Setup for generic calculation in n=1,2,3 dimensions and for none/exponential/linear growth

growing_domain/notes.tex

-\documentclass[11pt, reqno]{report}
-\usepackage{amssymb, amsmath, datetime, xcolor, soul, graphicx}
+\documentclass[11pt, reqno]{amsart}
+\usepackage{amssymb,amsmath,datetime,soul,graphicx,geometry}
+\usepackage[dvipsnames]{xcolor}
 \linespread{1.25}
-\usepackage[colorlinks=true, citecolor=blue, urlcolor=blue]{hyperref}
+\usepackage[colorlinks=true,citecolor=RoyalBlue,urlcolor=RoyalBlue]{hyperref}
 
 % shortcuts
 \newcommand{\vect}[1]{\boldsymbol{#1}}
@@ -13,243 +14,193 @@
 
 \begin{document}
 
-\title{Diffusion on a growing domains}
+\title{Diffusion in an 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{Diffusion on a growing disk}
-\subsection{Kinematics}
-
-We consider a linear reaction-diffusion process on a growing disk labelled by points $\vect{r} = (r, \varphi)$ with $0 < r < R(t)$ and $0 \leq \varphi < 2\pi$ where $R(t)$ is the increasing radius of the domain. Domain growth is associated with a \emph{radial} velocity field $\vect{v} = v(r,t) \, \uvec{r}$ which causes points on a circle of radius $r$ to move to a circle of radius $r + v(r,t) \, \tau $ in a short time $\tau$. Thus,
-\eqn{r \to r^{\prime} = r + v(r, t)\tau.\label{eq:rprime}} 
-And, similarly, 
-\eqn{r + \Delta r \to r^{\prime} + \Delta r^{\prime} = r + \Delta r + v(r + \Delta r, t)\tau .\label{eq:rprimeplusdeltarprime}} 
-Subtracting Eq(\ref{eq:rprime}) from Eq(\ref{eq:rprimeplusdeltarprime}),
-\begin{equation}
-\begin{split}
-\Delta r^{\prime} &= \Delta r + \tau[v(r + \Delta r, t) - v(r,t)] \\
-&\approx \Delta r \left( 1 + \tau \frac{\partial v}{\partial r}\right)
-\end{split}
-\end{equation}
-Integrating both sides, 
-\eq{
-R^{\prime}(t) = R(t) + \tau \int_0^{R(t)} \frac{\partial v}{\partial r} dr
+\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)
 }
-or
+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
-.
-\label{eq:growing_radius}
 }
-We consider uniform growth, i.e., $\partial_r v$ is independent of $r$ but potentially depends on time $t$,  so that we have $\partial_r v = \sigma(t)$. Combining this with \eqref{eq:growing_radius} gives
+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{
-\frac{\partial v}{\partial r} = \sigma(t) = \frac{1}{R(t)} \frac{dR(t)}{dt}.
+\sigma(t) = \frac{1}{R(t)} \frac{dR(t)}{dt}.
 \label{eq:growth_rate}
 }
-Assuming that the circular domain elongates with its center fixed, i.e., $v(0,t) = 0$, integrating \eqref{eq:growth_rate} gives
+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}.
+v(r,t) = \frac{r}{R(t)} \frac{dR(t)}{dt} = \sigma(t) \; r.
 }
 
-\subsection{Mass conservation}
+\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.
-
-We assume azimuthal symmetry $C(\vect{r}, t) = C(r, t)$, i.e., the dynamics is independent of $\varphi$. Then
+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} = D \, \frac{1}{r} \, \frac{\partial}{\partial r} \left(r \frac{\partial C}{\partial r} \right) - \frac{1}{r} \, \frac{\partial (r v C)}{\partial r}  + k\,C,
-\label{eq:azimuthal_symmetry}
+\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.
 }
-and 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}.
+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}.
 
