To begin this section with the following theorem.
Theorem 3
Let \((P,d,s,\le )\)be a complete partially ordered bmetric space with parameter \(s \ge 1\). Let \(S:P \rightarrow P\)be a continuous, nondecreasing mapping with regards to \(\le\)such that there exists \(\nu _0 \in P\)with \(\nu _0 \le S\nu _0\). Suppose that
$$\begin{aligned} \phi (sd(S\nu ,S\xi ))\le \phi (M(\nu ,\xi ))\psi (M(\nu ,\xi )) \end{aligned}$$
(3)
where \(\phi \in \Phi , \psi \in \Psi\), for any \(\nu ,\xi \in P\)with \(\nu \le \xi\)and
$$\begin{aligned} M(\nu ,\xi )=\max \left\{ \frac{d(\xi ,S\xi ) \left[ 1+d(\nu ,S\nu )\right] }{1+d(\nu ,\xi )},\frac{d(\xi ,S\nu ) \left[ 1+d(\nu ,S\xi )\right] }{1+d(\nu ,\xi )}, d(\nu ,\xi )\right\} . \end{aligned}$$
(4)
Then S has a fixed point in P.
Proof
If for some \(\nu _0 \in P\) such that \(S\nu _0=\nu _0\), then the proof is finished. Assume that \(\nu _0 < S\nu _0\), then construct a sequence \(\{\nu _n\} \subset P\) by \(\nu _{n+1}=S\nu _n\), for \(n\ge 0\). But S is nondecreasing then, we get the following by using mathematical induction
$$\begin{aligned} \nu _0 < S\nu _0=\nu _1\le S\nu _1=\nu _2\le .............\le S\nu _{n1}=\nu _n \le S\nu _n=\nu _{n+1}\le ........ \end{aligned}$$
(5)
If for some \(n_0\in {\mathbb {N}}\) such that \(\nu _{n_0}=\nu _{n_0+1}\) then from (5), \(\nu _{n_0}\) is a fixed point of S and we have nothing to prove. Suppose that \(\nu _n \ne \nu _{n+1}\), i.e., \(d(\nu _n, \nu _{n+1})>0\), for all \(n \ge 1\). Since \(\nu _n>\nu _{n1}\), for any \(n \ge 1\) and then by (3), we have
$$\begin{aligned} \begin{aligned} \phi (d(\nu _n,\nu _{n+1}))&= \phi (d(S\nu _{n1},S\nu _n))\le \phi (sd(S\nu _{n1},S\nu _n))\\ {}&\le \phi (M(\nu _{n1},\nu _n))\psi (M(\nu _{n1},\nu _n)), \end{aligned} \end{aligned}$$
(6)
where
$$\begin{aligned} \begin{aligned} M(\nu _{n1}, \nu _n)&=\max \left\{ \frac{d(\nu _n,S\nu _n) \left[ 1+d(\nu _{n1},S\nu _{n1})\right] }{1+d(\nu _{n1},\nu _n)}, \frac{d(\nu _n,S\nu _{n1})\left[ 1+d(\nu _{n1},S\nu _n)\right] }{1+d(\nu _{n1},\nu _n)},\right. \\&\quad \left. d(\nu _{n1},\nu _n)\right\} \\&= \max \{d(\nu _n,\nu _{n+1}),0, d(\nu _{n1},\nu _n)\}\\&= \max \{d(\nu _n,\nu _{n+1}),d(\nu _{n1},\nu _n)\} \end{aligned} \end{aligned}$$
(7)
which implies that
$$\begin{aligned} \phi (d(\nu _n,\nu _{n+1}))\le \phi (\max \{d(\nu _n,\nu _{n+1}),d(\nu _{n1},\nu _n)\})\psi (\max \{d(\nu _n,\nu _{n+1}),d(\nu _{n1},\nu _n)\}). \end{aligned}$$
(8)
If \(\max \{d(\nu _n,\nu _{n+1}), d(\nu _{n1},\nu _n)\}= d(\nu _n,\nu _{n+1})\) for some \(n \ge 1\), then from (8), we get
$$\begin{aligned} \phi (d(\nu _n,\nu _{n+1})) \le \phi (d(\nu _n,\nu _{n+1}))\psi (d(\nu _n,\nu _{n+1}))<\phi (d(\nu _n,\nu _{n+1})), \end{aligned}$$
(9)
which is a contradiction under (9). Thus, \(\max \{d(\nu _n,\nu _{n+1}), d(\nu _{n1},\nu _n)\}= d(\nu _{n1},\nu _n)\) for \(n \ge 1\) and hence from (8) again we have
$$\begin{aligned} \phi (d(\nu _n,\nu _{n+1})) \le \phi (d(\nu _n,\nu _{n1}))\psi (d(\nu _n,\nu _{n1}))<\phi (d(\nu _n,\nu _{n1})). \end{aligned}$$
(10)
Therefore, the sequence \(\{d(\nu _n,\nu _{n1})\}\) for \(n \ge 1\) is a monotone nonincreasing and bounded. From a result, we have
$$\begin{aligned} \lim \limits _{n \rightarrow +\infty }d(\nu _n,\nu _{n1})=\rho \ge 0. \end{aligned}$$
(11)
Now, taking the upper limit on both sides of (6), we obtain
$$\begin{aligned} \phi (\rho ) \le \phi (\rho )\lim \limits _{n \rightarrow +\infty } \inf \psi (d(\nu _n,\nu _{n1}))\le \phi (\rho )\psi (\rho )<\phi (\rho ) \end{aligned}$$
(12)
which is a contradiction under (12). Thus, \(\rho =0\). Hence, \(d(\nu _n,\nu _{n1})\rightarrow 0\) as \(n \rightarrow +\infty\).
Next, we prove that \(\{\nu _n\}\) is a Cauchy sequence in P. Assume opposite that \(\{\nu _n\}\) is not a Cauchy sequence. Then for some \(\epsilon >0\), we can get two subsequences \(\{\nu _{m_j}\}\) and \(\{\nu _{n_j}\}\) of \(\{\nu _n\}\), where \(n_j\) is the smallest index such that
$$\begin{aligned} n_j>m_j>j,~~~~~ d(\nu _{m_j},\nu _{n_j})\ge \epsilon \end{aligned}$$
(13)
and
$$\begin{aligned} d(\nu _{m_j},\nu _{{n_j}1}) < \epsilon . \end{aligned}$$
(14)
Applying triangular inequality in (13), we get
$$\begin{aligned} \begin{aligned} \epsilon&\le d(\nu _{m_j},\nu _{n_j})\le s d(\nu _{m_j},\nu _{{n_j}1})+sd(\nu _{n_j1},\nu _{n_j})\\&\le s^2d(\nu _{m_j},\nu _{{m_j}1})+s^2d(\nu _{{m_j}1}, \nu _{{n_j}1})+sd(\nu _{{n_j}1},\nu _{n_j}) \end{aligned} \end{aligned}$$
(15)
Furthermore,
$$\begin{aligned} \begin{aligned} d(\nu _{{m_j}1},\nu _{{n_j}1})\le s d(\nu _{{m_j}1},\nu _{m_j})+sd(\nu _{m_j},\nu _{{n_j}1}) \le s d(\nu _{{m_j}1},\nu _{{m_j}})+s \epsilon \end{aligned} \end{aligned}$$
(16)
Letting \(j \rightarrow +\infty\) in Eqs. (15), (16) and combining together, we obtain the following inequality
$$\begin{aligned} \frac{\epsilon }{s^2}\le \lim \limits _{j \rightarrow +\infty } \sup d(\nu _{{m_j}1},\nu _{{n_j}1}) \le s \epsilon . \end{aligned}$$
(17)
Similarly, we can get the following inequalities by using triangular inequality
$$\begin{aligned} \frac{\epsilon }{s^2}\le \lim \limits _{j \rightarrow +\infty } \inf d(\nu _{{m_j}1},\nu _{{n_j}1}) \le s \epsilon , \end{aligned}$$
(18)
and
$$\begin{aligned} \frac{\epsilon }{s}\le \lim \limits _{j \rightarrow +\infty } \sup d(\nu _{{m_j}1},\nu _{n_j}) \le s \epsilon ^2. \end{aligned}$$
(19)
Let
$$\begin{aligned} \begin{aligned} M(\nu _{{m_j}1}, \nu _{{n_j}1})&=\max \left\{ \frac{d(\nu _{{n_j}1},S\nu _{{n_j}1}) \left[ 1+d(\nu _{{m_j}1},S\nu _{{m_j}1})\right] }{1+d(\nu _{{m_j}1}, \nu _{{n_j}1})},\right. \\&\quad \left. \frac{d(\nu _{{n_j}1},S\nu _{{m_j}1}) ~\left[ 1+d(\nu _{{m_j}1},S\nu _{{n_j}1})\right] }{1+d(\nu _{{m_j}1}, \nu _{{n_j}1})}, d(\nu _{{m_j}1},\nu _{{n_j}1})\right\} \\&=\max \left\{ \frac{d(\nu _{{n_j}1},\nu _{n_j})\left[ 1+d(\nu _{{m_j}1},\nu _{m_j}) \right] }{1+d(\nu _{{m_j}1}, \nu _{{n_j}1}) },\right. \\&\quad \left. \frac{d(\nu _{{n_j}1},\nu _{m_j})\left[ 1+d(\nu _{{m_j}1},\nu _{n_j})\right] }{1+d(\nu _{{m_j}1},\nu _{{n_j}1})}, d(\nu _{{m_j}1}, \nu _{{n_j}1})\right\} . \end{aligned} \end{aligned}$$
(20)
From (20), we obtain the following inequalities
$$\begin{aligned} \frac{\epsilon }{s^2}\le \lim \limits _{j \rightarrow +\infty }\sup M (\nu _{{m_j}1}, \nu _{{n_j}1}) \le s \epsilon . \end{aligned}$$
(21)
and
$$\begin{aligned} \frac{\epsilon }{s^2}\le \lim \limits _{j \rightarrow +\infty } \inf M(\nu _{{m_j}1}, \nu _{{n_j}1}) \le s \epsilon . \end{aligned}$$
(22)
Form (5), we have \(\nu _{{m_j}1}<\nu _{{n_j}1}\), then
$$\begin{aligned} \phi (sd(\nu _{m_j},\nu _{n_j}))=\phi (sd(S\nu _{{m_j}1},S\nu _{{n_j}1})) \le \phi (M(\nu _{{m_j}1},\nu _{{n_j}1})) \psi (M(\nu _{{m_j}1},\nu _{{n_j}1})). \end{aligned}$$
(23)
Now, letting \(j \rightarrow +\infty\) and use of (21), (22), we obtain
$$\begin{aligned} \begin{aligned} \phi (s \epsilon )&\le \phi (s \lim \limits _{j \rightarrow +\infty } d(\nu _{m_j}, \nu _{n_j}))\\&\le \phi ( \lim \limits _{j \rightarrow +\infty } \sup M (\nu _{{m_j}1}, \nu _{{n_j}1})) \lim \limits _{j \rightarrow +\infty }\inf \psi ( M (\nu _{{m_j}1}, \nu _{{n_j}1}))\\&\le \phi (s \epsilon ) \psi (\lim \limits _{j \rightarrow +\infty }\inf M (\nu _{{m_j}1}, \nu _{{n_j}1}))\\&<\phi (s \epsilon ) \end{aligned} \end{aligned}$$
(24)
this is a contradiction under (24). Hence, \(\{\nu _n\}\) is a Cauchy sequence and converges to some \(\mu \in P\) as P is complete. Also, the continuity of S implies that
$$\begin{aligned} S\mu =S(\lim \limits _{n\rightarrow +\infty }\nu _n)=\lim \limits _{n\rightarrow +\infty }S\nu _n=\lim \limits _{n\rightarrow +\infty }\nu _{n+1}=\mu . \end{aligned}$$
(25)
Therefore, \(\mu\) is a fixed point of S in P. \(\square\)
By weakening the continuity property of a map S in Theorem 3, we have the following result.
Theorem 4
In Theorem 3, if P has a property that, the sequence \(\{\nu _n\}\)is a nondecreasing such that \(\nu _n\rightarrow v\), implies that \(\nu _n \le v\), for all \(n \in {\mathbb {N}}\), i.e., \(v=\sup \nu _n\)then a non continuous map S has a fixed point in P.
Proof
From Theorem 3, we take the same sequence \(\{\nu _n\}\) in P such that \(\nu _0 \le \nu _1\le \nu _2\le \nu _3\le ........\le \nu _n\le \nu _{n+1}\le ..........\), i.e., the sequence \(\{\nu _n\}\) is nondecreasing and converges to some v in P. Thus, from the hypotheses we have \(\nu _n \le v\), for any \(n \in {\mathbb {N}}\), implies that \(v=\sup \nu _n\).
Next, we prove that v is a fixed point of S in P, that is \(Sv=v\). Suppose \(Sv\ne v\), that is \(d(Sv,v)\ne 0\). Let
$$\begin{aligned} \begin{aligned} M(\nu _n, v)&=\max \left\{ \frac{d(v,Sv) \left[ 1+d(\nu _n,S\nu _n)\right] }{1+d(\nu _n, v)}, \frac{d(v,S\nu _n)\left[ 1+d(\nu _n,Sv)\right] }{1+d(\nu _n, v)}, d(\nu _n, v)\right\} \\&=\max \left\{ \frac{d(v,Sv) \left[ 1+d(\nu _n,\nu _{n+1})\right] }{1+d(\nu _n,v)}, \frac{d(v,\nu _{n+1})\left[ 1+d(\nu _n,Sv)\right] }{1+d(\nu _n,v)}, d(\nu _n,v)\right\} \end{aligned} \end{aligned}$$
(26)
Letting \(n\rightarrow +\infty\) and from \(\lim \limits _{n\rightarrow +\infty }\nu _n=v\), we get
$$\begin{aligned} \lim \limits _{n \rightarrow +\infty }M(\nu _n, v)= \max \{d(v,Sv),0,0\}=d(v,Sv). \end{aligned}$$
(27)
We know that \(\nu _n \le v\), for all n then from contraction condition (3), we get
$$\begin{aligned} \phi (d(\nu _{n+1}, Sv))=\phi (d(S\nu _n, Sv)\le \phi (s d(S\nu _n, Sv)\le \phi (M(\nu _n, v))\psi (M(\nu _n, v)) \end{aligned}$$
(28)
Letting \(n \rightarrow +\infty\) and use of (27), we get
$$\begin{aligned} \phi (d(v,Sv)) \le \phi (d(v,Sv))\psi (d(v,Sv))< \phi (d(v,Sv)) \end{aligned}$$
(29)
which is wrong under (29). Thus, \(Sv=v\), that is S has a fixed point v in P. \(\square\)
The uniqueness of, an existing fixed point in Theorems 3 and 4 can get, if P has the following property:
For any \(\nu\), \(\xi\)\(\in P\), there exists \(v \in P\) such that \(v \le \nu\) and \(v \le \xi\).
Theorem 5
If P satisfies the above mentioned condition in Theorem 3 (or Theorem 4) then S has a unique fixed point.
Proof
From Theorem 3 (or Theorem 4), we conclude that S has a nonempty set of fixed points. Suppose that \(\nu ^*\) and \(\xi ^*\) be two fixed points of S then, we claim that \(\nu ^*=\xi ^*\). Suppose that \(\nu ^*\ne \xi ^*\), then from the hypotheses we have
$$\begin{aligned} \begin{aligned} \phi (d(S\nu ^*, S\xi ^*))&\le \phi (sd(S\nu ^*, S\xi ^*)) \le \phi (M(\nu ^*, \xi ^*))\psi (M(\nu ^*, \xi ^*)) \end{aligned} \end{aligned}$$
(30)
where
$$\begin{aligned} \begin{aligned} M(\nu ^*, \xi ^*)&=\max \left\{ \frac{d(\xi ^*,S\xi ^*) \left[ 1+d(\nu ^*,S\nu ^*)\right] }{1+d(\nu ^*,\xi ^*)}, \frac{d(\xi ^*,S\nu ^*) \left[ 1+d(\nu ^*,S\xi ^*)\right] }{1+d(\nu ^*,\xi ^*)}, d(\nu ^*,\xi ^*)\right\} \\&= \max \left\{ \frac{d(\xi ^*,\xi ^*) \left[ 1+d(\nu ^*,\nu ^*)\right] }{1+d(\nu ^*,\xi ^*)},\frac{d(\xi ^*,\nu ^*) \left[ 1+d(\nu ^*,\xi ^*)\right] }{1+d(\nu ^*,\xi ^*)}, d(\nu ^*,\xi ^*)\right\} \\&= \max \{0,d(\xi ^*,\nu ^*), d(\nu ^*,\xi ^*) \}\\&=d(\nu ^*,\xi ^*). \end{aligned} \end{aligned}$$
(31)
Then by Eq. (30), we have
$$\begin{aligned} \begin{aligned} \phi (d(\nu ^*, \xi ^*))&=\phi (d(S\nu ^*, S\xi ^*)) \le \phi (d(\nu ^*, \xi ^*))\psi (d(\nu ^*, \xi ^*)) < \phi (d(\nu ^*, \xi ^*)) \end{aligned} \end{aligned}$$
(32)
which is a contradiction under (32). Hence, \(\nu ^*= \xi ^*\).
