclusion

Fixed point results show that rectangular cone b-metric space can be applied to obtain
the general existence results for implicit contractions subject to implicit ordered relation.
These results generalize many theorems in [1, 4, 34]. The obtained results can be applied
to obtain more general homotopy results and existence results for integral equations. This
idea can be further applied to obtain fixed point results in cone A-metric space. We refer
a cmparison between this paper and [44] for further studies.

We suggest the readers and interested researchers to compare the results presented in
this paper with the results appearing in [44] for further study.

Acknowledgements

The authors extend their appreciation to Taif University, Saudi Arabia, for supporting
this work through project number (TU-DSPP-2024-46)

9. Data Availability

No data were used to support this study.

Competing interests

The authors declare that they have no competing interests.

10. Author’s Contributions

All authors contributed equally to this work.

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