Let $\Omega$ be a precompact open subset of ${\mathbb R}^2$ with piecewise smooth boundary. Let $u$ be an eigenfunction of the Laplacian on $\Omega$ with either Dirichlet or Neumann conditions on $\partial \Omega$. Is it possible that some level set of $u$ has a `cusp'? By cusp, I mean a critical point $z$ of $u$ such that the level sets of $u$ topologically foliate a neighborhood of $z$.

The prototypical example of a cusp is the critical point $(0,0)$ of the function $f(x,y)=x^2+y^3$. A Mathematica generated sketch of the topological foliation is provided here:

enter image description here

S. Y. Cheng observed many years ago that if a critical point lies on the zero level set of a planar eigenfunction, then near the critical point, the zero level set is akin to that of a harmonic polynomial. It follows that the answer to the question is `no' if restricted to critical points that lie in the zero level set.

  • 3
    $\begingroup$ One way to try for a counterexample is to choose a region with an eigenvalue of large enough multiplicity that one might expect some eigenfunction to have a level set with the desired singularity. $\endgroup$ Jul 4, 2017 at 22:52
  • 1
    $\begingroup$ Thanks Noam. We've been looking at the square and its high dimensional eigenspaces, but nothing came up yet. $\endgroup$ Jul 4, 2017 at 22:59
  • 2
    $\begingroup$ The square was my first thought, but then I noticed you required smooth boundary. Is piecewise smooth good enough? $\endgroup$ Jul 4, 2017 at 23:17
  • 3
    $\begingroup$ @Noam. My apologies. I clumsily intended the word `say' to indicate the possibility of piecewise smooth. I will edit. Thanks! $\endgroup$ Jul 5, 2017 at 1:01

1 Answer 1


The question has been answered affirmatively by Olti Myrtaj and Kelly Chen as part of a Research Experiences for Undergraduates project. See here


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .