问题
I am trying to solve Poison equation with Dirichlet boundary condition for four sides of computational domain. As known that I should use FFTW_RODFT00 to satisfy the condition. However, the result is not correct.Could you please help me?
#include <stdio.h>
#include <math.h>
#include <cmath>
#include <fftw3.h>
#include <iostream>
#include <vector>
using namespace std;
int main() {
int N1=100;
int N2=100;
double pi = 3.141592653589793;
double L1 = 2.0;
double dx = L1/(double)(N1-1);
double L2= 2.0;
double dy=L2/(double)(N2-1);
double invL1s=1.0/(L1*L1);
double invL2s=1.0/(L2*L2);
std::vector<double> in1(N1*N2,0.0);
std::vector<double> in2(N1*N2,0.0);
std::vector<double> out1(N1*N2,0.0);
std::vector<double> out2(N1*N2,0.0);
std::vector<double> X(N1,0.0);
std::vector<double> Y(N2,0.0);
fftw_plan p, q;
int i,j;
p = fftw_plan_r2r_2d(N1,N2, in1.data(), out1.data(), FFTW_RODFT00, FFTW_RODFT00, FFTW_EXHAUSTIVE);
q = fftw_plan_r2r_2d(N1,N2, in2.data(), out2.data(), FFTW_RODFT00, FFTW_RODFT00, FFTW_EXHAUSTIVE);
int l=-1;
for(i = 0;i <N1;i++){
X[i] =-1.0+(double)i*dx ;
for(j = 0;j<N2;j++){
l=l+1;
Y[j] =-1.0+ (double)j*dy ;
in1[l]= sin(pi*X[i]) + sin(pi*Y[j]) ; // row major ordering
}
}
fftw_execute(p);
l=-1;
for ( i = 0; i < N1; i++){ // f = g / ( kx² + ky² )
for( j = 0; j < N2; j++){
l=l+1;
double fact=0;
in2[l]=0;
if(2*i<N1){
fact=((double)i*i)*invL1s;;
}else{
fact=((double)(N1-i)*(N1-i))*invL1s;
}
if(2*j<N2){
fact+=((double)j*j)*invL2s;
}else{
fact+=((double)(N2-j)*(N2-j))*invL2s;
}
if(fact!=0){
in2[l] = out1[l]/fact;
}else{
in2[l] = 0.0;
}
}
}
fftw_execute(q);
l=-1;
double erl1 = 0.;
for ( i = 0; i < N1; i++) {
for( j = 0; j < N2; j++){
l=l+1;
erl1 +=1.0/pi/pi*fabs( in1[l]- 0.25*out2[l]/((double)(N1-1))/((double)(N2-1)));
printf("%3d %10.5f %10.5f\n", l, in1[l], 0.25*out2[l]/((double)(N1-1))/((double)(N2-1)));
}
}
cout<<"error=" <<erl1 <<endl ;
fftw_destroy_plan(p); fftw_destroy_plan(q); fftw_cleanup();
return 0;
}
回答1:
I recognize a trick I provided you in Poisson equation using FFTW with rectanguar domain and the code I provided in my answer to Confusion testing fftw3 - poisson equation 2d test , which was adapted from the code of the asker @Charles_P ! Please consider adding links to these url in future questions !
The answer to Confusion testing fftw3 - poisson equation 2d test was devoted to the case of periodic boundary conditions. So here are a few modifications to solve the case of Dirichlet boundary conditions.
The fftw_plan_r2r_2d(N1,N2, in1.data(), out1.data(), FFTW_RODFT00, FFTW_RODFT00,FFTW_EXHAUSTIVE)
corresponds to a type I discrete sine transform as defined by the FFTW library:
It's meaning is well described in https://en.wikipedia.org/wiki/Discrete_sine_transform . If the size of the FFTW array is N1=4
and its values [a,b,c,d], the full array including boundaries is [0,a,b,c,d,0]. Hence the spatial step is:
And the frequencies f_k
of the type I DST are:
The inverse of the type I DST is the type I DST, except for a scale factor (see http://www.fftw.org/doc/1d-Real_002dodd-DFTs-_0028DSTs_0029.html#g_t1d-Real_002dodd-DFTs-_0028DSTs_0029), here 4.(N1+1).(N2+1)
.
Lastly, the test case must be adapted to the case of Dirichlet boundary conditions. Indeed, on the box of size L1,L2
the function
The source term
corresponds to a single frequency of the DST.The source term
is directly derived from the solution
Finally, here is a code solving the 2D Poisson equation using the type I DST of the FFTW library:
#include <stdio.h>
#include <math.h>
#include <cmath>
#include <fftw3.h>
#include <iostream>
#include <vector>
using namespace std;
int main() {
int N1=100;
int N2=200;
double pi = 3.141592653589793;
double L1 = 1.0;
double dx = L1/(double)(N1+1);//+ instead of -1
double L2= 5.0;
double dy=L2/(double)(N2+1);
double invL1s=1.0/(L1*L1);
double invL2s=1.0/(L2*L2);
std::vector<double> in1(N1*N2,0.0);
std::vector<double> in2(N1*N2,0.0);
std::vector<double> out1(N1*N2,0.0);
std::vector<double> out2(N1*N2,0.0);
std::vector<double> X(N1,0.0);
std::vector<double> Y(N2,0.0);
fftw_plan p, q;
int i,j;
p = fftw_plan_r2r_2d(N1,N2, in1.data(), out1.data(), FFTW_RODFT00, FFTW_RODFT00, FFTW_EXHAUSTIVE);
q = fftw_plan_r2r_2d(N1,N2, in2.data(), out2.data(), FFTW_RODFT00, FFTW_RODFT00, FFTW_EXHAUSTIVE);
int l=0;
for(i = 0;i <N1;i++){
for(j = 0;j<N2;j++){
X[i] =dx+(double)i*dx ;
Y[j] =dy+ (double)j*dy ;
//in1[l]= sin(pi*X[i])*sin(pi*Y[j]) ; // row major ordering
in1[l]=2*Y[j]*(L2-Y[j])+2*X[i]*(L1-X[i]);
l=l+1;
}
}
fftw_execute(p);
l=-1;
for ( i = 0; i < N1; i++){ // f = g / ( kx² + ky² )
for( j = 0; j < N2; j++){
l=l+1;
double fact=0;
fact=pi*pi*((double)(i+1)*(i+1))*invL1s;
fact+=pi*pi*((double)(j+1)*(j+1))*invL2s;
in2[l] = out1[l]/fact;
}
}
fftw_execute(q);
l=-1;
double erl1 = 0.;
for ( i = 0; i < N1; i++) {
for( j = 0; j < N2; j++){
l=l+1;
X[i] =dx+(double)i*dx ;
Y[j] =dy+ (double)j*dy ;
//double res=0.5/pi/pi*in1[l];
double res=X[i]*(L1-X[i])*Y[j]*(L2-Y[j]);
erl1 +=pow(fabs(res- 0.25*out2[l]/((double)(N1+1))/((double)(N2+1))),2);
printf("%3d %10.5g %10.5g\n", l, res, 0.25*out2[l]/((double)(N1+1))/((double)(N2+1)));
}
}
erl1=erl1/((double)N1*N2);
cout<<"error=" <<sqrt(erl1) <<endl ;
fftw_destroy_plan(p); fftw_destroy_plan(q); fftw_cleanup();
return 0;
}
It is compiled by g++ main.cpp -o main -lfftw3 -Wall
.
EDIT : How to compute the response to each frequency ?
The idea of FFT-based is to represent the solution as a linear combination of functions:
- In the case of periodic boundary conditions, the FFT is used. The base functions are:
In the case of Dirichlet Boundary conditions, the type-I DST is used. The base functions, being 0 at
x=0
andx=L1
, are:In the case of Neumann boundary conditions, the type-I DCT is used. The base functions are:
Their derivatives are null at x=0
and x=L1
.
Let's compute the second derivative of the componant f_k
of the type-I DST:
Hence, the componant k
of the DST of the solution corresponds to the componant k
of the DST of the source term, scaled by a factor
来源:https://stackoverflow.com/questions/35173102/fftw3-for-poisson-with-dirichlet-boundary-condition-for-all-side-of-computationa