和達三樹《物理用數學》習題2.1.1，向量加法的結合律

==問題==

證明結合律：$\bf A+(B+C) = (A+B)+C$。

==解答==

向量加法有兩種方法：平行四邊形法、三角形首尾相連法。證明關鍵在於使用三角形首尾相連法。如果用平行四邊形法會很難做。

$\bf A+(B+C)$

$\bf (A+B)+C$

由以上兩圖可知$\bf A+(B+C) = (A+B)+C$。

(解答結束)

2階Hermite矩陣的形式

==共軛矩陣的定義==

設${\rm A} = \left[ \begin{array}{cccc} z_{11} & z_{12} & \cdots & z_{1n} \\ z_{21} & z_{22} & \cdots & z_{2n} \\ \cdots & \cdots & \cdots & \cdots \\ z_{m1} & z_{m2} & \cdots & z_{mn} \end{array} \right] \in \mathbb{C}^{m \times n}$。定義${\rm A}$的共軛矩陣為${\rm A}^*$，
$${\rm A}^* = \left[ \begin{array}{cccc} \bar{z_{11}} & \bar{z_{12}} & \cdots & \bar{z_{1n}} \\ \bar{z_{21}} & \bar{z_{22}} & \cdots & \bar{z_{2n}} \\ \cdots & \cdots & \cdots & \cdots \\ \bar{z_{m1}} & \bar{z_{m2}} & \cdots & \bar{z_{mn}} \end{array} \right].$$

例子：${\rm A} = \left[ \begin{array}{cc} 1 & 1-2i \\ 1+2i & -5 \end{array} \right]$，則${\rm A}^* = \left[ \begin{array}{cc} \bar{1} & \overline{1-2i} \\ \overline{1+2i} & \overline{-5} \end{array} \right] = \left[ \begin{array}{cc} 1 & 1+2i \\ 1-2i & -5 \end{array} \right]$。

==Hermite共軛矩陣的定義==

設${\rm A}$為n階方陣，${\rm A} = \left[ \begin{array}{cccc} z_{11} & z_{12} & \cdots & z_{1n} \\ z_{21} & z_{22} & \cdots & z_{2n} \\ \cdots & \cdots & \cdots & \cdots \\ z_{n1} & z_{n2} & \cdots & z_{nn} \end{array} \right] \in \mathbb{C}^{n \times n}$。定義${\rm A}$的Hermite共軛為${\rm A}^{\dagger}$，

$${\rm A}^{\dagger} = \left( {\rm A}^* \right)^{\rm T} = \left( \left[ \begin{array}{cccc} \bar{z_{11}} & \bar{z_{12}} & \cdots & \bar{z_{1n}} \\ \bar{z_{21}} & \bar{z_{22}} & \cdots & \bar{z_{2n}} \\ \cdots & \cdots & \cdots & \cdots \\ \bar{z_{n1}} & \bar{z_{n2}} & \cdots & \bar{z_{nn}} \end{array} \right] \right)^{\rm T} = \left[ \begin{array}{cccc} \bar{z_{11}} & \bar{z_{21}} & \cdots & \bar{z_{n1}} \\ \bar{z_{12}} & \bar{z_{22}} & \cdots & \bar{z_{n2}} \\ \cdots & \cdots & \cdots & \cdots \\ \bar{z_{1n}} & \bar{z_{2n}} & \cdots & \bar{z_{nn}} \end{array} \right].$$

例子：${\rm A} = \left[ \begin{array}{cc} 1 & 1-2i \\ 1+2i & -5 \end{array} \right]$，則${\rm A}^{\dagger} = \left[ \begin{array}{cc} 1 & 1+2i \\ 1-2i & -5 \end{array} \right]^{\rm T} = \left[ \begin{array}{cc} 1 & 1-2i \\ 1+2i & -5 \end{array} \right]$。

==Hermite矩陣(Hermitian matrix)的定義==

設${\rm A}$為n階複方陣。若${\rm A}$滿足

$${\rm A}^{\dagger} = {\rm A},$$

則稱${\rm A}$為Hermite矩陣(Hermitian matrix)。

例子：${\rm A} = \left[ \begin{array}{cc} 1 & 1-2i \\ 1+2i & -5 \end{array} \right]$是Hermite矩陣。

==2階Hermite矩陣的形式==

設${\rm A} = \left[ \begin{array}{cc} z_{11} & z_{12} \\ z_{21} & z_{22} \end{array} \right]$，其中$z_{11}, z_{12}, z_{21}, z_{22} \in \mathbb{C}$。

