In the section, we denote by Ψ the class of functions \psi :{{\mathbb{R}}^{+}}^{3}\to {\mathbb{R}}^{+} satisfying the following conditions:
({\psi}_{1}) ψ is an increasing and continuous function in each coordinate;
({\psi}_{2}) for t\in {\mathbb{R}}^{+}, \psi (t,t,t)\le t, \psi (t,0,0)\le t and \psi (0,0,t)\le t.
Next, we denote by Θ the class of functions \phi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+} satisfying the following conditions:
({\phi}_{1}) φ is continuous and nondecreasing;
({\phi}_{2}) for t>0, \phi (t)>0 and \phi (0)=0.
And we denote by Φ the class of functions \varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+} satisfying the following conditions:
({\varphi}_{1}) ϕ is continuous;
({\varphi}_{2}) for t>0, \varphi (t)>0 and \varphi (0)=0.
We now state a new notion of cyclic \mathcal{CW}contractions in partial metric spaces as follows.
Definition 5 Let (X,p) be a partial metric space, m\in \mathbb{N}, {A}_{1},{A}_{2},\dots ,{A}_{m} be nonempty subsets of X and Y={\bigcup}_{i=1}^{m}{A}_{i}. An operator f:Y\to Y is called a cyclic \mathcal{CW}contraction if

(1)
{\bigcup}_{i=1}^{m}{A}_{i} is a cyclic representation of Y with respect to f;

(2)
for any x\in {A}_{i}, y\in {A}_{i+1}, i=1,2,\dots ,m,
\phi (p(fx,fy))\le \psi (\phi (p(x,y)),\phi (p(x,fx)),\phi (p(y,fy)))\varphi (M(x,y)),
(2.1)
where \psi \in \mathrm{\Psi}, \phi \in \mathrm{\Theta}, \varphi \in \mathrm{\Phi}, and M(x,y)=max\{p(x,y),p(x,fx),p(y,fy)\}.
Theorem 3 Let (X,p) be a complete partial metric space, m\in \mathbb{N}, {A}_{1},{A}_{2},\dots ,{A}_{m} be nonempty closed subsets of X and Y={\bigcup}_{i=1}^{m}{A}_{i}. Let f:Y\to Y be a cyclic \mathcal{CW}contraction. Then f has a unique fixed point z\in {\bigcap}_{i=1}^{m}{A}_{i}.
Proof Given {x}_{0} and let {x}_{n+1}=f{x}_{n}={f}^{n}{x}_{0} for n=0,1,2,\dots . If there exists {n}_{0}\in \mathbb{N} such that {x}_{{n}_{0}+1}={x}_{{n}_{0}}, then we finished the proof. Suppose that {x}_{n+1}\ne {x}_{n} for any n=0,1,2,\dots . Notice that for any n\ge 0, there exists {i}_{n}\in \{1,2,\dots ,m\} such that {x}_{n}\in {A}_{{i}_{n}} and {x}_{n+1}\in {A}_{{i}_{n}+1}.
Step 1. We will prove that
\underset{n\to \mathrm{\infty}}{lim}p({x}_{n},{x}_{n+1})=0,\phantom{\rule{1em}{0ex}}\text{that is},\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty}}{lim}{d}_{p}({x}_{n},{x}_{n+1})=0.
