Selçuk J. Appl. Math. Selçuk Journal of Vol. 13. No. 1. pp. 23-33, 2012 Applied Mathematics
Coupled Fixed Point Results in Ordered Partial Metric Spaces Hassen Aydi
Université de Sousse. Institut Supérieur d’Informatique et des Technologies de Communication de Hammam Sousse. Route GP1-4011, H. Sousse, Tunisie
e-mail: hassen.aydi@ isim a.rnu.tn
Received Date: May 23, 2011 Accepted Date: January 27, 2012
Abstract. In this paper, we prove some coupled fixed point theorems for map-pings having the mixed monotone property in partially ordered partial metric spaces. These results extend the main theorems of Bhaskar and Lakshmikan-tham [10] on ordered partial metric spaces.
Key words: Partially ordered; Partial metrics; Coupled fixed point; Mixed monotone property.
2000 Mathematics Subject Classification: 54H25, 47H10. 1. Introduction
Existence of a fixed point for contraction type mappings in partially ordered metric spaces and applications have been considered recently by many authors ( for details, see [1, 3, 5, 6, 10, 11, 12, 13, 14, 16, 17, 20, 24]). In [10], Bhaskar and Lakshmikantham proved some coupled fixed point theorems for mixed monotone mapping and discuss the existence and uniqueness of solution for periodic boundary value problem. In this paper, we extend their results to the class of ordered partial metric spaces. Also, we give an example illustrating our obtained results.
The concept of a partial metric space was introduced by Matthews [15]. In such spaces, the self-distance for any point need not be equal to zero. Specially, from the point of sequences, a convergent sequence need not have unique limit. First, we start by recalling some definitions and properties of partial metric spaces. For more details, we refer the reader to [2, 4, 7, 8, 9, 15, 18, 19, 21, 22, 23, 25, 26].
Definition 1. ([15].) A partial metric on a nonempty set X is a function p : X × X → R+ such that for all x, y, z ∈ X:
(p1) x = y ⇐⇒ p(x, x) = p(x, y) = p(y, y), (p2) p(x, x) ≤ p(x, y),
(p3) p(x, y) = p(y, x),
(p4) p(x, y) ≤ p(x, z) + p(z, y) − p(z, z).
A partial metric space is a pair (X, p) such that X is a nonempty set and p is a partial metric on X.
It is clear that, if p(x, y) = 0, then from (p1) and (p2), x = y. But if x = y, p(x, y) may not be 0. A basic example of a partial metric space is the pair (R+, p), where p(x, y) = max{x, y} for all x, y ∈ R+. Each partial metric p on
X generates a T0 topology τp on X which has as a base the family of open
p-balls {Bp(x, ε), x ∈ X, ε > 0}, where Bp(x, ε) = {y ∈ X : p(x, y) < p(x, x) + ε}
for all x ∈ X and ε > 0.
If p is a partial metric on X, then the function ps: X × X → R+ given by
(1) ps(x, y) = 2p(x, y) − p(x, x) − p(y, y), is a metric on X.
Definition 2. ([15].) Let (X, p) be a partial metric space and {xn} be a
se-quence in X. Then
(i) {xn} converges to a point x ∈ X if and only if p(x, x) = lim
n−→+∞p(x, xn),
(ii) {xn} is called a Cauchy sequence if there exists (and is finite) lim
n,m−→+∞p(xn, xm),
(iii) (X, p) is said to be complete if every Cauchy sequence {xn} in X converges,
with respect to τp, to a point x ∈ X, such that p(x, x) = lim
n,m−→+∞p(xn, xm).
Lemma 1. ([15].) Let (X, p) be a partial metric space. Then
(a) {xn} is a Cauchy sequence in (X, p) if and only if it is a Cauchy sequence
in the metric space (X, ps),
(b) (X, p) is complete if and only if the metric space (X, ps) is complete. Fur-thermore, lim n−→+∞p s(x n, x) = 0 if and only if p(x, x) = lim n−→+∞p(xn, x) =n,mlim−→+∞p(xn, xm).
Definition 3. ([2].) Suppose that (X, p) is a partial metric space. A mapping T : X → X is said to be continuous at x ∈ X, if for every ε > 0, there exists δ > 0 such that T (Bp(x, δ)) ⊆ Bp(T x, ε).
