2019年5月13日 星期一

C語言 例題5-6 四階 Runge-Kutta 解 ODE y'= -y + t^2 + 1 , 0<=t<=1 , y(0)=1 , 真實解 W(t)= -2e^(-t) + t ^2 - 2t + 3

C語言 例題5-6  四階 Runge-Kutta 解 ODE y'= -y + t^2 + 1 , 0<=t<=1 , y(0)=1 , 真實解 W(t)= -2e^(-t) + t ^2 - 2t + 3


//Code for RUNGE-KUTTA 4th ORDER METHOD in C Programming
// dy/dx = -y +x^2 +1  , 0<= x <=1 , y(0)=1

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

double rk4(double(*f)(double, double), double dx, double x, double y)
{
double k1 = dx * f(x, y),
k2 = dx * f(x + dx / 2, y + k1 / 2),
k3 = dx * f(x + dx / 2, y + k2 / 2),
k4 = dx * f(x + dx, y + k3);
return y + (k1 + 2 * k2 + 2 * k3 + k4) / 6;
}

double rate(double x, double y)
{
return (-y+ (x*x) +1);
}


int main(void)
{
double *y, x, y2;
double x0 = 0, x1 = 1, dx = .001;
int i, n = 1 + (x1 - x0)/dx;
y = (double *)malloc(sizeof(double) * n);

for (y[0] = 1, i = 1; i < n; i++)
y[i] = rk4(rate, dx, x0 + dx * (i - 1), y[i-1]);

printf("   x\t     y\t     real.        err.\n------------------------------------------\n");
for (i = 0; i < n; i++)
{
x = x0 + dx * i;
y2 = -2*exp(-x) + pow(x, 2) -2*x +3;
if (i%100==0)
    printf("%0.2lf\t%0.7lf\t%0.7lf\t%0.7lf\n", x, y[i], y2, (y[i]/y2 - 1));
}

return 0;
}


輸出畫面
   x      y      real.        err.
------------------------------------------
0.00 1.0000000 1.0000000 0.0000000
0.10 1.0003252 1.0003252 0.0000000
0.20 1.0025385 1.0025385 0.0000000
0.30 1.0083636 1.0083636 0.0000000
0.40 1.0193599 1.0193599 0.0000000
0.50 1.0369387 1.0369387 0.0000000
0.60 1.0623767 1.0623767 0.0000000
0.70 1.0968294 1.0968294 0.0000000
0.80 1.1413421 1.1413421 0.0000000
0.90 1.1968607 1.1968607 0.0000000
1.00 1.2642411 1.2642411 0.0000000

沒有留言:

張貼留言

8-QAM Signal 4 Phases 2 Amplitudes + 8PSK

 8-QAM Signal 4 Phases 2 Amplitudes + 8PSK import tkinter as tk from tkinter import messagebox import math import cmath # --- 8-QAM 參數設定 ---...