\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{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}),
&\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 $n \geq2$). 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
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,
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:
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
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}),
&\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
\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(t(\tau)) = \frac{R(t(\tau))^2}{D}\big[k - n \, \sigma(t(\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 $C(\xi, \tau)= u(\xi)\; w(\tau)$,
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{\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}
}
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}.
\subsection{Exact solution}
We first transform the spatial variable to a fixed domain $\xi=\dfrac{r}{R(t)}$. Then
\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
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
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,
