Meaning of degenerate PDE I am trying to understand the concept of degenerate PDE. The definitions I found depend on the type of equation but let say for elliptic (see here). Let's take an example:
$$x\partial_x^4u + \partial_y^4u + \partial_z^4 + u_x = 0 \qquad (1)$$
then $(1)$ is degenerate if $\exists \xi\in\mathbb R^3, \xi\neq 0$ such that
$$\sum_{\lvert \alpha \rvert = 4} \frac{\partial F(x,Du)}{\partial q_\alpha}\xi^\alpha=0.$$
So I have two questions:

*

*I don't understand the definition of the polynomial (what is $\frac{\partial}{\partial q_\alpha}$?), could someone explain how to apply it to equation $(1)$?

*What does it imply that an equation is degenerate? I have seen that the usual theory does not apply, but what parts exactly? Does it have consequences on the numerical solution?

 A: First, your equation is linear, and the definition of degenerate you cite is for a nonlinear PDE.
So let's start with linear: A linear pde on a domain in $\mathbb{R}^n$ is of the form
$$ \sum_{|\alpha|\le k} A_\alpha(x)\partial^\alpha u f. $$
Its principal symbol is defined to be
$$
\sigma(x,\xi) = \sum_{|\alpha|=k} A_\alpha(x)\xi^\alpha
$$
The PDE is elliptic if $\sigma(x,\xi) > 0$ for all $x$ and nonzero $\xi$. It is degenerate elliptic if $\sigma(x,\xi) \ge 0$ for all $x$ and $\xi$. You can use this definition to verify that your question is in fact degenerate elliptic.
A nonlinear PDE is of the form
$$ F(x,\partial^\alpha u) = 0 $$
In other words, it is given by a function $F(x,q^\alpha)$, where you set $q^\alpha = \partial^\alpha u$.
Its linearization at a function $u$ is defined to be the linear differential operator (acting on $v$)
$$ F'(x,\partial^\alpha u)(v) = \left.\frac{d}{dt}F(x,\partial^\alpha (u+tv))\right|_{t=0} = \frac{\partial F}{\partial q^\alpha}(x,\partial^\alpha u)\partial^\alpha v $$
The nonlinear PDE $F(x,\partial^\alpha u)$ is said to be (degenerate) elliptic at $u$, if the linearized operator is (degenerate) elliptic.
