# Cell cycle length and long-time behaviour of an age-size model

Katarzyna Pichór  K. Pichór, Institute of Mathematics, University of Silesia,
Bankowa 14, 40-007 Katowice, Poland<EMAIL_ADDRESS>and  Ryszard
Rudnicki  R. Rudnicki, Institute of Mathematics, Polish Academy of Sciences,
Bankowa 14, 40-007 Katowice, Poland<EMAIL_ADDRESS>
(Date: March 19th, 2021)

###### Abstract.

We consider an age-size structured cell population model based on the cell
cycle length. The model is described by a first order partial differential
equation with initial-boundary conditions. Using the theory of semigroups of
positive operators we establish new criteria for an asynchronous exponential
growth of solutions to such equations. We discuss the question of exponential
size growth of cells. We study in detail a constant size growth model and a
model with target size division. We also present versions of the model when
the population is heterogeneous.

###### Key words and phrases:

Cell cycle, size-age structured model, semigroup of operators, asynchronous
exponential growth

###### 2020 Mathematics Subject Classification:

Primary: 47D06; Secondary: 35F15, 45K05 92D25, 92C37

##  1\. Introduction

The cell cycle is a series of events that take place in a cell leading to its
replication. It is regulated by a complex network of protein interactions
[32]. Modern experimental techniques concerning the cell cycle [7, 20, 30, 33,
43, 44, 45, 48] allow us not only to understand processes inside single cells,
but also to build more precise cellular populations models.

Most of populations are usually heterogeneous. Thus it is important to
consider the distribution of the population according to some significant
parameters such as age, size, maturity, or proliferative state of cells.
Models of this type are called structured. This type of models are usually
represented by partial differential equations with some nonlocal perturbations
and specific boundary conditions. Knowing the length of cell cycle allows us
to predict the development of unicellular populations and tissues growth and
maintenance.

The aim of the paper is twofold. Firstly, to construct an age-size structured
model assuming that we know the growth of individual cells and the
distribution of the cell cycle length. Secondly, to study the long-time
behaviour of the solution of this model.

We consider a model which is based on the following assumptions. The
population grows in steady-state conditions. Cells can be described by their
age $a$ and size $x$ alone and reproduction occurs by fission into two equal
parts. The distribution of the cell cycle length depends only on the initial
size $x_{b}$ of a cell. The velocity of growth of an individual cell depends
only on its size $x$, i.e. $x^{\prime}(t)=g(x(t))$. We also assume that sizes
of cells and cell cycle durations are bounded above and bounded away from
zero. Moreover, we assume that the initial daughter cell sizes are distributed
in some interval which contains the mother initial size. We formulate a
mathematical model which describes the time evolution of distribution of
cellular age and size. The model consists of a partial differential equation
with an integral boundary condition and an initial condition. The novelty of
our model is that we use the distribution of the length of the cell cycle,
instead of a size dependent probability of division usually used in size-
structured models [6, 9, 10, 17, 18, 19]. Such probability is difficult to
measure experimentally in contrast to the length of the cell cycle.

We check that the solutions of our model generate a continuous semigroup of
operators $\\{U(t)\\}_{t\geq 0}$ on some $L^{1}$ space. Under additional
assumption that $g(2x)\neq 2g(x)$ for some $x$, we prove that the semigroup
$\\{U(t)\\}_{t\geq 0}$ has asynchronous exponential growth (AEG), i.e.

(1) $e^{-\lambda t}U(t)u_{0}(x_{b},a)\to Cv(x_{b},a)\quad\textrm{for
$t\to\infty$},$

where $\lambda$ is the Malthusian parameter and $v$ is a stable initial size
and age distribution, which does not depend on the initial distribution
$u_{0}$. The property AEG plays an important role in the study of structured
population models [4, 9, 17, 49], because we can expect that the real process
should be close to a stationary state and then it is easy to estimate
biological parameters [26].

The proof of AEG of $\\{U(t)\\}_{t\geq 0}$ is based on the reduction of the
problem to a stochastic (Markov) semigroup [24] by using the Perron
eigenvectors and on the theorem that a partially integral stochastic semigroup
having a unique invariant density is asymptotically stable [35]. A similar
technique was applied to study other population models [5, 34, 41] and to some
piecewise deterministic Markov processes [28, 36, 42]. We note that AEG
property can be proved by using known results on compact semigroups but it
seems to be difficult to check compactness and analyze the spectrum of the
generator of our semigroup. It is interesting that even nonlinear models of
cell population (cf. [27, 38]) can be reduced to stochastic semigroups.

The last two sections contain corollaries from our results (Section 6) and
some remarks concerning other models and experimental data (Section 7). One of
the main points of these sections is what can happen when $g(2x)=2g(x)$ for
all $x$ ? This is an important question because it is usually assumed that the
size (volume) of a cell grows exponentially, which means that $g(x)=\kappa x$
and in this case $g(2x)=2g(x)$. If we include in a model the assumption that
$g(x)=\kappa x$, then we can obtain some paradoxical results. For example, if
the size discrepancy between newborn cells is small, then the descendants of
one cell can have the same size at the same time and the size of the
population does not grow exponentially even in steady-state conditions. Of
course the law of exponential size growth is statistical in nature and we can
modify it by considering some random fluctuations in the growth rate. Another
problem considered in Section 7 is how to incorporate into our description
some models of the cell cycle: a constant $\Delta$ model and a model with
target size division. Finally, we present versions of the model when the
population is heterogeneous, e.g. with an asymmetric division or with fast and
slow proliferation.

##  2\. Model

We consider the following model of the cell cycle. Denote, respectively, by
$a$, $x_{b}$, and $x$ — the age, the initial size, and the size of a cell. We
assume that $\underline{x}_{b}$ and $\overline{x}_{b}$ are the minimum and
maximum sizes of newborn cells. We also assume that cells age with unitary
velocity and grow with a velocity $g(x)$, i.e. if a cell has the initial size
$x_{b}$, then the size at age $a$ satisfies the equation

(2) $x^{\prime}(a)=g(x(a)),\quad x(0)=x_{b}.$

We denote by $\pi_{a}x_{b}$ the solution of (2). The length $\tau$ of the cell
cycle is a random variable which depends on the initial cell size $x_{b}$; has
values in some interval $[\underline{a}(x_{b}),\overline{a}(x_{b})]$; and has
the probability density distribution $q(x_{b},a)$, i.e. the integral
$\int_{0}^{A}q(x_{b},a)\,da$ is the probability that $\tau\leq A$. According
to the definition of $q$, if a cell has the initial size $x_{b}$, then
$\Phi(x_{b},a)=\int_{a}^{\infty}q(x_{b},r)\,dr$ is its survival function, i.e.
$\Phi(x_{b},a)$ is the probability that a cell will not split before age $a$.
We assume that if the mother cell has size $x$ at the moment of division, then
the daughter cells have size $x/2$, i.e. if the initial size of the mother
cell is $x_{b}$ and $\tau=a$ is the length of its cell cycle, then the initial
size of the daughter cell is $S_{a}(x_{b})=\tfrac{1}{2}\pi_{a}x_{b}$.

Now we collect the assumptions concerning the functions $g$ and $q$ used in
the paper:

(A1) $g\colon[\underline{x}_{b},2\overline{x}_{b}]\to(0,\infty)$ is a
$C^{1}$-function,

(A2)
$q\colon[\underline{x}_{b},\overline{x}_{b}]\times[0,\infty)\to[0,\infty)$ is
a continuous function and for each $x_{b}$ the function $a\mapsto q(x_{b},a)$
is a probability density,

(A3) $0<\underline{a}(x_{b})<\overline{a}(x_{b})<\infty$, $q(x_{b},a)>0$ if
$a\in(\underline{a}(x_{b}),\overline{a}(x_{b}))$, and $q(x_{b},a)=0$ if
$a\notin(\underline{a}(x_{b}),\overline{a}(x_{b}))$ for each
$x_{b}\in[\underline{x}_{b},\overline{x}_{b}]$,

(A4) $x_{b}\mapsto\underline{a}(x_{b})$ and $x_{b}\mapsto\overline{a}(x_{b})$
are continuous functions,

(A5) $S_{\underline{a}(x_{b})}(x_{b})\geq\underline{x}_{b}$ and
$S_{\overline{a}(x_{b})}(x_{b})\leq\overline{x}_{b}$ for each
$x_{b}\in[\underline{x}_{b},\overline{x}_{b}]$,

(A6) $S_{\underline{a}(x_{b})}(x_{b})<x_{b}<S_{\overline{a}(x_{b})}(x_{b})$
for each $x_{b}\in(\underline{x}_{b},\overline{x}_{b})$.

Fig. 1 and Fig. 2 illustrate our assumptions. Only assumption (A6) needs some
explanation. We assume that a daughter cell can have the same initial size as
the initial size of a mother cell. In Section 5 we will add an extra
assumption (A7) which will be used only to show the long-time behaviour of the
distribution of $(x_{b},a)$.

$\underline{x}_{b}$$\overline{x}_{b}$$x_{b}$$\overline{a}(x_{b})$$\underline{a}(x_{b})$$a$

Figure 1. An example of functions $x_{b}\mapsto\underline{a}(x_{b})$ and
$x_{b}\mapsto\overline{a}(x_{b})$. The function $q$ is positive between the
graphs of these functions.

$\underline{x}_{b}$$\overline{x}_{b}$$x_{b}$$S_{\overline{a}(x_{b})}(x_{b})$$S_{\underline{a}(x_{b})}(x_{b})$$S_{a}(x_{b})$

Figure 2. The relation between the initial sizes of mother and daughter cells
(A5,A6).

Assume that a cell with initial size $x_{b}$ and age $a$ splits in the time
interval of the length $\Delta t$ with probability $p(x_{b},a)\Delta
t+o(\Delta t)$, i.e.

$p(x_{b},a)=\lim_{\Delta t\downarrow
0}\frac{\operatorname{P}(\tau\in[a,a+\Delta t]\mid\tau\geq a)}{\Delta t}.$

Since $\Phi(x_{b},a)=\exp\big{(}-\int_{0}^{a}p(x_{b},r)\,dr\big{)}$, an easy
computation shows that

$\displaystyle q(x_{b},a)$
$\displaystyle=p(x_{b},a)\exp\big{(}-\textstyle{\int_{0}^{a}}p(x_{b},r)\,dr\big{)},$
$\displaystyle\quad p(x_{b},a)$
$\displaystyle=\frac{q(x_{b},a)}{\int_{a}^{\infty}q(x_{b},r)\,dr}$

for $a<\overline{a}(x_{b})$. As $\Phi(x_{b},\overline{a}(x_{b}))=0$, we have
$\int_{0}^{\overline{a}(x_{b})}p(x_{b},a)\,da=\infty$.

In order to derive a master equation for the distribution of the population
with respect to $x_{b}$ and $a$ we need to introduce a family of Frobenius-
Perron operators which describe the relation between the initial sizes of
mother and daugther cells.

Let $f(x_{b})$ be the density of initial sizes of mother cells that have the
fixed length of cell cycle $\tau=a$ for some
$a\in(\underline{a},\overline{a})$, where
$\underline{a}=\min\underline{a}(x_{b})$ and
$\overline{a}=\max\overline{a}(x_{b})$. Denote by $P_{a}f(x_{b})$ the density
of initial sizes of daughter cells.

