Canonical Form - Second Order PDE

• Nov 22nd 2010, 06:40 AM
mukmar
Canonical Form - Second Order PDE
For the equation:

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

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

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

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

AND

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

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

$\eta u_{\xi\eta} + u_{\xi} = 0$
• Nov 24th 2010, 03:45 PM
zzzoak
To begin with

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

and the second $x$ derivative is

$
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}=
$

$
=(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}=
$

$
=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
$
\xi_x=-2 \frac{y}{x^3}
$

and
$
\eta_x=- \frac{y^2}{x^2},
$

also
$
\xi_{xx} \; and \; \eta_{xx}.
$

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

When we have everything then inserting this to initial de.