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\fancyhead[CE]{Mohammad  Hassan Saboori, Mahmoud Hassani,  Reza  Allahyari $^*$ and Mohammad Mehrabinezhad} 
\fancyhead[CO]{Fixed point theorems in $C^{*}$-algebra-valued $b_{v}(
s)$-metric spaces}



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{\noindent Journal of Mathematical Extension \\
Vol. XX, No. XX, (2014), pp-pp ()}\\
ISSN: 1735-8299\\
URL: http://www.ijmex.com\\
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{\Large \bf 
Fixed point theorems in $C^{*}$-algebra-valued $b_{v}(
s)$-metric spaces with application and numerical methods\\}
%{\bf Fixed point theorems in $C^{*}$-algebra-valued $b_{v}(
%s)$-metric spaces with application} 


\let\thefootnote\relax\footnote{\scriptsize Received: XXXX; Accepted: XXXX (Will be inserted by editor)}

{\bf Mohammad  Hassan Saboori}\vspace*{-2mm}\\
\vspace{2mm} {\small Department of Mathematics,  Mashhad Branch, Islamic Azad University, Mashhad, Iran.} \vspace{2mm}

{\bf  Mahmoud Hassani}\vspace*{-2mm}\\
\vspace{2mm} {\small  Department of Mathematics,  Mashhad Branch, Islamic Azad University, Mashhad, Iran.} \vspace{2mm}

{\bf  Reza  Allahyari$^*$\let\thefootnote\relax\footnote{$^*$
Corresponding Author}}\vspace*{-2mm}\\
\vspace{2mm} {\small   Department of Mathematics,  Mashhad Branch, Islamic Azad University, Mashhad, Iran.}\vspace{2mm}

{\bf  Mohammad Mehrabinezhad}\vspace*{-2mm}\\
\vspace{2mm} {\small   Department of Mathematics,  Mashhad Branch, Islamic Azad University, Mashhad, Iran.} \vspace{2mm}
\end{center}
{\footnotesize
\begin{quotation}
{\noindent \bf Abstract.} We first introduce a novel notion named $C^{*}$-algebra-valued $b_{v}(s)$-metric spaces. Then, we give proofs of the  Banach contraction principle, the expansion mapping theorem, and Jungck's theorem in $C^{*}$-algebra-valued $b_{v}(s)$-metric spaces.  As an application of our results, we establish a result for an integral equation in a $C^{*}$-algebra-valued $b_{v}(s)$-metric space.  Finally, a numerical method is presented to solve the proposed integral equation, and the convergence of this method is also studied.  Moreover, a numerical example is given to show applicability and accuracy of the numerical method and guarantee the theoretical results.
\end{quotation}
\begin{quotation}
\noindent{\bf AMS Subject Classification:}  47H10; 46L07

\noindent{\bf Keywords and Phrases:} $C^{*}$-algebra, $b_{v}(s)$-metric space, Fixed point theorem, Integral equation, Contractive mapping
\end{quotation}}

\section{Introduction and  preliminaries}
\label{intro} % It is advised to give each section and subsection a unique label.
 Banach contraction principle \cite{3} was introduced by Banach in 1922, and later it is called the fixed point theorem. The fixed point theorem is a strong tool for solving existence problems in many branches of mathematics and physics.\\
  Bakhtin \cite{2} introduced  $b$-metric spaces as a generalization of metric spaces and proved analogue of the Banach contraction principle in $b$-metric spaces. 
 In the paper \cite{5}, Branciari introduced the concept of $v$-generalized metric spaces. 
 Radenovi\'{c} and Mitrovi\'{c} \cite{12} introduced the concept $b_{v}(s)$-metric spaces as a generalization of metric spaces,  $b$-metric spaces,  and  $v$-generalized metric spaces. 
On the other hand, Ma et al. \cite{9} presented the concept of $C^{*}$-algebra-valued metric spaces. Later, this line of research was continued in \cite{4, 7, 8, 10, 13}, where several other fixed point results were obtained in the framework of $C^{*}$-algebra-valued metric, as well as (more general) $C^{*}$-algebra-valued $b$-metric spaces.  
Now,  in this paper, we introduce a new notion $C^{*}$-algebra-valued $b_{v}(s)$-metric spaces. Then, we  prove the  Banach contraction principle, expansion mapping theorem, and Jungck's theorem \cite{1} for $C^{*}$-algebra-valued $b_{v}(s)$-metric spaces.  Also,  we state a result for an integral equation in a $C^{*}$-algebra-valued $b_{v}(s)$-metric space, which demonstrates an application of our main theorem.  Finally,  we propose a numerical method for solving the  integral equation and investigate the convergence of this method.  Moreover,  to illustrate an application and accuracy, we present a  numerical example, which guarantees the theoretical results.\par 
We provide some auxiliary facts which will be used in the rest of the paper.
Throughout this paper,  $\Bbb{A}$ always denotes  a unital $C^{*}$-algebra with a unit $1_{\Bbb{A}}$. We call an element $a\in\Bbb{A}$ a \textit{positive element}, denoted $a\geq0_{\Bbb{A}}$, if $a\in\Bbb{A}_{h}$ and $\sigma(a)\subseteq\Bbb{R}_{+}=[0, \infty) $, where $\Bbb{A}_{h}=\{a\in\Bbb{A}$ :  $a^{*}=a\}$. The set $\Bbb{A}_{+}$ indicates the positive elements of $\Bbb{A}$.  Also,  $\Bbb{A'}=\{a\in\Bbb{A}: xa=ax,  \text{for all }x\in\Bbb{A}\}$. 
\begin{lemma} \cite{11}\label{in1} Let $\Bbb{A}$ be a unital $C^{*}$-algebra with unit $1_{\Bbb{A}}$.  
\begin{itemize}
\item[(1)] If $a,b\in\Bbb{A}_{h}$ with $a\leq b$  and $c\in\Bbb{A}$, then $c^*ac\leq c^*bc$.
\item[(2)] For all $a,b\in\Bbb{A}_{h}$,  if  ${0_{\Bbb{A}}}\leq a\leq b$, then $\|a\|\leq\|b\|$.
%\item[(3)] Suppose that $a,b\in\Bbb{A}_{+}$ and $ab=ba$ then $ab\geq0_{\Bbb{A}}$.
 \end{itemize}
 \end{lemma}
\begin{lemma} \cite{6, 11}\label{in2} Suppose that $\Bbb{A}$ is a unital $C^{*}$-algebra with unit $1_{\Bbb{A}}$. 
\begin{itemize}
\item[(1)] For any $x\in\Bbb{A}_{+}$, it follows that $x\leq 1_{\Bbb{A}}$ if and only if $\| x \|\leq1$.
\item[(2)] If $a\in\Bbb{A}_{+}$ with $\| a \|<\dfrac{1}{2}$, then $1_{\Bbb{A}}-a$ is invertible and $\| a(1_{\Bbb{A}}-a)^{-1}\|<1$.
\item[(3)] Suppose that $a,b\in\Bbb{A}_{+}$ with $ab=ba$; then $ab\geq0_{\Bbb{A}}$.
\item[(4)] For $a\in\Bbb{A'}$, if $b\geq c\geq0_{\Bbb{A}}$ and $1_{\Bbb{A}}-a\in\Bbb{A'_{+} }$ is an invertible element, then
 $(1_{\Bbb{A}}-a)^{-1}b\geq(1_{\Bbb{A}}-a)^{-1}c$.
\end{itemize}
\end{lemma}

\section{ Main results}
\label{sec:2}
\begin{definition} \label{bvsd} Let $X$ be a nonempty set  and let $\Bbb{A}$ be a $C^{*}$-algebra. The mapping $d:X\times X\rightarrow
\Bbb{A}_{+}$ is called \textit{$C^{*}$-algebra-valued $b_{v}(s)$-metric}, if there exists $s\in\Bbb{A'_{+}}$ with $\|s\|\geq1$ such that $d$ satisfies
\begin{itemize}
\item[(1)] $d(x,y)=0_\Bbb{A}$ if and only if $x=y$~~~ for all $x,y\in X$.
\item[(2)] $d(x,y)=d(y,x)$~~~ for all $x,y\in X$.
\item[(3)] $d(x,y)\leq s[d(x,u_{1})+d(u_{1},u_{2})+\cdots+d(u_{v-1},u_{v})+d(u_{v},y)]$, for all $x,y\in X$ and for all distinct elements $u_{1},u_{2},\ldots,u_{v}\in X-\{x,y\}$ in which $v\in\Bbb{N}$. 
\end{itemize}
\end{definition}
 \begin{definition}
 Suppose that $(X,\Bbb{A},d)$ is a $C^{*}$-algebra-valued $b_{v}(s)$-metric space. Then $T:X\rightarrow X$ is called a \textit{$C^{*}$-algebra-valued contractive mapping}, if there exists  $B\in\Bbb{A}$ with $\| B\|<1$ such that 
\begin{eqnarray}\label{cont1}
d(Tx,Ty)\leq B^{*}d(x,y)B\hspace{0.5cm} \text {for all }x,y\in X.
\end{eqnarray}
\end{definition}
\begin{example} Let $X=\ell^{p}=\{x=\{ x_{n}\}\subseteq\Bbb{R}:\sum_{n=1}^\infty| x_{n}|^{p}<+\infty\}, \  p\in(0,1)$, and let $\Bbb{A}=\Bbb{M}_{2\times2}(\Bbb{R})$. \\Define $d(x,y)=\text{diag}\Big({ (\sum_{n=1}^\infty|x_{n}-y_{n}|^{p})^\frac{1}{p}} ,(\sum_{n=1}^\infty|x_{n}-y_{n}|^{p})^\frac{1}{p}\Big)$ in which ``diag" denotes a diagonal matrix and $x,y\in X$. It is easy to verify that $d(.,.)$  is  a $C^{*}$-algebra-valued $b_{v}(s)$-metric.   For proving $(3)$  of Definition \ref{bvsd} with $v=2$,  we only need to use the following inequality:
\begin{footnotesize}
\begin{equation*}
{\Big(\displaystyle\sum_{n=1}^\infty\vert x_{n}-y_{n}\vert^{p}\Big)^\frac{1}{p}}\leq{ 2^{(\dfrac{2}{p})}}\Bigg[{\Big(\displaystyle\sum_{n=1}^\infty\vert x_{n}-u_{n}\vert^{p}\Big)^\frac{1}{p}}
 +{\Big(\displaystyle\sum_{n=1}^\infty\vert u_{n}-z_{n}\vert^{p}\Big)^\frac{1}{p}}+{\Big(\displaystyle\sum_{n=1}^\infty\vert z_{n}-y_{n}\vert^{p}\Big)^\frac{1}{p}}\Bigg],
 \end{equation*}
\end{footnotesize}
 which implies that $d(x,y)\leq s\Big[d(x,u)+d(u,z)+d(z,y)\Big]$, where $s={2^{(\dfrac{2}{p})}}I\in\Bbb{A'_{+}}$,  for all $x,y\in X$ and for all distinct elements  $u,z\in X-\{x,y\}$.
\end{example}
\begin{lemma} \label{bvsl1}Let $(X,\Bbb{A},d)$ be a $C^{*}$-algebra-valued $b_{v}(s)$-metric space and let $s\in\Bbb{A'_{+}}$. Then $(X,\Bbb{A},d)$ is a $C^{*}$-algebra-valued $b_{2v}(s^{2})$-metric space.
\end{lemma}
\begin{proof}
 Let $(X,\Bbb{A},d)$ be a $C^{*}$-algebra-valued $b_{v}(s)$-metric space. Let 
% $u_{1},u_{2},...,u_{v}$ are distinct and belong to $X-\{ x,y\} $ then there exists $A\in\Bbb{A'_{+}}$ with $\|A\|\geq1$ %such that for all $x,y\in X$ we have:\\ 
 \begin{eqnarray*}
d(x,y)\leq s\Big[d(x,u_{1})+d(u_{1},u_{2})+\cdots+d(u_{v},y)\Big]
 \end{eqnarray*}
for all  $x,y\in X$ and for all distinct elements $u_{1},u_{2},\ldots,u_{v}\in X-\{ x,y\} $.
Then, for different $s_{1},s_{2},\ldots,s_{v}\in X-\{x,y,u_{1},u_{2},\ldots,u_{v}\}$, we have
 \begin{eqnarray*}
d(u_{v},y)\leq s\Big[d(u_{v},s_{1})+d(s_{1},s_{2})+\cdots+d(s_{v},y)\Big].
\end{eqnarray*}
On the other hand,  for every $C^{*}$-algebra, if $a$ and $b$ are positive elements with $a\leq b$, and  in addition $s\in\Bbb{A'_{+}}$, then $sa\leq sb$.\\ 
Now, by the above inequality, we have
  \begin{eqnarray*}
sd(u_{v},y)\leq s\Big[s[d(u_{v},s_{1})+\cdots+d(s_{v},y)] \Big].
  \end{eqnarray*}  
Furthermore,  if $a,b\geq 0_{\Bbb{A}}$, then $a+b\geq 0_{\Bbb{A}}$. Hence we can write
  \begin{eqnarray*}
d(x,y)\leq s\Big[d(x,u_{1})+d(u_{1},u_{2})+\cdots+d(u_{v-1},u_{v})+s[d(u_{v},s_{1})+\cdots+d(s_{v},y)]\Big].
\end{eqnarray*}
Since $I\leq s$ and $s\in\Bbb{A'_{+}}$, so $s\leq s^2$ and $ sb\leq s^{2}b$ for all positive element $b\in\Bbb{A}$. Hence, for all positive elements $a,b,c\in\Bbb{A}$,  if $a\leq sb+s^{2}c$, then $a\leq s^{2}b+s^{2}c$. Thus, we get 
  \begin{eqnarray*}
d(x,y)\leq s^{2}\Big[d(x,u_{1})+d(u_{1},u_{2})+\cdots+d(u_{v},s_{1})+d(s_{1},s_{2})+\cdots+d(s_{v},y)\Big],
\end{eqnarray*}
which implies that  $(X,\Bbb{A},d)$ is a $C^{*}$-algebra-valued $b_{2v}(s^{2})$-metric space. 
\end{proof}
%%%%%%%%%%55555555555555555555555555555\\
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\begin{lemma}\label{bvsl2} Let $(X,\Bbb{A},d)$ be a $C^{*}$-algebra-valued $b_{v}(s)$-metric space, let $T:X\rightarrow X$, and let $\{x_{n}\}$ be a sequence in $X$ defined by $x_{0}\in X$ and $x_{n+1}=Tx_{n}$ such that $x_{n}\neq x_{n+1}$  $(n\geq0)$. If $T$ is a $C^*$-algebra-valued contractive mapping, then $x_{n}\neq x_{m}$  for all distinct numbers $m,n\in\Bbb{N}$.  
\end{lemma}
\begin{proof}
  Suppose, to the contrary, that  $x_{n}=x_{n+p}$ for some $n\geq0$ and  $p\geq1$.\\
Since $T$ is a $C^{*}$-algebra-valued contractive mapping, there exists $B\in\Bbb{A}$ with $\|B\|<1$ such that
 \begin{eqnarray*}
d(Tx,Ty)\leq B^{*}d(x,y)B,\hspace{0.5cm}\text{for all }x,y\in X.
\end{eqnarray*}
On the other hand,  we have  $Tx_{n}=Tx_{n+p}$,  and the assumptions imply $x_{n+1}=x_{n+p+1}$.\\
Now, we get
  \begin{eqnarray*}
d(x_{n+1},x_{n})=d(x_{n+p+1},x_{n+p})\leq B^{*}d(x_{n+p},x_{n+p-1})B.
\end{eqnarray*}
 Similarly,  
   \begin{eqnarray*}
d(x_{n+p},x_{n+p-1})\leq B^{*}d(x_{n+p-1},x_{n+p-2})B.
  \end{eqnarray*}
 Now, using Lemma \ref{in1},
we conclude 
\begin{eqnarray*}
 0_{\Bbb{A}}\leq d( x_{n+1},x_{n})= d(x_{n +p+1},x_{n+p })&\leq& B^{*} d(x_{n+p},x_{n+p-1})B\\ 
&\leq&( B^{*})^{2}d(x_{n+p-1},x_{n+p-2})B^{2}\\
&\vdots&\\
%\end{eqnarray*}
&\leq&( B^{*})^{p}d(x_{n+1},x_{n})B^{p}.
\end{eqnarray*}
Finally,  by applying  Lemma \ref{in1} again, we obtain
 %And since for every $C^{*}$-algebra for positive elements $a,b$ if $a \leq b$ then $\|a\|\leq\|b\|$,\\ therefore:\\
\begin{eqnarray*}
\|d(x_{n+1},x_{n})\|&\leq &\|( B^{*})^{p}d(x_{n+1},x_{n})B^{p}\|\\
%\end{eqnarray*}
%\\% Also, by submultiplicative of norm in $C^{*}$-algebra $\Bbb{A}$ and since $\|B\|<1$ we conclude:\\
%\begin{eqnarray*}
 & \leq &\|(B^{*})^{p}\|\|d(x_{n+1},x_{n}\|\|B^{p}\|\\ 
