和達三樹《物理用數學》習題1.4.3，極座標下的Laplace算子(Laplacian)

==問題==

$x=\rho \cos \varphi, y=\rho \sin \varphi$，試證
$$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{\partial^2 u}{\partial \rho^2} + \frac{1}{\rho} \frac{\partial u}{\partial \rho} + \frac{1}{\rho^2} \frac{\partial^2 u}{\partial \phi^2}.$$

==解答==

$u = u(x, y) = u(\rho \cos \varphi, \rho \sin \varphi)$。

\begin{align*}
\frac{\partial u}{\partial \rho} &= \frac{\partial u}{\partial x} \frac{\partial (\rho \cos \varphi)}{\partial \rho} + \frac{\partial u}{\partial y} \frac{\partial (\rho \sin \varphi)}{\partial \rho} \\
&= \frac{\partial u}{\partial x} \cos \varphi + \frac{\partial u}{\partial y} \sin \varphi.
\end{align*}
\begin{align*}
\frac{\partial^2 u}{\partial \rho^2} &= \frac{\partial}{\partial \rho} \left( \frac{\partial u}{\partial \rho} \right) \\
&= \frac{\partial}{\partial \rho} \left( \frac{\partial u}{\partial x} \cos \varphi + \frac{\partial u}{\partial y} \sin \varphi \right) \\
&= \frac{\partial}{\partial \rho} \left( \frac{\partial u}{\partial x} \cos \varphi  \right) + \frac{\partial}{\partial \rho} \left( \frac{\partial u}{\partial y} \sin \varphi \right) \\
&= \cos \varphi \frac{\partial}{\partial \rho} \frac{\partial u}{\partial x} + \sin \varphi \frac{\partial}{\partial \rho} \frac{\partial u}{\partial y} \\
&= \cos \varphi \left[ \frac{\partial}{\partial x} \frac{\partial u}{\partial x} \frac{\partial (\rho \cos \varphi)}{\partial \rho} \right] + \sin \varphi \left[ \frac{\partial}{\partial y} \frac{\partial u}{\partial x} \frac{\partial (\rho \sin \varphi)}{\partial \rho} \right] \\
&= \cos \varphi \left[ \frac{\partial^2 u}{\partial x^2} \cos \phi + \frac{\partial^2 u}{\partial y \partial x} \sin \varphi \right] + \sin \varphi \left[ \frac{\partial^2 u}{\partial x \partial y} \cos \varphi + \frac{\partial^2 u}{\partial y^2} \sin \varphi \right] \\
&= \frac{\partial^2 u}{\partial x^2} \cos^2 \varphi + \frac{\partial^2 u}{\partial y \partial x} \cos \varphi \sin \varphi + \frac{\partial^2 u}{\partial x \partial y} \sin \varphi \cos \varphi + \frac{\partial^2 u}{\partial y^2} \sin^2 \varphi \\
&= \frac{\partial^2 u}{\partial x^2} \cos^2 \varphi + \frac{\partial^2 u}{\partial x \partial y} \cos \varphi \sin \varphi + \frac{\partial^2 u}{\partial x \partial y} \cos \varphi \sin \varphi + \frac{\partial^2 u}{\partial y^2} \sin^2 \varphi \\
&= \frac{\partial^2 u}{\partial x^2} \cos^2 \varphi + 2\frac{\partial^2 u}{\partial x \partial y} \cos \varphi \sin \varphi  + \frac{\partial^2 u}{\partial y^2} \sin^2 \varphi.
\end{align*}
\begin{align*}
\frac{\partial u}{\partial \varphi} &= \frac{\partial u}{\partial x} \frac{\partial (\rho \cos \varphi)}{\partial \varphi} + \frac{\partial u}{\partial y} \frac{\partial (\rho \sin \varphi)}{\partial \varphi} \\
&= \frac{\partial u}{\partial x} (- \rho \sin \varphi) + \frac{\partial u}{\partial y} \rho \cos \varphi \\
