\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2010(2010), No. 47, pp. 1--12.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2010 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2010/47\hfil Well-posedness and asynchronous exponential growth] {Well-posedness and asynchronous exponential growth of solutions of a two-phase cell \\ division model} \author[M. Bai, S. Cui\hfil EJDE-2010/47\hfilneg] {Meng Bai, Shangbin Cui} % in alphabetical order \address{Meng Bai \newline Department of Mathematics, Sun Yat-Sen University, Guangzhou, Guangdong 510275, China} \email{baimeng.clare@yahoo.com.cn} \address{Shangbin Cui \newline Department of Mathematics, Sun Yat-Sen University, Guangzhou, Guangdong 510275, China} \email{cuisb3@yahoo.com.cn} \thanks{Submitted March 3, 2010. Published April 6, 2010.} \thanks{Supported by grant 10771223 from the National Natural Science Foundation of China} \subjclass[2000]{35L02, 35P99} \keywords{Cell division model; two-phase; well-posedness; \hfill\break\indent asynchronous exponential growth} \begin{abstract} In this article we study a two-phase cell division model. The cells of the two different phases have different growth rates. We mainly consider the model of equal mitosis. By using the semigroup theory, we prove that this model is well-posed in suitable function spaces and its solutions have the property of asynchronous exponential growth as time approaches infinity. The corresponding model of asymmetric mitosis is also studied and similar results are obtained. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \allowdisplaybreaks \section{Introduction} In the study of cell division, it has been recognized that the cell cycle can be divided into two major phase: The interphase and the \textbf{M} (mitosis) phase (cf. \cite{PE,JC}). In the interphase cells only increase their size and replicate their DNA, and do not undergo mitosis, whereas in the M phase the fully grown cells segregate the replicated chromosomes to opposite ends of the molecular scaffold (termed the spindle) and then cleave between them in a process known as cytokinesis to produce two daughter cells. The cells in the two phases are observably different. In this paper we study a mathematical model describing the proliferation of cells which are divided into two different phases: mitotic phase and non-mitotic phase. We refer these two phases as $m$-phase and $n$-phase, respectively. We denote by $m(t,x)$ and $n(t,x)$ the densities of $m$-phase cells and $n$-phase cells, respectively, of size $x$ (with a maximal size normalized to $x=1$) at time $t$. We assume that two daughter cells have equal sizes (i.e. \emph{equal mitosis}). In particular, we assume that the two phases have different growth rates $\gamma_1(x)$ and $\gamma_2(x)$, respectively. Then the model reads as follows: \label{e1.1} \begin{gathered} \begin{aligned} \frac{\partial m}{\partial t}+\frac{\partial (\gamma_1(x)m)}{\partial x} &=-B(x)m(t,x)-\nu(x)m(t,x)+\mu(x)n(t,x),\\ &\quad 00,\\ \frac{\partial n}{\partial t}+\frac{\partial (\gamma_2(x)n)}{\partial x} &=-\nu(x)n(t,x)-\mu(x)n(t,x)\\ &\quad +\begin{cases} 4B(2x)m(t,2x), & 0\leq x\leq\frac{1}{2},\; t>0 \\ 0, & \frac{1}{2}0, \end{cases} \end{aligned}\\ m(t,0)=0,\quad n(t,0)=0,\quad t>0,\\ m(0,x)=m_0(x),\quad n(0,x)=n_0(x),\quad 00$for almost all$x\in(0,1)$; \item[(H2)]$B$is a nonnegative and continuous function defined in$[0,1]$with$B(x)>0$for$x\in(0,1)$and$B(x)=0$for otherwise. \item[(H3)]$\gamma_1,\gamma_2\in C^{1}[0,1]$;$\gamma_1(x),\gamma_2(x)>0$for almost all$x\in [0,1]$; Moreover,$\gamma_1(x)\neq\gamma_2(x)$for$x\in [0,1]$and$\gamma_2(2x)\neq2\gamma_1(x)$for$x\in [0,1]$. \end{itemize} Our first main result considers well-posedness of \eqref{e1.1} and reads as follows. \begin{theorem} \label{thm1.1} For any pair of functions$(m_0,n_0)\in W^{1,1}(0,1)\times W^{1,1}(0,1)$such that$(m_0(0),n_0(0))=(0,0)$, problem \eqref{e1.1} has a unique solution $(m,n)\in C([0,\infty),W^{1,1}(0,1)\times W^{1,1}(0,1))\cap C^{1}([0,\infty),L^{1}[0,1]\times L^{1}[0,1]),$ and for any$T>0$, the mapping$(m_0,n_0)\mapsto(m,n)$from the space $$\{(m_0,n_0)\in W^{1,1}(0,1)\times W^{1,1}(0,1):(m_0(0),n_0(0))=(0,0)\}$$ to$C([0,T],W^{1,1}(0,1)\times W^{1,1}(0,1))\cap C^{1}([0,T],L^{1}[0,1]\times L^{1}[0,1])$is continuous. \end{theorem} The proof of this result will be given in Section 2. From the proof of this theorem we shall see that for any$(m_0,n_0)\in W^{1,1}(0,1)\times W^{1,1}(0,1)$we have$(m(t),n(t))= T(t)(m_0,n_0)$, for all$t\geq 0$, where$(T(t))_{t\geq 0}$is a strongly continuous semigroup in the space$X=L^{1}[0,1]\times L^{1}[0,1]$. Thus, for any$(m_0,n_0)\in X=L^{1}[0,1]\times L^{1}[0,1]$,$(m(t),n(t))= T(t)(m_0,n_0)$is well-defined for all$t\geq 0$, and$(m,n)\in C([0,\infty),X)$. As usual, for any$(m_0,n_0)\in X$we call the vector function$t\mapsto (m(t),n(t))= T(t)(m_0,n_0)$(for$t\geq 0$) a \emph{mild solution} of \eqref{e1.1}. Our second main result studies the asymptotic behavior of the solution of \eqref{e1.1}. Before stating this result, we introduce the eigenvalue problem \label{e1.2} \begin{gathered} (\gamma_1(x)\hat{m}(x))'+\lambda\hat{m}(x) =-B(x)\hat{m}(x)-\nu(x)\hat{m}(x)+\mu(x)\hat{n}(x),\quad 00,\\ U(0)=U_0, \end{gathered} where$U(t)=(m(t),n(t))$and$U_0=(m_0(x),n_0(x))$. Thus, to prove that \eqref{e1.1} is well-posed in$X$, we only need to show that the operator$L$generates a strongly continuous semigroup in$X$. \begin{lemma} \label{lem2.1} The operator$A+B$generates a strongly continuous semigroup \\$\{T_1(t)\}_{t\geq 0}$in$X$. \end{lemma} \begin{proof} Let$F\in X$and$U(t)=T_1(t)F$. We write$F=(f,g)$,$U(t)=(u(t,\cdot),v(t,\cdot))$. Then$(u,v)$is the solution of the problem \begin{gather*} \frac{\partial u}{\partial t}+\frac{\partial (\gamma_1(x)u)}{\partial x} =-a_1(x)u(t,x), \quad 0\leq x\leq 1,\;t>0, \\ \frac{\partial v}{\partial t}+\frac{\partial (\gamma_2(x)v)}{\partial x}=-a_2(x)v(t,x),\quad 0\leq x\leq 1,\;t>0, \\ u(t,0)=0,\quad v(t,0)=0,\quad t>0,\\ u(0,x)=f(x),\quad v(0,x)=g(x),\quad 0\leq x\leq 1, \end{gather*} where$a_1(x)=B(x)+\nu(x)$and$a_2(x)=\mu(x)+\nu(x)$. Let$S_1(t,x)$and$S_2(t,x)$be the solution of the following two equations $$\label{e2.2} \begin{gathered} \frac{dS_1}{dt}(t,x)=\gamma_{1}(S_1(t,x)),\quad S_1(0,x)=x,\\ \frac{dS_2}{dt}(t,x)=\gamma_{2}(S_2(t,x)),\quad S_2(0,x)=x \end{gathered}$$ Then $$\label{e2.3} S_1(t,x)=G_1^{-1}(t+G_1(x)),\quad S_2(t,x)=G_2^{-1}(t+G_2(x))$$ where $$\label{e2.4} G_1(x)=\int^{x}_{0}\frac{d\xi}{\gamma_{1}(\xi)},\quad G_2(x)=\int^{x}_{0}\frac{d\xi}{\gamma_{2}(\xi)}$$ By using the standard characteristic method, we obtain \begin{gather} u(t,x)=\begin{cases} \frac{E_1(x)}{\gamma_{1}(x)} \frac{\gamma_{1}(S_1(-t,x))}{E_1(S_1(-t,x))}f(S_1(-t,x)),& 00$, $\psi>0$, we have that $\langle T(t_{0})\varphi,\psi\rangle> 0$ for some $t_{0}> 0$, where $\langle\cdot,\cdot\rangle$ denotes the dual product between $X$ and $X^{\ast}$. We denote by $s(L)$ the \emph{spectral bound} of $L$; i.e., $$\label{e3.1} s(L)=\sup\{\mathop{\rm Re}\lambda:\lambda\in\sigma(L)\}.$$ If the above assertions on the semigroup $(T(t))_{t\geq 0}$ are proved, then by a well-known result in the theory of semigroups we see that $s(L)$ is a dominant eigenvalue of $L$ (i.e., $s(L)\in\sigma(L)$ and $\mathop{\rm Re}\lambda\max\{G_1(1),G_2(1)\}$. This particularly implies that $(T_1(t))_{t\geq 0}$ is norm continuous for $t>\max\{G_1(1),G_2(1)\}$. Thus, by \cite[Theorem III.1.16]{EN}, the desired assertion follows if we prove that the mapping $t\mapsto K(t)\in\mathcal{L}(X)$ is continuous for $t>0$, where $K(t)$ is the operator in $X$ defined by $$K(t)F=\int^{t}_{0}T_1(t-r)CT_1(r)Fdr \quad \text{for } F\in X.$$ Using the representations of $T_1(t)$ (given by \eqref{e2.5} and \eqref{e2.6}) and $C$ we see that for $F=(f,g)\in X$, \begin{align*} & T_1(t-r)CT_1(r)F\\ &= \Big(c_{1}(x,t,r)g(S_1(-t+r,S_2(-r,x))), c_{2}(x,t,r)f(S_2(-t+r,2S_1(-r,x)))\Big), \end{align*} where $c_{i}(x,t,r)$ ($i=1,2$) are continuous functions. Hence \begin{align*} K(t)F=\Big(&\int_{0}^{t} c_{1}(x,t,r)g(S_1(-t+r,S_2(-r,x)))dr,\\ &\int_{0}^{t} c_{2}(x,t,r)f(S_2(-t+r,2S_1(-r,x)))dr\Big), \end{align*} We substitute $\xi_{1}=S_1(-t+r,S_2(-r,x))$ for $r$ in the first term of $K(t)F$ and $\xi_{2}=S_2(-t+r,2S_1(-r,x))$ for $r$ in the second term of $K(t)F$. Because of the assumption (H3), We can find that \begin{gather*} \frac{d\xi_{1}}{dr}= \gamma_1(\xi_{1}) \Big(1-\frac{\gamma_2(S_2(-r,x))}{\gamma_1(S_2(-r,x))}\Big)\neq 0,\\ \frac{d\xi_{2}}{dr}= \frac{\gamma_2(\xi_2)}{\gamma_2(2S_1(-r,x))} (\gamma_2(2S_1(-r,x))-2\gamma_1(S_1(-r,x)))\neq 0. \end{gather*} Then we can easily verify that the mapping $t\mapsto K(t)$ from $(0,+\infty)$ to $\mathcal{L}(X)$ is continuous. Hence the desired assertion follows. This proves lemma 3.2. \end{proof} \begin{lemma} \label{lem3.3} The semigroup $(T(t))_{t\geq 0}$ is eventually compact. \end{lemma} \begin{proof} Since $R(\lambda,A)$ is compact and $B+C$ is the bounded operator, we conclude that $R(\lambda,L)=R(\lambda,A+B+C)$ is compact. Consequently, $R(\lambda,L)T(t)$ is compact for all $t>0$. Since $(T(t))_{t\geq 0}$ is eventually norm continuous, by \cite[Lemma II.4.28]{EN}, it follows that $(T(t))_{t\geq 0}$ is eventually compact. This completes the proof. \end{proof} \begin{lemma} \label{lem3.4} The semigroup $(T(t))_{t\geq 0}$ is irreducible. \end{lemma} \begin{proof} Since $R(\lambda,L)=\int^{+\infty}_{0} e^{-\lambda t}T(t)dt$, for all $Re\lambda >s(L)$ (see [\cite{EN},Lemma VI.1.9]), we have that for all $F=(f(x),g(x))\in X$, $\Psi=(\psi_1,\psi_2)\in X^{\ast}$, $F> 0$, $\Psi>0$, $$\langle\Psi,R(\lambda,L)F\rangle=\int^{+\infty}_{0} e^{-\lambda t}\langle\Psi,T(t)F\rangle dt\,.$$ If we prove that $\langle\Psi,R(\lambda,L)F\rangle>0$ for some $\lambda>0$, then from the above equation it follows that there exists a $t_{0}>0$ such that $\langle\Psi,T(t)F\rangle>0$, and the desired assertion then follows. Let $\pi_1$ and $\pi_2$ be the projections onto the first and second coordinates, respectively. We will prove that $\pi_1(R(\lambda,L)F)(x)>0$ and $\pi_2(R(\lambda,L)F)(x)>0$ for almost all $x\in[0,1]$. In the sequel we find the expression of $R(\lambda,L)$. For $F=(f(x),g(x))\in X$, we solve the equation $$\label{e3.2} (\lambda I-L)U=F.$$ By writing $U=(u(x),v(x))$ and $F=(f(x),g(x))$, we see that \eqref{e3.2} can be rewritten as \begin{gather*} (\gamma_1(x)u(x))'+\lambda u(x)+a_{1}(x)u(x)=f(x)+\mu(x)v(x)\quad \text{for } 00$such that$\|H_{\lambda}\|<1$for$\lambda\geq \lambda^{*}$. This implies$(I-H_{\lambda})^{-1}$exists for$\lambda\geq\lambda^{*}$. Then the resolvent of$L$is $$\label{e3.7} R(\lambda,L)F=(I-H_{\lambda})^{-1}S_{\lambda}F= \sum^{\infty}_{n=0}(H_{\lambda})^{n}S_{\lambda}F, \quad \text{for }\lambda>\lambda^{*}$$ For$0\leq(f(x),g(x))\in \mathbf{X}$and$(f(x),g(x))\neq 0$, without loss of generality, we can assume that$0\leq f\in L^{1}[0,1]$and$f(x)>0$for almost all$x\in [x_{0},x_{1}]$. Then \begin{gather*} \pi_1(S_{\lambda}(f,g))(x)>0, \quad \text{for } x\in[x_{0},1] \\ \pi_2(H_{\lambda}S_{\lambda}(f,g))(x)>0, \quad \text{for } x\in[\frac{x_{0}}{2},1] \\ \pi_1(H_{\lambda}H_{\lambda}S_{\lambda}(f,g))(x)>0, \quad \text{for } x\in[\frac{x_{0}}{2},1] \\ \pi_2(H_{\lambda}H_{\lambda}H_{\lambda}S_{\lambda}(f,g))(x)>0, \quad \text{for } x\in[\frac{x_{0}}{4},1] \end{gather*} Continuing in this way, we obtain$\pi_1(R(\lambda,L)F)(x)>0$and$\pi_2(R(\lambda,L)F)(x)>0$for almost all$x\in[0,1]$. If we assume that$g(x)>0$for almost all$x\in [x_{0},x_{1}]$, the result is the same. This completes the proof. \end{proof} \begin{corollary} \label{coro3.5}$\sigma(L)\neq\emptyset$. \end{corollary} \begin{proof} This follows from \cite[Theorem C-III.3.7]{AG}, which states that if a semigroup is irreducible, positive and eventually compact, then the spectrum of its generator is not empty. \end{proof} \begin{corollary} \label{coro3.6}$s(L)>-\infty$and$s(L)\in \sigma(L)$. \end{corollary} \begin{proof} The first assertion is an immediately consequence of Corollary \ref{coro3.5}. The second assertion follows from the positivity of the semigroup$(T(t))_{t\geq 0}$and the fact$s(L)>-\infty$; see \cite[Theorem VI.1.10]{EN}. \end{proof} By Lemmas \ref{lem3.1}--\ref{lem3.4}, Corollary \ref{coro3.6} and \cite[Corollary V.3.3]{EN}, we conclude that there exists an eigenvalue$\lambda$of$L$associated with a strictly positive eigenvector$(\hat{m},\hat{n})$such that $$\label{e3.8} \lim_{t\to+\infty}e^{-\lambda t}(m(t,x),n(t,x))=C(\hat{m}(x),\hat{n}(x))$$ where$\lambda=s(L)$. In the sequel we find the constant$C$. we know that$\lambda=s(L)$is the dominant eigenvalue of the eigenvalue problem \label{e3.9} \begin{gathered} (\gamma_1(x)\hat{m}(x))'+\lambda\hat{m}(x) =-B(x)\hat{m}(x)-\nu(x)\hat{m}(x)+\mu(x)\hat{n}(x),\quad 00$ and $\hat{n}(x)>0$ for all $00,\\ \frac{\partial n}{\partial t}+\frac{(\gamma_2(x)\partial n)}{\partial x} =-\nu(x)n(t,x)-\mu(x)n(t,x)+\int^{1}_{0}b(x,y)m(t,y)dy,\\ \quad 00, \\ m(t,0)=0,\quad n(t,0)=0,\\ m(0,x)=m_0(x),\quad n(0,x)=n_0(x),\quad 00$ for almost all $x\in [0,1]$ in this section. Besides, we assume that $b(\cdot,y)\in C[0,1]$ for any fixed $y\in [0,1]$. To establish well-posedness of \eqref{e4.1}, we redefine the operator $C_2$ in Section 2 as $$C_2(u,v)=\int^{1}_{0}b(x,y)u(y)dy\quad \text{for } (u,v)\in X,$$ and let $C_1(u,v)$ be as before. We note that the redefined operator $C(u,v)=(C_1(u,v),C_2(u,v))$ is bounded on $X$. Similar arguments as that in Section 2 yield that the redefined operator $L=A+B+C$ generates a strongly continuous semigroup $(T_{2}(t))_{t\geq 0}$ on $X$. Then we can obtain the same assertion as Theorem \ref{thm1.1} about model \eqref{e4.1}. Second, we will obtain the asynchronous exponential growth for \eqref{e4.1}. Note that the redefined operator $C$ is still positive on $X$. Then a similar argument as in the proof of Lemma \ref{lem3.1} shows that the semigroup $(T_{2}(t))_{t\geq 0}$ is positive. The proofs of the