In this section, we establish some coupled fixed point results by considering maps on metric spaces endowed with partial order.
Denote by Ψ the family of nondecreasing functions
such that
for all
, where
is the nth iterate of ψ satisfying (i)
, (ii)
for all
and (iii)
for all
.
Lemma 3.1
If
is nondecreasing and right continuous, then
as
for all
if and only if
for all
.
Definition 3.2 Let
be a partially ordered metric space and
be a mapping. Then a map
F is said to be

contractive if there exist two functions
and
such that
for all
with
and
.
Definition 3.3 Let
and
be two mappings. Then
F is said to be

admissible if
for all
.
Theorem 3.4
Let
be a partially ordered set and suppose there is a metric
d
on
X
such that
is a complete metric space.
Let
be a mapping having the mixed monotone property of
X.
Suppose that there exist
and
such that for
,
the following holds:
for all
and
.
Suppose also that
 (i)
F
is
admissible,
 (ii)
there exist
such that
 (iii)
If there exist
such that
and
,
then
F
has a coupled fixed point;
that is,
there exist
such that
Proof Let
be such that
and
and
(say) and
(say). Let
be such that
and
. Continuing this process, we can construct two sequences
and
in
X as follows:
for all
. We will show that
for all
. We will use the mathematical induction. Let
. Since
and
and as
and
, we have
and
. Thus, (3.2) hold for
. Now suppose that (3.2) hold for some fixed
n,
. Then, since
and
and by the mixed monotone property of
F, we have
From above, we conclude that
Thus, by the mathematical induction, we conclude that (3.2) hold for all
. If for some
n we have
, then
and
; that is,
F has a coupled fixed point. Now, we assumed that
for all
. Since
F is
admissible, we have
Thus, by the mathematical induction, we have
for all
. Using (3.1) and (3.3), we obtain
Adding (3.5) and (3.6), we get
Repeating the above process, we get
for all
. For
there exists
such that
Let
be such that
. Then, by using the triangle inequality, we have
This implies that
. Since
and hence
and
are Cauchy sequences in
. Since
is a complete metric space and hence
and
are convergent in
. Then there exist
such that
Since
F is continuous and
and
, taking limit
, we get
that is,
and
and hence F has a coupled fixed point. □
In the next theorem, we omit the continuity hypothesis of F.
for all
with
and
.
Suppose that
 (i)
conditions (i) and (ii) of Theorem 3.4 hold,
 (ii)
if
and
are sequences in
X
such that
for all
n
and
and
,
then
If there exist
such that
and
, then there exist
such that
and
; that is, F
has a coupled fixed point in
X.
Proof Proceeding along the same lines as in the proof of Theorem 3.4, we know that
and
are Cauchy sequences in the complete metric space
. Then there exist
such that
On the other hand, from (3.3) and hypothesis (ii), we obtain
for all
. Using the triangle inequality, (3.8) and the property of
for all
, we get
Similarly, using (3.9), we obtain
Taking the limit as
in the above two inequalities, we get
Hence,
and
. Thus, F has a coupled fixed point. □
In the following theorem, we will prove the uniqueness of the coupled fixed point. If
is a partially ordered set, then we endow the product
with the following partial order relation:
for all
.
and also assume that
is comparable to
and
. Then
F
has a unique coupled fixed point.
Proof From Theorem 3.4, the set of coupled fixed points is nonempty. Suppose
and
are coupled fixed points of the mappings
; that is,
,
and
,
. By assumption, there exists
in
such that
is comparable to
and
. Put
and
and choose
such that
and
. Thus, we can define two sequences
and
as
Since
is comparable to
, it is easy to show that
and
. Thus,
and
for all
. Since for every
, there exists
such that
Since
F is
admissible, so from (3.10), we have
Since
and
, we get
Therefore, by the mathematical induction, we obtain
for all
n∈ and similarly,
. From (3.10) and (3.11), we get
Adding (3.12) and (3.13), we get
for each
. Letting
in (3.14) and using Lemma 3.1, we get
Similarly, one can show that
From (3.15) and (3.16), we conclude that
and
. Hence, F has a unique coupled fixed point. □
Example 3.7 (Linear case)
Let
and
be a standard metric. Define a mapping
by
for all
. Consider a mapping
be such that
Since
holds for all
. Therefore, we have
Thus (3.1) holds for
for all
, and we also see that all the hypotheses of Theorem 3.4 are fulfilled. Then there exists a coupled fixed point of F. In this case,
is a coupled fixed point of F.
Example 3.8 (Nonlinear case)
Let
and
be a standard metric. Define a mapping
by
for all
. Consider a mapping
be such that
Therefore (3.1) holds for
for all
, and also the hypothesis of Theorem 3.4 is fulfilled. Then there exists a coupled fixed point of F. In this case,
is a coupled fixed point of F.