Generalized (σ,ξ)-Contractions and Related Fixed Point Results in a P.M.S

Abstract

In this paper, we present the concept of $\mathrm{\Theta }-{(\sigma ,\xi )}_{\mathrm{\Omega }}$-contraction mappings and we nominate some related fixed point results in ordered p-metric spaces. Our results extend several famous ones in the literature. Some examples and an application are given in order to validate our results.

1. Introduction

The Banach contraction principle (BCP) [1] is an applicable instrumentation to solve problems in nonlinear analysis. The BCP has been modified in variant procedures (see e.g., [2,3,4,5,6,7,8,9,10,11]).

Definition 1.

[12] The function$\xi :[0,+\infty )\to [0,+\infty )$verifying:

1.

ξ is non-decreasing and continuous;

2.

$\xi \left(t\right)=0$iff$t=0$,

is said to be an altering distance function.

Heretofore, many authors have concentrated on fixed point theorems depended on altering distance functions (see, e.g., [2,12,13,14,15,16,17,18,19]).

The concept of a b-metric space was nominated by Czerwik in [20]. Later, many interesting results about the existence of fixed points in b-metric spaces have been acquired (see, [2,21,22,23,24,25,26,27,28,29,30,31,32,33]).

Definition 2.

([20]) Let X be a (nonempty) set and$\varsigma \ge 1$be a real number. A function$d:X\times X\to R^{+}$is a b-metric if for all$\zeta ,\nu ,\mu \in X$,

(b₁)

$d(\zeta ,\nu )=0$iff$\zeta =\nu ;$

(b₂)

$d(\zeta ,\nu )=d(\nu ,\zeta );$

(b₃)

$d(\zeta ,\mu )\le \varsigma \left[d\right(\zeta ,\nu )+d(\nu ,\mu \left)\right].$

If $\varsigma =1$, the b-metric is a metric.

Let $\to$ be the set of strictly increasing continuous functions $\mathrm{\Omega }:[0,\infty )\to [0,\infty )$ such that $\mathrm{\Omega }\left(0\right)=0$ and $t\le \mathrm{\Omega }\left(t\right)$ for $t\ge 0$. Motivated by [20], we state the following.

Definition 3.

[34] Let X be a (nonempty) set. A function$\rho :X\times X\to R^{+}$is a p-metric iff there is$\mathrm{\Omega }\in \to$so that

(p₁)

$\rho (\zeta ,\nu )=0$iff$\zeta =\nu ,$

(p₂)

$\rho (\zeta ,\nu )=\rho (\nu ,\zeta ),$

(p₃)

$\rho (\zeta ,\mu )\le \mathrm{\Omega }\left(\rho \right(\zeta ,\nu )+\rho (\nu ,\mu \left)\right)$,

for all$\zeta ,\nu ,\mu \in X$.$(X,d)$is said to be a p.m.s. (or an extended b-metric space).

It should be mentioned that, the class of p-metric spaces is considerably comprehensive than the class of b-metric spaces. Note that a b-metric (with a coefficient $\varsigma \ge 1$) is a p-metric, when $\mathrm{\Omega }\left(t\right)=\varsigma t$. If $\mathrm{\Omega }\left(t\right)=t$, a p-metric is a metric.

Example 1.

[34] Let$(X,d)$be a metric space. Take$\rho (\zeta ,\nu )=e^{d(\zeta ,\nu )}-1$. Then ρ is a p-metric with$\mathrm{\Omega }\left(t\right)=e^t-1.$

The following example shows that a p-metric need not be a b-metric.

Example 2.

[34] Let$(X,d)$be a b-metric space (with a coefficient$\varsigma \ge 1$). Consider$\rho (\zeta ,\nu )=sinh\left[d\right(\zeta ,\nu \left)\right]$. Then ρ is a p-metric with$\mathrm{\Omega }\left(t\right)=sinh\left(\varsigma t\right)$,$t\ge 0$.

For$\varsigma =1$,$\zeta =2$,$\nu =-3$,$\mu =0$and$d(\zeta ,\nu )=|\zeta -\nu |$, we have

$$\rho (\zeta ,\nu )=sinh\left(5\right)>sinh\left(2\right)+sinh\left(3\right)=\rho (\zeta ,\mu )+\rho (\mu ,\nu ).$$

Definition 4.

[34] Let$(X,\rho )$be a p.m.s. A sequence$\left\{{\mu }_n\right\}$in X

(a) p-converges iff there is$\mu \in X$so that$\rho ({\mu }_n,\mu )\to 0$, as$n\to +\infty$. In this case, we write$\underset{n\to \infty }{lim}{\mu }_n=\mu ;$

(b) is p-Cauchy iff$\rho ({\mu }_n,{\mu }_m)\to 0$as$n,m\to +\infty .$

Note that a p.m.s$(X,\rho )$is p-complete if every p-Cauchy sequence in X is p-convergent.

Lemma 1.

Let$(X,\rho )$be a p.m.s. Suppose that$\left\{{\mu }_n\right\}$and$\left\{{\nu }_n\right\}$p-converge to$\mu ,\nu$, respectively. Then

$${\left({\mathrm{\Omega }}^2\right)}^{-1}\left(\rho (\mu ,\nu )\right)\le \underset{n⟶\infty }{lim\; inf}\rho ({\mu }_n,{\nu }_n)\le \underset{n⟶\infty }{lim\; sup}\rho ({\mu }_n,{\nu }_n)\le {\mathrm{\Omega }}^2\left(\rho (\mu ,\nu )\right).$$

Additionally, if$\mu =\nu ,$then${\displaystyle \underset{n⟶\infty }{lim}}\rho ({\mu }_n,{\nu }_n)=0.$Also, for any$z\in X$,

$${\mathrm{\Omega }}^{-1}\left(\rho (\mu ,z)\right)\le \underset{n⟶\infty }{lim\; inf}\rho ({\mu }_n,z)\le \underset{n⟶\infty }{lim\; sup}\rho ({\mu }_n,z)\le \mathrm{\Omega }\left(\rho (\mu ,z)\right).$$

The idea of $\mathrm{\Theta }$-contraction has been introduced by Jleli and Samet in [35] which provides an interesting generalization of BCP. Zhang and Song generalized the BCP using two altering distance functions [36]. Our approach provides a generalization of Zhang-Song result using the idea of $\mathrm{\Theta }$-contraction. In fact, we present the notion of generalized $\mathrm{\Theta }-{(\sigma ,\xi )}_{\mathrm{\Omega }}$-contractive mappings (where $\sigma$ and $\xi$ are altering distance functions) and we inaugurate some related fixed point results in complete ordered p-metric spaces. We also give some examples and an application.

2. Main Results

We first provide the notion of $\mathrm{\Theta }-{(\sigma ,\xi )}_{\mathrm{\Omega }}$-contractions.

