以y'=x+y,0<x<1,y(0)=1爲例,取步長h=0.1,已知精確值爲y=-x-1+2e^x,用來進行精度比較。
#include<stdio.h>
using namespace std;
double cor[10000];
double f(double x,double y)//改寫函數
{
return x+y;
}
double correctf(double x)//精確解函數
{
return -x-1+2*exp(x);
}
void Euler(double h,double l,double r,double *a,double *b,double tol)//歐拉法
{
double sum=0;
for(int i=1; i<=tol; i++)
{
b[i]=b[i-1]+h*f(a[i-1],b[i-1]);
sum+=fabs(b[i]-cor[i])/cor[i];
}
for(int i=1; i<=tol; i++)
printf("當x=%lf時,近似解爲:%lf,準確解爲:%lf\n",a[i],b[i],cor[i]);
printf("精度爲:%lf\n\n",sum/tol);
}
void improvedEuler(double h,double l,double r,double *a,double *b,double tol)//改進的歐拉法
{
double b1,sum=0;
for(int i=1; i<=tol; i++)
{
b1=b[i-1]+h*f(a[i-1],b[i-1]);
b[i]=b[i-1]+h/2*(f(a[i-1],b[i-1])+f(a[i],b1));
}
for(int i=1; i<=tol; i++)
printf("當x=%lf時,近似解爲:%lf,準確解爲:%lf\n",a[i],b[i],cor[i]);
printf("精度爲:%lf\n\n",sum/tol);
}
void RungeKutta(double h,double l,double r,double *a,double *b,double tol)//四階龍格庫塔法
{
double k1,k2,k3,k4,sum=0;
for(int i=1; i<=tol; i++)
{
k1=f(a[i-1],b[i-1]);
k2=f(a[i-1]+h/2,b[i-1]+h/2*k1);
k3=f(a[i-1]+h/2,b[i-1]+h/2*k2);
k4=f(a[i-1]+h,b[i-1]+h*k3);
b[i]=b[i-1]+h/6*(k1+2*k2+2*k3+k4);
}
for(int i=1; i<=tol; i++)
printf("當x=%lf時,近似解爲:%lf,準確解爲:%lf\n",a[i],b[i],cor[i]);
printf("精度爲:%lf\n\n",sum/tol);
}
int main()
{
double h,a[10000],b[10000],l,r;
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(cor,0,sizeof(cor));
printf("請輸入步長:");
scanf("%lf",&h);
printf("請輸入區間下限:");
scanf("%lf",&l);
printf("請輸入區間上限:");
scanf("%lf",&r);
printf("請賦予初始值:");
scanf("%lf",&b[0]);
double tol=(r-l)/h;
for(int i=0; i<=tol; i++)
a[i]=l+i*h;
for(int i=1; i<=tol; i++)
cor[i]=correctf(a[i]);
printf("以下爲歐拉法求解結果:\n");
Euler(h,l,r,a,b,tol);
printf("以下爲改進的歐拉法求解結果:\n");
improvedEuler(h,l,r,a,b,tol);
printf("以下爲四階龍格庫塔法求解結果:\n");
RungeKutta(h,l,r,a,b,tol);
return 0;
}
運行得到: