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section{极坐标变换下的Laplace算子}
对于函数$u=u(x,y)$,其中$(x,y)in D_{xy}subseteq mathbb R^2ackslash{(0,0)}$,构造极坐标变换
egin{equation}
x=r cos heta ,
end{equation}
egin{equation}
y=rsin heta,
end{equation}
其中$(r, heta)in D_{r heta}subseteq(0,+infty ) imesleft[ 0,2pi ight),$计算得雅可比行列式
$dfrac{partial (x,y)}{partial (r, heta)}=r>0$,因此（1）式和（2）式表示一个双射$T: (r, heta )mapsto (x,y)$，从而映射（函数）$(r, heta)mapsto u$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$rdfrac{partial r}{partial x}=x$，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$xsin heta=ycos heta$，两边对变量$x$求导得$sin heta+xdfrac{partial heta}{partial x}cos heta=-ydfrac{partial heta}{partial x}sin heta$，即$dfrac{partial heta}{partial x}=-dfrac{sin heta}{xcos heta+ysin heta}$，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$dfrac{partial heta}{partial x}=-dfrac{y}{r^2}$，同理可得$dfrac{partial heta}{partial y}=dfrac{x}{r^2}$.
为了计算$ abla^2 u$，用链式法则先求对变量$x$的一阶偏导数并代入上面的结论和化简得
[dfrac{partial u}{partial x}=dfrac{partial u}{partial r}dfrac{partial r}{partial x}+dfrac{partial u}{partial heta}dfrac{partial heta}{partial x}=dfrac{partial u}{partial r}dfrac{x}{r}-dfrac{partial u}{partial heta}dfrac{y}{r^2},]
[dfrac{partial u}{partial y}=dfrac{partial u}{partial r}dfrac{partial r}{partial y}+dfrac{partial u}{partial heta}dfrac{partial heta}{partial y}=dfrac{partial u}{partial r}dfrac{y}{r}+dfrac{partial u}{partial heta}dfrac{x}{r^2},]
运用求导的乘积法则和链式法则得
egin{equation}
dfrac{partial^2 u}{partial x^2}=dfrac{partial}{partial x}left(dfrac{partial u}{partial x} ight)=left(dfrac{partial^2 u}{partial r^2}dfrac{partial r}{partial x}+dfrac{partial^2 u}{partial rpartial heta}dfrac{partial heta}{partial x} ight)dfrac{partial r}{partial x}+dfrac{partial u}{partial r}dfrac{partial^2 r}{partial x^2}+left(dfrac{partial^2 u}{partial rpartial heta}dfrac{partial r}{partial x}+dfrac{partial^2 u}{partial heta^2}dfrac{partial heta}{partial x} ight)dfrac{partial heta}{partial x}+dfrac{partial u}{partial heta}dfrac{partial^2 heta}{partial x^2},
end{equation}
将$x$换成$y$得
egin{equation}
dfrac{partial^2 u}{partial y^2}=dfrac{partial}{partial y}left(dfrac{partial u}{partial y} ight)=left(dfrac{partial^2 u}{partial r^2}dfrac{partial r}{partial y}+dfrac{partial^2 u}{partial rpartial heta}dfrac{partial heta}{partial y} ight)dfrac{partial r}{partial y}+dfrac{partial u}{partial r}dfrac{partial^2 r}{partial y^2}+left(dfrac{partial^2 u}{partial rpartial heta}dfrac{partial r}{partial y}+dfrac{partial^2 u}{partial heta^2}dfrac{partial heta}{partial y} ight)dfrac{partial heta}{partial y}+dfrac{partial u}{partial heta}dfrac{partial^2 heta}{partial y^2}.
