\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2012 (2012), No. 222, pp. 1--21.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2012 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2012/222\hfil Transport equation] {Transport equation for growing bacterial populations (II)} \author[M. Boulanouar \hfil EJDE-2012/222\hfilneg] {Mohamed Boulanouar} % in alphabetical order \address{Mohamed Boulanouar \newline LMCM-RSA, Universite de Poitiers, 86000 Poitiers, France} \email{boulanouar@gmail.com} \thanks{Submitted September 3, 2012. Published December 4, 2012.} \thanks{Supported by LMCM-RSA} \subjclass[2000]{92C37, 82D75} \keywords{Cell population dynamic; asynchronous exponential growth; \hfill\break\indent semigroups; compactness; general boundary condition} \begin{abstract} This article studies the growing bacterial population. Each bacterium is distinguished by its degree of maturity and its maturation velocity. To complete the study in \cite{Boulanouar3}, we describe the bacterial profile of this population by proving that the generated semigroup possesses an asynchronous exponential growth property. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \allowdisplaybreaks \newcommand{\abs}[1]{|#1|} \newcommand{\set}[1]{\{#1\}} \newcommand{\norm}[1]{\|#1\|} \newcommand{\norme}[1]{|||#1|||} \newcommand{\parent}[1]{(#1)} \newcommand{\bracket}[1]{[#1]} \section{Introduction} This work studies a model for growing bacterial population, partially studied in \cite{Boulanouar3}, in which each bacteria is distinguished by its degree of maturity $0\le\mu\le1$ and its maturation velocity $v$. As each bacteria may not become less mature, then its maturation velocity must be positive $(0\le a0$, we have recently proved in \cite{Boulanouar1} that the strongly continuous semigroup $V_K=(V_K(t))_{t\ge0}$ is compact for a large time $t>\tfrac{2}{a}$ which led to an easy computation of the essential type ($\omega_{\rm ess}(V_K)=-\infty$). Biologically speaking, the case $a>0$ means that after a transitory phase, all bacteria will be divided or dead. In contrary to $a>0$, the case $a=0$ means that the maturation velocities can be arbitrary small and at any time there may be bacteria that are not yet divided. Consequently, the bacterial population never goes out of the transitory phase, which explains the non-compactness of the generated semigroup $V_K=(V_K(t))_{t\ge0}$ and therefore the computation of its essential type, $\omega_{\rm ess}(V_K)$, presents a lot of difficulties. In the sequel we organize the work as follows \begin{itemize} \item[(3)] Generation Theorem \item[(4)] Stability and Domination \item[(5)] Asynchronous Exponential Growth \end{itemize} In third Section, we recall some properties of the generated semigroup $V_K=(V_K(t))_{t\ge0}$ governing the model \eqref{MODEL:EQUATION}, \eqref{MODEL:BC}. We also complete some claims already proved in \cite{Boulanouar3}. In Fourth Section, we study the stability of the generated semigroup $V_K=(V_K(t))_{t\ge0}$ with respect to the boundary operator $K$. Domination result is also given. In fifth Section, we prove that the generated semigroup possesses Asynchronous Exponential Growth property as follows \begin{lemma}[{\cite[Theorems 9.10 and 9.11]{Clement}}] \label{AEG:LEM} Let $U=(U(t))_{t\ge0}$ be a positive and irreducible strongly continuous semigroup, on the Banach lattice space $X$, satisfying the inequality $\omega_{\rm ess}(U)<\omega_0(U)$. Then, there exist a rank one projector $\mathbb{P}$ into $X$ and an $\varepsilon>0$ such that: for any $\eta\in (0,\varepsilon)$, there exists $M(\eta)\ge1$ satisfying \begin{equation*} \norm{e^{-\omega_0(U)t}U(t)-\mathbb{P}}_{\mathcal{L}(X)} \le M(\eta)e^{-\eta t}\quad t\ge0. \end{equation*} \end{lemma} Thanks to \cite[Thereom 8.7]{Clement}, the rank one projection $\mathbb{P}$ can be written as follows: $\mathbb{P}\varphi=<\varphi,\varphi_0^*>\varphi_0^*$, where $\varphi_0^*\in(X^*)_+$ is a strictly positive functional. A strongly continuous semigroup $U=(U(t))_{t\ge0}$ satisfying Lemma~\ref{AEG:LEM} possesses the \emph{asynchronous exponential growth} with \emph{intrinsic growth density} $\varphi_0^*$. Lemma~\ref{AEG:LEM} describes the bacterial profile whose privileged direction is mathematically explained by the vector $\varphi_0^*$. This is what the biologist observes in his laboratory. Finally, note the novelty of this work. For all used theoretical results, we refer the reader to \cite{Clement} or \cite{Nagel}. \section{Mathematical preliminaries} This section deals with some useful mathematical tools that we will need in the sequel. These tools concern strongly continuous semigroups of linear operators in Banach spaces and Banach lattice spaces. Let $X$ be a Banach space and let $U=(U(t))_{t\ge0}$ be a strongly continuous semigroup of linear operators, on $X$. Following \cite[Chapter IV]{Nagel}, the \emph{type} $\omega_0(U)$ and the \emph{essential type} $\omega_{\rm ess}(U)$ of the semigroup $U=(U(t))_{t\ge0}$ are given by \begin{gather} \label{TYPE:DEF} \omega_0(U)=\lim_{t\to\infty} \frac{\ln\norm{U(t)}_{\mathcal{L}(X)}}{t}\\ \label{TYPE-ESS:DEF} \omega_{\rm ess}(U)=\lim_{t\to\infty} \frac{\ln\norm{U(t)}_{\rm ess}}{t}, \end{gather} where $\norm{\cdot}_{\rm ess}$ denotes the norm of Calkin algebra $\mathcal{L}(X)\slash\mathcal{K}(X)$ with $\mathcal{K}(X)$ stands for the two-sided closed ideal in $\mathcal{L}(X)$ of all compact operators. The types $\omega_{\rm ess}(U)$ and $\omega_{\rm ess}(U)$ are always ordered as follows \begin{equation}\label{TYPE:ORD} \omega_{\rm ess}(U)\le\omega_0(U). \end{equation} \begin{lemma}\label{COMPACIT} Let $U=(U(t))_{t\ge0}$ and $V=(V(t))_{t\ge0}$ be two strongly continuous semigroups, on $X$, and let $\lambda\in\rho(U(t_0))$ for some $t_0>0$. If \begin{equation*} \Big[\Big(V(t_0)-U(t_0)\Big)\Big(\lambda-U(t_0)\Big)^{-1}\Big]^n \end{equation*} is a compact operator for some integer $n>0$, then $\omega_{\rm ess}(U)=\omega_{\rm ess}(V)$ \end{lemma} \begin{proof} It suffices to apply \cite{Voigt} for the operators $V(t_0)-U(t_0)$ and $U(t_0)$. \end{proof} \begin{lemma}[\cite{Greiner}]\label{GREINER} Let $(\Omega, \Sigma, \mu)$ be a positive measure space and let $S$ and $T$ be linear and bounded operators on $L^1(\Omega, \mu)$. Then \begin{itemize} \item[(1)] The set of all weakly compact operators is norm-closed subset. \item[(2)] If $T$ is weakly compact and $0\le S\le T$ then $S$ is weakly compact too. \item[(3)] If $S$ and $T$ are weakly compact, then $ST$ is compact. \end{itemize} \end{lemma} \section{Generation Theorem} In this section, we recall briefly some properties of the model \eqref{MODEL:EQUATION}, \eqref{MODEL:BC} and its associated semigroup $V_K=(V_K(t))_{t\ge0}$ (see \cite{Boulanouar3}). Also, other needed properties will be proved. So, let us consider the functional framework $L^1(\Omega)$ whose norm is \begin{equation}\label{NORM:DEF} \norm{\varphi}_1=\int_\Omega\abs{\varphi(\mu,v)}\,d\mu\,dv, \end{equation} where $\Omega=(0,1)\times(0,\infty):=I\times J$. We also consider our regularity space \begin{equation*} W_1=\big\{\varphi\in L^1(\Omega)\;\, v\frac{\partial\varphi}{\partial \mu}\in L^1(\Omega)\quad\text{and}\quad v\varphi\in L^1(\Omega)\big\} \end{equation*} and the trace space $Y_1:=L^1(J,v\,dv)$ whose norms are \begin{equation*} \norm{\varphi}_{W_1}= \norm{v\frac{\partial\varphi}{\partial\mu}}_1+\norm{v\varphi}_1 \quad\text{and}\quad \norm{\psi}_{Y_1}=\int_0^\infty\abs{\psi(v)}v\,dv. \end{equation*} In this context, our recent result \cite[Theorem 2.2]{Boulanouar2} leads to \begin{lemma}\label{TRACES:LEM} The trace mappings $\gamma_0\varphi=\varphi(0,\cdot)$ and $\gamma_1\varphi=\varphi(1,\cdot)$ are linear bounded from $W_1$ into $Y_1$. \end{lemma} Let $T_0$ be the following unbounded operator \begin{equation}\label{T0:DEF} \begin{gathered} T_0\varphi=-v\frac{\partial \varphi}{\partial\mu}\quad\text{on the domain,}\\ D(T_0)=\{\varphi\in W_1\;\,\gamma_0\varphi=0\}, \end{gathered} \end{equation} corresponding to the model \eqref{MODEL:EQUATION}, \eqref{MODEL:BC} without bacterial division and without bacterial mortality. Some of its properties can be summarized as follows. \begin{lemma}\label{T0:LEM} The operator $T_0$ generates, on $L^1(\Omega)$, a strongly continuous positif semigroup $U_0=(U_0(t))_{t\ge0}$ of contraction given by \begin{equation}\label{U0:DEF} U_0(t)\varphi(\mu,v):=\chi(\mu,v,t)\varphi(\mu-tv,v), \end{equation} where \begin{equation}\label{CHI:DEF} \chi(\mu,v,t)= \begin{cases} 1 & \text{if } \mu\ge tv;\\ 0 & \text{if } \mu0\Longrightarrow\norm{\overline{K}_\lambda}<1. \end{equation} Next, if {\rm (Kc)} holds, then \begin{align*} \norm{\overline{K}_\lambda}_{\mathcal{L}(Y_1)} &=\sup_{\psi\in B}\norm{\theta_\lambda K\psi}_{Y_1}\\ &=\sup_{\varphi\in K(B)}\norm{\theta_\lambda\varphi}_{Y_1}\\ &\le\sup_{\varphi\in\overline{K(B)}} \norm{\theta_\lambda\varphi}_{Y_1}, \end{align*} where $B_0$ be the unit ball into $Y_1$. By virtue of the compactness of $\overline{K(B)}$, there exists $\varphi_0\in\overline{K(B)}$ such that \begin{equation*} \norm{\overline{K}_\lambda}_{\mathcal{L}(Y_1)}\le \norm{\theta_\lambda\varphi_0}_{Y_1} =\int_0^\infty e^{-\lambda/v}\abs{\varphi_0(v)}v\,dv \end{equation*} which leads to \begin{equation*} \lim_{\lambda\to\infty}\norm{\overline{K}_\lambda}_{\mathcal{L}(Y_1)} \le\lim_{\lambda\to\infty}\int_0^\infty e^{-\lambda/v}\abs{\varphi_0(v)}v\,dv=0 \end{equation*} and therefore \begin{equation}\label{KLAMBDA+BAR:E2} \norm{\overline{K}_\lambda}<1 \quad \text{for all large }\lambda. \end{equation} Now, by \eqref{KLAMBDA+BAR:E1} and \eqref{KLAMBDA+BAR:E2} we can infer that both hypotheses {\rm (Kb)} and {\rm (Kc)} imply that $\norm{\overline{K}_\lambda}<1$ for large $\lambda$. Suppose now that $K$ is positive. So, for all $\lambda\ge0$ we clearly get that \begin{equation}\label{KLAMBDA+BAR:E4} K_\lambda\le K\quad\text{and}\quad \overline{K}_\lambda\le K \end{equation} On the other hand, due to the obvious relation $K_\lambda K=K\overline{K}_\lambda$, it follows by induction that \begin{equation}\label{KLAMBDA+BAR:E3} K_\lambda^nK=K\overline{K}_\lambda^n\quad\text{for all integers $n\ge1$} \end{equation} which leads, by \eqref{KLAMBDA+BAR:E4}, to \begin{equation*} K_\lambda^{n+1}\le K\overline{K}_\lambda^n \quad\text{and}\quad \overline{K}_\lambda^{n+1}\le K_\lambda^nK \end{equation*} for all integers $n\ge1$ and therefore \begin{gather*} \norm{K_\lambda^{(n+1)}}^{\frac{1}{(n+1)}}\le\norm{K}^{\frac{1}{(n+1)}} \parent{\norm{\overline{K}_\lambda^n}^{\frac{1}{n}}}^\frac{n}{(n+1)}, \\ \norm{\overline{K}_\lambda^{(n+1)}}^{\frac{1}{(n+1)}} \le\parent{\norm{K_\lambda^n}^{\frac{1}{n}}}^{\frac{n}{(n+1)}}\norm{K}^{\frac{1}{(n+1)}}. \end{gather*} This easily leads to \eqref{KLAMBDA+BAR:R2} and completes the proof. \end{proof} Some useful properties of the unbounded operator \eqref{TK:DEF} are given next. \begin{lemma}\label{UK:LEM} Let $K$ be an admissible operator. {\rm(1)} The operator $T_K$ generates, on $L^1(\Omega)$, a strongly continuous semigroup $U_K=(U_K(t))_{t\ge0}$ satisfying \begin{equation}\label{UK:R1} U_K(t)\varphi(\mu,v)=U_0(t)\varphi(\mu,v)+\xi(\mu,v,t) K\parent{\gamma_1U_K\parent{t-\tfrac{\mu}{v}}\varphi}(v) \end{equation} for almost all $(\mu,v)\in\Omega$, where \begin{equation}\label{XI:DEF} \xi(\mu,v,t)= \begin{cases} 0 & \text{if } \mu\ge tv;\\ 1 & \text{if } \mu< tv. \end{cases} \end{equation} {\rm(2)} For all large $\lambda$, we have \begin{equation}\label{UK:R2} (\lambda-T_K)^{-1}\varphi= \varepsilon_\lambda K(I-\overline{K}_\lambda)^{-1}\gamma_1(\lambda - T_0)^{-1}\varphi +(\lambda -T_0)^{-1}\varphi, \end{equation} where $\varepsilon_\lambda(\mu,v)=e^{-\lambda\frac{\mu}{v}}$. {\rm(3)} If $K$ is positive, then $U_K=(U_K(t))_{t\ge0}$ is a positive semigroup and \begin{equation}\label{UK:R3} U_K(t)\ge U_0(t)\quad t\ge0. \end{equation} \end{lemma} \begin{proof} All the announced properties, but \eqref{UK:R2} and \eqref{UK:R3}, follow from \cite[Theorems 3.2 and 4.1 and Proposition 6.1]{Boulanouar3}. So, let us prove \eqref{UK:R2} and \eqref{UK:R3}. Let $\lambda$ be large and let $\varphi\in L^1(\Omega)$. Thanks to \cite[Proposition 3.1]{Boulanouar3} we infer that \begin{equation*} (\lambda-T_K)^{-1}= \varepsilon_\lambda(I- K_\lambda)^{-1} K\gamma_1(\lambda - T_0)^{-1} +(\lambda -T_0)^{-1} \end{equation*} which leads, by \eqref{KLAMBDA+BAR:R1} and \eqref{KLAMBDA+BAR:E3}, to \begin{align*} (\lambda-T_K)^{-1}\varphi-(\lambda -T_0)^{-1}\varphi &=\varepsilon_\lambda\Big(\sum_{n\ge0}K_\lambda^n\Big)K\gamma_1(\lambda - T_0)^{-1}\varphi\\ &=\varepsilon_\lambda K\Big(\sum_{n\ge0}\overline{K}_\lambda^n\Big)\gamma_1(\lambda - T_0)^{-1}\varphi\\ &=\varepsilon_\lambda K(I-\overline{K}_\lambda)^{-1}\gamma_1(\lambda - T_0)^{-1}\varphi \end{align*} and therefore \eqref{UK:R2} follows. Next, let $\varphi\in(L^1(\Omega))_+$. As $U_0=(U_0(t))_{t\ge0}$ is a positive semigroup (Lemma~\ref{T0:LEM}), it follows that $(\lambda -T_0)^{-1}\varphi$ is a positive function and therefore $\gamma_1(\lambda -T_0)^{-1}\varphi$ is a positive function too. So, the computation above clearly leads to \begin{equation*} (\lambda-T_K)^{-1}\varphi\ge(\lambda -T_0)^{-1}\varphi \end{equation*} because of the positivity of $K$ and therefore \begin{equation} \big[\lambda(\lambda-T_K)^{-1}\big]^n\varphi \ge \big[\lambda(\lambda-T_0)^{-1}\big]^n\varphi \end{equation} for all integers $n\ge1$. Putting now $\lambda=\frac{n}{t}$ $(t>0)$ and passing at the limit $n\to\infty$ we infer that \begin{equation} \lim_{n\to\infty}\Big[\frac{n}{t}\parent{\frac{n}{t}-T_K}^{-1}\Big]^n\varphi \ge \lim_{n\to\infty}\Big[\frac{n}{t}\parent{\frac{n}{t}-T_0}^{-1}\Big]^n\varphi \end{equation} which leads to \eqref{UK:R3} because of the exponential formula. \end{proof} Finally, let us consider the general model \eqref{MODEL:EQUATION}, \eqref{MODEL:BC} corresponding to the unbounded operator $L_K:=T_K+S$ on the domain $D(L_K)=D(T_K)$ and for which we have \begin{lemma}\label{VK:LEM} Let $K$ be an admissible operator. If the hypothesis \eqref{H-sigma} holds, then we have {\rm(1)} The operator $L_K$ generates, on $L^1(\Omega)$, a strongly continuous semigroup $V_K=(V_K(t))_{t\ge0}$. Furthermore \begin{equation}\label{VK:R1} V_K(t)=U_K(t)+\int_0^tU_K(t-s)SV_K(s)ds\quad t\ge0. \end{equation} {\rm(2)} Suppose that $K$ is positive. Then the semigroup $V_K=(V_K(t))_{t\ge0}$ is positive too and \begin{equation}\label{VK:R2} V_K(t)\ge V_0(t)\quad t\ge0. \end{equation} Moreover, if $K$ is irreducible then $V_K=(V_K(t))_{t\ge0}$ is also irreducible. {\rm(3)} If $K$ is a positive, irreducible and compact operator such that \begin{equation*} r(K_{\overline{\sigma}-\underline{\sigma}})>1 \end{equation*} then the type $\omega_0(V_K)$ of the semigroup $V_K=(V_K(t))_{t\ge0}$ satisfies to \begin{equation}\label{VK:R3} \omega_0(V_K)>-\underline{\sigma}, \end{equation} where $\underline{\sigma}$ is given by \eqref{INFSIGMA:DEF} and \begin{equation}\label{SUPSIGMA:DEF} \overline{\sigma}:=\operatorname{ess\,sup}_{(\mu,v)\in\Omega}\sigma(\mu,v). \end{equation} \end{lemma} \begin{proof} Almost all the announced properties follow from \cite[Theorem 5.1]{Boulanouar3} and \cite[Theorem 6.1]{Boulanouar3}. So, it only remains to prove \eqref{VK:R2} and \eqref{VK:R3}. \par Let $t>0$ and let $\varphi\in(L^1(\Omega))_+$. Then \eqref{UK:R3} clearly leads to \begin{equation*} \Big[e^{-\tfrac{t}{n}\sigma}U_K(\tfrac{t}{n})\Big]^n\varphi \ge \Big[e^{-\tfrac{t}{n}\sigma}U_0(\tfrac{t}{n})\Big]^n\varphi \quad\text{for all integers $n\in \mathbb{N}$.} \end{equation*} Passing at the limit $n\to\infty$, then \eqref{VK:R2} follows because of Trotter Formula. Finally, note that \eqref{VK:R3} follows from \cite[Th.7.1]{Boulanouar3} together with \eqref{KLAMBDA+BAR:R2}. \end{proof} We end this section by the following particular case \begin{corollary}\label{VK:COR} Let $K$ be a linear bounded operator from $Y_1$ into itself such that $\norm{K}<1$. If the hypothesis $(\mathrm{H}_\sigma)$ holds, then the semigroup $V_K=(V_K(t))_{t\ge0}$ satisfies \begin{equation}\label{VK:COR:R1} \norm{V_K(t)}_{\mathcal{L}(L^1(\Omega))}\le e^{-t\underline{\sigma}}\quad t\ge0. \end{equation} \end{corollary} The above corollary forllows from Lemma~\ref{VK:LEM} above together with \cite[Corollary.~5.1]{Boulanouar3}. \section{Stability and domination} In this section we are concerned with stability and domination results of the unperturbed semigroup $U_K=(U_K(t))_{t\ge0}$. That is one of the most useful results which will be used to insure Asynchronous Exponential Growth property for the semigroup $V_K=(V_K(t))_{t\ge0}$. Before we start, let us give the following useful result \begin{lemma}\label{STABILITY:LEM1} Let $K$ be an admissible operator and let $\lambda$ be large. Then, for all $\varphi\in L^1(\Omega)$, we have \begin{equation}\label{STABILITY:LEM1:R1} \int_0^\infty\int_0^\infty e^{-\lambda t}\abs{\gamma_1\big(U_K\parent{t}\varphi\big)(v)}v\,dt\,dv \le\frac{1}{1-\norm{\overline{K}_\lambda}}\norm{\varphi}_1, \end{equation} where $\overline{K}_\lambda$ is given by \eqref{KLAMBDA+BAR:DEF}. \end{lemma} \begin{proof} Let $\lambda$ be large and let $\varphi\in W_1$. Applying the trace mapping $\gamma_1$ to \eqref{UK:R1}, we infer that \begin{equation*} \gamma_1(U_K(t)\varphi)(v)=\gamma_1(U_0(t)\varphi)(v) +\xi(1,v,t) \left[K\gamma_1\parent{U_K\parent{t-\tfrac{1}{v}}\varphi}\right](v) \end{equation*} for all $t\ge0$ and for almost all $v\in(0,\infty)$. Integrating it, we obtain that \begin{equation}\label{STABILITY:LEM1:R10} \begin{aligned} &\int_0^\infty\int_0^\infty e^{-\lambda t}\abs{\gamma_1\big(U_K\parent{t}\varphi\big)(v)}v\,dt\,dv\\ &\le\int_0^\infty\int_0^\infty e^{-\lambda t}\abs{\gamma_1\big(U_0\parent{t}\varphi\big)(v)}v\,dt\,dv\\ &\quad +\int_0^\infty\int_0^\infty e^{-\lambda t}\xi(1,v,t) \abs{\parent{K\gamma_1\parent{U_K\parent{t-\tfrac{1}{v}}\varphi}}(v)}v\,dt\,dv\\ & :=I+J. \end{aligned} \end{equation} Thanks to Lemma~\ref{T0:LEM}, the term $I$ becomes \begin{align*} I&=\int_0^\infty\int_0^\infty e^{-\lambda t}\abs{\gamma_1(U_0(t)\varphi)(v)}v\,dvdt\\ &\le\int_0^\infty\int_0^\infty\abs{\chi(1,v,t)\varphi(1-tv,v)}vdt dv\\ &=\int_0^\infty\int_{1-tv}^1\abs{\varphi(\mu,v)}d\mu dt \end{align*} and therefore \begin{equation}\label{STABILITY:LEM1:R20} I\le\norm{\varphi}_1. \end{equation} For the term $J$ we have \begin{align*} J&=\int_0^\infty\int_0^\infty e^{-\lambda t} \xi(1,v,t) \abs{\parent{K\gamma_1\parent{U_K\parent{t-\tfrac{1}{v}}\varphi}}(v)}v\,dt\,dv\\ &=\int_0^\infty\int_0^\infty e^{-\lambda\parent{x+\frac{1}{v}}} \abs{\parent{K\gamma_1\parent{U_K(x)\varphi}}(v)}v\,dx\,dv\\ &=\int_0^\infty e^{-\lambda x}\Big[\int_0^\infty e^{-\lambda/v} \abs{\parent{K\gamma_1\parent{U_K(x)\varphi}}(v)}v\,dv\Big]dx\\ &=\int_0^\infty e^{-\lambda x}\Big[\int_0^\infty \abs{\parent{\overline{K}_\lambda\gamma_1\parent{U_K(x)\varphi}}(v)}v\,dv\Big]dx \end{align*} which leads, by the boundedness of $\overline{K}_\lambda$ (Lemma~\ref{KLAMBDA+BAR:LEM}), to \begin{equation}\label{STABILITY:LEM1:R30} J\le\norm{\overline{K}_\lambda}\int_0^\infty\int_0^\infty e^{-\lambda x}\abs{\gamma_1\parent{U_K(x)\varphi}(v)}v\,dx\,dv \end{equation} Now, \eqref{STABILITY:LEM1:R10} together with \eqref{STABILITY:LEM1:R20} and \eqref{STABILITY:LEM1:R30} clearly imply that \begin{equation*} \parent{1-\norm{\overline{K}_\lambda}}\int_0^\infty\int_0^\infty e^{-\lambda t}\abs{\gamma_1\big(U_K\parent{t}\varphi\big)(v)}v\,dt\,dv \le\norm{\varphi}_1 \end{equation*} and therefore \eqref{STABILITY:LEM1:R1} holds because of \eqref{KLAMBDA+BAR:R1}. Finally, the density of $W^1$ into $L^1(\Omega)$ achieves the proof. \end{proof} Now, we are ready to give the main result of this section. \begin{theorem}\label{STABILITY:THE} Let $K$ be an admissible operator and let $(K_n)_n$ be a sequence of admissible operators such that \begin{equation}\label{STABILITY:THE:R1} \lim_{n\to\infty}\norm{K_n-K}_{\mathcal{L}(Y_1)}=0. \end{equation} Then, for all $t\ge0$, we have \begin{equation}\label{STABILITY:THE:R2} \lim_{n\to\infty}\norm{U_{K_n}(t)-U_K(t)}_{\mathcal{L}\left(L^1(\Omega)\right)}=0. \end{equation} \end{theorem} \begin{proof} Let $\lambda$ be large and let $\varphi\in L^1(\Omega)$. In the sequel, we are going to divide this proof in two steps. \medskip \noindent\textbf{Step I}. If $H$ denotes another admissible operator, then by \eqref{UK:R1} it follows that \begin{equation}\label{STABILITY:E10} U_H(t)\varphi-U_K(t)\varphi :=A(t)\varphi+B(t)\varphi\quad t\ge0, \end{equation} where \begin{gather} A(t)\varphi(\mu,v)=\xi(\mu,v,t)(H-K)\gamma_1\parent{U_H\parent{t-\tfrac{\mu}{v}}\varphi}(v) \label{STABILITY:E20}\\ B(t)\varphi(\mu,v)=\xi(\mu,v,t)K\gamma_1 \Big(U_H\parent{t-\tfrac{\mu}{v}}\varphi-U_K \parent{t-\tfrac{\mu}{v}}\varphi\Big)(v) \label{STABILITY:E30} \end{gather} for almost all $(\mu,v)\in\Omega$. Furthermore, applying $\gamma_1$ to \eqref{STABILITY:E10} we obtain that \begin{equation}\label{STABILITY:E40} \gamma_1\Big(U_H(t)\varphi-U_K(t)\varphi\Big)= \gamma_1\big(A(t)\varphi\big)+\gamma_1\big(B(t)\varphi\big), \end{equation} where \begin{gather} \gamma_1\big(A(t)\varphi\big)(v) =\xi(1,v,t)(H-K)\gamma_1\parent{U_H\parent{t-\tfrac{1}{v}}\varphi}(v) \label{STABILITY:E50}\\ \gamma_1\big(B(t)\varphi\big)(v) =\xi(1,v,t)K\gamma_1 \Big(U_H\parent{t-\tfrac{1}{v}}\varphi-U_K\parent{t-\tfrac{1}{v}}\varphi\bigg)(v) \label{STABILITY:E60} \end{gather} for almost all $v\in(0,\infty)$. So, multiplying \eqref{STABILITY:E40} by $e^{-\lambda t}$ and integrating it over $\in(0,\infty)\times(0,\infty)$, we infer that \begin{equation}\label{STABILITY:E70} \begin{aligned} &\int_0^\infty\int_0^\infty e^{-\lambda t}\abs{\gamma_1\Big(U_H(t)\varphi-U_K(t)\varphi\Big)(v)}v\,dt\,dv\\ &\leq\int_0^\infty\int_0^\infty e^{-\lambda t}\abs{\gamma_1\big(A(t)\big)(v)}v\,dt\,dv +\int_0^\infty\int_0^\infty e^{-\lambda t}\abs{\gamma_1\big(B(t)\big)(v)}v\,dt\,dv\\ & :=I_1+I_2. \end{aligned} \end{equation} Firstly, thanks to \eqref{STABILITY:E50} the term $I_1$ becomes \begin{align*} I_1&=\int_0^\infty\int_0^\infty e^{-\lambda t}\xi(1,v,t)\abs{(H-K)\gamma_1 \parent{U_H\parent{t-\tfrac{1}{v}}\varphi}(v)}v\,dt\,dv\\ &=\int_0^\infty\int_0^\infty e^{-\lambda\parent{x+\frac{1}{v}}}\abs{(H-K)\gamma_1 \parent{U_H(x)\varphi}(v)}v\,dx\,dv\\ &=\int_0^\infty e^{-\lambda x}\bracket{\int_0^\infty \abs{(H-K)\gamma_1\parent{U_H(x)\varphi}(v)}v\,dv}dx\\ &\le\norm{H-K} \int_0^\infty\int_0^\infty e^{-\lambda x} \abs{\gamma_1\parent{U_H(x)\varphi}(v)}v\,dx\,dv \end{align*} which leads, by Lemma \ref{STABILITY:LEM1} (with $H$ instead $K$), to \begin{equation}\label{STABILITY:E80} I_1\le\frac{\norm{H-K}}{1-\norm{\overline{H}_\lambda}}\norm{\varphi}_1. \end{equation} Thanks to \eqref{STABILITY:E60}, the term $I_2$ becomes \begin{align*} I_2&=\int_0^\infty\int_0^\infty e^{-\lambda t}\abs{\gamma_1\big(B(t)\big)(v)}v\,dt\,dv\\ &=\int_0^\infty\int_0^\infty e^{-\lambda t}\xi(1,v,t) \big|K\gamma_1\Big(U_H\parent{t-\tfrac{1}{v}}\varphi-U_K \parent{t-\tfrac{1}{v}}\varphi\Big)(v)\big|v\,dt\,dv\\ &=\int_0^\infty\int_0^\infty e^{-\lambda\parent{x+\frac{1}{v}}} \big|K\gamma_1\Big(U_H(x)\varphi-U_K(x)\varphi\Big)(v)\big|v\,dx\,dv\\ &=\int_0^\infty e^{-\lambda x}\bracket{\int_0^\infty \abs{\overline{K}_\lambda\gamma_1\Big(U_H(x)\varphi-U_K(x)\varphi\Big)(v)}v\,dv}dx\\ &\le\norm{\overline{K}_\lambda}\int_0^\infty\int_0^\infty e^{-\lambda x} \abs{\gamma_1\Big(U_H(x)\varphi-U_K(x)\varphi\Big)(v)}v\,dx\,dv\\ \end{align*} which leads, by the boundedness of $\overline{K}_\lambda$ (Lemma~\ref{KLAMBDA+BAR:LEM}), to \begin{equation}\label{STABILITY:E90} I_2\le\norm{\overline{K}_\lambda}\int_0^\infty\int_0^\infty e^{-\lambda x} \abs{\gamma_1\Big(U_H(x)\varphi-U_K(x)\varphi\Big)(v)}v\,dx\,dv. \end{equation} Now, \eqref{STABILITY:E70} together with \eqref{STABILITY:E80} and \eqref{STABILITY:E90} clearly imply that \begin{equation*} \parent{1-\norm{\overline{K}_{\lambda}}}\int_0^\infty\int_0^\infty e^{-\lambda t}\abs{\gamma_1\Big(U_H(t)\varphi-U_K(t)\varphi\Big)(v)}v\,dt\,dv \le\frac{\norm{H-K}\norm{\varphi}_1} {1-\norm{\overline{H}_\lambda}} \end{equation*} and therefore, \eqref{KLAMBDA+BAR:R1} leads to \begin{equation}\label{STABILITY:E100} \int_0^\infty\int_0^\infty e^{-\lambda t}\abs{\gamma_1\Big(U_H(t)\varphi-U_K(t)\varphi\Big)(v)}v\,dt\,dv \le\frac{\norm{H-K}\norm{\varphi}_1} {\parent{1-\norm{\overline{K}_{\lambda}}}\parent{1-\norm{\overline{H}_\lambda}}}. \end{equation} On the other hand, by \eqref{STABILITY:E10} it follows that \begin{equation}\label{STABILITY:E110} \begin{aligned} \norm{U_H(t)\varphi-U_K(t)\varphi}_1 &\le\int_\Omega\abs{A(t)\varphi(\mu,v)}\,d\mu\,dv +\int_\Omega\abs{B(t)\varphi(\mu,v)}\,d\mu\,dv \\ &:=J_1+J_2. \end{aligned} \end{equation} for all $t\ge0$. So, using \eqref{STABILITY:E20} the term $J_1$ becomes \begin{equation*} \begin{aligned} J_1&=\int_\Omega\xi(\mu,v,t)\abs{(H-K)\gamma_1 \parent{U_H\parent{t-\tfrac{\mu}{v}}\varphi}(v)}\,d\mu\,dv\\ &\le\int_0^\infty\int_0^1e^{\lambda\frac{\mu}{v}}\xi(\mu,v,t) \abs{(H-K)\gamma_1\parent{U_H\parent{t-\tfrac{\mu}{v}}\varphi}(v)}\,d\mu\,dv\\ &\le e^{\lambda t}\int_0^\infty\int_0^\infty e^{-\lambda x}\abs{(H-K)\gamma_1\parent{U_H(x)\varphi}(v)}v\,dx\,dv\\ &\le e^{\lambda t}\int_0^\infty \Big[\int_0^\infty e^{-\lambda x}\abs{(H-K)\gamma_1\parent{U_H(x) \varphi}(v)}v\,dv\Big]dx\\ &\le e^{\lambda t}\norm{H-K} \int_0^\infty\int_0^\infty e^{-\lambda x} \abs{\gamma_1\parent{U_H(x)\varphi}(v)}v\,dx\,dv \end{aligned} \end{equation*} which leads, by Lemma \ref{STABILITY:LEM1} (with $H$ instead $K$), to \begin{equation}\label{STABILITY:E120} J_1\le e^{\lambda t}\frac{\norm{H-K}}{1-\norm{\overline{H}_\lambda}} \norm{\varphi}_1. \end{equation} Using \eqref{STABILITY:E30} the term $J_2$ becomes \begin{align*} J_2 &=\int_\Omega\xi(\mu,v,t)\Big|K\gamma_1 \Big(U_H\parent{t-\tfrac{\mu}{v}}\varphi-U_K \parent{t-\tfrac{\mu}{v}}\varphi\Big)(v)\Big|\,d\mu\,dv\\ &\le\int_0^\infty\int_0^1e^{\lambda\frac{\mu}{v}}\xi(\mu,v,t) \Big|K\gamma_1\Big(U_H\parent{t-\tfrac{\mu}{v}}\varphi-U_K \parent{t-\tfrac{\mu}{v}}\varphi\Big)(v)\Big|\,d\mu\,dv\\ &\le e^{\lambda t}\int_0^\infty\int_0^\infty e^{-\lambda x} \abs{K\gamma_1\Big(U_H(x)\varphi-U_K(x)\varphi\Big)(v)}v\,dx\,dv\\ &\quad\times e^{\lambda t}\int_0^\infty e^{-\lambda x} \Big[\int_0^\infty \abs{K\gamma_1\Big(U_H(x)\varphi-U_K(x)\varphi\Big)(v)}v\,dv\Big]dx\\ &\le e^{\lambda t}\norm{K} \int_0^\infty\int_0^\infty e^{-\lambda x} \big|\gamma_1\Big(U_H(x)\varphi-U_K(x)\varphi\Big)(v)\big|v\,dx\,dv \end{align*} which leads, by \eqref{STABILITY:E100}, to \begin{equation}\label{STABILITY:E130} J_2\le e^{\lambda t}\frac{\norm{K}\norm{H-K}} {\parent{1-\norm{\overline{K}_{\lambda}}} \parent{1-\norm{\overline{H}_\lambda}}}\norm{\varphi}_1. \end{equation} Now, \eqref{STABILITY:E110} together with \eqref{STABILITY:E120} and \eqref{STABILITY:E130} lead to \begin{equation*} \norm{U_H(t)\varphi-U_K(t)\varphi}_1 \le\frac{e^{\lambda t} \parent{\norm{K}+1}\norm{H-K}}{\parent{1-\norm{\overline{K}_{\lambda}}}\parent{1-\norm{\overline{H}_\lambda}}} \norm{\varphi}_1 \end{equation*} and therefore, for all $t\ge0$, we have \begin{equation}\label{STABILITY:E140} \norm{U_H(t)-U_K(t)}_{\mathcal{L}\parent{L^1(\Omega)}} \le\frac{e^{\lambda t}\parent{\norm{K}+1}\norm{H-K}} {\parent{1-\norm{\overline{K}_{\lambda}}} \parent{1-\norm{\overline{H}_\lambda}}}. \end{equation} \noindent\textbf{Step II}. Now, let $(K_n)_n\subset\mathcal{L}(Y_1)$ be a sequence of admissible operators such that \eqref{STABILITY:THE:R1} holds. As, we have \begin{equation*} \big|\norm{\overline{{K_n}}_{{}_\lambda}} -\norm{\overline{K}_{\lambda}}\big| \le\norm{\overline{{K_n}}_{{}_\lambda}-\overline{K}_{\lambda}} \le\norm{K_n-K} \end{equation*} for all integers $n\ge1$, it follows that \begin{equation}\label{STABILITY:E150} \lim_{n\to\infty}\norm{\overline{{K_n}}_{{}_\lambda}} =\norm{\overline{K}_{\lambda}}. \end{equation} On the other hand, putting $H=K_n$ into \eqref{STABILITY:E140} we obtain that \begin{equation}\label{STABILITY:E160} \norm{U_{K_n}(t)-U_K(t)}_{\mathcal{L}\parent{L^1(\Omega)}} \le\frac{e^{\lambda t}\parent{\norm{K}+1}\norm{K_n-K}} {\parent{1-\norm{\overline{K}_{\lambda}}} \parent{1-\norm{\overline{{K_n}}_{\lambda}}}} \end{equation} for all integers $n\ge1$. Passing now at the limit $n\to\infty$ into \eqref{STABILITY:E160} and using \eqref{STABILITY:THE:R1} and \eqref{STABILITY:E150}, we finally infer that \eqref{STABILITY:THE:R2}. The proof is now achieved. \end{proof} We complete this section by the following domination result. \begin{theorem}\label{DOMINATION:THE} Let $K$ and $H$ be two admissible operators. If $H$ is a positive operator and \begin{equation}\label{DOMINATION:THE:R1} \abs{K\psi}\le H\abs{\psi} \end{equation} for all $\psi\in Y_1$, then \begin{equation}\label{DOMINATION:THE:R2} \abs{\big[U_K(t)-U_0(t)\big]\varphi}\le \big[U_H(t)-U_0(t)\big]\abs{\varphi} \quad t\ge0 \end{equation} for all $\varphi\in L^1(\Omega)$. \end{theorem} \begin{proof} Let $\lambda$ be large. Firstly, note that \eqref{DOMINATION:THE:R1} obviously implies that $\abs{\overline{K}_\lambda\psi}\le\overline{H}_\lambda\abs{\psi}$ for all $\psi\in Y_1$ and by induction il follows that \begin{equation}\label{DOMINATION:THE:E10} \abs{\overline{K}_\lambda^n\psi}\le\overline{H}_\lambda^n\abs{\psi}. \end{equation} Next, let $\varphi\in L^1(\Omega)$. Due to \eqref{UK:R2} and \eqref{DOMINATION:THE:R1} we clearly get that \begin{align*} \abs{(\lambda-T_K)^{-1}\varphi} &\le \abs{\varepsilon_\lambda K(I-\overline{K}_\lambda)^{-1}\gamma_1(\lambda - T_0)^{-1}\varphi} +\abs{(\lambda -T_0)^{-1}\varphi}\\ &\le\varepsilon_\lambda H\abs{(I- \overline{K}_\lambda)^{-1}\gamma_1(\lambda - T_0)^{-1}\varphi} +\abs{(\lambda -T_0)^{-1}\varphi} \end{align*} which leads, by \eqref{KLAMBDA+BAR:R1} and \eqref{DOMINATION:THE:E10}, to \begin{align*} \abs{(\lambda-T_K)^{-1}\varphi} &\le\varepsilon_\lambda H \Big|\sum_{n\ge0}\overline{K}_\lambda^n\gamma_1(\lambda - T_0)^{-1}\varphi\Big| +\abs{(\lambda -T_0)^{-1}\varphi}\\ &\le\varepsilon_\lambda H \sum_{n\ge0}\abs{\overline{K}_\lambda^n\gamma_1(\lambda - T_0)^{-1}\varphi} +\abs{(\lambda -T_0)^{-1}\varphi}\\ &\le\varepsilon_\lambda H \sum_{n\ge0} \overline{H}_\lambda^n\gamma_1\abs{(\lambda -T_0)^{-1}\varphi} +\abs{(\lambda -T_0)^{-1}\varphi}\\ &=\varepsilon_\lambda H(I-\overline{H}_\lambda)^{-1}\gamma_1\abs{(\lambda -T_0)^{-1}\varphi} +\abs{(\lambda -T_0)^{-1}\varphi}. \end{align*} Thanks to the positivity of the semigroup $U_0=(U_0(t))_{t\ge0}$ (Lemma~\ref{T0:LEM}), we infer that of the operator $(\lambda -T_0)^{-1}$ and therefore \begin{align*} \abs{(\lambda-T_K)^{-1}\varphi} &\le \varepsilon_\lambda H(I-H_\lambda)^{-1}\gamma_1(\lambda -T_0)^{-1}\abs{\varphi} +(\lambda -T_0)^{-1}\abs{\varphi}\\ &=(\lambda-T_H)^{-1}\abs{\varphi}. \end{align*} This leads, by induction, to \begin{equation*} \abs{\parent{\lambda(\lambda-T_K)^{-1}}^n\varphi}\le \parent{\lambda(\lambda-T_H)^{-1}}^n\abs{\varphi} \end{equation*} for all integers $n$. Putting now $\lambda=\frac{n}{t}$ ($t>0$), we obtain that \begin{equation*} \abs{\bracket{\frac{n}{t}\parent{\frac{n}{t}-T_K}^{-1}}^n\varphi}\le \bracket{\frac{n}{t}\parent{\frac{n}{t}-T_H}^{-1}}^n\abs{\varphi} \end{equation*} and therefore \begin{equation}\label{DOMINATION:THE:E20} \abs{U_K(t)\varphi}\le U_H(t)\abs{\varphi}\quad t\ge0 \end{equation} because of Exponential formula. Finally, \eqref{UK:R1} together with \eqref{DOMINATION:THE:R1} and \eqref{DOMINATION:THE:E20} imply that \begin{align*} \abs{U_K(t)\varphi-U_0(t)\varphi}(\mu,v) &=\abs{\xi(\mu,v,t)K\parent{\gamma_1U_K\parent{t-\tfrac{\mu}{v}}\varphi}(v)}\\ &\le\xi(\mu,v,t)H\abs{\parent{\gamma_1U_K\parent{t-\tfrac{\mu}{v}}\varphi}(v)}\\ &\le\xi(\mu,v,t)H\gamma_1\abs{\parent{U_K\parent{t-\tfrac{\mu}{v}}\varphi}(v)}\\ &\le\xi(\mu,v,t)H\gamma_1\parent{U_H\parent{t-\tfrac{\mu}{v}}\abs{\varphi}}(v)\\ &=\Big(U_H(t)\abs{\varphi}-U_0(t)\abs{\varphi}\Big)(\mu,v) \end{align*} for almost all $(\mu,v)\in\Omega$. The proof is now achieved. \end{proof} \section{Asynchronous exponential growth} In this section, we prove that the generated semigroup $V_K=(V_K(t))_{t\ge0}$ possesses the asynchronous exponential growth property by applying Lemma~\ref{AEG:LEM}. However, one of the most difficulties to apply Lemma~\ref{AEG:LEM} is to compute the essential type $\omega_{\rm ess}(V_K)$ (given by \eqref{TYPE-ESS:DEF}) of the generated semigroup $V_K=(V_K(t))_{t\ge0}$ whose explicit form is unfortunately not available. In the sequel, we are going to circumvent this difficulty by proving some useful results. Before we start, let us recall that all rank one or finite rank