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?

  • 1
    $\begingroup$ The $p$-Laplacian is a non-linear operator. $\endgroup$
    – daw
    Sep 11, 2017 at 18:51
  • $\begingroup$ In other words^, it doesn't fit the supposed form with A, B, C, etc. $\endgroup$
    – Merkh
    Sep 12, 2017 at 14:21
  • $\begingroup$ @Merkh Good point. But how is it seen elliptic then? $\endgroup$
    – mavavilj
    Sep 12, 2017 at 15:04
  • 1
    $\begingroup$ 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. $\endgroup$
    – Merkh
    Sep 12, 2017 at 21:51
  • $\begingroup$ Also, there is a typo, it probably should be denoted as $\Delta_p$ rather than $\nabla_p.$ $\endgroup$
    – Merkh
    Sep 12, 2017 at 21:53

1 Answer 1


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 $y\neq0$.

  • $\begingroup$ For $p=1$, the assumption that $f$ is of class $C^2$ will not hold. Will the proof still remain the same ? $\endgroup$
    – creative
    Mar 10, 2020 at 4:30
  • $\begingroup$ Sir, I have asked this as a separate question. Can you spare some time to answer ? $\endgroup$
    – creative
    Mar 12, 2020 at 5:46
  • $\begingroup$ 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 $\endgroup$
    – Gio67
    Mar 13, 2020 at 17:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .