International Journal of Analysis and Applications
ISSN 2291-8639
Volume 13, Number 2 (2017), 185-197
http://www.etamaths.com

GENERALIZED MEIR-KEELER TYPE ψ-CONTRACTIVE MAPPINGS AND

APPLICATIONS TO COMMON SOLUTION OF INTEGRAL EQUATIONS

HÜSEYİN IŞIK1,∗, MOHAMMAD IMDAD2, DURAN TURKOGLU3 AND NAWAB HUSSAIN4

Abstract. The goal of the present article to introduce the notion of generalized Meir-Keeler type

ψ-contractions and prove some coupled common fixed point results for such type of contractions.
The theorems proved herein extend, generalize and improve some results of the existing literature.

Several examples and an application to integral equations are also given in order to illustrate the

genuineness of our results.

1. Introduction and Preliminaries

The Meir-Keeler contraction defined in 1969 by Meir and Keeler [13] is one of the most significant
generalizations of Banach contraction principle [1]. Owing to it’s utility, generality and effectiveness,
the result of Meir and Keeler [13] remains a novel result in metric fixed point theory. In recent years,
many authors extended and generalized this result in different ways and by now there exists extensive
literature on this theme. To mention a few, we recall [3, 7, 15, 16] and references cited therein.

Guo and Lakshmikantham [8] established some coupled fixed point theorems which has attracted the
attention of many researchers (e.g. [5–7,9–12] and references therein). Bhaskar and Lakshmikantham [2]
introduced the notion of mixed monotone mapping to prove results on coupled fixed points. As an
application, they also proved the existence and uniqueness of solution for a periodic boundary value
problem associated to a first order ordinary differential equation. Recently, Ding et al. [6] introduced
the notion of weakly increasing mappings with two variables and established several coupled common
fixed point theorems for these mappings in ordered metric spaces.

In this paper, we introduce the notion of generalized Meir-Keeler type ψ-contractions and prove
some coupled common fixed point theorems for such contractions using weakly increasing property
instead of mixed monotone property. Our results extend, generalize and improve several results of the
existing literature. Several interesting consequences of our theorems are derived besides furnishing an
example. As an application of the results presented herein, we discuss the existence of the common
solution for a system of integral equations.

We start by recalling some definitions and notions. In the sequel, the letters R,R+and N will denote
the set of all real numbers, the set of all non negative real numbers and the set of all natural numbers,
respectively.

Definition 1.1 ( [2]). An element (x,y) ∈ X2 is said to be a coupled fixed point of the mapping
F : X2 → X if x = F(x,y) and y = F(y,x).

Definition 1.2 ( [6]). An element (x,y) ∈ X2 is called a coupled common fixed point of mappings
F,G : X2 → X if F(x,y) = G(x,y) = x and F(y,x) = G(y,x) = y. We denote the set of all coupled
common points of F and G by F (F,G) .

Definition 1.3 ( [2]). Let (X,�) be a partially ordered set and F : X2 → X. We say that F has the
mixed monotone property if F(x,y) is monotone nondecreasing in x and monotone nonincreasing in
y, that is, for any x,y ∈ X,

x1,x2 ∈ X, x1 � x2 ⇒ F(x1,y) � F(x2,y)

2010 Mathematics Subject Classification. 47H10.
Key words and phrases. coupled common fixed point; Meir-Keeler contraction; weakly increasing mapping; integral

equation.

c©2017 Authors retain the copyrights of
their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

185



186 IŞIK, IMDAD, TURKOGLU AND HUSSAIN

and

y1,y2 ∈ X, y1 � y2 ⇒ F(x,y2) � F(x,y1).

Definition 1.4 ( [6]). Let (X,�) be a partially ordered set. Two mappings F,G : X2 → X are said
to be weakly increasing if

F(x,y) � G(F(x,y),F(y,x)) and G(x,y) � F(G(x,y),G(y,x)),

hold for all (x,y) ∈ X2.

Example 1.1. Let X = [1, +∞) be endowed with the usual ordering ≤ and F,G : X2 → X be given
by F(x,y) = x2 + 2y and G(x,y) = 2x + y2. Then, for all (x,y) ∈ X2,

F(x,y) = x2 + 2y ≤ G(F(x,y),F(y,x))
= G(x2 + 2y,y2 + 2x) = 2(x2 + 2y) + (y2 + 2x)2

and

G(x,y) = 2x + y2 ≤ F(G(x,y),G(y,x))

= F(2x + y2, 2y + x2) =
(
2x + y2

)2
+ 2

(
2y + x2

)
.

Thus, F and G are two weakly increasing mappings with respect to ≤.

On the lines of [4], we denote by Ψ the family of all functions ψ : R+
4

→ R+ such that
(ψ1) ψ is nondecreasing and continuous in each coordinate;
(ψ2) ψ (t,t,t, t) ≤ t for all t > 0;
(ψ3) ψ (t1, t2, t3, t4) = 0 iff t1 = t2 = t3 = t4 = 0.

In the rest of the paper, we denote by (X,�,d) an ordered metric space wherein � is a partial
order on the set X while d is a metric on X. In addition, we say that (x,y) ∈ X2 is comparable to
(u,v) ∈ X2 if x � u and y � v or u � x and v � y. For brevity, we show this with (x,y) � (u,v) or
(x,y) � (u,v) .

