我正在尝试使用具有简化假设和的 Runge-Kutta Dormand Prince 方法以数值方式求解宇宙学的背景方程。方程是
和
对于物质和辐射。
我应该得到一个关于物质。但是在这两种情况下，我都得到了相同的比例因子变化。我是初学者。我无法弄清楚我哪里出错了。请帮我。
我传递给我的代码的方程式如下
就事论事
void fcn(double t, double *y, double *f){
double p,hub;
p=0;
hub=y[1]/y[0];
f[0] = y[1];
f[1] = -(y[0]*y[2]+3*p*y[0])/6.0;
f[2] = -3*hub*y[2]-3*hub*p;
}
对于辐射
void fcn(double t, double *y, double *f){
/* a=y[0],adot=y[1],rho=y[2] */
double p,hub;
hub=y[1]/y[0];
p= (2.0/3.0)*y[2];
f[0] = y[1];
f[1] = -(y[0]*y[2]+3*p*y[0])/6.0;
f[2] = -3*hub*y[2]-3*hub*p;
}
我只是想看看我们得到的函数依赖类型是什么，而不是按照宇宙学来设置初始条件。这可能是因为初始条件错误吗？
另一件事是随时间的变化在两种情况下应该是相同的（）但我也得到了不同的结果。
vs情节如下
我的代码在求解其他耦合微分方程方面表现出色。
#include<stdio.h>
#include<math.h>
#include<stdbool.h>
#include <stdlib.h>
#define N 3
double ddopri5(void fcn(double, double *, double *),
double *y);
double alpha;
void fcn(double t, double *y, double *f);
double eps;
int main(void){
double y[N];
//eps = 1.e-9;
printf("Enter epsilon:\n");
scanf("%lg", &eps);
y[0]=1.0;
y[1]=12.00;
y[2]=30.00000567845;
ddopri5(fcn, y);
}
void fcn(double t, double *y, double *f){
double p,hub;
p=0;
hub=y[1]/y[0];
f[0] = y[1];
f[1] = -(y[0]*y[2]+3*p*y[0])/6.0;
f[2] = -3*hub*y[2]-3*hub*p;
}
double ddopri5(void fcn(double, double *, double *),
double *y){
double t, h, a, b, tw, chi;
double w[N], k1[N], k2[N], k3[N], k4[N], k5[N],
k6[N],k7[N], err[N], dy[N];
int i;
double errabs;
int iteration;
FILE *fpw;
fpw=fopen("case1.dat", "w");
iteration = 0;
//eps = 1.e-9;
h = 1.0e-6;
a = 0.0;
b = 30;
t = a;
while(t < b -eps){
fcn(t, y, k1);
tw = t+ (1.0/5.0)*h;
for(i = 0; i < N; i++){
w[i] = y[i] + h*(1.0/5.0)*k1[i];
}
fcn(tw, w, k2);
tw = t+ (3.0/10.0)*h;
for(i = 0; i < N; i++){
w[i] = y[i] + h*((3.0/40.0)*k1[i]+
(9.0/40.0)*k2[i]);
}
fcn(tw, w, k3);
tw = t+ (4.0/5.0)*h;
for(i = 0; i < N; i++){
w[i] = y[i] + h*((44.0/45.0)*k1[i] -n
(56.0/15.0)*k2[i] +
(32.0/9.0)*k3[i]);
}
fcn(tw, w, k4);
tw = t+ (8.0/9.0)*h;
for(i = 0; i < N; i++){
w[i] = y[i] + h*((19372.0/6561.0)*k1[i] -
(25360.0/2187.0)*k2[i] +
(64448.0/6561.0)*k3[i] -
(212.0/729.0)*k4[i]);
}
fcn(tw, w, k5);
tw = t + h;
for(i = 0; i < N; i++){
w[i] = y[i] + h*((9017.0/3168.0)*k1[i] -
(355.0/33.0)*k2[i] +
(46732.0/5247.0)*k3[i] +
(49.0/176.0)*k4[i] -
(5103.0/18656.0)*k5[i]) ;
}
fcn(tw, w, k6);
tw = t + h;
for(i = 0; i < N; i++){
w[i] = y[i] + h*((35.0/384.0)*k1[i] +
(500.0/1113.0)*k3[i] + (125.0/192.0)*k4[i] -
(2187.0/6784.0)*k5[i] + (11.0/84.0)*k6[i]);
}
fcn(tw, w, k7);
errabs = 0;
for(i = 0; i < N; i++){
dy[i] = h*((35.0/384.0)*k1[i] +
(500.0/1113.0)*k3[i] +
(125.0/192.0)*k4[i] -
(2187.0/6784.0)*k5[i] +
(11.0/84.0)*k6[i]);
err[i] = h*((71.0/57600.0)*k1[i] -
(71.0/16695.0)*k3[i]
+ (71.0/1920.0)*k4[i] -
(17253.0/339200.0)*k5[i] +
(22.0/525.0)*k6[i] - (1.0/40.0)*k7[i]);
errabs+=err[i]*err[i];
}
errabs = sqrt(errabs);
if( errabs < eps){
t+= h;
for(i = 0; i < N; i++){
y[i]+=dy[i];
}
}
chi=errabs/eps;
chi = pow(chi, (1.0/6.0));
if(chi > 10) chi = 10;
if(chi < 0.1) chi = 0.1;
h*= 0.95/chi;
if( t + h > b ) h = b - t;
iteration++;
fprintf(fpw ,"%.25lf %.25lf %.25lf %.25lf \n",
t,y[0],y[1],y[2]);
if(iteration > 30000) break;
}
fclose(fpw);
return 0;
}
我也试过 Scipy 。我在下面分享我的 Scipy 代码
from scipy.integrate import odeint
import numpy as np
import matplotlib.pyplot as plt
def odes(x,t):
#assign each ode to a vector
A=x[0]
B=x[1]
C=x[2]
#define each ODE
dAdt=B
dBdt=-(1/6)*A*C
dCdt=-(3*C*B)/A
return [dAdt,dBdt,dCdt]
#define initialcondition
x0=[1,5,13]
#declare a time vector
t=np.linspace(0,30,1000)
x=odeint(odes,x0,t)
A=x[:,0]
B=x[:,1]
C=x[:,2]
#plot the results
plt.semilogy(t,A)
plt.semilogy(t,B)
plt.semilogy(t,C)
plt.show()