\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 221, pp. 1--25.\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-2012/221\hfil Transport equation]
{Transport equation for growing bacterial populations (I)}

\author[M. Boulanouar \hfil EJDE-2012/221\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{Semigroups; positivity; irreducibility; spectral properties;
\hfill\break\indent cell population dynamics; general boundary condition}

\begin{abstract}
 This work deals with a mathematical study for growing a bacterial
 population. Each bacterium is distinguished by its degree of maturity
 and its maturation velocity.
 Here we study the limit case corresponding to infinite maturation
 velocities.  We show that this  model is governed by a strongly
 continuous semigroup. We also study the lattice and spectral
 properties of the generated semigroup and we compute its type.
\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}
\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 article deals with a bacterial population in which each bacteria
is distinguished by its own degree of maturity $\mu$ and its own maturation
velocity $v$.
At birth, the degree of maturity of a daughter bacteria is $\mu=0$.
At mitotic, the degree of maturity of a mother bacteria becomes $\mu=1$.
Between birth and division, the degree of maturity of each
bacterium is $0<\mu<1$.
As each bacterium may not become less mature,
then its maturation velocity
must be positive $(0\le a<v<b\le\infty)$.
So, if $f=f(t,\mu,v)$ denotes the bacterial density with
respect to the degree of maturity $\mu$ and the
maturation velocity $v$, at time $t$, then
\begin{equation}\label{MODEL:EQUATION}
\frac{\partial f}{\partial t}=-v\frac{\partial f}{\partial \mu}-\sigma f,
\end{equation}
where $\sigma=\sigma(\mu,v)$ denotes the rate of bacterial mortality or bacteria loss
due to causes other than division.

In most bacterial populations observed, there is often a correlation
$k=k(v,v')$ between the maturation
velocity of a bacteria mother $v'$ and that of its bacteria daughter $v$.
The bacterial mitotic obeys then to the biological
\emph{transition law} mathematically described by the following transition boundary condition
\begin{equation}\label{MODEL:TRANSITION}
vf(t,0,v)=p\int_a^bk(v,v')f(t,1,v')v'\,dv',
\end{equation}
where $p\ge0$ denotes the average number of bacteria daughter viable
per mitotic. To ensure the continuity of the bacterial flux for $p=1$,
the kernel of correlation $k$ must be positive and
fulfils the normalization condition
\begin{equation}\label{MODEL:NORMAL}
\int_a^bk(v,v')dv=1\quad\text{for all } v\in(a,b).
\end{equation}
The model \eqref{MODEL:EQUATION}-\eqref{MODEL:TRANSITION} was
introduced in \cite{Rotenberg}, where only a numerical study has been made.
The first theoretical studies of the model
\eqref{MODEL:EQUATION}-\eqref{MODEL:TRANSITION} were given
in \cite{Boulanouar1} and \cite{Boulanouar2},
where we have proved that the model \eqref{MODEL:EQUATION}-\eqref{MODEL:TRANSITION}
is governed by a strongly continuous semigroup provided that $b<\infty$.
However, if $b=\infty$, all claims of \cite{Boulanouar1}
and \cite{Boulanouar2} become no suitable
which leads to a serious mathematical difficulty.

To show the extent of this difficulty (i.e., $b=\infty$), we have studied
the model
\eqref{MODEL:EQUATION}-\eqref{MODEL:TRANSITION}
in the particular case $k(v,v')=\delta_v(v')$ which corresponds to the
\emph{perfect memory} law
mathematically described by the following boundary condition
\begin{equation} \label{MODEL:TRANSITION'} %\label{MODEL:PERFECT}
f(t,0,v)=p f(t,1,v)\quad\text{for all } v\in(a,b).
\end{equation}
We then proved, in \cite{Boulanouar3}, that the model
\eqref{MODEL:EQUATION}-\eqref{MODEL:TRANSITION'}
is well-posed if and only if $p\le1$. In other words,
there are no solutions for the most interesting case $p>1$
corresponding to an increasing number of bacteria.

In this work we  study the case $b=\infty$.
Furthermore, instead of using the biological transition law given
by \eqref{MODEL:TRANSITION},
we consider a general biological law
mathematically described by the following boundary condition
\begin{equation}\label{MODEL:BC}
f(t,0,v)=\bracket{Kf(t,1,\cdot)}(v) \quad\text{for all } v\in(a,b),
\end{equation}
where $K$ denotes a linear operator on suitable spaces (see Section 3).

According to $0<a<b=\infty$, we have recently proved in \cite{Boulanouar5}
that the general model \eqref{MODEL:EQUATION}, \eqref{MODEL:BC}
is governed by a strongly continuous semigroup.
However, all claims and computations unfortunately depend on $a>0$.
Therefore, we are naturally led to set the following question
\begin{quote}
What happens when $a=0$ and $b=\infty$?
\end{quote}
The purpose of this work is to answer the question above as follows
\begin{itemize}
\item[(2)] Mathematical preliminaries
\item[(3)] The unperturbed model (i.e., $\sigma=0$)
\item[(4)] Explicit form of the unperturbed semigroup
\item[(5)] Generation Theorem for the model \eqref{MODEL:EQUATION}, \eqref{MODEL:BC}
\item[(6)] Lattice property of the generated semigroup
\item[(7)] Spectral properties of the generated semigroup
\item[(8)] Application and comments
\end{itemize}
In the third Section, we consider the natural framework
of the general model \eqref{MODEL:EQUATION}, \eqref{MODEL:BC} which is
$L^1\parent{(0,1)\times(0,\infty)}$
whose norm
\begin{equation*}
\norm{f(t,\cdot,\cdot)}_1=\int_0^1\int_0^\infty\abs{f(t,\mu,v)}\,d\mu\,dv
\end{equation*}
denotes the bacteria number at time $t$.
We show then that the unperturbed model
\eqref{MODEL:EQUATION}, \eqref{MODEL:BC}
(i.e., without bacterial mortality ($\sigma=0$)) is governed by
a strongly continuous semigroup whose explicit form is given
in fourth Section. In fifth Section, we prove that the general
model \eqref{MODEL:EQUATION}, \eqref{MODEL:BC} is governed
by a strongly continuous semigroup whose
Lattice and Spectral properties are studied in sixth and seventh Sections.
The last section deals with an application to the model
\eqref{MODEL:EQUATION}--\eqref{MODEL:TRANSITION}.

Finally, note the novelty of this work and
for the used mathematical background,
we refer the reader to \cite{Clement} and \cite{Nagel}.

\section{Mathematical preliminaries}
\label{Mathematical preliminaries}

In this section, we are going to recall some useful mathematical tools
about strongly continuous semigroups of linear operators in a Banach space $X$.
So, the first one deals with the following known result

\begin{lemma}[{\cite[Theorem III.1.3]{Nagel}}] \label{PERTURBATION:LEM}
Let $T$ be the infinitesimal generator of a strongly continuous semigroup
$U=(U(t))_{t\ge0}$, on $X$, and
let $B$ be a bounded linear operator from $X$ into itself.
Then, the operator
$C:=T+B$ on the domain $D(C):=D(T)$
generates, on $X$, a strongly continuous semigroup $(V(t))_{t\ge0}$
given, for all $x\in X$, by
\begin{equation}\label{PERTURBATION:R1}
V(t)x=
\lim_{n\to\infty}\bracket{e^{-\frac{t}{n}B}U\parent{\tfrac{t}{n}}}^nx
\quad t\ge0.
\end{equation}
\end{lemma}

Let $U=(U(t))_{t\ge0}$
be a strongly continuous semigroup, on $X$, whose generator is $T$.
Following \cite[Chapter IV]{Nagel}, the {\sl type} $\omega_0(U)$ is
\begin{equation}\label{TYPE:DEF}
\omega_0(U)=\lim_{t\to\infty}
\frac{\ln\norm{U(t)}_{\mathcal{L}(X)}}{t}.
\end{equation}
The spectral bound $s(T)$ of the generator $T$ is given by
\begin{equation}\label{BOUND:DEF}
s(T)=
\begin{cases}
\sup\{\operatorname{Re}(\lambda),\,\lambda\in\sigma(T)\}
&\text{if $\sigma(T)\not =\emptyset$,}\\
-\infty &\text{if $\sigma(T)=\emptyset$}.
\end{cases}
\end{equation}
Generally, we have $\omega_0(U)\not=s(T)$.
However, if $X$ is an $L_1$ space then
\begin{equation}\label{BOUND:EQUALITY}
\omega_0(U)=s(T)
\end{equation}
because of \cite{Weis}.
Next, if $X$ denotes a Banach lattice space then, the positivity
and the irreducibility of the semigroup $U=(U(t))_{t\ge 0}$
are characterized as follows

\begin{lemma}[{\cite[Proposition 7.1 and 7.6]{Clement}}] \label{POS-IRRE:LEM}
\begin{enumerate}
\item
The semigroup $U=(U(t))_{t\ge 0}$ is
positive if and only if the resolvent operator $(\lambda-T)^{-1}$ is
positive for some great $\lambda$.

\item Suppose that the semigroup $U=(U(t))_{t\ge 0}$ is positive.
It is irreducible if and only if the resolvent operator
$(\lambda-T)^{-1}$ is irreducible for some great $\lambda$.
\end{enumerate}
\end{lemma}

\section{The unperturbed model (${\rm i.e.,}\;\sigma=0$)}
\label{The unperturbed model}

In this section, we are concerned with the unperturbed model
\eqref{MODEL:EQUATION}, \eqref{MODEL:BC}
(i.e., without bacterial mortality ($\sigma=0$)).
So, we are going to prove that this model is governed by
a strongly continuous semigroup $U_K=(U_K(t))_{t\ge0}$
which will be perturbed to infer the well posedness
of the general model \eqref{MODEL:EQUATION}, \eqref{MODEL:BC} (see Section 5).
Before we start, 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*}
Applying now \cite[Theorem 2.2]{Boulanouar7} to $\Omega=(0,1)\times(0,\infty)$
we infer that

\begin{lemma}\label{TRACES:LEM}
The trace mappings
$\gamma_0\varphi:=\varphi(0,\cdot)$ and $\gamma_1\varphi:=\varphi(1,\cdot)$
are linear continuous from $W_1$ into $Y_1$.
\end{lemma}

Next, let $T_0$ be the  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}
which makes sense because of Lemma~\ref{TRACES:LEM}.
This operator corresponds to the model
\eqref{MODEL:EQUATION}, \eqref{MODEL:BC} without bacterial mortality ($\sigma=0$)
and without bacterial division ($K=0$). Some of its useful properties
 can be summarized as follows

\begin{lemma}\label{T0:LEM} \quad
\begin{enumerate}
\item
The operator $T_0$ generates, on $L^1(\Omega)$,
a strongly continuous positif semigroup
$U_0=(U_0(t))_{t\ge0}$ of contractions 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 } \mu<tv.\\
\end{cases}
\end{equation}
\item
Let $\lambda>0$. Then $(\lambda-T_0)^{-1}$ is a positive operator from $L^1(\Omega)$
into itself. Furthermore, for all $g\in L^1(\Omega)$ we have
\begin{align}
\label{T0:R3}
\norm{(\lambda-T_0)^{-1}g}_1&\le\frac{\norm{g}_1}{\lambda}\\
\label{T0:R4}
\norm{v(\lambda-T_0)^{-1}g}_1&\le\norm{g}_1.
\end{align}
\item
Let $\lambda>0$. Then $\gamma_1(\lambda-T_0)^{-1}$
is a strictly positive operator from $L^1(\Omega)$ into $Y_1$.
\item
For all $\varphi\in W_1$, the following mapping
\begin{equation}\label{T0:R5}
t\to\gamma_1(U_0(t)\varphi)\in Y_1
\end{equation}
is continuous with respect to $t\ge0$.
\end{enumerate}
\end{lemma}

\begin{proof}
For all $\lambda>0$ and all $g\in L^1(\Omega)$, a simple computation
shows that
\begin{equation*}
(\lambda-T_0)^{-1}g(\mu,v)=\int_0^{\mu/v}e^{-\lambda s}g(\mu-sv,v)ds
\end{equation*}
which easily leads to the points (1) and (2) and (3).


(4) Let $t\ge0$. Firstly, for all $\varphi\in W_1$,
we have $\varphi=f+g$,
where $f:=\varepsilon(\gamma_0\varphi)$
and $g:=\varphi-f$ with $\varepsilon(\mu,v)=e^{-\frac{\mu}{v}}$.
Easy computations show that
\begin{equation*}
\norm{v\frac{\partial f}{\partial\mu}}_1=\norm{f}_1\le\norm{\gamma_0\varphi}_{Y_1}
\quad\text{and}\quad
\norm{vf}_1\le\norm{\gamma_0\varphi}_{Y_1}
\end{equation*}
which leads to $f\in W_1$ because of Lemma~\ref{TRACES:LEM}.
Furthermore
\begin{equation*}
\norm{g}_{W_1}\le\norm{\varphi}_{W_1}+\norm{f}_{W_1}<\infty
\quad\text{and}\quad\gamma_0g=\gamma_0\varphi-\gamma_0f=0
\end{equation*}
and therefore $g\in D(T_0)$.

Next, as $U_0=(U_0(t))_{t\ge0}$ is also a strongly continuous semigroup on
the domain $D(T_0)$ of its generator, then we can write
\begin{equation}\label{T0:E20}
\lim_{h\to0}\norm{U_0(t+h)g-U_0(t)g}_{D(T_0)}=0.
\end{equation}
By virtue of Lemma~\ref{TRACES:LEM} together with the fact
that $D(T_0)$ is a closed subspace of $W_1$, it follows that
\begin{equation*}
\begin{split}
\norm{\gamma_1U_0(t+h)g-\gamma_1U_0(t)g}_{Y_1}\le
\norm{\gamma_1}\norm{U_0(t+h)g-U_0(t)g}_{D(T_0)}
\end{split}
\end{equation*}
for all $h>0$ and therefore
\begin{equation}\label{T0:E30}
t\to\gamma_1U_0(t)g
\end{equation}
is a continuous mapping with respect to $t\ge0$,
because of \eqref{T0:E20}.
\par
On the other hand, due to \eqref{U0:DEF} and \eqref{CHI:DEF},
it is easy to check that
\begin{equation*}
\begin{split}
&\left\|\gamma_1U_0(t+h)f-\gamma_1U_0(t)f\right\|_{Y_1}\\
&=\int_0^\infty
\abs{\chi(1,v,t+h)e^{-\frac{(1-(t+h)v)}{v}}
-\chi(1,v,t)e^{-\frac{(1-tv)}{v}}}
\abs{\gamma_0\varphi(v)}v\,dv
\end{split}
\end{equation*}
for all $h>0$ and therefore
\begin{equation}\label{T0:E40}
t\to\gamma_1U_0(t)f
\end{equation}
is a continuous mapping with respect to $t\ge0$.
\par
Finally, writing \eqref{T0:R5} as follows
\begin{equation*}
t\to\gamma_1U_0(t)\varphi=\gamma_1U_0(t)f+\gamma_1U_0(t)g
\end{equation*}
we infer its continuity from those of the mappings
\eqref{T0:E30} and \eqref{T0:E40}.
\end{proof}

In the sequel we are going to study the model
\eqref{MODEL:EQUATION}, \eqref{MODEL:BC}, without bacterial mortality ($\sigma=0$),
modeled by the following unbounded operator
\begin{equation}\label{TK:DEF}
\begin{gathered}
T_K\varphi=-v\frac{\partial \varphi}{\partial\mu}\;\;
\text{on the domain,}\\
D(T_K)=\{\varphi\in W_1;\,
\gamma_0\varphi=K\gamma_1\varphi\}
\end{gathered}
\end{equation}
where $K$ denotes a linear operator from $Y_1$ into itself.
Note that \eqref{TK:DEF} has a sense because of Lemma~\ref{TRACES:LEM}.
So, in order to state the main goal of this section,
we are going to prove some preparative results.
The first one deals with the following operator
\begin{equation}\label{KLAMBDA:DEF}
K_\lambda\psi:=K(\theta_\lambda\psi),
\quad\text{where }
\theta_\lambda(v)=e^{-\lambda/v},\quad v\in J=(0,\infty)
\end{equation}
which is going to play
an important role in the sequel. So we have

\begin{lemma}\label{KLAMBDA:LEM}
Let $K$ be a linear operator from $Y_1$ into itself satisfying
one of the following hypotheses
\begin{enumerate}
\item[(Kb)] $K$ is bounded and $\norm{K}_{\mathcal{L}(Y_1)}<1$;
\item[(Kc)] $K$ is compact and $\norm{K}_{\mathcal{L}(Y_1)}\ge1$.
\end{enumerate}
Then, for all $\lambda\ge0$, the operator $K_\lambda$
is bounded linear from $Y_1$ into itself.
Furthermore, there exists a constant
\begin{equation}\label{OMEGAK:DEF}
\omega_K
\begin{cases}
=0,&\text{if {\rm (Kb)} holds}\\
>0,&\text{if {\rm (Kc)} holds},
\end{cases}
\end{equation}
such that
\begin{equation}\label{KLAMBDA:R1}
\lambda>\omega_K(\ln k)\Longrightarrow \norm{K_\lambda}_{\mathcal{L}(Y_1)}<1,
\end{equation}
where $k=\max\{1,\norm{K}\}$.
\end{lemma}
\begin{proof}


Firstly, note that the boundedness of $K_\lambda$ ($\lambda\ge0$) obviously
follows from
\begin{equation*}
\norm{K_\lambda\psi}_{Y_1}\le\norm{K}\norm{\psi}_{Y_1}\quad\text{for all }
\psi\in Y_1.
\end{equation*}
Therefore, if {\rm (Kb)} holds, then we clearly have
\begin{equation}\label{KLAMBDA:E1}
\lambda>0\Longrightarrow\norm{K_\lambda}<1.
\end{equation}

Suppose now {\rm (Kc)} holds and let $\omega\ge0$ be given.
So, the compactness of the operator $K$ obviously
leads to that of the operator $K\mathbb{I}_{\omega}$,
where $\mathbb{I}_{\omega}\in\mathcal{L}\parent{Y_1}$ is the following
characteristic operator
\begin{equation}\label{KLAMBDA:E05}
\mathbb{I}_{\omega}\psi(v)=
\begin{cases}
\psi(v)&\text{if } v>\omega;\\
0&\text{otherwise.}
\end{cases}
\end{equation}
Hence, there exists a finite sequence
$\parent{\psi_i}_{i=1}^{N_K}\subset B(0,1)\subset Y_1$
such that
\begin{equation}\label{KLAMBDA:E10}
K\mathbb{I}_{\omega}\Bigr(B(0,1)\Bigr)\subset
\cup_{i=1}^{N_K}B\Big(K\mathbb{I}_{\omega}\psi_i,\,\frac{1}{2}\Big),
\end{equation}
where $B(0,1)$ is the closed unit ball into $Y_1 $.

Now, for all $i\in\set{1,\cdots,N_K}$, we clearly have
\begin{equation*}
\norm{K\mathbb{I}_{\omega}\psi_i}_{Y_1}\le\norm{K}
\int_0^\infty\abs{\mathbb{I}_{\omega}\psi_i(v)}v\,dv
\end{equation*}
which implies that
\begin{equation*}
\lim_{\omega\to\infty}
\norm{K\mathbb{I}_{\omega}\psi_i}_{Y_1}
\le\norm{K}\lim_{\omega\to\infty}
\int_0^\infty\abs{\mathbb{I}_{\omega}\psi_i(v)}v\,dv=0
\end{equation*}
and therefore, there exists $\delta_{K,i}>0$ satisfying
\begin{equation*}
\norm{K\mathbb{I}_{\omega}\psi_i}_{Y_1}<\frac{1}{2}
\quad\text{for all } \omega>\delta_{K,i}.
\end{equation*}
Furthermore, if we set
\begin{equation}\label{KLAMBDA:E15}
\delta_K:=\max\set{\delta_{K,i};
\, i=1,\cdots, N_K}
\end{equation}
it follows that
\begin{equation}\label{KLAMBDA:E20}
\delta_K>0
\end{equation}
and
\begin{equation}\label{KLAMBDA:E25}
\max_{i\in\set{1,\cdots,N_K}}\norm{K\mathbb{I}_{\omega}\psi_i}_{Y_1}
<\frac{1}{2}
\quad\text{for all } \omega>\delta_K.
\end{equation}

Next, let $\omega>\delta_K$. For all $\psi\in B(0,1)\subset Y_1 $,
\eqref{KLAMBDA:E10} implies that there exists $i_0\in\set{1,\cdots,N_K}$ satisfying
\begin{equation*}
K\mathbb{I}_{\omega}\psi\in B\Big(K\mathbb{I}_{\omega}\psi_{i_0},\;\frac{1}{2}\Big)
\end{equation*}
which implies that
\begin{align*}
\norm{K\mathbb{I}_{\omega} \psi}_{Y_1}
&\le\norm{K\mathbb{I}_{\omega} \psi-K\mathbb{I}_{\omega} \psi_{i_0}}_{Y_1}
+\norm{K\mathbb{I}_{\omega} \psi_{i_0}}_{Y_1}\\
&\le\frac{1}{2}
+\max_{i\in\set{1,\cdots,N_K}}\norm{K\mathbb{I}_{\omega}\psi_i}_{Y_1}
\end{align*}
and therefore
\[
\norm{K\mathbb{I}_{\omega}}
=\sup_{\psi\in B(0,1)}\norm{K\mathbb{I}_{\omega} \psi}_{Y_1}
\le\frac{1}{2}+\max_{i\in\set{1,\cdots,N_K}}\norm{K\mathbb{I}_{\omega}\psi_i}_{Y_1}
<1
\]
because of \eqref{KLAMBDA:E25}.
Hence, we can say that
\begin{equation}\label{KLAMBDA:E30}
\norm{K\mathbb{I}_{\omega}}<1
\quad\text{for all } \omega>\delta_K.
\end{equation}
On the other hand, let $\omega>\delta_K$
and let $\overline{\mathbb{I}}_{\omega}\in\mathcal{L}\parent{Y_1}$ be the following
characteristic operator
\begin{equation}\label{KLAMBDA:E35}
\overline{\mathbb{I}}_{\omega}\psi(v):=
\begin{cases}
\psi(v)&\text{if } v\le\omega;\\
0&\text{otherwise}
\end{cases}
\end{equation}
for which we clearly have
\begin{equation}\label{KLAMBDA:E40}
\psi=\mathbb{I}_{\omega}\psi+\overline{\mathbb{I}}_{\omega}\psi
\quad\text{and}\quad
\mathbb{I}_{\omega}^2\psi=\mathbb{I}_{\omega}\psi.
\end{equation}
So, for all $\psi\in Y_1$ we have
\begin{equation*}
K_\lambda\psi=K_\lambda(\mathbb{I}_{\omega}\psi+\overline{\mathbb{I}}_{\omega}\psi)=
K_\lambda(\mathbb{I}_{\omega}^2\psi+\overline{\mathbb{I}}_{\omega}\psi)
=K\mathbb{I}_{\varepsilon}\parent{\mathbb{I}_{\varepsilon}\theta_\lambda\psi}
+K\overline{\mathbb{I}}_{\omega}\parent{\theta_\lambda\psi}
\end{equation*}
which implies that
\begin{align*}
\norm{K_\lambda\psi}_{Y_1}
&\le
\norm{K\mathbb{I}_{\omega}\parent{\mathbb{I}_{\omega}\theta_\lambda\psi}}_{Y_1}
+\norm{K\overline{\mathbb{I}}_{\omega}\parent{\theta_\lambda\psi}}_{Y_1}\\
&\le
\norm{K\mathbb{I}_{\omega}}\norm{\mathbb{I}_{\omega}\theta_\lambda\psi}_{Y_1}
+\norm{K}\norm{\overline{\mathbb{I}}_{\omega}\parent{\theta_\lambda\psi}}_{Y_1}\\
&\le
\norm{K\mathbb{I}_{\omega}}\norm{\mathbb{I}_{\omega}\psi}_{Y_1}
+e^{-\frac{\lambda}{\omega}}\norm{K}\norm{\overline{\mathbb{I}}_{\omega}\psi}_{Y_1}\\
&\le\max\set{\norm{K\mathbb{I}_{\omega}},\;e^{-\frac{\lambda}{\omega}}\norm{K}}
{\set{\norm{\mathbb{I}_{\omega}\psi}+\norm{\overline{\mathbb{I}}_{\omega}\psi}}}\\
&=\max\set{\norm{K\mathbb{I}_{\omega}},\;e^{-\frac{\lambda}{\omega}}\norm{K}}
\norm{\psi}_{Y_1}
\end{align*}
for all $\lambda\ge0$, and therefore
\begin{equation*}
\norm{K_\lambda}
\le\max\big\{\norm{K\mathbb{I}_{\omega}},\;e^{-\frac{\lambda}{\omega}}\norm{K}\big\}.
\end{equation*}
Now, \eqref{KLAMBDA:E30} clearly leads to
\begin{equation}\label{KLAMBDA:E45}
\lambda>\omega\ln\norm{K}\Longrightarrow\norm{K_\lambda}<1.
\end{equation}
Let $\lambda>\delta_K\ln\norm{K}$ be given. There exists $\omega$ such that
$\frac{\lambda}{\ln\norm{K}}>\omega>\delta_K$ which implies that $\lambda>\omega\ln\norm{K}$
and therefore $\norm{K_\lambda}<1$ because of \eqref{KLAMBDA:E45}.
Therefore, we can say that if {\rm (Kc)} holds, then
\begin{equation}\label{KLAMBDA:E55}
\lambda>\delta_K\ln\norm{K}\Longrightarrow\norm{K_\lambda}<1.
\end{equation}
Finally, by  \eqref{KLAMBDA:E20} we can set that
\begin{equation*}
\omega_K:=
\begin{cases}
0,& \text{if {\rm (Kb)} holds;}\\
\delta_K, &\text{if {\rm (Kc)} holds,}
\end{cases}
\end{equation*}
which obviously leads to \eqref{OMEGAK:DEF}.
Furthermore, \eqref{KLAMBDA:R1} clearly holds because of
\eqref{KLAMBDA:E1} and \eqref{KLAMBDA:E55}.
The proof is now achieved.
\end{proof}

\begin{remark}\label{ADM:REM} \rm
In the sequel, any linear operator $K$ from $Y_1$ into itself
is said to be \emph{admissible} if one of the following hypotheses holds
\begin{enumerate}
\item [{\rm (Kb)}] $K$ is bounded and
$\norm{K}_{\mathcal{L}(Y_1)}<1$;

\item[{\rm (Kc)}] $K$ is compact and
 $\norm{K}_{\mathcal{L}(Y_1)}\ge1$,
\end{enumerate}
The constant $\omega_K$, given by \eqref{OMEGAK:DEF}, is called
the \emph{abscissa} of the admissible operator $K$.
So, Lemma~\ref{KLAMBDA:LEM} means that \eqref{KLAMBDA:R1} holds for any
admissible operator $K$ whose abscissa is $\omega_K$.
\end{remark}

