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 ?

  • $\begingroup$ 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. $\endgroup$ Mar 1, 2013 at 6:08
  • $\begingroup$ thanks zev, i'll read up on it... I was wonder how people got the good looking formats. $\endgroup$
    – Neo
    Mar 1, 2013 at 6:10
  • $\begingroup$ Can you find $dr/dx_i$? Use the chain and product rules. $\endgroup$
    – anon
    Mar 1, 2013 at 6:22

1 Answer 1


$r^2 = ||<x_1,x_2, \dots x_n> ||^2 = x_1^2+x_2^2+ \cdots x_n^2$. 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.

  • $\begingroup$ i see, this works out well. I think I have got it, will look at it more. I'm glad there is a webpage like this... $\endgroup$
    – Neo
    Mar 1, 2013 at 6:32
  • $\begingroup$ great! Edwards is a lot of fun to work through. I've used it for a course I teach every so often. That problem usually needs a little push for most students. The chain rule is not understood by most students I run across. $\endgroup$ Mar 1, 2013 at 6:38
  • $\begingroup$ I think their is a typo in $r \frac{\partial r}{\partial x_j} = 2x_j$, there is an extra $2$. $\endgroup$
    – user10444
    Mar 11, 2014 at 2:04
  • $\begingroup$ @user10444 excellent comment. Thanks! Fixed. $\endgroup$ Mar 11, 2014 at 2:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .