FFT setting the boundary conditions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T08:07:34Z http://mathoverflow.net/feeds/question/44137 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/44137/fft-setting-the-boundary-conditions FFT setting the boundary conditions henryreed 2010-10-29T14:20:32Z 2010-10-31T09:58:02Z <p>I am using the FFT method to solve the two dimensional PDE on a $2\pi\times 2\pi$ square. I have boundary conditions that $u(x,0,t)=0$ and $u(x,2\pi,t)=0$. How do I go about setting these boundaries? I can't just expand in different basis functions because of the way my fft function works.</p> <p>Thanks in advance.</p> http://mathoverflow.net/questions/44137/fft-setting-the-boundary-conditions/44315#44315 Answer by Igor Khavkine for FFT setting the boundary conditions Igor Khavkine 2010-10-31T09:58:02Z 2010-10-31T09:58:02Z <p>If you don't mind wasting some resources, extend the $y$ coordinate to the interval $[0,4\pi]$. Then only look at solutions that are odd (anti-symmetric) under the flip $y\mapsto 4\pi-y$. In particular, if your PDE has source terms, they have to be anti-symmetrically mirrored from the $[0,2\pi]$ to the $[2\pi,4\pi]$ half of the extended interval. When restricted to $y\in[0,2\pi]$ a periodic solution on the extended interval will satisfy the desired boundary conditions.</p>