 
-\subsection{Exact solution}
+\section{Rescaling}
 
-We first transform the spatial variable to a fixed domain $\xi = \dfrac{r}{R(t)}$. Then
+We first transform the spatial variable to a fixed domain:
+\eqn{ r \to \xi = \frac{r}{R}.}
+This leads to
 \eqn{
-\frac{\partial C}{\partial t} \to \frac{\partial C}{\partial t} - \frac{v}{R(t)} \frac{\partial C}{\partial \xi}
+\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 r} \to \frac{1}{R(t)} \frac{\partial C}{\partial \xi}
+\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.
 }
-Then Eq(\ref{eq:azimuthal_symmetry}) becomes:
-\eqn{\frac{\partial C}{\partial t} = \frac{D}{\xi R^2(t)} \frac{\partial}{\partial \xi} \left( \xi \frac{\partial C}{\partial \xi} \right) + \left(k - 2 \sigma(t)\right) C}
-We now transform the variable $t: t \to T = \int_0^t \frac{D}{R^2(s)} ds$, giving
-\eqn{\frac{\partial C}{\partial T} = \frac{1}{\xi} \frac{\partial}{\partial \xi} \left( \xi \frac{\partial C}{\partial \xi} \right) + \frac{R^2(t)}{D}\left(k - 2 \sigma(t) \right) C}
-Finally, we express the second term on the RHS as a function of $T$ to write:
-\eqn{\frac{\partial C}{\partial T} = \frac{1}{\xi} \frac{\partial}{\partial \xi} \left( \xi \frac{\partial C}{\partial \xi} \right) + f(T) C}
-The above equation can be solved easily by separation of variables. Writing $C(\xi, T) = u(\xi) w(T)$,
-\eqn{\frac{1}{w} \frac{dw}{dt} = \frac{1}{u \xi} \frac{d}{d \xi} \left( \xi \frac{du}{d \xi} \right) + f(T)}
-Thus, 
-\eqn{\frac{1}{w} \frac{dw}{dT} - f(T) = \frac{1}{u \xi} \frac{d}{d \xi} \left( \xi \frac{du}{d \xi} \right) = -n^2}
-Solving the $w$ equation first:
-\eqn{w(T) = w(0) \exp\left[-n^2 T + \int_0^T f(T^{\star}) dT^\star \right]}
-The $u$ equation reads:
-\eqn{\xi \frac{d^2 u}{d \xi^2} + \frac{du}{d \xi} + n^2 u \xi = 0}
-This is the Bessel equation of zeroth order, having solutions
-\eqn{u(\xi) = c_1 J_0 (\xi n) + c_2 yY_0 (\xi n)}
-Thus, the exact solution in the $(\xi, T)$ coordinates is :
-\eqn{C(\xi, T) = \left[ a_1 J_0 (\xi n) + a_2 Y_0 (\xi n)\right] \exp\left[-n^2 T + \int_0^T f(T^\star) dT^\star \right]}
-Since we want the solution to be bounded at $\xi = 0$, we keep only the $J_0$ term:
-\eqn{C(\xi, T) = a_1 J_0 (\xi n) \exp\left[-n^2 T + \int_0^T f(T^\star) dT^\star \right]}  
-If we had Dirichlet boundary conditions at $\xi = 1$, $n$ would be the solution of 
