Fixed Point Results in Controlled Metric Spaces with Applications

Abstract

The aim of this paper is to obtain some common fixed point theorems for generalized contractions involving certain control functions in controlled metric space and derive some generalized fixed point results as a consequence of our main results. We also prove some common fixed point theorems in controlled metric spaces endowed with a graph. Our results will generalize and amend many famous results from the literature. We also provide an example to show the authenticity of the established results. As an application of our main result, we investigate the solution of integral equations.

1. Introduction

The manifest evolution of a metric space was fundamentally given by M. Frechet in 1906. Motivated by this contemporary idea, many researchers have generalized and extended this approach in recent years as: complex valued metric space, cone metric space, $\mathcal{F}$-metric space, $\mathrm{\Theta }$-metric space, orthogonal metric space, extended b-metric space, b-metric space and controlled metric space, etc.

Czerwik [1] gave the notion of a b-metric space as follows:

Definition 1.

(see. [1]) Let $\mathfrak{R}\ne \varnothing$ and $s\ge 1$ and $\tau :\mathfrak{R}\times \mathfrak{R}\to$$[0$,$\infty )$. If:

$\left(\tau 1\right)$$\tau (\mu ,\varsigma )=0$⇔$\mu =\varsigma ;$

$\left(\tau 2\right)$$\tau (\mu ,\varsigma )=\tau (\varsigma ,\mu )$for all$\mu ,\varsigma \in \mathfrak{R};$

$\left(\tau 3\right)$$\tau (\mu ,\omega )\le s\left[\tau \right(\mu ,\varsigma )+\tau (\varsigma ,\omega \left)\right]$for all$\mu ,\varsigma ,\omega \in \mathfrak{R}.$

then $(\mathfrak{R},\tau )$ is said to be a b-metric space.

Kamran et al. [2] defined the notion of a extended b-metric space in 2017:

Definition 2.

Let $\mathfrak{R}\ne \varnothing$ and $\sigma :\mathfrak{R}\times \mathfrak{R}\to [1,\infty )$ and $\tau :\mathfrak{R}\times \mathfrak{R}\to [0,\infty ).$ If

(i) $\tau (\mu ,\varsigma )=0$⇔$\mu =\varsigma$;

(ii) $\tau (\mu ,\varsigma )=\tau (\varsigma ,\mu )$;

(iii) $\tau (\mu ,\varsigma )\le \sigma (\mu ,\varsigma )\left[\tau \right(\mu ,\omega )+\tau (\varsigma ,\omega \left)\right].$

then $(\mathfrak{R},\tau )$ is said to be an extended b-metric space.

In 2018, a new type of extended b-metric space was given by Mlaiki et al. [3]:

Definition 3

([3]). Let $\mathfrak{R}\ne \varnothing$ and $\sigma :\mathfrak{R}\times \mathfrak{R}\to [1,\infty )$ and $\tau :\mathfrak{R}\times \mathfrak{R}\to [0,\infty ).$ If:

(i) $\tau (\mu ,\varsigma )=0$⇔$\mu =\varsigma$;

(ii) $\tau (\mu ,\varsigma )=\tau (\varsigma ,\mu )$;

(iii) $\tau (\mu ,\varsigma )\le \sigma (\mu ,\omega )\tau (\mu ,\omega )+\sigma (\varsigma ,\omega )\tau (\varsigma ,\omega ).$

then $(\mathfrak{R},\tau ,\sigma )$ is said to be a controlled metric space.

Example 1

([3]). Let $\mathfrak{R}=\left\{1,2,···\right\}.$ Take $\tau :\mathfrak{R}\times \mathfrak{R}\to [0,\infty )$ as

$$\tau (\mu ,\varsigma )=\left\{\begin{array}{c}0,\mathit{if}\mu =\varsigma \\ \frac{1}{\mu },\mathit{if}\mu \mathit{is}\mathit{even}\mathit{and}\varsigma \mathit{is}\mathit{odd}\\ \frac{1}{\varsigma },\mathit{if}\mu \mathit{is}\mathit{odd}\mathit{and}\varsigma \mathit{is}\mathit{even}\\ 1,\mathit{otherwise}.\end{array}\right.$$

Consider $\sigma :\mathfrak{R}\times \mathfrak{R}\to [1,\infty )$ as

$$\sigma (\mu ,\varsigma )=\left\{\begin{array}{c}\mu ,\mathit{if}\mu \mathit{is}\mathit{even}\mathit{and}\varsigma \mathit{is}\mathit{odd}\\ \varsigma ,\mathit{if}\mu \mathit{is}\mathit{odd}\mathit{and}\varsigma \mathit{is}\mathit{even}\\ 1,\mathit{otherwise}.\end{array}\right.$$

Then, τ is a controlled metric and $(\mathfrak{R},\tau ,\sigma )$ is a controlled metric space.

Theorem 1

([3]). Let $(\mathfrak{R},\sigma ,\tau )$ be a complete controlled metric space. Let $\mathcal{V}:\mathfrak{R}\to \mathfrak{R}$ be such that:

$$\tau (\mathcal{V}\mu ,\mathcal{V}\varsigma )\le \lambda \left(\tau \right(\mu ,\varsigma \left)\right)$$

∀$\mu ,\varsigma \in \mathfrak{R}$, where $\lambda \in [0,1).$ For ${\mu }_0\in \mathfrak{R},$ take ${\mu }_n={\mathcal{V}}^n{\mu }_0.$ Assume that:

$$\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\frac{\sigma ({\mu }_{i+1},{\mu }_{i+2})\sigma ({\mu }_{i+1},{\mu }_m)}{\sigma ({\mu }_i,{\mu }_{i+1})}<\frac{1}{\lambda }.$$

Furthermore, suppose that, $\forall \mu \in \mathfrak{R}$, we have ${lim}_{n\to \infty }\sigma ({\mu }_n,\mu )$ and ${lim}_{n\to \infty }\sigma (\mu ,{\mu }_n)$ which exist and are finite. Then, $\exists {\mu }^{*}\in \mathfrak{R}$ such that $\mathcal{V}{\mu }^{*}={\mu }^{*}$ which is unique.

Lateef [4] established Kannan [5] type fixed point theorems in the setting of a controlled metric space.

Theorem 2

([4]). Let $(\mathfrak{R},\sigma ,\tau )$ be a complete controlled metric space. Let $\mathcal{V}:\mathfrak{R}\to \mathfrak{R}$ be such that:

$$\tau (\mathcal{V}\mu ,\mathcal{V}\varsigma )\le \lambda \left(\tau \right(\mu ,\mathcal{V}\mu )+\tau (\varsigma ,\mathcal{V}\varsigma \left)\right)$$

∀$\mu ,\varsigma \in \mathfrak{R}$, where $\lambda \in (0,\frac{1}{2}).$ For ${\mu }_0\in \mathfrak{R},$ take ${\mu }_n={\mathcal{V}}^n{\mu }_0.$ Assume that:

$$\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\frac{\sigma ({\mu }_{i+1},{\mu }_{i+2})\sigma ({\mu }_{i+1},{\mu }_m)}{\sigma ({\mu }_i,{\mu }_{i+1})}<\frac{1}{\lambda }.$$

Furthermore, suppose that, ∀$\mu \in \mathfrak{R}$, we have: ${lim}_{n\to \infty }\sigma ({\mu }_n,\mu )$ and ${lim}_{n\to \infty }\sigma (\mu ,{\mu }_n)$ exist and are finite. Then, $\exists {\mu }^{*}\in \mathfrak{R}$ such that $\mathcal{V}{\mu }^{*}={\mu }^{*}$ which is unique.

Ahmad [6] established a Reich type fixed-point theorem in the setting of controlled metric space as follows.

Theorem 3.

Let $(\mathfrak{R},\sigma ,\tau )$ be a complete controlled metric space and $\mathcal{V}:\mathfrak{R}\to \mathfrak{R}$. If there exists $\alpha ,\beta ,\gamma \in (0,1)$ such that $\lambda =\alpha +\beta +\gamma <1$ and:

$$\tau (\mathcal{V}\mu ,\mathcal{V}\varsigma )\le \alpha \tau (\mu ,\varsigma )+\beta \tau (\mu ,\mathcal{V}\mu )+\gamma \tau (\varsigma ,\mathcal{V}\varsigma )$$

∀$\mu ,\varsigma \in \mathfrak{R}.$ For ${\mu }_0\in \mathfrak{R},$ take ${\mu }_n={\mathcal{V}}^n{\mu }_0.$ Suppose that:

$$\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\frac{\sigma ({\mu }_{i+1},{\mu }_{i+2})\sigma ({\mu }_{i+1},{\mu }_m)}{\sigma ({\mu }_i,{\mu }_{i+1})}<\frac{1}{\lambda }.$$

Furthermore, suppose that, for every $\mu \in \mathfrak{R}$, we have ${lim}_{n\to \infty }\sigma ({\mu }_n,\mu )$ and ${lim}_{n\to \infty }\sigma (\mu ,{\mu }_n)$ which exist and are finite. Then, $\exists {\mu }^{*}\in \mathfrak{R}$ such that $\mathcal{V}{\mu }^{*}={\mu }^{*}$ which is unique.

Later on, Abuloha et al. [7], Alamgir et al. [8], Abdeljawad et al. [9], Lateef [10], Hussain [11], Mlaiki et al. [12], Shatanawi et al. [13], Sezen et al. [14] and Tasneem et al. [15] studied controlled metric spaces and established different fixed-point results for self and multivalued mappings. For more details, in this direction, we refer the readers to [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17].

In this paper, we obtain some common fixed-point results for generalized contractions involving some certain control functions in the setting of controlled metric spaces. We also proved some common fixed-point theorems in controlled metric spaces endowed with a graph. We also provided an example to show the legitimacy of the established results. As an application of our main result, we investigate the solution of integral equations.

2. Main Results

We state our main result as follows.

Theorem 4.