Using (2.1), we have
\begin{array}{rcl}\phi (p({x}_{n},{x}_{n+1}))& =& \phi (p(f{x}_{n1},f{x}_{n}))\\ \le & \psi (\phi (p({x}_{n1},{x}_{n})),\phi (p({x}_{n1},f{x}_{n1})),\phi (p({x}_{n},f{x}_{n})))\varphi (M({x}_{n1},{x}_{n}))\\ =& \psi (\phi (p({x}_{n1},{x}_{n})),\phi (p({x}_{n1},{x}_{n})),\phi (p({x}_{n},{x}_{n+1})))\varphi (M({x}_{n1},{x}_{n})),\end{array}
where
\begin{array}{rl}M({x}_{n1},{x}_{n})& =max\{p({x}_{n1},{x}_{n}),p({x}_{n1},f{x}_{n1}),p({x}_{n},f{x}_{n})\}\\ =max\{p({x}_{n1},{x}_{n}),p({x}_{n1},{x}_{n}),p({x}_{n},{x}_{n+1})\}.\end{array}
If M({x}_{n1},{x}_{n})=p({x}_{n},{x}_{n+1}), then
\begin{array}{rcl}\phi (p({x}_{n},{x}_{n+1}))& \le & \psi (\phi (p({x}_{n},{x}_{n+1})),\phi (p({x}_{n},{x}_{n+1})),\phi (p({x}_{n},{x}_{n+1})))\varphi (p({x}_{n},{x}_{n+1}))\\ \le & \phi (p({x}_{n},{x}_{n+1}))\varphi (p({x}_{n},{x}_{n+1})),\end{array}
which implies that \varphi (p({x}_{n},{x}_{n+1}))=0, and hence p({x}_{n},{x}_{n+1})=0. This contradicts our initial assumption.
From the above argument, we have that for each n\in \mathbb{N},
\phi (p({x}_{n},{x}_{n+1}))\le \phi (p({x}_{n1},{x}_{n}))\varphi (p({x}_{n1},{x}_{n})),
(2.2)
and
p({x}_{n},{x}_{n+1})<p({x}_{n1},{x}_{n}).
And since the sequence \{p({x}_{n},{x}_{n+1})\} is decreasing, it must converge to some \eta \ge 0. Taking limit as n\to \mathrm{\infty} in (2.2) and by the continuity of φ and ϕ, we get
\phi (\eta )\le \phi (\eta )\varphi (\eta ),
and so we conclude that \varphi (\eta )=0 and \eta =0. Thus, we have
\underset{n\to \mathrm{\infty}}{lim}p({x}_{n},{x}_{n+1})=0.
(2.3)
By (p_{2}), we also have
\underset{n\to \mathrm{\infty}}{lim}p({x}_{n},{x}_{n})=0.
(2.4)
Since {d}_{p}(x,y)\le 2p(x,y)p(x,x)p(y,y) for all x,y\in X, using (2.3) and (2.4), we obtain that
\underset{n\to \mathrm{\infty}}{lim}{d}_{p}({x}_{n},{x}_{n+1})=0.
(2.5)
Step 2. We show that \{{x}_{n}\} is a Cauchy sequence in the metric space (Y,{d}_{p}). We claim that the following result holds.
Claim For every \epsilon >0, there exists n\in \mathbb{N} such that if r,q\ge n with rq=1modm, then {d}_{p}({x}_{r},{x}_{q})<\epsilon.
Suppose the above statement is false. Then there exists \u03f5>0 such that for any n\in \mathbb{N}, there are {r}_{n},{q}_{n}\in \mathbb{N} with {r}_{n}>{q}_{n}\ge n with {r}_{n}{q}_{n}=1modm satisfying
{d}_{p}({x}_{{q}_{n}},{x}_{{r}_{n}})\ge \u03f5.
Now, we let n>2m. Then corresponding to {q}_{n}\ge n use, we can choose {r}_{n} in such a way it is the smallest integer with {r}_{n}>{q}_{n}\ge n satisfying {r}_{n}{q}_{n}=1modm and {d}_{p}({x}_{{q}_{n}},{x}_{{r}_{n}})\ge \u03f5. Therefore, {d}_{p}({x}_{{q}_{n}},{x}_{{r}_{n}m})\le \u03f5 and
\begin{array}{rl}\u03f5& \le {d}_{p}({x}_{{q}_{n}},{x}_{{r}_{n}})\\ \le {d}_{p}({x}_{{q}_{n}},{x}_{{r}_{n}m})+\sum _{i=1}^{m}{d}_{p}({x}_{{r}_{ni}},{x}_{{r}_{ni+1}})\\ <\u03f5+\sum _{i=1}^{m}{d}_{p}({x}_{{r}_{ni}},{x}_{{r}_{ni+1}}).\end{array}
Letting n\to \mathrm{\infty}, we obtain that
\underset{n\to \mathrm{\infty}}{lim}{d}_{p}({x}_{{q}_{n}},{x}_{{r}_{n}})=\u03f5.
