\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2003(2003), No. 107, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2003 Texas State University-San Marcos.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE--2003/107\hfil
A mathematical model describing cellular division]
{A mathematical model describing cellular division with a
proliferating phase duration depending on the maturity of cells}

\author[Mostafa Adimy \& Laurent Pujo-Menjouet\hfil EJDE--2003/107\hfilneg]
{Mostafa Adimy \& Laurent Pujo-Menjouet} % in alphabetical order

\address{Mostafa Adimy \hfill\break
D\'{e}partement de Math\'{e}matiques Appliqu\'{e}es,
I.P.R.A., Universit\'{e} de Pau, Avenue de l'universit\'{e}, 64000
Pau, France}
\email{mostafa.adimy@univ-pau.fr}

\address{Laurent Pujo-Menjouet \hfill\break
Department of Physiology, Physics and
Mathematics, Centre for Nonlinear dynamics, McGill University,
3655 Drummond Street, Montreal, Quebec, Canada H3G 1Y6}
\email{pujo@cnd.mcgill.ca}

\date{}
\thanks{Submitted February 14, 2003. Published October 23, 2003.}
\thanks{L. P.-M. was partially  supported by MITACS, Canada.}
\subjclass[2000]{35F15, 35L60, 92C37, 92D25}
\keywords{Structured population, cell cycle, stem cells,
\hfill\break\indent
first order partial differential equation with delays,  non constant delay}

\begin{abstract}
 In this paper, we investigate a linear population model of cells
 that are capable of simultaneous proliferation and maturation.
 We consider the case when the time required for a cell to divide
 depends on its maturity. This model is described by first order
 partial differential system with a retardation of the maturation
 variable and a time delay depending on this maturity. Both delays
 are due to cell replication.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{thm}{Theorem}[section]
\newtheorem{lem}[thm]{Lemma}
\newtheorem{prop}[thm]{Proposition}

\section{Introduction}

Time-age and time-maturity structured models have been used, for
more than 40 years, to study cell replication (see for example Von
Foerster in 1959 \cite{vonfoerster1959}, Trucco in 1965
\cite{Trucco65}, \cite{Trucco65b}, in 1966 \cite{Trucco66},
Oldfield in 1966 \cite{oldfield66}, Nooney in 1967
\cite{nooney67}, Rubinow in 1968 \cite{rubinow68}, and Rubinow and
Lebowitz in 1975 \cite{rubinow75}).


More recently, in 1994 \cite{MR1} and in 1999 \cite{MR2}, Mackey
and Rudnicki considered a particular time-age-maturity structured
model to study the biological process of hematological cell
development in the bone marrow. This model is an extension of
models that have been considered previously both in the absence of
maturation (Mackey in 1978 \cite{Mackey78} and in 1979
\cite{Mackey79}) or with maturation and only one phase (Rey and
Mackey in 1992 \cite{rey92} and in 1993 \cite{rey93}). It consists
of a population of cells capable of both proliferating and
maturating. In this model, the period of life of each cell is
divided into a proliferating phase and a resting phase. In the
proliferating phase the cells are committed to undergo cell
division a time $\tau $ later. The position of a cell in the
proliferating phase is denoted by $a$ (cell age) which is assumed
to range from $a=0$ (the point of commitment) to $a=\tau $ (the
point of cytokinesis). In addition, each cell is characterized by
a maturation variable $m$, that is the concentration of what
composes a cell such as proteins, or other elements one can
measure experimentally. This maturity can be taken, without loss
of generality, from $m=0$ to $m=1$. Cells in the proliferating
phase can be lost at a rate $\gamma $. At age $a=\tau $ a cell
divides and gives two daughter cells, which enter directly the
resting phase. If the maturation of the mother cell at age $a=\tau
$ is $m$, the maturation of a daughter cell at birth is assumed to
be $g(m)$, with $g(m)\leq m$. In the resting phase, cells can
either return to the proliferating phase at a rate $\beta $ and
complete the cycle or die at a rate $\delta $ before ending the
cycle. A cell can remain in the resting phase indefinitely, and
then the cell age ranges from $a=0$, when the cell enters, to
$a=+\infty $. The maturation of a cell and the total number of
resting cells determine the capacity of this cell for entering the
next proliferating phase. We assume that cells of both types age
with unitary velocity, $da / dt=1$, and mature with a velocity
$V(m)$.

In their models, Mackey and Rudnicki \cite{MR1} and \cite{MR2},
Dyson, Villella-Bressan and Webb \cite{dyson96a}, \cite{dyson97},
\cite{dyson98}, \cite{dyson2000}, \cite{dyson2000a},
\cite{dyson99} and \cite{dyson96b}, Adimy and Pujo-Menjouet
\cite{adimypujo} and
 \cite{adimypujo2003}, and Pujo-Menjouet and
Rudnicki \cite{rudnickipujo}, assumed that the point of
cytokinesis $\tau $ is the same for all cells. This means that the
time required for a cell to divide does not depend on its
maturity, and in particular, the division duration of a cell with
small maturity (also called stem cell) is the same as one with a
higher maturity level. This assumption is not compatible with the
biological reality. It is commonly believed that a stem cell
proliferates more rapidly than a more mature cell. To our
knowledge, the hypothesis that $\tau $ depends on the maturity
variable, has been given for the first time by Mitchison in 1971
\cite{Mitchison1971}, and by John in 1981 \cite{john}, but never
been used in recent models.

We will assume in this paper that each cell entering the
proliferating phase with a maturity $m$ divides at age $\tau (m)$,
depending on this maturity, and we require that the mapping
\begin{equation}\label{For1}
m\to \int_{m}^{1}\frac{ds}{V(s)}-\tau(m)
\end{equation}
 is strictly decreasing on $(0,1] $. Note
that $ \int_{m}^{1}{\frac{ds}{V(s)}}$\ represents the time
required for a cell with maturity $m$ to reach the maximal
maturity $1$. The assumption (\ref{For1}) is satisfied in the
particular case when the mapping $m\mapsto \tau (m)$ is
increasing.

Our aim in this paper is to extend and to analyze the model of
Mackey and Rudnicki (\cite{MR1} and \cite{MR2}) taking into
account the new condition. We obtain two time-age-maturity
structured partial differential equations with two boundary
conditions. We integrate these equations with respect to age, and
we obtain two time-maturity structured partial differential
equations, in which there is a delay in the time variable as well
as in the maturation variable. The time delay depends on the
maturity. The model takes the form of a delay-differential
equation in a Banach space.

We prove in this model that if the cells have enough time in the
proliferating phase,
\begin{equation*}
\tau(m) >  \int_{m}^{g^{-1}(m)}\frac{ds}{V(s)},\quad
\mbox{for all }  m \in  \left(0,g(1)\right],
\end{equation*}
to increase sufficiently their maturity, $m>\Delta (m)$,
then the uniqueness of solutions depends, for a finite time, only
on cells with small maturity. In particular, if the initial
population of cells is $0$ for small maturity then the population
becomes extinct for a finite time. This result is a first step to
study the asymptotic behavior of solutions. We consider here only
the linear case. That is a simplification of a more general
nonlinear model. We defer to a further publication the use of our
approach in the nonlinear case $(\beta =\beta (N))$, to prove some
results on stability of solutions.


\section{Equations of the model}

Denote the density of resting cells by $n(t,m,a)$ and the
density of proliferating cells by $p(t,m,a)$,  the governing
equations of this model are
\begin{gather}\label{resting}
   \frac{\partial n}{\partial t}+\frac{\partial n}{\partial
a}+\frac{\partial \left( V(m)n\right)}{\partial m}=-\left( \delta
(m)+\beta (m)\right) n, \\
\label{prolif}
\frac{\partial p}{\partial t}+\frac{\partial p}{\partial
a}+\frac{\partial \left( V(m)p\right)}{\partial m} =-\gamma (m)p.
\end{gather}
 Throughout this paper, we shall require the following
hypothesis concerning the velocity of maturation $V$ in the two
phases.

\noindent  \textbf{(H1)} $V$ is continuously
differentiable on $[0,1] $, positive on $(0,1]$ and satisfies
$V(0)=0$, and
\begin{equation}\label{(H1).1}
\int_{0}^{m}\frac{ds}{V(s)}=+\infty, \;\quad\mbox{for } m\in (0,1] .
\end{equation}
Note that the integral $\int_{m_{1}}^{m_{2}}\frac{ds}{V(s)}$, $m_{1}<m_{2}$,
represents the time required for a cell to go from maturity
$m_{1}$ to maturity $m_{2}$. The condition (\ref{(H1).1})
describes the fact that the velocity of cells increases slowly for
a small maturity.
 As example, we have
\begin{equation*}
V(m)\underset{m\to 0}{\sim }\alpha_{1}m^{p},\quad \alpha _{1}>0,\;p\geq 1.
\end{equation*}

\noindent \textbf{(H2)}   The functions $\delta, \beta$ and $\gamma $ are
continuous and nonnegative on $[0,1]$.

 The total number of cells in the resting stage is given by
\begin{equation*}
N(t,m)=\int_{0}^{+\infty }n(t,m,a)da.
\end{equation*}
We assume that the quantity $N(t,m)$ determines the
reintroduction at a rate $\beta (m)$ of cells with maturity $m$
from the resting into the proliferating phase. This hypothesis is
given by the following boundary condition
\begin{equation}\label{bctauvariant1}
p(t,m,0)=\int_{0}^{+\infty }\beta (m)n(t,m,a)da=\beta
(m)N(t,m).
\end{equation}

In completing the formulation of this problem, we need a second
boundary condition which determines the transfer of cells from the
point of cytokinesis into the resting compartment.

We assume that a cell entering the proliferating phase
with a maturity $x_{0}\in [0,1] $ divides at age $\tau
(x_{0})>0$, and we require that

\noindent \textbf{(H3)} $\tau$ is a continuously
differentiable and positive function on $[0,1] $ such
that $\tau'(m)+\frac{1}{V(m)}>0$, for $m\in (0,1]$.

 Consider a cell in the proliferating phase at time $t$,
with maturity $x\in (0,1] $, age $a$ and initial
maturity $x_{0}$, i.e. at age $0$. Then, naturally we have
\begin{equation*}
x_{0}\leq x \quad\mbox{and}\quad
a=\int_{x_{0}}^{x}\frac{ds}{V(s)}\leq \tau (x_{0}).
\end{equation*}
 If $x$ is the maturity of the cell at the point of
cytokinesis, then there exists  $\Theta(x)\in(0,x) $ (the maturity
at the point of commitment) such that
\begin{equation}\label{intersection}
\int_{\Theta(x)}^{x}\frac{ds}{V(s)}=\tau (\Theta(x)).
\end{equation}
 This value $\Theta(x)$ is unique because the condition
$\rm\textbf{(H3)}$ implies that the function
\begin{equation*}
x_{0}\to \int_{x_{0}}^{x}\frac{ds}{V(s)}-\tau
(x_{0})
\end{equation*}
is continuous and strictly decreasing from $(0,x]$ into
$[-\tau (x),+\infty)$. Then we can define a function
$\Theta:(0,1]\to(0,1]$ such that, for each $x\in(0,1]$, $\Theta(x)$ is
the solution of Equation (\ref{intersection}).

 Remark that
$0< \Theta (x)<x$ for $x\in (0,1]$. This implies, in particular, that
\begin{equation*}
\lim_{x\to 0} \Theta (x)=0\quad\mbox{and}\quad
\lim_{x\to 0}\Big(\int_{\Theta (x)}^{x}\frac{ds}{V(s)}\Big) =\tau(0)<+\infty .
\end{equation*}
One can prove that $\Theta $ is continuously
differentiable on $(0,1] $ and satisfies
\begin{equation}\label{Form1}
 \Theta'(x)V(x)=\frac{V(\Theta (x))}{1+\tau'(\Theta
(x))V(\Theta (x))},\quad \mbox{for }  x\in (0,1].
\end{equation}
This implies, in particular, that
\begin{gather*}
  \Theta'(x)>0, \quad  \mbox{for }  x\in (0,1], \\
   \lim_{x\to 0}\big( \Theta '(x)V(x)\big) =0.
\end{gather*}
From a biological point of view, $\Theta (x)$ represents
the initial maturity of cells in the proliferating phase (the
point of commitment) that divide at maturity $x$ (the point of
cytokinesis). We deduce that the age of a cell with maturity $x$
at the point of cytokinesis is $\tau (\Theta (x))$. So, the total
number $P$ of proliferating cells of a given maturation level
$x\in [0,1] $ is given by
\begin{equation}\label{populationtotaleprolif}
P(t,x)=\int_{0}^{\tau (\Theta (x))}p(t,x,a)da.
\end{equation}
At the end of the proliferating phase, a cell with a
maturity $x$ divides into two daughter cells, with maturity
$g(x)$. These cells enter directly the resting phase with age
$a=0$. We assume that


\noindent \textbf{(H4)} $g:[0,1] \to[0,1] $ is a continuous function,
continuously differentiable on $[ 0,1)$ and such that $g(x)\leq x$,
for $x\in [0,1] $\ and$\;g'(x)>0$,\ for $x\in [0,1)$.


 We also assume, for technical reasons and without loss
of generality, that
\begin{equation*}
\lim_{x\to 1}g'(x)=+\infty .
\end{equation*}
We put $g^{-1}(m)=1$, for $m>g(1)$.
This implies that the function $g^{-1}:[0,1]
\to [0,1] $ is continuously differentiable and
satisfies
\begin{equation*}
(g^{-1})'(m)=0,\quad \mbox{ for } m>g(1).
\end{equation*}
Note that the maturity $m$ of the daughter cells just
after division is smaller than $g(1)$. Then, we can assume that
\begin{equation}\label{For2}
n(t,m,0)=0,\quad \mbox{for } m>g(1).
\end{equation}
The maturity of the mother cell at the point of
cytokinesis was $x=g^{-1}(m)$, its age was
$\tau (\Theta (g^{-1}(m)))$ and its maturity at the point of
commitment was $\Theta (g^{-1}(m))$. We set,
\begin{equation}\label{equationdelta}
\Delta (m)=\Theta \left( g^{-1}(m)\right) \quad \mbox{for } m\in [0,1] .
\end{equation}
 Then, we can give the second boundary condition
\begin{equation}\label{bctauvariant2}
n(t,m,0)=2(g^{-1})'(m)p\left( t,g^{-1}(m),\tau (\Delta
(m))\right) \quad \mbox{for } m\in [0,1],
\end{equation}
which includes also the condition (\ref{For2}). The
explanation of the multiplicative term $2(g^{-1})'(m)$ in
the condition (\ref{bctauvariant2}) is the following. The factor
$2$ accounts for the division of each cell at mitosis (doubling of
the cell). The term $(g^{-1})'(m)$ describes the fact
that the two new born cells with maturity on an interval
$(m,m+dm)$ come from a mother cell  with maturity on the interval
$\left(g^{-1}(m), g^{-1}(m)+(g^{-1})'(m)dm\right) $. In
fact, the boundary condition (\ref{bctauvariant2}) can be
reformulated, by considering the total population of new born
cells, as follows
\begin{equation*}
\int_{0}^{1} n(t,m,0)dm=  2 \int_{0}^{1} p\left( t,m,\tau (\Theta
(m))\right)dm.
\end{equation*}
From a biological point
of view, $\Delta $ gives the link between the maturity of a new
born cell and the maturity of its mother at the point of
commitment. This function $\Delta :[0,1] \to
[0,1] $ is continuous, continuously differentiable on
$(0,1] $ and satisfies the following properties:
\begin{gather*}
  \Delta (0)=0,    \\
  \Theta (m)\leq \Delta (m)\quad \mbox{and}\quad \Delta' (m)>0, \quad
  \mbox{for }  m\in ( 0,g(1)), \\
  \Delta (m)=\Theta (1), \quad  \mbox{for } m\in [ g(1),1].
\end{gather*}
Now, we specify initial conditions:
\begin{equation}\label{CI1}
\begin{gathered}
   p(0,m,a)=\Gamma (m,a),\quad \mbox{for }
 (m,a)\in [0,1] \times [0,\overline{\tau}] ,  \\
  n(0,m,a)=\mu (m,a), \quad\mbox{for }
(m,a)\in [0,1] \times [0,+\infty) ,
\end{gathered}
\end{equation}
where $\overline{\tau }:=\max_{x\in [0,1]}\tau (x)$, $\Gamma $ and $\mu $
are assumed to be continuous and the function
\begin{equation*}
 m\mapsto \int_{0}^{+\infty }\mu (m,a)da
 \end{equation*}
is continuous on $[0,1] $, (in
particular, $\lim_{a\to +\infty }\mu (m,a)=0$, for $m\in [0,1]$).



\section{Equations for the total population in the resting phase}

Before giving the equations for the total population in the
resting phase $N(t,m)$, we define the characteristic curves
$s\to \pi _{s}(m)$ through $(0,m)$, $m\in [0,1] $, given as follows:
$s\to \pi _{s}(m)$ is the solution of the
differential equation
\begin{equation}\label{Vitesse1}
\begin{gathered}
   \frac{d}{ds}u(s)=V(u(s)),  \quad s\in \mathbb{R},\\
  u(0)=m.
\end{gathered}
\end{equation}
Note that $\pi _{0}(m) =m$, for $m\in[0,1] $, and
$\pi _{s}\left( 0\right) =0$, for $s\in \mathbb{R}$.
 The expression $\pi _{s}(m) $ appears in our
model, only for $s\leq 0$. Furthermore, $\pi _{s}(m)$ is given
explicitly, for $s\leq 0$ and $m\in [0,1] $, by
\begin{equation}\label{explicitformflowlin}
 \pi _{s}(m) =h^{-1}\left(h(m)e^{s}\right),
\end{equation}
where $h:[0,1] \to [0,1] $ is defined as
\begin{equation}\label{h1}
h(m)=\begin{cases}
  \exp \big( - \int_{m}^{1}\frac{ds}{V(s)}\big),
 &  \mbox{if }  m\in (0,1] , \\
 0, & \mbox{if }  m=0.
\end{cases}
\end{equation}
 Let $m\in [0,1] $ and $t\geq 0$. We define
\begin{equation}\label{defksi}
\xi (m,t):=\exp \Big\{-\int_{0}^{t}\left( \gamma (\pi
_{-s}(m))+V'(\pi _{-s}(m))\right) ds\Big\}.
\end{equation}
 It is clear that
\begin{equation*}
\xi (m,t)=\frac{V(\pi _{-t}(m))}{V(m)}\exp \Big\{
-\int_{\pi _{-t}(m)}^{m}\frac{\gamma (y)}{V(y)}dy\Big\} .
\end{equation*}
Then we obtain the following result.

\begin{prop} \label{prop3.1}
Let $m\in [0,1] $ and $t \geq 0$. The total population $N(t,m)$ of
cells in the resting phase satisfies the following conditions:
\begin{enumerate}
\item If $0\leq t\leq \tau (\Delta (m))$, then
\begin{equation}\label{EqtauNvar1}
\frac{\partial }{\partial t}N(t,m)+\frac{\partial }{\partial
m}\left( V(m)N(t,m)\right) =-\left[ \delta (m)+\beta (m)\right]
N(t,m)+F(t,m),
\end{equation}
 with
\begin{equation}\label{equationf1}
F(t,m)=2(g^{-1})'(m)\xi (g^{-1}(m),t)\Gamma \left( \pi
_{-t}\left( g^{-1}(m)\right) ,\tau (\Delta (m))-t\right).
\end{equation}

\item If $\tau (\Delta (m))\leq t$, then
\begin{equation}\label{EqtauNvar2}
\begin{aligned}
&\frac{\partial }{\partial t}N(t,m)+\frac{\partial }{\partial
m}\left( V(m)N(t,m)\right)\\
&= -\left[ \delta (m)+\beta (m)\right]
N(t,m)+K(m)N\left( t-\tau (\Delta (m)),\Delta (m)\right) ,
\end{aligned}
\end{equation}
 with
\begin{equation}\label{Eq25}
K(m)=2(g^{-1})'(m)\xi \left( g^{-1}(m),\tau (\Delta
(m))\right) \beta \left( \Delta (m)\right) .
\end{equation}
\end{enumerate}
\end{prop}

\begin{proof} Equation (\ref{prolif}) can be solved
using the method of characteristics. First, we obtain the
following representation of solutions of (\ref{prolif}),
\begin{equation}\label{formp}
p(t,m,a)=\begin{cases}
   \xi (m,t)p(0,\pi _{-t}(m) ,a-t),&  \mbox{for } 0\leq t<a,\\
   \xi (m,a)p(t-a,\pi _{-a}(m),0), &  \mbox{for } a\leq t.
\end{cases}
\end{equation}
 The initial condition (\ref{CI1}) and the boundary
condition (\ref{bctauvariant1}) give
\begin{equation}\label{Caract1}
p(t,m,a)=\begin{cases}
   \xi (m,t)\Gamma (\pi _{-t}(m) ,a-t),&\mbox{for } 0\leq t<a, \\
  \xi (m,a)\beta (\pi_{-a}(m) )N(t-a,\pi _{-a}(m) ), &
  \mbox{for } a\leq t\,.
\end{cases}
\end{equation}
 By integrating (\ref{resting}) with respect to
the age between $0$ and $+\infty $, it follows that
\begin{equation*}
\frac{\partial }{\partial t}N(t,m)+\frac{\partial }{\partial
m}\left( V(m)N(t,m)\right) =-\left[ \delta (m)+\beta (m)\right]
N(t,m)+n(t,m,0).
\end{equation*}
Because of the second boundary condition (\ref{bctauvariant2}), we
obtain
\begin{align*}
 \frac{\partial }{\partial t}N(t,m)+\frac{\partial }{\partial
m}\left( V(m)N(t,m)\right) =&  -\left[ \delta (m)+\beta (m)\right]
N(t,m) \\
 &\quad+   2(g^{-1})'(m)p(t,g^{-1}(m),\tau (\Delta (m))).
\end{align*}
Furthermore, (\ref{Caract1}) implies
\begin{equation}\label{23}
\begin{aligned}
&p(t,g^{-1}(m),\tau (\Delta (m)))\\
&= \begin{cases}
  \xi (g^{-1}(m),t)\Gamma \left( \pi _{-t}\left( g^{-1}(m)\right)
,\tau (\Delta (m))-t\right) , & \mbox{for } 0\leq t<\tau (\Delta (m)),\\[3pt]
 \xi (g^{-1}(m),\tau (\Delta (m)))\beta \left( \pi _{-\tau (\Delta
(m))}\left( g^{-1}(m)\right) \right) \\
\times N\left( t-\tau (\Delta (m)),\pi _{-\tau (\Delta (m))}\left(
g^{-1}(m)\right)
\right) ,&  \mbox{for } \tau (\Delta (m))\leq t\,.
\end{cases}
\end{aligned}
\end{equation}
Set $x=\pi _{-\tau (\Delta (m))}\left( g^{-1}(m)\right)$
and remark that
\begin{equation*}
\frac{h(x)}{h(g^{-1}(m))}=e^{-\tau (\Theta(g^{-1}(m)))}.
\end{equation*}
Then
\begin{equation*}
\int_{x}^{g^{-1}(m)}\frac{ds}{V(s)}=\tau (\Theta (g^{-1}(m))).
\end{equation*}
According to the definition of $\Theta $, we deduce that
\begin{equation*}
x=\Theta (g^{-1}(m))=\Delta (m).
\end{equation*}
Then, the equations (\ref{EqtauNvar1}) and (\ref{EqtauNvar2}) follow
immediately from Equation (\ref{23}).
\end{proof}

We remark that the solutions of System
(\ref{EqtauNvar1})-(\ref{EqtauNvar2}) are independent of the
proliferating cells, and it is easy to prove (by steps) the
existence, uniqueness and regularity of these solutions.
We focus our study on Equation (\ref{EqtauNvar2}) for
$t\geq \overline{\tau }$, with an initial condition $\varphi $,
which is a solution of the system
(\ref{EqtauNvar1})-(\ref{EqtauNvar2}) on the set
$[0,\overline{\tau }] \times [0,1] $. We assume
that this function $\varphi $ is continuous ($\varphi \in C(
[ 0,\overline{\tau }] ,X)$, with $X=C[0,1]$).
We shall look for integrated solutions of Problem
(\ref{EqtauNvar2}), for $t\geq \overline{\tau }$, which are
continuous functions on $t$ with values in $X$. Therefore, we
rewrite Problem (\ref{EqtauNvar2}) as the following abstract delay
differential equation
\begin{equation}\label{For3}
\begin{gathered}
  \frac{d}{dt}u(t)=Au(t)+L(u_{t}^{\overline{\tau}}),\quad  t\geq
\overline{\tau }, \\
 u(t)=\varphi (t,.),\quad  t\in [0,\overline{\tau }] ,
\end{gathered}
\end{equation}
 where $u_{t}^{\overline{\tau}}:[ 0,\overline{\tau }] \mapsto X$ is the function defined for $t \geq
\overline{\tau}$  by
$u_{t}^{\overline{\tau}}(s)=u(t+s-\overline{\tau})$,
$A:D(A)\subseteq C[0,1] \to C[0,1] $ is the
unbounded closed linear operator defined by:
\begin{align*}
  D\left(A\right)=& \big\{ u\in C[0,1], u
\mbox{ differentiable on }(0,1], u'\in C(0,1],\\
   & \mbox{ and } \lim_{m\to 0} V(m)u'(m)=0 \big\},
\end{align*}
and
\[
Au(m)=\begin{cases}
-\left(\delta (m)+\beta (m)+V'(m)\right) u(m)-V(m)u'(m),&
\mbox{if }  m\in (0,1],\\
  -\left( \delta (0)+\beta (0)+V'(0)\right)u(0), &
 \mbox{if } m=0\,,
\end{cases}
\end{equation*}
and $L:C([ 0,\overline{\tau }],X) \to X$ is the linear operator defined for all
$\psi $ in $C([ 0,\overline{\tau }] ,X)$
by
\begin{equation*}
L(\psi )(m)=K(m)\psi \left( \overline{\tau}-\tau (\Delta
(m)),\Delta (m)\right) ,\quad m\in [0,1] .
\end{equation*}
We establish now the $C_{0}$-semigroup generated by the
operator $A$.

\begin{prop} \label{prop3.2}
Operator $A$ is the infinitesimal generator of
the $C_{0}$-semigroup $\left( S(t)\right) _{t\geq 0}$ defined on
$C[0,1] $ by
\begin{equation*}
\left( S(t)y\right) (m)=\overline{\xi }(m,t)y\left( \pi
_{-t}(m) \right) ,\quad \mbox{for } y\in C[0,1] \quad\mbox{and} \quad m\in [0,1],
\end{equation*}
where
\begin{equation*}
\overline{\xi }(m,t)=\frac{V\left( \pi _{-t}(m)
\right) }{V(m)}\exp \Big\{ -\int_{\pi _{-t}\left(
m\right) }^{m}\frac{\delta (s)+\beta (s)}{V(s)}ds\Big\}.
\end{equation*}
\end{prop}

The proof of this proposition is similar to the proof of
Proposition 1, in \cite{dyson96a}; therefore we omit it.

Let $\varphi \in C\left([ 0,\overline{\tau }] ,X\right) $ and
$t_{0}\in [ 0,\overline{\tau }]$. The following variation of
constants formula gives an integrated version of (\ref{For3}),
\begin{equation}\label{VCF1}
u(t)=\begin{cases}
  S(t-t_{0})\varphi (t_{0},.)+ \int_{t_{0}}^{t}S(t-s)L(u_{s}^{t_{0}})ds,
  & t\geq t_{0}, \\
 \varphi (t,.),&  t\in [ 0,t_{0}].
\end{cases}
\end{equation}
Then, for $ m\in [ 0,1]$ and $t\geq t_{0}$
\begin{equation}\label{VaOfCoFo1}
u(t)(m)=\overline{\xi }(m,t-t_{0})\varphi \left( t_{0},\pi
_{-(t-t_{0})}(m) \right)+\int_{t_{0}}^{t}\overline{\xi
}(m,t-s)L(u_{s}^{t_{0}})(\pi _{-(t-s)}(m))\,ds.
\end{equation}
Consequently, the integrated solutions of Equation
(\ref{EqtauNvar2})  for $m\in [0,1] $ and
$t\geq \overline{\tau }$ are given by
\begin{equation}\label{VCF2}
\begin{aligned}
&N(t,m)\\
&= \overline{\xi }(m,t-\overline{\tau })\varphi \left( \overline{\tau
},\pi _{-(t-\overline{\tau })}(m)\right)\\
&\quad+ \int_{\overline{\tau }}^{t}\overline{\xi }(m,t-s)
K(\pi_{-(t-s)}(m))N\left( s-\tau \left( \Delta (\pi_{-(t-s)}(m))\right) ,
\Delta (\pi _{-(t-s)}(m))\right) ds,
\end{aligned}
\end{equation}
and $N(t,m)=\varphi (t,m)$, for $(t,m)\in [ 0,\overline{\tau }] \times [0,1] $.

\begin{prop} \label{prop3.3}
There exists a unique continuous solution $N$ of
(\ref{VCF2}) with initial condition
$\varphi \in C([0,\overline{\tau }] \times [0,1])$.
\end{prop}

The proof of this proposition follows immediately using standard techniques.

The maturity of the daughter cells just after division is smaller
than $g(1)$. Then, we can first focus our study on the resting
phase for the maturity interval $[ 0,g(1)] $, which is
the source of the cells production.
Suppose that

\noindent \textbf{(H5)} the mapping
\begin{equation*}
m\mapsto\int_{m}^{g^{-1}(m)}\frac{ds}{V(s)},\;\;m\in
[0,g(1)],
\end{equation*}
is bounded on a neighborhood of $0$.

This condition means that the time required for
a cell to reach the maturity of its mother is bounded. As a simple
mathematical example we have
\begin{equation*}
V(m)\underset{m\to 0}{\sim}\alpha _{1}m,\quad \alpha _{1}>0.
\end{equation*}
Then, Condition (H5) is satisfied if and
only if $g'(0)>0$. We also assume that

\noindent \textbf{(H6)} $\tau (m)>\int_{m}^{g^{-1}(m)}\frac{ds}{V(s)}$,
 for all $m\in (0,g(1)]$.

This assumption means that, in the proliferating phase,
cells have enough time to reach the maturity of their mother. With
the conditions $\rm \textbf{(H5)}$ and $\rm \textbf{(H6)}$, we
obtain more than uniqueness. Before reformulating this result, we
need the following lemmas.


\begin{lem}\label{lemma4}
Let $m\in [0,1] $. Then $\Theta (m)$ is the unique solution of the equation
\begin{equation*}
x=\pi _{-\tau (x)}(m)\quad\mbox{and}\quad x\leq m.
\end{equation*}
\end{lem}

\begin{proof} The equation
\begin{equation*}
\int_{x}^{m}\frac{ds}{V(s)}=\tau (x)\quad \mbox{with } x\leq m
\end{equation*}
is equivalent to
$x=\pi _{-\tau (x)}(m)$  and $x\leq m$,
which completes the proof.
\end{proof}

\begin{lem}\label{lemma1}
If the conditions
(H5)  and  (H6) are satisfied, then for all
$m\in ( 0,g(1)]$, we have $\Delta(m)<m$.
\end{lem}

\begin{proof} First, recall that for all $m\in ( 0,g(1)]$,
\begin{equation*}
\int_{\Delta(m)}^{g^{-1}(m)}\frac{ds}{V(s)}=\tau (\Delta(m))\, .
\end{equation*}
Then, from Condition (H6), it follows that
\begin{equation*}
\tau(m)-\int_{m}^{1}\frac{ds}{V(s)}>\tau
(\Delta(m))-\int_{\Delta(m)}^{1}\frac{ds}{V(s)}.
\end{equation*}
On the other hand, Condition (H3)
implies that the mapping
\begin{equation*}
m\mapsto \tau(m)-\int_{m}^{1}\frac{ds}{V(s)}
\end{equation*}
is continuous and strictly increasing on $(0,g(1)]$. Then
$m>\Delta (m)$ which completes the proof.
\end{proof}

We remark that Lemma \ref{lemma1} implies, in particular, that
\begin{equation*}
\Theta(1):=\Delta (g(1))<g(1).
\end{equation*}

Let $0<b<\Theta (1)$ be fixed. The transmission of the
maturity from a generation $n$ to the generation $(n+1)$ can be
defined by the sequence
\begin{equation}\label{suitebn}
b_{n+1}=\begin{cases}
  \Delta^{-1}(b_{n}), & \mbox{ if }  b_{n}\in [ 0,\Theta(1)] , \\
  g(1),&  \mbox{if } b_{n}\in [\Theta (1),g(1)] ,
\end{cases}
\end{equation}
and $b_{0}=b$.
Note that the sequence $\left( b_{n}\right) _{n\in \mathbb{N}}$ is increasing.
The following result is immediate.

\begin{lem}\label{lemma2}
There exists $N\in \mathbb{N}$ such that $b_{N}<\Theta (1)\leq b_{N+1}\leq g(1)$.
\end{lem}

We give now our main result,
which emphasizes the strong link between the process of production
of cells and the population of stem cells.


\begin{thm}\label{theoprincipal}
Let $N(t,m)$ be a solution of  (\ref{VCF2}) with an initial condition
$\varphi \in C\left( [ 0,\overline{\tau }] \times [0,1]\right) $.
Suppose that there exists $0<b<\Theta (1)$ such that
$\varphi (t,m)=0,$\ for $t\in [ 0,\overline{\tau }] $
and $m\in [0,b]$. Then, there exists
$\overline{t}\geq \overline{\tau }$ such that $N(t,m)=0$, for
$m\in [0,g(1)] $ and $t\geq \overline{t}$, where
$\overline{t}$ can be chosen to be
\begin{equation*}
\overline{t}=\ln \left[ \frac{h\left(
g(1)\right) }{h\left( b\right) }\right] +(N+2)\overline{\tau},
\end{equation*}
$N\in \mathbb{N}$ is given by Lemma \ref{lemma2}, and
$N(t,m)=0$, for $m\in \left[ g(1),1\right] $ and
\newline{}$t\geq \overline{t}+\overline{\tau }-\ln \left( h\left(
g(1)\right) \right) =(N+3)\overline{\tau }-\ln \left( h\left(
b\right) \right) $.
\end{thm}

\begin{proof} First, we prove by induction that
$N(t,m)=0$,  for $m\in [ 0,b]$ and $t\geq 0$.
We know that
$N(t,m)=0$ for $(t,m)\in [ 0,\overline{\tau }] \times \left[ 0,b\right]$.
Let $m\in [ 0,b] $ and $t\geq \overline{\tau}$. We have
\begin{equation*}
\pi_{-(t-\overline{\tau })}(m)\leq m\leq b.
\end{equation*}
So $\varphi \left( \overline{\tau },\pi_{-(t-\overline{\tau })}(m)\right) =0$.
 We suppose that $t\in \left[ \overline{\tau
},\overline{\tau }+\underline{\tau }\right] $, with
$\underline{\tau }:=\underset{x\in [0,1] }{\min }\tau
(x)>0$, and $s\in \left[ \overline{\tau },t\right] $. Then $s-\tau
\left( \Delta (\pi _{-(t-s)}(m))\right) \in [ 0,\overline{\tau }] $. Because $\Delta $ is an increasing
function and $\pi _{-(t-s)}(m)\leq m$, then
\begin{equation*}
\Delta (\pi_{-(t-s)}(m))\leq \Delta (m).
\end{equation*}
 From lemma \ref{lemma1}, we deduce that
$\Delta (\pi_{-(t-s)}(m))<m\leq b$.
Consequently,
\begin{equation*}
N\left( s-\tau \left( \Delta (\pi
_{-(t-s)}(m))\right) ,\Delta (\pi _{-(t-s)}(m))\right) =0.
\end{equation*}
Then, (\ref{VCF2}) implies that
$N(t,m)=0$ for $m\in[ 0,b]$  and
$t\in \left[0,\overline{\tau }+\underline{\tau }\right]$.
 By steps on the intervals
 $[ 0,\overline{\tau}+n\underline{\tau }]$, $n=1,2,\dots $, we conclude that
\begin{equation*}
N(t,m)=0,\quad\mbox{for }  m\in [ 0,b] \mbox{  and } t\geq 0.
\end{equation*}
Let us now reconsider the sequence $(b_{n})_{n\geq 0}$
given by (\ref{suitebn}), and the sequence $(t_{n})_{n\in \mathbb{N}}$
defined by
\begin{equation}\label{recurr1}
\begin{gathered}
  t_{0}  =  0, \\
   t_{n+1} =  t_{n}+\ln
[h( b_{n+1})/h( b_{n})] +\overline{\tau }.
\end{gathered}
\end{equation}
 Then,
\begin{equation}\label{recurr2}
t_{n}=\ln \big[ \frac{h\left(
b_{n}\right) }{h(b)}\big] +n\overline{\tau}.
\end{equation}
We recall that the sequence $(b_{n}) _{n\in
\mathbb{N}}$ is increasing. Then, the sequence $(t_{n})_{n\in \mathbb{N}}$ is also
increasing.

We will prove, by induction, the following result

\noindent $\mathbf{(H_n)}$  $N(t,m)=0$ for $m\in[ 0,b_{n}]$ and $t\geq t_{n}$.

We remark that for $m\in ( 0,g(1)]$,
\begin{equation*}
\int_{m}^{g^{-1}(m)}\frac{ds}{V(s)}
=\ln \Big(\frac{h\left( g^{-1}(m)\right) }{h(m)}\Big).
\end{equation*}
First, $( H_{0}) $ is true.
Let us  suppose that $( H_{n})$ is true.
Let $m\in [ 0,b_{n+1}]$ and $t\geq t_{n+1}$. Then, because of (\ref{recurr1}),
\begin{equation*}
t_{n+1}\geq t_{n}+\overline{\tau }.
\end{equation*}
Using the variation of constant formula
(\ref{VaOfCoFo1}), the solutions of (\ref{VCF2}) can be
reformulated, for $t\geq t_{n+1}$, as
\begin{align*}
   N(t,m) & =  \overline{\xi
}(m,t-t_{n}-\overline{\tau })N\big( t_{n}+\overline{\tau },
 \pi_{-(t-t_{n}-\overline{\tau })}(m))\big) \\
&\quad +  \int_{t_{n}+\overline{\tau}}^{t}\overline{\xi }
   (m,t-s)K(\,\pi_{-(t-s)}(m))\\
&\quad \times N\left( s-\tau (\Delta (\pi _{-(t-s)}(m))),
  \Delta (\pi _{-(t-s)}(m))\right) ds.
\end{align*}
We remark that
$\pi_{-(t-t_{n}-\overline{\tau })}(m))=h^{-1}\big(
h(m)e^{-(t-t_{n}-\overline{\tau })}\big)$ and
\begin{equation*}
  e^{-(t-t_{n}-\overline{\tau
})}  =  \frac{h( b_{n}) }{h( b_{n+1})}e^{-(t-t_{n+1})}
 \leq   \frac{h( b_{n})}{h( b_{n+1}) }.
\end{equation*}
Then, we deduce that
\begin{equation*}
 \pi_{-(t-t_{n}-\overline{\tau })}(m))
\leq  h^{-1}\big(h(m)\frac{h( b_{n}) }{h( b_{n+1}) }\big)
\leq  h^{-1}\big( h(b_{n+1})\frac{h(b_{n}) }{h( b_{n+1})}\big)
=b_{n}.
\end{equation*}
Hence, $(H_{n})$ implies
\begin{equation*}
N\left( t_{n}+\overline{\tau },\pi
_{-(t-t_{n}-\overline{\tau })}(m))\right) =0.
\end{equation*}
Furthermore, for $t_{n}+\overline{\tau }\leq s\leq t$, we have
\begin{equation*}
s-\tau (\Delta (\pi _{-(t-s)}(m)))\geq
\left( t_{n}+\overline{\tau }\right) -\tau (\Delta (\pi
_{-(t-s)}(m))\geq t_{n},
\end{equation*}
and
$\Delta (\pi _{-(t-s)}(m))\leq \Delta (m)\leq \Delta (b_{n+1})=b_{n}$.
Consequently,
\begin{equation*}
N\left( s-\tau (\Delta (\pi
_{-(t-s)}(m)))\;,\;\Delta (\pi _{-(t-s)}(m)\right) =0.
\end{equation*}
Then, we deduce that
$\left(H_{n+1}\right)$: $N(t,m)=0$ for $m\in[ 0,b_{n+1}]$ and
$t\geq t_{n+1}$.
Thanks to Lemma \ref{lemma2}, we have
$b_{N}<\Theta (1)\leq b_{N+1}\leq g(1)$,
which implies
\[
N(t,m)=0,\quad\mbox{for }m\in [ 0,\Theta (1)], \quad
t\geq t_{N+1}=\ln \big[ \frac{h( b_{n+1})
}{h( b) }\big] +(N+1)\overline{\tau }.
\]

