# Coupled and Tripled Coincidence Point Results with Application to Fredholm Integral Equations

Authors: Marwan Amin Kutbi; Nawab Hussain; Jamal Rezaei Roshan; Vahid Parvaneh
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

The aim of this paper is to define weakα-ψ-φ-contractive mappings and to establish coupled and tripled coincidence point theorems for such mappings defined on G
b-metric spaces using the concept of rectangular G-α-admissibility. As an application, we derive new coupled and tripled coincidence point results for weak ψ-φ-contractive mappings in partially ordered G
b-metric spaces. Our results are generalizations and extensions of some recent results in the literature. We also present an example as well as an application to nonlinear Fredholm integral equations in order to illustrate the effectiveness of our results.


---

## Body

## 1. Introduction and Mathematical Preliminaries

The concept of generalized metric space, or aG-metric space, was introduced by Mustafa and Sims.Definition 1 (G-metric space [1]).
LetX be a nonempty set and let G
:
X
×
X
×
X
→
R
+ be a function satisfying the following properties:(G1)
G
(
x
,
y
,
z
)
=
0 if and only if x
=
y
=
z;
(G2)
0
<
G
(
x
,
x
,
y
), for all x
,
y
∈
X with x
≠
y;
(G3)
G
(
x
,
x
,
y
)
≤
G
(
x
,
y
,
z
), for all x
,
y
,
z
∈
X with y
≠
z;
(G4)
G
(
x
,
y
,
z
)
=
G
(
x
,
z
,
y
)
=
G
(
y
,
z
,
x
)
=
⋯ (symmetry in all three variables);
(G5)
G
(
x
,
y
,
z
)
≤
G
(
x
,
a
,
a
)
+
G
(
a
,
y
,
z
), for all x
,
y
,
z
,
a
∈
X (rectangle inequality).
Then, the functionG is called a G-metric on X and the pair (
X
,
G
) is called a G-metric space.Recently, Aghajani et al. in [2] motivated by the concept of b-metric [3] introduced the concept of generalized b-metric spaces (G
b-metric spaces) and then they presented some basic properties of G
b-metric spaces.The following is their definition ofG
b-metric spaces.Definition 2 (see [2]).
LetX be a nonempty set and let s
≥
1 be a given real number. Suppose that a mapping G
:
X
×
X
×
X
→
R
+ satisfies the following:(G
b1)
G
(
x
,
y
,
z
)
=
0 if x
=
y
=
z,
(G
b2)
0
<
G
(
x
,
x
,
y
) for all x
,
y
∈
X with x
≠
y,
(G
b3)
G
(
x
,
x
,
y
)
≤
G
(
x
,
y
,
z
) for all x
,
y
,
z
∈
X with y
≠
z,
(G
b4)
G
(
x
,
y
,
z
)
=
G
(
p
{
x
,
y
,
z
}
), where p is a permutation of x
,
y
,
z (symmetry),
(G
b5)
G
(
x
,
y
,
z
)
≤
s
[
G
(
x
,
a
,
a
)
+
G
(
a
,
y
,
z
)
] for all x
,
y
,
z
,
a
∈
X (rectangle inequality).
ThenG is called a generalized b-metric and the pair (
X
,
G
) is called a generalized b-metric space or a G
b-metric space.EachG-metric space is a G
b-metric space with s
=
1.Example 3 (see [2]).
Let(
X
,
G
) be a G-metric space and G
*
(
x
,
y
,
z
)
=
G
(
x
,
y
,
z
)
p, where p
>
1 is a real number. Then G
* is a G
b-metric with s
=
2
p
-
1.Example 4 (see [4]).
LetX
=
R and d
(
x
,
y
)
=
|
x
-
y
|
2. We know that (
X
,
d
) is a b-metric space with s
=
2. Let G
(
x
,
y
,
z
)
=
d
(
x
,
y
)
+
d
(
y
,
z
)
+
d
(
z
,
x
), it is easy to see that (
X
,
G
) is not a G
b-metric space. Indeed, (G
b
3
) is not true for x
=
0, y
=
2, and z
=
1. However, G
(
x
,
y
,
z
)
=
max
⁡
{
d
(
x
,
y
)
,
d
(
y
,
z
)
,
d
(
z
,
x
)
} is a G
b-metric on R with s
=
2.Definition 5 (see [2]).
AG
b-metric G is said to be symmetric if G
(
x
,
y
,
y
)
=
G
(
y
,
x
,
x
), for all x
,
y
∈
X.Proposition 6 (see [2]).
LetX be a G
b-metric space. Then for each x
,
y
,
z
,
a
∈
X it follows that(1)
ifG
(
x
,
y
,
z
)
=
0, then x
=
y
=
z,
(2)
G
(
x
,
y
,
z
)
≤
s
(
G
(
x
,
x
,
y
)
+
G
(
x
,
x
,
z
)
),
(3)
G
(
x
,
y
,
y
)
≤
2
s
G
(
y
,
x
,
x
),
(4)
G
(
x
,
y
,
z
)
≤
s
(
G
(
x
,
a
,
z
)
+
G
(
a
,
y
,
z
)
).Definition 7 (see [2]).
LetX be a G
b-metric space. One defines d
G
(
x
,
y
)
=
G
(
x
,
y
,
y
)
+
G
(
x
,
x
,
y
), for all x
,
y
∈
X. It is easy to see that d
G defines a b-metric d on X, which one calls the b-metric associated with G.Definition 8 (see [2]).
LetX be a G
b-metric space. A sequence {
x
n
} in X is said to be(1)
G
b-Cauchy if, for each ɛ
>
0, there exists a positive integer n
0 such that, for all m
,
n
,
l
≥
n
0, G
(
x
n
,
x
m
,
x
l
)
<
ɛ;
(2)
G
b-convergent to a point x
∈
X if, for each ɛ
>
0, there exists a positive integer n
0 such that, for all m
,
n
≥
n
0, G
(
x
n
,
x
m
,
x
)
<
ɛ.Proposition 9 (see [2]).
LetX be a G
b-metric space. Then the following are equivalent:(1)
the sequence
{
x
n
}
is
G
b
-Cauchy,
(2)
for any
ɛ
>
0
there exists
n
0
∈
N
such that
G
(
x
n
,
x
m
,
x
m
)
<
ɛ
for all
m
,
n
≥
n
0.Proposition 10 (see [2]).
LetX be a G
b-metric space. The following are equivalent.(1)
{
x
n
}
is
G
b
-convergent to
x.
(2)
G
(
x
n
,
x
n
,
x
)
→
0
, as
n
→
+
∞.
(3)
G
(
x
n
,
x
,
x
)
→
0
, as
n
→
+
∞.Definition 11 (see [2]).
AG
b-metric space X is called G
b-complete if every G
b-Cauchy sequence is G
b-convergent in X.Proposition 12.
Let(
X
,
G
) and (
X
′
,
G
′
) be two G
b-metric spaces. Then a function f
:
X
→
X
′ is G
b-continuous at a point x
∈
X if and only if it is G
b-sequentially continuous at x; that is, whenever {
x
n
} is G
b-convergent to x, {
f
(
x
n
)
} is G
b
′-convergent to f
(
x
).Proposition 13.
Let(
X
,
G
) be a G
b-metric space. A mapping F
:
X
×
X
→
X is said to be continuous if, for any two G
b-convergent sequences {
x
n
} and {
y
n
} converging to x and y, respectively, {
F
(
x
n
,
y
n
)
} is G
b-convergent to F
(
x
,
y
).In general, aG
b-metric function G
(
x
,
y
,
z
) for s
>
1 is not jointly continuous in all its variables. The following is an example of a discontinuous G
b-metric.Example 14 (see [4]).
LetX
=
N
∪
{
∞
} and let D
:
X
×
X
→
R be defined by
(1)
D
(
m
,
n
)
=
{
0
,
if
m
=
n
,
|
1
m
-
1
n
|
,
if
one
of
m
,
n
is
even
and
the
other
is
even
or
∞
,
5
,
if
one
of
m
,
n
is
odd
and
the
other
is
odd
(
and
m
≠
n
)
or
∞
,
2
,
otherwise
.
Then it is easy to see that, for all m
,
n
,
p
∈
X, we have
(2)
D
(
m
,
p
)
≤
5
2
(
D
(
m
,
n
)
+
D
(
n
,
p
)
)
.
Thus, (
X
,
D
) is a b-metric space with s
=
5
/
2 (see [5]).
LetG
(
x
,
y
,
z
)
=
max
⁡
{
D
(
x
,
y
)
,
D
(
y
,
z
)
,
D
(
z
,
x
)
}. It is easy to see that G is a G
b-metric with s
=
5
/
2 which is not a continuous function.We will need the following simple lemma about theG
b-convergent sequences in the proof of our main results.Lemma 15 (see [4]).
Let(
X
,
G
) be a G
b-metric space with s
>
1 and suppose that {
x
n
},{
y
n
}, and {
z
n
} are G
b-convergent to x, y, and z, respectively. Then one has
(3)
1
s
3
G
(
x
,
y
,
z
)
≤
lim
⁡
inf
⁡
n
→
∞
G
(
x
n
,
y
n
,
z
n
)
≤
lim
⁡
sup
⁡
n
→
∞
⁡
G
(
x
n
,
y
n
,
z
n
)
≤
s
3
G
(
x
,
y
,
z
)
.
In particular, if x
=
y
=
z, then we have lim
⁡
n
→
∞
⁡
G
(
x
n
,
y
n
,
z
n
)
=
0.The existence of fixed points, coupled fixed points, and tripled fixed points for contractive type mappings in partially ordered metric spaces has been considered recently by several authors (see [6–28], etc.)Lakshmikantham and Ćirić [17] introduced the notions of mixed g-monotone mapping and coupled coincidence point and proved some coupled coincidence point and common coupled fixed point theorems in partially ordered complete metric spaces.Definition 16 (see [17]).
Let(
X
,
⪯
) be a partially ordered set and let F
:
X
×
X
→
X and g
:
X
→
X be two mappings. F has the mixed g-monotone property, if F is monotone g-nondecreasing in its first argument and is monotone g-nonincreasing in its second argument; that is, for all x
1
,
x
2
∈
X, g
x
1
⪯
g
x
2 implies F
(
x
1
,
y
)
⪯
F
(
x
2
,
y
) for any y
∈
X and for all y
1
,
y
2
∈
X, g
y
1
⪯
g
y
2 implies F
(
x
,
y
1
)
⪰
F
(
x
,
y
2
) for any x
∈
X.Definition 17 (see [7, 17]).
An element(
x
,
y
)
∈
X
×
X is called(1)
a coupled fixed point of mappingF
:
X
×
X
→
X if x
=
F
(
x
,
y
) and y
=
F
(
y
,
x
),
(2)
a coupled coincidence point of mappingsF
:
X
×
X
→
X and g
:
X
→
X if g
(
x
)
=
F
(
x
,
y
) and g
(
y
)
=
F
(
y
,
x
),
(3)
a common coupled fixed point of mappingsF
:
X
×
X
→
X and g
:
X
→
X if x
=
g
(
x
)
=
F
(
x
,
y
) and y
=
g
(
y
)
=
F
(
y
,
x
).Definition 18 (see [17]).
LetX be a nonempty set. We say that the mappings F
:
X
×
X
→
X and g
:
X
→
X are commutative if g
(
F
(
x
,
y
)
)
=
F
(
g
x
,
g
y
) for all x
,
y
∈
X.Choudhury and Maity [10] have established some coupled fixed point results for mappings with mixed monotone property in partially ordered G-metric spaces. They obtained the following results.Theorem 19 (see [10, Theorem 3.1]).
Let(
X
,
⪯
) be a partially ordered set and let G be a G-metric on X such that (
X
,
G
) is a complete G-metric space. Let F
:
X
×
X
→
X be a continuous mapping having the mixed monotone property on X. Assume that there exists k
∈
[
0,1
) such that
(4)
G
(
F
(
x
,
y
)
,
F
(
u
,
v
)
,
F
(
w
,
z
)
)

≤
k
2
[
G
(
x
,
u
,
w
)
+
G
(
y
,
v
,
z
)
]
,
for all x
⪯
u
⪯
w and y
⪰
v
⪰
z, where either u
≠
w or v
≠
z.
If there existx
0
,
y
0
∈
X such that x
0
⪯
F
(
x
0
,
y
0
) and y
0
⪰
F
(
y
0
,
x
0
), then F has a coupled fixed point in X; that is, there exist x
,
y
∈
X such that x
=
F
(
x
,
y
) and y
=
F
(
y
,
x
).Theorem 20 (see [10, Theorem 3.2]).
If, in the above theorem, in place of the continuity ofF, one assumes the following conditions, namely,(i)
if a nondecreasing sequence{
x
n
}
→
x, then x
n
⪯
x for all n,
(ii)
if a nonincreasing sequence{
y
n
}
→
y, then y
n
⪰
y for all n,then F has a coupled fixed point.Definition 21 (see [29]).
Let(
X
,
⪯
) be a partially ordered set and let G be a G-metric on X. One says that (
X
,
G
,
⪯
) is regular if the following conditions hold.(i)
If{
x
n
} is a nondecreasing sequence with x
n
→
x, then x
n
⪯
x for all n
∈
N.
(ii)
If{
x
n
} is a nonincreasing sequence with x
n
→
x, then x
n
⪰
x for all n
∈
N.Definition 22 (see [10]).
Let(
X
,
G
) be a generalized b-metric space. Mappings f
:
X
2
→
X and g
:
X
→
X are called compatible if
(5)
lim
⁡
n
→
∞
G
(
g
f
(
x
n
,
y
n
)
,
f
(
g
x
n
,
g
y
n
)
,
f
(
g
x
n
,
g
y
n
)
)
=
0
,
lim
⁡
n
→
∞
G
(
g
f
(
y
n
,
x
n
)
,
f
(
g
y
n
,
g
x
n
)
,
f
(
g
y
n
,
g
x
n
)
)
=
0
hold whenever {
x
n
} and {
y
n
} are sequences in X such that
(6)
lim
⁡
n
→
∞
f
(
x
n
,
y
n
)
=
lim
⁡
n
→
∞
g
x
n
,
lim
⁡
n
→
∞
f
(
y
n
,
x
n
)
=
lim
⁡
n
→
∞
g
y
n
.On the other hand, Berinde and Borcut [25] introduced the concept of tripled fixed point and obtained some tripled fixed point theorems for contractive type mappings in partially ordered metric spaces. For a survey of tripled fixed point theorems and related topics we refer the reader to [25–28, 30].Definition 23 (see [25, 26]).
Let(
X
,
⪯
) be a partially ordered set, f
:
X
3
→
X, and g
:
X
→
X.(1)
An element(
x
,
y
,
z
)
∈
X
3 is called a tripled fixed point of f if f
(
x
,
y
,
z
)
=
x, f
(
y
,
x
,
y
)
=
y, and f
(
z
,
y
,
x
)
=
z.
(2)
An element(
x
,
y
,
z
)
∈
X
3 is called a tripled coincidence point of the mappings f and g if f
(
x
,
y
,
z
)
=
g
x, f
(
y
,
x
,
y
)
=
g
y, and f
(
z
,
y
,
x
)
=
g
z.
(3)
An element(
x
,
y
,
z
)
∈
X
3 is called a tripled common fixed point of f and g if x
=
g
(
x
)
=
f
(
x
,
y
,
z
), y
=
g
(
y
)
=
f
(
y
,
x
,
y
), and z
=
g
(
z
)
=
f
(
z
,
y
,
x
).
(4)
One says thatf has the mixed g-monotone property if f
(
x
,
y
,
z
) is g-nondecreasing in x, g-nonincreasing in y, and g-nondecreasing in z; that is, if, for any x
,
y
,
z
∈
X,
(7)
x
1
,
x
2
∈
X
,
g
x
1
⪯
g
x
2
⟹
f
(
x
1
,
y
,
z
)
⪯
f
(
x
2
,
y
,
z
)
,
y
1
,
y
2
∈
X
,
g
y
1
⪯
g
y
2
⟹
f
(
x
,
y
1
,
z
)
⪰
f
(
x
,
y
2
,
z
)
,
z
1
,
z
2
∈
X
,
g
z
1
⪯
g
z
2
⟹
f
(
x
,
y
,
z
1
)
⪯
f
(
x
,
y
,
z
2
)
.Definition 24 (see [28]).
LetX be a nonempty set. One says that the mappings f
:
X
3
→
X and g
:
X
→
X commute if g
(
f
(
x
,
y
,
z
)
)
=
f
(
g
x
,
g
y
,
g
z
), for all x
,
y
,
z
∈
X.In [26], Borcut obtained the following.Theorem 25 (see [26, Corollary 1]).
Let(
X
,
⪯
) be a partially ordered set and suppose there is a metric d on X such that (
X
,
d
) is a complete metric space. Let f
:
X
3
→
X and g
:
X
→
X be such that f has the g-mixed monotone property. Assume that there exists k
∈
[
0,1
) such that
(8)
d
(
f
(
x
,
y
,
z
)
,
f
(
u
,
v
,
w
)
)
≤
k
max
⁡
⁡
{
d
(
g
x
,
g
u
)
,
d
(
g
y
,
g
v
)
,
d
(
g
z
,
g
w
)
}
for all x
,
y
,
z
,
u
,
v
,
w
∈
X with g
x
⪯
g
u, g
y
⪰
g
v, and g
z
⪯
g
w. Suppose f
(
X
3
)
⊆
g
(
X
) and g is continuous and commutes with f and also suppose either(a)
f is continuous, or
(b)
X has the following properties:
(i)
if a nondecreasing sequencex
n
→
x, then x
n
⪯
x for all n,
(ii)
if a nonincreasing sequencey
n
→
y, then y
n
⪰
y for all n.
If there existx
0
,
y
0
,
z
0
∈
X such that g
x
0
⪯
f
(
x
0
,
y
0
,
z
0
), g
y
0
⪰
f
(
y
0
,
x
0
,
z
0
), and g
z
0
⪯
f
(
z
0
,
y
0
,
x
0
), then f and g have a tripled coincidence point.Definition 26.
Let(
X
,
G
) be a generalized b-metric space. Mappings f
:
X
3
→
X and g
:
X
→
X are called compatible if
(9)
lim
⁡
n
→
∞
G
(
g
f
(
x
n
,
y
n
,
z
n
)
,
f
(
g
x
n
,
g
y
n
,
g
z
n
)
,
f
(
g
x
n
,
g
y
n
,
g
z
n
)
)
=
0
,
lim
⁡
n
→
∞
G
(
g
f
(
y
n
,
x
n
,
y
n
)
,
f
(
g
y
n
,
g
x
n
,
g
y
n
)
,
f
(
g
y
n
,
g
x
n
,
g
y
n
)
)
=
0
,
lim
⁡
n
→
∞
G
(
g
f
(
z
n
,
y
n
,
x
n
)
,
f
(
g
z
n
,
g
y
n
,
g
x
n
)
,
f
(
g
z
n
,
g
y
n
,
g
x
n
)
)
=
0
hold whenever {
x
n
},{
y
n
}, and {
z
n
} are sequences in X such that
(10)
lim
⁡
n
→
∞
f
(
x
n
,
y
n
,
z
n
)
=
lim
⁡
n
→
∞
g
x
n
,
lim
⁡
n
→
∞
f
(
y
n
,
x
n
,
y
n
)
=
lim
⁡
n
→
∞
g
y
n
,
lim
⁡
n
→
∞
f
(
z
n
,
y
n
,
x
n
)
=
lim
⁡
n
→
∞
g
z
n
.Letψ
:
[
0
,
+
∞
)
→
[
0
,
+
∞
)satisfies the following:(i)
ψ is continuous and nondecreasing,
(ii)
ψ
(
t
)
=
0 if and only if t
=
0.That is,ψ is an altering distance function.In this paper, we obtain some coupled and tripled coincidence point theorems for nonlinear(
ψ
,
φ
) weakly contractive mappings which are G-α-admissible with respect to another function in partially ordered G
b-metric spaces. These results generalize and modify several comparable results in the literature.


## 2. Main Results

Samet et al. [31] defined the notion of α-admissible mapping as follows.Definition 27.
LetT be a self-mapping on X and let α
:
X
×
X
→
[
0
,
+
∞
) be a function. One says that T is an α-admissible mapping if
(11)
x
,
y
∈
X
,
α
(
x
,
y
)
≥
1
⟹
α
(
T
x
,
T
y
)
≥
1
.Definition 28 (see [32]).
Let(
X
,
G
) be a G-metric space, let T be a self-mapping on X, and let α
:
X
3
→
[
0
,
+
∞
) be a function. One says that T is an G-α-admissible mapping if
(12)
x
,
y
,
z
∈
X
,
α
(
x
,
y
,
z
)
≥
1
⟹
α
(
T
x
,
T
y
,
T
z
)
≥
1
.Following the recent work in [33–35] we present the following definition in the setting of G-metric spaces.Definition 29.
Let(
X
,
G
) be a G-metric space and let f
,
g
:
X
→
X and α
:
X
3
→
[
0
,
+
∞
). One says that f is a rectangular G-α-admissible mapping with respect to g if(R1)
α
(
g
x
,
g
y
,
g
z
)
≥
1
implies
α
(
f
x
,
f
y
,
f
z
)
≥
1, x
,
y
,
z
∈
X,
(R2)
{
α
(
g
x
,
g
y
,
g
y
)
≥
1
,
α
(
g
y
,
g
z
,
g
z
)
≥
1
}implies α
(
g
x
,
g
z
,
g
z
)
≥
1
,
x
,
y
,
z
∈
X.Lemma 30.
Letf be a rectangular G-α-admissible mapping with respect to g such that f
(
X
)
⊆
g
(
X
). Assume that there exists x
0
∈
X such that α
(
g
x
0
,
f
x
0
,
f
x
0
)
≥
1. Define sequence {
y
n
} by y
n
=
g
x
n
=
f
x
n
-
1. Then
(13)
α
(
y
n
,
y
m
,
y
m
)
≥
1
∀
n
,
m
∈
N
w
i
t
h
n
<
m
.Now, we prove the following coincidence point result.Theorem 31.
Let(
X
,
G
) be a generalized b-metric space and let f
,
g
:
X
→
X satisfy the following condition:
(14)
α
(
g
x
,
g
y
,
g
z
)
ψ
(
s
G
(
f
x
,
f
y
,
f
z
)
)
≤
ψ
(
G
(
g
x
,
g
y
,
g
z
)
)
-
φ
(
G
(
g
x
,
g
y
,
g
z
)
)
for all x
,
y
,
z
∈
X, where ψ
,
φ
:
[
0
,
∞
)
→
[
0
,
∞
) are two altering distance mappings, α
:
X
3
→
[
0
,
+
∞
), and f is a rectangular G-α-admissible mapping with respect to g.
Then, mapsf and g have a coincidence point if(i)
f
(
X
)
⊆
g
(
X
),
(ii)
there existsx
0
∈
X such that α
(
g
x
0
,
f
x
0
,
f
x
0
)
≥
1,
(iii)
f and g are continuous and compatible and (
X
,
G
) is complete,
(iii′)
one off
(
X
) or g
(
X
) is complete and assume that whenever {
x
n
} in X is a sequence such that α
(
x
n
,
x
n
+
1
,
x
n
+
1
)
≥
1 for all n
∈
N
∪
{
0
} and x
n
→
x as n
→
+
∞, we have α
(
x
n
,
x
,
x
)
≥
1 for all n
∈
N
∪
{
0
}.Proof.
Letx
0
∈
X be such that α
(
g
x
0
,
f
x
0
,
f
x
0
)
≥
1. According to (i) one can define the sequence {
y
n
} as y
n
+
1
=
g
x
n
+
1
=
f
x
n for all n
=
0,1
,
2
,
….
Asα
(
g
x
0
,
g
x
1
,
g
x
1
)
=
α
(
g
x
0
,
f
x
0
,
f
x
0
)
≥
1 and since f is an G-α-admissible mapping with respect to g, then α
(
y
1
,
y
2
,
y
2
)
=
α
(
f
x
0
,
f
x
1
,
f
x
1
)
≥
1. Continuing this process, we get α
(
y
n
,
y
n
+
1
,
y
n
+
1
)
≥
1 for all n
∈
N
∪
{
0
}.
Ify
n
=
y
n
+
1, then x
n is a coincidence point of f and g.
Now, assume thaty
n
≠
y
n
+
1 for all n; that is,
(15)
G
(
y
n
,
y
n
+
1
,
y
n
+
2
)
>
0
,
for all n. Let G
n
=
G
(
y
n
,
y
n
+
1
,
y
n
+
2
). Then, from (14) we obtain that
(16)
ψ
(
s
G
(
y
n
+
1
,
y
n
+
2
,
y
n
+
3
)
)
≤
α
(
y
n
,
y
n
+
1
,
y
n
+
2
)
ψ
(
s
G
(
y
n
+
1
,
y
n
+
2
,
y
n
+
3
)
)
=
α
(
g
x
n
,
g
x
n
+
1
,
g
x
n
+
2
)
ψ
(
s
G
(
f
x
n
,
f
x
n
+
1
,
f
x
n
+
2
)
)
≤
ψ
(
G
(
y
n
,
y
n
+
1
,
y
n
+
2
)
)
-
φ
(
G
(
y
n
,
y
n
+
1
,
y
n
+
2
)
)
.
We prove that G
n
+
1
≤
G
n for each n
∈
N. If G
n
+
1
>
G
n for some n
∈
N, then from (16) we have ψ
(
G
n
+
1
)
≤
ψ
(
G
n
+
1
)
-
φ
(
G
n
) which implies that G
n
=
0, a contradiction to (15).
Hence, we have0
<
G
n
+
1
≤
G
n for each n
∈
N. Thus, the sequence {
G
n
} is nonincreasing and so there exists r
≥
0 such that lim
⁡
n
→
∞
⁡
G
n
=
r
≥
0.
Suppose thatr
>
0. Then from (16), taking the limit as n
→
∞ implies that
(17)
ψ
(
r
)
≤
ψ
(
r
)
-
φ
(
r
)
a contradiction. Hence,
(18)
lim
⁡
n
→
∞
G
n
=
lim
⁡
n
→
∞
G
(
y
n
,
y
n
+
1
,
y
n
+
2
)
=
0
.
Since y
n
+
1
≠
y
n
+
2 for every n, so by property (G
b3) we obtain
(19)
G
(
y
n
,
y
n
+
1
,
y
n
+
1
)
≤
G
(
y
n
,
y
n
+
1
,
y
n
+
2
)
.
Hence,
(20)
lim
⁡
n
→
∞
G
(
y
n
,
y
n
+
1
,
y
n
+
1
)
=
0
.
Also, by part (3) of Proposition 6 we have
(21)
lim
⁡
n
→
∞
G
(
y
n
,
y
n
,
y
n
+
1
)
=
0
.
Now, we prove that{
y
n
} is a G
b-Cauchy sequence. Assume on contrary that {
y
n
} is not a G
b-Cauchy sequence. Then there exists ɛ
>
0 for which we can find subsequences {
y
m
k
} and {
y
n
k
} of {
y
n
} such that m
k is the smallest index for which m
k
>
n
k
>
k and
(22)
G
(
y
n
k
,
y
m
k
,
y
m
k
)
≥
ɛ
.
This means that
(23)
G
(
x
n
k
,
x
m
k
-
1
,
x
m
k
-
1
)
<
ɛ
.
Since f is a rectangular G-α-admissible mapping with respect to g, then from Lemma 30  α
(
y
n
k
,
y
m
k
-
1
,
y
m
k
-
1
)
≥
1. Now, from (14) we have
(24)
ψ
(
s
G
(
y
n
k
+
1
,
y
m
k
,
y
m
k
)
)
≤
α
(
y
n
k
,
y
m
k
-
1
,
y
m
k
-
1
)
ψ
(
s
G
(
y
n
k
+
1
,
y
m
k
,
y
m
k
)
)
=
α
(
g
x
n
k
,
g
x
m
k
-
1
,
g
x
m
k
-
1
)
ψ
(
s
G
(
f
x
n
k
,
f
x
m
k
-
1
,
f
x
m
k
-
1
)
)
≤
ψ
(
G
(
y
n
k
,
y
m
k
-
1
,
y
m
k
-
1
)
)
-
φ
(
G
(
y
n
k
,
y
m
k
-
1
,
y
m
k
-
1
)
)
.
Using (G
b5) we obtain that
(25)
G
(
y
n
k
,
y
m
k
,
y
m
k
)
≤
s
G
(
y
n
k
,
y
n
k
+
1
,
y
n
k
+
1
)
+
s
G
(
y
n
k
+
1
,
y
m
k
,
y
m
k
)
.
Taking the upper limit as k
→
∞ and using (20) and (23) we obtain that
(26)
limsup
⁡
k
→
∞
⁡
G
(
y
n
k
+
1
,
y
m
k
,
y
m
k
)
≥
ɛ
s
.
Using (G
b
5) we obtain that
(27)
G
(
y
n
k
,
y
m
k
,
y
m
k
)
≤
s
G
(
y
n
k
,
y
m
k
-
1
,
y
m
k
-
1
)
+
s
G
(
y
m
k
-
1
,
y
m
k
,
y
m
k
)
.
Taking the upper limit as k
→
∞ and using (20) and (23) we obtain that
(28)
lim
⁡
inf
⁡
k
→
∞
G
(
y
n
k
,
y
m
k
-
1
,
y
m
k
-
1
)
≥
ɛ
s
.
Taking the upper limit as k
→
∞ in (24) and using (23) and (26) we obtain that
(29)
ψ
(
ϵ
)
≤
ψ
(
s
limsup
⁡
k
→
∞
⁡
G
(
y
n
k
+
1
,
y
m
k
,
y
m
k
)
)
≤
ψ
(
limsup
⁡
k
→
∞
⁡
G
(
y
n
k
,
y
m
k
-
1
,
y
m
k
-
1
)
)
-
liminf
⁡
k
→
∞
⁡
φ
(
G
(
y
n
k
,
y
m
k
-
1
,
y
m
k
-
1
)
)
≤
ψ
(
ϵ
)
-
φ
(
liminf
⁡
k
→
∞
⁡
G
(
y
n
k
,
y
m
k
-
1
,
y
m
k
-
1
)
)
which implies that
(30)
φ
(
lim
⁡
k
→
∞
inf
⁡
⁡
G
(
y
n
k
,
y
m
k
-
1
,
y
m
k
-
1
)
)
=
0
,
so li
m
k
→
∞
⁡
inf
⁡
⁡
G
(
y
n
k
,
y
m
k
-
1
,
y
m
k
-
1
)
=
0, a contradiction to (28). It follows that {
y
n
} is a G
b-Cauchy sequence in X.
Suppose first that (iii) holds. Then there exists(31)
lim
⁡
n
→
∞
f
x
n
=
lim
⁡
n
→
∞
g
x
n
=
z
∈
X
.
Further, since f and g are continuous and compatible, we get that
(32)
lim
⁡
n
→
∞
f
g
x
n
=
f
z
,
lim
⁡
n
→
∞
g
f
x
n
=
g
z
,
lim
⁡
n
→
∞
G
(
f
g
x
n
,
f
g
x
n
,
g
f
x
n
)
=
0
.
We will show that f
z
=
g
z. Indeed, we have
(33)
G
(
f
z
,
g
z
,
g
z
)
≤
s
[
G
(
f
z
,
f
g
x
n
,
f
g
x
n
)
+
G
(
f
g
x
n
,
g
z
,
g
z
)
]
≤
s
G
(
f
z
,
f
g
x
n
,
f
g
x
n
)
+
s
2
[
G
(
f
g
x
n
,
g
f
x
n
,
g
f
x
n
)
+
G
(
g
f
x
n
,
g
z
,
g
z
)
]
⟶
s
·
0
+
s
2
·
0
+
s
2
·
0
=
0
as
(
n
→
∞
)
,
and it follows that f
z
=
g
z. It means that f and g have a coincidence point.
In the case (iii′), if we assume thatg
(
X
) is G
b-complete, then
(34)
lim
⁡
n
→
∞
f
x
n
=
lim
⁡
n
→
∞
g
x
n
=
g
u
=
z
for some u
∈
X. Also, from (iii′) we have α
(
g
x
n
,
g
u
,
g
u
)
≥
1. Applying (14) with x
=
x
n and y
=
u, we have
(35)
ψ
(
G
(
f
x
n
,
f
u
,
f
u
)
)
≤
α
(
g
x
n
,
g
u
,
g
u
)
ψ
(
s
G
(
f
x
n
,
f
u
,
f
u
)
)
≤
ψ
(
G
(
g
x
n
,
g
u
,
g
u
)
)
-
φ
(
G
(
g
x
n
,
g
u
,
g
u
)
)
.
It follows that G
(
f
x
n
,
f
u
,
f
u
)
→
0 when n
→
∞; that is, f
x
n
→
f
u. Uniqueness of the limit yields that f
u
=
z
=
g
u. Hence, f and g have a coincidence point u
∈
X.Theorem 32.
Let(
X
,
G
,
⪯
) be an ordered generalized b-metric space and let f
,
g
:
X
→
X satisfy the following condition:
(36)
ψ
(
s
G
(
f
x
,
f
y
,
f
z
)
)
≤
ψ
(
G
(
g
x
,
g
y
,
g
z
)
)
-
φ
(
G
(
g
x
,
g
y
,
g
z
)
)
for all x
,
y
,
z
∈
X, with g
x
⪯
g
y
⪯
g
z, where ψ
,
φ
:
[
0
,
∞
)
→
[
0
,
∞
) are two altering distance functions.
Then, mapsf and g have a coincidence point if(i)
f is g-nondecreasing with respect to ⪯ and f
(
X
)
⊆
g
(
X
);
(ii)
there existsx
0
∈
X such that g
x
0
⪯
f
x
0;
(iii)
f and g are continuous and compatible and (
X
,
G
) is complete, or
(iii′)
(
X
,
G
,
⪯
) is regular and one of f
(
X
) or g
(
X
) is complete.Proof.
Defineα
:
X
×
X
×
X
→
[
0
,
+
∞
) by
(37)
α
(
x
,
y
,
z
)
=
{
1
,
if
x
⪯
y
⪯
z
,
0
,
otherwise
.
First, we prove that f is a rectangular G-α-admissible mapping with respect to g. Assume that α
(
g
x
,
g
y
,
g
z
)
≥
1. Therefore, we have g
x
⪯
g
y
⪯
g
z. Since f is g-nondecreasing with respect to ⪯, we get f
x
⪯
f
y
⪯
f
z; that is, α
(
f
x
,
f
y
,
f
z
)
≥
1. Also, let α
(
x
,
z
,
z
)
≥
1 and α
(
z
,
y
,
y
)
≥
1, and then x
⪯
z and z
⪯
y. Consequently, we deduce that x
⪯
y
⪯
y; that is, α
(
x
,
y
,
y
)
≥
1. Thus, f is a rectangular G-α-admissible mapping with respect to g. Since
(38)
ψ
(
s
G
(
f
x
,
f
y
,
f
z
)
)
≤
ψ
(
G
(
g
x
,
g
y
,
g
z
)
)
-
φ
(
G
(
g
x
,
g
y
,
g
z
)
)
for all x
,
y
,
z
∈
X, with g
x
⪯
g
y
⪯
g
z, then
(39)
α
(
g
x
,
g
y
,
g
z
)
ψ
(
s
G
(
f
x
,
f
y
,
f
z
)
)
≤
ψ
(
G
(
g
x
,
g
y
,
g
z
)
)
-
φ
(
G
(
g
x
,
g
y
,
g
z
)
)
.
Moreover, from (ii) there exists x
0
∈
X such that g
x
0
⪯
f
x
0
⪯
f
x
0; that is, α
(
g
x
0
,
f
x
0
,
f
x
0
)
≥
1. Hence, all the conditions of Theorem 31 are satisfied and therefore f and g have a coincidence point.Ifα
(
x
,
y
,
z
)
=
1 for all x
,
y
,
z
∈
X in Theorem 31, then we obtain the following coincidence point result.Theorem 33.
Let(
X
,
G
) be a generalized b-metric space and let f
,
g
:
X
→
X satisfy the following condition:
(40)
ψ
(
s
G
(
f
x
,
f
y
,
f
z
)
)
≤
ψ
(
G
(
g
x
,
g
y
,
g
z
)
)
-
φ
(
G
(
g
x
,
g
y
,
g
z
)
)
for all x
,
y
,
z
∈
X, where ψ
,
φ
:
[
0
,
∞
)
→
[
0
,
∞
) are two altering distance functions.
Then, mapsf and g have a coincidence point if(i)
f
(
X
)
⊆
g
(
X
),
(ii)
f and g are continuous and compatible and (
X
,
G
) is complete, or
(ii′)
one off
(
X
) or g
(
X
) is complete.