Let Y be a self-map on the ordered p.m.s $(X,⪯,\rho )$. Consider

$$P(x,y)=max\left\{\rho (x,y),\rho (x,\mathrm{Y}x),\rho (y,\mathrm{Y}y),\frac{{\mathrm{\Omega }}^{-1}[\rho (x,\mathrm{Y}y)+\rho (y,\mathrm{Y}x)]}{2}\right\}.$$

Motivated by [35], denote by $\Delta$ the set of functions $\mathrm{\Theta }:[0,\infty )\to [1,\infty )$ so that

$\left({\mathrm{\Theta }}_1\right)$

$\mathrm{\Theta }$ is continuous and non-decreasing;

$\left({\mathrm{\Theta }}_2\right)$

for any $\left\{t_n\right\}\subseteq (0,\infty )$, $\underset{n\to \infty }{lim}\mathrm{\Theta }\left(t_n\right)=1$ iff $\underset{n\to \infty }{lim}{t}_n=0$.

Definition 5.

Let$(X,⪯,\rho )$be an ordered p.m.s. The mapping$\mathrm{Y}:X\to X$is an ordered$\mathrm{\Theta }-{(\sigma ,\xi )}_{\mathrm{\Omega }}$-contraction if there are$\mathrm{\Theta }\in \Delta$,$\mathrm{\Omega }\in \omega$and two altering distance functions σ and ξ, so that

$$\mathrm{\Theta }\left(\sigma \left({\mathrm{\Omega }}^2\left(\rho (\mathrm{Y}x,\mathrm{Y}y)\right)\right)\right)\le \frac{\mathrm{\Theta }\left(\sigma \right(P(x,y)\left)\right)}{\mathrm{\Theta }\left(\xi \right(P(x,y)\left)\right)}$$

(1)

for all comparable elements$x,y\in X$.

Our first result is

Theorem 1.

Let$(X,⪯,\rho )$be an ordered p-complete p.m.s. Suppose that$\mathrm{Y}:X\to X$is an ordered non-decreasing continuous$\mathrm{\Theta }-{(\sigma ,\xi )}_{\mathrm{\Omega }}$-contractive mapping. If there is$r_0\in X$such that$r_0⪯\mathrm{Y}{r}_0$, then Y admits a fixed point.

Proof. 

Let $r_0\in X$ satisfy $r_0⪯\mathrm{Y}{r}_0$. Consider a sequence $\left(r_n\right)$ in X so that $r_{n+1}=\mathrm{Y}{r}_n$ for each $n\ge 0$. Since $r_0⪯\mathrm{Y}{r}_0=r_1$ and $\mathrm{Y}$ is non-decreasing, we have $r_1=\mathrm{Y}{r}_0⪯r_2=\mathrm{Y}{r}_1$. Inductively, we have

$$r_0⪯r_1⪯\cdots ⪯r_n⪯r_{n+1}⪯\cdots .$$

If $r_k=r_{k+1}$ for some $k\in \mathbb{N}$, so $r_k$ is a fixed point of $\mathrm{Y}$. Suppose that $r_n\ne r_{n+1}$ for each $n\ge 0$. According to (1) and the fact $\mathrm{\Omega }\in \to$, we have

$$\begin{array}{ll}\mathrm{\Theta }\left(\sigma \left(\rho (r_n,r_{n+1})\right)\right)& \le \mathrm{\Theta }\left(\sigma \left({\mathrm{\Omega }}^2\left(\rho (r_n,r_{n+1})\right)\right)\right)\\ & =\mathrm{\Theta }\left(\sigma \left({\mathrm{\Omega }}^2\left(\rho (\mathrm{Y}{r}_{n-1},\mathrm{Y}{r}_n)\right)\right)\right)\\ & \le \frac{\mathrm{\Theta }\left(\sigma \left(P(r_{n-1},r_n)\right)\right)}{\mathrm{\Theta }\left(\xi \left(P(r_{n-1},r_n)\right)\right)},\end{array}$$

(2)

where

$$\begin{array}{ll}P(r_{n-1},r_n)& =max\left\{\rho (r_{n-1},r_n),\rho (r_{n-1},\mathrm{Y}{r}_{n-1}),\rho (r_n,\mathrm{Y}{r}_n),\frac{{\mathrm{\Omega }}^{-1}[\rho (r_{n-1},\mathrm{Y}{r}_n)+\rho (r_n,\mathrm{Y}{r}_{n-1})]}{2}\right\}\\ & \le max\left\{\rho (r_{n-1},r_n),\rho (r_n,r_{n+1})\right\}.\end{array}$$

(3)

From (2) to (3) and the assumptions on $\sigma$ and $\xi$, we deduce that

$$\begin{array}{ll}\mathrm{\Theta }\left(\sigma \left(\rho (r_n,r_{n+1})\right)\right)& \le \frac{\mathrm{\Theta }\left(\sigma \left(max\left\{\rho (r_{n-1},r_n),\rho (r_n,r_{n+1})\right\}\right)\right)}{\mathrm{\Theta }\left(\xi \left(P(r_{n-1},r_n)\right)\right)}\\ & <\mathrm{\Theta }\left(\sigma \left(max\left\{\rho (r_{n-1},r_n),\rho (r_n,r_{n+1})\right\}\right)\right).\end{array}$$

(4)

If for some n,

$$max\left\{\rho (r_{n-1},r_n),\rho (r_n,r_{n+1})\right\}=\rho (r_n,r_{n+1}),$$

then by (4) we have

$$\begin{array}{ll}\mathrm{\Theta }\left(\sigma \left(\rho (r_n,r_{n+1})\right)\right)& \le \frac{\mathrm{\Theta }\left(\sigma \left(\rho (r_n,r_{n+1})\right)\right)}{\mathrm{\Theta }\left(\xi \left(P(r_{n-1},r_n)\right)\right)}\\ & <\mathrm{\Theta }\left(\sigma \left(\rho (r_n,r_{n+1})\right)\right),\end{array}$$

which gives a contradiction. Thus,

$$max\left\{\rho (r_{n-1},r_n),\rho (r_n,r_{n+1})\right\}=\rho (r_{n-1},r_n),\mathrm{f}\mathrm{o}\mathrm{r} \mathrm{e}\mathrm{a}\mathrm{c}\mathrm{h}n\ge 0.$$

Therefore, (4) yields that

$$\mathrm{\Theta }\left(\sigma \left(\rho (r_n,r_{n+1})\right)\right)\le \frac{\mathrm{\Theta }\left(\sigma \left(\rho (r_n,r_{n-1})\right)\right)}{\mathrm{\Theta }\left(\xi \left(P(r_{n-1},r_n)\right)\right)}<\mathrm{\Theta }\left(\sigma \left(\rho (r_n,r_{n-1})\right)\right),\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{h}n\ge 0.$$

