notes added

thesis_notes.tex

@@ -5,33 +5,68 @@
 \usepackage{amssymb}
 \usepackage{amssymb}
 \usepackage{graphicx}
 \usepackage{graphicx}
 \usepackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry}
 \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}
 \begin{document}
-\chapter{A minimal model}
-We model a 1-D tissue as a visco-elastic medium. We write hydrodynamic equations for the evolution of a displacement field $\boldsymbol u(x,t)$, a density field $\rho(x,t)$ and a concentration field for the signaling morphogen $c(x,t)$ that tells the cells when to divide.
-THe equation for $\boldsymbol u$ is a force balance equation (divergence of the total stress = force density). We use the Kelvin Voigt model to be able to get long-time leastic behaviour:
-\begin{equation*}
-\gamma \dot u = \partial_x \sigma
-\end{equation*}
+
+\author{Jigyasa Watwani}
+\title{Physics of Growth Regulation in Cells and Tissues}
+\maketitle
+\tableofcontents
+\chapter{Size regulation in living systems}
+Evidence for size regulation: Transplantation experiments in salamander limbs and mouse spleens \cite{Bryant1984-pa}
+\chapter{Cell size regulation}
+\section{Evidence}
+Probability distributions of quantities like cell length at division, added length in a generation, generation time, cell elongation rate, etc are sharply peaked \cite{Jun2018-wz}, indicating that single cells robustly control their sizes.
+\section{Cell size control mechanisms: Sizers, Adders and Timers}
+Definitions, correlations between growth and size at birth, time taken (in generations) to achieve size homeostasis for the three mechanisms (\cite{RHIND2021R1414})
+\section{Modeling sizers, adders and timers}
+Largely following \cite{RHIND2021R1414}
+\subsection*{Division protein model}
+\subsection*{Diluted inhibitor sizer}
+\subsection*{Accumulating activator sizer}
+\chapter{Tissue size regulation}
+\section{Existing models}
+\subsection{Signaling-based models}
+\subsection{Mechanical models}
+\cite{AEGERTERWILMSEN2007318}, \cite{Shraiman2005-wl}, \cite{Hufnagel2007-nt}, \cite{PhysRevLett.129.048102}
+\chapter{A minimal mechanical model}
+We model a 1-D tissue as a visco-elastic medium. The relevant fields are a displacement field $\boldsymbol u(x,t)$, a density field $\rho(x,t)$ and a concentration field for the signaling morphogen $c(x,t)$ that tells the cells when to divide.
+\begin{figure}[h]
+\includegraphics[scale=0.3]{fields}
+\end{figure}
+
+The equation for $\boldsymbol u$ is a force balance equation $\boldsymbol \nabla \cdot \boldsymbol \sigma = \boldsymbol f$, where $\boldsymbol \sigma$ is the total stress in the tissue and $\boldsymbol f$ is the force density. We use the Kelvin Voigt model [\ref{kelvin_voigt}] to be able to get long-time eleastic behaviour. Thus, 
 \begin{equation*}
 \begin{equation*}
-\sigma = \sigma_{e} + \sigma_{v} + \sigma_{a} = K \partial_x u + \eta \partial_x \dot u + \lambda \partial_x f(c,\rho)
+\boldsymbol \sigma = \boldsymbol {\sigma_{e}} + {\boldsymbol \sigma_{v}} + {\boldsymbol \sigma_{a}}
 \end{equation*}
 \end{equation*}
-Thus, 
+The elastic and viscous stresses can be written in terms of their irreducible components as:
+\begin{equation}
+\boldsymbol{\sigma_e} = -K \frac{1}{d} (\boldsymbol \nabla \cdot \boldsymbol u)\boldsymbol g - 2\mu \left[\frac{1}{2}[\boldsymbol \nabla \boldsymbol u + (\boldsymbol \nabla \boldsymbol u)^T] - \frac{1}{d} (\boldsymbol \nabla \cdot \boldsymbol u)\boldsymbol g \right] 
+\end{equation}
+\begin{equation}
+\boldsymbol{\sigma_v} = -\zeta \frac{1}{d} (\boldsymbol \nabla \cdot \boldsymbol v)\boldsymbol g - 2\eta \left[\frac{1}{2}[\boldsymbol \nabla \boldsymbol v + (\boldsymbol \nabla \boldsymbol v)^T] - \frac{1}{d} (\boldsymbol \nabla \cdot \boldsymbol v)\boldsymbol g \right] 
+\end{equation}
+where $K, \mu, \zeta, \eta$ are the bulk modulus, shear modulus, bulk viscosity and shear viscosity of the visco-elastic material.