Let $(\mathfrak{R},\sigma ,\tau )$ be a complete controlled metric space and $\mathcal{V}$,$\mathcal{U}:\mathfrak{R}\to \mathfrak{R}$. If there exists $\alpha ,\beta :\mathfrak{R}\to [0,1)$ such that:

(i) $\alpha \left(\mathcal{V}\mu \right)\le \alpha \left(\mu \right)$ and $\beta \left(\mathcal{V}\mu \right)\le \beta \left(\mu \right);$

(ii) $\alpha \left(\mathcal{U}\mu \right)\le \alpha \left(\mu \right)$ and $\beta \left(\mathcal{U}\mu \right)\le \beta \left(\mu \right);$

(iii) ($\alpha +\beta \left)\right(\mu )<1;$

(iv)

$$\tau (\mathcal{V}\mu ,\mathcal{U}\varsigma )\le \alpha \left(\mu \right)\tau (\mu ,\varsigma )+\beta \left(\mu \right)\frac{\tau (\mu ,\mathcal{V}\mu )\tau (\varsigma ,\mathcal{U}\varsigma )}{1+\tau (\mu ,\varsigma )}$$

(1)

for all $\mu ,\varsigma \in \mathfrak{R}$. For ${\mu }_0\in \mathfrak{R},$ we set $\frac{\alpha \left({\mu }_0\right)}{1-\beta \left({\mu }_0\right)}=\lambda .$ Suppose that:

$$\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\frac{\sigma ({\mu }_{i+1},{\mu }_{i+2})\sigma ({\mu }_{i+1},{\mu }_m)}{\sigma ({\mu }_i,{\mu }_{i+1})}<\frac{1}{\lambda }$$

(2)

where ${\mu }_{2n+1}=\mathcal{V}{\mu }_{2n}$ and ${\mu }_{2n+2}=\mathcal{U}{\mu }_{2n+1}$ for each $n\ge 0.$ In addition, assume that, for every $\mu \in \mathfrak{R}$, we have ${lim}_{n\to \infty }\sigma ({\mu }_n,\mu )$ and ${lim}_{n\to \infty }\sigma (\mu ,{\mu }_n)$ which exist and are finite. Then, $\mathcal{V}$ and $\mathcal{U}$ have a unique common fixed point.

Proof. 

Let ${\mu }_0\in$ $\mathfrak{R}$. We construct $\left\{{\mu }_n\right\}$ in $\mathfrak{R}$ by ${\mu }_{2n+1}=\mathcal{V}{\mu }_{2n}$ and ${\mu }_{2n+2}=\mathcal{U}{\mu }_{2n+1}$ for each $n\ge 0.$ From hypothesis and (1) we obtain:

$$\begin{array}{ccc}\hfill \tau ({\mu }_{2n+1},{\mu }_{2n+2})& =& \tau (\mathcal{V}{\mu }_{2n},\mathcal{U}{\mu }_{2n+1})\hfill \\ & \le & \alpha \left({\mu }_{2n}\right)\tau ({\mu }_{2n},{\mu }_{2n+1})+\beta \left({\mu }_{2n}\right)\frac{\tau ({\mu }_{2n},\mathcal{V}{\mu }_{2n})\tau ({\mu }_{2n+1},\mathcal{U}{\mu }_{2n+1})}{1+\tau ({\mu }_{2n},{\mu }_{2n+1})}\hfill \\ & =& \alpha \left({\mu }_{2n}\right)\tau ({\mu }_{2n},{\mu }_{2n+1})+\beta \left({\mu }_{2n}\right)\frac{\tau ({\mu }_{2n},{\mu }_{2n+1})\tau ({\mu }_{2n+1},{\mu }_{2n+2})}{1+\tau ({\mu }_{2n},{\mu }_{2n+1})}\hfill \\ & \le & \alpha \left({\mu }_{2n}\right)\tau ({\mu }_{2n},{\mu }_{2n+1})+\beta \left({\mu }_{2n}\right)\tau ({\mu }_{2n+1},{\mu }_{2n+2})\hfill \\ & =& \alpha \left(\mathcal{U}{\mu }_{2n-1}\right)\tau ({\mu }_{2n},{\mu }_{2n+1})+\beta \left(\mathcal{U}{\mu }_{2n-1}\right)\tau ({\mu }_{2n+1},{\mu }_{2n+2})\hfill \\ & \le & \alpha \left({\mu }_{2n-1}\right)\tau ({\mu }_{2n},{\mu }_{2n+1})+\beta \left({\mu }_{2n-1}\right)\tau ({\mu }_{2n+1},{\mu }_{2n+2})\hfill \\ & =& \alpha \left(\mathcal{V}{\mu }_{2n-2}\right)\tau ({\mu }_{2n},{\mu }_{2n+1})+\beta \left(\mathcal{V}{\mu }_{2n-2}\right)\tau ({\mu }_{2n+1},{\mu }_{2n+2})\hfill \\ & \le & \alpha \left({\mu }_{2n-2}\right)\tau ({\mu }_{2n},{\mu }_{2n+1})+\beta \left({\mu }_{2n-2}\right)\tau ({\mu }_{2n+1},{\mu }_{2n+2})\hfill \\ & & ⋮\hfill \\ & \le & \alpha \left({\mu }_0\right)\tau ({\mu }_{2n},{\mu }_{2n+1})+\beta \left({\mu }_0\right)\tau ({\mu }_{2n+1},{\mu }_{2n+2})\hfill \end{array}$$

which implies that:

$$\tau ({\mu }_{2n+1},{\mu }_{2n+2})\le \left(\frac{\alpha \left({\mu }_0\right)}{1-\beta \left({\mu }_0\right)}\right)\tau ({\mu }_{2n},{\mu }_{2n+1})=.$$

Similarly:

$$\begin{array}{ccc}\hfill \tau ({\mu }_{2n+2},{\mu }_{2n+3})& =& \tau (\mathcal{U}{\mu }_{2n+1},\mathcal{V}{\mu }_{n+2})\hfill \\ & =& \tau (\mathcal{V}{\mu }_{2n+2},\mathcal{U}{\mu }_{2n+1})\hfill \\ & \le & \alpha \left({\mu }_{2n+2}\right)\tau ({\mu }_{2n+2},{\mu }_{2n+1})+\beta \left({\mu }_{2n+2}\right)\frac{\tau ({\mu }_{2n+2},\mathcal{V}{\mu }_{2n+2})\tau ({\mu }_{2n+1},\mathcal{V}{\mu }_{2n+1})}{1+\tau ({\mu }_{2n+2},{\mu }_{2n+1})}\hfill \\ & =& \alpha \left({\mu }_{2n+2}\right)\tau ({\mu }_{2n+2},{\mu }_{2n+1})+\beta \left({\mu }_{2n+2}\right)\frac{\tau ({\mu }_{2n+2},{\mu }_{2n+3})\tau ({\mu }_{2n+1},{\mu }_{2n+2})}{1+\tau ({\mu }_{2n+2},{\mu }_{2n+1})}\hfill \\ & =& \alpha \left({\mu }_{2n+2}\right)\tau ({\mu }_{2n+2},{\mu }_{2n+1})+\beta \left({\mu }_{2n+2}\right)\tau ({\mu }_{2n+2},{\mu }_{2n+3})\hfill \\ & =& \alpha \left(\mathcal{U}{\mu }_{2n+1}\right)\tau ({\mu }_{2n+1},{\mu }_{2n+2})+\beta \left(\mathcal{U}{\mu }_{2n+1}\right)\tau ({\mu }_{2n+2},{\mu }_{2n+3})\hfill \\ & \le & \alpha \left({\mu }_{2n+1}\right)\tau ({\mu }_{2n+1},{\mu }_{2n+2})+\beta \left({\mu }_{2n+1}\right)\tau ({\mu }_{2n+2},{\mu }_{2n+3})\hfill \\ & =& \alpha \left(\mathcal{V}{\mu }_{2n}\right)\tau ({\mu }_{2n+1},{\mu }_{2n+2})+\beta \left(\mathcal{V}{\mu }_{2n}\right)\tau ({\mu }_{2n+2},{\mu }_{2n+3})\hfill \\ & \le & \alpha \left({\mu }_{2n}\right)\tau ({\mu }_{2n+1},{\mu }_{2n+2})+\beta \left({\mu }_{2n}\right)\tau ({\mu }_{2n+2},{\mu }_{2n+3})\hfill \\ & & ⋮\hfill \\ & \le & \alpha \left({\mu }_0\right)\tau ({\mu }_{2n+1},{\mu }_{2n+2})+\beta \left({\mu }_0\right)\tau ({\mu }_{2n+2},{\mu }_{2n+3})\hfill \end{array}$$

which implies that:

$$\tau ({\mu }_{2n+2},{\mu }_{2n+3})\le \left(\frac{\alpha \left({\mu }_0\right)}{1-\beta \left({\mu }_0\right)}\right)\tau ({\mu }_{2n+1},{\mu }_{2n+2})=\lambda \tau ({\mu }_{2n+1},{\mu }_{2n+2}).$$

pursuing in this direction, we obtain:

$$\begin{array}{ccc}\hfill \tau ({\mu }_n,{\mu }_{n+1})& \le & \lambda \tau ({\mu }_{n-1},{\mu }_n)\hfill \\ & \le & {\lambda }^2\tau ({\mu }_{n-2},{\mu }_{n-1})\hfill \\ & \le & ...\le {\lambda }^n\tau ({\mu }_0,{\mu }_1).\hfill \end{array}$$

Thus:

$$\tau ({\mu }_n,{\mu }_{n+1})\le {\lambda }^n\tau ({\mu }_0,{\mu }_1).$$

(3)

