牛顿下山法求解非线性方程（组）（C实现）

1、算法描述

（1）符号说明与基本假设

对于非线性方程组：


（1）
引入向量：



可将（1）式改写为


（2）
通常考虑方程（2）只有唯一解的情形。
（2）牛顿下山算法

引入下山因子λ，产生一下计算格式：



下山因子λ一般依次取1、1/2、1/4、1/8、……
其中



计算步骤为：



2、C语言实现
newton.h头文件：

#ifndef __NEWTON_H__
#define __NEWTON_H__
// 牛顿下山法求解非线性方程（组）

int newton( double** X0
, int n
, double lmada
, double eps_x
, double eps_f
, void   (*f)( double** X, int n)
, void   (*df)(double** X, int n)
);

#endif

newton.c文件

#include "newton.h"
#include <stdlib.h>
#include <string.h>

// 计算矩阵的逆
static int inverse( double** dfX, int n )
{
int i, j,k, p;
double maxV, tmp;
double* A = *dfX;
double* T = (double*)malloc(sizeof(double) * n * n);
double* tArr = (double*)malloc(sizeof(double) * n);
int row_size = sizeof(double) * n;

memset( T, 0, sizeof(double)*n*n);
for ( i = 0; i < n; ++i )
{
T[i*n+i] = 1.0;
}

for ( j = 0; j < n; ++j )
{
p = j;
tmp = A[j*n+j];
maxV = (tmp>=0)?(tmp):(-tmp);
for ( i = j +1; i < n; ++i )
{
tmp = A[i*n+j];
if ( tmp < 0 ) tmp = -tmp;
if ( maxV < tmp )
{
p = i;
maxV = tmp;
}
}

if ( maxV < 1e-20 )
{
return -1;
}

if ( j != p )
{
memcpy(  tArr, A+j*n, row_size);
memcpy( A+j*n, A+p*n, row_size);
memcpy( A+p*n,  tArr, row_size);
memcpy(  tArr, T+j*n, row_size);
memcpy( T+j*n, T+p*n, row_size);
memcpy( T+p*n,  tArr, row_size);
}

tmp = A[j*n+j];
for ( i = j; i < n; ++i ) A[j*n+i] /= tmp;
for ( i = 0; i < n; ++i ) T[j*n+i] /= tmp;
for ( i = 0; i < n; ++i )
{
if ( i != j )
{
tmp = A[i*n+j];
for ( k = j; k < n; ++k )
A[i*n+k] -= tmp * A[j*n+k];
for ( k = 0; k < n; ++k )
T[i*n+k] -= tmp * T[j*n+k];
}
}
}
memcpy( A, T, row_size * n );
free( T );
free( tArr );

return 0;
}

// 计算步长dx
static void calc_dx( double** dx
, double*  df
, double*  dfx
, double   lamda
, int      n
)
{
int i, j;
double* x = *dx;
memset( x, 0, sizeof(double) * n);
for ( i = 0; i < n; ++i )
{
for ( j = 0; j < n; ++j )
{
x[i] += -lamda * df[j] * dfx[i*n+j];
}
}
}

// 计算向量的无穷范数
static double norm_inf( double* A, int n )
{
int i;
double t = A[0];
double ret = t;

if ( t < 0 )
{
ret = -t;
}

for ( i = 1; i < n; ++i )
{
t = A[i];
if ( t < 0 ) t = -t;
if ( ret < t ) ret = t;
}

return ret;
}

// 牛顿下山法求解非线性方程组，求解成功返回0，失败返回-1
int  newton  ( double** X0                     // 迭代起始点
, int n                           // 方程组维数
, double lamda                    // 起始下山因子
, double eps_x                    // 阈值
, double eps_f                    // 阈值
, void   (*f)( double** X, int n) // 带求解非线性方程组函数
, void   (*df)(double** X, int n) // 带求解非线性方程组的导函数（Jacobi矩阵）
)
{
int i, ret = 0;
int row_size = sizeof(double) * n;
double*   X = *X0;
double*  dx = (double*)malloc( row_size );
double*  fX = (double*)malloc( row_size );
double* dfX = (double*)malloc( n * row_size );
double max_f, max_f1;

memcpy( fX, X, row_size );
f ( &fX, n );
for ( ;; )
{
memcpy( dfX, X, row_size);
df( &dfX, n );

ret = inverse( &dfX, n ); // 计算逆

if ( ret < 0 ) // Jacobi矩阵不可逆
{
goto end;
}
calc_dx( &dx, fX, dfX, lamda, n); // 计算步长

max_f = norm_inf( fX, n );
for ( i = 0; i < n; ++i )
{
X[i] += dx[i];
}

memcpy( fX, X, row_size);
f( &fX, n );
max_f1 = norm_inf( fX, n );

if ( max_f1 < max_f )
{
if ( norm_inf( dx, n ) < eps_x )
{
break;
}
else
{
continue;
}
}
else
{
if ( max_f1 < eps_f )
{
break;
}
else
{
lamda /= 2.0;
}
}
}

end:
free( dx );
free( fX );
free( dfX);

return ret;
}

3、测试

考虑用牛顿下山法求解以下非线性方程组：



求解以上方程的主程序：

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <assert.h>

#include "newton.h"
#define  MATH_E   2.7182818285
#define  MATH_PI  3.1415926536

void f( double** X, int n )
{
assert( n == 3 );
double* A = *X;
double x = A[0];
double y = A[1];
double z = A[2];

A[0] = 3*x-cos(y*z)-0.5;
A[1] = x*x-81*(y+0.1)*(y+0.1)+sin(z)+1.06;
A[2] = pow( MATH_E, -x*y)+20*z + 10*MATH_PI/3.0-1;
}

void df( double** X, int n )
{
assert( n == 3 );
double* A = *X;
double x = A[0];
double y = A[1];
double z = A[2];

A[0] = 3.0;
A[1] = z * sin(y*z);
A[2] = y * sin(y*z);
A[3] = 2*x;
A[4] = -162.0*(y+0.1);
A[5] = cos(z);
A[6] = -y * pow( MATH_E, -x*y);
A[7] = -x * pow( MATH_E, -x*y);
A[8] = 20.0;
}

int main()
{
int n = 3;
double* X = (double*)malloc(sizeof(double)*n);
X[0] = 1.0;
X[1] = 1.0;
X[2] = 1.0;

double eps_x = 1e-14;
double eps_f = eps_x;
double lamda = 1.0;

newton( &X, n, lamda, eps_x, eps_f, f, df);
printf("%f\t%f\t%f", X[0], X[1], X[2]);

free( X );
return 1;
}

计算结果：

x = 0.500000
y = -0.000000
z = -0.523599