Let (X, p) be a partial metric. We endow X × X with the partial metric ν defined for (x, y), (u, v) ∈ X × X by
In view of Definition 3, a mapping F : X × X → X is said to be continuous at (x, y) ∈ X ×X, if for every ε > 0, there exists δ > 0 such that F (Bν((x, y), δ)) ⊆
Bp(F (x, y), ε).
The following result is easy to check.
Lemma 2. Let (X, p) be a partial metric space. F : X × X → X is continuous if and only if given a sequence {(xn, yn)}n∈N and (x, y) in X × X such that
ν((x, y), (x, y)) = lim
n→+∞ν((x, y), (xn, yn)), then
p(F (x, y), F (x, y)) = lim
n→+∞p(F (x, y), F (xn, yn)).
The aim of this paper is to give the analogous of the results of Bhashkar and Lakshmikantham [10] on ordered partial metric spaces.
2. Main Results
We start by recalling some known definitions.
Definition 4. ([10].) An element (x, y) ∈ X × X is called a coupled fixed point of mapping F : X × X → X if x = F (x, y) and y = F (y, x).
Definition 5. ([15].) Let (X, ≤X) be a partially ordered set and F : X × X →
X. The mapping F is said to has the mixed monotone property if for any x, y ∈ X
(2) x1, x2∈ X, x1≤Xx2=⇒ F (x1, y) ≤XF (x2, y) for any y ∈ X,
and
(3) y1, y2∈ X, y1≤Xy2=⇒ F (x, y1) ≥X F (x, y2) for any x ∈ X.
Our first result is the following
Theorem 1. Let (X, ≤X) be a partially ordered set and suppose there is a
partial metric p on X such that (X, p) is a complete partial metric space. Let F : X × X → X be a continuous mapping having the mixed monotone property on X. Assume that there exists a k ∈ [0, 1) with
(4)
p(F (x, y), F (u, v)) ≤ k2(p(x, u) + p(y, v)) for each x ≥X u and y ≤X v.
If there exist x0, y0∈ X such that x0≤X F (x0, y0) and y0 ≥X F (y0, x0), then
Proof. Since x0 ≤X F (x0, y0) = x1 (say) and y0 ≥X F (y0, x0) = y1 (say),
letting x2= F (x1, y1) and y2= F (y1, x1), we denote
F2(x0, y0) = F (F (x0, y0), F (y0, x0)) = F (x1, y1) = x2
F2(y0, x0) = F (F (y0, x0), F (x0, y0)) = F (y1, x1) = y2.
With this notation, we now have, due to the mixed monotone property of F , x2 = F2(x0, y0) = F (x1, y1) ≥X F (x0, y0) = x1 and
y2 = F2(y0, x0) = F (y1, x1) ≤X F (y0, x0) = y1.
Further, for n = 1, 2, ..., we let,
xn+1= Fn+1(x0, y0) = F (Fn(x0, y0), Fn(y0, x0)),
and
yn+1= Fn+1(y0, x0) = F (Fn(y0, x0), Fn(x0, y0)).
We can easily verify that
x0≤X F (x0, y0) = x1≤XF2(x0, y0) = x2≤X ... ≤XFn+1(x0, y0) = xn+1,
y0≥X F (y0, x0) = y1≥XF2(y0, x0) = y2≥X... ≥XFn+1(y0, x0) = yn+1.
Now, we claim that, for n ∈ N,
(5) p(xn+1, xn) ≤ kn 2 [p(x1, x0) + p(y1, y0)] (6) p(yn+1, yn) ≤ kn 2 [p(y1, y0) + p(x1, x0)].
Indeed, for n = 1, using x1 = F (x0, y0) ≥X x0 and y1 = F (y0, x0) ≤X y0, we
get p(x2, x1) =p(F2(x0, y0), F (x0, y0)) =p(F (x1, y1), F (x0, y0)) ≤k2[p(x1, x0) + p(y1, y0)]. Similarly p(y2, y1) = p(F (y1, x1), F (y0, x0)) ≤ k 2[p(y1, y0) + p(x1, x0)].
Now, assume that (5) and (6) hold. Using xn+1≥X xn and yn+1≤X yn, we get
p(xn+2, xn+1) =p(Fn+2(x0, y0), Fn+1(x0, y0)) =p(F (xn+1, yn+1), F (xn, yn)) ≤k 2[p(xn+1, xn) + p(yn+1, yn)] ≤k n+1 2 [p(x1, x0) + p(y1, y0)].
