我最近刚刚为学校选择了编程，我遇到了涉及快速傅里叶变换包FFTW的代码问题。
在我的代码中，我从一个初始函数开始（在这种情况下u（x，t = 0）= sin（x）+ sin（3 * x）并将使用RK4来尝试求解热方程的U_t对于那些有FFTW包经验的人，我发送一个大小为N的数组并对其进行转换，得到两个导数，然后采用逆变换来求解热方程的U_xx。
#include <iostream>
#include <cmath>
#include <fstream>
#include <fftw3.h>
using namespace std;
const double Pi= 3.141592653589;
const int N = 2048;
void Derivative(double& num1, double& num2, int J){
//takes two derivatives
num1 = (-1)*J*J*num1;
num2 = (-1)*J*J*num2;
}
double Function(double X){ // Initial function
double value = sin(X) + sin(3*X);
return value;
}
void FFT(double *in, double *Result){
double *input, *outI;
input = new double [N];
outI = new double [N];
for(int i=0; i<N; i++){
input[i] = in[i];
}
//Declaring transformed matricies;
fftw_complex *out;
out= (fftw_complex*) fftw_malloc(sizeof(fftw_complex)*((N/2)+1));
//Plans for execution
fftw_plan p;
p= fftw_plan_dft_r2c_1d(N,input, out, FFTW_ESTIMATE);
fftw_execute(p);
//Derivative
for(int i=0; i<(N/2+1);i++){
Derivative(out[i][0],out[i][1],i);
}
fftw_plan pI;
//Execution of Inverse FFT
pI = fftw_plan_dft_c2r_1d(N,out,outI,FFTW_PRESERVE_INPUT);
fftw_execute(pI);
for(int i=0;i<N; i++){//Dividing by the size of the array
Result[i] = (1.0/N)*outI[i];
}
fftw_free(outI); fftw_free(out);
fftw_free(input);
fftw_destroy_plan(pI); fftw_destroy_plan(p);
}
int main()
{
//Creating and delcaring function variables
double *initial,*w1,*w2,*w3,*w4,*y,*holder1,*holder2, *holder3;
double dx= 2*Pi/N, x=0, t=0;
double h = pow(dx,2), tmax= 100*h;
//creating arrays for the RK4 procedure
initial = new double [N];
w1 = new double [N]; w2 = new double [N]; y= new double [N];
w3 = new double [N]; w4 = new double [N]; holder1 = new double [N];
holder2 = new double [N]; holder3 = new double [N];
double *resultArray=new double[N];
int j =0;
//Output files
ofstream sendtofileINITIAL("HeatdataINITIAL.dat");
ofstream sendtofile5("Heatdata5.dat");
ofstream sendtofile25("Heatdata25.dat");
ofstream sendtofile85("Heatdata85.dat");
ofstream sendtofileFINAL("HeatdataFINAL.dat");
//Initial data
for(int i=0; i<N; i++){
y[i] = Function(x);
sendtofileINITIAL << i*2*Pi/N <<" "<< y[i] << endl;
x+=dx;
}
//RK4
// y[i+1] = y[i] + (h/2)*(w1 + 2*w2 + 2*w3 + w4)
// where the w1,w2,w3,w4 is the standard RK4 calculations
// w1= f(y[i]), w2= f(y[i]+(h/2)*w1[i])
// w3= f(y[i] + (h/2)*w2[i]) and w4 = f(y[i]+h*w3[i])
while(t<=tmax){
FFT(y,resultArray);
for(int i=0; i<N;i++){ //Calculating w1
w1[i] = resultArray[i];
}
for(int i=0; i<N; i++){
holder1[i] = y[i] + (h/2)*w1[i];}
FFT(holder1,resultArray);
for(int i=0; i<N; i++){ //Calculating w2
w2[i] = resultArray[i];
}
for(int i=0; i<N; i++){