(X2,δ) is a metric space under the following metric:

δ ((x,y) , (u,v)) := d (x,u) + d (y,v) ,

for all (x,y) , (u,v) ∈ X2. By the definition of δ, it is obvious that

δ ((x,y) , (u,v)) = δ ((y,x) , (v,u)) .

Definition 1.5. Let (X,�,d) be an ordered metric space and F,G : X2 → X be two mappings. We
say that (F,G) is a generalized Meir-Keeler type ψ-contraction pair if, for all ε > 0, there exists η > 0
such that, for all comparable pairs (x,y) , (u,v) ∈ X2,

ε ≤
1

2
M (x,y,u,v) ≤ ε + η (ε) ⇒ d (F (x,y) ,G (u,v)) < ε, (1.1)

where

M (x,y,u,v) = ψ




δ ((x,y) , (u,v)) ,δ ((x,y) , (F (x,y) ,F (y,x))) ,
δ ((u,v) , (G (u,v) ,G (v,u))) ,

1
2

[
δ ((x,y) , (G (u,v) ,G (v,u)))

+δ ((u,v) , (F (x,y) ,F (y,x)))

]



with ψ ∈ Ψ.

If we take F = G in the above definition, then we have:

Definition 1.6. Let (X,�,d) be an ordered metric space and F : X2 → X be a mapping. We say
that F is a generalized Meir-Keeler type ψ-contraction if, for all ε > 0, there exists η > 0 such that,
for all comparable pairs (x,y) , (u,v) ∈ X2,

ε ≤
1

2
MF (x,y,u,v) ≤ ε + η (ε) ⇒ d (F (x,y) ,F (u,v)) < ε,



GENERALIZED MEIR-KEELER CONTRACTIONS 187

where

MF (x,y,u,v) = ψ




δ ((x,y) , (u,v)) ,δ ((x,y) , (F (x,y) ,F (y,x))) ,
δ ((u,v) , (F (u,v) ,F (v,u))) ,

1
2

[
δ ((x,y) , (F (u,v) ,F (v,u)))

+δ ((u,v) , (F (x,y) ,F (y,x)))

]



with ψ ∈ Ψ.

Remark 1.1. Let (X,�,d) be an ordered metric space and let F,G : X2 → X be two given mappings.
If (F,G) is a generalized Meir-Keeler type ψ-contraction pair, then

d (F (x,y) ,G (u,v)) <
1

2
M (x,y,u,v) ,

for all comparable pairs (x,y) , (u,v) ∈ X2 when M (x,y,u,v) > 0. Also, if M (x,y,u,v) = 0, then
d (F (x,y) ,G (u,v)) = 0, that is,

d (F (x,y) ,G (u,v)) ≤
1

2
M (x,y,u,v) ,

for all comparable pairs (x,y) , (u,v) ∈ X2.

2. Main Results

Our main result is stated as follows:

Theorem 2.1. Let (X,�,d) be a complete ordered metric space, F,G : X2 → X be two weakly
increasing mappings with respect to � and (F,G) be a generalized Meir-Keeler type ψ-contraction pair.
If either F or G is continuous, then F and G have a coupled common fixed point.

Proof. Let x0,y0 ∈ X. Define two sequences {xn} and {yn} in X as follows:

x2n+1 = F(x2n,y2n), x2n+2 = G(x2n+1,y2n+1),

and

y2n+1 = F(y2n,x2n), y2n+2 = G(y2n+1,x2n+1),

for all n ∈ N0 where N0 = N∪{0} . Since F and G are weakly increasing, we have

x1 = F (x0,y0) � G (F (x0,y0) ,F (y0,x0)) = G (x1,y1)
= x2 � F (G (x1,y1) ,G (y1,x1)) = F (x2,y2) = x3 � ··· ,

and

y1 = F (y0,x0) � G (F (y0,x0) ,F (x0,y0)) = G (y1,x1)
= y2 � F (G (y1,x1) ,G (x1,y1)) = F (y2,x2) = y3 � ··· ,

so that the sequences {xn} and {yn} are nondecreasing.
Step 1. Now, we show that {xn} and {yn} are Cauchy sequences in (X,�,d).
Case 1. If for some n ∈ N0, xn = xn+1 and yn = yn+1, then xn+1 = xn+2 and yn+1 = yn+2. If not,
then for n = 2m where m ∈ N, by Remark 1.1, we have

d (xn+1,xn+2) = d (x2m+1,x2m+2)

= d (F(x2m,y2m),G (x2m+1,y2m+1))

<
1

2
M (x2m,y2m,x2m+1,y2m+1) ,



188 IŞIK, IMDAD, TURKOGLU AND HUSSAIN

where

M (x2m,y2m,x2m+1,y2m+1)

= ψ




δ ((x2m,y2m) , (x2m+1,y2m+1)) ,
δ ((x2m,y2m) , (F (x2m,y2m) ,F (y2m,x2m))) ,

δ ((x2m+1,y2m+1) , (G (x2m+1,y2m+1) ,G (y2m+1,x2m+1))) ,

1
2

[
δ ((x2m,y2m) , (G (x2m+1,y2m+1) ,G (y2m+1,x2m+1)))

+δ ((x2m+1,y2m+1) , (F (x2m,y2m) ,F (y2m,x2m)))

]



= ψ




δ ((x2m,y2m) , (x2m+1,y2m+1)) ,δ ((x2m,y2m) , (x2m+1,y2m+1)) ,
δ ((x2m+1,y2m+1) , (x2m+2,y2m+2)) ,

1
2

[
δ ((x2m,y2m) , (x2m+2,y2m+2))

+δ ((x2m+1,y2m+1) , (x2m+1,y2m+1))

]



= ψ

{
0, 0,δ ((x2m+1,y2m+1) , (x2m+2,y2m+2)) ,

1

2
[δ ((x2m,y2m) , (x2m+2,y2m+2))]

}
.

