TITLE: How to transform a PDE into canonical form
QUESTION [5 upvotes]: Question: $$4U_{xx} +12U_{xy} +9U_{yy}=0$$
I would like to transform this pde into canonical form. I know that the pde is a parabolic type but I am unsure how to proceed with rewriting it without cross-derivatives.

REPLY [2 votes]: (I'm going to use lowercase $u$'s for the PDE (instead of capital, like you did), then change to uppercase when I change variables, if that's okay.)
First, I'll show you how to do it, then I'll summarize why:
Set $\xi=y-\frac{3}{2}x$, $\eta=x.$ Making the change of variables $u(x(\xi,\eta),y(\xi,\eta))=U(\xi,\eta),$ we can use the chain rule to calculate
\begin{align*}
u_{xx}&=\frac{9}{4}U_{\xi\xi}-3U_{\xi\eta}+U_{\eta\eta}\\
u_{yy}&=U_{\xi\xi}\\
u_{xy}&=-\frac{3}{2}U_{\xi\xi}+U_{\xi\eta}
\end{align*}
So, $$0=4u_{xx}+12u_{xy}+9u_{yy}=4U_{\eta\eta}.$$ That is, our PDE is in the canonical form $${U_{\eta\eta}=0.}$$
This comes from looking at the PDE $$au_{xx}+2bu_{xy}+cu_{yy}+du_x+eu_y+fu=g$$ (with $a,b,c,d,e,f,g$ functions of $x,y$), with Cauchy data $u|_{\Gamma}=u_0$ and $\partial_\nu u|_{\Gamma}=u_1$, from which one can set up the characteristic-type equation
$$\frac{dy}{dx}=\frac{b\pm\sqrt{b^2-ac}}{a}$$ (which in the parabolic case has only one characteristic), then setting $\xi$ to be the constant of integration in the solution (we want it constant on characteristic curves). Finally, we just pick $\eta$ so that the change of variables matrix is invertible.
See, e.g., https://math.la.asu.edu/~dajones/class/476/canonical.pdf
for more details on how to find the change of variables for different classes of PDEs.