-\eqn{J_0 (n) = 0}
-i.e $n$'s are the roots of the zeroth bessel function.
-Thus, the most general solution is:
-\eqn{C(\xi, T) = \sum_{n} a_n J_0(\eta_n \xi) \exp\left[-\eta_n^2 T + \int_0^T f(T^\star) dT^\star \right]
-\label{general_sol}}
-where $\eta_n$ is the $n$th root of the zeroth Bessel function for Dirichlet boundary conditions. \\$a_n$ is determined by the initial conditions. Let us take 
-\eqn{C(\xi, 0) = J_0(\eta_m \xi) = \sum_{n} a_n J_0(\eta_n \xi)}
-where the last equality follows by putting $T=0$ in Eq(\ref{general_sol}). Now using the orthogonality relation
-\eq{\int_0^a J_v \left( \eta_{\nu p}\frac{\xi}{a}\right)J_v \left( \eta_{\nu q}\frac{\xi}{a}\right) \xi d \xi = \frac{a^2}{2}[J_{\nu+1} (\eta_{\nu p})]^2 \delta_{pq}}
-where $\eta_{\nu p}$ and $\eta_{\nu q}$ are the $p$th and $q$th roots of the $\nu$th bessel function, we get 
-\eqn{a_{n=m} = 1, a_{n \neq m} = 0}.
-In summary, the solution to our problem for the initial condition $C(\xi, 0) = J_0(\eta_m \xi)$ where $\eta_m$ is the mth root of $j_0$ is:
-\eqn{C(\xi, T) = J_0(\eta_{m} \xi) \exp \left[-\eta_m^2 T + \int_0^T f(T^\star)dT^\star \right]}.                                                                        
-\subsection{Specific examples}
-\subsubsection*{No growth}
-In this case, $R(t) = R, \sigma(t) = 0, f(T^\star) = R^2 k/D$.
-This gives \eq{C(\xi, T) =  J_0 (\eta_m \xi) e^{-\eta_m^2 T} e^{R^2 k T/D}}
-Plugging in $T = D t/R^2 $ and $\xi = r/R$,
-\eqn{C(r,t) = J_0 (\eta_m r/R) e^{-\frac{\eta_m^2 D t}{R^2}}e^{kt}}
-\subsubsection{Exponential domain growth}
-In this case, $R(t) = R_0 e^{\alpha t}, \sigma(t)=\alpha$
-\eq{ T = \frac{D(1 - e^{-2 \alpha t})}{2 \alpha R_0^2}}
-And \eq{f(	T^\star) = \frac{R_0^2 (k - 2 \alpha)}{D - 2 \alpha R_0^2 T^{\star}}} 
-Then,
-\eq{C(\xi, T) = J_0(\eta_m \xi) e^{-\eta_m^2 T} \left(\frac{D}{D - 2\alpha R_0^2 T}\right)^{(k-2\alpha)/2\alpha}}
-and \eqn{C(r,t) = J_0\left(\eta_m \frac{r}{R(t)}\right) \exp\left( -\frac{\eta_m^2 D(1 - e^{-2 \alpha t})}{2 \alpha R_0^2}\right) e^{(k - 2\alpha)t}}
-\subsubsection{Linear domain growth}
-In this case, $R(t)=R_0 + bt, \sigma(t) = b/R(t)$.\\
-Further, \eq{T = \frac{D t}{R(t) R_0}}, or, 
-\eq{t = \frac{R_0^2 T}{D - R_0 b T}}
-And \eq{f(T^\star) = \frac{R_0}{(D - R_0 bT^\star)^2} (k D R_0 - 2 b D + 2bR_0T^\star)}
