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dxstatic 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-1int  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  3.1415926536void 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