# 拉普拉斯方程

## 定義

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

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

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

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

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

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

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

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

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

## 二維拉普拉斯方程

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

### 解析函數

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

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

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

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

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

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

φ 滿足拉普拉斯方程意味著 ψ 滿足可積條件：

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

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

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

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

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

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

### 流體動力學

${\displaystyle u}$${\displaystyle v}$ 分別為滿足定常不可壓縮無旋條件的流體速度場的 ${\displaystyle x}$${\displaystyle y}$ 方向分量（這裡僅考慮二維流場），那麼不可壓縮條件為：[3]:99-101

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

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

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

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

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

### 靜電學

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

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

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

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

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

## 三維拉普拉斯方程

### 基本解

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

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

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

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

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

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

### 格林函數

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

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

u在邊界S上取值為g，那麼我們可以應用格林定理（是高斯散度定理的一個推論），得到[1]: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分別代表兩個函數在邊界S上的法向導數。考慮到uG滿足的條件，可將這滿足狄利克雷邊界條件的公式化簡為

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

### 圓球殼案例

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

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

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

## 參考文獻

1. Boas, Mary. Mathematical Methods in the Physical Sciences 3rd. Wiley. 2005. ISBN 978-0471198260.
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
• 嚴鎮軍編，《數學物理方程》，第二版，中國科學技術大學出版社，合肥，2002，ISBN 978-7-312-00799-6/O•177
• L.C. Evans, Partial Differential Equations, American Mathematical Society, Providence, 1998. ISBN 978-0-8218-0772-9
• I. G. Petrovsky, Partial Differential Equations, W. B. Saunders Co., Philadelphia, 1967.
• A. D. Polyanin, Handbook of Linear Partial Differential Equations for Engineers and Scientists, Chapman & Hall/CRC Press, Boca Raton, 2002. ISBN 978-1-58488-299-2
• A. Sommerfeld, Partial Differential Equations in Physics, Academic Press, New York, 1949.
• Pijush K.Kundu, Fluid Mechanics, Academic Press, 2002.