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\keywords{Controlled metric space; Fixed point; Fredholm integral equation}
\subjclass[2000]{46T99; 47H10; 54H25}

\begin{document}
\title[ ]{\bfseries A
 Fredholm integral equation via a fixed point result in controlled metric spaces}


\author[]
{Nayab Alamgir, Quanita Kiran, Hassen Aydi, Vahid Parvaneh}



\address{Nayab Alamgir,  \newline
 School of Natural Sciences, \texttt{National University of  Sciences and Technology (NUST)}, Sector H-$12$, Islamabad-Pakistan.}
\email{\href{mailto:nayab@sns.nust.edu.pk}{nayab@sns.nust.edu.pk}}


\address{Quanita Kiran,  \newline
 School of Electrical Engineering and Computer Science (SEECS), National University of Sciences and Technology (NUST), Sector H-$12$, Islamabad-Pakistan.}
\email{\href{mailto:quanita.kiran@seecs.edu.pk}{quanita.kiran@seecs.edu.pk}}

\address{Hassen Aydi,  \newline
Institut Sup\'erieur d'Informatique et des Techniques de Communication, Universit\'e de Sousse,
 Hammam~Sousse~4000, Tunisia.}
\email{\href{mailto:hassen.aydi@isima.rnu.tn}{hassen.aydi@isima.rnu.tn}}

\address{Vahid Parvaneh,  \newline
 Department of Mathematics, Gilan-E-Gharb  Branch, Islamic Azad University,\\
Gilan-E-Gharb, Iran.}
\email{\href{mailto:zam.dalahoo@gmail.com}{zam.dalahoo@gmail.com}}



\begin{abstract}
In this article, we establish various fixed point results in the structure of controlled metric spaces. By the help of that results, we give efficient solution of a Fredholm type integral equation. Moreover, some examples are given to show the conceptual depth within this approach.
\end{abstract}

\maketitle

\section{Introduction}
\noindent More than a century ago in $1903$, Fredholm
\cite{FE} did wonders in the theory of non linear integral equations
by presenting an integration expression with fixed limits:

 \begin{equation}\label{equnntt1} \textsl{k}(\omega) = \Upsilon(\omega) + \lambda \int_\textsl{p}^\textsl{q} \! \Phi(\omega, \textsl{t})\textsl{k}(\textsl{t}) \, \mathrm{d}\textsl{t}.\end{equation}

 \noindent In this expression, $\textsl{p}$ and $\textsl{q}$ are constants and $ \lambda$ is a parameter. Given  $\Upsilon(\omega)$ and the kernel function $\Phi(\omega, \textsl{t})$, the problem is typically to find the function $\textsl{k}(\omega)$. Integral equations of Fredholm have been widely used in different fields of mathematics, e.g, physical mathematics, computational mathematics and approximation theory. Motivated by\textcolor{white}{o}his\textcolor{white}{o}great work, many\textcolor{white}{o}researchers have\textcolor{white}{o}focused their\textcolor{white}{o}work on\textcolor{white}{o}solving the\textcolor{white}{o}Fredholm integral equation by applying the technique of fixed point theory.\textcolor{white}{o}Rus in \cite{RMD} used\textcolor{white}{o}Krasnoselskii's fixed point\textcolor{white}{o}theorem on\textcolor{white}{o}cones in order to\textcolor{white}{o}obtain positive\textcolor{white}{o}continuous solutions of the following\textcolor{white}{o}Fredholm\textcolor{white}{o}integral\textcolor{white}{o}equation

\begin{equation}\label{equnntt2} \textsl{k}(\omega) = \Upsilon(\omega) + \int_0^\textsl{T} \! \Phi(\omega, \textsl{t})\textsl{f}(\textsl{k}(\textsl{t})) \, \mathrm{d}\textsl{t},\qquad \omega \in [0, \textsl{T}],\qquad \textsl{T} > 0.\end{equation}

 \noindent using\textcolor{white}{o}limit type\textcolor{white}{o}conditions for $\textsl{f}$ in $(0,\infty).$ In \cite{BMG}, \textcolor{white}{o}Berenguer et al. established a method of\textcolor{white}{o } a numerical\textcolor{white}{o}approximation of the fixed\textcolor{white}{o}point of an\textcolor{white}{o}operator, especially the\textcolor{white}{o}integral one\textcolor{white}{o}associated with a non\textcolor{white}{o}linear\textcolor{white}{o}Fredholm integral\textcolor{white}{o}equation, that uses the\textcolor{white}{o}properties of a\textcolor{white}{o}classical\textcolor{white}{o}Schauder basis in the\textcolor{white}{o}Banach space\textcolor{white}{o}$\mathbb{C}([a, b] \times [a, b]).$ In the same\textcolor{white}{o}direction, \textcolor{white}{o}Pathak et al. in \cite{PKT}\textcolor{white}{o}proved a fixed\textcolor{white}{o}point result for a\textcolor{white}{o}pair of\textcolor{white}{o}weakly\textcolor{white}{o}compatible mappings, and by the use of that result they proved the\textcolor{white}{o}existence of a \textcolor{white}{o}solution of\textcolor{white}{o} a system of non\textcolor{white}{o}linear\textcolor{white}{o}integral\textcolor{white}{o}equations. For more related works, you can see \cite{Kar, RSH}. Recently, Karapinar et al. in \cite{KKL} proved fixed point results over extended $b$-metric spaces and applied that results to find a solution of the equation \eqref{equnntt1}. The in-depth analysis of a non linear expression has resulted one of the best principle, called the Banach contraction principle (BCP). Researchers have been greatly benefited from generalizations of this principle. BCP has been widely utilized by many researchers in their vital researches. In nut shell, many portion of fixed point theory contains BCP generalization. Below, we recall some basic generalizations of it.

 \begin{definition}[\cite{KSV}]\label{def11}
 A mapping $\textsl{T} : \mathbf{\textsl{S}}_{1} \cup \mathbf{\textsl{S}}_{2} \rightarrow \mathbf{\textsl{S}}_{1} \cup \mathbf{\textsl{S}}_{2}$ is called cyclic, if $\textsl{T}(\mathbf{\textsl{S}}_{1}) \subset \mathbf{\textsl{S}}_{2}$ and $\textsl{T}(\mathbf{\textsl{S}}_{2}) \subset \mathbf{\textsl{S}}_{1}$, where $\mathbf{\textsl{S}}_{1}$, $\mathbf{\textsl{S}}_{2}$ are nonempty subsets of a metric space $(\textsl{W}, \textsl{d}).$
 \end{definition}

 \begin{definition}[\cite{KSV}]\label{101}
 A cyclic map $\textsl{T} : \mathbf{\textsl{S}}_{1} \cup \mathbf{\textsl{S}}_{2} \rightarrow \mathbf{\textsl{S}}_{1} \cup \mathbf{\textsl{S}}_{2}$ is called cyclic contraction, if for all $\omega_{1} \in \mathbf{\textsl{S}}_{1}$, $\omega_{2} \in \mathbf{\textsl{S}}_{2}$, there exists $\kappa \in (0, 1)$ such that
 $$\textsl{d}(\textsl{T}\omega_{1}, \textsl{T}\omega_{2}) \leq \kappa \textsl{d}(\omega_{1}, \omega_{2}).$$
 \end{definition}

 \begin{definition}[\cite{KAS}]\label{def12}
 A cyclic map $\textsl{T} : \mathbf{\textsl{S}}_{1} \cup \mathbf{\textsl{S}}_{2} \rightarrow \mathbf{\textsl{S}}_{1} \cup \mathbf{\textsl{S}}_{2}$ is called a cyclic orbital contraction, if for all $\omega_{1} \in \mathbf{\textsl{S}}_{1}$, there exists $\kappa \in (0, 1)$ such that

\begin{equation}\label{equ11} \textsl{d}(\textsl{T}^{2n}\omega_{1}, \textsl{T}\omega_{2}) \leq \kappa \textsl{d}(\textsl{T}^{2n - 1}\omega_{1}, \omega_{2}),\end{equation}

where $\omega_{2} \in \mathbf{\textsl{S}}_{1}$, $n \in \mathbb{N}$ and $\mathbf{\textsl{S}}_{1}$, $\mathbf{\textsl{S}}_{2}$ are closed subsets of $\textsl{W}.$
 \end{definition}

 \begin{definition}[\cite{DW}]\label{def13}
 A mapping $\textsl{T} : \textsl{W} \rightarrow \textsl{W}$ on a metric space $(\textsl{W}, \textsl{d})$ is called $\mathbf{\textsl{F}}$-contraction, if there exists $\Omega > 0$ such that

\begin{equation}\label{eque2} \Omega + \mathbf{\textsl{F}}(\textsl{d}(\textsl{T}\omega_{1}, \textsl{T}\omega_{2})) \leq \mathbf{\textsl{F}}(\textsl{d}(\omega_{1}, \omega_{2})), \end{equation}

\noindent for all $\omega_{1}, \omega_{2} \in \textsl{W}$ with
$d(\omega_{1}, \omega_{2}) > 0,$ where $\mathbf{\textsl{F}} : [0,
\infty) \rightarrow \mathbb{R}$ is function satisfying the following
axioms:
\begin{itemize}
\item [\rm {(I)}] $\mathbf{\textsl{F}}$ is strictly increasing;

\item[\rm {(II)}] for each sequence $\{a_{n}\} \subset (0, \infty)$ of positive real numbers, $\lim_{n \rightarrow \infty} a_{n} = 0$ if and only if $\lim_{n \rightarrow \infty} \mathbf{\textsl{F}}(a_{n}) = -\infty;$

\item[\rm{(III)}] for each sequence $\{a_{n}\} \subset (0, \infty)$ of positive real numbers, $\lim_{n \rightarrow \infty} a_{n} = 0$, there exists $\kappa \in (0, 1)$ such that $\lim_{n \rightarrow \infty} (a_{n})^{\kappa} \mathbf{\textsl{F}}(a_{n}) = 0;$
\end{itemize}

\noindent We denote by $\mathcal{F}$ the family of all functions
$\mathbf{\textsl{F}}$ satisfying $(I)-(III)$.
\end{definition}

\noindent Recently in \cite{MN}, Mlaiki et al. introduced controlled type metric space as a generalization of a $b$-metric space \cite{SC1, BI}, which
is different from extended $b$-metric spaces \cite{TK1}.

\begin{definition}[\cite{MN}]\label{def15}
Let $\textsl{W}$ be a nonempty set and $\alpha : \textsl{W} \times
\textsl{W} \rightarrow [1, \infty).$ Then a mapping
$\textsl{d}_{\alpha} : \textsl{W} \times \textsl{W} \rightarrow [0,
\infty)$ is called a controlled metric, if for all $\omega_{1},
\omega_{2}, \omega_{3} \in \textsl{W}$, it satisfies the following
axioms:

\begin{itemize}
\item[\rm (i)] $\textsl{d}_{\alpha}(\omega_{1}, \omega_{2}) = 0$~~iff~~$\omega_{1} =
\omega_{2},$
\item[\rm (ii)] $\textsl{d}_{\alpha}(\omega_{1}, \omega_{2}) = \textsl{d}_{\alpha}(\omega_{2}, \omega_{1}),$
\item[\rm (iii)] $\textsl{d}_{\alpha}(\omega_{1}, \omega_{3}) \leq \alpha(\omega_{1}, \omega_{2})\textsl{d}_{\alpha}(\omega_{1}, \omega_{2}) + \alpha(\omega_{2}, \omega_{3}) \textsl{d}_{\alpha}(\omega_{2}, \omega_{3})$.
\end{itemize}

\noindent The pair $(\textsl{W}, \textsl{d}_{\alpha})$ is called a
controlled  metric space.
\end{definition}

\textbf{Remark 1.1} Every $b$-metric space is a controlled metric
space, if we take $\alpha(\omega_{1}, \omega_{2}) = s \geq 1$ for
all $\omega_{1}, \omega_{2} \in \textsl{W}.$ Generally, a
 controlled metric space is
not an extended $b$-metric space, if we take same functions $\alpha
= \mathcal{\theta}$ as follows:

\begin{example}[\cite{MN}]

Let $\textsl{W} = \{1, 2, \ldots\ldots\}.$ Define $\textsl{d}_{\alpha} : \textsl{W} \times \textsl{W} \rightarrow [0, \infty)$ as:

$$\textsl{d}_{\alpha}(\omega_{1}, \omega_{2}) = \left\{
             \begin{array}{ll}
               0, & \hbox{if $\omega_{1} = \omega_{2}$;} \\

               \frac{1}{\omega_{1}}, & \hbox{if $\omega_{1}$ is even and $\omega_{2}$ is odd;}\\

               \frac{1}{\omega_{2}}, & \hbox{if $\omega_{1}$ is odd and $\omega_{2}$ is even;}\\

               1,           &\hbox{otherwise.}
              \end{array}
            \right.$$
Hence $(\textsl{W}, \textsl{d}_{\alpha})$ is a controlled metric
space, where $\alpha : \textsl{W} \times \textsl{W} \rightarrow [1,
\infty)$ is defined as:

$$ \alpha(\omega_{1}, \omega_{2}) = \left\{
             \begin{array}{ll}
               \omega_{1}, & \hbox{if $\omega_{1}$ is even and $\omega_{2}$ is odd;}\\
               \omega_{2}, & \hbox{if $\omega_{1}$ is odd and $\omega_{2}$ is even;}\\
               1,  &\hbox{otherwise.}
              \end{array}
            \right.$$
Clearly, $\textsl{d}_{\alpha}$ is not an extended $b$-metric for the
same function $\alpha = \mathcal{\theta}$.
\end{example}

\noindent In this paper, we introduce and establish new sort of
contractions,  called controlled cyclic orbital contractions and
controlled cyclic orbital $F$-contractions in the setting of
controlled metric spaces. By the help of that results, we are in the position to consider
real-life applications in a very general structure such as simple
and efficient solution of a Fredholm integral equation in the setting
of a controlled metric space. Moreover, some examples are given to
show the conceptual depth within this approach.



\section{Main results}

\noindent In this section, we introduce the notion of controlled
cyclic orbital contractions.

\begin{definition}\label{def11}
Let $\mathbf{\textsl{S}}_{1}, \mathbf{\textsl{S}}_{2}$ be nonempty
subsets of a controlled metric space $(\textsl{W},
\textsl{d}_{\alpha})$. Then a cyclic map $\textsl{T} :
\mathbf{\textsl{S}}_{1} \cup \mathbf{\textsl{S}}_{2} \rightarrow
\mathbf{\textsl{S}}_{1} \cup \mathbf{\textsl{S}}_{2}$ is called a
controlled cyclic orbital contraction, if for some $\omega_{1} \in
\mathbf{\textsl{S}}_{1}$, there exists $\kappa \in (0, 1)$ such that

\begin{equation}\label{equ11} \textsl{d}_{\alpha}(\textsl{T}^{2n}\omega_{1}, \textsl{T}\omega_{2}) \leq \kappa \textsl{d}_{\alpha}(\textsl{T}^{2n - 1}\omega_{1}, \omega_{2}),\end{equation}

where $\omega_{2} \in \mathbf{\textsl{S}}_{1}$ and $n \in \mathbb{N}.$
\end{definition}

\begin{theorem}\label{thm11}
Let $\mathbf{\textsl{S}}_{1}$ and $\mathbf{\textsl{S}}_{2}$ be
nonempty closed subsets of a  complete controlled metric space
$(\textsl{W}, \textsl{d}_{\alpha})$ such that $\textsl{d}_{\alpha}$
is a continuous functional. Let $\textsl{T} :
\mathbf{\textsl{S}}_{1} \cup \mathbf{\textsl{S}}_{2} \rightarrow
\mathbf{\textsl{S}}_{1} \cup \mathbf{\textsl{S}}_{2}$ be a
controlled cyclic orbital contraction. Also suppose that for any
$\omega_{0} \in \textsl{W}$,
\begin{equation}\label{c1}
  \sup_{m\geq1}\lim_{i\rightarrow\infty}\frac{\alpha(\omega_{i+1},\omega_{i+2})}{\alpha(\omega_{i},\omega_{i+1})}\alpha(\omega_{i+1},\omega_m)<\frac{1}{k},
 \end{equation}
%$\lim_{n, m \rightarrow \infty} \alpha(\omega_{n}, \omega_{m})\kappa < 1,$
where $\omega_{n} = \textsl{T}^{n} \omega_{0}$, $n = 1, 2, \ldots.$
Then $\mathbf{\textsl{S}}_{1} \cap \mathbf{\textsl{S}}_{2}$ is
nonempty and $\textsl{T}$ has a unique fixed point.
\end{theorem}

\begin{proof}
Suppose there exists  $\omega_{0} \in \mathbf{\textsl{S}}_{1}$
satisfying equation \eqref{equ11}. Define an iterative sequence
$\{\omega_{n}\}$ starting from $\omega_{0}$ as follows:

$$\omega_{1} = \textsl{T}\omega_{0},~\omega_{2} = \textsl{T}\omega_{1} = \textsl{T}(\textsl{T}\omega_{0}) = \textsl{T}^{2}\omega_{0},~\ldots, \omega_{n} = \textsl{T}^{n}\omega_{0}, \ldots$$

