拉格朗日乘子法的简单数学推导

作者: andyhacker | 来源:发表于2022-08-06 15:17 被阅读0次

拉格朗日乘子法公式

结论

原问题

$min f( \mathbf{x}) \\ st. \mathbf{G(x)} = 0 \tag{1}$

转换问题
$min\mathbf{F(x)}\tag{2}$
其中
$\mathbf{F(x)} = f( \mathbf{x})+\lambda\mathbf{G(x)}\tag{3}$

推导过程

一、隐函数

将自变量 $\mathbf{x}$ 展开成向量形式
$\mathbf{x}=(x_0, x_1, x_2, ..., x_n)$
则等式 $\mathbf{G(x)} = 0$ 存在隐函数使得
$x_0=g(x_1,x_2,x_3, ..., x_n) \tag{4}$
令
$\mathbf{x'}=(x_1,x_2,x_3, ..., x_n) \tag{5}$
隐函数偏导数
对于等式（方程） $\mathbf{G(x)} = 0$ 有式 $(4)$ 的隐函数，对其两边同时进行求导得
$\frac{\partial {\mathbf{G_{x'}}}}{\partial x_1} = \frac{\partial {\mathbf{G}} }{\partial x_1}+ \frac{\partial {\mathbf{G}} }{\partial x_0} \frac{\partial {{g}} }{\partial x_1} = 0 \\ \frac{\partial {\mathbf{G_{x'}}}}{\partial x_2} = \frac{\partial {\mathbf{G}} }{\partial x_2}+ \frac{\partial {\mathbf{G}} }{\partial x_0} \frac{\partial {{g}} }{\partial x_2} = 0 \\ \ \\ ... \\ \ \\ \frac{\partial {\mathbf{G_{x'}}}}{\partial x_n} = \frac{\partial {\mathbf{G}} }{\partial x_n}+ \frac{\partial {\mathbf{G}} }{\partial x_0} \frac{\partial {{g}} }{\partial x_n} = 0 \tag{6}$

二、原问题的转换

原问题 $(1)$ 结合等式 $(4)$ 可以等价为
$\begin{align} min\mathbf{F(x)} &= min f(x_0,x_1, x_2,...,x_n)\\ &= minf(g(x_1,x_2,x_3,...,x_n),x_1,x_2,x_3,...x_n)\\ \tag{7}\end{align}$
对式 $(7)$ 求解，即为
$\frac{\partial {\mathbf{F_{x'}}}}{\partial x_1} = \frac{\partial {\mathbf{F}} }{\partial x_1}+ \frac{\partial {\mathbf{F}} }{\partial x_0} \frac{\partial {{g}} }{\partial x_1} = 0 \\ \frac{\partial {\mathbf{F_{x'}}}}{\partial x_2} = \frac{\partial {\mathbf{F}} }{\partial x_2}+ \frac{\partial {\mathbf{F}} }{\partial x_0} \frac{\partial {{g}} }{\partial x_2} = 0 \\ \ \\ ... \\ \ \\ \frac{\partial {\mathbf{F_{x'}}}}{\partial x_n} = \frac{\partial {\mathbf{F}} }{\partial x_n}+ \frac{\partial {\mathbf{F}} }{\partial x_0} \frac{\partial {{g}} }{\partial x_n} = 0 \tag{8}$
观察 $(6)$ 式，等式中含有共同项 $\frac{\partial {\mathbf{G}} }{\partial x_0}$ ，式子两侧同除以共同项，可以变换为
$\frac{\partial {{g}} }{\partial x_1} = -\frac{\frac{\partial {\mathbf{G}} }{\partial x_1}}{\frac{\partial {\mathbf{G}} }{\partial x_0}} \\ \frac{\partial {{g}} }{\partial x_2} = -\frac{\frac{\partial {\mathbf{G}} }{\partial x_2}}{\frac{\partial {\mathbf{G}} }{\partial x_0}} \\ \ \\ ... \\ \ \\ \frac{\partial {{g}} }{\partial x_n} = -\frac{\frac{\partial {\mathbf{G}} }{\partial x_n}}{\frac{\partial {\mathbf{G}} }{\partial x_0}} \tag{9}$
将 $(9)$ 式依次带入 $(8)$ 式，得
$\frac{\partial {\mathbf{F_{x'}}}}{\partial x_1} = \frac{\partial {\mathbf{F}} }{\partial x_1}+ \frac{\partial {\mathbf{F}} }{\partial x_0} (-\frac{\frac{\partial {\mathbf{G}} }{\partial x_1}}{\frac{\partial {\mathbf{G}} }{\partial x_0}})= \frac{\partial {\mathbf{F}} }{\partial x_1}+ (-\frac{\frac{\partial {\mathbf{F}} }{\partial x_0}}{{\frac{\partial {\mathbf{G}} }{\partial x_0}}})\frac{\partial {\mathbf{G}} }{\partial x_1}= 0 \\ \frac{\partial {\mathbf{F_{x'}}}}{\partial x_2} = \frac{\partial {\mathbf{F}} }{\partial x_2}+ \frac{\partial {\mathbf{F}} }{\partial x_0} (-\frac{\frac{\partial {\mathbf{G}} }{\partial x_2}}{\frac{\partial {\mathbf{G}} }{\partial x_0}})= \frac{\partial {\mathbf{F}} }{\partial x_2}+ (-\frac{\frac{\partial {\mathbf{F}} }{\partial x_0}}{{\frac{\partial {\mathbf{G}} }{\partial x_0}}})\frac{\partial {\mathbf{G}} }{\partial x_2}= 0 \\ \ \\ ... \\ \ \\ \frac{\partial {\mathbf{F_{x'}}}}{\partial x_n} = \frac{\partial {\mathbf{F}} }{\partial x_n}+ \frac{\partial {\mathbf{F}} }{\partial x_0} (-\frac{\frac{\partial {\mathbf{G}} }{\partial x_n}}{\frac{\partial {\mathbf{G}} }{\partial x_0}})= \frac{\partial {\mathbf{F}} }{\partial x_n}+ (-\frac{\frac{\partial {\mathbf{F}} }{\partial x_0}}{{\frac{\partial {\mathbf{G}} }{\partial x_0}}})\frac{\partial {\mathbf{G}} }{\partial x_n}= 0 \tag{10}$
令 $\lambda=-\frac{\frac{\partial {\mathbf{F}} }{\partial x_0}}{{\frac{\partial {\mathbf{G}} }{\partial x_0}}}\tag{11}$
代入 $(10)$ 得
$\frac{\partial {\mathbf{F}} }{\partial x_1}+ \lambda\frac{\partial {\mathbf{G}} }{\partial x_1}= 0 \\ \frac{\partial {\mathbf{F}} }{\partial x_2}+ \lambda\frac{\partial {\mathbf{G}} }{\partial x_2}= 0 \\ \ \\ ... \\ \ \\ \frac{\partial {\mathbf{F}} }{\partial x_n}+ \lambda\frac{\partial {\mathbf{G}} }{\partial x_n}= 0 \tag{12}$
同时，式 $(11)$ 可变换为
$\frac{\partial {\mathbf{F}} }{\partial x_0}+ \lambda\frac{\partial {\mathbf{G}} }{\partial x_0}= 0 \tag{13}$
结合式 $(12)(13)$ ，即可等价于
$\nabla {\mathbf{F}} +\lambda \nabla {\mathbf{G}}=0 \tag{14}$
意其即为式 $(3)$ 的最优解