## 3. Coupled Fixed Point Results

We will use the following simple lemma in proving our next results. A similar case in the context ofb-metric spaces can be found in [24].Lemma 34.
Let(
X
,
G
) be a generalized b-metric space (with the parameter s) and let f
:
X
2
→
X and g
:
X
→
X. Suppose that F
:
X
2
→
X
2 is given by
(41)
F
X
=
(
f
(
x
,
y
)
,
f
(
y
,
x
)
)
,
X
=
(
x
,
y
)
∈
X
2
and H
:
X
2
→
X
2 is defined by
(42)
H
X
=
(
g
x
,
g
y
)
,
X
=
(
x
,
y
)
∈
X
2
.
(a)
If a mappingΩ
2
m
:
X
2
×
X
2
×
X
2
→
R
+ is given by
(43)
Ω
2
m
(
X
,
U
,
A
)
=
max
⁡
⁡
{
G
(
x
,
u
,
a
)
,
G
(
y
,
v
,
b
)
}
,
X
=
(
x
,
y
)
,
U
=
(
u
,
v
)
,
A
=
(
a
,
b
)
∈
X
2
,
then (
X
2
,
Ω
2
m
) is a generalized b-metric space (with the same parameter s). The space (
X
2
,
Ω
2
m
) is G
b-complete if and only if (
X
,
G
) is G
b-complete.
(b)
Iff is continuous from (
X
2
,
Ω
2
m
) to (
X
,
G
), then F is continuous in (
X
2
,
Ω
2
m
).
(c)
Iff and g are compatible, then F and H are compatible.
(d)
The mappingf
:
X
2
→
X is G-α-admissible with respect to g; that is,
(44)
x
,
y
,
u
,
v
,
a
,
b
∈
X
,
α
(
(
g
x
,
g
y
)
,
(
g
u
,
g
v
)
,
(
g
a
,
g
b
)
)
≥
1
⟹
α
(
f
(
x
,
y
)
,
f
(
y
,
x
)
)
,
(
f
(
u
,
v
)
,
f
(
v
,
u
)
)
,
(
(
f
(
a
,
b
)
,
f
(
b
,
a
)
)
)
≥
1
if and only if the mapping F
:
X
2
→
X
2 is G-α-admissible with respect to H; that is,
(45)
X
,
U
,
A
∈
X
2
,
α
(
H
X
,
H
U
,
H
A
)
≥
1
⟹
α
(
F
X
,
F
U
,
F
A
)
≥
1
,
where α
:
X
2
×
X
2
×
X
2
→
[
0
,
∞
) is a function.
(e)
The statement (d) holds if we replace theG-α-admissibility by rectangular G-α-admissibility.Let(
X
,
G
) be a generalized b-metric space, f
:
X
2
→
X, and g
:
X
→
X. In the rest of this paper unless otherwise stated, for all x
,
y
,
u
,
v
,
z
,
w
∈
X, let
(46)
N
f
m
(
x
,
y
,
u
,
v
,
z
,
w
)
=
max
⁡
⁡
{
G
(
f
(
x
,
y
)
,
f
(
u
,
v
)
,
f
(
z
,
w
)
)
,
G
(
f
(
y
,
x
)
,
f
(
v
,
u
)
,
f
(
w
,
z
)
)
}
,
N
g
m
(
x
,
y
,
u
,
v
,
z
,
w
)
=
max
⁡
⁡
{
G
(
g
x
,
g
u
,
g
z
)
,
G
(
g
y
,
g
v
,
g
w
)
}
.Now, we have the following coupled coincidence point result.Theorem 35.
Let(
X
,
G
) be a generalized b-metric space with the parameter s and let f
:
X
2
→
X and g
:
X
→
X. Assume that
(47)
α
(
(
g
x
,
g
y
)
,
(
g
u
,
g
v
)
,
(
g
z
,
g
w
)
)
×
ψ
(
s
N
f
m
(
x
,
y
,
u
,
v
,
z
,
w
)
)
≤
ψ
(
N
g
m
(
x
,
y
,
u
,
v
,
z
,
w
)
)
-
φ
(
N
g
m
(
x
,
y
,
u
,
v
,
z
,
w
)
)
,
for all x
,
y
,
u
,
v
,
z
,
w
∈
X, where ψ
,
φ
:
[
0
,
∞
)
→
[
0
,
∞
) are altering distance functions, α
:
(
X
2
)
3
→
[
0
,
∞
), and f is a rectangular G-α-admissible mapping with respect to g.
Assume also that(1)
f
(
X
2
)
⊆
g
(
X
);
(2)
there existx
0
,
y
0
∈
X such that
(48)
α
(
(
g
x
0
,
g
y
0
)
,
(
f
(
x
0
,
y
0
)
,
f
(
y
0
,
x
0
)
)
,
f
(
x
0
,
y
0
)
,
f
(
y
0
,
x
0
)
)
≥
1
,
α
(
(
g
y
0
,
g
x
0
)
,
(
f
(
y
0
,
x
0
)
,
f
(
x
0
,
y
0
)
)
,
f
(
y
0
,
x
0
)
,
f
(
x
0
,
y
0
)
)
≥
1
.Also, suppose that either(a)
f and g are continuous, the pair (
f
,
g
) is compatible, and (
X
,
G
) is G
b-complete, or
(b)
(
g
(
X
)
,
G
) is G
b-complete and assume that whenever {
x
n
} and {
y
n
} in X are sequences such that
(49)
α
(
(
x
n
,
y
n
)
,
(
x
n
+
1
,
y
n
+
1
)
,
(
x
n
+
1
,
y
n
+
1
)
)
≥
1
,
α
(
(
y
n
,
x
n
)
,
(
y
n
+
1
,
x
n
+
1
)
,
(
y
n
+
1
,
x
n
+
1
)
)
≥
1for all n
∈
N
∪
{
0
} and x
n
→
x, y
n
→
y as n
→
+
∞, we have
(50)
α
(
(
x
n
,
y
n
)
,
(
x
,
y
)
,
(
x
,
y
)
)
≥
1
,
α
(
(
y
n
,
x
n
)
,
(
y
,
x
)
,
(
y
,
x
)
)
≥
1
for all n
∈
N
∪
{
0
}.
Then,f and g have a coupled coincidence point in X.Proof.
LetΩ
2
m be the generalized b-metric on X
2 defined in Lemma 34. Also, define the mappings F
,
H
:
X
2
→
X
2 by F
X
=
(
f
(
x
,
y
)
,
f
(
y
,
x
)
) and H
X
=
(
g
x
,
g
y
), X
=
(
x
,
y
) as in Lemma 34. Then, (
X
2
,
Ω
2
m
) is a generalized b-metric space (with the same parameter s as X), such that F
(
X
2
)
⊆
H
(
X
2
). Moreover, the contractive condition (47) implies that
(51)
α
(
H
X
,
H
U
,
H
A
)
ψ
(
s
Ω
2
m
(
F
X
,
F
U
,
F
A
)
)
≤
ψ
(
Ω
2
m
(
H
X
,
H
U
,
H
A
)
)
-
φ
(
Ω
2
m
(
H
X
,
H
U
,
H
A
)
)
holds for all X
,
U
,
A
∈
X
2. Also, one can show that all conditions of Theorem 31 are satisfied for F and H and we have proved in Theorem 31 that, under these conditions, it follows that F and H have a coincidence point X
¯
=
(
x
¯
,
y
¯
)
∈
X
2 which is obviously a coupled coincidence point of f and g.In the following theorem, we give a sufficient condition for the uniqueness of the common coupled fixed point (see also [23]).Theorem 36.
In addition to the hypotheses of Theorem35, suppose that f and g are commutative and that, for all (
x
,
y
) and (
x
*
,
y
*
)
∈
X
2, there exists (
u
,
v
)
∈
X
2, such that α
(
(
g
x
,
g
y
)
,
(
g
u
,
g
v
)
,
(
g
u
,
g
v
)
)
≥
1 and α
(
(
g
x
*
,
g
y
*
)
,
(
g
u
,
g
v
)
,
(
g
u
,
g
v
)
)
≥
1. Then, f and g have a unique common coupled fixed point of the form (
a
,
a
).Proof.
We will use the notations as in the proof of Theorem35. It was proved in this theorem that the set of coupled coincidence points of f and g; that is, the set of coincidence points of F and H in X
2 is nonempty. We will show that if X and X
* are coincidence points of F and H, that is,
(52)
H
X
=
F
X
,
H
X
*
=
F
X
*
,
then H
X
=
H
X
*.
Choose an elementU
=
(
u
,
v
)
∈
X
2 such that α
(
(
g
x
,
g
y
)
,
(
g
u
,
g
v
)
,
(
g
u
,
g
v
)
)
≥
1 and α
(
(
g
x
*
,
g
y
*
)
,
(
g
u
,
g
v
)
,
(
g
u
,
g
v
)
)
≥
1. Let U
0
=
U and choose U
1
∈
X
2 so that H
U
1
=
F
U
0. Then, we can inductively define a sequence {
H
U
n
} such that H
U
n
+
1
=
F
U
n.
Asα
(
(
g
x
,
g
y
), (
g
u
,
g
v
), (
g
u
,
g
v
)
)
≥
1 and f is rectangular G-α-admissible with respect to g, then α
(
(
f
(
x
,
y
)
,
f
(
y
,
x
)
)
,
(
f
(
u
,
v
)
,
f
(
v
,
u
)
)
,
(
f
(
u
,
v
)
,
f
(
v
,
u
)
)
)
≥
1; that is, α
(
H
X
,
H
U
0
,
H
U
0
)
≥
1 yields that
(53)
α
(
F
X
,
F
U
,
F
U
)
=
α
(
F
X
,
F
U
0
,
F
U
0
)
=
α
(
H
X
,
H
U
1
,
H
U
1
)
≥
1
.
Therefore, by the mathematical induction, we obtain that α
(
H
X
,
H
U
n
,
H
U
n
)
≥
1, for all n
≥
0.
Applying (47), one obtains that
(54)
ψ
(
s
Ω
2
m
(
F
X
,
H
U
n
+
1
,
H
U
n
+
1
)
)
≤
α
(
H
X
,
H
U
n
,
H
U
n
)
ψ
(
s
Ω
2
m
(
F
X
,
F
U
n
,
F
U
n
)
)
≤
ψ
(
Ω
2
m
(
H
X
,
H
U
n
,
H
U
n
)
)
-
φ
(
Ω
2
m
(
H
X
,
H
U
n
,
H
U
n
)
)
≤
ψ
(
Ω
2
m
(
H
X
,
H
U
n
,
H
U
n
)
)
.
From the properties of ψ, we deduce that the sequence {
Ω
2
m
(
H
X
,
H
U
n
,
H
U
n
)
} is nonincreasing. Hence, if we proceed as in Theorem 31, we can show that
(55)
lim
⁡
n
→
∞
Ω
2
m
(
H
X
,
H
U
n
,
H
U
n
)
=
0
;
that is, {
H
U
n
} is G
b-convergent to H
X.
Similarly, we can show that{
H
U
n
} is G
b-convergent to H
X
*. Since the limit is unique, it follows that H
X
=
H
X
*.
The compatibility off and g yields that F and H are compatible, and hence F and H are weak compatible. Since H
X
=
F
X, we have H
H
X
=
H
F
X
=
F
H
X. Let H
X
=
A. Then, H
A
=
F
A. Thus, A is another coincidence point of F and H. Then, A
=
H
X
=
H
A. Therefore, A
=
(
a
,
b
) is a coupled common fixed point of f and g.
To prove the uniqueness, assume thatP is another common fixed point of F and H. Then, P
=
H
P
=
F
P and also H
P
=
H
A. Thus, P
=
H
P
=
H
A
=
A. Hence, the coupled common fixed point is unique. Also, if (
a
,
b
) is a common coupled fixed point of f and g, then (
b
,
a
) is also a common coupled fixed point of f and g. Uniqueness of the common coupled fixed point yields that a
=
b.LetΩ
2
a
:
X
2
×
X
2
×
X
2
→
R
+ be given by
(56)
Ω
2
a
(
X
,
U
,
A
)
=
G
(
x
,
u
,
a
)
+
G
(
y
,
v
,
b
)
2
,
X
=
(
x
,
y
)
,
U
=
(
u
,
v
)
,
A
=
(
a
,
b
)
∈
X
2
,
and then (
X
2
,
Ω
2
a
) is a generalized b-metric space (with the same parameter s).Let(
X
,
G
) be a generalized b-metric space, f
:
X
2
→
X, and g
:
X
→
X. For all x
,
y
,
u
,
v
,
z
,
w
∈
X, let
(57)
N
f
a
(
x
,
y
,
u
,
v
,
z
,
w
)
=
G
(
f
(
x
,
y
)
,
f
(
u
,
v
)
,
f
(
z
,
w
)
)
2
+
G
(
f
(
y
,
x
)
,
f
(
v
,
u
)
,
f
(
w
,
z
)
)
2
,
N
g
a
(
x
,
y
,
u
,
v
,
z
,
w
)
=
G
(
g
x
,
g
u
,
g
z
)
+
G
(
g
y
,
g
v
,
g
w
)
2
.Remark 37.
The result of Theorems35 and 36 holds, if we replace Ω
2
m, N
f
m, and N
g
m by Ω
2
a, N
f
a, and N
g
a, respectively.


