# DE Hyperbolic change of scale

• January 12th 2011, 03:10 PM
dwsmith
DE Hyperbolic change of scale
$x=\mu\xi \ \ \ y=\nu\eta$

I don't understand what the book is getting at and how this is done.

The final step is a "change of scale," where mu and nu are chosen so that in the transformed equation the coefficients of $\omega_{\xi\xi}, \ \omega_{\eta\eta}, \ \mbox{and} \ \omega$ are equal in absolute value. We have

$\displaystyle\frac{\partial^2}{\partial x^2}=\frac{1}{\mu^2}\frac{\partial^2}{\partial\xi^ 2}, \ \frac{\partial^2}{\partial y^2}=\frac{1}{\nu^2}\frac{\partial}{\partial\eta^2 },$

and equation $\omega_{xx}-4\omega_{yy}-\frac{5}{4}\omega=0$.

The codition

$\displaystyle\frac{1}{\mu^2}=\frac{4}{\nu^2}=\frac {5}{4}$

will be satisfied if $\mu=\frac{2}{\sqrt{5}}, \ \nu=\frac{4}{\sqrt{5}}.$
• January 13th 2011, 05:03 AM
Jester
What the book is getting at is to transform your PDE to

$\omega_{\xi \xi}-\omega_{\eta \eta}-\omega=0\;\;\;(*)$

Under the change of variables we obtain

$\dfrac{1}{\mu^2}\omega_{\xi \xi}-4\dfrac{1}{\nu^2}\omega_{\eta \eta}-\dfrac{5}{4}\omega=0$.

To hit the target (*) choose

$\dfrac{1}{\mu^2} = \dfrac{5}{4},\;\;\;4\dfrac{1}{\nu^2} = \dfrac{5}{4}.$