Similarly, one can show that p(yn+2, yn+1) ≤
kn+1
2 [p(y1, y0) + p(x1, x0)]. Using the definition of ps given by (1) in (5) and (6), we find (7) ps(xn+1, xn) ≤ 2p(xn+1, xn) ≤ kn[p(x1, x0) + p(y1, y0)] (8) ps(yn+1, yn) ≤ 2p(yn+1, yn) ≤ kn[p(y1, y0) + p(x1, x0)]. Letting m > n, we have ps(xm, xn) ≤ps(xm, xm−1) + ... + ps(xn+1, xn) ≤(km−1+ ... + kn)[p(x1, x0) + p(y1, y0)] =k n− km 1 − k [p(x1, x0) + p(y1, y0)] ≤ k n 1 − k[p(x1, x0) + p(y1, y0)]. (9)
Similarly, we can verify
(10) ps(ym, yn) ≤
kn
1 − k[p(x1, x0) + p(y1, y0)].
Since k ∈ [0, 1), so from (9) and (10), the sequences {xn} and {yn} are Cauchy in
the metric space (X, ps). The partial metric space (X, p) is complete, then from
Lemma 1, the sequences {xn} and {yn} converge in the metric space (X, ps),
say lim n→+∞p s(x n, x) = 0 and lim n→+∞p s(y
n, y) = 0. Again from Lemma 1, we have
(11) p(x, x) = lim
n→+∞p(xn, x) = limn→+∞p(xn, xn)
(12) p(y, y) = lim
n→+∞p(yn, y) =n→+∞lim p(yn, yn).
By (p2) we have p(xn, xn) ≤ p(xn+1, xn), so letting n → +∞ in (5), we get
lim
n→+∞p(xn, xn) = 0 because k ∈ [0, 1). Therefore, from (11), we have
p(x, x) = lim
n→+∞p(xn, x) = limn→+∞p(xn, xn) = 0.
Similarly, we have
p(y, y) = lim
n→+∞p(yn, y) = limn→+∞p(yn, yn) = 0.
Now, we claim that F (x, y) = x and F (y, x) = y.
Let ε > 0. Since F is continuous at (x, y), hence there exists δ > 0 such that if (u, v) ∈ X × X verifying ν((x, y), (u, v)) < ν((x, y), (x, y)) + δ, meaning that
we have
p(F (x, y), F (u, v)) < p(F (x, y), F (x, y)) +ε 2. Since lim
n→+∞p(xn, x) =n→+∞lim p(yn, y) = 0, for α = min( δ 2,
ε
2) > 0, there exist
n0, m0∈ N such that, for n ≥ n0, m ≥ m0,
p(xn, x) < α and p(ym, y) < α.
Then, for n ∈ N, n ≥ max(n0, m0), we have p(xn, x) + p(yn, y) < 2α < δ. Now,
for any n ≥ max(n0, m0),
p(F (x, y), x) ≤p(F (x, y), xn+1) + p(xn+1, x)
=p(F (x, y), F (xn, yn)) + p(xn+1, x)
<p(F (x, y), F (x, y)) +ε 2+ α <p(F (x, y), F (x, y)) + ε. On the other hand, by (4)
p(F (x, y), F (x, y)) ≤ k2(p(x, x) + p(y, y)) = 0, then p(F (x, y), F (x, y)) = 0, so for any ε > 0
p(F (x, y), x) < ε.
This implies that F (x, y) = x. Similarly, we can show that F (y, x) = y. The proof of Theorem 1 is completed.
In the following theorem we remove the continuity of F . But, we add a condition on X.
Theorem 2. Let (X, ≤X) be a partially ordered set and suppose there is a
partial metric p on X such that (X, p) is a complete partial metric space. Assume that X satisfies the following property:
(i) if a non-decreasing sequence {xn} converges to x in (X, p), then xn ≤X x,
(ii) if a non-increasing sequence {yn} converges to y in (X, p), then yn≥Xy.
Let F : X × X → X be a mapping having the mixed monotone property on X. Suppose that there exists a k ∈ [0, 1) such that (4) holds for each x ≥X u
and y ≤X v. If there exist x0, y0 ∈ X such that x0 ≤X F (x0, y0) and y0 ≥X
F (y0, x0), then F has a coupled fixed point.
Proof. Following the proof of Theorem 1, we only have to show that F (x, y) = x and F (y, x) = y. Since lim
n→+∞p(xn, x) = limn→+∞p(ym, y) = 0, there exist n1∈ N,
n2∈ N such that, for all n ≥ n1 and m ≥ n2, we have
p(xn, x) <
ε
3, p(xm, y) < ε 3.
Since {xn} and {yn} converge respectively to x and y in (X, p), hence taking
n ∈ N, n ≥ max{n1, n2}, we have xn≤X x and yn≥Xy. Therefore, from (4)
p(F (x, y), x) ≤p(F (x, y), xn+1) + p(xn+1, x)
=p(F (x, y), F (xn, yn)) + p(xn+1, x)
≤k
2[p(x, xn) + p(y, yn)] + p(xn+1, x) ≤p(x, xn) + p(y, yn) + p(xn+1, x) < ε.
This implies that F (x, y) = x. Similarly, we can show that p(F (y, x), y) < ε, implying that F (y, x) = y. This completes the proof of Theorem 2.
Now we shall prove the uniqueness of a coupled fixed point. Note that if (X, ≤X)
is a partially ordered set, then we endow the product X × X with the following partial order
for (x, y), (u, v) ∈ X × X, (x, y) ≤ (u, v) ⇐⇒ x ≤Xu, y ≥X v.
Theorem 3. In addition to the hypotheses of Theorem 1, suppose that for every (x, y), (y∗, x∗) ∈ X×X, there exists a (u, v) ∈ X×X such that (F (u, v), F (v, u))
is comparable to (F (x, y), F (y, x)) and (F (x∗, y∗), F (y∗, x∗)). Then F has a
unique coupled fixed point.
Proof. Let (x, y) be a coupled fixed of F . If (x∗, y∗) is another coupled fixed
point, then it suffices to show that
ν((x, y), (x∗, y∗)) =: p(x, x∗) + p(y, y∗) = 0, where lim
n→+∞p(xn, x) = limn→+∞p(yn, y) = 0. We consider two cases
• If (x, y) is comparable to (x∗, y∗), with respect to the ordering in X ×X, then,
for every n = 0, 1, 2, ..., we have (Fn(x, y), Fn(y, x)) = (x, y) is comparable to
(Fn(x∗, y∗), Fn(y∗, x∗)) = (x∗, y∗). Also
ν((x, y), (x∗, y∗)) =p(x, x∗) + p(y, y∗)
=p(Fn(x, y), Fn(x∗, y∗)) + p(Fn(y, x), Fn(y∗, x∗)) ≤kn(p(x, x∗) + p(y, y∗))
=knν((x, y), (x∗, y∗)). This implies that ν((x, y), (x∗, y∗)) = 0.
• If (x, y) is not comparable to (x∗, y∗), then there exists a (u, v) ∈ X × X such
that (F (u, v), F (v, u)) is comparable to (F (x, y), F (y, x)) and (F (x∗, y∗), F (y∗, x∗)).
(Fn(x, y), Fn(y, x)) = (x, y) and (Fn(x∗, y∗), Fn(y∗, x∗)) = (x∗, y∗). Therefore, ν((x, y), (x∗, y∗)) =ν((Fn(x, y), Fn(y, x)), (Fn(x∗, y∗), Fn(y∗, x∗)))
≤ν((Fn(x, y), Fn(y, x)), (Fn(u, v), Fn(v, u))) +ν((Fn(u, v), Fn(v, u)), (Fn(x∗, y∗), Fn(y∗, x∗))) ≤kn{p(x, u) + p(y, v) + p(u, x∗) + p(v, y∗)} →0 as n → +∞
so that ν((x, y), (x∗, y∗)) = 0.
Alternatively, if we know that the elements x0, y0are such that x0≤Xy0, then
we can also demonstrate that the components x and y of the coupled fixed point are indeed the same.
Theorem 4. In addition to the hypothesis of Theorem 1 (resp. Theorem 2), suppose that x0, y0 in X are comparable. Then x = y.
Proof. Recall that x0∈ X is such that x0≤X F (x0, y0). Now, if x0≤Xy0, we
claim that, for all n ∈ N,
xn ≤X yn.
Indeed, by the mixed monotone property of F ,
x1= F (x0, y0) ≤X F (y0, x0) = y1.
Assume that xn≤Xyn for some n. Now, consider,
xn+1=Fn+1(x0, y0) = F (Fn(x0, y0), Fn(y0, x0))
=F (xn, yn)
≤XF (yn, xn) = yn+1.