For all $n,m\in \mathbb{N}(n<m)$, we have:

$$\begin{array}{ccc}\hfill \tau ({\mu }_n,{\mu }_m)& \le & \sigma ({\mu }_n,{\mu }_{n+1})\tau ({\mu }_n,{\mu }_{n+1})+\sigma ({\mu }_{n+1},{\mu }_m)\tau ({\mu }_{n+1},{\mu }_m)\hfill \\ & \le & \sigma ({\mu }_n,{\mu }_{n+1})\tau ({\mu }_n,{\mu }_{n+1})+\sigma ({\mu }_{n+1},{\mu }_m)\sigma ({\mu }_{n+1},{\mu }_{n+2})\tau ({\mu }_{n+1},{\mu }_{n+2})\hfill \\ & & +\sigma ({\mu }_{n+1},{\mu }_m)\sigma ({\mu }_{n+2},{\mu }_m)\tau ({\mu }_{n+2},{\mu }_m)\hfill \\ & \le & \sigma ({\mu }_n,{\mu }_{n+1})\tau ({\mu }_n,{\mu }_{n+1})+\sigma ({\mu }_{n+1},{\mu }_m)\sigma ({\mu }_{n+1},{\mu }_{n+2})\tau ({\mu }_{n+1},{\mu }_{n+2})\hfill \\ & & +\sigma ({\mu }_{n+1},{\mu }_m)\sigma ({\mu }_{n+2},{\mu }_m)\sigma ({\mu }_{n+2},{\mu }_{n+3})\tau ({\mu }_{n+2},{\mu }_{n+3})\hfill \\ & & +\sigma ({\mu }_{n+1},{\mu }_m)\sigma ({\mu }_{n+2},{\mu }_m)\sigma ({\mu }_{n+3},{\mu }_m)\tau ({\mu }_{n+3},{\mu }_m)\hfill \\ & \le & ···\hfill \\ & \le & \sigma ({\mu }_n,{\mu }_{n+1})\tau ({\mu }_n,{\mu }_{n+1})+\sum _{i=n+1}^{m-2}\left(\prod _{j=n+1}^i\sigma ({\mu }_j,{\mu }_m)\right)\sigma ({\mu }_i,{\mu }_{i+1})\tau ({\mu }_i,{\mu }_{i+1})\hfill \\ & & +\prod _{i=n+1}^{m-1}\sigma ({\mu }_i,{\mu }_m)\tau ({\mu }_{m-1},{\mu }_m)\hfill \end{array}$$

which further implies that:

$$\begin{array}{ccc}\hfill \tau ({\mu }_n,{\mu }_m)& \le & \sigma ({\mu }_n,{\mu }_{n+1})\tau ({\mu }_n,{\mu }_{n+1})+\sum _{i=n+1}^{m-2}\left({\displaystyle \prod _{j=n+1}^i\sigma ({\mu }_j,{\mu }_m)}\right)\sigma ({\mu }_i,{\mu }_{i+1})\tau ({\mu }_i,{\mu }_{i+1})\hfill \\ & & +\left(\prod _{i=n+1}^{m-1}\sigma ({\mu }_i,{\mu }_m)\right)\sigma ({\mu }_{m-1},{\mu }_m)\tau ({\mu }_{m-1},{\mu }_m)\hfill \\ & \le & \sigma ({\mu }_n,{\mu }_{n+1}){\lambda }^n\tau ({\mu }_0,{\mu }_1)+\sum _{i=n+1}^{m-2}\left(\prod _{j=n+1}^i\sigma ({\mu }_j,{\mu }_m)\right)\sigma ({\mu }_i,{\mu }_{i+1}){\lambda }^i\tau ({\mu }_0,{\mu }_1)\hfill \\ & & +\left(\prod _{i=n+1}^{m-1}\sigma ({\mu }_i,{\mu }_m)\right)\sigma ({\mu }_{m-1},{\mu }_m){\lambda }^{m-1}\tau ({\mu }_0,{\mu }_1)\hfill \\ & =& \sigma ({\mu }_n,{\mu }_{n+1}){\lambda }^n\tau ({\mu }_0,{\mu }_1)+\sum _{i=n+1}^{m-1}\left(\prod _{j=n+1}^i\sigma ({\mu }_j,{\mu }_m)\right)\sigma ({\mu }_i,{\mu }_{i+1}){\lambda }^i\tau ({\mu }_0,{\mu }_1).\hfill \end{array}$$

Thus:

$$\tau ({\mu }_n,{\mu }_m)\le \sigma ({\mu }_n,{\mu }_{n+1}){\lambda }^n\tau ({\mu }_0,{\mu }_1)+\sum _{i=n+1}^{m-1}\left(\prod _{j=n+1}^i\sigma ({\mu }_j,{\mu }_m)\right)\sigma ({\mu }_i,{\mu }_{i+1}){\lambda }^i\tau ({\mu }_0,{\mu }_1).$$

(4)

Let:

$$S_l=\sum _{i=0}^l\left(\prod _{j=0}^i\sigma ({\mu }_j,{\mu }_m)\right)\sigma ({\mu }_i,{\mu }_{i+1}){\lambda }^i\tau ({\mu }_0,{\mu }_1).$$

From (4), we obtain:

$$\tau ({\mu }_n,{\mu }_m)\le \tau ({\mu }_0,{\mu }_1)[{\lambda }^n\sigma ({\mu }_n,{\mu }_{n+1})+(S_{m-1}-S_n)].$$

(5)

Now, using the fact that $\sigma (\mu ,\varsigma )\ge 1$, and by using the ratio test, ${lim}_{n\to \infty }{S}_n$ exists. Hence, $\left\{S_n\right\}$ is Cauchy. Eventually, taking $n,m\to \infty$ in (5), we obtain that:

$$\underset{n,m\to \infty }{lim}\tau ({\mu }_n,{\mu }_m)=0.$$

(6)

Hence, $\left\{{\mu }_n\right\}$ is a Cauchy $(\mathfrak{R},\tau ,\sigma )$. Thus, ∃${\mu }^{*}\in \mathfrak{R}$ such that

$$\underset{n\to \infty }{lim}\tau ({\mu }_n,{\mu }^{*})=0$$

(7)

that is ${\mu }_n\to {\mu }^{*}$ as $n\to \infty .$ Now, by (1) and condition (iii), we obtain:

$$\begin{array}{ccc}\hfill \tau ({\mu }^{*},\mathcal{V}{\mu }^{*})& \le & \sigma ({\mu }^{*},{\mu }_{2n+2})\tau ({\mu }^{*},{\mu }_{2n+2})+\sigma ({\mu }_{2n+2},\mathcal{V}{\mu }^{*})\tau ({\mu }_{2n+2},\mathcal{V}{\mu }^{*})\hfill \\ & =& \sigma ({\mu }^{*},{\mu }_{2n+2})\tau ({\mu }^{*},{\mu }_{2n+2})+\sigma ({\mu }_{2n+2},\mathcal{V}{\mu }^{*})\tau (\mathcal{U}{\mu }_{2n+1},\mathcal{V}{\mu }^{*})\hfill \\ & =& \sigma ({\mu }^{*},{\mu }_{2n+2})\tau ({\mu }^{*},{\mu }_{2n+2})+\sigma ({\mu }_{2n+2},\mathcal{V}{\mu }^{*})\tau (\mathcal{V}{\mu }^{*},\mathcal{U}{\mu }_{2n+1})\hfill \\ & =& \sigma ({\mu }^{*},{\mu }_{2n+2})\tau ({\mu }^{*},{\mu }_{2n+2})+\sigma ({\mu }_{2n+2},\mathcal{V}{\mu }^{*})\left[\begin{array}{c}\alpha \left({\mu }^{*}\right)\tau ({\mu }^{*},{\mu }_{2n+1})\\ +\beta \left({\mu }^{*}\right)\frac{\tau ({\mu }^{*},\mathcal{V}{\mu }^{*})\tau ({\mu }_{2n+1},{\mu }_{2n+2})}{1+({\mu }^{*},{\mu }_{2n+1})}\end{array}\right]\hfill \\ & =& \sigma ({\mu }^{*},{\mu }_{2n+2})\tau ({\mu }^{*},{\mu }_{2n+2})+\sigma ({\mu }_{2n+2},\mathcal{V}{\mu }^{*})\left[\begin{array}{c}\alpha \left({\mu }^{*}\right)\tau ({\mu }^{*},{\mu }_{2n+1})\\ +\beta \left({\mu }^{*}\right)\frac{\tau ({\mu }^{*},\mathcal{V}{\mu }^{*})\tau ({\mu }_{2n+1},{\mu }_{2n+2})}{1+({\mu }^{*},{\mu }_{2n+1})}\end{array}\right]\hfill \end{array}$$

Letting $n\to \infty$ and using (7), we obtain a contradiction to $\tau ({\mu }^{*},\mathcal{V}{\mu }^{*})>0$. Thus, $\tau ({\mu }^{*},\mathcal{V}{\mu }^{*})=0$. This implies that ${\mu }^{*}=\mathcal{V}{\mu }^{*}$. It follows similarly that ${\mu }^{*}=\mathcal{U}{\mu }^{*}.$ Therefore, ${\mu }^{*}$ is a common fixed point of $\mathcal{V}$ and $\mathcal{U}$. Eventually, we show that ${\mu }^{*}$ is a unique common fixed point of $\mathcal{V}$ and $\mathcal{U}$. Assume that there exists another common fixed point ${\mu }^{/}$ that is ${\mu }^{/}=\mathcal{V}{\mu }^{/}=\mathcal{U}{\mu }^{/}.$ It follows from:

$$\begin{array}{ccc}\hfill \tau ({\mu }^{*},{\mu }^{/})& =& \tau (\mathcal{V}{\mu }^{*},\mathcal{U}{\mu }^{/})\le \alpha \left({\mu }^{*}\right)\tau ({\mu }^{*},{\mu }^{/})+\beta \left({\mu }^{*}\right)\frac{\tau ({\mu }^{*},\mathcal{V}{\mu }^{*})\tau (\varsigma ,\mathcal{U}{\mu }^{/})}{1+\tau ({\mu }^{*},{\mu }^{/})}\hfill \\ & =& \alpha \left({\mu }^{*}\right)\tau ({\mu }^{*},{\mu }^{/}).\hfill \end{array}$$

Since $\alpha \left({\mu }^{*}\right)\in [0,1),$ so we have $\tau ({\mu }^{*},{\mu }^{/}).$ Therefore, we have ${\mu }^{*}={\mu }^{/}$ and thus ${\mu }^{*}$ is a unique common fixed point of $\mathcal{V}$ and $\mathcal{U}$. □

Corollary 1.

