Basic HTML version of Foils prepared April 22 2000

Foil 4 Solving Poisson's Equation with the FFT I

From Parallel FFT and use in PDE Solvers Computational Science Class CPS615 -- Winter Semester 2000. by Geoffrey C. Fox
1 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)
2 The inverse of this is: ?jk = 4 ?(x,y) sin(p jx) sin(p ky)
3 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)
4 This implies PDE can be solved exactly algebraically, ?jk = fjk / (-p2j2 - p2k2)
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