Since ψ is nondecreasing, we deduce

M (x2m,y2m,x2m+1,y2m+1)

≤ ψ




δ ((x2m+1,y2m+1) , (x2m+2,y2m+2)) ,δ ((x2m+1,y2m+1) , (x2m+2,y2m+2)) ,
δ ((x2m+1,y2m+1) , (x2m+2,y2m+2)) ,

1
2

[δ ((x2m,y2m) , (x2m+1,y2m+1)) + δ ((x2m+1,y2m+1) , (x2m+2,y2m+2))]




≤ψ
{
δ ((x2m+1,y2m+1) , (x2m+2,y2m+2)) ,δ ((x2m+1,y2m+1) , (x2m+2,y2m+2)) ,
δ ((x2m+1,y2m+1) , (x2m+2,y2m+2)) ,δ ((x2m+1,y2m+1) , (x2m+2,y2m+2))

}
≤ δ ((x2m+1,y2m+1) , (x2m+2,y2m+2)) = δ ((xn+1,yn+1) , (xn+2,yn+2)) .

Hence, it follows that

d (xn+1,xn+2) <
1

2
δ ((xn+1,yn+1) , (xn+2,yn+2)) . (2.1)

Similarly, we can also show that

d (yn+1,yn+2) <
1

2
δ ((xn+1,yn+1) , (xn+2,yn+2)) . (2.2)

Thus, from (2.1) and (2.2)

δ ((xn+1,yn+1) , (xn+2,yn+2)) < δ ((xn+1,yn+1) , (xn+2,yn+2)) ,

which is a contradiction. Hence we must have xn+1 = xn+2 and yn+1 = yn+2 when n is even. By
similar arguments we can show that this equality holds also when n is odd. Therefore, in any case for
all those n for which xn = xn+1 and yn = yn+1 holds, we always obtain xn+1 = xn+2 and yn+1 = yn+2.
Repeating above process inductively, one obtains xn = xn+k and yn = yn+k for all k ∈ N. Hence {xn}
and {yn} are constant sequences so that {xn} and {yn} are Cauchy sequences in (X,�,d).
Case 2. Suppose that xn 6= xn+1 and yn 6= yn+1 for all n ∈ N0. Then, for n = 2m + 1, using Remark
1.1, we have

d (xn,xn+1) = d (x2m+1,x2m+2)

= d (F(x2m,y2m),G (x2m+1,y2m+1))

<
1

2
M (x2m,y2m,x2m+1,y2m+1) , (2.3)

where

M (x2m,y2m,x2m+1,y2m+1)

= ψ




δ ((x2m,y2m) , (x2m+1,y2m+1)) ,δ ((x2m,y2m) , (x2m+1,y2m+1)) ,
δ ((x2m+1,y2m+1) , (x2m+2,y2m+2)) ,

1
2

[δ ((x2m,y2m) , (x2m+2,y2m+2)) + δ ((x2m+1,y2m+1) , (x2m+1,y2m+1))]


 .

Since,

δ ((x2m+1,y2m+1) , (x2m+1,y2m+1)) = d (x2m+1,x2m+1) + d (y2m+1,y2m+1) = 0,



GENERALIZED MEIR-KEELER CONTRACTIONS 189

and

δ ((x2m,y2m) , (x2m+2,y2m+2))

= d (x2m,x2m+2) + d (y2m,y2m+2)

≤ d (x2m,x2m+1) + d (x2m+1,x2m+2) + d (y2m,y2m+1) + d (y2m+1,y2m+2)
= δ ((x2m,y2m) , (x2m+1,y2m+1)) + δ ((x2m+1,y2m+1) , (x2m+2,y2m+2)) ,

so we get

M (x2m,y2m,x2m+1,y2m+1)

≤ ψ




δ ((x2m,y2m) , (x2m+1,y2m+1)) ,δ ((x2m,y2m) , (x2m+1,y2m+1)) ,
δ ((x2m+1,y2m+1) , (x2m+2,y2m+2)) ,

1
2

[δ ((x2m,y2m) , (x2m+1,y2m+1)) + δ ((x2m+1,y2m+1) , (x2m+2,y2m+2))]


 . (2.4)

If δ ((x2m,y2m) , (x2m+1,y2m+1)) ≤ δ ((x2m+1,y2m+1) , (x2m+2,y2m+2)) for some m, since ψ is nonde-
creasing, we obtain

M (xn−1,yn−1,xn,yn)

= M (x2m,y2m,x2m+1,y2m+1)

≤ ψ




δ ((x2m,y2m) , (x2m+1,y2m+1)) ,δ ((x2m,y2m) , (x2m+1,y2m+1)) ,
δ ((x2m+1,y2m+1) , (x2m+2,y2m+2)) ,

1
2
[δ ((x2m,y2m) , (x2m+1,y2m+1)) + δ ((x2m+1,y2m+1) , (x2m+2,y2m+2))]




