c - 用于查找初始值的牛顿法实现，使用 Dormand Prince 求解 C 中的微分方程

Question

下面的代码可以很好地求解其中的微分方程组(代码中的 fcn 函数)，并具有正确的初始值。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该值是否为正确值，因为根据定义，ddopri 5 的输出应将 y_2(1) 和 y_3(1) 设为 0。如何针对此问题实现 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("%lg\n", 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 = %.15lf\n", 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 = %.25lf\n", 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] = %.15lf\n", i, y[i]);
    }*/

    return 0;
}

Answer 1

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

可以使用有限差分获得雅可比行列式，即每次将 Y 的元素向上和向下碰撞，计算 F，进行有限差分。

明确一点，矩阵J的元素(i,j)为dF_i/dY0_j