# 柯西－黎曼方程

(1a)     ${ \partial u \over \partial x } = { \partial v \over \partial y }$

(1b)    ${ \partial u \over \partial y } = -{ \partial v \over \partial x } .$

## 注释和其他表述

### 共形映射

(2)    ${ i { \partial f \over \partial x } } = { \partial f \over \partial y } .$

$\begin{pmatrix} a & -b \\ b & \;\; a \end{pmatrix},$

### 複共轭的独立性

(3)    $\frac{\partial f}{\partial\bar{z}} = 0$

$\frac{\partial}{\partial\bar{z}} = \frac{1}{2}\left(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y}\right).$

### 複可微性

$f(z) = u(z) + i v(z)$

$\lim_{\underset{h\in\mathbb{C}}{h\to 0}} \frac{f(z_0+h)-f(z_0)}{h} = f'(z_0)$

$\lim_{\underset{h\in\mathbb{R}}{h\to 0}} \frac{f(z_0+h)-f(z_0)}{h} = \frac{\partial f}{\partial x}(z_0).$

$\lim_{\underset{ih\in i\mathbb{R}}{h\to 0}} \frac{f(z_0+ih)-f(z_0)}{ih} = \lim_{\underset{ih\in i\mathbb{R}}{h\to 0}} -i\frac{f(z_0+ih)-f(z_0)}{h} =-i\frac{\partial f}{\partial y}(z_0).$

f沿着两个轴的导数相同也即

$\frac{\partial f}{\partial x}(z_0)=-i\frac{\partial f}{\partial y}(z_0),$

### 物理解释

$\bar{f} = \begin{bmatrix}u\\ -v\end{bmatrix}$

$\frac{\partial (-v)}{\partial x} - \frac{\partial u}{\partial y} = 0.$

$\frac{\partial u}{\partial x} + \frac{\partial (-v)}{\partial y}=0.$

### 其它解释

$\frac{\partial u}{\partial s} = \frac{\partial u}{\partial n},\quad \frac{\partial u}{\partial n} = -\frac{\partial u}{\partial s}$

${ \partial u \over \partial r } = {1 \over r}{ \partial v \over \partial \theta},\quad{ \partial v \over \partial r } = -{1 \over r}{ \partial u \over \partial \theta}.$

${\partial f \over \partial r} = {1 \over i r}{\partial f \over \partial \theta}.$

## 非齐次方程

$\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y} = \alpha(x,y)$
$\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x} = \beta(x,y)$

$\frac{\partial f}{\partial\bar{z}} = \phi(z,\bar{z})$

$f(\zeta,\bar{\zeta}) = \frac{1}{2\pi i}\iint_D \phi(z,\bar{z})\frac{dz\wedge d\bar{z}}{z-\zeta}$

## 推广

### Goursat定理及其推广

f = u+iv为复函数，作为函数f : R2R2可微。则柯西积分定理（柯西－古尔萨定理）断言f在开域Ω上解析当且仅当它在该域上满足柯西-黎曼方程（Rudin 1966，Theorem 11.2）。特别是，f不需假定为连续可微（Dieudonné 1969，§9.10, Ex. 1）。

f在整个域Ω上满足柯西-黎曼方程是要点。可以构造在一点满足柯西-黎曼方程的连续函数，但它不在该点解析（譬如，f(z) = z5/|z|4）。只滿足柯西-黎曼方程也是不夠的，（需額外滿足连续性），下面的例子表明了这一点：（Looman 1923，p.107）

$f(z) = \begin{cases}\exp(-z^{-4})&\mathrm{if\ }z\not=0\\ 0&\mathrm{if\ }z=0 \end{cases}$

• f(z)在开域Ω⊂C上局部可积，并以弱形式满足柯西-黎曼方程，则f和Ω上的一个解析函数几乎处处相等。

### 多变量的情况

$\bar{\partial}$

${\partial f \over \partial \bar z} = 0$,

${\partial f \over \partial \bar z} = {1 \over 2}\left({\partial f \over \partial x} - {1 \over i}{\partial f \over \partial y}\right).$