International Journal of Analysis and Applications
Volume 18, Number 6 (2020), 1083-1107
URL: https://doi.org/10.28924/2291-8639
DOI: 10.28924/2291-8639-18-2020-1083
COUPLED COINCIDENCE POINT FOR f(ψ,ϕ)−CONTRACTIONS VIA
GENERALIZED α−ADMISSIBLE MAPPINGS WITH AN APPLICATION
DHEKRA M. ALBAQERI1, HASANEN A. HAMMAD2,∗, MANUEL DE LA SEN3
1Department of Mathematics, Faculty of Education, Sanaa University, Sanaa 1247, Yemen
2Department of Mathematics, Faculty of Science, Sohag University, Sohag 82524, Egypt
3Institute of Research and Development of Processes University of the Basque Country 48940- Leioa
(Bizkaia), Spain
∗Corresponding author: hassanein hamad@science.sohag.edu.eg
Abstract. The main objective of this manuscript is to discuss some coupled coincidence point (ccp) results
for generalized α− admissible mappings which are f(ψ,ϕ)− contractions in the context of b−metric spaces
(b-ms). Also, an example to support the obtained theoretical theorems is derived. Ultimately, an analytical
solution for nonlinear integral equation (nie) is discussed as an application.
1. Introduction and elementary discussions
Fixed point techniques plays an enormous role in many applications of mathematics. During the past
thirty years various extension of a metric space have been discussed. The Banach contraction principle is a
popular tool helps to solve problems in nonlinear analysis. A number of publications are interested to the
study and solutions of many practical and theoretical problems by using this principle [1–8].
Bakhtin [9] in 1993 and Czerwik [10] in 1998 introduced the concept of (b-ms). Since then, several papers
have been published on the fixed point theory of both classes of single-valued and multi-valued operators in
(b-ms). [11], [12], [13–16].
Received September 13th, 2020; accepted October 15th, 2020; published November 4th, 2020.
2010 Mathematics Subject Classification. 46N40, 47H10, 46T99.
Key words and phrases. coupled coincidence point; generalized α− admissible mapping; b−metric spaces; nonlinear integral
equations.
©2020 Authors retain the copyrights
of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.
1083
https://doi.org/10.28924/2291-8639
https://doi.org/10.28924/2291-8639-18-2020-1083
Int. J. Anal. Appl. 18 (6) (2020) 1084
Definition 1.1. [10] Let Γ be a nonempty set and s ≥ 1 be a given real number. A function νb : Γ × Γ →
[0,∞) is a b-metric (b-m) iff, for all e,r,ζ ∈ Υ , the stipulations below are fulfilled:
(b1) νb(e,r) = 0 ⇔ e = r;
(b2) νb(e,r) = νb(r,e);
(b3) νb(e,r) ≤ s[νb(e,ζ) + νb(ζ,r)].
The pair (Γ,νb) is called a (b-ms) with a constant s ≥ 1.
Example 1.1. Let Γ = [0,∞). Define the function νb : Υ2 → [0,∞) by νb(e,r) = (e−r)2.
Then (Γ,νb) is a (b-ms) with a constant s = 2.
Definition 1.2. [10] Suppose that (Γ,νb) is a (b-ms). So a sequence {en} in Γ is called:
(i) convergent if there is e ∈ Γ so that νb(en,e) → 0 as n →∞.
(ii) Cauchy sequence iff limm,n→∞νb(em,en) → 0 as n,m →∞.
(iii) the pair (Γ,νb) is called a complete iff every Cauchy sequence {en} in Γ converges to e ∈ Γ.
Lemma 1.1. [11] Let (Γ,νb) be a (b-ms) with a coefficient s ≥ 1, {en} and {rn} be a convergent to points
e,r ∈ Γ, respectively. Then we have
1
s2
νb(e,r) ≤ lim inf
n→∞
νb(en,rn) ≤ lim sup
n→∞
νb(en,rn) ≤ s2νb(e,r).
In particular, if e = r, then limn→∞νb(en,rn) = 0. Moreover, for each δ ∈ Γ, we have
1
s
νb(e,δ) ≤ lim inf
n→∞
νb(en,δ) ≤ lim sup
n→∞
νb(en,δ) ≤ sνb(e,δ).
Lemma 1.2. [12] Let {en} be a sequence in a (b-ms) (Γ,νb) so that
νb(en,en+1) ≤ λνb(en−1,en),
for some λ, 0 < λ < 1
s
, and for each n ∈ N. Then {en} is a Cauchy sequence in Γ.
The idea of coupled fixed point initiated and studied by Guo and Lakshmikantham [17]. After that, the
monotone property is studied by Bhaskar and Lakshmikantham [18]. Many works are made to generalized
this concept in various spaces under certain conditions, the reader can shed light on [19–23, 25, 26].
Definition 1.3. [18] An element (e,r) ∈ Υ × Υ is called a (ccp) of the mappings Υ : Γ × Γ → Γ and
Λ : Γ → Γ if Υ(e,r) = Λe and Υ(r,e) = Λr.
Definition 1.4. [27] An element (e,r) ∈ Γ × Γ is called a (ccp) of mappings Υ, Λ : Γ × Γ → Γ if
Υ(e,r) = Λ(e,r) and Υ(r,e) = Λ(r,e).
Example 1.2. Let Υ, Λ : R×R → R be defined by Υ(e,r) = Λr and Λ(e,r) = 2
3
(e + r) for all (e,r) ∈ Γ×Γ
. Note that (0, 0), (1, 2) and (2, 1) are (ccp) of Υ and Λ .
Int. J. Anal. Appl. 18 (6) (2020) 1085
Definition 1.5. [27] Let Υ, Λ : Γ × Γ → Γ. We say that the pair (Υ, Λ) is generalized compatible if
νb(Υ(Λ(en,rn), Λ(rn,en)), Λ(Υ(en,rn), Υ(rn,en))) → 0 as n →∞,
νb(Υ(Λ(rn,en), Λ(en,rn)), Λ(Υ(rn,en), Υ(en,rn))) → 0 as n →∞,
whenever {en} and {rn} are sequences in Γ such that for all t1, t2 ∈ Γ, we have
lim
n→∞
Υ(en,rn) = lim
n→∞
Λ(en,rn) = t1,
lim
n→∞
Υ(rn,en) = lim
n→∞
Λ(rn,en) = t2.
Definition 1.6. [28] Let Υ : Γ → Γ and α : Γ×Γ → [0, +∞). We say that Υ is an α− admissible mapping
if α(e,r) ≥ 1 implies α(Υe, Υr) ≥ 1 for all e,r ∈ Υ.
Ansari [29] initiated the remarkable of C−class functions. This contribution covers a large class of con-
tractive conditions. Here, we denote C-class functions as C.
Definition 1.7. [29] A C-class function is a continuous mapping f : [0,∞)2 → R which fulfills the
stipulations below:
(1) f(e,r) ≤ e,
(2) f(e,r) = e implies that either e = 0 or r = 0 for all e,r ∈ [0,∞).
Example 1.3. The functions below f : [0,∞)2 → R are elements of C, for all e,r ∈ [0,∞) :
(1) f(e,r) = e−r;
(2) f(e,r) = λe, 0 < λ < 1;
(3) f(e,r) = e
(1+r)σ
; σ ∈ (0,∞);
(4) f(e,r) =
log(r+ae)
(1+r)
,a > 1.
Here in this manuscript, we refers to:
• Ψ = {ψ : ψ : [0,∞) → [0,∞) is a strictly nondecrasing and continuous function, ψ(t) = 0 ⇔ t = 0}.
• An ultra altering distance function Φ = {ϕ : ϕ : [0,∞) → [0,∞)} is a continuous, non-decreasing mapping
such that ϕ(t) > 0 for t > 0 and ϕ(0) ≥ 0.