-Then
-\eq{C(\xi, T) = J_0(\eta_m \xi) e^{-\eta_m^2 T} e^{\frac{(kR_0^2 - 2b + 2 R_0)T}{D - R_0 bT}}\left(\frac{D - R_0 bT}{D}\right)^\frac{2 R_0}{b}}
-\eqn{C(r, t) = J_0\left(\eta_{m} \frac{n r}{R(t)}\right)e^{-\frac{\eta_m^2 D t}{R(t) R_0}} \left(\frac{R_0}{R(t)}\right)^{2R_0/b} e^{kt} e^{\frac{2( R_0-b)t}{R_0^2}}}
-
-\section{Diffusion in a growing sphere}
-\subsection{Kinematics}
-We consider a linear reaction-diffusion process on a growing sphere labelled by points $\vect{r} = (r, \theta, \varphi)$ with $0 < r < R(t)$, $0 \leq \varphi < 2\pi$ and $0 \leq \theta \leq \pi$ where $R(t)$ is the increasing radius of the domain. Domain growth is associated with a \emph{radial} velocity field $\vect{v} = v(r,t) \, \uvec{r}$ which causes points on the surface of the sphere of radius $r$ to move to a circle of radius $r + v(r,t) \, \tau $ in a short time $\tau$. Thus,
-\eqn{r \to r^{\prime} = r + v(r, t)\tau.\label{eq:rprime}} 
-And, similarly, 
-\eqn{r + \Delta r \to r^{\prime} + \Delta r^{\prime} = r + \Delta r + v(r + \Delta r, t)\tau .\label{eq:rprimeplusdeltarprime}} 
-Subtracting Eq(\ref{eq:rprime}) from Eq(\ref{eq:rprimeplusdeltarprime}),
-\begin{equation}
-\begin{split}
-\Delta r^{\prime} &= \Delta r + \tau[v(r + \Delta r, t) - v(r,t)] \\
-&\approx \Delta r \left( 1 + \tau \frac{\partial v}{\partial r}\right)
-\end{split}
-\end{equation}
-Integrating both sides, 
-\eq{
-R^{\prime}(t) = R(t) + \tau \int_0^{R(t)} \frac{\partial v}{\partial r} dr
-}
-or
+Next, we transform the variable:
 \eqn{
-\frac{dR(t)}{dt} = \int_0^{R(t)} \frac{\partial v}{\partial r} dr
-.
-\label{eq:growing_radius}
+t \to \tau  = \int_0^t ds \; \frac{D}{R^2(s)}.
 }
-We consider uniform growth, i.e., $\partial_r v$ is independent of $r$ but potentially depends on time $t$,  so that we have $\partial_r v = \sigma(t)$. Combining this with \eqref{eq:growing_radius} gives
+Then
 \eqn{
-\frac{\partial v}{\partial r} = \sigma(t) = \frac{1}{R(t)} \frac{dR(t)}{dt}.
-\label{eq:growth_rate}
+\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}
 }
-Assuming that the circular domain elongates with its center fixed, i.e., $v(0,t) = 0$, integrating \eqref{eq:growth_rate} gives
+where
 \eqn{
-v(r,t) = \frac{r}{R(t)} \frac{dR(t)}{dt}.
+f(\tau) = \frac{R(\tau)^2}{D} \big[k - n \, \sigma(\tau) \big].
 }
 