&= - \rho \frac{\partial u}{\partial x} \sin \varphi + \rho \frac{\partial u}{\partial y} \cos \varphi.
\end{align*}
\begin{align*} \frac{\partial^2 u}{\partial \varphi^2} &= \frac{\partial}{\partial \varphi} \left( \frac{\partial u}{\partial \varphi} \right)  \\ &= \frac{\partial}{\partial \varphi} \left( - \rho \frac{\partial u}{\partial x} \sin \varphi + \rho \frac{\partial u}{\partial y} \cos \varphi \right)  \\ &= \frac{\partial}{\partial \varphi} \left( - \rho \frac{\partial u}{\partial x} \sin \varphi \right) + \frac{\partial}{\partial \varphi} \left( \rho \frac{\partial u}{\partial y} \cos \varphi \right)  \\ &= -\rho \frac{\partial}{\partial \varphi} \left( \frac{\partial u}{\partial x} \sin \varphi \right) + \rho \frac{\partial}{\partial \varphi} \left( \frac{\partial u}{\partial y} \cos \varphi \right)  \\ &= -\rho \left[ \left( \frac{\partial}{\partial \varphi} \frac{\partial u}{\partial x} \right) \sin \varphi + \frac{\partial u}{\partial x} \left( \frac{\partial}{\partial \varphi} \sin \varphi \right) \right] + \rho \left[ \left( \frac{\partial}{\partial \varphi} \frac{\partial u}{\partial y} \right) \cos \varphi + \frac{\partial u}{\partial y} \left( \frac{\partial}{\partial \varphi} \cos \varphi \right) \right] \\ &= -\rho \left\{ \left[ \frac{\partial}{\partial x} \frac{\partial u}{\partial x} \frac{\partial (\rho \cos \varphi)}{\partial \varphi} + \frac{\partial}{\partial y} \frac{\partial u}{\partial x} \frac{\partial (\rho \sin \varphi)}{\partial \varphi} \right] \sin \varphi + \frac{\partial u}{\partial x} \frac{\partial (\sin \varphi)}{\partial \varphi} \right\} \\ & \quad +\rho \left\{ \left[ \frac{\partial}{\partial x} \frac{\partial u}{\partial y} \frac{\partial (\rho \cos \varphi)}{\partial \varphi} + \frac{\partial}{\partial y} \frac{\partial u}{\partial y} \frac{\partial (\rho \sin \varphi)}{\partial \varphi} \right] \cos \varphi + \frac{\partial u}{\partial y} \frac{\partial (\cos \varphi)}{\partial \varphi} \right\} \\ &= -\rho \left\{ \left[ \frac{\partial^2 u}{\partial x^2} (- \rho \sin \varphi) + \frac{\partial^2 u}{\partial y \partial x} \rho \cos \varphi \right] \sin \varphi + \frac{\partial u}{\partial x} \cos \varphi \right\} \\ & \quad + \rho \left\{ \left[ \frac{\partial^2 u}{\partial x \partial y} (- \rho \sin \varphi) + \frac{\partial^2 u}{\partial y^2} \rho \cos \varphi \right] \cos \varphi + \frac{\partial u}{\partial y} (- \sin \varphi) \right\} \\ &= \frac{\partial^2 u}{\partial x^2} \rho^2 \sin^2 \varphi - \frac{\partial^2 u}{\partial y \partial x} \rho^2 \cos \varphi \sin \varphi - \frac{\partial u}{\partial x} \rho \cos \varphi - \frac{\partial^2 u}{\partial x \partial y} \rho^2 \sin \varphi \cos \varphi + \frac{\partial^2 u}{\partial y^2} \rho^2 \cos^2 \varphi - \frac{\partial u}{\partial y} \rho \sin \varphi \\ &= \frac{\partial^2 u}{\partial x^2} \rho^2 \sin^2 \varphi - \frac{\partial^2 u}{\partial x \partial y} \rho^2 \cos \varphi \sin \varphi - \frac{\partial u}{\partial x} \rho \cos \varphi - \frac{\partial^2 u}{\partial x \partial y} \rho^2 \cos \varphi \sin \varphi + \frac{\partial^2 u}{\partial y^2} \rho^2 \cos^2 \varphi - \frac{\partial u}{\partial y} \rho \sin \varphi \\ &= \frac{\partial^2 u}{\partial x^2} \rho^2 \sin^2 \varphi - 2 \frac{\partial^2 u}{\partial x \partial y} \rho^2 \cos \varphi \sin \varphi + \frac{\partial^2 u}{\partial y^2} \rho^2 \cos^2 \varphi - \frac{\partial u}{\partial x} \rho \cos \varphi - \frac{\partial u}{\partial y} \rho \sin \varphi.
\end{align*}
\begin{align*} \frac{\partial^2 u}{\partial \rho^2} + \frac{1}{\rho} \frac{\partial u}{\partial \rho} + \frac{1}{\rho^2} \frac{\partial^2 u}{\partial \phi^2} &= \left( \frac{\partial^2 u}{\partial x^2} \cos^2 \varphi + 2\frac{\partial^2 u}{\partial x \partial y} \cos \varphi \sin \varphi  + \frac{\partial^2 u}{\partial y^2} \sin^2 \varphi \right) \\ & \quad +\frac{1}{\rho} \left( \frac{\partial u}{\partial x} \cos \varphi + \frac{\partial u}{\partial y} \sin \varphi \right) \\ & \qquad +\frac{1}{\rho^2} \left( \frac{\partial^2 u}{\partial x^2} \rho^2 \sin^2 \varphi - 2 \frac{\partial^2 u}{\partial x \partial y} \rho^2 \cos \varphi \sin \varphi + \frac{\partial^2 u}{\partial y^2} \rho^2 \cos^2 \varphi \right. \\ &\qquad \qquad \qquad \left. - \frac{\partial u}{\partial x} \rho \cos \varphi - \frac{\partial u}{\partial y} \rho \sin \varphi \right) \\ &= \frac{\partial^2 u}{\partial x^2} \cos^2 \varphi + 2\frac{\partial^2 u}{\partial x \partial y} \cos \varphi \sin \varphi  + \frac{\partial^2 u}{\partial y^2} \sin^2 \varphi + \frac{\partial u}{\partial x} \frac{1}{\rho} \cos \varphi + \frac{\partial u}{\partial y} \frac{1}{\rho} \sin \varphi + \frac{\partial^2 u}{\partial x^2}  \sin^2 \varphi \\ & \quad - 2 \frac{\partial^2 u}{\partial x \partial y} \cos \varphi \sin \varphi + \frac{\partial^2 u}{\partial y^2} \cos^2 \varphi - \frac{\partial u}{\partial x} \frac{1}{\rho} \cos \varphi - \frac{\partial u}{\partial y} \frac{1}{\rho} \sin \varphi \\ &= \frac{\partial^2 u}{\partial x^2} \cos^2 \varphi + \frac{\partial^2 u}{\partial y^2} \sin^2 \varphi + \frac{\partial^2 u}{\partial x^2}  \sin^2 \varphi + \frac{\partial^2 u}{\partial y^2} \cos^2 \varphi \\
&= \left( \frac{\partial^2 u}{\partial x^2} \cos^2 \varphi + \frac{\partial^2 u}{\partial x^2}  \sin^2 \varphi \right) + \left( \frac{\partial^2 u}{\partial y^2} \sin^2 \varphi + \frac{\partial^2 u}{\partial y^2} \cos^2 \varphi \right)  \\
&= \frac{\partial^2 u}{\partial x^2} \left( \cos^2 \varphi + \sin^2 \varphi \right) + \frac{\partial^2 u}{\partial y^2} \left( \sin^2 \varphi + \cos^2 \varphi \right)  \\
&= \frac{\partial^2 u}{\partial x^2} \cdot 1 + \frac{\partial^2 u}{\partial y^2} \cdot 1 \\ &= \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}.
\end{align*}

