Canonical form of the PDE $u_{xx}+2u_{xy}+2u_{yz}+u_{zz}=0$

Question

Find the canonical form of the PDE:

$$u_{xx}+2u_{xy}+2u_{yz}+u_{zz}=0$$

I know how to do that in the "normal" way: finding the the actual transformation using the eigenvectors of the appropriate matrix.

However, I think that there should be another (easier) way using the fact that $u_{yy}$ is not in the equation. Any ideas?

First think how to factorize $x^2+2xy+2yz+z^2$ in $\mathbb{C}$ . — doraemonpaul, Nov 22 '13 at 03:53

score 0 · Answer 1 · answered May 29 '18 at 06:25

0

Actual transformation is

$$\xi=x,\;\eta=x-y,\;\zeta=\frac{x-y-z}{\sqrt2}.$$

We get hyperbolic equation

$$u_{\xi\xi}+u_{\zeta\zeta}-u_{\eta\eta}=0.$$

answered May 29 '18 at 06:25

Aleksas Domarkas

1 Answers1