(5)

Since $\mathrm{\Theta }\in \Delta$ and $\sigma$ is non-decreasing, the positive sequence $\left\{\rho (r_n,r_{n+1})\right\}$ is non-increasing. Thus, there is $r\ge 0$ so that

$$\underset{n\to \infty }{lim}\rho (r_n,r_{n+1})=r.$$

Taking $n\to \infty$ in (5), we get

$$\mathrm{\Theta }\left(\sigma \left(r\right)\right)\le \frac{\mathrm{\Theta }\left(\sigma \right(r\left)\right)}{\mathrm{\Theta }\left(\xi \left({lim}_{n\to \infty }P(r_{n-1},r_n)\right)\right)}\le \mathrm{\Theta }\left(\sigma \left(r\right)\right).$$

Therefore, $\mathrm{\Theta }\left(\xi \left({lim}_{n\to \infty }P(r_{n-1},r_n)\right)\right)=1$ which supplies that $\xi \left({lim}_{n\to \infty }P(r_{n-1},r_n)\right)=0$, and so $r=0$, that is,

$$\underset{n\to \infty }{lim}\rho (r_n,r_{n+1})=0.$$

(6)

Next, we demonstrate that $\left\{r_n\right\}$ is a p-Cauchy sequence in X. By contradiction, there is $\epsilon >0$ for which we can gain $\left\{r_{m_i}\right\}$ and $\left\{r_{n_i}\right\}$ of $\left\{r_n\right\}$ so that

$$n_i>m_i>i,\rho (r_{m_i},r_{n_i})\ge \epsilon$$

(7)

and

$$\rho (r_{m_i},r_{n_i-1})<\epsilon .$$

(8)

The p-triangular inequality leads to

$$\begin{array}{ll}\epsilon & \le \rho (r_{m_i},r_{n_i})\\ & \le \mathrm{\Omega }(\rho (r_{m_i},r_{m_i-1})+\rho (r_{m_i-1},r_{n_i}))\\ & \le \mathrm{\Omega }(\rho (r_{m_i},r_{m_i-1})+\mathrm{\Omega }(\rho (r_{m_i-1},r_{n_i-1})+\rho (r_{n_i-1},r_{n_i}))).\end{array}$$

Exploiting (6), (7) and (8), we have

$${\left({\mathrm{\Omega }}^2\right)}^{-1}\left(\epsilon \right)\le \underset{i⟶\infty }{lim\; inf}\rho (r_{m_i-1},r_{n_i-1}).$$

Likewise,

$$\rho (r_{m_i-1},r_{n_i-1})\le \mathrm{\Omega }(\rho (r_{m_i-1},r_{m_i})+\rho (r_{m_i},r_{n_i-1})).$$

Handling (6) and (8), we have

$$\underset{i⟶\infty }{lim\; sup}\rho (r_{m_i-1},r_{n_i-1})\le \mathrm{\Omega }\left(\epsilon \right),$$

(9)

Moreover,

$$\rho (r_{m_i},r_{n_i})\le \mathrm{\Omega }(\rho (r_{m_i},r_{n_i-1})+\rho (r_{n_i-1},r_{n_i})).$$

Appling (5) and (8), we have

$$\underset{i⟶\infty }{lim\; sup}\rho (r_{m_i},r_{n_i-1})\ge {\mathrm{\Omega }}^{-1}\left(\epsilon \right),$$

In addition,

$$\rho (r_{m_i-1},r_{n_i})\le \mathrm{\Omega }(\rho (r_{m_i-1},r_{m_i})+\mathrm{\Omega }[\rho (r_{m_i},r_{n_i-1})+\rho (r_{n_i-1},r_{n_i})]).$$

Using (6) and (8), we have

$$\underset{i⟶\infty }{lim\; sup}\rho (r_{m_i},r_{n_i-1})\le {\mathrm{\Omega }}^2\left(\epsilon \right).$$

Moreover,

$$\rho (r_{m_i},r_{n_i})\le \mathrm{\Omega }(\rho (r_{m_i},r_{m_i-1})+\rho (r_{m_i-1},r_{n_i})).$$

Appling (6) and (8), we get

$$\underset{i⟶\infty }{lim\; sup}\rho (r_{m_i-1},r_{n_i})\ge {\mathrm{\Omega }}^{-1}\left(\epsilon \right).$$

From (1),

$$\begin{array}{l}\begin{array}{ll}\mathrm{\Theta }\left(\sigma \left({\mathrm{\Omega }}^2\left(\rho (r_{m_i},r_{n_i})\right)\right)\right)& =\mathrm{\Theta }\left(\sigma \left({\mathrm{\Omega }}^2\left(\rho (\mathrm{Y}{r}_{m_i-1},\mathrm{Y}{r}_{n_i-1})\right)\right)\right)\\ & \le \frac{\mathrm{\Theta }\left(\sigma \left(P(r_{m_i-1},r_{n_i-1})\right)\right)}{\mathrm{\Theta }\left(\xi \left(P(r_{m_i-1},r_{n_i-1})\right)\right)},\end{array}\end{array}$$

(10)

where

$$\begin{array}{l}\begin{array}{l}P(r_{m_i-1},r_{n_i-1})]\\ =max\left\{\rho (r_{m_i-1},r_{n_i-1}),\rho (r_{m_i-1},\mathrm{Y}{r}_{m_i-1}),\rho (r_{n_i-1},\mathrm{Y}{r}_{n_i-1}),\frac{{\mathrm{\Omega }}^{-1}[\rho (r_{m_i-1},\mathrm{Y}{r}_{n_i-1})+\rho (r_{n_i-1},\mathrm{Y}{r}_{m_i-1})]}{2}\right\}\\ =max\left\{\rho (r_{m_i-1},r_{n_i-1}),\rho (r_{m_i-1},r_{m_i}),\rho (r_{n_i-1},r_{n_i}),\frac{{\mathrm{\Omega }}^{-1}[\rho (r_{m_i-1},r_{n_i})+\rho (r_{n_i-1},r_{m_i})]}{2}\right\}.\end{array}\end{array}$$

(11)

Taking $i\to \infty$ in (11) and using (6), we achieve that,

$$\begin{array}{l}\underset{i⟶\infty }{lim\; sup}P(r_{m_i-1},r_{n_i-1})\\ =max\{\underset{i⟶\infty }{lim\; sup}\rho (r_{m_i-1},r_{n_i-1}),0,0,\underset{i⟶\infty }{lim\; sup}\rho (r_{m_i},r_{n_i-1})\}\le {\mathrm{\Omega }}^2\left(\epsilon \right).\end{array}$$

(12)

