exact solution

growing_domain/notes.tex

@@ -22,7 +22,22 @@ Largely following \href{https://journals.plos.org/plosone/article?id=10.1371/jou
 
 
 \section{Kinematics}
 \section{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 $\tau$. By considering the expansion of an element of initial size $\Delta r$, we can derive an expression relating $R(t)$ and $v(r,t)$ which can be written as
+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
+}
+or
 \eqn{
 \eqn{
 \frac{dR(t)}{dt} = \int_0^{R(t)} \frac{\partial v}{\partial r} dr
 \frac{dR(t)}{dt} = \int_0^{R(t)} \frac{\partial v}{\partial r} dr
 .
 .
@@ -59,8 +74,36 @@ and impose zero diffusive flux conditions $\partial_r C = 0$ at $r=0$ and at $r=
 
 
 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 $\xi = \dfrac{r}{R(t)}$. Then
 \eqn{
 \eqn{
-\frac{\partial C}{\partial t} =
+\frac{\partial C}{\partial t} \to \frac{\partial C}{\partial t} - \frac{v}{R(t)} \frac{\partial C}{\partial \xi}
 }
 }
-
-
-\end{document}
+\eqn{
+\frac{\partial C}{\partial r} \to \frac{1}{R(t)} \frac{\partial C}{\partial \xi}
+}
\ No newline at end of file
+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) = -k^2}
+Solving the $v$ equation first:
+\eqn{v(T) = v(0) \exp\left[-k^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} + k^2 u = 0}
+This is the Bessel equation of zeroth order, having solutions
+\eqn{u(\xi) = c_1 j_0 (k \xi) + c_2 y_0 (k \xi)}
+where $j_0, y_0$ are spherical bessel functions given by 
+\eq{j_0 (k \xi) = \frac{\sin(k \xi)}{k \xi}}
+\eq{y_0 (k \xi) = -\frac{\cos(k \xi)}{k \xi}}
+Thus, the exact solution in the $(\xi, T)$ coordinates is :
+\eqn{C(\xi, T) = \left[ a_1 j_0 (k \xi) + a_2 y_0 (k \xi)\right] \exp\left[-k^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 (k \xi) \exp\left[-k^2 T + \int_0^T f(T^\star) dT^\star \right]}  
+And
+\eqn{\frac{\partial C}{\partial \xi} = a_1 \left(\frac{1}{\xi} \cos(k \xi) - \frac{1}{k \xi^2} \sin(k \xi) \right) \exp\left[-k^2 T + \int_0^T f(T^\star) dT^\star \right].\label{neumann}}
+Demanding Eq(\ref{neumann}) to be zero at $\xi = 1$ (Neumann boundary conditions at $\xi = 1$), one can get $k$ to be the solution of:
+\eqn{\tan (k)= k} 
+\end{document}