English

Reverse Faber-Krahn inequality for a truncated laplacian operator

Analysis of PDEs 2020-03-30 v1

Abstract

In this paper we prove a reverse Faber-Krahn inequality for the principal eigenvalue μ1(Ω)\mu_1(\Omega) of the fully nonlinear eigenvalue problem \labeleq{λN(D2u)=μuin Ω,u=0on Ω. \label{eq} \left\{\begin{array}{r c l l} -\lambda_N(D^2 u) & = & \mu u & \text{in }\Omega, \\ u & = & 0 & \text{on }\partial \Omega. \end{array}\right. Here λN(D2u) \lambda_N(D^2 u) stands for the largest eigenvalue of the Hessian matrix of uu. More precisely, we prove that, for an open, bounded, convex domain ΩRN\Omega \subset \mathbb{R}^N, the inequality μ1(Ω)π2[diam(Ω)]2=μ1(Bdiam(Ω)/2), \mu_1(\Omega) \leq \frac{\pi^2}{[\text{diam}(\Omega)]^2} = \mu_1(B_{\text{diam}(\Omega)/2}), where diam(Ω)\text{diam}(\Omega) is the diameter of Ω\Omega, holds true. The inequality actually implies a stronger result, namely, the maximality of the ball under a diameter constraint. Furthermore, we discuss the minimization of μ1(Ω)\mu_1(\Omega) under different kinds of constraints.

Keywords

Cite

@article{arxiv.2003.12107,
  title  = {Reverse Faber-Krahn inequality for a truncated laplacian operator},
  author = {Enea Parini and Julio Rossi and Ariel Salort},
  journal= {arXiv preprint arXiv:2003.12107},
  year   = {2020}
}

Comments

11 pages, 1 figure