operators are compact which leads to their admissibility because of Definition~\ref{DEF:ADM}. Therefore, all the semigroups of this work exist. \begin{lemma}\label{ONERANK:LEM} Let $K$ be the following rank one operator in $Y_1$; i.e., \begin{equation*} K\psi=h\int_0^\infty k(v')\psi(v')v'dv', \quad h\in Y_1,\quad k\in L^\infty(0,\infty). \end{equation*} Then we have \begin{equation*}\label{E:SVM} U_K(t)=U_0(t)+\sum_{m=1}^\infty U_m(t)\quad t\ge0, \end{equation*} where $U_0(t)$ is given by \eqref{U0:DEF} and $U_m(t)$ is defined~by \begin{align*} &U_1(t)\varphi(\mu,v)\\ &=\xi(\mu,v,t)h(v)\int_0^\infty k(v_1) \chi \left(1,v_1,t-\tfrac{\mu}{v}\right) \varphi\left(1-\parent{t-\tfrac{\mu}{v}}v_1,v_1\right)v_1dv_1 \end{align*} and, for $m\ge2$, by \begin{align*} &U_m(t)\varphi(\mu,v)\\ &= \xi(\mu,v,t)h(v) \underbrace{\int_0^\infty\cdots\int_0^\infty}_\text{$m$ times} \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_mdv_1\cdots dv_m. \end{align*} for all $\varphi\in L^1(\Omega)$. Furthermore, for all $t\ge0$ we have \begin{equation}\label{ONERANK:LEM:R1} \lim_{N\to\infty} \norm{U_K(t)-U_0(t)-\sum_{m=1}^N U_m(t)}_{\mathcal{L}(L^1(\Omega))}=0. \end{equation} \end{lemma} \begin{proof} Let $\varphi\in L^1(\Omega)$. By \eqref{UK:R1}, it is easy to check, by induction, that for all integer $N\ge1$ we have \begin{equation}\label{ONERANK:E10} U_K(t)=U_0(t)+\sum_{m=1}^NU_m(t)+R_N(t) \end{equation} where the rest $R_{N+1}(t)$ is given by \begin{align*} &R_N(t)\varphi(\mu,v)\\ &=\xi(\mu,v,t)h(v) \underbrace{\int_0^\infty\cdots\int_0^\infty}_\text{$(N+1)$ times}\\ &\quad\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_1v_2\cdots v_{N+1}dv_1\cdots dv_{N+1}. \end{align*} Now, let us prove \eqref{ONERANK:LEM:R1}. Let $\lambda$ be large. Then we have \begin{align*} &\norm{R_N(t)\varphi}_1\\ &\le\int_\Omega\abs{R_{N+1}(t)\varphi(\mu,v)} e^{\lambda\frac{\mu}{v}}\,d\mu\,dv\\ &=\int_\Omega\Big|\xi(\mu,v,t)h(v) \underbrace{\int_0^\infty\cdots\int_0^\infty}_\text{$(N+1)$ times} e^{\lambda\frac{\mu}{v}}\prod_{j=1}^N h(v_j) \prod_{j=1}^{N+1}k(v_j) \\ &\quad \times \xi\Big(1,v_N,t-\frac{\mu}{v} -\sum_{i=1}^{(N-1)}\frac{1}{v_i}\Big) \gamma_1\Big(U_K\Big(t-\frac{\mu}{v}- \sum_{i=1}^N\frac{1}{v_i}\Big)\varphi\Big)(v_{N+1}) \\ &\quad\times v_1v_2\cdots v_{N+1}dv_1\cdots dv_{N+1} \vphantom{\underbrace{\int_a^b}}\Big|\,d\mu\,dv. \end{align*} By the change of variables $x=t-\frac{\mu}{v}-\sum_{i=1}^N\frac{1}{v_i}$ and $vdx=-d\mu$, we infer that \begin{align*} &\norm{R_N(t)\varphi}_1\\ &\le\int_0^\infty\int_0^t \Big| h(v) \underbrace{\int_0^\infty\cdots\int_0^\infty}_\text{$(N+1)$ times} e^{\lambda\parent{t-x-\sum_{i=1}^N\frac{1}{v_i}}}\\ &\quad\times \prod_{j=1}^Nh(v_j) \prod_{j=1}^{N+1}k(v_j)\gamma_1\left(U_K(x)\varphi\right)(v_{N+1}) v_1v_2\cdots v_{N+1}dv_1\cdots dv_{N+1} \vphantom{\underbrace{\int_a^b}}\Big| v\,dx\,dv \end{align*} which leads to \begin{align*} &\norm{R_N(t)\varphi}_1\le e^{\lambda t} \Big[\Big(\int_0^\infty\abs{h(v)}v\,dv\Big) \Big(\operatorname{ess\,sup}_{v\in(0,\infty)}\abs{k(v)}\Big) \Big]\\ &\quad\times\Big[\Big(\int_0^\infty e^{-\lambda/v}\abs{h(v)}v\,dv\Big) \Big(\operatorname{ess\,sup}_{v\in(0,\infty)}\abs{k(v)}\Big)\Big]^N\\ &\quad\times \int_0^\infty\int_0^t e^{-\lambda x}\abs{\gamma_1\left(U_K(x)\varphi\right)(v_{N+1})}v_{N+1} \,dx\, dv_{N+1}. \end{align*} Hence \begin{equation*} \norm{R_N(t)\varphi}_1 \le e^{\lambda t}\norm{K} \norm{\overline{K}_\lambda}^N \int_0^\infty\int_0^\infty e^{-\lambda x} \abs{\gamma_1(U_K(x)\varphi)(v)}v\,dx\,dv \end{equation*} which implies, by \eqref{STABILITY:LEM1:R1}, that \begin{equation}\label{ONERANK:E20} \norm{R_N(t)\varphi}_1 \le\frac{e^{\lambda t}\norm{K} \norm{\overline{K}_\lambda}^N}{1-\norm{\overline{K}_\lambda}}\norm{\varphi}_1 \end{equation} and therefore \begin{equation*} \lim_{N\to\infty}\norm{R_N(t)}_{\mathcal{L}(L^1(\Omega))}=0 \end{equation*} because of \eqref{KLAMBDA+BAR:R1}. Now the proof is complete. \end{proof} The second useful result concerns the linear operator \begin{equation}\label{UK(T,S):DEF} \mathbb{U}_K(t,s):=\Big[U_K(t)-U_0(t)\Big]U_0(s)\Big[U_K(t)-U_0(t)\Big] \end{equation} which is clearly bounded from $L^1(\Omega)$ into itself for all $t\ge0$ and all $s\ge0$ because of Lemmas~\ref{T0:LEM} and \ref{UK:LEM}. One of the most important properties of this operator is as follows. \begin{proposition}\label{KCOMPACT:PROP} Let $K$ be a compact operator from $Y_1$ into itself. Then $\mathbb{U}_K(t,s)$ is a weakly compact operator into $L^1(\Omega)$ for all $t>0$ and all $s\ge0$. \end{proposition} \begin{proof} Let $\varphi\in L^1(\Omega)$, $t>0$ and $s\ge0$. We divide this proof in several steps. \medskip \noindent \textbf{Step 1.} Let us consider the boundary operator \begin{equation}\label{KCOMPACT:PROP:E10} K\psi=h\int_0^\infty k(v')\psi(v')v'dv' \quad h\in C_c(J)\; k\in L^\infty(0,\infty). \end{equation} By Lemma~\ref{ONERANK:LEM}, the operator $U_K(t)$ is expressed as follows \begin{equation}\label{KCOMPACT:PROP:E20} U_K(t)=U_0(t)+\sum_{m=1}^\infty U_m(t). \end{equation} Let us show that $U_m(t)$ is weakly compact into $L^1(\Omega)$ for all $m\ge2$. So, as $h\in C_c(J)$, there exist $a$ and $b$ $(00$ and all $s\ge0$. \end{quote} \medskip \noindent\textbf{Step 2.} Let us consider now the rank one boundary operator \begin{equation}\label{KCOMPACT:PROP:E70} K\psi=h\int_0^\infty k(v')\psi(v')v'dv' \quad h\in Y_1\; k\in L^\infty(0,\infty). \end{equation} As $h\in Y_1$, there exists a sequence $(h_n)_n$ of $C_c(J)$ converging to $h$ into $Y_1$. So, let $K_n$ be the following operator \begin{equation*} K_n\psi=h_n\int_0^\infty k(v')\psi(v')v'dv' \end{equation*} obviously of the form \eqref{KCOMPACT:PROP:E10} and for which $\mathbb{U}_{K_n}(t,s)$ is a weakly compact operator into $L^1(\Omega)$ because of the step I. Furthermore, it is easy to check that \begin{equation}\label{KNKNM} \lim_{n\to\infty}\norm{K_n-K}_{\mathcal{L}(Y_1)}=0 \end{equation} which leads, by Theorem~\ref{STABILITY:THE}, to \begin{equation}\label{CONV} \lim_{n\to\infty}\norm{U_{K_n}(t)-U_K(t)}_{\mathcal{L}(L^1(\Omega))}=0 \end{equation} and therefore, $\big(\norm{U_{K_n}(t)}\big)_n$ is a bounded sequence; i.e., \begin{equation}\label{BOUNDEDNESS10} \norm{U_{K_n}(t)}\le M_t\quad\text{for all integer $n$.} \end{equation} On the other hand, writing \begin{equation}\label{BOUNDEDNESS20} \begin{aligned} \mathbb{U}_{K_n}(t,s)-\mathbb{U}_K(t,s) &=\Big[U_{K_n}(t)-U_K(t)\Big]U_0(s)\Big[U_{K_n}(t)-U_0(t)\Big]\\ &\quad +\Big[U_K(t)-U_0(t)\Big]U_0(s)\Big[U_{K_n}(t)-U_K(t)\Big] \end{aligned} \end{equation} it follows that \begin{equation}\label{BOUNDEDNESS30} \begin{aligned} \norm{\mathbb{U}_{K_n}(t,s)-\mathbb{U}_K(t,s)} &\leq\Big[M_t+\norm{U_0(t)}\Big]\norm{U_0(s)}\norm{U_{K_n}(t)-U_K(t)} \\ &\quad +\Big[\norm{U_K(t)}+\norm{U_0(s)}\Big]\norm{U_0(s)} \norm{U_{K_n}(t)-U_K(t)} \end{aligned} \end{equation} which leads, by \eqref{CONV}, to \begin{equation}\label{BOUNDEDNESS40} \lim_{n\to\infty}\norm{\mathbb{U}_{K_n}(t,s)-\mathbb{U}_K(t,s)}_{\mathcal{L}(L^1(\Omega))}=0 \end{equation} and therefore $\mathbb{U}_K(t,s)$ is a weakly compact operator because of the first point of Lemma~\ref{GREINER}. Now, we can say that: \begin{quote} for any rank one operator $K$, the operator $\mathbb{U}_K(t,s)$ is weakly compact into $L^1(\Omega)$ for all $t>0$ and all $s\ge0$. \end{quote} \medskip \noindent\textbf{Step 3.} Let us consider now the finite rank operator \begin{equation*} K\psi=\sum_{i=1}^{M_K}h_i\int_0^\infty k_i(v')\psi(v')v'dv', \quad h_i\in Y_1,\quad k_i\in L^\infty(0,\infty),\quad i=1,\dots, M_K. \end{equation*} So, if we set \begin{equation*} h:=\max_{i=1,\dots,M_K}\abs{h_i}\in Y_1\quad\text{and} \quad k:=\sum_{i=1}^{M_K}\abs{k_i}\in L^\infty(0,\infty) \end{equation*} it follows that \begin{equation*} H\psi=h\int_0^\infty k(v')\psi(v')v'dv' \end{equation*} is obviously a positive operator of the form \eqref{KCOMPACT:PROP:E70} and therefore $\mathbb{U}_H(t,s)$ is a weakly compact operator into $L^1(\Omega)$ because of the step 2. Furthermore, for all $\psi\in Y_1$ we have \begin{align*} \abs{K\psi} &\le \sum_{i=1}^{M_K}\abs{h_i}\int_0^\infty\abs{k_i(v')}\abs{\psi(v')}v'dv'\\ &\le \Big[\max_{i=1,..,M_K}\abs{h_i}\Big] \int_0^\infty \Big[\sum_{i=1}^{M_K}\abs{k_i(v')}\Big]\abs{\psi(v')}v'dv'\\ &\le H\abs{\psi} \end{align*} which leads, by Theorem~\ref{DOMINATION:THE} and the positivity of $U_0(s)$ (Lemma~\ref{T0:LEM}), to \begin{align*} \abs{\mathbb{U}_K(t,s)\varphi} &=\abs{\Big[U_K(t)-U_0(t)\Big]U_0(s)\Big[U_K(t)-U_0(t)\Big]\varphi}\\ &\le\Big[U_H(t)-U_0(t)\Big]\abs{U_0(s)\Big[U_K(t)-U_0(t)\Big]\varphi}\\ &\le\Big[U_H(t)-U_0(t)\Big]U_0(s)\abs{\Big[U_K(t)-U_0(t)\Big]\varphi}\\ &\le\Big[U_H(t)-U_0(t)\Big]U_0(s)\Big[U_H(t)-U_0(t)\Big]\abs{\varphi}\\ \end{align*} and therefore \begin{equation*} \abs{\mathbb{U}_K(t,s)\varphi}\le\mathbb{U}_H(t,s)\abs{\varphi} \end{equation*} for all $\varphi\in L^1(\Omega)$. This implies \begin{equation*} 0\le \mathbb{U}_K(t,s)+\mathbb{U}_H(t,s)\le 2\mathbb{U}_H(t,s) \end{equation*} and therefore $\mathbb{U}_K(t,s)+\mathbb{U}_H(t,s)$ is a weakly compact operator into $L^1(\Omega)$ because of the second point of Lemma \ref{GREINER}. Writing now \begin{equation*} \mathbb{U}_K(t,s)= \Big(\mathbb{U}_K(t,s)+\mathbb{U}_H(t,s)\Big)-\mathbb{U}_H(t,s) \end{equation*} we can say that: \begin{quote} for any finite rank operator $K$, the operator $\mathbb{U}_K(t,s)$ is weakly compact into $L^1(\Omega)$ for all $t>0$ and all $s\ge0$. \end{quote} \medskip \noindent\textbf{Step 4.} Let $K$ be a compact operator into $Y_1$. Thanks to \cite[Corollary 5.3, p.276]{Edmunds}, there exists a sequence $(K_n)_n$ of finite rank operators satisfying \begin{equation*} \lim_{n\to\infty}\norm{K_n-K}_{\mathcal{L}(Y_1)}=0 \end{equation*} and for which $\mathbb{U}_{K_n}(t,s)$ is a weakly compact operator into $L^1(\Omega)$ because of the step IV. Furthermore, Theorem~\ref{STABILITY:THE} leads to \begin{equation*} \lim_{n\to\infty}\norm{U_{K_n}(t)-U_K(t)}=0. \end{equation*} On the other hand, preceding as before by using \eqref{BOUNDEDNESS10}, \eqref{BOUNDEDNESS20} and \eqref{BOUNDEDNESS30}, we infer \eqref{BOUNDEDNESS40} and therefore $\mathbb{U}_K(t,s)$ is a weakly compact operator into $L^1(\Omega)$ because of the first point of Lemma~\ref{GREINER}. The proof is now achieved. \end{proof} Now, we are finally able to compute the essential type $\omega_{\rm ess}(V_K)$ of the semigroup $V_K=(V_K(t))_{t\ge0}$ as follows. \begin{theorem}\label{TYPEESSE:THE} Let $K$ be a positive compact operator from $Y_1$ into itself. If the hypothesis \eqref{H-sigma} holds, then we have \begin{equation}\label{TYPEESSEVK:THE:R1} \omega_{\rm ess}(V_K)\le-\underline{\sigma}, \end{equation} where $\underline{\sigma}$ is given by \eqref{INFSIGMA:DEF}. \end{theorem} \begin{proof} Let $t>0$ be fixed. We divide this proof into two steps. \medskip \noindent\textbf{Step 1.} Let $s\ge0$ be given. First, due to \eqref{VK:R1} and \eqref{V0:R2} we obtain that \begin{align*} V_K(t)-V_0(t) &=U_K(t)-U_0(t)+\int_0^tU_K(t-s)SV_K(s)ds\\ &\quad +\int_0^tU_0(t-s)(-S)V_0(s)ds. \end{align*} As $-S$, given by \eqref{S:DEF}, is clearly a positive operator, then \eqref{UK:R3} together with the positivity of the semigroup $V_0=(V_0(t))_{t\ge0}$ (Lemma~\ref{T0:LEM}) imply that \begin{align*} &V_K(t)-V_0(t)\\ &\le U_K(t)-U_0(t)+\int_0^tU_K(t-s)SV_K(s)ds +\int_0^tU_K(t-s)(-S)V_0(s)ds\\ &\le U_K(t)-U_0(t)+\int_0^tU_K(t-s)S\Big[V_K(s)-V_0(s)\Big]ds \end{align*} which leads, by \eqref{VK:R2}, to \begin{equation}\label{TYPEESSE:THE:R1} 0\le V_K(t)-V_0(t)\le U_K(t)-U_0(t). \end{equation} On the other hand, due to \eqref{V0:R2} together with the non-positivity of the operator $S$, we easily infer that $0\le V_0(s)\le U_0(s)$. So, this together with \eqref{TYPEESSE:THE:R1} lead to \begin{equation*} 0\le \Big[V_K(t)-V_0(t)\Big]V_0(s)\Big[V_K(t)-V_0(t)\Big]\le\mathbb{U}_K(t,s), \end{equation*} where $\mathbb{U}_K(t,s)$ is given by \eqref{UK(T,S):DEF}. Now, Proposition~\ref{KCOMPACT:PROP} together with the second point of Lemma~\ref{GREINER}, imply that \begin{equation*} \Big[V_K(t)-V_0(t)(t)\Big]V_0(s)\Big[V_K(t)-V_0(t)\Big] \end{equation*} is a weakly compact operator into $L^1(\Omega)$, for all $t>0$ and all $s\ge0$. \medskip \noindent\textbf{Step 2.