Let $(\mathfrak{R},\sigma ,\tau )$ be a complete controlled metric space and $\mathcal{V}$,$\mathcal{U}:\mathfrak{R}\to \mathfrak{R}$. If $\mathcal{V}$,$\mathcal{U}$ satisfy:

$$\tau (\mathcal{V}\mu ,\mathcal{U}\varsigma )\le \alpha \tau (\mu ,\varsigma )+\beta \frac{\tau (\mu ,\mathcal{V}\mu )\tau (\varsigma ,\mathcal{U}\varsigma )}{1+\tau (\mu ,\varsigma )}$$

for all $\mu ,\varsigma \in \mathfrak{R}$. For ${\mu }_0\in \mathfrak{R},$ we set $\frac{\alpha }{1-\beta }=\lambda .$ Suppose that:

$$\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\frac{\sigma ({\mu }_{i+1},{\mu }_{i+2})\sigma ({\mu }_{i+1},{\mu }_m)}{\sigma ({\mu }_i,{\mu }_{i+1})}<\frac{1}{\lambda }$$

where ${\mu }_{2n+1}=\mathcal{V}{\mu }_{2n}$ and ${\mu }_{2n+2}=\mathcal{U}{\mu }_{2n+1}$ for each $n\ge 0.$ Moreover, assume that ∀$\mu \in \mathfrak{R}$, we have ${lim}_{n\to \infty }\sigma ({\mu }_n,\mu )$ and ${lim}_{n\to \infty }\sigma (\mu ,{\mu }_n)$ which exist and are finite. Then, $\mathcal{V}$ and $\mathcal{U}$ have a unique common fixed point.

Proof. 

We can prove this result by applying Theorem 4 by setting $\alpha \left(\mu \right)=\alpha$ and $\beta \left(\mu \right)=\beta .$ □

Corollary 2.

Let $(\mathfrak{R},\sigma ,\tau )$ be a complete controlled metric space and $\mathcal{U}:\mathfrak{R}\to \mathfrak{R}$. If there exists $\alpha ,\beta :\mathfrak{R}\to [0,1)$ such that:

(i) $\alpha \left(\mathcal{U}\mu \right)\le \alpha \left(\mu \right)$ and $\beta \left(\mathcal{U}\mu \right)\le \beta \left(\mu \right);$

(ii) ($\alpha +\beta \left)\right(\mu )<1;$

(iii)

$$\tau (\mathcal{U}\mu ,\mathcal{U}\varsigma )\le \alpha \left(\mu \right)\tau (\mu ,\varsigma )+\beta \left(\mu \right)\frac{\tau (\mu ,\mathcal{U}\mu )\tau (\varsigma ,\mathcal{U}\varsigma )}{1+\tau (\mu ,\varsigma )}$$

(8)

for all $\mu ,\varsigma \in \mathfrak{R}$. For ${\mu }_0\in \mathfrak{R},$ we set $\frac{\alpha \left({\mu }_0\right)}{1-\beta \left({\mu }_0\right)}=\lambda .$ Assume that:

$$\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\frac{\sigma ({\mu }_{i+1},{\mu }_{i+2})\sigma ({\mu }_{i+1},{\mu }_m)}{\sigma ({\mu }_i,{\mu }_{i+1})}<\frac{1}{\lambda }$$

(9)

where ${\mu }_{n+1}=\mathcal{U}{\mu }_n$ for each $n\ge 0.$ Furthermore, assume that ∀$\mu \in \mathfrak{R}$, we have ${lim}_{n\to \infty }\sigma ({\mu }_n,\mu )$ and ${lim}_{n\to \infty }\sigma (\mu ,{\mu }_n)$ which exist and are finite. Then, $\mathcal{U}$ has a unique fixed point.

Proof. 

We can prove this result by applying Theorem 4 with $\mathcal{V}=$$\mathcal{U}$. □

Example 2.

Let $\mathfrak{R}=\left[0,1\right].$ Now, we define $\tau :\mathfrak{R}\times \mathfrak{R}\to [0,\infty )$ by

$$\tau (\mu ,\varsigma )={\left(\mu +\varsigma \right)}^2$$

where $\sigma (\mu ,\varsigma )=2+\mu +\varsigma ,$$\forall \mu ,\varsigma \in \mathfrak{R}.$ Now, we define $\mathcal{V}$,$\mathcal{U}:\mathfrak{R}\to \mathfrak{R}$ by

$$\mathcal{V}\mu =\frac{\mu }{4}\mathit{and}\mathcal{U}\mu =\frac{\mu }{5}$$

for $\mu \in \mathfrak{R}.$ Choose $\alpha ,\beta :\mathfrak{R}\to [0,1)$ by

$$\alpha \left(\mu \right)=\frac{25+\mu }{400}\mathit{and}\beta \left(\mu \right)=\frac{23+\mu }{400}.$$

Take ${\mu }_0=0,$ so (2) is satisfied. Let $\mu ,\varsigma \in \mathfrak{R}.$ Then:

$$\begin{array}{ccc}\hfill \tau (\mathcal{V}\mu ,\mathcal{U}\varsigma )& =& \frac{{\left(5\mu +4\varsigma \right)}^2}{400}\le \frac{{\left(5\mu +5\varsigma \right)}^2}{400}\hfill \\ & \le & \frac{25+\mu }{400}{\left(\mu +\varsigma \right)}^2+\frac{23+\mu }{400}\frac{{\left(\frac{5\mu }{4}\right)}^2{\left(\frac{6\varsigma }{5}\right)}^2}{1+{\left(\mu +\varsigma \right)}^2}\hfill \\ & =& \alpha \left(\mu \right)\tau (\mu ,\varsigma )+\beta \left(\mu \right)\frac{\tau (\mu ,\mathcal{V}\mu )\tau (\varsigma ,\mathcal{U}\varsigma )}{1+\tau (\mu ,\varsigma )}.\hfill \end{array}$$

Note that all the assumptions of Theorem 4 are satisfied. Thus, ∃μ$\in \mathfrak{R}$ such that $\mathcal{V}\mu$ = μ, which is $\mu =0$.

Corollary 3.

Let $(\mathfrak{R},\sigma ,\tau )$ be a complete controlled metric space and $\mathcal{U}:\mathfrak{R}\to \mathfrak{R}$. If there exists $\alpha :\mathfrak{R}\to [0,1)$ such that:

(i) $\alpha \left(\mathcal{U}\mu \right)\le \alpha \left(\mu \right);$

(ii)

$$\tau (\mathcal{U}\mu ,\mathcal{U}\varsigma )\le \alpha \left(\mu \right)\tau (\mu ,\varsigma )$$

for all $\mu ,\varsigma \in \mathfrak{R}$. For ${\mu }_0\in \mathfrak{R}.$ Suppose that:

$$\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\frac{\sigma ({\mu }_{i+1},{\mu }_{i+2})\sigma ({\mu }_{i+1},{\mu }_m)}{\sigma ({\mu }_i,{\mu }_{i+1})}<\frac{1}{\alpha \left({\mu }_0\right)}$$

where ${\mu }_{n+1}=\mathcal{U}{\mu }_n$ for each $n\ge 0.$ In addition to these, assume that, $\forall \mu \in \mathfrak{R}$, we have ${lim}_{n\to \infty }\sigma ({\mu }_n,\mu )$ and ${lim}_{n\to \infty }\sigma (\mu ,{\mu }_n)$ which exist and are finite. Then, $\mathcal{U}$ has a unique fixed point.

Proof. 

Taking $\beta (·)=\beta =0$ in Corollary 2. □

Corollary 4.

Let $(\mathfrak{R},\sigma ,\tau )$ be a complete controlled metric space and $\mathcal{U}:\mathfrak{R}\to \mathfrak{R}$. If $\mathcal{U}$ satisfy:

$$\tau (\mathcal{U}\mu ,\mathcal{U}\varsigma )\le \alpha \tau (\mu ,\varsigma )+\beta \frac{\tau (\mu ,\mathcal{U}\mu )\tau (\varsigma ,\mathcal{U}\varsigma )}{1+\tau (\mu ,\varsigma )}$$

for all $\mu ,\varsigma \in \mathfrak{R}$. For ${\mu }_0\in \mathfrak{R},$ we set $\frac{\alpha }{1-\beta }=\lambda .$ Suppose that:

$$\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\frac{\sigma ({\mu }_{i+1},{\mu }_{i+2})\sigma ({\mu }_{i+1},{\mu }_m)}{\sigma ({\mu }_i,{\mu }_{i+1})}<\frac{1}{\lambda }$$

where ${\mu }_{n+1}=\mathcal{U}{\mu }_n$ for each $n\ge 0.$ Moreover, assume that ∀$\mu \in \mathfrak{R}$, we have ${lim}_{n\to \infty }\sigma ({\mu }_n,\mu )$ and ${lim}_{n\to \infty }\sigma (\mu ,{\mu }_n)$ which exist and are finite. Then, $\mathcal{U}$ has a unique fixed point.

Proof. 

We can prove this result by applying Corollary 2 by setting $\alpha \left(\mu \right)=\alpha$ and $\beta \left(\mu \right)=\beta .$ □

Corollary 5.

(See. [3]) Let $(\mathfrak{R},\sigma ,\tau )$ be a complete controlled metric space and $\mathcal{U}:\mathfrak{R}\to \mathfrak{R}$. If $\mathcal{U}$ satisfies:

$$\tau (\mathcal{U}\mu ,\mathcal{U}\varsigma )\le \alpha \tau (\mu ,\varsigma )$$

for all $\mu ,\varsigma \in \mathfrak{R}$. For ${\mu }_0\in \mathfrak{R},$ we set $\alpha <1.$ Suppose that:

$$\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\frac{\sigma ({\mu }_{i+1},{\mu }_{i+2})\sigma ({\mu }_{i+1},{\mu }_m)}{\sigma ({\mu }_i,{\mu }_{i+1})}<\frac{1}{\alpha }$$

where ${\mu }_{n+1}=\mathcal{U}{\mu }_n$ for each $n\ge 0.$ Moreover, assume that ∀$\mu \in \mathfrak{R}$, we have ${lim}_{n\to \infty }\sigma ({\mu }_n,\mu )$ and ${lim}_{n\to \infty }\sigma (\mu ,{\mu }_n)$ which exist and are finite. Then, $\mathcal{U}$ has a unique fixed point.