&\leq&\|B^*\|^{p}\|d(x_{n+1},x_{n})\|\|B\|^{p}\\
&=&\|B\|^{2p}\|d(x_{n+1},x_{n})\|\\
&<&\|d(x_{n+1},x_{n})\|,
\end{eqnarray*}
which is a contradiction.
\end{proof}
%%%%%%%%%%5555555555555555555555555\\
%%%%%%5555555555555555555555555555555555\\
\begin{definition}
   Let $(X,\Bbb{A},d)$ be a $C^{*}$-algebra-valued $b_{v}(s)$-metric space. Suppose that $\{x_{n}\}\subset X$ and $x\in X$.
   If, for any $\varepsilon>0$, there is a natural number $N$ such that  $\|d({x_{n}},x)\|\leq \varepsilon$ for all $n>N$, then $\{x_{n}\}$ is said to be \textit{convergent with respect to $\Bbb{A}$}, also \textit{$\{x_{n}\}$ converges to $x$}, or \textit{ $x$ is the limit of $\{x_{n}\}$}. We denote it by $\displaystyle{\lim_{n\rightarrow\infty}}x_{n}=x$.\\
 For any $\varepsilon>0$, if there is  a natural number $N$ such that  $\|d(x_{n},x_{m})\|\leq \varepsilon$ for all $n,m>N$, then $\{x_{n}\}$ is called a \textit{Cauchy sequence with respect to $\Bbb{A}$}.\\
   We say $(X,\Bbb{A},d)$ is a \textit{complete $C^*$-algebra-valued $b_{v}(s)$-metric space} if every Cauchy sequence with respect to $\Bbb{A}$ is convergent.
 %  Then\\
%   \begin{itemize}
%\item[(1)] The sequence $\{x_{n}\}\subset{X}$ is called to be convergent to a point $x\in X$ with respect to $\Bbb{A}$ if and only if for any  $\varepsilon>0$, there exists an $N\in\Bbb{N}$ such that for all $n>N$, $\|d(x_{n},x)\|\leq\varepsilon$, it means that 
%$ \displaystyle{\lim_{n\rightarrow\infty}}x_{n}=x$.
 %\item[(2)]The sequence $\{x_{n}\}\subset{X}$ is called Cauchy sequence with respect to $\Bbb{A}$ if for any $\varepsilon>0$, there exists $N\in\Bbb{N}$ such that for all $m,n>N$, $\|d(x_{m},x_{n})\|<\varepsilon$.\\
%\item[(3)]The space $(X,\Bbb{A},d)$ is said to complete $C^*$-algebra-valued $b_{v}(s)$-metric space if every Cauchy sequence in $X$ is convergent with respect to $\Bbb{A}$.\\
%  \end{itemize} 
\end{definition} 
%\end{Theorem} %%%%%55555555555555555555555555555\\
%%%5555555555555555555555555555555555555555555555555\\ Section 4 \\
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\begin{theorem}\label{bvs}
Suppose that $(X,\Bbb{A},d)$ is a complete $C^{*}$-algebra-valued $b_{v}(s)$-metric space with coefficient $s$. Let  $T:X\rightarrow X$ be a $C^{*}$-algebra-valued contractive mapping with constant $B$.  If   there exists a  natural number $n_{0}$ such that $s(B^*)^{n_{0}}B^{n_{0}}< 1_{\Bbb{A}}$ and $B^{n_{0}}\in\Bbb{A'}$, then $T$ has a unique fixed point in $X$.
 \end{theorem}  
\begin{proof}
It is clear that if $B=0_{\Bbb{A}}$, then $T$ maps  $X$ into a single point. Thus without loss of generality, one can suppose that $B\neq 0_{\Bbb{A}}$.\\
Choose $x_{0}\in X$, and  set $\{x_{n}\}$ by $x_{n+1}=Tx_{n}=T^{n+1}x_{0}, n=0,1,2,\ldots $. If $x_{n}=x_ {n+1}$  for some $n\geq 0$, then $T$ has a unique fixed point in $X$.  Otherwise,  we consider $x_{n}\neq x_ {n+1}$
 $(n\geq0)$.  Using Lemma \ref{bvsl2} implies that $x_{n}\neq x_ {m}$ for all distinct numbers $n, m\in\Bbb{N}$.  On the other hand,  notice that   $s[d(x,u_{1})+d(u_{1},u_{2})+\cdots+d(u_{v-1},u_{v})+d(u_{v},y)]$,  $s\in\Bbb{A'_{+}}$, is also a positive element. Now, by Lemma \ref{in1} and  the  condition (\ref{cont1}) on $T$, it follows that
\begin{eqnarray*}
d(x_{n+1}, x_{n})=d(Tx_{n}, Tx_{n-1})&\leq&B^*d(x_{n}, x_{n-1})B\\&\leq&(B^*)^{2}d(x_{n-1}, x_{n-2})B^{2}\\
&\vdots&\\
&\leq&(B^*)^{n}d(x_{1}, x_{0})B^{n}.
\end{eqnarray*} 
We consider the following two cases:
\begin{itemize}
\item[(1)] $v\geq2$
\item[(2)] $v=1$.
\end{itemize}
Let $v\geq2$ . Also,  suppose that $m>n$; then the triangle inequality for  the $b_{v}(s)$-metric $d$ implies that
\begin{eqnarray*}
 d(x_{n},x_{m})&\leq& s
\Big[d(x_{n},x_{n+1})+d(x_{n+1},x_{n+2})+\cdots+d(x_{n+v-3},x_{n+v-2})\\&&\quad+ 
          d(x_{n+v-2},x_{n+n_{0}})+d(x_{n+n_{0}},x_{m+n_{o}})
          +d(x_{m+n_{o}},x_{m})\Big]\\
          &\leq & s\Big[(B^*)^{n}d(x_{0},x_{1})B^{n}+(B^*)^{n+1}d(x_{0},x_{1})B^{n+1}+\cdots\\&&
          \quad+(B^*)^{n+v-3}d(x_{0},x_{1})B^{n+v-3}+(B^*)^{n}d(x_{v-2},x_{n_{0}})B^{n}\\&&\quad+
         ( B^*)^{n_{0}}d(x_{n},x_{m})B^{n_{0}}+(B^*)^{m}d(x_{n_{0}},x_{0})B^{m}\Big].
 \end{eqnarray*}
  So,
  \begin{eqnarray*}
   d(x_{n},x_{m})-s(B^*)^{n_{0}}d(x_{n},x_{m})
  B^{n_{0}}&\leq & s(B^*)^{n}d(x_{0},x_{1})B^{n}+s(B^*)^{n+1}d(x_{0},x_{1})B^{n+1}\\&&+\cdots+
 s(B^*)^{n+v-3}d(x_{0},x_{1})B^{n+v-3}\\&&+s(B^*)^{n}d(x_{v-2},x_{n_{0}})B^{n}\\&&+
  s(B^*)^{m}d(x_{n_{0}},x_{0})B^{m}.
\end{eqnarray*}
 On the other hand,  by Lemma \ref{in1}, we have
  \begin{footnotesize} 
\begin{eqnarray*}
d(x_{n},x_{m})(1_{\Bbb{A}}-s{(B^*)^{n_{0}}}B^{n_{0}})&\leq&
      \|d(x_{n},x_{m})(1_{\Bbb{A}}-s{(B^*)^{n_{0}}}B^{n_{0}})\|1_{\Bbb{A}} \\&\leq& \| s{(B^*)^{n}}d(x_{0},x_{1})B^{n}+ s{(B^*)^{n+1}}d(x_{0},x_{1})B^{n+1}+ \cdots\\&&+s{(B^*)^{n+v-3}}d(x_{0},x_{1})B^{n+v-3}+s{(B^*)^{n}}d(x_{v-2},x_{n_{0}})B^{n}\\&&+s{(B^*)^{m}}d(x_{n_{0}},x_{0} )B^{m}\|1_{\Bbb{A}} \\
     &\leq& \|s{(B^*)^{n}} d(x_{0},x_{1})B^{n}\|1_{\Bbb{A}}+\|s{(B^*)^{n+1}}d(x_{0},x_{1})B^{n+1}\|1_{\Bbb{A}}+\cdots\\&
     &+ \|s{(B^*)^{n+v-3}}d(x_{0},x_{1})B^{n+v-3}\|1_{\Bbb{A}}+\|s{(B^*)^{n}}d(x_{v-2},x_{n_{0}})B^{n}\|1_{\Bbb{A}} \\
     &&+\|s{(B^*)^{m}}d(x_{n_{0}},x_{0})B^{m}\|1_{\Bbb{A}}\\
     &\leq&\|s\|\|(B^*)^{n}\|\|d(x_{0},x_{1})\|\|B^{n}\|1_{\Bbb{A}}+\|s\|\|(B^*)^{n+1}\|\|d(x_{0},x_{1})\|\|B^{n+1}\|1_{\Bbb{A}}\\
     &&+\|s\|\|(B^*)^{n+v-3}\|\|d(x_{0},x_{1})\|\|B^{n+v-3}\|1_{\Bbb{A}}\\
     &&+\|s\|\|(B^*)^{n}\|\|d(x_{v-2},x_{n_{0}})\|\|B^{n}\|1_{\Bbb{A}}\\
     &&+\|s\|\|(B^*)^{m}\|\|d(x_{n_{0}},x_{0})\|\|B^{m}\|1_{\Bbb{A}}\\
     &\leq&\|s\|\|B^*\|^{n}\|d(x_{0},x_{1})\|\|B\|^{n}1_{\Bbb{A}}+\|s\|\|B^*\|^{n+1}\|d(x_{0},x_{1})\|\|B\|^{n+1}1_{\Bbb{A}}\\
     &&+\|s\|\|B^*\|^{n+v-3}\|d(x_{0},x_{1})\|\|B\|^{n+v-3}1_{\Bbb{A}}
     +\|s\|\|B^*\|^{n}\|d(x_{v-2},x_{n_{0}})\|\|B\|^{n}1_{\Bbb{A}}\\
     &&+\|s\|\|B^*\|^{m}\|d(x_{n_{0}},x_{0})\|\|B\|^{m}1_{\Bbb{A}}.
     \end{eqnarray*}
\end{footnotesize}
 Now, since  $ (1_{\Bbb{A}}-s{(B^*)^{n_{0}}}B^{n_{0}})\in\Bbb{A'_{+}}$ and it is invertible.  Hence,   by Lemma \ref{in2}, we have
   
