What does $\Delta_p$ denote in $p$-Laplace equation?

Question

What does $\Delta_p$ denote in $p$-Laplace equation?

Why isn't it $\Delta_{p-2}$?

The $p$-Laplace equation is:

$$- \Delta_p u = - \text{div} ( | \nabla u |^{p-2} \nabla u) = 0$$

Answer 1

The operator $\triangle_p$ is related to $L^p$-space instead of $L^{p-2}$ space. For instance, $\triangle_p$ is a maximal monotone operator from $W_0^{1,p}(\Omega)$ into $(W_0^{1,p}(\Omega))'$. If $p=2$, $\triangle_p$ is a "natural" operator in $L^2$ space.

Answer 2

The $p$-Laplace equation is the Euler–Lagrange equation for the $p$-energy $$ E_p[u] = \int |\nabla u|^p $$ That is, a function that minimizes $E_p$ subject to some boundary conditions is a solution of the $p$-Laplace equation. So it makes sense that $p$ is the number associated to it. It's also the exponent of the natural domain of definition of $\Delta_p$, which is the Sobolev space $W^{1,p}$. Indeed, being in this space is what one needs to have $E_p$ finite.