## 4. Coupled Fixed Point Results in Partially Ordered Generalizedb-Metric Spaces

We will use the following simple lemma in proving our results.Lemma 38.
Let(
X
,
G
b
,
⪯
) be an ordered generalized b-metric space (with the parameter s) and let f
:
X
2
→
X and g
:
X
→
X. Let F
:
X
2
→
X
2 be given by
(58)
F
X
=
(
f
(
x
,
y
)
,
f
(
y
,
x
)
)
,
X
=
(
x
,
y
)
∈
X
2
and H
:
X
2
→
X
2 is defined by
(59)
H
X
=
(
g
x
,
g
y
)
,
X
=
(
x
,
y
)
∈
X
2
.
(a) If a relation⊑
2 is defined on X
2 by
(60)
X
⊑
2
U
⟺
x
⪯
u
∧
y
⪰
v
,
X
=
(
x
,
y
)
,
U
=
(
u
,
v
)
∈
X
2
,
then (
X
2
,
Ω
2
m
,
⊑
2
) and (
X
2
,
Ω
2
a
,
⊑
2
) are ordered generalized b-metric spaces (with the same parameter s).
(b) If the mappingf has the g-mixed monotone property, then the mapping F
:
X
2
→
X
2 is G-nondecreasing with respect to ⊑
2; that is,
(61)
H
X
⊑
2
H
U
⟹
F
X
⊑
2
F
U
.Theorem 39.
Let(
X
,
G
b
,
⪯
) be a partially ordered generalized b-metric space with the parameter s and let f
:
X
2
→
X and g
:
X
→
X. Assume that
(62)
ψ
(
s
N
f
m
(
x
,
y
,
u
,
v
,
z
,
w
)
)
≤
ψ
(
N
g
m
(
x
,
y
,
u
,
v
,
z
,
w
)
)
-
φ
(
N
g
m
(
x
,
y
,
u
,
v
,
z
,
w
)
)
,
for all x
,
y
,
u
,
v
,
z
,
w
∈
X with g
x
⪯
g
u
⪯
g
z and g
y
⪰
g
v
⪰
g
w, where ψ
,
φ
:
[
0
,
∞
)
→
[
0
,
∞
) are altering distance functions.
Assume also that(1)
f has the mixed g-monotone property and f
(
X
2
)
⊆
g
(
X
);
(2)
there existx
0
,
y
0
∈
X such that g
x
0
⪯
f
(
x
0
,
y
0
) and g
y
0
⪰
f
(
y
0
,
x
0
).
Also, suppose that either(a)
f and g are continuous, the pair (
f
,
g
) is compatible, and (
X
,
G
) is G
b-complete, or
(b)
(
X
,
G
b
) is regular and (
g
(
X
)
,
G
) is G
b-complete.Then, f and g have a coupled coincidence point in X.Proof.
By Lemma38, (
X
2
,
Ω
2
m
,
⊑
2
) is an ordered generalized b-metric space (with the same parameter s).
Defineα
:
X
2
×
X
2
×
X
2
→
[
0
,
+
∞
) by
(63)
α
(
(
x
,
y
)
,
(
u
,
v
)
,
(
a
,
b
)
)
=
{
1
,
if
(
x
,
y
)
⊑
2
(
u
,
v
)
⊑
2
(
a
,
b
)
,
0
,
otherwise
.
First, we prove that F is a rectangular G-α-admissible mapping with respect to H. Hence, we assume that α
(
H
X
,
H
U
,
H
A
)
≥
1, where X
=
(
x
,
y
),U
=
(
u
,
v
), and A
=
(
a
,
b
). Therefore, we have H
X
⊑
2
H
U
⊑
2
H
A. Since f has the mixed g-monotone property, then from Lemma 38, the mapping F
:
X
2
→
X
2 is G-nondecreasing with respect to ⊑
2; that is,
(64)
F
X
⊑
2
F
U
⊑
2
F
A
;
that is, α
(
F
X
,
F
U
,
F
A
)
≥
1. Also, let α
(
X
,
A
,
A
)
≥
1 and α
(
A
,
U
,
U
)
≥
1; then X
⊑
2
A and A
⊑
2
U. Consequently, we deduce that X
⊑
2
U; that is, α
(
X
,
U
,
U
)
≥
1. Thus, F is a rectangular G-α-admissible mapping with respect to H.
From (62) and the definition of α and ⊑
2,
(65)
ψ
(
s
Ω
2
m
(
F
X
,
F
U
,
F
A
)
)
≤
ψ
(
Ω
2
m
(
H
X
,
H
U
,
H
A
)
)
-
φ
(
Ω
2
m
(
H
X
,
H
U
,
H
A
)
)
,
for all X
,
U
,
A
∈
X
2 with H
X
⊑
2
H
U
⊑
2
H
A. Moreover, from (2) there exists (
x
0
,
y
0
)
∈
X
2 such that
(66)
H
(
x
0
,
y
0
)
=
(
g
x
0
,
g
y
0
)
⊑
2
(
(
x
0
,
y
0
)
,
f
(
y
0
,
x
0
)
)
=
F
(
x
0
,
y
0
)
.
Hence, all the conditions of Theorem 32 are satisfied and so F and H have a coincidence point X
¯
=
(
x
¯
,
y
¯
)
∈
X
2 which is a coupled coincidence point of f and g.In the following theorem, we give a sufficient condition for the uniqueness of the common coupled fixed point (see also [25, 28, 30]).Theorem 40.
In addition to the hypotheses of Theorem39, suppose that, for all (
x
,
y
) and (
x
*
,
y
*
)
∈
X
2, there exists (
u
,
v
)
∈
X
2, such that (
g
u
,
g
v
) is comparable with (
g
x
,
g
y
) and (
g
x
*
,
g
y
*
). Then, f and g have a unique common coupled fixed point of the form (
a
,
a
).Proof.
It was proved in Theorem39 that the set of coupled coincidence points of f and g, that is, the set of coincidence points of F and H in X
2, is nonempty. We will show that if X and X
* are coincidence points of F and H, that is,
(67)
H
X
=
F
X
,
H
X
*
=
F
X
*
,
then H
X
=
H
X
*. There exists (
u
,
v
)
∈
X
2, such that (
g
u
,
g
v
) is comparable with (
g
x
,
g
y
) and (
g
x
*
,
g
y
*
). Without any loss of generality, we may assume that (
g
x
,
g
y
)
⊑
2
(
g
u
,
g
v
) and (
g
x
*
,
g
y
*
)
⊑
2
(
g
u
,
g
v
). According to the definition of α in the above theorem, α
(
(
g
x
,
g
y
)
,
(
g
u
,
g
v
)
,
(
g
u
,
g
v
)
)
≥
1 and α
(
(
g
x
*
,
g
y
*
)
,
(
g
u
,
g
v
)
,
(
g
u
,
g
v
)
)
≥
1.
Now, following the proof of Theorem36, one can obtain that H
X
=
H
X
*. The remainder part of proof is analogous to the proof of Theorem 36 and so we omit it.Remark 41.
In Theorem39, we can replace the contractive condition (62) by the following:
(68)
α
(
(
g
x
,
g
y
)
,
(
g
u
,
g
v
)
,
(
g
z
,
g
w
)
)
×
ψ
(
s
N
f
a
(
x
,
y
,
u
,
v
,
z
,
w
)
)
≤
ψ
(
N
g
a
(
x
,
y
,
u
,
v
,
z
,
w
)
)
-
φ
(
N
g
a
(
x
,
y
,
u
,
v
,
z
,
w
)
)
.Remark 42.
Theorem39 provides conclusions of Theorems 3.1 and 3.2 of [4] for more general pair of compatible maps.In Theorem39, if we take ψ
(
t
)
=
t for all t
∈
[
0
,
∞
), we obtain the following result.Corollary 43.
Let(
X
,
G
,
⪯
) be a partially ordered G
b-complete generalized b-metric space with the parameter s
≥
1 and let f
:
X
2
→
X. Assume that
(69)
G
(
f
(
x
,
y
)
,
f
(
u
,
v
)
,
f
(
z
,
w
)
)
2
+
G
(
f
(
y
,
x
)
,
f
(
v
,
u
)
,
f
(
w
,
z
)
)
2
≤
1
s
G
(
g
x
,
g
u
,
g
z
)
+
G
(
g
y
,
g
v
,
g
w
)
2
-
1
s
φ
(
G
(
g
x
,
g
u
,
g
z
)
+
G
(
g
y
,
g
v
,
g
w
)
2
)
,
for all x
,
y
,
u
,
v
,
z
,
w
∈
X with x
⪯
u
⪯
z and y
⪰
v
⪰
w, where φ
:
[
0
,
∞
)
→
[
0
,
∞
) is an altering distance function.
Assume also that(1)
f has the mixed g-monotone property and f
(
X
2
)
⊆
g
(
X
);
(2)
there existx
0
,
y
0
∈
X such that g
x
0
⪯
f
(
x
0
,
y
0
) and g
y
0
⪰
f
(
y
0
,
x
0
).
Also, suppose that either(a)
f is continuous, or
(b)
(
X
,
G
) is regular.
Then,f and g have a coupled coincidence point in X.
In addition, suppose that, for all(
x
,
y
) and (
x
*
,
y
*
)
∈
X
2, there exists (
u
,
v
)
∈
X
2, such that (
u
,
v
) is comparable with (
x
,
y
) and (
x
*
,
y
*
). Then,fandghave a unique coupled fixed point of the form (
a
,
a
).In Corollary43, if we take φ
(
t
)
=
(
1
-
k
)
t for all t
∈
[
0
,
∞
), where k
∈
[
0,1
), we obtain the following extension of the results by Choudhury and Maity (Theorems 19 and 20).Corollary 44.
Let(
X
,
G
,
⪯
) be a partially ordered G
b-complete generalized b-metric space with the parameter s
≥
1 and let f
:
X
2
→
X. Assume that
(70)
G
(
f
(
x
,
y
)
,
f
(
u
,
v
)
,
f
(
z
,
w
)
)
2
+
G
(
f
(
y
,
x
)
,
f
(
v
,
u
)
,
f
(
w
,
z
)
)
2
≤
k
s
G
(
x
,
u
,
z
)
+
G
(
y
,
v
,
w
)
2
,
for all x
,
y
,
u
,
v
,
z
,
w
∈
X with x
⪯
u
⪯
z and y
⪰
v
⪰
w, or z
⪯
u
⪯
x and w
⪰
v
⪰
y, where k
∈
[
0,1
).
Assume also that(1)
f has the mixed monotone property;
(2)
there existx
0
,
y
0
∈
X such that x
0
⪯
f
(
x
0
,
y
0
) and y
0
⪰
f
(
y
0
,
x
0
).
Also, suppose that either(a)
f is continuous, or
(b)
(
X
,
G
) is regular.
Then,f has a coupled fixed point in X.
In addition, suppose that, for all(
x
,
y
) and (
x
*
,
y
*
)
∈
X
2, there exists (
u
,
v
)
∈
X
2, such that (
u
,
v
) is comparable with (
x
,
y
) and (
x
*
,
y
*
). Then, f has a unique coupled fixed point of the form (
a
,
a
).Now, we present an example to illustrate Theorem39 and Remark 41.Example 45.
LetX
=
R be endowed with the usual ordering and let G
b-metric G on X be given by G
(
x
,
y
,
z
)
=
max
⁡
{
|
x
-
y
|
2
,
|
y
-
z
|
2
,
|
x
-
z
|
2
}, where s
=
2. Define F
:
X
×
X
→
X as
(71)
F
(
x
,
y
)
=
x
-
y
9
,
(
x
,
y
)
∈
X
×
X
.
We define ψ
,
φ
:
[
0
,
∞
)
→
[
0
,
∞
) by
(72)
ψ
(
t
)
=
ln
⁡
⁡
(
t
+
1
)
,
φ
(
t
)
=
ln
⁡
⁡
(
t
+
1
c
t
+
1
)
,
where c
=
8
/
81. Also, F has mixed monotone property and satisfies the condition (68). Indeed, for all x
,
y
,
u
,
v
,
a
,
b
∈
X with x
≤
u
≤
a and y
≥
v
≥
b, we have
(73)
ψ
(
2
(
G
(
F
(
x
,
y
)
,
F
(
u
,
v
)
,
F
(
a
,
b
)
)
)
)
=
ln
⁡
(
2
max
⁡
{
(
x
-
y
9
-
u
-
v
9
)
2
,
(
u
-
v
9
-
a
-
b
9
)
2
,
(
a
-
b
9
-
x
-
y
9
)
2
}
+
1
)
=
ln
⁡
(
2
max
⁡
{
(
x
-
u
+
v
-
y
9
)
2
,
(
u
-
a
+
b
-
v
9
)
2
,
(
a
-
x
+
y
-
b
9
)
2
}
+
1
)
=
ln
⁡
(
2
81
max
⁡
{
(
x
-
u
+
v
-
y
)
2
,
(
u
-
a
+
b
-
v
)
2
,
(
a
-
x
+
y
-
b
)
2
}
+
1
2
81
)
≤
ln
⁡
(
4
81
max
⁡
{
(
x
-
u
)
2
+
(
v
-
y
)
2
,
(
u
-
a
)
2
+
(
b
-
v
)
2
,
(
a
-
x
)
2
+
(
y
-
b
)
2
}
+
1
4
81
)
≤
ln
⁡
(
4
81
[
max
⁡
⁡
{
(
x
-
u
)
2
,
(
u
-
a
)
2
,
(
a
-
x
)
2
}
+
max
⁡
⁡
{
(
v
-
y
)
2
,
(
b
-
v
)
2
,
(
y
-
b
)
2
}
]
+
1
4
81
)
=
ln
⁡
(
8
81
G
(
x
,
u
,
a
)
+
G
(
y
,
v
,
b
)
2
+
1
)
=
ln
⁡
(
c
G
(
x
,
u
,
a
)
+
G
(
y
,
v
,
b
)
2
+
1
)
=
ln
⁡
(
G
(
x
,
u
,
a
)
+
G
(
y
,
v
,
b
)
2
+
1
)
-
ln
⁡
(
(
(
G
(
x
,
u
,
a
)
+
G
(
y
,
v
,
b
)
)
/
2
)
+
1
c
(
(
G
(
x
,
u
,
a
)
+
G
(
y
,
v
,
b
)
)
/
2
)
+
1
)
=
ψ
(
G
(
x
,
u
,
a
)
+
G
(
y
,
v
,
b
)
2
)
-
φ
(
G
(
x
,
u
,
a
)
+
G
(
y
,
v
,
b
)
2
)
.
Similarly,
(74)
ψ
(
2
(
G
(
F
(
y
,
x
)
,
F
(
v
,
u
)
)
,
F
(
b
,
a
)
)
)
≤
ψ
(
G
(
x
,
u
,
a
)
+
G
(
y
,
v
,
b
)
2
)
-
φ
(
G
(
x
,
u
,
a
)
+
G
(
y
,
v
,
b
)
2
)
.
So,
(75)
ψ
(
2
(
G
(
F
(
x
,
y
)
,
F
(
u
,
v
)
,
F
(
a
,
b
)
)
2
+
G
(
F
(
y
,
x
)
,
F
(
v
,
u
)
,
F
(
b
,
a
)
)
2
)
)
≤
ψ
(
G
(
x
,
u
,
a
)
+
G
(
y
,
v
,
b
)
2
)
-
φ
(
G
(
x
,
u
,
a
)
+
G
(
y
,
v
,
b
)
2
)
.Finally, there are obviouslyx
0
,
y
0
∈
X such that x
0
≤
F
(
x
0
,
y
0
) and y
0
≥
F
(
y
0
,
x
0
). Thus, we conclude that the mapping F has a coupled fixed point (which is (
0,0
)).Consider now the same example, but with theG-metric
(76)
G
(
x
,
y
,
z
)
=
max
⁡
⁡
{
|
x
-
y
|
,
|
y
-
z
|
,
|
x
-
z
|
}
on X
=
R. The respective contractive condition
(77)
ψ
(
G
(
F
(
x
,
y
)
,
F
(
u
,
v
)
,
F
(
a
,
b
)
)
)
≤
ψ
(
G
(
x
,
u
,
a
)
+
G
(
y
,
v
,
b
)
2
)
-
φ
(
G
(
x
,
u
,
a
)
+
G
(
y
,
v
,
b
)
2
)
does not hold for all x
,
y
,
u
,
v
,
a
,
b
∈
X such that x
≥
u
≥
a and y
≤
v
≤
b. Indeed, for x
=
1, y
=
u
=
v
=
a
=
b
=
0 it reduces to
(78)
ψ
(
G
(
F
(
1,0
)
,
F
(
0,0
)
,
F
(
0,0
)
)
)
=
ψ
(
1
9
)
=
0.10536051565
≰
0.4054651081
-
0.35726300629
=
ψ
(
1
2
)
-
φ
(
1
2
)
=
ψ
(
G
(
1,0
,
0
)
+
G
(
0,0
,
0
)
2
)
-
φ
(
G
(
1,0
,
0
)
+
G
(
0,0
,
0
)
2
)
.
We conclude that, using a G
b-metric instead of the standard one, one has more possibilities for choosing a control function in order to get a coupled fixed point result.Remark 46.
Theorems 3.5, 3.6, and 3.7 of [4] are special cases of Theorem 39.