The proof is completed. \(\square\)
Now, we have the results below, which are the generalizations of Theorems2.1 and 2.2 of [38] and the Corollaries2.1 and 2.2 of [39] in the space.
Corollary 1
Let \((P,d,s,\le )\)be a partially ordered bmetric space with a parameter s. Suppose \(S,f: P \rightarrow P\)are continuous mappings such that
 \((C_1)\).:

for some \(\psi \in \Psi\)and \(\phi \in \Phi\)with
$$\begin{aligned} \phi (sd(S\nu ,S\xi ))\le \phi (M_f(\nu ,\xi ))\psi (M_f(\nu ,\xi )) \end{aligned}$$
(33)
for any \(\nu\), \(\xi\)\(\in P\)such that \(f\nu \le f\xi\)and
$$\begin{aligned} M_f(\nu ,\xi )=\max \left\{ \frac{d(f\xi ,S\xi ) \left[ 1+d(f\nu ,S\nu )\right] }{1+d(f\nu ,f\xi )},\frac{d(f\xi ,S\nu ) \left[ 1+d(f\nu ,S\xi )\right] }{1+d(f\nu ,f\xi )}, d(f\nu ,f\xi )\right\} , \end{aligned}$$
(34)
 \((C_2)\):

\(SP \subset fP\)and fP is a complete subspace of P,
 \((C_3)\):

S is a monotone fnon decreasing mapping,
 \((C_4)\):

S and f are compatible.
If for some \(\nu _0 \in P\)such that \(f\nu _0 \le S\nu _0\), then S and f have a coincidence point in P.
Proof
By using lemma 1, we obtain a complete subspace fE of P, where \(E \subset P\) and f is onetoone self mapping on P. By Corollary 2.1 of [39], we have a sequence \(\{f\nu _n\} \subset fE\) for some \(\nu _0 \in E\) with \(f\nu _{n+1}=S\nu _n=g(f\nu _n)\), for \(n \ge 0\), where g is a selfmapping on fE with \(g(f\nu )=S\nu\), \(\nu \in E\). So, from the hypotheses we have
$$\begin{aligned} \phi (sd(g(f\nu ),g(f\xi )))\le \phi (M_f(\nu ,\xi ))\psi (M_f(\nu ,\xi )) \end{aligned}$$
(35)
for all \(\nu\), \(\xi\)\(\in P\) with \(f\nu \le f\xi\) and,
$$\begin{aligned} M_f(\nu ,\xi )=\max \left\{ \frac{d(f\xi ,g(f\xi )) \left[ 1+d(f\nu ,g(f\nu ))\right] }{1+d(f\nu ,f\xi )}, \frac{d(f\xi ,g(f\nu ))\left[ 1+d(f\nu ,g(f\xi ))\right] }{1+d(f\nu ,f\xi )}, d(f\nu ,f\xi )\right\} . \end{aligned}$$
(36)
From the same argument in Theorem 3, \(\{f\nu _n\}\) is a Cauchy sequence and which converges for some \(v \in fE\). Thus, by the compatibility of S and f, we obtain
$$\begin{aligned} \lim \limits _{n \rightarrow +\infty }d(f(S\nu _n), S(f\nu _n))=0. \end{aligned}$$
(37)
Further, by use of triangular inequality,
$$\begin{aligned} d(Sv,fv)\le sd(Sv,S(f\nu _n))+s^2 d(S(f\nu _n), f(S\nu _n))+s^2d(f(S\nu _n), fv). \end{aligned}$$
(38)
Finally, we arrive at \(d(Sv,fv)=0\) as \(n \rightarrow +\infty\) in (38). Therefore, v is a coincidence point for S and f in P. \(\square\)
Replace the condition, weakly compatible instead of (C4) in Corollary 1, we obtain the following result.
