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前期是分享了matlab下面实现四阶龙格库塔（Runge-Kutta）求解微分方程，这期分享一下C++、C、Java、Python下面的四阶龙格库塔（Runge-Kutta）求解微分方程。
前文传送门：matlab代码实现四阶龙格库塔求解微分方程
C++方法
#include <bits/stdc++.h>using namespace std;
// "dy/dx = (x - y)/2"float dydx(float x, float y){ return((x - y)/2);}
// Finds value of y for a given x using step size h// and initial value y0 at x0.float rungeKutta(float x0, float y0, float x, float h){ // Count number of iterations using step size or // step height h int n = (int)((x - x0) / h);
float k1, k2, k3, k4, k5;
// Iterate for number of iterations float y = y0; for (int i=1; i<=n; i++) { // Apply Runge Kutta Formulas to find // next value of y k1 = h*dydx(x0, y); k2 = h*dydx(x0 + 0.5*h, y + 0.5*k1); k3 = h*dydx(x0 + 0.5*h, y + 0.5*k2); k4 = h*dydx(x0 + h, y + k3);
// Update next value of y y = y + (1.0/6.0)*(k1 + 2*k2 + 2*k3 + k4);;
// Update next value of x x0 = x0 + h; }
return y;}
// Driver Codeint main(){ float x0 = 0, y = 1, x = 2, h = 0.2; cout << "The value of y at x is : " << rungeKutta(x0, y, x, h);
return 0;}
C：
// C program to implement Runge Kutta method#include<stdio.h>
// A sample differential equation "dy/dx = (x - y)/2"float dydx(float x, float y){ return((x - y)/2);}
// Finds value of y for a given x using step size h// and initial value y0 at x0.float rungeKutta(float x0, float y0, float x, float h){ // Count number of iterations using step size or // step height h int n = (int)((x - x0) / h);
float k1, k2, k3, k4, k5;
// Iterate for number of iterations float y = y0; for (int i=1; i<=n; i++) { // Apply Runge Kutta Formulas to find // next value of y k1 = h*dydx(x0, y); k2 = h*dydx(x0 + 0.5*h, y + 0.5*k1); k3 = h*dydx(x0 + 0.5*h, y + 0.5*k2); k4 = h*dydx(x0 + h, y + k3);
// Update next value of y y = y + (1.0/6.0)*(k1 + 2*k2 + 2*k3 + k4);;
// Update next value of x x0 = x0 + h; }
return y;}
// Driver methodint main(){ float x0 = 0, y = 1, x = 2, h = 0.2; printf("\nThe value of y at x is : %f", rungeKutta(x0, y, x, h)); return 0;}
Java：
// Java program to implement Runge Kutta methodimport java.io.*;class differential{ double dydx(double x, double y) { return ((x - y) / 2); } // Finds value of y for a given x using step size h // and initial value y0 at x0. double rungeKutta(double x0, double y0, double x, double h) { differential d1 = new differential(); // Count number of iterations using step size or // step height h int n = (int)((x - x0) / h);
double k1, k2, k3, k4, k5;
// Iterate for number of iterations double y = y0; for (int i = 1; i <= n; i++) { // Apply Runge Kutta Formulas to find // next value of y k1 = h * (d1.dydx(x0, y)); k2 = h * (d1.dydx(x0 + 0.5 * h, y + 0.5 * k1)); k3 = h * (d1.dydx(x0 + 0.5 * h, y + 0.5 * k2)); k4 = h * (d1.dydx(x0 + h, y + k3));
// Update next value of y y = y + (1.0 / 6.0) * (k1 + 2 * k2 + 2 * k3 + k4); // Update next value of x x0 = x0 + h; } return y; } public static void main(String args[]) { differential d2 = new differential(); double x0 = 0, y = 1, x = 2, h = 0.2; System.out.println("\nThe value of y at x is : " + d2.rungeKutta(x0, y, x, h)); }}
Python3
# Python program to implement Runge Kutta method# A sample differential equation "dy / dx = (x - y)/2"def dydx(x, y): return ((x - y)/2)
# Finds value of y for a given x using step size h# and initial value y0 at x0.def rungeKutta(x0, y0, x, h): # Count number of iterations using step size or # step height h n = (int)((x - x0)/h) # Iterate for number of iterations y = y0 for i in range(1, n + 1): "Apply Runge Kutta Formulas to find next value of y" k1 = h * dydx(x0, y) k2 = h * dydx(x0 + 0.5 * h, y + 0.5 * k1) k3 = h * dydx(x0 + 0.5 * h, y + 0.5 * k2) k4 = h * dydx(x0 + h, y + k3)
# Update next value of y y = y + (1.0 / 6.0)*(k1 + 2 * k2 + 2 * k3 + k4)
# Update next value of x x0 = x0 + h return y
# Driver methodx0 = 0y = 1x = 2h = 0.2print ('The value of y at x is:', rungeKutta(x0, y, x, h))
The value of y at x is: 1.1036393232374955