Now, we compute the resolvent operator of $T_K$ as follows

\begin{proposition}\label{RTK:PRO}
Let $K$ be an admissible operator whose abscissa is $\omega_K$.
Then
\begin{equation}\label{RTK:R1}
\Big(\omega_K\ln k;\,\infty\Big)\subset\rho(T_K).
\end{equation}
Furthermore, if $\lambda>\omega_K(\ln k)$ $(k=\max\{1,\norm{K}\})$ then we have
\begin{equation}\label{RTK:R2}
(\lambda-T_K)^{-1}g=
\varepsilon_\lambda(I- K_\lambda)^{-1}K\gamma_1(\lambda - T_0)^{-1}g
+(\lambda -T_0)^{-1}g
\end{equation}
for all $g\in L^1(\Omega)$,
where $\varepsilon_\lambda(\mu,v)=e^{-\lambda\frac{\mu}{v}}$.
\end{proposition}

\begin{proof}
Let $\lambda>\omega_K(\ln k)$ and let $g\in L^1(\Omega)$.
So, the general solution of the following equation
\begin{equation}\label{RTK:E10}
\lambda\varphi=-v\frac{\partial \varphi}{\partial \mu}+g
\end{equation}
is given by
\begin{equation}\label{RTK:E20}
\varphi=\varepsilon_\lambda\psi+(\lambda-T_0)^{-1}g
\end{equation}
where $\psi$ is any function of the variable $v\in J=(0,\infty)$.
When $\psi\in Y_1$,  we claim that $\varphi$ belongs to $W_1$.
Indeed, integrating \eqref{RTK:E20} and using
\eqref{T0:R3} we infer that
\begin{equation*}
\norm{\varphi}_1
\le\frac{1}{\lambda}\norm{\psi}_{Y_1}+
\frac{1}{\lambda}\norm{g}_1<\infty
\end{equation*}
which leads, by virtue of \eqref{RTK:E10}, to
\begin{equation*}
\norm{v\frac{\partial \varphi}{\partial \mu}}_1
=\norm{-\lambda\varphi+g}_1
\le\lambda\norm{\varphi}_1+\norm{g}_1<\infty.
\end{equation*}
Once more, integrating \eqref{RTK:E20} and using \eqref{T0:R4} we obtain that
\begin{equation*}
\begin{aligned}
\norm{v \varphi}_1
&=\norm{v \varepsilon_\lambda\psi}_1
+\norm{v (\lambda-T_0)^{-1}g}_1\\
&\le\norm{\psi}_{Y_1}+\norm{g}_1<\infty.
\end{aligned}
\end{equation*}
Hence, $\varphi\in W_1$. Furthermore, $\varphi$ belongs to $D(T_K)$ if
$\gamma_0\varphi=K\gamma_1\varphi$.
Namely, $\psi$ satisfies
\begin{equation*}
\psi=K_\lambda\psi+K\gamma_1(\lambda-T_0)^{-1}g
\end{equation*}
which admits, by virtue of \eqref{KLAMBDA:R1}, the following unique solution
\begin{equation}\label{RTK:E40}
\psi=(I-K_\lambda)^{-1}K\gamma_1(\lambda-T_0)^{-1}g\in Y_1.
\end{equation}
In order to achieve the proof, it suffices to put \eqref{RTK:E40} in \eqref{RTK:E20}.
\end{proof}

Now we are able to state the main result of this section
as follows.

\begin{theorem}\label{UK:THE1}
Let $K$ be an admissible operator whose abscissa is $\omega_K$.
Then, 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}
\norm{U_K(t)\varphi}_1
\le k e^{\omega_K(\ln k)t}\norm{\varphi}_1\quad t\ge0
\end{equation}
for all $\varphi\in L^1(\Omega)$, where $k=\max\set{1;\,\norm{K}}$.
\end{theorem}

\begin{remark} \rm
Actually, the admissibility concept that we have gave in Remark~\ref{ADM:REM},
is a particular case of a general and theoretical concept already defined in \cite{Boulanouar9}.
Accordingly, Theorem~\ref{UK:THE1} can be inferred from \cite[Theorem 3.1]{Boulanouar9}.
However, taking into account to the practical and biological aspect of this work, we prefer to give a slightly different proof for the reader.
\end{remark}

\begin{proof}[Proof of Theorem \ref{UK:THE1}]
Firstly, let $\omega>\omega_K$ be given and let
\begin{equation}\label{UK:E10}
\norme{\varphi}_1=\int_0^\infty\int_0^1\abs{\varphi(\mu,v)} h_\omega(\mu,v)\,d\mu\,dv
\end{equation}
be another norm on $L^1(\Omega)$ where
\begin{equation*}
 h_\omega(\mu,v)=k^{\min\set{\omega\frac{\mu}{v};\,1}}.
\end{equation*}
The norms \eqref{NORM:DEF} and \eqref{UK:E10} are equivalent because,
for all $\varphi\in L^1(\Omega)$ we have
\begin{equation}\label{EQUIV}
\norm{\varphi}_1\le\norme{\varphi}_1\le k\norm{\varphi}_1.
\end{equation}

Next, let $\lambda>\omega(\ln k)$ and let $g\in L^1(\Omega)$.
Proposition~\ref{RTK:PRO} means that
$\varphi=(\lambda-T_K)^{-1}g\in D(T_K)$ is the unique solution of
the following system
\begin{align}
\label{UK:E30}
\lambda\varphi&=-v\frac{\partial \varphi}{\partial \mu}+g\\
\label{UK:E31}
\gamma_0\varphi&=K\gamma_1\varphi.
\end{align}
So, multiplying \eqref{UK:E30} by $(\operatorname{sgn}\varphi) h_\omega$
and integrating it over $\Omega$, we obtain that
\begin{equation}\label{UK:E40}
\begin{aligned}
&\lambda\norme{\varphi}_1\\
&=-\int_\Omega
\parent{v\frac{\partial\abs{\varphi}}{\partial\mu}}(\mu,v) h_\omega(\mu,v)\,d\mu\,dv
 +\int_\Omega\operatorname{sgn}\varphi(\mu,v)( h_\omega g)(\mu,v)\,d\mu\,dv\\
&\le-\int_\Omega
\parent{v\frac{\partial\abs{\varphi}}{\partial\mu}}(\mu,v) h_\omega(\mu,v)\,d\mu\,dv
+\norme{g}_1\\
&:=A+\norme{g}_1.
\end{aligned}
\end{equation}
Integrating $A$ by parts and using \eqref{UK:E31} we infer that
\begin{equation}\label{UK:E50}
\begin{aligned}
A&=\int_0^\infty\abs{\gamma_0( h_\omega\varphi)(v)}v\,dv
-\int_0^\infty\abs{\gamma_1( h_\omega\varphi)(v)}v\,dv\\
&\quad +\int_\Omega
\parent{v\frac{\partial h_\omega}{\partial\mu}}
(\mu,v)\abs{\varphi(\mu,v)}\,d\mu\,dv\\
&=\int_0^\infty\abs{\gamma_0\varphi(v)}v\,dv
-\int_0^\infty\abs{\gamma_1( h_\omega\varphi)(v)}v\,dv\\
&\quad +\int_\Omega
\parent{v\frac{\partial h_\omega}{\partial\mu}}
(\mu,v)\abs{\varphi(\mu,v)}\,d\mu\,dv\\
&=\int_0^\infty\abs{K\gamma_1\varphi(v)}v\,dv
-\int_0^\infty\abs{\gamma_1( h_\omega\varphi)(v)}v\,dv\\
&\quad +\int_\Omega
\parent{v\frac{\partial h_\omega}{\partial\mu}}
(\mu,v)\abs{\varphi(\mu,v)}\,d\mu\,dv\\
&:=A_1-A_2+A_3.
\end{aligned}
\end{equation}
Applying \eqref{KLAMBDA:E40} together with \eqref{KLAMBDA:E05}
and \eqref{KLAMBDA:E35} for $\psi=\gamma_1\varphi\in Y_1$,
it follows that
\begin{align*}
A_1&=\int_0^\infty\abs{K\Bigr(\mathbb{I}_{\omega}(\gamma_1\varphi)
+\overline{\mathbb{I}}_{\omega}(\gamma_1\varphi)\Bigr)(v)}v\,dv\\
&\le\int_0^\infty\abs{K\Bigr(\mathbb{I}_{\omega}(\gamma_1\varphi)\Bigr)(v)}v\,dv
+\int_0^\infty\abs{K\Bigr(\overline{\mathbb{I}}_{\omega}(\gamma_1\varphi)\Bigr)(v)}v\,dv\\
&\le\int_0^\infty\abs{K\Bigr(\mathbb{I}_{\omega}(\gamma_1\varphi)\Bigr)(v)}v\,dv
+\norm{K}\int_0^\infty\abs{\Bigr(\overline{\mathbb{I}}_{\omega}(\gamma_1\varphi)\Bigr)(v)}v\,dv\\
&=\norm{K\Bigr(\mathbb{I}_{\omega}\Bigr(\theta_{\lambda'}(\gamma_1 h_\omega)\Bigr)(\gamma_1\varphi)\Bigr)}_{Y_1}
+\norm{K}\norm{\overline{\mathbb{I}}_{\omega}(\gamma_1\varphi)}_{Y_1}
\end{align*}
because of $\Bigr(\theta_{\lambda'}(\gamma_1 h_\omega)\Bigr)(v)=1$
for all $v\in(\omega,\infty)$, where $\theta_{\lambda'}$ is given by
\eqref{KLAMBDA:DEF} for $\lambda'=\omega(\ln k)$.
Hence
\begin{align*}
A_1&\le\norm{K\theta_{\lambda'}\Bigr(\mathbb{I}_{\omega}\gamma_1( h_\omega\varphi)\Bigr)}_{Y_1}
+\norm{K}\norm{\overline{\mathbb{I}}_{\omega}(\gamma_1\varphi)}_{Y_1}\\
&=\norm{K_{\lambda'}\Bigr(\mathbb{I}_{\omega}\gamma_1( h_\omega\varphi)\Bigr)}_{Y_1}
+\norm{K}\norm{\overline{\mathbb{I}}_{\omega}(\gamma_1\varphi)}_{Y_1}\\
&\le\norm{K_{\lambda'}}\norm{\mathbb{I}_{\omega}\gamma_1( h_\omega\varphi)}_{Y_1}
+\norm{K}\norm{\overline{\mathbb{I}}_{\omega}(\gamma_1\varphi)}_{Y_1}
\end{align*}
and therefore
\begin{equation}\label{UK:E60}
A_1\le\norm{\mathbb{I}_{\omega}\gamma_1( h_\omega\varphi)}_{Y_1}
+\norm{K}\norm{\overline{\mathbb{I}}_{\omega}(\gamma_1\varphi)}_{Y_1}
\end{equation}
because of \eqref{KLAMBDA:R1}.
Once more, applying \eqref{KLAMBDA:E40} together with \eqref{KLAMBDA:E05} and \eqref{KLAMBDA:E35}
for $\psi=\abs{\gamma_1( h_\omega\varphi)}\in Y_1$,
we infer that
\begin{align*}
A_2&=\int_0^\infty\Bigr(\mathbb{I}_{\omega}\abs{\gamma_1( h_\omega\varphi)}
+\overline{\mathbb{I}}_{\omega}\abs{\gamma_1( h_\omega\varphi)}\Bigr)(v)v\,dv\\
&=\int_0^\infty\abs{\mathbb{I}_{\omega}\gamma_1( h_\omega\varphi)(v)}v\,dv
+\int_0^\infty\abs{\overline{\mathbb{I}}_{\omega}(\gamma_1 h_\omega)(\gamma_1\varphi)(v)}v\,dv\\
&=\int_0^\infty\abs{\mathbb{I}_{\omega}\gamma_1( h_\omega\varphi)(v)}v\,dv
+k\int_0^\infty\abs{\overline{\mathbb{I}}_{\omega}(\gamma_1\varphi)(v)}v\,dv
\end{align*}
and therefore
\begin{equation}\label{UK:E70}
A_2=\norm{\mathbb{I}_{\omega}\gamma_1( h_\omega\varphi)}_{Y_1}+
k\norm{\overline{\mathbb{I}}_{\omega}(\gamma_1\varphi)}_{Y_1}.
\end{equation}
Next, for almost all $(\mu,v)\in\Omega$ we have
\begin{equation*}
\begin{aligned}
\parent{v\frac{\partial h_\omega}{\partial\mu}}(\mu,v)
&=v\frac{\partial}{\partial\mu}
\parent{
\begin{cases}
k^{\omega\frac{\mu}{v}}&\text{if  $\omega\frac{\mu}{v}\le1$}\\
k&\text{if  $\omega\frac{\mu}{v}>1$}
\end{cases}
}\\
&=\begin{cases}
\omega(\ln k)k^{\omega\frac{\mu}{v}}&\text{if  $\omega\frac{\mu}{v}\le1$}\\
0&\text{if  $\omega\frac{\mu}{v}>1$}
\end{cases}
\\
&
\le\omega(\ln k)
\begin{cases}
k^{\omega\frac{\mu}{v}}&\text{if  $\omega\frac{\mu}{v}\le1$}\\
k&\text{if  $\omega\frac{\mu}{v}>1$}
\end{cases}
\end{aligned}
\end{equation*}
which leads to
\begin{equation*}
\parent{v\frac{\partial h_\omega}{\partial\mu}}
\le\omega(\ln k) h_\omega.
\end{equation*}
Hence
\begin{equation}\label{UK:E80}
A_3\le\omega(\ln k)\norme{\varphi}_1.
\end{equation}
Replacing  now \eqref{UK:E60} and \eqref{UK:E70} and \eqref{UK:E80}
into \eqref{UK:E50} we infer that
\begin{equation*}
A\le\omega(\ln k)\norme{\varphi}_1
\end{equation*}
which we put into \eqref{UK:E40} to finally get that
\begin{equation*}
\norme{\varphi}_1=\norme{(\lambda-T_K)^{-1}g}_1\le
\frac{\norme{g}_1}
{\left(\lambda-\omega(\ln k)\right)}.
\end{equation*}
On the other hand, \eqref{RTK:R1} obviously leads to
$\rho(T_K)\not=\emptyset$ and therefore $T_K$ is a closed operator.
Furthermore, $T_K$ is densely defined because of
$\mathcal{C}_c(\Omega)\subset D(T_K)\subset L^1(\Omega)$.
Now, thanks to Hille-Yosida Theorem,
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:E100}
\norme{U_K(t)\varphi}_1\le
e^{\omega(\ln k)t}
\norme{\varphi}_1\quad t\ge0
\end{equation}
for all $\varphi\in L^1(\Omega)$.
As $\omega$ $(\omega>\omega_K)$ is arbitrary chosen, then passing
at the limit $\omega\to\omega_K$ in \eqref{UK:E100} we obtain that
\begin{equation}\label{UK:E110}
\norme{U_K(t)\varphi}_1\le
e^{\omega_K(\ln k)t}
\norme{\varphi}_1\quad t\ge0.
\end{equation}
Finally, in order to archives the proof, it suffices to infer \eqref{UK:R1}
from \eqref{UK:E110} together with \eqref{EQUIV}.
\end{proof}