Corollary 2
If P has the property in Corollary 1instead of the compatibility for S and f that, for any nondecreasing sequence \(\{f\nu _n\}\subset P\)such that \(\lim \limits _{n \rightarrow +\infty } f\nu _n=f\nu\)implies that \(f\nu _n \le f\nu\)for all \(n \in {\mathbb {N}}\), that is \(f\nu =\sup f\nu _n\). Then S and f have a common fixed point in P, if for some coincidence point \(\mu\)of S and f with \(f\mu \le f(f\mu )\). Furthermore, the set of common fixed points of S and f is well ordered if and only if S and f have one and only one common fixed point.
Proof
It is obvious from Corollary 1 and Theorem 4 that S and f have a coincidence point in P, as \(f\mu =g(f\mu )=S\mu\) for some \(\mu\) in P.
Next, assume that a pair of mappings (S, f) is weakly compatible and let \(\vartheta\) be an element in P such that \(\vartheta =S\mu =f\mu\). Then, \(S\vartheta =S(f\mu )=f(S\mu )=f\vartheta\). Let
$$\begin{aligned} \begin{aligned} M(\mu ,\vartheta )&=\max \left\{ \frac{d(f\vartheta ,S\vartheta ) \left[ 1+d(f\mu ,S\mu )\right] }{1+d(f\mu ,f\vartheta )},\frac{d(f\vartheta ,S\mu ) \left[ 1+d(f\mu ,S\vartheta )\right] }{1+d(f\mu ,f\vartheta )}, d(f\mu ,f\vartheta )\right\} \\&=\max \{0,d(S\mu ,S\vartheta )\}\\&=d(S\mu ,S\vartheta ). \end{aligned} \end{aligned}$$
(39)
then from contraction condition, we have
$$\begin{aligned} \begin{aligned} \phi (d(S\mu ,S\vartheta )) \le \phi (M(\mu ,\vartheta ))\psi (M(\mu ,\vartheta )) \le \phi (d(S\mu ,S\vartheta ))\psi (d(S\mu ,S\vartheta )). \end{aligned} \end{aligned}$$
(40)
Hence, we get \(d(S\mu ,S\vartheta )=0\) by the property of \(\psi\). Therefore, \(S\vartheta =f\vartheta =\vartheta\).
Eventually, by following Theorem 5, we deduce that S and f have one and only one common fixed point if and only if the set of common fixed points of S and f is well ordered. \(\square\)
We illustrate the usefulness of the obtained results in different cases such as continuity and discontinuity of a metric d in a space P.
Example 1
Define a metric \(d:P \rightarrow P\) as below and \(\le\) be an usual order in P, where \(P=\{1,2,3,4,5\}\)
$$\begin{aligned}&d(\nu , \xi )=d(\xi ,\nu )=0, ~\text {if}~ \nu , \xi =1,2,3,4,5 ~\text {and}~ \nu = \xi .\\&d(\nu , \xi )=d(\xi ,\nu )=1, ~if~ \nu ,\xi =1,2,3,4 ~\text {and}~ \nu \ne \xi .\\&d(\nu , \xi )=d(\xi ,\nu )=6, ~if~ \nu =1,2,3 ~\text {and}~ \xi =5.\\&d(\nu , \xi )=d(\xi ,\nu )=12, ~if~ \nu =4 ~\text {and}~ \xi =5. \end{aligned}$$
Define a map \(S:P \rightarrow P\) by \(S1=S2=S3=S4=1, S5=3\) and let \(\phi (t)=\frac{t}{2}\), \(\psi (t)=\frac{t}{3}\) for \(t \in [0,+\infty )\). Then S has a fixed point in P.
