# Laplacian(F) = (n-1/r)g'(r) + g''(r)

I got one more problem from my self reading of Methods of Advanced Calculus by Edwards, hints and solutions are equally appreciated:

If f(x) = g(r), r= |x|, and n>=3, show that Laplace(f) = d^2f/dx1^2 +.....+d^2f/dx2^2 = (n-1/r)g'(r) + g''(r)

and using this result show that if Lapace(f) =0 then f(x) = a/|x|^n-2 + b when x does not = 0

Does the n-1 come from the definition of the derivative ?

• You can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own TeX.SE site. Mar 1, 2013 at 6:08
• thanks zev, i'll read up on it... I was wonder how people got the good looking formats.
– Neo
Mar 1, 2013 at 6:10
• Can you find $dr/dx_i$? Use the chain and product rules.
– anon
Mar 1, 2013 at 6:22

$r^2 = ||<x_1,x_2, \dots x_n> ||^2 = x_1^2+x_2^2+ \cdots x_n^2$. Consequently, $r \frac{\partial r}{\partial x_j} = x_j$. Consider then, by the chain-rule and the result above: $$(\nabla g)_j = \frac{\partial }{\partial x_j} g(r) = g'(r)\frac{\partial r }{\partial x_j} = g'(r) \frac{x_j}{r}$$ Now, differentiate once more: $$\frac{\partial}{\partial x_j}(\nabla g)_j = \frac{\partial }{\partial x_j} \left( g'(r) \frac{x_j}{r} \right) = g''(r)\frac{x_j^2}{r^2}+\frac{\partial }{\partial x_j}\frac{x_j}{r} = g''(r)\frac{x_j^2}{r^2}+ g'(r)\frac{r-x_j\frac{x_j}{r}}{r^2}$$ cleaning it up a bit:
$$\frac{\partial}{\partial x_j}(\nabla g)_j= g''(r)\frac{x_j^2}{r^2}+ g'(r)\frac{r^2-x_j^2}{r^3}$$ Finally, to find the Laplacian sum the formula above from $j=1,2, \dots n$. It's obvious where the $g''(r)$ term arises. On the other hand, you get $n$ copies of $r^2$ less one copy of $r^2$ from the $g'(r)$ term. Think on this for a bit, you'll see it. Welcome to the MSE.
• I think their is a typo in $r \frac{\partial r}{\partial x_j} = 2x_j$, there is an extra $2$. Mar 11, 2014 at 2:04