## 5. Tripled Coincidence Point Results

In this section we prove some tripled coincidence and tripled common fixed point results.Lemma 47.
Let(
X
,
G
,
⪯
) be an ordered generalized b-metric space (with the parameter s) and let f
:
X
3
→
X and g
:
X
→
X.
(a) If a relation⊑
3 is defined on X
3 by
(79)
X
⊑
3
U
⟺
x
⪯
u
∧
y
⪰
v
∧
z
⪯
w
,
X
=
(
x
,
y
,
z
)
,
U
=
(
u
,
v
,
w
)
∈
X
3
,
and a mapping Ω
3
m
:
X
3
×
X
3
×
X
3
→
R
+ is given by
(80)
Ω
3
m
(
X
,
U
,
A
)
=
max
⁡
⁡
{
G
(
x
,
u
,
a
)
,
G
(
y
,
v
,
b
)
,
G
(
z
,
w
,
c
)
}
,
for all X
=
(
x
,
y
,
z
), U
=
(
u
,
v
,
w
), and A
=
(
a
,
b
,
c
)
∈
X
3, then (
X
3
,
Ω
3
m
,
⊑
3
) is an ordered generalized b-metric space (with the same parameter s). The space (
X
3
,
Ω
3
m
) is G
b-complete if and only if (
X
,
G
) is G
b-complete.
(b) If the mappingf has the g-mixed monotone property, then the mapping F
:
X
3
→
X
3 given by
(81)
F
X
=
(
f
(
x
,
y
,
z
)
,
f
(
y
,
x
,
y
)
,
f
(
z
,
y
,
x
)
)
,
X
=
(
x
,
y
,
z
)
∈
X
3
is G-nondecreasing with respect to ⊑
3; that is,
(82)
H
X
⊑
3
H
U
⟹
F
X
⊑
3
F
U
,
where H
:
X
3
→
X
3 is defined by
(83)
H
X
=
(
g
x
,
g
y
,
g
z
)
,
X
=
(
x
,
y
,
z
)
∈
X
3
.
(c) Iff is continuous from (
X
3
,
Ω
3
m
) to (
X
,
G
), then F is continuous in (
X
3
,
Ω
3
m
).
(d) Iff and g are compatible, then F and H are compatible.
(e) The mappingf
:
X
3
→
X is G-α-admissible with respect to g; that is,
(84)
x
,
y
,
z
,
u
,
v
,
w
,
a
,
b
,
c
∈
X
,
α
(
(
g
x
,
g
y
,
g
z
)
,
(
g
y
,
g
x
,
g
y
)
,
(
g
z
,
g
y
,
g
x
)
)
≥
1
⟹
α
(
(
f
(
x
,
y
,
z
)
,
f
(
y
,
x
,
y
)
,
f
(
z
,
y
,
x
)
)
,
(
f
(
u
,
v
,
w
)
,
f
(
v
,
u
,
v
)
,
f
(
w
,
v
,
u
)
)
,
(
f
(
a
,
b
,
c
)
,
f
(
b
,
a
,
b
)
,
f
(
c
,
b
,
a
)
)
)
≥
1
if and only if the mapping F
:
X
3
→
X
3 is G-α-admissible with respect to H; that is,
(85)
X
,
U
,
A
∈
X
3
,
α
(
H
X
,
H
U
,
H
A
)
≥
1
⟹
α
(
F
X
,
F
U
,
F
A
)
≥
1
,
where α
:
X
3
×
X
3
×
X
3
→
[
0
,
∞
) is a function.Let(
X
,
G
,
⪯
) be an ordered generalized b-metric space, f
:
X
3
→
X, and g
:
X
→
X. For all x
,
y
,
z
,
u
,
v
,
w
,
a
,
b
,
c
∈
X, let
(86)
M
f
m
(
x
,
y
,
z
,
u
,
v
,
w
,
a
,
b
,
c
)
=
max
⁡
⁡
{
G
(
f
(
x
,
y
,
z
)
,
f
(
u
,
v
,
w
)
,
f
(
a
,
b
,
c
)
)
,
G
(
f
(
y
,
x
,
y
)
,
f
(
v
,
u
,
v
)
,
f
(
b
,
a
,
b
)
)
,
G
(
f
(
z
,
y
,
x
)
,
f
(
w
,
v
,
u
)
,
f
(
c
,
b
,
a
)
)
}
,
M
g
m
(
x
,
y
,
z
,
u
,
v
,
w
,
a
,
b
,
c
)
=
max
⁡
⁡
{
G
(
g
x
,
g
u
,
g
a
)
,
G
(
g
y
,
g
v
,
g
b
)
,
G
(
g
z
,
g
w
,
g
c
)
}
.Theorem 48.
Let(
X
,
G
) be a generalized b-metric space with the parameter s and let f
:
X
3
→
X and g
:
X
→
X. Assume that
(87)
α
(
(
g
x
,
g
y
,
g
z
)
,
(
g
u
,
g
v
,
g
w
)
,
(
g
a
,
g
b
,
g
c
)
)
×
ψ
(
s
M
f
m
(
x
,
y
,
z
,
u
,
v
,
w
,
a
,
b
,
c
)
)
≤
ψ
(
M
g
m
(
x
,
y
,
z
,
u
,
v
,
w
,
a
,
b
,
c
)
)
-
φ
(
M
g
m
(
x
,
y
,
z
,
u
,
v
,
w
,
a
,
b
,
c
)
)
,
for all x
,
y
,
z
,
u
,
v
,
w
,
a
,
b
,
c
∈
X where ψ
,
φ
:
[
0
,
∞
)
→
[
0
,
∞
) are altering distance functions and α
:
(
X
3
)
3
→
[
0
,
∞
) is a mapping such that f is a rectangular G-α-admissible mapping with respect to g.
Assume also that(1)
f
(
X
3
)
⊆
g
(
X
);
(2)
there existx
0
,
y
0
,
z
0
∈
X such that
(88)
α
(
(
g
x
0
,
g
y
0
,
g
z
0
)
,
(
f
(
x
0
,
y
0
,
z
0
)
,
f
(
y
0
,
x
0
,
y
0
)
,
f
(
z
0
,
y
0
,
x
0
)
)
,
(
f
(
x
0
,
y
0
,
z
0
)
,
f
(
y
0
,
x
0
,
y
0
)
,
f
(
z
0
,
y
0
,
x
0
)
)
)
≥
1
,
α
(
(
g
y
0
,
g
x
0
,
g
y
0
)
,
(
f
(
y
0
,
x
0
,
y
0
)
,
f
(
x
0
,
y
0
,
z
0
)
,
f
(
y
0
,
x
0
,
y
0
)
)
,
(
f
(
y
0
,
x
0
,
y
0
)
,
f
(
x
0
,
y
0
,
z
0
)
,
f
(
y
0
,
x
0
,
y
0
)
)
)
≥
1
,
α
(
(
g
z
0
,
g
y
0
,
g
x
0
)
,
(
f
(
z
0
,
y
0
,
x
0
)
,
f
(
y
0
,
x
0
,
y
0
)
,
f
(
x
0
,
y
0
,
z
0
)
)
,
(
f
(
z
0
,
y
0
,
x
0
)
,
f
(
y
0
,
x
0
,
y
0
)
,
f
(
x
0
,
y
0
,
z
0
)
)
)
≥
1
.
Also, suppose that either(a)
f and g are continuous, the pair (
f
,
g
) is compatible, and (
X
,
G
) is G
b-complete, or
(b)
(
g
(
X
)
,
G
) is G
b-complete and assume that whenever {
x
n
},{
y
n
},{
z
n
} in X are sequences such that
(89)
α
(
(
x
n
,
y
n
,
z
n
)
,
(
x
n
+
1
,
y
n
+
1
,
z
n
+
1
)
,
(
x
n
+
1
,
y
n
+
1
,
z
n
+
1
)
)
≥
1
,
α
(
(
y
n
,
x
n
,
y
n
)
,
(
y
n
+
1
,
x
n
+
1
,
y
n
+
1
)
,
(
y
n
+
1
,
x
n
+
1
,
y
n
+
1
)
)
≥
1
,
α
(
(
z
n
,
y
n
,
x
n
)
,
(
z
n
+
1
,
y
n
+
1
,
x
n
+
1
)
,
(
z
n
+
1
,
y
n
+
1
,
x
n
+
1
)
)
≥
1 for all n
∈
N
∪
{
0
} and x
n
→
x, y
n
→
y, and z
n
→
z as n
→
+
∞, we have
(90)
α
(
(
x
n
,
y
n
,
z
n
)
,
(
x
,
y
,
z
)
,
(
x
,
y
,
z
)
)
≥
1
,
α
(
(
y
n
,
x
n
,
y
n
)
,
(
y
,
x
,
y
)
,
(
y
,
x
,
y
)
)
≥
1
,
α
(
(
z
n
,
y
n
,
x
n
)
,
(
z
,
y
,
x
)
,
(
z
,
y
,
x
)
)
≥
1
for all n
∈
N
∪
{
0
}.
Then,f and g have a tripled coincidence point in X.Theorem 49.
In addition to the hypotheses of Theorem48, suppose that f and g are commutative and that, for all (
x
,
y
,
z
) and (
x
*
,
y
*
,
z
*
)
∈
X
3, there exists (
u
,
v
,
w
)
∈
X
3, such that
(91)
α
(
(
g
x
,
g
y
,
g
z
)
,
(
g
u
,
g
v
,
g
w
)
,
(
g
u
,
g
v
,
g
w
)
)
≥
1
,
α
(
(
g
x
*
,
g
y
*
,
g
z
*
)
,
(
g
u
,
g
v
,
g
w
)
,
(
g
u
,
g
v
,
g
w
)
)
≥
1
.
Then, f and g have a unique common tripled fixed point of the form (
a
,
a
,
a
).Let(92)
Ω
3
a
(
X
,
U
,
A
)
=
G
(
x
,
u
,
a
)
+
G
(
y
,
v
,
b
)
+
G
(
z
,
w
,
c
)
3
,
X
=
(
x
,
y
,
z
)
,
U
=
(
u
,
v
,
w
)
,
A
=
(
a
,
b
,
c
)
∈
X
3
,
M
f
a
(
x
,
y
,
z
,
u
,
v
,
w
,
a
,
b
,
c
)
=
G
(
f
(
x
,
y
,
z
)
,
f
(
u
,
v
,
w
)
,
f
(
a
,
b
,
c
)
)
3
+
G
(
f
(
y
,
x
,
y
)
,
f
(
v
,
u
,
v
)
,
f
(
b
,
a
,
b
)
)
3
+
G
(
f
(
z
,
y
,
x
)
,
f
(
w
,
v
,
u
)
,
f
(
c
,
b
,
a
)
)
3
,
M
g
a
(
x
,
y
,
z
,
u
,
v
,
w
,
a
,
b
,
c
)
=
G
(
g
x
,
g
u
,
g
a
)
+
G
(
g
y
,
g
v
,
g
b
)
+
G
(
g
z
,
g
w
,
g
c
)
3
.Remark 50.
In Theorem48, we can replace the contractive condition (87) by the following:
(93)
α
(
(
g
x
,
g
y
,
g
z
)
,
(
g
u
,
g
v
,
g
w
)
,
(
g
a
,
g
b
,
g
c
)
)
×
ψ
(
s
M
f
a
(
x
,
y
,
z
,
u
,
v
,
w
,
a
,
b
,
c
)
)
≤
ψ
(
M
g
a
(
x
,
y
,
z
,
u
,
v
,
w
,
a
,
b
,
c
)
)
-
φ
(
M
g
a
(
x
,
y
,
z
,
u
,
v
,
w
,
a
,
b
,
c
)
)
.The following tripled fixed point results in ordered metric spaces can be obtained.Theorem 51.
Let(
X
,
G
,
⪯
) be a partially ordered generalized b-metric space with the parameter s and let f
:
X
3
→
X and g
:
X
→
X. Assume that
(94)
ψ
(
s
M
f
m
(
x
,
y
,
z
,
u
,
v
,
w
,
a
,
b
,
c
)
)
≤
ψ
(
M
g
m
(
x
,
y
,
z
,
u
,
v
,
w
,
a
,
b
,
c
)
)
-
φ
(
M
g
m
(
x
,
y
,
z
,
u
,
v
,
w
,
a
,
b
,
c
)
)
,
for all x
,
y
,
z
,
u
,
v
,
w
,
a
,
b
,
c
∈
X with g
x
⪯
g
u
⪯
g
a, g
y
⪰
g
v
⪰
g
b, and g
z
⪯
g
w
⪯
g
c, where ψ
,
φ
:
[
0
,
∞
)
→
[
0
,
∞
) are altering distance functions.
Assume also that(1)
f has the mixed g-monotone property and f
(
X
3
)
⊆
g
(
X
);
(2)
there existx
0
,
y
0
,
z
0
∈
X such that g
x
0
⪯
f
(
x
0
,
y
0
,
z
0
), g
y
0
⪰
f
(
y
0
,
x
0
,
y
0
), and g
z
0
⪯
f
(
z
0
,
y
0
,
x
0
).
Also, suppose that either(a)
f and g are continuous, the pair (
f
,
g
) is compatible, and (
X
,
G
) is G
b-complete, or
(b)
(
X
,
G
) is regular and (
g
(
X
)
,
G
) is G
b-complete.
Then,f and g have a tripled coincidence point in X.Theorem 52.
In addition to the hypotheses of Theorem48, suppose that f and g are commutative and that, for all (
x
,
y
,
z
) and (
x
*
,
y
*
,
z
*
)
∈
X
3, there exists (
u
,
v
,
w
)
∈
X
3, such that (
g
u
,
g
v
,
g
w
) is comparable with (
g
x
,
g
y
,
g
z
) and (
g
x
*
,
g
y
*
,
g
z
*
). Then, f and g have a unique common tripled fixed point of the form (
a
,
a
,
a
).Remark 53.
In Theorem51, we can replace the contractive condition (94) by the following:
(95)
ψ
(
s
M
f
a
(
x
,
y
,
z
,
u
,
v
,
w
,
a
,
b
,
c
)
)
≤
ψ
(
M
g
a
(
x
,
y
,
z
,
u
,
v
,
w
,
a
,
b
,
c
)
)
-
φ
(
M
g
a
(
x
,
y
,
z
,
u
,
v
,
w
,
a
,
b
,
c
)
)
.Remark 54.
Theorem51 extends Theorem 2.1 of [36] to a compatible pair.Remark 55.
Corollaries 2.2, 2.3, 2.6, 2.7,  and 2.8 of [36] are special cases of Theorem 51.Remark 56.
Theorem25 is a special case of Theorems 51.