\noindent From equation \eqref{equ11}, we have

$$\textsl{d}_{\alpha}(\textsl{T}^{2}\omega_{0}, \textsl{T}\omega_{0}) \leq \kappa \textsl{d}_{\alpha}(\textsl{T}\omega_{0}, \omega_{0}).$$

Recursively, we have

\begin{align*} \textsl{d}_{\alpha}(\textsl{T}^{3}\omega_{0}, \textsl{T}^{2}\omega_{0}) & \leq \kappa \textsl{d}_{\alpha}(\textsl{T}^{2}\omega_{0}, \textsl{T}\omega_{0})\\ & \leq \kappa^{2} \textsl{d}_{\alpha}(\textsl{T}\omega_{0}, \omega_{0}).\end{align*}

\noindent As for any $n \in \mathbb{N}$, either $n$ or $n + 1$ is
even, so we have

\begin{align*} \textsl{d}_{\alpha}(\textsl{T}^{n + 1}\omega_{0}, \textsl{T}^{n}\omega_{0}) \leq \kappa^{n} \textsl{d}_{\alpha}(\textsl{T}\omega_{0}, \omega_{0}),\end{align*}

that is,

\begin{equation}\label{equ12} \textsl{d}_{\alpha}(\omega_{n + 1}, \omega_{n}) \leq \kappa^{n} \textsl{d}_{\alpha}(\omega_{1}, \omega_{0}).\end{equation}

\noindent From the triangle inequality and equation \eqref {equ12}
for $m > n,$ we have

\begin{align*} \textsl{d}_{\alpha}(\omega_{n}, \omega_{m}) &\leq \alpha(\omega_{n}, \omega_{n + 1})\textsl{d}_{\alpha}(\omega_{n}, \omega_{n + 1}) + \alpha(\omega_{n + 1}, \omega_{m})\textsl{d}_{\alpha}(\omega_{n + 1}, \omega_{m})\\ & \leq \alpha(\omega_{n}, \omega_{n + 1})\textsl{d}_{\alpha}(\omega_{n}, \omega_{n + 1}) + \alpha(\omega_{n}, \omega_{m})\alpha(\omega_{n + 1}, \omega_{n + 2})\textsl{d}_{\alpha}(\omega_{n + 1}, \omega_{n + 2})\\& + \alpha(\omega_{n}, \omega_{m})\alpha(\omega_{n + 2}, \omega_{m})\textsl{d}_{\alpha}(\omega_{n + 2}, \omega_{m})\\ &~~\vdots \\ &\leq \alpha(\omega_{n}, \omega_{n + 1})\textsl{d}_{\alpha}(\omega_{n}, \omega_{n + 1}) + \sum_{i = n + 1}^{m - 2}\left(\prod_{j = 1}^{i}\alpha(\omega_{j}, \omega_{m})\right) \alpha(\omega_{i}, \omega_{i + 1})\textsl{d}_{\alpha}(\omega_{i}, \omega_{i + 1})\\& + \prod_{j = n + 1}^{m - 1} \alpha(\omega_{j}, \omega_{m}) \alpha(\omega_{m - 1}, \omega_{m}) \textsl{d}_{\alpha}(\omega_{m - 1}, \omega_{m})\\ & \leq \alpha(\omega_{n}, \omega_{n + 1})\textsl{d}_{\alpha}(\omega_{n}, \omega_{n + 1}) + \sum_{i = n + 1}^{m - 1}\left(\prod_{j = 1}^{i}\alpha(\omega_{j}, \omega_{m})\right) \alpha(\omega_{i}, \omega_{i + 1})\textsl{d}_{\alpha}(\omega_{i}, \omega_{i + 1})\\ & \leq [\alpha(\omega_{n}, \omega_{n + 1})\kappa^{n} + \sum_{i = n + 1}^{m - 1}\left(\prod_{j = 1}^{i}\alpha(\omega_{j}, \omega_{m})\right) \alpha(\omega_{i}, \omega_{i + 1})\kappa^{i}]\textsl{d}_{\alpha}(\omega_{1}, \omega_{0}).\end{align*}

\noindent Using (\ref{c1}) % $\lim_{n, m \rightarrow \infty} \alpha(\omega_{n}, \omega_{m})\kappa < 1$,
 for any $\omega_{0} \in \mathbf{\textsl{S}}_{1},$ we get the series $\sum_{i = 1}^{\infty}\left(\prod_{j = 1}^{i}\alpha(\omega_{j}, \omega_{m})\right) \alpha(\omega_{i}, \omega_{i + 1})\kappa^{i}$ converges by the ratio test for each $m \in \mathbb{N}$. Let $S = \sum_{i = 1}^{\infty}\left(\prod_{j = 1}^{i}\alpha(\omega_{j}, \omega_{m})\right) \alpha(\omega_{i}, \omega_{i + 1})\kappa^{i}$ and
 $$S_{n} = \sum_{p = 1}^{n}\left(\prod_{q = 1}^{p}\alpha(\omega_{q}, \omega_{m})\right) \alpha(\omega_{p}, \omega_{p + 1})\kappa^{p}.$$ Thus, for $m > n$, we have
%Note that
\begin{equation}
\label{c2}
\textsl{d}_{\alpha}(\omega_{n}, \omega_{m}) \leq [\alpha(\omega_{n}, \omega_{n + 1})\kappa^{n} + (S_{m - 1} - S_{n})]\textsl{d}_{\alpha}(\omega_{1}, \omega_{0}).
\end{equation}
Condition (\ref{c1}) implies that $\lim_{n\rightarrow\infty}S_n$ exists and hence the real sequence $\{S_n\}$ is Cauchy.
Finally, if we take limit in  the inequality (\ref{c2}) as $n, m\rightarrow \infty,$ we deduce that
\begin{equation}\label{h0}
\lim_{n,m\rightarrow\infty}\omega(\omega_{n},\omega_{m}) =0 ,
\end{equation}
%\noindent By letting $n \rightarrow \infty$, we conclude that
that is, $\{\omega_{n}\}$ is a Cauchy sequence. Hence as a result,
there exists $\rho \in \mathbf{\textsl{S}}_{1} \cup
\mathbf{\textsl{S}}_{2}$ such that $\omega_{n} \rightarrow \rho$.
Now, note that $\{\omega_{2n}\} = \{T^{2n}\omega_{0}\}$ is a
sequence in $\mathbf{\textsl{S}}_{1}$ and $\{\omega_{2n - 1}\} =
\{T^{2n - 1}\omega_{0}\}$ is a sequence in $\mathbf{\textsl{S}}_{2}$
and both converge to $\rho$. As the sets $\mathbf{\textsl{S}}_{1}$
and $\mathbf{\textsl{S}}_{2}$ are closed in $\textsl{W}$ and $\rho
\in \mathbf{\textsl{S}}_{1} \cap \mathbf{\textsl{S}}_{2}$, hence
$\mathbf{\textsl{S}}_{1} \cap \mathbf{\textsl{S}}_{2}$ is nonempty.
Next, we prove that $\rho$ is a fixed point of $\textsl{T}.$ The
continuity of $\textsl{d}_{\alpha}$ yields that

\begin{align*} \textsl{d}_{\alpha}(\rho, \textsl{T}\rho) &= \lim_{n \rightarrow \infty} \textsl{d}_{\alpha}(\textsl{T}^{2n}\omega, \textsl{T}\rho)\\& \leq \kappa \lim_{n \rightarrow \infty} \textsl{d}_{\alpha}(\textsl{T}^{2n - 1}\omega, \rho)\\& = 0.\end{align*}

\noindent Thus $\rho$ is a fixed point of $\textsl{T}.$ For the
uniqueness, assume there exists $\varrho \in \mathbf{\textsl{S}}_{1}
\cap \mathbf{\textsl{S}}_{2}$, $\rho \neq \varrho$ such that
$\textsl{T}\varrho = \varrho.$ Now,

\begin{align*} \textsl{d}_{\alpha}(\rho, \varrho) &= \textsl{d}_{\alpha}(\textsl{T}\rho, \textsl{T}\varrho)\\& \leq\kappa \textsl{d}_{\alpha}(\rho, \varrho) \\& < \textsl{d}_{\alpha}(\rho, \varrho).\end{align*}

\noindent Which is a contradiction. Hence, $\rho = \varrho$ and $\rho$
is the unique fixed point of $\textsl{T}.$
\end{proof}
\begin{remark}
\label{R2.1}
%\textbf{Remark 2.1}
\begin{itemize}
\item[\rm (1)] For $\alpha(\omega_{1}, \omega_{2}) = s \geq 1$, the above theorem reduces to the $b$-metric space.

\item[\rm (2)] For $\alpha(\omega_{1}, \omega_{2}) = 1$, the above theorem reduces to the main result of Karpagam et al. \cite{KAS}.
\end{itemize}
\end{remark}
\begin{example}
Let $\textsl{W} = \mathbb{R}$. Define $\textsl{d}_{\alpha} :
\textsl{W} \times \textsl{W} \rightarrow [0, \infty)$ by
$\textsl{d}_{\alpha}(\omega_{1}, \omega_{2}) = (\omega_{1} -
\omega_{2})^{2}.$ Then, clearly $(\textsl{W}, \textsl{d}_{\alpha})$
is a complete controlled metric space with $\alpha: \textsl{W}
\times \textsl{W} \rightarrow [1, \infty)$ defined as
$\alpha(\omega_{1}, \omega_{2}) = 3\omega_{1} + 2\omega_{2} + 5$.
Let $\mathbf{\textsl{S}}_{1} = [0, \frac{1}{3}]$ and
$\mathbf{\textsl{S}}_{2} = [\frac{1}{4}, 1].$ Define a mapping
$\textsl{T} : \mathbf{\textsl{S}}_{1} \cup \mathbf{\textsl{S}}_{2}
\rightarrow \mathbf{\textsl{S}}_{1} \cup \mathbf{\textsl{S}}_{2}$ as

$$\textsl{T}\omega_{1} = \left\{
             \begin{array}{ll}
              \frac{1}{3}, & \hbox{$0 \leq \omega_{1} \leq \frac{1}{4}$;}\\

               \frac{1}{2}(1 - \omega_{1}), & \hbox{$\frac{1}{4} < \omega_{1} \leq 1$.}
              \end{array}
            \right.$$

\noindent First, we have to show that $\textsl{T}$ is a cyclic map.
\begin{itemize}
\item[\rm (i)] If $\omega_{1} = 0 \in \mathbf{\textsl{S}}_{1}$, then $\textsl{T}0 = \frac{1}{3} \in \mathbf{\textsl{S}}_{2}.$

\item[\rm (ii)] If $\omega_{1} = \frac{1}{3} \in \mathbf{\textsl{S}}_{1}$, then $\textsl{T}\frac{1}{3} = \frac{1}{3} \in \mathbf{\textsl{S}}_{2}.$

\item[\rm (iii)] If $\omega_{1} = \frac{1}{4} \in \mathbf{\textsl{S}}_{2}$, then $\textsl{T}\frac{1}{4} = \frac{1}{3} \in \mathbf{\textsl{S}}_{1}.$

\item[\rm (iv)] If $\omega_{1} = 1 \in \mathbf{\textsl{S}}_{2}$, then $\textsl{T}1 = 0 \in \mathbf{\textsl{S}}_{1}.$
\end{itemize}

\noindent Hence $\textsl{T}(\mathbf{\textsl{S}}_{1}) \subseteq
\mathbf{\textsl{S}}_{2}$, $\textsl{T}(\mathbf{\textsl{S}}_{2})
\subseteq \mathbf{\textsl{S}}_{1}$ and $\textsl{T}$ is a cyclic map.
Next fix any $\omega_{1} \in \mathbf{\textsl{S}}_{1}$. Let
$\omega_{1} = 0$, then we have

$$\textsl{T}\omega_{1} = \frac{1}{3},~ \textsl{T}^{2}\omega_{1} = \textsl{T}(\textsl{T}\omega_{1}) = \frac{1}{3},~\ldots.$$

\noindent Thus $\textsl{T}^{n}\omega_{1} = \frac{1}{3},$ therefore $\textsl{T}^{2n}\omega_{1} = \frac{1}{3}$ and $\textsl{T}^{2n - 1}\omega_{1} = \frac{1}{3}.$ For $\omega_{2}$, we will take the following cases:

\begin{itemize}
\item[\rm Case 1:] If $\omega_{2} = 0$, $\textsl{T}0 = \frac{1}{3},$ then $\textsl{d}_{\alpha}(\textsl{T}^{2n}\omega_{1}, \textsl{T}\omega_{2}) = \textsl{d}_{\alpha}(\frac{1}{3}, \frac{1}{3}) = 0$ and $\textsl{d}_{\alpha}(\textsl{T}^{2n - 1}\omega_{1}, \omega_{2}) = \textsl{d}_{\alpha}(\frac{1}{3}, 0) = \frac{1}{9}.$ Hence, for $\kappa \in (0, 1)$
    $$\textsl{d}_{\alpha}(\textsl{T}^{2n}\omega_{1}, \textsl{T}\omega_{2}) \leq \kappa \textsl{d}_{\alpha}(\textsl{T}^{2n - 1}\omega_{1}, \omega_{2}).$$


\item[\rm Case 2:] If $\omega_{2} = \frac{1}{3}$, $\textsl{T}\frac{1}{3} = \frac{1}{3},$ then $\textsl{d}_{\alpha}(\textsl{T}^{2n}\omega_{1}, \textsl{T}\omega_{2}) = \textsl{d}_{\alpha}(\frac{1}{3}, \frac{1}{3}) = 0$ and $\textsl{d}_{\alpha}(\textsl{T}^{2n - 1}\omega_{1}, \omega_{2}) = \textsl{d}_{\alpha}(\frac{1}{3}, \frac{1}{3}) = 0.$ Hence, for $\kappa \in (0, 1)$
    $$\textsl{d}_{\alpha}(\textsl{T}^{2n}\omega_{1}, \textsl{T}\omega_{2}) \leq \kappa \textsl{d}_{\alpha}(\textsl{T}^{2n - 1}\omega_{1}, \omega_{2}).$$

\item[\rm Case 3:] If $0 < \omega_{2} < \frac{1}{3}$, we will take subcases:
\begin{itemize}
\item[\rm Subcase A:]  If $0 < \omega_{2} \leq \frac{1}{4}$, $\textsl{T}\omega_{2} = \frac{1}{3},$ then $\textsl{d}_{\alpha}(\textsl{T}^{2n}\omega_{1}, \textsl{T}\omega_{2}) = \textsl{d}_{\alpha}(\frac{1}{3}, \frac{1}{3}) = 0$ and $\textsl{d}_{\alpha}(\textsl{T}^{2n - 1}\omega_{1}, \omega_{2}) = \textsl{d}_{\alpha}(\frac{1}{3}, \omega_{2}) = (\frac{1}{3} - \omega_{2})^{2}.$ Hence for $\kappa \in (0, 1)$
    $$\textsl{d}_{\alpha}(\textsl{T}^{2n}\omega_{1}, \textsl{T}\omega_{2}) \leq \kappa \textsl{d}_{\alpha}(\textsl{T}^{2n - 1}\omega_{1}, \omega_{2}).$$