Now, let us infer some interesting Corollaries.

\begin{corollary}\label{UK:COR1}
Let $K$ be a bounded linear operator from $Y_1$ into itself such that
$\norm{K}<1$. Then the operator $T_K$
generates, on $L^1(\Omega)$, a strongly continuous
semigroup $U_K=(U_K(t))_{t\ge0}$ of contractions; i.e.,
\begin{equation*}
\norm{U_K(t)\varphi}_1
\le \norm{\varphi}_1 \quad t\ge0
\end{equation*}
for all $\varphi\in L^1(\Omega)$.
\end{corollary}

\begin{proof}
Thanks to Remark~\ref{ADM:REM}, the hypothesis {\rm (Kb)}
holds and therefore $K$ is an admissible operator whose
abscissa $\omega_K=0$ because of \eqref{OMEGAK:DEF}.
Now, it suffices to apply Theorem~\ref{UK:THE1}
for $\omega_K =0$ and $k=1$.
\end{proof}

\begin{remark}\label{UK:REM} \rm
According to Corollary~\ref{UK:COR1} we infer that
\begin{equation*}
\norm{U_K(t)\varphi}_1=\norm{U_K(t-s)U_K(s)\varphi}_1
\le\norm{U_K(s)\varphi}_1
\end{equation*}
for all initial data $\varphi\in L^1(\Omega)$,
where $t$ and $s$ ($t>s$) are two arbitrary times.
Namely, the unperturbed model \eqref{MODEL:EQUATION}, \eqref{MODEL:BC}
(without bacterial mortality ($\sigma=0$)),
corresponding to Corollary~\ref{UK:COR1}
is biologically uninteresting because the bacteria number is obviously decreasing.
\end{remark}

In contrary to Remark~\ref{UK:REM}, we can say that $\norm{K}>1$ corresponds
to an increasing bacteria number during each mitotic. Hence we have

\begin{corollary}\label{UK:COR2}
Let $K$ be a linear compact operator from $Y_1$ into itself
such that $\norm{K}>1$.
Then, the operator $T_K$ generates, on $L^1(\Omega)$,
a strongly continuous semigroup $U_K=(U_K(t))_{t\ge0}$ satisfying
\begin{equation*}
\norm{U_K(t)\varphi}_1
\le\norm{K}\norm{K}^{t\omega_K}\norm{\varphi}_1\quad t\ge0
\end{equation*}
for all $\varphi\in L^1(\Omega)$, where $\omega_K>0$ is abscissa of the operator $K$.
\end{corollary}
\begin{proof}
By virtue of Remark~\ref{ADM:REM}, we infer that the hypothesis {\rm (Kc)}
holds and therefore $K$ is an admissible operator.
Furthermore, its abscissa $\omega_K>0$ because of \eqref{OMEGAK:DEF}.
Now, Theorem~\ref{UK:THE1} together with $k=\norm{K}$
achieve the proof.
\end{proof}

\section{Explicit form of the unperturbed semigroup}
\label{Explicit form}

The purpose of this section is to find
the explicit form of the semigroup $U_K=(U_K(t))_{t\ge0}$
which will be very useful to describe the asynchronous exponential growth
related to the model \eqref{MODEL:EQUATION}, \eqref{MODEL:BC}.

\begin{theorem}\label{UK:THE2}
Let $K$ be an admissible operator whose abscissa is $\omega_K$.
Then, for all $\varphi\in L^1(\Omega)$,
we have
\begin{equation}\label{UK:THE2:R1}
U_K(t)\varphi=U_0(t)\varphi+A_K(t)\varphi\quad t\ge0,
\end{equation}
where the operator $A_K(t)$ is defined by
\begin{equation}\label{UK:THE2:R2}
A_K(t)\varphi(\mu,v):=\xi(\mu,v,t)
K\parent{\gamma_1U_K\parent{t-\frac{\mu}{v}}\varphi}(v)
\end{equation}
with
\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}
\end{theorem}

\begin{proof}
Let $\lambda>\omega_K(\ln k)$ be fixed, where $k=\max\{1,\norm{K}\}$.
In the sequel, we are going to divide the proof in several steps.
\smallskip

\noindent\textbf{Step I.}
Let $L_\lambda^1:=L^1\Bigr(\Delta,\; e^{\lambda x/v}\Bigr)$
be the weighted Banach space whose norm is
\begin{equation*}
\norm{f}_{L_\lambda^1}=\int_{\Delta}\abs{f(x,v)}e^{\lambda x/v}\,dx\,dv,
\end{equation*}
where $\Delta=(-\infty,0)\times(0,\infty)$.
Let $H_K$ and $V_K$ be the following linear operators
\begin{gather*}
H_Kf(x,v)
:=K\Bigr(\xi(1,\cdot,-xv^{-1})f(1+xv^{-1}\cdot,\cdot)\Bigr)(v)\\
V_K\varphi(x,v)
:=K\Bigr(\gamma_1U_0(-xv^{-1})\varphi\Bigr)(v).
\end{gather*}
So, for all  $f\in L_\lambda^1$,
\begin{align*}
\norm{H_Kf}_{L_\lambda^1}
&=\int_{-\infty}^0\int_0^{\infty}
\abs{K\Bigr(\xi(1,\cdot,-xv^{-1})f(1+xv^{-1}\cdot,\cdot)\Bigr)(v)}
e^{\lambda x/v}\,dx\,dv\\
&=\int_0^\infty\Big[\int_0^{\infty}
\abs{K\Bigr(\xi(1,\cdot,t)f(1-t\cdot,\cdot)\Bigr)(v)}v\,dv\Big]
e^{-\lambda t}dt\\
&=\int_0^\infty
\Big[\int_0^{\infty}\abs{
K\theta_{\lambda}\Bigr(\theta_{-\lambda}\xi(1,\cdot,t)f(1-t\cdot,\cdot)\Bigr)(v)}
v\,dv\Big]e^{-\lambda t}dt\\
&=\int_0^\infty
\Big[\int_0^{\infty}\abs{
K_\lambda\Bigr(\theta_{-\lambda}\xi(1,\cdot,t)f(1-t\cdot,\cdot)\Bigr)(v)}
v\,dv\Big]e^{-\lambda t}dt
\end{align*}
because of \eqref{KLAMBDA:DEF}. Due to the boundedness
of the operator $K_\lambda$ (see Lemma~\ref{KLAMBDA:LEM}) we infer that
\begin{equation*}
\begin{aligned}
\norm{H_Kf}_{L_\lambda^1}
&\le\norm{K_\lambda}_{\mathcal{L}(Y_1)}\int_0^\infty\int_0^{\infty}
e^{\frac{\lambda}{v}}\xi(1,v,t)\abs{f(1-tv,v)}e^{-\lambda t}v\,dvdt\\
&=\norm{K_\lambda}_{\mathcal{L}(Y_1)}\int_0^\infty
\int_{-\infty}^1
e^{\frac{\lambda}{v}}\xi(1,v,\tfrac{(1-x)}{v})\abs{f(x,v)}
e^{-\lambda\frac{(1-x)}{v}}\,dx\,dv\\
&=\norm{K_\lambda}_{\mathcal{L}(Y_1)}\int_0^\infty
\int_{-\infty}^0\abs{f(x,v)}
e^{\lambda x/v}dx dv\\
&=\norm{K_\lambda}_{\mathcal{L}(Y_1)}\norm{f}_{L_\lambda^1}
\end{aligned}
\end{equation*}
and therefore
\begin{equation}\label{UK:THE2:E1}
\norm{H_K}_{\mathcal{L}(L_\lambda^1)}
\le\norm{K_\lambda}_{\mathcal{L}(Y_1)}.
\end{equation}
On the other hand, for all $\varphi\in L^1(\Omega)$ we have
\begin{equation*}
\begin{aligned}
\norm{V_K\varphi}_{L_\lambda^1}
&=\int_0^\infty\int_{-\infty}^0
\abs{K\Bigr(\gamma_1U_0(-xv^{-1})\varphi\Bigr)(v)}e^{\lambda x/v}\,dx\,dv\\
&\le\int_0^\infty\bracket{\int_0^\infty
\abs{K\Bigr(\gamma_1U_0(t)\varphi\Bigr)(v)}v\,dv}dt\\
&\le\norm{K}_{\mathcal{L}(Y_1)}\int_0^\infty\int_0^\infty
\abs{\Bigr(\gamma_1U_0(t)\varphi\Bigr)(v)}vdtdv
\end{aligned}
\end{equation*}
which leads, by virtue of \eqref{U0:DEF} and \eqref{CHI:DEF}, to
\begin{equation*}
\begin{aligned}
\norm{V_K\varphi}_{L_\lambda^1}
&\le\norm{K}_{\mathcal{L}(Y_1)}\int_0^\infty\int_0^\infty
\abs{\chi(1,v,t)\varphi(1-tv,v)}vdtdv\\
&=\norm{K}_{\mathcal{L}(Y_1)}\int_0^\infty\int_{-\infty}^1
\abs{\chi(1,v,\tfrac{(1-\mu)}{v})\varphi(\mu,v)}\,d\mu\,dv\\
&=\norm{K}_{\mathcal{L}(Y_1)}\int_0^\infty\int_0^1
\abs{\varphi(\mu,v)}\,d\mu\,dv
\end{aligned}
\end{equation*}
and therefore
\begin{equation}\label{UK:THE2:E1BIS}
\norm{V_K\varphi}_{L_\lambda^1}
\le\norm{K}_{\mathcal{L}(Y_1)}\norm{\varphi}_1.
\end{equation}
Now, \eqref{KLAMBDA:R1} together with \eqref{UK:THE2:E1} and \eqref{UK:THE2:E1BIS} imply that the problem
\begin{equation}\label{UK:THE2:E2}
f=H_Kf+V_K\varphi
\end{equation}
admits, for all $\varphi\in L^1(\Omega)$, the unique solution
\begin{equation}\label{UK:THE2:E3}
f_\varphi^K=(I-H_K)^{-1}V_K\varphi\in L_\lambda^1
\end{equation}
satisfying
\begin{equation}\label{UK:THE2:E4}
\norm{f_\varphi^K}_{L_\lambda^1}\le\frac{\norm{K}_{\mathcal{L}(Y_1)}}
{1-\norm{K_\lambda}_{\mathcal{L}(Y_1)}}
\norm{\varphi}_1.
\end{equation}
Furthermore
\begin{equation}\label{UK:THE2:E5}
\varphi\in L^1(\Omega)\to f_\varphi^K\in L_\lambda^1
\end{equation}
is a linear mapping because of those of the operators $H_K$ and $V_K$.

Now we can say that :
\emph{If $\lambda>\omega_K(\ln k)$ then,
for all $\varphi\in L^1(\Omega)$
the problem \eqref{UK:THE2:E2} admits the unique solution \eqref{UK:THE2:E3}
satisfying \eqref{UK:THE2:E4}.
Moreover \eqref{UK:THE2:E5} is a linear continuous mapping
from $L^1(\Omega)$ into $L_\lambda^1$.}
\smallskip

\noindent\textbf{Step II.}
Thanks to the step I, we can define the following operator
\begin{equation*}
B_K(t)\varphi(\mu,v):=\xi(\mu,v,t)f_\varphi^K(\mu-tv,v)\quad t\ge0.
\end{equation*}
Note that the linearity of the operator $B_K(t)$ follows from that of
\eqref{UK:THE2:E5}.
\par
First, let $t\ge0$. For all $\varphi\in L^1(\Omega)$ we have
\begin{equation*}
\begin{aligned}
\norm{B_K(t)\varphi}_1
&=\int_\Omega
\xi(\mu,v,t)\abs{f_\varphi^K(\mu-tv,v)}\,d\mu\,dv\\
&\le\int_0^\infty\int_0^{tv}
\abs{f_\varphi^K(\mu-tv,v)}
e^{\lambda\frac{\mu}{v}}\,d\mu\,dv\\
&=\int_0^\infty\int_{-tv}^0
\abs{f_\varphi^K(x,v)}
e^{\lambda\parent{\frac{x}{v}+t}}\,dx\,dv
\end{aligned}
\end{equation*}
which implies that
\begin{equation}\label{UK:THE2:E5BIS}
\norm{B_K(t)\varphi}_1
\le e^{\lambda t}\int_0^\infty\int_{-tv}^0
\abs{f_\varphi^K(x,v)}
e^{\lambda x/v}\,dx\,dv
\end{equation}
and therefore
\begin{equation*}
\norm{B_K(t)\varphi}_1\le e^{\lambda t}\norm{f_\varphi^K}_{L_\lambda^1}
\le e^{\lambda t}\frac{\norm{K}_{\mathcal{L}(Y_1)}}
{1-\norm{K_\lambda}_{\mathcal{L}(Y_1)}}
\norm{\varphi}_1
\end{equation*}
because of \eqref{UK:THE2:E4}.
Hence, $B_K(t)$ is a bounded operator from $L^1(\Omega)$ into itself.
Furthermore, \eqref{UK:THE2:E5BIS} obviously leads to
\begin{equation}\label{UK:THE2:E6}
\lim_{t\to0_+}\norm{B_K(t)\varphi}_1=0
\quad\text{and}\quad B_K(0)=0.
\end{equation}
\par
Next. For all $\varphi\in W_1$ we have
\begin{align*}
&\int_0^\infty\int_0^\infty\abs{\gamma_0B_K(t)\varphi(v)-
K\gamma_1B_K(t)\varphi(v)-K\gamma_1U_0(t)\varphi(v)}vdtdv\\
&=\int_0^\infty\int_0^\infty\abs{f_\varphi^K(-tv,v)-
H_Kf_\varphi^K(-tv,v)-V_K\varphi(-tv,v)}vdtdv\\
&=\int_{\Delta}
\abs{f_\varphi^K(x,v)-H_Kf_\varphi^K(x,v)-V_K\varphi(x,v)}\,dx\,dv
=0
\end{align*}
because $f_\varphi^K$ is the unique solution of
the problem \eqref{UK:THE2:E2}
and therefore
\begin{equation}\label{E:AAA}
\gamma_0B_K(t)\varphi-
K\gamma_1B_K(t)\varphi=K\gamma_1U_0(t)\varphi
\quad\text{a.e. } t\in\mathbb{R}_+.
\end{equation}
Due to the continuity of the mapping \eqref{T0:R5}, it follows that
\eqref{E:AAA} holds for all $t\ge0$; i.e.,
\begin{equation}\label{E:BB1}
\gamma_0B_K(t)\varphi-
K\gamma_1B_K(t)\varphi=K\gamma_1U_0(t)\varphi
\quad\text{for all } t\in\mathbb{R}_+.
\end{equation}
Now we can say that : \emph{for all $t\ge0$, $B_K(t)$ is a
bounded linear operator from $L^1(\Omega)$ into itself satisfying
\eqref{UK:THE2:E6} and \eqref{E:BB1}.}
\smallskip

\noindent\textbf{Step III.}
Thanks to the step II together with
Lemma~\ref{T0:LEM}, we can define the following operator
\begin{equation}\label{UK:THE2:E7}
S_K(t):=U_0(t)+B_K(t)\quad t\ge0
\end{equation}
which is clearly linear and bounded from $L^1(\Omega)$ into itself.

First, let $\varphi\in L^1(\Omega)$.
By virtue Lemma~\ref{T0:LEM}(1) together with \eqref{UK:THE2:E6}
we infer that
\begin{equation}\label{UK:THE2:E8}
S_K(0)=U_0(0)+B_K(0)=U_0(0)=I,
\end{equation}
where $I$ is the identity operator into $L^1(\Omega)$, and
\begin{equation}\label{UK:THE2:E9}
\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.
\end{equation}
\par
Next, let $t\ge0$ and let $\varphi\in W_1 $.
Due to \eqref{UK:THE2:E7} we obtain
\begin{equation} \label{UK:THE2:E10}
\begin{aligned}
&\xi(\mu,v,t)\gamma_0\left(S_K\parent{t-\frac{\mu}{v}}\varphi\right)(v) \\
&=\xi(\mu,v,t)
\gamma_0\left(U_0\parent{t-\frac{\mu}{v}}\varphi\right)(v)
+\xi(\mu,v,t)\gamma_0\left(B_K\parent{t-\frac{\mu}{v}}\varphi\right)(v)
\end{aligned}
\end{equation}
 for almost all $(\mu,v)\in\Omega$.
However, \eqref{U0:DEF} together with \eqref{CHI:DEF} and \eqref{XI:DEF} lead to
\begin{equation*}
\xi(\mu,v,t)\gamma_0\left(U_0\parent{t-\frac{\mu}{v}}\varphi\right)(v)=0
\end{equation*}
for almost all $(\mu,v)\in \Omega$, and therefore \eqref{UK:THE2:E10} becomes
\begin{align*}
\xi(\mu,v,t)\gamma_0\left(S_K\parent{t-\frac{\mu}{v}}\varphi\right)(v)
&=\xi(\mu,v,t)\gamma_0\left(B_K\parent{t-\frac{\mu}{v}}\varphi\right)(v)\\
&=\xi(\mu,v,t)B_K\parent{t-\frac{\mu}{v}}\varphi(0,v)\\
&=\xi(\mu,v,t)f_\varphi^K(\mu-tv,v).
\end{align*}
Hence
\begin{equation}\label{UK:THE2:E11}
\xi(\mu,v,t)\gamma_0\left(S_K\parent{t-\frac{\mu}{v}}\varphi\right)(v)=
B_K(t)\varphi(\mu,v)
\end{equation}
for almost all $(\mu,v)\in \Omega$.

On the other hand, \eqref{E:BB1} and \eqref{UK:THE2:E7} imply that
\begin{align*}
\gamma_0S_K(t)\varphi-K\gamma_1S_K(t)\varphi
&=\gamma_0U_0(t)\varphi+\gamma_0B_K(t)\varphi
 -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
=0
\end{align*}
for $t\ge0$ and therefore
\begin{equation}\label{UK:THE2:E12}
\xi(\mu,v,t)
\gamma_0\left(S_K\parent{t-\tfrac{\mu}{v}}\varphi\right)(v)=
\xi(\mu,v,t)
\left(K\gamma_1S_K\parent{t-\tfrac{\mu}{v}}\varphi\right)(v)
\end{equation}
for almost all $(\mu,v)\in \Omega$.
Hence
\begin{equation}\label{UK:THE2:E13}
B_K(t)\varphi(\mu,v)=\xi(\mu,v,t)
\left(K\gamma_1S_K\parent{t-\tfrac{\mu}{v}}\varphi\right)(v)\quad t\ge0
\end{equation}
because of \eqref{UK:THE2:E11} and \eqref{UK:THE2:E12}.
Moreover, the density of $W_1$ in $L^1(\Omega)$ implies that
\eqref{UK:THE2:E13} holds too for all $\varphi\in L^1(\Omega)$.

Now we can say that :
\emph{for all $t\ge0$, $S_K(t)$ given by \eqref{UK:THE2:E7},
is a bounded linear operator from $L^1(\Omega)$
into itself satisfying \eqref{UK:THE2:E8} and \eqref{UK:THE2:E9}.
Furthermore, \eqref{UK:THE2:E13} holds for all $\varphi\in L^1(\Omega)$.}
\smallskip

\noindent \textbf{Step IV.}
In order to prove that $(S_K(t))_{t\ge0}$ is
a strongly continuous semigroup it remains, by virtue of the step III,
to show only that
\begin{equation*}
G(t,s):=S_K(t)S_K(s)-S_K(t+s)=0
\quad \text{for all $t\ge0$ and all $s\ge0$}.
\end{equation*}
So, let $t\ge0$ and $s\ge0$ and let $\varphi\in L^1(\Omega)$.
By virtue of \eqref{U0:DEF} and \eqref{UK:THE2:E7} and \eqref{UK:THE2:E13}, a simple computation
leads to
\begin{equation}\label{UK:THE2:E15}
G(t,s)\varphi(\mu,v)
=\xi(\mu,v,t)\left(K\gamma_1G\parent{t-\tfrac{\mu}{v},s}\varphi\right)(v)
\end{equation}
for almost all $(\mu,v)\in \Omega$.

First, applying the trace mapping $\gamma_1$ to \eqref{UK:THE2:E15}
and integrating it,
we obtain that
\begin{align*}
&\int_0^\infty\int_0^{\infty}
e^{\lambda\parent{\frac{1}{v}-t}}
\abs{\gamma_1G(t,s)\varphi(v)}vdtdv\\
&=\int_0^\infty\int_0^{\infty}
e^{\lambda\parent{\frac{1}{v}-t}}\xi(1,v,t)\abs{
\Big(K\gamma_1G(t-\tfrac{1}{v},s)\varphi\Big)(v)} vdtdv\\
&\le\int_0^\infty e^{-\lambda x}
\Big[\int_0^{\infty}
\abs{K\Big(\gamma_1G(x,s)\varphi\Big)(v)}v\,dv\Big]dx\\
&=\int_0^\infty e^{-\lambda x}
\Big[\int_0^{\infty}
\abs{K_\lambda\Big(\theta_{-\lambda}\gamma_1G(x,s)\varphi\Big)(v)}v\,dv\Big]dx
\end{align*}
because of \eqref{KLAMBDA:DEF}. Due to the boundedness
of the operator $K_\lambda$ (see Lemma~\ref{KLAMBDA:LEM}), it follows that
\begin{align*}
&\int_0^\infty\int_0^{\infty}
e^{\lambda\parent{\frac{1}{v}-t}}
\abs{\gamma_1G(t,s)\varphi(v)}vdtdv\\
&\leq\norm{K_\lambda} \int_0^\infty e^{-\lambda x}
\Big[\int_0^{\infty}
\big|\theta_{-\lambda}\big(\gamma_1G(x,s)\varphi\big)(v)\big|v\,dv\Big]dx\\
&=\norm{K_\lambda}
\int_0^\infty\int_0^{\infty}
e^{\lambda\parent{\frac{1}{v}-x}}\big|\gamma_1G(x,s)\varphi(v)\big|v\,dx\, dv
\end{align*}
which leads, by virtue of \eqref{KLAMBDA:R1}, to
\begin{equation}\label{UK:THE2:E16}
\gamma_1G(t,s)=0\quad\text{for all $t\ge0$ and all $s\ge0$}.
\end{equation}
On the other hand, integrating \eqref{UK:THE2:E15} we obtain that
\begin{align*}
\int_\Omega\abs{G(t,s)\varphi(\mu,v)}\,d\mu\,dv
&=\int_\Omega\xi(\mu,v,t)\abs{
K\Big(\gamma_1G(t-\tfrac{\mu}{v},s)\varphi\Big)(v)}
\,d\mu\,dv\\
&=\int_0^\infty\int_0^{tv}\abs{
K\Big(\gamma_1G(t-\tfrac{\mu}{v},s)\varphi\Big)(v)}  \,d\mu\,dv\\
&=\int_0^\infty\int_0^t\abs{
K\Big(\gamma_1G(x,s)\varphi\Big)(v)}v\,dx\,dv\\
&=0
\end{align*}
because of \eqref{UK:THE2:E16} and therefore
$G(t,s)=0$ for all $t\ge0$ and all $s\ge0$.

Now we can say that: \emph{ the family operators $(S_{K}(t))_{t\ge0}$
is a strongly continuous semigroup on $L^1(\Omega)$.}
\smallskip

\noindent\textbf{Step V.}
To achieve the proof, it suffices to show that
the semigroups $(S_K(t))_{t\ge0}$ and $(U_K(t))_{t\ge0}$
are equal.
So, let us suppose that $B$ denotes the generator
of the semigroup $(S_K(t))_{t\ge0}$.
\par
First, let $\varphi\in L^1(\Omega)$.
Due to \eqref{UK:THE2:E7} and \eqref{UK:THE2:E13} we infer that
\begin{align*}
&\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)
\left[K\gamma_1S_K(t-\tfrac{\mu}{v})\varphi\right](v)dt\\
&=\int_0^\infty e^{-\lambda t}U_0(t)\varphi(\mu,v)dt+
e^{-\lambda\frac{\mu}{v}}
\int_0^\infty e^{-\lambda t}\left[K\gamma_1S_K(t)\varphi\right](v)dt\\
&=\int_0^\infty e^{-\lambda t}U_0(t)\varphi(\mu,v)dt+
\varepsilon_\lambda(\mu,v) K\gamma_1
 \Big[\int_0^\infty e^{-\lambda t}S_K(t)\varphi dt\Big](v)
\end{align*}
for all almost $(\mu,v)\in\Omega$, and therefore
\begin{equation}\label{UK:THE2:E17}
(\lambda-B)^{-1}\varphi=
\varepsilon_\lambda K\gamma_1(\lambda-B)^{-1}\varphi+(\lambda-T_0)^{-1}\varphi.
\end{equation}
Applying $\theta_\lambda^{-1}\gamma_1$ to both hand side of
\eqref{UK:THE2:E17} we obtain
\begin{align*}
\theta_\lambda^{-1}\gamma_1(\lambda-B)^{-1}\varphi
&=\theta_\lambda^{-1}\gamma_1\Big(\varepsilon_\lambda K\gamma_1(\lambda-B)^{-1}\varphi\Big)
 +\theta_\lambda^{-1}\gamma_1(\lambda-T_0)^{-1}\varphi\\
&=\theta_\lambda^{-1}\theta_\lambda
K\gamma_1(\lambda-B)^{-1}\varphi
+\theta_\lambda^{-1}\gamma_1(\lambda-T_0)^{-1}\varphi\\
&=K\gamma_1(\lambda-B)^{-1}\varphi
+\theta_\lambda^{-1}\gamma_1(\lambda-T_0)^{-1}\varphi\\
&=K\theta_\lambda\theta_\lambda^{-1}\gamma_1(\lambda-B)^{-1}\varphi
+\theta_\lambda^{-1}\gamma_1(\lambda-T_0)^{-1}\varphi
\end{align*}
which leads, by  \eqref{KLAMBDA:DEF}, to
\begin{equation*}
\theta_\lambda^{-1}\gamma_1(\lambda-B)^{-1}\varphi=
K_\lambda\theta_\lambda^{-1}\gamma_1(\lambda-B)^{-1}\varphi
+\theta_\lambda^{-1}\gamma_1(\lambda-T_0)^{-1}\varphi
\end{equation*}
and therefore
\begin{equation}\label{UK:THE2:E18}
\gamma_1(\lambda-B)^{-1}\varphi
=\theta_\lambda(I-K_\lambda)^{-1}\theta_\lambda^{-1}\gamma_1(\lambda-T_0)^{-1}\varphi
\end{equation}
because of \eqref{KLAMBDA:R1}. Putting now \eqref{UK:THE2:E18} into
\eqref{UK:THE2:E17} we infer that
\begin{align*}
(\lambda-B)^{-1}\varphi
&=\varepsilon_\lambda K\theta_\lambda(I-K_\lambda)^{-1}\theta_\lambda^{-1}\gamma_1(\lambda-T_0)^{-1}\varphi
+(\lambda-T_0)^{-1}\varphi\\
&=\varepsilon_\lambda K_\lambda(I-K_\lambda)^{-1}\theta_\lambda^{-1}\gamma_1(\lambda-T_0)^{-1}\varphi
+(\lambda-T_0)^{-1}\varphi\\
&=\varepsilon_\lambda (I-K_\lambda)^{-1}K_\lambda\theta_\lambda^{-1}\gamma_1(\lambda-T_0)^{-1}\varphi
+(\lambda-T_0)^{-1}\varphi.
\end{align*}
Hence
\begin{equation}\label{UK:THE2:E19}
(\lambda-B)^{-1}\varphi=
\varepsilon_\lambda (I-K_\lambda)^{-1}K\gamma_1(\lambda-T_0)^{-1}\varphi
+(\lambda-T_0)^{-1}\varphi.
\end{equation}
Finally, \eqref{RTK:R2} and \eqref{UK:THE2:E19} obviously imply that
$(\lambda-T_K)^{-1}=(\lambda-B)^{-1}$ and therefore
\begin{equation}\label{UK:THE2:E20}
U_K(t)\varphi=S_K(t)\varphi\quad t\ge 0
\end{equation}
because of the uniqueness of the generated semigroup.
Now, in order to achieves the proof, it suffices to infer
\eqref{UK:THE2:R1} and \eqref{UK:THE2:R2} from
\eqref{UK:THE2:E7} and \eqref{UK:THE2:E13}
together with \eqref{UK:THE2:E20}.
\end{proof}