若${\rm A}$為Hermite矩陣，則意味著${\rm A}$滿足
$${\rm A}^{\dagger} = {\rm A}.$$
其中${\rm A}^{\dagger} = \left( \left[ \begin{array}{cc} z_{11} & z_{12} \\ z_{21} & z_{22} \end{array} \right]^* \right)^{\rm T} = \left( \left[ \begin{array}{cc} \bar{z_{11}} & \bar{z_{12}} \\ \bar{z_{21}} & \bar{z_{22}} \end{array} \right] \right)^{\rm T} = \left[ \begin{array}{cc} \bar{z_{11}} & \bar{z_{21}} \\ \bar{z_{12}} & \bar{z_{22}} \end{array} \right]$，於是有
$$\left[ \begin{array}{cc} \bar{z_{11}} & \bar{z_{21}} \\ \bar{z_{12}} & \bar{z_{22}} \end{array} \right] = \left[ \begin{array}{cc} z_{11} & z_{12} \\ z_{21} & z_{22} \end{array} \right].$$
從而得到
$$\bar{z_{11}} = z_{11}, \bar{z_{21}} = z_{12}, \bar{z_{12}} = z_{21}, \bar{z_{22}} = z_{22}.$$
所以$z_{11}, z_{22}$皆為實數，而$z_{12}$與$z_{21}$互為共軛複數(因此實部相同)。

總結上述，我們可以得到：${\rm A} = \left[ \begin{array}{cc} z_{11} & z_{12} \\ z_{21} & z_{22} \end{array} \right]$為Hermite矩陣，若且唯若${\rm A}$形如
$${\rm A} = \left[ \begin{array}{cc} a & p+qi \\ p-qi & b \end{array} \right], a, b, p, q \in \mathbb{R}.$$
(主對角線元素為實數，副對角線元素為共軛複數對)

(證明終了)

和達三樹《物理用數學》習題2.3.3，Pauli矩陣

==問題==

令$\sigma_1 = \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right], \sigma_2 = \left[ \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right], \sigma_3 = \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right]$，求證：
(i) $\sigma_1, \sigma_2, \sigma_3$為Hermite矩陣。
(ii) $\sigma_1 \sigma_1 = I, \sigma_2 \sigma_2 = I, \sigma_3 \sigma_3 = I$。
(iii) $\sigma_1 \sigma_2 = i \sigma_3, \sigma_2 \sigma_3 = i \sigma_1, \sigma_3 \sigma_1 = i \sigma_2$。
$\sigma_1, \sigma_2, \sigma_3$稱為Pauli矩陣。

==解答==

(i) 首先確認2階Hermite矩陣的形式。

設${\rm A} = \left[ \begin{array}{cc} z_{11} & z_{12} \\ z_{21} & z_{22} \end{array} \right]$，其中$z_{11}, z_{12}, z_{21}, z_{22} \in \mathbb{C}$。

若${\rm A}$為Hermite矩陣，則意味著${\rm A}$滿足
$${\rm A}^{\dagger} = {\rm A}.$$
其中${\rm A}^{\dagger} = \left( \left[ \begin{array}{cc} z_{11} & z_{12} \\ z_{21} & z_{22} \end{array} \right]^* \right)^{\rm T} = \left( \left[ \begin{array}{cc} \bar{z_{11}} & \bar{z_{12}} \\ \bar{z_{21}} & \bar{z_{22}} \end{array} \right] \right)^{\rm T} = \left[ \begin{array}{cc} \bar{z_{11}} & \bar{z_{21}} \\ \bar{z_{12}} & \bar{z_{22}} \end{array} \right]$，於是有
$$\left[ \begin{array}{cc} \bar{z_{11}} & \bar{z_{21}} \\ \bar{z_{12}} & \bar{z_{22}} \end{array} \right] = \left[ \begin{array}{cc} z_{11} & z_{12} \\ z_{21} & z_{22} \end{array} \right].$$
從而得到
$$\bar{z_{11}} = z_{11}, \bar{z_{21}} = z_{12}, \bar{z_{12}} = z_{21}, \bar{z_{22}} = z_{22}.$$
所以$z_{11}, z_{22}$皆為實數，而$z_{12}$與$z_{21}$互為共軛複數(因此實部相同)。

總結上述，我們可以得到：${\rm A} = \left[ \begin{array}{cc} z_{11} & z_{12} \\ z_{21} & z_{22} \end{array} \right]$為Hermite矩陣，若且唯若${\rm A}$形如
$${\rm A} = \left[ \begin{array}{cc} a & p+qi \\ p-qi & b \end{array} \right], a, b, p, q \in \mathbb{R}.$$
(主對角線元素為實數，副對角線元素為共軛複數對)

現在回頭來判定Pauli矩陣是否為Hermite矩陣。

顯然$\sigma_1, \sigma_2, \sigma_3$都是Hermite矩陣。

(ii) 直接計算。

$\sigma_1 \sigma_1 = \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right] \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right] = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right] = I.$

注意對矩陣$\left[ \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right]$左乘$\sigma_1 = \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right]$的效果相當於「交換1st row與2nd row」，因此可以利用此性質立刻寫出$\sigma_1 \sigma_1$的乘積。

$\sigma_2 \sigma_2 = \left[ \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right] \left[ \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right] = i \left[ \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right] \left[ \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right] = i \left[ \begin{array}{cc} -i & 0 \\ 0 & -i \end{array} \right] = \left[ \begin{array}{cc} -i^2 & 0 \\ 0 & -i^2 \end{array} \right] = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right] = I.$

這裡一樣使用了基本矩陣(Elementary matrix)的性質。對矩陣$\left[ \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right]$左乘$\left[ \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right]$的效果相當於「先將2nd row的元素添上負號後，再與1st row交換」。