≤ ψ
{
δ ((x2m+1,y2m+1) , (x2m+2,y2m+2)) ,δ ((x2m+1,y2m+1) , (x2m+2,y2m+2)) ,
δ ((x2m+1,y2m+1) , (x2m+2,y2m+2)) ,δ ((x2m+1,y2m+1) , (x2m+2,y2m+2))

}
≤ δ ((x2m+1,y2m+1) , (x2m+2,y2m+2)) = δ ((xn,yn) , (xn+1,yn+1)) ,

and so from (2.3)

d (xn,xn+1) <
1

2
δ ((xn,yn) , (xn+1,yn+1)) . (2.5)

Similarly, we can show that

d (yn,yn+1) <
1

2
δ ((xn,yn) , (xn+1,yn+1)) . (2.6)

Thus, from (2.5) and (2.6), we get

δ ((xn,yn) , (xn+1,yn+1)) < δ ((xn,yn) , (xn+1,yn+1)) ,

which is a contradiction. Hence, it must be

δ ((x2m+1,y2m+1) , (x2m+2,y2m+2)) = δ ((xn,yn) , (xn+1,yn+1))

< δ ((xn−1,yn−1) , (xn,yn))

= δ ((x2m,y2m) , (x2m+1,y2m+1)) ,

for all n ∈ N0. Set δn := {δ ((xn,yn) , (xn+1,yn+1))} , then the sequence {δn} is decreasing and
bounded below. Thus, there exists r ≥ 0 such that limn→∞δn = r. Notice that

r = inf {δn : n ∈ N0} . (2.7)
We now prove that r = 0. If not, then by (2.4), we deduce that

lim
n→∞

M (xn−1,yn−1,xn,yn) = r.

Then there exists a positive integer p such that

ε ≤
1

2
M (xp−1,yp−1,xp,yp) < ε + η (ε) ,

where ε = r/2. Owing to the condition (1.1), we have

d (F(xp−1,yp−1),G (xp,yp)) < ε,

which implies

d (xp,xp+1) < ε.



190 IŞIK, IMDAD, TURKOGLU AND HUSSAIN

Similarly, we can obtain that

d (yp,yp+1) < ε.

Summing the two foregoing inequalities, we get

δ ((xp,yp) , (xp+1,yp+1)) < 2ε = r,

which contradicts (2.7) for n = p. Thus, ε = r/2 = 0, that is,

lim
n→∞

δn = lim
n→∞

[d (xn,xn+1) + d (yn,yn+1)] = 0. (2.8)

Now, we prove that {xn} and {yn} are Cauchy sequences. It is sufficient to show that {x2n} and {y2n}
are Cauchy sequences in (X,d). Suppose, to the contrary, that at least one of {x2n} or {y2n} is not
Cauchy sequence. Then, there exists an ε > 0 for which we can find subsequences {x2mk},{x2nk} of
{x2n} and {y2mk},{y2nk} of {y2n} such that nk is the smallest index for which nk > mk > k and

d(x2nk,x2mk ) + d(y2nk,y2mk ) ≥ ε and d(x2nk−1,x2mk ) + d(y2nk−1,y2mk ) < ε. (2.9)

Using the triangular inequality and (2.9), we get

ε ≤ d(x2nk,x2mk ) + d(y2nk,y2mk )
≤ d(x2mk,x2nk−1) + d(x2nk−1,x2nk ) + d(y2mk,y2nk−1) + d(y2nk−1,y2nk )
< ε + δ2nk−1.

Taking k →∞ in the above inequality and using (2.8), we deduce

lim
k→∞

[d(x2nk,x2mk ) + d(y2nk,y2mk )] = ε. (2.10)

Again, by the triangle inequality, we get

d(x2nk,x2mk ) + d(y2nk,y2mk ) ≤ d(x2nk,x2nk+1) + d(x2nk+1,x2mk−1) + d(x2mk−1,x2mk )
+ d(y2nk,y2nk+1) + d(y2nk+1,y2mk−1) + d(y2mk−1,y2mk )

≤ δ2nk + δ2mk−1 + d(x2nk+1,x2mk ) + d(x2mk,x2mk−1)
+ d(y2nk+1,y2mk ) + d(y2mk,y2mk−1)

≤ δ2nk + 2δ2mk−1 + d(x2nk+1,x2nk ) + d(x2nk,x2mk )
+ d(y2nk+1,y2nk ) + d(y2nk,y2mk )

= 2δ2nk + 2δ2mk−1 + d(x2nk,x2mk ) + d(y2nk,y2mk ).

Letting k →∞ in the above inequality besides using (2.8) and (2.10), we have

limk→∞ [d(x2nk+1,x2mk−1) + d(y2nk+1,y2mk−1)] = ε, and
limk→∞ [d(x2nk+1,x2mk ) + d(y2nk+1,y2mk )] = ε.

