Linear partial differential eqn.

Posted September 22nd, 2007 by vic777

Analysis questions

How can I reduce the following equation to canonical form?
$2u_{xx} - 4u_{xy} +2u_{yy} + 3u = 0$

Comments

Answer

On September 22nd, 2007 Anonymous says:

You have to start with characteristic equation (Keep in mind you have to change the sign of the mixed coefficient)
$2y'^2+4y'+2=0$ or $2(y'+1)^2=0$ As the solution is double the equation is of parabolic type. Integrating the equation y'+1=0 you get $y+x=c$
As there are double characteristics you have to take any independent other function to reduce to canonic form.
Write $u(x,y)=v(x+y,x)$
You have
$\frac{\partial u}{\partial x}(x,y)=\frac{\partial v}{\partial \xi}(x+y,x)+\frac{\partial v}{\partial \psi}(x+y,x)$
$\frac{\partial u}{\partial y}(x,y)=\frac{\partial v}{\partial \xi}(x+y,x)$
$\frac{\partial ^2u}{\partial x^2}(x,y)=\frac{\partial ^2 v}{\partial \xi^2}(x+y,x)+2\frac{\partial ^2v}{\partial \psi}(x+y,x)+\frac{\partial ^2 v}{\partial \psi^2}(x+y,x)$
$\frac{\partial ^2u}{\partial x \partial y}(x,y)=\frac{\partial ^2 v}{\partial \xi^2}(x+y,x)+\frac{\partial ^2v}{\partial \psi}(x+y,x)$
$\; \frac{\partial ^2u}{\partial y^2}(x,y)=\frac{\partial ^2 v}{\partial \xi^2}(x+y,x)$

Now to replace into equation you may multiply by 2,-4 and 2 and add up down

$2\frac{\partial ^2u}{\partial x^2}(x,y)=2\frac{\partial ^2 v}{\partial \xi^2}(x+y,x)+4\frac{\partial ^2v}{\partial \psi}(x+y,x)+2\frac{\partial ^2 v}{\partial \psi^2}(x+y,x)$
$-4\frac{\partial ^2u}{\partial x \partial y}(x,y)=-4\frac{\partial ^2 v}{\partial \xi^2}(x+y,x)-4\frac{\partial ^2v}{\partial \psi}(x+y,x)$
$2\frac{\partial ^2u}{\partial y^2}(x,y)=2\frac{\partial ^2 v}{\partial \xi^2}(x+y,x)$

You have $2\frac{\partial ^2 v}{\partial \psi^2}(x+y,x)+3v(x+y,x)=0$
or
$2\frac{\partial ^2 v}{\partial \psi^2}(\xi,\psi)+3v(\xi,\psi)=0$
This is the canonical form

Great help.Thank-you!

On September 22nd, 2007 vic777 says:

Great help.Thank-you!