eventual compactness and the irreducibility of this semigroup have some differences from those given in Lemmas \ref{lem3.3} and \ref{lem3.4}. We thus give them in the following two lemmas. \begin{lemma} \label{lem4.1} The semigroup $(T_{2}(t))_{t\geq 0}$ is eventually norm continuous and eventually compact. \end{lemma} \begin{proof} In view of the Fr\'{e}chet-Kolmogorov compactness criterion in $L^{1}$ we conclude from \label{e4.6} \begin{aligned} \Big|\int^{1}_{0}b(x,y)u(y)dy-\int^{1}_{0}b(x',y)u(y)dy\Big| &\leq \int^{1}_{0}|b(x,y)-b(x',y)||u(y)|dy\\ &\leq \|b(x,y)-b(x',y)\|_{\infty}\|u(y)\|_{L^{1}} \end{aligned} and the continuity of $b$ that the operator $C$ is compact. From \eqref{e2.5} and \eqref{e2.6}, we can easily see that the semigroup $(T_{1}(t))_{t\geq 0}$ generated by the operator $A+B$ is compact for $t>\max\{G_{1}(1),G_{2}(1)\}$. Hence, the semigroup $(T_{2}(t))_{t\geq 0}$ is compact for $t>\max\{G_{1}(1),G_{2}(1)\}$; see \cite[Lemma III.1.14]{EN}. By \cite[Lemma II.4.22]{EN}, the semigroup $(T_{2}(t))_{t\geq 0}$ is norm continuous for $t>\max\{G_{1}(1),G_{2}(1)\}$. That completes the proof. \end{proof} \begin{lemma} \label{lem4.2} The semigroup $(T_2(t))_{t\geq 0}$ is irreducible. \end{lemma} \begin{proof} The proof of this lemma is similar as that in Lemma \ref{lem3.4} except for the definition of the operators $H_{\lambda}$. Here for each $\lambda\in \mathbb{C}$, we define \begin{align*} H_{\lambda}(f_1(x),f_2(x)) &=\Big( \int^{x}_{0}\frac{\varepsilon_{1\lambda}(x)\mu(s)f_2(s)} {\varepsilon_{1\lambda}(s)\gamma_{1}(s)}ds, \int^{x}_{0}\frac{\varepsilon_{2\lambda}(x)} {\varepsilon_{2\lambda}(s)\gamma_{2}(s)}\int^{1}_{s}b(s,y)f_1(y)\,dy\,ds\Big) \\ &=\Big(\int^{x}_{0}\frac{\varepsilon_{1\lambda}(x)\mu(s)f_2(s)} {\varepsilon_{1\lambda}(s)\gamma_{1}(s)}ds, \int^{x}_{0}f_1(y)\int^{y}_{0}\frac{\varepsilon_{2\lambda}(x)b(s,y)} {\varepsilon_{2\lambda}(s)\gamma_{2}(s)}\,ds\,dy\\ &\quad +\int^{1}_{x}f_1(y)\int^{x}_{0}\frac{\varepsilon_{2\lambda}(x)b(s,y)} {\varepsilon_{2\lambda}(s)\gamma_{2}(s)}\,ds\,dy\Big), \end{align*} where $\varepsilon_{1\lambda}(x)$ and $\varepsilon_{2\lambda}(x)$ are correspondingly the same with those appearing in the proof of Lemma \ref{lem3.4}. We also note that \eqref{e4.2}, \eqref{e4.3} and the assumption on $B(x)$ play a important role to obtain that $\pi_1(R(\lambda,L)F)(x)>0$ and $\pi_2(R(\lambda,L)F)(x)>0$ for almost all $x\in[0,1]$. \end{proof} We also have that there exist an eigenvalue $\lambda$ of the redefined operator $L$ (which is also the spectral bound of the redefined operator $L$) and the strictly positive associated eigenvector $(\hat{m}(x),\hat{n}(x))$; i.e., \label{e4.7} \begin{gathered} (\gamma_1(x)\hat{m}(x))'+\lambda\hat{m}(x) =-\nu(x)\hat{m}(x)-B(x)\hat{m}(x)+\mu(x)\hat{n}(x),\quad 0