# physics of the 1-dimensional wave equation

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.

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 $$t=11$$.
• ooh, so considering positive time the wave is observable in the time interval $[9,11]$ Sep 17, 2020 at 19:52