Finding the first nodal line of a convex domain
Let \(\Omega\) be a convex open set in \(\mathbb{R}^2\). Then the Dirichlet problem has eigenvalues \(0<\lambda_1 <\lambda_2 \leq\lambda_3 \leq\dots\) Let \(u\) be the second eigenfunction: \(\Delta u= -\lambda_2u\) on \(\Omega\) and \(u=0\) on \(\partial\Omega\). The set \(\Lambda= \{z\in \Omega: u(z)=0\}\) is called the first nodal line, and it is known that \(\Omega \setminus \Lambda\) consists of two components and that \(\overline\Lambda \cap\partial \Omega \neq \emptyset\). The paper under review is concerned with the case where \(\Omega\) is long and thin. Let \((x,y)\) be Cartesian coordinates and suppose that \(\Omega\) is rotated and dilated so that its projection onto the \(y\)-axis has length 1 and this is the shortest of all projections of \(\Omega\). The main result says that \(\Lambda\) lies in a strip \([x_0-C,\;x_0+C] \times\mathbb{R}\). Here \(x_0\) is such that the domains \(\{(x,y)\in \Omega: x<x_0\}\) and \(\{(x,y)\in \Omega: x>x_0\}\) have equal first eigenvalue for the Dirichlet problem, and \(C\) is an absolute constant. The main ideas of the proof are presented in a discursive style with the aid of several examples and diagrams. The details of the proof are given in another paper of the author's [Ann. Math., II. Ser. 141, 1-33 (1995; Zbl 0831.35115)].
scientific article; zbMATH DE number 1115820