######  Lemma 1.

Let $x_{a}$ be the minimum initial size of cells which can split at age $a$.
Then

(3)
$P_{a}f(x_{b})=\frac{2g(\pi_{-a}(2x_{b}))}{g(2x_{b})}f(\pi_{-a}(2x_{b}))\mathbf{1}_{[S_{a}(x_{a}),\overline{x}_{b}]}(x_{b}).$

###### Proof.

Observe that $x_{a}=\underline{x}_{b}$ if
$S_{a}(\underline{x}_{b})\geq\underline{x}_{b}$ or
$x_{a}=S_{a}^{-1}(\underline{x}_{b})$ otherwise. It is clear that
$q(x_{b},a)=0$ for $x_{b}<x_{a}$. We have

$\int_{S_{a}(x_{a})}^{S_{a}(y)}P_{a}f(x_{b})\,dx_{b}=\int_{x_{a}}^{y}f(x_{b})\,dx_{b}\quad\textrm{for
$y\geq x_{a}$}$

or, equivalently,

(4)
$\int_{S_{a}(x_{a})}^{x}P_{a}f(x_{b})\,dx_{b}=\int_{x_{a}}^{S_{a}^{-1}(x)}f(x_{b})\,dx_{b}\quad\textrm{for
$x\geq S_{a}(x_{a})$}.$

From (4) it follows that

$P_{a}f(x_{b})=\frac{d}{dx_{b}}\big{(}S_{a}^{-1}(x_{b})\big{)}f\big{(}S_{a}^{-1}(x_{b})\big{)}\mathbf{1}_{[S_{a}(x_{a}),\overline{x}_{b}]}(x_{b}).$

Using the formula $S_{a}^{-1}(x_{b})=\pi_{-a}(2x_{b})$ we check that

(5)
$\frac{d}{dx_{b}}\big{(}S_{a}^{-1}(x_{b})\big{)}=\frac{2g(\pi_{-a}(2x_{b}))}{g(2x_{b})}.$

In order to show (5) we introduce two functions:

$\varphi(x_{b},a)=\pi_{-a}(2x_{b})\quad\textrm{and}\quad\psi(x_{b},a)=\frac{\partial\varphi}{\partial
x_{b}}(x_{b},a).$

From $\varphi(x_{b},0)=2x_{b}$ we obtain $\psi(x_{b},0)=2$. Since
$\frac{\partial\varphi}{\partial a}(x_{b},a)=-g(\varphi(x_{b},a))$, we have

$\displaystyle\frac{\partial\psi}{\partial a}(x_{b},a)$
$\displaystyle=\frac{\partial}{\partial a}\frac{\partial\varphi}{\partial
x_{b}}(x_{b},a)=\frac{\partial}{\partial x_{b}}\frac{\partial\varphi}{\partial
a}(x_{b},a)$ $\displaystyle=\frac{\partial}{\partial
x_{b}}\big{(}-g(\varphi(x_{b},a))\big{)}=-g^{\prime}(\varphi(x_{b},a))\psi(x_{b},a).$

We have received the linear equation $\partial\psi/\partial
a=-g^{\prime}(\varphi(x_{b},a))\psi$ with the initial condition
$\psi(x_{b},0)=2$ which has the solution

$\psi(x_{b},a)=2\exp\bigg{(}-\int_{0}^{a}g^{\prime}(\varphi(x_{b},r))\,dr\bigg{)}.$

Substituting $y=\varphi(x_{b},r)$ we receive $dy/dr=-g(y)$ and

$\psi(x_{b},a)=2\exp\bigg{(}\int_{2x_{b}}^{\pi_{-a}(2x_{b})}\frac{g^{\prime}(y)}{g(y)}\,dy\bigg{)}=\frac{2g(\pi_{-a}(2x_{b}))}{g(2x_{b})}.\qed$

The formula (3) defines a family of operators $P_{a}\colon
L^{1}[\underline{x}_{b},\overline{x}_{b}]\to
L^{1}[\underline{x}_{b},\overline{x}_{b}]$, $a\geq 0$. The operators $P_{a}$
are well defined for $a\in(\underline{a},\overline{a})$ but we extend the
definition of $P_{a}$ setting $P_{a}f\equiv 0$ for others $a$’s. For each $a$
the operator $P_{a}$ is linear and positive, i.e. if $f\geq 0$, then
$P_{a}f\geq 0$. Moreover $\|P_{a}f\|_{L^{1}}\leq\|f\|_{L^{1}}$. The adjoint
operator of $P_{a}$ acts on the space
$L^{\infty}[\underline{x}_{b},\overline{x}_{b}]$ and it is given by
$P_{a}^{*}f(x_{b})=f(S_{a}(x_{b}))=f(\frac{1}{2}\pi_{a}x_{b})$ for $x_{b}\geq
x_{a}$ and $P_{a}^{*}f(x_{b})=0$ for $x_{b}<x_{a}$.

We denote by $u(t,x_{b},a)$ the number of individuals in a population having
initial size $x_{b}$ and age $a$ at time $t$. Then, according to our
assumptions concerning the model, $p(x_{b},a)u(t,x_{b},a)\Delta t$ is the
number of cells of initial size $x_{b}$ and age $a$ which split in a time
interval of the length $\Delta t$. It means that $2\Delta
t\int_{0}^{\infty}\Big{(}P_{a}\big{(}p(\cdot,a)u(t,\cdot,a)\big{)}\Big{)}(x_{b})\,da$
is the number of new born cells in this time interval. It should be noted that
the operator $P_{a}$ in the last integral acts on the function
$\psi(x_{b})=p(x_{b},a)u(t,x_{b},a)$ at fixed values $t$ and $a$. If there are
no limitations concerning the growth of the population and all cells split,
then the function $u$ satisfies the following initial-boundary problem:

(6) $\displaystyle\frac{\partial u}{\partial t}(t,x_{b},a)+\frac{\partial
u}{\partial a}(t,x_{b},a)=-p(x_{b},a)u(t,x_{b},a),\quad
a<\overline{a}(x_{b}),$ (7) $\displaystyle
u(t,x_{b},0)=2\int_{0}^{\infty}\Big{(}P_{a}\big{(}p(\cdot,a)u(t,\cdot,a)\big{)}\Big{)}(x_{b})\,da,$
(8) $\displaystyle u(0,x_{b},a)=u_{0}(x_{b},a).$

We assume that $u_{0}$ is a nonnegative measurable function such that

(9)
$\int_{\underline{x}_{b}}^{\overline{x}_{b}}\int_{0}^{\infty}u_{0}(x_{b},a)\Psi(x_{b},a)\,da\,dx_{b}<\infty,$

where
$\Psi(x_{b},a)=\exp\big{(}\int_{0}^{a}p(x_{b},\bar{a})\,d\bar{a}\big{)}$. In
(9) we have assumed that the initial condition $u_{0}$ is integrable with
weight $\Psi(x_{b},a)=\Phi(x_{b},a)^{-1}$ because $\Phi(x_{b},a)$ is the
fraction of cells which will survive beyond age $a$. Since
$\overline{a}(x_{b})$ is the maximum age of a cell with initial size $x_{b}$,
it is reasonable to consider variables $x_{b}$ and $a$ only from the set

$X=\\{(x_{b},a)\colon\underline{x}_{b}\leq
x_{b}\leq\overline{x}_{b},\,\,\,0\leq a\leq\overline{a}(x_{b})\\}$

(see Fig. 3).

$\underline{x}_{b}$$\overline{x}_{b}$$x_{b}$$\overline{a}(x_{b})$$a$$X$

Figure 3. The set $X$.

Though we consider $a\leq\overline{a}(x_{b})$, it will be convenient to keep
the notation of integral $\int_{0}^{\infty}$ with respect to $a$ as in formula
(7) assuming that $u(t,x_{b},a)=0$ for $a>\overline{a}(x_{b})$.

Let $\mathcal{B}(X)$ be the $\sigma$-algebra of Borel subsets of $X$, $\ell$
be the Lebesgue measure on $X$, and $E$ be the space
$L^{1}(X)=L^{1}(X,\mathcal{B}(X),\ell)$. By $\|\cdot\|_{E}$ we denote the norm
in $E$.

The main purpose of the paper is to show that the solutions of the system
(6)–(8) have asynchronous exponential growth (1). The AEG property can be
written in the following way:

(10) $\lim_{t\to\infty}e^{-\lambda t}u(t,x_{b},a)=\alpha(u_{0})v(x_{b},a),$

where the limit is in the space $E$, $\alpha$ is a linear and bounded
functional on $E$, and $v\in E$ does not depend on $u_{0}$ (see Theorem 18).
We prove this fact under an additional assumption that $g(2x)\neq 2g(x)$ for
some $x$. The schedule of the proof is the following. In Section 3 we replace
the system (6)–(8) by one in which the first equation of the system has zero
on the right-hand side. Then we construct a $C_{0}$-semigroup of positive
operators $\\{T(t)\\}_{t\geq 0}$ on the space $L^{1}(X)$ corresponding to the
new system. In Section 4 we prove that the infinitesimal generator
$\mathcal{A}$ of the semigroup $\\{T(t)\\}_{t\geq 0}$ and the adjoint operator
$\mathcal{A}^{*}$ have positive eigenvectors, $f_{i}$ and $v$, respectively,
for some eigenvalue $\lambda$. In Section 5 we introduce a semigroup
$\\{P(t)\\}_{t\geq 0}$ given by $P(t)f=e^{-\lambda t}T(t)f$ defined on the
space $E_{1}=L^{1}(X,\mathcal{B}(X),\mu)$ with the measure $\mu$ given by
$d\mu=v\,dx_{b}da$. We check that $\\{P(t)\\}_{t\geq 0}$ is a stochastic
semigroup on $E_{1}$ and that $f_{i}$ is the unique invariant density of
$\\{P(t)\\}_{t\geq 0}$. We also formulate some general theorem concerning
asymptotic stability of stochastic semigroups. We apply this theorem to the
semigroup $\\{P(t)\\}_{t\geq 0}$ and prove its asymptotic stability. We
translate this result in terms of the semigroup $\\{T(t)\\}_{t\geq 0}$.
Finally we return to the semigroup $\\{U(t)\\}_{t\geq 0}$ generated by the
system (6)–(8) and we show that it has the AEG property.

Most of experiments concerning microorganisms are conducted in chemostats,
where cells can be grown in a physiological steady state under constant
environmental conditions. Then cells are removed from the system with the
outflow with rate $D$. In this case we add to the right-hand side of equation
(6) the term $-Du(t,x_{b},a)$. Similarly, if cells die with rate
$d(t,x_{b},a)$, then we add to the right-hand side of (6) the term
$-d(t,x_{b},a)u(t,x_{b},a)$. One can consider more advanced models with
cellural competition, but in this case all functions $q$, $g$, and $d$ can
depend also on the total number of cells and we do not investigate such
models.

##  3\. A semigroup approach

It is convenient to substitute $z(t,x_{b},a)=u(t,x_{b},a)\Psi(x_{b},a)$ and
$z_{0}(x_{b},a)=u_{0}(x_{b},a)\Psi(x_{b},a)$ in (6)–(8). Then the system
(6)–(8) takes the form

(11) $\displaystyle\frac{\partial z}{\partial t}(t,x_{b},a)+\frac{\partial
z}{\partial a}(t,x_{b},a)=0,\quad a<\overline{a}(x_{b}),$ (12) $\displaystyle
z(t,x_{b},0)=2\int_{0}^{\infty}\Big{(}P_{a}\big{(}q(\cdot,a)z(t,\cdot,a)\big{)}\Big{)}(x_{b})\,da,$
(13) $\displaystyle z(0,x_{b},a)=z_{0}(x_{b},a).$

Observe that $z_{0}$ is an integrable function. Since the expression on the
right-hand side of (12) will be used quite often, so instead of it, we will
use the shortened notation $\mathcal{P}z(t,x_{b})$. Thus, equation (12) takes
the form $z(t,x_{b},0)=\mathcal{P}z(t,x_{b})$. We also use a simplified
notation $\mathcal{P}f(x_{b})$ for the expression
$2\int_{0}^{\infty}\Big{(}P_{a}\big{(}q(\cdot,a)f(\cdot,a)\big{)}\Big{)}(x_{b})\,da$.

We consider the solutions of (11)–(13) as continuous functions
$z\colon[0,\infty)\to E$ defined by $z(t)(x_{b},a)=z(t,x_{b},a)$, where $z(t)$
is the solution of the evolution

$z^{\prime}(t)=\mathcal{A}z(t),\quad z(0)=z_{0},$

with the operator

$\mathcal{A}f(x_{b},a)=-\frac{\partial f}{\partial a}(x_{b},a)$

which has the domain

$\mathcal{D}(\mathcal{A})=\Big{\\{}f\in E,\quad\frac{\partial f}{\partial
a}\in E,\quad f(x_{b},0)=\mathcal{P}f(x_{b})\Big{\\}}.$

Since a function $f\in E$ is only almost everywhere defined we need to clarify
the formula for $f(x_{b},0)$. Consider the (partial) Sobolev space

$W_{1}(X)=\Big{\\{}f\in E\colon\frac{\partial f}{\partial a}\in E\Big{\\}}$

with the norm $\|f\|_{W_{1}(X)}=\|f\|_{E}+\Big{\|}\dfrac{\partial f}{\partial
a}\Big{\|}_{E}$. In the space $W_{1}(X)$ we introduce a trace operator
$\mathcal{T}\colon W_{1}(X)\to L^{1}[\underline{x}_{b},\overline{x}_{b}]$,
$\mathcal{T}f(x_{b})=f(x_{b},0)$, in the following way. First we show that
there exists a constant $c>0$ such that for each function $f\in W_{1}(X)\cap
C(X)$ we have

$\|\mathcal{T}f\|_{L^{1}[\underline{x}_{b},\overline{x}_{b}]}\leq
c\|f\|_{W_{1}(X)}$

(see [11] Theorem 1, Chapter 5.5). Since the set $W_{1}(X)\cap C(X)$ is dense
in $W_{1}(X)$ we can extend $\mathcal{T}$ uniquely to a linear bounded
operator on the whole space $W_{1}(X)$.

######  Proposition 2.

The operator $\mathcal{A}$ with the domain

$\mathcal{D}(\mathcal{A})=\big{\\{}f\in
W_{1}(X)\colon\mathcal{T}f=\mathcal{P}f\big{\\}}$

generates a positive $C_{0}$-semigroup $\\{T(t)\\}_{t\geq 0}$ on $E$.

We recall that a family $\\{T(t)\\}_{t\geq 0}$ of linear operators on a Banach
space $E$ is a $C_{0}$-semigroup or strongly continuous semigroup if it
satisfies the following conditions:

  1. (a)

$T(0)=I$, i.e., $T(0)f=f$ for $f\in E$,

  2. (b)

$T(t+s)=T(t)T(s)\quad\textrm{for}\quad s,\,t\geq 0$,

  3. (c)

for each $f\in E$ the function $t\mapsto T(t)f$ is continuous.

The proof of this result is based on a perturbation method related to
operators with boundary conditions developed in [15] and an extension of this
method to unbounded perturbations in $L^{1}$ space in [16] (see Theorem 3
below).

######  Theorem 3.

Let $(\Gamma,\Sigma,m)$, $(\Gamma_{\partial},\Sigma_{\partial},m_{\partial})$
be $\sigma$-finite measure spaces and let $L^{1}=L^{1}(\Gamma,\Sigma,m)$ and
$L_{\partial}^{1}=L^{1}(\Gamma_{\partial},\Sigma_{\partial},m_{\partial})$.
Let $\mathcal{D}$ be a linear subspace of $L^{1}$. We assume that
$A\colon\mathcal{D}\to L^{1}$ and $\Upsilon_{0},\Upsilon\colon\mathcal{D}\to
L_{\partial}^{1}$ are linear operators satisfying the following conditions:

  1. (1)

for each $\lambda>0$, the operator $\Upsilon_{0}\colon\mathcal{D}\to
L^{1}_{\partial}$ restricted to the nullspace $\mathcal{N}(\lambda
I-A)=\\{f\in\mathcal{D}\colon\lambda f-Af=0\\}$ has a positive right inverse
$\Upsilon(\lambda)\colon L^{1}_{\partial}\to\mathcal{N}(\lambda I-A)$, i.e.
$\Upsilon_{0}\Upsilon(\lambda)f_{\partial}=f_{\partial}$ for $f_{\partial}\in
L^{1}_{\partial}$;

  2. (2)

the operator $\Upsilon\colon\mathcal{D}\to L^{1}_{\partial}$ is positive and
there is $\omega>0$ such that $\|\Upsilon\Upsilon(\lambda)\|<1$ for
$\lambda>\omega$;

  3. (3)

the operator $A_{0}=A\big{|}_{\mathcal{D}(A_{0})}$, where
$\mathcal{D}(A_{0})=\\{f\in\mathcal{D}\colon\Upsilon_{0}f=0\\}$, generates a
positive $C_{0}$-semigroup on $L^{1}$;

  4. (4)

$\int_{\Gamma}Af(x)\,m(dx)\leq\int_{\Gamma_{\partial}}\Upsilon_{0}f(x_{\partial})\,m_{\partial}(dx_{\partial})$
for $f\in\mathcal{D}_{+}=\\{f\in\mathcal{D}\colon f\geq 0\\}$.

Then $A$ with the domain
$\mathcal{D}(A)=\\{f\in\mathcal{D}\colon\Upsilon_{0}f=\Upsilon f\\}$ generates
a positive semigroup on $L^{1}$.

###### Proof of Proposition 2.

First we translate our notation to that from Theorem 3. Let $\Gamma=X$,
$\Gamma_{\partial}=[\underline{x}_{b},\overline{x}_{b}]$, $L^{1}=E$,
$L_{\partial}^{1}=L^{1}[\underline{x}_{b},\overline{x}_{b}]$,
$\Upsilon_{0}=\mathcal{T}$, $\Upsilon=\mathcal{P}$, $Af=-\frac{\partial
f}{\partial a}$ and $\mathcal{D}=\Big{\\{}f\in E\colon\,\,\frac{\partial
f}{\partial a}\in E\\}$.

(1): Since $Af=-\frac{\partial f}{\partial a}$, the nullspace
$\mathcal{N}(\lambda I-A)$ is the set of functions $f\in\mathcal{D}$
satisfying equation

$\frac{\partial f}{\partial a}+\lambda f=0.$

Solving this equation we obtain that $f(x_{b},a)=f(x_{b},0)e^{-\lambda a}$.
Thus the operator $\Upsilon_{0}$ restricted to $\mathcal{N}(\lambda I-A)$ is
invertible and the inverse operator $\Upsilon(\lambda)\colon
L^{1}[\underline{x}_{b},\overline{x}_{b}]\to\mathcal{N}(\lambda I-A)$ given by
$\Upsilon(\lambda)f(x_{b},a)=f(x_{b})e^{-\lambda a}$ is positive.

(2): Since $\Upsilon=\mathcal{P}$ we check that
$\|\mathcal{P}\Upsilon(\lambda)\|<1$ for $\lambda>\omega=\ln 2/\underline{a}$.
Take $f\in L^{1}[\underline{x}_{b},\overline{x}_{b}]$, $f\geq 0$, and let
$\Theta_{\lambda}(x_{b},a)=2e^{-\lambda a}q(x_{b},a)$. Then

$\displaystyle\int_{\underline{x}_{b}}^{\overline{x}_{b}}(\mathcal{P}\Upsilon(\lambda)f)(x_{b})\,dx_{b}$
$\displaystyle=\int_{\underline{x}_{b}}^{\overline{x}_{b}}\int_{0}^{\infty}P_{a}(f(\cdot)\Theta_{\lambda}(\cdot,a))(x_{b})\,da\,dx_{b}$
$\displaystyle\leq\int_{\underline{x}_{b}}^{\overline{x}_{b}}\int_{0}^{\infty}f(x_{b})\Theta_{\lambda}(x_{b},a)\,da\,dx_{b}.$

Since

$\int_{0}^{\infty}\Theta_{\lambda}(x_{b},a)\,da=\int_{\underline{a}}^{\overline{a}}2e^{-\lambda
a}q(x_{b},a)\,da\leq 2e^{-\lambda\underline{a}}<1$

for $\lambda>\omega=\ln 2/\underline{a}$, we have
$\|\mathcal{P}\Upsilon(\lambda)\|<1$ for $\lambda>\omega$.

(3): The operator $A_{0}$ generates a positive $C_{0}$-semigroup
$\\{T_{0}(t)\\}_{t\geq 0}$ on $E$ given by

$T_{0}(t)f(x_{b},a)=\begin{cases}f(x_{b},a-t)\quad\textrm{for $a>t$,}\\\
0\quad\textrm{for $a<t$}.\end{cases}$

(4): If $f\in\mathcal{D}_{+}$ then

$\int_{X}Af(x_{b},a)\,dx_{b}\,da=-\int_{X}\frac{\partial f}{\partial
a}(x_{b},a)\,da\,dx_{b}=\int_{\underline{x}_{b}}^{\overline{x}_{b}}\mathcal{T}f(x_{b})\,dx_{b}.$

Since $\Upsilon_{0}=\mathcal{T}$, $\Gamma=X$, and
$\Gamma_{\partial}=[\underline{x}_{b},\overline{x}_{b}]$ we have

$\int_{X}Af(x_{b},a)\,dx_{b}\,da-\int_{\underline{x}_{b}}^{\overline{x}_{b}}\Upsilon_{0}f(x_{b},a)\,dx_{b}=0.\qed$

######  Remark 4.

It is not difficult to check that the resolvent
$R(\lambda,\mathcal{A})=(\lambda I-\mathcal{A})^{-1}$ exists for sufficiently
large $\lambda>0$ and it is given by the formula

(14) $R(\lambda,\mathcal{A})=(I-\mathcal{P}_{\lambda})^{-1}R(\lambda,A_{0}),$

where

$\mathcal{P}_{\lambda}f(x_{b},a)=e^{-\lambda a}\mathcal{P}f(x_{b}),\quad
R(\lambda,A_{0})f(x_{b},a)=\int_{0}^{a}f(x_{b},r)e^{\lambda(r-a)}\,dr$

for $f\in E$. Another proof of Proposition 2 can be done directly using
formula (14) and the Hille–Yosida theorem.

######  Remark 5.

If $z_{0}\in\mathcal{D}(\mathcal{A})$ and $z(t)=T(t)z_{0}$, then
$z(t)\in\mathcal{D}(\mathcal{A})$, $z^{\prime}(t)$ exists and
$z^{\prime}(t)=\mathcal{A}z(t)$. Assume that the function $z_{0}$ and the
derivative $\dfrac{\partial z_{0}}{\partial a}$ are continuous bounded
functions and the consistency condition
$z_{0}(x_{b},0)=\mathcal{P}z_{0}(x_{b})$ holds. Then the problem (11)–(13) has
a unique classical solution. By the classical solution we understand a
continuous function $z$, which has continuous derivatives $\dfrac{\partial
z}{\partial a}$ and $\dfrac{\partial z}{\partial t}$ outside the set
$\mathcal{Z}=\\{(t,x_{b},a)\colon a=t,\,(x_{b},a)\in X\\}$, $z$ satisfies (11)
outside $\mathcal{Z}$, and $z$ satisfies conditions (12)–(13). In this case
$z_{0}\in\mathcal{D}(\mathcal{A})$ and $z(t,x_{b},a)=T(t)z_{0}(x_{b},a)$, i.e.
classical and “semigroup” solutions are identical.

##  4\. Eigenvectors of $\mathcal{A}$ and $\mathcal{A}^{*}$

Our aim is to study the long-time behaviour of the semigroup
$\\{T(t)\\}_{t\geq 0}$. The strategy is the following. First we check that the
adjoint operator $\mathcal{A}^{*}$ of $\mathcal{A}$ has a positive eigenvector
$v=v(x_{b},a)$ corresponding to some positive eigenvalue $\lambda$. Then we
introduce the semigroup $\\{P(t)\\}_{t\geq 0}$ given by $P(t)f=e^{-\lambda
t}T(t)f$ defined on the space $E_{1}=L^{1}(X,\mathcal{B}(X),\mu)$ with the
measure $\mu$ given by $d\mu=v\,d\ell$. Then we prove that semigroup
$\\{P(t)\\}_{t\geq 0}$ has an invariant density $f_{i}$ and
$\lim_{t\to\infty}P(t)f=f_{i}$ for each density $f$. Finally, we translate
this result in terms of the semigroup $\\{T(t)\\}_{t\geq 0}$.

We first study some properties of the adjoint operator of $\mathcal{A}$.
Denote by $H$ the operator $H\colon E\to
L^{1}[\underline{x}_{b},\overline{x}_{b}]$ defined by
$Hf(x_{b})=\mathcal{P}f(x_{b})$. Then the operator $H^{*}\colon
L^{\infty}[\underline{x}_{b},\overline{x}_{b}]\to E^{*}$ is given by

(15)
$H^{*}f(x_{b},a)=2q(x_{b},a)P^{*}_{a}f(x_{b})=2q(x_{b},a)f(S_{a}(x_{b})).$

It should be noted that we omit in the last product the factor
$\mathbf{1}_{[x_{a},\overline{x}_{b}]}(x_{b})$ because $q(x_{b},a)=0$ for
$x_{b}\leq x_{a}$.

######  Lemma 6.

Let

$\mathcal{D}^{\odot}=\\{f\in C(X),\,\,\,\frac{\partial f}{\partial a}\in
C(X),\,\,\,f(x_{b},\overline{a}(x_{b}))=0,\,\,\,\frac{\partial f}{\partial
a}(x_{b},\overline{a}(x_{b}))=0\\}.$

Then $\mathcal{D}^{\odot}\subset\mathcal{D}(\mathcal{A}^{*})$ and

(16) $\mathcal{A}^{*}f=\frac{\partial f}{\partial
a}+H^{*}{\tilde{f}}\quad\textrm{for $f\in\mathcal{D}^{\odot}$},$

where $\tilde{f}(x_{b})=f(x_{b},0)$.

