Suppose $u$ is a solution of
$$ \begin{cases}u_{tt} = \Delta u & \textrm{ in } \mathbb R \times \mathbb R^n, \\ u(0,\cdot) = 0, \\u_t(0,\cdot) = g(x), \end{cases} $$
with $g$ compactly supported. Immediately, we may note that $u$ is odd in time with respect to $t = 0$ (i.e $u(t,x) = -u(-t,x)$ for all $t,x$).
My question is this:
can there be an open set $E \subset \mathbb R^n$ and time $t_1 \neq 0$ so that $u(t_1,\cdot)$ vanishes on $E$, but $u_t(t_1,\cdot)$ does not?
In some cases, the answer is clear.
- $E$ cannot be all of $\mathbb R^n$; if that were the case, then $u$ would be odd in time (for all $x$) with respect to $t = t_1$ as well, and thus $u$ is actually periodic in time for all $x$ (see answers to my question here). This will violate local energy decay (for most reasonable wave speeds, including the constant one above, energy of a wave in a bounded set must decrease as $t \to \infty$).
In general, if $u(t_1,\cdot) \equiv 0$ in $E$, we can only conclude that it is odd w.r.t. $t$ in the light cone determined by $E$. In some cases, this is enough to leverage "oddness" again; for example,
- If $E$ is "on the edge" of the natural support of $u$: suppose $n = 1$ and $g$ is supported in $(-1,1)$, and $t_1 = 1$. We know by finite speed of propagation, $u(t_1,\cdot)$ and $u_t(t_1,\cdot)$ vanish outside $(-2,2)$. If we suppose $u(t_1,\cdot)$ vanishes on $(1,2)$ (without assuming anything about whether or not the derivative vanishes), then $u$ will be odd with respect to $t_1$ in the light cone determined by $(1,\infty)$. In particular, at a point like $x = 2.1$, where we already knew that $u(t,2.1)$ vanishes for $t \in (-1.1,t_1)$, we can reflect, and conclude $u(t,2.1) = 0$ for $t \in (t_1,t_1 + 1.1)$ as well. Thus, $u$ (and its normal derivative) will vanish on $-1.1 \leq t \leq 2.1$, and we can then use unique continuation (Holmgren's theorem) to determine that $u(t_1,\cdot)$ and its time derivative vanish on $(1,2)$.
What about in general though? Is it possible for it to vanish on some other set $E$ which is nestled further inside the natural support of the wave? Also, what is true if you replace "open" by "positive measure"?
Yes, this can happen. In one space dimension, let $$g(x) = (\sin x) \chi_{[-2\pi,2\pi]} $$ The solution of $u_{tt}=u_{xx}$ with $u(x,0)=0$, $u_t(x,0)=g(x)$ is $$ u(x,t) = \frac12\Big ( \cos \max(x-t,2\pi) - \cos \min(x+t,2\pi) \Big) $$ In particular, $u(x,\pi)=0$ for $|x|<\pi$. Also, $$ u_t(x,t) = \frac12\Big( \sin \max(x-t,2\pi) + \sin \min(x+t,2\pi) \Big) $$ satisfies $u_t(x,\pi)= -\sin x$ for $|x|<\pi$.
The function $g$ can be smoothed out near $\pm 2\pi$ to create a $C^\infty$ example.
To create an example in higher dimensions, let $g$ depend on one coordinate as above, and multiply it by smooth cutoff function that is $\equiv 1$ in a suitably large neighborhood of the origin.