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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

user88595
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2 Answers2

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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.

Harald Hanche-Olsen
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  • That looks like a good idea so I guess the solution for $u$ is just a damped solution for $v$ in time. The solution for the wave equation goes right and left on the graph, do you know if the telegraph equation does that as well? – user88595 Dec 17 '13 at 12:07
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    As far as I understand, the telegraph equation doesn't have proper wave solutions, but you do get damped waves traveling in the two directions. I think it's a little bit more complicated than so, but that should be the gist of it. For short wavelengths, the derivatives dominate, so you get the wave equation. But for longer wavelengths, something altogether different happens. I am not sure what; I have never studied this equation myself. – Harald Hanche-Olsen Dec 17 '13 at 16:41
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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 $v_t(x,0)=\begin{cases}\dfrac{c^2}{2D}&|x|<L\\0&\text{otherwise}\end{cases}$

doraemonpaul
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