Laplacian of spherically symmetric function

hbghlyj

When a scalar field $ϕ$ depends only on the magnitude of the position vector $r = \left|\bf r\right|$, we have
$$\nabla^{2} \phi(r)=\phi^{\prime \prime}(r)+\frac{2 \phi^{\prime}(r)}{r}$$page 3

Czhang271828

一般地, $\mathbb R^n$ 中径向对称函数 Laplace 算子一般形式为
$$
\begin{align*}
\Delta f(r)&=\sum_{k=1}^n \partial_{x_kx_k}f(r)\\
&=\sum_{k=1}^n \partial_{x_k}(f'(r)\cdot \partial_{x_k} r)\\
&=\sum_{k=1}^n \partial_{x_k}(f'(r)\cdot (x_k/r))\\
&=\sum_{k=1}^n [f''(r)\cdot (x_k/r)^2+f'(r)\cdot (1/r-x_k^2/r^3)]\\
&=f''(r)+\dfrac{n-1}{r}\cdot f'(r).
\end{align*}
$$

hbghlyj

$n=2$,极坐标系$(r,\theta)$:(又见这里)$$∇^2f=\frac{1}{r^2} \frac{\partial^2 f}{\partial \theta^2} + \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial f}{\partial r}\right)$$
$f=ϕ(r)$⇒$\nabla^{2} \phi(r)=\phi^{\prime \prime}(r)+\frac{\phi^{\prime}(r)}{r}$
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$n=3$,球坐标系$(r,\theta,\varphi)$:(又见这里)$$∇^2f={1 \over r^{2}}{\partial  \over \partial r}\!\left(r^{2}{\partial f \over \partial r}\right)\!+\!{1 \over r^{2}\!\sin \theta }{\partial  \over \partial \theta }\!\left(\sin \theta {\partial f \over \partial \theta }\right)\!+\!{1 \over r^{2}\!\sin ^{2}\theta }{\partial ^{2}f \over \partial \varphi ^{2}}$$
$f=ϕ(r)$⇒$\nabla^{2} \phi(r)=\phi^{\prime \prime}(r)+\frac{2 \phi^{\prime}(r)}{r}$

Czhang271828

回复 3# hbghlyj

嗯, 即 "仅与模长有关" 之意. 正式写作中会用数学语言加以说明.

hbghlyj

Kelvin transform
Given a fixed sphere $S(0,R)$ with centre $0$ and radius $R$, the inversion of a point $x$ in $\Bbb R^n$ is defined to be\[ x^{*}={\frac {R^{2}}{|x|^{2}}}x.\]If $D$ is an open subset of Rn which does not contain 0, then for any function $f$ defined on $D$, the Kelvin transform $f^*$ of $f$ with respect to the sphere $S(0,R)$ is
\[f^{*}(x^{*})={\frac {|x|^{n-2}}{R^{2n-4}}}f(x)={\frac {1}{|x^{*}|^{n-2}}}f(x)={\frac {1}{|x^{*}|^{n-2}}}f\left({\frac {R^{2}}{|x^{*}|^{2}}}x^{*}\right).\]
Let $D$ be an open subset in $\Bbb R^n$ which does not contain the origin 0. Then a function $u$ is harmonic, subharmonic or superharmonic in $D$ if and only if the Kelvin transform $u^*$ with respect to the sphere $S(0,R)$ is harmonic, subharmonic or superharmonic in $D^*$.

This follows from the formula
\[\Delta u^{*}(x^{*})={\frac {R^{4}}{|x^{*}|^{n+2}}}(\Delta u)\left({\frac {R^{2}}{|x^{*}|^{2}}}x^{*}\right).\]