\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2010(2010), No. 144, pp. 1--20.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2010 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2010/144\hfil Transport equation] {Transport equations in cell population dynamics I } \author[M. Boulanouar \hfil EJDE-2010/144\hfilneg] {Mohamed Boulanouar} \address{Mohamed Boulanouar \newline LMCM-RMI, Universite de Poitiers\\ 86000 Poitiers, 86000 Poitiers, France} \email{boulanouar@free.fr} \thanks{Submitted May 23, 2010. Published October 12, 2010.} \thanks{Supported by LMCM-RMI} \subjclass[2000]{92C37, 82D75} \keywords{Semigroups; operators; boundary value problem; \hfill\break\indent cell population dynamic; general boundary condition} \begin{abstract} In this article, we study a cell proliferating model, where each cell is characterized by its degree of maturity and its maturation velocity. The boundary conditions in this model generalize the known biological rules. We consider also the degenerate case corresponding to infinite maturation velocity. Then we show that this model is governed by a strongly continuous semigroup and give its explicit expression. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{definition}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \newcommand{\abs}[1]{|#1|} \newcommand{\norm}[1]{\|#1\|} \newcommand{\norme}[1]{|||#1|||} \newcommand{\pare}[1]{\left(#1\right)} \newcommand{\cro}[1]{\left[#1\right]} \section{Introduction} We consider a cell population in which each cell is distinguished by its degree of maturity $\mu$ and its maturation velocity $v$. At the birth, the degree of maturity of each (daughter) cell is null ($\mu=0$) and at the division, the degree of maturity of each (mother) cell becomes $\mu=1$. Between the birth and the division of each cell, its degree of maturity is $0<\mu<1$. As each cell may not become less mature, then its maturation velocity $v$ must be positive $(0\le a1$. When $01$ corresponding to an increasing cell density. In this work, we are concerned with the case $0\omega$ we have $\lambda\in\rho(A)$ and $$\norm{(\lambda-A)^{-n}}_{\mathcal{L}(X)} \le M(\lambda-\omega)^{-n}$$ for all $n\in\mathbb{N}$. \end{enumerate} \end{lemma} \begin{lemma}[{\cite[Theorem II.3.15]{Nagel}}] \label{L:PHI-LUM} Let $(A,D(A))$ be a densely defined linear operator on a Banach space $X$. If $A$ is dissipative and the range $rg(\lambda-A)=X$ for some $\lambda>0$, then $A$ generates a strongly continuous semigroup of contractions. \end{lemma} Section~\ref{S:CONS} deals with the explicit expression of the unperturbed semigroup. This expression will be very useful to describe the asymptotic behavior which is the main goal of \cite{Boulanouar3}. The end of this work concerns the generation theorem for the perturbed model \eqref{E:MODEL}-\eqref{E:CAL}, where we have applied the following perturbation results \begin{lemma}[{\cite[Theorem III.1.3]{Nagel}}] \label{L:PER} Let $(A,D(A))$ be the infinitesimal generator of a strongly continuous semigroup $(T(t))_{t\ge0}$ on a Banach space $X$ and let $B$ be a linear bounded operator from $X$ into itself. Then, the operator $C:=A+B$ on the domain $D(C):=D(A)$ generates a strongly continuous semigroup $(S(t))_{t\ge0}$ given by Trotter's formula $$\label{E:TRO} S(t)x= \lim_{n\to\infty}\cro{e^{-\frac{t}{n}B}T\pare{\tfrac{t}{n}}}^nx \quad t\ge0,$$ for all $x\in X$. \end{lemma} \begin{lemma}[{\cite[Theorem II.2.7]{Nagel}}] \label{L:PER1} Let $(A,D(A))$ be the generator of a strongly continuous semigroup $(T(t))_{t\ge0}$ of contractions on a Banach space $X$ and let $B$ be a dissipative operator satisfying $D(A)\subset D(B)$ and $$\norm{Bx}\le a\norm{Ax}+b\norm{x}$$ for all $x\in D(A)$, where, $0\le a <1$ and $b \ge 0$. Then, $A+B$ is the infinitesimal generator of a strongly continuous semigroup of contractions. \end{lemma} Finally, some of these results were announced in \cite{Boulanouar4} and here we explicitly state the detailed conditions and outline all the proofs. For all theoretical results used here, we refer the reader to \cite{Nagel}. \section{The unperturbed model ($r=0$)}\label{S:N-PER} In this section, we are going to study the unperturbed model \eqref{E:MODEL}-\eqref{E:CAL} (i.e. $r=0$). So, let us consider the following functional framework $L^1(\Omega)$ whose natural norm is $$\label{NORM} \norm{\varphi}_1= \int_\Omega\abs{\varphi(\mu,v)}\,d\mu\,dv$$ where, $\Omega=(0,\;1)\times(a,\;\infty):=I\times J$. We emphasize that we are only concerned with the following important assumption $$\label{A>0} a>0$$ until the end of this work. We consider also the partial Sobolev space $W^1(\Omega)=\big\{\varphi\in L^1(\Omega),\; v\frac{\partial\varphi}{\partial \mu}\in L^1(\Omega) \text{ and } v\varphi\in L^1(\Omega)\big\}$ whose norm is $\norm{\varphi}_{W^1(\Omega)}=\big\|v\frac{\partial\varphi}{\partial\mu}\big\|_1 +\norm{v\varphi}_1.$ Finally, we consider the trace space $Y_1:=L^1(J, vdv)$ endowed with the norm $\norm{\psi}_{Y_1}= \int_a^\infty\abs{\psi(v)}vdv.$ \begin{lemma}[\cite{Boulanouar0}]\label{TRACES:LEM} The trace mappings $\gamma_0\varphi=\varphi(0,\cdot)$ and $\gamma_1\varphi=\varphi(1,\cdot)$ are linear bounded from $W^1(\Omega)$ into $Y_1$. \end{lemma} Thanks to Lemma above, it is easy to check the following useful lemma. \begin{lemma}\label{A0:LEM} Let $A_0$ be the following unbounded operator $$\label{A0:R1} \begin{gathered} A_0\varphi=-v\frac{\partial \varphi}{\partial\mu}\text{ on the domain}\\ D(A_0)= \{\varphi\in W^1(\Omega),\; \gamma_0\varphi=0\}. \end{gathered}$$ \begin{enumerate} \item The operator $A_0$ generates, on $L^1(\Omega)$, a positive strongly continuous semigroup $(U_0(t))_{t\ge0}$ of contractions given by $$\label{A0:R2} U_0(t)\varphi(\mu,v)=\chi(\mu,v,t)\varphi(\mu-tv,v)\quad t\ge0,$$ where, \label{A0:R3} \chi(\mu,v,t)= \begin{cases} 1 & \text{if}\quad \mu\ge tv,\\ 0 & \text{if}\quad \mu0$. Then, for all$g\in L^1(\Omega)$, we have $$\label{A0:R5} \norm{v(\lambda-A_0)^{-1}g}_1\le \norm{g}_1.$$ \item$t\to\gamma_1U_0(t)\varphi\in Y_1$is a continuous mapping with respect to$t\ge0$. \end{enumerate} \end{lemma} Now, let us consider a linear boundary operator$K$from$Y_1$into itself until the end of this work. This leads to write the boundary condition \eqref{E:CAL} as $$\label{BC:DEF} \gamma_0\varphi=K\gamma_1\varphi$$ and allows us to give a sense, by Lemma~\ref{TRACES:LEM}, to the following unbounded operator $$\label{AK:DEF} \begin{gathered} A_K\varphi=-v\frac{\partial \varphi}{\partial\mu}\text{ on the domain}\\ D(A_K)= \{\varphi\in W^1(\Omega), \text{ satisfying }\gamma_0\varphi=K\gamma_1\varphi\} \end{gathered}$$ To study$A_K$, let us define the operator $$\label{KLAMBDA:DEF} K_\lambda:=\theta_\lambda K, \quad\text{where}\quad\theta_\lambda(\cdot)=e^{-\frac{\lambda}{\cdot}}$$ which plays an important role in the sequel. \begin{lemma}\label{AK:LEM1} Suppose$K$is bounded satisfying$\norm{K}_{\mathcal{L}(Y_1)}<1$. Then \begin{enumerate} \item For all$\lambda\ge0$,$K_\lambda$is a linear bounded operator from$Y_1$into itself satisfying$\norm{K_\lambda}_{\mathcal{L}(Y_1)}<1$. \item For all$\lambda>0$, the resolvent operator of \eqref{AK:DEF} is given by $$\label{AK:R1} (\lambda-A_K)^{-1}g= \varepsilon_\lambda K(I- K_\lambda)^{-1}\gamma_1(\lambda - A_0)^{-1}g +(\lambda -A_0)^{-1}g$$ for all$g\in L^1(\Omega)$, where$\varepsilon_\lambda(\mu,v)=e^{-\lambda\frac{\mu}{v}}$. \item The operator defined by \eqref{AK:DEF} generates, on$L^1(\Omega)$, a strongly continuous semigroup$(U_K(t))_{t\ge0}$satisfying $$\label{AK:R2} \norm{U_K(t)\varphi}_1 \le\norm{\varphi}_1 \quad t\ge0,$$ for all$\varphi\in L^1(\Omega)$. \end{enumerate} \end{lemma} \begin{proof} (1) This point clearly follows from$\norm{K_\lambda}_{\mathcal{L}(Y_1)}\le\norm{K}_{\mathcal{L}(Y_1)}<1$for all$\lambda\ge0$. (2) Let$\lambda>0$and$g\in L^1(\Omega)$. The general solution of $$\label{AK:E1} \lambda\varphi=-v\frac{\partial \varphi}{\partial \mu}+g$$ is given by $$\label{AK:E2} \varphi=\varepsilon_\lambda\psi+(\lambda-A_0)^{-1}g$$ where$\psi$is any function of the variable$v$. If$\psi\in Y_1$then, by \eqref{A0:R5} we clearly have $\norm{v\varphi}_1 \le\norm{v\varepsilon_\lambda\psi}_1 +\norm{v(\lambda-A_0)^{-1}g}_1 \leq \norm{\psi}_{Y_1}+\norm{g}_1<\infty$ which implies, by \eqref{AK:E1}, that $\big\|v\frac{\partial \varphi}{\partial \mu}\varphi\big\|_1 \le\lambda\norm{\varphi}_1+\norm{g}_1 \le\frac{\lambda}{a}\norm{v\varphi}_1+\norm{g}_1<\infty$ and therefore$\varphi\in W^1(\Omega)$. Now,$\varphi\in D(A_K)$if and only if$\varphi$satisfies \eqref{BC:DEF}. This means that$\psi$is the unique solution of the following system $$\label{AK:E3} \begin{gathered} \psi =K\gamma_1\varphi\\ \gamma_1\varphi =K_\lambda\gamma_1\varphi+\gamma_1(\lambda-A_0)^{-1}g. \end{gathered}$$ As$\lambda>0$, then the first point implies that$(I-K_\lambda)$is invertible into$Y_1$and therefore $$\psi=K(I-K_\lambda)^{-1}\gamma_1(\lambda-A_0)^{-1}g$$ which we put in \eqref{AK:E2} to get \eqref{AK:R1}. (3) Let$\varphi\in D(A_K). Then, we have \begin{align*} \big\langle\operatorname{sgn}\varphi,\; A_K\varphi\big\rangle &=-\int_\Omega\pare{\operatorname{sgn}\varphi(\mu,v)} \Big(v\frac{\partial\varphi}{\partial\mu}(\mu,v)\Big)\,d\mu\,dv\\ &=-\int_0^1\int_a^\infty v\frac{\partial\abs{\varphi}}{\partial\mu}(\mu,v)\,d\mu\,dv\\ &=\int_a^\infty\abs{\varphi(0,v)}vdv -\int_a^\infty\abs{\varphi(1,v)}vdv\\ &=\norm{\gamma_0\varphi}_{Y_1}-\norm{\gamma_1\varphi}_{Y_1}. \end{align*} By \eqref{BC:DEF}, it follows that $$\label{AK:E4} \langle\operatorname{sgn}\varphi, A_K\varphi\rangle \le \pare{\norm{K}_{\mathcal{L}(Y_1)}-1}\norm{\gamma_1\varphi}_{Y_1}\le0$$ because of\norm{K}_{\mathcal{L}(Y_1)}<1$and hence,$A_K$is a dissipative operator. Furthermore,$A_K$is densely defined because of$\mathcal{C}_c(\Omega)\subset D(A_K)\subset L^1(\Omega)$. Now, Lemma \ref{L:PHI-LUM} completes the proof. \end{proof} Clearly, Lemma above does not hold for the case$\norm{K}_{\mathcal{L}(Y_1)}\ge1$because \eqref{AK:E4} can not be satisfied and therefore other ways are needed. So, according to the compactness of the boundary operator$K$, we have the following result. \begin{lemma}\label{AK:LEM2} Suppose$K$is compact satisfying$\norm{K}_{\mathcal{L}(Y_1)}\ge1$. Then \begin{enumerate} \item There exists$\lambda_0=\lambda_0(K)$, with$\lambda_0>0$if$\norm{K}_{\mathcal{L}(Y_1)}>1$and$\lambda_0\ge0$if$\norm{K}_{\mathcal{L}(Y_1)}=1$, satisfying$\norm{K_\lambda}_{\mathcal{L}(Y_1)}<1$for all$\lambda>\lambda_0$and$\norm{K_{\lambda_0}}_{\mathcal{L}(Y_1)}\le1$. \item For all$\lambda>\lambda_0$, the resolvent operator of \eqref{AK:DEF} is given by \eqref{AK:R1}. \item The operator defined by \eqref{AK:DEF} generates, on$L^1(\Omega)$, a strongly continuous semigroup$(U_K(t))_{t\ge0}$satisfying $$\label{AK:R5} \norm{U_K(t)\varphi}_1 \le e^{\frac{\lambda_0}{a}}e^{t\lambda_0}\norm{\varphi}_1 \quad t\ge0,$$ for all$\varphi\in L^1(\Omega)$. \end{enumerate} \end{lemma} \begin{proof} (1) Let$\lambda, \eta\ge0$and let$B$be the unit ball in$Y_1. So we have \begin{align*} \norm{K_\lambda-K_{\eta}}_{\mathcal{L}(Y_1)} &=\sup_{\psi\in B}\norm{K_\lambda\psi-K_{\eta}\psi}_{Y_1}\\ &=\sup_{\varphi\in K\pare{B}}\norm{\theta_\lambda\varphi-\theta_\eta\varphi}_{Y_1}\\ &\le\sup_{\varphi\in\overline{K\pare{B}}} \norm{\theta_\lambda\varphi-\theta_\eta\varphi}_{Y_1}. \end{align*} By the compactness of the set\overline{K(B)}$, then there exists$\varphi_0\in \overline{K(B)}$(so independent of$\lambda$and$\mu$) satisfying $\norm{K_\lambda-K_{\eta}}_{\mathcal{L}(Y_1)}\le \norm{\theta_\lambda\varphi_0-\theta_\eta\varphi_0}_{Y_1}$ which implies $\lim_{\lambda\to\eta}\norm{K_\lambda-K_{\eta}}_{\mathcal{L}(Y_1)} \le \lim_{\lambda\to\eta} \norm{\theta_\lambda\varphi_0-\theta_\eta\varphi_0}_{Y_1} =0$ and therefore, the continuity of the mapping $$\label{LEM2:E1} \lambda \to \norm{K_\lambda}_{\mathcal{L}(Y_1)}$$ follows, for$\lambda\ge0$. Furthermore, proceeding as previously we obtain $$\norm{K_\lambda}_{\mathcal{L}(Y_1)} =\sup_{\psi\in B}\norm{K_\lambda\psi}_{Y_1} \le\sup_{\varphi\in \overline{K\pare{B}}} \norm{\theta_\lambda\varphi}_{Y_1} =\norm{\theta_\lambda\varphi_0}_{Y_1}$$ and therefore $$\label{LEM2:E2} \lim_{\lambda\to\infty}\norm{K_\lambda}_{\mathcal{L}(Y_1)}=0.