(2.6)
On the other hand, we can conclude that
\begin{array}{rl}\u03f5& \le {d}_{p}({x}_{{q}_{n}},{x}_{{r}_{n}})\\ \le {d}_{p}({x}_{{q}_{n}},{x}_{{q}_{n+1}})+{d}_{p}({x}_{{q}_{n+1}},{x}_{{r}_{n+1}})+{d}_{p}({x}_{{r}_{n+1}},{x}_{{r}_{n}})\\ \le {d}_{p}({x}_{{q}_{n}},{x}_{{q}_{n+1}})+{d}_{p}({x}_{{q}_{n+1}},{x}_{{q}_{n}})+{d}_{p}({x}_{{q}_{n}},{x}_{{r}_{n}})+{d}_{p}({x}_{{r}_{n}},{x}_{{r}_{n+1}})+{d}_{p}({x}_{{r}_{n+1}},{x}_{{r}_{n}}).\end{array}
Letting n\to \mathrm{\infty}, we obtain that
\underset{n\to \mathrm{\infty}}{lim}{d}_{p}({x}_{{q}_{n+1}},{x}_{{r}_{n+1}})=\u03f5.
(2.7)
Since {d}_{p}(x,y)=2p(x,y)p(x,x)p(y,y) and using (2.4), (2.6) and (2.7), we have that
\underset{n\to \mathrm{\infty}}{lim}p({x}_{{q}_{n}},{x}_{{r}_{n}})=\frac{\u03f5}{2},
(2.8)
and
\underset{n\to \mathrm{\infty}}{lim}p({x}_{{q}_{n+1}},{x}_{{r}_{n+1}})=\frac{\u03f5}{2}.
(2.9)
Since {x}_{{q}_{n}} and {x}_{{r}_{n}} lie in different adjacently labeled sets {A}_{i} and {A}_{i+1} for certain 1\le i\le m, by using the fact that f is a cyclic \mathcal{CW}contraction, we have
\begin{array}{rcl}\phi (p(f{x}_{{q}_{n}+1},f{x}_{{r}_{n}+1}))& =& \phi (p(f{x}_{{q}_{n}},f{x}_{{r}_{n}}))\\ \le & \psi (\phi (p({x}_{{q}_{n}},{x}_{{r}_{n}})),\phi (p({x}_{{q}_{n}},f{x}_{{q}_{n}})),\phi (p({x}_{{r}_{n}},f{x}_{{r}_{n}})))\\ \varphi (M({x}_{{q}_{n}},{x}_{{r}_{n}}))\\ =& \psi (\phi (p({x}_{{q}_{n}},{x}_{{r}_{n}})),\phi (p({x}_{{q}_{n}},{x}_{{q}_{n}+1})),\phi (p({x}_{{r}_{n}},{x}_{{r}_{n}+1})))\\ \varphi (M({x}_{{q}_{n}},{x}_{{r}_{n}})),\end{array}
where
M({x}_{{q}_{n}},{x}_{{r}_{n}})=max\{p({x}_{{q}_{n}},{x}_{{r}_{n}}),p({x}_{{q}_{n}},{x}_{{q}_{n}+1}),p({x}_{{r}_{n}},{x}_{{r}_{n}+1})\}.
Thus, letting n\to \mathrm{\infty}, we can conclude that
\phi \left(\frac{\u03f5}{2}\right)\le \psi (\phi \left(\frac{\u03f5}{2}\right),\phi (0),\phi (0))\varphi \left(\frac{\u03f5}{2}\right)\le \phi \left(\frac{\u03f5}{2}\right)\varphi \left(\frac{\u03f5}{2}\right),
which implies \varphi (\frac{\u03f5}{2})=0, that is, \u03f5=0. So, we get a contradiction. Therefore, our claim is proved.
In the sequel, we will show that \{{x}_{n}\} is a Cauchy sequence in the metric space (Y,{d}_{p}). Let \epsilon >0 be given. By our claim, there exists {n}_{1}\in \mathbb{N} such that if r,q\ge {n}_{1} with rq=1modm, then
{d}_{p}({x}_{r},{x}_{q})\le \frac{\epsilon}{2}.
Since {lim}_{n\to \mathrm{\infty}}{d}_{p}({x}_{n},{x}_{n+1})=0, there exists {n}_{2}\in \mathbb{N} such that
{d}_{p}({x}_{n},{x}_{n+1})\le \frac{\epsilon}{2m}
for any n\ge {n}_{2}.
Let r,q\ge max\{{n}_{1},{n}_{2}\} and r>q. Then there exists k\in \{1,2,\dots ,m\} such that rq=kmodm. Therefore, rq+j=1modm for j=mk+1, and so we have
\begin{array}{rl}{d}_{p}({x}_{q},{x}_{r})& \le {d}_{p}({x}_{q},{x}_{r+j})+{d}_{p}({x}_{r+j},{x}_{r+j1})+\cdots +{d}_{p}({x}_{r1},{x}_{r})\\ \le \frac{\epsilon}{2}+j\times \frac{\epsilon}{2m}\\ \le \frac{\epsilon}{2}+m\times \frac{\epsilon}{2m}=\epsilon .\end{array}
Thus, \{{x}_{n}\} is a Cauchy sequence in the metric space (Y,{d}_{p}).
Step 3. We show that f has a fixed point ν in {\bigcap}_{i=1}^{m}{A}_{i}.
Since Y is closed, the subspace (Y,p) is complete. Then from Lemma 1, we have that (Y,{d}_{p}) is complete. Thus, there exists \nu \in X such that
\underset{n\to \mathrm{\infty}}{lim}{d}_{p}({x}_{n},\nu )=0.
And it follows from Lemma 1 that we have
p(\nu ,\nu )=\underset{n\to \mathrm{\infty}}{lim}p({x}_{n},\nu )=\underset{n,m\to \mathrm{\infty}}{lim}p({x}_{n},{x}_{m}).
(2.10)
On the other hand, since the sequence \{{x}_{n}\} is a Cauchy sequence in the metric space (Y,{d}_{p}), we also have
\underset{n\to \mathrm{\infty}}{lim}{d}_{p}({x}_{n},{x}_{m})=0.
Since {d}_{p}(x,y)=2p(x,y)p(x,x)p(y,y), we can deduce that
\underset{n\to \mathrm{\infty}}{lim}p({x}_{n},{x}_{m})=0.
(2.11)
Since Y={\bigcup}_{i=1}^{m}{A}_{i} is a cyclic representation of X with respect to f, the sequence \{{x}_{n}\} has infinite terms in each {A}_{i} for i\in \{1,2,\dots ,m\}. Now, for all i=1,2,\dots ,m, we may take a subsequence \{{x}_{{n}_{k}}\} of \{{x}_{n}\} with {x}_{{n}_{k}}\in {A}_{i1} and also all converge to ν. Using (2.10) and (2.11), we have
p(\nu ,\nu )=\underset{n\to \mathrm{\infty}}{lim}p({x}_{n},\nu )=\underset{n\to \mathrm{\infty}}{lim}p({x}_{{n}_{k}},\nu )=0.
By (2.1),
\begin{array}{rl}\phi (p({x}_{{n}_{k+1}},f\nu ))& =\phi (p(f{x}_{{n}_{k}},f\nu ))\\ \le \psi (\phi (p({x}_{{n}_{k}},\nu )),\phi (p({x}_{{n}_{k}},f{x}_{{n}_{k}})),\phi (p(\nu ,f\nu )))\varphi (M({x}_{{n}_{k}},\nu ))\\ =\psi (\phi (p({x}_{{n}_{k}},\nu )),\phi (p({x}_{{n}_{k}},{x}_{{n}_{k}+1})),\phi (p(\nu ,f\nu )))\varphi (M({x}_{{n}_{k}},\nu )),\end{array}
where
M({x}_{{n}_{k}},\nu )=max\{p({x}_{{n}_{k}},\nu ),p({x}_{{n}_{k}},{x}_{{n}_{k}+1}),p(\nu ,f\nu )\}.
