标签：right partial dfrac 坐标 rho 算子 theta 极坐标 left

\documentclass{article}

\usepackage{amsmath}

\usepackage{amsthm}

\usepackage{amsfonts}

\usepackage{ctex}

\usepackage{mathrsfs}

\begin{document}

\section{极坐标变换下的Laplace算子}

对于函数$u=u(x,y)$,其中$(x,y)\in D_{xy}\subseteq \mathbb R^2\backslash\{(0,0)\}$,构造极坐标变换

\begin{equation}

x=r \cos \theta ,

\end{equation}

\begin{equation}

y=r\sin\theta,

\end{equation}

其中$(r,\theta)\in D_{r\theta}\subseteq(0,+\infty )\times\left[ 0,2\pi\right),$计算得雅可比行列式

$\dfrac{\partial (x,y)}{\partial (r,\theta)}=r>0$,因此（1）式和（2）式表示一个双射$T: (r,\theta )\mapsto (x,y)$，从而映射（函数）$(r,\theta)\mapsto u$存在。为了方便，我们还假设$u$是二阶连续可微的,使得$u$对$x$,$y$的混合偏导数与求导顺序无关。

 

由（1）（2）易得$r^2=x^2+y^2$，两边对变量$x$求导得$r\dfrac{\partial r}{\partial x}=x$，所以$\dfrac{\partial r}{\partial x}=\dfrac{x}{r}$，同理可得$\dfrac{\partial r}{\partial y}=\dfrac{y}{r}$。由（1）（2）也易得$x\sin\theta=y\cos\theta$，两边对变量$x$求导得$\sin\theta+x\dfrac{\partial \theta}{\partial x}\cos\theta=-y\dfrac{\partial \theta}{\partial x}\sin\theta$，即$\dfrac{\partial\theta}{\partial x}=-\dfrac{\sin\theta}{x\cos\theta+y\sin\theta}$，为了使得表达式简洁，我们在分子分母都乘以非零的$r$并将（1）（2）分别代入式中的$x$,$y$得$\dfrac{\partial\theta}{\partial x}=-\dfrac{y}{r^2}$，同理可得$\dfrac{\partial\theta}{\partial y}=\dfrac{x}{r^2}$.

 

为了计算$\nabla^2 u$，用链式法则先求对变量$x$的一阶偏导数并代入上面的结论和化简得

\[\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial r}\dfrac{\partial r}{\partial x}+\dfrac{\partial u}{\partial \theta}\dfrac{\partial \theta}{\partial x}=\dfrac{\partial u}{\partial r}\dfrac{x}{r}-\dfrac{\partial u}{\partial \theta}\dfrac{y}{r^2},\]

\[\dfrac{\partial u}{\partial y}=\dfrac{\partial u}{\partial r}\dfrac{\partial r}{\partial y}+\dfrac{\partial u}{\partial \theta}\dfrac{\partial \theta}{\partial y}=\dfrac{\partial u}{\partial r}\dfrac{y}{r}+\dfrac{\partial u}{\partial \theta}\dfrac{x}{r^2},\]

运用求导的乘积法则和链式法则得

\begin{equation}

\dfrac{\partial^2 u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial x}\right)=\left(\dfrac{\partial^2 u}{\partial r^2}\dfrac{\partial r}{\partial x}+\dfrac{\partial^2 u}{\partial r\partial \theta}\dfrac{\partial \theta}{\partial x}\right)\dfrac{\partial r}{\partial x}+\dfrac{\partial u}{\partial r}\dfrac{\partial^2 r}{\partial x^2}+\left(\dfrac{\partial^2 u}{\partial r\partial \theta}\dfrac{\partial r}{\partial x}+\dfrac{\partial^2 u}{\partial \theta^2}\dfrac{\partial \theta}{\partial x}\right)\dfrac{\partial \theta}{\partial x}+\dfrac{\partial u}{\partial \theta}\dfrac{\partial^2 \theta}{\partial x^2},

