exact solution for time-dependent alpha

thesis_notes.tex

@@ -101,7 +101,7 @@ Putting $\frac{\partial c}{\partial t} = D \frac{\partial^2 c}{\partial x^2} - \
 \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}
+\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:
@@ -185,6 +185,56 @@ 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}
+\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{\pi x}{L_0} \exp(-\alpha t_c)\right)\exp\left(\frac{-\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} = -k^2 c_k
+\end{equation}
+\begin{equation}
+\implies c_k(t) = c_k (t_c) \exp[-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}{2 \alpha L_0^2} \pi^2 (1 - e^{-2 \alpha t_c})} \left[\delta(k - \pi/L) + \delta (k+ \pi/L)\right]\right]
+\end{equation}
+where $L = L_0 \exp(\alpha t_c)$
+Thus, 
+\begin{equation}
+c_k(t) = \exp\left[-k^2(t-t_c)\right] 2 \pi e^{-\alpha t_c} \left[\delta(k) + 0.1 e^{-\frac{D}{2 \alpha L_0^2} \pi^2 (1 - e^{-2 \alpha t_c})} \left[\delta(k - \pi/L) + \delta (k+ \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{\pi x}{L}\right) \exp \left( -\frac{\pi^2 (t - t_c)}{L^2} - \frac{D \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}