Letting k\to \mathrm{\infty}, we have
\begin{array}{rl}\phi (p(\nu ,f\nu ))& \le \psi (\phi (0),\phi (0),\phi (p(\nu ,f\nu )))\varphi (p(\nu ,f\nu ))\\ \le \phi (p(\nu ,f\nu ))\varphi (p(\nu ,f\nu )),\end{array}
which implies \varphi (p(\nu ,f\nu ))=0, that is, p(\nu ,f\nu )=0. So, \nu =f\nu.
Step 4. Finally, to prove the uniqueness of the fixed point, suppose that μ, ν are fixed points of f. Then using the inequality (2.1), we obtain that
\begin{array}{rcl}\phi (p(\mu ,\nu ))& =& \phi (p(f\mu ,f\nu ))\le \psi (\phi (p(\mu ,\nu )),\phi (p(\mu ,f\mu )),\phi (p(\nu ,f\nu )))\\ \varphi (M(\mu ,\nu )),\end{array}
where
M(\mu ,\nu )=max\{p(\mu ,\nu ),p(\mu ,f\mu ),p(\nu ,f\nu )\}=p(\mu ,\nu ).
So, we also deduce that
\begin{array}{rl}\phi (p(\mu ,\nu ))& \le \psi \left(\phi (p(\mu ,\nu ),0,0)\right)\\ \le \phi (p(\mu ,\nu ))\varphi (p(\mu ,\nu )),\end{array}
which implies that \varphi (p(\mu ,\nu ))=0, and hence p(\mu ,\nu )=0, that is, \mu =\nu. So, we complete the proof. □
The following provides an example for Theorem 3.
Example 1 Let X=[0,1] and A=[0,1], B=[0,\frac{1}{2}], C=[0,\frac{1}{4}]. We define the partial metric p on X by
p(x,y)=max\{x,y\}\phantom{\rule{1em}{0ex}}\text{for all}x,y\in X,
and define the function f:X\to X by
f(x)=\frac{{x}^{2}}{1+x}\phantom{\rule{1em}{0ex}}\text{for all}x\in X.
Now, we let \phi ,\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+} and \psi :{{\mathbb{R}}^{+}}^{3}\to {\mathbb{R}}^{+} be
\phi (t)=2t,\phantom{\rule{2em}{0ex}}\varphi (t)=\frac{2t}{5(1+t)}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\psi (t)=\frac{4}{5}\cdot max\{{t}_{1},{t}_{2},{t}_{3}\}.
Then f is a cyclic \mathcal{CW}contraction and 0 is the unique fixed point.
Proof We claim that f is a cyclic \mathcal{CW}contraction.

(1)
Note that f(A)=[0,\frac{1}{2}]\subset B, f(B)=[0,\frac{1}{6}]\subset C and f(C)=[0,\frac{1}{20}]\subset A. Thus, A\cup B\cup C is a cyclic representation of X with respect to f;

(2)
For x\in A and y\in B (or, x\in B and y\in C), without loss of generality, we may assume that x\ge y, then we have
and
Since
\frac{2{x}^{2}}{1+x}\le \frac{8x}{5}\frac{2x}{5(1+x)},
we have
\begin{array}{rcl}\phi (p(fx,fy))& \le & \psi (\phi (p(x,y)),\phi (p(x,fx)),\phi (p(y,fy)))\\ \varphi (max\{p(x,y),p(x,fx),p(y,fy)\}).\end{array}
On the other hand, for x\in C and y\in A, without loss of generality, we may assume that x\le y, then it is easy to get the above inequality.
Note that Example 1 satisfies all of the hypotheses of Theorem 3, and we get that 0 is the unique fixed point. □