# 拉普拉斯方程

## 定义

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

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

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

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

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

${\displaystyle \nabla ^{2}\varphi =0}$

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

${\displaystyle \Delta \varphi =0}$

${\displaystyle \Delta \varphi =f}$

## 二维拉普拉斯方程

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

### 解析函数

${\displaystyle f(z)=u(x,y)+iv(x,y)}$

${\displaystyle u_{x}=v_{y},\quad v_{x}=-u_{y}}$

${\displaystyle u_{yy}=(-v_{x})_{y}=-(v_{y})_{x}=-(u_{x})_{x}}$

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

${\displaystyle \psi _{x}=-\varphi _{y},\quad \psi _{y}=\varphi _{x}}$

${\displaystyle d\psi =-\varphi _{y}\,dx+\varphi _{x}\,dy}$

φ 满足拉普拉斯方程意味着 ψ 满足可积条件：

${\displaystyle \psi _{xy}=\psi _{yx}}$

${\displaystyle \varphi =\log r}$

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

${\displaystyle f(z)=\sum _{n=0}^{\infty }c_{n}z^{n}}$

${\displaystyle c_{n}=a_{n}+ib_{n}}$

${\displaystyle f(z)=\sum _{n=0}^{\infty }\left[a_{n}r^{n}\cos n\theta -b_{n}r^{n}\sin n\theta \right]+i\sum _{n=1}^{\infty }\left[a_{n}r^{n}\sin n\theta +b_{n}r^{n}\cos n\theta \right]}$

### 流体动力学

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

${\displaystyle u_{x}+v_{y}=0}$

${\displaystyle v_{x}-u_{y}=0}$

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

${\displaystyle \psi _{x}=-v,\quad \psi _{y}=u}$

${\displaystyle \varphi _{x}=u,\quad \varphi _{y}=v}$

### 静电学

${\displaystyle \nabla \times (u,v)=v_{x}-u_{y}=0}$

${\displaystyle \nabla \cdot (u,v)=\rho }$

${\displaystyle d\varphi =-u\,dx-v\,dy}$

${\displaystyle \varphi _{x}=-u,\quad \varphi _{y}=-v}$

${\displaystyle \varphi _{xx}+\varphi _{yy}=-\rho }$

## 三维拉普拉斯方程

### 基本解

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

${\displaystyle \iiint _{V}\nabla \cdot \nabla udV=-1}$

${\displaystyle -1=\iiint _{V}\nabla \cdot \nabla u\,dV=\iint _{S}u_{r}dS=4\pi a^{2}u_{r}(a)}$

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

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

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

### 格林函数

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

${\displaystyle \nabla \cdot \nabla u=-f}$

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:652-659

${\displaystyle \iiint _{V}\left[G\,\nabla \cdot \nabla u-u\,\nabla \cdot \nabla G\right]\,dV=\iiint _{V}\nabla \cdot \left[G\nabla u-u\nabla G\right]\,dV=\iint _{S}\left[Gu_{n}-uG_{n}\right]\,dS}$

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，可将这满足狄利克雷边界条件的公式化简为

${\displaystyle u(x',y',z')=\iiint _{V}Gf\,dV-\iint _{S}G_{n}g\,dS}$

### 圆球壳案例

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

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

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