Hence, xn ≤X yn for all n. Since lim
n→+∞p(xn, x) =n→+∞lim p(yn, y) = 0, then for
a given ε > 0, there exists n0∈ N such that for all n ≥ n0
p(x, xn) <
ε
4, p(y, yn) < ε 4. Now for all n ≥ n0,
p(x, y) ≤p(x, xn+1) + p(xn+1, y) ≤p(x, xn+1) + p(xn+1, yn+1) + p(yn+1, y) <ε 4+ p(F (xn, yn), F (yn, xn)) + ε 4 ≤ε2+ kp(yn, xn) ≤ε2+ k[p(yn, y) + p(y, x) + p(x, xn)] <ε + kp(y, x).
This implies that (1 − k)p(x, y) < ε, which in turn leads to p(x, y) = 0, since ε > 0 is arbitrary and hence we have x = y.
Similarly, if x0 ≥X y0, then it is possible to show xn ≥X yn for all n and that
p(x, y) = 0.
Now, we give an example illustrating our result. Here, neither the continuity of the mapping F is satisfied, neither the conditions (i) and (ii) given in Theorem 2 hold, but we still obtain a coupled fixed point result.
Example 1. Let X = [0, 1] endowed with the natural ordering of real numbers ≤. Let p(x, y) = max(x, y). Obviously, for any x, y ∈ X, ps(x, y) = |x − y|. Then, (X, ps) is a complete metric space, and so for (X, p). Consider k =25. let F : X × X → X be defined as
F (x, y) = (x−3y
5 si x ≥ 3y
0 si x < 3y. Also, there are x0= 0 and y0= 0 in X such that
x0= 0 ≤ F (0, 0) = F (x0, y0) and y0= 0 ≥ F (0, 0) = F (y0, x0).
Obviously, F has the mixed monotone property. Letting x ≥ u and y ≤ v, then p(F (x, y), F (u, v)) = max{F (x, y), F (u, v)}
≤15x = 1
5max{x, u} ≤15(p(x, u) + p(y, v)) =k
2(p(x, u) + p(y, v)), which is the contraction (4).
The mapping F is not continuous on X × X, since it isn’t continuous at (x, y) = (2, 1). Indeed, let us take xn = 3n+14 and yn = 1n for any n ∈ N∗. For a large
n, we have ν((xn, yn), (x, y)) =p( 3 n+ 1 4, 2) + p( 1 n, 1)
=2 + 1 = p(x, x) + p(y, y) = ν((x, y), (x, y)).
Thus, (xn, yn) → (2, 1) in (X × X, ν). On the other hand, x = 2 < 3y = 3 and
xn≥ 3yn, so F (x, y) = 0 and F (xn, yn) =15(xn− 3yn). Then, p(F (xn, yn), F (x, y)) = F (xn, yn) = 1 5(xn− 3yn) = 1 20 6= 0 = F (x, y) = p(F (x, y), F (x, y)), giving that F is not continuous at (2, 1).
Moreover, the condition (ii) given in Theorem 2 doesn’t hold. Indeed, the sequence {yn} = {n1} is non-increasing and converges for example to 4 in (X, p),
but yn ≤ 4 for a large n.
Even the continuity of F and the conditions (i) and (ii) given in Theorem 2 don’t hold, we have obtained a coupled fixed point result, that is F has (0, 0) as a coupled fixed point.
Now, we will show that the coupled fixed point result of Bhashkar and Laksh-mikantham [10] is not applicable in this case. Suppose that
d(F (x, y), F (u, v)) ≤ k0
2 (d(x, u) + d(y, v))
for each k0 ∈ [0, 1) and all x ≥ u, y ≤ v, where d is the standard metric given
by d(x, y) = |x − y|. We have
|F (x, y) − F (u, v)| ≤ k0
2 ((x − u) + (y − v)). Taking x = u = 9, v = 2 and y = 1, we get F (x, y) = 6
5 and F (u, v) = 3 5. It follows that 3 5≤ k0 2, so k0 ≥6 5, that is a contradiction. References
1. Agarwal, R.P., El-Gebeily, M.A. and O’Regan, D.(1978): Generalized contractions in partially ordered metric spaces, Appl. Anal. 87, 1-8.
2. Altun, I. and Erduran, A. (2011): Fixed point theorems for monotone mappings on partial metric spaces, Fixed Point Theory Appl. Article ID 508730, 10 pages. 3. Altun, I. and Simsek, H.(2010): Some fixed point theorems on ordered metric spaces and application, Fixed Point Theory Appl. Article ID 621469, 17 pages.
4. Altun, I., Sola, F. and Simsek, H.(2010): Generalized contractions on partial metric spaces, Topology and its Appl. 157, no. 18, 2778-2785.
5. Amini-Harandi, A. and Emami, H.(2010): A fixed point theorem for contraction type maps in partially ordered metric spaces and application to ordinary differential equations, Nonlinear Anal. 72, 2238-2242.
6. Aydi, H. (2011): Coincidence and common fixed point results for contraction type maps in partially ordered metric spaces, Int. J. Math. Anal. 5, no. 13, 631-642. 7. Aydi, H.(2011): Some fixed point results in ordered partial metric spaces, J. Non-linear Sci. Appl, 4, no. 3, 210-217.
8. Aydi, H.(2011): Some coupled fixed point results on partial metric spaces, Interna-tional J. Math. Math. Sciences, Article ID 647091, 11 pages.
9. Aydi, H.(2011): Fixed point results for weakly contractive mappings in ordered partial metric spaces, J. Adv. Math. Studies, 4, no. 2, 1-12.
10. Bhashkar, T.G. and Lakshmikantham, V.(2006): Fixed point theorems in partially ordered cone metric spaces and applications, Nonlinear Anal. 65, no. 7, 825—832.
11. ´Ciri´c, Lj., Caki´c, N., Rajovi´c, M. and Ume, J.S.(2008): Monotone generalized nonlinear contractions in partially ordered metric spaces, Fixed Point Theory Appl. Article ID 131294, 17 pages.
12. Harjani, J. and Sadarangani, K.(2010): Generalized contractions in partially or-dered metric spaces and applications to ordinary differential equations, Nonlinear Anal. 72, 1188-1197.
13. Lakshmikantham, V. and ´Ciri´c, Lj.(2009): Coupled fixed point theorems for non-linear contractions in partially ordered metric space, Nonnon-linear Anal. 70, 4341—4349. 14. Van Luong, N. and Xuan Thuan, N.(2010): Coupled fixed point theorems in partially ordered metric spaces, Bulletin Math. Anal. Appl. 2, no. 4, 16-24.
15. Matthews, S.G.(1994): Partial metric topology, in: Proc. 8th Summer Conference on General Topology and Applications, in: Ann. New York Acad. Sci. 728, 183-197. 16. Nieto, J.J. and Rodríguez-López, R.(2005): Contractive mapping theorems in partially ordered sets and applications to ordinary differential equation, Order 22, 223-239.
17. Nashine, N.K. and Samet, B.(2010): Fixed point results for mapping satisfying (ψ, ϕ)-weakly contraction in partially ordered metric spaces, Nonlinear Anal. 74, 2201-2209.
18. Oltra, S. and Valero, O.(2004): Banach’s fixed point theorem for partial metric spaces, Rend. Istit. Mat. Univ. Trieste, 36, 17-26.
19. O’ Neill, S.J.(1996): Partial metrics, valuations and domain theory, in: Proc. 11th Summer Conference on General Topology and Applications, in: Ann. New York Acad. Sci. 806, 304-315.
20. Ran, A.C.M and Reurings, M.C.B.(2004): A fixed point theorem in partially ordered sets and some applications to matrix equations, Proc. Amer. Math. Soc. 132, 1435-1443.
21. Romaguera, S.(2010): A Kirk type characterization of completeness for partial metric spaces, Fixed Point Theory Appl. Article ID 493298, 6 pages.
22. Romaguera, S. and Schellekens, M.(2005): Partial metric monoids and semivalu-ation spaces, Topology and its Appl. 153, 948-962.
23. Romaguera, S. and Valero, O.(2009): A quantitative computational model for complete partialmetric spaces via formal balls, Mathematical Structures in Computer Science 19, no. 3, 541-563.
24. Samet, B.(2010): Coupled fixed point theorems for a generalized Meir-Keeler contraction in partially ordered metric spaces, Nonlinear Anal. 72, 4508-4517. 25. Schellekens, M.(2004): The correspondence between partial metrics and semival-uations, Theoret. Comput. Sci. 315, 135-149.
26. Valero, O. (2005): On Banach fixed point theorems for partial metric spaces, Appl. Gen. Topol. 6, no. 2, 229-240.