Theorem 5.

Let $(\mathfrak{R},\sigma ,\tau )$ be a complete controlled metric space and $\mathcal{U}:\mathfrak{R}\to \mathfrak{R}$. If there exists $\alpha ,\beta :\mathfrak{R}\to [0,1)$ such that for all $\mu ,\varsigma \in \mathfrak{R}$ and for some $n\in N,$ we have:

(i) $\alpha \left({\mathcal{U}}^n\mu \right)\le \alpha \left(\mu \right)$ and $\beta \left({\mathcal{U}}^n\mu \right)\le \beta \left(\mu \right);$

(ii) ($\alpha +\beta \left)\right(\mu )<1;$

(iii)

$$\tau ({\mathcal{U}}^n\mu ,{\mathcal{U}}^n\varsigma )\le \alpha \left(\mu \right)\tau (\mu ,\varsigma )+\beta \left(\mu \right)\frac{\tau (\mu ,{\mathcal{U}}^n\mu )\tau (\varsigma ,{\mathcal{U}}^n\varsigma )}{1+\tau (\mu ,\varsigma )}.$$

For ${\mu }_0\in \mathfrak{R},$ we set $\frac{\alpha \left({\mu }_0\right)}{1-\beta \left({\mu }_0\right)}=\lambda .$ Suppose that:

$$\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\frac{\sigma ({\mu }_{i+1},{\mu }_{i+2})\sigma ({\mu }_{i+1},{\mu }_m)}{\sigma ({\mu }_i,{\mu }_{i+1})}<\frac{1}{\lambda }$$

where ${\mu }_{n+1}=\mathcal{U}{\mu }_n$ for each $n\ge 0.$ Moreover, assume that ∀$\mu \in \mathfrak{R}$, we have ${lim}_{n\to \infty }\sigma ({\mu }_n,\mu )$ and ${lim}_{n\to \infty }\sigma (\mu ,{\mu }_n)$ which exist and are finite. Then, $\mathcal{U}$ has a unique fixed point.

Proof. 

From Corollary 2, we obtain ${\mathcal{U}}^n$ which has a unique fixed point ${\mu }^{*}$. It follows from:

$${\mathcal{U}}^n\left(\mathcal{U}{\mu }^{*}\right)=\mathcal{U}\left({\mathcal{U}}^n{\mu }^{*}\right)=\mathcal{U}{\mu }^{*}$$

that $\mathcal{U}{\mu }^{*}$ is a fixed point of ${\mathcal{U}}^n$. Therefore, $\mathcal{U}{\mu }^{*}$ = ${\mu }^{*}$ by the uniqueness of a fixed point of ${\mathcal{U}}^n$ and then ${\mu }^{*}$ is also a fixed point of $\mathcal{U}$. Since the fixed point of $\mathcal{U}$ is also a fixed point of ${\mathcal{U}}^n$, the fixed point of $\mathcal{U}$ is unique. □

Corollary 6.

Let $(\mathfrak{R},\sigma ,\tau )$ be a complete controlled metric space and $\mathcal{U}:\mathfrak{R}\to \mathfrak{R}$. If $\mathcal{U}$ satisfies:

$$\tau ({\mathcal{U}}^n\mu ,{\mathcal{U}}^n\varsigma )\le \alpha \tau (\mu ,\varsigma )+\beta \frac{\tau (\mu ,{\mathcal{U}}^n\mu )\tau (\varsigma ,{\mathcal{U}}^n\varsigma )}{1+\tau (\mu ,\varsigma )}$$

for all $\mu ,\varsigma \in \mathfrak{R}$. For ${\mu }_0\in \mathfrak{R},$ we set $\frac{\alpha }{1-\beta }=\lambda .$ Suppose that:

$$\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\frac{\sigma ({\mu }_{i+1},{\mu }_{i+2})\sigma ({\mu }_{i+1},{\mu }_m)}{\sigma ({\mu }_i,{\mu }_{i+1})}<\frac{1}{\lambda }$$

where ${\mu }_{n+1}=\mathcal{U}{\mu }_n$ for each $n\ge 0.$ Moreover, assume that, $\forall \mu \in \mathfrak{R}$, we have ${lim}_{n\to \infty }\sigma ({\mu }_n,\mu )$ and ${lim}_{n\to \infty }\sigma (\mu ,{\mu }_n)$ which exist and are finite. Then, $\mathcal{U}$ have a unique fixed point.

Proof. 

We can prove this result by applying the Theorem 5 by setting $\alpha \left(\mu \right)=\alpha$ and $\beta \left(\mu \right)=\beta .$ □

Example 3.

Let $\mathfrak{R}=\{0,1,2\}$. Define $\sigma :\mathfrak{R}\times \mathfrak{R}\to [1,\infty )$ and $\tau :\mathfrak{R}\times \mathfrak{R}\to [1,\infty )$ as $\sigma (\mu ,\varsigma )=1+\mu \varsigma$ and:

$$\begin{array}{ccc}\hfill \tau (2,2)& =& \tau (0,0)=\tau (1,1)=0\hfill \\ \hfill \tau (2,0)& =& \tau (0,2)=5,\tau (0,1)=\tau (0,1)=10\hfill \\ \hfill \tau (1,2)& =& \tau (2,1)=15.\hfill \end{array}$$

Now, define:

$$\mathcal{V}:\mathfrak{R}\to \mathfrak{R}$$

by

$$\mathcal{V}0=\mathcal{V}2=0,\mathcal{V}1=2$$

and $\alpha ,\beta :\mathfrak{R}\to [0,1)$:

$$\alpha \left(0\right)=\frac{1}{2},\alpha \left(1\right)=\frac{3}{4},\alpha \left(2\right)=\frac{2}{3}$$

$$\beta \left(0\right)=\frac{1}{6},\beta \left(1\right)=\frac{1}{5},\beta \left(2\right)=\frac{2}{11}.$$

It is very simple to show that (8) holds. By result 2, there exists a unique μ such that $\mathcal{V}\mu =\mu ,$ that is, $\mu =0.$

3. Fixed-Point Results for Graphs

Let $(\mathfrak{R},\sigma ,\tau )$ be a controlled metric space and $\Delta$ denote the diagonal of $\mathfrak{R}\times \mathfrak{R}$. A directed graph is represented by G with $V\left(G\right)$ as its vertices coincide with $\mathfrak{R}$ and $\Delta$ is contained in $E\left(G\right),$ where $E\left(G\right)$ is the set of edges of the graph. Additionally suppose that G has no parallel edges and thus, one can recognize G with the pair $\left(V\left(G\right),E\left(G\right)\right).$ In this paper, we shall represent G as a graph satisfying the above conditions. Let us represent by $G^{-1}$ the graph attained from G by reversing the direction of edges. Thus:

$$E\left(G^{-1}\right)=\left\{\left(\mu ,\varsigma \right)\in \mathfrak{R}\times \mathfrak{R}:\left(\varsigma ,\mu \right)\in E\left(G\right)\right\}.$$

Definition 4.

An element $\mu \in$$\mathfrak{R}$ is called a common fixed-point of the pair ($\mathcal{V},\mathcal{U}$), if $\mathcal{V}\left(\mu \right)=\mathcal{U}\left(\mu \right)=$$\mu .$ We shall represent by CFix($\mathcal{V},\mathcal{U}$) the collection of all common fixed points of the pair ($\mathcal{V},\mathcal{U}$), meaning that:

$$CFix(\mathcal{V},\mathcal{U})=\left\{\mu \in \mathfrak{R}:\mathcal{V}\left(\mu \right)=\mathcal{U}\left(\mu \right)=\mu \right\}.$$

Definition 5.

Suppose that $\mathcal{V},\mathcal{U}$$:\mathfrak{R}\to \mathfrak{R}$ are two mappings on controlled metric space $(\mathfrak{R},\tau ,\sigma )$ endowed with a directed graph G. We say that the pair ($\mathcal{V},\mathcal{U}$) is G-orbital pair, if for any $\mu \in$$\mathfrak{R}$:

$$\left(\mu ,\mathcal{V}\mu \right)\in E\left(G\right)⟹\left(\mathcal{V}\mu ,\mathcal{U}\left(\mathcal{V}\mu \right)\right)\in E\left(G\right)$$

$$\left(\mu ,\mathcal{U}\mu \right)\in E\left(G\right)⟹\left(\mathcal{U}\mu ,\mathcal{V}\left(\mathcal{U}\mu \right)\right)\in E\left(G\right).$$

Let us consider the following sets:

$$\begin{array}{ccc}\hfill {\mathfrak{R}}^{\mathcal{V}}& =& \left\{\mu \in \mathfrak{R}:\left(\mu ,\mathcal{V}\mu \right)\in E\left(G\right)\right\}\hfill \\ \hfill {\mathfrak{R}}^{\mathcal{U}}& =& \left\{\mu \in \mathfrak{R}:\left(\mu ,\mathcal{U}\mu \right)\in E\left(G\right)\right\}.\hfill \end{array}$$

Remark 1.

If the pair ($\mathcal{V},\mathcal{U}$) is a G-orbital-cyclic pair, then ${\mathfrak{R}}^{\mathcal{V}}\ne ⌀⟺{\mathfrak{R}}^{\mathcal{U}}\ne ⌀.$

Proof. 

Let ${\mu }_0\in {\mathfrak{R}}^{\mathcal{V}}.$ Then, $\left({\mu }_0,\mathcal{V}{\mu }_0\right)\in E\left(G\right)⟹\left(\mathcal{V}{\mu }_0,\mathcal{U}\left(\mathcal{V}{\mu }_0\right)\right)\in E\left(G\right).$ If we represent by ${\mu }_1=\mathcal{V}{\mu }_0,$ then we obtain that $\left({\mu }_1,\mathcal{U}\left({\mu }_1\right)\right)\in E\left(G\right),$ thus ${\mathfrak{R}}^{\mathcal{U}}\ne ⌀.$ □

We now state our main result.

Theorem 6.

Let $(\mathfrak{R},\sigma ,\tau )$ be a complete controlled metric space endowed with a directed graph G. Let $\mathcal{V},\mathcal{U}$$:\mathfrak{R}\to \mathfrak{R}$ be a self mapping such that the pair ($\mathcal{V},\mathcal{U}$) forms a G-orbital pair. Suppose that:

(i). ${\mathfrak{R}}^{\mathcal{V}}\ne ⌀,$

(ii) For all $\mu \in {\mathfrak{R}}^{\mathcal{V}}$ and $\varsigma \in {\mathfrak{R}}^{\mathcal{U}}$ and $\alpha ,\beta ,\gamma >0$, with $\alpha +\beta +\gamma <1:$

$$\tau (\mathcal{V}\mu ,\mathcal{U}\varsigma )\le \alpha \tau (\mu ,\varsigma )+\beta \tau (\mu ,\mathcal{V}\mu )+\gamma \tau (\varsigma ,\mathcal{U}\varsigma )$$

(10)

(iii) For any sequence (${\mu }_n$)${}_{n\in \mathbb{N}}\subset \mathfrak{R}$, with (${\mu }_n,{\mu }_{n+1})\in E\left(G\right):$

$$\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\frac{\sigma ({\mu }_{i+1},{\mu }_{i+2})\sigma ({\mu }_{i+1},{\mu }_m)}{\sigma ({\mu }_i,{\mu }_{i+1})}<\frac{1}{\lambda },$$

(11)

where $\lambda =max\left\{\frac{\alpha +\beta }{1-\gamma },\frac{\alpha +\gamma }{1-\beta }\right\};$

(iv) $\mathcal{V}$ and $\mathcal{U}$ are continuous, or for any sequence (${\mu }_n$)${}_{n\in \mathbb{N}}\subset \mathfrak{R}$, with ${\mu }_n\to \mu$ as $n\to \infty ,$ and (${\mu }_n,{\mu }_{n+1})\in E\left(G\right)$ for $n\in \mathbb{N},$ we have $\mu \in {\mathfrak{R}}^{\mathcal{V}}\cap {\mathfrak{R}}^{\mathcal{U}}.$ Under these conditions, CFix$\left(\mathcal{V},\mathcal{U}\right)\ne \varnothing ,$

(v) For every $\mu \in \mathfrak{R}$, we have ${lim}_{n\to \infty }\sigma ({\mu }_n,\mu )$ and ${lim}_{n\to \infty }\sigma (\mu ,{\mu }_n)$ which exist and are finite,

(vi) If (${\mu }^{*},{\mu }^{/})\in$ CFix$\left(\mathcal{V},\mathcal{U}\right)$ implies ${\mu }^{*}\in {\mathfrak{R}}^{\mathcal{V}}$ and ${\mu }^{/}\in {\mathfrak{R}}^{\mathcal{U}},$ then the pair ($\mathcal{V},\mathcal{U}$) has a unique common fixed point.

Proof. 

Let ${\mu }_0\in {\mathfrak{R}}^{\mathcal{V}}$. Thus, (${\mu }_0,\mathcal{V}{\mu }_0)\in E\left(G\right).$ As the pair ($\mathcal{V},\mathcal{U}$) is G-orbital-cyclic, we have ($\mathcal{V}{\mu }_0,\mathcal{UV}{\mu }_0)\in E\left(G\right).$ If we denote by ${\mu }_1=\mathcal{V}{\mu }_0,$ we have (${\mu }_1,\mathcal{U}{\mu }_1)\in E\left(G\right)$ and from here ($\mathcal{U}{\mu }_1,\mathcal{VU}{\mu }_1)\in E\left(G\right).$ Denoting by ${\mu }_2=\mathcal{U}{\mu }_1,$ we have (${\mu }_2,\mathcal{V}{\mu }_2)\in E\left(G\right).$ Continuing in this way, we construct a sequence (${\mu }_n$)${}_{n\in \mathbb{N}}$ with ${\mu }_{2n}=\mathcal{U}{\mu }_{2n-1}$ and ${\mu }_{2n+1}=\mathcal{V}{\mu }_{2n},$ such that $\left({\mu }_{2n},{\mu }_{2n+1}\right)\in E\left(G\right).$ We shall suppose that ${\mu }_n\ne {\mu }_{n+1}.$ If, there exists $n_0\in \mathbb{N},$ such that ${\mu }_{n_0}={\mu }_{n_0+1},$ then because $\Delta \subset E\left(G\right),$$\left({\mu }_{n_0},{\mu }_{n_0+1}\right)\in E\left(G\right)$ and ${\mu }^{*}={\mu }_{n_0}$ is a fixed point of $\mathcal{V}$.

In order to show that ${\mu }^{*}\in$ CFix$\left(\mathcal{V},\mathcal{U}\right),$ we shall consider two cases for $n_0.$

If $n_0$ is even, then $n_0=2n.$ In this way, ${\mu }_{2n}={\mu }_{2n+1}=\mathcal{V}{\mu }_{2n}$ and thus, ${\mu }_{2n}$ is a fixed point of $\mathcal{V}$. Suppose that $\tau (\mathcal{V}{\mu }_{2n},\mathcal{U}{\mu }_{2n+1})>0,$ and let $\mu ={\mu }_{2n}\in {\mathfrak{R}}^{\mathcal{V}}$ and $\varsigma ={\mu }_{2n+1}\in {\mathfrak{R}}^{\mathcal{U}}.$ Thus:

$$\begin{array}{ccc}\hfill 0& <& \tau ({\mu }_{2n+1},{\mu }_{2n+2})=\tau (\mathcal{V}{\mu }_{2n},\mathcal{U}{\mu }_{2n+1})\hfill \\ & \le & \alpha \tau ({\mu }_{2n},{\mu }_{2n+1})+\beta \tau ({\mu }_{2n},\mathcal{V}{\mu }_{2n})+\gamma \tau ({\mu }_{2n+1},\mathcal{U}{\mu }_{2n+1})\hfill \\ & =& \gamma \tau ({\mu }_{2n+1},{\mu }_{2n+2})\hfill \end{array}$$

which is a contradiction. Thus, ${\mu }_{2n}$ is also a fixed point of $\mathcal{U}$.

Similarly, if $n_0$ is odd, then there exists a common fixed point of $\mathcal{V}$ and $\mathcal{U}$. We therefore suppose that ${\mu }_n\ne {\mu }_{n+1}$ for all $n\in \mathbb{N}.$

We shall now prove that (${\mu }_n$)${}_{n\in \mathbb{N}}$ is a Cauchy sequence. In order to do this, we shall consider two possible cases:

Case 1. $\mu ={\mu }_{2n}\in {\mathfrak{R}}^{\mathcal{V}}$ and $\varsigma ={\mu }_{2n+1}\in {\mathfrak{R}}^{\mathcal{U}}:$

$$\begin{array}{ccc}\hfill 0& <& \tau ({\mu }_{2n+1},{\mu }_{2n+2})=\tau (\mathcal{V}{\mu }_{2n},\mathcal{U}{\mu }_{2n+1})\hfill \\ & \le & \alpha \tau ({\mu }_{2n},{\mu }_{2n+1})+\beta \tau ({\mu }_{2n},\mathcal{V}{\mu }_{2n})+\gamma \tau ({\mu }_{2n+1},\mathcal{U}{\mu }_{2n+1})\hfill \\ & =& \alpha \tau ({\mu }_{2n},{\mu }_{2n+1})+\beta \tau ({\mu }_{2n},{\mu }_{2n+1})+\gamma \tau ({\mu }_{2n+1},{\mu }_{2n+2})\hfill \end{array}$$

that is:

$$(1-\gamma )\tau ({\mu }_{2n+1},{\mu }_{2n+2})\le (\alpha +\beta )\tau ({\mu }_{2n},{\mu }_{2n+1})$$

which implies:

$$\tau ({\mu }_{2n+1},{\mu }_{2n+2})\le \frac{\alpha +\beta }{1-\gamma }\tau ({\mu }_{2n},{\mu }_{2n+1}).$$

(12)

Case 2. $\mu ={\mu }_{2n}\in {\mathfrak{R}}^{\mathcal{V}}$ and $\varsigma ={\mu }_{2n-1}\in {\mathfrak{R}}^{\mathcal{U}}:$

$$\begin{array}{ccc}\hfill 0& <& \tau ({\mu }_{2n+1},{\mu }_{2n})=\tau (\mathcal{V}{\mu }_{2n},\mathcal{U}{\mu }_{2n-1})\hfill \\ & \le & \alpha \tau ({\mu }_{2n},{\mu }_{2n-1})+\beta \tau ({\mu }_{2n},\mathcal{V}{\mu }_{2n})+\gamma \tau ({\mu }_{2n-1},\mathcal{U}{\mu }_{2n-1})\hfill \\ & =& \alpha \tau ({\mu }_{2n},{\mu }_{2n-1})+\beta \tau ({\mu }_{2n},{\mu }_{2n+1})+\gamma \tau ({\mu }_{2n-1},{\mu }_{2n})\hfill \end{array}$$

that is:

$$(1-\beta )\tau ({\mu }_{2n+1},{\mu }_{2n})\le (\alpha +\gamma )\tau ({\mu }_{2n},{\mu }_{2n-1})$$

which implies:

$$\tau ({\mu }_{2n},{\mu }_{2n+1})\le \frac{\alpha +\gamma }{1-\beta }\tau ({\mu }_{2n-1},{\mu }_{2n})=\frac{\alpha +\beta }{1-\gamma }\tau ({\mu }_{2n-1},{\mu }_{2n}).$$

(13)

Since $\lambda =max\left\{\frac{\alpha +\beta }{1-\gamma },\frac{\alpha +\gamma }{1-\beta }\right\},$ so we have:

$$\tau ({\mu }_n,{\mu }_{n+1})\le \lambda \tau ({\mu }_{n-1},{\mu }_n).$$

(14)

Thus, we have:

$$\begin{array}{ccc}\hfill \tau ({\mu }_n,{\mu }_{n+1})& \le & \lambda \tau ({\mu }_{n-1},{\mu }_n)\hfill \\ & \le & {\lambda }^2\tau ({\mu }_{n-2},{\mu }_{n-1})\hfill \\ & \le & \\ & & ⋮\hfill \\ & \le & {\lambda }^n\tau ({\mu }_0,{\mu }_1).\hfill \end{array}$$