(解答結束)

和達三樹《物理用數學》習題1.4.1

==問題==

$f(x)$與$g(x)$為任意函數，試證明
$$u(a, t) = f(x+at) + g(x-at)$$
可看作是
$$\frac{\partial^2 u}{\partial t^2} = a^2 \frac{\partial^2 u}{\partial x^2}.$$

==解答==

\begin{align*}
\frac{\partial}{\partial t} u(x, t) &= \frac{\partial}{\partial t} \left[ f(x+at) + g(x-at) \right] \\
&= \frac{\partial}{\partial t} f(x+at) + \frac{\partial}{\partial t} g(x-at) \\
&= f'(x+at) \cdot a + g'(x-at) \cdot (-a) \\
&= af'(x+at) - ag'(x-at) \\
\frac{\partial^2}{\partial t^2} u(x, t) &= \frac{\partial}{\partial t} \left[ \frac{\partial}{\partial t} u(x, t) \right] \\
&= \frac{\partial}{\partial t} \left[ af'(x+at) - ag'(x-at) \right] \\
&= \frac{\partial}{\partial t} \left[ af'(x+at) \right] - \frac{\partial}{\partial t} \left[ ag'(x-at) \right] \\
&= a \frac{\partial}{\partial t} f'(x+at) - a \frac{\partial}{\partial t} g'(x-at) \\
&= a f''(x+at) \cdot a - a g''(x-at) \cdot (-a) \\
&= a^2 \left[ f''(x+at) + g''(x-at) \right] \\
\frac{\partial}{\partial x} u(x, t) &=  \frac{\partial}{\partial x} \left[ f(x+at) + g(x-at) \right] \\
&= \frac{\partial}{\partial x} f(x+at) + \frac{\partial}{\partial x} g(x-at) \\
&= f'(x+at) \cdot 1 + g'(x-at) \cdot 1 \\
&= f'(x+at) + g'(x-at) \\
\frac{\partial^2}{\partial x^2} u(x, t)  &= \frac{\partial}{\partial x} \left[ \frac{\partial}{\partial x} u(x, t)  \right] \\
&= \frac{\partial}{\partial x} \left[ f'(x+at) + g'(x-at) \right] \\
&= \frac{\partial}{\partial x} f'(x+at) + \frac{\partial}{\partial x} g'(x-at) \\
&= f''(x+at) \cdot 1 + g''(x-at) \cdot 1 \\
&= f''(x+at) + g''(x-at) \\
\frac{\partial^2 u}{\partial t^2} &= a^2 \left[ f''(x+at) + g''(x-at) \right] \\
&= a^2 \cdot \frac{\partial^2}{\partial x^2} u(x, t).
\end{align*}

(解答結束)

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。

複合函數的求導，鏈鎖律，Chain Rule

==定理==(鏈鎖律，Chain Rule)

若$g( \xi ) = \eta , g'( \xi ) = t, f'(\eta) = \tau$，則$f \left( g(x) \right)$在$\xi$可導，且導數為$\tau t$。

==證明==

由於$f$在$\eta$可導，意味著存在正數$p$使得$f$在$(\eta - p, \eta + p)$上有定義，且差商$\frac{f( \eta + k) - f( \eta)}{k}$在$k \rightarrow 0$時有極限$\tau$。

定義函數$\varphi : (-p, p) \rightarrow \mathbb{R}$，
$$\varphi (k) = \left\{ \begin{array}{ll} \frac{f( \eta + k) - f( \eta)}{k}, & k \neq 0 \\ \tau, & k = 0 \end{array} \right.$$
此函數的幾何意義為「割線與切線斜率變化函數」。

