1 球坐标变换

极坐标变换：$\phi:(r,\theta)\mapsto(x,y)$（$\mathbb R^+,\theta\in[0,2\pi)$），求 $J_\phi,J_{\phi^{-1}}$

$$
J_\phi=\begin{pmatrix}
\frac{\partial x}{\partial r}&\frac{\partial x}{\partial\theta}\
\frac{\partial y}{\partial r}&\frac{\partial y}{\partial\theta}
\end{pmatrix}=\begin{pmatrix}
\cos\theta & -r\sin\theta\
\sin\theta & r\cos\theta
\end{pmatrix}
$$

$$
\phi^{-1}=\begin{cases}
r=\sqrt{x^2+y^2}\
\theta=\arccos\frac x{\sqrt{x^2+y^2}}
\end{cases}(\theta\in(0,\pi),y>0)
$$

计算其逆变换的雅各布矩阵时，须要考虑怎么表达这个逆变换。

$$
\frac{\partial r}{\partial x}=\frac x{\sqrt{x^2+y^2}},\ \frac{\partial r}{\partial y}=\frac y{\sqrt{x^2+y^2}}
$$

$$
\begin{aligned}
\frac{\partial\theta}{\partial x}&=-\frac1{\sqrt{1-\frac{x^2}{x^2+y^2}}}\frac{\sqrt{x^2+y^2}-x\frac{x}{\sqrt{x^2+y^2}}}{x^2+y^2}=-\frac y{x^2+y^2}\
\frac{\partial\theta}{\partial y}&=-\frac x{x^2+y^2}
\end{aligned}
$$

从而

$$
J_{\phi^{-1}}=\begin{pmatrix}
\frac x{\sqrt{x^2+y^2}} & \frac y{\sqrt{x^2+y^2}}\
-\frac y{x^2+y^2} & -\frac x{x^2+y^2}
\end{pmatrix}=\begin{pmatrix}
\cos\theta & \sin\theta\
-\sin\theta/r & \cos\theta/r
\end{pmatrix}=(J_\phi\circ\phi^{-1})^{-1}
$$

$\phi:(r,\theta,\varphi)\mapsto(x,y,z)$

$\mathbb R^3$ 上球极坐标变换

虽然矩阵直接看是互逆的，但其自变量是不一样的，不能直接 $J_\phi^{-1}=J_{\phi^{-1}}$。

$$
\begin{cases}
x=r\sin\theta\cos\varphi\
y=r\sin\theta\sin\varphi\
z=r\cos\theta
\end{cases}
$$

$$
J_\phi=\begin{pmatrix}
\sin\theta\cos\varphi & r\cos\varphi\cos\theta & -r\sin\theta\sin\varphi\
\sin\theta\sin\varphi & r\sin\varphi\cos\theta & r\sin\theta\cos\varphi\
\cos\theta & -r\sin\theta & 0
\end{pmatrix}
$$

$$
\begin{cases}
r=\sqrt{x^2+y^2+z^2}\
\theta=\arccos(z/r)\
\varphi=\arccos\left(x/\sqrt{r^2-z^2}\right)
\end{cases}
$$

$$
\begin{aligned}
J_{\phi^{-1}}&=\begin{pmatrix}
\frac x{\sqrt{x^2+y^2+z^2}} & \frac y{\sqrt{x^2+y^2+z^2}} & \frac z{\sqrt{x^2+y^2+z^2}}\
\frac{xz}{(x^2+y^2+z^2)\sqrt{x^2+y^2}} & \frac{yz}{(x^2+y^2+z^2)\sqrt{x^2+y^2}} & -\frac{\sqrt{x^2+y^2}}{x^2+y^2+z^2}\
-\frac y{x^2+y^2} & -\frac x{x^2+y^2} & 0
\end{pmatrix}\
&=\begin{pmatrix}
\sin\theta\cos\varphi & \sin\theta\sin\varphi & \cos\theta\
\cos\theta\cos\varphi/r & \cos\theta\sin\varphi/r & \sin\theta/r\
-\sin\varphi/r\sin\theta & \cos\varphi/r\sin\theta & 0
\end{pmatrix}
\end{aligned}
$$

$\mathbb R^n$ 上的球坐标变换 $(r,\varphi_1,\cdots,\varphi_{n-1})\mapsto(x_1,\cdots,x_n)$

$$
g:\mathbb R^n\to\mathbb R^m,\ f:\mathbb R^m\to\mathbb R^l
$$

2 复合函数求导

$$
\begin{cases}
x_1&=r\cos\varphi_1\
x_2&=r\sin\varphi_1\cos\varphi_2\
x_3&=r\sin\varphi_1\sin\varphi_2\cos\varphi_3\
\vdots\
x_{n-1}&=r\sin\varphi_1\cdots\sin\varphi_{n-2}\cos\varphi_{n-1}\
x_n&=r\sin\varphi_1\cdots\sin\varphi_{n-2}\sin\varphi_{n-1}
\end{cases}
$$

$$
J_{f\circ g}(x_0)=J_{f}(g(x_0))\cdot J_g(x_0)
$$

例子

$$
\mathrm{id}=\phi\circ\phi^{-1},\ \ x_0=J_{\mathrm{id}}(x_0)=J_{\phi}(\phi^{-1}(x_0))\cdot J_{\phi^{-1}}(x_0)
$$

