The Courant nodal domain theorem (for Neumann boundary conditions) says that the $n$-th eigenfunction has at most $n$ nodal domains (connected components where the eigenfunction has the same sign. However, Chavel in Eigenvalues in Riemannian Geometry seems to point out (P23) that Divergence theorem is used, so the regularity of the nodal set matters, whose proof by Cheng in dimension $\ge 3$ is incomplete. My question is: does it really affect the proof of the nodal domain theorem? More generally, does it affect the application of divergence theorem to a nodal domain (say to functions that are restrictions of $C^{\infty}(\mathbb R^n)$.
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$\begingroup$ Courant nodal domain theorem only involves the max-min characterization of eigenvalues. $\endgroup$– Kelei WangCommented Nov 29, 2013 at 1:58
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$\begingroup$ @kelei no you are probably thinking of the courant-fischer-weyl min-max principle en.wikipedia.org/wiki/Min-max_theorem. The nodal domain theorem is about the characterization of zeros of eigenfunctions, not so much about the characterization of eigenvalues. $\endgroup$– guestCommented Jan 5, 2014 at 18:18
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3$\begingroup$ But the argument proceeds as follows. Assume the number of nodal domains of the $k$th eigenfunction $u_k$ is larger than $k$, then you can find a function vanishing on a nodal domain, orthogonal to the first $k-1$ eigenfuncions and attaining the $k$th eigenvalue. Thus by the min-max principle it's an eigenfunction. But it vanishes on an open set, thus must be $0$ by the unique continuation principle. This argument does not involve any regularity of the nodal sets. $\endgroup$– Kelei WangCommented Jan 7, 2014 at 4:26
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$\begingroup$ Could you elaborate on your comment that Cheng's proof is incomplete (or provide a reference)? $\endgroup$– Graham CoxCommented Feb 8, 2017 at 22:11
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$\begingroup$ @Kelei Wang As I understand the construction, the constructed function is not necessarily differentiable on the nodal set (at the border between nodal domains). How can it then obey the unique continuation principle? (In Courant's proof that step is more complicated. Can it be as simple as you say?) $\endgroup$– steworiCommented Mar 9, 2023 at 13:58
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2 Answers
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Not an precise answer : As far as I remember this is not really an issue. One has to replace by an other version of divergence theorem ( for less regular domain) using geometric measure theory . The key point I think is that the set where the nodal lines are not regular (often called the singular set $\{u=0\}\cap\{\nabla u=0\}$ is of dimension at most $n-2$ so it has measure 0 for the $H^{n-1}$ measure. But this is quite specific to solutions of elliptic pde and I don't think that smoothness is enough to guarantee any kind of regularity of the nodal set.
A more precise answer : Take a look (for example) at the proof of the nice Sogge-Zelditch formula which precisely uses this type of div theroem (formula (8) in the paper).
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In P. BÉRARD's and D. MEYER's paper "Inégalités isopérimétriques et applications" ("Annales scientifiques de l'École Normale Supérieure", Serie 4, Volume 15 (1982) no. 3, pp. 513-541), the proof is given for the case where the nodal domain is a non-normal domain.