The goal of this paper is to obtain some new (ccp) results for a certain class of f(ψ,ϕ)- contractive via
generalized α− admissible mappings in (b-ms). Ultimately, to support our work we present an example and
application to find an analytical solution to the (nie).
2. Main results
We begin this part with the definition below:
Int. J. Anal. Appl. 18 (6) (2020) 1086
Definition 2.1. Let Υ, Λ : Γ2 → Γ and α : Γ2×Γ2 → R+ be given mappings. We say that Υ is a generalized
α-admissible with respect to (w.r.t.) Λ if
α((Λ(e,r), Λ(r,e)), (Λ(µ,κ), Λ(κ,µ))) ≥ 1 implies α((Υ(e,r), Υ(r,e)), (Υ(µ,κ), Υ(κ,µ))) ≥ 1,
for all (e,r), (µ,κ) ∈ Γ2.
Now, we present the first main theorem:
Theorem 2.1. Let (Γ,νb) be a complete (b-ms) (with parameter s > 1), and Υ, Λ : Γ
2 → Γ be two generalized
compatible mappings such that Υ is a generalized α−admissible mapping w.r.t. Λ and Λ is continuous. Let
there is f ∈ C, ψ ∈ Ψ,ϕ ∈ Φ so that the stipulation below holds
(ψ (sσνb(Υ(e,r), Υ(µ,κ)) + ρ)
α
(Λ(e,r), Λ(r,e)),
(Υ(e,r), Υ(r,e))
α
(Λ(µ,κ), Λ(κ,µ)),
(Υ(µ,κ), Υ(κ,µ))
≤ f
ψ(νb(Λ(e,r),Λ(µ,κ))+νb(Λ(r,e),Λ(κ,µ))2 ),
ϕ(
νb(Λ(e,r),Λ(µ,κ))+νb(Λ(r,e),Λ(κ,µ))
2
)
+ρ,(2.1)
for all e,r,µ,κ ∈ Γ,σ,ρ > 0, and α : (Γ2 × Γ2) → [0,∞). Assume that
(i) Υ(Γ2) ⊆ Λ(Γ2),
(ii) there is e0,r0 ∈ Γ so that
α((Λ(e0,r0), Λ(r0,e0)), (Υ(e0,r0), Υ(r0,e0))) ≥ 1,
α((Λ(r0,e0), Λ(e0,r0)), (Υ(r0,e0), Υ(e0,r0))) ≥ 1.
Also, suppose either
(iv) Υ is continuous, or
(v) {en},{rn} are two sequences in Γ so that
α((en+1,rn+1), (en,rn)) ≥ 1 and α((rn,en), (rn+1,en+1)) ≥ 1.
for all n ∈ N∪{0}, and en → e,rn → r as n →∞, e,r ∈ Γ, we have
α((e,r), (en,rn)) ≥ 1 and α((rn,en), (r,e)) ≥ 1.
Then Υ and Λ have a (ccp).
Proof. Let e0,r0 ∈ Γ, so by condition (ii), we have
α((Λ(e0,r0), Λ(r0,e0)), (Υ(e0,r0), Υ(r0,e0))) ≥ 1,
α((Λ(r0,e0), Λ(e0,r0)), (Υ(r0,e0), Υ(e0,r0))) ≥ 1.
Int. J. Anal. Appl. 18 (6) (2020) 1087
According to (i), define two sequences {en},{rn} in Γ by
Υ(en,rn) = Λ(en+1,rn+1), Υ(rn,en) = Λ(rn+1,en+1),∀n = 0, 1, 2, ... .
Since Υ(Γ2) ⊆ Λ(Γ2), then we can write
α((Λ(e0,r0), Λ(r0,e0)), (Λ(e1,r1), Λ(r1,e1)))
= α((Λ(e0,r0), Λ(r0,e0)), (Υ(e0,r0), Υ(r0,e0))) ≥ 1,
α((Λ(r0,e0), Λ(e0,r0)), (Λ(r1,e1), Λ(e1,r1)))
= α((Λ(r0,e0), Λ(e0,r0)), (Υ(r0,e0), Υ(e0,r0))) ≥ 1.
Again, since Υ is a generalized α-admissible mapping w.r.t. Λ, then we have that
α((Υ(e0,r0), Υ(r0,e0)), (Υ(e1,r1), Υ(r1,e1))) ≥ 1,
and
α((Υ(r0,e0), Υ(e0,r0)), (Υ(r1,e1), Υ(e1,r1))) ≥ 1.
By induction, we get for all n ∈ N∪{0},
α((Λ(en+1,rn+1), Λ(rn+1,en+1)), (Υ(en+1,rn+1), Υ(rn+1,en+1)))
= α((Υ(en,rn), Υ(rn,en)), (Υ(en+1,rn+1), Υ(rn+1,en+1))) ≥ 1,
and α((Λ(rn+1,en+1), Λ(en+1,rn+1)), (Υ(rn+1,en+1), Υ(en+1,rn+1)))
= α((Υ(rn,en), Υ(en,rn)), (Υ(rn+1,en+1), Υ(en+1,rn+1))) ≥ 1.
(2.2)
Denote
λn = νb(Λ(en,rn), Λ(en+1,rn+1)) + νb(Λ(rn,en), Λ(rn+1,en+1)),n ∈ N∪{0}.
We suppose that λn > 0,∀n ∈ N because if not, (en,rn) will be a (ccp) and the proof is finished.
We claim that ψ(sσλn+1) ≤ ψ(λn). Using (2.2) and letting e = en,r = rn,µ = en+1, and κ = rn+1 in (2.1),
Int. J. Anal. Appl. 18 (6) (2020) 1088
we get
ψ (sσνb(Λ(en+1,rn+1), Λ(en+2,rn+2)) + ρ
= ψ (sσνb(Υ(en,rn), Υ(en+1,rn+1)) + ρ
≤ (ψ (sσνb(Υ(en,rn), Υ(en+1,rn+1)) + ρ)
µn ,
≤ f
ψ
(
νb(Λ(en,rn),Λ(en+1,rn+1))+νb(Λ(rn,en),Λ(rn+1,en+1))
2
)
,
ϕ
(
νb(Λ(en,rn),Λ(en+1,rn+1))+νb(Λ(rn,en),Λ(rn+1,en+1))
2
)
+ ρ
= f
(
ψ(
λn
2
),ϕ(
λn
2
)
)
+ ρ,(2.3)
where
µn = α((Λ(en,rn), Λ(rn,en)), (Υ(en,rn), Υ(rn,en))) ×
α((Λ(en+1,rn+1), Λ(rn+1,en+1)), (Υ(en+1,rn+1), Υ(rn+1,en+1))).
Similarly, we have
ψ (sσνb(Λ(rn+2,en+2), Λ(rn+1,en+1)) + ρ
= ψ (sσνb(Υ(rn+1,en+1), Υ(rn,en)) + ρ
≤ (ψ (sσνb(Υ(rn+1,en+1), Υ(rn,en)) + ρ)
µn ,
≤ f
ψ
(
νb(Λ(rn+1,en+1),Λ(rn,en))+νb(Λ(en+1,rn+1),Λ(en,rn))
2
)
,
ϕ
(
νb(Λ(rn+1,en+1),Λ(rn,en))+νb(Λ(en+1,rn+1),Λ(en,rn))
2
)
+ ρ
= f
(
ψ(
λn
2
),ϕ(
λn
2
)
)
+ ρ,(2.4)
where
µn = α((Λ(rn+1,en+1), Λ(en+1,rn+1)), (Υ(rn+1,en+1), Υ(en+1,rn+1)) ×
α((Λ(rn,en), Λ(en,rn)), (Υ(rn,en), Υ(en,rn)).
