Let
- $d\in\mathbb N$;
- $\Omega_i\subseteq\mathbb R^d$ be open and $$P_i\varphi:=-\Delta\varphi\;\;\;\text{for }\varphi\in C_c^\infty(\Omega_1);$$
- $U:L^2(\Omega_1)\to L^2(\Omega_2)$ be an isometry$^1$
How can we show that $UP_1=P_2U$ in the weak sense?
Let $f_1\in L^2(\Omega_1)$. We need to show that $f_2:=Uf_1\in L^2(\Omega_2)$ is a weak solution of $P_2f_2=g_2:=UP_1f_1\in ^2(\Omega_2)$, i.e. $$\langle f_2,P_2\varphi\rangle_{L^2(\Omega_2)}=\langle g_2,\varphi\rangle_{L^2(\Omega_2)}\;\;\;\text{for all }\varphi\in C_c^\infty(\Omega_2)\tag1.$$
How can we do that?
Even when I impose the stronger assumption of $U$ being an unitary linear operator (from which we know that $U$ is bijective and $U^\ast=U^{-1}$), I only obtain $$\langle f_2,P_2\varphi\rangle_{L^2(\Omega_2)}=\langle f_1,U^{-1}P_2\varphi\rangle_{L^2(\Omega_1)}\tag2$$ and don't know how to proceed from there ...
EDIT: I've read that "it is very easy to verify that if $Ω_1$ and $Ω_2$ are isometric, then the Dirichlet eigenvalues of the laplacian on $Ω_1$ and $Ω_2$ coincide" ... If the assertion, as stated in my question, is not true, how can that claim be stated and proven correctly?
$^1$ I'm not sure, but maybe we need to assume that $U$ is linear and surjective as well. Phrased differently, $U$ is then an unitary linear operator from $L^2(\Omega_1)$ to $L^2(\Omega_2)$.