\section{Generation theorem for the model
\eqref{MODEL:EQUATION}, \eqref{MODEL:BC}}
\label{The global model}

The main goal of this section is to prove that
the general model \eqref{MODEL:EQUATION}, \eqref{MODEL:BC}
is governed by a strongly continuous semigroup $V_K=(V_K(t))_{t\ge0}$
as a linear perturbation of the unperturbed semigroup
$U_K=(U_K(t))_{t\ge0}$ already studied.
To this end, we suppose that the rate of bacterial mortality
fulfills the  hypothesis
\begin{itemize}
\item[$(H_\sigma)$]
$\sigma\in \big(L^\infty(\Omega)\big)_+$
\end{itemize}
and we denote
\begin{equation}\label{SUPINFSIGMA:DEF}
\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).
\end{equation}
Thanks to the hypothesis ${\rm(H_\sigma)}$, the perturbation operator
\begin{equation*}
S\varphi(\mu,v):=-\sigma(\mu,v)\varphi(\mu,v)\quad (\mu,v)\in\Omega
\end{equation*}
is obviously linear and bounded from $L^1(\Omega)$ into itself.
So, let $L_K$ be the unbounded operator
\begin{gather*}
L_K:=T_K+S\\
D(L_K)=D(T_K)
\end{gather*}
closely related to the model \eqref{MODEL:EQUATION}, \eqref{MODEL:BC},
and for which we finally have

\begin{theorem}\label{VK:THE}
Let $K$ be an admissible operator whose abscissa is $\omega_K$.
If the hypothesis $(\mathrm{H}_\sigma)$ holds, then the operator $L_K$ generates,
on $L^1(\Omega)$,
a strongly continuous semigroup $V_K=(V_K(t))_{t\ge0}$ satisfying
\begin{equation*}
\norm{V_K(t)\varphi}_1
\le k e^{t(\omega_K\ln k-\underline{\sigma})}
\norm{\varphi}_1\quad t\ge0
\end{equation*}
for all $\varphi\in L^1(\Omega)$, where $k=\max\set{1;\;\norm{K}}$.
\end{theorem}

\begin{proof}
As $L_K=T_K+S$ is a bounded linear perturbation of the generator
$T_K$, it follows by virtue of Lemma~\ref{PERTURBATION:LEM}
that $L_K$ is a generator of a strongly continuous semigroup
denoted $V_K=(V_K(t))_{t\ge0}$ satisfying
\begin{equation*}\label{}
V_K(t)\varphi=
\lim_{n\to\infty}\bracket{e^{-\sigma\frac{t}{n}}U_K\parent{\frac{t}{n}}}^n
\varphi\quad t\ge0
\end{equation*}
for all $\varphi\in L^1(\Omega)$.
Using the norm \eqref{UK:E10} together with \eqref{UK:E110}
and the hypothesis $(\mathrm{H}_\sigma)$ we infer
\begin{equation*}
\norme{V_K(t)\varphi}_1\le\lim_{n\to\infty}
\Big[e^{-\underline{\sigma}\tfrac{t}{n}}e^{\frac{t}{n}\omega_K(\ln k)}\Big]^n
\norme{\varphi}_1
=e^{t(\omega_K\ln k-\underline{\sigma})}\norme{\varphi}_1.
\end{equation*}
Now \eqref{EQUIV} completes the proof.
\end{proof}

Let us end this section with some interesting Corollaries.

\begin{corollary}\label{VK:COR1}
Let $K$ be a bounded linear operator from $Y_1$ into itself such that
$\norm{K}<1$.
If the hypothesis $(\mathrm{H}_\sigma)$ holds, then the operator $L_K$ generates,
on $L^1(\Omega)$, a strongly continuous semigroup
$V_K=(V_K(t))_{t\ge0}$ satisfying
\begin{equation*}
\norm{V_K(t)\varphi}_1
\le e^{-t\underline{\sigma}}
\norm{\varphi}_1\quad t\ge0
\end{equation*}
for all $\varphi\in L^1(\Omega)$.
\end{corollary}

The proof of the above corollary is similar to that of Corollary \ref{UK:COR1},
and is omitted.

\begin{remark}\label{VK:REM} \rm
Corollary~\ref{VK:COR1} means that the general model \eqref{MODEL:EQUATION},
\eqref{MODEL:BC} corresponding to the case $\norm{K}<1$
is biologically uninteresting because the bacteria number is decreasing.
Indeed, if $t$ and $s$ ($t>s$) are two arbitrary times, then we have
\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
<\norm{V_K(s)\varphi}_1
\end{equation*}
for all initial data $\varphi\in L^1(\Omega)$.
\end{remark}

In contrary to Remark~\ref{VK:REM}, we understand that $\norm{K}>1$
is closely related to an increasing bacteria number during each mitotic.
This is the most observed and biologically interesting case for which we have

\begin{corollary}\label{VK:COR2}
Let $K$ be a linear compact operator from $Y_1$ into itself
such that $\norm{K}>1$.
If the hypothesis $(\mathrm{H}_\sigma)$ holds, then the operator $L_K$ generates,
on $L^1(\Omega)$,
a strongly continuous semigroup $V_K=(V_K(t))_{t\ge0}$ satisfying
\begin{equation*}
\norm{V_K(t)\varphi}_1
\le\norm{K}\norm{K}^{t(\omega_K-\underline{\sigma})}
\norm{\varphi}_1\quad t\ge0
\end{equation*}
for all $\varphi\in L^1(\Omega)$, where $\omega_K>0$ is abscissa of the operator $K$.
\end{corollary}

The proof of the above is similar to that of Corollary \ref{UK:COR2}, and
it is omitted.
PAGE 18

\section{Lattice property of the generated semigroup}
\label{Lattice property}

In this section we are concerned with the lattice properties of the 
generated semigroup $V_K=(V_K(t))_{t\ge0}$. These properties can be
 inferred from those of the  linear operator
\begin{equation}\label{KLAMBDABAR:DEF}
\overline{K}_\lambda\psi:=\theta_\lambda K\psi,
\quad\text{where }
\theta_\lambda(v)=e^{-\lambda/v},\quad v\in J=(0,\infty)
\end{equation}
which is going to play an important role in the sequel.
So, before we start, let us note that $\overline{K}_\lambda$ ($\lambda\ge0$)
is clearly a bounded operator from $Y_1$ into itself because,
for all $\psi\in Y_1$ we have
\begin{equation*}
\norm{\overline{K}_\lambda\psi}_{Y_1}\le\norm{K\psi}_{Y_1}
\le\norm{K}\norm{\psi}_{Y_1}.
\end{equation*}

Despite the obvious difference between
\eqref{KLAMBDABAR:DEF} and \eqref{KLAMBDA:DEF},
both operators are related by the following result.

\begin{lemma}\label{KLAMBDABAR:LEM}
Let $K$ be an admissible operator whose abscissa is $\omega_K$.
If $K$ is positive, then
\begin{enumerate}
\item $\overline{K}_\lambda$ $(\lambda\ge0)$ is positive too.
Moreover, if $\lambda>\omega_K(\ln k)$ then
\begin{equation}\label{KLAMBDABAR:R1}
(I-K_\lambda)^{-1}K\ge\overline{K}_\lambda^n
\quad\text{for all integers } n\ge1,
\end{equation}
where $k=\max\{1,\norm{K}\}$.

\item Furthermore, if $K$ is irreducible, then
 $\overline{K}_\lambda$ is also irreducible.
\end{enumerate}
\end{lemma}

\begin{proof}
(1). Let $\lambda\ge0$. Firstly, it is easy to see that the positivity of
$\overline{K}_\lambda$ follows from that $K$. Furthermore, we obviously have
\begin{equation}\label{KLAMBDABAR:E5}
K\ge\overline{K}_\lambda.
\end{equation}
Next, as we clearly have $K_\lambda K=K\overline{K}_\lambda$,
it follows by induction that
\begin{equation}\label{KLAMBDABAR:E10}
K_\lambda^nK=K\overline{K}_\lambda^n
\quad\text{for all integers}\quad n\ge1.
\end{equation}
Therefore, if $\lambda>\omega_K(\ln k)$,
then \eqref{KLAMBDA:R1}, \eqref{KLAMBDABAR:E10} and \eqref{KLAMBDABAR:E5}
lead to
\begin{equation*}
(I-K_\lambda)^{-1}K=\sum_{m\ge0}K_\lambda^m K =\sum_{m\ge0}K\overline{K}_\lambda^m
\ge K\overline{K}_\lambda^{n-1}\ge\overline{K}_\lambda^n
\end{equation*}
for all integers $n\ge1$.

(2).  Firstly, let $M$ be a closed ideal in $Y_1$ such that
\begin{equation}\label{KLAMBDABAR:E100}
K_\lambda(M)\subset M.
\end{equation}
By virtue of the characterization of closed ideals in $L_1$-spaces
(see \cite[pp.309]{Nagel}), there exists $\Delta\subset J$ such that
\begin{equation*}
M=\{\psi\in Y_1;\, \psi(v)=0\text{ a.e. } v\in \Delta\}.
\end{equation*}
So, for all $\varphi\in K(M)$, there exists
$\psi\in M$ such that $\varphi=K\psi$. This implies that
\begin{equation*}
\theta_\lambda\varphi=\theta_\lambda K\psi=\overline{K}_\lambda\psi\in M
\end{equation*}
which leads to $\varphi\in M$ and therefore $K(M)\subset M$.
Now, by virtue of the irreducibility of $K$,
we obviously infer that
\begin{equation}\label{KLAMBDA-IRRE:E200}
M=\emptyset\quad\text{or}\quad M=Y_1
\end{equation}
and therefore, $\overline{K}_\lambda$ is irreducible
because \eqref{KLAMBDABAR:E100} holds only for \eqref{KLAMBDA-IRRE:E200}.
\end{proof}

Now, the lattice properties of the
semigroup $U_K=(U_K(t))_{t\ge0}$ are given as follows.

\begin{proposition}\label{UK-POS-IRR:THE}
Let $K$ be an admissible operator whose abscissa is $\omega_K$.
If $K$ is positive, then
\begin{enumerate}
\item The semigroup $U_K=(U_K(t))_{t\ge0}$ is also positive.

\item Furthermore, if $K$ is irreducible then
$U_K=(U_K(t))_{t\ge0}$ is also irreducible.
\end{enumerate}
\end{proposition}

\begin{proof}
(1) Let $\lambda>\omega_K(\ln k)$ and let $\varphi\in(L^1(\Omega))_+$.
Thanks to the second and the third point of Lemma~\ref{T0:LEM},
we infer that $(\lambda - T_0)^{-1}\varphi\ge0$
and $\gamma_1(\lambda - T_0)^{-1}\varphi\ge0$.
This together with \eqref{RTK:R2} and \eqref{KLAMBDABAR:R1} imply
\begin{equation}\label{UK-POS-IRR:E10}
(\lambda-T_K)^{-1}\varphi\ge\varepsilon_\lambda
\overline{K}_\lambda^n\gamma_1(\lambda - T_0)^{-1}\varphi
\quad\text{for all integers } n\ge1
\end{equation}
and therefore $(\lambda-T_K)^{-1}\varphi\ge0$
because of the positivity of the operator $\overline{K}_\lambda$
 (see Lemma~\ref{KLAMBDABAR:LEM}).
Now, the positivity of $U_K=(U_K(t))_{t\ge0}$
follows from the first point of Lemma~\ref{POS-IRRE:LEM}.