For all $n,m\in \mathbb{N}(n<m)$, we have:

$$\begin{array}{ccc}\hfill \tau ({\mu }_n,{\mu }_m)& \le & \sigma ({\mu }_n,{\mu }_{n+1})\tau ({\mu }_n,{\mu }_{n+1})+\sigma ({\mu }_{n+1},{\mu }_m)\tau ({\mu }_{n+1},{\mu }_m)\hfill \\ & \le & \sigma ({\mu }_n,{\mu }_{n+1})\tau ({\mu }_n,{\mu }_{n+1})+\sigma ({\mu }_{n+1},{\mu }_m)\sigma ({\mu }_{n+1},{\mu }_{n+2})\tau ({\mu }_{n+1},{\mu }_{n+2})\hfill \\ & & +\sigma ({\mu }_{n+1},{\mu }_m)\sigma ({\mu }_{n+2},{\mu }_m)\tau ({\mu }_{n+2},{\mu }_m)\hfill \\ & \le & \sigma ({\mu }_n,{\mu }_{n+1})\tau ({\mu }_n,{\mu }_{n+1})+\sigma ({\mu }_{n+1},{\mu }_m)\sigma ({\mu }_{n+1},{\mu }_{n+2})\tau ({\mu }_{n+1},{\mu }_{n+2})\hfill \\ & & +\sigma ({\mu }_{n+1},{\mu }_m)\sigma ({\mu }_{n+2},{\mu }_m)\sigma ({\mu }_{n+2},{\mu }_{n+3})\tau ({\mu }_{n+2},{\mu }_{n+3})\hfill \\ & & +\sigma ({\mu }_{n+1},{\mu }_m)\sigma ({\mu }_{n+2},{\mu }_m)\sigma ({\mu }_{n+3},{\mu }_m)\tau ({\mu }_{n+3},{\mu }_m)\hfill \\ & \le & ···\hfill \\ & \le & \sigma ({\mu }_n,{\mu }_{n+1})\tau ({\mu }_n,{\mu }_{n+1})+\sum _{i=n+1}^{m-2}\left(\prod _{j=n+1}^i\sigma ({\mu }_j,{\mu }_m)\right)\sigma ({\mu }_i,{\mu }_{i+1})\tau ({\mu }_i,{\mu }_{i+1})\hfill \\ & & +\prod _{i=n+1}^{m-1}\sigma ({\mu }_i,{\mu }_m)\tau ({\mu }_{m-1},{\mu }_m)\hfill \end{array}$$

which further implies that:

$$\begin{array}{ccc}\hfill \tau ({\mu }_n,{\mu }_m)& \le & \sigma ({\mu }_n,{\mu }_{n+1})\tau ({\mu }_n,{\mu }_{n+1})+\sum _{i=n+1}^{m-2}\left(\prod _{j=n+1}^i\sigma ({\mu }_j,{\mu }_m)\right)\sigma ({\mu }_i,{\mu }_{i+1})\tau ({\mu }_i,{\mu }_{i+1})\hfill \\ & & +\left(\prod _{i=n+1}^{m-1}\sigma ({\mu }_i,{\mu }_m)\right)\sigma ({\mu }_{m-1},{\mu }_m)\tau ({\mu }_{m-1},{\mu }_m)\hfill \\ & \le & \sigma ({\mu }_n,{\mu }_{n+1}){\lambda }^n\tau ({\mu }_0,{\mu }_1)+\sum _{i=n+1}^{m-2}\left(\prod _{j=n+1}^i\sigma ({\mu }_j,{\mu }_m)\right)\sigma ({\mu }_i,{\mu }_{i+1}){\lambda }^i\tau ({\mu }_0,{\mu }_1)\hfill \\ & & +\left(\prod _{i=n+1}^{m-1}\sigma ({\mu }_i,{\mu }_m)\right)\sigma ({\mu }_{m-1},{\mu }_m){\lambda }^{m-1}\tau ({\mu }_0,{\mu }_1)\hfill \\ & =& \sigma ({\mu }_n,{\mu }_{n+1}){\lambda }^n\tau ({\mu }_0,{\mu }_1)+\sum _{i=n+1}^{m-1}\left(\prod _{j=n+1}^i\sigma ({\mu }_j,{\mu }_m)\right)\sigma ({\mu }_i,{\mu }_{i+1}){\lambda }^i\tau ({\mu }_0,{\mu }_1).\hfill \end{array}$$

(15)

Let:

$$S_l=\sum _{i=0}^l\left({\displaystyle \prod _{j=0}^i\sigma ({\mu }_j,{\mu }_m)}\right)\sigma ({\mu }_i,{\mu }_{i+1}){\lambda }^i\tau ({\mu }_0,{\mu }_1).$$

From (15), we obtain:

$$\tau ({\mu }_n,{\mu }_m)\le \tau ({\mu }_0,{\mu }_1)[{\lambda }^n\sigma ({\mu }_n,{\mu }_{n+1})+(S_{m-1}-S_n)].$$

(16)

Now, by using the fact that $\sigma (\mu ,\varsigma )\ge 1$, and by using a ratio test, ${lim}_{n\to \infty }{S}_n$ exists. Thus, $\left\{S_n\right\}$ is Cauchy. Eventually, if in (16), we take $n,m\to \infty$, we obtain that:

$$\underset{n,m\to \infty }{lim}\tau ({\mu }_n,{\mu }_m)=0.$$

(17)

Hence, $\left\{{\mu }_n\right\}$ is a Cauchy in $(\mathfrak{R},\tau )$. So ∃${\mu }^{*}\in X$ such that:

$$\underset{n\to \infty }{lim}\tau ({\mu }_n,{\mu }^{*})=0$$

(18)

that is ${\mu }_n\to {\mu }^{*}$ as $n\to \infty .$ It is obvious that:

$$\underset{n\to \infty }{lim}{\mu }_{2n}=\underset{n\to \infty }{lim}{\mu }_{2n+1}={\mu }^{*}.$$

(19)

As $\mathcal{V}$ and $\mathcal{U}$ are continuous, so we have:

$$\begin{array}{ccc}\hfill {\mu }^{*}& =& \underset{n\to \infty }{lim}{\mu }_{2n+1}=\underset{n\to \infty }{lim}\mathcal{V}\left({\mu }_{2n}\right)=\mathcal{V}\left({\mu }^{*}\right)\hfill \\ \hfill {\mu }^{*}& =& \underset{n\to \infty }{lim}{\mu }_{2n+2}=\underset{n\to \infty }{lim}\mathcal{U}\left({\mu }_{2n+1}\right)=\mathcal{U}\left({\mu }^{*}\right).\hfill \end{array}$$

Now, let $\mu ={\mu }^{*}\in {\mathfrak{R}}^{\mathcal{V}}$ and $\varsigma ={\mu }_{2n+2}\in {\mathfrak{R}}^{\mathcal{U}}$, we have:

$$\begin{array}{ccc}\hfill 0& <& \tau (\mathcal{V}{\mu }^{*},{\mu }_{2n+2})=\tau (\mathcal{V}{\mu }^{*},\mathcal{U}\left({\mu }_{2n+1}\right))\hfill \\ & \le & \alpha \tau ({\mu }^{*},{\mu }_{2n+1})+\beta \tau ({\mu }^{*},\mathcal{V}{\mu }^{*})+\gamma \tau ({\mu }_{2n+1},\mathcal{U}\left({\mu }_{2n+1}\right))\hfill \\ & =& \alpha \tau ({\mu }^{*},{\mu }_{2n+1})+\beta \tau ({\mu }^{*},\mathcal{V}{\mu }^{*})+\gamma \tau ({\mu }_{2n+1},{\mu }_{2n+2}).\hfill \end{array}$$

Taking the limit as $n\to \infty$ and using (19), we can easily deduce that $\tau ({\mu }^{*},\mathcal{V}{\mu }^{*})=0$. This yields that ${\mu }^{*}=\mathcal{V}{\mu }^{*}$. Similarly, let $\mu ={\mu }_{2n+1}\in {\mathfrak{R}}^{\mathcal{V}}$ and $\varsigma ={\mu }^{*}\in {\mathfrak{R}}^{\mathcal{U}}$, we have:

$$\begin{array}{ccc}\hfill 0& <& \tau ({\mu }_{2n+2},\mathcal{U}{\mu }^{*})=\tau (\mathcal{V}\left({\mu }_{2n}\right),\mathcal{U}{\mu }^{*})\hfill \\ & \le & \alpha \tau ({\mu }_{2n},{\mu }^{*})+\beta \tau ({\mu }_{2n},\mathcal{V}\left({\mu }_{2n}\right))+\gamma \tau ({\mu }^{*},\mathcal{U}{\mu }^{*})\hfill \\ & =& \alpha \tau ({\mu }_{2n},{\mu }^{*})+\beta \tau ({\mu }_{2n},{\mu }_{2n+1})+\gamma \tau ({\mu }^{*},\mathcal{U}{\mu }^{*}).\hfill \end{array}$$

Taking the limit as $n\to \infty$ and using (19), we can easily deduce that $\tau ({\mu }^{*},\mathcal{U}{\mu }^{*})=0$. This yields that ${\mu }^{*}=\mathcal{U}{\mu }^{*}$. □

Corollary 7.

