Reverse Faber-Krahn inequality for a truncated laplacian operator

preprint2020arXivOpen access

Abstract

In this paper we prove a reverse Faber-Krahn inequality for the principal eigenvalue μ1(Ω)μ_1(Ω) of the fully nonlinear eigenvalue problem \[ \label{eq} \left\{\begin{array}{r c l l} -λ_N(D^2 u) & = & μu & \text{in }Ω, \\ u & = & 0 & \text{on }\partial Ω. \end{array}\right. \] Here λN(D2u) λ_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 Ω⊂RNΩ\subset \mathbb{R}^N, the inequality μ1(Ω)≤π2[diam(Ω)]2=μ1(Bdiam(Ω)/2), μ_1(Ω) \leq \frac{π^2}{[\text{diam}(Ω)]^2} = μ_1(B_{\text{diam}(Ω)/2}), where diam(Ω)\text{diam}(Ω) is the diameter of ΩΩ, 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)μ_1(Ω) under different kinds of constraints.

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