Summing (2.3), (2.4), and since ψ is nondecreasing, we get
(2.5) ψ (sσλn+1) ≤ f
(
ψ(
λn
2
),ϕ(
λn
2
)
)
≤ ψ(
λn
2
),
since ψ is nondecreasing. By inequality (2.5), one can write
λn+1 ≤
1
sσ
λn.
Hence, by Lemma 1.2, the sequence λn is b−Cauchy in Γ and then {Λ(en,rn)} and {Λ(rn,en)} are also
Cauchy sequences in Γ. By the completeness of Γ, there exist e,r ∈ Γ so that
(2.6) lim
n→∞
Λ(en,rn) = Υ(en,rn) = Λ, and lim
n→∞
Λ(rn,en) = Υ(rn,en) = r.
Int. J. Anal. Appl. 18 (6) (2020) 1089
Since the pair (Υ, Λ) satisfies the generalized compatible, by (2.6), we can write
(2.7) lim
n→∞
νb(Υ(Λ(en,rn), Λ(rn,en)), Λ(Υ(en,rn), Υ(rn,en))) = 0,
(2.8) lim
n→∞
νb(Υ(Λ(rn,en), Λ(en,rn)), Λ(Υ(rn,en), Υ(en,rn))) = 0.
Now, if (i) holds, that is Υ is continuous and Λ is already continuous from the hypothesis of the theorem,
then in view of triangle inequality, we have
νb(Λ(e,r), Υ(Λ(en,rn), Λ(rn,en))) ≤ s[νb(Λ(e,r), Λ(Υ(en,rn), Υ(rn,en)))
+ νb(Λ(Υ(en,rn), Υ(rn,en)), Υ(Λ(en,rn), Λ(rn,en)))].
Passing n → ∞ in the above inequality and using (2.6), (2.7), and the continuity of Υ and Λ, we get
Λ(e,r) = Υ(e,r). Similarly, by (2.6), (2.8), we can show that Λ(r,e) = Υ(r,e).
Next, assume that (ii) holds. Since the pair Υ,Λ satisfies the generalized compatible and Λ is continuous,
we have
lim
n→∞
Λ(Λ(en,rn), Λ(rn,en)) = Λ(e,r)
= lim
n→∞
Λ(Υ(en,rn), Υ(rn,en))
= lim
n→∞
Υ(Λ(en,rn), Λ(rn,en)),(2.9)
and
lim
n→∞
Λ(Λ(rn,en), Λ(en,rn)) = Λ(r,e)
= lim
n→∞
Λ(Υ(rn,en), Υ(en,rn))
= lim
n→∞
Υ(Λ(rn,en), Λ(en,rn)).(2.10)
Then, we have
α((Λ(Λ(en,rn),e(rn,en)), Λ(Λ(rn,en), Λ(en,rn))), (Λ(e,r), Λ(r,e))) ≥ 1,
and
α((Λ(Λ(rn,en), Λ(en,rn)), Λ(Λ(en,rn), Λ(rn,en))), (Λ(r,e), Λ(e,r))) ≥ 1.
Int. J. Anal. Appl. 18 (6) (2020) 1090
Applying (2.1), we get
ψ(νb(Υ(e,r),e(e,r))) = lim
n→∞
ψ(νb(Υ(e,r),e(Υ(en,rn), Υ(rn,en))))
≤ lim
n→∞
ψ(νb(Υ(e,r), Υ(e(en,rn), Λ(rn,en)))) + ρ
≤ lim
n→∞
(ψ(sσνb(Υ(e,r), Υ(e(en,rn), Λ(rn,en)))) + ρ)
µn,
≤ lim
n→∞
f
ψ
(
νb(Λ(e,r),Λ(Λ(en,rn),Λ(rn,en)))+νb(Λ(r,e),Λ(Λ(rn,en),Λ(enrn)))
2
)
,
ϕ
(
νb(Λ(e,r),Λ(Λ(en,rn),Λ(rn,en)))+νb(Λ(r,e),Λ(Λ(rn,en),Λ(enrn)))
2
)
+ ρ,
where
µn = α(Λ(e,r), Λ(r,e)), (Υ(e,r), Υ(r,e)) ×
α
(Λ(Λ(rn,en), Λ(en,rn)), Λ(Λ(en,rn), Λ(rn,en))),
(Υ(Λ(rn,en)), Λ(en,rn), Υ(Λ(en,rn)), Λ(rn,en))
.
Using (2.9),(2.10), we get ψ(νb(Υ(e,r), Λ(e,r))) = 0 implies that Υ(e,r) = Λ(e,r). Similarly, we can prove
that Υ(r,e) = Λ(r,e). �
Theorem 2.2. Let (Γ,νb) be a complete (b-ms) (with parameter s > 1), and Υ, Λ : Γ
2 → Γ be two generalized
compatible mappings so that Υ is a generalized α−admissible mapping w.r.t. Λ and Λ is continuous. Let
there is f ∈ C and ψ ∈ Ψ,ϕ ∈ Φ so that the stipulation below holds:
α
(Λ(e,r), Λ(r,e)),
(Υ(e,r), Υ(r,e))
α
(Λ(µ,κ), Λ(κ,µ)),
(Υ(µ,κ), Υ(κ,µ))
+1
ψ(sσνb(Υ(e,r),Υ(µ,κ))
≤ 2f
(
ψ(
νb(Λ(e,r),Λ(µ,κ))+νb(Λ(r,e),Λ(κ,µ))
2
),ϕ(
νb(Λ(e,r),Λ(µ,κ))+νb(Λ(r,e),Λ(κ,µ))
2
)
)
,(2.11)
for all e,r,µ,κ ∈ Γ,ρ,σ > 0, and α : (Γ2 × Γ2) → [0,∞). Assume that
(i) Υ(Γ2) ⊆ Λ(Γ2),
(ii) there is e0,r0 ∈ Γ so that
α((Λ(e0,r0), Λ(r0,e0)), (Υ(e0,r0), Υ(r0,e0))) ≥ 1,
α((Λ(r0,e0), Λ(e0,r0)), (Υ(r0,e0), Υ(e0,r0))) ≥ 1.
Also, suppose either
(iv) Υ is continuous, or
(v) {en},{rn} are two sequences in Γ such that
α((en+1,rn+1), (en,rn)) ≥ 1 and α((rn,en), (rn+1,en+1)) ≥ 1.
Int. J. Anal. Appl. 18 (6) (2020) 1091
for all n ∈ N∪{0}, and en → e,rn → r as n →∞, e,r ∈ Γ, we have
α((e,r), (en,rn)) ≥ 1 and α((rn,en), (r,e)) ≥ 1.
Then Υ and Λ have a (ccp).
Proof. As in Theorem (2.1), we can conclude that for all n ∈ N∪{0},
α((Λ(en+1,rn+1), Λ(rn+1,en+1)), (Υ(en+1,rn+1), Υ(rn+1,en+1)))
= α((Υ(en,rn), Υ(rn,en)), (Υ(en+1,rn+1), Υ(rn+1,en+1))) ≥ 1,
and α((Λ(rn+1,en+1), Λ(en+1,rn+1)), (Υ(rn+1,en+1), Υ(en+1,rn+1)))
= α((Υ(rn,en), Υ(en,rn)), (Υ(rn+1,en+1), Υ(en+1,rn+1))) ≥ 1.
(2.12)
Denote
λn = νb(Λ(en,rn), Λ(en+1,rn+1)) + νb(Λ(rn,en), Λ(rn+1,en+1)),n ∈ N∪{0}.
We suppose that λn > 0,∀n ∈ N because if not, (en,rn) will be a (ccp) and the proof is finished.
