234阶龙格库塔.c

数值分析的几个C实现,比较小.

#include<stdlib.h> 
#include<stdio.h> 
/*n表示几等分，n+1表示他输出的个数*/ 
int RungeKutta(double y0,double a,double b,int n,double *x,double *y,int style,double (*function)(double,double)) 
{ 
double h=(b-a)/n,k1,k2,k3,k4; 
int i; 
 /*x=(double*)malloc((n+1)*sizeof(double));
y=(double*)malloc((n+1)*sizeof(double)); */
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; 
default: 
return 0; 
} 
return 1; 
} 
double function(double x,double y) 
{ 
return y-2*x/y; 
} 
/*例子求y'=y-2*x/y(0<x<1);y0=1;*/
int main() 
{int i;
double x[6],y[6]; 
printf("用二阶龙格-库塔方法\n");
RungeKutta(1,0,1,5,x,y,2,function); 
for(i=0;i<6;i++)
printf("x[%d]=%f,y[%d]=%f\n",i,x[i],i,y[i]); 
printf("用三阶龙格-库塔方法\n");
RungeKutta(1,0,1,5,x,y,3,function); 
for(i=0;i<6;i++) 
printf("x[%d]=%f,y[%d]=%f\n",i,x[i],i,y[i]); 
printf("用四阶龙格-库塔方法\n");
RungeKutta(1,0,1,5,x,y,4,function); 
for(i=0;i<6;i++) 
printf("x[%d]=%f,y[%d]=%f\n",i,x[i],i,y[i]); 
 getch();
}