Thread: Canonical Form - Second Order PDE

1. Canonical Form - Second Order PDE

For the equation:

$\displaystyle 2x^2u_{xx} + 5xyu_{xy} + 2y^2u_{yy} + 8xu_{x} + 5yu_y = 0$

I'm just trying to understand how to put it in canonical form.

I've understood till the point where the characteristics are:

$\displaystyle \xi = \frac{y}{x^2}$

AND

$\displaystyle \eta = \frac{y^2}{x}$

I'm just not sure how that leads to the canonical form:

$\displaystyle \eta u_{\xi\eta} + u_{\xi} = 0$

2. To begin with

$\displaystyle u_x=u_\xi \; \xi_x+u_\eta \; \eta_x$

and the second $\displaystyle x$ derivative is

$\displaystyle u_{xx}=(u_\xi \; \xi_x)_x+(u_\eta \; \eta_x)_x=u_{\xi x}\; \xi_x+u_\xi \; \xi_{xx}+u_{\eta x}\; \eta_x+u_\eta \; \eta_{xx}=$

$\displaystyle =(u_{\xi \xi} \; \xi_x+u_{\xi \eta} \; \eta_x) \; \xi_x+u_\xi \; \xi_{xx}+(u_{\eta \xi} \xi_x+u_{\eta \eta} \eta_x) \; \eta_x+u_\eta \; \eta_{xx}=$

$\displaystyle =u_{\xi \xi} \; \xi_x^2+2u_{\xi \eta} \; \xi_x \eta_x+u_{\eta \eta} \eta_x^2+u_\xi \; \xi_{xx}+u_\eta \; \eta_{xx}.$

Known functions are
$\displaystyle \xi_x=-2 \frac{y}{x^3}$
and
$\displaystyle \eta_x=- \frac{y^2}{x^2},$
also
$\displaystyle \xi_{xx} \; and \; \eta_{xx}.$

In the same way with
$\displaystyle u_{xy} \; \; and \; \; u_{yy}.$

When we have everything then inserting this to initial de.