$$ \par On the other hand, if$\eta>\lambda\ge0$then for all$\psi\in Y_1$we have $\abs{K_{\eta}\psi}=\theta_\eta\abs{K\psi}= \theta_{\eta-\lambda}\theta_\lambda\abs{K\psi} =\theta_{\eta-\lambda}\abs{K_\lambda\psi}<\abs{K_\lambda\psi}$ which implies $$\norm{K_{\eta}}_{\mathcal{L}(Y_1)}\le \norm{K_\lambda}_{\mathcal{L}(Y_1)}$$ and therefore the mapping \eqref{LEM2:E1} is decreasing. Now, if$\norm{K}_{\mathcal{L}(Y_1)}>1$, then$\norm{K_0}_{\mathcal{L}(Y_1)}=\norm{K}_{\mathcal{L}(Y_1)}>1$together with \eqref{LEM2:E2} imply that the equation$\norm{K_\lambda}_{\mathcal{L}(Y_1)}=1$has at least one strictly positive solution ($\lambda_1>0$) and therefore the closed set $$\label{LEM2:E3} E:=\{\lambda\ge0,\; \norm{K_\lambda}_{\mathcal{L}(Y_1)}=1\}$$ is not empty and bounded. Now, it suffices to set $$\lambda_0:=\max E\ge\lambda_1>0$$ Next. If$\norm{K}_{\mathcal{L}(Y_1)}=1$then$\norm{K_0}_{\mathcal{L}(Y_1)}=\norm{K}_{\mathcal{L}(Y_1)}=1$implies that$\lambda=0$is obviously a solution of$\norm{K_\lambda}_{\mathcal{L}(Y_1)}=1$and therefore the closed set \eqref{LEM2:E3} is not empty and bounded. Now, it suffices again to set $$\lambda_0:=\max E\ge0.$$ Finally, in both cases we have$\norm{K_\lambda}_{\mathcal{L}(Y_1)}<1$if$\lambda>\lambda_0$and$\norm{K_{\lambda_0}}_{\mathcal{L}(Y_1)}\le1$. (2) Following the proof of the second point of Lemma~\ref{AK:LEM1}, we have only to solve \eqref{AK:E3}. This clearly follows from the first point, and therefore$(I-K_\lambda)$is an invertible operator into$Y_1$. (3) First, let us introduce on,$L^1(\Omega)$, the following norm $$\label{NORME} \norme{\varphi}_1= \int_\Omega\abs{\varphi(\mu,v)}h(\mu,v)\,d\mu\,dv$$ where$h(\mu,v)=e^{-\lambda_0\frac{(1-\mu)}{v}}.$The norms \eqref{NORME} and \eqref{NORM} are clearly equivalent because of $$\label{LEM2:E4} e^{-\lambda_0/a}\norm{\varphi}_1 \le \norme{\varphi}_1\le \norm{\varphi}_1$$ for all$\varphi\in L^1(\Omega)$. \par Next, let$\lambda>\lambda_0$and$g\in L^1(\Omega)$. So, the second point means that \eqref{AK:E2} is the unique solution of \eqref{AK:E1} satisfying \eqref{BC:DEF}. Multiplying \eqref{AK:E1} by$(\operatorname{sgn}\varphi)h$and integrating it over$\Omega$, we obtain $$\label{LEM2:E5} \lambda\norme{\varphi}_1 =-\int_\Omega v\frac{\partial\abs{\varphi}}{\partial\mu}h(\mu,v)\,d\mu\,dv +\int_\Omega(\operatorname{sgn}\varphi) hg(\mu,v)\,d\mu\,dv =I+J.$$ For the term$J$, we obviously have $$\label{LEM2:E7} J\le\norme{g}_1.$$ Integrating by parts and using \eqref{BC:DEF} and \eqref{KLAMBDA:DEF}, the term$Ibecomes %\label{e:I1} \begin{align*} I&=\int_a^\infty e^{-\frac{\lambda_0}{v}}\abs{\varphi(0,v)}vdv -\int_a^\infty \abs{\varphi(1,v)}vdv +\lambda_0\int_\Omega\abs{\varphi h}(\mu,v)\,d\mu\,dv \\ &=\int_a^\infty e^{-\frac{\lambda_0}{v}}\abs{K\gamma_1\varphi(v)}vdv -\int_a^\infty \abs{\gamma_1\varphi(v)}vdv+\lambda_0\norme{\varphi}_1\\ &=\norm{K_{\lambda_0}\gamma_1\varphi}_{Y_1} -\norm{\gamma_1\varphi}_{Y_1} +\lambda_0\norme{\varphi}_1\\ &\le \pare{\norm{K_{\lambda_0}}_{\mathcal{L}(Y_1)}-1} \norm{\gamma_1\varphi}_{Y_1} +\lambda_0\norme{\varphi}_1 \end{align*} and by the first point, we are led to $$\label{LEM2:E6} I\le \lambda_0\norme{\varphi}_1.$$ Putting \eqref{LEM2:E6} and \eqref{LEM2:E7} in \eqref{LEM2:E5}, we obtain $\norme{\varphi}_1 =\norme{(\lambda-A_K)^{-1}g}_1\le \frac{\norme{g}_1} {\left(\lambda-\lambda_0\right)}.$ Moreover,A_K$is a closed operator (because of$\rho(A_K)\not=\emptyset$) and densely defined (because of$\mathcal{C}_c(\Omega)\subset D(A_K)\subset L^1(\Omega)$). Therefore, Lemma~\ref{L:HIL-YOS} leads to the existence of a strongly continuous semigroup$(U_K(t))_{t\ge0}$satisfying $$\label{LEM2:E8} \norme{U_K(t)g}_1\le e^{t\lambda_0}\norme{g}_1,\quad t\ge0.$$ Then \eqref{LEM2:E4} completes the proof. \end{proof} Lemmas~\ref{AK:LEM1} and \ref{AK:LEM2} suggest to set the following definition. \begin{definition}\label{D:ADMISSIBLE} \rm$K$is said to be an \emph{admissible} operator if ($K$is bounded and$\norm{K}_{\mathcal{L}(Y_1)}<1$) or ($K$is compact and$\norm{K}_{\mathcal{L}(Y_1)}\ge1$). In this case, the number $$\label{OMEGA0} \omega_0= \begin{cases} 0,&\text{if K bounded and \norm{K}_{\mathcal{L}(Y_1)}<1;}\\ \lambda_0,&\text{if K compact and \norm{K}_{\mathcal{L}(Y_1)}\ge1.} \end{cases}$$ is called the \emph{abscissa} of the admissible operator$K$. \end{definition} Lemmas~\ref{AK:LEM1} and \ref{AK:LEM2} together with the definition above clearly lead to the main result of this section. \begin{theorem}\label{AK:THE} Let$K$be an admissible operator whose abscissa is$\omega_0$. Then \begin{enumerate} \item$\norm{K_\lambda}_{\mathcal{L}(Y_1)}<1$for all$\lambda>\omega_0$and$\norm{K_{\omega_0}}_{\mathcal{L}(Y_1)}\le1$. \item For all$\lambda>\omega_0$, the resolvent operator of \eqref{AK:DEF} is given by \eqref{AK:R1}. \item The operator defined by \eqref{AK:DEF} generates, on$L^1(\Omega)$, a strongly continuous semigroup$(U_K(t))_{t\ge0}$satisfying $$\label{AK:R8} \norm{U_K(t)\varphi}_1 \le e^{\frac{\omega_0}{a}}e^{t\omega_0}\norm{\varphi}_1, \quad t\ge0,$$ for all$\varphi\in L^1(\Omega)$. \end{enumerate} \end{theorem} \section{Construction of the unperturbed semigroup ($r=0$)}\label{S:CONS} In this section, we are going to give the expression of the unperturbed semigroup$(U_K(t))_{t\ge0}$. This expression is very useful to describe the behavior asymptotic which is the main goal of \cite{Boulanouar3}. So, let us consider the Banach space$Z_\omega^1:=L^1((-\infty,0)\times J,h_\omega)$($\omega\ge0$) whose norm is $$\norm{f}_{Z_\omega^1}=\int_a^\infty\int_{-\infty}^0 \abs{f(x,v)}e^{-\omega\frac{(1-x)}{v}}\,dx\,dv.$$ In this context we have the following result. \begin{lemma}\label{BK:LEM} Let$K$be an admissible operator whose abscissa is$\omega_0$and let$B_K(t)$be the operator $$\label{BK:R0} B_K(t)\varphi(\mu,v)=\xi(\mu,v,t)(I-H_K)^{-1}V_K\varphi(\mu-tv,v)\quad t\ge0$$ for almost all$(\mu,v)\in\Omega$, where, the operators$H_K$and$V_K$are defined as \begin{gather*} H_Kf(x,v)=\Big(K\big(\xi(1,\cdot,-xv^{-1}) f(1+xv^{-1}\cdot,\cdot)\big)\Big)(v),\\ V_K\varphi(x,v)=\big(K\pare{\gamma_1U_0(-xv^{-1})\varphi}\big)(v), \end{gather*} with $$\label{BK:R3} \xi(\mu,v,t)= \begin{cases} 1 & \text{if } \mu< tv;\\ 0 & \text{if } \mu\ge tv.\\ \end{cases}$$ Then \begin{enumerate} \item$H_K$and$V_K$are bounded operators respectively from$Z_\omega^1(\omega\ge0)$and$L^1(\Omega)$into$Z_\omega^1$. Furthermore, we have \begin{gather}\label{BK:R1} \norm{H_K}_{\mathcal{L}(Z_\omega^1)} \le \norm{K_\omega}_{\mathcal{L}(Y_1)},\\ \label{BK:R2} \norm{V_K}_{\mathcal{L}(L^1(\Omega), Z_\omega^1)} \le\norm{K_\omega}_{\mathcal{L}(Y_1)}. \end{gather} \item Let$\omega>\omega_0$. Then, for all$t\ge0$, the operator$B_K(t)$is linear and bounded from$L^1(\Omega)$into itself. \item For all$\varphi\in L^1(\Omega)$, the mapping$t\in\mathbb{R}_+\to B_K(t)\varphi$is continuous at :$t=0_+$and$B_K(0)=0$. \item For all$\varphi\in W^1(\Omega)$and for all$t\ge0$we have $$\label{BK:R5} \gamma_0B_K(t)\varphi -K\gamma_1B_K(t)\varphi = K\gamma_1U_0(t)\varphi.