Similarly,

$${\left({\mathrm{\Omega }}^2\right)}^{-1}\left(\epsilon \right)\le \underset{i⟶\infty }{lim\; inf}P(r_{m_i-1},r_{n_i-1}).$$

(13)

Now, taking $i\to \infty$ in (10) and using (7) and (12),

$$\begin{array}{ll}\mathrm{\Theta }(\sigma ({\mathrm{\Omega }}^2(\epsilon )))& \le \mathrm{\Theta }(\sigma ({\mathrm{\Omega }}^2(\underset{i⟶\infty }{lim\; sup}\rho (r_{m_i},r_{n_i}))))\\ & \le \frac{\mathrm{\Theta }(\sigma (\underset{i⟶\infty }{lim\; sup}P(r_{m_i-1},r_{n_i-1})))}{\mathrm{\Theta }(\underset{i⟶\infty }{lim\; inf}\xi (P(r_{m_i-1},r_{n_i-1})))}\\ & \le \frac{\mathrm{\Theta }(\sigma ({\mathrm{\Omega }}^2(\epsilon )))}{\mathrm{\Theta }(\xi (\underset{i⟶\infty }{lim\; inf}P(r_{m_i-1},r_{n_i-1})))}.\end{array}$$

It yields that

$$\xi \left(\underset{i⟶\infty }{lim\; inf}P(r_{m_i-1},r_{n_i-1})\right)=0,$$

so, $\underset{i⟶\infty }{lim\; inf}P(r_{m_i-1},r_{n_i-1})=0,$ a contradiction to (13). Thus, $\{r_{n+1}=\mathrm{Y}{r}_n\}$ is a p-Cauchy sequence in the p-complete space X, so there is $u\in X$ so that $r_n\to u$. According to the continuity of $\mathrm{Y}$,

$$\underset{n\to \infty }{lim}{r}_{n+1}=\underset{n\to \infty }{lim}\mathrm{Y}{r}_n=\mathrm{Y}u.$$

(14)

The p-triangular inequality leads to

$$\begin{array}{ll}\rho (u,\mathrm{Y}u)\le & \mathrm{\Omega }(\rho (u,\mathrm{Y}{r}_n)+\rho (\mathrm{Y}{r}_n,\mathrm{Y}u))\\ & =\mathrm{\Omega }(\rho (u,r_{n+1})+\rho (\mathrm{Y}{r}_n,\mathrm{Y}u))\\ & \le \mathrm{\Omega }[\mathrm{\Omega }(\rho (u,r_n)+\rho (r_n,r_{n+1}))+\rho (\mathrm{Y}{r}_n,\mathrm{Y}u)].\end{array}$$

The continuity of $\mathrm{\Omega }$ together with and (14) imply that

$$\rho (u,\mathrm{Y}u)\le \mathrm{\Omega }[\mathrm{\Omega }(\underset{n\to \infty }{lim}\rho (u,r_n)+\underset{n\to \infty }{lim}\rho (r_n,r_{n+1}))+\underset{n\to \infty }{lim}\rho (\mathrm{Y}{r}_n,\mathrm{Y}u)]=0.$$

We find that $\mathrm{Y}u=u$. ☐

The continuity of $\mathrm{Y}$ in Theorem 1 can be substituted by the following reservation:

An ordered p.m.s $(X,⪯,p)$ possesses the sequential limit comparison property (s.l.c.p) if for each nondecreasing sequence $\left\{r_n\right\}$ in X, converging to some $x\in X$, we have $r_n⪯x$ for each $n\in \mathbb{N}$.

Theorem 2.

Having the same assumptions of Theorem 1, by replacing the continuity of Y with the s.l.c.p. property of $(X,⪯,\rho )$, Y encompasses a fixed point.

Proof. 

Reviewing the lines of the proof of Theorem 1, we have that $\left\{r_n\right\}$ is an increasing sequence in X so that $r_n\to u$, for $u\in X$. Using the $s.l.c.p.$ obligation on X, we have $r_n⪯u$, for any $n\in \mathbb{N}$. We claim that $\mathrm{Y}u=u$. By (1),

$$\begin{array}{l}\begin{array}{ll}\mathrm{\Theta }\left(\sigma \left({\mathrm{\Omega }}^2\left(\rho (r_{n+1},\mathrm{Y}u)\right)\right)\right)& =\mathrm{\Theta }\left(\sigma \left({\mathrm{\Omega }}^2\left(\rho (\mathrm{Y}{r}_n,\mathrm{Y}u)\right)\right)\right)\\ & \le \frac{\mathrm{\Theta }\left(\sigma \left(P(r_n,u)\right)\right)}{\mathrm{\Theta }\left(\xi \left(P(r_n,u)\right)\right)},\end{array}\end{array}$$

(15)

where

$$\begin{array}{l}\begin{array}{l}P(r_n,u)\\ =max\left\{\rho (r_n,u),\rho (r_n,\mathrm{Y}{r}_n),\rho (u,\mathrm{Y}u),\frac{{\mathrm{\Omega }}^{-1}[\rho (r_n,\mathrm{Y}u)+\rho (u,\mathrm{Y}{r}_n)]}{2}\right\}\\ =max\left\{\rho (r_n,u),\rho (r_n,r_{n+1}),\rho (u,\mathrm{Y}u),\frac{{\mathrm{\Omega }}^{-1}[\rho (r_n,\mathrm{Y}u)+\rho (u,r_{n+1})]}{2}\right\}.\end{array}\end{array}$$

(16)

Making $n\to \infty$ in (16) and using Lemma 1, we get

$$\begin{array}{c}\begin{array}{c}\underset{n⟶\infty }{lim\; sup}P(r_n,u)=\rho (u,\mathrm{Y}u).\end{array}\end{array}$$

(17)

Likely, we can obtain

$$\underset{n⟶\infty }{lim\; inf}P(r_n,u)=\rho (u,\mathrm{Y}u).$$

(18)

The the upper limit as $n\to \infty$ in (15) together with Lemma 1 and (17) imply that

$$\begin{array}{ll}\mathrm{\Theta }(\sigma (\rho (u,\mathrm{Y}u)))& =\mathrm{\Theta }(\sigma (\mathrm{\Omega }({\mathrm{\Omega }}^{-1}(\rho (u,\mathrm{Y}u))))\\ & \le \mathrm{\Theta }(\sigma ({\mathrm{\Omega }}^2(\underset{n⟶\infty }{lim\; sup}\rho (r_{n+1},\mathrm{Y}u))))\\ & \le \frac{\mathrm{\Theta }(\sigma (\underset{n⟶\infty }{lim\; sup}P(r_n,u)))}{\mathrm{\Theta }(\underset{n⟶\infty }{lim\; inf}\xi (P(r_n,u)))}\\ & \le \frac{\mathrm{\Theta }(\sigma (\rho (u,\mathrm{Y}u)))}{\mathrm{\Theta }(\xi (\underset{n⟶\infty }{lim\; inf}P(r_n,u)))}.\end{array}$$

Therefore, $\xi \left({\displaystyle \underset{n⟶\infty }{lim\; inf}}P(r_n,u)\right)\to 0,$ equivalently, ${\displaystyle \underset{n⟶\infty }{lim\; inf}}P(r_n,u)=0.$ Thus, from (18) we get $u=\mathrm{Y}u$ and hereupon u is a fixed point of Y. ☐

Remark 1.

