Solving Poisson’s Equation with the FFT I
Express any 2D function defined in 0 ? x,y ? 1 as a series ?(x,y) = Sj Sk ?jk sin(p jx) sin(p ky)
- Here ?jk are called Fourier coefficient of ?(x,y)
The inverse of this is: ?jk = 4 ?(x,y) sin(p jx) sin(p ky)
Poisson’s equation ?2 ? /? x2 + ? 2 ? /? y2 = f(x,y) becomes
Sj Sk (-p2j2 - p2k2) ?jk sin(p jx) sin(p ky)
= Sj Sk fjk sin(p jx) sin(p ky)
- where fjk are Fourier coefficients of f(x,y)
- and f(x,y) = Sj Sk fjk sin(p jx) sin(p ky)
This implies PDE can be solved exactly algebraically, ?jk = fjk / (-p2j2 - p2k2)