基于反函数定理的偏导数等式证明及相关疑问咨询

嗨，我来帮你梳理这个问题的思路，一步步解决你的困惑：

关于反函数表述的小纠正 #

你提到的 u=f^{-1}(x,y), v=g^{-1}(x,y) 表述不太准确哦。根据反函数定理，我们有一个从 $(u,v)$ 到 $(x,y)$ 的映射：$(u,v) \mapsto (x,y)=(f(u,v),g(u,v))$，它的逆映射是一个整体，应该是存在定义在 $(x,y)$ 邻域内的二元函数 $U(x,y)$ 和 $V(x,y)$，满足 $u=U(x,y)$、$v=V(x,y)$，并且代入原映射后成立：$x=f(U(x,y),V(x,y))$，$y=g(U(x,y),V(x,y))$。简单说，逆映射是把 $(x,y)$ 整体映射回 $(u,v)$，不是分别对$f$和$g$单独取逆~

证明目标等式的核心思路 #

我们可以结合复合函数求导法则和反函数定理的雅可比矩阵性质来推导，步骤很清晰：

第一步：对恒等式求偏导 #

首先有两个恒等式（因为逆映射存在，所以代入后等式恒成立）：

$x = f(u(x,y), v(x,y))$
$y = g(u(x,y), v(x,y))$

1. 对第一个恒等式两边关于$x$求偏导（注意$u$、$v$都是$x,y$的函数，要用链式法则）：
   $$\frac{\partial x}{\partial x} = \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x}$$
   左边是1，而$\frac{\partial f}{\partial u}$就是$\frac{\partial x}{\partial u}$，整理得：
   $$1 = \frac{\partial x}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial x}{\partial v} \cdot \frac{\partial v}{\partial x} \tag{1}$$

2. 对第二个恒等式两边关于$y$求偏导：
   $$\frac{\partial y}{\partial y} = \frac{\partial g}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial g}{\partial v} \cdot \frac{\partial v}{\partial y}$$
   左边是1，$\frac{\partial g}{\partial u}=\frac{\partial y}{\partial u}$，$\frac{\partial g}{\partial v}=\frac{\partial y}{\partial v}$，整理得：
   $$1 = \frac{\partial y}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial y}{\partial v} \cdot \frac{\partial v}{\partial y} \tag{2}$$

3. 对第一个恒等式关于$y$求偏导（左边为0）：
   $$0 = \frac{\partial x}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial x}{\partial v} \cdot \frac{\partial v}{\partial y} \tag{3}$$

4. 对第二个恒等式关于$x$求偏导（左边为0）：
   $$0 = \frac{\partial y}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial y}{\partial v} \cdot \frac{\partial v}{\partial x} \tag{4}$$

第二步：用克莱姆法则求解偏导关系 #

原映射的雅可比矩阵$J$是：
$$J = \begin{pmatrix}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}
\end{pmatrix}$$
根据反函数定理，$J$可逆，所以它的行列式 $\det J = \frac{\partial x}{\partial u}\frac{\partial y}{\partial v} - \frac{\partial x}{\partial v}\frac{\partial y}{\partial u} \neq 0$，这是我们能解方程组的前提。

推导第一个等式 $\frac{\partial x}{\partial u}\cdot \frac{\partial u}{\partial x}= \frac{\partial y}{\partial v}\cdot \frac{\partial v}{\partial y}$

把(1)(4)看成关于$\frac{\partial u}{\partial x}$和$\frac{\partial v}{\partial x}$的线性方程组，用克莱姆法则解$\frac{\partial u}{\partial x}$：
$$\frac{\partial u}{\partial x} = \frac{\begin{vmatrix}1 & \frac{\partial x}{\partial v} \ 0 & \frac{\partial y}{\partial v}\end{vmatrix}}{\det J} = \frac{\frac{\partial y}{\partial v}}{\det J}$$
两边乘$\frac{\partial x}{\partial u}$得：
$$\frac{\partial x}{\partial u} \cdot \frac{\partial u}{\partial x} = \frac{\frac{\partial x}{\partial u}\cdot\frac{\partial y}{\partial v}}{\det J} \tag{5}$$

再把(3)(2)看成关于$\frac{\partial u}{\partial y}$和$\frac{\partial v}{\partial y}$的线性方程组，解$\frac{\partial v}{\partial y}$：
$$\frac{\partial v}{\partial y} = \frac{\begin{vmatrix}\frac{\partial x}{\partial u} & 0 \ \frac{\partial y}{\partial u} & 1\end{vmatrix}}{\det J} = \frac{\frac{\partial x}{\partial u}}{\det J}$$
两边乘$\frac{\partial y}{\partial v}$得：
$$\frac{\partial y}{\partial v} \cdot \frac{\partial v}{\partial y} = \frac{\frac{\partial x}{\partial u}\cdot\frac{\partial y}{\partial v}}{\det J} \tag{6}$$

对比(5)(6)，显然两者相等，第一个等式得证。

推导第二个等式 $\frac{\partial x}{\partial v}\cdot \frac{\partial v}{\partial x}= \frac{\partial y}{\partial u}\cdot \frac{\partial u}{\partial y}$

同样用克莱姆法则，从(1)(4)解$\frac{\partial v}{\partial x}$：
$$\frac{\partial v}{\partial x} = \frac{\begin{vmatrix}\frac{\partial x}{\partial u} & 1 \ \frac{\partial y}{\partial u} & 0\end{vmatrix}}{\det J} = \frac{-\frac{\partial y}{\partial u}}{\det J}$$
两边乘$\frac{\partial x}{\partial v}$得：
$$\frac{\partial x}{\partial v} \cdot \frac{\partial v}{\partial x} = \frac{-\frac{\partial x}{\partial v}\cdot\frac{\partial y}{\partial u}}{\det J} \tag{7}$$

从(3)(2)解$\frac{\partial u}{\partial y}$：
$$\frac{\partial u}{\partial y} = \frac{\begin{vmatrix}0 & \frac{\partial x}{\partial v} \ 1 & \frac{\partial y}{\partial v}\end{vmatrix}}{\det J} = \frac{-\frac{\partial x}{\partial v}}{\det J}$$
两边乘$\frac{\partial y}{\partial u}$得：
$$\frac{\partial y}{\partial u} \cdot \frac{\partial u}{\partial y} = \frac{-\frac{\partial x}{\partial v}\cdot\frac{\partial y}{\partial u}}{\det J} \tag{8}$$

对比(7)(8)，两者相等，第二个等式得证。

对你思路的补充说明 #

你不需要单独计算每一个偏导的具体值，核心是利用恒等式的链式求导得到线性方程组，再结合反函数定理保证雅可比矩阵可逆，从而通过克莱姆法则建立不同偏导乘积之间的等式关系。

备注：内容来源于stack exchange，提问作者Algo