We claim that ψ(sσλn+1) ≤ ψ(λn). Using (2.12), letting e = en,r = rn,µ = en+1, and κ = rn+1 in (2.11),
we have
2ψ(s
σνb(Λ(en+1,rn+1),Λ(en+2,rn+2)) = 2ψ(s
σνb(Υ(en,rn),Υ(en+1,rn+1))
≤ (µn + 1)ψ(s
σνb(Υ(en,rn),Υ(en+1,rn+1))
≤ 2
f
ψ
(
νb(Λ(en,rn),Λ(en+1,rn+1))+νb(Λ(rn,en),Λ(rn+1,en+1))
2
)
,
ϕ
(
νb(Λ(en,rn),Λ(en+1,rn+1))+νb(Λ(rn,en),Λ(rn+1,en+1))
2
)
= 2f(ψ(
λn
2
),ϕ(
λn
2
)),(2.13)
where
µn = α((Λ(en,rn), Λ(rn,en)), (Υ(en,rn), Υ(rn,en))) ×
α((Λ(en+1,rn+1), Λ(rn+1,en+1)), (Υ(en+1,rn+1), Υ(rn+1,en+1))).
Thus, we get
ψ (νb(Λ(en+1,rn+1), Λ(en+2,rn+2)) ≤ f
(
ψ(
λn
2
),ϕ(
λn
2
)
)
.
Int. J. Anal. Appl. 18 (6) (2020) 1092
Similarly, we have
2ψ(s
σνb(Λ(rn+2,en+2),Λ(rn+1,en+1)) = 2ψ(s
σνb(Υ(rn+1,en+1),Υ(rn,en))
≤ (µn + 1)ψ(s
σνb(Υ(rn+1,en+1),Υ(rn,en))
≤ 2
f
ψ
(
νb(Λ(rn+1,en+1),Λ(rn,en))+νb(Λ(en+1,rn+1),Λ(en,rn))
2
)
,
ϕ
(
νb(Λ(rn+1,en+1),Λ(rn,en))+νb(Λ(en+1,rn+1),Λ(en,rn))
2
)
= 2f(ψ(
λn
2
),ϕ(
λn
2
)),(2.14)
where
µn = α((Λ(rn+1,en+1), Λ(en+1,rn+1)), (Υ(rn+1,en+1), Υ(en+1,rn+1))) ×
α((Λ(rn,en), Λ(en,rn)), (Υ(rn,en), Υ(en,rn))).
Thus, we get
(2.15) ψ (sσνb(Λ(rn+2,en+2), Λ(rn+1,en+1)) ≤ f
(
ψ(
λn
2
),ϕ(
λn
2
)
)
.
Summing both inequalities (2), (2.15), and since ψ is nondecreasing, we have
(2.16) ψ (sσλn+1) ≤ f
(
ψ(
λn
2
),ϕ(
λn
2
)
)
≤ ψ(
λn
2
),
since ψ is nondecreasing and from inequality (2.16), one can get
λn+1 ≤
1
sσ
λn.
Hence, by Lemma 1.2, the sequence λn is b−Cauchy in Γ and then {Λ(en,rn)} and {Λ(rn,en)} are also
Cauchy sequences in Γ. By the completeness of Γ, there exist e,r ∈ Γ such that
(2.17) lim
n→∞
e(en,rn) = Υ(en,rn) = e, and lim
n→∞
e(rn,en) = Υ(rn,en) = r.
Since the pair (Υ,e) satisfies the generalized compatible condition, then by (2.17), one can write
(2.18) lim
n→∞
νb(Υ(Λ(en,rn), Λ(rn,en)), Λ(Υ(en,rn), Υ(rn,en))) = 0,
(2.19) lim
n→∞
νb(Υ(Λ(rn,en), Λ(en,rn)), Λ(Υ(rn,en), Υ(en,rn))) = 0.
Now, if (i) holds, we apply the same steps of Theorem (2.1), and we get Λ(r,e) = Υ(r,e).
Now, assume that (ii) holds. Since the pair Υ,Λ satisfies the generalized compatible and Λ is continuous, we
Int. J. Anal. Appl. 18 (6) (2020) 1093
have
lim
n→∞
Λ(Λ(en,rn),e(rn,en)) = Λ(e,r)
= lim
n→∞
Λ(Υ(en,rn), Υ(rn,en))
= lim
n→∞
Υ(Λ(en,rn), Λe(rn,en)),(2.20)
and
lim
n→∞
Λ(Λ(rn,en), Λ(en,rn)) = Λ(r,e)
= lim
n→∞
Λ(Υ(rn,en), Υ(en,rn))
= lim
n→∞
Υ(Λ(rn,en), Λ(en,rn)).(2.21)
Then, we can write
α((Λ(Λ(en,rn), Λ(rn,en)), Λ(Λ(rn,en), Λ(en,rn))), (Λ(e,r), Λ(r,e))) ≥ 1,
and
α((Λ(Λ(rn,en), Λ(en,rn)), Λ(Λ(en,rn), Λ(rn,en))), (Λ(r,e), Λ(e,r))) ≥ 1.
Applying (2.11), we get
2ψ(νb(Υ(e,r),Λ(e,r))) = lim
n→∞
2ψ(νb(Υ(e,r),e(Υ(en,rn),Υ(rn,en))))
= lim
n→∞
2ψ(νb(Υ(e,r),Υ(e(en,rn),Λ(rn,en))))
≤ lim
n→∞
(µn + 1)
ψ(sσνb(Υ(e,r),Υ(e(en,rn),Λ(rn,en)))),
≤ lim
n→∞
2f(ψ(χn),ϕ(χn)),
where
µn = α(Λ(e,r), Λ(r,e)), (Υ(e,r), Υ(r,e)) ×
α
Λ(Λ(rn,en), Λ(en,rn)), Λ(Λ(en,rn), Λ(rn,en))),
(Υ(Λ(rn,en), Λ(en,rn)), Υ(Λ(en,rn), Λ(rn,en)))
,
and
χn =
νb(Λ(e,r), Λ(e(en,rn), Λ(rn,en))) + νb(Λ(r,e), Λ(Λ(rn,en), Λ(enrn)))
2
,
Using (2.20),(2.21), we get ψ(νb(Υ(e,r), Λ(e,r))) = 0 this leads to Υ(e,r) = Λ(e,r). Similarly, we can prove
that Υ(r,e) = Λ(r,e). �
Int. J. Anal. Appl. 18 (6) (2020) 1094
Theorem 2.3. Let (Γ,νb) be a complete (b-ms) (with parameter s > 1), and Υ, Λ : Γ
2 → Γ be two generalized
compatible mappings such that Υ is a generalizedα−admissible mapping w.r.t. Λ and Λ is continuous. Let
there is f ∈ C, ψ ∈ Ψ,ϕ ∈ Φ so that the stipulation below holds:
α
(Λ(e,r), Λ(r,e)),
(Υ(e,r), Υ(r,e))
α
(Λ(µ,κ), Λ(κ,µ)),
(Υ(µ,κ), Υ(κ,µ)))
ψ(sσνb(Υ(e,r), Υ(µ,κ)))
≤ f
ψ(νb(Λ(e,r),Λ(µ,κ))+νb(Λ(r,e),Λ(κ,µ))2 ),
ϕ(
νb(Λ(e,r),Λ(µ,κ))+νb(Λ(r,e),Λ(κ,µ))
2
)
,(2.22)
for all e,r,µ,κ ∈ Γ,ρ,σ > 0,, and α : (Γ2 × Γ2) → [0,∞). Assume that
(i) Υ(Γ2) ⊆ Λ(Γ2)
(ii) there is e0,r0 ∈ Γ so that
α((Λ(e0,r0), Λ(r0,e0)), (Υ(e0,r0), Υ(r0,e0))) ≥ 1,
α((Λ(r0,e0), Λ(e0,r0)), (Υ(r0,e0), Υ(e0,r0))) ≥ 1.
Also, suppose either
(iv) Υ is continuous, or
(v) {en},{rn} are two sequences in Γ so that
α((en+1,rn+1), (en,rn)) ≥ 1,
α((rn,en), (rn+1,en+1)) ≥ 1.
for all n ∈ N∪{0}, and en → e,rn → r as n →∞, e,r ∈ Γ, we have
α((e,r), (en,rn)) ≥ 1,
α((rn,en), (r,e)) ≥ 1.
Then Υ and Λ have a (ccp).
Proof. Again, as in Theorem (2.1), we can conclude that for all n ∈ N∪{0},
α((Λ(en+1,rn+1), Λ(rn+1,en+1)), (Υ(en+1,rn+1), Υ(rn+1,en+1)))
= α((Υ(en,rn), Υ(rn,en)), (Υ(en+1,rn+1), Υ(rn+1,en+1))) ≥ 1,and
α((Λ(rn+1,en+1), Λ(en+1,rn+1)), (Υ(rn+1,en+1), Υ(en+1,rn+1)))
= α((Υ(rn,en), Υ(en,rn)), (Υ(rn+1,en+1), Υ(en+1,rn+1))) ≥ 1.
(2.23)
Int. J. Anal. Appl. 18 (6) (2020) 1095
Denote
λn = νb(Λ(en,rn), Λ(en+1,rn+1)) + νb(Λ(rn,en), Λ(rn+1,en+1)),n ∈ N∪{0}
.
We suppose that λn > 0,∀n ∈ N because if not, (en,rn) will be a (ccp) and the proof is finished.
We claim that ψ(sσλn+1) ≤ ψ(λn). Using (2.12), letting e = en,r = rn,µ = en+1, and κ = rn+1 in (2.22),
we have
ψ (sσνb(Λ(en+1,rn+1), Λ(en+2,rn+2))
= ψ (sσνb(Υ(en,rn), Υ(en+1,rn+1))
≤ µnψ (sσνb(Υ(en,rn), Υ(en+1,rn+1))
≤ f
ψ
(
νb(Λ(en,rn),Λ(en+1,rn+1))+νb(Λ(rn,en),Λ(rn+1,en+1))
2
)
,
ϕ
(
νb(Λ(en,rn),Λ(en+1,rn+1))+νb(Λ(rn,en),Λ(rn+1,en+1))
2
)
= f
(
ψ(
λn
2
),ϕ(
λn
2
)
)
,(2.24)
where
µn = α((Λ(en,rn), Λ(rn,en)), (Υ(en,rn), Υ(rn,en))) ×
α((Λ(en+1,rn+1), Λ(rn+1,en+1)), (Υ(en+1,rn+1), Υ(rn+1,en+1))).
Thus, we get
ψ (sσνb(Λ(en+1,rn+1), Λ(en+2,rn+2)) ≤ f
(
ψ(
λn
2
),ϕ(
λn
2
)
)
.(2.25)
Similarly, we have
ψ (sσνb(Λ(rn+2,en+2), Λ(rn+1,en+1))
= ψ (sσνb(Υ(rn+1,en+1), Υ(rn,en))
≤ µnψ (sσνb(Υ(rn+1,en+1), Υ(rn,en))
≤ f
ψ
(
νb(Λ(rn+1,en+1),Λ(rn,en))+νb(Λ(en+1,rn+1),Λ(en,rn))
2
)
,
ϕ
(
νb(Λ(rn+1,en+1),Λ(rn,en))+νb(Λ(en+1,rn+1),Λ(en,rn))
2
)
= f
(
ψ(
λn
2
),ϕ(
λn
2
)
)
,(2.26)
where
µn = α((Λ(rn+1,en+1), Λ(en+1,rn+1)), (Υ(rn+1,en+1), Υ(en+1,rn+1))) ×
α((Λ(rn,en), Λ(en,rn)), (Υ(rn,en), Υ(en,rn))).
Int. J. Anal. Appl. 18 (6) (2020) 1096
Thus, we get
ψ (sσνb(Λ(rn+2,en+2), Λ(rn+1,en+1)) ≤ f
(
ψ(
λn
2
),ϕ(
λn
2
))
)
.(2.27)
By inequalities (2.25), (2.27), and since ψ is non-decreasing, we have
ψ (sσλn+1) ≤ f
(
ψ(
λn
2
),ϕ(
λn
2
)
)
≤ ψ(
λn
2
).(2.28)
since ψ is nondecreasing and by inequality (2.28), we get
λn+1 ≤
1
sσ
λn.
Thus, by Lemma 1.2, the sequence λn is b-Cauchy in Γ and then {Λ(en,rn)} and {Λ(rn,en)} are also Cauchy
sequences in Γ. By the completeness of Γ, there exist e,r ∈ Γ such that
lim
n→∞
Λ(en,rn) = Υ(en,rn) = e, and lim
n→∞
Λ(rn,en) = Υ(rn,en) = r.(2.29)
Since the pair (Υ,e) satisfies the generalized compatible condition, then by (2.29), one can write
lim
n→∞
νb(Υ(Λ(en,rn), Λ(rn,en)), Λ(Υ(en,rn), Υ(rn,en))) = 0,(2.30)
lim
n→∞
νb(Υ(Λ(rn,en), Λ(en,rn)), Λ(Υ(rn,en), Υ(en,rn))) = 0.(2.31)
Now, if (i) holds, we apply the same steps of Theorem (2.1), and we get Λ(r,e) = Υ(r,e).
Now, assume that (ii) holds. Since the pair Υ, Λ satisfies the generalized compatible condition and Λ is
continuous, we have
lim
n→∞
Λ(Λ(en,rn), Λ(rn,en)) = Λ(e,r)
= lim
n→∞
Λ(Υ(en,rn), Υ(rn,en))
= lim
n→∞
Υ(Λ(en,rn), Λ(rn,en)),(2.32)
and
lim
n→∞
Λ(Λ(rn,en), Λ(en,rn)) = Λ(r,e)
= lim
n→∞
Λ(Υ(rn,en), Υ(en,rn))
= lim
n→∞
Υ(Λ(rn,en), Λ(en,rn)).(2.33)
Then, we have
α((Λ(Λ(en,rn), Λ(rn,en)), Λ(Λ(rn,en), Λ(en,rn))), (Λ(e,r), Λ(r,e))) ≥ 1,
and
α((Λ(Λ(rn,en), Λ(en,rn)), Λ(Λ(en,rn), Λ(rn,en))), (Λ(r,e), Λ(e,r))) ≥ 1.
Int. J. Anal. Appl. 18 (6) (2020) 1097
Applying (2.22), we get
ψ(νb(Υ(e,r), Λ(e,r))) = lim
n→∞
ψ(νb(Υ(e,r), Λ(Υ(en,rn), Υ(rn,en))))
= lim
n→∞
ψ(νb(Υ(e,r), Υ(Λ(en,rn), Λ(rn,en))))
≤ lim
n→∞
µnψ(νb(Υ(e,r), Υ(Λ(en,rn), Λ(rn,en)))),
≤ lim
n→∞
f (ψ(χn),ϕ(χn)) ,
where
µn = α(Λ(e,r), Λ(r,e)), (Υ(e,r), Υ(r,e)) ×
α
(Λ(Λ(en,rn), Λ(rn,en)), Λ(Λ(rn,en), Λ(en,rn))),
(Υ(Λ(en,rn), Λ(rn,en)), Υ(Λ(rn,en), Λ(en,rn)))
.
and
χn =
νb(Λ(e,r), Λ(Λ(en,rn), Λ(rn,en))) + νb(Λ(r,e), Λ(Λ(rn,en), Λ(enrn)))
2
.