-\subsection{Mass conservation}
+\section{Exact solutions}
 
-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
+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{
-\partial_t C = D \nabla^2 C - \nabla \cdot (\vect{v} C) + k \, C,
-\label{eq:general_reaction_diffusion}
+\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],
 }
-where $D>0$ and $k$ are the diffusion constant and reaction rate respectively.
-
-We assume azimuthal symmetry $C(\vect{r}, t) = C(r, t)$, i.e., the dynamics is independent of $\varphi$. Then
+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 need $\omega=m\pi$ 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(z) = \frac{\sin z}{z},
+\qquad \mathrm{and} \qquad 
+y_0(z) = -\frac{\cos z}{z}.} 
+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{
-\frac{\partial C}{\partial t} = D \, \frac{1}{r^2} \, \frac{\partial}{\partial r} \left(r^2 \frac{\partial C}{\partial r} \right) - \frac{1}{r^2} \, \frac{\partial (r^2 v C)}{\partial r}  + k \, C,
-\label{eq:azimuthal_symmetry}
+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).
 }
-and 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}.
+%
+\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{Exact solution}
+\subsection{Exponential domain growth}
+\hl{Should check from here onwards....}
 
-We first transform the spatial variable to a fixed domain $\xi = \dfrac{r}{R(t)}$. Then
-\eqn{
-\frac{\partial C}{\partial t} \to \frac{\partial C}{\partial t} - \frac{v}{R(t)} \frac{\partial C}{\partial \xi}
-}
-\eqn{
-\frac{\partial C}{\partial r} \to \frac{1}{R(t)} \frac{\partial C}{\partial \xi}
-}
-Then Eq(\ref{eq:azimuthal_symmetry}) becomes:
-\eqn{\frac{\partial C}{\partial t} = \frac{D}{\xi^2 R^2(t)} \frac{\partial}{\partial \xi} \left( \xi^2 \frac{\partial C}{\partial \xi} \right) + \left(k - 3 \sigma(t)\right) C}
-We now transform the variable $t: t \to T = \int_0^t \frac{D}{R^2(s)} ds$, giving
-\eqn{\frac{\partial C}{\partial T} = \frac{1}{\xi^2} \frac{\partial}{\partial \xi} \left( \xi^2 \frac{\partial C}{\partial \xi} \right) + \frac{R^2(t)}{D}\left(k - 3 \sigma(t) \right) C}
-Finally, we express the second term on the RHS as a function of $T$ to write:
-\eqn{\frac{\partial C}{\partial T} = \frac{1}{\xi^2} \frac{\partial}{\partial \xi} \left( \xi^2 \frac{\partial C}{\partial \xi} \right) + f(T) C}
-The above equation can be solved easily by separation of variables. Writing $C(\xi, T) = u(\xi) v(T)$,
-\eqn{\frac{1}{v} \frac{dv}{dt} = \frac{1}{u \xi^2} \frac{d}{d \xi} \left( \xi^2 \frac{du}{d \xi} \right) + f(T)}
-Thus, 
-\eqn{\frac{1}{v} \frac{dv}{dT} - f(T) = \frac{1}{u \xi^2} \frac{d}{d \xi} \left( \xi^2 \frac{du}{d \xi} \right) = -n^2}
-Solving the $v$ equation first:
-\eqn{v(T) = v(0) \exp\left[-n^2 T + \int_0^T f(T^{\star}) dT^\star \right]}
-And the $u$ equation reads:
-\eqn{\frac{d^2 u}{d \xi^2} + \frac{2}{\xi} \frac{du}{d \xi} + n^2 u = 0}
-This is the Bessel equation of zeroth order, having solutions
-\eqn{u(\xi) = c_1 j_0 (n \xi) + c_2 y_0 (n \xi)}
-where $j_0, y_0$ are spherical bessel functions given by 
-\eq{j_0 (n \xi) = \frac{\sin(n \xi)}{n \xi}}
-\eq{y_0 (n \xi) = -\frac{\cos(n \xi)}{n \xi}}
-Thus, the exact solution in the $(\xi, T)$ coordinates is :
-\eqn{C(\xi, T) = \left[ a_1 j_0 (n \xi) + a_2 y_0 (n \xi)\right] \exp\left[-n^2 T + \int_0^T f(T^\star) dT^\star \right]}
-Since we want the solution to be bounded at $\xi = 0$, we keep only the $j_0$ term:
-\eqn{C(\xi, T) = a_1 j_0 (n \xi) \exp\left[-n^2 T + \int_0^T f(T^\star) dT^\star \right]}  
-If we have Dirichlet boundary conditions at $\xi =0, 1$ then $n$ can be determined to be the solution of $j_0(n) = 0$, i.e.
-\eq{n = m \pi, m \in \mathbb{Z}}
-Thus, the most general solution is:
-\eq{C(\xi, T) = \sum_{m} a_m j_0(m \pi \xi) \exp\left[-(m \pi)^2 T + \int_0^T f(T^\star) dT^\star \right]}
-We now determine $a_m$ by the initial conditions. Let us take 
-\eqn{C(\xi, 0) = j_0(p \pi \xi) = \sum_m a_m j_0(m \pi \xi)}
-Using the orthogonality relation
-\eqn{\int_0^a j_{\nu}\left(\eta_{\nu p}\frac{\xi}{a}\right) j_{\nu}\left(\eta_{\nu q}\frac{\xi}{a}\right) \xi^2 d \xi = \frac{a^3}{2} [j_{\nu + 1}(\eta_{\nu p})]^2 \delta_{pq}}
-where $\eta_{\nu p}$ and $\eta_{\nu q}$ are the $p$th and $q$th roots of the $\nu$th bessel function, we get
-\eqn{a_{m=p}=1, a_{m \neq p} = 0}
-And thus the solution for initial condition $C(\xi, 0) = j_0(p \pi \xi)$ and Dirichlet boundaries at $\xi = 0,1$ is:
-\eqn{C(\xi, T) = j_0(p \pi \xi) \exp\left[-(p \pi)^2 T + \int_0^T f(T^\star) dT^\star\right]}
-\subsection{Specific examples}
-\subsubsection*{No growth}
-In this case, $R(t) = R, \sigma(t) = 0, f(T^\star) = R^2 k/D$.
-This gives \eq{C(\xi, T) = j_0(p \pi \xi) e^{-p^2 \pi^2 T} e^{R^2 k T/D}}
-Plugging in $T = D t/R^2 $ and $\xi = r/R$,
-\eqn{C(r,t) = j_0\left(p \pi \frac{r}{R(t)}\right) e^{-\frac{p^2 \pi^2 D t}{R^2}}e^{kt}}
-\subsubsection*{Exponential domain growth}
-In this case, $R(t) = R_0 e^{\alpha t}, \sigma(t)=\alpha$
-\eq{ T = \frac{D(1 - e^{-2 \alpha t})}{2 \alpha R_0^2}}
-And \eq{f(	T^\star) = \frac{R_0^2 (k - 3 \alpha)}{D - 2 \alpha R_0^2 T^{\star}}} 
-Then,
-\eq{C(\xi, T) = j_0(p \pi \xi) e^{-(p \pi)^2 T} \left(\frac{D}{D - 2 \alpha R_0^2 T}\right)^{(k-3\alpha)/2\alpha}}
-and \eqn{C(r,t) = j_0\left(p \pi \frac{r}{R(t)}\right) \exp\left[-\frac{(p \pi)^2 D(1 - e^{-2 \alpha t})}{2 \alpha R_0^2} \right] e^{(k-3\alpha)t}}
-\subsubsection*{Linear domain growth}
-In this case, $R(t)=R_0 + bt, \sigma(t) = b/R(t)$.\\
-Further, \eq{T = \frac{D t}{R(t) R_0}}, or, 
-\eq{t = \frac{R_0^2 T}{D - R_0 b T}}
-And \eq{f(T^\star) = \frac{R_0}{(D - R_0 bT^\star)^2} (k D R_0 - 3 b D + 3bR_0T^\star)}
-Then
-\eq{C(\xi, T) = j_0(p \pi \xi) e^{-p^2 \pi^2 T} e^{\frac{(k R_0 + 3 - 3b) R_0 T}{D - R_0 b T}}\left(\frac{D - R_0 b T}{D}\right)^{3/b}}
-\eqn{C(r, t) = \left(\frac{R_0}{R(t)} \right)^{3/b} j_0\left(\frac{p \pi r}{R(t)}\right)e^{-\frac{p^2 \pi^2 D t}{R(t) R_0}} e^{kt} e^{\frac{3(1-b)t}{R_0}} }
-\end{document}
+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 f= \frac{R_0^2 \; (k-n\alpha) \; e^{2\alpha t}}{D} .}
+Hence $\tau \to D/(2\alpha R_0^2)$ as $t\to\infty$. Thus
+\begin{itemize}
+%
+\item $n=1$
+
+%
+\item $n=2$
+
+%
+\item $n=3$
+
+\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 f= \frac{(R_0 + \alpha t)^2}{D} \; \left(k-\frac{n\alpha}{R_0 + \alpha t} \right).}
+Hence $\tau \to D/(\alpha R_0)$ as $t\to\infty$. Thus
+\begin{itemize}
+%
+\item $n=1$
+
+%
+\item $n=2$
+
+%
+\item $n=3$
+
+\end{itemize}
 
 \end{document}