龙格库塔法

经典四阶龙格库塔法

龙格-库塔(Runge-Kutta)方法是一种在工程上应用广泛的高精度单步算法,经常被称为“RK4”或者就是“龙格库塔法”。

令初值问题表述如下。

对于该问题的RK4由如下方程给出：

其中：

RK4法是四阶方法，也就是说每步的误差是h⁵阶，而总积累误差为h⁴阶。

 

RK4计算程序如下：

//[email protected]
#include <iostream>
using namespace std;

#include <vector>
#include <math.h>
using namespace std;
class RK  
{
public:
    class DiffFunc
    {
    public:
        double operator()(double x,double y)
        {
            // y'=cos(x)            return cos(x);
            // y'=x*y-1
            //    return x*y-1;        }
    } m_df;
    
    RK(double xend,double x0,double y0,double h=1e-3)
    {
        m_max_x = xend;
        m_x0 = x0;
        m_y0 = y0;
        m_h = h;
        m_half_h=h/2.0;
    }
    
    void Solve()
    {
        double yn=m_y0,xstart=m_x0;
        while( xstart<m_max_x)
        {
            double y=yn+K(xstart,yn)*m_h;//y(n+1)=y(n)+h*y'            m_x.push_back(xstart);
            m_y.push_back(yn);
            cout<<xstart<<""<<yn<<endl;
            yn=y;
            xstart+=m_h;
        }    
    }
    //求解后的点    std::vector<double> m_x,m_y;
    //步长h    double m_h;
    //初始点x0,y0    double m_x0,m_y0;
    //x范围    double m_max_x;
    private:
        double m_half_h;
        double m_ptx,m_pty;
        double  K1(double x,double y)
        {
            double v=m_df(x,y);
            m_ptx=x;
            m_pty=y;
            return v;
        }
        double  K2(double _k1)
        {
            double v=m_df(m_ptx+m_half_h,m_pty+m_half_h*_k1);
            return v;
        }
        double  K3(double _k2)
        {
            
            double v=m_df(m_ptx+m_half_h,m_pty+m_half_h*_k2);
            return v;
        }
        double  K4(double _k3)
        {
            double v=m_df(m_ptx+m_h,m_pty+m_h*_k3);
            return v;
        }
        double  K(double x,double y)
        {
            double _k1=K1(x,y),_k2=K2(_k1),_k3=K3(_k2),_k4=K4(_k3);
            return  (_k1+2.0*_k2+2.0*_k3+_k4)/6.0;
        }
        
};

main()
{
    RK s1(1,0,0);
    s1.Solve();
}

 

计算实例：

y'=cos(x)   y(0)=0,   ->   solution  y=sin(x)

 

y'=x*cos(x)   y(0)=0

 

y'=1.0/sqrt(x*x+y*y),y(0)=0.1

http://www.cnblogs.com/JustHaveFun-SAN/archive/2012/02/09/2330655.html