Suppose now that $m\in [ \Theta (1),g(1)]$
and $t\geq t_{N+1}+\ln [ \frac{h( g(1)) }{h( b_{n+1}) }]
+\overline{\tau }=\overline{t}$.
Since $\Delta ^{-1}(g(1))=\Theta (1)$, using the same technique as in the
first part of this proof, we obtain
$N(t,m)=0$ for $m\in [0,g(1)]$ and $t\geq \overline{t}$.
Finally, take $m\in [ g(1),1]$ and $t\geq \overline{t}+\overline{\tau }$.
We can write
\begin{align*}
N(t,m) & =  \overline{\xi}(m,t-\overline{t}-\overline{\tau })N\left(
\overline{t}+\overline{\tau },\pi_{-(t-\overline{t}-\overline{\tau
})}(m))\right) \\
 &\quad +  \int_{\overline{t}+\overline{\tau }}^{t}\overline{\xi }(m,t-s)
 K(\,\pi_{-(t-s)}(m))\\
&\quad \times N\left( s-\tau(\Delta (\pi _{-(t-s)}(m))),\Delta (\pi _{-(t-s)}(m))
\right) ds.
\end{align*}
Let $\overline{t}+\overline{\tau }\leq s\leq t$. Then
\begin{equation*}
s-\tau (\Delta (\pi_{-(t-s)}(m)))\geq \left( \overline{t}+\overline{\tau }\right)
-\tau (\Delta (\pi _{-(t-s)}(m))\geq \overline{t}.
\end{equation*}
Consequently, if $\pi _{-(t-s)}(m)\leq g(1)$, then,
$\Delta (\pi_{-(t-s)}(m))\leq g(1)$.
This implies that
\begin{equation*}
N\left( s-\tau (\Delta (\pi_{-(t-s)}(m))),\Delta (\pi _{-(t-s)}(m))\right) =0.
\end{equation*}
On the other hand, if $ \pi_{-(t-s)}(m)>g(1)$,
then, by the definition of $K$, we have
$K(\pi _{-(t-s)}(m))=0$.
We conclude that, for $m\in [ g(1),1] $ and
$t\geq \overline{t}+\overline{\tau }$,
\begin{equation*}
N(t,m)=\overline{\xi}(m,t-\overline{t}-\overline{\tau })
N\left(\overline{t}+\overline{\tau },
\pi_{-(t-\overline{t}-\overline{\tau })}(m))\right) .
\end{equation*}
We remark that $\ln (h(m))<0$, for all $m\in (0,g(1)] $. Then,
for $m\in [ g(1),1]$ and
$t\geq (N+3)\overline{\tau }-\ln \left( h( b) \right)
=\overline{t}+\overline{\tau }-\ln (h(g(1)))$, we have
\begin{equation*}
\pi_{-(t-\overline{t}-\overline{\tau })}(m)
= h^{-1}\big(h(m)e^{-(t-\overline{t}-\overline{\tau })}\big)
\leq h^{-1}\big(h(m)h(g(1))\big)
\leq h^{-1}( h(g(1))) =g(1).
\end{equation*}
Hence,
\begin{equation*}
N\left( \overline{t}+\overline{\tau
},\pi _{-(t-\overline{t}-\overline{\tau })}(m))\right) =0.
\end{equation*}
This completes the proof of the theorem.
\end{proof}

This result proves that the production of cells depends strongly
on the state of the population of stem cells. It describes, in
particular, the destruction of the cell population in the resting
phase when its starting value is defective. It is believed that
the pathology of aplastic anemia is due to injury or destruction
of a common pluripotential stem cell. This result is a first step
to prove stability or instability results in the nonlinear case
$\beta =\beta (N)$ (see further publication).
In the next section, we prove that the proliferating
cells also depends strongly on the state of the population of stem
cells.

\section{Equations for the total population in the proliferating phase}

To complete the formulation of our model, we give the equations satisfied
by the total population
\begin{equation*}
P(t,m)=\int_{0}^{\tau(\Theta (m))}p(t,m,a)da
\end{equation*}
of proliferating cells for a given maturation level
$m\in [0,1] $.

\begin{prop} \label{prop4.1}
Let $m\in [0,1] $ and $t\geq 0$. Then $P(t,m)$
satisfies the following equations:
\begin{enumerate}
\item If $0\leq t\leq \tau (\Theta (m))$, then
\begin{equation}\label{EqtauPvar1Bis}
\begin{aligned}
&\frac{\partial}{\partial t}P(t,m)+\frac{\partial }{\partial m}\left(
V(m)P(t,m)\right) \\
&=-\gamma(m)P(t,m)+\beta(m)N(t,m)-G(t,m),
\end{aligned}
\end{equation}
 with
\begin{equation}\label{f1}
G(t,m)=\frac{1}{1+\tau'(\Theta
(m))V(\Theta (m))}\xi (m,t)\Gamma \left( \pi _{-t}(m),\tau (\Theta
(m))-t\right) ,
\end{equation}
where $\xi $ is given by (\ref{defksi}) and
$\Gamma $ is the initial data given by (\ref{CI1}).

\item If $t\geq \tau (\Theta (m))$, then
\begin{equation}\label{EqtauPvar2}
\begin{aligned}
&\frac{\partial}{\partial t}P(t,m)+\frac{\partial }{\partial m}\left(
V(m)P(t,m)\right)\\
&=  -\gamma(m)P(t,m)+\beta (m)N(t,m)- H(m)N\left( t-\tau (\Theta (m)),
\Theta (m)\right),
\end{aligned}
\end{equation}
 where
\begin{equation}\label{conditionK2}
H(m)=\frac{\xi \left( m,\tau (\Theta
(m))\right) \beta \left( \Theta (m)\right) }{1+\tau ^{\prime
}(\Theta (m))V(\Theta (m))}.
\end{equation}
\end{enumerate}
\end{prop}

\begin{proof}
 Taking $a=\tau (\Theta (m))$ in (\ref{Caract1}), we obtain
\begin{equation}\label{equationproliftaudeteta}
p(t,m,\tau(\Theta (m)))
=\begin{cases}
\xi (m,t)\Gamma (\pi _{-t}(m) ,\tau (\Theta (m))-t),&
\mbox{for } 0\leq t<\tau (\Theta(m)), \\[3pt]
 \xi (m,\tau (\Theta (m)))\beta
\left( \pi_{-\tau (\Theta (m))}(m) \right)\\
\times N(t-\tau (\Theta(m)),\pi _{-\tau (\Theta (m))}(m) ), &
\mbox{for } \tau (\Theta (m))\leq t.
\end{cases}
\end{equation}
According to Lemma \ref{lemma4}, we deduce that
$\pi_{-\tau (\Theta (m))}(m) =\Theta (m)$ for   all $m \in [0,1]$.
On the other hand, we have
\begin{align*}
&\frac{\partial}{\partial m}\left( V(m)P(t,m)\right) \\
&=\frac{\partial }{\partial m}\Big(\int_{0}^{\tau (\Theta (m))}
   V(m)p(t,m,a)da\Big),\\
&=  \int_{0}^{\tau(\Theta (m))}\frac{\partial }{\partial m}\left(
V(m)p(t,m,a)\right) da+\tau'(\Theta (m))\Theta '(m)V(m)p(t,m,\tau (\Theta (m))).
\end{align*}
Consequently, by integrating (\ref{prolif})
with respect to the age between $0$ and $\tau (\Theta (m))$, we
obtain the following time-maturation structured equation,
\begin{align*}
& \frac{\partial}{\partial t}P(t,m)+\frac{\partial }{\partial m}
\left(V(m)P(t,m)\right) \\
&=-\gamma(m)P(t,m)+p(t,m,0)-p(t,m,\tau (\Theta (m)))\\
&\quad +\tau'(\Theta (m))\Theta '(m)V(m)p(t,m,\tau
 (\Theta(m))), \\
&=-\gamma(m)P(t,m)+p(t,m,0) -\left[1-\tau'(\Theta (m))\Theta'(m)V(m)\right]
p(t,m,\tau (\Theta (m))).
\end{align*}
Using (\ref{bctauvariant1}), (\ref{Form1}) and
(\ref{equationproliftaudeteta}) in this last equation, we obtain
(\ref{EqtauPvar1Bis}) and (\ref{EqtauPvar2}).
\end{proof}

It is not difficult to prove, by steps, the existence, uniqueness
and regularity of solutions of System (\ref{EqtauPvar1Bis})--(\ref{EqtauPvar2}).
Let $\psi \in C\left( [ 0,\overline{\tau }] \times [0,1] \right)$
be an integrated solution of this system on the set
$[ 0,\overline{\tau }] \times [0,1]$. As for the resting phase, an integrated
expression of Equation (\ref{EqtauPvar2}), for $t\geq \overline{\tau }$ and
$m\in [0,1] $, is given by
\begin{equation}\label{Eq27}
\begin{aligned}
&P(t,m)\\
&=\xi (m,t-\overline{\tau })\psi \left( \overline{\tau },\pi
_{-(t-\overline{\tau })}(m)\right) \\
&\quad  + \int_{\overline{\tau }}^{t}\xi (m,t-s)\beta
(\pi _{-(t-s)}(m))N\left( s,\pi_{-(t-s)}(m)\right) ds \\
&\quad + \int_{\overline{\tau }}^{t}\xi (m,t-s)H(\pi _{-(t-s)}(m))
N \left( s-\tau \left( \Theta\left( \pi _{-(t-s)}(m)\right) \right) ,
\Theta \left( \pi_{-(t-s)}(m)\right)\right)  ds,
\end{aligned}
\end{equation}
and $P(t,m)=\psi \left( t,m\right) $, for
$(t,m)\in [ 0,\overline{\tau }] \times [0,1]$.
Then we have the following result.

\begin{prop} \label{prop4.2}
Under the assumptions in Theorem \ref{theoprincipal} and  $\psi (t,m)=0$
for $t\in [ 0,\overline{\tau }] $, $m\in [ 0,b] $,
the solution $P$ of \eqref{Eq27} is equal to $0$, for $t\geq
\overline{t}$, $m\in [0,g(1)] $, and for $t\geq
\overline{t}+\overline{\tau }-\ln (h(g(1)))$, $m\in [g(1),1]$.
\end{prop}

\begin{proof} In the proof of Theorem \ref{theoprincipal}, we obtained that
$N(t,m)=0$ for $m\in \left[
0,b_{n}\right] $ and $t\geq t_{n}$. Then (\ref{Eq27}) becomes
\begin{equation*}
P(t,m)=\xi (m,t-t_{n})P\left( t_{n},\pi _{-(t-t_{n})}(m)\right)
\quad \mbox{for }  t\geq t_{n}\; m\in [0,b_{n}].
\end{equation*}
Then, by steps, we prove that
$P(t,m)=0$  for $t\geq t_{n}$ and $m\in [ 0,b_{n}]$.
This leadds to
\begin{equation*}
P(t,m)=0,\quad \mbox{for }  t\geq \overline{t},\; m\in [0,g(1)].
\end{equation*}
Using the same argument as in the proof of Theorem
\ref{theoprincipal}, we also obtain, for $m\in [0,1] $
and $t\geq \overline{t}+\overline{\tau }-\ln (h(g(1)))$,
\begin{equation*}
P(t,m)=\xi (m,t-\overline{t}-\overline{\tau })P\left(
\overline{t}+\overline{\tau },\pi_{-(t-\overline{t}-\overline{\tau })}(m)\right).
\end{equation*}
Then $P(t,m)=0$ for $m\in [ g(1),1]$ and
$t\geq \overline{t}+\overline{\tau }-\ln(h(g(1)))$,
which completes the proof.
\end{proof}

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\end{document}