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growing_domain/notes.blg

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growing_domain/notes.out

-\BOOKMARK [1][-]{section.0.1}{Diffusion on a growing disk}{}% 1
-\BOOKMARK [2][-]{subsection.0.1.1}{Kinematics}{section.0.1}% 2
-\BOOKMARK [2][-]{subsection.0.1.2}{Mass conservation}{section.0.1}% 3
-\BOOKMARK [2][-]{subsection.0.1.3}{Exact solution}{section.0.1}% 4
-\BOOKMARK [2][-]{subsection.0.1.4}{Specific examples}{section.0.1}% 5
-\BOOKMARK [1][-]{section.0.2}{Diffusion in a growing sphere}{}% 6
-\BOOKMARK [2][-]{subsection.0.2.1}{Kinematics}{section.0.2}% 7
-\BOOKMARK [2][-]{subsection.0.2.2}{Mass conservation}{section.0.2}% 8
-\BOOKMARK [2][-]{subsection.0.2.3}{Exact solution}{section.0.2}% 9
-\BOOKMARK [2][-]{subsection.0.2.4}{Specific examples}{section.0.2}% 10

growing_domain/notes.tex

-\documentclass[11pt, reqno]{amsart}
-\usepackage{amssymb,amsmath,datetime,soul,graphicx,geometry}
-\usepackage[dvipsnames]{xcolor}
-\linespread{1.25}
-\usepackage[colorlinks=true,citecolor=RoyalBlue,urlcolor=RoyalBlue]{hyperref}
-
-% shortcuts
-\newcommand{\vect}[1]{\boldsymbol{#1}}
-\newcommand{\tens}[1]{\mathsf{#1}}
-\newcommand{\uvec}[1]{\hat{\boldsymbol{#1}}}
-\newcommand{\eqn}[1]{\begin{align}#1\end{align}}
-\newcommand{\eq}[1]{\begin{align*}#1\end{align*}}
-
-
-\begin{document}
-
-\title{Diffusion in isotropically growing domains}
-\date{\currenttime, \today}
-\maketitle
-
-Largely following \href{https://journals.plos.org/plosone/article?id=10.1371/journal.pone.0117949}{Exact Solutions of Linear Reaction-Diffusion Processes on a Uniformly Growing Domain: Criteria for Successful Colonization}, M J Simpson, PLoS One (2015).
-
-\section{Kinematics}
-
-We consider a linear reaction-diffusion process on a growing $n-$dimensional domain labelled by points $\vect{r} = (r, \uvec{n})$ where $0 < r < R(t)$ with $R(t)$ is the increasing radius of the domain and $\uvec{n}$ is a unit-vector indicating a direction ($\uvec{n}$ only exists for $n \geq 2$). Domain growth is associated with a \emph{radial} velocity field $\vect{v} = v(r,t) \, \uvec{r}$ which causes points at a distance $r$ from the origin to move to a distance $r + v(r,t) \, \Delta t $ in a short time $\Delta t$. Thus points at distances $r$ and $r + \Delta r$, respectively, move to
-\eqn{r \to r^{\prime} &= r + v(r, t) \;\tau,
-\\
-r + \Delta r \to r^{\prime} + \Delta r^{\prime} &= r + \Delta r + v(r + \Delta r, t) \; \Delta t.} 
-Thus
-\eqn{
-\Delta r^{\prime} = \Delta r + \Delta t \; \big[ v(r + \Delta r, t) - v(r,t) \big] 
-\approx 
-\Delta r \left( 1 + \Delta t \frac{\partial v}{\partial r}\right)
-}
-implying
-\eqn{
-R^{\prime}(t) = R(t) + \Delta t \int_0^{R(t)} \frac{\partial v}{\partial r} dr
-\qquad
-\implies
-\qquad
-\frac{dR(t)}{dt} = \int_0^{R(t)} \frac{\partial v}{\partial r} dr
-}
-We consider uniform growth, i.e., $\partial_r v \equiv \sigma(t)$ is independent of $r$ but could potentially depend on time $t$. Thus
-\eqn{
-\sigma(t) = \frac{1}{R(t)} \frac{dR(t)}{dt}.
