There are a lot of fixed and common fixed point results in different types of spaces. For example, metric spaces, fuzzy metric spaces, and uniform spaces. One of the most interesting is partial metric space, which is defined by Matthews [1]. In partial metric spaces, the distance of a point in the self may not be zero. After the definition of partial metric space, Matthews proved the partial metric version of Banach fixed point theorem. Then, Valero [2], Oltra and Valero [3], and Altun et al. [4] gave some generalizations of the result of Matthews. Again, Romaguera [5] proved the Caristi type fixed point theorem on this space.

First, we recall some definitions of partial metric spaces and some properties of theirs. See [1–3, 5–7] for details.

A partial metric on a nonempty set is a function such that for all

(p₁),

(p₂),

(p₃),

(p₄).

A partial metric space is a pair such that is a nonempty set and is a partial metric on . It is clear that if , then from (p₁) and (p₂) . But if , may not be 0. A basic example of a partial metric space is the pair , where for all . Other examples of partial metric spaces, which are interesting from a computational point of view, may be found in [1, 8].

Each partial metric on generates a topology on , which has as a base the family open -balls , where for all and .

If is a partial metric on , then the function given by

is a metric on .

Let be a partial metric space, then we have the following.

(i)A sequence in a partial metric space converges to a point if and only if .

(ii)A sequence in a partial metric space is called a Cauchy sequence if there exists (and is finite) .

(iii) A partial metric space is said to be complete if every Cauchy sequence in converges, with respect to , to a point such that .

(iv) A mapping is said to be continuous at , if for every , there exists such that .

Lemma 1.1 (see [1, 3]).

Let be a partial metric space.

(a) is a Cauchy sequence in if and only if it is a Cauchy sequence in the metric space .

(b)A partial metric space is complete if and only if the metric space is complete. Furthermore, if and only if

On the other hand, existence of fixed points in partially ordered sets has been considered recently in [9], and some generalizations of the result of [9] are given in [10–15] in a partial ordered metric spaces. Also, in [9], some applications to matrix equations are presented; in [14, 15], some applications to ordinary differential equations are given. Also, we can find some results on partial ordered fuzzy metric spaces and partial ordered uniform spaces in [16–18], respectively.

The aim of this paper is to combine the above ideas, that is, to give some fixed point theorems on ordered partial metric spaces.

Theorem 2.1.

Let be partially ordered set, and suppose that there is a partial metric on such that is a complete partial metric space. Suppose is a continuous and nondecreasing mapping such that

for all with , where is a continuous, nondecreasing function such that is convergent for each . If there exists an with , then there exists such that . Moreover, .

Proof.

From the conditions on , it is clear that for and . If , then the proof is finished, so suppose . Now, let for . If for some , then it is clear that is a fixed point of . Thus, assume for all . Notice that since and is nondecreasing, we have

Now, since , we can use the inequality (2.1) for these points, then we have

since

and is nondecreasing. Now, if

for some , then from (2.3) we have

which is a contradiction since . Thus

for all . Therefore, we have

and so

On the other hand, since

then from (2.9) we have

Therefore,

This shows that . Now, we have

Since is convergent for each , then is a Cauchy sequence in the metric space . Since is complete, then, from Lemma 1.1, the sequence converges in the metric space , say . Again, from Lemma 1.1, we have

Moreover, since is a Cauchy sequence in the metric space , we have , and, from (2.11), we have , thus, from definition , we have . Therefore, from (2.14), we have

Now, we claim that . Suppose . Since is continuous, then, given , there exists such that . Since , then there exists such that for all . Therefore, we have for all . Thus, , and so for all . This shows that . Now, we use the inequality (2.1) for , then we have

Therefore, we obtain

and letting , we have

which is a contradiction since . Thus, , and so .

In the following theorem, we remove the continuity of . Also, The contractive condition (2.1) does not have to be satisfied for , but we add a condition on .

Theorem 2.2.

Let be a partially ordered set, and suppose that there is a partial metric on such that is a complete partial metric space. Suppose is a nondecreasing mapping such that

for all with (i.e., and ), where is a continuous, nondecreasing function such that is convergent for each . Also, the condition

holds. If there exists an with , then there exists such that . Moreover, .

Proof.

As in the proof of Theorem 2.1, we can construct a sequence in by for . Also, we can assume that the consecutive terms of are different. Otherwise we are finished. Therefore, we have

Again, as in the proof of Theorem 2.1, we can show that is a Cauchy sequence in the metric space , and, therefore, there exists such that

Now, we claim that . Suppose . Since the condition (2.20) is satisfied, then we can use (2.19) for . Therefore, we obtain

using the continuity of and letting , we have . Therefore, we obtain

which is a contradiction. Thus, , and so .

Example 2.3.

Let and , then it is clear that is a complete partial metric space. We can define a partial order on as follows:

Let ,

and , . Therefore, is continuous and nondecreasing. Again we can show by induction that , and so we have that is convergent. Also, is nondecreasing with respect to , and for , we have

that is, the condition (2.19) of Theorem 2.2 is satisfied. Also, it is clear that the condition (2.20) is satisfied, and for , we have . Therefore, all conditions of Theorem 2.2 are satisfied, and so has a fixed point in . Note that if and , then

This shows that the contractive condition of Theorem  1 of [4] is not satisfied.

Theorem 2.4.

If one uses the following condition instead of (2.1) in Theorem 2.1, one has the same result.

for all with .

In what follows, we give a sufficient condition for the uniqueness of the fixed point in Theorem 2.4, this condition is

In [15], it was proved that condition (2.30) is equivalent to

Theorem 2.5.

Adding condition (2.31) to the hypotheses of Theorem 2.4, one obtains uniqueness of the fixed point of .

Proof.

Suppose that there exists are different fixed points of , then . Now, we consider the following two cases.

(i) If and are comparable, then and are comparable for . Therefore, we can use the condition (2.1), then we have

which is a contradiction.

(ii) If and are not comparable, then there exists comparable to and . Since is nondecreasing, then is comparable to and for . Moreover,

Now, if for some , then we have

which is a contradiction. Thus, for all , and so

This shows that is a nonnegative and nondecreasing sequence and so has a limit, say . From the last inequality, we can obtain

hence . Similarly, it can be proven that, . Finally,

and taking limit , we have . This contradicts .

Consequently, has no two fixed points.