(2) Let $\lambda>\omega_K(\ln k)$ and let $\varphi\in (L^1(\Omega))_+$ 
be such that $\varphi\not=0$.
Thanks to the third point of Lemma~\ref{T0:LEM},
we infer that $\gamma_1(\lambda-T_0)^{-1}\varphi$
is a strictly positive function.
As $\overline{K}_\lambda$ is an irreducible operator 
(see Lemma~\ref{KLAMBDABAR:LEM})
then, there exists an integer $m\ge1$ such that
\begin{equation}\label{UK-POS-IRR:E30}
\overline{K}_\lambda^m\gamma_1(\lambda-T_0)^{-1}\varphi(v)>0
\quad\text{a.e. $v\in(0,\infty)$}.
\end{equation}
Putting now $n=m$ into \eqref{UK-POS-IRR:E10} we infer that
\begin{equation*}
(\lambda-T_K)^{-1}\varphi(\mu,v)\ge\varepsilon_\lambda(\mu,v)
\overline{K}_\lambda^m\gamma_1(\lambda - T_0)^{-1}\varphi(v)
\quad\text{a.e. $(\mu,v)\in \Omega$}
\end{equation*}
which leads, by virtue of \eqref{UK-POS-IRR:E30}, to
\begin{equation*}
(\lambda-T_K)^{-1}\varphi(\mu,v)>0\quad\text{a.e. $(\mu,v)\in \Omega$}
\end{equation*}
and therefore the irreducibility of $(\lambda-T_K)^{-1}$ follows.
Finally, the second point of Lemma \ref{POS-IRRE:LEM} leads 
to the irreducibility of $U_K=(U_K(t))_{t\ge0}$.
\end{proof}

Now, the main resutl of this section is as follows.

\begin{theorem}\label{VK-POS-IRR:THE}
Let $K$ be an admissible operator whose abscissa is $\omega_K$
and suppose that $(\mathrm{H}_\sigma)$ holds.
If $K$ is positive, then
\begin{enumerate}
\item
The semigroup $V_K=(V_K(t))_{t\ge0}$ is positive satisfying
\begin{equation}\label{VK-POS-IRR:R1}
e^{-t\overline{\sigma}}U_K(t)\le V_K(t),\quad t\ge0
\end{equation}
where $\overline{\sigma}$ is given by \eqref{SUPINFSIGMA:DEF}.
\item
Furthermore, if $K$ is irreducible, then
$V_K=(V_K(t))_{t\ge0}$ is also irreducible.
\end{enumerate}
\end{theorem}

\begin{proof}
(1) Let $t>0$ and let $\varphi \in (L^1(\Omega))_+$.
By  the first point of Proposition~\ref{UK-POS-IRR:THE}
we obtain the positivity of the semigroup $U_K=(U_K(t))_{t\ge0}$
which leads to
\begin{equation*}
\Big[e^{-\tfrac{t}{n}\sigma}U_K(t)\Big]^n\varphi \in (L^1(\Omega))_+
\quad\text{for all integers }n\in \mathbb{N},
\end{equation*}
and therefore
\begin{equation*}
e^{-t\overline{\sigma}}U_K(t)\varphi
\le\Big[e^{-\tfrac{t}{n}\sigma} U_K(\tfrac{t}{n})\Big]^n\varphi
\quad\text{for all integers }n\in \mathbb{N}.
\end{equation*}
Passing at the limit $n\to\infty$ and using \eqref{PERTURBATION:R1}, we infer that
\begin{equation*}
e^{-t\overline{\sigma}}U_K(t)\varphi
\le V_K(t)\varphi
\end{equation*}
and therefore the positivity
$V_K=(V_K(t))_{t\ge0}$ and \eqref{VK-POS-IRR:R1} follow.

(2) The irreducibility of the semigroup
$V_K=(V_K(t))_{t\ge0}$ obviously follows from that of the semigroup
$U_K=(U_K(t))_{t\ge0}$ (Proposition~\ref{UK-POS-IRR:THE})
together with \eqref{VK-POS-IRR:R1}.
\end{proof}

\section{Spectral properties of the generated semigroup}
\label{Spectral properties}

The purpose of this section is to compute the
type $\omega_0(V_K)$ of the semigroup $V_K=(V_K(t))_{t\ge0}$.
This will be obtained through spectral properties of the operator
$\overline{K}_\lambda$ given by \eqref{KLAMBDABAR:DEF}.

In the sequel, we  suppose that the operator
$K$ is compact from $Y_1$  into itself.
Thanks to Remark~\ref{ADM:REM}, $K$ is then an admissible operator whose abscissa is denoted by $\omega_K$.
Therefore, all results of this work hold.
So, let us start by the following preparative result.

\begin{lemma}\label{SIGMATK}
Let $K$ be a compact operator from $Y_1$  into itself and let
$\lambda\in\mathbb{C}$ be such that $\operatorname{Re}\lambda\ge0$.
Then we have
\begin{equation*}
\lambda\in\sigma(T_K)\Longrightarrow1\in\sigma_p\left(\overline{K}_\lambda\right).
\end{equation*}
\end{lemma}

\begin{proof}
Let $\lambda\in\mathbb{C}$ be such that $\operatorname{Re}\lambda\ge0$ and
let $g\in L^1(\Omega)$.
If $1\in\rho(\overline{K}_\lambda)$, then the  equation
\begin{equation} \label{LEMMA7L2:E1}
\psi=\overline{K}_\lambda\psi+\gamma_1(\lambda-T_0 )^{-1}g\\
\end{equation}
admits a unique solution $\psi\in Y_1$.
So, let $\varphi$ be the  function
\begin{equation}\label{LEMMA7L2:E2}
\varphi=\varepsilon_\lambda K\psi+(\lambda-T_0 )^{-1}g.
\end{equation}
On one hand, simple computations together with \eqref{T0:R3} 
and \eqref{T0:R4} infer that
\begin{equation}\label{LEMMA7L2:E3}
\norm{\varphi}_1
\le\frac{1}{\lambda}\norm{K}\norm{\psi}_{Y_1}+
\frac{1}{\lambda}\norm{g}_1<\infty
\end{equation}
and
\begin{equation}\label{LEMMA7L2:E4}
\begin{aligned}
\norm{v \varphi}_1
&=\norm{K}\norm{\psi}_{Y_1}
+\norm{v(\lambda-T_0)^{-1}g}_1\\
&\le\norm{\psi}_{Y_1}+\norm{g}_1<\infty.
\end{aligned}
\end{equation}
Moreover, we have
\begin{align*}
\lambda\varphi+v\frac{\partial\varphi}{\partial\mu}&=
\lambda\varphi+v\frac{\partial}{\partial\mu}\parent{\varepsilon_\lambda K\psi+(\lambda-T_0)^{-1}g}\\
&=\lambda\varphi-\lambda\varepsilon_\lambda K\psi-\lambda(\lambda-T_0)^{-1}g+g
=g
\end{align*}
and therefore
\begin{equation}\label{LEMMA7L2:E5}
\norm{v\frac{\partial\varphi}{\partial \mu}}_1
=\norm{-\lambda\varphi+g}_1
\le\lambda\norm{\varphi}_1+\norm{g}_1<\infty.
\end{equation}
So, we have $\varphi\in W_1$ because of \eqref{LEMMA7L2:E3},
\eqref{LEMMA7L2:E4} and \eqref{LEMMA7L2:E5}.
\par
On the other hand, \eqref{LEMMA7L2:E1} and \eqref{LEMMA7L2:E2}
lead to
\begin{align*}
\gamma_0\varphi=K\psi&=K\left[\overline{K}_\lambda\psi+\gamma_1(\lambda-T_0 )^ {-1}g\right]\\
&=K\gamma_1\left[\varepsilon_\lambda K\psi+(\lambda-T_0 )^ {-1}g\right]\\
&=K\gamma_1\varphi
\end{align*}
which implies that $\varphi\in D(T_K)$
and therefore $(\lambda-T_K)$ is an operator with bounded inverse.
Hence, we have $\lambda\in \rho(T_K)$.
\end{proof}

Now we are ready to state the main result of this section.

\begin{theorem}\label{VK-TYPE:THE}
Let $K$ be a positive, irreducible and compact operator
from $Y_1$  into itself such that
\begin{equation}\label{VK-TYPE:R1}
r(\overline{K}_{\overline{\sigma}-\underline{\sigma}})>1.
\end{equation}
If $(\mathrm{H}_\sigma)$ holds,
then the type $\omega_0(V_K)$ of the semigroup $V_K=(V_K(t))_{t\ge0}$
satisfies to
\begin{equation}\label{SPECTRAL:E2}
\omega_0(V_K)>-\underline{\sigma},
\end{equation}
where $\underline{\sigma}$ and $\overline{\sigma}$ are given 
by \eqref{SUPINFSIGMA:DEF}.
\end{theorem}

\begin{proof}
We  divide this proof into several steps.
\smallskip

\noindent\textbf{Step I}.
Let $\lambda\ge 0$. Due to Lemma~\ref{KLAMBDABAR:LEM},
we infer the positivity and
the irreducibility of the operator $\overline{K}_\lambda$.
Furthermore, its compactness follows from that of the operator $K$.
So, thanks to \cite{Pagter} we infer that $r(\overline{K}_\lambda)>0$
and there exists
a quasi-interior vector $\psi_\lambda$ of $(Y_1)_+$
and a strictly positive functional $\psi_\lambda^*\in (Y_1^*)_+$
such that
\begin{equation}\label{SPECTRAL:E3}
\overline{K}_\lambda\psi_\lambda=r(\overline{K}_\lambda)\psi_\lambda
\quad\text{and}\quad
\overline{K}_\lambda^*\psi_\lambda^*=r(\overline{K}_\lambda)\psi_\lambda^*
\end{equation}
with $\norm{\psi_\lambda}_{Y_1}=\norm{\psi_\lambda^*}_{Y_1^*}=1$,
where $\overline{K}_\lambda^*$ is the adjoint operator of $\overline{K}_\lambda$.
Now, we claim that
\begin{equation}\label{SPECTRAL:E5}
\lambda\ge0\to r(\overline{K}_\lambda)
\end{equation}
is a continuous and strictly decreasing mapping.
So, let $\lambda>\eta\ge0$.
First, for all $\psi\in(Y_1)_+$ we have
\begin{equation*}
\overline{K}_\lambda\psi=\theta_\lambda K\psi=
\theta_{\lambda-\eta}\theta_\eta K\psi
<\theta_\eta K\psi
=\overline{K}_\eta\psi
\end{equation*}
and therefore
\begin{equation}\label{SPECTRAL:E6}
\overline{K}_\eta\psi-\overline{K}_\lambda\psi>0.
\end{equation}
Using \eqref{SPECTRAL:E3} for $\lambda$ and for $\eta$ we obtain that
\begin{equation*}
\begin{aligned}
r(\overline{K}_\eta)-r(\overline{K}_\lambda)&=\frac{\langle \overline{K}_\eta^*\psi_\eta^*,\psi_\lambda\rangle}{\langle\psi_\eta^*,
\psi_\lambda\rangle}-r(\overline{K}_\lambda)\\
&=\frac{\langle\psi_\eta^*, \overline{K}_\eta\psi_\lambda\rangle}{\langle\psi_\eta^*,
\psi_\lambda\rangle}-r(\overline{K}_\lambda)\\
&=\frac{\langle\psi_\eta^*,\overline{K}_\lambda\psi_\lambda\rangle}{\langle\psi_\eta^*,\psi_\lambda\rangle}
+\frac{\langle\psi_\eta^*,(\overline{K}_\eta-\overline{K}_\lambda)\psi_\lambda\rangle}{\langle\psi_\eta^*,\psi_\lambda\rangle}-r(\overline{K}_\lambda)\\
&=r(\overline{K}_\lambda)+\frac{\langle\psi_\eta^*,(\overline{K}_\eta-\overline{K}_\lambda)\psi_\lambda\rangle}{\langle\psi_\eta^*,\psi_\lambda\rangle}-r(\overline{K}_\lambda)
\end{aligned}
\end{equation*}
which leads, by  \eqref{SPECTRAL:E6}, to
\begin{equation}\label{SPECTRAL:E8}
r(\overline{K}_\eta)-r(\overline{K}_\lambda)
=\frac{\langle\psi_\eta^*,(\overline{K}_\eta-\overline{K}_\lambda)\psi_\lambda\rangle}
{\langle\psi_\eta^*,\psi_\lambda\rangle}>0
\end{equation}
because $\psi_\eta^*$ is a strictly positive functional on $(Y_1)_+$ 
and therefore \eqref{SPECTRAL:E5} is a strictly decreasing mapping.
In particular, we infer that 
$r(\overline{K}_0)\ge r(\overline{K}_{\overline{\sigma}-\underline{\sigma}})$
which leads to
\begin{equation}\label{SPECTRAL:E9}
r(\overline{K}_0)>1
\end{equation}
because \eqref{VK-TYPE:R1}.
On the other hand, \eqref{SPECTRAL:E8} implies that
\begin{align*}
\abs{r(\overline{K}_\eta)-r(\overline{K}_\lambda)}&\le\frac{\norm{\psi_\eta^*}_{Y_1^*}}
{\langle\psi_\eta^*,\psi_\lambda\rangle}
\norm{(\overline{K}_\eta- \overline{K}_\lambda)\psi_\lambda}_{Y_1}\\
&\le\frac{1}{\langle\psi_\eta^*,\psi_\lambda\rangle}
\sup_{\psi\in B}\norm{(\overline{K}_\eta- \overline{K}_\lambda)\psi}_{Y_1}\\
&=\frac{1}{\langle\psi_\eta^*,\psi_\lambda\rangle}
\sup_{\psi\in B}\norm{(\theta_\eta-\theta_\lambda)K\psi}_{Y_1}\\
&\le\frac{1}{\langle\psi_\eta^*,\psi_\lambda\rangle}
\sup_{\varphi\in K\parent{B}}\norm{(\theta_\eta-\theta_\lambda)\varphi}_{Y_1}\\
&\le\frac{1}{\langle\psi_\eta^*,\psi_\lambda\rangle}
\sup_{\varphi\in \overline{K\parent{B}}}
\norm{(\theta_\eta-\theta_\lambda)\varphi}_{Y_1},
\end{align*}
where $B$ is the unit ball into $Y_1$.
Since $\overline{K(B)}$ is a compact set, then
there exists $\varphi_0\in\overline{K(B)}$ such that
\begin{equation*}
\left|r(\overline{K}_\eta)-r(\overline{K}_\lambda)\right|\le
\frac{1}{\langle\psi_\eta^*,\psi_\lambda\rangle}
\norm{(\theta_\eta-\theta_\lambda)\varphi_0}_{Y_1}
\end{equation*}
which leads to
\begin{equation*}
\lim_{\mu\to\lambda}\left|r(\overline{K}_\mu)-r(\overline{K}_\lambda)\right|
\le
\lim_{\mu\to\lambda}
\frac{1}{\langle\psi_\eta^*,\psi_\lambda\rangle}
\lim_{\mu\to\lambda}\norm{(\theta_\eta-\theta_\lambda)\varphi_0}_{Y_1}
=0
\end{equation*}
and therefore \eqref{SPECTRAL:E5} is a continuous mapping.
Note that a similar computation leads to
\begin{equation*}
r(\overline{K}_\lambda)\le\norm{\overline{K}_\lambda}_{\mathcal{L}(Y_1)}
\le\sup_{\varphi\in \overline{K\parent{B}}}
\norm{\theta_\lambda\varphi}_{Y_1}=
\norm{\theta_\lambda\varphi_0}_{Y_1}
\end{equation*}
and therefore
\begin{equation}\label{SPECTRAL:E10}
\lim_{\lambda\to\infty}r(\overline{K}_\lambda)=0.
\end{equation}
Finally, as \eqref{SPECTRAL:E5} is a continuous and strictly decreasing mapping, 
then by \eqref{SPECTRAL:E9} and \eqref{SPECTRAL:E10},
there exists a unique $\lambda_0$ such that
\begin{equation}\label{SPECTRAL:E12}
\lambda_0>0\quad\text{and}\quad r(\overline{K}_{\lambda_0})=1.
\end{equation}
\smallskip

\noindent \textbf{Step II}.
In this step we prove that
$\lambda_0=\omega_0(U_K)$, where $\omega_0(U_K)$ is the type of the semigroup
$U_K=(U_K(t))_{t\ge0}$.
So, let $\lambda\in \sigma(T_K)$ such that $\operatorname{Re}(\lambda)\ge0$.
By virtue of Lemma~\ref{SIGMATK}, there exists $\psi\not=0$ such
 that $\overline{K}_\lambda\psi=\psi$.
This clearly leads to
\begin{equation*}
\abs{\psi}=\abs{\overline{K}_\lambda\psi}
\le \abs{\theta_\lambda}K\abs{\psi}\\
=\theta_{_{\operatorname{Re}{\lambda}}}K\abs{\psi}=\overline{K}_{\operatorname{Re}{\lambda}}\abs{\psi}
\end{equation*}
which implies that
$\parent{\overline{K}_{\operatorname{Re}{\lambda}}}^n\abs{\psi}\ge\abs{\psi} $
for all integers $n$ and therefore
$r\parent{\overline{K}_{\operatorname{Re}{\lambda}}}\ge 1$.
This together with \eqref{SPECTRAL:E12} lead to
$\operatorname{Re}{\lambda}\le\lambda_0$ because \eqref{SPECTRAL:E5}
is a strictly decreasing mapping and therefore \eqref{BOUND:DEF} leads to
\begin{equation}\label{SPECTRAL:E13}
s(T_K)\le\lambda_0.
\end{equation}

Conversely. Applying \eqref{SPECTRAL:E3} to $\lambda_0$,
it follows that $\overline{K}_{\lambda_0}\psi_{\lambda_0}=\psi_{\lambda_0}$
with $\psi_{\lambda_0}\not=0$.
Following the proof of Lemma~\ref{SIGMATK} (put $g=0$ in \eqref{LEMMA7L2:E1} 
and \eqref{LEMMA7L2:E2})
we easily infer that $\varphi:=\varepsilon_{\lambda_0}K\psi_{\lambda_0}$ satisfies to
\begin{equation*}
\varphi\in W_1\quad\text{and}\quad
-v\frac{\partial\varphi}{\partial \mu}=\lambda_0\varphi\quad\text{and}\quad
\gamma_0\varphi=K\gamma_1\varphi
\end{equation*}
which implies that
$T_K\varphi=\lambda_0\varphi$
and therefore $\lambda_0\in \sigma_p(T_K)\subset\sigma(T_K)$.
Now, \eqref{BOUND:DEF} leads to
\begin{equation}\label{SPECTRAL:E14}
\lambda_0\le s(T_K).
\end{equation}
Finally, \eqref{SPECTRAL:E12} together with \eqref{SPECTRAL:E13} 
and \eqref{SPECTRAL:E14} and \eqref{BOUND:EQUALITY} imply that
\begin{equation}\label{SPECTRAL:E15}
\omega_0(U_K)=\lambda_0>0\quad\text{and}\quad
r(\overline{K}_{\lambda_0})=1.
\end{equation}
\smallskip

\noindent\textbf{Step III}.
On one hand, \eqref{VK-TYPE:R1} and \eqref{SPECTRAL:E15} lead to
\begin{equation}\label{SPECTRAL:E17}
\overline{\sigma}-\underline{\sigma}<\omega_0(U_K)
\end{equation}
because \eqref{SPECTRAL:E5} is a strictly decreasing mapping.
On the other hand, Proposition~\ref{UK-POS-IRR:THE} and Theorem~\ref{VK-POS-IRR:THE} imply the positivity of the semigroups
$U_K=(U_K(t))_{t\ge0}$ and $V_K=(V_K(t))_{t\ge0}$
which leads, by virtue of \eqref{VK-POS-IRR:R1}, to
\begin{equation*}
e^{-t\overline{\sigma}}\norm{U_K(t)}_{\mathcal{L}(L^1(\Omega))}
\le\norm{V_K(t)}_{\mathcal{L}(L^1(\Omega))}
\end{equation*}
for all $t\ge0$ and therefore
\begin{equation*}
-\overline{\sigma}+\lim_{t\to\infty}
\frac{\ln \norm{U_K(t)}_{\mathcal{L}(L^1(\Omega))}}{t}
\le
\lim_{t\to\infty}\frac{\ln \norm{V_K(t)}_{\mathcal{L}(L^1(\Omega))}}{t}.
\end{equation*}
Hence, we have
\begin{equation}\label{SPECTRAL:E18}
-\overline{\sigma}+\omega_0(U_K)\le\omega_0(V_K)
\end{equation}
because of \eqref{TYPE:DEF}.
Now, \eqref{SPECTRAL:E17} and \eqref{SPECTRAL:E18} achieve the proof.
\end{proof}

\begin{remark} \rm
Note that the choice of the functional framework $L^1(\Omega)$
was natural because $\norm{V_K(t)\varphi}_1$ denotes the bacteria number at time $t$;
nevertheless, according to a lot of modification,
all the results of this work still hold into $L^p(\Omega)$ $(p>1)$.
\end{remark}

\section{Application and comments}
\label{APPLICATION}

Taking now the particular model \eqref{MODEL:EQUATION}-\eqref{MODEL:TRANSITION} 
that is
\begin{equation} \label{eR}
\begin{gathered}
\frac{\partial f}{\partial t}=-v\frac{\partial f}{\partial \mu}
-\sigma f \quad t\ge0\\
vf(t,0,v)=p\int_0^\infty k(v,v')f(t,1,v')v'dv' \quad t\ge0\\
f(0,\cdot,\cdot)=\varphi\in L^1(\Omega),
\end{gathered}
\end{equation}
where $p\ge1$ denotes the average number of daughter bacteria viable 
per mitotic.
T ensure the continuity of the bacterial flux for $p=1$,
the kernel of correlation $k$ must be positive and
fulfils the  normalization condition
\begin{equation*}
\int_0^\infty k(v,v')dv=1\quad\text{for all } v\in(0,\infty).
\end{equation*}
If $K$ denotes  the  transition operator
\begin{equation*}
K\psi(v)=\frac{p}{v}\int_0^\infty k(v,v')\psi(v')v'dv'
\end{equation*}
then for all $\psi\in(Y_1)_+$ we have
\begin{equation*}
\begin{aligned}
\norm{K\psi}_{Y_1}&=\int_0^\infty vK\psi(v)dv=p\int_0^\infty\parent{\int_0^\infty k(v,v')dv}\psi(v')v'dv'=p\norm{\psi}_{Y_1}
\end{aligned}
\end{equation*}
which leads to $\norm{K}_{\mathcal{L}(Y_1)}=p$
and therefore $K$ is a bounded linear operator from $Y_1$
into itself. Furthermore, if $k$ is a continuous kernel, then
$K$ becomes compact which leads to its admissibility because of
Remark~\ref{ADM:REM}.
Now, thanks to Theorems~\ref{VK:THE} and \ref{VK-POS-IRR:THE},
we can say that the model {\rm (R)}
is well posed and admits, for all initial data $\varphi\in (L^1(\Omega))_+$,
the following positive solution
\begin{equation*}
f(t,\cdot,\cdot)=V_K(t)\varphi\quad t\ge0.
\end{equation*}

\begin{remark}\rm
In \cite[p. 475]{Proto}, there is an incorrect study of the model \eqref{eR}.
Indeed, the authors claim that there exists a unique solution $f$
belonging to $L^1((0,1)\times(0,\infty))$
when $1<p<2$ and they have proceeded as follows:

In order to use the easy case $0<p<1$,
the authors consider the change
\begin{equation*}
\widetilde{f}=q^\mu f
\end{equation*}
and say that the model \eqref{eR} becomes
\begin{equation} \label{eR'}
\begin{gathered}
\frac{\partial\widetilde{f}}{\partial t}
=-v\frac{\partial \widetilde{f}}{\partial \mu}
-\sigma f+(\ln q)\widetilde{f}\\
\widetilde{f}(t,0,v)=\frac{p}{qv}\int_0^\infty k(v,v')\widetilde{f}(t,1,v')v'dv'\\
\widetilde{f}(0,\cdot,\cdot)=q^\mu\varphi\in L^1(\Omega),
\end{gathered}
\end{equation}
where $q>p$ is fixed.  So, after some computations, the authors infer that
the well posedness of the model \eqref{eR'} follows from
the boundedness of the multiplicative operator
\begin{equation*}
\widetilde{f}\to(\ln q)\widetilde{f}
\end{equation*}
from $L^1(\Omega)$ into itself.
Actually, the previous model \eqref{eR'} is incorrectly computed
and the correct model is
\begin{equation} \label{eR''}
\begin{gathered}
\frac{\partial \widetilde{f}}{\partial t}=-v\frac{\partial \widetilde{f}}{\partial \mu}
-\sigma f+(v\ln q)\widetilde{f}\\
\widetilde{f}(t,0,v)=\frac{p}{qv}\int_0^\infty k(v,v')\widetilde{f}(t,1,v')v'dv'\\
\widetilde{f}(0,\cdot,\cdot)=q^\mu\varphi\in L^1((0,1)\times(0,\infty)).
\end{gathered}
\end{equation}
Unfortunately, the operator
$$ 
\widetilde{f}\to(v\ln q)\widetilde{f}
$$
is obviously not bounded or dissipative into
$L^1((0,1)\times(0,\infty))$ and therefore we cannot, by this way, infer
any well posedness of the model \eqref{eR}.
\end{remark}

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\section*{Addendum posted on June 24, 2013.} % Vol.  2012, No. 221}

The author would like to make the following changes:

\noindent (1) Lemma~3.3 and Remark~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) Relation~(3.13) must be replaced by
\begin{equation}
\omega_K=
\begin{cases}
0 &\text{if {\rm (Kb)} holds};\\
\delta_K:=\inf\{\omega>0: \|K\mathbb{I}_{\omega}\|<1\}&
\text{if {\rm (Kc)} holds}.
\end{cases}
\tag{3.13}
\end{equation}

\noindent (3) Proof of Lemma~3.3: from 
``{\it So, the compactness...}''(Page 6 line 5) to 
``{\it because of (3.20)}'' (Page 7 line 1)
and from ``{\it Finally, by (3.19)...}'' (Page 7 line 24) to the end the proof, 
must be deleted.

\noindent(4) Corollaries~3.10~and~5.4: 
``{\it Let $K$ be a linear compact}'' 
must be replaced by \\
``{\it Let $K$ be a linear admissible}''.
The proof must be replaced by : ``{\it Proof: Obvious.}''

\noindent(5) Page 20: from ``{\it In the sequel...to...result }''
(Lines -7 to -4) must be deleted.

\noindent (6) Theorem 7.2: 
``{\it $K$ admissible}'' must be inserted  in the preamble.

\noindent(7) Page 24: 
``{\it T ensure...$v\in(0,\infty)$}'' (Lines 10 to 12) 
must be deleted. 
``{\it Furthermore...Ramark~3.4}'' (Lines 18 and 19) must be replaced by\\
``{\it If ${p \sup_{v'\geq \omega}\int_0^\infty| k(v,v')| dv<1}$
for some $\omega>0$, then $K$ is an admissible operator.}''

End of addendum.


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