Substituting$\mathrm{\Theta }\left(t\right)=e^t$in (1), we obtain the following contractive condition:

$$\sigma \left({\mathrm{\Omega }}^2\left(\rho (\mathrm{Y}x,\mathrm{Y}y)\right)\right)\le \sigma \left(P(x,y)\right)-\xi \left(P(x,y)\right)$$

which is the Zhang-Song contractive condition in a p-metric space.

Corollary 1.

Let$(X,⪯,\rho )$be an ordered p-complete p.m.s. Let$\mathrm{Y}:X\to X$be an ordered non-decreasing mapping. Assume there is$k\in [0,1)$so that

$$\begin{array}{cc}{\mathrm{\Omega }}^2\left(\rho (\mathrm{Y}x,\mathrm{Y}y)\right)& \le kmax\left\{\rho (x,y),\rho (x,\mathrm{Y}x),\rho (y,\mathrm{Y}y),\frac{{\mathrm{\Omega }}^{-1}[\rho (x,\mathrm{Y}y)+\rho (y,\mathrm{Y}x)]}{2}\right\},\end{array}$$

for all comparable elements$x,y\in X$. If there is$r_0\in X$so that$r_0⪯\mathrm{Y}{r}_0$, then Y admits a fixed point provided that either Y is continuous, or $(X,⪯,p)$ enjoys the $s.l.c.p.$

Proof. 

It follows using Theorems 1 and 2 by taking $\mathrm{\Theta }\left(t\right)=e^t$, $\sigma \left(t\right)=t$ and $\xi \left(t\right)=(1-k)t$. ☐

Corollary 2.

Let$(X,⪯,\rho )$be an ordered p-complete p.m.s. Let$\mathrm{Y}:X\to X$be an ordered non-decreasing mapping. Assume that there are$\alpha ,\beta ,\gamma ,\delta \in [0,1)$with$\alpha +\beta +\gamma +\delta \in [0,1)$so that

$$\begin{array}{c}{\mathrm{\Omega }}^2\left(\rho (\mathrm{Y}x,\mathrm{Y}y)\right)\le \alpha \rho (x,y)+\beta \rho (x,\mathrm{Y}x)+\gamma \rho (y,\mathrm{Y}y)+\delta \frac{{\mathrm{\Omega }}^{-1}[\rho (x,\mathrm{Y}y)+\rho (y,\mathrm{Y}x)]}{2},\end{array}$$

for all comparable elements$x,y\in X$. If there is$r_0\in X$such that$r_0⪯\mathrm{Y}{r}_0$, then Y has a fixed point provided that either Y is continuous, or $(X,⪯,p)$ possesses the $s.l.c.p.$

The following corollary is an enlargement of BCP in a p.m.s., where $\rho (x,y)=e^{d(x,y)}-1$.

Corollary 3.

Let Y be a non-decreasing self-mapping on an ordered p-complete p.m.s $(X,⪯,\rho )$. Assume that there is $\alpha \in [0,1)$ such that

$$\begin{array}{c}{e}^{\rho (\mathrm{Y}x,\mathrm{Y}y)}-1\le \alpha \rho (x,y),\end{array}$$

for all comparable elements $x,y\in X$. If there is $r_0\in X$ such that $r_0⪯\mathrm{Y}{r}_0$, then Y has a fixed point provided that either Y is continuous, or $(X,⪯,p)$ enjoys the $s.l.c.p.$

Remark 2.

A subset W in an ordered set X is well ordered if each two elements of W are comparable. In Theorems 1 and 2, Y admits a unique fixed point whenever the fixed points of Y are comparable.

Remark 3.

For any p-metric space$(X,\rho )$, the conclusion of Theorems 1 and 2 remains true if$\sigma ,\xi$are only non-decreasing on$diam\left(X\right)={sup}_{x,y\in X}\rho (x,y)$.

Corollary 4.

Let$(X,⪯,\rho )$be a partially ordered p-complete p-metric space. Let$\mathrm{Y}:X\to X$be an ordered non-decreasing mapping. Suppose that there exists$k\in [0,1)$such that

$$\begin{array}{cc}{\mathrm{\Omega }}^2\left(\rho (\mathrm{Y}x,\mathrm{Y}y)\right)& \le kmax\left\{\rho (x,y),\rho (x,\mathrm{Y}x),\rho (y,\mathrm{Y}y),\frac{{\mathrm{\Omega }}^{-1}[\rho (x,\mathrm{Y}y)+\rho (y,\mathrm{Y}x)]}{2}\right\},\end{array}$$

for all comparable elements$x,y\in X$. If there is$r_0\in X$such that$r_0⪯\mathrm{Y}{r}_0$, then Y has a fixed point provided that either Y is continuous, or $(X,⪯,p)$ enjoys the $s.l.c.p.$

Proof. 

It follows from Theorems 1 and 2, by taking $\mathrm{\Theta }\left(t\right)=e^t$, $\sigma \left(t\right)=t$ and $\xi \left(t\right)=(1-k)t$ for each $t\in [0,+\infty )$. ☐

Corollary 5.

Let$(X,⪯,\rho )$be a partially ordered p-complete p-metric space. Let$\mathrm{Y}:X\to X$be an ordered non-decreasing mapping. Suppose that there are$\alpha ,\beta ,\gamma ,\delta \in [0,1)$with$\alpha +\beta +\gamma +\delta \in [0,1)$such that

$$\begin{array}{cc}{\mathrm{\Omega }}^2\left(\rho (\mathrm{Y}x,\mathrm{Y}y)\right)& \le \alpha \rho (x,y)+\beta \rho (x,\mathrm{Y}x)+\gamma \rho (y,\mathrm{Y}y)+\delta \frac{{\mathrm{\Omega }}^{-1}[\rho (x,\mathrm{Y}y)+\rho (y,\mathrm{Y}x)]}{2},\end{array}$$

for all comparable elements$x,y\in X$. If there is$r_0\in X$such that$r_0⪯\mathrm{Y}{r}_0$, then Y has a fixed point provided that either Y is continuous, or $(X,⪯,p)$ enjoys the $s.l.c.p.$

Example 3.