Using (2.32),(2.33), we get ψ(νb(Υ(e,r), Λ(e,r))) = 0 implies that Υ(e,r) = Λ(e,r). Similarly, we can prove
that Υ(r,e) = Λ(r,e). �
Theorem 2.4. Suppose that all requirements of Theorems (2.1) or (2.2) or (2.3) are fulfilled. In addition,
let the stipulation below holds:
(vi) If Λ(e,r) = Υ(e,r) and Λ(r,e) = Υ(r,e)
then
α((Λ(e,r), Λ(r,e)), (Υ(e,r), Υ(r,e))) ≥ 1,
and
α((Λ(r,e), Λ(e,r)), (Υ(r,e), Υ(e,r))) ≥ 1.
Then Λ and Υ have a unique (ccp).
Proof. From Theorem (2.1) or (2.2) or (2.3), we know that the set of (ccp) of Λ and Υ is nonempty. Let
(e,r) and (e∗,r∗) are (ccp) of Λ and Υ, this yields
Λ(e,r) = Υ(e,r), Λ(r,e) = Υ(r,e),
and
Λ(e∗,r∗) = Υ(e∗,r∗), Λ(r∗,e∗) = Υ(r∗,e∗).
Int. J. Anal. Appl. 18 (6) (2020) 1098
We will prove that Λ(e,r) = Λ(e∗,r∗) and Λ(r,e) = Λ(r∗,e∗).
It follows from (vi) that
α((Λ(e,r), Λ(r,e)), (Υ(e,r), Υ(r,e))) ≥ 1
α((Λ(r,e), Λ(e,r)), (Υ(r,e), Υ(e,r))) ≥ 1,
and
α ((Λ(e∗,r∗), Λ(r∗,e∗)), (Υ(e∗,r∗), Υ(r∗,e∗))) ≥ 1
α((Λ(r∗,e∗), Λ(e∗,r∗)), (Υ(r∗,e∗), Υ(e∗,r∗))) ≥ 1,
From Theorem 2.1, using above inequalities, we have
ψ (νb(Λ(e,r), Λ(e
∗,r∗)) + ρ
≤ ψ (sσνb(Υ(e,r), Υ(e∗,r∗)) + ρ
≤ (ψ (sσνb(Υ(e,r), Υ(e∗,r∗)) + ρ)
µn
≤ f
ψ(νb(Λ(e,r),Λ(e∗,r∗))+νb(Λ(r,e),Λ(r∗,e∗))2 ),
ϕ(
νb(Λ(e,r),Λ(e
∗,r∗))+νb(Λ(r,e),Λ(r
∗,e∗))
2
)
+ ρ,(2.34)
where
µn = α((Λ(e,r), Λ(r,e)), (Υ(e,r), Υ(r,e))),α((Λ(e
∗,r∗), Λ(r∗,e∗)), (Υ(e∗,r∗), Υ(r∗,e∗))).
From Theorem 2.2, we have
2ψ(νb(Λ(e,r),Λ(e
∗,r∗))) ≤ 2ψ(s
σνb(Υ(e,r),Υ(e
∗,r∗)))
≤ (µn + 1)ψ(s
σνb(Υ(e,r),Υ(e
∗,r∗)))
≤ 2f
(
ψ(
νb(Λ(e,r),Λ(e
∗,r∗))+νb(Λ(r,e),Λ(r
∗,e∗))
2
),ϕ(
νb(Λ(e,r),Λ(e
∗,r∗))+νb(Λ(r,e),Λ(r
∗,e∗))
2
)
)
.(2.35)
From Theorem 2.3, we have
ψ(νb(Λ(e,r), Λ(e
∗,r∗))) ≤ ψ(sσνb(Υ(e,r), Υ(e∗,r∗)))
≤ (µn + 1)ψ(sσνb(Υ(e,r), Υ(e∗,r∗)))
≤ f
ψ(νb(Λ(e,r),Λ(e∗,r∗))+νb(Λ(r,e),Λ(r∗,Λ∗))2 ),
ϕ(
νb(Λ(e,r),Λ(e
∗,r∗))+νb(Λ(r,e),Λ(r
∗,e∗))
2
)
.(2.36)
From (2.34), (2.35) and (2.36), we have
f (ψ(νb(Λ(e,r), Λ(e
∗,r∗)),ϕ(νb(Λ(e,r), Λ(e
∗,r∗)))) = ψ(νb(Λ(e,r), Λ(e
∗,r∗)).
Int. J. Anal. Appl. 18 (6) (2020) 1099
By the hypotheses of f,ψ,ϕ, we get either ψ(νb(Λ(e,r), Λ(e
∗,r∗)) = 0 or ϕ(νb(Λ(e,r), Λ(e
∗,r∗))) = 0. Thus
we have Λ(e,r) = Λ(e∗,r∗). Similarly, we can prove that Λ(r,e) = Λ(r∗,e∗). �
An example below support Theorems 2.1, 2.2, 2.3, and 2.4.
Example 2.1. Let f(e,r) = τe, 0 < τ < 1 and Γ = [0,∞) endowed with
νb(e,r) = (e−r)
2
for all e,r ∈ Γ. It is clear that (Γ,νb) is a complete b−ms with a coefficient s = 2. Assume that Υ, Λ : Γ×Γ →
Γ by
Υ(e,r) =
e−r e ≥ r0 otherwise and Λ(e,r) =
e + r e ≥ r0 otherwise ,
It is obvious that Υ(Γ2) ⊆ Λ(Γ2) and the mapping Υ is continuous. Define α : Γ2 × Γ2 → [0, +∞) by
α ((e,r), (µ,κ)) =
2, e ≥ µ, r ≤ κ0, otherwise .
Then for each e◦,r◦ ∈ Γ, we find that
α [(Λ (e◦,r◦) , Λ (r◦,e◦)) , (Υ (e◦,r◦) , Υ (r◦,e◦))] = 2 > 1,
α [(Λ (r◦,e◦) , Λ (e◦,r◦)) , (Υ (r◦,e◦) , Υ (e◦,r◦))] = 2 > 1.
For all n ∈ N, let en = nn+1 and rn =
1
n
be two sequences such that
α [(Λ (en+1,rn+1) , Λ (rn,en))] ≥ 1,
α [(Λ (rn,en) , Λ (en+1,rn+1))] ≥ 1.
Then, limn→∞en = 1 and limn→∞rn = 0. Certainly, 0, 1 ∈ Γ and
α [(Λ (0, 1) , Λ (rn,en))] = 2 > 1,
α [(Λ (1, 0) , Λ (en,rn))] = 2 > 1.
Under this sequences, we can write
lim
n→∞
νb (Υ [Λ (en,rn) , Λ (rn,en)] , Λ [Υ (en,rn) , Υ (rn,en)])
= νb (Υ [Λ (1, 0) , Λ (0, 1)] , Λ [Υ (1, 0) , Υ (0, 1)])
= νb (Υ (1, 0) , Λ (1, 0)) = νb (1, 1) = 0,
similarly, one can prove that
lim
n→∞
νb (Υ [Λ (rn,en) , Λ (en,rn)] , Λ [Υ (rn,en) , Υ (en,rn)]) = 0.
Int. J. Anal. Appl. 18 (6) (2020) 1100
whenever en,rn ∈ Γ, such that
lim
n→∞
Υ (en,rn) = Υ (1, 0) = 1 = Λ (1, 0) = lim
n→∞
Λ (en,rn) ,
lim
n→∞
Υ (rn,en) = Υ (0, 1) = 0 = Λ (0, 1) = lim
n→∞
Λ (rn,en) .
Therefore, Υ and Λ are generalized compatible.
