Canonical form for parabolic PDE? I'm having trouble reducing this parabolic equation to canonical form.
$$\frac{\partial^2 u}{\partial x^2} + 2\frac{\partial^2 u}{\partial x \partial y} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial u}{\partial x} - \frac{\partial u}{\partial y} = 0$$
I know it's parabolic because I checked: $B^2 - AC0$,
$$\begin{align}
A = 1,\\
B = 1,\\
C = 1,\\
B^2 - AC = 1 - (1)(1) = 0\end{align}$$ so it's parabolic
I'm really not sure where to go from here.  I know a change of variables is involved but I'm not sure how to reduce this to canonical form.  I appreciate any help.  Thanks in advance!
 A: Set
$$
\frac{\partial}{\partial t}=\frac{\partial}{\partial y}-\frac{\partial}{\partial x}\quad
\quad\text{and}\quad \frac{\partial}{\partial z}=\frac{\partial}{\partial x}+\frac{\partial}{\partial y},
$$
and you have that
$$
\frac{\partial^2 u}{\partial x^2} + 2\frac{\partial^2 u}{\partial x \partial y} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial u}{\partial x} - \frac{\partial u}{\partial y}=\frac{\partial}{\partial z^2}-\frac{\partial}{\partial t}.
$$
A: Edit:
The PDE is parabolic and the characteristics are to be found from the equation:
$$\xi_x^2 + 2 \xi_x \xi_y + \xi_y^2 = (\xi_x + \xi_y)^2 = 0.$$
You can solve for $\xi_x/\xi_y$ as follows:
$$\frac{\xi_x}{\xi_y} + 1 = 0 \Rightarrow \frac{\xi_x}{\xi_y} = -1 =-\frac{dy}{dx},$$
and hence you have information of only one characteristic since the solution of the equation above is double:
$$\begin{align}
\xi = & x-y, \\
\eta = & f(x,y).
\end{align}$$
Beeing $f(x,y)$ any function so $\eta$ is not a linear combination of $\xi$. Use now the chain rule, remembering that:
$$u_x = u_\xi \xi_x + u_\eta \eta_x, \quad u_y = u_\xi \xi_y + u_\eta \eta_y, \quad u_{xx} = \ldots$$
I'm sure you can work from here. Let us know about your progress.
I hope this is useful to you.
Cheers!