注意$\lim \limits_{k \rightarrow 0} \varphi (k) = \lim \limits_{k \rightarrow 0} \frac{f( \eta + k) - f( \eta)}{k} = f'(\eta) = \tau = \varphi (0)$，所以$\varphi$在$0$連續。

由$\varphi$的定義，在$(- p, p)$上，恆有
$$f(\eta + k) - f(\eta) = \varphi (k) \cdot k.$$

要確認$f \left( g(x) \right)$在$\xi$是否可導，必須計算極限式
$$\lim \limits_{h \rightarrow 0} \frac{f \left( g( \xi + h) \right) - f \left( g( \xi ) \right)}{h}.$$
將$g ( \xi ) = \eta$代入，式子變為
$$\lim \limits_{h \rightarrow 0} \frac{f \left( g( \xi + h) \right) - f (\eta)}{h}.$$

但我們目前只知道
$$\lim \limits_{k \rightarrow 0} \frac{f (\eta + k) - f (\eta)}{k} = \tau.$$
將兩個極限式的差商相比較：
$$\frac{f \left( g( \xi + h) \right) - f (\eta)}{h} \text{與} \frac{f (\eta + k) - f (\eta)}{k},$$
注意到其中兩者分子的被減項$f \left( g( \xi + h) \right)$與$f (\eta + k)$形式上有所不同。

$g( \xi + h) = \eta + k$？

$g( \xi + h) = g ( \xi ) + k$？

變量$k$受到$h$影響！

定義$k(h) = g ( \xi + h) - g (\xi )$。

$k(h)$的定義域呢？

由於$g$在$\xi$可導，所以存在正數$q$使得$g$在$(\xi - q, \xi + q)$上有定義。於是$k(h)$的定義域就是$(-q, q)$。

而因$g$在$\xi$可導，所以$g$在$\xi$連續，即有$\lim \limits_{h \rightarrow 0} g( \xi + h ) = g ( \xi )$，於是$\lim \limits_{h \rightarrow 0} k(h) = \lim \limits_{h \rightarrow 0} [g ( \xi + h) - g ( \xi )] = g ( \xi ) - g( \xi ) = 0$。

這意味著當$h$足夠小(夠接近$0$)，那麼$k(h)$也會充分接近$0$。

所以對於正數$p$，存在正數$q'$使得當$h \in (-q', q')$時，有$k(h) \in (-p, p)$。

因此，當$h \in (-q', q')$時，
$$k(h) = g(\xi + h) - g(\xi) \in (-p, p),$$
然後
\begin{align}
f \left( g(\xi + h) \right) - f \left( g(\xi) \right) &= f \left( g(\xi) + k(h) \right) - f\left( g(\xi) \right) \notag \\
&= f \left( \eta + k(h) \right) - f(\eta) \notag \\
&= \varphi \left( k(h) \right) \cdot k(h) \notag
\end{align}
所以
\begin{align}
\lim \limits_{h \rightarrow 0} \frac{f \left( g(\xi + h) \right) - f \left( g(\xi) \right)}{h}
&= \lim \limits_{h \rightarrow 0} \frac{\varphi \left( k(h) \right) \cdot k(h)}{h} \notag \\
&= \lim \limits_{h \rightarrow 0} \varphi \left( k(h) \right) \cdot \frac{k(h)}{h} \notag \\
&= \lim \limits_{h \rightarrow 0} \varphi \left( k(h) \right) \cdot \frac{g ( \xi + h) - g (\xi )}{h} \notag \\
&= \left[ \lim \limits_{h \rightarrow 0} \varphi \left( k(h) \right) \right] \cdot \left[ \lim \limits_{h \rightarrow 0} \frac{g ( \xi + h) - g (\xi )}{h}\right] \notag \\
&= \varphi (0) \cdot t \notag \\
&= \tau \cdot t \notag \\
&= \tau t \notag
\end{align}

(證明終了)