$$ \item Let$K'$be an admissible operator whose abscissa is$\omega_0'$. Then, for all$\omega>\max\{\omega_0,\;\omega_0'\}$, we have $$\label{BK:R6} \norm{B_K(t)-B_{K'}(t)}_{\mathcal{L}\left(L^1(\Omega)\right)} \le \frac{e^{\omega(\frac{1}{a}+t)}\norm{K-K'}_{\mathcal{L}(Y_1)}} {(1-\norm{K_\omega}_{\mathcal{L}(Y_1)})(1-\norm{K'_\omega}_{\mathcal{L}(Y_1)})}.$$ \end{enumerate} \end{lemma} \begin{proof} (1) For all$f\in Z_\omega^1(\omega\ge0), we have \begin{align*} \norm{H_Kf}_{\mathcal{L}(Z_\omega^1)}&=\int_a^\infty \int_{-\infty}^0 \abs{ \left[K\xi\big(1,\cdot,-\tfrac{x}{v}\big)f\big(1+\tfrac{x}{v}\cdot,\cdot\big)\right](v)} e^{-\omega\frac{(1-x)}{v}} \,dx\,dv\\ &=\int_a^\infty\int_0^{\infty}\abs{ \left[K\xi(1,\cdot,t)f(1-t\cdot,\cdot)\right](v)} e^{-\omega(\frac{1}{v}+t)}v\,dt\,dv \end{align*} which leads, by \eqref{KLAMBDA:DEF}, to \begin{align*} \norm{H_Kf}_{\mathcal{L}(Z_\omega^1)} &=\int_0^\infty\cro{\int_a^{\infty}\abs{ \left[K_\omega\xi(1,\cdot,t)f(1-t\cdot,\cdot)\right](v)} vdv}e^{-\omega t}dt\\ &\le\norm{K_\omega}_{\mathcal{L}(Y_1)} \int_0^\infty\Big[\int_a^\infty\abs{ \xi(1,v,t)f(1-tv,v)}vdv\Big]e^{-\omega t}dt\\ &\le\norm{K_\omega}_{\mathcal{L}(Y_1)} \int_a^\infty\int_{-\infty}^0\abs{ \xi\left(1,v,\tfrac{1-x}{v}\right)f(x,v)}e^{-\omega\frac{(1-x)}{v}}\,dx\,dv\\ &\le\norm{K_\omega}_{\mathcal{L}(Y_1)} \norm{f}_{Z_\omega^1} \end{align*} and therefore \eqref{BK:R1} holds. A similar calculation implies that $\norm{V_K\varphi}_{Z_\omega^1} \le\norm{K_\omega}_{\mathcal{L}(Y_1)}\norm{\varphi}_1$ for all\varphi\in L^1(\Omega)$and therefore \eqref{BK:R2} holds. (2) Let$\omega>\omega_0$and$t\ge0$. Due to the first point of Theorem~\ref{AK:THE} together with \eqref{BK:R1}, we obtain$(I-H_K)$is an invertible operator into$Z_\omega^1$. So, the operator$B_K(t)$given by \eqref{BK:R0} is well defined and its linearity follows from those of$(I-H_K)^{-1}$and$V_K. For its boundedness, we have \begin{align*} \norm{B_K(t)\varphi}_1 &=\int_\Omega\xi(\mu,v,t) \abs{(I-H_K)^{-1}V_K\varphi(\mu-tv,v)}\,d\mu\,dv\\ &\le\int_a^\infty\int_0^{tv} e^{\omega\frac{\mu}{v}} \abs{(I-H_K)^{-1}V_K\varphi(\mu-tv,v)}\,d\mu\,dv\\ &=\int_a^\infty\int_{-tv}^0e^{\omega\pare{\frac{x}{v}+t}} \abs{(I-H_K)^{-1}V_K\varphi(x,v)}\,dx\,dv \end{align*} for all\varphi\in L^1(\Omega)and therefore $$\label{BK:E3} \norm{B_K(t)\varphi}_1 \le e^{\omega\pare{\frac{1}{a}+t}} \int_a^\infty \int_{-tv}^0 e^{-\omega\frac{(1-x)}{v}} \abs{(I-H_K)^{-1}V_K\varphi(x,v)} \,dx\,dv.$$ This implies $$\label{BK:E4} \norm{B_K(t)\varphi}_1 \le e^{\omega\pare{\frac{1}{a}+t}}\norm{(I-H_K)^{-1}V_K\varphi}_{Z_\omega^1}$$ and by \eqref{BK:R1} and \eqref{BK:R2}, we clearly have \begin{align*} \norm{B_K(t)\varphi}_1&\le e^{\omega\pare{\frac{1}{a}+t}}\norm{(I-H_K)^{-1}V_K\varphi}_{Z_\omega^1}\\ &=e^{\omega\pare{\frac{1}{a}+t}} \norm{\sum_{n\ge0}H_K^n}_{Z_\omega^1} \norm{V_K\varphi}_{Z_\omega^1}\\ &\le e^{\omega\pare{\frac{1}{a}+t}}\frac{1} {1-\norm{K_\omega}_{\mathcal{L}(Y_1)}} \norm{K_\omega}_{\mathcal{L}(Y_1)}\norm{\varphi}_1 \end{align*} which leads to the boundedness ofB_K(t)$from$L^1(\Omega)$into itself. (2) This point obviously follows from \eqref{BK:E3}. (3) Let$\varphi\in W^1(\Omega). Using \eqref{BK:R0}, we obtain \begin{align*} &\int_a^\infty\int_0^\infty\big\vert K\gamma_1U_0(t)\varphi(v) -\gamma_0B_K(t)\varphi(v)- K\gamma_1B_K(t)\varphi(v)\big\vert v\,dt\,dv\\ &=\int_a^\infty\int_0^\infty\Big\vert V_K\varphi(-tv,v) -(I-H_K)^{-1}V_K(-tv,v)\\ &\quad -K\cro{\xi(1,\cdot,t)(I-H_K)^{-1}V_K\varphi(1-t\cdot,\cdot)}(v)\Big\vert v\,dt\,dv. \end{align*} The changex=-tv$with$dx=-vdtinfers that \begin{align*} &\int_a^\infty\int_0^\infty\big\vert K\gamma_1U_0(t)\varphi(v)-\gamma_0B_K(t)\varphi(v)- K\gamma_1B_K(t)\varphi(v)\big\vert v\,dt\,dv\\ &=\int_a^\infty\int_{-\infty}^0 \Big\vert V_K\varphi(x,v)-(I-H_K)^{-1}V_K(x,v)\\ &\quad -K\cro{\xi(1,\cdot,-xv^{-1})(I-H_K)^{-1}V_K\varphi(1+xv^{-1}\cdot,\cdot)}(v)\Big\vert \,dx\,dv\\ &=\int_a^\infty\int_{-\infty}^0 \Big\vert V_K\varphi(x,v)-(I-H_K)^{-1}V_K(x,v)\\ &\quad -H_K(I-H_K)^{-1}V_K\varphi(x,v)\Big\vert \,dx\,dv =0 \end{align*} and therefore \eqref{BK:R5} holds for almost allt\in \mathbb{R}_+$. Now, thanks to the third point of Lemma~\ref{A0:LEM}, we obtain \eqref{BK:R5} holds for all$t\ge0$. (4) Let$\omega>\sup\{\omega_0,\;\omega_0'\}. A simple calculation, like the proof of \eqref{BK:E4}, infers that $\norm{B_K(t)\varphi-B_{K'}(t)\varphi}_1\le e^{\omega\pare{\frac{1}{a}+t}}\norm{(I-H_K)^{-1}V_K\varphi -(I-H_{K'})^{-1}V_{K'}\varphi}_{Z_\omega^1}$ As \begin{align*} &(I-H_K)^{-1}V_K-(I-H_{K'})^{-1}V_{K'}\\ &=(I-H_{K'})^{-1}[V_K-V_{K'}] +(I-H_K)^{-1}[H_K-H_{K'}](I-H_{K'})^{-1}V_K, \end{align*} we clearly have \begin{align*} &\norm{B_K(t)\varphi-B_{K'}(t)\varphi}_1\\ &\le e^{\omega\pare{\frac{1}{a}+t}} \norm{(I-H_{K'})^{-1}} \norm{V_K\varphi-V_{K'}\varphi}\\ &\quad +e^{\omega\pare{\frac{1}{a}+t}}\norm{(I-H_K)^{-1}} \norm{H_K-H_{K'}}\norm{(I-H_{K'})^{-1}}\norm{V_K\varphi} \end{align*} and therefore, by \eqref{BK:R1} and \eqref{BK:R2}, we are lead to \begin{align*} &\norm{B_K(t)\varphi-B_{K'}(t)\varphi}_1\\ &\le e^{\omega\pare{\frac{1}{a}+t}} \frac{1}{1-\norm{K'_\omega}_{\mathcal{L}(Y_1)}} \norm{V_K\varphi-V_{K'}\varphi}_{Z_\omega^1}\\ &\quad + e^{\omega\pare{\frac{1}{a}+t}} \frac{1}{(1-\norm{K_\omega}_{\mathcal{L}(Y_1)})} \norm{H_K-H_{K'}}\frac{1}{(1-\norm{K'_\omega}_{\mathcal{L}(Y_1)})} \norm{K_\omega}_{\mathcal{L}(Y_1)}\norm{\varphi}_1. \end{align*} Thanks to the obvious linearity of the mappingsK\to H_K$and$K\to V_K, \eqref{BK:R1} and \eqref{BK:R2} imply \begin{align*} \norm{B_K(t)\varphi-B_{K'}(t)\varphi}_1 &\le e^{\omega\pare{\frac{1}{a}+t}} \frac{\norm{K_\omega-K'_\omega}_{\mathcal{L}(Y_1)} \norm{\varphi}_1}{1-\norm{K'_\omega}_{\mathcal{L}(Y_1)}}\\ &\quad +e^{\omega\pare{\frac{1}{a}+t}} \frac{\norm{K_\omega-K'_\omega}_{\mathcal{L}(Y_1)} \norm{K_\omega}_{\mathcal{L}(Y_1)}\norm{\varphi}_1} {(1-\norm{K_\omega}_{\mathcal{L}(Y_1)}) (1-\norm{K'_\omega}_{\mathcal{L}(Y_1)})}\\ &\le e^{\omega\pare{\frac{1}{a}+t}} \frac{\norm{K_\omega-K'_\omega}_{\mathcal{L}(Y_1)}}{(1-\norm{K_\omega}_{\mathcal{L}(Y_1)}) (1-\norm{K'_\omega}_{\mathcal{L}(Y_1)})}\norm{\varphi}_1\\ &\le e^{\omega\pare{\frac{1}{a}+t}} \frac{\norm{K-K'}_{\mathcal{L}(Y_1)}}{(1-\norm{K_\omega}_{\mathcal{L}(Y_1)}) (1-\norm{K'_\omega}_{\mathcal{L}(Y_1)})}\norm{\varphi}_1 \end{align*} and therefore \eqref{BK:R6} holds. The proof is complete. \end{proof} \begin{theorem}\label{UK:THE} LetK$be an admissible operator whose abscissa is$\omega_0$. Then, the semigroup$(U_K(t))_{t\ge0}$is given by $$\label{UK:R1} U_K(t)\varphi=U_0(t)\varphi+B_K(t)\varphi\quad t\ge0,$$ for all$\varphi\in L^1(\Omega)$. Furthermore, the operator$B_K(t)$satisfies $$\label{UK:R2} B_K(t)\varphi(\mu,v)= \xi(\mu,v,t)K\gamma_1\big(U_K\pare{t -\tfrac{\mu}{v}}\varphi\big)(v),\quad t\ge0,$$ for almost all$(\mu,v)\in\Omega$. \end{theorem} \begin{proof} Let$t\ge0$and$\varphi\in L^1(\Omega)$and let$S_K(t)$be the following operator $$\label{UK:E1} S_K(t)=U_0(t)+B_K(t),\quad t\ge0,$$ where,$B_K(t)$is given by \eqref{BK:R0}. In the sequel, we are going to prove that$(S_K(t))_{t\ge0}$is a strongly continuous semigroup into$L^1(\Omega)$. At the end of this proof, we will show that$U_K(t)=S_K(t)$for all$t\ge0$. Then, let us divide the proof in several steps. \textbf{Step one.} This step deals with a useful expression of the operator$B_K(t)$like \eqref{UK:R2}. So, for all$\varphi\in W^1(\Omega)$and for almost all$(\mu,v)\in \Omega, \eqref{UK:E1} implies that $S_K\pare{t-\tfrac{\mu}{v}}\varphi(0,v)= U_0\pare{t-\tfrac{\mu}{v}}\varphi(0,v) +B_K\pare{t-\tfrac{\mu}{v}}\varphi(0,v)$ which leads, by \eqref{A0:R2}, to $\xi(\mu,v,t)S_K\pare{t-\tfrac{\mu}{v}}\varphi(0,v)= \xi(\mu,v,t)B_K\pare{t-\tfrac{\mu}{v}}\varphi(0,v).$ Using \eqref{BK:R0} we obtain \begin{align*} &\xi(\mu,v,t)S_K\pare{t-\tfrac{\mu}{v}}\varphi(0,v)\\ &=\xi(\mu,v,t)\xi\pare{0,v,\pare{t-\tfrac{\mu}{v}}} (I-H_K)^{-1}V_K\varphi\pare{0-\pare{t-\tfrac{\mu}{v}}v,v}\\ &=\xi(\mu,v,t)(I-H_K)^{-1}V_K\varphi(\mu-tv,v) \end{align*} and therefore $$\label{UK:E5} \xi(\mu,v,t)\gamma_0\pare{S_K\pare{t-\tfrac{\mu}{v}}\varphi}(v)= B_K(t)\varphi(\mu,v).$$ On the other hand, \eqref{UK:E1} implies \begin{align*} \gamma_0S_K(t)\varphi-K\gamma_1S_K(t)\varphi&= \gamma_0U_0(t)\varphi+\gamma_0B_K(t)\varphi\\ &\quad -K\gamma_1U_0(t)\varphi-K\gamma_1B_K(t)\varphi\\ &=\gamma_0B_K(t)\varphi -K\gamma_1U_0(t)\varphi-K\gamma_1B_K(t)\varphi. \end{align*} According to \eqref{BK:R5}, we obtain $\gamma_0S_K(t)\varphi=K\gamma_1S_K(t)\varphi$ and hence $$\label{UK:E7} \xi(\mu,v,t) \gamma_0 \pare{ S_K\pare{t-\tfrac{\mu}{v}}\varphi}(v) = \xi(\mu,v,t) K\gamma_1\pare{S_K\pare{t-\tfrac{\mu}{v}}\varphi}(v).$$ Now, combining \eqref{UK:E5} and \eqref{UK:E7}, we obtain the following useful relation $$\label{UK:E9} B_K(t)\varphi(\mu,v)=\xi(\mu,v,t) K\gamma_1\pare{S_K\pare{t-\tfrac{\mu}{v}}}\varphi(v).$$ Finally, the density ofW^1(\Omega)$in$L^1(\Omega)$, implies that \eqref{UK:E9} holds again for all$\varphi\in L^1(\Omega)$. \textbf{Step two.} In this step, we are going to prove that$(S_K(t))_{t\ge0}$is a strongly continuous semigroup into$L^1(\Omega)$. So, Lemma~\ref{A0:LEM} together with the second and the third points of Lemma~\ref{BK:LEM} clearly lead to the linearity and the boundedness of the operator$S_K(t)$from$L^1(\Omega)$into itself and$S_K(0)=U_0(0)+B_K(0)=I+0=I$and $$\lim_{t\to0_+}\norm{S_K(t)\varphi-\varphi}_1\le \lim_{t\to0_+}\norm{U_0(t)\varphi-\varphi}_1 +\lim_{t\to0_+}\norm{B_K(t)\varphi}_1=0.$$ Now, in order to state that the family operators$(S_K(t))_{t\ge0}$is a strongly continuous semigroup into$L^1(\Omega)$, it suffices to prove that $$\label{UK:E3} G(t,s):=S_K(t)S_K(s)-S_K(t+s)=0$$ for all$t\ge0$and all$s\ge0. So, by \eqref{UK:E1} and \eqref{UK:E9}, a simple calculation shows that \begin{align*} &G(t,s)\varphi(\mu,v)\\ &=\xi(\mu,v,t)K\gamma_1\pare{U_K\pare{t-\tfrac{\mu}{v}} S_K(s)\varphi}(v)\\ &\quad + \Big( \chi(\mu,v,t)\xi\pare{t+s-\tfrac{\mu}{v}}- \xi\pare{t+s-\tfrac{\mu}{v}} \Big) K\gamma_1\pare{S_K\pare{t+s-\tfrac{\mu}{v},s}\varphi}(v) \end{align*} for almost all(\mu,v)\in \Omega$. Furthermore, \eqref{A0:R3} and \eqref{BK:R3} allow to reduce the relation above to $$\label{UK:E11} G(t,s)\varphi(\mu,v) =\xi(\mu,v,t)K\gamma_1\pare{G(t-\tfrac{\mu}{v},s)\varphi}(v).$$ Next, let$\omega>\omega_0. On one hand, \eqref{UK:E11} implies \begin{align*} \int_0^{\infty}e^{-\omega t}\norm{\gamma_1G(t,s)\varphi}_{Y_1}dt &=\int_0^{\infty}e^{-\omega t}\int_a^\infty \abs{\gamma_1G(t,s)\varphi(v)}v\,dv\,dt\\ &=\int_a^\infty\int_0^{\infty}e^{-\omega t} \xi(1,v,t)\abs{K\gamma_1\pare{G(t-\tfrac{1}{v},s)\varphi}(v)} v\,dt\,dv\\ &=\int_a^\infty\int_{\tfrac{1}{v}}^{\infty}e^{-\omega t} \abs{K\gamma_1\pare{G(t-\tfrac{1}{v},s)\varphi}(v)} v\,dt\,dv\\ &=\int_a^\infty\int_0^{\infty} e^{-\omega\pare{x+\tfrac{1}{v}}} \abs{K\gamma_1G(x,s)\varphi(v)} v\,dx\,dv \end{align*} which leads, by \eqref{KLAMBDA:DEF}, to \begin{align*} \int_0^{\infty}e^{-\omega t}\norm{\gamma_1G(t,s)\varphi}_{Y_1}dt& = \int_0^{\infty} \int_a^\infty e^{-\omega x} \abs{K_\omega\gamma_1G(x,s)\varphi(v)}vdvdx\\ &=\int_0^{\infty} e^{-\omega x} \norm{K_\omega\gamma_1G(x,s)\varphi}_{Y_1}dx \end{align*} and therefore, $$\label{UK:E13} \int_0^{\infty}e^{-\omega t}\norm{\gamma_1G(t,s)\varphi}_{Y_1}dt \le\norm{K_\omega} \int_0^{\infty}e^{-\omega t}\norm{\gamma_1G(t,s)\varphi}_{Y_1}dt.$$ This obviously means that $$\label{UK:E15} \int_0^{\infty}e^{-\omega t}\norm{\gamma_1G(t,s)\varphi}_{Y_1}dt=0$$ because of the first point of Theorem~\ref{AK:THE}. On the other hand, \eqref{UK:E11} implies \begin{align*} \norm{G(t,s)\varphi}_1 &=\int_\Omega\xi(\mu,v,t) \abs{K\gamma_1\pare{G\pare{t-\tfrac{\mu}{v},s}\varphi}(v)}\,d\mu\,dv\\ &=\int_a^\infty\int_0^{tv} \abs{K\gamma_1\pare{G\pare{t-\tfrac{\mu}{v},s}\varphi}(v)}\,d\mu\,dv\\ &=\int_a^\infty\int_0^t\abs{ K\gamma_1G(x,s)\varphi(v)} v\,dx\,dv\\ &=\int_0^t\norm{K\gamma_1G(x,s)\varphi}_{Y_1}dx\\ &\le\norm{K}_{\mathcal{L}(Y_1)}\int_0^t\norm{\gamma_1G(x,s)\varphi}_{Y_1}dx \end{align*} and therefore, $$\label{UK:E17} \norm{G(t,s)\varphi}_1 \le\norm{K}_{\mathcal{L}(Y_1)} \int_0^\infty\norm{\gamma_1G(x,s)\varphi}_{Y_1}dx.$$ Now, \eqref{UK:E15} and \eqref{UK:E17} obviously implyG(t,s)=0$and therefore \eqref{UK:E3} holds for all$t\ge0$and all$s\ge0$. Hence,$(S_K(t))_{t\ge0}$is well a strongly continuous semigroup satisfying $$\label{UK:E19} S_K(t)\varphi(\mu,v)=U_0(t)\varphi(\mu,v)+ \xi(\mu,v,t) K\gamma_1\big(S_K\pare{t-\tfrac{\mu}{v}}\varphi\big)(v)$$ because of \eqref{UK:E1} and \eqref{UK:E9}. \textbf{Step three.} Let$\lambda>\omega_0$and$\varphi\in L^1(\Omega)$and let$B$be the generator of the semigroup$(S_K(t))_{t\ge0}. Then, by \eqref{UK:E19} we obtain \begin{align*} &(\lambda-B)^{-1}\varphi(\mu,v)\\ &=\int_0^\infty e^{-\lambda t}S_K(t)\varphi(\mu,v) dt\\ &=\int_0^\infty e^{-\lambda t}U_0(t)\varphi(\mu,v)dt+ \int_0^\infty e^{-\lambda t}\xi(\mu,v,t) K\gamma_1\pare{S_K\pare{t-\tfrac{\mu}{v}}\varphi}(v)dt\\ &=(\lambda-A_0)^{-1}\varphi(\mu,v)+ e^{-\lambda\tfrac{\mu}{v}} \int_0^\infty e^{-\lambda t}K\gamma_1S_K(t)\varphi(v)dt\\ &=(\lambda-A_0)^{-1}\varphi(\mu,v)+ e^{-\lambda\tfrac{\mu}{v}}K\gamma_1 \Big[\int_0^\infty e^{-\lambda t}S_K(t)\varphi dt\Big](v)\\ &=(\lambda-A_0)^{-1}\varphi(\mu,v)+ e^{-\lambda\tfrac{\mu}{v}}K\gamma_1(\lambda-B)^{-1}\varphi(v) \end{align*} for almost all(\mu,v)\in\Omega$, and therefore $$\label{UK:E21} (\lambda-B)^{-1}\varphi= \varepsilon_\lambda K\gamma_1(\lambda-B)^{-1}\varphi+(\lambda-A_0)^{-1}\varphi$$ where,$\varepsilon_\lambda=e^{-\lambda\tfrac{\mu}{v}}$. Applying$\gamma_1$to \eqref{UK:E21}, we infer that $\gamma_1(\lambda-B)^{-1}\varphi =K_\lambda\gamma_1(\lambda-B)^{-1}\varphi+\gamma_1(\lambda-A_0)^{-1}\varphi$ and thanks to the first point of Theorem~\ref{AK:THE}, it follows that $$\label{UK:E23} \gamma_1(\lambda-B)^{-1}\varphi =(I-K_\lambda)^{-1}\gamma_1(\lambda-A_0)^{-1}\varphi$$ because of$\lambda>\omega_0$. Now, putting \eqref{UK:E23} in \eqref{UK:E21}, we finally obtain $$\label{UK:E25} (\lambda-B)^{-1}\varphi= \varepsilon_\lambda K(I-K_\lambda)^{-1}\gamma_1(\lambda-A_0)^{-1}\varphi +(\lambda-A_0)^{-1}\varphi.$$ \textbf{Step four.} By \eqref{AK:R1} and \eqref{UK:E25}, we have$(\lambda-A_K)^{-1}=(\lambda-B)^{-1}$and hence,$U_K(t)=S_K(t)$for all$t\ge0$because of the uniqueness of the generated semigroup. Moreover, \eqref{UK:R2} holds because of \eqref{UK:E9}. Now, the proof is complete. \end{proof} \begin{corollary}\label{INEQUALITY:LEM} Let$K$be an admissible operator whose abscissa is$\omega_0$. Then we have $$\label{INEQUALITY:R1} \int_a^\infty\int_0^te^{-\omega x} \abs{\gamma_1\left(U_K(x)\varphi\right)(v)}v\,dx\,dv \le\frac{\norm{\varphi}_1}{1-\norm{K_\omega}_{\mathcal{L}(Y_1)}},\quad t\ge0,$$ for all$\varphi\in L^1(\Omega)$. \end{corollary} \begin{proof} The corollary is obvious for$t=0$. So, let$t>0$and$\omega>\omega_0$be a given real and let$\psi\in W^1(\Omega)$. Applying$\gamma_1$to \eqref{UK:R1} and \eqref{UK:R2} we obtain $$\gamma_1(U_K(x)\psi)(v)=\gamma_1(U_0(x)\psi)(v) +\xi(1,v,x) \left[K\gamma_1\pare{U_K\pare{x-\tfrac{1}{v}}\psi}\right](v)$$ for all$x\ge0$and for almost all$v\in J$. Multiplying the relation above by$e^{-\omega x}$and integrating it over$(0,t)\times J, we infer that \begin{align*} &\int_a^\infty\int_0^te^{-\omega x}\abs{\gamma_1(U_K(x)\psi)(v)}v\,dx\,dv\\ &\le\int_a^\infty\int_0^te^{-\omega x}\chi(1,v,x)\abs{\psi(1-xv,v)}v\,dx\,dv\\ &\quad +\int_a^\infty\int_0^te^{-\omega x} \xi(1,v,x)\abs{K\gamma_1\pare{U_K\pare{x-\tfrac{1}{v}}\psi}(v)}v\,dx\,dv \end{align*} where we have used \eqref{A0:R2}. A suitable change of variables leads to \begin{align*} &\int_a^\infty\int_0^te^{-\omega x}\abs{\gamma_1(U_K(x)\psi)(v)}v\,dx\,dv\\ &\le\int_a^\infty\int_{1-tv}^1 e^{-\omega\tfrac{(1-\mu)}{v}} \chi\pare{1,v,\tfrac{1-\mu}{v}}\abs{\psi(\mu,v)}\,d\mu\,dv\\ &\quad + \int_a^\infty\int_0^te^{-\omega y}e^{-\omega\frac{1}{v}} \abs{K\gamma_1\pare{U_K(y)\psi}(v)}vdydv\\ &\leq\int_\Omega\abs{\psi(\mu,v)}\,d\mu\,dv +\int_0^te^{-\omega x} \norm{K_\omega\gamma_1\left(U_K(x)\psi\right)}_{Y_1}dx \end{align*} which implies \begin{align*} &\int_a^\infty\int_0^te^{-\omega x} \abs{\gamma_1(U_K(x)\psi)(v)}v\,dx\,dv\\ &\le\norm{\psi}_1 +\norm{K_\omega}_{\mathcal{L}(Y_1)} \int_a^\infty\int_0^te^{-\omega x} \abs{\gamma_1(U_K(x)\psi)(v)}v\,dx\,dv. \end{align*} Therefore, $\int_a^\infty\int_0^te^{-\omega x} \abs{\gamma_1(U_K(x)\psi)(v)}v\,dx\,dv\le \frac{1}{1-\norm{K_\omega}_{\mathcal{L}(Y_1)}}\norm{\psi}_1$ because of the first point of Theorem\ref{AK:THE}. Now, the density ofW^1(\Omega)$in$L^1(\Omega)$leads to \eqref{INEQUALITY:R1} for all$\varphi\in L^1(\Omega)$. \end{proof} Note that a rank one or a compact boundary operator is admissible and therefore Theorem~\ref{AK:THE} holds. Accordingly, we give three important results very useful for the results in \cite{Boulanouar3}. The first one is as follows. \begin{theorem}\label{STAB:THE} Let$K$and$K'$be two compact operators. Then, we have $$\label{STAB:R1} \norm{U_K(t)-U_{K'}(t)}_{\mathcal{L}\left(L^1(\Omega)\right)} \le 4e^{\omega(\frac{1}{a}+t)}\norm{K-K'}_{\mathcal{L}(Y_1)},\quad t\ge0$$ for all$\omega$big enough. \end{theorem} \begin{proof} Let$\omega$be a positive real and let$t\ge0$. First, \eqref{UK:R1} and \eqref{BK:R6} clearly lead to $$\label{STAB:E1} \norm{U_K(t)-U_{K'}(t)}_{\mathcal{L}\left(L^1(\Omega)\right)} \le \frac{e^{\omega(\frac{1}{a}+t)}\norm{K-K'}_{\mathcal{L}(Y_1)}} {(1-\norm{K_\omega}_{\mathcal{L}(Y_1)})(1-\norm{K'_\omega}_{\mathcal{L}(Y_1)})}$$ Next, let$B$be the unit ball in$Y_1$. So we have $\norm{K_\omega}_{\mathcal{L}(Y_1)} =\sup_{\psi\in B}\norm{K_\omega\psi}_{Y_1} =\sup_{\varphi\in K\pare{B}}\norm{\theta_\omega\varphi}_{Y_1} \le\sup_{\varphi\in\overline{K\pare{B}}} \norm{\theta_\omega\varphi}_{Y_1}.$ By the compactness of the set$\overline{K(B)}$, then there exists$\varphi_0\in \overline{K(B)}$(independent of$\omega$) satisfying $\norm{K_\omega}_{\mathcal{L}(Y_1)} \le\norm{\theta_\omega\varphi_0}_{Y_1}$ and hence $\lim_{\omega\to\infty}\norm{K_\omega}_{\mathcal{L}(Y_1)}= \lim_{\omega\to\infty}\norm{\theta_\omega\varphi_0}_{Y_1}=0.$ Therefore, there exists$\omega_1>0$such that $$\label{STAB:E2} \omega>\omega_1\Longrightarrow\norm{K_\omega}_{\mathcal{L}(Y_1)}<\frac{1}{2}.$$ The same calculation above holds for the compact operator$K'$and therefore there exists$\omega_1'>0$such that $$\label{STAB:E3}\ \omega>\omega_1'\Longrightarrow\norm{K_\omega'}_{\mathcal{L}(Y_1)} <\frac{1}{2}.$$ Finally, if$\omega>\sup\{\omega_1,\omega_1'\}$then \eqref{STAB:E1} and \eqref{STAB:E2} and \eqref{STAB:E3} clearly lead to \eqref{STAB:R1}. The proof is complete. \end{proof} Let us end this section by the following result. \begin{lemma}\label{1RANK:LEM} Let$K$be a rank one operator in$Y_1$; i.e., $K\psi=h\int_a^\infty k(v')\psi(v')v'dv', \quad h\in Y_1,\quad k\in L^\infty(J).$ Then, for all$\varphi\in L^1(\Omega)$, we have $%\label{E:S5.L1.E1} U_K(t)\varphi=\sum_{m=0}^\infty U_m(t)\varphi,\quad t\ge0,$ where,$U_0(t)$is given by \eqref{A0:R2} and $%\label{E:V1} U_1(t)\varphi(\mu,v) = \xi(\mu,v,t)h(v) \int_a^\infty k(v_1) \chi \left( 1,v_1,t-\tfrac{\mu}{v} \right) \varphi \left( 1- \pare{ t-\tfrac{\mu}{v}}v_1 , v_1 \right) v_1dv_1$ and, for$m\ge2, by %\label{E:VM} \begin{align*} U_m(t)\varphi(\mu,v) &=\xi(\mu,v,t)h(v) \underbrace{\int_a^\infty\cdots\int_a^\infty}_\text{mtimes} \prod_{j=1}^{m-1}h(v_{j})\prod_{j=1}^mk(v_j)\\ &\quad \times \xi\Big(1,v_{m-1},t-\frac{\mu}{v}-\sum_{i=1}^{(m-2)}\frac{1}{v_i}\Big) \chi\Big(1,v_m,t-\frac{\mu}{v}-\sum_{i=1}^{(m-1)}\frac{1}{v_i}\Big)\\ &\quad\times \varphi\Big(1-\Big(t-\frac{\mu}{v} -\sum_{i=1}^{m-1}\frac{1}{v_i}\Big) v_m,v_m\Big)v_1v_2\cdots v_m\,dv_1\cdots dv_m. \end{align*} Furthermore, for allt\ge0$, $$\label{1RANK:R1} \lim_{N\to\infty} \big\|U_K(t)-\sum_{m=0}^N U_m(t)\big\|_{\mathcal{L}(L^1(\Omega))}=0.$$ \end{lemma} \begin{proof} Let$\varphi\in L^1(\Omega)$and let$\omega$be a large real. By Theorem~\ref{UK:THE}, it is easy to check, by induction, that for all integer$N\ge1$we have $% U_K(t)=U_0(t)+\sum_{m=1}^N U_m(t)+R_N(t)$ where$R_N(t)is given by \begin{align*} & R_N(t)\varphi(\mu,v)\\ &=\xi(\mu,v,t)h(v)\underbrace{\int_a^\infty\cdot\cdot\int_a^\infty}_\text{(N+1)times} \prod_{j=1}^Nh(v_j)\prod_{j=1}^{N+1}k(v_j) \xi \Big( 1,v_N,t - \frac{\mu}{v} - \sum_{i=1}^{(N-1)} \frac{1}{v_i} \Big)\\ &\quad\times \gamma_1 \Big( U_K \Big(t - \frac{\mu}{v} - \sum_{i=1}^N \frac{1}{v_i}\Big) \varphi \Big) (v_{N+1}) v_1 \cdot\cdot v_{N+1}\,dv_1 \cdot\cdot dv_{N+1}. \end{align*} As1=e^{\omega\frac{(1-\mu)}{v}}e^{-\omega\frac{(1-\mu)}{v}} \le e^{\omega/a}e^{-\omega\frac{(1-\mu)}{v}}$for all$(\mu,v)\in\Omega, then \begin{align*} &\norm{R_N(t)\varphi}_1\\ &\le e^{\omega/a}\int_\Omega \Big| e^{-\omega\frac{(1-\mu)}{v}} \xi(\mu,v,t)h(v) \underbrace{\int_a^\infty\cdots\int_a^\infty}_\text{(N+1)times} \\ &\quad\times \prod_{j=1}^N h(v_j) \prod_{j=1}^{N+1}k(v_j) \xi\Big(1,v_N,t-\frac{\mu}{v}-\sum_{i=1}^{(N-1)}\frac{1}{v_i}\Big) \\ &\quad\times \gamma_1\Big(U_K\Big(t-\frac{\mu}{v}- \sum_{i=1}^N\frac{1}{v_i}\Big)\varphi\Big)(v_{N+1}) v_1v_2\cdots v_{N+1}\,dv_1\cdots dv_{N+1} \Big|\,d\mu\,dv. \end{align*} By the change of variablesx=t-\frac{\mu}{v}-\sum_{i=1}^N\frac{1}{v_i}$with$vdx=-d\mu, we infer that \begin{align*} &\norm{R_N(t)\varphi}_1\\ &\le e^{\omega/a}\int_a^\infty \int_0^t \Big| e^{-\omega\pare{x-t+\frac{1}{v}}}h(v) \underbrace{\int_a^\infty\cdots\int_a^\infty}_\text{(N+1)times} e^{-\omega\pare{\sum_{i=1}^N\frac{1}{v_i}}}\prod_{j=1}^Nh(v_j) \\ &\quad\times \prod_{j=1}^{N+1}k(v_j)\gamma_1\big(U_K(x)\varphi\big)(v_{N+1}) v_1v_2\cdots v_Ndv_1\cdots dv_{N+1}\Big| v\,dx\,dv\\ &\le e^{\omega/a}e^{\omega t} \Big[\int_a^\infty e^{-\frac{\omega}{v}}\abs{h(v)}vdv\Big] \Big[\int_a^\infty e^{-\frac{\omega}{v}}\abs{h(v)}\abs{k(v)}vdv\Big]^N\\ &\quad\times \int_0^te^{-\omega x}\int_a^\infty \abs{k(v_{N+1})}\abs{\gamma_1\left(U_K(x)\varphi\right)(v_{N+1}) v_{N+1}}\,dv_{N+1}\,dx. \end{align*} Ask\in L^\infty(J), then we obtain \begin{align*} \norm{R_N(t)\varphi}_1 &\le e^{\omega/a}e^{\omega t} \Big(\norm{k}_\infty\int_a^\infty e^{-\frac{\omega}{v}}\abs{h(v_i)}v\,dv \Big)^{N+1}\\ &\quad\times \int_0^te^{-\omega x}\int_a^\infty \abs{\gamma_1\left(U_K(x)\varphi\right)(v_{N+1})}v_{N+1}\,dv_{N+1}\,dx\\ &=e^{\omega/a}e^{\omega t}\norm{K_\omega}_{\mathcal{L}(Y_1)}^{N+1} \int_a^\infty \int_0^t e^{-\omega x}\abs{\gamma_1\left(U_K(x)\varphi\right) (v_{N+1})}v_{N+1}\,dx\,dv_{N+1} \end{align*} which, by \eqref{INEQUALITY:R1}, implies $$\norm{R_N(t)\varphi}_1 \le e^{\omega/a}e^{\omega t}\frac{ \norm{K_\omega}_{\mathcal{L}(Y_1)}^{N+1}}{1-\norm{K_\omega}_{\mathcal{L}(Y_1)}} \norm{\varphi}_1$$ and therefore $$\lim_{N\to\infty}\|U_K(t)-\sum_{m=0}^N U_m(t)\|_{\mathcal{L}(L^1(\Omega))} =\lim_{N\to\infty}\norm{R_N(t)}_{\mathcal{L}(L^1(\Omega))}=0$$ because of the first point of Theorem~\ref{AK:THE}. The proof is now complete. \end{proof} \section{The perturbed semigroup \eqref{E:MODEL}-\eqref{E:CAL}} \label{S:PER} In this section, we are going to prove that the perturbed model \eqref{E:MODEL}-\eqref{E:CAL} is governed by a strongly continuous semigroup like a linear perturbation of the unperturbed semigroup(U_K(t))_{t\ge0}$already studied. So, let us define the following two perturbation operators \begin{gather*} R\varphi (\mu,v)=\int_a^\infty r(\mu,v,v')\varphi (\mu,v')dv',\\ S\varphi(\mu,v)=-\sigma (\mu,v)\varphi(\mu,v), \end{gather*} where$\sigma$is given by \eqref{E:SIGMA}. Let us impose the following hypothesis \begin{itemize} \item[(H1)]$r$is measurable positive, and$\sigma\in L^\infty(\Omega)$. \end{itemize} Denoting $$\underline{\sigma}:=\operatorname{ess\,inf}_{(\mu,v)\in\Omega}\sigma(\mu,v) \quad\text{and}\quad \overline{\sigma}:=\operatorname{ess\,sup}_{(\mu,v)\in\Omega}\sigma(\mu,v),$$ we have the following result. \begin{lemma}\label{S+R:LEM} Suppose that {\rm (H1)} holds. Then,$S$and$R$are linear bounded operators from$L^1(\Omega)$into itself. Furthermore,$S+R$is a dissipative operator. \end{lemma} \begin{proof} Let$\varphi\in L^1(\Omega)$. The boundedness of the operators$S$and$Rclearly follows from \begin{align*} \norm{R\varphi}_1 &\le\int_0^1\int_a^\infty\int_a^\infty r(\mu,v,v')\abs{\varphi(\mu,v')}dv'dvd\mu\\ &=\int_0^1\int_a^\infty \Big[\int_a^\infty r(\mu,v,v')dv\Big]\abs{\varphi(\mu,v')}dv'd\mu\\ &=\int_0^1\int_a^\infty \sigma(\mu,v')\abs{\varphi(\mu,v')}dv'd\mu =\norm{S\varphi}_1 \end{align*} and $\norm{S\varphi}_1=\int_\Omega\sigma(\mu,v)\abs{\varphi(\mu,v)}\,d\mu\,dv \le\overline{\sigma}\norm{\varphi}_1.$ Furthermore, we have \begin{align*} &\langle\operatorname{sgn}\varphi, (S+R)\varphi\rangle\\ &= \int_\Omega\operatorname{sgn}\varphi(\mu,v)\pare{R\varphi(\mu,v)+S\varphi(\mu,v)}\,d\mu\,dv\\ &\le\int_0^1 \int_a^\infty \Big[\int_a^\infty r(\mu,v,v')dv\big] \abs{\varphi(\mu,v')}dv'd\mu - \int_\Omega \sigma(\mu,v)\abs{\varphi(\mu,v)}\,d\mu\,dv\\ &=\int_\Omega\sigma(\mu,v')\abs{\varphi(\mu,v')}\,d\mu\,dv' -\int_\Omega\sigma(\mu,v)\abs{\varphi(\mu,v)}\,d\mu\,dv \end{align*} and therefore $\langle\operatorname{sgn}\varphi, (S+R)\varphi\rangle\le0.