differential equation of a harmonic function Let $v$ be a smooth harmonic function on $R^n$. If $r^2=\sum_{i=1}^{n}|x_i|^2$, where $x=(x_1,x_2,....x_n) \in R^n$ and if $v$ is a radial function i.e $v(x)=v(r)$, write down the differential equation satisfied by $v$.
 A: Wealll, 
let's just stay away from that nasty point $r = 0$, i.e., from $(0, 0, . . . , 0)$, and see what we get:
first off, $v$ harmonic means that $v$ satisfies
$\sum_1^n \dfrac{\partial^2 v}{\partial x_i^2} = 0, \tag{1}$
so we just start grinding, using $v = v(r)$ and $r = \sqrt{\sum_1^n x_i^2} = (\sum_1^n x_i^2)^{1/2}$:
$\dfrac{\partial v}{\partial x_i} = \dfrac{\partial v}{\partial r} \dfrac{\partial r}{\partial x_i}, \tag{3}$
and 
$\dfrac{\partial r}{\partial x_i} = \dfrac{\partial (\sum_1^n x_i^2)^{1/2}}{\partial x_i} = \dfrac{1}{2}(\sum_1^n x_i^2)^{-1/2}(2x_i) = \dfrac{x_i}{r}, \tag{4}$
so
$\dfrac{\partial v}{\partial x_i} =  \dfrac{\partial v}{\partial r} \dfrac{x_i}{r}; \tag{5}$
keep grinding!  Thus:
$\dfrac{\partial^2 v}{\partial x_i^2} = \dfrac{\partial}{\partial x_i}(\dfrac{\partial v}{\partial x_i}) = \dfrac{\partial}{\partial x_i}(\dfrac{\partial v}{\partial r} \dfrac{x_i}{r}) = \dfrac{\partial}{\partial x_i}(\dfrac{\partial v}{\partial r})(\dfrac{x_i}{r}) + (\dfrac{\partial v}{\partial r})\dfrac{\partial}{\partial x_i}(\dfrac{x_i}{r}); \tag{6}$
and thus:
$\dfrac{\partial}{\partial x_i}(\dfrac{\partial v}{\partial r}) = \dfrac{\partial^2v}{\partial^2 r}\dfrac{\partial r}{\partial x_i} = \dfrac{\partial^2v}{\partial^2 r}(\dfrac{x_i}{r}), \tag{7}$
in a manner parallel to (3), again using (4), and thus:
$\dfrac{\partial}{\partial x_i}(\dfrac{x_i}{r}) = \dfrac{1}{r} + x_i \dfrac{\partial r^{-1}}{\partial x_i} = \dfrac{1}{r} - x_i r^{-2} \dfrac{\partial r}{\partial x_i} = \dfrac{1}{r} - \dfrac{x_i^2}{r^{3}}, \tag{8}$
where (4) has been used yet once again.  Using (7) and (8), (6) yields
$\dfrac{\partial^2 v}{\partial x_i^2} = \dfrac{\partial^2v}{\partial^2 r}(\dfrac{x_i}{r})^2 + \dfrac{\partial v}{\partial r}(\dfrac{1}{r} - \dfrac{x_i^2}{r^{3}}), \tag{9}$
and so
$\sum_1^n \dfrac{\partial^2 v}{\partial x_i^2} = \sum_1^n (\dfrac{\partial^2v}{\partial^2 r}(\dfrac{x_i}{r})^2 + \dfrac{\partial v}{\partial r}(\dfrac{1}{r} - \dfrac{x_i^2}{r^{3}})) = \dfrac{\partial^2v}{\partial^2 r} \sum_1^n (\dfrac{x_i}{r})^2 + \dfrac{\partial v}{\partial r} \sum_1^n (\dfrac{1}{r} - \dfrac{x_i^2}{r^{3}}). \tag{10}$
It is easy to see that
$\sum_1^n (\dfrac{x_i}{r})^2 = 1 \tag{11}$
and this implies
$\sum_1^n (\dfrac{1}{r} - \dfrac{x_i^2}{r^{3}})= \dfrac{1}{r}\sum_1^n(1 - \dfrac{x_i^2}{r^{2}}) = \dfrac{n - 1}{r}; \tag{12}$
pulling together (1), (10), (11), and (12) we finally arrive at
$\dfrac{\partial^2v}{\partial^2 r} + \dfrac{n - 1}{r}\dfrac{\partial v}{\partial r} = 0, \tag{13}$
in accord with the expression given in this Wikipedia page.
Hope this helps!  Cheerio,
and as always,
Fiat Lux!!!