(2.11)

On the other hand, we also obtain

d(x2nk,x2mk−1) + d(y2nk,y2mk−1) ≤ d(x2nk,x2mk ) + d(x2mk,x2mk−1)
+d(y2nk,y2mk ) + d(y2mk,y2mk−1)

= d(x2nk,x2mk ) + d(y2nk,y2mk ) + δ2mk−1,

which implies

lim
k→∞

[d(x2nk,x2mk−1) + d(y2nk,y2mk−1)] ≤ ε. (2.12)

Since (x2nk,y2nk ) � (x2mk−1,y2mk−1) for nk > mk, using Remark 1.1, we have

d (x2nk+1,x2mk ) = d (F(x2nk,y2nk ),G (x2mk−1,y2mk−1))

<
1

2
M (x2nk,y2nk,x2mk−1,y2mk−1) , (2.13)



GENERALIZED MEIR-KEELER CONTRACTIONS 191

where

M (x2nk,y2nk,x2mk−1,y2mk−1)

= ψ




δ ((x2nk,y2nk ) , (x2mk−1,y2mk−1)) ,
δ ((x2nk,y2nk ) , (F(x2nk,y2nk ),F(y2nk,x2nk ))) ,

δ ((x2mk−1,y2mk−1) , (G (x2mk−1,y2mk−1) ,G (y2mk−1,x2mk−1))) ,

1
2

[
δ ((x2nk,y2nk ) , (G (x2mk−1,y2mk−1) ,G (y2mk−1,x2mk−1)))

+δ ((x2mk−1,y2mk−1) , (F(x2nk,y2nk ),F(y2nk,x2nk )))

]



= ψ




δ ((x2nk,y2nk ) , (x2mk−1,y2mk−1)) ,δ ((x2nk,y2nk ) , (x2nk+1,y2nk+1)) ,
δ ((x2mk−1,y2mk−1) , (x2mk,y2mk )) ,

1
2

[δ ((x2nk,y2nk ) , (x2mk,y2mk )) + δ ((x2mk−1,y2mk−1) , (x2nk+1,y2nk+1))]


 .

By a similar method, we can also show that

d (y2nk+1,y2mk ) <
1

2
M (x2nk,y2nk,x2mk−1,y2mk−1) . (2.14)

Summing the inequalities (2.13) and (2.14), we get

d (x2nk+1,x2mk ) + d (y2nk+1,y2mk ) < M (x2nk,y2nk,x2mk−1,y2mk−1) .

Now, using (2.8), (2.10), (2.11) and (2.12) as k →∞ in the above inequality, we deduce

ε < lim
k→∞

M (x2nk,y2nk,x2mk−1,y2mk−1)

= ψ{ lim
k→∞

δ ((x2nk,y2nk ) , (x2mk−1,y2mk−1)) , 0, 0,ε}

≤ ψ{ε, 0, 0,ε}≤ ε,

which is a contradiction. Therefore, {xn} and {yn} are Cauchy sequences in X.
Step 2. Now, we prove the existence of coupled common fixed point of F and G.
Owing to the completeness of (X,d) , there exist x,y ∈ X such that

lim
n→∞

xn = x and lim
n→∞

yn = y. (2.15)

Without loss of generality, we assume that F is continuous. Now, we have

x = lim
n→∞

x2n+1 = lim
n→∞

F(x2n,y2n) = F
(

lim
n→∞

x2n, lim
n→∞

y2n

)
= F (x,y) .

and

y = lim
n→∞

y2n+1 = lim
n→∞

F(y2n,x2n) = F
(

lim
n→∞

y2n, lim
n→∞

x2n

)
= F (y,x) .

We now assert that d (x,G (x,y)) = d (y,G (y,x)) = 0. To establish the claim, assume that d (x,G (x,y)) >
0 and d (y,G (y,x)) > 0. Since (x,y) ∈ X2 is comparable to its own, making use of Remark 1.1, we
obtain

d (x,G (x,y)) = d (F (x,y) ,G (x,y)) <
1

2
M (x,y,x,y) , (2.16)

where

M (x,y,x,y) = ψ




δ ((x,y) , (x,y)) ,δ ((x,y) , (F(x,y),F(y,x))) ,
δ ((x,y) , (G (x,y) ,G (y,x))) ,

1
2

[δ ((x,y) , (G (x,y) ,G (y,x))) + δ ((x,y) , (F(x,y),F(y,x)))]




= ψ

{
0, 0,δ ((x,y) , (G (x,y) ,G (y,x))) ,

1

2
δ ((x,y) , (G (x,y) ,G (y,x)))

}
≤ δ ((x,y) , (G (x,y) ,G (y,x))) .

Similarly, we have

d (y,G (y,x)) <
1

2
M (x,y,x,y) . (2.17)



192 IŞIK, IMDAD, TURKOGLU AND HUSSAIN

Thus, it follows from (2.16) and (2.17) that

d (x,G (x,y)) + d (y,G (y,x)) < M (x,y,x,y)

≤ δ ((x,y) , (G (x,y) ,G (y,x)))
= d (x,G (x,y)) + d (y,G (y,x)) ,

which implies d (x,G (x,y)) = d (y,G (y,x)) = 0. Hence, x = F (x,y) = G (x,y) and y = F (y,x) =
G (y,x) . �

Now, we furnish the following example which illustrates the results of Theorem 2.1.

Example 2.1. Let X = [0, 1] be equipped with the usual metric and the partial order defined by

x � y ⇐⇒ y ≤ x.

Define the function ψ : R+
4

→ R+ by ψ (t1, t2, t3, t4) = max{t1, t2, t3, t4} and the mappings F,G :
X2 → X by F (x,y) = x+y

7
and G (x,y) = x+y

6
. Then, it is easy to see that F and G are weakly

increasing with respect to � .
Also, (F,G) is a generalized Meir-Keeler type ψ-contraction. Indeed, for all comparable (x,y) , (u,v) ∈

X2

d (F (x,y) ,G (u,v)) =

∣∣∣∣x + y7 − u + v6
∣∣∣∣

≤
1

7
(|x−u| + |y −v|)

=
1

7
δ ((x,y) , (u,v)) ≤

1

7
M (x,y,u,v)

<
1

7
· 2 (ε + η (ε)) < ε,

which holds if we choose η (ε) < 5
2
ε. Thus, it can easily see that all the hypotheses of Theorem 2.1 are

fulfilled. Therefore, F and G have a coupled common fixed point which is (0, 0) .