   \begin{eqnarray*}
  d (x_{n},x_{m})&\leq& \Big(\|s\|\|B^*\|^{n}\|d(x_{0},x_{1})\|\|B\|^{n}+\|s\|\|B^*\|^{n+1}\|d(x_{0},x_{1})\|\|B\|^{n+1}\\
     &&+\|s\|\|B^*\|^{n+v-3}\|d(x_{0},x_{1})\|\|B\|^{n+v-3}
     +\|s\|\|B^*\|^{n}\|d(x_{v-2},x_{n_{0}})\|\|B\|^{n}\\
     &&+\|s\|\|B^*\|^{m}\|d(x_{n_{0}},x_{0})\|\|B\|^{m}\Big)(1_{\Bbb{A}}-s{(B^*)^{n_{0}}}B^{n_{0}})^{-1}\\
    & \longrightarrow& 0_{\Bbb{A}} \hspace{0.25cm} ( as~~ m, n\rightarrow\infty). 
\end{eqnarray*}
Therefore $\{x_{n}\}$ is a Cauchy sequence with respect to $\Bbb{A}$.
If $v=1$, then the proof follows from Lemma \ref{bvsl1}.  
By completeness of  $(X,\Bbb{A},d)$, there exists $x^*\in X$ such that $\displaystyle{\lim_{n\rightarrow\infty}}x_{n}=\displaystyle{\lim_{n\rightarrow\infty}}Tx_{n-1}
=x^{*}$.
Since 
\begin{eqnarray*}
d(Tx^{*},x^{*})&\leq & s\Big[d(Tx^{*},x_{n+1})+d(x_{n+1},x_{n+2})+\cdots +d(x_{n+v},x^{*})\Big]\\&
=&s\Big[d(Tx^{* },Tx_{n}) +d(Tx_{n},Tx_{n+1})+\cdots +d(x_{n+v},x^{*})\Big]\\
&\leq& s\Big[B^{*}d(x^{*},x_{n})B+(B^*)^{n}d(x_{0},x_{1})B^{n}+ \cdots +d(Tx_{n+v-1},x^{*}) \Big],
\end{eqnarray*}
it follows that
%Since right and left ineqaulity are positive elements of $C^{*}$-algebra $\Bbb{A}$ we have:
\begin{eqnarray*}
\| d(Tx^{*},x^{*})\| &\leq&\| s[B^*d(x^{*},x_{n})B+(B^*)^{n}d(x_{0},x_{1})B^{n}+\cdots+d(Tx_{n+v-1},x^{*})\| \\
&\leq&\| s\| \|B^{*}\| \| d(x^{*},x_{n})\| \| B \| +\|s\|\|(B^*)^{n}\|\|d(x_{0},x_{1})\|\| B^{n}\|+\cdots\\&
&+\|s\|\|(B^*)^{n+v-2}\|\|d(x_{0},x_{1})\|\|B^{n+v-2}\|+\|d(x_{n+v},x^{*})\|\\&\leq&\|s\|\|B^{*}\|\|d(x^{*},x_{n})\|\|B\|+\|s\|\|B^*\|^{n}\|d(x_{0},x_{1})\|\|B\|^{n}+\cdots\\&
&+\|s\|\|B^*\|^{n+v-2}\|d(x_{0},x_{1})\|\|B\|^{n+v-2}+\|d(x_{n+v},x^{*})\|\\
&\longrightarrow &0\hspace{0.5cm} (as~~n\rightarrow\infty),
\end{eqnarray*}
%But since $\displaystyle{\lim_{n\rightarrow\infty}}x_{n}=x^{*}$ hence $\|d(x^{*},x_{n+1})\|$ and $\|d(x_{n+v},x^{*})\|$ are less of zero.\\
which shows that $Tx^*=x^*$.\\
 To prove that $x^*$ is the unique fixed point, we suppose that $y^*(\neq x^*)$ is another fixed point of $T$. Then by applying condition (\ref{cont1}), we have  
%$\|d(Tx^{*},x^{*})\|\leq0$ and property of norm implies that $d(Tx^{*},x^{*})=0_{\Bbb{A}}$ and finally conclude $Tx^{*}=x^{*}$, it means that $x^{*}$ is a fixed point for $T$.\\
%It remains that uniqueness of $x^{*}$.\\
%By contradiction, suppose $y^{*}$ is another fixed point for $T$.We show that $x^{*}=y^{*}$.
\begin{eqnarray*}
0_{\Bbb{A}} \leq d(x^{*},y^{*})=d(Tx^{*},Ty^{*})\leq B^{*}d(x^{*},y^{*})B.
\end{eqnarray*}
%Since each both of inequality are positive element we have :
Using the norm of $\Bbb{A}$, we have
\begin{eqnarray*}
0&\leq& \|d(x^{*},y^{*})\|=\|d(Tx^{*},Ty^{*})\|\\&\leq&\| B^{*}\|\|d(x^{*},y^{*})\|\|B\|\\&=&\|B\|^2\|d(x^*,y^*)\|\\
 &<&\|d(x^{*},y^{*})\|,
 \end{eqnarray*}
which is  impossible. So $d(x^{*},y^{*})=0_{\Bbb{A}}$ and  $x^{*}=y^{*}$, which implies that  the  fixed point is unique.
   \end{proof}
%%%%%%%%5555555555555555555555555555\\
 %%%%%%%555555555555555555555555555555555 \\
 %%%%%%%%%%5555555555555555555555555555555\\
 \begin{definition} \cite{9} \label{exd}  let $X$ be a nonempty set. We call a mapping $T$ is a \textit{$C^{*}$-algebra-valued expansion mapping on $X$}, if $T:X\rightarrow X$ satisfies
\begin{itemize}
\item[(1)] $T(X)=X$;
\item[(2)] $d(Tx,Ty)\geq B^{*}d(x,y)B,\hspace{0.5cm}\text{for all }x,y\in X$,
\end{itemize}
 where $B\in\Bbb{A}$ is an invertible element and $\|B^{-1} \| < 1$.
\end{definition}
 %%%%%%%%%%%%%%%%%%%%% 
\begin{theorem} Consider a complete $C^{*}$-algebra-valued $b_{v}(s)$-metric space $(X,\Bbb{A},d)$  with coefficient $s$.  Let $T: X\rightarrow X $  be a $C^{*}$-algebra-valued  expansion mapping with constant $B$.  If there exists a natural number $n_{0}$ such that $(B^{-1})^{n_{0}}\in \Bbb{A'}$ and $s((B^{-1})^{*})^{n_{0}}(B^{-1})^{n_{0}}<1_{\Bbb{A}}$, then $T$  has a unique fixed point in $X$. 
\end{theorem}
\begin{proof} First, we show that $T$ is invertible. Since by condition (1) of Definition \ref{exd}, $T$ is surjective, it is enough  to show that $T$ is injective. Indeed, for any $x,y\in X$ with $x\neq y$, if $T(x)=T(y)$, we have
 \begin{eqnarray*}
 0_{\Bbb{A}}=d(Tx,Ty)\geq B^{*}d(x,y)B.
 \end{eqnarray*}
 Since $B^{*}d(x,y)B\in\Bbb{A_{+}}$,  therefore   $B^{*}d(x,y)B=0_{\Bbb{A}}$.   On the other hand,  $B$ is invertible, then  $d(x,y)=0_{\Bbb{A}}$, which is impossible. Thus $T$ is injective.\\
 
%\begin{eqnarray*}
%0_{\Bbb{A}}=d(Tx,Ty)\geq B^{*}d(x,y)B,\hspace*{0.5cm}   \forall x,y\in X,
 %\end{eqnarray*}
%and since $B$ is invertible so $B\neq 0_{A}}$. Therefore it is necessery $d(x,y)=0_{\Bbb{A}}$ and so $x=y$ this is contradiction, hence $T$ is invertible.\\
Next, we will show that $T$ has a unique fixed point in $X$. In fact, since $T$ is invertible, for any  $x,y\in X$, it follows that
\begin{eqnarray*}
d(Tx,Ty)\geq B^{*}d(x,y)B.
\end{eqnarray*}
 In the above formula, we replace $x$ and $y$ by $T^{-1}(x)$ and $T^{-1}(y)$, respectively, and we get
\begin{eqnarray*}
d(x,y)\geq B^{*}d(T^{-1}x,T^{-1}y)B.
 \end{eqnarray*}
 Now by part (1) of Lemma \ref{in1}, we have 
\begin{eqnarray*}
 (B^{-1})^{*}d(x,y)B^{-1}& \geq& (B^{-1})^{*} B^{*}d(T^{-1}x,T^{-1}y)B B^{-1}\\
 &=& (B^{*})^{-1} B^{*}d(T^{-1}x,T^{-1}y)B B^{-1}\\
 &=&d(T^{-1}x,T^{-1}y).
 \end{eqnarray*}
% Finally we get:\\
%\begin{eqnarray*}
%d(T^{-1}x,T^{-1}y)\leq (B^{-1})^{*}d(x,y)B^{-1}.
% \end{eqnarray*}
Using Theorem \ref{bvs}, there exists unique $x^*\in X$ such that $T^{-1}x^{*}=x^{*}$, which means that there is a unique fixed point $x^*\in X$ such that $Tx^{*}=x^*$.  %$TT^{-1}x^{*}= it means $T$ has unique fixed point in $X$.\blacksquare
 \end{proof} 
\\
 In the following theorem, we prove Jungck's theorem in $ C^{*} $-algebra-valued $b_{v}(s)$-metric spaces.
\begin{theorem}\label{j}
Consider $(X,\Bbb{A},d)$ is a complete $ C^{*} $-algebra-valued $b_{v}(s)$-metric space with coefficient $s$. Let $T$ and $I$ be commuting mappings of $X$ into itself such that    the range of $I$ contains the range of $T$ and $I$ is continuous and satisfies the inequality
\begin{eqnarray}\label{j1}
d(Tx,Ty)\leq B^{*}d(Ix,Iy)B\hspace{0.5cm}    \text{for all }x,y\in X,
\end{eqnarray}
where $B\in\Bbb{A}$ with $\|B\|<1$. If   there exists a  natural number $n_{0}$ such that $s(B^*)^{n_{0}}B^{n_{0}}< 1_{\Bbb{A}}$ and $B^{n_{0}}\in\Bbb{A'}$.
  Then $T$ and $I$ have a unique common fixed point.
\end{theorem}
\begin{proof}
Let $x_{0}\in X$ be arbitrary. Then $Tx_{0}$ and $Ix_{0}$ are well-defined. Since  $Tx_{0}\in I(X)$, there is $x_{1}\in X$ such that $Ix_{1}=Tx_{0}$. 
 In general,  if $x_{n}$ is chosen, then we choose a point $x_{n+1}$ in $X$  such that $I x_{n+1}=Tx_{n}$.
 Now, we show that $\{I x_{n}\}$ is Cauchy. From (\ref{j1}), for all $m,n\in\Bbb{N}$, we have
\begin{eqnarray}\label{j2}
 d(Ix_{m},Ix_{n})=d(Tx_{m-1},Tx_{n-1})\leq B^{*}d(Ix_{m-1},Ix_{n-1})B.
\end{eqnarray}
Now, we have the following two cases.\\
\textbf{Case 1} If $Ix_{n}=Ix_{n+1}$ for some $n\geq0$, then $Ix_{n}=Ix_{n+1}=Tx_{n}=\omega$. We show that $\omega$ is a unique common fixed point of $T$ and $I$.  Since $T$ and $I$ commute, thus $I\omega=I(Tx_{n})=T(Ix_{n})=T\omega$.\\
Now, let $d(T\omega,\omega)>0_{\Bbb{A}}$. Hence
\begin{eqnarray*}
d(T\omega,\omega)=d(T\omega,Tx_{n})\leq B^{*}d(I\omega,Ix_{n})B= B^{*}d(T\omega,\omega)B.    
\end{eqnarray*}
%Since right and left above inequality are positive elements and $\|B\|=\|B^{*}\|<1$ we can write:\\
Using the norm of $\Bbb{A}$, we have
\begin{eqnarray*}
\|d(T\omega,\omega)\|<\|d(T\omega,\omega)\|.
\end{eqnarray*}
 This is a contradiction. Thus $\|d(T\omega,\omega)\|=0$,   $d(T\omega,\omega)=0_{\Bbb{A}}$, and     $T\omega=\omega=I\omega$. By  condition (\ref{j1}),  $\omega$ is a unique common fixed point of $T$ and $I$.\\
\textbf{Case 2} Now suppose that $Ix_{n}\neq Ix_{n+1}$ for all $n\geq0$. From Lemma \ref{bvsl2} and  inequality (\ref{j2}), we have  $Ix_{n}\neq Ix_{n+p}$ for all $n\geq0$ and $ p\geq1$. With a similar argument used in the proof of Theorem \ref{bvs}, we can prove that  the sequence  $\{Ix_{n}\}$ is Cauchy. Since the $ C^{*} $-algebra-valued $b_{v}(s)$-metric space $(X,\Bbb{A},d)$ is complete,  so  $\{Ix_{n}\}$  converges to $u\in X$ such that
\begin{eqnarray*}
 \displaystyle{\lim_{n\rightarrow\infty}}Ix_{n}={\lim_{n\rightarrow\infty}}Tx_{n-1}=u. 
 \end{eqnarray*}
Since $I$ is continuous, inequality (\ref{j1}) implies that both $I$ and  $T$ are continuous.
  Since $T$ and $I$ commute, we obtain
 \begin{eqnarray*}
 Iu=I(\displaystyle{\lim_{n\rightarrow\infty}}Tx_{n-1})=I({\lim_{n\rightarrow\infty}}Tx_{n})=
 \displaystyle{\lim_{n\rightarrow\infty}}ITx_{n}
 =\displaystyle{\lim_{n\rightarrow\infty}}TIx_{n}=T({\lim_{n\rightarrow\infty}}Ix_{n})=Tu.\\
\end{eqnarray*}
 Let $Tu=Iu=\nu$. Thus $T\nu=TIu=ITu=I\nu$.\\
If $Tu\neq T\nu$, then from (\ref{j1}), we get
\begin{eqnarray*}
\|d(Tu,T\nu)\|&\leq&\|B^{*}d(Iu,I\nu)B\|=\|B^*d(Tu,T\nu)B\|\\
&\leq&\| B^{*}\|\|d(Tu,T\nu)\|\|B\|\\
&<& \|d(Tu,T\nu)\|.
\end{eqnarray*}
This is a contradiction. So $\|d(Tu,T\nu)\|=0$,   $d(Tu,T\nu)=0_{\Bbb{A}}$, and    $Tu=T\nu$.  Thus,  we obtain $T\nu=I\nu=\nu$.\\
Now, we claim $\nu$ is the unique common fixed point for $T$ and $I$.\\
Let $\nu^{*}(\neq \nu)$ be another fixed point  for $T$ and $I$.  By inequality (\ref{j1}), we have
\begin{eqnarray*}
d(\nu,\nu^{*})=d(T\nu,T\nu^{*})\leq B^{*} d(I\nu,I\nu^{*})B.
\end{eqnarray*}
Now, by using the norm of $\Bbb{A}$, we have
\begin{eqnarray*}
\|d(\nu,\nu^{*})\|=\|d(T\nu,T\nu^{*})\|&\leq&\|B^{*} d(I\nu,I\nu^{*})B\|\\
&\leq&\|B^{*}\|\ d(I\nu,I\nu^{*})\|\|B\|\\
&<&\|d(I\nu,I\nu^{*})\|=\|d(\nu,\nu^{*})\|.
\end{eqnarray*}
This is a contradiction,  which  implies that $\nu=\nu^{*}$.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{remark}  In Theorem \ref{j}, if $I$ is the identity map on $X$,  then, Theorem \ref{bvs} holds.
\end{remark}
\section{Application}
In this section, we give an existence theorem for a solution of the following
integral equation.
\begin{equation}\label{ap1}
x(t)=\int _{E}K(t,s,x(s))ds+g(t), ~~~t\in E,
\end{equation} 
where $K:E\times E\times\Bbb{R}\rightarrow \Bbb{R}$ and $g\in C_{\Bbb{R}}(E)$.\\
Let $X=C_{\Bbb{R}}(E)$ be the set of all real valued continuous functions on E, where E is a nonempty Lebesgue measurable
compact set in $\Bbb{R}_{+}$. Also,  $\Bbb{A}=L(H)$ is the set of all bounded linear operators on $H=L^{2}(E)$ with usual operator norm. We define $d':X\times X\rightarrow\Bbb{R}_{+}$ by $d'(x,y)=\displaystyle\sup_{t\in E}(x(t)-y(t))^{2}$ for all $x,y\in X$.  Then,  $(X, d')$ is a complete $b_{2}(3)$-metric space.
 Moreover, $\Pi_{\gamma}:H\rightarrow H$ is defined by $\Pi_{\gamma}(h)=\gamma .h$ for all $\gamma\in\Bbb{C}$ and $h\in H$. Now,   define  $d:X\times X\rightarrow\Bbb{A}_
{+}$ by $d(x,y)=\Pi_{d'(x,y)}$. It is clear that $(X,\Bbb{A},d)$ is a complete  $C^{*}$-algebra-valued $b_{v}(s)$-metric space with $v=2$  and  $s=3I$.
We assume that the following
conditions are satisfied:\\
$(i)$ There exists a continuous function $ f :E\times E\rightarrow \Bbb{R}$ such that 
$$|K(t,s,u)-K(t,s,v)|\leq \alpha|f(t,s)(u-v)|,$$ 
for $t,s\in E,$ $\alpha\in(0,1)$  and $u,v\in\Bbb{R}$.
\\$(ii)$  It follows that $\displaystyle{\sup_{t\in E}}\int_{E}|f(t,s)|ds\leq1 
\hspace{0.3cm}$ for any $t,s\in E$.
\begin{theorem}\label{apT}
Under the assumptions $(i)$ and $(ii)$ equation (\ref{ap1}) has a unique solution in $X$
\end{theorem}
\begin{proof}
Let $T :X \rightarrow X$ be defined by $Tx(t)=\int _{E}K(t,s,x(s))ds+g(t)$, $t\in E$.  Then 
\begin{eqnarray*}
\|d(Tx,Ty)\|&=&\|\Pi_{d'(Tx,Ty)}\|=\displaystyle{\sup_{\|h\|=1}}\big\langle\Pi_{d'(Tx,Ty)}( h ) , h \big\rangle  ;h\in H\\
&=&\displaystyle{\sup_{\|h\|= 1}}\int_{E}d'(Tx,Ty)h(u)\overline{h}(u)d(u)    ;  u\in E\\
&=&\displaystyle{\sup_{\|h\|=1}}\int_{E} {\sup_{t\in E}}\Big[Tx(t)-Ty(t)\Big]^{2} h(u)\overline{h}(u)d(u)   ;  u\in E\\
&=&\displaystyle{\sup_{\|h\|=1}}\int_{E} \sup_{t\in E}\Big[\int_{E}\big [K\big(t,s,x(s)\big)-K\big(t,s,y(s)\big)\big]ds\Big]^{2}|h(u)|^{2}du   ;  u\in E\\
&\leq&\displaystyle{\sup_{\|h\|=1}}\int_{E}\sup_{t\in E}\Big[\int_{E}\alpha| f(t,s)|\big(x(s)-y(s)\big)ds\Big]^{2} |h(u)|^{2}du  ;  u\in E\\
&=&\alpha^{2}d'(x,y)\displaystyle{\sup_{\|h\|=1}}\int_{E}{\sup_{t\in E}}\Big[\int_{E}|f(t,s)|ds\Big]^{2} |h(u)|^{2}du ;   u\in E\\
&\leq& \alpha^{2}d'(x,y) \displaystyle\sup_{\|h\|=1}\int_{E} |h(u)|^{2}du  ;  u\in E\\
&=&\alpha ^{2} \displaystyle\sup_{\|h\|=1}\int_{E}d'(x,y) |h(u)|^{2}du  ;  u\in E\\
&=&\alpha^{2}\|d(x,y)\|.
\end{eqnarray*} 
 By  take   $B=\alpha1_{\Bbb{A}}$, then $\|B\|<1$.   Using Theorem \ref{bvs},  the  integral equation (\ref{ap1}) has a  unique solution in $X$.
 \end{proof}
 \begin{example}\label{exam11} Consider the following functional integral equation:
% \begin{eqnarry}
 \begin{equation}
 x(t)=\int _{0}^{1}\frac{4e^{-(t+1)s}}{3({(t+1)^2}+2)}\hspace{0.1cm}\frac{|x(s)|}{1+|x(s)|} ds+t
\end{equation}
for $t\in E=[0,1]$. Observe that this equation is a special case of (\ref{ap1}) with
\begin{itemize}
%\begin{equation*}
\item[] $K(t,s,x(s))=\frac{4e^{-(t+1)s}}{3({(t+1)^2}+2)}\hspace{0.1cm}\frac{|x(s)|}{1+|x(s)|}$,
\item[] $f(t,s)=\frac{4e^{-(t+1)s}}{(t+1)^{2}+2}$,
\item[] $g(t)=t$.
\end{itemize}
Notice that, for arbitrary fixed numbers $u, v\in \Bbb{R}$ and  $t, s\in E=[0,1]$, we have\\
\begin{eqnarray*}
|K(t,s,u)-K(t,s,v)|&=&\big|\frac{4e^{-(t+1)s}}{3({(t+1)^2}+2)}\ \frac{|u|}{1+|u|}-\frac{4e^{-(t+1)s}}{3({(t+1)^2}+2)}\ \frac{|v|}{1+|v|}\big|\\
&\leq&\frac{1}{3} |\frac{4e^{-(t+1)s}}{(t+1)^{2}+2}||u-v|.\end{eqnarray*}
Thus the function $K$ satisfies the assumption $(i)$ with $\alpha=\dfrac{1}{3}$.\\
Also, we have\\
\begin{footnotesize}
$\displaystyle{\sup_{ 0\leq t\leq1}}\int_{0}^{1}|f(t,s)|ds=\displaystyle{\sup_{0\leq t\leq1}}\int_{0}^{1}|\frac{4e^{-(t+1)s}}{(t+1)^{2}+2}|ds=\displaystyle{\sup_{ 0\leq t\leq1}}\frac{4}{(t+1)^{2}+2}\int_{0}^{1}e^{-(t+1)s}ds<1$.
\end{footnotesize}
This shows that the assumption $(ii)$  holds. Consequently, all the conditions of Theorem \ref{apT} are satisfied. Hence the integral equation (\ref{exam11}) has a unique solution in $C_{\Bbb{R}}(E)$.   
 \end{example}
 \section{Iterative method for   solving  integral equation}
 \begin{theorem}
 
 Consider the integral equation (\ref{ap1}). The following iteration process leads to the fixed point (function) solution  of (\ref{ap1})
 \begin{equation}\label{ap4}
 x_{n+1}(t)=\int_{E}K(t, s, x_{n}(s))ds+g(t),\quad t\in E,
 \end{equation}
 where the initial guess $x_{0}(t)$ can be any arbitrary function such as $0,\ 1,$ or  $ t$. 
 \end{theorem}
 \begin{proof}
 Assume that the exact solution of (\ref{ap1}) is $ \tilde{x}(t)$.\\
 We have
 \begin{eqnarray*}
| x_{1}(t)-\tilde{x}(t)|&=&|\int_{E}\big(K(t, s, x_{0}(s)-K(t, s, \tilde{x}(s)\big)ds|\\&\leq& \int_{E}\alpha|f(t,s)||x_{0}(s)-\tilde{x}(s)|ds\\&\leq&\alpha M,
 \end{eqnarray*}
 where $M=\max|x_{0}(s)-\tilde{x}(t)|,\hspace{0.1cm}t\in E$. One can show similarly  that
 \begin{eqnarray*}
 |x_{2}(t)-\tilde{x}(t)|&\leq&\alpha\int_{E}|f(t,s)||x_{1}(s)-\tilde{x}(s)|ds\\&\leq&\alpha^{2}M\int_{E}|f(t, s)|ds\\&\leq&\alpha^{2}M.
 \end{eqnarray*}  Finally, 
 \begin{equation*}
 |x_{n+1}(t)-\tilde{x}(t)|\leq\alpha^{n+1}M.
 \end{equation*}
 It is clear that when $n$ tends to infinity, $x_{n+1}(t)$ tends to the exact solution $\tilde{x}(t)$.
 \end{proof}
\\ Consider the integral equation (\ref{ap4}), we set 
 \begin{equation*}
 H(x_{n}(t))=\int_{E}K\big(t,s,x_{n}(s)\big)ds+g(t),
 \end{equation*}
   so the integral equation (\ref{ap4}) can be rewritten as follows:
   \begin{equation*}
   x_{n+1}(t)=H(x_{n}(t)).
   \end{equation*}
 It is clear that the exact solution $\tilde{x}(t)$ satisfies
 \begin{equation*}
 \tilde{x}(t)=H(\tilde{x}(t))
 \end{equation*}
 and $|\tilde{x}(t)-H(\tilde{x}(t))|=0$.\\
 Now in order to start the iterations for Example \ref{exam11}, we consider $x_{0}(t)=0$ and do four iterations  according to relation (\ref{ap4}) to obtain
 $x_{4}(t)$. we have used Maple 2018 to plot
 \begin{equation*}
| x_{4}(t)-x_{3}(t)|=|H(x_{3}(t))-x_{3}(t)|
 \end{equation*}
 in Figure \ref{fig1}, which shows  small errors between $x_{3}(t)$ and $x_{4}(t)$, and it can be considered as a good approximation for the exact solution $\tilde{x}(t)$.  In Figure \ref{fig2}, we have plotted $x_{4}(t)$ in the interval $[0,1]$.
 \newpage
 \begin{figure}[t]
  \centering
  \includegraphics[width=2.75in,height=1.75in]{fig1}
  \caption{graph of  $| x_{4}(t)-x_{3}(t)|$}\label{fig1}
\end{figure}
\begin{figure}[t]
  \centering
  \includegraphics[width=2.75in,height=1.75in]{fig2}
  \caption{graph of  $x_{4}(t)$}\label{fig2}
\end{figure}
 

