# Fourier transform for Neumann boundary condition

I need to solve system of two coupled partial differential equations numerically.

$\frac{\partial x_1}{\partial t} = c_1\nabla ^2 x_1 + f_1(x_1,x_2) \\$

$\frac{\partial x_2}{\partial t} = c_2\nabla ^2 x_2 + K\frac{\partial x_1}{\partial t}$

The domain of system is a square region.

Boundary condition:

$x = constant \implies \frac{\partial x_1}{\partial x} = \frac{\partial x_2}{\partial x} = 0$

$y = constant \implies \frac{\partial x_1}{\partial y} = \frac{\partial x_2}{\partial y} = 0$

I tried to solve this system with Fourier transform. Solution becomes unstable after few iterations. I have solved this system earlier with finite difference scheme and it worked well so I know that constants of system are perfectly fine.

My question is can Fourier transform be used to solve these equations? I read somewhere that it because of Neumann boundary condition one cannot apply Fourier transform. Is this correct? If yes what is alternative?(I have read that cosine transform should be used but want to confirm). Thanks.

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How do you define $f_1(x_1,x_2)$ ? Is $f_1$ a periodic function ? – aberration Sep 10 '12 at 16:35
$f_1(x_1,x_2) = P(x_1)arctan(x_2)$, where P(x) is a polynomial. – chatur Sep 10 '12 at 19:36
Maybe the numerical unstability come from the function $f_1$. If $P$ is not equal to zero at the limite of the domain you can have a jump at the boundary condition when reconstructing the periodic signal. – aberration Sep 10 '12 at 20:07
@aberration, Thanks for comment but do not understand what it means. Probably I should study FFT more thoroughly. But If I am solving system in all the changes are limited to center part only(roughly) so that values at the boundary are not affected can I use FFT here? – chatur Sep 11 '12 at 6:01
Well maybe what I said is wrong, I think it's may work with some smooth function $f_1$ but anyway it's will give probably a wrong solution because near boundary process will not be solved properly. As you said you probably need to use a cosine modes or Chebyshev modes. If you really want to use Fourier transform you can use some kind of penalization method. – aberration Sep 11 '12 at 16:08