rtt proof, moving domain notes

thesis_notes.tex

@@ -8,7 +8,100 @@
 \author{Jigyasa Watwani}
 \author{Jigyasa Watwani}
 \title{Physics of Growth Regulation in Cells and Tissues - Notes}
 \title{Physics of Growth Regulation in Cells and Tissues - Notes}
 \begin{document}
 \begin{document}
-\chapter{Exact solution for advection diffusion equation on a moving domain}
+\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 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}
 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}.
 \bibliographystyle{plain}.
 The equation we intend to solve is:
 The equation we intend to solve is:
@@ -92,5 +185,73 @@ Putting in the initial condition,
 \begin{equation}
 \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]
 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}
 \end{equation}
+\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}
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 \end{document}
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