---

$$
f(x,y)\mapsto f(r,\theta),\ \ J_f(r,\theta)=J_f(\phi(x,y))\cdot J_\phi(r,\theta)
$$

$$
[\partial f/\partial r,\partial f/\partial\theta]=[\partial f/\partial x,\partial f/\partial y]\begin{pmatrix}
\cos\theta & -r\sin\theta\
\sin\theta & r\cos\theta
\end{pmatrix}
$$

---

$f(x_1,\cdots,x_n),\ \ x_i=g_i(u_1,\cdots,u_m)$

那么

$$
\frac{\partial f}{\partial u_i}=\sum_{j=1}^n\frac{\partial f}{\partial x_j}\cdot\frac{\partial x_j}{\partial u_i}
$$

---

对于齐次函数有欧拉恒等式：

$f(x_1,\cdots,x_n)$ 称为齐次函数，如果 $\forall t\in\mathbb R$ 有 $f(tx_1,\cdots,tx_n)=t^df(x_1,\cdots,x_n)$，$d\in\mathbb N$ 称为次数。

欧拉向量组

$$
\langle\nabla f(x),x\rangle=d\cdot f(x)
$$

$$
\boxed{\frac d{dt}f(tx_1,\cdots,tx_n)=\sum_{i=1}^n\frac{\partial f}{\partial x_i}\frac{\partial (tx_i)}{\partial t}}=\sum_{i=1}^n\frac{\partial f}{\partial x_i}\cdot x_i=\langle\nabla f(x),x\rangle
$$

证明：固定 $x$，对恒等式 $f(tx_1,\cdots,tx_n)=t^df(x_1,\cdots,x_n)$ 两边求导：

画框的步骤是上面链式求导的直接推论。

3 拉普拉斯算子和调和函数

$\mathbb R^2$ 上，拉普拉斯算子 $\Delta=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}$。

调和函数：$\Delta f\equiv 0$ 的函数。

$\Delta$ 用极坐标的指出：

$$
\begin{aligned}
\Delta&=\frac1r\frac\partial{\partial r}\left(r\frac\partial{\partial r}\right)+\frac1{r^2}\frac{\partial^2}{\partial\theta^2}\
\Delta f&=\frac1r\frac\partial{\partial r}\left(r\frac{\partial f}{\partial r}\right)+\frac1{r^2}\frac{\partial^2 f}{\partial\theta^2}
\end{aligned}
$$
证明：

$$
\begin{aligned}
\frac\partial{\partial x}&=\frac\partial{\partial r}\frac{\partial r}{\partial x}+\frac\partial{\partial\theta}\frac{\partial\theta}{\partial x}\
&=\cos\theta\cdot\frac\partial{\partial r}-\frac{\sin\theta}r\frac\partial{\partial\theta}\

\frac{\partial^2}{\partial x^2}&=\frac\partial{\partial x}\left(\frac\partial{\partial x}\right)
=\cos\theta\cdot\frac\partial{\partial r}\left(\frac\partial{\partial x}\right)-\frac{\sin\theta}r\frac\partial{\partial\theta}\left(\frac\partial{\partial x}\right)\
&=\cos\theta\cdot\frac\partial{\partial r}\left[\cos\theta\cdot\frac\partial{\partial r}-\frac{\sin\theta}r\frac\partial{\partial\theta}\right]\
&-\frac{\sin\theta}r\frac\partial{\partial\theta}\left[\cos\theta\cdot\frac\partial{\partial r}-\frac{\sin\theta}r\frac\partial{\partial\theta}\right]\
&=\cos^2\theta\frac{\partial^2}{\partial r^2}-\frac{\sin\theta\cos\theta}r\frac{\partial^2}{\partial r\partial\theta}+\frac{\sin\theta\cos\theta}{r^2}\frac\partial{\partial\theta}\
&+\frac{\sin^2\theta}r\frac\partial{\partial r}-\frac{\sin\theta\cos\theta}r\frac{\partial^2}{\partial r\partial\theta}+\frac{\sin\theta\cos\theta}{r^2}\frac\partial{\partial\theta}+\frac{\sin^2\theta}{r^2}\frac{\partial^2}{\partial\theta^2}
\end{aligned}
$$

同理有

从而

$$
\begin{aligned}
\frac{\partial^2}{\partial y^2}&=\sin^2\theta\frac{\partial^2}{\partial r^2}+\frac{\cos^2\theta}r\frac\partial{\partial r}+\frac{\cos^2\theta}{r^2}\frac{\partial^2}{\partial\theta^2}\
&-\frac{2\sin\theta\cos\theta}{r^2}\frac\partial{\partial\theta}+\frac{2\sin\theta\cos\theta}r\frac{\partial^2}{\partial r\partial\theta}
\end{aligned}
$$

数分习题课笔记（2）