Take$X=\{0,1,2,3\}$. Define on X the partial order ⪯:

$$⪯:=\left\{\right(0,0),(1,1),(2,2),(3,3),(1,2),(0,1),(0,2\left)\right\}.$$

Define the metric

$$d(x,y)=\left\{\begin{array}{cc}0,& ifx=y,\\ x+y,& ifx\ne y\end{array}\right.$$

and let$\rho (x,y)=sinh\left[d\right(x,y\left)\right].$Note that$(X,\rho )$is a p-complete p-metric space [Here,$\mathrm{\Omega }\left(t\right)=sinh\left(t\right)$for$t\ge 0$].

Define the self-map Y by

$$\begin{array}{c}\mathrm{Y}=\left(\begin{array}{cccc}0& 1& 2& 3\\ 0& 0& 1& 2\end{array}\right).\end{array}$$

We see that Y is an ordered increasing mapping and $(X,⪯,\rho )$ enjoys the $s.l.c.p.$ Define $\sigma \left(t\right)=\sqrt{t}$ cand $\xi \left(t\right)={\displaystyle \frac{t^2}{15+t^2}}$ and $\mathrm{\Theta }\left(t\right)=1+t^2$. We show that Y is an ordered non-decreasing $\mathrm{\Theta }-{(\sigma ,\xi )}_{\mathrm{\Omega }}$-contractive mapping. Indeed, let $x,y\in X$ with $x⪯y$. If $(x,y)\in \left\{\right(0,0),(1,1),(2,2),(3,3),(0,1\left)\right\}$, then we have nothing to prove. Thus, we need to only check the following cases:

Case 1.$(x,y)=(1,2).$Here,

$$\begin{array}{ll}\sigma \left({\mathrm{\Omega }}^2\left(\rho (\mathrm{Y}x,\mathrm{Y}y)\right)\right)& =\sqrt{{sinh}^3(\mathrm{Y}1+\mathrm{Y}2)}\\ & =\sqrt{{sinh}^3(0+1)}\\ & =1.623,\end{array}$$

$$\sigma \left(P(x,y)\right)=\sqrt{P(x,y)}=\sqrt{sinh3}=3.16,$$

$$\xi \left(P(x,y)\right)=\frac{P{(x,y)}^2}{15+P{(x,y)}^2}=\frac{{(sinh3)}^2}{15+{(sinh3)}^2}=0.86,$$

$$\begin{array}{ll}\mathrm{\Theta }\left(\sigma \left({\mathrm{\Omega }}^2\left(\rho (\mathrm{Y}x,\mathrm{Y}y)\right)\right)\right)& =\mathrm{\Theta }\left(1.623\right)\\ & =3.63\le 6.31=\frac{{3.16}^2+1}{{0.86}^2+1}\\ & =\frac{\mathrm{\Theta }\left(\sigma \right(P(x,y)\left)\right)}{\mathrm{\Theta }\left(\xi \right(P(x,y)\left)\right)}\end{array}$$

Case 2.$(x,y)=(0,2).$We have

$$\begin{array}{ll}\sigma \left({\mathrm{\Omega }}^2\left(\rho (\mathrm{Y}x,\mathrm{Y}y)\right)\right)& =\sqrt{{sinh}^3(\mathrm{Y}0+\mathrm{Y}2)}\\ & =\sqrt{{sinh}^3(0+1)}\\ & =1.623,\end{array}$$

$$\sigma \left(P(x,y)\right)=\sqrt{P(x,y)}=\sqrt{sinh2}=1.904,$$

$$\xi \left(P(x,y)\right)=\frac{P{(x,y)}^2}{15+P{(x,y)}^2}=\frac{{(sinh2)}^2}{15+{(sinh2)}^2}=0.467,$$

$$\begin{array}{cc}\mathrm{\Theta }\left(\sigma \left({\mathrm{\Omega }}^2\left(\rho (\mathrm{Y}x,\mathrm{Y}y)\right)\right)\right)& =\mathrm{\Theta }\left(1.623\right)=3.634\le 3.797=\frac{{1.904}^2+1}{{0.467}^2+1}=\frac{\mathrm{\Theta }\left(\sigma \right(P(x,y)\left)\right)}{\mathrm{\Theta }\left(\xi \right(P(x,y)\left)\right)}.\end{array}$$

Also, any two fixed points of Y are comparable. Thus, all of the conditions of Theorem 2 are satisfied, and so Y has a unique fixed point, which is, 0.

Remark 4.

Taking$(x,y)=(0,2)$in Example 3, we have

$$\sigma \left({\mathrm{\Omega }}^2\left(\rho (\mathrm{Y}x,\mathrm{Y}y)\right)\right)=1.623>1.437=1.904-0.467=\sigma \left(P(x,y)\right)-\xi \left(P(x,y)\right).$$

Thus, we can not apply the main result of Roshan et al. [30]. Also, we have$|\mathrm{Y}1-\mathrm{Y}2|=|0-1|=|1-2|$and$1⪯2$. Thus, Y is neither a Banach contraction, nor an ordered Banach contraction, with the usual metric. This example shows that our result is a real generalization of the similar results in literature in the setting of b-metric spaces and metric spaces.

Corollary 6.

Let$(X,⪯,\rho )$be an ordered p-complete p-metric space. Let$\mathrm{Y}:X\to X$be an ordered non-decreasing continuous ordered mapping and suppose that there exist altering distance functions$\sigma ,\xi$satisfying

$$1+ln(1+(\sigma \left({\mathrm{\Omega }}^2\left(\rho (\mathrm{Y}x,\mathrm{Y}y)\right)\right))\le \frac{1+ln(1+(\sigma \left(P\right(x,y\left)\right))}{1+ln(1+(\xi \left(P\right(x,y\left)\right))}.$$

(19)

If there is$r_0\in X$such that$r_0⪯\mathrm{Y}{r}_0$, then Y has a fixed point. Moreover if any two fixed points of Y are comparable, then the fixed point of Y is unique and for any $r_0\in X$, the iterated sequence ${\left\{{\mathrm{Y}}^n\left(r_0\right)\right\}}_{n\in \mathbb{N}}$ converges to the fixed point.

In much the same way as in Theorem 2 we can prove:

Theorem 3.

Let$(X,⪯,\rho )$be an ordered p-complete p-metric space. Let$\mathrm{Y}:X\to X$be an ordered continuous non-decreasing mapping satisfying

$$\mathrm{\Theta }\left(\sigma \left(\mathrm{\Omega }\left(\rho (\mathrm{Y}x,\mathrm{Y}y)\right)\right)\right)\le \frac{\mathrm{\Theta }\left(\sigma \right(\rho (x,y)\left)\right)}{\mathrm{\Theta }\left(\xi \right(\rho (x,y)\left)\right)}$$

(20)

for all$x,y\in X$with$x⪯y$. If there is$r_0\in X$such that$r_0⪯\mathrm{Y}{r}_0$, then Y has a fixed point. Moreover, if any two fixed points of Y are comparable, then the fixed point of Y is unique and for any $r_0\in X$, the iterated sequence ${\left\{{\mathrm{Y}}^n\left(r_0\right)\right\}}_{n\in \mathbb{N}}$ converges to the fixed point.