## 6. Application to Integral Equations

As an application of the (coupled) fixed point theorems established in Section4, we study the existence and uniqueness of a solution to a Fredholm nonlinear integral equation.In order to compare our results to the ones in [37, 38] we will consider the same integral equation; that is,
(96)
x
(
t
)
=
∫
a
b
‍
(
K
1
(
t
,
s
)
+
K
2
(
t
,
s
)
)
(
f
(
s
,
x
(
s
)
)
+
g
(
s
,
x
(
s
)
)
)
d
s
x
(
t
)
=
+
h
(
t
)
,
where t
∈
I
=
[
a
,
b
].LetΘ denote the set of all functions θ
:
[
0
,
∞
)
→
[
0
,
∞
) satisfying the following.(
i
θ
)
θ is nondecreasing and (
θ
(
r
)
)
p
≤
θ
(
r
p
), for all p
≥
1.
(
i
i
θ
)
There existsφ
∈
Φ such that 2
θ
(
r
)
=
(
r
/
2
)
-
φ
(
r
/
2
), for all r
∈
[
0
,
∞
).
Θ
is nonempty, asθ
1
(
r
)
=
2
k
r with 0
≤
4
k
<
1 is an element of Θ.Like in [38], we assume that the functions K
1, K
2, f, and g fulfill the following conditions.Assumption 57.
Consider the following:(i)
K
1
(
t
,
s
)
≥
0 and K
2
(
t
,
s
)
≤
0, for all t
,
s
∈
I;
(ii)
there exist two positive numbersλ and μ and θ
∈
Θ such that, for all x
,
y
∈
R with x
≥
y, the following Lipschitzian type conditions hold:
(97)
0
≤
f
(
t
,
x
)
-
f
(
t
,
y
)
≤
λ
θ
(
x
-
y
)
,
-
μ
θ
(
x
-
y
)
≤
g
(
t
,
x
)
-
g
(
t
,
y
)
≤
0
;
(iii)
(98)
(
max
⁡
⁡
{
λ
p
,
μ
p
}
)
·
sup
⁡
t
∈
I
[
(
∫
a
b
‍
K
1
(
t
,
s
)
d
s
)
p
+
(
∫
a
b
‍
-
K
2
(
t
,
s
)
d
s
)
p
]
≤
1
2
3
p
-
3
.Definition 58 (see [38]).
A pair(
α
,
β
)
∈
X
2 with X
=
C
(
I
,
R
) is called a coupled lower-upper solution of (96) if, for all t
∈
I,
(99)
α
(
t
)
≤
∫
a
b
‍
K
1
(
t
,
s
)
[
f
(
s
,
α
(
s
)
)
+
g
(
s
,
β
(
s
)
)
]
d
s
+
∫
a
b
‍
K
2
(
t
,
s
)
[
f
(
s
,
β
(
s
)
)
+
g
(
s
,
α
(
s
)
)
]
d
s
+
h
(
t
)
,
β
(
t
)
≥
∫
a
b
‍
K
1
(
t
,
s
)
[
f
(
s
,
β
(
s
)
)
+
g
(
s
,
α
(
s
)
)
]
d
s
+
∫
a
b
‍
K
2
(
t
,
s
)
[
f
(
s
,
α
(
s
)
)
+
g
(
s
,
β
(
s
)
)
]
d
s
+
h
(
t
)
.Theorem 59.
Consider the integral equation (96) with
(100)
K
1
,
K
2
∈
C
(
I
×
I
,
R
)
,
h
∈
C
(
I
,
R
)
.
Suppose that there exists a coupled lower-upper solution (
α
,
β
) of (96) with α
≤
β and that Assumption 57 is satisfied. Then the integral equation (96) has a unique solution in C
(
I
,
R
).Proof.
Consider onX
=
C
(
I
,
R
) the natural partial order relation; that is, for x
,
y
∈
C
(
I
,
R
),
(101)
x
≤
y
⟺
x
(
t
)
≤
y
(
t
)
,
∀
t
∈
I
.
It is well known that X is a complete metric space with respect to the sup metric:
(102)
d
(
x
,
y
)
=
sup
⁡
t
∈
I
|
x
(
t
)
-
y
(
t
)
|
,
x
,
y
∈
C
(
I
,
R
)
.
Now forp
≥
1 we define
(103)
ρ
(
x
,
y
)
=
d
(
x
,
y
)
p
=
(
sup
⁡
t
∈
I
|
x
(
t
)
-
y
(
t
)
|
)
p
=
sup
⁡
t
∈
I
|
x
(
t
)
-
y
(
t
)
|
p
,
x
,
y
∈
C
(
I
,
R
)
.
Define
(104)
G
(
x
,
y
,
z
)
=
max
⁡
⁡
{
ρ
(
x
,
y
)
,
ρ
(
y
,
z
)
,
ρ
(
z
,
x
)
}
.
It is easy to see that (
X
,
G
) is a complete G
b-metric space with s
=
2
p
-
1 (see Example 3).
Now define onX
2 the following partial order:
for all(
x
,
y
)
,
(
u
,
v
)
∈
X
2
(105)
(
x
,
y
)
≤
(
u
,
v
)
⟺
x
(
t
)
≤
u
(
t
)
,
y
(
t
)
≥
v
(
t
)
∀
t
∈
I
.
Obviously, for any (
x
,
y
)
,
(
z
,
t
)
∈
X
2, the element (
max
⁡
{
F
(
x
,
y
)
,
F
(
z
,
t
)
}
,
min
⁡
{
F
(
y
,
x
)
,
F
(
t
,
z
)
}
) is comparable with (
F
(
x
,
y
)
,
F
(
y
,
x
)
) and (
F
(
z
,
t
)
,
F
(
t
,
z
)
).
Define now the mappingF
:
X
×
X
→
X by
(106)
F
(
x
,
y
)
(
t
)
=
∫
a
b
‍
K
1
(
t
,
s
)
[
f
(
s
,
x
(
s
)
)
+
g
(
s
,
y
(
s
)
)
]
d
s
+
∫
a
b
‍
K
2
(
t
,
s
)
[
f
(
s
,
y
(
s
)
)
+
g
(
s
,
x
(
s
)
)
]
d
s
+
h
(
t
)
,
∀
t
∈
I
.
It is not difficult to prove, like in [38], that F has the mixed monotone property. Now for x
,
y
,
u
,
v
,
a
,
b
∈
X with x
≥
u
≥
a and y
≤
v
≤
b, we have
(107)
ρ
(
F
(
x
,
y
)
,
F
(
u
,
v
)
)
=
sup
⁡
t
∈
I
|
F
(
x
,
y
)
(
t
)
-
F
(
u
,
v
)
(
t
)
|
p
.
Let us first evaluate the expression in the right hand side. According to the computations done by Berinde in [37],
(108)
F
(
x
,
y
)
(
t
)
-
F
(
u
,
v
)
(
t
)
=
∫
a
b
‍
K
1
(
t
,
s
)
[
f
(
s
,
x
(
s
)
)
+
g
(
s
,
y
(
s
)
)
]
d
s
+
∫
a
b
‍
K
2
(
t
,
s
)
[
f
(
s
,
y
(
s
)
)
+
g
(
s
,
x
(
s
)
)
]
d
s
-
∫
a
b
‍
K
1
(
t
,
s
)
[
f
(
s
,
u
(
s
)
)
+
g
(
s
,
v
(
s
)
)
]
d
s
-
∫
a
b
‍
K
2
(
t
,
s
)
[
f
(
s
,
v
(
s
)
)
+
g
(
s
,
u
(
s
)
)
]
d
s
=
∫
a
b
‍
K
1
(
t
,
s
)
[
f
(
s
,
x
(
s
)
)
-
f
(
s
,
u
(
s
)
)
+
g
(
s
,
y
(
s
)
)
=
∫
a
b
‍
K
1
(
t
,
s
)
-
g
(
s
,
v
(
s
)
)
]
d
s
+
∫
a
b
‍
K
2
(
t
,
s
)
[
f
(
s
,
y
(
s
)
)
-
f
(
s
,
v
(
s
)
)
+
g
(
s
,
x
(
s
)
)
=
∫
a
b
‍
K
1
(
t
,
s
)
-
g
(
s
,
u
(
s
)
)
]
d
s
=
∫
a
b
‍
K
1
(
t
,
s
)
[
(
f
(
s
,
x
(
s
)
)
-
f
(
s
,
u
(
s
)
)
)
=
∫
a
b
‍
K
1
(
t
,
s
)
-
(
g
(
s
,
v
(
s
)
)
-
g
(
s
,
y
(
s
)
)
)
]
d
s
-
∫
a
b
‍
K
2
(
t
,
s
)
[
(
f
(
s
,
y
(
s
)
)
-
f
(
s
,
v
(
s
)
)
)
=
∫
a
b
‍
K
1
(
t
,
s
)
-
(
g
(
s
,
u
(
s
)
)
-
g
(
s
,
x
(
s
)
)
)
]
d
s
≤
∫
a
b
‍
K
1
(
t
,
s
)
[
λ
θ
(
x
(
s
)
-
u
(
s
)
)
+
μ
θ
(
v
(
s
)
-
y
(
s
)
)
]
d
s

-
∫
a
b
‍
K
2
(
t
,
s
)
[
λ
θ
(
y
(
s
)
-
v
(
s
)
)
+
μ
θ
(
u
(
s
)
-
x
(
s
)
)
]

d
s
.
Since the function θ is nondecreasing and x
≥
u and y
≤
v, we have
(109)
θ
(
x
(
s
)
-
u
(
s
)
)
≤
θ
(
sup
⁡
t
∈
I
|
x
(
t
)
-
u
(
t
)
|
)
=
θ
(
d
(
x
,
u
)
)
,
θ
(
v
(
s
)
-
y
(
s
)
)
≤
θ
(
sup
⁡
t
∈
I
|
y
(
t
)
-
v
(
t
)
|
)
=
θ
(
d
(
y
,
v
)
)
.
Hence, by (108), in view of the fact that K
2
(
t
,
s
)
≤
0, we obtain that
(110)
|
F
(
x
,
y
)
(
t
)
-
F
(
u
,
v
)
(
t
)
|
≤
∫
a
b
‍
K
1
(
t
,
s
)
[
λ
θ
(
d
(
x
,
u
)
)
+
μ
θ
(
d
(
y
,
v
)
)
]
d
s
-
∫
a
b
‍
K
2
(
t
,
s
)
[
λ
θ
(
d
(
y
,
v
)
)
+
μ
θ
(
d
(
x
,
u
)
)
]
d
s
,
as all quantities in the right hand side of (108) are nonnegative.
Now from (108) we have
(111)
|
F
(
x
,
y
)
(
t
)
-
F
(
u
,
v
)
(
t
)
|
p
≤
(
∫
a
b
‍
K
1
(
t
,
s
)
[
λ
θ
(
d
(
x
,
u
)
)
+
μ
θ
(
d
(
y
,
v
)
)
]
d
s
-
∫
a
b
‍
K
2
(
t
,
s
)
[
λ
θ
(
d
(
v
,
y
)
)
+
μ
θ
(
d
(
x
,
u
)
)
]
d
s
)
p
≤
2
p
-
1
(
(
∫
a
b
‍
K
1
(
t
,
s
)
d
s
)
p
(
λ
θ
(
d
(
x
,
u
)
)
+
μ
θ
(
d
(
v
,
y
)
)
)
p
+
(
-
∫
a
b
‍
K
2
(
t
,
s
)
d
s
)
p
×
(
λ
θ
(
d
(
v
,
y
)
)
+
μ
θ
(
d
(
x
,
u
)
)
)
p
+
(
-
∫
a
b
‍
K
2
(
t
,
s
)
d
s
)
p
)
≤
2
p
-
1
(
(
∫
a
b
‍
K
1
(
t
,
s
)
d
s
)
p
≤
2
p
-
1
×
2
p
-
1
(
λ
p
θ
(
d
(
x
,
u
)
)
p
+
μ
p
θ
(
d
(
v
,
y
)
)
p
)
≤
2
p
-
1
+
(
-
∫
a
b
‍
K
2
(
t
,
s
)
d
s
)
p
≤
2
p
-
1
×
2
p
-
1
(
λ
p
θ
(
d
(
v
,
y
)
)
p
+
μ
p
θ
(
d
(
x
,
u
)
)
p
)
(
∫
a
b
‍
K
1
(
t
,
s
)
d
s
)
p
)
≤
2
p
-
1
(
(
∫
a
b
‍
K
1
(
t
,
s
)
d
s
)
p
≤
2
p
-
1
×
2
p
-
1
(
λ
p
θ
(
ρ
(
x
,
u
)
)
+
μ
p
θ
(
ρ
(
v
,
y
)
)
)
≤
2
p
-
1
+
(
-
∫
a
b
‍
K
2
(
t
,
s
)
d
s
)
p
≤
2
p
-
1
×
2
p
-
1
(
λ
p
θ
(
ρ
(
v
,
y
)
)
+
μ
p
θ
(
ρ
(
x
,
u
)
)
)
(
∫
a
b
‍
K
1
(
t
,
s
)
d
s
)
p
)
≤
2
2
p
-
2
[
(
λ
p
(
∫
a
b
‍
K
1
(
t
,
s
)
d
s
)
p
≤
2
p
-
1
+
μ
p
(
-
∫
a
b
‍
K
2
(
t
,
s
)
d
s
)
p
)
θ
ρ
(
x
,
u
)
≤
2
p
-
1
+
(
μ
p
(
∫
a
b
‍
K
1
(
t
,
s
)
d
s
)
p
≤
2
p
-
1
+
λ
p
(
-
∫
a
b
‍
K
2
(
t
,
s
)
d
s
)
p
)
θ
ρ
(
v
,
y
)
(
λ
p
(
∫
a
b
‍
K
1
(
t
,
s
)
d
s
)
p
+
μ
p
(
-
∫
a
b
‍
K
2
(
t
,
s
)
d
s
)
p
)
]
.
So, we have
(112)
|
F
(
x
,
y
)
(
t
)
-
F
(
u
,
v
)
(
t
)
|
p
≤
2
2
p
-
2
[
(
λ
p
(
∫
a
b
‍
K
1
(
t
,
s
)
d
s
)
p
≤
2
2
p
-
2
+
μ
p
(
-
∫
a
b
‍
K
2
(
t
,
s
)
d
s
)
p
)
θ
ρ
(
x
,
u
)
≤
2
2
p
-
2
+
(
μ
p
(
∫
a
b
‍
K
1
(
t
,
s
)
d
s
)
p
≤
2
2
p
-
2
+
λ
p
(
-
∫
a
b
‍
K
2
(
t
,
s
)
d
s
)
p
)
θ
ρ
(
y
,
v
)
(
λ
p
(
∫
a
b
‍
K
1
(
t
,
s
)
d
s
)
p
+
μ
p
(
-
∫
a
b
‍
K
2
(
t
,
s
)
d
s
)
p
)
]
.
Similarly, one can obtain that
(113)
|
F
(
u
,
v
)
(
t
)
-
F
(
z
,
t
)
(
t
)
|
p
≤
2
2
p
-
2
[
(
λ
p
(
∫
a
b
‍
K
1
(
t
,
s
)
d
s
)
p
≤
2
2
p
-
2
+
μ
p
(
-
∫
a
b
‍
K
2
(
t
,
s
)
d
s
)
p
)
θ
ρ
(
u
,
z
)
≤
2
2
p
-
2
+
(
μ
p
(
∫
a
b
‍
K
1
(
t
,
s
)
d
s
)
p
≤
2
2
p
-
2
+
λ
p
(
-
∫
a
b
‍
K
2
(
t
,
s
)
d
s
)
p
)
θ
ρ
(
v
,
t
)
(
λ
p
(
∫
a
b
‍
K
1
(
t
,
s
)
d
s
)
p
+
μ
p
(
-
∫
a
b
‍
K
2
(
t
,
s
)
d
s
)
p
)
]
,
|
F
(
z
,
t
)
(
t
)
-
F
(
x
,
y
)
(
t
)
|
p
≤
2
2
p
-
2
[
(
λ
p
(
∫
a
b
‍
K
1
(
t
,
s
)
d
s
)
p
≤
2
2
p
-
2
+
μ
p
(
-
∫
a
b
‍
K
2
(
t
,
s
)
d
s
)
p
)
θ
ρ
(
x
,
z
)
≤
2
2
p
-
2
+
(
μ
p
(
∫
a
b
‍
K
1
(
t
,
s
)
d
s
)
p
≤
2
2
p
-
2
+
λ
p
(
-
∫
a
b
‍
K
2
(
t
,
s
)
d
s
)
p
)
θ
ρ
(
y
,
t
)
(
λ
p
(
∫
a
b
‍
K
1
(
t
,
s
)
d
s
)
p
]
.
Taking the supremum with respect to t and using (98) we get
(114)
G
(
F
(
x
,
y
)
,
F
(
u
,
v
)
,
F
(
z
,
t
)
)
=
max
⁡
{
sup
⁡
t
∈
I
|
F
(
x
,
y
)
(
t
)
-
F
(
u
,
v
)
(
t
)
|
p
,
sup
⁡
t
∈
I
|
F
(
u
,
v
)
(
t
)
-
F
(
z
,
t
)
(
t
)
|
p
,
sup
⁡
t
∈
I
|
F
(
z
,
t
)
(
t
)
-
F
(
x
,
y
)
(
t
)
|
p
}
≤
2
2
p
-
2
(
max
⁡
{
λ
p
,
μ
p
}
)
×
sup
⁡
t
∈
I
[
(
∫
a
b
‍
K
1
(
t
,
s
)
d
s
)
p
+
(
∫
a
b
‍
-
K
2
(
t
,
s
)
d
s
)
p
]
×
max
⁡
⁡
{
θ
(
ρ
(
x
,
u
)
)
+
θ
(
ρ
(
y
,
v
)
)
,
θ
(
ρ
(
u
,
z
)
)
+
θ
(
ρ
(
v
,
t
)
)
,
θ
(
ρ
(
x
,
z
)
)
+
θ
(
ρ
(
y
,
t
)
)
}
≤
2
2
p
-
2
(
max
⁡
{
λ
p
,
μ
p
}
)
×
sup
⁡
t
∈
I
[
(
∫
a
b
‍
K
1
(
t
,
s
)
d
s
)
p
+
(
∫
a
b
‍
-
K
2
(
t
,
s
)
d
s
)
p
]
×
θ
(
G
(
x
,
u
,
z
)
)
+
θ
(
G
(
y
,
v
,
t
)
)
≤
2
2
p
-
2
2
3
p
-
3
[
θ
(
G
(
x
,
u
,
z
)
)
+
θ
(
G
(
y
,
v
,
t
)
)
]
=
1
2
p
-
1
[
θ
(
G
(
x
,
u
,
z
)
)
+
θ
(
G
(
y
,
v
,
t
)
)
]
.
Now, since θ is nondecreasing, we have
(115)
θ
(
G
(
x
,
u
,
z
)
)
≤
θ
(
G
(
x
,
u
,
z
)
+
G
(
y
,
v
,
t
)
)
,
θ
(
G
(
y
,
v
,
t
)
)
≤
θ
(
G
(
x
,
u
,
z
)
+
G
(
y
,
v
,
t
)
)
and so
(116)
θ
(
G
(
x
,
u
,
z
)
)
+
θ
(
G
(
y
,
v
,
t
)
)
≤
2
θ
(
G
(
x
,
u
,
z
)
+
G
(
y
,
v
,
t
)
)
=
G
(
x
,
u
,
z
)
+
G
(
y
,
v
,
t
)
2
-
φ
(
G
(
x
,
u
,
z
)
+
G
(
y
,
v
,
t
)
2
)
,
by the definition of θ. Finally, from (114) we get that
(117)
G
(
F
(
x
,
y
)
,
F
(
u
,
v
)
,
F
(
z
,
t
)
)
≤
1
2
p
-
1
G
(
x
,
u
,
z
)
+
G
(
y
,
v
,
t
)
2
-
1
2
p
-
1
φ
(
G
(
x
,
u
,
z
)
+
G
(
y
,
v
,
t
)
2
)
.
Similarly, we can obtain that
(118)
G
(
F
(
y
,
x
)
,
F
(
v
,
u
)
,
F
(
t
,
z
)
)
≤
1
2
p
-
1
G
(
x
,
u
,
z
)
+
G
(
y
,
v
,
t
)
2
-
1
2
p
-
1
φ
(
G
(
x
,
u
,
z
)
+
G
(
y
,
v
,
t
)
2
)
,
which is just the contractive condition (69) in Corollary 43.
Now, let(
α
,
β
)
∈
X
2 be a coupled upper-lower solution of (96). Then we have
(119)
α
(
t
)
≤
F
(
α
,
β
)
(
t
)
,
β
(
t
)
≥
F
(
β
,
α
)
(
t
)
,
for all t
∈
I, which show that all hypotheses of Corollary 43 are satisfied.
This proves thatF has a coupled fixed point (
x
*
,
y
*
) in X
2.
Sinceα
≤
β, by Corollary 43 it follows that x
*
=
y
*; that is,
(120)
x
*
=
F
(
x
*
,
x
*
)
,
and therefore x
*
∈
C
(
I
,
R
) is the solution of the integral equation (96).



---

*Source: 568718-2014-05-18.xml*