###### Proof.

If $f\in\mathcal{D}^{\odot}$ and $\varphi\in\mathcal{D}(\mathcal{A})$ then

$\displaystyle\langle f,\mathcal{A}\varphi\rangle$
$\displaystyle=-\int_{\underline{x}_{b}}^{\overline{x}_{b}}\int_{0}^{\overline{a}(x_{b})}f(x_{b},a)\frac{\partial\varphi}{\partial
a}(x_{b},a)\,da\,dx_{b}$
$\displaystyle=\int_{\underline{x}_{b}}^{\overline{x}_{b}}f(x_{b},0)\varphi(x_{b},0)\,dx_{b}+\Big{\langle}\frac{\partial
f}{\partial a},\varphi\Big{\rangle}$
$\displaystyle=\int_{\underline{x}_{b}}^{\overline{x}_{b}}f(x_{b},0)(H\varphi)(x_{b})\,dx_{b}+\Big{\langle}\frac{\partial
f}{\partial a},\varphi\Big{\rangle}$ $\displaystyle=\langle
H^{*}\tilde{f},\varphi\rangle+\Big{\langle}\frac{\partial f}{\partial
a},\varphi\Big{\rangle}=\Big{\langle}\frac{\partial f}{\partial
a}+H^{*}\tilde{f},\varphi\Big{\rangle}.$

Thus $\mathcal{D}^{\odot}\subset\mathcal{D}(\mathcal{A}^{*})$ and (16) holds.
∎

Now we will check that the operator $\mathcal{A}^{*}$ has a positive
eigenvector $v\in\mathcal{D}^{\odot}$ for some eigenvalue $\lambda>0$. Let
$\mathcal{A}^{*}v=\lambda v$. From (16) it follows that

(17) $\lambda v-\frac{\partial v}{\partial a}=H^{*}{\tilde{v}}.$

The problem is that the adjoint semigroup $\\{T^{*}(t)\\}_{t\geq 0}$ is not
continuous. Instead of this semigroup we may use the sun dual semigroup, but
it will be more convenient to consider a little modification of the sun dual
semigroup. Consider a semigroup $\\{T^{\odot}(t)\\}_{t\geq 0}$ defined on the
space

$\widetilde{C}(X)=\\{f\in C(X)\colon\,\,\,f(x_{b},\overline{a}(x_{b}))=0\\}$

with the infinitesimal generator $\mathcal{A}^{\odot}$ with the domain
$\mathcal{D}(\mathcal{A}^{\odot})=D^{\odot}$ and given by the same formula as
$\mathcal{A}^{*}$. Let $\mathcal{A}^{\odot}_{0}f=\frac{\partial f}{\partial
a}$ and
$\mathcal{D}(\mathcal{A}^{\odot}_{0})=\mathcal{D}(\mathcal{A}^{\odot})$. Then
$\mathcal{A}_{0}^{\odot}$ is the infinitesimal generator of a
$C_{0}$-semigroup $\\{T^{\odot}_{0}(t)\\}_{t\geq 0}$ on the space
$\widetilde{C}(X)$ given by the formula

$T^{\odot}_{0}(t)f(x_{b},a)=\begin{cases}f(x_{b},a+t)&\textrm{for
$a\leq\overline{a}(x_{b})-t$},\\\ 0&\textrm{for
$a>\overline{a}(x_{b})-t$}.\end{cases}$

Let $H^{\odot}\colon C[\underline{x}_{b},\overline{x}_{b}]\to\widetilde{C}(X)$
be given by $H^{\odot}f(x_{b},a)=2q(x_{b},a)f(S_{a}(x_{b}))$. Then
$\mathcal{A}^{\odot}f=\mathcal{A}^{\odot}_{0}f+H^{\odot}\tilde{f}$ for
$f\in\mathcal{D}(\mathcal{A}^{\odot})$. Denote by
$R(\lambda,\mathcal{A}^{\odot}_{0})$ the resolvent of the operator
$\mathcal{A}^{\odot}_{0}$.

######  Lemma 7.

Let $K_{\lambda}\colon C[\underline{x}_{b},\overline{x}_{b}]\to
C[\underline{x}_{b},\overline{x}_{b}]$, $\lambda\geq 0$, be the integral
operator given by

(18)
$K_{\lambda}\tilde{v}(x_{b})=\int_{\underline{a}(x_{b})}^{\overline{a}(x_{b})}2e^{-\lambda
a}q(x_{b},a)\tilde{v}(S_{a}(x_{b}))\,da.$

If $\tilde{v}$ is a function such that $K_{\lambda}\tilde{v}=\tilde{v}$, then
the function

(19)
$v(x_{b},a)=\int_{a}^{\infty}H^{\odot}\tilde{v}(x_{b},s)e^{-\lambda(s-a)}\,ds.$

satisfies (17) and $v\in\mathcal{D}(\mathcal{A}^{\odot})$.

###### Proof.

If $v\in\widetilde{C}(X)$ satisfies the equation

(20) $v=R(\lambda,\mathcal{A}^{\odot}_{0})H^{\odot}{\tilde{v}}$

then $v$ also satisfies (17). Since
$R(\lambda,\mathcal{A}^{\odot}_{0})f=\int_{0}^{\infty}e^{-\lambda
s}T^{\odot}_{0}(s)f\,ds$ we have

$R(\lambda,\mathcal{A}^{\odot}_{0})f(x_{b},a)=\int_{a}^{\infty}f(x_{b},s)e^{-\lambda(s-a)}\,ds.$

Now (20) can be written as the integral equation (19). Observe that in order
to find $v(x_{b},a)$ it is enough to solve (19) for $a=0$. Equation (19) for
$a=0$ takes the form

$v(x_{b},0)=\int_{0}^{\infty}H^{\odot}\tilde{v}(x_{b},s)e^{-\lambda s}\,ds.$

In the above formula we replace $s$ by $a$ and apply (15). Then

$v(x_{b},0)=\int_{\underline{a}(x_{b})}^{\overline{a}(x_{b})}2e^{-\lambda
a}q(x_{b},a)v(S_{a}(x_{b}),0)\,da.$

Thus if $K_{\lambda}\tilde{v}=\tilde{v}$, then $v$ given by (19) satisfies
(17). Since $v$ belongs to the range of resolvent
$R(\lambda,\mathcal{A}^{\odot}_{0})$, we have
$v\in\mathcal{D}(\mathcal{A}^{\odot}_{0})=\mathcal{D}(\mathcal{A}^{\odot})$. ∎

We want to prove that there exists a constant $\lambda>0$ and a positive
function $\tilde{v}\in C[\underline{x}_{b},\overline{x}_{b}]$ such that
$K_{\lambda}\tilde{v}=\tilde{v}$. We split the proof of this fact into two
lemmae.

######  Lemma 8.

For each $\lambda\geq 0$ the spectral radius $r(K_{\lambda})$ of $K_{\lambda}$
is a positive, isolated and simple eigenvalue of $K_{\lambda}$ associated with
a positive eigenfunction $\tilde{v}_{\lambda}\in
C[\underline{x}_{b},\overline{x}_{b}]$.

###### Proof.

In order to check this property we write the operator $K_{\lambda}$ in the
standard integral form. We substitute $y(a)=S_{a}(x_{b})$ in (18). Then
$da=2\,dy/g(2y)$ and we find that the operator $K_{\lambda}$ can be written in
the form

$K_{\lambda}\tilde{v}(x_{b})=\int_{\underline{x}_{b}}^{\overline{x}_{b}}k_{\lambda}(x_{b},y)\tilde{v}(y)\,dy,\quad
k_{\lambda}(x_{b},y)=\frac{4e^{-\lambda
a(y;x_{b})}q(x_{b},a(y;x_{b}))}{g(2y)},$

where

$a(y;x_{b})=\int_{x_{b}}^{2y}\frac{dr}{g(r)}.$

The expression $a(y;x_{b})$ has the following interpretation. If $x_{b}$ is
the initial size of a mother cell and it splits at the age $a(y;x_{b})$, then
$y$ is the initial size of its daughter cells. Since the function $g$ is
continuous and positive, the kernel $k_{\lambda}$ is continuous and
nonnegative. Moreover, $k_{\lambda}(x_{b},x_{b})>0$ for all
$x_{b}\in(\underline{x}_{b},\overline{x}_{b})$. Indeed,
$k_{\lambda}(x_{b},x_{b})>0$ if and only if $q(x_{b},a(x_{b};x_{b}))>0$. The
last inequality follows from (A6) (see Fig. 2). This implies that the spectral
radius $r(K_{\lambda})$ is a positive, isolated and simple eigenvalue of
$K_{\lambda}$ associated with an eigenfunction $\tilde{v}_{\lambda}\in
C[\underline{x}_{b},\overline{x}_{b}]$ such that
$\tilde{v}_{\lambda}(x_{b})>0$ for
$x_{b}\in(\underline{x}_{b},\overline{x}_{b})$ (see comments after the proof
of Theorem 7.4 of [2]). Observe that $\tilde{v}_{\lambda}$ is also positive at
$\underline{x}_{b}$ and $\overline{x}_{b}$, because $\tilde{v}_{\lambda}(y)>0$
for $y\in(\underline{x}_{b},\overline{x}_{b})$ and the functions $y\mapsto
q(\underline{x}_{b},a(y;\underline{x}_{b}))$ and $y\mapsto
q(\overline{x}_{b},a(y;\overline{x}_{b}))$ are positive on some nontrivial
intervals. ∎

######  Lemma 9.

There exists $\lambda>0$ such that $r(K_{\lambda})=1$.

###### Proof.

First we check that the function $\lambda\mapsto K_{\lambda}$ is continuous
with respect to the operator norm. Indeed, let $\lambda_{1}\leq\lambda_{2}$
and $T_{\lambda_{1},\lambda_{2}}=K_{\lambda_{1}}-K_{\lambda_{2}}$. Since $\int
q(x_{b},a)\,da=1$ and
$e^{-\lambda_{1}a}-e^{-\lambda_{2}a}\leq\overline{a}(\lambda_{2}-\lambda_{1})$,
we have

$\|T_{\lambda_{1},\lambda_{2}}\|\leq\max\limits_{\underline{x}_{b}\leq
x_{b}\leq\overline{x}_{b}}\int_{\underline{a}(x_{b})}^{\overline{a}(x_{b})}2\big{(}e^{-\lambda_{1}a}-e^{-\lambda_{2}a}\big{)}q(x_{b},a)\,da\leq
2\overline{a}(\lambda_{2}-\lambda_{1}).$

Since the function $k_{\lambda}$ is continuous, the operator
$K_{\lambda}\colon C[\underline{x}_{b},\overline{x}_{b}]\to
C[\underline{x}_{b},\overline{x}_{b}]$ is compact for each $\lambda\geq 0$.
The spectral radius mapping restricted to compact linear bounded operators on
any Banach space is continuous with respect to the operator norm (see e.g.
Theorem 2.1 of [8]). Thus the function $\lambda\mapsto r(K_{\lambda})$ is
continuous. Observe that, $r(K_{0})=2$, because $\|K_{0}\|=2$ and
$K_{0}\mathbf{1}_{[\underline{x}_{b},\overline{x}_{b}]}=2\cdot\mathbf{1}_{[\underline{x}_{b},\overline{x}_{b}]}$.
Now, let $\bar{\lambda}>0$ be a constant such that
$e^{-\bar{\lambda}\underline{a}}\leq 1/4$. Then

$K_{\bar{\lambda}}\tilde{v}(x_{b})\leq\frac{1}{2}\int_{\underline{a}(x_{b})}^{\overline{a}(x_{b})}q(x_{b},a)\tilde{v}(S_{a}(x_{b}))\,da\leq\frac{1}{2}\|\tilde{v}\|\quad\textrm{for
$\tilde{v}\geq 0$,}$

and consequently
$r(K_{\bar{\lambda}})\leq\|K_{\bar{\lambda}}\|\leq\frac{1}{2}$. From the
continuity of the function $\lambda\mapsto r(K_{\lambda})$ it follows that
there exists a $\lambda\in(0,\bar{\lambda})$ such that $r(K_{\lambda})=1$. ∎

Now we apply formula (19) to find a nonnegative eigenfunction $v(x_{b},a)$ of
the operator $\mathcal{A}^{*}$.

######  Proposition 10.

The operator $\mathcal{A}^{*}$ has an eigenvalue $\lambda>0$ and a
corresponding eigenfunction $v$ such that

(21) $c_{1}\Phi(x_{b},a)\leq v(x_{b},a)\leq c_{2}\Phi(x_{b},a)$

for some positive constants $c_{1}$ and $c_{2}$ independent of $x_{b}$ and
$a$.

###### Proof.

According to Lemma 9 there exists $\lambda>0$ such that $r(K_{\lambda})=1$.
Let $\tilde{v}_{\lambda}$ be a positive fixed point of $K_{\lambda}$. Then
from formulae (15) and (19) it follows that

$v(x_{b},a)=\int_{a}^{\infty}2q(x_{b},s)\tilde{v}_{\lambda}(S_{s}(x_{b}))e^{-\lambda(s-a)}\,ds$

is the eigenfunction of the operator $\mathcal{A}^{*}$ corresponding to
$\lambda$. Since the functions $\tilde{v}_{\lambda}(S_{s}(x_{b}))$ and
$e^{-\lambda(s-a)}$ are bounded above and bounded away from zero, there exist
positive constants $c_{1}$ and $c_{2}$ such that

$c_{1}\int_{a}^{\infty}q(x_{b},s)\,ds\leq v(x_{b},a)\leq
c_{2}\int_{a}^{\infty}q(x_{b},s)\,ds$

and (21) follows from the definition of $\Phi$. ∎

From now on $\lambda$ denotes the eigenvalue from Proposition 10.

######  Lemma 11.

A function $f_{i}(x_{b},a)$ is an eigenvector of $\mathcal{A}$ corresponding
to $\lambda$ if and only if

(22) $f_{i}(x_{b},a)=e^{-\lambda a}f_{i}(x_{b},0)\quad\textrm{for}\quad
a\leq\overline{a}(x_{b})$

and the function $f(x_{b})=f_{i}(x_{b},0)$ satisfies the equation $Jf=f$,
where the operator $J\colon L^{1}[\underline{x}_{b},\overline{x}_{b}]\to
L^{1}[\underline{x}_{b},\overline{x}_{b}]$ is given by the formula

(23) $Jf(x_{b})=\int_{\underline{x}_{b}}^{2x_{b}}2e^{-\lambda
a(x_{b};y)}q(y,a(x_{b};y))f(y)\,dy.$

###### Proof.

If a function $f_{i}(x_{b},a)$ is an eigenvector of $\mathcal{A}$
corresponding to $\lambda$, then the function $z(t,x_{b},a)=e^{\lambda
t}f_{i}(x_{b},a)$ is a solution of (11)–(13). Substituting $z=e^{\lambda
t}f_{i}$ into (11)–(12) we obtain

(24) $\lambda f_{i}(x_{b},a)+\frac{\partial f_{i}}{\partial
a}(x_{b},a)=0,\quad f_{i}(x_{b},0)=\mathcal{P}f_{i}(x_{b}).$

From the first of equations (24) it follows that the function
$(x_{b},a)\mapsto e^{\lambda a}f_{i}(x_{b},a)$ has zero partial derivative
with respect to $a$. Thus $e^{\lambda a}f_{i}(x_{b},a)=f_{i}(x_{b},0)$, where
$f_{i}(x_{b},0)$ is the value of the trace operator $\mathcal{T}$ on $f_{i}$.
Therefore (22) holds and the function $x_{b}\mapsto f_{i}(x_{b},0)$ satisfies
the following integral equation

(25) $f_{i}(x_{b},0)=\int_{0}^{\infty}2e^{-\lambda
a}\Big{(}P_{a}\big{(}q(\cdot,a)f_{i}(\cdot,0)\big{)}\Big{)}(x_{b})\,da.$

Since

$\Big{(}P_{a}\big{(}q(\cdot,a)f(\cdot)\big{)}\Big{)}(x_{b})=\frac{d}{dx_{b}}\big{(}S_{a}^{-1}(x_{b})\big{)}q\big{(}S_{a}^{-1}(x_{b}),a\big{)}f\big{(}S_{a}^{-1}(x_{b})\big{)},$

the substitution $y=S_{a}^{-1}(x_{b})$ to (25) gives (23). ∎

From (23) it follows that $J$ is an integral operator with a continuous
kernel. In particular $Jf\in C[\underline{x}_{b},\overline{x}_{b}]$ for $f\in
L^{1}[\underline{x}_{b},\overline{x}_{b}]$ and the operator $J$ restricted to
$C[\underline{x}_{b},\overline{x}_{b}]$ is a continuous and positive.

The operators $J$ and $K_{\lambda}$ are adjoint, i.e.

$\int_{\underline{x}_{b}}^{\overline{x}_{b}}g(x_{b})Jf(x_{b})\,dx_{b}=\int_{\underline{x}_{b}}^{\overline{x}_{b}}K_{\lambda}g(x_{b})f(x_{b})\,dx_{b}$

for $f\in L^{1}[\underline{x}_{b},\overline{x}_{b}]$, $g\in
L^{\infty}[\underline{x}_{b},\overline{x}_{b}]$.

######  Lemma 12.

There exists $\tilde{f}_{i}\in C[\underline{x}_{b},\overline{x}_{b}]$ such
that $J\tilde{f}_{i}=\tilde{f}_{i}$ and $\tilde{f}_{i}(x_{b})>0$ for
$x_{b}\in(\underline{x}_{b},\overline{x}_{b})$. The function $\tilde{f}_{i}$
is the unique, up to a multiplicative constant, fixed point of $J$.

###### Proof.

Using the same arguments as in Lemma 8 we prove that there exists an
eigenfunction $\tilde{f}_{i}\in C[\underline{x}_{b},\overline{x}_{b}]$ of $J$
such that $\tilde{f}_{i}(x_{b})>0$ for
$x_{b}\in(\underline{x}_{b},\overline{x}_{b})$. This eigenfunction is indeed a
fixed point of $J$ because $\langle
J\tilde{f}_{i},\tilde{v}_{\lambda}\rangle=\langle\tilde{f}_{i},K_{\lambda}\tilde{v}_{\lambda}\rangle=\langle\tilde{f}_{i},\tilde{v}_{\lambda}\rangle$.
Since $r=1$ is an isolated and simple eigenvalue of $J$, the function
$\tilde{f}_{i}$ is the unique, up to a multiplicative constant, fixed point of
$J$. ∎

######  Remark 13.

It is generally not true that $\tilde{f}_{i}(\underline{x}_{b})>0$ and
$\tilde{f}_{i}(\overline{x}_{b})>0$. If we assume additionally that
$S_{\underline{a}(x_{b})}(x_{b})=\underline{x}_{b}$ for
$x_{b}\in[\underline{x}_{b},\underline{x}_{b}+\delta]$, $\delta>0$, then
$\tilde{f}_{i}(\underline{x}_{b})>0$ because a mother cell with the initial
size $x_{b}\in[\underline{x}_{b},\underline{x}_{b}+\delta]$ can have a
daughter cell with the initial size $\underline{x}_{b}$. Analogously, if
$S_{\overline{a}(x_{b})}(x_{b})=\overline{x}_{b}$ for
$x_{b}\in[\overline{x}_{b}-\delta,\overline{x}_{b}]$, $\delta>0$, then
$\tilde{f}_{i}(\overline{x}_{b})>0$.

From Lemma 12 and from formulae (22) and (25) we have

######  Proposition 14.

Let $\tilde{f}_{i}$ be the function from Lemma 12. If
$f_{i}(x_{b},a)=e^{-\lambda a}\tilde{f}_{i}(x_{b})$, then
$\mathcal{A}f_{i}=\lambda f_{i}$. The function $f_{i}$ is the unique, up to a
multiplicative constant, eigenfunction of $\mathcal{A}$ corresponding to the
eigenvalue $\lambda$.

##  5\. Asymptotic behaviour

We precede the formulation of the main result of this section by some
definitions and some general theorem concerning asymptotic stability of
stochastic semigroups.

Let a triple $(X,\Sigma,m)$ be a $\sigma$-finite measure space. Denote by $D$
the subset of the space $L^{1}=L^{1}(X,\Sigma,m)$ which contains all densities

$D=\\{f\in\,L^{1}\colon\,f\geq 0,\,\,\|f\|=1\\}.$

A $C_{0}$-semigroup $\\{P(t)\\}_{t\geq 0}$ of linear operators on $L^{1}$ is
called stochastic semigroup or Markov semigroup if $P(t)(D)\subseteq D$ for
each $t\geq 0$.

A stochastic semigroup $\\{P(t)\\}_{t\geq 0}$ is asymptotically stable if
there exists a density $f_{i}$ such that

(26) $\lim_{t\to\infty}\|P(t)f-f_{i}\|=0\quad\text{for}\quad f\in D.$

From (26) it follows immediately that $f_{i}$ is invariant  with respect to
$\\{P(t)\\}_{t\geq 0}$, i.e. $P(t)f_{i}=f_{i}$ for each $t\geq 0$.

A stochastic semigroup $\\{P(t)\\}_{t\geq 0}$ is called partially integral if
there exists a measurable function $k\colon(0,\infty)\times X\times
X\to[0,\infty)$, called a kernel, such that

$P(t)f(x)\geq\int_{X}k(t,x,y)f(y)\,m(dy)$

for every density $f$ and

$\int_{X}\int_{X}k(t,x,y)\,m(dx)\,m(dy)>0$

for some $t>0$. The following result was proved in [35].

######  Theorem 15.

Let $\\{P(t)\\}_{t\geq 0}$ be a partially integral stochastic semigroup.
Assume that the semigroup $\\{P(t)\\}_{t\geq 0}$ has a unique invariant
density $f_{i}$. If $f_{i}>0$ a.e., then the semigroup $\\{P(t)\\}_{t\geq 0}$
is asymptotically stable.

New results concerning positive operators on Banach lattices similar in spirit
to Theorem 15 may be found in [13, 14].

Investigation of the long-time behaviour of the semigroup $\\{T(t)\\}_{t\geq
0}$ can be reduced to the study of asymptotic stability of some stochastic
semigroup. Let $\lambda$ and $v$ be the eigenvalue and the eigenfunction from
Proposition 10. We define a semigroup $\\{P(t)\\}_{t\geq 0}$ as the extension
of semigroup $\\{e^{-\lambda t}T(t)\\}_{t\geq 0}$ to the space
$E_{1}=L^{1}(X,\mathcal{B}(X),\mu)$ with measure $\mu$ given by
$d\mu=v\,d\ell$. Observe that we can indeed extend the semigroup
$\\{e^{-\lambda t}T(t)\\}_{t\geq 0}$ to a stochastic semigroup on $E_{1}$.
Since $\mathcal{A}^{*}v=\lambda v$, we have $T^{*}(t)v=e^{\lambda t}v$. If
$f\in E$ then $P(t)f=e^{-\lambda t}T(t)f$ and

$\displaystyle\iint\limits_{X}P(t)f(x_{b},a)\,\mu(dx_{b},da)=\iint\limits_{X}e^{-\lambda
t}T(t)f(x_{b},a)v(x_{b},a)\,dx_{b}\,da$
$\displaystyle=\iint\limits_{X}f(x_{b},a)e^{-\lambda
t}T^{*}(t)v(x_{b},a)\,dx_{b}\,da=\iint\limits_{X}f(x_{b},a)v(x_{b},a)\,dx_{b}\,da$
$\displaystyle=\iint\limits_{X}f(x_{b},a)\,\mu(dx_{b},da).$

Since the function $v$ is bounded and positive almost everywhere, $E$ is dense
in $E_{1}$. If $f\in E_{1}$, we choose a sequence $(f_{n})$ from $E$ such that
$f_{n}\to f$ in $E_{1}$ and define $P(t)f=\lim\limits_{n\to\infty}P(t)f_{n}$
in $E_{1}$. Since the operators $P(t)$ are positive and preserve the integral
with respect to $\mu$, this extension is uniquely defined and
$\\{P(t)\\}_{t\geq 0}$ is a stochastic semigroup on $E_{1}$.

In order to prove asymptotic stability of the semigroup $\\{P(t)\\}_{t\geq 0}$
we need to add an additional assumption concerning function $g$:

(A7) there exists $x\in(\underline{x}_{b},\overline{x}_{b})$ such that
$g(2x)\neq 2g(x)$.

We precede the formulation of a theorem on asymptotic stability of
$\\{P(t)\\}_{t\geq 0}$ by the following lemma.

######  Lemma 16.

Assume (A1)–(A7). Then the semigroup $\\{T(t)\\}_{t\geq 0}$ is partially
integral.

###### Proof.

Observe that the operator $T(t)$ has the kernel $k(t,x,y)$ if and only if the
operator $T^{*}(t)$ has the kernel $k^{*}(t,x,y)=k(t,y,x)$. Thus, in order to
prove that the semigroup $\\{T(t)\\}_{t\geq 0}$ is partially integral it is
sufficient to check that the semigroup $\\{T^{\odot}(t)\\}_{t\geq 0}$ has the
similar property. The semigroup $\\{T^{\odot}(t)\\}_{t\geq 0}$ is given by the
Dyson-Phillips expansion

(27) $T^{\odot}(t)f=\sum_{n=0}^{\infty}T^{\odot}_{n}(t)f,$

where

$T^{\odot}_{n+1}f(t)=\int_{0}^{t}T^{\odot}_{0}(\tau)\mathcal{H}T^{\odot}_{n}(t-\tau)f\,d\tau,\quad
n\geq 0,$

where $\mathcal{H}f(x_{b},a)=2q(x_{b},a)f(S_{a}(x_{b}),0)$ and
$T^{\odot}_{0}(t)f(x_{b},a)=f(x_{b},a+t)$ for $a\leq\overline{a}(x_{b})-t$.
Since
$\mathcal{H}T^{\odot}_{0}(t-\tau)f(x_{b},a)=2q(x_{b},a)f(S_{a}(x_{b}),t-\tau)$,
we have

$\displaystyle T^{\odot}_{1}f(t)(x_{b},a)$
$\displaystyle=\int_{0}^{t}T^{\odot}_{0}(\tau)\mathcal{H}T^{\odot}_{0}(t-\tau)f(x_{b},a)\,d\tau$
$\displaystyle=\int_{0}^{t}2q(x_{b},a+\tau)f(S_{a+\tau}(x_{b}),t-\tau)\,d\tau.$

Analogously, since

$T^{\odot}_{1}(t-\tau_{1})f(x_{b},a)=\int_{0}^{t-\tau_{1}}2q(x_{b},a+\tau)f(S_{a+\tau}(x_{b}),t-\tau_{1}-\tau)\,d\tau,$

we have

$\mathcal{H}T^{\odot}_{1}(t-\tau_{1})f(x_{b},a)=2q(x_{b},a)\int_{0}^{t-\tau_{1}}2q(S_{a}(x_{b}),\tau)f(S_{\tau}(S_{a}(x_{b})),t-\tau_{1}-\tau)\,d\tau,$

and finally

$\displaystyle T^{\odot}_{2}f(t)(x_{b},a)$
$\displaystyle=\int_{0}^{t}T^{\odot}_{1}(\tau_{1})\mathcal{H}T^{\odot}_{1}(t-\tau_{1})f(x_{b},a)\,d\tau_{1}$
$\displaystyle=\int_{0}^{t}2q(x_{b},a+\tau_{1})\int_{0}^{t-\tau_{1}}2q(S_{a+\tau_{1}}(x_{b}),\tau)$
$\displaystyle\hskip 85.35826pt{}\cdot
f(S_{\tau}(S_{a+\tau_{1}}(x_{b})),t-\tau_{1}-\tau)\,d\tau\,d\tau_{1}.$

We substitute in the last integral $\tilde{x}=S_{\tau}(S_{a+\tau_{1}}(x_{b}))$
and $\tilde{a}=t-\tau_{1}-\tau$. Then

$\displaystyle\frac{\partial\tilde{x}}{\partial\tau}$
$\displaystyle=\frac{1}{2}g\big{(}S_{\tau}(S_{a+\tau_{1}}(x_{b}))\big{)},$
$\displaystyle\frac{\partial\tilde{x}}{\partial\tau_{1}}$
$\displaystyle=\frac{1}{2}\frac{g(\pi_{\tau}S_{a+\tau_{1}}(x_{b}))}{g(S_{a+\tau_{1}}(x_{b}))}\cdot\frac{1}{2}g\big{(}S_{a+\tau_{1}}(x_{b})\big{)}=\frac{1}{4}g(\pi_{\tau}S_{a+\tau_{1}}(x_{b})),$
$\displaystyle\frac{\partial\tilde{a}}{\partial\tau}$
$\displaystyle=\frac{\partial\tilde{a}}{\partial\tau_{1}}=-1.$

Let $\mathcal{J}_{\tau,\tau_{1}}$ be the Jacobian matrix of the transformation
$(x_{b},a)\mapsto(\tilde{x},\tilde{a})$. Then

$\det\mathcal{J}_{\tau,\tau_{1}}(x_{b},a)=\frac{1}{4}g(\pi_{\tau}S_{a+\tau_{1}}(x_{b}))-\frac{1}{2}g\big{(}S_{\tau}(S_{a+\tau_{1}}(x_{b}))\big{)}.$

According to (A7) there exists $x\in(\underline{x}_{b},\overline{x}_{b})$ such
that $g(2x)\neq 2g(x)$. We find
$x^{1}_{b}\in(\underline{x}_{b},\overline{x}_{b})$ and $\tau^{0}>0$ such that
$q(x_{b}^{1},\tau^{0})>0$ and $S_{\tau^{0}}(x_{b}^{1})=x$, i.e. $x$ and
$x_{b}^{1}$ are the initial sizes of daughter and mother cells. Next we find
$x^{0}_{b}\in(\underline{x}_{b},\overline{x}_{b})$, $a^{0}>0$, and
$\tau_{1}^{0}>0$ such that $q(x_{b}^{0},a^{0}+\tau_{1}^{0})>0$ and
$S_{a^{0}+\tau_{1}^{0}}(x_{b}^{0})=x^{1}_{b}$. We also choose $t>0$ such that
the point $(x,t-\tau^{0}-\tau_{1}^{0})$ lies in the interior of the set $X$.
Then we find a neighbourhood $\mathcal{U}$ of the point
$(\tau^{0},\tau_{1}^{0},x_{b}^{0},a^{0})$ such that
$\det\mathcal{J}_{\tau,\tau_{1}}(x_{b},a)\neq 0$, $q(x_{b},a+\tau_{1})>0$,
$q(S_{a+\tau_{1}}(x_{b}),\tau)>0$,
$(S_{\tau}(S_{a+\tau_{1}}(x_{b})),t-\tau-\tau_{1})\in X$ for
$(\tau,\tau_{1},x_{b},a)\in\mathcal{U}$. Thus there exist neighbourhoods
$V_{1}$ and $V_{2}$ of the points $(x_{b}^{0},a^{0})$ and
$(x,t-\tau_{1}^{0}-\tau^{0})$ and there exist $\varepsilon>0$ and a
nonnegative kernel $k(t,x_{b},a,\tilde{x},\tilde{a})$ such that
$k(t,x_{b},a,\tilde{x},\tilde{a})\geq\varepsilon$ for
$(x_{b},a,\tilde{x},\tilde{a})\in V_{1}\times V_{2}$ and

$T^{\odot}_{2}f(t)(x_{b},a)\geq\iint\limits_{X}k(t,x_{b},a,\tilde{x},\tilde{a})f(\tilde{x},\tilde{a})\,d\tilde{x}\,d\tilde{a}.$

From (27) it follows that the semigroups $\\{T^{\odot}(t)\\}_{t\geq 0}$ and
$\\{T(t)\\}_{t\geq 0}$ are partially integral. ∎

######  Theorem 17.

Assume (A1)–(A7). Then the semigroup $\\{P(t)\\}_{t\geq 0}$ is asymptotically
stable. The eigenfunction of $\mathcal{A}$ from Proposition 14 is the
invariant density of $\\{P(t)\\}_{t\geq 0}$.

###### Proof.

We need to check that the semigroup $\\{P(t)\\}_{t\geq 0}$ satisfies
assumptions of Theorem 15. Since the semigroup $\\{T(t)\\}_{t\geq 0}$ is
partially integral and $P(t)f=e^{-\lambda t}T(t)f$ for $f\in E$, the same
property has the semigroup $\\{P(t)\\}_{t\geq 0}$. According to Proposition 14
there exists a function $f_{i}$ such that $\mathcal{A}f_{i}=\lambda f_{i}$ and
$f_{i}>0$ a.e. As $\mu(X)<\infty$ and $f_{i}$ is bounded, $f_{i}$ is
integrable with respect to $\mu$. Since the eigenfunction is determined up to
a multiplicative constant, we may assume that $\int_{X}f_{i}\,d\mu=1$. Also
according to Proposition 14 the function $f_{i}$ is the unique invariant
density of $\\{P(t)\\}_{t\geq 0}$. ∎

######  Theorem 18.

For every $u_{0}\in E$ we have

(28) $\lim_{t\to\infty}e^{-\lambda t}U(t)u_{0}=\Phi
f_{i}\iint\limits_{X}u_{0}(x_{b},a)\Psi(x_{b},a)v(x_{b},a)\,dx_{b}\,da\quad\textrm{in
$E$}.$

Moreover, $\Phi f_{i}$ and $\Psi v$ are eigenfunctions of the semigroups
$\\{U(t)\\}_{t\geq 0}$ and $\\{U^{*}(t)\\}_{t\geq 0}$ corresponding to the
eigenvalue $\lambda$.

###### Proof.

Condition of asymptotic stability of the semigroup $\\{P(t)\\}_{t\geq 0}$ can
be written in the following equivalent form: for every $f\in E_{1}$ we have

(29)
$\lim_{t\to\infty}P(t)f=f_{i}\iint\limits_{X}f(x_{b},a)v(x_{b},a)\,dx_{b}\,da\quad\textrm{in
$E_{1}$}.$

We can extend the semigroup $\\{T(t)\\}_{t\geq 0}$ to a $C_{0}$-semigroup on
$E_{1}$ setting $T(t)f=e^{\lambda t}P(t)f$ for $f\in E_{1}$. From (29) it
follows that

(30) $\lim_{t\to\infty}e^{-\lambda
t}T(t)f=f_{i}\iint\limits_{X}f(x_{b},a)v(x_{b},a)\,dx_{b}\,da\quad\textrm{in
$E_{1}$}.$

Now we return to the problem (6)–(8). We recall that after substitution
$z(t,x_{b},a)=u(t,x_{b},a)\Psi(x_{b},a)$ and
$z_{0}(x_{b},a)=u_{0}(x_{b},a)\Psi(x_{b},a)$ we have replaced this problem by
the system (11)–(13) and the semigroup $\\{T(t)\\}_{t\geq 0}$ describes the
evolution of the solutions of this system. Since $z=u\Psi$ and
$z_{0}=u_{0}\Psi$, we have $u(t)=\Phi T(t)(u_{0}\Psi)$ because $\Phi=1/\Psi$.
Thus we can define a semigroup $\\{U(t)\\}_{t\geq 0}$ corresponding to (6)–(8)
by

(31) $U(t)u_{0}=\Phi T(t)(u_{0}\Psi).$

From inequalities (21) it follows that $u_{0}\Psi\in E_{1}$ if and only if
$u_{0}\in E$ and $\\{U(t)\\}_{t\geq 0}$ is a $C_{0}$-semigroup on the space
$E$. It should be noted that we consider solutions of (6)–(8) for a wider
class of initial conditions because we do not assume that $u_{0}$ satisfies
inequality (9). From (30) it follows that

$\lim_{t\to\infty}e^{-\lambda t}\Psi
U(t)u_{0}=f_{i}\iint\limits_{X}u_{0}(x_{b},a)\Psi(x_{b},a)v(x_{b},a)\,dx_{b}\,da\quad\textrm{in
$E_{1}$}.$

Using again inequalities (21) we finally obtain (28). ∎

Property (28) is called the asynchronous exponential growth of the semigroup
$\\{U(t)\\}_{t\geq 0}$. Precisely, we say that a semigroup $\\{U(t)\\}_{t\geq
0}$ on a Banach space $\mathbb{X}$ has asynchronous (or balanced) exponential
growth if there exist $\lambda\in\mathbb{C}$, a nonzero $x_{i}\in\mathbb{X}$,
and a nonzero linear functional $\alpha\colon\mathbb{X}\to\mathbb{C}$ such
that

$\lim_{t\to\infty}e^{-\lambda t}U(t)x=x_{i}\alpha(x)\quad\textrm{for
$x\in\mathbb{X}$}.$

It should be mentioned that one can find in literature, e.g. [50], a more
general definition of asynchronous exponential growth, where it is only
assumed that $e^{-\lambda t}U(t)x$ converges to a nonzero finite rank
operator.

##  6\. Remarks

###  6.1. Chemostat

In Section 2 we have mentioned that if we consider experiment in a chemostat,
then we need to add to the right-hand side of equation (6) the term
$-Du(t,x_{b},a)$. In this case we substitute
$u(t,x_{b},a)=e^{-Dt}\bar{u}(t,x_{b},a)$ and then we check that the function
$\bar{u}$ satisfies the system (6)–(8). From Theorem 18 we deduce that

(32) $\lim_{t\to\infty}e^{(D-\lambda)t}U(t)u_{0}=\Phi
f_{i}\iint\limits_{X}u_{0}(x_{b},a)\Psi(x_{b},a)v(x_{b},a)\,dx_{b}\,da\quad\textrm{in
$E$}.$

From (32) it follows that in order to grow cells under constant environmental
conditions, cells should be removed from the system with rate $D=\lambda$.

###  6.2. Age-size structured model

Now we consider an age-size structured model consistent with our biological
description. Let $\bar{p}(x,a)\Delta t$ be the probability that a cell with
size $x$ and age $a$ splits in the time interval of the length $\Delta t$.
Since such a cell had the initial size $x_{b}=\pi_{-a}x$, we see that

$\bar{p}(x,a)=p(\pi_{-a}x,a)=q(\pi_{-a}x,a)\big{/}\textstyle{\int_{a}^{\infty}}q(\pi_{-a}x,r)\,dr$

for $a<\overline{a}(\pi_{-a}x)$. We set $\bar{p}(x,a)=0$ for
$a\geq\overline{a}(\pi_{-a}x)$. Let $w(t,x,a)$ be the number of cells having
size $x$ and age $a$ at time $t$. Then the function $w$ satisfies the
following initial-boundary problem:

(33) $\displaystyle\frac{\partial w}{\partial t}(t,x,a)+\frac{\partial
w}{\partial a}(t,x,a)+\frac{\partial(gw)}{\partial
x}(t,x,a)=-\bar{p}(x,a)w(t,x,a),$ (34) $\displaystyle
w(t,x,0)=4\int_{0}^{\infty}\bar{p}(2x,a)w(t,2x,a)\,da,$ (35) $\displaystyle
w(0,x,a)=w_{0}(x,a).$

We have the following relationship between solutions of the systems (6)–(8)
and (33)–(35):

(36)
$\int_{0}^{a}\int_{0}^{x}u(t,x_{b},r)\,dx_{b}\,dr=\int_{0}^{a}\int_{0}^{\pi_{r}x}w(t,y,r)\,dy\,dr.$

Differentiating both sides of (36) with respect to $a$ and $x$ we obtain

$u(t,x,a)=\frac{\partial(\pi_{a}x)}{\partial
x}w(t,\pi_{a}x,a)=\frac{g(\pi_{a}x)}{g(x)}w(t,\pi_{a}x,a).$

Using the above formula and Theorem 18 we get

$e^{-\lambda t}\frac{g(\pi_{a}x)}{g(x)}w(t,\pi_{a}x,a)\to(\Phi
f_{i})(x,a)\iint\limits_{X}w_{0}(\pi_{a}x_{b},a)h(x_{b},a)\,dx_{b}\,da$

in $E$ as $t\to\infty$, where
$h(x_{b},a)=\Psi(x_{b},a)v(x_{b},a)g(\pi_{a}x_{b})/g(x_{b})$. Finally we
conclude that

$e^{-\lambda t}w(t,x,a)\to
h_{i}(x,a)\iint\limits_{X}w_{0}(\pi_{a}x_{b},a)h(x_{b},a)\,dx_{b}\,da$

in $L^{1}$, where $h_{i}(x,a)=(\Phi f_{i})(\pi_{-a}x,a)g(\pi_{-a}x)/g(x)$.

###  6.3. Solutions with values in the space of measures

If we study the dynamics of population growth of microorganisms starting from
a single cell, then initial distribution of the population is described by a
singular measure, precisely with a delta Dirac measure. Thus it is natural to
consider a model which describes the evolution of measures instead of $L^{1}$
functions. We can introduce such a model by considering weak solutions. Let
$\\{U(t)\\}_{t\geq 0}$ be the semigroup introduced in Section 5 and let
$\\{U^{\odot}(t)\\}_{t\geq 0}$ be the “dual semigroup” given by
$U^{\odot}(t)u_{0}=\Psi T^{\odot}(t)(\Phi u_{0})$ (see formula (31)). Denote
by $\mathcal{M}(X)$ the space of all finite Borel measures on $X$. For any
measure $\nu_{0}\in\mathcal{M}(X)$ we define the weak solution of the problem
(6)–(8) as a function $u\colon[0,\infty)\to\mathcal{M}(X)$, $u(t)=\nu_{t}$,
where the measures $\nu_{t}$ satisfy the condition

(37)
$\iint\limits_{X}f(x_{b},a)\,\nu_{t}(dx_{b},da)=\iint\limits_{X}U^{\odot}(t)f(x_{b},a)\,\nu_{0}(dx_{b},da)$

for all $f\in C(X)$. Since the set $X$ is compact, the existence and
uniqueness of the measures $\nu_{t}$ is a simple consequence of the Riesz
representation theorem.

One can ask about the long-time behaviour of the measures $\nu_{t}$. We are
interested in convergence of measures in the total variation norm. We denote
by $d(\nu,\bar{\nu})_{TV}$ the distance between $\nu$ and $\bar{\nu}$ in the
total variation norm in $\mathcal{M}(X)$. We recall that

$d(\nu,\bar{\nu})_{TV}=(\nu-\bar{\nu})^{+}(X)+(\nu-\bar{\nu})^{-}(X),$

where the symbols $\nu^{+}$ and $\nu^{-}$ denote the positive and negative
part of a signed measure $\nu$. We can formulate Theorem 18 in a slightly
stronger form:

######  Proposition 19.

Assume that conditions (A1)–(A7) hold. Let $\nu_{0}\in\mathcal{M}(X)$ and let
$\nu_{\infty}$ be the measure given by

$\nu_{\infty}(A)=\int\limits_{A}\Phi(x_{b},a)f_{i}(x_{b},a)\,dx_{b}\,da\cdot\iint\limits_{X}\Psi(x_{b},a)v(x_{b},a)\,\nu_{0}(dx_{b},\,da)$

for $A\in\mathcal{B}(X)$. Then

(38) $\lim_{t\to\infty}d(e^{-\lambda t}\nu_{t},\nu_{\infty})_{TV}=0.$

We only give some idea of the proof of Proposition 19. We consider weak
solutions connected with the semigroup $\\{P^{\odot}(t)\\}_{t\geq 0}$, i.e. we
replace in (37) the semigroup $\\{U^{\odot}(t)\\}_{t\geq 0}$ by
$\\{P^{\odot}(t)\\}_{t\geq 0}$, where $P^{\odot}(t)=e^{-\lambda
t}T^{\odot}(t)$. It is enough to check that if $\nu_{0}$ is a probability
measure then $\lim_{t\to\infty}d(\nu_{t},\mu_{*})_{TV}=0$, where
$d\mu_{*}=f_{i}d\mu$. We write $\nu_{t}$ as a sum $\nu_{t}^{a}+\nu_{t}^{s}$,
where $\nu_{t}^{a}$ is the absolutely continuous part of $\nu_{t}$ with
respect to the Lebesgue measure and $\nu_{t}^{s}$ is the singular part of
$\nu_{t}$. We deduce from conditions (A6) and (A7) that there exist $t_{0}>0$
and $\varepsilon>0$ independent of $\nu_{0}$ such that $\nu^{s}_{t_{0}}(X)\leq
1-\varepsilon$. The proof of this part is very technical but it uses similar
arguments as the proof of Lemma 16. From the last inequality it follows that
$\nu^{s}_{t}(X)\leq(1-\varepsilon)^{n}$ for $t\geq nt_{0}$. Fix $\eta>0$ and
let $t_{1}>0$ be such that $\nu^{s}_{t_{1}}(X)\leq\eta$. Let
$f=d\nu_{t_{1}}^{a}/d\mu$. Then $\lim_{t\to\infty}\|P(t)f-f_{i}\int
f\,d\mu\|_{E_{1}}=0$. Since $\int f\,d\mu\geq 1-\eta$ and
$\nu^{s}_{t_{1}}(X)\leq\eta$, we have $d(\nu_{t},\mu_{*})_{TV}\leq 2\eta$. As
$\eta>0$ can be chosen arbitrary small we finally obtain
$\lim_{t\to\infty}d(\nu_{t},\mu_{*})_{TV}=0$.

It should be noted that a similar result can be obtained by using the theory
of positive recurrent and aperiodic Harris processes (see Theorem 13.3.3 of
[31]), but to apply this theorem we need to formulate the problem properly in
the language of stochastic processes. First, we construct a family of Markov
processes corresponding to the semigroup $\\{P(t)\\}_{t\geq 0}$. In
particular, we need to define additionally the processes started from points
$(x_{b},\overline{a}(x_{b}))$. Since Theorem 13.3.3 applies to discrete-time
processes, we consider this Markov family for times $t=0,1,2,\dots$ and check
assumptions of this theorem. As a result we obtain that
$\lim_{n\to\infty}d(\nu_{n},\mu_{*})_{TV}=0$. Finally, we pass from discrete
time convergence to continuous time convergence.

###  6.4. Case $g(2x)=2g(x)$

A function $g$ satisfying condition $g(2x)=2g(x)$ for all
$x\in[\underline{x}_{b},\overline{x}_{b}]$ can be constructed in the following
way. Let $g\colon[\underline{x}_{b},2\underline{x}_{b}]\to(0,\infty)$ be a
given $C^{1}$-function such that $g(2\underline{x}_{b})=2g(\underline{x}_{b})$
and $g^{\prime}(2\underline{x}_{b})=g^{\prime}(\underline{x}_{b})$. Then we
define $g(x)=2^{n}g(2^{-n}x)$ for
$x\in[2^{n}\underline{x}_{b},2^{n+1}\underline{x}_{b}]$.

Observe that if $g(2x)=2g(x)$ for all
$x\in[\underline{x}_{b},\overline{x}_{b}]$, then the semigroup
$\\{U(t)\\}_{t\geq 0}$ has no asynchronous exponential growth. Indeed,
consider a cell with initial size $x_{b}$. Fix time $t>0$ and assume that the
cell splits at age $a\leq t$. Then the daughter cells at time $t$ have size

$x(a)=\pi_{t-a}(\tfrac{1}{2}\pi_{a}x_{b}).$

Since

$x^{\prime}(a)=-g(\pi_{t-a}(\tfrac{1}{2}\pi_{a}x_{b}))+\frac{g(\pi_{t-a}(\tfrac{1}{2}\pi_{a}x_{b}))}{g(\tfrac{1}{2}\pi_{a}x_{b})}\cdot\frac{1}{2}g(\pi_{a}x_{b})=0,$

the function $x$ is constant and $x(a)=\tfrac{1}{2}\pi_{t}x_{b}$. Thus the
size of all daughter cells is exactly twice smaller than the size of the
mother cell. If $x_{n}(t)$ is the size of a cell from the $n$th generation
then its mother, grandmother, etc. cells have sizes $2x_{n}(t)$, $4x_{n}(t)$,
$\dots$ But since cells have minimum and maximum size $\underline{x}_{b}$ and
$2\overline{x}_{b}$, the maximum number of existing generations at a given
time $t$ is not greater than $2+\log_{2}(\overline{x}_{b}/\underline{x}_{b})$.
Moreover, if $x_{1}\in(\underline{x}_{b},2\underline{x}_{b})$ and
$f_{1}(x_{b},a)=\mathbf{1}_{(\underline{x}_{b},x_{1})}(x_{b})$,
$f_{2}(x_{b},a)=\mathbf{1}_{(x_{1},2\underline{x}_{b})}(x_{b})$, then
$U(t)f_{1}\cdot U(t)f_{2}\equiv 0$ for all $t\geq 0$. Consequently, the
semigroup $\\{U(t)\\}_{t\geq 0}$ has no asynchronous exponential growth.

######  Remark 20.

One can check that the semigroup $\\{U(t)\\}_{t\geq 0}$ is irreducible i.e.
$\int_{0}^{\infty}U(t)f\,dt>0$ a.e. even if (A7) does not hold. Our example
shows that a semigroup can be irreducible but not overlapping supports. A
stochastic semigroup $\\{P(t)\\}_{t\geq 0}$ is called overlapping supports if
$P(t)f_{1}\cdot P(t)f_{2}\neq 0$ for any two densities $f_{1}$ and $f_{2}$ and
some $t=t(f_{1},f_{2})$. Another simple example of irreducible stochastic
semigroup which does not overlap supports is the rotation semigroup. If
$X=S^{1}$ is a unit circle on the complex plain with centre $z_{0}=0$,
$\,\Sigma=\mathcal{B}(X)$ is the $\sigma$-algebra of Borel subsets of $X$ and
$m$ is the arc-Lebesgue measure on $X$, the rotation semigroup
$\\{P(t)\\}_{t\geq 0}$ is given by $P(t)f(z)=f(ze^{it})$.

Now we consider a special case when $g(x)=\kappa x$, $\kappa>0$. We start at
time $t=0$ with a single cell with size $x$. Cells from the $n$th generation
have size $2^{-n}e^{\kappa t}x$ at time $t$. Then $\bar{p}(2^{-n}e^{\kappa
t}x,a)\Delta t$ is the probability that a cell from the $n$th generation with
age $a$ splits in the time interval of the length $\Delta t$. This observation
allows us to describe the evolution of the population using discrete
parameters. Denote by $w_{n}(t,a)$ the number of cells from the $n$th
generation with age $a$ at time $t$. Then the functions $w_{n}$ satisfy the
following infinite system of partial differential equations with boundary
conditions:

$\displaystyle\frac{\partial w_{n}}{\partial t}(t,a)+\frac{\partial
w_{n}}{\partial a}(t,a)=-\bar{p}\big{(}2^{-n}e^{\kappa
t}x,a\big{)}w_{n}(t,a),$ $\displaystyle
w_{n}(t,0)=2\int_{0}^{\infty}\bar{p}\big{(}2^{1-n}e^{\kappa
t}x,a\big{)}w_{n-1}(t,a)\,da.$

It should be noted that it is not easy to find a direct formula for the
eigenvector $f_{i}(x_{b},a)$ of the operator $\mathcal{A}$ even in the case
$g(x)=\kappa x$. Indeed, we have $\pi_{-a}(2x_{b})=2e^{-\kappa a}x_{b}$ and

$\displaystyle P_{a}f(x_{b})$
$\displaystyle=\frac{2g(\pi_{-a}(2x_{b}))}{g(2x_{b})}f(\pi_{-a}(2x_{b}))=\frac{2\pi_{-a}(2x_{b})}{2x_{b}}f(\pi_{-a}(2x_{b}))$
$\displaystyle=2e^{-\kappa a}f(2e^{-\kappa a}x_{b}).$

Then $f_{i}(x_{b},a)=e^{-\lambda a}f_{i}(x_{b},0)$, where $\lambda$ and
$f_{i}(x_{b},0)$ should be found by solving the following equation

$f_{i}(x_{b},0)=\int_{0}^{\infty}4e^{-(\lambda+\kappa)a}q(2e^{-\kappa
a}x_{b},a)f_{i}(2e^{-\kappa a}x_{b},0)\,da,$

which is not a simple task.

##  7\. Comparison with experimental data and other models

Modern experimental techniques enable studies of individual cells growth in
well-controlled environments. Especially interesting are experimental results
concerning rod-shaped bacteria, for example E. coli, C. crescentus and B.
subtilis [7, 20, 43, 48], because they change only their length. Although such
bacteria have similar shape there are variety of distinct models of cell cycle
and cell division. For example, we consider models with symmetric or
asymmetric divisions, with different velocities of proliferation,
deterministic or stochastic growth of individuals or models based on special
assumptions as fixed cell length extension or models with target size
division. We give a short review of such models and show how to incorporate
them to our model.

###  7.1. Models with exponential growth

Since experimental data suggest that cells grow exponentially, one can find a
number of models with the assumption $g(x)=\kappa x$ but with various
descriptions of the cell cycle length.

In [7, 17, 43, 47] it is considered an additive model (or a constant $\Delta$
model), where it is assumed that the difference $\Delta(x_{b})=x_{d}-x_{b}$
between the size at division $x_{d}$ and the initial size $x_{b}$ of a cell is
a random variable independent of $x_{b}$. From this assumption it follows that

$\tau(x_{b})=\kappa^{-1}\ln((x_{b}+\Delta)/x_{b}).$

If $h(x)$ is the density distribution of $\Delta$, then

(39) $q(x_{b},a)=\kappa x_{b}e^{\kappa a}h\big{(}x_{b}e^{\kappa
a}-x_{b}\big{)}$

is the density of $\tau(x_{b})$. We obtain a special case of our model if the
density distribution of $\Delta$ is positive on the interval
$(\underline{x}_{b},\overline{x}_{b})$. According to experimental data from
[43] the coefficient of variation $c_{v}$ of $\Delta$ for E. coli is in the
range of $0.17$ to $0.28$ depending on the different growth conditions. We
recall that $c_{v}=\sigma/\mu$, where $\sigma$ is the standard deviation, and
$\mu$ is the mean.

In [1, 21] it is assumed that a cell with initial size $x_{b}$ attempts to
divide at a target size $x_{d}=f(x_{b})$. Then the expected length of the cell
cycle is $\tau_{0}(x_{b})=\kappa^{-1}\ln(f(x_{b})/x_{b})$, but
$\tau_{0}(x_{b})$ is additively perturbed by a symmetric random variable
$\xi$, and finally $\tau(x_{b})=\tau_{0}(x_{b})+\xi$. If $h(a)$ is the density
distribution of $\xi$, then $q(x_{b},a)=h(a-\tau_{0}(x_{b}))$ is the density
of $\tau(x_{b})$. The authors assume in these papers that $h$ has a normal
distribution but in this case $\tau(x_{b})$ can be negative therefore a
truncated normal distribution located in some interval
$[-\varepsilon,\varepsilon]$ seems to be more suitable. They also assume that
$f(x_{b})=2x_{b}^{1-\alpha}x_{0}^{\alpha}$, $\alpha\in[0,1]$ and $x_{0}>0$. If
$\alpha>0$, $\underline{x}_{b}=x_{0}e^{-\kappa\varepsilon/\alpha}$ and
$\overline{x}_{b}=x_{0}e^{\kappa\varepsilon/\alpha}$, then we obtain a
particular case of our model. If $\alpha=0$, then
$\tau_{0}\equiv\kappa^{-1}\ln 2$ and the length of cell cycle does not depend
on $x_{b}$. In this case a daughter cell size is distributed in some
neighbourhood of the initial mother cell size, so there is no minimum
$\underline{x}_{b}$ and maximum size $\overline{x}_{b}$.

###  7.2. Paradoxes of exponential growth

Models with exponential growth law can lead to some odd mathematical results.
If the population starts with a single cell of size $x$, cells from $n$th
generation have size $x_{n}(t)=2^{-n}e^{\kappa t}x$ at time $t$. Since
$\underline{x}_{b}\leq x_{n}(t)\leq\overline{x}_{b}$, population consists of a
few generations at each time and all cells in each generation have the same
size. Usually the quotient $\overline{x}_{b}/\underline{x}_{b}$ is not too
large. The initial size for E. coli under steady-growth conditions is
$x_{b}=2.32\,\pm\,0.38\,\,\mu m$ (mean $\pm$ SD) [7]. Thus we can assume that
in this case $\overline{x}_{b}/\underline{x}_{b}<2$. Then it is easy to check
that if

$t\in\left(\frac{n+\log_{2}(\overline{x}_{b}/x)}{\kappa\log_{2}e},\frac{1+n+\log_{2}(\overline{x}_{b}/x)}{\kappa\log_{2}e}\right),$

the population consists only of cells from the $n$th generation, thus all
cells have the same size and they cannot split in this time interval.
Consequently the size of the population never reaches an exponential
(balanced) growth. On this point we also observe that the large quotient
$\overline{x}_{b}/\underline{x}_{b}$ helps the population to stabilize its
growth, which explains why in the model with target size division [21] it
takes the population a longer time to reach its balanced growth for greater
$\alpha$, because
$\overline{x}_{b}/\underline{x}_{b}=e^{2\kappa\varepsilon/\alpha}$.

The exponential growth law of cells should be a little bit modified in order
to achieve AEG. For example it is enough to assume that $\kappa$ depends on
the initial size $x_{b}$. But according to the experimental results, the
average growth rate does not depend on the initial size of cells. On the other
hand, even if a population grows under perfect conditions the individual cells
have different growth rates: the standard deviation of the growth rate is 15%
of their respective means [43]. Thus there is other factor called maturity,
which decides about the growth rate of an individual cell. The mathematical
models based on the concept of maturity were formulated in the late sixties
[25, 40]. In such models the growth rate is identified with maturation
velocity $v$ which is constant during the life of cell and is inherited in a
random way from mother to daughter cells.

Rotenberg [39] considered a version of maturity models with random jumps of
$v$ during the cell cycle. If we replace random jumps of $v$ by stochastic
fluctuations of $\kappa$, we obtain a cell growth model described by a
stochastic equation considered in the next subsection.

We consider here a simple generalization of our model assuming that $\kappa$
is a random variable with the distribution dependent on $x_{b}$. Let the
function $r\mapsto k(r|x_{b})$ be the density of $\kappa$. The question is how
to describe the joint distribution of age and initial size in this case.
Equations (6) and (8) remain the same and it is enough to derive a version of
the boundary condition (7). Denote by $f(x;x_{b},a)$ the density distribution
of the random variable $\xi_{a}^{x_{b}}=x_{b}e^{\kappa a}$. Then (7) takes the
form

(40)
$u(t,x,0)=4\iint\limits_{X}f(2x;x_{b},a)p(x_{b},a)u(t,x_{b},a)\,dx_{b}\,da.$

It remains to find the function $f(x;x_{b},a)$. We have

$\operatorname{Prob}(x_{b}e^{\kappa a}\leq
x)=\operatorname{Prob}\big{(}\kappa\leq
a^{-1}\ln(x/x_{b})\big{)}=\int_{0}^{a^{-1}\ln(x/x_{b})}k(r|x_{b})\,dr.$

Hence

$f(x;x_{b},a)=\frac{1}{ax}k(a^{-1}\ln(x/x_{b})|x_{b}).$

At first glance formulae (7) and (40) differ significantly, but if we replace
in (40) the term $f(2x;x_{b},a)$ by the delta Dirac $\delta_{S_{a}(x_{b})}(x)$
we will receive (7).

###  7.3. Stochastic growth of $x$

The size of a cell having initial size $x_{b}$ grows according to Itô
stochastic differential equation

(41)
$d\xi_{t}^{x_{b}}=\kappa\xi_{t}^{x_{b}}\,dt+\sigma(\xi_{t}^{x_{b}})\,dB_{t},$

where $B_{t}$, $t\geq 0$, is a one dimensional Wiener process (Brownian
motion), and $\kappa>0$. In [20, 37] the authors assume that
$\sigma(x)=\sqrt{D}x^{\gamma}$, where $D>0$ and $\gamma\in(0,1)$. The great
strength of this formula is that equation (41) was intensively studied for
such $\sigma$ and we can solve (41) and find various properties of solutions.
But there is one weak point: the size can go to zero and even solutions can be
absorbed at zero. To omit this problem we propose to assume that
$\sigma\colon[\underline{x}_{b},\infty)\to\mathbb{R}$ is a $C^{1}$-function
and $\sigma(\underline{x}_{b})=0$. Then $\xi_{t}^{x_{b}}>\underline{x}_{b}$
for $t>0$. It should be noted that solutions can decrease at some moments,
i.e. a cell can shrink, but if the diffusion coefficient $\sigma$ is small, we
observe exponential growth with small stochastic noise. If $f(x;x_{b},a)$ is
the density distribution of the random variable $\xi_{a}^{x_{b}}$, then the
joint distribution of age and initial size $u(t,x_{b},a)$ satisfies equations
(6), (8), (40).

###  7.4. Models with asymmetric division and with slow-fast proliferation

A lot of cellular populations are heterogeneous. For example, C. crescentus
has an asymmetric cell division; B. subtilis occasionally produces minicells;
melanoma cells have slowly and quickly proliferating cells [33]; and
precursors of blood cells replicate and maturate going through the levels of
morphological development [29]. It is difficult to find one universal model of
the evolution of heterogeneous populations. Now we present a model of the
distribution of heterogeneous population based on similar assumptions as the
model presented in Section 2. We divide the population into a number of
subpopulations. We assume that cells in the $i$th subpopulation grow according
to the equation $x^{\prime}=g_{i}(x)$ and their length of the cell cycle has
the probability density distribution $q_{i}(x_{b},a)$. We also assume that
$r_{ij}$ is the probability that a daughter of a cell from the $i$th
subpopulation belongs to the $j$th subpopulation and the daughter has initial
size $\beta_{ij}x$, where $x$ is the size of the mother cell at division. As
in Section 2 we introduce the function

$p_{i}(x_{b},a)=\frac{q_{i}(x_{b},a)}{\int_{a}^{\infty}q_{i}(x_{b},r)\,dr}$

and operators $P^{ij}_{a}$ which describe the relation between the density of
the initial sizes of mother and daughter cells satisfying the equation:

$\int_{\underline{x}_{b}}^{\beta_{ij}\pi^{i}_{a}y}P^{ij}_{a}f(x_{b})\,dx_{b}=r_{ij}\int_{\underline{x}_{b}}^{y}f(x_{b})\,dx_{b}.$

Then

$P^{ij}_{a}f(x_{b})=\frac{r_{ij}}{\beta_{ij}}\frac{g(\pi^{i}_{-a}(x_{b}/\beta_{ij}))}{g(x_{b}/\beta_{ij})}f(\pi^{i}_{-a}(x_{b}/\beta_{ij})).$

We denote by $u_{i}(t,x_{b},a)$ the number of individuals in the $i$th
population having initial size $x_{b}$ and age $a$ at time $t$. Then the
system (6)–(8) will be replaced by the following one

(42) $\displaystyle\frac{\partial u_{i}}{\partial t}(t,x_{b},a)+\frac{\partial
u_{i}}{\partial a}(t,x_{b},a)=-p_{i}(x_{b},a)u_{i}(t,x_{b},a),$ (43)
$\displaystyle
u_{j}(t,x_{b},0)=2\sum\limits_{i}\int_{0}^{\infty}P^{ij}_{a}(p_{i}(x_{b},a)u_{i}(t,x_{b},a))\,da,$
(44) $\displaystyle u_{i}(0,x_{b},a)=u_{i0}(x_{b},a).$

In some cases of asymmetric division the size of daughter cells is not
strictly determined and it is better to consider a model where the density
$k(x_{b}|x_{d})$ describes the distribution of the initial size of a daughter
cell $x_{b}$ if the mother cell has the size $x_{d}$, see e.g. [3, 18, 19, 22,
41].

As an example of application of the model (42)–(44) we consider C. crescentus
which has asymmetric cell division into a ”stalked” cell which can replicate
and a mobile ”swarmer” cell which differentiates into a stalked cell after a
short period of motility. Thus we have two subpopulations: the first – stalked
cells and the second – swarmer cells. Then $r_{ij}=1/2$ for $i=1,2$ and
$j=1,2$. The stalked daughter has length of $0.56\,\pm\,0.04$ (mean $\pm$ SD)
of the mother cell [7]. Hence we can assume that $\beta_{11}=\beta_{21}=0.56$
and $\beta_{12}=\beta_{22}=0.44$. If we assume that both stalked and swarmer
cells have the same growth rate $\kappa$, i.e. $g_{i}(x)=\kappa x$, then
$q_{2}(x_{b},a)=q_{1}(x_{b},a-\rho)$, where $\rho$ satisfies the formula
$e^{\kappa\rho}=0.56/0.44$ and $q_{1}=q$ is given by (39).

A model for the growth B. subtilis should be more advanced. B. subtilis can
divide symmetrically to make two daughter cells (binary fission), but some
mutants split asymmetrically, producing a single endospore, which can
differentiate to a ”typical” cell. Assume that the first population consists
of typical cells and the second of minicells. If $\mathfrak{p}$ is the
probability of asymmetric fission, then
$r_{11}=r_{21}=1-\mathfrak{p}+\mathfrak{p}/2=1-\mathfrak{p}/2$ and
$r_{12}=r_{22}=\mathfrak{p}/2$. Some information on the size of minicells can
be found in [23].

In a model which describes slowly and quickly proliferating cells we should
assume that the length of the cell cycle of slowly proliferating cells is
longer than in quickly proliferating cells and slowly proliferating cells also
grow slower. Thus the sensible assumptions are: $g_{1}(x)<g_{2}(x)$ and

$\int_{0}^{a}q_{1}(x_{b},r)\,dr<\int_{0}^{a}q_{2}(x_{b},r)\,da\quad\text{for
$a<\overline{a}_{1}(x_{b})$}.$

We should also assume that there is some transition between both
subpopulations. Other model of the growth of the population with slowly and
quickly proliferating cells was recently studied in [46].

## Acknowledgments

This research was partially supported by the National Science Centre (Poland)
Grant No. 2017/27/B/ST1/00100.

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