%\vspace{4mm}\noindent{\bf Acknowledgements}\\
%\noindent The authors wish to thank the anonymous referee(s) for reading the article so carefully and for the comments made on this article.

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\begin{thebibliography}{99} % Enter references in alphabetical order and according to the following format.
\bibitem{1} S. Aleksi\'{c}, Z.D. Mitrovi\'{c}, S. Radenovi\'{c}, \textit{A fixed point theorem of Jungck in $b_{v}$(s)-metric spaces}, Period. Math. Hung. 77, \textbf{(2018)}, 224--231.
\bibitem{2} I.A. Bakhtin,  \textit{The contraction mapping principle in quasimetric spaces}, Funct. Anal. Unianowsk Gos. Ped. Inst. 30,  \textbf{(1989)}, 26--37.
\bibitem {3} S.  Banach,   
\textit{ Sur les operations dans les ensembles abstraits et leurs applications aux equations integrals}, Fundam. Math. 3, \textbf{(1922)}, 133--181.  
\bibitem{4} S. Batul, T. Kamran, \textit{$C^*$-valued contractive type mapping}, Fixed Point Theory Appl. 2015, 142 \textbf{(2015)}, 9 pp.
\bibitem{5} A. Branciari, \textit{A fixed point theorem of Banach-Caccioppoli type on a class of generalized metric spaces},
Publ. Math. Debrecen, 57, \textbf{(2000)}, 31--37. 
\bibitem {6} RG. Douglas, \textit{Banach Algebra Techniques in Operator Theory}., Springer, Berlin \textbf{(1998)}.
\bibitem{7} T. Kamran, M. Postolache, A. Ghiura, S. Batul, R. Ali, \textit{The Banach contraction principle in $C^*$-algebra-valued $b$-metric spaces with application},  Fixed Point Theory Appl. 2016, 10 \textbf{(2016)}, 1--7.
\bibitem{8} C. Klin-eama, P. Kaskasemay, \textit{Fixed point theorems for cyclic contractions in $C^*$-algebra-valued $b$-metric spaces}, J. Funct. Spaces,  2016, Article ID 7827040,  \textbf{(2016)}, 16 pp.
%\bibitem{3} S.  Cezerwc,k ; \textit{Contraction mapping in b-metric spaces,}
%Acta. Math. Inform Univ. Ostrav, \textbf{(1993)}, 1,5-11
\bibitem{10} Z. Ma, L. Jiang,  \textit{$C^*$-algebra-valued $b$-metric spaces and related fixed point theorems}, Fixed Point Theory Appl. 2015, 222 \textbf{(2015)}, 1--12.
\bibitem {9} Z.  Ma, L. Jiang, H. Sun, \textit{$C^{*}$-algebra-valued metric spaces and related fixed point Theorems},                       Fixed  Point Theory  Appl. 2014, 206 \textbf{(2014)}, 1--11.