holder2[i] = y[i] + (h/2)*w2[i];
}
FFT(holder2,resultArray);
for(int i=0; i<N; i++){ //Calculating w3
w3[i] = resultArray[i];
}
for(int i=0; i<N; i++){
holder3[i]= y[i] + h*w3[i];
}
FFT(holder3,resultArray);
for(int i=0; i<N; i++){ //Calculating w4
w4[i] = resultArray[i];
}
for(int i=0; i<N; i++){
y[i] += (h/6.0)*(w1[i] + 2*w2[i] + 2*w3[i] +w4[i]);
//Outputing data at certain time periods
if(j==5)
sendtofile5 << i*2*Pi/N <<" "<< y[i] << endl;
if(j==25)
sendtofile25 << i*2*Pi/N <<" "<< y[i] << endl;
if(j==85)
sendtofile85 << i*2*Pi/N <<" "<< y[i] << endl;
if(j==100)
sendtofileFINAL << i*2*Pi/N <<" "<< y[i] << endl;
}
j++;// counter
t+=h;// time counter
}
sendtofileINITIAL.close();
sendtofile5.close();
sendtofile25.close();
sendtofile85.close();
sendtofileFINAL.close();
return 0;
}
当我在t的几次迭代之后绘制y [i]的图形时，这些值非常快速地爆炸并且在正面和负面之间交替。
如果有人对可能导致这些奇怪结果的语法问题有任何建议，或者如果在我的数学方法中发现错误，我会非常感谢任何信息。谢谢。
答案 0 :(得分：2)
您是否知道如何在不使用FFT的情况下解决此PDE？我建议你这样做。在开始之前，有助于了解答案。
你说它是热/扩散方程，你给出了初始温度分布，但你没有说到边界条件是什么。既然你什么都没说，假设两个端点都是绝缘的（U = x = 0，x = 0，x = L），这是正确的吗？或者它们的价值是固定的吗？如果你知道它们是什么会有所帮助 - 没有适当的边界条件就没有解决方案。
答案 1 :(得分：1)
这是我在这里的第一个答案，因此请避免任何愚蠢的错误。大约7年前就问过这个问题，但是我在这段代码中发现了问题。希望这会帮助某人。
稳定性标准不被认为是结果，数值解也没有像#duffymo所述那样收敛。因此，根据稳定性标准，我添加了用于规范网格大小的域的长度，并减少了空间中的点数。
傅立叶空间中的频率定义不正确。我在代码中添加了此部分，现在此代码可以正常工作。
我将修改后的代码附有此注释。
#include <iostream>
#include <cmath>
#include <fstream>
#include <fftw3.h>
using namespace std;
const double Pi = 3.141592653589;
const int N = 64;
const double L = 5.0;
void Derivative(double& num1, double& num2, double J) {
//takes two derivatives
num1 = -1 * pow(J, 2)*num1;
num2 = -1 * pow(J, 2)*num2;
}
double Function(double X) { // Initial function
double value = 1.0 / cosh(3 * (X - M_PI));//sin(X) + sin(3*X);
return value;
}
void FFT(double *in, double *Result) {
double *input, *outI, *kx;
input = new double[N];
outI = new double[N];
kx = new double[N];
for (int i = 0; i<N; i++) {
input[i] = in[i];
}
//Declaring transformed matricies;
fftw_complex *out;
out = (fftw_complex*)fftw_malloc(sizeof(fftw_complex)*(N));
//Plans for execution
fftw_plan p;
p = fftw_plan_dft_r2c_1d(N, input, out, FFTW_ESTIMATE);
fftw_execute(p);
//----This part calculate frequencies--------//
//Frequencies
for (int i = 0; i< N / 2; ++i) {
kx[i] = i / L;
}
kx[N / 2] = 0.00;
for (int i = 0; i < ((N / 2) - 1); ++i) {
kx[i + 1 + N / 2] = -kx[N / 2 - i - 1];
}
//Derivative
for (int i = 0; i<N; i++) {
//the frequencies is passed to this function to calculate
//the derivatives
Derivative(out[i][0], out[i][1], kx[i]);
}
fftw_plan pI;
//Execution of Inverse FFT
pI = fftw_plan_dft_c2r_1d(N, out, outI, FFTW_PRESERVE_INPUT);
