1
$\begingroup$

I was given the task to solve the 1-dimensional wave equation $$ \partial_{tt}u-\partial_{xx}u=0 $$ with the conditions \begin{align} u(x,0)&=f(x), \\ \partial_tu(x,0)&=0, \end{align} where $\mathrm{supp}f\subseteq[-1,1]$.

By factorizing the equation into $$ (\partial_t-\partial_x)(\partial_t+\partial_x)u=0 $$ and using new variables, and then applying the initial conditions, I get the general solution $$ u(x,t)=\frac{1}{2}(f(x+t)+f(x-t)). $$

Now I'm asked to interpret the solution the following way: If I'm sitting on the real line at point $x=10$, at what time will I be able to see the wave for the first time and how long am I able to observe it?

My guess would be that I don't see the wave at all, since $f\equiv0$ outside of $[-1,1]$ but maybe I'm misinterpreting that here.

$\endgroup$

1 Answer 1

1
$\begingroup$

The wave is observable when at least one of $x\pm t$ is in $f$'s support, i.e. from $t=-11$ to $t=-9$, then again from $t=9$ to $t=11$.

$\endgroup$
1
  • $\begingroup$ ooh, so considering positive time the wave is observable in the time interval $[9,11]$ $\endgroup$
    – maryam
    Sep 17, 2020 at 19:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .