c++牛顿迭代法求解二元非线性方程
时间: 2023-09-16 16:10:23 浏览: 96
牛顿迭代法可以用于求解二元非线性方程组。假设我们要求解如下方程组:
f1(x1, x2) = 0
f2(x1, x2) = 0
其中,x1和x2是未知量,f1和f2是已知的非线性函数。
牛顿迭代法的思路是,从一个初始点(x1^0, x2^0)开始,通过不断迭代,使得每一步迭代后的点(x1^k, x2^k)都更加接近方程组的解。具体迭代公式如下:
x1^(k+1) = x1^k - [J^-1(x1^k, x2^k) * F(x1^k, x2^k)]1
x2^(k+1) = x2^k - [J^-1(x1^k, x2^k) * F(x1^k, x2^k)]2
其中,J是雅可比矩阵,F是非线性方程组的函数向量,[J^-1(x1^k, x2^k) * F(x1^k, x2^k)]表示矩阵J的逆与向量F的乘积。
具体的C++代码如下:
```c++
#include <iostream>
#include<cmath>
using namespace std;
double f1(double x1, double x2)
{
return pow(x1, 2) + pow(x2, 2) - 4;
}
double f2(double x1, double x2)
{
return pow(x1, 2) - pow(x2, 2) - 1;
}
double df1_dx1(double x1, double x2)
{
return 2 * x1;
}
double df1_dx2(double x1, double x2)
{
return 2 * x2;
}
double df2_dx1(double x1, double x2)
{
return 2 * x1;
}
double df2_dx2(double x1, double x2)
{
return -2 * x2;
}
void newton(double &x1, double &x2)
{
double eps = 1e-8;
int maxIter = 1000;
int iter = 0;
while (iter < maxIter)
{
double J[2][2];
J[0][0] = df1_dx1(x1, x2);
J[0][1] = df1_dx2(x1, x2);
J[1][0] = df2_dx1(x1, x2);
J[1][1] = df2_dx2(x1, x2);
double F[2];
F[0] = f1(x1, x2);
F[1] = f2(x1, x2);
double detJ = J[0][0] * J[1][1] - J[0][1] * J[1][0];
double invJ[2][2];
invJ[0][0] = J[1][1] / detJ;
invJ[0][1] = -J[0][1] / detJ;
invJ[1][0] = -J[1][0] / detJ;
invJ[1][1] = J[0][0] / detJ;
double dx1 = invJ[0][0] * F[0] + invJ[0][1] * F[1];
double dx2 = invJ[1][0] * F[0] + invJ[1][1] * F[1];
x1 -= dx1;
x2 -= dx2;
double err = sqrt(dx1 * dx1 + dx2 * dx2);
if (err < eps)
{
break;
}
iter++;
}
cout << "x1 = " << x1 << endl;
cout << "x2 = " << x2 << endl;
}
int main()
{
double x1 = 1.0;
double x2 = 1.0;
newton(x1, x2);
return 0;
}
```
在上述代码中,我们定义了两个非线性函数f1和f2,并且分别计算了它们对x1和x2的偏导数。在newton函数中,我们实现了牛顿迭代法的迭代公式,并且将计算结果输出。最后,在main函数中,我们初始化了x1和x2,并且调用newton函数求解方程组的解。