 Research
 Open Access
 Published:
Some unique fixed point theorems for rational contractions in partially ordered metric spaces
Journal of Inequalities and Applications volume 2013, Article number: 248 (2013)
Abstract
In this paper, we prove some unique fixed point results for an operator T satisfying certain rational contraction condition in a partially ordered metric space. Our results generalize the main result of Jaggi (Indian J. Pure Appl. Math. 8(2):223230, 1977). We give several examples to show that our results are proper generalization of the existing one.
MSC:47H10, 54H25, 46J10, 46J15.
1 Introduction
Fixed point theory is one of the famous and traditional theories in mathematics and has a broad set of applications. In this theory, contraction is one of the main tools to prove the existence and uniqueness of a fixed point. Banach’s contraction principle, which gives an answer on the existence and uniqueness of a solution of an operator equation Tx=x, is the most widely used fixed point theorem in all of analysis. This principle is constructive in nature and is one of the most useful tools in the study of nonlinear equations. There are many generalizations of Banach’s contraction mapping principle in the literature [1–6]. These generalizations were made either by using the contractive condition or by imposing some additional conditions on an ambient space. There have been a number of generalizations of metric spaces such as rectangular metric spaces, pseudo metric spaces, fuzzy metric spaces, quasi metric spaces, quasi semimetric spaces, probabilistic metric spaces, Dmetric spaces and cone metric spaces
The basic topological properties of ordered sets were discussed by Wolk [7] and Monjardet [8]. The existence of fixed points in partially ordered metric spaces was considered by Ran and Reurings [9]. After this paper, Nieto et al. [10–12] published some new results. Recently, many papers have been reported on partially ordered metric spaces (see, e.g., [9–19] and also [8, 20–33]).
The triple (X,d,\u2aaf) is called partially ordered metric spaces (POMS) if (X,\u2aaf) is a partially ordered set and (X,d) is a metric space. Further, if (X,d) is a complete metric space, the triple (X,d,\le ) is called partially ordered complete metric spaces (POCMS). Throughout the manuscript, we assume that X\ne \mathrm{\varnothing}. A partially ordered metric space (X,d,\u2aaf) is called ordered complete (OC) if for each convergent sequence {\{{x}_{n}\}}_{n=0}^{\mathrm{\infty}}\subset X, the following condition holds: either

if \{{x}_{n}\} is a nonincreasing sequence in X such that {x}_{n}\to {x}^{\ast} implies {x}^{\ast}\u2aaf{x}_{n}\mathrm{\forall}n\in \mathbb{N}, that is, {x}^{\ast}=inf\{{x}_{n}\}, or

if \{{x}_{n}\} is a nondecreasing sequence in X such that {x}_{n}\to {x}^{\ast} implies {x}_{n}\u2aaf{x}^{\ast}\mathrm{\forall}n\in \mathbb{N}, that is, {x}^{\ast}=sup\{{x}_{n}\}.
In this manuscript, we prove that an operator T satisfying certain rational contraction condition has a fixed point in a partially ordered metric space. Our results generalize the main result of Jaggi [34].
2 Main results
We start this section with the following definition.
Definition 1 Let (X,d,\u2aaf) be a partially ordered metric space. A selfmapping T on X is called an almost Jaggi contraction if it satisfies the following condition:
for any distinct x,y\in X with x\u2aafy, where L\ge 0 and \alpha ,\beta \in [0,1) with \alpha +\beta <1.
Theorem 2 Let (X,d,\u2aaf) be a complete partially ordered metric space. Suppose that a selfmapping T is an almost Jaggi contraction, continuous and nondecreasing. Suppose there exists {x}_{0}\in X with {x}_{0}\u2aafT{x}_{0}. Then T has a unique fixed point.
Proof Let {x}_{0}\in X and set {x}_{n+1}=T{x}_{n}. If {x}_{{n}_{0}}={x}_{{n}_{0}+1} for some {n}_{0}\in \mathbb{N}, then T has a fixed point. In particular, {x}_{{n}_{0}} is a fixed point of T. So, we assume that {x}_{n}\ne {x}_{n+1} for all n. Since {x}_{0}\u2aafT{x}_{0}, then
Now
which implies that
By the triangle inequality, for m\ge n we have
where k=\frac{\beta}{1\alpha}. Letting n\to \mathrm{\infty} in the inequality (3), we get d({x}_{n},{x}_{m})=0. Thus, the sequence \{{x}_{n}\} is Cauchy. Since X is complete, there exists a point z\in X such that {x}_{n}\to z. Furthermore, the continuity of T in X implies that
Therefore, z is a fixed point of T in X. Now, if there exists another point w\ne z in X such that Tw=w, then
a contradiction. Hence u is a unique fixed point of T in X. □
Example 3 Let X=[0,1] with the usual metric and usual order ≤. We define an operator T:X\to X as follows:
Then T is continuous and nondecreasing. Take \beta =\frac{1}{4}. Then, for any \alpha \in [0,1) with \alpha +\beta <1, we have the result. Let us examine in detail. Without loss of generality, we assume that y\u2aafx.
Case 01. If x,y\in [0,\frac{1}{2}], then
holds for any L\ge 1 and any \alpha \in [0,1) with \alpha +\beta <1. Thus, all the conditions of Theorem 2 are satisfied.
Case 02. If x,y\in (\frac{1}{2},1], then
holds for any L\ge 0 and any \alpha \in [0,1) with \alpha +\beta <1. Hence, all the conditions of Theorem 2 are satisfied.
Case 03. If x\in (\frac{1}{2},1] and y\in [0,\frac{1}{2}], then we can easily evaluate that \frac{1}{20}2x1\le \frac{1}{20}. Further, we have \frac{9}{20}\le d(x,Ty)=x\frac{y}{10}\le 1 and \frac{3}{20}\le d(y,Tx)=\frac{x}{5}\frac{1}{20}y\le \frac{9}{20}. By the help of these observations, we derive that
Notice that 0\in X is the fixed point of T.
Definition 4 Let (X,d,\u2aaf) be a partially ordered metric space. A selfmapping T on X is called a Jaggi contraction if it satisfies the following condition:
for any distinct x,y\in X with x\u2aafy, where \alpha ,\beta \in [0,1) with \alpha +\beta <1.
Corollary 5 Let (X,d,\u2aaf) be a complete partially ordered metric space. Suppose that a selfmapping T is a Jaggi contraction, continuous and nondecreasing. Suppose that there exists {x}_{0}\in X with {x}_{0}\u2aafT{x}_{0}. Then T has a fixed point.
Proof Set L=0 in Theorem 2. □
Example 6 Let X=[0,\mathrm{\infty}), d:X\times X\to {\mathbb{R}}_{+} be defined by
Then (X,d) is a complete metric space. Let T:X\to X be defined by
Also, x\u2aafy iff x\le y. Clearly, T is an increasing and continuous selfmapping on X. We shall prove that conditions of Corollary 5 hold and T has a fixed point.
Proof For the proof of this example, we have the following cases.

Let 0\le x<y\le 2. Then
\begin{array}{rl}d(Tx,Ty)& =max\{Tx,Ty\}=\frac{y}{8(1+y)}\le \frac{1}{4}(x+y)=\frac{1}{4}(\frac{xy}{y}+y)\\ =\frac{1}{4}[\frac{max\{x,\frac{x}{8(1+x)}\}max\{y,\frac{y}{8(1+y)}\}}{max\{x,y\}}+max\{x,y\}]\\ =\frac{1}{4}(\frac{d(x,Tx)d(y,Ty)}{d(x,y)}+d(x,y)),\end{array}
that is,

Let 2<x<y. Then
\begin{array}{rcl}d(Tx,Ty)& =& max\{Tx,Ty\}=\frac{y}{12}\le \frac{1}{4}(x+y)=\frac{1}{4}(\frac{xy}{y}+y)\\ =& \frac{1}{4}[\frac{max\{x,\frac{x}{12}\}max\{y,\frac{y}{12}\}}{max\{x,y\}}+max\{x,y\}]\\ =& \frac{1}{4}(\frac{d(x,Tx)d(y,Ty)}{d(x,y)}+d(x,y)),\end{array}
that is,

Let 0\le x\le 2 and 2<y. Then
\begin{array}{rcl}d(Tx,Ty)& =& max\{Tx,Ty\}=\frac{y}{12}\le \frac{1}{4}(x+y)=\frac{1}{4}(\frac{xy}{y}+y)\\ =& \frac{1}{4}[\frac{max\{x,\frac{x}{8(1+x)}\}max\{y,\frac{y}{12}\}}{max\{x,y\}}+max\{x,y\}]\\ =& \frac{1}{4}(\frac{d(x,Tx)d(y,Ty)}{d(x,y)}+d(x,y)),\end{array}
that is,
Then conditions of Corollary 5 hold and T has a fixed point (here, x=0 is a fixed point of T). □
In the next theorem, we establish the existence of a unique fixed point of a map T by assuming only the continuity of some iteration of T.
Theorem 7 Let (X,d,\u2aaf) be a complete partially ordered metric space. Suppose that a selfmapping T is nondecreasing and an almost Jaggi contraction. Suppose there exists {x}_{0}\in X with {x}_{0}\u2aafT{x}_{0}. If the operator {T}^{p} is continuous for some positive integer p, then T has a unique fixed point.
Proof As in Theorem 2, we define a sequence {x}_{n} and conclude that the sequence {x}_{n} converges to some point z\in X. Thus its subsequence {x}_{{n}_{k}} ({n}_{k}=kp) also converges to z. Also,
Therefore z is a fixed point of {T}^{p}. We now show that Tz=z. Let m be the smallest positive integer such that {T}^{m}z=z but {T}^{q}\ne z (q=1,2,\dots ,m1). If m>1, then
which implies that
Regarding (1), we have
Inductively, we get
where k=\frac{\beta}{1\alpha}. Notice that k<1. Therefore,
a contradiction. Hence Tz=z. The uniqueness of z follows as in Theorem 2. □
Corollary 8 Let (X,d,\u2aaf) be a complete partially ordered metric space. Suppose that a selfmapping T is nondecreasing and a Jaggi contraction. Suppose there exists {x}_{0}\in X with {x}_{0}\u2aafT{x}_{0}. If the operator {T}^{p} is continuous for some positive integer p, then T has a unique fixed point.
Proof Set L=0 in Theorem 7. □
The following theorem generalizes Theorem 2.
Theorem 9 Let (X,d,\u2aaf) be a complete partially ordered metric space and let T be a nondecreasing selfmapping defined on X. Suppose that for some positive integer m, selfmapping T satisfies the following condition:
for any distinct x,y\in X with x\u2aafy and for some \alpha ,\beta \in [0,1) with \alpha +\beta <1 and L\ge 0. Suppose there exists {x}_{0}\in X with {x}_{0}\u2aaf{T}^{m}{x}_{0}. If {T}^{m} is continuous, then T has a unique fixed point.
Proof Due to Theorem 2, we conclude that {T}^{m} has a unique fixed point, say z\in X. Consider now
Thus, Tz is also a fixed point of {T}^{m}. But, by Theorem 2, we know that {T}^{m} has a unique fixed point z. It follows that z=Tz. Hence, z is the unique fixed point of T. □
Corollary 10 Let (X,d,\u2aaf) be a complete partially ordered metric space and let T be a nondecreasing selfmapping defined on X. Suppose that for some positive integer m, the selfmapping T satisfies the following condition:
for all distinct x,y\in X and for some \alpha ,\beta \in [0,1) with \alpha +\beta <1. Suppose there exists {x}_{0}\in X with {x}_{0}\u2aaf{T}^{m}{x}_{0}. If {T}^{m} is continuous, then T has a unique fixed point.
Proof Set L=0 in Theorem 9. □
Now, we give the following example.
Example 11 Let X=[0,1] with the usual metric and usual order ≤. We define an operator T:X\leftarrow X as follows:
It can be easily seen that T is discontinuous and does not satisfy (1) for any \alpha ,\beta \in [0,1) with \alpha +\beta \prec 1 when x=\frac{1}{3}, y=1. Now {T}^{2}(x)=0 for all x\in [0,1]. It can be verified that {T}^{2} satisfies the conditions of Theorem 9 and 0 is a unique fixed point of {T}^{2}.
Theorem 12 Let (X,d,\u2aaf) be a complete partially ordered metric space and let T be a nondecreasing selfmapping defined on X. Suppose that a selfmapping T on X satisfies the condition
for any points x,y\in X with x\u2aafy, and for some \alpha ,\beta \in [0,1) with \alpha +\beta <1 and L\ge 0. Suppose there exists {x}_{0}\in X with {x}_{0}\u2aafT{x}_{0}. Then T has a fixed point.
Proof Define sequences {x}_{n} as in Theorem 2. If {x}_{{n}_{0}}={x}_{{n}_{0}+1} for some {n}_{0}\in \mathbb{N} then T has a fixed point. In particular, {x}_{{n}_{0}} is a fixed point of T. Therefore, we assume that
Due to (7), we have
which implies that
Recursively, we obtain that
As in Theorem 2, we prove that \{{x}_{n}\} is a Cauchy sequence. Indeed, by the triangle inequality, we have for m\ge n,
where k=\frac{\beta}{1\alpha}<1. Letting n\to \mathrm{\infty}, then the righthand side of the inequality (9) tends to 0. Thus, the sequence \{{x}_{n}\} is Cauchy.
Since X is complete, there exists a z\in X such that
Consider (7)
Letting n\to \mathrm{\infty} in (11), we get
which is possible only if d(Tz,z)=0. Thus, Tz=z.
Now, we show that z is the unique fixed point of T. Assume, on the contrary, that the operator T has another fixed point u\ne z. Keeping (7) in mind, we obtain that
a contradiction. Hence z is a unique fixed point of T in X. □
Corollary 13 Let (X,d,\u2aaf) be a complete partially ordered metric space and let T be a nondecreasing selfmapping defined on X. Suppose that a selfmapping T on X satisfies the condition
for any points x,y\in X with x\u2aafy, and for some \alpha ,\beta \in [0,1) with \alpha +\beta <1. Then T has a fixed point.
Proof Set L=0 in Theorem 12. □
Example 14 Let X=[0,1], d:X\times X\to {\mathbb{R}}_{+} be defined by
Then (X,d) is a complete metric space. Let T:X\to X be defined by
Also, x\u2aafy iff x\le y. Suppose that \beta =\frac{1}{3} and \alpha \in [0,1) such that \alpha +\beta <1. Clearly, T is an injective, continuous and sequentially convergent mapping on X. We shall prove that conditions of Corollary 8 hold and T has a fixed point.
Proof For the proof of this example, we have the following cases.
Let x,y\in [0,1]. Then
That is,
Hence, conditions of Corollary 8 hold and T has a fixed point (here x=0 is a fixed point of T). □
3 Further results
Theorem 15 Let (X,d,\u2aaf) be a complete partially ordered metric space and let T be a nondecreasing, continuous selfmapping defined on X. Suppose that a selfmapping T satisfies the following condition:
for all x,y\in X with y\u2aafx, where A=d(y,Tx)+d(x,Ty) and λ, μ are nonnegative reals such that \lambda +\mu <1. If there exists {x}_{0}\in X with {x}_{0}\u2aafT{x}_{0}, then T has a fixed point.
Proof By assumption, there exists {x}_{0}\in X with {x}_{0}\u2aafT{x}_{0}. If {x}_{0}=T{x}_{0}, then the proof is finished. So, we suppose that {x}_{0}\prec T{x}_{0}. Since T is a nondecreasing mapping, we get
by iteration. Put {x}_{n+1}=T{x}_{n}. If there exists {n}_{0}\in N such that {x}_{{n}_{0}}={x}_{{n}_{0}+1}, then from {x}_{{n}_{0}}={x}_{{n}_{0}+1}=T{x}_{{n}_{0}}, we get {x}_{{n}_{0}} is a fixed point, and the proof is finished. Suppose that {x}_{n}\ne {x}_{n+1} for n\in N. Since the points {x}_{n} and {x}_{n1} are comparable for all n\in N due to (14), we have the following two cases.
Case 1. If A=d({x}_{n1},T{x}_{n})+d({x}_{n},T{x}_{n1})\ne 0, then using the contractive condition (13), we get
Hence, we derive that
where h=(\lambda +\mu )<1. Moreover, by the triangular inequality, we have, for m\ge n,
and this proves that d({x}_{m},{x}_{n})\to 0 as m,n\to \mathrm{\infty}.
So, \{{x}_{n}\} is a Cauchy sequence and, since X is a complete metric space, there exists z\in X such that {lim}_{n\to \mathrm{\infty}}{x}_{n}=z. Further, the continuity of T implies
Thus z is a fixed point.
Case 2. If A=d({x}_{n1},T{x}_{n})+d({x}_{n},T{x}_{n1})=0, then d({x}_{n+1},{x}_{n})=0. This implies that {x}_{n}={x}_{n+1}, a contradiction. Thus there exists a fixed point z of T. □
Example 16 Let X=[0,1] with the usual metric and usual order ≤. We define an operator T:X\to X in the following way:
It is clear that T is continuous on [0,1]. Now, for \lambda =\frac{16}{25} and any \mu \in [0,1) such that \lambda +\mu <1. Without loss of generality, we assume that x\le y. So, we have
for all x,y\in X. Also, there exists {x}_{0}=0\in X such that
is satisfied. This shows that conditions of Theorem 15 hold and T has a fixed point \frac{1}{2}\in [0,1].
We may remove the continuity criteria on T in Theorem 15 as follows.
Theorem 17 Let (X,d,\u2aaf) be a complete partially ordered metric space and let T be a nondecreasing selfmapping defined on X. Suppose that a selfmapping T satisfies the following condition:
for all x,y\in X with y\u2aafx, where A=d(x,Tx)+d(y,Ty) and λ, μ are nonnegative reals with \lambda +\mu <1. And also suppose that X has the (OC) property. If there exists {x}_{0}\in X with {x}_{0}\u2aafT{x}_{0}, then T has a fixed point.
Proof We only have to check that z=Tz. As \{{x}_{n}\}\subset X is a nondecreasing sequence and {x}_{n}\to z\in X, then z=sup\{{x}_{n}\} for all n\in N. Since T is a nondecreasing mapping, then T{x}_{n}\u2aafTz for all n\in \mathbb{N} or, equivalently, {x}_{n+1}\u2aafTz for all n\in N. Moreover, as {x}_{0}\prec {x}_{1}\u2aafTz and z=sup\{{x}_{n}\}, we get z\u2aafTz. Suppose that z\prec Tz. Using a similar argument as that in the proof of Theorem 15 for {x}_{0}\u2aafT{x}_{0}, we obtain that \{{T}^{n}z\} is a nondecreasing sequence and {lim}_{n\to \mathrm{\infty}}{T}^{n}z=y for certain y\in X. Again, using (OC), we have that y=sup\{{T}^{n}z\}. Moreover, from {x}_{0}\u2aafz, we get {x}_{n}={T}^{n}{x}_{0}\u2aaf{T}^{n}z for n\ge 1 and {x}_{n}\prec {T}^{n}z for n\ge 1 because {x}_{n}\u2aafz\prec Tz\u2aaf{T}^{n}z for n\ge 1 as {x}_{n} and {T}^{n}z are comparable and distinct for n\ge 1.
Case 1. If d({T}^{n}z,T{x}_{n})+d({x}_{n},{T}^{n+1}z)\ne 0, then applying the contractive condition (16), we get
Making n\to \mathrm{\infty} in the above inequality, we obtain
As \lambda <1, d(x,z)=0, thus z=y. Particularly, z=y=sup\{{T}^{n}z\} and consequently, Tz\u2aafz which is a contradiction. Hence, we conclude that Tz=z.
Case 2. If A=d({T}^{n}z,T{x}_{n})+d({x}_{n},{T}^{n+1}z)=0, then d({x}_{n+1},{T}^{n+1}z)=0. Taking the limit as n\to \mathrm{\infty}, we get d(z,y)=0. Then z=y=sup\{{T}^{n}z\}, which implies that Tz\u2aafz, a contradiction. Thus Tz=z. □
Now we prove the sufficient condition for the uniqueness of the fixed point in Theorem 15 and Theorem 17, that is,
U: for any y,z\in X, there exists x\in X which is comparable to y and z.
Theorem 18 Adding the above mentioned condition to the hypothesis of Theorem 15 (or Theorem 17), one obtains the uniqueness of the fixed point of T.
Proof We distinguish two cases.
Case 1. If y and z are comparable and y\ne z. Now we have two subcases that are as follows:

(i)
If d(z,Ty)+d(y,Tz)\ne 0, then using the contractive condition, we have
\begin{array}{rcl}d(y,z)& =& d(Ty,Tz)\le \lambda d(y,z)+\mu \frac{d(y,Ty)d(y,Tz)+d(z,Ty)d(z,Tz)}{d(z,Ty)+d(y,Tz)}\\ \le & \lambda d(y,z)+\mu \frac{d(y,y)d(y,z)+d(z,y)d(z,z)}{d(z,y)+d(y,z)}=\lambda d(y,z).\end{array}
As \lambda <1, so by the last inequality, we have a contradiction. Thus y=z.

(ii)
If d(z,Ty)+d(y,Tz)=0, then d(y,z)=0, a contradiction. Thus y=z.
Case 2. If y and z are not comparable, then by a given condition there exists x\in X comparable to y and z. Monotonicity implies that {T}^{n}x is comparable to {T}^{n}y=y and {T}^{n}z=z for n=0,1,2,\dots .
If there exists {n}_{0}\ge 1 such that {T}^{{n}_{0}}x=y, then as y is a fixed point, the sequence \{{T}^{n}x:n\ge {n}_{0}\} is constant, and consequently {lim}_{n\to \mathrm{\infty}}{T}^{n}x=y. On the other hand, if {T}^{n}x\ne y for n\ge 1. Now we have two subcases as follows:

(i)
If d({T}^{n1}y,{T}^{n}x)+d({T}^{n1}x,{T}^{n}y)\ne 0, then using the contractive condition, we obtain, for n\ge 2,
\begin{array}{rcl}d({T}^{n}x,y)& =& d({T}^{n}x,{T}^{n}y)\\ \le & \lambda d({T}^{n1}x,{T}^{n1}y)\\ +\mu \frac{d({T}^{n1}x,{T}^{n}x)d({T}^{n1}x,{T}^{n}y)+d({T}^{n1}y,{T}^{n}x)d({T}^{n1}y,{T}^{n}y)}{d({T}^{n1}y,{T}^{n}x)+d({T}^{n1}x,{T}^{n}y)}\\ \le & \lambda d({T}^{n1}x,y)+\mu \frac{d({T}^{n1}x,{T}^{n}x)d({T}^{n1}x,y)+d(Ty,{T}^{n}x)d(y,y)}{d(y,{T}^{n}x)+d({T}^{n1}x,y)}\\ \le & \lambda d({T}^{n1}x,y)+\mu \frac{d({T}^{n1}x,{T}^{n}x)d({T}^{n1}x,y)}{d(y,{T}^{n}x)+d({T}^{n1}x,y)}.\end{array}
This implies that
By induction we get
Taking limit as n\to \mathrm{\infty} in the above inequality, we get
as \lambda +\mu <1. Using a similar argument, we can prove that
Now, the uniqueness of the limit gives that y=z.

(ii)
If d({T}^{n1}y,{T}^{n}x)+d({T}^{n1}x,{T}^{n}y)=0, then d({T}^{n}x,y)=0. Then
\underset{n\to \mathrm{\infty}}{lim}{T}^{n}x=y.
Using a similar argument, we can prove that
Now, the uniqueness of the limit gives that y=z. This completes the proof. □
Remark 19 If in Theorem 15Theorem 18 \mu =0, then we obtain Theorem 2.1Theorem 2.3 of [10].
We get the following fixed point theorem in partially ordered metric spaces if we take \lambda =0 in the theorems of Section 3.
Theorem 20 Let (X,d,\u2aaf) be a complete partially ordered metric space and let T be a nondecreasing selfmapping defined on X. Suppose that a selfmapping T satisfies the following condition:
for all x,y\in X with y\u2aafx, where A=d(y,Tx)+d(x,Ty) and μ is a nonnegative real with 0\le \mu <1. Suppose also that either T is continuous or X satisfies the condition (OC). If there exists {x}_{0}\in X with {x}_{0}\u2aafT{x}_{0}, then T has a fixed point.
If (X,\u2aaf) satisfies the condition used in Theorem 18, then the uniqueness of a fixed point can be proved.
4 Applications
In this section we state some applications of the main results. The first result is the consequence of Theorem 2.
Corollary 21 Let (X,d,\u2aaf) be a Torbitally complete partially ordered metric space and let T be a nondecreasing selfmapping defined on X. Suppose that a selfmapping T satisfies the following condition:
for all distinct x,y\in X with x\u2aafy and for \alpha ,\beta \in [0,1) with \alpha +\beta <1, where L\ge 0. If there exists {x}_{0}\in X with {x}_{0}\u2aafT{x}_{0} then T has at least one fixed point.
Similarly, the following result is the consequence of Corollary 5.
Corollary 22 Let T be a continuous, nondecreasing selfmap defined on a complete partially ordered metric space (X,d,\u2aaf). Suppose that T satisfies the following condition:
for any distinct x,y\in X with x\u2aafy, where \alpha ,\beta \in [0,1) with \alpha +\beta <1. Suppose there exists {x}_{0}\in X with {x}_{0}\u2aafT{x}_{0}. Then T has a fixed point.
The following result is the consequence of Theorem 12.
Corollary 23 Let (X,d,\u2aaf) be a partially ordered metric space. Let T:X\to X be a nondecreasing, continuous mapping. Suppose that a selfmapping T satisfies
for any x,y\in X with x\u2aafy, and for some \alpha ,\beta \in [0,1) with \alpha +\beta <1 and L\ge 0. Suppose that there exists {x}_{0}\in X with {x}_{0}\u2aafT{x}_{0}. Then T has a fixed point.
References
Edelstein M: On fixed points and periodic points under contraction mappings. J. Lond. Math. Soc. 1962, 37: 74–79.
Hardy GC, Rogers T: A generalization of fixed point theorem of S. Reich. Can. Math. Bull. 1973, 16: 201–206. 10.4153/CMB19730360
Kannan R: Some results on fixed points  II. Am. Math. Mon. 1969, 76: 71–76.
Reich S: Some remarks concerning contraction mappings. Can. Math. Bull. 1971, 14: 121–124. 10.4153/CMB19710249
Smart DR: Fixed Point Theorems. Cambridge University Press, Cambridge; 1974.
Wong CS: Common fixed points of two mappings. Pac. J. Math. 1973, 48: 299–312. 10.2140/pjm.1973.48.299
Wolk ES: Continuous convergence in partially ordered sets. Gen. Topol. Appl. 1975, 5: 221–234. 10.1016/0016660X(75)900227
Monjardet B: Metrics on partially ordered sets  a survey. Discrete Math. 1981, 35: 173–184. 10.1016/0012365X(81)902065
Ran ACM, Reurings MCB: A fixed point theorem in partially ordered sets and some application to matrix equations. Proc. Am. Math. Soc. 2004, 132: 1435–1443. 10.1090/S0002993903072204
Nieto JJ, Lopez RR: Contractive mapping theorems in partially ordered sets and applications to ordinary differential equations. Order 2005, 22: 223–239. 10.1007/s1108300590185
Nieto JJ, Lopez RR: Existence and uniqueness of fixed point in partially ordered sets and applications to ordinary differential equation. Acta Math. Sin. Engl. Ser. 2007, 23(12):2205–2212. 10.1007/s1011400507690
Nieto JJ, Pouso L, RodríguezLópez R: Fixed point theorems in ordered abstract spaces. Proc. Am. Math. Soc. 2007, 135: 2505–2517. 10.1090/S0002993907087291
Agarwal RP, ElGebeily MA, O’Regan D: Generalized contractions in partially ordered metric spaces. Appl. Anal. 2008, 87: 1–8. 10.1080/00036810701714164
Bhaskar TG, Lakshmikantham V: Fixed point theory in partially ordered metric spaces and applications. Nonlinear Anal., Theory Methods Appl. 2006, 65: 1379–1393. 10.1016/j.na.2005.10.017
Choudhury BS, Kundu A: A coupled coincidence point result in partially ordered metric spaces for compatible mappings. Nonlinear Anal., Theory Methods Appl. 2010, 73: 2524–2531. 10.1016/j.na.2010.06.025
Hong S: Fixed points of multivalued operators in ordered metric spaces with applications. Nonlinear Anal., Theory Methods Appl. 2010, 72: 3929–3942. 10.1016/j.na.2010.01.013
Lakshmikantham V, Ćirić LB: Couple fixed point theorems for nonlinear contractions in partially ordered metric spaces. Nonlinear Anal., Theory Methods Appl. 2009, 70: 4341–4349. 10.1016/j.na.2008.09.020
Ozturk M, Basarir M: On some common fixed point theorems with rational expressions on cone metric spaces over a Banach algebra. Hacet. J. Math. Stat. 2012, 41(2):211–222.
