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I've just read a review paper about Yau's conjecture on nodal sets of the eigenfunctions for the Laplace operator on manifolds. Briefly, if $\phi_\lambda$, $\lambda$ are an eigenpair for the Laplace-Beltrami operator on a manifold $M$, i.e.

$$-\Delta\phi_\lambda = \lambda\phi,$$

then as $\lambda \to \infty$, there are constants $c$, $C$ such that

$$c\sqrt\lambda \le \text{area}(\{\phi_\lambda = 0\}) \le C\sqrt\lambda.$$

I've seen several results for manifolds without boundary of varying degrees of regularity. For example, Donnelly and Feffermann proved that Yau's conjecture is true for real analytic manifolds, but as I understand, the conjecture hasn't been proven yet for $C^\infty$ manifolds.

Are there extensions of the Yau conjecture to manifolds with boundary? Most of the work I've found (mainly other papers by Malinnikova and Logunov) considers only manifolds without boundary.

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    $\begingroup$ Is it fair to assume you are interested in Dirichlet/Neumann boundary conditions? $\endgroup$ – Josiah Park Oct 22 '19 at 22:13
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Yau's conjecture was proved for the real analytic manifolds by Donelly and Fefferman, first for manifolds without boundary, then for manifolds with boundary in the paper Nodal Sets of Eigenfunctions: Riemannian Manifolds With Boundary, Analysis, et Cetera (1990), 251--262. The main result there is stated below.

Theorem 1.2 On any real analytic Riemannian manifold $M$ with boundary, the $n-1$ dimensional Hausdorff measure, $\mathcal{H}^{n-1}(N)$, of the nodal set $N$ of eigenfunction $f$, $\Delta f = -\lambda f$, (with Dirichlet or Neumann conditions on the boundary of $M$) satisfies $$c_1\sqrt{\lambda}\leq \mathcal{H}^{n-1}(N) \leq c_2 \sqrt{\lambda},$$ for some positive constants $c_1$, $c_2$.

There are other papers that handle more general boundary conditions, many of which reference the above paper.

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  • $\begingroup$ Theorem 1.2 in that paper is only for real analytic manifolds. Theorem 1.1, which applies to $C^\infty$ manifolds, only says that eigenfunctions vanish to order at most $c\sqrt{\lambda}$. $\endgroup$ – user7868 Apr 16 '20 at 10:21
  • $\begingroup$ @user7868 Thanks for the correction. $\endgroup$ – Josiah Park Apr 17 '20 at 10:25

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