Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Problem in English (original problem 7 on page 813 here)

Suppose $f(|\bar{x}|)=\sqrt{x_{1}^2+...+x_{n}^{2}}$. For what kind of real $f$ it holds that $f$ is harmonic everywhere but not in origin? If $f$ is harmonic, then $\triangle f=0$.


The "real function" apparently here means some $g$ such that $g: \mathbb R^{n}\to\mathbb R$, not vector in the co-domain but scalar (can be realized by looking at the norm) but $\bar{x}\in\mathbb R^n$ (please verify).

I am not sure whether this problem is just a brute-force calculation -practise or some clever trick, below some of my calculations for one term, not summing it up because it is a messy.

enter image description here

share|cite|improve this question
Work out what it means for $\Delta f=0$. In other words, you are summing $\frac{\partial^2 f}{\partial x_i^2}$ which you can use the chain rule on to get in terms of $f''(|x|)$ and everything else that is in your above expression. When you sum that's when you will get 0. Do not expect that each individual term gives you 0. – Alex R. Mar 20 '12 at 20:13
I can't read the PDF because it is not in English. Could you please post the full question? Are you looking for all harmonic functions on the punctured plane $\mathbb{R}^2-\{0\}$? – user7530 Mar 20 '12 at 21:49
up vote 1 down vote accepted

First, if $f(r)$ is a scalar function and $g: \mathbb{R}^n\to \mathbb{R}$, we have $$\Delta f(g) = \nabla \cdot \nabla f(g) = \nabla \cdot (f'(g) \nabla g) = f''(g)\nabla g\cdot\nabla g + f'(g)\Delta g.$$ When $g(\mathbf{x}) = \|\mathbf{x}\|$, we have, by direct computation, $\nabla g = \hat{\mathbf{x}}$ and $\Delta g = (n-1)/\|\mathbf{x}\|$. Therefore $$\Delta f(g) = f''(g) + f'(g)(n-1)/g,$$ which vanishes whenever $f$ satisfies the ordinary differential equation $$rf''+(n-1)f'=0.$$ My ODE chops are not up to finding a general solution (I'm sure someone will post an answer completing this step), but as a sanity check note that the expected $f(r) = \log r$ does work when $n=2$.

share|cite|improve this answer
...yes this is the way to do it, then the last part of this problem is to solve the second-degree differential-eq where $f=f(r)$ (so cannot use the Euler here but $g(r)=y'(r)$ and that way to solve it as a first degree). One can show that the above holds by writing things such as $\triangledown=\bar{i} \partial_x+\bar{j} \partial_y+\bar{k} \partial_z$ open. Also, notice that the $r$ is the distance from origin.Thanks, I think I am now on the right track about this problem, +1. – hhh Mar 21 '12 at 11:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.