Under which domain transformations are the eigenvalues of $-\Delta$ invariant?

Question

Let $d\in\mathbb N$, $\Omega\subseteq\mathbb R^d$ be bounded and open, $$\mathfrak a(u,v):=\langle\nabla u,\nabla v\rangle_{L^2(\Omega,\:\mathbb C^d)}\;\;\;\text{for }u,v\in H_0^1(\Omega)$$ and $$P:=-\Delta:C_c^\infty(\Omega)\to C_c^\infty(\Omega).$$ By the Lax-Milgram theorem, $$\forall f\in L^2(\Omega):\exists !v_f\in H_0^1(\Omega):\forall u\in H_0^1(\Omega):\mathfrak a(u,v_f)=\mathcal l_f(u):=\langle u,f\rangle_{L^2(\Omega)}.$$ The map $$S:L^2(\Omega)\to H_0^1(\Omega)\;,\;\;\;f\mapsto v_f$$ is

1. a bounded linear operator from $L^2(\Omega)$ to $H_0^1(\Omega))$;
2. a compact linear operator on $L^2(\Omega)$;
3. a self-adjoint bounded linear operator on $L^2(\Omega)$.

Remember that if $f\in L^2(\Omega)$, then $u\in L^2(\Omega)$ is a weak solution of the symbolic equation $Pu=f$ if $$\langle u,P\varphi\rangle_{L^2(\Omega)}=\langle f,\varphi\rangle_{L^2(\Omega)}\;\;\;\text{for all }\varphi\in C_c^\infty(\Omega).\tag1$$ It's easy to see that $u\in H_0^1(\Omega)$ is a weak solution of $Pu=f$ iff $$Sf=u\tag2.$$

By the spectral theorem, $$S=\sum_{i\in I}\mu_ie_i\otimes e_i,$$ where the sum converges in $\mathfrak L(L^2(\Omega))$ and $I:=\mathbb N\cap[0,\operatorname{rank}S]$, for some $(\mu_i)_{i\in I}\subseteq(0,\infty)$ nonincreasing and some orthonormal basis $(e_i)_{i\in I}$ of $\overline{\mathcal R(S)}^{L^2(\Omega)}$. Moreover, if $$\mu_n:=0\;\;\;\text{for }n\in\mathbb N\setminus I,$$ then $$\mu_n\xrightarrow{n\to\infty}0\tag3.$$ As usual, $(e_i)_{i\in I}$ can be supplemented to an orthonormal basis $(e_n)_{n\in\mathbb N}$ of $L^2(\Omega)$ by an orthonormal basis $(e_n)_{n\in\mathbb N\setminus I}$ of $\mathcal N(S)$.

If $$\lambda_n:=\left.\begin{cases}\displaystyle\frac1{\mu_i}&\text{, if }\mu_i\ne 0\\\infty&\text{, otherwise}\end{cases}\right\}\;\;\;\text{for }n\in\mathbb N,$$ we immediately see that $e_i\in H_0^1(\Omega)$ and $e_i$ is the unique weak solution of $Pu=\lambda_ie_i$ for all $i\in I$.

Let $\tilde\Omega\subseteq\mathbb R^d$ be bounded and open. Assume $\tilde\Omega=\iota(\Omega)$ for some $\iota:\Omega\to\tilde\Omega$. Under which assumption on $\iota$ will the eigenvalues $\tilde{\lambda_i}$ of $$\tilde P:=-\Delta:C_c^\infty(\tilde\Omega)\to C_c^\infty(\tilde\Omega),$$ obtained in the same way as the $\lambda_i$, coincide with the eigenvalues $\lambda_i$ of $P$?

Clearly, if $\iota$ is a $C^1$-diffeomorphism with $$\left|\det{\rm D}\iota\right|=1\tag4,$$ we easily see that $$\tilde e_n:=e_n\circ\iota^{-1}\;\;\;\text{for }n\in\mathbb N$$ is an orthonormal basis of $L^2(\tilde\Omega)$. I guess that's the least we need to assume.

But what do we need to assume further? Do we need $\iota$ to extend to a linear isometry on $\mathbb R^d$ or something like that?

Answer 1

The question when two domains have the same Laplacian eigenvalues has been popularized as "Can one hear the shape of a drum" by Kac and has received a lot of research interest.

Some (increasingly less) simple observations:

• It is clearly sufficient that $\iota$ is an isometry, because in this case $f\mapsto f\circ \iota$ is a unitary operator from $L^2(\tilde\Omega)$ to $L^2(\Omega)$.
• You cannot expect to extract too much information on the map $\iota$ itself, because the input data only fix the range of $\iota$. The better question would be when there exists a map between $\Omega$ and $\tilde \Omega$ with nice properties.
• To get an interesting result, you should make at least some minimal assumptions on the boundaries of the domains. Otherwise, you can always just remove a set of capacity $0$ from $\Omega$ without changing the eigenvalues of $\Delta$. For example, the Laplacian on the disk and the Laplacian on the disk with countably many points removed have the same eigenvalues. In this case, the domains are not even homeomorphic.
• Even for Lipschitz domains, there does not need to exist an isometry between $\Omega$ and $\tilde \Omega$. The existence of isospectral domains that are not isometric was first proven by Gordon, Webb and Wolpert building on previous work most notably by Sunada. You can find pictures of a pair of such domains in the wikipedia article linked above.
• To the best of my knowledge, the question whether isospectral domains are are necessarily isometric is open for some very natural classes such as convex domains and domains with $C^\infty$-boundary.
• $\Delta$ and $\tilde \Delta$ have the same eigenvalues if and only if there exists a unitary $U\colon L^2(\Omega)$ to $L^2(\tilde \Omega)$ such that $U\Delta=\tilde \Delta U$. The reason why this question often has a negative answer is that there are just too many isometries of $L^2$. If in contrast $V\Delta_p=\tilde \Delta_p V$ for the Laplacians acting on $L^p$ and a surjective isometry $V\colon L^p(\Omega)\to L^p(\tilde \Omega)$, then $\Omega$ and $\tilde \Omega$ must be isometric under minor regularity assumptions by work of Arendt.