Int. J. Anal. Appl. (2023), 21:7

Existence Fixed Point Solutions for C-Class Functions in Bipolar Metric Spaces With

Applications

G. Upender Reddy1, C. Ushabhavani1,2,∗, B. Srinuvasa Rao3

1Department of Mathematics, Mahatma Gandhi University, Nalgonda, Telangana, India
2Department of Humanities & Basic Science, Sreechaitanya College of Engineering, Thimmapur,

Karimnagar-505001,Telangana, India
3Department of Mathematics, Dr.B.R.Ambedkar University, Srikakulam, Etcherla-532410, Andhra

Pradesh, India

∗Corresponding author: n.ushabhavani@gmail.com

Abstract. In this study, the idea of C-class functions is introduced in the process of building a bi-polar

metric space, along with often coupled fixed point theorems for these mappings in complete bi-polar

metric spaces that associate altering distance function and ultra-altering distance function. Further-

more, we provide applications to integral equations as well as homotopy and we give an interpretation

that demonstrates the relevance of the results obtained.

1. Introduction

Fixed point theory is a crucial topic of non-linear analysis. Numerous types of equations that exist

in natural, biological, social, engineering, and other branches of science and technology are studied in

order to understand their underlying relevance. Examining the situations in which single or multi-valued

mappings have solutions is a common application of this technique.

Coupled fixed points was originally understood by Guo and Lakshmikantham [1] in 1987. Bhaskar

and Lakshmikantham [2] developed a novel fixed point theorem for mixed monotone mapping in a

metric space with partial ordering after using a weak contractivity condition. For further information

Received: Dec. 28, 2022.

2020 Mathematics Subject Classification. 54H25, 47H10, 54E50.
Key words and phrases. complete bipolar metric space; ω-compatible mappings; C-class functions; common coupled

fixed point.

https://doi.org/10.28924/2291-8639-21-2023-7
ISSN: 2291-8639

© 2023 the author(s).

https://doi.org/10.28924/2291-8639-21-2023-7


2 Int. J. Anal. Appl. (2023), 21:7

on coupled fixed point outcomes, see the study results ( [3], [4], [5], [6], [7], [8], [9]) and relevant

references.

In 2014, Ansari [10] proposed the idea of C-class functions and the proofs of unique fixed point

theorems for specific contractive mappings. This marked the beginning of a significant amount of

work in this area (See.( [11], [12], [13], [14], [15], [16], [17] ).

In addition to providing variant-related (coupled) fixed point solutions for co-variant and contravari-

ant contractive mappings, Muttu and Gurdal [18] recently developed the concept of bi-polar metric

spaces. Later, we proved some fixed point theorems in our earlier papers (see. [19], [20], [21], [22],

[23], [24]).

The purpose of this article is to propose a coupled common fixed point theorem for a covariant

mappings of C-class functions in relation to bi-polar metric spaces. Examples that are appropriate and

relevant applications to integral equations along with homotopy are also provided.

What follows is In our subsequent conversations, we compile a few suitable definitions.

2. Preliminaries

Definition 2.1. ( [18]) The mapping d : S×T → [0,∞) is said to be a Bipolar-metric on pair of non
empty sets (S,T ).If

(B1) d(µ,ν) = 0 implies that µ = ν;

(B2) µ = ν implies that d(µ,ν) = 0;

(B3) if (µ,ν) ∈ (S,T ), then d(µ,ν) = d(ν,µ);
(B4) d(µ1,ν2) ≤ d(µ1,ν1) + d(µ2,ν1) + d(µ2,ν2),

for all µ,µ1,µ2 ∈S and ν,ν1,ν2 ∈T , and the triple (S,T ,d) is called a Bipolar-metric space.

Example 2.1. ( [18]) Let d : S×T → [0, +∞) be defined as d(ψ,a) = ψ(a), for all (ψ,a) ∈ (S,T )
where S = {ψ/ψ : R→ [1, 3]} be the set of all functions and T = R. Then the triple (S,T ,d) is a
disjoint Bipolar-metric space.

Definition 2.2. ( [18]) Let Ω : S1 ∪T1 →S2 ∪T2 be a function defined on two pairs of sets (S1,T1)
and (S2,T2) is said to be

(i) covariant if Ω(S1) ⊆S2 and Ω(T1) ⊆T2. This is denoted as
Ω : (S1,T1) ⇒ (S2,T2);

(ii) contravariant if Ω(S1) ⊆T2 and Ω(T1) ⊆S2. It is denoted as
Ω : (S1,T1) � (S2,T2).

Particularly, if d1 is bipolar metrics on (S1,T1) and d2 is bipolar metrics on (S2,T2), we often write
Ω : (S1,T1,d1) ⇒ (S2,T2,d2) and
Ω : (S1,T1,d1) � (S2,T2,d2) respectively.



Int. J. Anal. Appl. (2023), 21:7 3

Definition 2.3. ( [18]) In a bipolar metric space (S,T ,d) for any ξ ∈S ∪T is left point if ξ ∈S, is
right point if ξ ∈T and is central point if ξ ∈S∩T .

Also, {αi} in S and {βi} in T are left and right sequence respectively. In a bipolar metric space,
we call a sequence, a left or a right one. A sequence {ξi} is said to be convergent to ξ iff either {ξi}
is a left sequence, ξ is a right point and lim

i→∞
d(ξi,ξ) = 0, or {ξi} is a right sequence, ξ is a left point

and lim
i→∞

d(ξ,ξi ) = 0. The bisequence ({αi},{βi}) on (S,T ,d) is a sequence on S×T . In the case
where {αi} and {βi} are both convergent, then ({αi},{βi}) is convergent.
The bi-sequence ({αi},{βi}) is a Cauchy bisequence if lim

i,j→∞
d(αi,βj) = 0.

Note that every convergent Cauchy bisequence is biconvergent. The bipolar metric space is com-

plete, if each Cauchy bisequence is convergent (and so it is biconvergent).

Definition 2.4. ( [22]) Let (S,T ,d) be a bipolar metric space and a pair (℘,$) is called

(a) coupled fixed point of covariant mapping Ω :
(
S2,T 2

)
⇒ (S,T )

if Ω (℘,$) = ℘, Ω ($,℘) = $ for (℘,$) ∈S2 ∪T 2 ;
(b) coupled coincident point of Ω :

(
S2,T 2

)
⇒ (S,T ) and Λ : (S,T ) ⇒ (S,T )

if F (℘,$) = Λ℘, Ω ($,℘) = Λ$;

(c) coupled common point of Ω :
(
S2,T 2

)
⇒ (S,T ) and Λ : (S,T ) ⇒ (S,T )

if Ω (℘,$) = Λ℘ = ℘, Ω ($,℘) = Λ$ = $;

(d) the pair (Ω, Λ) is weakly compatible if Λ(Ω(℘,$)) = Ω(Λ℘, Λ$) and

Λ(Ω($,℘)) = Ω(Λ$, Λ℘) whenever Ω (℘,$) = Λ℘, Ω ($,℘) = Λ$

Definition 2.5. ( [10]) Let C = {∆/∆ : [0, +∞)×[0.+∞) → R} be a family of continuous functions
is called a C-class function if for all s∗,t∗ ∈ [0,∞),

(a) ∆(s∗,t∗) ≤ s?;
(b) ∆(s∗,t∗) = s∗ ⇒ s∗ = 0 or t∗ = 0.

Example 2.2. ( [10]) Each of the functions ∆ : [0, +∞) × [0. + ∞) → R defined below are elements
of C.

(a) ∆(s∗,t∗) = s? − t?;
(b) ∆(s∗,t∗) = ms∗ where m ∈ (0, 1).
(c) ∆(s∗,t∗) = s

∗

(1+t?)r
where r ∈ (0,∞).

(d) ∆(s∗,t∗) = s?η(s?) where η : [0,∞) → [0,∞) is continuous function.
(e) ∆(s∗,t∗) = s? −ϕ(s?) for all s∗,t∗ ∈ [0, +∞) where, the continuous function

ϕ : [0,∞) → [0,∞) such that ϕ(s?) = 0 ⇔ s? = 0.
(f ) ∆(s∗,t∗) = sΩ(s?,t?) for all s∗,t∗ ∈ [0, +∞) where, the continuous function

Ω : [0,∞)2 → [0,∞) such that Ω(s?,t?) < 1.

Khan et al. [25] and A. H. Ansari et al. [11] both addressed a new category of contractive fixed

point outcomes. The idea of an altering distance function and ultra altering distance functions, which



4 Int. J. Anal. Appl. (2023), 21:7

are control functions that vary the distance between two locations in a metric space, were introduced

in their work.

We say F = {ψ?/ψ? : [0,∞) → [0,∞)} and G = {φ?/φ? : [0,∞) → [0,∞)} be the class of all
altering distance and ultra altering distance functions satisfying the following condition:

(ψ0) ψ? is nondecreasing and continuous;

(ψ1) ψ?(t) = 0 if and only if t = 0.

(ψ2) ψ?(t) is subadditivity, ψ?(a + b) ≤ ψ?(a) + ψ?(b);
(φ0) φ? is continuous;

(φ1) φ?(t) > 0, t > 0 and φ?(0) ≥ 0.

3. Main Results

In this section, two covariant mappings that meet new type contractive criteria in bipolar metric

spaces are given some common coupled fixed point theorems via C-class functions.

Theorem 3.1. Let (S,T ,d) be a complete bipolar metric space. Suppose that Γ :
(
S2,T 2

)
⇒ (S,T )

and Λ : (S,T ) ⇒ (S,T ) be two covariant mappings satiesfies

ψ? (d(Γ(u,v), Γ(p,q))) ≤ ∆ (ψ? (M(u,v,p,q)) ,φ? (M(u,v,p,q))) (3.1)

where, M(u,v,p,q) = ` max
{
d (Λu, Λp) ,d (Λv, Λq)

}
for all u,v ∈ S and p,q ∈ T and ∆ ∈ C,

ψ? ∈ F, φ? ∈ G with ` ∈ (0, 1)

(ξ0) Γ(S2 ∪T 2) ⊆ Λ(S∪T ) and Λ(S∪T ) is a complete subspace of S∪T ,
(ξ1) pair (Γ, Λ) is ω-compatible.

Then there is a unique common coupled fixed point of Γ and Λ in S∪T .

Proof. Let x0,y0 ∈ S and p0,q0 ∈ T be arbitrary, and from (ξ0), we construct the bisequences
({ακ} ,{ζκ}), ({βκ} ,{ηκ}) in (S,T ) as

Γ (xκ,yκ) = Λxκ+1 = ακ, Γ (pκ,qκ) = Λpκ+1 = ζκ

Γ (yκ,yκ) = Λyκ+1 = βκ, Γ (qκ,pκ) = Λqκ+1 = ηκ

where κ = 0, 1, 2, . . . .

Then from (3.1), we can get

ψ? (d(ακ,ζκ+1)) = ψ? (d(Γ (xκ,yκ) , Γ (pκ+1,qκ+1)))

≤ ∆ (ψ? (M(xκ,yκ,pκ+1,qκ+1)) ,φ? (M(xκ,yκ,pκ+1,qκ+1)))

(3.2)



Int. J. Anal. Appl. (2023), 21:7 5

where,

M(xκ,yκ,pκ+1,qκ+1) = ` max
{
d (Λxκ, Λpκ+1) ,d (Λyκ, Λqκ+1)

}
= ` max

{
d (ακ−1,ζκ) ,d (βκ−1,ηκ)

}
From (3.2), deduce that

ψ? (d(ακ,ζκ+1)) ≤ ∆

(
ψ?

(
` max

{
d (ακ−1,ζκ) ,

d (βκ−1,ηκ)

})
,φ?

(
` max

{
d (ακ−1,ζκ) ,

d (βκ−1,ηκ)

}))

≤ ψ?
(
` max

{
d (ακ−1,ζκ) ,d (βκ−1,ηκ)

})
By using (ψ0), we have

d(ακ,ζκ+1) ≤ ` max
{
d (ακ−1,ζκ) ,d (βκ−1,ηκ)

}
(3.3)

Similarly, we can prove

d (βκ,ηκ+1) ≤ ` max
{
d (ακ−1,ζκ) ,d (βκ−1,ηκ)

}
(3.4)

Combining (3.3) and (3.4), we have

max
{
d (ακ,ζκ+1) ,d (βκ,ηκ+1)

}
≤ ` max

{
d (ακ−1,ζκ) ,d (βκ−1,ηκ)

}
≤ `2 max

{
d (ακ−2,ζκ−1) ,d (βκ−2,ηκ−1)

}
...

≤ `κ max
{
d (α0,ζ1) ,d (β0,η1)

}
→ 0 as κ →∞.

(3.5)

On the other hand, we have

ψ? (d(ακ+1,ζκ)) = ψ? (d(Γ (xκ+1,yκ+1) , Γ (pκ,qκ)))

≤ ∆ (ψ? (M(xκ+1,yκ+1,pκ,qκ)) ,φ? (M(xκ+1,yκ+1,pκ,qκ)))

≤ ψ?
(
` max

{
d (ακ,ζκ−1) ,d (βκ,ηκ−1)

})
By using (ψ0), we have

d(ακ+1,ζκ) ≤ ` max
{
d (ακ,ζκ−1) ,d (βκ,ηκ−1)

}
(3.6)

Because of

M(xκ+1,yκ+1,pκ,qκ) = ` max
{
d (Λxκ+1, Λpκ) ,d (Λyκ+1, Λqκ)

}
= ` max

{
d (ακ,ζκ−1) ,d (βκ,ηκ−1)

}



6 Int. J. Anal. Appl. (2023), 21:7

Similarly, we can prove

d (βκ+1,ηκ) ≤ ` max
{
d (ακ,ζκ−1) ,d (βκ,ηκ−1)

}
(3.7)

Combining (3.6) and (3.7), we have

max
{
d (ακ+1,ζκ) ,d (βκ+1,ηκ)

}
≤ ` max

{
d (ακ,ζκ−1) ,d (βκ,ηκ−1)

}
≤ `2 max

{
d (ακ−1,ζκ−2) ,d (βκ−1,ηκ−2)

}
...

≤ `κ max
{
d (α1,ζ0) ,d (β1,η0)

}
→ 0 as κ →∞.

(3.8)

Moreover,

ψ? (d(ακ,ζκ)) = ψ? (d(Γ (xκ,yκ) , Γ (pκ,qκ)))

≤ ∆ (ψ? (M(xκ,yκ,pκ,qκ)) ,φ? (M(xκ,yκ,pκ,qκ)))

≤ ψ?
(
` max

{
d (ακ−1,ζκ−1) ,d (βκ−1,ηκ−1)

})
By using (ψ0), we have

d(ακ,ζκ) ≤ ` max
{
d (ακ−1,ζκ−1) ,d (βκ−1,ηκ−1)

}
(3.9)

Because of

M(xκ,yκ,pκ,qκ) = ` max
{
d (Λxκ, Λpκ) ,d (Λyκ, Λqκ)

}
= ` max

{
d (ακ−1,ζκ−1) ,d (βκ−1,ηκ−1)

}
Similarly, we can prove

d (βκ,ηκ) ≤ ` max
{
d (ακ−1,ζκ−1) ,d (βκ−1,ηκ−1)

}
(3.10)

Combining (3.9) and (3.10), we have

max
{
d (ακ,ζκ) ,d (βκ,ηκ)

}
≤ ` max

{
d (ακ−1,ζκ−1) ,d (βκ−1,ηκ−1)

}
≤ `2 max

{
d (ακ−2,ζκ−2) ,d (βκ−2,ηκ−2)

}
...

≤ `κ max
{
d (α0,ζ0) ,d (β0,η0)

}
→ 0 as κ →∞.

(3.11)



Int. J. Anal. Appl. (2023), 21:7 7

For each κ,δ ∈N with κ < δ. Then, from (3.5), (3.8), (3.11) and using property (B4), we have

d (ακ,ζδ) + d (βκ,ηδ) ≤ (d (ακ,ζκ+1) + d (βκ,ηκ+1))

+ (d (ακ+1,ζκ+1) + d (βκ+1,ηκ+1))

+ · · · + (d (αδ−1,ζδ−1) + d (βδ−1,ηδ−1))

+ (d (αδ−1,ζδ) + d (βδ−1,ηδ))

≤ 2
(
`κ + `κ+1 + · · · + `δ−1

)
max

{
d (α0,ζ1) ,d (β0,η1)

}
+2
(
`κ+1 + `κ+2 + · · · + `δ−1

)
max

{
d (α0,ζ0) ,d (β0,η0)

}
≤

2`κ

1 − `
max

{
d (α0,ζ1) ,d (β0,η1)

}
+

2`κ+1

1 − `
max

{
d (α0,ζ1) ,d (β0,η1)

}
→ 0 as κ →∞.

Similarly, we can prove that (d (αδ,ζκ) + d (βδ,ηκ)) → 0 as κ,δ →∞. Then the bisequence (ακ,ζδ)
and (βκ,ηδ) are Cauchy bisequences in (S,T ). Suppose Λ(S∪T ) is complete subspace of (S,T ,d),
then the sequences {ακ} ,{βκ} and {ζκ} ,{ηκ} ⊆ f (S ∪ T ) are convergence in complete bipolar
metric spaces (Λ(S), Λ(T ),d). Therefore, there exist a,b ∈ Λ(S) and l,m ∈ Λ(T ) such that

lim
κ→∞

ακ = l lim
κ→∞

βκ = m lim
κ→∞

ζκ = a lim
κ→∞

ηκ = b. (3.12)

Since Λ : S ∪T → S ∪T and a,b ∈ Λ(S) and l,m ∈ Λ(T ), there exist x,y ∈ S and p,q ∈ T such
that Λx = a, Λy = b and Λp = l, Λq = m. Hence

lim
κ→∞

ακ = l = Λp lim
κ→∞

βκ = m = Λq lim
κ→∞

ζκ = a = Λx lim
κ→∞

ηκ = b = Λy.

Claim that Γ(x,y) = l, Γ(y,x) = m and Γ(p,q) = a, Γ(q,p) = b.

By using (3.1), (B4), (ψ1) and (ψ2), we have

ψ? (d(Γ(x,y), l)) ≤ ψ? (d(Γ(x,y),ζκ+1)) + ψ? (d(ακ+1,ζκ+1)) + ψ? (d(ακ+1, l))

≤ ψ? (d(Γ(x,y), Γ(pκ+1,qκ+1))) + ψ? (d(ακ+1,ζκ+1)) + ψ? (d(ακ+1, l))

≤ ∆ (ψ? (M(x,y,pκ+1,qκ+1)) ,φ? (M(x,y,pκ+1,qκ+1)))

+ψ? (d(ακ+1,ζκ+1)) + ψ? (d(ακ+1, l))

≤ ψ?
(
` max

{
d (Λx,ζκ) ,d (Λy,ηκ)

})
+ψ? (d(ακ+1,ζκ+1)) + ψ? (d(ακ+1, l)) → 0 as κ →∞.

It follows that ψ? (d(Γ(x,y), l)) = 0 implies that d(Γ(x,y), l) = 0, which deduce that Γ(x,y) = l.

Similarly, we can prove that Γ(y,x) = m and Γ(p,q) = a, Γ(q,p) = b. Therefore, it follows that

Γ(x,y) = l = Λp, Γ(y,x) = m = Λq and Γ(p,q) = a = Λx, Γ(q,p) = b = Λy.



8 Int. J. Anal. Appl. (2023), 21:7

Since {Γ, Λ} is ω-compatible pair, we have Γ(l,m) = Λl, Γ(m,l) = Λm and Γ(a,b) = Λa,
Γ(b,a) = Λb. Now we prove that Λl = l, Λm = m and Λa = a, Λb = b. Now we have

ψ? (d(Λa,ζκ)) ≤ ψ? (d(Γ(a,b), Γ(pκ,qκ)))

≤ ∆ (ψ? (M(a,b,pκ,qκ)) ,φ? (M(a,b,pκ,qκ)))

≤ ψ?
(
` max

{
d (Λa,ζκ−1) ,d (Λb,ηκ−1)

})
By using (ψ0), we have

d(Λa,ζκ) ≤ ` max
{
d (Λa,ζκ−1) ,d (Λb,ηκ−1)

}
Letting κ →∞, we have

d(Λa,a) ≤ ` max
{
d (Λa,a) ,d (Λb,b)

}
and

ψ? (d(Λb,ηκ)) ≤ ψ? (d(Γ(b,a), Γ(qκ,pκ)))

≤ ∆ (ψ? (M(b,a,qκ,pκ)) ,φ? (M(b,a,qκ,pκ)))

≤ ψ?
(
` max

{
d (Λb,ηκ−1) ,d (Λa,ζκ−1)

})
By using (ψ0), we have

d(Λb,ηκ) ≤ ` max
{
d (Λb,ηκ−1,d (Λa,ζκ−1))

}
Letting κ →∞, we have

d(Λb,b) ≤ ` max
{
d (Λb,b) ,d (Λa,a)

}
Therefore,

max
{
d (Λa,a) ,d (Λb,b)

}
≤ ` max

{
d (Λa,a) ,d (Λb,b)

}
which implies that d (Λa,a) = 0 and d (Λb,b) = 0 and hence Λa = a and Λb = b. Therefore,

Γ(a,b) = Λa = a, Γ(b,a) = Λb = b.

Similarly, we can prove Γ(l,m) = Λl = l, Γ(m,l) = Λm = m. Therefore,

Γ(p,q) = Λx = a = Λa = Γ(a,b) Γ(x,y) = Λp = l = Λl = Γ(l,m)

Γ(q,p) = Λy = b = Λb = Γ(b,a) Γ(y,x) = Λq = m = Λm = Γ(m,l)

On the other hand, from (3.12), we get

d (Λp, Λx) = d(l,a) = d
(

lim
κ→∞

ακ, lim
κ→∞

ζκ

)
= lim
κ→∞

d(ακ,ζκ) = 0

and

d (Λq, Λy) = d(m,b) = d
(

lim
κ→∞

βκ, lim
κ→∞

ηκ

)
= lim
κ→∞

d(βκ,ηκ) = 0.



Int. J. Anal. Appl. (2023), 21:7 9

Thus a = l,b = m. Therefore, (a,b) ∈ S2 ∩T 2 is a common coupled fixed point of Γ and Λ. In
the following we will show the uniqueness. Assume that there is another coupled fixed point (a′,b′)

of Γ, Λ. Then from (3.1), we have

ψ?
(
d(a,a′)

)
= ψ?

(
d(Γ(a,b), Γ(a′,b′)

)
≤ ∆

(
ψ?
(
M(a,b,a′,b′)

)
,φ?

(
M(a,b,a′,b′)

))
≤ ψ?

(
` max

{
d (Λa, Λa′) ,d (Λb, Λb′)

})
≤ ψ?

(
` max

{
d (a,a′) ,d (b,b′)

})
by the property of (ψ0), we have

d(a,a′) ≤ ` max
{
d (a,a′) ,d (b,b′)

}
Therefore, we have

max
{
d (a,a′) ,d (b,b′)

}
≤ ` max

{
d (a,a′) ,d (b,b′)

}
hence, we get a = a′,b = b′. Therefore, (a,b) is a unique common coupled fixed point of Γ and Λ.

Finally we will show a = b.

ψ? (d(a,b)) = ψ? (d(Γ(a,b), Γ(b,a))

≤ ∆ (ψ? (M(a,b,b,a)) ,φ? (M(a,b,b,a)))

≤ ψ?
(
` max

{
d (Λa, Λb) ,d (Λb, Λa)

})
≤ ψ?

(
` max

{
d (a,b) ,d (b,a)

})
by the property of (ψ0), we have

d(a,b) ≤ ` max
{
d (a,b) ,d (b,a)

}
Therefore, we have

max
{
d (a,b) ,d (b,a)

}
≤ ` max

{
d (a,b) ,d (b,a)

}
hence, we get a = b. Which means that Γ and Λ have a unique common fixed point of the form

(a,a). �

Corollary 3.1. Let (S,T ,d) be a complete bipolar metric space. Suppose that Γ :
(
S2,T 2

)
⇒ (S,T )

be a covariant mapping satisfy

ψ? (d(Γ(u,v), Γ(p,q))) ≤ ∆

(
ψ?

(
` max

{
d (u,p) ,

d (v,q)

})
,φ?

(
` max

{
d (u,p) ,

d (v,q)

}))
for all u,v ∈ S and p,q ∈ T and ∆ ∈ C, ψ? ∈ F, φ? ∈ G with ` ∈ (0, 1) Then there is a unique
coupled fixed point of Γ in S∪T .



10 Int. J. Anal. Appl. (2023), 21:7

Corollary 3.2. Let (S,T ,d) be a complete bipolar metric space. Suppose that
Γ : (S×T ,T ×S) ⇒ (S,T ) be a covariant mapping satisfy

ψ? (d(Γ(u,p), Γ(q,v))) ≤ ∆

(
ψ?

(
` max

{
d (u,q) ,

d (v,p)

})
,φ?

(
` max

{
d (u,q) ,

d (v,p)

}))

for all u,v ∈ S and p,q ∈ T and ∆ ∈ C, ψ? ∈ F, φ? ∈ G with ` ∈ (0, 1) Then there is a unique
coupled fixed point of Γ in S∪T .

Example 3.1. Let S = Un(R) and T = Ln(R) be the set of all n × n upper and lower triangular

matrices over R. Define d : S×T → [0,∞) as d(X,Y ) =
κ∑

i,j=1

|αij −βij|

for all X = (αij)n×n ∈ Un(R) and Y = (βij)n×n ∈ Ln(R). Then obviously (S,T ,d) is a Bipolar-
metric space. And define Γ : S2 ∪T 2 →S∪T as
Γ(A,B) = (

aij−bij
10

)n×n where (A = (aij)n×n,B = (bij)n×n) ∈ Un(R)2 ∪Ln(R)2 and define
Λ : S ∪T → S ∪T as `(A) = (aij

2
)n×n and let ∆ : [0, +∞) × [0. + ∞) → R as ∆(s∗,t∗) = s∗ − t∗,

also define ψ? : [0,∞) → [0,∞), φ? : [0,∞) → [0,∞) as ψ?(t∗) = t∗ and φ?(t∗) = t
∗

2
respectively.

Then obviously, Γ(S2 ∪T 2) ⊆ Λ(S∪T ) and the pairs (Γ, Λ) is ω-compatible.
In fact, we have

ψ? (d(Γ(A,B), Γ(X,Y ))) = d(Γ(A,B), Γ(X,Y ))

=

κ∑
i,j=1

|
aij −bij

10
−
xij −yij

10
|

≤
1

4


 κ∑
i,j=1

|
aij
2
−
xij
2
| +

κ∑
i,j=1

|
bij
2
−
yij
2
|




≤
1

4
(d(ΛA, ΛX) + d(ΛB, ΛY ))

≤
1

2

(
1

2
max{d(ΛA, ΛX),d(ΛB, ΛY )}

)

≤ ∆

(
ψ?

(
` max

{
d(ΛA, ΛX),

d(ΛB, ΛY )

})
,φ?

(
` max

{
d(ΛA, ΛX),

d(ΛB, ΛY )

}))

Thus all the conditions of the theorem (3.1) are satisfied and (On×n,On×n) is unique coupled fixed

point.

3.1. Application to the existence of solutions of integral equations.

Let S = C (L∞(E1)) ,T = C (L∞(E2)) be the set of essential bounded measurable continuous
functions on E1 and E2 where E1,E2 are two Lebesgue measurable sets with m(E1 ∪ E2) < ∞.
Define d : S×T → R+ as d(`,σ) = ||`−σ|| for all ` ∈S,σ ∈T . Therefore, (S,T ,d) is a complete
bipolar metric space.

In this section, we apply our theorem (3.1) to establish the existence and uniqueness solution of



Int. J. Anal. Appl. (2023), 21:7 11

nonlinear integral equation defined by:

x(t) = f (t) + κ

∫
E1∪E2

Ω(t,`, (x,y))d`. (3.13)

where x,y ∈ C (L∞(E1) ∪L∞(E2)), κ ∈ R and t,` ∈ E1 ∪E2,
Ω : E21 ∪E

2
2 ×L

∞(E1)
2 ∪L∞(E2)2 → R and f : E1 ∪E2 → R are given continuous functions

Theorem 3.2. Assume that the following conditions are fulfilled

(i) Define, ∆ : [0, +∞) × [0. + ∞) → R as ∆(s∗,t∗) = θs∗ where θ ∈ (0, 1),
let ψ? : [0,∞) → [0,∞) as ψ?(t∗) = t∗. Let Λ : S ∪ T → S ∪ T as Λ(x) = x and
Γ : S2 ∪T 2 →S∪T by Γ(x,y)(t) = f (t) + κ

∫
E1∪E2

Ω(t,`, (x,y))d`

(ii) There exists a continuous function χ : E21 ∪E
2
2 → R

+ such that for all x,y ∈ S,p,q ∈ T ,
κ ∈ R and t,` ∈ E1 ∪E2, we get that
||Ω(t,`, (x,y)) − Ω(t,`, (p,q))|| ≤ χ(t,`)M(x,y,p,q) where,
M(x,y,p,q) = λ max{d(Λx, Λp),d(Λy, Λq)} where λ ∈ (0, 1)

(iii) ||κ||
∫

E1∪E2
χ(t,`)d` ≤ θ

(iv) Γ
(
S2 ∪T 2

)
⊆ Λ(S∪T ), Λ(S∪T ) is closed and the pair (Γ, Λ) is weakly compatible.

Then there exists unique solution in C (L∞(E1) ∪L∞(E2)) for the initial value problem 3.13.

Proof. The existence of a solution of (3.13) is equivalent to the existence of a common fixed point

of Γ and Λ. Obviously, Γ
(
S2 ∪T 2

)
⊆ Λ(S ∪T ), Λ(S ∪T ) is closed and the pair (Γ, Λ) is weakly

compatible. Using the inequalities, (i), (ii) and (iii), we have

ψ? (d(Γ(x,y), Γ(p,q))) = d(Γ(x,y), Γ(p,q))

= ||κ
∫

E1∪E2

(Ω(t,`, (x,y)))d`−κ
∫

E1∪E2

(Ω(t,`, (p,q)))d`||

≤ ||κ||
∫

E1∪E2

||Ω(t,`, (x,y)) − Ω(t,`, (p,q))||d`

≤ ||κ||
∫

E1∪E2

χ(t,`)M(x,y,p,q)d`

≤ ||κ||


 ∫
E1∪E2

χ(t,`)d`


M(x,y,p,q)

≤ θM(x,y,p,q)

≤ ∆

(
ψ?

(
λ max

{
d(Λx, Λp),

d(Λy, Λq)

})
,φ?

(
λ max

{
d(Λx, Λp),

d(Λy, Λq)

}))
Hence, all the conditions of Theorem (3.1) hold, we conclude that Γ and Λ have a unique solution in

S∪T to the integral equation (3.13). �



12 Int. J. Anal. Appl. (2023), 21:7

3.2. Application to the existence of solutions of Homotopy.

In this part, we examine the possibility that homotopy theory has a unique solution.

Theorem 3.3. Let (S,T ,d) be complete bipolar metric space, (P,Q) and (P,Q) be an open and
closed subset of (S,T ) such that (P,Q) ⊆ (P,Q). Suppose
H :

(
P ×Q

)
∪
(
Q×P

)
× [0, 1] →S∪T be an operator with following conditions are satisfying,

`0) ℘ 6= H(℘,$,s), $ 6= H($,℘,s), for each ℘ ∈ ∂P,$ ∈ ∂Q and s ∈ [0, 1] (Here ∂P ∪ ∂Q is
boundary of P ∪Q in S∪T );
`1) for all ℘,$ ∈P, ı,  ∈Q, s ∈ [0, 1] and ψ? ∈ F,φ? ∈ G ∆ ∈ C and ` ∈ (0, 1) such that

ψ? (d (H(℘,ı,s),H(,$,s))) ≤ ∆

(
ψ?

(
` max

{
d (℘,) ,

d ($,ı)

})
,φ?

(
` max

{
d (℘,) ,

d ($,ı)

}))

`2) ∃ M ≥ 0 3 d(H(℘,ı,s),H(,$,t)) � M|s − t|
for every ℘,$ ∈P, ı,  ∈Q and s,t ∈ [0, 1].
Then H(., 0) has a coupled fixed point ⇐⇒ H(., 1) has a coupled fixed point.

Proof. Let the set

Θ =
{
s ∈ [0, 1] : H(℘,ı,s) = ℘,H(ı,℘,s) = ı for some ℘ ∈P, ı ∈Q

}
.

Υ =
{
t ∈ [0, 1] : H(,$,t) = ,H($,,t) = $ for some $ ∈P,  ∈Q

}
.

Suppose that H(., 0) has a coupled fixed point in (P ×Q) ∪ (Q×P), we have that
(0, 0) ∈ (Θ × Υ) ∩ (Υ × Θ). Now we show that (Θ × Υ) ∩ (Υ × Θ) is both closed and open in
[0, 1] and hence by the connectedness Θ = Υ = [0, 1]. As a result, H(., 1) has a coupled fixed point
in (Θ × Υ) ∩ (Υ × Θ). First we show that (Θ × Υ) ∩ (Υ × Θ) closed in [0, 1]. To see this, Let
(
{
a
p

}∞
p=1

,
{
x
p

}∞
p=1

) ⊆ (Θ, Υ) and (
{
y
p

}∞
p=1

,
{
b
p

}∞
p=1

) ⊆ (Υ, Θ) with (ap,xp) → (α,β), (yp,bp) →
(β,α) ∈ [0, 1] as p →∞. We must show that (α,β) ∈ (Θ × Υ) ∩ (Υ × Θ).
Since (ap,xp) ∈ (Θ, Υ), (yp,bp) ∈ (Υ, Θ) for p = 0, 1, 2, 3, · · · , there exists sequences ({℘p} ,{$p})
and ({ıp} ,{p}) with ℘p+1 = H(℘p,$p,ap), $p+1 = H($p,℘p,xp) and ıp+1 = H(ıp, p,yp),
p+1 = H(p, ıp,bp)
Consider

ψ? (d(℘p, p+1)) = ψ? (d (H(℘p−1,$p−1,ap−1),H(p, ıp,bp)))

≤ ∆

(
ψ?

(
` max

{
d (℘p−1, p) ,

d (ıp,$p−1)

})
,φ?

(
` max

{
d (℘p−1, p) ,

d (ıp,$p−1)

}))

≤ ψ?

(
` max

{
d (℘p−1, p) ,

d (ıp,$p−1)

})

By using (ψ0), we have



Int. J. Anal. Appl. (2023), 21:7 13

d(℘p, p+1) ≤ ` max

{
d (℘p−1, p) ,

d (ıp,$p−1)

}

Similar lines we can prove that

d(ıp+1,$p) ≤ ` max

{
d (℘p−1, p) ,

d (ıp,$p−1)

}

Therefore, we get

max

{
d (℘p, p+1) ,

d (ıp+1,$p)

}
≤ ` max

{
d (℘p−1, p) ,

d (ıp,$p−1)

}

≤ `2 max

{
d (℘p−2, p−1) ,

d (ıp−1,$p−2)

}
...

≤ `p max

{
d (℘0, 1) ,

d (ı1,$0)

}
(3.14)

Similarly, we can prove

max

{
d (℘p+1, p) ,

d (ıp,$p+1)

}
≤ `p max

{
d (℘1, 0) ,

d (ı0,$1)

}
(3.15)

and

max

{
d (℘p, p) ,

d (ıp,$p)

}
≤ `p max

{
d (℘0, 0) ,

d (ı0,$0)

}
(3.16)

For each p,q ∈N with p < q. Then, from (3.14), (3.15), (3.16) and using property (B4), we have

d (℘p, q) + d (ıp,$q)

≤ (d (℘p, p+1) + d (ıp,$p+1)) + (d (℘p+1, p+1) + d (ıp+1,$p+1))

+ · · · + (d (℘q−1, q−1) + d (ıq−1,$q−1)) + (d (℘q−1, q) + d (ıq−1,$q))

≤ (M|ap−1 −bp| + M|xp −yp−1|) + · · · + (M|aq−2 −bq−1| + M|xq−1 −yq−2|)

+2
(
`p+1 + `p+2 + · · · + `q−1

)
max

{
d (℘0, 0) ,

d (ı0,$0)

}
≤ (M|ap−1 −bp| + M|xp −yp−1|) + · · · + (M|aq−2 −bq−1| + M|xq−1 −yq−2|)

+
2`p+1

1 − `
max

{
d (℘0, 0) ,

d (ı0,$0)

}
→ 0 as p,q →∞.



14 Int. J. Anal. Appl. (2023), 21:7

It follows that lim
p,q→∞

(d (℘p, q) + d (ıp,$q)) = 0. Similarly, we can prove that

lim
p,q→∞

(d (℘q, p) + d (ıq,$p)) = 0. Therefore, ({℘p} ,{$p}) and ({ıp} ,{p}) are Cauchy bi-
sequences in (P,Q). By completeness, there exist (a,x) ∈P ×Q and (y,b) ∈Q×P with

lim
p→∞

℘p+1 = x lim
p→∞

ıp+1 = y lim
p→∞

$p+1 = a lim
p→∞

p+1 = b (3.17)

we have

d (H(b,y,α),x) ≤ d (H(b,y,α), p+1) + d(℘p+1, p+1) + d(℘p+1,x)

≤ d (H(b,y,α),H(p, ıp,bp)) + M|ap −bp| + d(℘p+1,x)

Letting p →∞ in the above inequality and ψ? is continuous and non-decreasing, we have

ψ? (d (H(b,y,α),x)) ≤ ψ? (d (H(b,y,α),H(p, ıp,bp)))

≤ ∆

(
ψ?

(
` max

{
d (b, p) ,

d (ıp,y)

})
,φ?

(
` max

{
d (b, p) ,

d (ıp,y)

}))

≤ ψ?

(
` max

{
d (b, p) ,

d (ıp,y)

})

By using (ψ0) and letting as p → ∞, we get that d (H(b,y,α),x) = 0 implies that H(b,y,α) = x.
Similarly, we can prove that H(y,b,β) = a and H(x,a,α) = y, H(a,x,β) = b. On the other hand,
from (3.17), we get

d (a,y) = d

(
lim
p→∞

$p, lim
p→∞

ıp

)
= lim
p→∞

d(ıp,$p) = 0

and

d (b,x) = d

(
lim
p→∞

p, lim
p→∞

℘p

)
= lim
p→∞

d(℘p, p) = 0.

Therefore, a = y and b = x and hence (α,β) ∈ (Θ × Υ) ∩ (Υ × Θ).
Clearly (Θ×Υ)∩(Υ×Θ) is closed in [0, 1]. Let (α0,β0) ∈ Θ×Υ, there exists bisequences (℘0,$0)
and (ı0, 0) with ℘0 = H(℘0,$0,α0), $0 = H($0,℘0,β0) and ı0 = H(ı0, 0,β0), 0 = H(0, ı0,α0).
Since (P×Q) ∪ (Q×P) is open, then there exist δ > 0 such that Bd(℘0,δ) ⊆ (P×Q) ∪ (Q×P),
Bd($0,δ) ⊆ (P×Q)∪(Q×P), Bd(ı0,δ) ⊆ (P×Q)∪(Q×P) and Bd(0,δ) ⊆ (P×Q)∪(Q×P).
Choose α ∈ (α0 − �,α0 + �), β ∈ (β0 − �,β0 + �) such that |α−α0| ≤ 1Mp <

�
2
, |β −β0| ≤ 1Mp <

�
2

and |α0 −β0| ≤ 1Mp <
�
2
.

Then for,  ∈ BP∪Q(℘0,δ) = {, 0 ∈Q/d(℘0, ) ≤ d(℘0, 0) + δ},
ı ∈ BP∪Q(δ,$0) = {ı, ı0 ∈P/d(ı,$0) ≤ d(ı0,$0) + δ}
℘ ∈ BP∪Q(δ, 0) = {℘,℘0 ∈P/d(℘,0) ≤ d(℘0, 0) + δ}
$ ∈ BP∪Q(ı0,δ) = {$,$0 ∈Q/d(ı0,$) ≤ d(ı0,$0) + δ}



Int. J. Anal. Appl. (2023), 21:7 15

d (H(℘,$,α), 0)) = d (H(℘,$,α),H(0, ı0,α0))

≤ d (H(℘,$,α),H(, ı,α0)) + d (H(℘0,$0,α),H(, ı,α0))

+d (H(℘0,$0,α),H(0, ı0,α0))

≤ 2M|α−α0| + d (H(℘0,$0,α),H(, ı,α0))

≤
2

Mp−1
+ d (H(℘0,$0,α),H(, ı,α0))

Letting p →∞ and using (ψ0), then we have

ψ? (d (H(℘,$,α), 0))) ≤ ψ? (d (H(℘0,$0,α),H(, ı,α0)))

≤ ∆

(
ψ?

(
` max

{
d (℘0, ) ,

d (ı,$0)

})
,φ?

(
` max

{
d (℘0, ) ,

d (ı,$0)

}))

≤ ψ?

(
` max

{
d (℘0, ) ,

d (ı,$0)

})
Using the property of ψ?, we get

d (H(℘,$,α), 0)) ≤ ` max

{
d (℘0, ) ,

d (ı,$0)

}
Similarly we can prove

d (ı0,H($,℘,β))) ≤ ` max

{
d (℘0, ) ,

d (ı,$0)

}
Therefore,

max

{
d (H(℘,$,α), 0)) ,
d (ı0,H($,℘,β)))

}
≤ ` max

{
d (℘0, ) ,

d (ı,$0)

}

≤ ` max

{
d (℘0, 0) + δ,

d (ı0,$0) + δ

}

Thus, d (H(℘,$,α), 0)) ≤ d (℘0, 0) + δ and d (ı0,H($,℘,β))) ≤ d (ı0,$0) + δ.
Similarly, we can prove

d (H(ı, ,β),$0)) ≤ d (ı0,$0) + δ and d (℘0,H(, ı,α))) ≤ d (℘0, 0) + δ.
On the other hand,

d(℘0,$0) = d (H(℘0,$0,α0),H($0,℘0,β0)) ≤ M|α0 −β0| <
1

Mp−1
→ 0 as p →∞.

and

d(ı0, 0) = d (H(ı0, 0,β0),H(0, ı0,α0)) ≤ M|α0 −β0| <
1

Mp−1
→ 0 as p →∞.



16 Int. J. Anal. Appl. (2023), 21:7

So ℘0 = $0 and ı0 = 0 and hence α = β. Thus for each fixed α ∈ (α0 − �,α0 + �), H(.,α) :
BΘ∪Υ(℘0,δ) → BΘ∪Υ(℘0,δ) and H(.,α) : BΘ∪Υ(ı0,δ) → BΘ∪Υ(ı0,δ). Thus, we conclude that
H(.,α) has a coupled fixed point in (P×Q)∩(Q×P). But this must be in (P×Q)∪(Q×P). Therefore,
(α,α) ∈ (Θ × Υ) ∩ (Υ × Θ) for α ∈ (α0 −�,α0 + �).Hence, (α0 −�,α0 + �) ⊆ (Θ × Υ) ∩ (Υ × Θ).
Clearly, (Θ×Υ)∩(Υ×Θ) is open in [0, 1]. For the reverse implication, we use the same strategy. �

Theorem 3.4. Let (S,T ,d) be complete bipolar metric space, (P,Q) and (P,Q) be an open and
closed subset of (S,T ) such that (P,Q) ⊆ (P,Q). Suppose
H :

(
P2 ∪Q2

)
× [0, 1] →S∪T be an operator with following conditions are satisfying,

`0) ℘ 6= H(℘,$,s), $ 6= H($,℘,s), for each ℘,$ ∈ ∂P ∪ ∂Q and s ∈ [0, 1] (Here ∂P ∪ ∂Q is
boundary of P ∪Q in S∪T );
`1) for all ℘,$ ∈P, ı,  ∈Q, s ∈ [0, 1] and ψ? ∈ F,φ? ∈ G ∆ ∈ C and ` ∈ (0, 1) such that

ψ? (d (H(℘,$,s),H(ı, ,s))) ≤ ∆

(
ψ?

(
` max

{
d (℘,ı) ,

d ($,)

})
,φ?

(
` max

{
d (℘,ı) ,

d ($,)

}))
`2) ∃ M ≥ 0 3 d(H(℘,$,s),H(ı, , t)) � M|s − t|
for every ℘,$ ∈P, ı,  ∈Q and s,t ∈ [0, 1].
Then H(., 0) has a coupled fixed point ⇐⇒ H(., 1) has a coupled fixed point.

CONCLUSION

We ensured the existence and uniqueness of a common coupled fixed point for two covariant mappings

in the class of complete bipolar metric spaces with examples via C-class functions. Two illustrated

application has been provided.

Conflicts of Interest: The authors declare that there are no conflicts of interest regarding the publi-

cation of this paper.

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	1. Introduction
	2. Preliminaries
	3. Main Results
	3.1.  Application to the existence of solutions of integral equations 
	3.2.  Application to the existence of solutions of Homotopy 

	References