Non homogeneous PDE

• Apr 13th 2010, 01:26 PM
kushagra452
Non homogeneous PDE
d^2 f/dx^2 - d^f/dy^2 = x

This is a non homogeneous PDE. I first tried to find the solution to the homogeneous PDE associated with it.

d^2 f/dx^2 - d^f/dy^2 = 0

I equated d^2 f/dx^2 = d^f/dy^2 = k (a constant)

Now integrated both the sides to get f = k [(x^2)/2 + (y^2)/2]

Is this correct?
• Apr 14th 2010, 05:26 AM
Jester
Quote:

Originally Posted by kushagra452
d^2 f/dx^2 - d^f/dy^2 = x

This is a non homogeneous PDE. I first tried to find the solution to the homogeneous PDE associated with it.

d^2 f/dx^2 - d^f/dy^2 = 0

I equated d^2 f/dx^2 = d^f/dy^2 = k (a constant)

Now integrated both the sides to get f = k [(x^2)/2 + (y^2)/2]

Is this correct?

The solution will consist of two parts - a particular solution and the general solution of the homeogenous problem (which is $f_{xx} - f_{yy} = 0$). For the particular solution, any one will work. Say $f_p = \frac{1}{6}x^3$. The homogenous equation

$f_{xx} - f_{yy} = 0$ is the standard wave equation that has the general solution

$f = F(x+y) + G(x-y)$ where F and G are arbitrary.

Thus, the general solution is $f = F(x+y) + G(x-y) + \frac{1}{6}x^3$ .

As for your solution is doesn't satisfy the homogenous equation.
• Apr 14th 2010, 05:36 AM
kushagra452

How did you solve the homogeneous equation f(xx)-f(yy) = 0? I could not figure out how to use the change of variable. x = u+v , y= u-v.

Thanks again!

Cheers!
• Apr 14th 2010, 06:34 AM
Jester
Quote:

Originally Posted by kushagra452

How did you solve the homogeneous equation f(xx)-f(yy) = 0? I could not figure out how to use the change of variable. x = u+v , y= u-v.

Thanks again!

Cheers!

Yes, that's it! If you use $x = u+v$, $y = u-v$, then

$
f_{xx}-f_{yy} = f_{uv} = 0,
$

which you can integrate twice.