Definition 2.1. Let (X,�,d) be an ordered metric space. We say that (X,�,d) is regular if for
non-decreasing sequence {xn} with d(xn,x) → 0 implies that xn � x for all n.

In our next theorem, we replace the continuity of F or G in Theorem 2.1 with the regularity of
(X,�,d).

Theorem 2.2. Let (X,�,d) be a complete ordered metric space, F,G : X2 → X be two weakly
increasing mappings with respect to � and (F,G) be a generalized Meir-Keeler type ψ-contraction pair.
If (X,�,d) is regular, then F and G have a coupled common fixed point.

Proof. We define sequences {xn} and {yn} as in Theorem 2.1. Proceeding on the lines of the proof of
Theorem 2.1, we can show that the sequences {xn} and {yn} are non-decreasing and limn→∞xn = x
and limn→∞yn = y. Then, since (X,�,d) is regular, we conclude that (xn,yn) is comparable to (x,y)
for all n ∈ N0. Now, by Remark 1.1, we obtain

d (x2n+1,G (x,y)) = d (F(x2n,y2n),G (x,y)) <
1

2
M (x2n,y2n,x,y) , (2.18)

where

M (x2n,y2n,x,y)

= ψ




δ ((x2n,y2n) , (x,y)) ,δ ((x2n,y2n) , (F(x2n,y2n),F(y2n,x2n))) ,
δ ((x,y) , (G (x,y) ,G (y,x))) ,

1
2

[
δ ((x2n,y2n) , (G (x,y) ,G (y,x)))

+δ((x,y) , (F(x2n,y2n),F(y2n,x2n)))

]



= ψ




δ ((x2n,y2n) , (x,y)) ,δ ((x2n,y2n) , (x2n+1,y2n+1)) ,
δ ((x,y) , (G (x,y) ,G (y,x))) ,

1
2

[δ ((x2n,y2n) , (G (x,y) ,G (y,x))) + δ((x,y) , (x2n+1,y2n+1))]


 .



GENERALIZED MEIR-KEELER CONTRACTIONS 193

Similarly, we have

d (y2n+1,G (y,x)) = d (F(y2n,x2n),G (y,x)) <
1

2
M (x2n,y2n,x,y) . (2.19)

Thus it follows from (2.18) and (2.19) that

d (x2n+1,G (x,y)) + d (y2n+1,G (y,x)) < M (x2n,y2n,x,y) .

Letting n →∞ in the above inequality, we get

d (x,G (x,y)) + d (y,G (y,x)) < ψ

{
0, 0,δ ((x,y) , (G (x,y) ,G (y,x))) ,

1
2
δ ((x,y) , (G (x,y) ,G (y,x)))

}
≤ δ ((x,y) , (G (x,y) ,G (y,x)))
= d (x,G (x,y)) + d (y,G (y,x)) ,

which implies d(x,G (x,y)) = 0 and d (y,G (y,x)) = 0 that is, x = G (x,y) and y = G (y,x) .
Since (x,y) ∈ X2 is comparable to its own, using of Remark 1.1, we obtain

d (F (x,y) ,x) = d (F (x,y) ,G (x,y)) <
1

2
M (x,y,x,y) , (2.20)

where

M (x,y,x,y) = ψ




δ ((x,y) , (x,y)) ,δ ((x,y) , (F(x,y),F(y,x))) ,
δ ((x,y) , (G (x,y) ,G (y,x))) ,

1
2

[δ ((x,y) , (G (x,y) ,G (y,x))) + δ ((x,y) , (F(x,y),F(y,x)))]




= ψ

{
0,δ ((x,y) , (F(x,y),F(y,x))) , 0,

1

2
δ ((x,y) , (F(x,y),F(y,x)))

}
≤ δ ((x,y) , (F(x,y),F(y,x))) .

Similarly, we have

d (F (y,x) ,y) <
1

2
M (x,y,x,y) . (2.21)

Thus, it follows from (2.20) and (2.21) that

d (F (x,y) ,x) + d (F (y,x) ,y) < M (x,y,x,y)

≤ δ ((x,y) , (F(x,y),F(y,x)))
= d (x,F (x,y)) + d (y,F (y,x)) ,

which implies d (x,F (x,y)) = d (y,F (y,x)) = 0. Therefore, x = F (x,y) = G (x,y) and y = F (y,x) =
G (y,x) . �

Definition 2.2. Let (X,�,d) be an ordered metric space and F,G : X2 → X be two mappings. We
say that (F,G) is a generalized Meir-Keeler type contraction if, for all ε > 0, there exists η > 0 such
that, for all comparable (x,y) , (u,v) ∈ X2,

ε ≤
1

2
Mmax (x,y,u,v)≤ ε + η (ε)⇒ d (F (x,y) ,G (u,v)) < ε, (2.22)

where

Mmax (x,y,u,v) = max




δ ((x,y) , (u,v)) ,δ ((x,y) , (F (x,y) ,F (y,x))) ,
δ((u,v) , (G (u,v) ,G (v,u))),

1
2

[
δ ((x,y) , (G (u,v) ,G (v,u)))
+δ (u,v) , (F (x,y) ,F (y,x))

]

 .

If we take ψ (t1, t2, t3, t4) = max{t1, t2, t3, t4} in Theorem 2.1, we have the following result.

Corollary 2.1. Let (X,�,d) be a complete ordered metric space, F,G : X2 → X be weakly increasing
mappings with respect to � and (F,G) be a generalized Meir-Keeler type contraction. Assume that the
following conditions are satisfied:

(a) F (or G) is continuous or
(b) (X,�,d) is regular.

Then F and G have a coupled common fixed point.



194 IŞIK, IMDAD, TURKOGLU AND HUSSAIN

Proposition 2.1. Let (X,�,d) be an ordered metric space and let F,G : X2 → X be two given
mappings. If the following contraction is satisfied, then (F,G) is a generalized Meir-Keeler type con-
traction:

d (F (x,y) ,G (u,v)) ≤
k

2
Mmax (x,y,u,v) , k ∈ [0, 1).

Proof. Assume that the above inequality holds. Then, for all ε > 0, we can easily show that (2.22) is
satisfied with η (ε) =

(
1
k
− 1
)
ε. �

Definition 2.3. Let (X,�) be an ordered set and F : X2 → X. We say that F is nondecreasing if,
for any x,y ∈ X,

x1,x2 ∈ X, x1 � x2 ⇒ F(x1,y) � F(x2,y)
and

y1,y2 ∈ X, y1 � y2 ⇒ F(x,y1) � F(x,y2).

If we choose F = G in Theorem 2.1, we have the following corollary.

Corollary 2.2. Let (X,�,d) be a complete ordered metric space, F : X2 → X be a nondecreasing
mapping and F be a generalized Meir-Keeler type ψ-contraction. Assume that the following conditions
hold:

(a) F is continuous or
(b) (X,�,d) is regular.

Then F has a coupled fixed point.

We denote by Φ the family of all functions φ : R+ → R+ such that
(φ1) φ is nondecreasing and right continuous;
(φ2) φ (0) = 0 and φ (t) > 0 for any t > 0.

Theorem 2.3. Let (X,�,d) be a complete ordered metric space and F,G : X2 → X be two weakly
increasing mappings with respect to �. Assume that, for all ε > 0, there exists η > 0 such that, for all
comparable (x,y) , (u,v) ∈ X2,

ε ≤ φ
(

1

2
M (x,y,u,v)

)
≤ ε + η (ε)⇒ φ (d (F (x,y) ,G (u,v))) < ε, (2.23)

where φ ∈ Φ. Then (F,G) is a generalized Meir-Keeler type ψ-contraction.

Proof. Fix ε > 0. Since φ (ε) > 0, there exists γ > 0 such that, for all comparable (u,v) , (w,z) ∈ X2,

φ (ε) ≤ φ
(

1

2
M (u,v,w,z)

)
≤ φ (ε) + γ ⇒ φ (d (F (x,y) ,G (u,v))) < φ (ε) . (2.24)

Due to the right continuity of φ, there exists η > 0 such that φ (ε + η) < φ (ε) +γ. For any comparable
(x,y) , (u,v) ∈ X2, such that

ε ≤
1

2
M (x,y,u,v) ≤ ε + η.

Since φ is nondecreasing, we have:

φ (ε) ≤ φ
(

1

2
M (x,y,u,v)

)
< φ (ε + η) < φ (ε) + γ,

By (2.24), we obtain φ (d (F (x,y) ,G (u,v))) < φ (ε) which implies d (F (x,y) ,G (u,v)) < ε, as φ is
nondecreasing. This completes the proof. �

Corollary 2.3. Let (X,�,d) be a complete ordered metric space and F,G : X2 → X be two weakly
increasing mappings with respect to �. Assume that, for all ε > 0, there exists η > 0 such that, for
every comparable pair (x,y) , (u,v) ∈ X2,

ε ≤
∫ 1

2
M(x,y,u,v)

0

θ (t) dt ≤ ε + η (ε)⇒
∫ d(F (x,y),G(u,v))

0

θ (t) dt < ε,

where θ is a locally integrable function from R+ into itself satisfying
∫ s

0
θ (t) dt > 0 for all s > 0.

Suppose also that the following conditions are satisfied:



GENERALIZED MEIR-KEELER CONTRACTIONS 195

(a) F (or G) is continuous or
(b) (X,�,d) is regular.

Then F and G have a coupled common fixed point.

Corollary 2.4. Let (X,�,d) be a complete ordered metric space and F,G : X2 → X be two weakly
increasing mappings with respect to �. Assume that, for all comparable (x,y) , (u,v) ∈ X2,∫ d(F (x,y),G(u,v))

0

θ (t) dt ≤ k
∫ 1

2
M(x,y,u,v)

0

θ (t) dt,

where k ∈ (0, 1) and the function θ is defined as in Corollary 2.3. Suppose also that the following
conditions are satisfied:

(a) F (or G) is continuous or
(b) (X,�,d) is regular.

Then F and G have a coupled common fixed point.

