complete analytical solutions in 2D/3D with specified initial and boundary conditions

growing_domain/notes.tex

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+\documentclass[11pt, reqno]{report}
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@@ -13,14 +13,14 @@
 
 
 \begin{document}
 \begin{document}
 
 
-\title{Diffusion on growing domains}
+\title{Diffusion on a growing domains}
 \date{\currenttime, \today}
 \date{\currenttime, \today}
 \maketitle
 \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).
 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}
+\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,
 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}} 
 \eqn{r \to r^{\prime} = r + v(r, t)\tau.\label{eq:rprime}} 
@@ -53,7 +53,122 @@ Assuming that the circular domain elongates with its center fixed, i.e., $v(0,t)
 v(r,t) = \frac{r}{R(t)} \frac{dR(t)}{dt}.
 v(r,t) = \frac{r}{R(t)} \frac{dR(t)}{dt}.
 }
 }
 
 
-\section{Mass conservation}
+\subsection{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
+\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}
+}
+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
+\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 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
+\eqn{
+\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
+\eqn{
+\frac{\partial v}{\partial r} = \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
+\eqn{
+v(r,t) = \frac{r}{R(t)} \frac{dR(t)}{dt}.
+}
+
+\subsection{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
 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{
 \eqn{
@@ -70,7 +185,7 @@ We assume azimuthal symmetry $C(\vect{r}, t) = C(r, t)$, i.e., the dynamics is i
 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}.
 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}.
 
 
 
 
-\section{Exact solution}
+\subsection{Exact solution}
 
 
 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{
@@ -102,31 +217,39 @@ 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]}
 \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:
 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]}  
 \eqn{C(\xi, T) = a_1 j_0 (n \xi) \exp\left[-n^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(n \xi) - \frac{1}{n \xi^2} \sin(n \xi) \right) \exp\left[-n^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 $n$ to be the solution of:
-\eqn{\tan (n)= n} 
+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:
 Thus, the most general solution is:
-\eq{C(\xi, T) = \sum_{n} a_n j_0(n \xi) \exp\left[-n^2 T + \int_0^T f(T^\star) dT^\star \right]}
-where the sum is over solutions of $\tan(n) = n$ and $a_n$ is determined by the initial conditions.
-\section{No growth}
+\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, $R(t) = R, \sigma(t) = 0, f(T^\star) = R^2 k/D$.
 In this case, $R(t) = R, \sigma(t) = 0, f(T^\star) = R^2 k/D$.
-This gives \eq{C(\xi, T) = \sum_n a_n \left( \frac{\sin (n \xi)}{n \xi}\right) e^{-n^2 T} e^{R^2 k T/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 $\xi = r/R$,
 Plugging in $T = D t/R^2 $ and $\xi = r/R$,
-\eqn{C(r,t) = \sum_n a_n \left(\frac{\sin(nr/R)}{nr/R}\right) e^{-\frac{n^2 D t}{R^2}}e^{kt}}
-\section{Exponential domain growth}
+\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, $R(t) = R_0 e^{\alpha t}, \sigma(t)=\alpha$
 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}}
 \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}}} 
 And \eq{f(	T^\star) = \frac{R_0^2 (k - 3 \alpha)}{D - 2 \alpha R_0^2 T^{\star}}} 
 Then,
 Then,
-\eq{C(\xi, T) = \sum_n a_n j_0(n \xi) e^{-n^2 T} \frac{k - 3 \alpha}{2 \alpha} \left(1 - \frac{2 \alpha R_0^2 T}{D}\right)}
-and \eqn{C(r,t) = \frac{k - 3 \alpha}{2 \alpha} e^{- 2 \alpha t} \sum_n a_n j_0 \left( \frac{n r}{R(t)}\right) e^{-\frac{n^2 D(1 - e^{-2 \alpha t})}{2 \alpha R_0^2}}}
-\section{Linear domain growth}
+\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, $R(t)=R_0 + bt, \sigma(t) = b/R(t)$.\\
 In this case, $R(t)=R_0 + bt, \sigma(t) = b/R(t)$.\\
 Further, \eq{T = \frac{D t}{R(t) R_0}}, or, 
 Further, \eq{T = \frac{D t}{R(t) R_0}}, or, 
 \eq{t = \frac{R_0^2 T}{D - R_0 b T}}
 \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)}
 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
 Then
-\eq{C(\xi, T) = e^{kt} \left(\frac{R_0}{R(t)}\right)^3 \sum_n a_n j_0(n \xi) e^{-n^2 T} e^{\frac{k R_0^2 T}{D - R_0 b T}}\left(\frac{D - R_0 b T}{D}\right)^3}
-\eqn{C(r, t) = \sum_n a_n j_0\left(\frac{n r}{R(t)}\right)e^{-\frac{n^2 D t}{R(t) R_0}} }
+\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}
+
 \end{document}
 \end{document}