Now, for e = κ = 1 and r = µ = 0, if
α [(Λ (e,r) , Λ (r,e)) , (Λ (µ,κ) , Λ (κ,µ))] > 1,
this implies that
α [(Υ (e,r) , Υ (r,e)) , (Υ (µ,κ) , Υ (κ,µ))]
= α [(Υ (1, 0) , Υ (0, 1)) , (Υ (0, 1) , Υ (1, 0))] = α [(1, 0) , (0, 1)] = 2 > 1
for all (1, 0), (0, 1) ∈ Γ2. This prove that Υ is a generalized α−admissible w.r.t. Λ.
Finally, we will try to verify the contractive conditions (2.1),(2.11) and (2.22) of Theorems 2.1, 2.2, and
2.3 respectively. Take ψ(θ) = θ
3
and φ(θ) = θ for all θ ∈ [0,∞). Put σ = ρ = 1, τ = 1
2
, e = κ = 1
2
and
r = µ = 1
4
, then we have
α [(Λ (e,r) , Λ (r,e)) , (Υ (e,r) , Υ (r,e))]
×α [(Λ (µ,κ) , Λ (κ,µ)) , (Υ (µ,κ) , Υ (κ,µ))]
= α
[(
Λ
(
1
2
,
1
4
)
, Λ
(
1
4
,
1
2
))
,
(
Υ
(
1
2
,
1
4
)
, Υ
(
1
4
,
1
2
))]
×α
[(
Λ
(
1
4
,
1
2
)
, Λ
(
1
2
,
1
4
))
,
(
Υ
(
1
4
,
1
2
)
, Υ
(
1
2
,
1
4
))]
= α
[(
3
4
, 0
)
,
(
0,
1
4
)]
×α
[(
0,
3
4
)
,
(
0,
1
4
)]
= 2 × 0 = 0,(2.37)
ψ
(
νb (Λ (e,r) , Λ (µ,κ)) + νb (Λ (r,e) , Λ (κ,µ))
2
)
= ψ
(
νb
(
Λ
(
1
2
, 1
4
)
, Λ
(
1
4
, 1
2
))
+ νb
(
Λ
(
1
4
, 1
2
)
, Λ
(
1
2
, 1
4
))
2
)
= ψ
(
νb
(
3
4
, 0
)
+ νb
(
0, 3
4
)
2
)
= ψ
(
9
16
)
=
3
16
,(2.38)
(2.39) φ
(
νb (Λ (e,r) , Λ (µ,κ)) + νb (Λ (r,e) , Λ (κ,µ))
2
)
= φ
(
9
16
)
=
9
16
,
Int. J. Anal. Appl. 18 (6) (2020) 1101
(2.40) sσνb (Υ (e,r) , Υ (µ,κ)) = 2νb
(
Υ
(
1
2
,
1
4
)
, Υ
(
1
4
,
1
2
))
= 2νb
(
1
4
, 0
)
=
1
8
.
It follows from definition of f and (2.37)-(2.40) that
(ψ (sσνb (Υ (e,r) , Υ (µ,κ))) + ρ)
α
(Λ (e,r) , Λ (r,e)) ,
(Υ (e,r) , Υ (r,e))
×α
(Λ (µ,κ) , Λ (κ,µ)) ,
(Υ (µ,κ) , Υ (κ,µ))
=
(
ψ
(
1
4
)
+ 1
)0
= 1
≤
35
32
=
1
2
×
3
16
+ 1
= f
(
3
16
,
9
16
)
+ 1
= f
ψ
(
νb(Λ(e,r),Λ(µ,κ))+νb(Λ(r,e),Λ(κ,µ))
2
)
,
φ
(
νb(Λ(e,r),Λ(µ,κ))+νb(Λ(r,e),Λ(κ,µ))
2
)
+ ρ.(2.41)
Also, we can write
α
(Λ (e,r) , Λ (r,e)) ,
(Υ (e,r) , Υ (r,e))
×α
(Λ (µ,κ) , Λ (κ,µ)) ,
(Υ (µ,κ) , Υ (κ,µ))
+1
ψ(sσνb(Υ(e,r),Υ(µ,κ)))
= (0 + 1)ψ(
1
8
) = 1
< 2
3
32 = 2
3
16
×1
2 = 2f(
3
16
, 9
16 )
= 2
f
(
ψ
(
νb(Λ(e,r),Λ(µ,κ))+νb(Λ(r,e),Λ(κ,µ))
2
)
,φ
(
νb(Λ(e,r),Λ(µ,κ))+νb(Λ(r,e),Λ(κ,µ))
2
))
.(2.42)
Additionally, the contractive condition (2.22) of Theorem 2.3 is directly hold. So, by (2.41)-(2.42) all hy-
potheses of Theorems 2.1, 2.2, and 2.3 are fulfilled, so by Theorem 2.4 the mappings Υ and e have a unique
(ccp), here it is (0, 0) ∈ Γ2.
If we put Λ = I, (where I is the identity mapping) in Theorem 2.1, we get the important result below.
Corollary 2.1. Let (Γ,νb) be a complete (b-ms) (with parameter s > 1), and Υ,I : Γ
2 → Γ be two
generalized compatible mappings such that Υ is generalized α−admissible mapping w.r.t. I. Let there is
Int. J. Anal. Appl. 18 (6) (2020) 1102
f ∈ C, ψ ∈ Ψ,ϕ ∈ Φ so that the stipulation below holds
(ψ (sσνb(Υ(e,r), Υ(µ,κ)) + ρ)
α
((e,r), (r,e)),
(Υ(e,r), Υ(r,e))
α
((µ,κ), (κ,µ)) ,
(Υ(µ,κ), Υ(κ,µ)
≤ f
ψ
(
νb((e,r),(µ,κ))+νb((r,e),(κ,µ))
2
)
,
ϕ
(
νb((e,r),(µ,κ))+νb((r,e),(κ,µ))
2
)
+ρ,(2.43)
for all e,r,µ,κ ∈ Γ,ρ,σ > 0, and α : (Γ2 × Γ2) → [0,∞). Assume that
(i) Υ(Γ2) ⊆ Γ2,
(ii) there is e0,r0 ∈ Γ so that
α(((e0,r0), (r0,e0)), (Υ(e0,r0), Υ(r0,e0))) ≥ 1,
α(((r0,e0), (e0,r0)), (Υ(r0,e0), Υ(e0,r0))) ≥ 1.
Also, suppose either
(iv) Υ is continuous, or
(v) {en},{rn} are two sequences in Γ so that
α((en+1,rn+1), (en,rn)) ≥ 1,
α((rn,en), (rn+1,en+1)) ≥ 1.
for all n ∈ N∪{0}, and en → e,rn → r as n →∞, e,r ∈ Γ, we have
α((e,r), (en,rn)) ≥ 1,
α((rn,en), (r,e)) ≥ 1.
Then Υ has a (ccp).
3. An important application
This part is very important in this paper, where the existence solution to a (nie) using Corollary 2.1 is
presented.
Here, we refers to χ by the class functions κ : [0,∞) → [0,∞) so that κ is an increasing function and
there is ψ ∈ Ψ, φ ∈ Φ, and f ∈ C such that κ(κ) = 1
2
f(ψ(κ),φ(κ)) for all κ ∈ [0,∞). Assume the problem
below:
(3.1) j($) =
v∫
u
(a1($,`) + a2($,`)) (k1(`,j(`)) + k2(`,j(`))) d`,
for all $ ∈ [u,v]. Suppose that a1,a2,k1,k2 are continuous functions which satisfy the hypotheses below:
Int. J. Anal. Appl. 18 (6) (2020) 1103
(i) for all $,` ∈ [u,v], a1($,`),a2($,`) ≥ 0,
(ii) for all y,z ∈ R with y ≥ z, there is f,U so that
0 ≤ k1(`,y) −k1(`,z) ≤ fκ(y −z),
0 ≤ k2(`,y) −k2(`,z) ≤Uκ(y −z),
(iii) we get
max{f2,U2}
sup$∈[u,v]
v∫
u
(a1($,`) + a2($,`))d`
2
≤ 1.