Proof. For any ε > 0, choosing η (ε) =
(

1
k
− 1
)
ε and applying Corollary 2.3, the desired result is

obtained. �

3. An Application

Consider the following integral equations:

x (s) =
∫ b
a
H1 (s,r,x (r) ,y (r)) dr,

y (s) =
∫ b
a
H1 (s,r,y (r) ,x (r)) dr,

(3.1)

and
x (s) =

∫ b
a
H2 (s,r,x (r) ,y (r)) dr,

y (s) =
∫ b
a
H2 (s,r,y (r) ,x (r)) dr,

(3.2)

where s ∈ I = [a,b] , H1,H2 : I × I ×R×R → R and b > a ≥ 0.
In this section, we present an existence theorem for a common solution to (3.1) and (3.2) that

belongs to X := C(I,R) (the set of continuous functions defined on I) by using the obtained result in
Corollary 2.1.

We consider the operators F,G : X2 → X given by

F (x,y) (s) =

∫ b
a

H1 (s,r,x (r) ,y (r)) dr, x,y ∈ X, s ∈ I,

and

G (x,y) (s) =

∫ b
a

H2 (s,r,x (r) ,y (r)) dr, x,y ∈ X, s ∈ I.

Then the existence of a common solution to (3.1) and (3.2) is equivalent to the existence of a coupled
common fixed point of F and G.

It is well known that X endowed with the metric d defined by d(x,y) = sups∈I|x (s) −y (s) | for all
x,y ∈ X, forms a complete metric space. Also, equip X with the partial order � given by

x,y ∈ X, x � y ⇔ x(s) ≤ y(s), ∀s ∈ I.
Recall that in [14], it is proved that (X,�,d) is regular.
Suppose that the following conditions hold.

(A) H1,H2 : I × I ×R×R → R are continuous;
(B) for all s,r ∈ I and x,y ∈ X, we have

H1 (s,r,x (r) ,y (r))≤H2
(
s,r,

∫ b
a
H1 (r,τ,x (τ) ,y (τ)) dτ,

∫ b
a
H1 (r,τ,y (τ) ,x (τ)) dτ

)
and

H2 (s,r,x (r) ,y (r))≤H1
(
s,r,

∫ b
a
H2 (r,τ,x (τ) ,y (τ)) dτ,

∫ b
a
H2 (r,τ,y (τ) ,x (τ)) dτ

)
(C) for all comparable (x,y) , (u,v) ∈ X2 and for every s,r ∈ I, we have



196 IŞIK, IMDAD, TURKOGLU AND HUSSAIN

|H1 (s,r,x (r) ,y (r)) −H2 (s,r,u (r) ,v (r))|
2 ≤

k2

4
γ (s,r) [|x (r) −u (r)| + |y (r) −v (r)|]2 ,

where k ∈ [0, 1) and γ : I2 → R+ is a continuous function satisfying sups∈I
∫ b
a
γ (s,r) ≤ 1/ (b−a) .

Theorem 3.1. Assume that conditions (A)-(C) are satisfied. Then, integral equations (3.1) and (3.2)
have a common solution in X.

Proof. From condition (B), the mappings F and G are weakly increasing with respect to �.
Let (x,y) is comparable to (u,v) . Then, by (C), for all s ∈ I, we deduce

|F (x,y) (s) −G (u,v) (s)|2 ≤

(∫ b
a

|H1 (s,r,x (r) ,y (r)) −H2 (s,r,u (r) ,v (r))|dr

)2

≤
∫ b
a

12dr

∫ b
a

|H1 (s,r,x (r) ,y (r)) −H2 (s,r,u (r) ,v (r))|
2
dr

≤ (b−a)
∫ b
a

k2

4
γ (s,r) [|x (r) −u (r)| + |y (r) −v (r)|]2 dr

≤
k2

4
(b−a)

∫ b
a

γ (s,r) [d (x,u) + d (y,v)]
2
dr

≤
k2

4
(b−a) sup

s∈I

(∫ b
a

γ (s,r) dr

)
[δ ((x,y) , (u,v))]

2

≤
[
k

2
δ ((x,y) , (u,v))

]2
≤
[
k

2
Mmax (x,y,u,v)

]2
,

where

Mmax (x,y,u,v) = max




δ ((x,y) , (u,v)) ,δ ((x,y) , (F (x,y) ,F (y,x))) ,
δ((u,v) , (G (u,v) ,G (v,u))),

1
2

[
δ ((x,y) , (G (u,v) ,G (v,u)))
+δ (u,v) , (F (x,y) ,F (y,x))

]

 .

Therefore, we obtain (
sup
s∈I
|F (x,y) (s) −G (u,v) (s)|

)2
≤
[
k

2
Mmax (x,y,u,v)

]2
,

and so

d (F (x,y) ,G (u,v)) ≤
k

2
Mmax (x,y,u,v) . (3.3)

Hence, by Proposition 2.1, (F,G) is a generalized Meir-Keeler type contraction. Therefore, from
Corollary 2.1, F and G have a coupled common fixed point, that is, integral equations (3.1) and (3.2)
have a common solution in X. �

References

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3 (1922), 133-181.

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1Department of Mathematics, Faculty of Science and Arts, Muş Alparslan University, Muş 49100, Turkey

2Department of Mathematics, Aligarh Muslim University, Aligarh, Uttar Pradesh, 202002, India

3Department of Mathematics, Faculty of Science, University of Gazi, 06500-Teknikokullar, Ankara,
Turkey

4Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah, 21589, Saudi Arabia

∗Corresponding author: isikhuseyin76@gmail.com


	1. Introduction and Preliminaries
	2. Main Results
	3. An Application
	References