\end{equation}

将$x$换成$y$得

\begin{equation}

\dfrac{\partial^2 u}{\partial y^2}=\dfrac{\partial}{\partial y}\left(\dfrac{\partial u}{\partial y}\right)=\left(\dfrac{\partial^2 u}{\partial r^2}\dfrac{\partial r}{\partial y}+\dfrac{\partial^2 u}{\partial r\partial \theta}\dfrac{\partial \theta}{\partial y}\right)\dfrac{\partial r}{\partial y}+\dfrac{\partial u}{\partial r}\dfrac{\partial^2 r}{\partial y^2}+\left(\dfrac{\partial^2 u}{\partial r\partial \theta}\dfrac{\partial r}{\partial y}+\dfrac{\partial^2 u}{\partial \theta^2}\dfrac{\partial \theta}{\partial y}\right)\dfrac{\partial \theta}{\partial y}+\dfrac{\partial u}{\partial \theta}\dfrac{\partial^2 \theta}{\partial y^2}.

\end{equation}

注意由于假设二阶连续可微，所以两个混合偏导数用同一个记号表示。下面计算所需的四个二阶偏导数。$\dfrac{\partial r}{\partial x}=\dfrac{x}{r}$的两边求对变量$x$的偏导数得\[\dfrac{\partial^2 r}{\partial x^2}=\dfrac{r^2-rx\dfrac{\partial r}{\partial x}}{r^3}=\dfrac{y^2}{r^3},\]

这里利用了一个小技巧，为了能够通过关系式$r^2=x^2+y^2$来化简最后的结果，分子分母同时乘以非零的$r$。把$x$和$y$互换即得\[\dfrac{\partial^2 r}{\partial y^2}=\dfrac{x^2}{r^3};\]

$\dfrac{\partial\theta}{\partial x}=-\dfrac{y}{r^2}$的两边求对变量$x$的偏导数得\[\dfrac{\partial^2 \theta}{\partial x^2}=\dfrac{2y}{r^3}\dfrac{\partial r}{\partial x}=\frac{2xy}{r^4}\],同理可得

\[\dfrac{\partial^2 \theta}{\partial y^2}=-\frac{2xy}{r^4},\]

将这些结果代入（3）（4）式得

\[\dfrac{\partial^2 u}{\partial x^2}=\dfrac{1}{r^4}\left[ r^2\dfrac{\partial^2 u}{\partial r^2}x^2+\left(r\dfrac{\partial u}{\partial r}+\dfrac{\partial^2 u}{\partial \theta^2}\right)y^2+2\left(\dfrac{\partial u}{\partial \theta}-r\dfrac{\partial^2 u}{\partial r\partial\theta}\right)xy\right],\]

\[\dfrac{\partial^2 u}{\partial y^2}=\dfrac{1}{r^4}\left[ \left(r\dfrac{\partial u}{\partial r}+\dfrac{\partial^2 u}{\partial \theta^2}\right)x^2+r^2\dfrac{\partial^2 u}{\partial r^2}y^2+2\left(r\dfrac{\partial^2 u}{\partial r\partial \theta}-\dfrac{\partial u}{\partial \theta}\right)xy\right],\]

两式相加得

\[\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=\dfrac{1}{r^4}\left(r^2\dfrac{\partial^2 u}{\partial r^2}+r\dfrac{\partial u}{\partial r}+\dfrac{\partial^2 u}{\partial \theta^2}\right)\left(x^2+y^2\right),\]

注意到$r^2=x^2+y^2$，所以在极坐标变换下二维Laplace算子的表达式为

\[\boxed{\nabla^2 =\dfrac{\partial^2 }{\partial r^2}+\dfrac{1}{r}\dfrac{\partial }{\partial r}+\dfrac{1}{r^2}\dfrac{\partial^2 }{\partial \theta^2}}\]

\section{柱面坐标变换下的Laplace算子}

函数$u=u(x,y,z)$的柱面坐标变换是指

\[x=\rho \cos\phi,\]

\[y=\rho \sin\phi,\]

\[z=z,\]