$\sigma_3 \sigma_3 = \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right] \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right] = \left[ \begin{array}{cc} 1 & 0 \\ -0 & -(-1) \end{array} \right] = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right] = I.$

這裡一樣使用了基本矩陣(Elementary matrix)的性質。對矩陣$\left[ \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right]$左乘$\left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right]$的效果相當於「將2nd row的元素添上負號」。

(iii) 與(ii)相同，直接計算，但一樣使用基本矩陣的性質來化簡計算量。

$\sigma_1 \sigma_2 = \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right] \left[ \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right] = \left[ \begin{array}{cc} i & 0 \\ 0 & -i \end{array} \right] = i \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right] = i \sigma_3$。

$\sigma_2 \sigma_3 = \left[ \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right] \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right] = i \left[ \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right] \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right] = i \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right] = i \sigma_1$。

$\sigma_3 \sigma_1 = \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right] \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right] = \left[ \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right] = \left[ \begin{array}{cc} 0 & -i^2 \\ i^2 & 0 \end{array} \right] = i \left[ \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right] = i \sigma_2$。

(解答結束)

==附註==

維基百科關於Pauli矩陣的介紹：https://zh.wikipedia.org/wiki/%E6%B3%A1%E5%88%A9%E7%9F%A9%E9%99%A3

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1.4.2，Maxwell熱力學關係式

==問題==

令溫度為T，壓力為p，體積為V，熵為S，根據熱力學公式
\begin{align*}
dU&=TdS-pdV &\quad& (U\text{為內能})\\
dH&=TdS+Vdp &\quad& (H\text{為焓})\\
dF&=-SdT-pdV &\quad& (F\text{為亥姆霍茲(Helmholtz)自由能})\\
dG&=-SdT+Vdp &\quad& (G\text{為吉布斯(Gibbs)自由能})
\end{align*}
證明Maxwell關係式
\begin{align*}
\left( \frac{\partial p}{\partial S} \right)_V = -\left( \frac{\partial T}{\partial V} \right)_S, & \left( \frac{\partial V}{\partial S} \right)_p = \left( \frac{\partial T}{\partial p} \right)_S \\
\left( \frac{\partial S}{\partial V} \right)_T = \left( \frac{\partial p}{\partial T} \right)_V, & \left( \frac{\partial S}{\partial p} \right)_T = -\left( \frac{\partial V}{\partial T} \right)_p
\end{align*}

==解答==

由全微分(total differential)
$$df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy.$$
對比
\begin{align*}
dU&=TdS-pdV,\\
dH&=TdS+Vdp,\\
dF&=-SdT-pdV,\\
dG&=-SdT+Vdp.
\end{align*}

可得

\begin{align*} &\frac{\partial U}{\partial S} = T, \frac{\partial U}{\partial V} = -p.\\ &\frac{\partial H}{\partial S} = T, \frac{\partial H}{\partial p} = V.\\ &\frac{\partial F}{\partial T} = -S, \frac{\partial F}{\partial V} = -p.\\ &\frac{\partial G}{\partial T} = -S, \frac{\partial G}{\partial p} = V. \end{align*}

於是
\begin{align*} &\frac{\partial p}{\partial S} = \frac{\partial}{\partial S}p = \frac{\partial}{\partial S} \left( -\frac{\partial U}{\partial V} \right) = - \frac{\partial^2 U}{\partial S \partial V} = - \frac{\partial^2 U}{\partial V \partial S} = - \frac{\partial}{\partial V} \left( \frac{\partial U}{\partial S} \right) = - \frac{\partial}{\partial V} T = - \frac{\partial T}{\partial V}, \\ &\frac{\partial V}{\partial S} = \frac{\partial}{\partial S} V = \frac{\partial}{\partial S} \left( \frac{\partial H}{\partial p} \right) = \frac{\partial^2 H}{\partial S \partial p} = \frac{\partial^2 H}{\partial p \partial S} = \frac{\partial}{\partial p} \left( \frac{\partial H}{\partial S} \right) = \frac{\partial}{\partial p} T = \frac{\partial T}{\partial p}, \\ &\frac{\partial S}{\partial V} = \frac{\partial}{\partial V} S = \frac{\partial}{\partial V} \left( -\frac{\partial F}{\partial T} \right) = - \frac{\partial^2 F}{\partial V \partial T} = - \frac{\partial^2 F}{\partial T \partial V} = - \frac{\partial}{\partial T} \left( \frac{\partial F}{\partial V} \right) = - \frac{\partial}{\partial T} (-p) = \frac{\partial p}{\partial T}, \\ &\frac{\partial S}{\partial p} = \frac{\partial}{\partial p} S = \frac{\partial}{\partial p} \left( -\frac{\partial G}{\partial T} \right) = - \frac{\partial^2 G}{\partial p \partial T} = - \frac{\partial^2 G}{\partial T \partial p} = - \frac{\partial}{\partial T} \left( \frac{\partial G}{\partial p} \right) = - \frac{\partial}{\partial T} V = - \frac{\partial V}{\partial T}. \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。