\item[\rm Subcase B:] If $\frac{1}{4} < \omega_{2} \leq \frac{1}{3}$, $\textsl{T}\omega_{2} = \frac{1}{2}(1 - \omega_{2}),$ then $\textsl{d}_{\alpha}(\textsl{T}^{2n}\omega_{1}, \textsl{T}\omega_{2}) = \textsl{d}_{\alpha}(\frac{1}{3}, \frac{1}{2}(1 - \omega_{2})) = (\frac{1}{3} - \frac{1}{2} + \frac{\omega_{2}}{2})^{2}$. This implies that $\textsl{d}_{\alpha}(\textsl{T}^{2n}\omega_{1}, \textsl{T}\omega_{2}) = (\frac{-1}{6} + \frac{\omega_{2}}{2})^{2} = \frac{1}{4}(\omega_{2} - \frac{1}{3})^{2}$, and $\textsl{d}_{\alpha}(\textsl{T}^{2n - 1}\omega_{1}, \omega_{2}) = \textsl{d}_{\alpha}(\frac{1}{3}, \omega_{2}) = (\frac{1}{3} - \omega_{2})^{2}.$ Hence for $\kappa = \frac{1}{2}\in (0, 1)$
    $$\frac{1}{4}(\omega_{2} - \frac{1}{3})^{2} \leq \frac{1}{2}(\omega_{2} - \frac{1}{3})^{2}.$$
\end{itemize}
\end{itemize}

\noindent Hence, in all cases, the  conditions of Theorem
\ref{thm11} are  satisfied, and $\frac{1}{3}$ is the
 unique fixed point of
$\textsl{T}.$

\end{example}


\begin{definition}\label{def21}
Let $\mathbf{\textsl{S}}_{1}, \mathbf{\textsl{S}}_{2}$ be nonempty
subsets of a controlled metric space $(\textsl{W},
\textsl{d}_{\alpha})$. Then a cyclic map $\textsl{T} :
\mathbf{\textsl{S}}_{1} \cup \mathbf{\textsl{S}}_{2} \rightarrow
\mathbf{\textsl{S}}_{1} \cup \mathbf{\textsl{S}}_{2}$ is called a
controlled cyclic orbital $\mathbf{\textsl{F}}$-contraction, if for
some $\omega_{1} \in \mathbf{\textsl{S}}_{1}$, there exists $\Omega
>
0$ such that for all $\omega_{1}, \omega_{2} \in \textsl{W}$ with
$\textsl{d}_{\alpha}(\textsl{T}\omega_{1}, \textsl{T}\omega_{2}) >
0$, the following inequality holds:

\begin{equation}\label{equ21} \Omega + \mathbf{\textsl{F}}(\textsl{d}_{\alpha}(\textsl{T}^{2n}\omega_{1}, \textsl{T}\omega_{2})) \leq \mathbf{\textsl{F}}(\textsl{d}_{\alpha}(\textsl{T}^{2n - 1}\omega_{1}, \omega_{2})),\end{equation}

where $\omega_{2} \in \mathbf{\textsl{S}}_{1}$ and $n \in
\mathbb{N}.$
\end{definition}

\begin{theorem}\label{thm22}
Let $\mathbf{\textsl{S}}_{1}$ and $\mathbf{\textsl{S}}_{2}$ be two  nonempty subsets of a  complete controlled metric space $(\textsl{W}, \textsl{d}_{\alpha})$.% such that $\textsl{d}_{\alpha}$ is a continuous functional.
Let $\textsl{T} : \mathbf{\textsl{S}}_{1} \cup
\mathbf{\textsl{S}}_{2} \rightarrow \mathbf{\textsl{S}}_{1} \cup
\mathbf{\textsl{S}}_{2}$ be a continuous controlled cyclic orbital
$\mathbf{\textsl{F}}$-contraction such that for any $\omega_{0} \in
\textsl{W}$,
\begin{equation}\label{c3}
  \sup_{m\geq1}\lim_{i\rightarrow\infty}\frac{\alpha(\omega_{i+1},\omega_{i+2})}{\alpha(\omega_{i},\omega_{i+1})}\alpha(\omega_{i+1},\omega_m)<1,
 \end{equation}
%$\lim_{n, m \rightarrow \infty} \alpha(\omega_{n}, \omega_{m})\kappa < 1,$
where $\omega_{n} = \textsl{T}^{n} \omega_{0}$, $n = 1, 2, \ldots.$
Then $\mathbf{\textsl{S}}_{1} \cap \mathbf{\textsl{S}}_{2}$ is
nonempty and $\textsl{T}$ has a unique fixed point.
\end{theorem}

\begin{proof}
let us consider an arbitrary $\omega_{0} \in
\mathbf{\textsl{S}}_{1}$ satisfying equation \eqref{equ21}. As for
any $n \in \mathbb{N}$, either $n$ or $n + 1$ is even, so we have

\begin{equation}\label{thrm1} \mathbf{\textsl{F}}(\textsl{d}_{\alpha}(\textsl{T}^{n + 1}\omega, \textsl{T}^{n}\omega)) \leq \mathbf{\textsl{F}}(\textsl{d}_{\alpha}(\textsl{T}\omega, \omega)) - n\Omega.\end{equation}

\noindent By taking limit $n \rightarrow \infty$ in equation
\eqref{thrm1}, we get $\lim_{n \rightarrow \infty}
\mathbf{\textsl{F}}(\textsl{d}_{\alpha}(\omega_{n}, \omega_{n + 1}))
= -\infty$. Thus from condition $(II)$ of Definition \ref{def13}, we
have
$$\lim_{n \rightarrow \infty} \textsl{d}_{\alpha}(\omega_{n}, \omega_{n + 1}) = 0.$$
Also from condition $(III)$, there exists $l \in (0, 1)$ such that

$$\lim_{n \rightarrow \infty} \textsl{d}_{\alpha}(\omega_{n}, \omega_{n + 1})^{l} \mathbf{\textsl{F}}(\textsl{d}_{\alpha}(\omega_{n}, \omega_{n + 1})) = 0.$$

From equation \eqref{thrm1}, \mbox{for all} $n \in \mathbb{N}$, the
following holds:

\begin{equation}\label{eque11} \lim_{n \rightarrow \infty} \textsl{d}_{\alpha}(\omega_{n}, \omega_{n + 1})^{l} \mathbf{\textsl{F}}(\textsl{d}_{\alpha}(\omega_{n}, \omega_{n + 1})) - \lim_{n \rightarrow \infty} \textsl{d}_{\alpha}(\omega_{n}, \omega_{n + 1})^{l} \mathbf{\textsl{F}}(\mathcal{A}_{0}) \leq \lim_{n \rightarrow \infty} -\textsl{d}_{\alpha}(\omega_{n}, \omega_{n + 1})^{l} n\Omega \leq 0.\end{equation}

By letting $n \rightarrow \infty$ in \eqref{eque11}, we obtain

\begin{equation}\label{eque12} \lim_{n \rightarrow \infty} n \textsl{d}_{\alpha}(\omega_{n}, \omega_{n + 1})^{l} = 0. \end{equation}

From equation \eqref{eque12}, there exists $n_{1} \in \mathbb{N}$
such that $n \textsl{d}_{\alpha}(\omega_{n}, \omega_{n + 1})^{l}
\leq 1$ for all $n \geq n_{1}.$ Thus \mbox{for all} $n \geq n_{1},$
we have

\begin{equation}\label{eque13} \textsl{d}_{\alpha}(\omega_{n}, \omega_{n + 1}) \leq \frac{1}{n^{\frac{1}{l}}}.\end{equation}

\noindent From the triangle inequality and equation \eqref {eque13}
for $m > n \geq n_{1},$ we have

\begin{align*} \textsl{d}_{\alpha}(\textsl{T}^{n}\omega, \textsl{T}^{m}\omega) &= \textsl{d}_{\alpha}(\omega_{n}, \omega_{m})\\&\leq \alpha(\omega_{n}, \omega_{n + 1})\textsl{d}_{\alpha}(\omega_{n}, \omega_{n + 1}) + \alpha(\omega_{n + 1}, \omega_{m})\textsl{d}_{\alpha}(\omega_{n + 1}, \omega_{m})\\ & \leq \alpha(\omega_{n}, \omega_{n + 1})\textsl{d}_{\alpha}(\omega_{n}, \omega_{n + 1}) + \alpha(\omega_{n}, \omega_{m})\alpha(\omega_{n + 1}, \omega_{n + 2})\textsl{d}_{\alpha}(\omega_{n + 1}, \omega_{n + 2})\\& + \alpha(\omega_{n}, \omega_{m})\alpha(\omega_{n + 2}, \omega_{m})\textsl{d}_{\alpha}(\omega_{n + 2}, \omega_{m})\\ & \leq \ldots\\ &\leq \alpha(\omega_{n}, \omega_{n + 1})\textsl{d}_{\alpha}(\omega_{n}, \omega_{n + 1}) + \sum_{i = n + 1}^{m - 2}\left(\prod_{j = 1}^{i}\alpha(\omega_{j}, \omega_{m})\right) \alpha(\omega_{i}, \omega_{i + 1})\textsl{d}_{\alpha}(\omega_{i}, \omega_{i + 1})\\& + \prod_{j = n + 1}^{m - 1} \alpha(\omega_{j}, \omega_{m}) \alpha(\omega_{m - 1}, \omega_{m}) \textsl{d}_{\alpha}(\omega_{m - 1}, \omega_{m})\\ & \leq \alpha(\omega_{n}, \omega_{n + 1})\frac{1}{n^\frac{1}{l}} + \sum_{i = n + 1}^{m - 1}\left(\prod_{j = 1}^{i}\alpha(\omega_{j}, \omega_{m})\right) \alpha(\omega_{i}, \omega_{i + 1})\frac{1}{i^\frac{1}{l}}\\& \leq \alpha(\omega_{n}, \omega_{n + 1})\frac{1}{n^\frac{1}{l}} + \sum_{i = n + 1}^{\infty}\left(\prod_{j = 1}^{i}\alpha(\omega_{j}, \omega_{m})\right) \alpha(\omega_{i}, \omega_{i + 1})\frac{1}{i^\frac{1}{l}}.\end{align*}

