Let $u(x,0) = f(x), u_t(x,0) = g(x),$ and $C(x,t)$ all be smooth functions supported in $B_R(0),$ the ball of radius $R$ centered at the origin. Prove that for $u$ solving $$u_{tt} - u_{xx} + C(x,t)u = 0,$$ and having the initial conditions as previously stated, we have that $u(x,t) = 0$ for $|x| > R+t.$

My attempt:

Let $u(x,0) = u_t(x,0) = 0$ in $B(x_0,t_0),$ where $|x_0| > R.$ Then show that the appropriate energy is decreasing, $$ e(t) = \frac{1}{2} \int_{B(x_0, t_0 - t)} u_t^2 + u_x^2 \; dx \leq e(0) = 0. $$ Calculate, \begin{align} \frac{de}{dt} &= \int_B u_tu_{tt} + u_xu_{xt} \; dx + \int_{\partial B} -\frac{1}{2} u_t^2 - \frac{1}{2} u_x^2 \; ds\\ &= \int_B u_t(u_{tt} - u_{xx}) \; dx + \int_{\partial B} u_xu_t -\frac{1}{2} u_t^2 - \frac{1}{2} u_x^2 \; ds\\ &\leq \int_B - u_t u C(x,t)\; dx = 0 \end{align} since $|u_xu_t| \leq \frac{1}{2} u_t^2 + \frac{1}{2}u_x^2$ and $C(x,t)$ is only supported in a ball of size $R.$ I believe from this, it can be concluded that information has finite propagation speed (speed 1) and thus the result.

Could anyone confirm if this is correct? Thanks!

  • $\begingroup$ Wouldn't this follow from the dispersion relation associated with the wave equation? I.e. insert a general Fourier mode solution $u = \exp{(\kappa x - \omega t)}$ to find that the propagation speed of the solution is 1 outside $B_R(0)$. It would seem that the finite support of the initial data implies the stated assertion. $\endgroup$
    – ekkilop
    Jul 19, 2017 at 15:50
  • $\begingroup$ Thanks for the comment. I am not sure if this constitutes proof. I am a little confused: To me, this type of argument seems to also gives us that the speed of propagation inside $B_R(0)$ is greater than 1 since $\omega = \pm \sqrt{k^2 + C(x,t)}.$ How do we know that a given mode hasn't travelled outside of $B_{R+t}(0)?$ I'm also not sure if assuming $u$ is a Fourier mode is a valid method of proof to begin with. $\endgroup$
    – Merkh
    Jul 19, 2017 at 16:03
  • $\begingroup$ No, this was rather intended as an argument around which one maybe could build a proof. Anyway, it may work better if $C$, $f$ and $g$ are compactly supported (are they?) in which case I think $u$ would have a Fourier series representation. $C$ vanishes outside $B_R(0)$ ($R$ is constant right?) so how could the mode have reached further than $B_{R+t}(0)$? $\endgroup$
    – ekkilop
    Jul 19, 2017 at 16:58
  • $\begingroup$ True true, it makes sense now. I was imagining something else. The mode has to slow its speed to 1 after leaving $B_R(0);$ I was imagining that it would maintain its speed after leaving the ball. $\endgroup$
    – Merkh
    Jul 19, 2017 at 18:23
  • $\begingroup$ Ah, I see. Well, there are clearly some details to fill in here, but perhaps the dispersion argument could be the way to go... $\endgroup$
    – ekkilop
    Jul 19, 2017 at 19:35


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.