下面的代码可以用正确的初始值来求解其中的微分方程组(代码中的fcn函数(。然而，任务的重点是用一些随机值替换初始值y_1(0(和y_2(0(，并实现一些迭代方法来找到正确的初始值来求解方程。我已经知道如何检查这个值是否是正确的值，因为根据定义，ddopri 5的输出应该给出y_2(1(和y_3(1(为0。对于这个问题，我如何实现Newton-Raphson？
#include<stdio.h>
#include<math.h>
#include<stdbool.h>
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[4];
//eps = 1.e-9;
printf("Enter alpha:n");
scanf("%lg", &alpha);
printf("Enter epsilon:n");
scanf("%lg", &eps);
y[0]=1.0;//x1(0)
y[1]=-1.22565282791;//x2(0)
y[2]=-0.274772807644;//p1(0)
y[3]=0.0;//p2(0)
ddopri5(fcn, y);
}
void fcn(double t, double *y, double *f){
/* double h = 0.25;*/
f[0] = y[1];
f[1] = y[3] - sqrt(2)*y[0]*exp(-alpha*t);
f[2] = sqrt(2)*y[3]*exp(-alpha*t) + y[0];
f[3] = -y[2];
}
double ddopri5(void fcn(double, double *, double *), double *y){
double t, h, a, b, tw, chi;
double w[4], k1[4], k2[4], k3[4], k4[4], k5[4], k6[4], k7[4], err[4], dy[4];
int i;
double errabs;
int iteration;
iteration = 0;
//eps = 1.e-9;
h = 0.1;
a = 0.0;
b = 1;//3.1415926535;
t = a;
while(t < b -eps){
printf("%lgn", eps);
fcn(t, y, k1);
tw = t+ (1.0/5.0)*h;
for(i = 0; i < 4; i++){
/*printf("k1[%i] = %.15lf n", i, k1[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 < 4; i++){
/*printf("k2[%i] = %.15lf n", i, k2[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 < 4; i++){
/*printf("k3[%i] = %.15lf n", i, k3[i]);*/
w[i] = y[i] + h*((44.0/45.0)*k1[i] - (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 < 4; i++){
/*printf("k4[%i] = %.15lf n", i, k4[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 < 4; i++){
/*printf("k5[%i] = %.15lf n", i, k5[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 < 4; i++){
/*printf("k6[%i] = %.15lf n", i, k6[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 < 4; i++){
/* printf("k7[%i] = %.15lf n", i, k7[i]);*/
/* dy[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]);*/
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])*/;
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]);
/*printf("err[%i] = %.15lf n", i, err[i]);*/
errabs+=err[i]*err[i];
}
errabs = sqrt(errabs);
printf("errabs = %.15lfn", errabs);
if( errabs < eps){
t+= h;
printf(" FROM IF t t = %.25lf, n h = %.25lf, n errabs = %.25lf, n iteration = %i . n", t, h, errabs, iteration);
for(i = 0; i < 4; i++){
y[i]+=dy[i];
}
}
/*Avtomaticheskiy vibor shaga*/
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;
/* for(i = 0; i < 4; i++){
printf("y[%i] = %.15lf n", i, y[i]);
}*/
iteration++;
printf("t = %.25lf t h = %.25lfn", t, h);
/*if(iteration > 5) break;*/
printf("end n");
for(i = 0; i < 4; i++){
printf("y[%i] = %.15lf n", i, y[i]);
}
if(iteration > 30000) break;
}
/* for(i = 0; i < 4; i++){
printf("y[%i] = %.15lfn", i, y[i]);
}*/
return 0;
}
试试这个：
Y0=initial_guess
while (true) {
F=ddopri(Y0);
Error=F-F_correct
if (Error small enough)
break;
J=jacobian(ddopri, Y0) // this is the matrix dF/dY0
Y0=Y0-J^(-1)*Error // here you have to solve a linear system
雅可比可以使用有限差来获得，即一次向上和向下碰撞Y的元素一个，计算F，取有限差。
需要明确的是，矩阵j的元素(i，j(是dF_i/dY0_j