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The $p$-Laplacian eigenvalue problem is formulated as

\begin{align} -\Delta_p u &= \lambda_p^p |u|^{p-2}u && \text{in } \Omega\\ u &=0 && \text{on }\partial\Omega. \end{align}

Now for a function $\phi \in C^2(\Omega)$ we can rewrite the PDE as $$ -(p-2)|\nabla \phi(x)|^{p-4}\left[ {|\nabla \phi(x)|^2 \over p-2}\Delta \phi(x) + \Delta_{\infty}\phi(x) \right] = \lambda_p^p|\phi|^{p-2}u $$

Taking the approach of viscosity sub/supersolutions, in short, we find that the limit problem is

\begin{align} \min\{ |\nabla u(x)| - \Lambda_\infty u(x), -\Delta_\infty u(x)\} &= 0 &&\text{in }\Omega\\ u&=0 && \text{on }\partial \Omega, \end{align}

where $\Lambda_\infty = \lim_{p \to \infty} \lambda_p$.

I guess what I don't understand is why we are taking the new problem as the minimum of two equations. It seems a bit random to notice that $|\nabla u(x)| - \Lambda_\infty u(x) = 0$ and that $-\Delta_\infty u(x) = 0$ and then take the minimum of these two. Why exactly is this?

tulio
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    Are you familiar with The $\infty$-Eigenvalue Problem? In Theorem 1.21 they show that solutions of the $p$-laplacian eigenvalue problem converge to a solution of $\infty$-laplacian eigenvalue problem written with this minimum – Raul Fernandes Horta Mar 29 '25 at 16:52
  • @RaulFernandesHorta I've seen that, I just don't get the logic behind combining those two equations into one with this minimum (instead of a system of equations for example) – tulio Mar 29 '25 at 17:53

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