Let $(\mathfrak{R},\sigma ,\tau )$ be a complete controlled metric space endowed with a directed graph G. Let $\mathcal{U}$$:\mathfrak{R}\to \mathfrak{R}$ be a self mapping such that:

(i) ${\mathfrak{R}}^{\mathcal{U}}\ne ⌀,$

(ii) for all $\mu \in {\mathfrak{R}}^{\mathcal{U}}$ and $\varsigma \in {\mathfrak{R}}^{\mathcal{U}}$ and $\alpha ,\beta ,\gamma >0$, with $\alpha +\beta +\gamma <1:$

$$\tau (\mathcal{U}\mu ,\mathcal{U}\varsigma )\le \alpha \tau (\mu ,\varsigma )+\beta \tau (\mu ,\mathcal{U}\mu )+\gamma \tau (\varsigma ,\mathcal{U}\varsigma )$$

(iii) for any sequence (${\mu }_n$)${}_{n\in \mathbb{N}}\subset \mathfrak{R}$, with (${\mu }_n,{\mu }_{n+1})\in E\left(G\right):$

$$\underset{m\ge 1}{sup}\underset{i\to \infty }{lim}\frac{\sigma ({\mu }_{i+1},{\mu }_{i+2})\sigma ({\mu }_{i+1},{\mu }_m)}{\sigma ({\mu }_i,{\mu }_{i+1})}<\frac{1}{\lambda },$$

where $\lambda =max\left\{\frac{\alpha +\beta }{1-\gamma },\frac{\alpha +\gamma }{1-\beta }\right\};$

(iv) $\mathcal{U}$ is continuous, or for any sequence (${\mu }_n$)${}_{n\in \mathbb{N}}\subset \mathfrak{R}$, with ${\mu }_n\to \mu$ as $n\to \infty ,$ and (${\mu }_n,{\mu }_{n+1})\in E\left(G\right)$ for $n\in \mathbb{N},$ we have $\mu \in {\mathfrak{R}}^{\mathcal{U}}.$ Under these conditions, Fix$\left(\mathcal{U}\right)\ne \varnothing ,$

(v) For every $\mu \in \mathfrak{R}$, we have ${lim}_{n\to \infty }\sigma ({\mu }_n,\mu )$ and ${lim}_{n\to \infty }\sigma (\mu ,{\mu }_n)$ exist and finite,

(vi) If (${\mu }^{*},{\mu }^{/})\in$ Fix$\left(\mathcal{U}\right)$ implies that ${\mu }^{*}\in {\mathfrak{R}}^{\mathcal{U}}$ and ${\mu }^{/}\in {\mathfrak{R}}^{\mathcal{U}},$ then the $\mathcal{U}$ has a unique fixed point.

Example 4.

Let $\mathfrak{R}=\left\{0,1,2,3,4\right\}.$ We define the controlled metric:

$$\tau (\mu ,\varsigma )={\left|\mu -\varsigma \right|}^2$$

for all $\mu ,\varsigma \in \mathfrak{R},$ where $\sigma (\mu ,\varsigma )=1+\mu \varsigma .$ Then, $(\mathfrak{R},\sigma ,\tau )$ is a complete controlled metric space. We now define $\mathcal{U}$$:\mathfrak{R}\to \mathfrak{R}$ by

$$\mathcal{U}\mu =0,\mathrm{for}\mu \in \{0,1\}$$

and:

$$\mathcal{U}\mu =1,\mathrm{for}\mu \in \{2,3\}.$$

We now define the directed graph as G=$\left\{(0,1),(0,2),(2,3),(0,0),(1,1),(2,2),(3,3)\right\}.$ Then, with $\alpha =\frac{1}{3},\beta =0,\gamma =\frac{1}{3},$ all assumptions of result 7 are satisfied and $\mu =0$ is the unique fixed point.

4. Application

Theorem 7.

Let $\mathfrak{R}=C\left(\left[0,1\right]\right).$ Now, we define $\tau :\mathfrak{R}\times \mathfrak{R}\to [$0,$\infty ):$

$$\tau \left(\mu ,\varsigma \right)=\underset{t\in \left[0,1\right]}{max}{\left(\mu \left(t\right)-\varsigma \left(t\right)\right)}^2.$$

Then, $(\mathfrak{R},\sigma ,\tau )$ is a complete controlled metric space with $\sigma \left(\mu ,\varsigma \right)=\sigma \left(\varsigma ,\omega \right)=2.$

Consider the integral equations:

$$\mu \left(t\right)={\int }_0^1{K}_1\left(t,s,\mu \left(s\right)\right)ds+g\left(t\right)$$

(20)

$$\mu \left(t\right)={\int }_0^1{K}_2\left(t,s,\mu \left(s\right)\right)ds+h\left(t\right)$$

(21)

for $t\in \left[0,1\right],$ where $\mu ,g,h\in \mathfrak{R}.$

Suppose that $K_1,K_2:\left[0,1\right]\times \left[0,1\right]$→$\mathbb{R}$ are such that ${\vartheta }_{\mu }\left(t\right),{\theta }_{\mu }\left(t\right)\in \mathfrak{R}$ for each $\mu \in \mathfrak{R},$ where:

$${\vartheta }_{\mu }\left(t\right)={\int }_0^1{K}_1\left(t,s,\mu \left(s\right)\right)ds$$

and

$${\theta }_{\mu }\left(t\right)={\int }_0^1{K}_2\left(t,s,\mu \left(s\right)\right)ds$$

for all $t\in [0,1].$ If ∃$\alpha$, $\beta :\mathfrak{R}\to [0,1)$ such that these assertions hold:

(i) $\alpha ({\vartheta }_{\mu }+g\left(t\right))\le \alpha \left(\mu \right)$ and $\beta ({\vartheta }_{\mu }+g\left(t\right))\le \beta \left(\mu \right);$

(ii) $\alpha ({\theta }_{\mu }+g\left(t\right))\le \alpha \left(\mu \right)$ and $\beta ({\theta }_{\mu }+g\left(t\right))\le \beta \left(\mu \right);$

(iii) ($\alpha +\beta \left)\right(\mu )<1;$

(iv)

$${∥{\vartheta }_{\mu }\left(t\right)-{\theta }_{\varsigma }\left(t\right)+g\left(t\right)-h\left(t\right)∥}^2\le \alpha \left(\mu \right){\aleph }_1(\mu ,\varsigma )\left(t\right)+\beta \left(\mu \right){\aleph }_2(\mu ,\varsigma )\left(t\right)$$

$\forall \mu ,\varsigma \in \mathfrak{R},$ where:

$${\aleph }_1(\mu ,\varsigma )\left(t\right)={∥\mu \left(t\right)-\varsigma \left(t\right)∥}^2,$$

$${\aleph }_2(\mu ,\varsigma )\left(t\right)=\frac{{∥{\vartheta }_{\mu }\left(t\right)+\theta \left(t\right)-\mu \left(t\right)∥}^2{∥{\theta }_{\varsigma }\left(t\right)+h\left(t\right)-\varsigma \left(t\right)∥}^2}{1+{∥\mu \left(t\right)-\varsigma \left(t\right)∥}^2}$$

then, the system of integral equations (20) and (21) has a unique common solution.

Proof. 

Define $\mathcal{V}$,$\mathcal{U}:\mathfrak{R}\to \mathfrak{R}$ by

$$\mathcal{V}\mu ={\vartheta }_{\mu }+g$$

and:

$$\mathcal{U}\mu ={\theta }_{\mu }+h.$$

Then:

$$\tau \left(\mathcal{V}\mu ,\mathcal{U}\varsigma \right)=\underset{t\in \left[0,1\right]}{max}\left({∥{\vartheta }_{\mu }\left(t\right)-{\theta }_{\varsigma }\left(t\right)+g\left(t\right)-h\left(t\right)∥}^2\right)$$

$$\tau \left(\mu ,\mathcal{V}\mu \right)=\underset{t\in \left[0,1\right]}{max}\left({∥{\vartheta }_{\mu }\left(t\right)+g\left(t\right)-\mu \left(t\right)∥}^2\right)$$

$$\tau \left(\varsigma ,\mathcal{U}\varsigma \right)=\underset{t\in \left[0,1\right]}{max}\left({∥{\theta }_{\varsigma }\left(t\right)+h\left(t\right)-\varsigma \left(t\right)∥}^2\right).$$

It is simple to show that:

(i) $\alpha \left(\mathcal{V}\mu \right)\le \alpha \left(\mu \right)$ and $\beta \left(\mathcal{V}\mu \right)\le \beta \left(\mu \right);$

(ii) $\alpha \left(\mathcal{U}\mu \right)\le \alpha \left(\mu \right)$ and $\beta \left(\mathcal{U}\mu \right)\le \beta \left(\mu \right);$

(iii) ($\alpha +\beta \left)\right(\mu )<1;$

(iv)

$$\tau \left(\mathcal{V}\mu ,\mathcal{U}\varsigma \right)=\alpha \left(\mu \right)\tau \left(\mu ,\varsigma \right)+\beta \left(\mu \right)\frac{\tau \left(\mu ,\mathcal{V}\mu \right)\tau \left(\varsigma ,\mathcal{U}\varsigma \right)}{1+\tau \left(\mu ,\varsigma \right)}$$

$\forall \mu ,\varsigma \in \mathfrak{R}.$ Thus, by Theorem 4, we obtain that $\mathcal{V}$ and $\mathcal{U}$ have a common fixed point. Thus, there exists a unique point $\mu \in \mathfrak{R}$ such that $\mu =\mathcal{V}\mu =\mathcal{U}\mu$. Now, we have:

$$\mu =\mathcal{V}\mu ={\vartheta }_{\mu }+g$$

and:

$$\mu =\mathcal{U}\mu ={\theta }_{\mu }+h$$

that is:

$$\mu \left(t\right)={\int }_0^1{K}_1\left(t,s,\mu \left(s\right)\right)ds+g\left(t\right)$$

$$\mu \left(t\right)={\int }_0^1{K}_2\left(t,s,\mu \left(s\right)\right)ds+h\left(t\right).$$

Therefore, we can conclude that the integral equations (20) and (21) have a unique common solution. □

5. Conclusions

We summarize our conclusions as follows:

(1) To generalize the main result of Mlaiki et al. [3], we defined a new contractive condition in a controlled metric space by employing two control functions $\alpha ,\beta :\mathfrak{R}\to [0,1)$ to the right-hand side of the inequality. Moreover, we have used a certain rational expression in the contractive condition;

(2) We have taken two self mappings instead of one self mapping in the contractive condition of our main results;

(3) Some examples are given to illustrate the validity of our main results;

(4) We obtained some common fixed-point results in controlled metric spaces endowed with a graph;

(5) We also investigated integral equations as an application of our main results.

6. Future Work

In this area, our future work will focus on studying the fixed points of multi-valued and fuzzy mappings in controlled metric spaces, with fractional differential inclusion problems as applications.