Theorem 4.

Let$(X,⪯,\rho )$be an ordered p-complete p-metric space. Let$\mathrm{Y}:X\to X$be a non-decreasing mapping satisfying

$$\mathrm{\Theta }\left(\sigma \left(\mathrm{\Omega }\left(\rho (\mathrm{Y}x,\mathrm{Y}y)\right)\right)\right)\le \frac{\mathrm{\Theta }\left(\sigma \right(\rho (x,y)\left)\right)}{\mathrm{\Theta }\left(\xi \right(\rho (x,y)\left)\right)}$$

(21)

for all$x,y\in X$with$x⪯y$. Assume that$(X,⪯,p)$enjoys the$s.l.c.p.$If there is$r_0\in X$so that$r_0⪯\mathrm{Y}{r}_0$, then Y has a fixed point. Moreover, if any two fixed points of Y are comparable, then the fixed point of Y is unique and for any $r_0\in X$, ${\left\{{\mathrm{Y}}^n\left(r_0\right)\right\}}_{n\in \mathbb{N}}$ converges to the fixed point.

Example 4.

Let$X=[5,6]$. Given the p-metric$\rho (\zeta ,\nu )=e^{|\zeta -\nu |}-1$(Here,$\mathrm{\Omega }\left(t\right)=e^t-1).$

Consider on X:$\zeta ⪯\nu$iff$\nu \le \zeta$. Given$\mathrm{Y}:X\to X$as

$$\mathrm{Y}\zeta =3ln(1+\zeta )$$

Take$\sigma \left(t\right)=2ln(1+t)$and$\xi \left(t\right)=2ln(1+t)-0.9t$for each$t\ge 0$. Now, we show that Y is an ordered $\mathrm{\Theta }-{(\sigma ,\xi )}_{\mathrm{\Omega }}$-contractive mapping with $\mathrm{\Theta }\left(t\right)=1+{[ln(t+1)]}^2$.

Let$\zeta ⪯\nu$, that is$\nu \le \zeta$. The mean value theorem for$s⟼3ln(1+s)$yields that

$$\begin{array}{ll}\sigma (\mathrm{\Omega }(\rho (\mathrm{Y}\zeta ,\mathrm{Y}\nu )))& =2ln(\mathrm{\Omega }(\rho (\mathrm{Y}\zeta ,\mathrm{Y}\nu ))+1))\\ & =2\rho (\mathrm{Y}\zeta ,\mathrm{Y}\nu )\\ & =2(e^{\frac{|\mathrm{Y}\zeta -\mathrm{Y}\nu |}{2}}-1)\\ & =2(e^{\frac{3ln(1+\zeta )-3ln(1+\nu )}{2}}-1)\\ & =2(e^{\frac{3}{(1+c(\zeta ,\nu ))}(\zeta -\nu )}-1)\\ & \le 2(e^{\frac{1}{2}(\zeta -\nu )}-1)\\ & \le (e^{(\zeta -\nu )}-1).\end{array}$$

Therefore,

$$\begin{array}{ll}\mathrm{\Theta }\left(\sigma \right(\mathrm{\Omega }\left(\rho \right(F\zeta ,F\nu \left)\right)\left)\right)& \le \mathrm{\Theta }(e^{|\zeta -\nu |}-1)\\ & ={1+|\zeta -\nu |}^2\le \frac{{1+[ln(2|\zeta -\nu |+1\left)\right]}^2}{1+[ln(2|\zeta -\nu |-0.9(e^{|\zeta -\nu |}-1)+1{\left)\right]}^2}\\ & =\frac{\mathrm{\Theta }\left(\sigma \right(\rho (\zeta ,\nu ))}{\mathrm{\Theta }\left(\xi \right(\rho (\zeta ,\nu )\left)\right)}\end{array}$$

where$c(\zeta ,\nu )$is a constant dependent on$\zeta ,\nu$, obtained from mean value theorem such that$3ln(1+\zeta )-3ln(1+\nu )=\frac{3}{1+c(\zeta ,\nu )}(\zeta -\nu )$. So, we conclude that Y is a $\mathrm{\Theta }-{(\sigma ,\xi )}_{\mathrm{\Omega }}$-contractive mapping. Thus, all of the hypotheses of Theorem 3 are verified and hence Y has a fixed point in $[5,7]$. Moreover, since any two elements of $[5,7]$ are comparable, the fixed point of Y is unique and for any $r_0\in X$, the iterated sequence ${\left\{{\mathrm{Y}}^n\left(r_0\right)\right\}}_{n\in \mathbb{N}}$ is convergent to the fixed point.

Note that we can not apply the main result of Roshan et al. [30]. Indeed, for $\zeta =5$ and $\nu =6$, we get

$$\begin{array}{ll}\sigma \left(\mathrm{\Omega }\right(\rho (\mathrm{Y}\zeta ,\mathrm{Y}\nu )\left)\right)& =2ln\left(\mathrm{\Omega }\right(\rho (\mathrm{Y}\zeta ,\mathrm{Y}\nu ))+1))\\ & =2\rho (\mathrm{Y}\zeta ,\mathrm{Y}\nu )\\ & =2(e^{|\mathrm{Y}\zeta -\mathrm{Y}\nu |}-1)\\ & =2(e^{|3ln(6)-3ln(7\left)\right|}-1)\\ & =2(\frac{7^3}{6^3}-1)\\ & =1.175\\ & >1.546\\ & =0.9(e-1)\\ & =2ln(e-1+1)-(2ln(e-1+1)-0.9(e-1\left)\right)\\ & =2ln(e^{|\zeta -\nu |}-1+1)-(2ln(e^{|\zeta -\nu |}-1+1)-0.9(e^{|\zeta -\nu |}-1))\\ & =\sigma \left(\rho \right(\zeta ,\nu \left)\right)-\xi \left(\rho \right(\zeta ,\nu \left)\right).\end{array}$$

3. Application

For $T>0$, consider

$$\zeta \left(s\right)=p\left(s\right)+{\int }_0^T\lambda (s,r)f(r,\zeta \left(r\right))dr,s\in I=[0,T]$$

(22)

Here, we give an existence theorem for a solution of (22) in $X=C(I,[0,ln\left(\frac{20}{9}\right)])$ using Theorem 2. Take

$$\rho (\zeta ,\nu )=e^{\left|\right|\zeta -{\nu \left|\right|}_{\infty }}-1$$

for all $\zeta ,\nu \in X$. Note that X is a p-complete p-metric space with $\mathrm{\Omega }\left(s\right)=e^s-1$, where ${\left|\right|\zeta \left|\right|}_{\infty }={sup}_{q\in I}\left|\zeta \left(q\right)\right|$.

X is endowed with the partial order ⪯:

$$\zeta ⪯\nu ⟺\zeta \left(s\right)\le \nu \left(s\right),$$

for each $s\in I.$ Note that $(X,⪯,\rho )$ is regular. Assume that

(i)

$f:I\times [0,ln\left(\frac{20}{9}\right)]\to [0,ln\left(\frac{20}{9}\right)]$ and $p:I\to [0,ln\left(\frac{20}{9}\right)]$ are continuous;

($ii$)

$\lambda :I\times I\to [0,\infty )$ is continuous;

($iii$)

For all $\zeta ,\nu$ with $\zeta ⪯\nu$

$$0\le f(r,\nu )-f(r,\zeta )\le \nu -\zeta .$$

($iv$)

${max}_{s\in I}{\int }_0^T\mid \lambda (s,r)\mid dr\le \frac{1}{2}$;

(v)

There exists a continuous function $\alpha :[0,T]\to [0,ln\left(\frac{20}{9}\right)]$ so that

$$\alpha \left(s\right)\le p\left(s\right)+{\int }_0^T\lambda (s,r)f(r,\alpha \left(r\right))dr.$$

Theorem 5.

Under the conditions (i)-(v), (22) has a solution in$X=C([0,T],ln\left(\frac{20}{9}\right)]$.

Proof. 

Take $F:X\to X$ as

$$F\left(\zeta \left(s\right)\right)=p\left(s\right)+{\int }_0^T\lambda (s,r)f(r,\eta \left(r\right))dr.$$

For $\zeta ⪯\nu$,

$$f(s,\zeta )\le f(s,\nu ),$$

the operator F is ordered increasing. Having that $\lambda (s,r)>0$, so

$$F\left(\zeta \left(s\right)\right)=p\left(s\right)+{\int }_0^T\lambda (s,r)f(r,\zeta \left(r\right))dr\le p\left(s\right)+{\int }_0^T\lambda (s,r)f(r,\nu \left(r\right))dr=F\left(\nu \left(s\right)\right).$$

Now, take $\mathrm{\Theta }\left(s\right)=1+{[ln(s+1)]}^2$, $\sigma \left(s\right)=2ln(1+s)$ and $\xi \left(s\right)=2ln(1+s)-0.9s$. Note that $\xi$ is increasing iff $0\le s\le \frac{11}{9}$. For $\zeta ,\nu \in X$, we have $0\le \left|\right|\zeta -{\nu \left|\right|}_{\infty }\le ln(20/9)$, hence $0\le \rho (\zeta ,\nu )=e^{\left|\right|\zeta -{\nu \left|\right|}_{\infty }}-1\le 11/9$. Thus, $diam\left(X\right)={sup}_{\zeta ,\nu \in X}\rho (\zeta ,\nu )=\frac{11}{9}$.

Now,

$$\begin{array}{ll}\sigma \left(\mathrm{\Omega }\right(\rho (F\zeta ,F\nu )\left)\right)& =2ln(\mathrm{\Omega }(e^{\left|\right|F\zeta -{F\nu \left|\right|}_{\infty }}-1)+1)\\ & =2ln(e^{e^{\left|\right|F\zeta -{F\nu \left|\right|}_{\infty }}-1}-1+1)\\ & =2(e^{\left|\right|F\zeta -{F\nu \left|\right|}_{\infty }}-1)\\ & \le 2(e^{{max}_{s\in I}\left|{\int }_0^T\lambda (s,r)[f(r,\zeta \left(r\right))-f(r,\nu \left(r\right))]dr\right|}-1)\\ & \le 2(e^{({max}_{s\in I}{\int }_0^T\left|\lambda (s,r)\right|dr\left)\right||\zeta -{\nu \left|\right|}_{\infty }}-1)\\ & \le 2(e^{\frac{\left|\right|\zeta -{\nu \left|\right|}_{\infty }}{2}}-1)\\ & \le e^{\left|\right|\zeta -{\nu \left|\right|}_{\infty }}-1.\end{array}$$

Therefore,

$$\begin{array}{ll}\mathrm{\Theta }\left(\sigma \right(\mathrm{\Omega }\left(\rho \right(F\zeta ,F\nu \left)\right)\left)\right)& \le \mathrm{\Theta }(e^{\left|\right|\zeta -{\nu \left|\right|}_{\infty }}-1)\\ & =1+\left|\right|\zeta -{\nu \left|\right|}^2\le \frac{1+[ln(2\left|\right|\zeta -\nu \left|\right|+{1\left)\right]}^2}{1+[ln(2\left|\right|\zeta -\nu \left|\right|-0.9(e^{\left|\right|\zeta -{\nu \left|\right|}_{\infty }}-1)+1{\left)\right]}^2}\\ & =\frac{\mathrm{\Theta }\left(\sigma \right(\rho (\zeta ,\nu ))}{\mathrm{\Theta }\left(\xi \right(\rho (\zeta ,\nu )\left)\right)}.\end{array}$$

Due to assumption (v),

$$\alpha ⪯F\left(\alpha \right).$$

By Theorem 4, there is $\zeta \in X$ such that $\zeta =F\left(\zeta \right)$, which is a solution of (22). ☐

Note that we can not apply the theorem of Roshan et al. [30] to have a solution of (22). Indeed,

$$\begin{array}{ll}{e}^{\left|\right|\zeta -{\nu \left|\right|}_{\infty }}-1>& 2\left|\right|\zeta -\nu \left|\right|-\left(2\right||\zeta -\nu ||-0.9(e^{\left|\right|\zeta -{\nu \left|\right|}_{\infty }}-1))\\ & =2ln(e^{\left|\right|\zeta -{\nu \left|\right|}_{\infty }}-1+1)-(2ln(e^{\left|\right|\zeta -{\nu \left|\right|}_{\infty }}-1+1)-0.9(e^{\left|\right|\zeta -{\nu \left|\right|}_{\infty }}-1))\\ & =\sigma \left(\rho \right(\zeta -\nu \left)\right)-\xi \left(\rho \right(\zeta -\nu \left)\right).\end{array}$$

4. Conclusions

We introduced contraction type mappings by intervening $\mathrm{\Theta }$-contractions of Jleli and Samet [35] and some control functions including altering distance functions. We gave some fixed point theorems related to above mappings in the class of p-metric spaces. The obtained results have been illustrated by some concrete examples and an application on integral equations.