time dependent alpha solution for mth mode cosine initial condition

thesis_notes.tex

@@ -181,9 +181,9 @@ This is exactly the same as Eq(11) in the reference with $f(T^\star)=\alpha L_0^
 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,
+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{\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{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:
@@ -212,27 +212,27 @@ L_0 \exp(\alpha t_c), \quad t>t_c
 \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]
+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} = -k^2 c_k
+\frac{d c_k}{dt} = -D k^2 c_k
 \end{equation}
 \begin{equation}
-\implies c_k(t) = c_k (t_c) \exp[-k^2(t-t_c)]
+\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}{2 \alpha L_0^2} \pi^2 (1 - e^{-2 \alpha t_c})} \left[\delta(k - \pi/L) + \delta (k+ \pi/L)\right]\right]
+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[-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]
+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{\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]
+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}