于是由上一节的讨论易得Laplace算子在柱面坐标变换下的表示为

\[\boxed{\nabla^2 =\dfrac{\partial^2 }{\partial \rho^2}+\dfrac{1}{\rho}\dfrac{\partial }{\partial \rho}+\dfrac{1}{\rho^2}\dfrac{\partial^2 }{\partial \phi^2}+\dfrac{\partial^2 }{\partial z^2}}\]

\section{球面坐标变换下的Laplace算子}

对于函数$u=u(x,y,z)$,其中$(x,y,z)\in \Omega_{xyz}\subseteq \mathbb R^3\backslash\{(0,0,0)\}$,构造球面坐标变换

\[x=r\sin\theta\cos\phi,\]

\[y=r\sin\theta\sin\phi,\]

\[z=r\cos\theta,\]

其中$(r,\theta,\phi)\in \Omega_{r\theta\phi}\subseteq(0,+\infty)\times[0,\pi]\times[0,2\pi)$.类似于极坐标变换的情形，以上三式给出了一个双射$S:(r,\theta,\phi)\mapsto (x,y,z)$,从而函数$(r,\theta,\phi)\mapsto u$存在。为了方便我们仍然假设$u$是二阶连续可微的。

 

为了能够利用第一节的结论，我们令$\rho=r\sin\theta$（这实际上是引入了柱面坐标$(\rho,\phi,z)$），于是$x=\rho \cos\phi$，$y=\rho\sin\phi$,对$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}$可以利用极坐标变换的结论得

\begin{equation}

\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=\dfrac{\partial^2 u }{\partial \rho^2}+\dfrac{1}{\rho}\dfrac{\partial u}{\partial \rho}+\dfrac{1}{\rho^2}\dfrac{\partial^2 u}{\partial \phi^2}

\end{equation}

同理，由$\rho=r\sin\theta$，$z=r\cos\theta$得

\begin{equation}

\dfrac{\partial^2 u}{\partial \rho^2}+\dfrac{\partial^2 u}{\partial z^2}=\dfrac{\partial^2 u }{\partial r^2}+\dfrac{1}{r}\dfrac{\partial u}{\partial r}+\dfrac{1}{r^2}\dfrac{\partial^2 u}{\partial \theta^2}

\end{equation}

（5）和（6）相加得

\begin{equation}

\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}+\dfrac{\partial^2 u}{\partial z^2}=\dfrac{1}{\rho}\dfrac{\partial u}{\partial \rho}+\dfrac{1}{\rho^2}\dfrac{\partial^2 u}{\partial \phi^2}+\dfrac{\partial^2 u }{\partial r^2}+\dfrac{1}{r}\dfrac{\partial u}{\partial r}+\dfrac{1}{r^2}\dfrac{\partial^2 u}{\partial \theta^2}

\end{equation}

我们还需计算$\dfrac{\partial u}{\partial \rho}$,注意到$(z,\rho)\mapsto (r,\theta)$是一个极坐标变换，于是由第一节的讨论可知

\[\dfrac{\partial u}{\partial \rho}=\dfrac{\partial u}{\partial r}\dfrac{\rho}{r}+\dfrac{\partial u}{\partial \theta}\dfrac{z}{r^2},\]

将$z=r\cos\theta$和$\rho=r\sin\theta$代入得

\[\dfrac{\partial u}{\partial \rho}=\dfrac{\partial u}{\partial r}\sin\theta+\dfrac{1}{r}\dfrac{\partial u}{\partial \theta}\cos\theta,\]

代入（7）式得到Laplace算子在球面坐标变换下的表示为

\[\boxed{\nabla^2=\dfrac{\partial^2}{\partial r^2}+\dfrac{2}{r}\dfrac{\partial}{\partial r}+\dfrac{1}{r^2}\left(\dfrac{\partial^2}{\partial \theta^2}+\cot\theta\dfrac{\partial}{\partial\theta}+\csc^2 \theta \dfrac{\partial^2}{\partial \phi^2}\right)}\]

\end{document}

标签：right,partial,dfrac,坐标,rho,算子,theta,极坐标,left
来源： https://www.cnblogs.com/Eufisky/p/15005858.html

本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享；
2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关；
3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关；
4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除；
5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。