Can a wave vanish at two times?

Question

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"?

Answer 1

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.