I am having trouble explaining/understanding the following phenomenon: To test fftw3 i am using the 2d poisson test case: laplacian(f(x,y)) = - g(x,y) with periodic boundary conditions. After applying the fourier transform to the equation we obtain : F(kx,ky) = G(kx,ky) /(kx² + ky²) (1) if i take g(x,y) = sin (x) + sin(y) , (x,y) \in [0,2 \pi] i have immediately f(x,y) = g(x,y) which is what i am trying to obtain with the fft : i compute G from g with a forward Fourier transform From this i can compute the Fourier transform of f with (1). Finally, i compute f with the backward Fourier transform (without forgetting to normalize by 1/(nx*ny)). In practice, the results are pretty bad? (For instance, the amplitude for N = 256 is twice the amplitude obtained with N = 512) Even worse, if i try g(x,y) = sin(x)*sin(y) , the curve has not even the same form of the solution. (note that i must change the equation; i divide by two the laplacian in this case : (1) becomes F(kx,ky) = 2*G(kx,ky)/(kx²+ky²) Here is the code: /* * fftw test -- double precision */ #include <iostream> #include <stdio.h> #include <stdlib.h> #include <math.h> #include <fftw3.h> using namespace std; int main() { int N = 128; int i, j ; double pi = 3.14159265359; double *X, *Y ; X = (double*) malloc(N*sizeof(double)); Y = (double*) malloc(N*sizeof(double)); fftw_complex *out1, *in2, *out2, *in1; fftw_plan p1, p2; double L = 2.*pi; double dx = L/((N - 1)*1.0); in1 = (fftw_complex*) fftw_malloc(sizeof(fftw_complex)*(N*N) ); out2 = (fftw_complex*) fftw_malloc(sizeof(fftw_complex)*(N*N) ); out1 = (fftw_complex*) fftw_malloc(sizeof(fftw_complex)*(N*N) ); in2 = (fftw_complex*) fftw_malloc(sizeof(fftw_complex)*(N*N) ); p1 = fftw_plan_dft_2d(N, N, in1, out1, FFTW_FORWARD,FFTW_MEASURE ); p2 = fftw_plan_dft_2d(N, N, in2, out2, FFTW_BACKWARD,FFTW_MEASURE); for(i = 0; i < N; i++){ X[i] = -pi + (i*1.0)*2.*pi/((N - 1)*1.0) ; for(j = 0; j < N; j++){ Y[j] = -pi + (j*1.0)*2.*pi/((N - 1)*1.0) ; in1[i*N + j][0] = sin(X[i]) + sin(Y[j]) ; // row major ordering //in1[i*N + j][0] = sin(X[i]) * sin(Y[j]) ; // 2nd test case in1[i*N + j][1] = 0 ; } } fftw_execute(p1); // FFT forward for ( i = 0; i < N; i++){ // f = g / ( kx² + ky² ) for( j = 0; j < N; j++){ in2[i*N + j][0] = out1[i*N + j][0]/ (i*i+j*j+1e-16); in2[i*N + j][1] = out1[i*N + j][1]/ (i*i+j*j+1e-16); //in2[i*N + j][0] = 2*out1[i*N + j][0]/ (i*i+j*j+1e-16); // 2nd test case //in2[i*N + j][1] = 2*out1[i*N + j][1]/ (i*i+j*j+1e-16); } } fftw_execute(p2); //FFT backward // checking the results computed double erl1 = 0.; for ( i = 0; i < N; i++) { for( j = 0; j < N; j++){ erl1 += fabs( in1[i*N + j][0] - out2[i*N + j][0]/N/N )*dx*dx; cout<< i <<" "<< j<<" "<< sin(X[i])+sin(Y[j])<<" "<< out2[i*N+j][0]/N/N <<" "<< endl; // > output } } cout<< erl1 << endl ; // L1 error fftw_destroy_plan(p1); fftw_destroy_plan(p2); fftw_free(out1); fftw_free(out2); fftw_free(in1); fftw_free(in2); return 0; } I can't find any (more) mistakes in my code (i installed the fftw3 library last week) and i don't see a problem with the maths either but i don't think it's the fft's fault. Hence my predicament. I am all out of ideas and all out of google as well. Any help solving this puzzle would be greatly appreciated. note : compiling : g++ test.cpp -lfftw3 -lm executing : ./a.out output and i use gnuplot in order to plot the curves : (in gnuplot ) splot "output" u 1:2:4 ( for the computed solution )