-\label{eq:growth_rate}
-}
-We impose the boundary condition $v(0,t) = 0$ using which, on integrating \eqref{eq:growth_rate}, we find
-\eqn{
-v(r,t) = \frac{r}{R(t)} \frac{dR(t)}{dt} = \sigma(t) \; r.
-}
-
-\section{Mass conservation}
-
-We now consider the conservation of a mass density $C(\vect{r}, t)$, assuming that it evolves according to a linear reaction-diffusion mechanism. The associated conservation statement on the growing domain can be written as
-\eqn{
-\partial_t C = D \nabla^2 C - \nabla \cdot (\vect{v} C) + k \, C,
-\label{eq:general_reaction_diffusion}
-}
-where $D>0$ and $k$ are the diffusion constant and reaction rate respectively. Assuming a solution that only depends on the growth direction, i.e., $C(\vect{r}, t) = C(r, t)$, we get
-\eqn{
-\frac{\partial C}{\partial t} = \frac{D}{r^{n-1}} \, \frac{\partial}{\partial r} \left(r^{n-1} \frac{\partial C}{\partial r} \right) - \frac{1}{r^{n-1}} \, \frac{\partial (r^{n-1} v C)}{\partial r}  + k\,C,
-\\
-= \frac{D}{r^{n-1}} \, \frac{\partial}{\partial r} \left(r^{n-1} \frac{\partial C}{\partial r} \right) - \frac{\sigma(t)}{r^{n-1}} \, \frac{\partial (r^{n} C)}{\partial r}  + k\,C.
-\label{eq:azimuthal_symmetry}
-}
-We impose zero diffusive flux conditions $\partial_r C = 0$ at $r=0$ and at $r=R(t)$. With some given initial conditions $C(r, 0)$ our aim is to find the exact solution of \eqref{eq:azimuthal_symmetry}.
-
-
-\section{Rescaling}
-
-We first transform the spatial variable to a fixed domain:
-\eqn{ r \to \xi = \frac{r}{R}.}
-This leads to
-\eqn{
-\frac{\partial}{\partial t} \to \frac{\partial}{\partial t} - \sigma \; \xi \frac{\partial}{\partial \xi},
-\qquad \qquad
-\frac{\partial}{\partial r} \to \frac{1}{R} \frac{\partial}{\partial \xi}
-}
-which implies
-\eqn{
-\frac{\partial C}{\partial t} = 
-\frac{D }{\xi^{n-1} R^2} \, \frac{\partial}{\partial \xi} \left(\xi^{n-1} \frac{\partial C}{\partial \xi} \right)  + (k - n \, \sigma) C.
-}
-Next, we transform the variable:
-\eqn{
-t \to \tau  = \int_0^t ds \; \frac{D}{R^2(s)}.
-}
-Then
-\eqn{
-\frac{\partial C}{\partial \tau} = 
-\frac{1}{\xi^{n-1}} \, \frac{\partial}{\partial \xi} \left(\xi^{n-1} \frac{\partial C}{\partial \xi} \right)  + f(\tau) C
-\label{eq:nondim_eqn}
-}
-where
-\eqn{
-f(t(\tau)) = \frac{R(t(\tau))^2}{D} \big[k - n \, \sigma(t(\tau)) \big].
-}
-
-\section{Exact solutions}
-
-We use separation of variables to solve \eqref{eq:nondim_eqn}. Writing $C(\xi, \tau) = u(\xi) \; w(\tau)$,
-\eqn{\frac{1}{w} \frac{dw}{d\tau} - f(\tau)= \frac{1}{u \, \xi^{n-1}} \frac{d}{d \xi} \left( \xi^{n-1} \frac{du}{d \xi} \right).}
-With $-\omega^2$ as a separation constant, we get
-\eqn{
-\frac{dw}{d\tau} = \big[ f(\tau) - \omega^2 \big] w
-\qquad
-\implies
-\qquad
-w(\tau) = w(0) \exp\left[-\omega^2 \tau + \int_0^\tau ds\; f(s) \right],
-}
-and
-\eqn{
-\frac{d^2 u}{d\xi^2} + \frac{(n-1)}{\xi} \frac{du}{d\xi} + \omega^2 u = 0.
-\label{eq:u_eqn}
-}
-We need to find solutions of \eqref{eq:u_eqn} with the boundary conditions that $u'(\xi) = 0$ at both $\xi=0$ and $\xi=1$.
-\begin{itemize}
-\item When $n=1$, we get
-\eqn{u'' + \omega^2 u = 0,}
-and thus we need $\omega=m\pi, m \in \mathbb{Z}$ to satisfy the boundary conditions. The general solution is thus
-\eqn{u(\xi, \tau) = \sum_{m=0}^{\infty} a_m \cos\left(m \pi \xi \right) \; \exp\left[-m^2 \pi^2 \tau + \int_0^\tau ds\; f(s) \right].}
-%
-\item When $n=2$, we get
-\eqn{\xi \, u'' + u' + \omega^2 \, \xi \, u = 0,}
-the solutions of which are the zeroth-order Bessel functions $J_0(\omega\xi)$ and $Y_0(\omega\xi)$. Requiring $u$ to be bounded at $\xi=0$, we neglect the $Y_0$ solution. Imposing the boundary condition
-\eqn{\frac{dJ_0(z)}{dz}\bigg\vert_{z=\omega}=0 \qquad \implies J_1(\omega)=0,
-\label{eq:bc_n2}}
-determines the allowed values of $\omega$ as the zeroes of the 1st-order Bessel function. The general solution is thus
-\eqn{u(\xi, \tau) = \sum_{\omega} a_{\omega}  \, J_0(\omega \xi) \; \exp\left[-\omega^2 \tau + \int_0^\tau ds\; f(s) \right],}
-where the summation is over those values of $\omega$ satisfying \eqref{eq:bc_n2}.
-%
-\item When $n=3$, we get
-\eqn{\xi \, u'' + 2 \, u' + \omega^2 \, \xi \, u = 0,}
-the solutions of which are the zeroth-order \emph{spherical} Bessel functions
-\eqn{
-j_0(\omega \xi) = \frac{\sin (\omega \xi)}{\omega \xi},
-\qquad \mathrm{and} \qquad 
-y_0(\omega \xi) = -\frac{\cos (\omega \xi)}{\omega \xi}.} 
-Requiring $u$ to be bounded at $\xi=0$, we neglect the $y_0$ solution. Imposing the boundary condition
-\eqn{\frac{dj_0(z)}{dz}\bigg\vert_{z=\omega}=0 \qquad \implies \qquad
-\tan \omega = \omega,
-\label{eq:bc_n3}}
-determines the allowed values of $\omega$. The general solution is thus
-\eqn{u(\xi, \tau) = \sum_{\omega} a_{\omega} \, \frac{\sin (\omega \xi)}{\omega \xi} \; \exp\left[-\omega^2 \tau + \int_0^\tau ds\; f(s) \right],}
-where the summation is over those values of $\omega$ satisfying \eqref{eq:bc_n3}.
-\end{itemize}
-
-\section{Three cases}
-
-\subsection{Non-growing domain}
-With $R(t) = R_0$ in this case, we get 
-\eqn{\sigma(t)=0, \qquad \tau=\frac{Dt}{R_0^2}, \qquad f=\frac{R_0^2k}{D}.} 
-Hence $\tau \to\infty$ as $t\to\infty$. Thus
-\begin{itemize}
-%
-\item $n=1$
-\eqn{
-C(r,t) = \sum_{m=0}^{\infty} a_m \cos\left(\frac{m \pi r}{R_0} \right) \; \exp\left(-\frac{m^2 \pi^2 D t}{R_0^2} + k t \right),
-\qquad \mathrm{with} \quad m \in \mathbb{Z}
-}
-%
-\item $n=2$
-\eqn{C(r, t) = \sum_{\omega} a_{\omega}  \, J_0\left(\frac{\omega r}{R_0} \right) \; \; \exp\left(-\frac{\omega^2 D t}{R_0^2} + k t \right),
-\qquad \mathrm{with} \quad J_1(\omega)=0.