\noindent By (\ref{c3}), %Since $\lim_{n, m \rightarrow \infty}\alpha(\omega_{n}, \omega_{m})\kappa < 1$,
the series $\sum_{n = 1}^{\infty}\frac{1}{n^\frac{1}{l}}\prod_{i =
1}^{n}\alpha(\omega_{i}, \omega_{m})\alpha(\omega_{i}, \omega_{i +
1})$ converges by the ratio test for each $m \in \mathbb{N}.$ Let
$$ S = \sum_{i = 1}^{\infty}\frac{1}{i^\frac{1}{l}}\prod_{j = 1}^{i}\alpha(\omega_{j},
\omega_{m})\alpha(\omega_{i}, \omega_{i + 1}),~~S_{n}~~= ~~\sum_{p =
1}^{n}\frac{1}{p^\frac{1}{l}}\prod_{q = 1}^{p}\alpha(\omega_{q},
\omega_{m})\alpha(\omega_{p}, \omega_{p + 1}).$$ Therefore for $m >
n$, we have
\begin{equation}\label{c4}
\textsl{d}_{\alpha}(\omega_{n}, \omega_{m}) \leq \alpha(\omega_{n}, \omega_{n + 1})\frac{1}{n^\frac{1}{l}} + S_{m - 1} - S_{n}.
\end{equation}
Condition (\ref{c3}) yields that $\lim_{n\rightarrow\infty}S_n$ exists and hence the real sequence $\{S_n\}$ is Cauchy.
Taking $n,m\rightarrow \infty,$ in  the inequality (\ref{c4}), we get that
\begin{equation}\label{c5}
\lim_{n,m\rightarrow\infty}\omega(\omega_{n},\omega_{m})=0.
\end{equation}
%\noindent By letting $n \rightarrow \infty$,
We conclude that $\{\omega_{n}\}$ is a Cauchy sequence. Hence, as a
result, there exists  $\rho \in \mathbf{\textsl{S}}_{1} \cup
\mathbf{\textsl{S}}_{2}$ such that $\omega_{n} \rightarrow \rho$.
Now, note that $\{\omega_{2n}\} = \{T^{2n}\omega_{0}\}$ is a
sequence in $\mathbf{\textsl{S}}_{1}$ and $\{\omega_{2n - 1}\} =
\{T^{2n - 1}\omega_{0}\}$ is a sequence in $\mathbf{\textsl{S}}_{2}$
and both converge to $\rho$. As the sets $\mathbf{\textsl{S}}_{1}$
and $\mathbf{\textsl{S}}_{2}$ are closed in $\textsl{W}$ and $\rho
\in \mathbf{\textsl{S}}_{1} \cap \mathbf{\textsl{S}}_{2}$. Hence,
$\mathbf{\textsl{S}}_{1} \cap \mathbf{\textsl{S}}_{2}$ is nonempty.
Next, we prove that $\rho$ is a fixed point of $\textsl{T}.$ Let
$\rho \neq \textsl{T}\rho$, then from the triangle inequality

\begin{equation}\label{equ211} \textsl{d}_{\alpha}(\rho, \textsl{T}\rho) \leq \alpha(\rho, \textsl{T}^{2n}\omega_{0})\textsl{d}_{\alpha}(\rho, \textsl{T}^{2n}\omega_{0}) + \alpha(\textsl{T}^{2n}\omega_{0}, \textsl{T}\rho)\textsl{d}_{\alpha}(\textsl{T}^{2n}\omega_{0}, \textsl{T}\rho).\end{equation}

\noindent Since $\textsl{T}^{2n - 1}\omega_{0} \rightarrow \rho$ as
$n \rightarrow \infty$, and from the continuity of $\textsl{T}$, we
have $\lim_{n \rightarrow
\infty}\textsl{d}_{\alpha}(\textsl{T}^{2n}\omega_{0},
\textsl{T}\rho) = 0.$ Hence, from equation \eqref{equ211}, we get
$\textsl{d}_{\alpha}(\rho, \textsl{T}\rho) = 0$ as $n \rightarrow
\infty.$ For the uniqueness, assume there exists $\varrho \in
\mathbf{\textsl{S}}_{1} \cap \mathbf{\textsl{S}}_{2}$, $\rho \neq
\varrho$ such that $\textsl{T}\varrho = \varrho.$ From equation
\eqref{equ21}, we have

\begin{align*} &\Omega + \mathbf{\textsl{F}}(\textsl{d}_{\alpha}(\textsl{T}^{2}\rho, \textsl{T}\rho)) \leq \mathbf{\textsl{F}}(\textsl{d}_{\alpha}(\textsl{T}\rho, \rho))\\& \Omega + \mathbf{\textsl{F}}(\textsl{d}_{\alpha}(\textsl{T}\rho, \rho)) \leq \mathbf{\textsl{F}}(\textsl{d}_{\alpha}(\textsl{T}\rho, \rho))\\& \Omega \leq \mathbf{\textsl{F}}(\textsl{d}_{\alpha}(\textsl{T}\rho, \rho)) - \mathbf{\textsl{F}}(\textsl{d}_{\alpha}(\textsl{T}\rho, \rho)) = 0.\end{align*}

\noindent It is a contradiction. Hence, $\rho = \varrho$ and $\rho$
is the unique fixed point of $\textsl{T}.$

\end{proof}
\begin{remark}
\label{R2.2}
%\textbf{Remark 2.2}
\begin{itemize}
\item[\rm (a)] For $\alpha(\omega_{1}, \omega_{2}) = s \geq 1$, Theorem \ref{thm22} reduces to the $b$-metric space.

\item[\rm (b)] For $\alpha(\omega_{1}, \omega_{2}) = 1$, Theorem \ref{thm22} reduces to the metric space.
\end{itemize}
\end{remark}
\begin{example}
Let $\textsl{W} = \{\frac{1}{2^{n}} : n \in \mathbb{N}\} \cup
\{0\}.$  Define $\textsl{d}_{\alpha} : \textsl{W} \times \textsl{W}
\rightarrow [0, \infty)$ by $\textsl{d}_{\alpha}(\omega_{1},
\omega_{2}) = (\omega_{1} - \omega_{2})^{2}.$ Then, clearly
$(\textsl{W}, \textsl{d}_{\alpha})$ is a complete controlled metric
space with $\alpha: \textsl{W} \times \textsl{W} \rightarrow [1,
\infty)$ defined as $\alpha(\omega_{1}, \omega_{2}) = 3\omega_{1} +
2\omega_{2} + 5$. Let $\mathbf{\textsl{S}}_{1} = \{\frac{1}{2^{2n -
1}} : n \in \mathbb{N}\} \cup \{0\}$ and $\mathbf{\textsl{S}}_{2} =
\{\frac{1}{2^{2n}} : n \in \mathbb{N}\} \cup \{0\}.$ Define a
mapping $\textsl{T} : \mathbf{\textsl{S}}_{1} \cup
\mathbf{\textsl{S}}_{2} \rightarrow \mathbf{\textsl{S}}_{1} \cup
\mathbf{\textsl{S}}_{2}$ as

$$\textsl{T}\omega_{1} = \left\{
             \begin{array}{ll}
              \frac{1}{2^{n + 1}}, & \hbox{if $\omega_{1} \in \{\frac{1}{2^{n}} : n \in \mathbb{N}\}$;}\\