Proof
It is apparent that, \((P,d,s,\le )\) is a complete partially ordered bmetric space for \(s=2\). Consider the possible cases for \(\nu\), \(\xi\) in P:
Case 1 Suppose \(\nu , \xi \in \{1,2,3,4\}\) and \(\nu <\xi\), thence,
$$\begin{aligned} \phi (2d(S\nu ,S\xi ))=0 \le \phi (M(\nu ,\xi ))\psi (M(\nu ,\xi )). \end{aligned}$$
Case 2 Suppose that \(\nu \in \{1,2,3,4\}\) and \(\xi =5\), then \(d(S\nu ,S\xi )=d(1,3)=1\), \(M(4,5)=12\) and \(M(\nu ,5)=6\), for \(\nu \in \{1,2,3\}\). Therefore, we have the following inequality,
$$\begin{aligned} \phi (2d(S\nu ,S\xi )) \le \frac{M(\nu ,\xi )}{6} =\phi (M(\nu ,\xi ))\psi (M(\nu ,\xi )). \end{aligned}$$
Thus, condition (3) of Theorem 3 holds. Furthermore, the remaining assumptions in Theorem 3 are fulfilled. Hence, S has a fixed point in P as Theorem 3 is appropriate to \(S, \phi , \psi\) and \((P,d,s,\le )\). \(\square\)
Example 2
A metric \(d:P \rightarrow P\), where \(P=\{0, 1, \frac{1}{2},\frac{1}{3},\frac{1}{4},........\frac{1}{n},.....\}\) with usual order \(\le\) is defined as follows
$$\begin{aligned} \begin{aligned} d(\nu ,\xi )= {\left\{ \begin{array}{ll} &{} ~0 ~~~~~~~~~,~~ if~ \nu =\xi \\ &{} ~1 ~~~~~~~~~,~~ if~ \nu \ne \xi \in \{0,1\} \\ &{} ~\nu \xi ~~~,~~ if~ \nu ,\xi \in \left\{ 0, \frac{1}{2n},\frac{1}{2m}: n \ne m \ge 1\right\} \\ &{} ~5 ~~~~~~~~~~,~~ otherwise. \end{array}\right. } \end{aligned} \end{aligned}$$
A map \(S: P \rightarrow P\) be such that \(S0=0, S\frac{1}{n}=\frac{1}{9n}\) for all \(n\ge 1\) and let \(\phi (t)=t\), \(\psi (t)=\frac{3t}{4}\) for \(t \in [0,+\infty )\). Then S has a fixed point in P.
Proof
It’s obvious that for \(s=\frac{9}{4}\), \((P,d,s,\le )\) is a complete partially ordered bmetric space and also by definition, d is discontinuous bmetric space. Now, for \(\nu ,\xi \in P\) with \(\nu <\xi\) then consider the following cases:
Case 1 If \(\nu =0\) and \(\xi =\frac{1}{n}\), \(n \ge 1\), then \(d(S\nu ,S\xi )=d(0,\frac{1}{9n})=\frac{1}{9n}\) and \(M(\nu ,\xi )=\frac{1}{n}\) or \(M(\upsilon ,\xi )= \{1,5\}\). Therefore, we have
$$\begin{aligned} \phi \left( \frac{9}{4}d(S\nu ,S\xi )\right) \le \frac{M(\nu ,\xi )}{4} =\phi (M(\nu ,\xi ))\psi (M(\nu ,\xi )). \end{aligned}$$
Case 2 If \(\nu =\frac{1}{m}\) and \(\xi =\frac{1}{n}\) with \(m>n\ge 1\), then
$$\begin{aligned} d(S\nu ,S\xi )=d\left( \frac{1}{9m},\frac{1}{9n}\right) ~\text {and}~ M(\nu ,\xi )\ge \frac{1}{n}\frac{1}{m}~ \text {or}~ M(\upsilon ,\xi )=5. \end{aligned}$$
Therefore,
$$\begin{aligned} \phi \left( \frac{9}{4}d(S\nu ,S\xi )\right) \le \frac{M(\nu ,\xi )}{4} =\phi (M(\nu ,\xi ))\psi (M(\nu ,\xi )). \end{aligned}$$
Hence, condition (3) of Theorem 3 and remaining assumptions are satisfied. Thus, S has a fixed point in P. \(\square\)