和達三樹《物理用數學》習題1.4.4，偏導數相乘無消去律

==問題==

$x, y, z$之間具有函數關係，即$f(x, y, z) = 0$，試證明下式成立
$$\left( \frac{\partial x}{\partial y} \right) \cdot \left( \frac{\partial y}{\partial z} \right) \cdot \left( \frac{\partial z}{\partial x} \right)  = -1.$$

==解答==

首先看看最簡單的例子。

設$f: \mathbb{R^3} \rightarrow \mathbb{R}$定義為

$$f(x, y, z) = ax + by + cz + d,$$

其中$a, b, c, d$皆為實數，且$abc \neq 0$。

考慮方程式$f(x, y, z) = 0$，即$ax + by + cz + d = 0$，此時決定了三個函數關係$x = g_1 (y, z), y = g_2 (x, z), z = g_3 (x, y)$，具體來說則是

$$\begin{align*} x = g_1 (y, z) &= \frac{-1}{a} (by + cz + d), \\ y = g_2 (x, z) &= \frac{-1}{b} (ax + cz + d), \\ z = g_3 (x, y) &= \frac{-1}{c} (ax + by + d). \end{align*}$$

於是乎得到

$$\begin{align*} \frac{\partial x}{\partial y} = \frac{\partial g_1}{\partial y} = \frac{-b}{a}, \\ \frac{\partial y}{\partial z} = \frac{\partial g_2}{\partial z} = \frac{-c}{b}, \\ \frac{\partial z}{\partial x} = \frac{\partial g_3}{\partial x} = \frac{-a}{c}. \end{align*}$$

因此

$$\left( \frac{\partial x}{\partial y} \right) \cdot \left( \frac{\partial y}{\partial z} \right) \cdot \left( \frac{\partial z}{\partial x} \right)  = \frac{-b}{a} \cdot \frac{-c}{b} \cdot \frac{-a}{c} = -1.$$

以下均假定函數有足夠多次的可導階數(足夠光滑)，同時再假定$\frac{\partial f}{\partial x} \cdot \frac{\partial f}{\partial y} \cdot \frac{\partial f}{\partial z} \neq 0$。

根據題設，方程式$f(x, y, z) = 0$決定了$x, y, z$彼此之間的函數關係，所以我們可以假設三個函數關係為$x = g_1 (y, z), y = g_2 (x, z), z = g_3 (x, y)$。此時有

$$\begin{align*} f\left( g_1(y, z), y, z \right) = 0, \\ f\left( x, g_2(x, z), z \right) = 0, \\ f\left( x, y, g_3(x, y) \right) = 0. \end{align*}$$

分別對此三式取偏微商$\frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}, \frac{\partial f}{\partial x}$，得

$$\begin{align*} \frac{\partial f}{\partial x} \cdot \frac{\partial g_1}{\partial y} + \frac{\partial f}{\partial y} = 0, \\ \frac{\partial f}{\partial y} \cdot \frac{\partial g_2}{\partial z} + \frac{\partial f}{\partial z} = 0, \\ \frac{\partial f}{\partial x} + \frac{\partial f}{\partial z} \cdot \frac{\partial g_3}{\partial x} = 0. \end{align*}$$

解出

$$\begin{align*} \frac{\partial g_1}{\partial y} = \frac{- \frac{\partial f}{\partial y}}{\frac{\partial f}{\partial x}}, \\ \frac{\partial g_2}{\partial z} = \frac{- \frac{\partial f}{\partial z}}{\frac{\partial f}{\partial y}}, \\ \frac{\partial g_3}{\partial x} = \frac{- \frac{\partial f}{\partial x}}{\frac{\partial f}{\partial z}}. \end{align*}$$

於是

$$\left( \frac{\partial x}{\partial y} \right) \cdot \left( \frac{\partial y}{\partial z} \right) \cdot \left( \frac{\partial z}{\partial x} \right)  =  \frac{\partial g_1}{\partial y} \cdot \frac{\partial g_2}{\partial z} \cdot \frac{\partial g_3}{\partial x} = \frac{- \frac{\partial f}{\partial y}}{\frac{\partial f}{\partial x}} \cdot \frac{- \frac{\partial f}{\partial z}}{\frac{\partial f}{\partial y}} \cdot \frac{- \frac{\partial f}{\partial x}}{\frac{\partial f}{\partial z}} = -1.$$

(解答結束)

==註記==

原書在書末提供的解答過於「物理」，完全無法理解何以任意取值，不具備數學之嚴密性。本題其實只是「隱函數定理」的簡單應用。在單變數函數的微分學中，我們有

$$\frac{dy}{dx} \cdot \frac{dx}{dy} = 1.$$

看似可以用分數運算的消去律約去$dy, dx$。但透過本題可發現多變數函數之微分學與單變數函數的微分學有著本質上的差異，不能將單變數函數的結果隨便地照搬到多變數函數來。