               0, & \hbox{if $\omega_{1} = 0$.}
              \end{array}
            \right.$$

\noindent Clearly $\textsl{T}(\mathbf{\textsl{S}}_{1}) \subseteq
\mathbf{\textsl{S}}_{2}$, $\textsl{T}(\mathbf{\textsl{S}}_{2})
\subseteq \mathbf{\textsl{S}}_{1}$ and $\textsl{T}$ is a cyclic map.
Next, fix any $\omega_{1} \in \mathbf{\textsl{S}}_{1}$. Let
$\omega_{1} = \frac{1}{2^{2n - 1}}$, then we have

$$\textsl{T}\omega_{1} = \frac{1}{2^{2n}},~ \textsl{T}^{2}\omega_{1} = \textsl{T}(\textsl{T}\omega_{1}) = \frac{1}{2^{2n + 1}},~\ldots.$$

\noindent Thus, $\textsl{T}^{2n}\omega_{1} = \frac{1}{2^{2n + 2n
-1}} = \frac{1}{2^{4n -1}}$ and $\textsl{T}^{2n - 1}\omega_{1} =
\frac{1}{2^{4n - 2}}.$  For $\omega_{2}$, we will take the following
cases:

\begin{itemize}
\item[\rm Case 1:] If $\omega_{2} \in \mathbf{\textsl{S}}_{1}/\{0, 1\}$, let

$$\omega_{2} = \frac{1}{2^{2m - 1}},~~(m > n \geq 1).$$

\noindent We have $\textsl{T}\omega_{2} = \frac{1}{2^{2m}}.$ Also,

\begin{align*} \textsl{d}_{\alpha}(\textsl{T}^{2n}\omega_{1}, \textsl{T}\omega_{2}) &= \textsl{d}_{\alpha}\left(\frac{1}{2^{4n -1}}, \frac{1}{2^{2m}}\right)\\& = \left(\frac{1}{2^{4n -1}} - \frac{1}{2^{2m}}\right)^{2}\\& = \left(\frac{2^{2m} - 2^{4n - 1}}{2^{4n + 2m - 1}}\right)^{2},\end{align*}

\noindent and

\begin{align*} \textsl{d}_{\alpha}(\textsl{T}^{2n - 1}\omega_{1}, \omega_{2}) & = \textsl{d}_{\alpha}\left(\frac{1}{2^{4n - 2}}, \frac{1}{2^{2m - 1}}\right)\\& = \left(\frac{1}{2^{4n -2}} - \frac{1}{2^{2m - 1}}\right)^{2}\\& = \left(\frac{2^{2m - 1} - 2^{4n - 2}}{2^{4n + 2m - 3}}\right)^{2}.\end{align*}

\noindent Define the function $\mathbf{\textsl{F}} : [0, \infty)
\rightarrow \mathbb{R}$ by $\mathbf{\textsl{F}}(t) = \ln t$, for all
$t \in [0, \infty)$ and $\Omega > 0.$

\begin{align*} \mathbf{\textsl{F}}(\textsl{d}_{\alpha}(\textsl{T}^{2n}\omega_{1}, \omega_{2})) - \mathbf{\textsl{F}}(\textsl{d}_{\alpha}(\textsl{T}^{2n - 1}\omega_{1}, \omega_{2}))& = 2 \left(\ln \frac{2^{2m} - 2^{4n - 1}}{2^{4n + 2m - 1}})^{2} - \ln \frac{2^{2m - 1} - 2^{4n - 2}}{2^{4n + 2m - 3}} \right)\\& = 2\ln \left(\frac{2^{2m} - 2^{4n - 1}}{2^{4n + 2m - 1}})^{2} \times \frac{2^{4n + 2m - 3}}{2^{2m - 1} - 2^{4n - 2}} \right)\\& = 2\ln \left(\frac{2^{2m} - 2^{4n - 1}}{2^{2m - 1} - 2^{4n - 2}} \times 2^{-2} \right)\\& = 2\ln \left(\frac{2^{2m} - 2^{4n - 1}}{2^{2m + 1} - 2^{4n}} \right)\\& = 2\ln \left(\frac{2^{2m} - 2^{4n - 1}}{2(2^{2m - 1} - 2^{4n - 1})} \right)\\& = 2\ln \frac{1}{2}\\& < -1.\end{align*}


\item[\rm Case 2:] If $\omega_{2} = 0$, $\textsl{T}0 = 0,$ then

\begin{align*} \textsl{d}_{\alpha}(\textsl{T}^{2n}\omega_{1}, \textsl{T}\omega_{2}) &= \textsl{d}_{\alpha}\left(\frac{1}{2^{4n -1}}, 0\right)\\& = \left(\frac{1}{2^{4n -1}}\right)^{2},\end{align*}

\noindent and

\begin{align*} \textsl{d}_{\alpha}(\textsl{T}^{2n - 1}\omega_{1}, \omega_{2}) & = \textsl{d}_{\alpha}\left(\frac{1}{2^{4n - 2}}, 0\right)\\& = \left(\frac{1}{2^{4n -2}}\right)^{2}.\end{align*}

\noindent Thus, we have

\begin{align*} \mathbf{\textsl{F}}(\textsl{d}_{\alpha}(\textsl{T}^{2n}\omega_{1}, \omega_{2})) - \mathbf{\textsl{F}}(\textsl{d}_{\alpha}(\textsl{T}^{2n - 1}\omega_{1}, \omega_{2}))& = 2 \left(\ln \frac{1}{2^{4n -1}} - \ln \frac{1}{2^{4n -2}}\right) \\& = 2 \left(\frac{1}{2^{4n -1}} \times \ln \frac{2^{4n -2}}{1}\right) \\&= 2\ln \left(2^{-1} \right)\\& = 2\ln \frac{1}{2}\\& < -1.\end{align*}
\end{itemize}

\noindent Hence, for $\Omega = 1$, $\textsl{T}$ is a controlled
cyclic orbital $F$-contraction. Thus,  all the conditions of Theorem
\ref{thm22} are  satisfied, and $0$ is the unique fixed point of
$\textsl{T}.$


\end{example}


\section{ Applications to the existence of a solution of a Fredholm integral equation }

In this section, we ensure the existence of solution of a Fredholm
integral equation.

\begin{theorem}\label{thrm31}
Let $\textsl{W} = \mathbb{C}([a, b], \mathbb{R})$ be the space of all continuous
real valued functions defined on closed interval $[a, b]$ with a
controlled  metric given as

\begin{equation}\label{31} \textsl{d}_{\alpha}(\omega_{1}, \omega_{2}) = \sup_{t \in [a, b]} |\omega_{1}(t) - \omega_{2}(t)|^{2}.\end{equation}

\noindent Clearly, $(\textsl{W}, \textsl{d}_{\alpha})$ is a complete
controlled metric space with $\alpha: \textsl{W} \times \textsl{W}
\rightarrow [1, \infty)$ defined as $\alpha(\omega_{1}, \omega_{2})
= 3|\omega_{1}(t)| + 2|\omega_{2}(t)| + 5$. Let us consider the
 Fredholm integral equation:

\begin{equation}\label{32}\omega(t) = \int_a^b \! \mathcal{M}(t, r,\omega(r)) \, \mathrm{d}r + \mathbf{v}(t), \end{equation}

\noindent where $t, r \in [a, b],$ $\mathbf{v} : [a, b] \rightarrow
\mathbb{R}$ and $\mathcal{M} : [a, b] \times [a, b] \times
\mathbb{R} \rightarrow \mathbb{R}$ both are continuous functions.
Let $\mathbf{\textsl{S}}_{1} = \mathbf{\textsl{S}}_{2} = \textsl{W} = (\mathbb{C}([a, b]), \mathbb{R})$. Clearly,
$\mathbf{\textsl{S}}_{1}$, $\mathbf{\textsl{S}}_{2}$ are closed
subsets of $\textsl{W}.$ Define $\textsl{T} :
\mathbf{\textsl{S}}_{1} \cup \mathbf{\textsl{S}}_{2} \rightarrow
\mathbf{\textsl{S}}_{1} \cup \mathbf{\textsl{S}}_{2}$ by

\begin{equation}\label{33}\omega(t) = \int_a^b \! \mathcal{M}(t, r,\omega(r)) \, \mathrm{d}r + \mathbf{v}(t),~~~for~~t,~r \in [a, b], \end{equation}

\noindent where both $\mathbf{v}$ and $\mathcal{M}$ are continuous
functions. Clearly, $\textsl{T}(\mathbf{\textsl{S}}_{1}) \subset
\mathbf{\textsl{S}}_{2}$ and $\textsl{T}(\mathbf{\textsl{S}}_{2})
\subset \mathbf{\textsl{S}}_{1}$. Thus,  $\textsl{T}$ is a cyclic
map on $\mathbf{\textsl{S}}_{1} \cup \mathbf{\textsl{S}}_{2}.$
Moreover, let us consider that for $\omega \in \textsl{W}$ and $t, r
\in [a, b]$, the following condition holds:

\begin{equation}\label{34} |\mathcal{M}(t, r, \omega(r)) - \mathcal{M}(t, r, \textsl{T}\omega(r))| \leq
\frac{1}{2} |\omega(r) - \textsl{T}\omega(r)|.\end{equation}

\noindent Then, the integral equation \eqref{32} has a solution.
Now, we will show that $\textsl{T}$ satisfies all the conditions of
Theorem \ref{thm11}. For $\omega \in \textsl{W}$, consider

\begin{align*} |\textsl{T}^{2}\omega(t) - \textsl{T}\omega(t)|^{2} & = |\textsl{T}(\textsl{T}\omega(t)) - \textsl{T}\omega(t)|^{2}\\& = |\int_a^b \! [\mathcal{M}(t, r, \textsl{T}\omega(r)) - \mathcal{M}(t, r, \omega(r)) \,
\mathrm{d}r|^{2} \\& \leq \int_a^b \! \frac{1}{4}|\textsl{T}\omega(r) - \omega(r)|^{2} \, \mathrm{d}r \\ & \leq \frac{1}{4} \textsl{d}_{\alpha}(\textsl{T}\omega, \omega). \end{align*}

\noindent This implies that

$$\textsl{d}_{\alpha}(\textsl{T}^{2}\omega, \textsl{T}\omega) \leq \frac{1}{4} \textsl{d}_{\alpha}(\textsl{T}\omega, \omega),$$

\noindent where $\kappa = \frac{1}{4} \in (0, 1).$ Thus, all the
conditions of Theorem \ref{thm11} follow by the hypothesis. Hence,
$\textsl{T}$ has a fixed point, that is, the Fredholm integral
equation \eqref{32} has a solution.
\end{theorem}


\begin{theorem}\label{thrm32}
Let $\textsl{W} = \mathbb{C}([a, b], \mathbb{R})$ be the space of all continuous
real valued functions defined on the closed interval $[a, b]$ with
a metric given by

\begin{equation}\label{35} \textsl{d}_{\alpha}(\omega_{1}, \omega_{2}) = \sup_{t \in [a, b]} |\omega_{1}(t) - \omega_{2}(t)|^{2}.\end{equation}

\noindent Clearly, $(\textsl{W}, \textsl{d}_{\alpha})$ is a complete
controlled metric space with $\alpha: \textsl{W} \times \textsl{W}
\rightarrow [1, \infty)$ defined as $\alpha(\omega_{1}, \omega_{2})
= 3|\omega_{1}(t)| + 2|\omega_{2}(t)| + 5$. Let us consider  the
Fredholm integral equation

\begin{equation}\label{36} \omega(t) = \int_a^b \! \mathcal{M}(t, r, \omega(r)) \, \mathrm{d}r + \mathbf{v}(t), \end{equation}

\noindent where $t, r \in [a, b],$ $\mathbf{v} : [a, b] \rightarrow
\mathbb{R}$ and $\mathcal{M} : [a, b] \times [a, b] \times
\mathbb{R} \rightarrow \mathbb{R}$ both are continuous functions.
Let $\mathbf{\textsl{S}}_{1} = \mathbf{\textsl{S}}_{2} = \textsl{W} = (\mathbb{C}([a, b]), \mathbb{R})$. clearly
$\mathbf{\textsl{S}}_{1}$, $\mathbf{\textsl{S}}_{2}$ are closed
subsets of $\textsl{W}.$ Define $\textsl{T} :
\mathbf{\textsl{S}}_{1} \cup \mathbf{\textsl{S}}_{2} \rightarrow
\mathbf{\textsl{S}}_{1} \cup \mathbf{\textsl{S}}_{2}$ by

\begin{equation}\label{37} \omega(t) = \int_a^b \! \mathcal{M}(t, r, \omega(r)) \, \mathrm{d}r + \mathbf{v}(t),~~~for~~t,~r \in [a, b], \end{equation}

\noindent where both $\mathbf{v}$ and $\mathcal{M}$ are continuous
functions. Clearly, $\textsl{T}(\mathbf{\textsl{S}}_{1}) \subset
\mathbf{\textsl{S}}_{2}$ and $\textsl{T}(\mathbf{\textsl{S}}_{2})
\subset \mathbf{\textsl{S}}_{1}$. Thus, $\textsl{T}$ is a cyclic map
on $\mathbf{\textsl{S}}_{1} \cup \mathbf{\textsl{S}}_{2}.$
Moreover, let us consider that for $\omega \in \textsl{W}$, $\Omega
> 0$ and $t, r \in [a, b]$,
the following condition holds:

\begin{equation}\label{38} |\mathcal{M}(t, r, \omega(r)) - \mathcal{M}(t, r, \textsl{T}\omega(r))| \leq
|e^{-\frac{\Omega}{2}}(\omega(r) - \textsl{T}\omega(r))|.\end{equation}

\noindent Then, the integral equation \eqref{36} has a solution.
Now, we will show that $\textsl{T}$ satisfies all the conditions of
Theorem \ref{thm22}. For $\omega \in \textsl{W}$, consider

\begin{align*} |\textsl{T}^{2}\omega(t) - \textsl{T}\omega(t)|^{2} & = |\textsl{T}(\textsl{T}\omega(t)) - \textsl{T}\omega(t)|^{2}\\& = |\int_a^b \! [\mathcal{M}(t, r, \textsl{T}\omega(r)) - \mathcal{M}(t, r, \omega(r)) \,
\mathrm{d}r|^{2} \\& \leq \int_a^b \! |e^{-\frac{\Omega}{2}} (\textsl{T}\omega(r) - \omega(r))|^{2} \, \mathrm{d}r \\ & \leq e^{-\Omega}\textsl{d}_{\alpha}(\textsl{T}\omega, \omega). \end{align*}

\noindent It implies that

$$\textsl{d}_{\alpha}(\textsl{T}^{2}\omega, \textsl{T}\omega) \leq e^{-\Omega}\textsl{d}_{\alpha}(\textsl{T}\omega, \omega).$$

\noindent Thus,

\begin{align*} \ln(\textsl{d}_{\alpha}(\textsl{T}^{2}\omega, \textsl{T}\omega)) \leq \ln(e^{-\Omega}) + \ln(\textsl{d}_{\alpha}(\textsl{T}\omega, \omega)).
\end{align*}
That is,
%\ln(\textsl{d}_{\alpha}(\textsl{T}^{2}\omega, \textsl{T}\omega)) \leq -\Omega + \ln(\textsl{d}_{\alpha}(\textsl{T}\omega, \omega))\\&
$$\Omega + \ln(\textsl{d}_{\alpha}(\textsl{T}^{2}\omega, \textsl{T}\omega)) \leq \ln(\textsl{d}_{\alpha}(\textsl{T}\omega, \omega)).$$%\end{align*}

\noindent Hence, all the conditions of Theorem \ref{thm22}  are
satisfied for $\mathbf{\textsl{F}}(t_{1}) = \ln t_{1}$, $t_{1} > 0$.
Hence, $\textsl{T}$ has a fixed point, that is, the Fredholm
integral equation \eqref{36} has a solution.
\end{theorem}
















   \section{ Data Availability Statement:}

No data were used to support this study.



   \section{Competing Interest}
   The authors declare that there is not any competing interest regarding the publication of this manuscript.

\section{Author contributions} All authors contributed equally to the writing of this paper. All authors read and
   approved the final manuscript.







   \section{Funding}
   This research received no external funding.




\section{Acknowledgements}

Not applicable.








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