<s>[INST] Reduction of a quadratic form to a canonical form I'm supposed to reduce following polynomial to its canonical form. But my result differs from the one given in my book, so I'm not sure if it's correct too.
$$
q = u_{xx} - u_{xy} - 2 u_{yy} + u_x + u_y = 0
$$
So, the characteristic quadratic polynomial is
$$
x^2 - xy -2y^2
$$
Here I'm using Lagrange's reduction method for quadratic polynomial:
\begin{align}
x^2 - xy -2y^2 &= (x^2 - xy + \frac{1}{4}y^2) - \frac{1}{4}y^2 - 2y^2\\
&= (x-\frac{1}{2}y)^2 - \frac{9}{4}y^2\\
&= \xi^2 - \frac{9}{4}\eta^2\\\\
&\xi = x-\frac{1}{2}y\\
&\eta = y
\end{align}
Now I can define the function $u$ like this:
$$
u(x,y) = u(\xi(x,y), \eta(x,y))
$$
Now I need to deduce $u_x$ and $u_y$ too:
\begin{align}
u_x &= u_\xi \cdot \xi_x + u_\eta \cdot \eta_x = u_\xi\\
u_y &= u_\xi \cdot \xi_y + u_\eta \cdot \eta_y = -\frac{1}{2}u_\xi + u_\eta
\end{align}
So, my canonical form looks like this:
\begin{align}
q &= u_{\xi\xi} - \frac{9}{4}u_{\eta\eta} + u_\xi - \frac{1}{2}u_\xi + u_\eta\\
&= \underline{\underline{u_{\xi\xi} - \frac{9}{4}u_{\eta\eta} + \frac{1}{2}u_\xi + u_\eta}}
\end{align}
Is my solution correct or is there a mistake?
 [/INST] We reduce second order PDE with constant coefficients to canonic form
$ u_{xx} - u_{xy} - 2 u_{yy} + u_x + u_y = 0$
Quadratic form:
$Q=x^2 - xy -2y^2=\left(x-\frac{y}{2}\right)^2-\left(\frac{3y}{2}\right)^2$
We get transform
$\xi=x,\quad\eta=\frac{x}{3}+\frac{2y}{3}$
and canonical form of PDE:
$u_{\xi\xi}-u_{\eta\eta}+u_\xi+u_\eta=0$
see:
V.S.Vladimirov, A Collection of Problems on the Equations
of Mathematical Physics, Springer, 1986

In our case
$$B=\begin{pmatrix}1 & \frac{1}{3}\\
0 & \frac{2}{3}\end{pmatrix}$$
</s>