Confusion testing fftw3 - poisson equation 2d test - c++

Question

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);
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*dx ;
for(j = 0; j < N; j++){
Y[j] = -pi + j*dx ;
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 )

Answer 1

Here are some little points to be modified :
You need to account for all small frequencies, including the negative ones ! Index i corresponds to the frequency 2PI i/N but also to the frequency 2PI (i-N)/N. In the Fourier space, the end of the array matters as much as the beginning ! In our case, we keep the smallest frequency : it's 2PI i/N for the first half of the array, and 2PI(i-N)/N on the second half.
Of course, as Paul said, N-1 should be Nin double dx = L/(N - 1); => double dx = L/(N ); N-1 does not correspond to a continious periodic signal. It woud be hard to use it as a test case...
Scaling...I did it empirically
The result i obtain is closer to the expected one, for both cases. 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 );
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*dx ;
for(j = 0; j < N; j++){
Y[j] = -pi + j*dx ;
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++){
double fact=0;
in2[i*N + j][0]=0;
in2[i*N + j][1]=0;
if(2*i<N){
fact=((double)i*i);
}else{
fact=((double)(N-i)*(N-i));
}
if(2*j<N){
fact+=((double)j*j);
}else{
fact+=((double)(N-j)*(N-j));
}
if(fact!=0){
in2[i*N + j][0] = out1[i*N + j][0]/fact;
in2[i*N + j][1] = out1[i*N + j][1]/fact;
}else{
in2[i*N + j][0] = 0;
in2[i*N + j][1] = 0;
}
//in2[i*N + j][0] = out1[i*N + j][0];
//in2[i*N + j][1] = out1[i*N + j][1];
// in2[i*N + j][0] = out1[i*N + j][0]*(1.0/(i*i+1e-16)+1.0/(j*j+1e-16)+1.0/((N-i)*(N-i)+1e-16)+1.0/((N-j)*(N-j)+1e-16))*N*N;
// in2[i*N + j][1] = out1[i*N + j][1]*(1.0/(i*i+1e-16)+1.0/(j*j+1e-16)+1.0/((N-i)*(N-i)+1e-16)+1.0/((N-j)*(N-j)+1e-16))*N*N;
//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<< 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;
}
This code is far from being perfect, it is neither optimized nor beautiful. But it gives almost what is expected.
Bye,