To discuss the existence of a unique solution of the problem (3.1), we formulate the theorem below:
Theorem 3.1. Under the assumptions (i)-(iii) with a1,a2 ∈ C([u,v]× [u,v],R) and k1,k2 ∈ C([u,v]×R×
R), the problem (3.1) has a solution in C([u,v],R).
Proof. Let Γ = C([u,v],R) be the set of real continuous functions on [u,v] endowed with the distance
νb(e,r) = sup
$∈[u,v]
(|e($) −r($)|)2 , ∀e,r ∈ Γ.
It’s obvious that, the pair (Γ,νb) is a complete b−ms with a coefficient s = 2.
Define mappings Υ : Γ × Γ → Γ and α : Γ2 × Γ2 → R+ by
Υ(e,r)($) =
v∫
u
a1($,`) (k1(`,e(`)) + k2(`,r(`))) d`
+
q∫
p
a2($,`) (k1(`,e(`)) + k2(`,r(`))) d`,
and
α ((e,r), (µ,κ)) =
1, e ≥ µ, r ≤ κ0, otherwise .
for all $ ∈ [u,v], (e,r), (µ,κ) ∈ Γ2. If the mapping Υ has a (ccp) in Γ, then it is a solution of the problem
(3.1).
Since, for each e,r,µ,κ ∈ Γ, α [((e,r) , (r,e)) , (I (µ,κ) ,I (κ,µ))] = 1 and
α [(Υ (e,r) , Υ (r,e)) , (Υ (µ,κ) , Υ (κ,µ))] = 1,
we conclude that Υ is a generalized α−admissible w.r.t. I and by the continuity of a1,a2,k1, and k2, we
have Υ is a continuous mapping. Also, for any two sequences {en} and {rn} in Γ, suppose that
lim
n→∞
νb (Υ [I (en) ,I (rn)] ,I [Υ (en,rn) , Υ (rn,en)]) = 0,
lim
n→∞
νb (Υ [I (rn) ,I (en)] ,I [Υ (rn,en) , Υ (en,rn)]) = 0.
Int. J. Anal. Appl. 18 (6) (2020) 1104
Thus, we have
lim
n→∞
Υ (en,rn) = lim
n→∞
Ien = lim
n→∞
en, lim
n→∞
Υ (rn,en) = lim
n→∞
rn.
Therefore, the pair (Υ,I) is generalized compatible. Again, it follows from the definition of α that if
α [(I (e) ,I (r)) , (I (µ) ,I (κ))] = α (e,r,µ,κ) = 1,
this implies that
α [(Υ (e,r) , Υ (r,e)) , (Υ (µ,κ) , Υ (κ,µ))] = 1,
and
(3.2) α [(e,r) , (Υ (e,r) , Υ (r,e))] ×α [(µ,κ) , (Υ (µ,κ) , Υ (κ,µ))] = 1.
Now we are going to verify the hypothesis (2.43) of Corollary 2.1, for all e,r,µ,κ ∈ Γ,
νb (Υ(e,r), Υ(µ,κ))
= sup
$∈[u,v]
(|Υ(e,r)($) − Υ(µ,κ)($)|)2
= sup
$∈[u,v]
∣∣∣∣ v∫
u
a1($,`) (k1(`,e(`)) + k2(`,r(`))) d` +
v∫
u
a2($,`) (k1(`,e(`)) + k2(`,r(`))) d`
−
v∫
u
a1($,`) (k1(`,µ(`)) + k2(`,κ(`))) d`−
v∫
u
a2($,`) (k1(`,µ(`)) + k2(`,κ(`))) d`
∣∣∣∣
2
= sup
$∈[u,v]
∣∣∣∣ v∫
u
a1($,`) [(k1(`,e(`)) −k1(`,µ(`))) + (k2(`,r(`)) −k2(`,κ(`)))] d`
+
v∫
u
a2($,`) [(k1(`,e(`)) −k1(`,µ(`))) + (k2(`,r(`)) −k2(`,κ(`)))] d`
∣∣∣∣
2
.
Applying assumption (ii), one can get
νb (Υ(e,r), Υ(µ,κ))
≤ sup
$∈[u,v]
∣∣∣∣ v∫
u
a1($,`) [fκ(e(`) −µ(`))) + Uκ(r(`) −κ(`))] d`
+
v∫
u
a2($,`) [fκ(e(`) −µ(`))) + Uκ(r(`) −κ(`))] d`
∣∣∣∣
2
≤ max{f2,U2}×
sup
$∈[u,v]
∣∣∣∣∣∣
v∫
u
(a1($,`) + a2($,`)) [κ(|e(`) −µ(`)|) + κ(|r(`) −κ(`)|)] d`
∣∣∣∣∣∣
2 .(3.3)
By using the definition of κ and the distance νb, we have
(3.4) κ |e(`) −µ(`)|2 ≤ κνb(e,µ) and κ |r(`) −κ(`)|
2 ≤ κνb(r,κ), ∀$ ∈ [u,v].
Int. J. Anal. Appl. 18 (6) (2020) 1105
It follows from (3.3), (3.4) and assumption (iii) that
νb (Υ(e,r), Υ(µ,κ))
≤ max{f2,U2}× [κ2νb(e,µ) + κ2νb(r,κ)] ×
sup$∈[u,v]
v∫
u
(a1($,`) + a2($,`)) d`
2
≤ κ2νb(e,µ) + κ2νb(r,κ)
= 2π2
(
νb(e,µ) + νb(r,κ)
2
)
≤ 2 ×
1
4
f
(
ψ
(
νb(e,µ) + νb(r,κ)
2
)
,φ
(
νb(e,µ) + νb(r,κ)
2
))
=
1
2
f
(
ψ
(
νb(e,µ) + νb(r,κ)
2
)
,φ
(
νb(e,µ) + νb(r,κ)
2
))
.
Thus, for all e,r,µ,κ ∈ Γ, we
21νb (Υ(e,r), Υ(µ,κ)) ≤ f
(
ψ
(
νb(e,µ) + νb(r,κ)
2
)
,φ
(
νb(e,µ) + νb(r,κ)
2
))
.
Add ρ > 0 to the both sides, we have
(
21νb (Υ(e,r), Υ(µ,κ)) + ρ
)
≤ f
(
ψ
(
νb(e,µ) + νb(r,κ)
2
)
,φ
(
νb(e,µ) + νb(r,κ)
2
))
+ ρ.
Put ψ(κ) = κ, for all κ ∈ [0,∞), s = 2, σ = 1, and using (3.2), we get
ψ(sσνb (Υ(e,r), Υ(µ,κ)) + ρ)
α[(e,r),(Υ(e,r),Υ(r,e))]×α[(µ,κ),(Υ(µ,κ),Υ(κ,µ))]
≤ f
(
ψ
(
νb(e,µ) + νb(r,κ)
2
)
,φ
(
νb(e,µ) + νb(r,κ)
2
))
+ ρ.
Therefore all stipulations of Corollary 2.1 are fulfilled. Then the mapping Υ has a (ccp) which is a solution
of the system (3.1) in Γ. �
Availability of data and material: Not applicable.
Funding: This work was supported in part by the Basque Government under Grant IT1207-19.
Author Contributions: All authors contributed equally and significantly in writing this article. All
authors read and approved the final manuscript.
Acknowledgments: The authors are grateful to the Spanish Government and the European Commission
for Grant IT1207-19.
Conflicts of Interest: The author(s) declare that there are no conflicts of interest regarding the publication
of this paper.
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1. Introduction and elementary discussions
2. Main results
3. An important application
References