-}
-%
-\item $n=3$
-\eqn{C(r, t) = \sum_{\omega} a_{\omega}  \, \frac{R_0}{\omega r} \; \sin \left(\frac{\omega r}{R_0}\right) \; \; \exp\left(-\frac{\omega^2 D t}{R_0^2} + k t \right),
-\qquad \mathrm{with} \quad \tan \omega = \omega.}
-\end{itemize}
-
-
-\subsection{Exponential domain growth}
-
-With $R(t) = R_0 e^{\alpha t}$ in this case, we get, 
-\eqn{\sigma(t)=\alpha, \qquad \tau=\frac{D \; (1-e^{-2\alpha t})}{2\alpha R_0^2} , \qquad \int_0^\tau f(s) ds = \ln\left(\frac{D}{D - 2 \alpha R_0^2 \tau}\right)^{(k-n\alpha)/2\alpha} .}
-Hence $\tau \to D/(2\alpha R_0^2)$ as $t\to\infty$. Thus
-\begin{itemize}
-%
-\item $n=1$
-\eqn{C(r,t) = \sum_{m=0}^\infty a_m \cos\left( \frac{m \pi r}{R(t)}\right) \exp\left[{-\frac{m^2 \pi^2 D}{2 \alpha R_0^2} (1 - e^{-2\alpha t})+t(k-\alpha)}\right],
-\qquad \mathrm{with} \quad m \in \mathbb{Z}}
-%
-\item $n=2$
-\eqn{C(r,t) = \sum_{\omega}^\infty a_\omega J_0\left( \frac{\omega r}{R(t)}\right) \exp\left[{-\frac{\omega^2 D}{2 \alpha R_0^2} (1 - e^{-2\alpha t})+t(k-2\alpha)}\right],
-\qquad \mathrm{with} \quad J_1(\omega)=0.}
-%
-\item $n=3$
-\eqn{C(r,t) = \sum_{\omega}^\infty a_\omega \frac{ R(t)}{\omega r} \sin\left( \frac{\omega r}{R(t)}\right) \exp\left[{-\frac{\omega^2 D}{2 \alpha R_0^2} (1 - e^{-2\alpha t})+t(k-3\alpha)}\right],
-\qquad \mathrm{with} \quad \tan \omega = \omega.}
-\end{itemize}
-
-
-\subsection{Linear domain growth}
-With $R(t) = R_0 + \alpha t$ in this case, we get 
-\eqn{\sigma(t) = \frac{\alpha}{R_0 + \alpha t}, \qquad\tau=\frac{D \; t}{R_0 (R_0 + \alpha t)} , \qquad \int_0^\tau f(s) ds = \frac{R_0^2 k \tau}{(D - \alpha R_0 \tau)} + \ln \left( \frac{D - R_0 \alpha \tau}{D}\right)^{n}.}
-Hence $\tau \to D/(\alpha R_0)$ as $t\to\infty$. Thus
-\begin{itemize}
-%
-\item $n=1$
-\eqn{C(r,t) = \sum_{m=0}^\infty a_m \cos\left( \frac{m \pi r}{R(t)}\right) \exp\left[{-\frac{m^2 \pi^2 D t}{R(t) R_0} + tk}\right] \left(\frac{R_0}{R(t)}\right),
-\qquad \mathrm{with} \quad m \in \mathbb{Z}}
-%
-\item $n=2$
-\eqn{C(r,t) = \sum_{\omega}^\infty a_\omega J_0\left( \frac{\omega r}{R(t)}\right) \exp\left[{-\frac{\omega^2 D t}{R(t) R_0} + tk}\right] \left(\frac{R_0}{R(t)}\right)^2,
-\qquad \mathrm{with} \quad J_1(\omega)=0.}
-%
-\item $n=3$
-\eqn{C(r,t) = \sum_{\omega}^\infty a_\omega \frac{ R(t)}{\omega r} \sin\left( \frac{\omega r}{R(t)}\right) \exp\left[{-\frac{\omega^2 D t}{R(t) R_0} + tk} \right] \left(\frac{R_0}{R(t)}\right)^3,
-\qquad \mathrm{with} \quad \tan \omega = \omega.}
-\end{itemize}
-
-\end{document}