# How to see that the $p$-Laplacian is elliptic?

How to see that the $p$-Laplacian is elliptic?

The $p$-Laplacian can be formulated as:

$$\Delta_p u = |\nabla u|^{p-2} [(p-1)u_{vv}+(n-1)Hu_v]$$

where $H$ is a sort of sign function.

In the case of a linear 2nd order elliptic PDE one could try finding $A, B$ and $C$:

Since any linear 2nd order PDE can be written as:

$$A u_{xx}+2 B u_{xy}+C u_{yy} + D u_x + E u_y + Fu + G=0$$

But since the $p$-Laplacian is nonlinear (or quasilinear), then what techniques display if it's elliptic?

• The $p$-Laplacian is a non-linear operator.
– daw
Sep 11, 2017 at 18:51
• In other words^, it doesn't fit the supposed form with A, B, C, etc. Sep 12, 2017 at 14:21
• @Merkh Good point. But how is it seen elliptic then? Sep 12, 2017 at 15:04
• Elliptic operators really only make sense as linear operators. However, that isn't to say that people don't call a class of nonlinear operators "elliptic". If your operator is written as $F(x, u, Du, D^2u, \dots),$ and there exists a first order Taylor expansion of $F$ at whatever point you are interested in, if the first order expansion is an elliptic operator, then we say your nonlinear op is elliptic at that point. Sep 12, 2017 at 21:51
• Also, there is a typo, it probably should be denoted as $\Delta_p$ rather than $\nabla_p.$ Sep 12, 2017 at 21:53

The $p$-laplacian equation $\Delta_{p}u=0$ is the Euler-Lagrange equation of the functional $$F(u)=\int_{\Omega}\frac{1}{p}|\nabla u|^{p}dx.$$ The main feature of this functional is that the function $f(\xi)=\frac{1} {p}|\xi|^{p}$ is convex. In general if you have a convex function $f:\mathbb{R}^{N}\rightarrow\mathbb{R}$ and you consider the the Euler-Lagrange equation of the functional $$F(u)=\int_{\Omega}f(\nabla u)\,dx,$$ if $f$ is sufficiently smooth, say $C^{1}$, you find that critical points $u$ satisfy $$\int_{\Omega}\nabla_{\xi}f(\nabla u)\cdot\nabla v\,dx=0$$ for all $v\in C_{c}^{\infty}(\Omega)$, or integrating by parts, $$\int_{\Omega}v\operatorname{div}(\nabla_{\xi}f(\nabla u))\,dx=0$$ for all $v\in C_{c}^{\infty}(\Omega)$, which lead to the equation $$\operatorname{div}(\nabla_{\xi}f(\nabla u))=\sum_{i=1}^{N}\frac{\partial }{\partial x_{i}}\left( \frac{\partial f}{\partial\xi_{i}}(\nabla u)\right) =0.$$ This is the canonical example of an elliptic equation in divergence form.
If $f$ and $u$ are of class $C^{2}$ you can use the chain rule to write the previous equation as $$\sum_{i=1}^{N}\sum_{j=1}^{N}\frac{\partial^{2}f}{\partial\xi_{j}\partial \xi_{i}}(\nabla u)\frac{\partial^{2}u}{\partial x_{j}\partial x_{i}}=0.$$ Now for a convex function $f$ of class $C^{2}$, the Hessian matrix $\left( \frac{\partial^{2}f}{\partial\xi_{j}\partial\xi_{i}}(\xi)\right) _{i,j=1}% ^{N}$ is positive semi-definite, meaning that all the eigenvalues are nonnegative, that is $$\sum_{i=1}^{N}\sum_{j=1}^{N}\frac{\partial^{2}f}{\partial\xi_{j}\partial \xi_{i}}(\xi)y_{i}y_{j}\geq0$$ for all $y\in\mathbb{R}^{N}$. An even stronger condition is the Hessian matrix $\left( \frac{\partial^{2}f}{\partial\xi_{j}\partial\xi_{i}}(\xi)\right) _{i,j=1}^{N}$ is positive definite, meaning that all the eigenvalues are positive, that is $$\sum_{i=1}^{N}\sum_{j=1}^{N}\frac{\partial^{2}f}{\partial\xi_{j}\partial \xi_{i}}(\xi)y_{i}y_{j}>0$$ for all $y\in\mathbb{R}^{N}$, $y\neq0$. This latter condition is used to define elliptic equations of second order. Precisely, if you have an equation of the form $$G(x,u,\nabla u,\nabla^{2}u)=0,$$ where $G=G(x,u,\xi,P)$, with $P=\left( p_{i,j}\right) _{i,j=1}^{N}% \in\mathbb{R}^{N\times N}$ the space of square matrices, is of class $C^{2}$ you say that this equation is elliptic in a subset $\Gamma$ of $\Omega \times\mathbb{R}\times\mathbb{R}^{N}\times\mathbb{R}^{N\times N}$ if the matrix $\left( \frac{\partial G}{\partial p_{i,j}}(x,u,\xi,P)\right) _{i,j=1}^{N}$ is positive definite for every $(x,u,\xi,P)\in\Gamma$. This is the standard definition (see for example Chapter 17 in the book Gilbarg-Trudinger). For equations coming from a convex functional you have $$G(\xi,P)=\sum_{i=1}^{N}\sum_{j=1}^{N}\frac{\partial^{2}f}{\partial\xi _{j}\partial\xi_{i}}(\xi)p_{i,j}%$$ and so you have ellipticity at every point $(\xi,P)$ such that $\left( \frac{\partial^{2}f}{\partial\xi_{j}\partial\xi_{i}}(\xi)\right) _{i,j=1}% ^{N}$ is positive definite (and not just positive semi-definite).
In particular for the $p$-laplacian equation for $\xi\neq0$\ you get \begin{align*} \frac{\partial f}{\partial\xi_{i}}(\xi) & =|\xi|^{p-2}\xi_{i}=(\xi_{1}% ^{2}+\cdots+\xi_{N}^{2})^{\frac{p-2}{2}}\xi_{i},\\ \frac{\partial^{2}f}{\partial\xi_{j}\partial\xi_{i}}(\xi) & =(p-2)(\xi _{1}^{2}+\cdots+\xi_{N}^{2})^{\frac{p-2}{2}-1}\xi_{i}\xi_{j}+(\xi_{1}% ^{2}+\cdots+\xi_{N}^{2})^{\frac{p-2}{2}}\delta_{i,j}\\ & =|\xi|^{p-4}\left[ (p-2)\xi_{i}\xi_{j}+|\xi|^{2}\delta_{i,j}\right] , \end{align*} which is a positive definite matrix since \begin{align*} \sum_{i=1}^{N}\sum_{j=1}^{N}\frac{\partial^{2}f}{\partial\xi_{j}\partial \xi_{i}}(\xi)y_{i}y_{j} & =\sum_{i=1}^{N}\sum_{j=1}^{N}|\xi|^{p-4}\left[ (p-2)\xi_{i}\xi_{j}y_{i}y_{j}+|\xi|^{2}\delta_{i,j}y_{i}y_{j}\right] \\ & =|\xi|^{p-4}\left[ (p-2)\sum_{i=1}^{N}\sum_{j=1}^{N}\xi_{i}\xi_{j}% y_{i}y_{j}+|\xi|^{2}\sum_{i=1}^{N}\sum_{j=1}^{N}\delta_{i,j}y_{i}y_{j}\right] \\ & =|\xi|^{p-4}\left[ (p-2)(\xi\cdot y)^{2}+|\xi|^{2}|y|^{2}\right] >0 \end{align*} for $y\neq0$.
• For $p=1$, the assumption that $f$ is of class $C^2$ will not hold. Will the proof still remain the same ? Mar 10, 2020 at 4:30
• Hi Hirak, I deleted my answer to your question because it was misleading. I don't know how to do it for $p=1$. Try looking at this paper math.berkeley.edu/~evans/brezis.pdf Mar 13, 2020 at 17:21