PDE : Mixture of Wave and Heat equations

Question

Today I was given the following equation : $$\frac{1}{c^2}u_{tt} + \frac{1}{D}u_t = u_{xx}$$ with initial conditions : $u(x,0) = 1$ if $|x|<L$ and $0$ otherwise, $u_t(x,0) = 0$. So fairly simple initial conditions.

I can see that there is a bit of wave and heat equation so I first solved each case but I couldn't "glue" the answers together. If $c$ gets large, then the equation will behave like a heat equation and similarly, if $D$ is large then it will behave like a wave equation.

Using dimensional analysis I deduced that if $\frac{c^2}{LD}$ is the criterion to say if $c$ and $D$ are "large".

I know that I can solve the equation using separation of variable but what would be a way to be able see how the solution behaves without solving it? Like being able to sketch a solution for varying $t$ would be really nice.

Cheers

Answer 1

A substitution of the form $u=e^{-at}v$ with $a=c^2/(2D)$ transforms the equation into the telegraph equation $$\frac1{c^2}v_{tt}-v_{xx}=bv$$ with $b=a/(2D)$. The telegraph equation is a much studied equation.

Answer 2

Hint:

Let $u=e^{-\frac{c^2t}{2D}}v$ ,

Then $u_t=e^{-\frac{c^2t}{2D}}v_t-\dfrac{c^2}{2D}e^{-\frac{c^2t}{2D}}v$

$u_{tt}=e^{-\frac{c^2t}{2D}}v_{tt}-\dfrac{c^2}{2D}e^{-\frac{c^2t}{2D}}v_t-\dfrac{c^2}{2D}e^{-\frac{c^2t}{2D}}v_t+\dfrac{c^4}{4D^2}e^{-\frac{c^2t}{2D}}v=e^{-\frac{c^2t}{2D}}v_{tt}-\dfrac{c^2}{D}e^{-\frac{c^2t}{2D}}v_t+\dfrac{c^4}{4D^2}e^{-\frac{c^2t}{2D}}v$

$u_x=e^{-\frac{c^2t}{2D}}v_x$

$u_{xx}=e^{-\frac{c^2t}{2D}}v_{xx}$

$\therefore\dfrac{1}{c^2}\left(e^{-\frac{c^2t}{2D}}v_{tt}-\dfrac{c^2}{D}e^{-\frac{c^2t}{2D}}v_t+\dfrac{c^4}{4D^2}e^{-\frac{c^2t}{2D}}v\right)+\dfrac{1}{D}\left(e^{-\frac{c^2t}{2D}}v_t-\dfrac{c^2}{2D}e^{-\frac{c^2t}{2D}}v\right)=e^{-\frac{c^2t}{2D}}v_{xx}$ with $v(x,0)=\begin{cases}1&|x|<L\\0&\text{otherwise}\end{cases}$ and $v_t(x,0)=\begin{cases}\dfrac{c^2}{2D}&|x|<L\\0&\text{otherwise}\end{cases}$

$\dfrac{1}{c^2}v_{tt}-\dfrac{1}{D}v_t+\dfrac{c^2}{4D^2}v+\dfrac{1}{D}v_t-\dfrac{c^2}{2D^2}v=v_{xx}$ with $v(x,0)=\begin{cases}1&|x|<L\\0&\text{otherwise}\end{cases}$ and $v_t(x,0)=\begin{cases}\dfrac{c^2}{2D}&|x|<L\\0&\text{otherwise}\end{cases}$

$\dfrac{1}{c^2}v_{tt}=\dfrac{c^2}{4D^2}v+v_{xx}$ with $v(x,0)=\begin{cases}1&|x|<L\\0&\text{otherwise}\end{cases}$ and $v_t(x,0)=\begin{cases}\dfrac{c^2}{2D}&|x|<L\\0&\text{otherwise}\end{cases}$