public class D {
static double[] x=new double[6];
static double[] y=new double[6];
public static double function(double a,double b)
{
return b-2*a/b;
}
/*n表示幾等分，n+1表示他輸出的個數*/
public static void RungeKutta(double y0,double a,double b,int n,int style)
{
double h=(b-a)/n,k1,k2,k3,k4;
int i;
x[0]=a;
y[0]=y0;
switch(style)
{
case 2:
for(i=0;i<n;i++)
{
x[i+1]=x[i]+h;
k1=function(x[i],y[i]);
k2=function(x[i]+h/2,y[i]+h*k1/2);
y[i+1]=y[i]+h*k2;
}
break;
case 3:
for(i=0;i<n;i++)
{
x[i+1]=x[i]+h;
k1=function(x[i],y[i]);
k2=function(x[i]+h/2,y[i]+h*k1/2);
k3=function(x[i]+h,y[i]-h*k1+2*h*k2);
y[i+1]=y[i]+h*(k1+4*k2+k3)/6;
}
break;
case 4:
for(i=0;i<n;i++)
{
x[i+1]=x[i]+h;
k1=function(x[i],y[i]);
k2=function(x[i]+h/2,y[i]+h*k1/2);
k3=function(x[i]+h/2,y[i]+h*k2/2);
k4=function(x[i]+h,y[i]+h*k3);
y[i+1]=y[i]+h*(k1+2*k2+2*k3+k4)/6;
}
break;
}
}
public static void main(String[] args) {
//例子求y'=y-2*x/y(0<x<1);y0=1;
System.out.println("用二階龍格-庫塔方法");
RungeKutta(1,0,1,5,2);
for(int i=0;i<6;i++)
System.out.printf("x[%d]=%f,y[%d]=%f\n",i,x[i],i,y[i]);
System.out.println("用三階龍格-庫塔方法");
RungeKutta(1,0,1,5,3);
for(int i=0;i<6;i++)
System.out.printf("x[%d]=%f,y[%d]=%f\n",i,x[i],i,y[i]);
System.out.println("用四階龍格-庫塔方法");
RungeKutta(1,0,1,5,4);
for(int i=0;i<6;i++){
System.out.printf("x[%d]=%f,y[%d]=%f\n",i,x[i],i,y[i]);
}
}
}
運行結果:
用二階龍格-庫塔方法
x[0]=0.000000,y[0]=1.000000
x[1]=0.200000,y[1]=1.183636
x[2]=0.400000,y[2]=1.342656
x[3]=0.600000,y[3]=1.485014
x[4]=0.800000,y[4]=1.615225
x[5]=1.000000,y[5]=1.736182
用三階龍格-庫塔方法
x[0]=0.000000,y[0]=1.000000
x[1]=0.200000,y[1]=1.183244
x[2]=0.400000,y[2]=1.341729
x[3]=0.600000,y[3]=1.483408
x[4]=0.800000,y[4]=1.612727
x[5]=1.000000,y[5]=1.732472
用四階龍格-庫塔方法
x[0]=0.000000,y[0]=1.000000
x[1]=0.200000,y[1]=1.183229
x[2]=0.400000,y[2]=1.341667
x[3]=0.600000,y[3]=1.483281
x[4]=0.800000,y[4]=1.612514
x[5]=1.000000,y[5]=1.732142