fftw_execute(pI);
for (int i = 0; i<N; i++) {//Dividing by the size of the array
Result[i] = (1.0 / N)*outI[i];
}
fftw_free(outI); fftw_free(out);
fftw_free(input);
fftw_destroy_plan(pI); fftw_destroy_plan(p);
}
int main()
{
//Creating and delcaring function variables
double *initial, *w1, *w2, *w3, *w4, *y, *holder1, *holder2, *holder3;
double dx = 2 * Pi / N, x = 0, t = 0, dt = 0.01;
double h = pow(dx, 2), tmax = 100 * h;
//creating arrays for the RK4 procedure
initial = new double[N];
w1 = new double[N]; w2 = new double[N]; y = new double[N];
w3 = new double[N]; w4 = new double[N]; holder1 = new double[N];
holder2 = new double[N]; holder3 = new double[N];
double *resultArray = new double[N];
double *resultnext = new double[N];
int j = 1;
// The stability parameter are declared but not used
float alpha = 0.1, CFL = alpha * dt / (10 * pow(dx, 2));
//Output files
ofstream sendtofileINITIAL("HeatdataINITIAL.dat");
ofstream sendtofile5("Heatdata5.dat");
ofstream sendtofile25("Heatdata25.dat");
ofstream sendtofile85("Heatdata85.dat");
ofstream sendtofileFINAL("HeatdataFINAL.dat");
//Initial data
for (int i = 0; i < N; i++) {
y[i] = Function(x);
sendtofileINITIAL << i * 2 * Pi / N << " " << y[i] << endl;
x += dx;
}
//RK4
// y[i+1] = y[i] + (h/2)*(w1 + 2*w2 + 2*w3 + w4)
// where the w1,w2,w3,w4 is the standard RK4 calculations
// w1= f(y[i]), w2= f(y[i]+(h/2)*w1[i])
// w3= f(y[i] + (h/2)*w2[i]) and w4 = f(y[i]+h*w3[i])
while (t <= tmax) {
FFT(y, resultArray);
for (int i = 0; i < N; i++) { //Calculating w1
w1[i] = resultArray[i];
}
for (int i = 0; i < N; i++) {
holder1[i] = y[i] + (h / 2)*w1[i];
}
FFT(holder1, resultArray);
for (int i = 0; i < N; i++) { //Calculating w2
w2[i] = resultArray[i];
}
for (int i = 0; i < N; i++) {
holder2[i] = y[i] + (h / 2)*w2[i];
}
FFT(holder2, resultArray);
for (int i = 0; i < N; i++) { //Calculating w3
w3[i] = resultArray[i];
}
for (int i = 0; i < N; i++) {
holder3[i] = y[i] + h * w3[i];
}
FFT(holder3, resultArray);
for (int i = 0; i < N; i++) { //Calculating w4
w4[i] = resultArray[i];
}
for (int i = 0; i < N; i++) {
y[i] += (h / 6.0)*(w1[i] + 2 * w2[i] + 2 * w3[i] + w4[i]);
//Outputing data at certain time periods
if (j == 5)
sendtofile5 << i * 2 * Pi / N << " " << y[i] << endl;
if (j == 25)
sendtofile25 << i * 2 * Pi / N << " " << y[i] << endl;
if (j == 85)
sendtofile85 << i * 2 * Pi / N << " " << y[i] << endl;
if (j == 100)
sendtofileFINAL << i * 2 * Pi / N << " " << y[i] << endl;
}
j++;// counter
t += h;// time counter
}
sendtofileINITIAL.close();
sendtofile5.close();
sendtofile25.close();
sendtofile85.close();
sendtofileFINAL.close();
return 0;
}