# A PDE transformation

• Sep 22nd 2010, 05:49 AM
Silverflow
A PDE transformation
Hi there,
In the notes I'm reading, they have stated the following PDE:
\$\displaystyle u_{xx}+u_{yy}+2au_x+2bu_y+cu=d\$.
Then, by defining:
\$\displaystyle v(x,y)=e^{ax+by}u(x,y)\$
The PDE becomes, in terms of v:
\$\displaystyle v_{xx}+v_{yy}+(c-(a^2+b^2))v=e^{ax+by}d\$
I don't see how they made this transformation. Should I taking partial derivatives of u as a function of v?

• Sep 22nd 2010, 06:04 AM
Ackbeet
Quote:

Should I taking partial derivatives of u as a function of v?
I would write u as a function of v and the exponential function there. That is,

\$\displaystyle u(x,y)=e^{-ax-by}\,v(x,y).\$

Then start taking the partial derivatives you need to assemble the ingredients of your first equation. Then plug everything in and simplify.
• Sep 22nd 2010, 06:21 AM
Silverflow
Cheers.
• Sep 22nd 2010, 06:23 AM
Ackbeet
• Sep 22nd 2010, 01:33 PM
Silverflow
Yeah, took a bit, but I got it out. Thanks for your advice.
• Sep 22nd 2010, 04:19 PM
Ackbeet
You're welcome. Have a good one!