exact solution diffusion advection moving domain

thesis_notes.tex

+\documentclass[10pt,a4paper]{report}
+\usepackage[utf8]{inputenc}
+\usepackage{amsmath}
+\usepackage{amsfonts}
+\usepackage{amssymb}
+\usepackage{graphicx}
+\usepackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry}
+\author{Jigyasa Watwani}
+\title{Physics of Growth Regulation in Cells and Tissues - Notes}
+\begin{document}
+\chapter{Exact solution for advection diffusion equation on a moving domain}
+We checked if our numerical solution for the advection-diffusion equation on a moving domain matches the analytical expression in \cite{Simpson2015-xe}
+\bibliographystyle{plain}.
+The equation we intend to solve is:
+\begin{equation} \label{diffusion_advection_equation}
+\frac{\partial C}{\partial t} = D\frac{\partial^2 C}{\partial x^2} - \frac{\partial(vC)}{\partial x}
+\end{equation}
+Here, $C$ is the concentration of the morphogen diffusing and advecting in a $1$-D domain of size $L(t)$. 
+
+Let the fixed domain be parameterized by $s$ and the moving domain be parameterized by $x$. We define a map $x(s,t)$. The velocity of a point $s$ at time $t$ is given by 
+\begin{equation} \label{vel_def}
+v=\frac{\partial x(s,t)}{\partial t} 
+\end{equation}
+
+
+Eventually, we want to scale $x(t)$ by $L(t)$. So we need to find $L(t)$.
+Now, 
+\begin{equation} \label{length_def}
+L(t) = \int_0^{L_0} \frac{\partial x(s,t)}{\partial s} ds
+\end{equation}
+Let us choose a map $x(s,t) = s \exp(\alpha t) $. The velocity of the domain is then given by (from Eq \ref{vel_def}): 
+\begin{equation}
+v(s,t) = \alpha s \exp(\alpha t)
+\end{equation}
+or, in the moving domain:
+\begin{equation}
+v(x,t) = \alpha x
+\end{equation}
+Thus, for our chosen map, the advection velocity is $v = \alpha x$. We also move the domain by this velocity. 
+And finally, the length of the domain is given by (from Eq \ref{length_def} and the definition of our chosen map):
+\begin{equation}
+L(t) = L_0 \exp(\alpha t)
+\end{equation}
+We now scale variables $x \to \xi = x/L(t)$. Note then that the derivatives transform as: 
+\begin{equation}
+\frac{\partial C}{\partial x} \to \frac{1}{L(t)} \frac{\partial C}{\partial \xi} 
+\end{equation}
+\begin{equation}
+\frac{\partial^2 C}{\partial x^2} \to \frac{1}{L(t)^2} \frac{\partial^2 C}{\partial \xi^2} 
+\end{equation}
+and 
+\begin{equation}
+\frac{\partial C}{\partial t} \to \frac{\partial C}{\partial t} - \alpha \xi \frac{\partial C}{\partial \xi} 
+\end{equation}
+
+Converting Eq(\ref{diffusion_advection_equation}) in terms of $\{\zeta, t\}$
+
+\begin{equation} \label{rescaled1}
+\frac{\partial C}{\partial t} = \frac{D}{L(t)^2}\frac{\partial^2 C}{\partial \zeta^2} - \alpha C
+\end{equation}
+We again scale $t \to T = \int_0^t \frac{D}{L(s)^2}ds $, i.e, 
+\begin{equation}
+T = \frac{D}{2 \alpha L_0^2}(1- \exp(-2 \alpha t))
+\end{equation}
+or, 
+\begin{equation}
+\exp(-2 \alpha t) = 1-\frac{2 \alpha L_0^2 T}{D} \implies \exp(2 \alpha t) = \frac{D}{D-2\alpha L_0^2 T} 
+\end{equation}
+and again changing variables in Eq(\ref{rescaled1}) from $\{\zeta, t\} \to \{\zeta, T\}$, the derivatives transform as:
+\begin{equation}
+\frac{\partial C}{\partial t} \to \frac{D-2\alpha L_0^2 T}{L_0^2} \frac{\partial C}{\partial T}
+\end{equation}
+This implies Eq(\ref{rescaled1}) becomes:
+\begin{equation}
+\frac{\partial C}{\partial T} + \frac{\alpha C L_0^2}{D - 2 \alpha L_0^2 T}= \frac{\partial^2 C}{\partial \zeta^2}
+\end{equation}
+This can be solved by separation of variables to give the most general solution:
+\begin{equation}
+C(\zeta, T) = \exp(-k^2 T) \sqrt{\frac{D - 2\alpha L_0^2 T}{D}}[c_1 \sin(k\zeta) + c_2 \cos(k\zeta)]
+\end{equation}
+Using Neumann boundary conditions $\partial C/\partial \zeta = 0$ at $\zeta = 0$ and $\zeta = 1$, we get $c_1=0,k=n\pi$.
+Thus, for any initial condition, we have,
+\begin{equation}
+C(\zeta, T) = \sum_n a_n \exp\left(-n^2 \pi^2 T \right) \cos\left(n \pi\zeta\right) \sqrt{\frac{D - 2 \alpha L_0^2 T}{D}}
+\end{equation}
+This is exactly the same as Eq(11) in the reference with $f(T^\star)=\alpha L_0^2/(2\alpha L_0^2T^\star-D)$. All that is left now is to convert back to the old variables and put in an initial condition:
+\begin{equation}
+C(x,t) = \sum_n a_n \cos\left(\frac{n \pi x}{L(t)}\right)\exp\left(\frac{-n^2 \pi^2 D(1-\exp(-2\alpha t))}{2 \alpha L_0^2}\right) \exp(-\alpha t)
+\end{equation}
+This matches the result (13) of the paper.
+Putting in the initial condition,
+\begin{equation}
+C(x,t) = \exp(-\alpha t)\left[1+ 0.2\cos\left(\frac{\pi x}{L_0} \exp(-\alpha t)\right)\exp\left(\frac{-\pi^2D}{2 \alpha L_0^2}(1-\exp(-2 \alpha t))\right)\right]
+\end{equation}
+\bibliography{ref}
+\end{document}
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