end{equation}
注意由于假设二阶连续可微，所以两个混合偏导数用同一个记号表示。下面计算所需的四个二阶偏导数。$dfrac{partial r}{partial x}=dfrac{x}{r}$的两边求对变量$x$的偏导数得[dfrac{partial^2 r}{partial x^2}=dfrac{r^2-rxdfrac{partial r}{partial x}}{r^3}=dfrac{y^2}{r^3},]
这里利用了一个小技巧，为了能够通过关系式$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};]
$dfrac{partial heta}{partial x}=-dfrac{y}{r^2}$的两边求对变量$x$的偏导数得[dfrac{partial^2 heta}{partial x^2}=dfrac{2y}{r^3}dfrac{partial r}{partial x}=frac{2xy}{r^4}],同理可得
[dfrac{partial^2 heta}{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^2dfrac{partial^2 u}{partial r^2}x^2+left(rdfrac{partial u}{partial r}+dfrac{partial^2 u}{partial heta^2} ight)y^2+2left(dfrac{partial u}{partial heta}-rdfrac{partial^2 u}{partial rpartial heta} ight)xy ight],]
[dfrac{partial^2 u}{partial y^2}=dfrac{1}{r^4}left[ left(rdfrac{partial u}{partial r}+dfrac{partial^2 u}{partial heta^2} ight)x^2+r^2dfrac{partial^2 u}{partial r^2}y^2+2left(rdfrac{partial^2 u}{partial rpartial heta}-dfrac{partial u}{partial heta} ight)xy ight],]
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^2dfrac{partial^2 u}{partial r^2}+rdfrac{partial u}{partial r}+dfrac{partial^2 u}{partial heta^2} ight)left(x^2+y^2 ight),]
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
[oxed{ abla^2 =dfrac{partial^2 }{partial r^2}+dfrac{1}{r}dfrac{partial }{partial r}+dfrac{1}{r^2}dfrac{partial^2 }{partial heta^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= ho cosphi,]
[y= ho sinphi,]
[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
[oxed{ abla^2 =dfrac{partial^2 }{partial ho^2}+dfrac{1}{ ho}dfrac{partial }{partial ho}+dfrac{1}{ ho^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$(x,y,z)in Omega_{xyz}subseteq mathbb R^3ackslash{(0,0,0)}$,构造球面坐标变换
[x=rsin hetacosphi,]
[y=rsin hetasinphi,]
[z=rcos heta,]
其中$(r, heta,phi)in Omega_{r hetaphi}subseteq(0,+infty) imes[0,pi] imes[0,2pi)$.类似于极坐标变换的情形，以上三式给出了一个双射$S:(r, heta,phi)mapsto (x,y,z)$,从而函数$(r, heta,phi)mapsto u$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$ ho=rsin heta$（这实际上是引入了柱面坐标$( ho,phi,z)$），于是$x= ho cosphi$，$y= hosinphi$,对$dfrac{partial^2 u}{partial x^2}+dfrac{partial^2 u}{partial y^2}$可以利用极坐标变换的结论得
egin{equation}
dfrac{partial^2 u}{partial x^2}+dfrac{partial^2 u}{partial y^2}=dfrac{partial^2 u }{partial ho^2}+dfrac{1}{ ho}dfrac{partial u}{partial ho}+dfrac{1}{ ho^2}dfrac{partial^2 u}{partial phi^2}
end{equation}
同理，由$ ho=rsin heta$，$z=rcos heta$得
egin{equation}
dfrac{partial^2 u}{partial ho^2}+dfrac{partial^2 u}{partial z^2}=dfrac{partial^2 u }{partial r^2}+dfrac{1}{r}dfrac{partial u}{partial r}+dfrac{1}{r^2}dfrac{partial^2 u}{partial heta^2}
end{equation}
（5）和（6）相加得
egin{equation}
dfrac{partial^2 u}{partial x^2}+dfrac{partial^2 u}{partial y^2}+dfrac{partial^2 u}{partial z^2}=dfrac{1}{ ho}dfrac{partial u}{partial ho}+dfrac{1}{ ho^2}dfrac{partial^2 u}{partial phi^2}+dfrac{partial^2 u }{partial r^2}+dfrac{1}{r}dfrac{partial u}{partial r}+dfrac{1}{r^2}dfrac{partial^2 u}{partial heta^2}
end{equation}
我们还需计算$dfrac{partial u}{partial ho}$,注意到$(z, ho)mapsto (r, heta)$是一个极坐标变换，于是由第一节的讨论可知
[dfrac{partial u}{partial ho}=dfrac{partial u}{partial r}dfrac{ ho}{r}+dfrac{partial u}{partial heta}dfrac{z}{r^2},]
将$z=rcos heta$和$ ho=rsin heta$代入得
[dfrac{partial u}{partial ho}=dfrac{partial u}{partial r}sin heta+dfrac{1}{r}dfrac{partial u}{partial heta}cos heta,]
代入（7）式得到Laplace算子在球面坐标变换下的表示为
[oxed{ abla^2=dfrac{partial^2}{partial r^2}+dfrac{2}{r}dfrac{partial}{partial r}+dfrac{1}{r^2}left(dfrac{partial^2}{partial heta^2}+cot hetadfrac{partial}{partial heta}+csc^2 heta dfrac{partial^2}{partial phi^2} ight)}]
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