# 拉普拉斯方程

$\int_M \mathrm{d}\omega = \oint_{\partial M} \omega$

## 定義

$\Delta f = \frac{\partial^2 f}{\partial x^2 } + \frac{\partial^2 f}{\partial y^2 } + \frac{\partial^2 f}{\partial z^2 } = 0$

$\Delta f=\frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial f}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 f}{\partial \phi^2} + \frac{\partial^2 f}{\partial z^2} =0$

$\Delta f = \frac{1}{\rho^2}\frac{\partial}{\partial \rho} \left(\rho^2 \frac{\partial f}{\partial \rho}\right) + \frac{1}{\rho^2 \sin\theta} \frac{\partial}{\partial \theta} \left(\sin\theta \frac{\partial f}{\partial \theta}\right) + \frac{1}{\rho^2 \sin^2\theta} \frac{\partial^2 f}{\partial \varphi^2} =0$

$\Delta f =\frac{\partial}{\partial \xi^j}\left(\frac{\partial f}{\partial \xi^k}g^{ki}\right) + \frac{\partial f}{\partial \xi^j} g^{jm}\Gamma^n_{mn} =0,$

$\Delta f = \frac{1}{\sqrt{|g|}} \frac{\partial}{\partial \xi^i}\!\left(\sqrt{|g|}g^{ij} \frac{\partial f}{\partial \xi^j}\right) =0, \qquad (g=\mathrm{det}\{g_{ij}\})$

$\nabla^2 \varphi = 0$

$\operatorname{div}\,\operatorname{grad}\,\varphi = 0$

$\Delta \varphi = 0$

$\Delta \varphi = f$

## 二维拉普拉斯方程

$\frac{\partial^2\psi}{\partial x^2} + \frac{\partial^2\psi}{\partial y^2} \equiv \psi_{xx} + \psi_{yy} = 0.$

### 解析函数

$f(z) = u(x,y) + iv(x,y)$

$u_x = v_y, \quad v_x = -u_y$

$u_{yy} = (-v_x)_y = -(v_y)_x = -(u_x)_x$

$f(z) = \varphi(x,y) + i \psi(x,y)$

$\psi_x = -\varphi_y, \quad \psi_y = \varphi_x$

$d \psi = -\varphi_y\, dx + \varphi_x\, dy$

φ 满足拉普拉斯方程意味着 ψ 满足可积条件：

$\psi_{xy} = \psi_{yx}$

$\varphi = \log r$

$f(z) = \log z = \log r + i\theta$

$f(z) = \sum_{n=0}^\infty c_n z^n$

$c_n = a_n + i b_n$

$f(z) = \sum_{n=0}^\infty \left[ a_n r^n \cos n \theta - b_n r^n \sin n \theta\right] + i \sum_{n=1}^\infty \left[ a_n r^n \sin n\theta + b_n r^n \cos n \theta\right]$

### 流體動力學

$u$$v$ 分别为满足定常不可压缩无旋条件的流体速度场的 $x$$y$ 方向分量（这里仅考虑二维流场），那么不可压缩条件为：[3]:99-101

$u_x + v_y=0$

$v_x - u_y =0$

$d \psi = -v\, dx + u\, dy$

$\psi_x = -v, \quad \psi_y=u$

$\varphi_x=u, \quad \varphi_y=v$

### 靜電學

$\nabla \times (u,v) = v_x -u_y =0$

$\nabla \cdot (u,v) = \rho$

$d \varphi = -u\, dx -v\, dy$

$\varphi_x = -u, \quad \varphi_y = -v$

$\varphi_{xx} + \varphi_{yy} = -\rho$

## 三维拉普拉斯方程

### 基本解

$\nabla \cdot \nabla u = u_{xx} + u_{yy} + u_{zz} = -\delta(x-x',y-y',z-z')$

$\iiint_V \nabla \cdot \nabla u dV =-1$

$-1= \iiint_V \nabla \cdot \nabla u \, dV = \iint_S u_r dS = 4\pi a^2 u_r(a)$

$u_r(r) = -\frac{1}{4\pi r^2}$

$u = \frac{1}{4\pi r}$

$u = \frac{-\ln r}{2\pi}$

### 格林函数

$\nabla \cdot \nabla G = -\delta(x-x',y-y',z-z') \quad \hbox{in} \quad V$
$G = 0 \quad \hbox{if} \quad (x,y,z) \quad \hbox{on} \quad S$

$\nabla \cdot \nabla u = -f$

u在边界S上取值为g，那么我们可以应用格林定理（是高斯散度定理的一个推论），得到[1]:652-659

$\iiint_V \left[ G \, \nabla \cdot \nabla u - u \, \nabla \cdot \nabla G \right]\, dV = \iiint_V \nabla \cdot \left[ G \nabla u - u \nabla G \right]\, dV = \iint_S \left[ G u_n -u G_n \right] \, dS$

unGn分别代表两个函数在边界S上的法向导数。考虑到uG满足的条件，可将這滿足狄利克雷边界条件的公式化简为

$u(x',y',z') = \iiint_V G f \, dV - \iint_S G_n g \, dS$

### 圓球殼案例

$\rho' = \frac{a^2}{\rho}$

$G= \frac{1}{4 \pi R} - \frac{a}{4 \pi \rho R'}$

$u(P) = \frac{1}{4\pi} a^3\left( 1 - \frac{\rho^2}{a^2} \right) \iint \frac{g(\theta',\varphi') \sin \theta' \, d\theta' \, d\varphi'}{(a^2 + \rho^2 - 2 a \rho \cos \Theta)^{3/2} }$

## 参考文献

1. ^ 1.0 1.1 1.2 1.3 1.4 1.5 Boas, Mary. Mathematical Methods in the Physical Sciences 3rd. Wiley. 2005. ISBN 978-0471198260.
2. ^ 2.0 2.1 2.2 Jackson, John David, Classical Electrodynamic 3rd., USA: John Wiley & Sons, Inc., 1999, ISBN 978-0-471-30932-1
3. ^ Batchelor, George. An Introduction to Fluid Dynamics. Cambridge University Press. ISBN 978-0521663960.
4. ^ Griffiths, David J., Introduction to Electrodynamics (3rd ed.), Prentice Hall, 1998, ISBN 0-13-805326-X
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