deleting redundant files

thesisnotes/thesis_notes.tex

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-\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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