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?
 A: 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 $y\neq0$.
