How to transform a PDE into canonical form 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. 
 A: Here's a method I learnt for PDE's of the form $a_{11}u_{xx}+2a_{12}u_{xy}+a_{22}u_{yy}=0$.

If you write your PDE as a problem $\mathcal{L} u=0$, we have that $\mathcal{L}$ is equal to the differential operator
$$\mathcal{L}=4\partial_x^2+12\partial_x \partial_y+9\partial_y^2=(2\partial_x+3\partial_y)^2$$
We then define new operators:
$$\partial_{\xi}=2\partial_x+3\partial_y,\quad \partial_{\eta}=\partial_{y} \tag{1}$$
We then have that your PDE reduces to the form:
$$u_{\xi \xi}=0 \tag{2}$$
To find the variable transformation, we rewrite $(1)$ as:
$$\begin{pmatrix} \partial_{\xi} \\ \partial_{\eta} \end{pmatrix}=B\begin{pmatrix} \partial_x \\ \partial_y \end{pmatrix},\qquad B=\begin{pmatrix} 2 & 3 \\ 0 & 1 \end{pmatrix}$$
Then, the transformation can be found by taking the transpose of $B$:
$$\begin{pmatrix} x \\ y \end{pmatrix}=B^T \begin{pmatrix} \xi \\ \eta \end{pmatrix}$$
Hence, in our case, we obtain:
$$x=2\xi,\qquad y=3\xi+\eta$$
so the inverse transform is:
$$\xi=x/2,\qquad \eta=y-\frac{3x}{2}$$
You can check that using this change of variable leads to the PDE in $(2)$. Note that this method doesn't just work for parabolic PDE's, in general what you should do is complete the square on $\mathcal{L}$ and conveniently define the new operators so that you get the desired canonical form. Then you can proceed in the same way as I have done with your problem.
A: (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.
