\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:
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
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}
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 $L(t)$.
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:
\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:
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
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
