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.
[Math] How to transform a PDE into canonical form
linear-pdeparabolic pdepartial differential equations
Best Answer
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.