# Hyperbolic DE

• Jan 12th 2011, 01:25 PM
dwsmith
Hyperbolic DE
I don't understand the second step in the change of variables for this hyperbolic DE.

After a change of variables and some algebra, we have

$\mbox{2.3.11} \ \ u_{xx}-4u_{yy}+3u_{x}+u=0$

I am then told that the second step is a change of dependent variable

$\mbox{2.3.12} \ \ u=\exp{(\beta x)}\omega$,

where beta is chosen so that in the transformed equation the coefficient of $\omega_{x}$ vanishes. Differentiating 2.3.12 and substituting in 2.3.11, we obtain for $\omega$ the equation

$\omega_{xx}-4\omega_{yy}+(2\beta+3)\omega_{x}+(\beta^2+3\beta+ 1)\omega=0$

I can't obtain this equation (directly above) when I follow the directions.
• Jan 12th 2011, 02:30 PM
zzzoak
If

$
\beta = -3/2 \; ?
$

$
u_x=w_x+\beta w
$

$
u_{xx}=w_{xx}+2 \beta w_x +\beta^2w
$

$
u_{yy}=w_{yy}
$
• Jan 12th 2011, 02:37 PM
dwsmith
Quote:

Originally Posted by zzzoak
If

$
\beta = -3/2 \; ?
$

The book has that but I can't get the omega equation which is the first priority.
• Jan 12th 2011, 02:38 PM
dwsmith
Quote:

Originally Posted by zzzoak

$
u_x=w_x+\beta w
$

$
u_{xx}=w_{xx}+2 \beta w_x +\beta^2w
$

$
u_{yy}=w_{y}
$

How did you get that?
• Jan 12th 2011, 02:41 PM
zzzoak
Sorry, I write all derivatives without exp term.
• Jan 12th 2011, 02:45 PM
zzzoak
$
u=w(x,y) \; exp(\beta x)
$

$
\frac{d}{dx}u=(\frac{d}{dx}w)exp(\beta x)+w(\frac{d}{dx}exp(\beta x))
$