+We choose
+\begin{equation}
+\boldsymbol {\sigma_a} = \lambda \boldsymbol g(c, \rho, \boldsymbol u)
+\end{equation}
+The equation for the velocity field then becomes, in the overdamped limit:
 \begin{equation}
 \begin{equation}
-\gamma \dot u = K \partial_x^2 u + \eta \partial_x^2 \dot u + \lambda \partial_x f 
+\gamma \boldsymbol v = \boldsymbol \nabla \cdot (\boldsymbol{\sigma_e} + \boldsymbol{\sigma_v} + \boldsymbol{\sigma_a})
 \end{equation}
 \end{equation}
+
 The equation for the cell density is a diffusion advection equation, with terms added to account for cell birth and death:
 The equation for the cell density is a diffusion advection equation, with terms added to account for cell birth and death:
 \begin{equation}
 \begin{equation}
-\dot \rho = D_\rho \partial_x^2 \rho - \partial_x(\dot u \rho) - k_{\rho}(\rho - \rho_0) + F(u, c_i, \rho, t)
+\dot \rho = D_\rho \boldsymbol \nabla^2 \rho - \boldsymbol \nabla \cdot (\boldsymbol v \rho) - k_{\rho}(\rho - \rho_0) + F(\boldsymbol u, c_i, \rho, t)
 \end{equation} 
 \end{equation} 
 The $c-$dependence of $F$ is the sizer/adder-like behaviour that changes cell density/number based on the concentration field.
 The $c-$dependence of $F$ is the sizer/adder-like behaviour that changes cell density/number based on the concentration field.
+
 The dynamics for concentration of the $i-$th signaling morphogen is:
 The dynamics for concentration of the $i-$th signaling morphogen is:
 \begin{equation}
 \begin{equation}
-\dot c_i = D_{c_i} \partial_x^2 c_i - \partial_x(\dot {u} c_i) + G(u, c_i, \rho, t)
+\dot c_i = D_{c_i} \boldsymbol \nabla^2 c_i - \boldsymbol \nabla \cdot(\boldsymbol v c_i) + G(\boldsymbol u, c_i, \rho, t)
 \end{equation}
 \end{equation}
 \section{No signaling on fixed boundaries}
 \section{No signaling on fixed boundaries}
-Choosing $f = \frac{\rho}{\rho + \rho_s}$,
+Working in 1-D, ignoring the couplings $F, G$ and choosing $g(u,\rho,t) = \frac{\rho}{\rho + \rho_s}$,
 \begin{eqnarray}
 \begin{eqnarray}
 \dot u = v \\
 \dot u = v \\
 \gamma \dot u = K \partial_x^2 u + \eta \partial_x^2 \dot u + \lambda \partial_x \left( \frac{\rho}{\rho + \rho_s}\right)\\
 \gamma \dot u = K \partial_x^2 u + \eta \partial_x^2 \dot u + \lambda \partial_x \left( \frac{\rho}{\rho + \rho_s}\right)\\
@@ -48,6 +83,7 @@ The boundary conditions are:
 \begin{equation}
 \begin{equation}
 \rho \big|_{\partial \Omega} = 0 \qquad \sigma \big|_{\partial \Omega} = 0
 \rho \big|_{\partial \Omega} = 0 \qquad \sigma \big|_{\partial \Omega} = 0
 \end{equation}
 \end{equation}
+At every timestep, we move the boundary by $vdt$.
 \section{Controlling growth}
 \section{Controlling growth}
 A generic control problem goes as follows: Given a differential equation $\dot x = f(x,u)$ with the initial condition $x(0)=0$, find the control path $u(t)$ that minimizes a cost (which we will construct) subject to the equation $\dot x = f(x,u)$ being satisfied.
 A generic control problem goes as follows: Given a differential equation $\dot x = f(x,u)$ with the initial condition $x(0)=0$, find the control path $u(t)$ that minimizes a cost (which we will construct) subject to the equation $\dot x = f(x,u)$ being satisfied.
 The cost $J$ involves a running cost, for the path reaching a final state and a terminal cost.
 The cost $J$ involves a running cost, for the path reaching a final state and a terminal cost.
@@ -59,7 +95,16 @@ Going back to oyr problem, we tackle the control aspects in two steps:
 \item What form of active stress would minimize the cost function $J = \Phi[L(T)] + \int_0^T h(L(t)) dt$ subject to $\gamma \dot u = \partial_x \sigma$
 \item What form of active stress would minimize the cost function $J = \Phi[L(T)] + \int_0^T h(L(t)) dt$ subject to $\gamma \dot u = \partial_x \sigma$
 \item Use the optimum active stress obtained in step 1 to get appropriate reaction terms/boundary conditions on $c_i$.
 \item Use the optimum active stress obtained in step 1 to get appropriate reaction terms/boundary conditions on $c_i$.
 \end{enumerate}
 \end{enumerate}
-\chapter{Model on fixed bouundaries}
+\chapter{Model on fixed boundaries}
+On fixed boundaries, our model can explain pattern formation in the desnity field of cells and/or signaling morphogens. 
+
+In the absence of the signaling morphogen, we first look at patterns in the density field. We obtained homogeneous, patterned and oscillatory states in different regions of the phase space:
+\begin{figure}[h]
+\includegraphics[scale=0.2]{phaseplot}
+\end{figure}
+\begin{figure}
+\includegraphics[scale=0.3]{patterns}
+\end{figure}
 \chapter{Model on moving boundaries}
 \chapter{Model on moving boundaries}
 Note:
 Note:
 \section{Diffusion on fixed boundaries becomes diffusion-advection on moving boundaries}
 \section{Diffusion on fixed boundaries becomes diffusion-advection on moving boundaries}
@@ -235,7 +280,8 @@ The final solution is the inverse fourier transform of this, viz,
 c(x,t) = \exp(-\alpha t_c) \left[ 1 + 0.2 \cos\left(\frac{m \pi x}{L}\right) \exp \left( -\frac{D m^2 \pi^2 (t - t_c)}{L^2} - \frac{D m^2 \pi^2 (1 - e^{-2 \alpha t_c})}{2 \alpha L_0^2}\right) \right]
 c(x,t) = \exp(-\alpha t_c) \left[ 1 + 0.2 \cos\left(\frac{m \pi x}{L}\right) \exp \left( -\frac{D m^2 \pi^2 (t - t_c)}{L^2} - \frac{D m^2 \pi^2 (1 - e^{-2 \alpha t_c})}{2 \alpha L_0^2}\right) \right]
 \end{equation}
 \end{equation}
 where again $L = L_0 \exp(\alpha t_c)$
 where again $L = L_0 \exp(\alpha t_c)$
-\chapter*{Appendix: Visco-elastic models}
+\appendix
+\chapter{Visco-elastic models}
 As the name suggests, viscoelastic materials have an elastic component and a viscous component. For the elastic component, the stress $\sigma_e$ is related to the strain $\epsilon_e$ as:
 As the name suggests, viscoelastic materials have an elastic component and a viscous component. For the elastic component, the stress $\sigma_e$ is related to the strain $\epsilon_e$ as:
 \begin{equation}
 \begin{equation}
 \sigma_e = E \epsilon_e
 \sigma_e = E \epsilon_e
@@ -270,7 +316,7 @@ As $t \to \infty$,
 \sigma(t) = \eta \dot \epsilon
 \sigma(t) = \eta \dot \epsilon
 \end{equation}
 \end{equation}
 That is, at long times, a Maxwell viscoelastic material behaves like a fluid.
 That is, at long times, a Maxwell viscoelastic material behaves like a fluid.
-\subsection{Kelvin-Voigt model}
+\subsection{Kelvin-Voigt model} \label{kelvin_voigt}
 In this model, the elastic and viscous components(spring and dashpot, respectively) are in parallel with each other. We subject both of them to equal strain $\epsilon = \epsilon_e = \epsilon_v$. The total stress of the combination is 
 In this model, the elastic and viscous components(spring and dashpot, respectively) are in parallel with each other. We subject both of them to equal strain $\epsilon = \epsilon_e = \epsilon_v$. The total stress of the combination is 
 \begin{equation}
 \begin{equation}
 \sigma = \sigma_e + \sigma_v
 \sigma = \sigma_e + \sigma_v
@@ -287,7 +333,7 @@ As $t \to \infty$,
 \epsilon(t) = \frac{\sigma}{\epsilon}
 \epsilon(t) = \frac{\sigma}{\epsilon}
 \end{equation}
 \end{equation}
 That is, at long times, a Kelvin-Voigt viscoelastic material behaves like a solid.
 That is, at long times, a Kelvin-Voigt viscoelastic material behaves like a solid.
-\chapter*{Appendix: Proof of the Reynolds Transport Theorem}
+\chapter{Proof of the Reynolds Transport Theorem}
 From Leibniz theorem,
 From Leibniz theorem,
 \begin{equation}
 \begin{equation}
 \begin{split}
 \begin{split}
@@ -303,5 +349,5 @@ Thus,
 \begin{equation}
 \begin{equation}
 \frac{d}{dt} \int_{\alpha(t)}^{\beta(t)} c(x,t) dx = \int_{\alpha(t)}^{\beta(t)} \left[ \frac{\partial c}{\partial t} + \frac{\partial}{\partial x}(cv) \right]dx
 \frac{d}{dt} \int_{\alpha(t)}^{\beta(t)} c(x,t) dx = \int_{\alpha(t)}^{\beta(t)} \left[ \frac{\partial c}{\partial t} + \frac{\partial}{\partial x}(cv) \right]dx
 \end{equation}
 \end{equation}
-\bibliography{ref}
+\bibliography{references}
 \end{document}
 \end{document}
\ No newline at end of file

thesisnotes/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{A minimal model}
+We model a 1-D tissue as a visco-elastic medium. We write hydrodynamic equations for the evolution of a displacement field $\boldsymbol u(x,t)$, a density field $\rho(x,t)$ and a concentration field for the signaling morphogen $c(x,t)$ that tells the cells when to divide.
+THe equation for $\boldsymbol u$ is a force balance equation (divergence of the total stress = force density). We use the Kelvin Voigt model to be able to get long-time leastic behaviour:
+\begin{equation*}
+\gamma \dot u = \partial_x \sigma
+\end{equation*}
+\begin{equation*}
+\sigma = \sigma_{e} + \sigma_{v} + \sigma_{a} = K \partial_x u + \eta \partial_x \dot u + \lambda \partial_x f(c,\rho)
+\end{equation*}
+Thus, 
+\begin{equation}
+\gamma \dot u = K \partial_x^2 u + \eta \partial_x^2 \dot u + \lambda \partial_x f 
+\end{equation}
+The equation for the cell density is a diffusion advection equation, with terms added to account for cell birth and death:
+\begin{equation}
+\dot \rho = D_\rho \partial_x^2 \rho - \partial_x(\dot u \rho) - k_{\rho}(\rho - \rho_0) + F(u, c_i, \rho, t)
+\end{equation} 
+The $c-$dependence of $F$ is the sizer/adder-like behaviour that changes cell density/number based on the concentration field.
+The dynamics for concentration of the $i-$th signaling morphogen is:
+\begin{equation}
+\dot c_i = D_{c_i} \partial_x^2 c_i - \partial_x(\dot {u} c_i) + G(u, c_i, \rho, t)
+\end{equation}
+\section{No signaling on fixed boundaries}
+Choosing $f = \frac{\rho}{\rho + \rho_s}$,
+\begin{eqnarray}
+\dot u = v \\
+\gamma \dot u = K \partial_x^2 u + \eta \partial_x^2 \dot u + \lambda \partial_x \left( \frac{\rho}{\rho + \rho_s}\right)\\
+\dot \rho = D_{\rho} \partial_x^2 \rho - \partial_x (\dot u \rho) - k_{\rho}(\rho - \rho_0)
+\end{eqnarray}
+The boundary conditions are:
+\begin{equation}
+u\big|_{\partial \Omega} = 0 \qquad v\big|_{\partial \Omega} = 0 \qquad
+J_{\rho}\big|_{\partial \Omega} = 0 \implies (v \rho - \partial_X \rho)\big|_{\partial \Omega} = 0
+\end{equation}
+\section{Moving boundaries}
+The equations are the same.
