Consider the wave equation in $\mathbb{R}^{3+1}$ dimensions with the following boundary conditions:
$$\begin{align} \partial_t^2 \psi &= \Delta \psi \ \text{ in }\ \mathbb{R} \times B(0,1) \\ \psi = \psi_t &=0 \ \text{ on } \ \mathbb{R} \times \partial B(0,1) \end{align}$$
where $B(0,1)$ is the ball of radius $1$ in $3$ dimensions and $\partial B(0,1)$ is its boundary. I am asked to show the solution vanishes inside $\partial \mathbb{R} \times B(0,1)$
The proof given in my course is:
I'm confused as to why we need an open closed argument and I believe I am getting something wrong about the size of the ball on which the solution has to vanish.
Assume we have already proved that $\psi =0$ on $\mathbb{R} \times \{x : 1-\rho \le |x| \le 1\}.$ Then for each $x\in \mathbb{R} \times \partial B(0,1-\rho),$ there exists $\epsilon(x)>0$ such that $\psi$ vanishes on $B(x,\epsilon(x)).$ Furthermore, we can take an uniform $\epsilon$ for $x\in \mathbb{R} \times \partial B(0,1-\rho),$ as by Cauchy-Kovalevskaya theorem, $\epsilon(x)$ depends only on the coefficients of the pde and the values of $\partial^\alpha(\psi)(x)$ with $|\alpha|\le 2$. Note that both the coefficients and the values of $\partial^\alpha(\psi)(x)$ are constant in $\mathbb{R} \times \partial B(0,1-\rho)$.
At this point, we proved $\epsilon(x)$ is a function of $|x|$, as its value can be taken uniformly on $\mathbb{R} \times \partial B(0,|x|)$.
The problem is that $\epsilon(\rho)$ could shrink very fast as $\rho$ decreases.
Nonetheless, note that the the coefficients and the values of $\partial^\alpha(\psi)(x)$ are constant on $ \mathbb{R} \times \{x : \rho<|x| <1\}$ , hence i claim $\epsilon$ can be chosen uniformly on this set, i.e. we can take $\epsilon(\rho) = \epsilon(1).$ As $\epsilon(1)>0,$ we will cover the entire cylinder in finite time.
The latter has the advantage of not using an open closed argument, but I am not etirely sure about its correctness.
Also, how are the translation and rotation invariance remarks related to this solution?