$ The proof is complete. \end{proof} Let us define the perturbed operatorsL_K$and$T_K$as follows $$\label{LK:DEF} \begin{gathered} L_K:=A_K+S,\\ D(L_K)=D(A_K) \end{gathered}$$ and $$\label{TK:DEF} \begin{gathered} T_K:=L_K+R=A_K+S+R,\\ D(T_K)=D(A_K) \end{gathered}$$ for which we have the following generation results. \begin{lemma}\label{VK-TK:LEM1} Assume {\rm (H1)} and let$K$be a bounded operator with$\norm{K}_{\mathcal{L}(Y_1)}<1$. Then \begin{enumerate} \item The operator defined by \eqref{LK:DEF} generates, on$L^1(\Omega)$, a strongly continuous semigroup$(V_K(t))_{t\ge0}$satisfying $$\label{VK:R1} \norm{V_K(t)\varphi}_1 \le e^{-t\underline{\sigma}} \norm{\varphi}_1,\quad t\ge0,$$ for all$\varphi\in L^1(\Omega)$. \item The operator defined by \eqref{TK:DEF} generates, on$L^1(\Omega)$, a strongly continuous semigroup$(W_K(t))_{t\ge0}$satisfying $$\label{TK:R1} \norm{W_K(t)\varphi}_1 \le\norm{\varphi}_1,\quad t\ge0,$$ for all$\varphi\in L^1(\Omega)$. \end{enumerate} \end{lemma} \begin{proof} (1). First,$L_K=A_K+S$is a bounded linear perturbation of the generator$A_K$and therefore, Lemma~\ref{L:PER} implies that$L_K$is a generator of a strongly continuous semigroup which we denote as$(V_K(t))_{t\ge0}$. Next, Trotter's formula \eqref{E:TRO} implies $$\label{VK:E1} V_K(t)\varphi= \lim_{t\to\infty}\Big[e^{-\sigma t/n}U_K\big(\frac{t}{n}\big) \Big]^n\varphi , \quad t\ge0,$$ for all$\varphi\in L^1(\Omega)$. By \eqref{AK:R2}, we obtain $\norm{V_K(t)\varphi}_1\le \lim_{t\to\infty} \Big[e^{-\underline{\sigma}t/n}\cdot 1\Big]^n\norm{\varphi}_1 \le e^{-t\underline{\sigma}}\norm{\varphi}_1 \quad t\ge0,$ and therefore \eqref{VK:R1} holds. (2) By the third point of Lemma~\ref{AK:LEM1}, the operator$A_K$generates, on$L^1(\Omega)$, a strongly continuous semigroup of contractions. Furthermore, Lemma~\ref{S+R:LEM} implies that$S+R$is a bounded and dissipative operator. As we have,$D(A_K)\subset L^1(\Omega)=D(S+R)$and $$\norm{(S+R)\varphi}_1\leq\norm{S+R}\norm{\varphi}_1 =0.\norm{A_K\varphi}_1 +\norm{S+R}\norm{\varphi}_1$$ for all$\varphi\in D(A_K)$then, all conditions of Lemma~\ref{L:PER1} are clearly satisfied. The proof is complete. \end{proof} \begin{lemma}\label{VK-TK:LEM2} Assume {\rm (H1)} and let$K$be a compact operator with$\norm{K}_{\mathcal{L}(Y_1)}\ge1$. \begin{enumerate} \item The operator defined by \eqref{LK:DEF} generates, on$L^1(\Omega)$, a strongly continuous semigroup$(V_K(t))_{t\ge0}$satisfying $$\label{VK:R2} \norm{V_K(t)\varphi}_1 \le e^\frac{\lambda_0}{a}e^{t(\lambda_0-\underline{\sigma})}\norm{\varphi}_1 \quad t\ge0,$$ for all$\varphi\in L^1(\Omega)$. \item The operator defined by \eqref{TK:DEF} generates, on$L^1(\Omega)$, a strongly continuous semigroup$(W_K(t))_{t\ge0}$. \end{enumerate} \end{lemma} \begin{proof} (1) Following the proof of Lemma \ref{VK-TK:LEM1}, it suffices to show \eqref{VK:R2}. So, applying the norm \eqref{NORME} to Trotter's formula \eqref{VK:E1}, we obtain $\norme{V_K(t)\varphi}_1\le \lim_{t\to\infty} \Big[e^{-\underline{\sigma}t/n}e^{\frac{t}{n}\lambda_0}\Big]^n \norme{\varphi}_1 \le e^{t(\lambda_0-\underline{\sigma})}\norme{\varphi}_1, \quad t\ge0$ for all$\varphi\in L^1(\Omega)$, where we have used \eqref{LEM2:E8}. Now, \eqref{LEM2:E4} completes this part of the proof. (2) Clearly,$T_K=L_K+R$is a bounded linear perturbation of the generator$L_K$and therefore, Lemma~\ref{L:PER} implies that$T_K$is a generator too. The proof is now complete. \end{proof} We can summarize Lemmas~\ref{VK-TK:LEM1} and \ref{VK-TK:LEM2} as follows. \begin{theorem}\label{VK-TK:THE} Suppose that {\rm (H1)} holds and let$K$be an admissible operator whose abscissa is$\omega_0$. \begin{enumerate} \item The operator defined by \eqref{LK:DEF} generates, on$L^1(\Omega)$, a strongly continuous semigroup$(V_K(t))_{t\ge0}$satisfying $$\label{VK:R3} \norm{V_K(t)\varphi}_1 \le e^\frac{\omega_0}{a}e^{t(\omega_0-\underline{\sigma})}\norm{\varphi}_1 \quad t\ge0,$$ for all$\varphi\in L^1(\Omega)$. \item The operator defined by \eqref{TK:DEF} generates, on$L^1(\Omega)$, a strongly continuous semigroup$(W_K(t))_{t\ge0}$. \end{enumerate} \end{theorem} Let us finish this work with the following Remarks \begin{remark} \rm Inequality$a>0$has been used in many places in this work. So the open question is: What happens when$a=0$? \end{remark} \begin{remark} \rm According to Theorem \ref{VK-TK:THE}, we can say that the model \eqref{E:MODEL}-\eqref{E:CAL} is well-posed. However, the case corresponding to$\norm{K}_{\mathcal{L}(Y_1)}<1$in Lemma \ref{VK-TK:LEM1} is biologically uninteresting because the cell density is decreasing. Indeed, for all$t$and$s$with$t>s$we have $\norm{W_K(t)\varphi}_1 =\norm{W_K(t-s)W_K(s)\varphi}_1 \le e^{-(t-s)\underline{\sigma}}\norm{W_K(s)\varphi}_1 \le \norm{W_K(s)\varphi}_1$ for all initial data$\varphi\in L^1(\Omega)$. However, the case corresponding to$\norm{K}_{\mathcal{L}(Y_1)}>1\$ in Lemma~\ref{VK-TK:LEM2} means that the cell density is increasing during each mitotic. This corresponds to the most observed and biologically interesting case for which we ask the following natural question: What happens when the cell density is increasing? The answer is given in \cite{Boulanouar3}. \end{remark} \begin{thebibliography}{99} \bibitem{Boulanouar0} M. Boulanouar. \emph{New results for neutronic equations}. C. R. Acad. Sci. Paris, S\'erie I, 347, 623--626, 2009. \bibitem{Boulanouar1} M. Boulanouar. \emph{A Transport Equation in Cell Population Dynamics.} Diff and Int Equa., 13, 125--144, 2000. \bibitem{Boulanouar2} M. Boulanouar. \emph{Un mod\ele de Rotenberg avec la loi \a m\'emoire parfaite.} C. R. Acad. Sci. Paris, S\'erie I, 327, 955--958, 1998. \bibitem{Boulanouar3} M. Boulanouar. \emph{A Transport Equation in Cell Population Dynamics (II)}. Submitted. \bibitem{Boulanouar4} M. Boulanouar. \emph{Sur une \'equation de transport dans la dynamique des populations.} Preprint. \bibitem{Nagel} K. Engel and R. Nagel. \emph{One-Parameter Semigroups for Linear Evolution Equations.} Graduate texts in mathematics, 194, Springer-Verlag, New York, Berlin, Heidelberg, 1999. \bibitem{Rotenberg} M. Rotenberg \emph{Transport theory for growing cell populations.} J. Theor. Biol., 103, 181--199, 1983. \end{thebibliography} \end{document}