 \bibitem {12} Z.D. Mitrovic, S. Radenovic,   \textit{The Banach and Reich contractions in $b_{v}$(s)-metric spaces}. J. Fixed Point Theory Appl. 19. \textbf{(2017)}, 3087--3095. 
 \bibitem{11} G.J.  Murphy,  \textit{$C^{*}$-Algebras and Operator Theory}, Academic Press, London  \textbf{(1990)}.
% \bibitem{12} Z.  Ma,  L. Jiang, \textit{$C^{*}$-algebra-valued metric spaces and related fixed point Theorems}. Fixed   Point Theory  Appl. 2015:222 \textbf{(2015)}   
 
\bibitem{13} D. Shehwar, T. Kamran, \textit{$C^*$-valued $G$-contractions and fixed points}. J. Inequal. Appl. 2015, 304 \textbf{(2015)}, 1--8.


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\end{center}



{\small

\noindent{\bf Mohammad  Hassan Saboori}

\noindent Department of Mathematics

\noindent Graduated Phd. of Mathematics 

\noindent  Mashhad Branch, Islamic Azad University, Mashhad, Iran.

%\noindent Mashhad, Iran.

\noindent E-mail: mhs72859@gmail.com}\\

{\small
\noindent{\bf  Mahmoud Hassani  }

\noindent  Department of Mathematics

\noindent Associate Professor of Mathematics

\noindent Mashhad Branch, Islamic Azad University, Mashhad, Iran.


%\noindent Mashhad, Iran.


\noindent E-mail: hassani@mshdiau.ac.ir}\\

{\small
\noindent{\bf  Reza  Allahyari}

\noindent  Department of Mathematics

\noindent Assistant Professor of Mathematics

\noindent Mashhad Branch, Islamic Azad University, Mashhad, Iran.


%\noindent Mashhad, Iran.


\noindent E-mail: rezaallahyari@mshdiau.ac.ir}\\

{\small
\noindent{\bf  Mohammad Mehrabinezhad}

\noindent  Department of Mathematics

\noindent Assistant Professor of Mathematics

\noindent Mashhad Branch, Islamic Azad University, Mashhad, Iran.


%\noindent Mashhad, Iran.


\noindent E-mail: mmehrabinezhad@gmail.com}\\




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