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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 $\triangle f=0$.

Definitions

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.

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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

1 Answer 1

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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 $n=2$.

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...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

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