Rouzkard F, Imdad M: Some common fixed point theorems on complex valued metric spaces. Comput. Math. Appl. 2012. doi:10.1016/j.camwa.2012.02.063
Ahmad J, Arshad M, Vetro C: On a theorem of Khan in a generalized metric space. Int. J. Anal. 2013., 2013: Article ID 852727
Altun I, Damjanovic B, Djoric D: Fixed point and common fixed point theorems on ordered cone metric spaces. Appl. Math. Lett. 2010, 23: 310–316. 10.1016/j.aml.2009.09.016
AminiHarandi A, Emami H: A fixed point theorem for contraction type maps in partially ordered metric spaces and application to ordinary differential equations. Nonlinear Anal., Theory Methods Appl. 2010, 72: 2238–2242. 10.1016/j.na.2009.10.023
Arshad M, Azam A, Vetro P: Some common fixed results in cone metric spaces. Fixed Point Theory Appl. 2009., 2009: Article ID 493965
Arshad M, Ahmad J, Karapınar E: Some common fixed point results in rectangular metric spaces. Int. J. Anal. 2013., 2013: Article ID 307234
Aydi H, Karapınar E, Shatanawi W:Coupled fixed point results for (\psi ,\phi )weakly contractive condition in ordered partial metric spaces. Comput. Math. Appl. 2011, 62(12):4449–4460. 10.1016/j.camwa.2011.10.021
Azam A, Fisher B, Khan M: Common fixed point theorems in complex valued metric spaces. Numer. Funct. Anal. Optim. 2011, 32(3):243–253. 10.1080/01630563.2011.533046
Beg I, Butt AR: Fixed point for setvalued mappings satisfying an implicit relation in partially ordered metric spaces. Nonlinear Anal. 2009, 71: 3699–3704. 10.1016/j.na.2009.02.027
Dricia Z, McRaeb FA, Devi JV: Fixedpoint theorems in partially ordered metric spaces for operators with PPF dependence. Nonlinear Anal., Theory Methods Appl. 2007, 67: 641–647. 10.1016/j.na.2006.06.022
Karapınar E: Couple fixed point on cone metric spaces. Gazi Univ. J. Sci. 2011, 24(1):51–58.
Karapınar E, Luong NV: Quadruple fixed point theorems for nonlinear contractions. Comput. Math. Appl. 2012, 64(6):1839–1848. 10.1016/j.camwa.2012.02.061
Luong NV, Thuan NX: Coupled fixed points in partially ordered metric spaces and application. Nonlinear Anal., Theory Methods Appl. 2011, 74: 983–992. 10.1016/j.na.2010.09.055
Samet B: Coupled fixed point theorems for a generalized MeirKeeler contraction in partially ordered metric spaces. Nonlinear Anal. 2010, 74(12):4508–4517.
Zhang X: Fixed point theorems of multivalued monotone mappings in ordered metric spaces. Appl. Math. Lett. 2010, 23: 235–240. 10.1016/j.aml.2009.06.011
Jaggi DS: Some unique fixed point theorems. Indian J. Pure Appl. Math. 1977, 8(2):223–230.
Acknowledgements
The authors express their gratitude to the anonymous referees for constructive and useful remarks, comments and suggestions.
Author information
Authors and Affiliations
Corresponding author
Additional information
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.
Rights and permissions
Open Access This article is distributed under the terms of the Creative Commons Attribution 2.0 International License (https://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
About this article
Cite this article
Arshad, M., Karapınar, E. & Ahmad, J. Some unique fixed point theorems for rational contractions in partially ordered metric spaces. J Inequal Appl 2013, 248 (2013). https://doi.org/10.1186/1029242X2013248
Received:
Accepted:
Published:
DOI: https://doi.org/10.1186/1029242X2013248
Keywords
 fixed point
 rational contractions
 partially ordered metric spaces