} Thanks to the step above, we infer that \begin{equation*} \Big[V_K(t)-V_0(t)(t)\Big]V_0(nt)\Big[V_K(t)-V_0(t)\Big] \end{equation*} is a weakly compact operator into $L^1(\Omega)$, for all integers $n\ge0$. Therefore, for all integers $N\ge1$, the following finite sum \begin{align*} \mathbb{V}_N(t) &:=\frac{1}{2}\sum_{n=0}^N \frac{1}{2^n}\Big[V_K(t)-V_0(t)\Big]V_0(nt)\Big[V_K(t)-V_0(t)\Big]\\ &=\frac{1}{2}\Big[V_K(t)-V_0(t)\Big]\Big[\sum_{n=0}^{N}\frac{1}{2^n}V_0(nt)\Big] \Big[V_K(t)-V_0(t)\Big] \end{align*} is also a weakly compact operator into $L^1(\Omega)$. Du to \eqref{V0:R1}, it follows that $2\in\rho(V_0(t))$ which implies that \begin{equation*} \lim_{N\to\infty} \Big\|\mathbb{V}_N(t)-\Big[V_K(t)-V_0(t)\Big](2-V_0(t))^{-1} \Big[V_K(t)-V_0(t)\Big]\Big\|=0 \end{equation*} and therefore \begin{equation*} \Big[V_K(t)-V_0(t)\Big](2-V_0(t))^{-1}\Big[V_K(t)-V_0(t)\Big] \end{equation*} is a weak compact operator into $L^1(\Omega)$ because of the first point of Lemma~\ref{GREINER}. Hence, the following operator \begin{equation*} \Big(\Big[V_K(t)-V_0(t)\Big](2-V_0(t))^{-1}\Big)^4 \end{equation*} is compact into $L^1(\Omega)$ because of the third point of Lemma~\ref{GREINER}, which leads, by Lemma~\ref{COMPACIT}, to \begin{equation}\label{TYPEESSEVK:E10} \omega_{\rm ess}(V_K)=\omega_{\rm ess}(V_0). \end{equation} Finally, du to \eqref{V0:R1} together with \eqref{TYPE:DEF} and \eqref{TYPE:ORD} and \eqref{TYPEESSEVK:E10}, we clearly infer that $\omega_{\rm ess}(V_K)=\omega_{\rm ess}(V_0)\le\omega(V_0)\le-\underline{\sigma}$ and therefore \eqref{TYPEESSEVK:THE:R1} easily follows. The proof is now achieved. \end{proof} Now, we are able to prove that the semigroup $V_K=(V_K(t))_{t\ge0}$, governing the general model \eqref{MODEL:EQUATION}, \eqref{MODEL:BC}, possesses Asynchronous Exponential Growth property. Before we start, note that in the case $\norm{K}<1$, the model \eqref{MODEL:EQUATION}, \eqref{MODEL:BC} is biologically uninteresting because the bacterial density is decreasing. Indeed, for all $t$ and all $s$ with $t>s$, \eqref{VK:COR:R1} implies that \begin{equation*} \norm{V_K(t)\varphi}_1=\norm{V_K(t-s)V_K(s)\varphi}_1 \le e^{-(t-s)\underline{\sigma}}\norm{V_K(s)\varphi}_1 \le\norm{V_K(s)\varphi}_1 \end{equation*} for all initial data $\varphi\in L^1(\Omega)$. Therefore, we well understand that $\norm{K}>1$ is closely related to an increasing number of bacteria during each mitotic. This situation is the most biologically observed for which Asynchronous Exponential Growth property is given by \begin{theorem} Suppose that \eqref{H-sigma} holds and let $K$ be a positive, irreducible and compact operator in $Y_1$ such that \begin{equation*} r(K_{\overline{\sigma}-\underline{\sigma}})>1, \end{equation*} where $\underline{\sigma}$ and $\overline{\sigma}$ are given by \eqref{INFSIGMA:DEF} and \eqref{SUPSIGMA:DEF}. Then, there exist a rank one projector $\mathbb{P}$ in $L^1(\Omega)$ and an $\varepsilon>0$ such that for every $\eta\in( 0,\varepsilon)$, there exist $M(\eta)\ge 1$ satisfying \begin{equation*} \norm{e^{-\omega_0(V_K)t}V_K(t)-\mathbb{P}}_{\mathcal{L}(L^1(\Omega))} \le M(\eta)e^{-\eta t} \quad t\ge0. \end{equation*} \end{theorem} \begin{proof} Thanks to Lemma~\ref{VK:LEM}, it follows that $V_K=(V_K(t))_{t\ge0}$ is a positive and irreducible semigroup. Furthermore, \eqref{TYPEESSEVK:THE:R1} and \eqref{VK:R3} obviously lead to $\omega_{\rm ess}(V_K)<\omega(V_K)$. Now, all the conditions of Lemma~\ref{AEG:LEM} are satisfied. \end{proof} When there is no bacterial mortality or bacteria loss due to causes other than division (i.e., $\sigma=0$), then Asynchronous Exponential Growth property of the $U_K=(U_K(t))_{t\ge0}$ can be described as follows. \begin{corollary} Let $K$ be a positive, irreducible and compact operator in $Y_1$ with $r(K)>1$. Then, there exist a rank one projector $\mathbb{P}$ in $L^1(\Omega)$ and an $\varepsilon>0$ such that for every $\eta\in( 0,\varepsilon)$, there exist $M(\eta)\ge 1$ satisfying \begin{equation*} \norm{e^{-\omega_0(U_K)t}U_K(t)-\mathbb{P}}_{\mathcal{L}(L^1(\Omega))} \le M(\eta)e^{-\eta t} \quad t\ge0. \end{equation*} \end{corollary} The proof follows from Theorem above because $\overline{\sigma}=\underline{\sigma}=0$. \begin{remark} \rm According to a lot of modifications, we claim that all the results of this work still hold into $L^p(\Omega)$ $(p>1)$. However, the norm of such space have no biological meaning. \end{remark} \begin{thebibliography}{99} \bibitem{Boulanouar1} M. Boulanouar; \emph{Transport Equation in Cell Population Dynamics (II).} Elec. Journ. Diff. Eqns., N. 145, 1-20, 2010. \bibitem{Boulanouar2} M. Boulanouar; \emph{New trace theorem for neutronic function spaces.} Trans. Theor. Stat. Phys., Vol., 38, 228-242, 2009. \bibitem{Boulanouar3} M. Boulanouar; \emph{Transport equation for growing bacterial population (I).} Electron. J. Diff. Equ., Vol. 2012 (2012), No. 221, 1-25. \bibitem{Clement} Ph. Clement, H. J. A. M. Heijmans, S. Angenent, C. J. van Duijn, B. de Pagter; \emph{One-parameter semigroups}. CWI Monographs, 5. North-Holland Publishing Co., Amsterdam, 1987. \bibitem{Edmunds} D. E. Edmunds, W. D. Evans; \emph{Spectral Theory and Differential Operators}, Oxford Science Publications, 1987. \bibitem{Greiner} G. Greiner; \emph{Spectral Properties and Asymptotic Behaviour of Linear Transport Equation.} Math. Z., (185), 167--177. 1984. \bibitem{Nagel} K. Engel, R. Nagel; \emph{One-Parameter Semigroups for Linear Evolution Equations.} Graduate texts in mathematics, 194. Springer-Verlag, New York, Berlin, Heidelberg, 1999. \bibitem{Voigt} J. Voigt; \emph{A Perturbation Theorem for the Essential Spectral Radius of Strongly Continuous Semigroups}. Mh. Math. 90. pp.153--161. 1980. \end{thebibliography} \section*{Addendum posted on June 24, 2013.} % Vol. 2012, No. 222. The author would like to make the following changes: \noindent (1) Definition~3.4: ${\textrm (K\!c)}$ must be replaced by: ``{\it $\|K\mathbb{I}_{\omega}\|<1$ for some $\omega>0$ and $\|K\|\geq 1$. $\mathbb{I}_{\omega}$ denotes the characteristic operator of the set $(\omega,\infty)$.}'' \noindent (2) Lemmas~3.5: ``{\it $K$ compact}'' must be inserted in the preamble. Line~10 of the proof: ``{\it Next, if ${\textrm (K\!c)}$ holds}'' must be replaced by ``{\it As $K$ is compact}''. \noindent (3) Lemmas~3.5, 3.6, 3.7, 4.1 and Theorems~4.2, 4.3: ``{\it compact}'' must be inserted in the preamble (for the operators $K$, $H$ and $K_n$). \noindent(4) Page~13: ``{\it In the sequel...to...exist}'' (Lines 7 to 10) must be deleted. \noindent (5) Lemma~5.1 to Corollary~5.5: ``{\it $K$ admissible}'' must be inserted in the preamble. \noindent (6) Lemma~3.7(3) and Theorem~5.4: ``$r(K_{\overline{\sigma}-\underline{\sigma}}>$'' must be replaced by ``$r(\overline K_{\overline{\sigma}-\underline{\sigma}})>1$.'' \noindent (7) Proof of Proposition~5.2 must start by: ``{\it In the sequel, all one or finite rank operators must be admissible like the operator $K$ in the preamble. As $$ \|K_n\mathbb{I}_{\omega}\|\leq \|K_n\mathbb{I}_{\omega} -K\mathbb{I}_{\omega}\| +\|K\mathbb{I}_{\omega}\| \leq \|K_n-K\|+\|K\mathbb{I}_{\omega}\|, $$ we infer then that all approximation operators $K_n$ of $K$ are also admissible for large integer $n$.}'' End of addendum. \end{document}