+The boundary conditions are:
+\begin{equation}
+\rho \big|_{\partial \Omega} = 0 \qquad \sigma \big|_{\partial \Omega} = 0
+\end{equation}
+\section{Controlling growth}
+A generic control problem goes as follows: Given a differential equation $\dot x = f(x,u)$ with the initial condition $x(0)=0$, find the control path $u(t)$ that minimizes a cost (which we will construct) subject to the equation $\dot x = f(x,u)$ being satisfied.
+The cost $J$ involves a running cost, for the path reaching a final state and a terminal cost.
+\begin{equation}
+J = \int_0^T V(x,u)dt + \Phi(x(T))
+\end{equation}
+Going back to oyr problem, we tackle the control aspects in two steps:
+\begin{enumerate}
+\item What form of active stress would minimize the cost function $J = \Phi[L(T)] + \int_0^T h(L(t)) dt$ subject to $\gamma \dot u = \partial_x \sigma$
+\item Use the optimum active stress obtained in step 1 to get appropriate reaction terms/boundary conditions on $c_i$.
+\end{enumerate}
+\chapter{Model on fixed bouundaries}
+\chapter{Model on moving boundaries}
+Note:
+\section{Diffusion on fixed boundaries becomes diffusion-advection on moving boundaries}
+
+\begin{equation}
+\frac{d}{dt}\int_{\Omega(t)} c(x,t) dV = - \int_{\Omega(t)} \boldsymbol J \cdot \boldsymbol {dS} = - \int_{\Omega(t)} \boldsymbol \nabla \cdot \boldsymbol J dV 
+\end{equation}
+where $\boldsymbol J=-D\boldsymbol \nabla c$ is the diffusiove flux.
+In 1-D, this becomes, 
+\begin{equation}
+\frac{d}{dt}\int_{\Omega(t)} c(x,t) dx = D \int_{\Omega(t)} \frac{\partial^2 c}{\partial x^2} dx
+\end{equation}
+\begin{equation}
+\implies \frac{d}{dt} \int_{\Omega(t)} \left[ \frac{\partial c}{\partial t} + \frac{\partial (vc)}{\partial x}\right]dx = D \int_{\Omega(t)} \frac{\partial^2 c}{\partial x^2} dx
+\end{equation}
+where the LHS follows from Reynolds TRansport Theorem in 1-D and $v$ is the velocity of the moving domain.\\
+Since the above is true for any arbitrary interval and any arbitrarry $\Omega(t)$, equality of integrals implies equality of integrands:
+\begin{equation}
+\implies \frac{\partial c}{\partial t} + \frac{\partial (vc)}{\partial x} = D \frac{\partial^2 c}{\partial x^2} 
+\end{equation}
+That is, a simple diffusion equation has become a diffusion-advection equation on a moving domain.
+\section{Conservation laws}
+Claim: If neumann conditions are satified at all times, total concentration on the moving domain is conserved in time, that is 
+\begin{equation}
+\frac{d}{dt} \int_0^{L(t)} c(x,t)dx =0
+\end{equation}
+Proof: From RTT, \begin{equation} 
+\begin{split}
+\frac{d}{dt} \int_0^{L(t)} c(x,t)dx = \int_0^{L(t)} \left[ \frac{\partial c}{\partial t} + \frac{\partial}{\partial x} (vc)\right]dx
+\end{split}
+\end{equation}
+Putting $\frac{\partial c}{\partial t} = D \frac{\partial^2 c}{\partial x^2} - \frac{\partial (vc)}{\partial x}$ in the above,
+\begin{equation}
+\begin{split}
+\frac{d}{dt} \int_0^{L(t)} c(x,t)dx &= \int_0^{L(t)} D \frac{\partial^2 c}{\partial x^2} dx\\
+&= D \frac{\partial c}{\partial x} \bigg{|}_{0}^{L(t)}\\
+&= 0
+\end{split}
+\end{equation}
+The last step follows from the requirement that neumann conditions are satified at all times.
+
+\section{Exact solution for advection diffusion equation on a domain growing exponentially}
+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 $C(x,0) = 1 + 0.2 \cos\left(\frac{m \pi x}{L_0}\right)$
+\begin{equation}
+C(x,t) = \exp(-\alpha t)\left[1+ 0.2\cos\left(\frac{m \pi x}{L_0} \exp(-\alpha t)\right)\exp\left(\frac{-m^2 \pi^2D}{2 \alpha L_0^2}(1-\exp(-2 \alpha t))\right)\right]
+\end{equation}
+\section{Exact solution for advection diffusion equation on a domain growing exponentially till $t=t_c$}
+In this case, the map is:
+\begin{equation}
+x(s,t)=
+\begin{cases}
+s \exp(\alpha t), \quad t \leq t_c \\
+s \exp(\alpha t_c), \quad t>t_c
+\end{cases}
+\end{equation}
+The velocity is:
+\begin{equation}
+v(x,t) = 
+\begin{cases}
+\alpha x, \quad t \leq t_c \\
+0, \quad t>t_c
+\end{cases}
+\end{equation}
+The domain length is:
+\begin{equation}
+L(t) = 
+\begin{cases}
+L_0 \exp(\alpha t), \quad t \leq t_c \\
+L_0 \exp(\alpha t_c), \quad t>t_c
+\end{cases}
+\end{equation}
+The solution for $t \leq t_c$ is the same as obtained in the previous section. For $t > t_c$, the problem is one of solving the diffusion equation on a fixed domain of length $L = L_0 \exp(\alpha t_c)$ with the initial condition to be:
+\begin{equation} \label{ic}
+c(x, t_c) = \exp(-\alpha t_c)\left[1+ 0.2\cos\left(\frac{m \pi x}{L_0} \exp(-\alpha t_c)\right)\exp\left(\frac{-m^2 \pi^2D}{2 \alpha L_0^2}(1-\exp(-2 \alpha t_c))\right)\right]
+\end{equation}
+Going to the fourier domain, the diffusion equation is:
+\begin{equation}
+\frac{d c_k}{dt} = -D k^2 c_k
+\end{equation}
+\begin{equation}
+\implies c_k(t) = c_k (t_c) \exp[-D k^2(t-t_c)]
+\end{equation}
+Now, $c_k(t_c)$ is just the Fourier transform of Eq(\ref{ic}). Thus, 
+\begin{equation}
+c_k(t_c) = 2 \pi e^{-\alpha t_c} \left[\delta(k) + 0.1 e^{-\frac{D m^2}{2 \alpha L_0^2} \pi^2 (1 - e^{-2 \alpha t_c})} \left[\delta(k - m \pi/L) + \delta (k+ m \pi/L)\right]\right]
+\end{equation}
+where $L = L_0 \exp(\alpha t_c)$
+Thus, 
+\begin{equation}
+c_k(t) = \exp\left[-D k^2(t-t_c)\right] 2 \pi e^{-\alpha t_c} \left[\delta(k) + 0.1 e^{-\frac{D m^2}{2 \alpha L_0^2} \pi^2 (1 - e^{-2 \alpha t_c})} \left[\delta(k - m \pi/L) + \delta (k+ m \pi/L)\right]\right]
+\end{equation}
+The final solution is the inverse fourier transform of this, viz,
+\begin{equation}
+c(x,t) = \exp(-\alpha t_c) \left[ 1 + 0.2 \cos\left(\frac{m \pi x}{L}\right) \exp \left( -\frac{D m^2 \pi^2 (t - t_c)}{L^2} - \frac{D m^2 \pi^2 (1 - e^{-2 \alpha t_c})}{2 \alpha L_0^2}\right) \right]
+\end{equation}
+where again $L = L_0 \exp(\alpha t_c)$
+\chapter*{Appendix: Visco-elastic models}
+As the name suggests, viscoelastic materials have an elastic component and a viscous component. For the elastic component, the stress $\sigma_e$ is related to the strain $\epsilon_e$ as:
+\begin{equation}
+\sigma_e = E \epsilon_e
+\end{equation}
+where $E$ is the elastic modulus.\\
+For the viscous component, the stress $\sigma_v$ is related to the strain rate $\dot \epsilon_v$ as:
+\begin{equation}
+\sigma_v = \eta \dot \epsilon_v
+\end{equation}
+where $\eta$ is the viscosity.
+\section{Linear visco-elasticity}
+Linear viscoelasticity is when the stress is separable in its elastic and viscous responses:
+\begin{equation}
+\sigma(t) = E \epsilon(t) + \int_0^t F(t-t^\prime) \epsilon(t^\prime) d t^\prime
+\end{equation}
+\section{Models of linear viscoleasticity}
+\subsection{Maxwell model}
+In this model, the elastic and viscous components(spring and dashpot, respectively) are in series with each other. We subject both of them to a stress $\sigma(t)$. The total strain of the combination is 
+\begin{equation}
+\epsilon = \epsilon_e + \epsilon_v
+\end{equation}
+\begin{eqnarray}
+\implies \dot \epsilon = \dot \epsilon_e + \dot \epsilon_v\\
+= \frac{\dot \sigma}{E} + \frac{\sigma}{\eta}
+\end{eqnarray}
+Solving this for $\sigma(t)$,
+\begin{equation}
+\sigma(t) = \sigma(0) \exp(-Et/\eta) + \eta \dot \epsilon [1 - \exp(-Et/\eta)]
+\end{equation}
+As $t \to \infty$, 
+\begin{equation}
+\sigma(t) = \eta \dot \epsilon
+\end{equation}
+That is, at long times, a Maxwell viscoelastic material behaves like a fluid.
+\subsection{Kelvin-Voigt model}
+In this model, the elastic and viscous components(spring and dashpot, respectively) are in parallel with each other. We subject both of them to equal strain $\epsilon = \epsilon_e = \epsilon_v$. The total stress of the combination is 
+\begin{equation}
+\sigma = \sigma_e + \sigma_v
+\end{equation}
+\begin{eqnarray}
+\implies \sigma = E \epsilon_ + \eta \dot \epsilon \\
+\end{eqnarray}
+Solving this for $\epsilon(t)$,
+\begin{equation}
+\epsilon(t) = \epsilon(0) \exp(-Et/\eta) + \frac{\sigma}{E}[1 - \exp(-Et/\eta)]
+\end{equation}
+As $t \to \infty$, 
+\begin{equation}
+\epsilon(t) = \frac{\sigma}{\epsilon}
+\end{equation}
+That is, at long times, a Kelvin-Voigt viscoelastic material behaves like a solid.
+\chapter*{Appendix: Proof of the Reynolds Transport Theorem}
+From Leibniz theorem,
+\begin{equation}
+\begin{split}
+\frac{d}{dt} \int_{\alpha(t)}^{\beta(t)} c(x,t) dx&= \int_{\alpha(t)}^{\beta(t)} \frac{\partial c}{\partial t} dx + c(\beta, t) \dot \beta - c(\alpha, t) \dot \alpha\\
+&= \int_{\alpha(t)}^{\beta(t)} \frac{\partial c}{\partial t} dx + c(\beta, t) v(\beta,t) - c(\alpha, t) v(\alpha, t)
+\end{split}
+\end{equation}
+Now, \begin{equation}
+c(\beta, t) v(\beta,t) - c(\alpha, t) v(\alpha, t) = \int_{\alpha(t)}^{\beta(t)} \left[c \frac{\partial v}{\partial x} + v \frac{\partial c}{\partial x}\right] dx = \int_{\alpha(t)}^{\beta(t)} \frac{\partial}{\partial x}(cv) dx
+\end{equation}
+The first equality above can be seen just by integrating the expression on the RHS by parts.
+Thus, 
+\begin{equation}
+\frac{d}{dt} \int_{\alpha(t)}^{\beta(t)} c(x,t) dx = \int_{\alpha(t)}^{\beta(t)} \left[ \frac{\partial c}{\partial t} + \frac{\partial}{\partial x}(cv) \right]dx
+\end{equation}
+\bibliography{ref}
+\end{document}
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