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The function $f : \mathbb{R}^n \rightarrow \mathbb{R}$ given by $f(x) = \|x\|^{2-n}$, where $\|~\|$ denotes the Euclidean norm, is harmonic. This is just a simple computation.

My question is: why should we expect this function to be harmonic? In other words, if someone asked you whether this function is harmonic or not, what kind of reasoning could lead you to suspect that this is the case, before actually computing its laplacian?

I'm just trying to understand the intuition behind harmonic functions. Since this can be a little vague, I'm willing to accept alternative proofs of this fact as an answer, even if they are overkill.

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There was good reason to try to find radially symmetric Green's functions for the Laplacian en.wikipedia.org/wiki/… –  Will Jagy Nov 22 '12 at 5:05
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Harmonic except at the origin, if $n > 2$. –  Robert Israel Nov 22 '12 at 5:08

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up vote 3 down vote accepted

As I commented, if $n \ge 2$ this function is not harmonic at the origin. By the Divergence Theorem, the flux of the gradient of $f$ outward through the boundary of a bounded region with smooth boundary is $0$ if $f$ is harmonic in the region. In particular this should be true for the region between two spheres centred at the origin. For $f(x) = \|x\|^{2-n}$ the gradient is directed towards the origin (and thus normal to the surface), with magnitude proportional to $\|x\|^{1-n}$. Since the $n-1$-dimensional area of the sphere of radius $r$ is proportional to $r^{n-1}$, that means the flux through the sphere is the same for every $r$, so the flux out of the region between the two spheres is $0$, as it should be.

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I like this specially because it explains the exponent $2-n$, which was rather mysterious to me. Thanks! –  student Nov 22 '12 at 21:39

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