\documentclass[11pt,twoside]{article} \usepackage{amsmath, amsthm, amscd, amsfonts, amssymb, graphicx, color} \usepackage[bookmarksnumbered, colorlinks]{hyperref} \usepackage{float} \usepackage{lipsum} \usepackage{afterpage} \usepackage[labelfont=bf]{caption} \usepackage[nottoc,notlof,notlot]{tocbibind} %\renewcommand\bibname{References} \def\bibname{\Large \bf References} \usepackage{lipsum} \usepackage{fancyhdr} \pagestyle{fancy} \fancyhf{} \renewcommand{\headrulewidth}{0pt} \fancyhead[LE,RO]{\thepage} \thispagestyle{empty} %\afterpage{\lhead{new value}} \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} %\topmargin=-1.6cm \textheight 17.5cm% \textwidth 12cm % \topmargin 8mm % \oddsidemargin 20mm % \evensidemargin 20mm % \footskip=24pt % \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{xca}[theorem]{Exercise} %\theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \renewenvironment{proof}{{\bfseries \noindent Proof.}}{~~~~$\square$} \makeatletter \def\th@newremark{\th@remark\thm@headfont{\bfseries}} \makeatletter %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % If you want to insert other packages. Insert them here %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %\long\def\symbolfootnote[#1]#2{\begingroup% %\def\thefootnote{\fnsymbol{footnote}}\footnote[#1]{#2}\endgroup} \def \thesection{\arabic{section}} \begin{document} %\baselineskip 9mm %\setcounter{page}{} \thispagestyle{plain} {\noindent Journal of Mathematical Extension \\ Vol. XX, No. XX, (2014), pp-pp ()}\\ ISSN: 1735-8299\\ URL: http://www.ijmex.com\\ %\vspace*{9mm} \begin{center} {\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\\ %%%%%%%%%%%%5555555555555555555555555555555555555\\ %%%%%%%%%%%%%%5555555555555555555555555555555555555555555\\ \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 \\ %%%%555555555555555555555555555555555555555555555555555555555\\ \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. % BibTeX users please use one of %\bibliographystyle{spbasic} % basic style, author-year citations %\bibliographystyle{spmpsci} % mathematics and physical sciences %\bibliographystyle{spphys} % APS-like style for physics %\bibliography{} % name your BibTeX data base % Non-BibTeX users please use \begin{center} \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. % Format for books % etc \end{thebibliography} \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}\\ \end{document}