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Is there a non brute force way to show $v(x)=|x|^{2-n}u\left(\frac{x}{|x|^2}\right)$ is harmonic if $u$ is harmonic on a domain of $\mathbb{R}^n$?

I have tested this for the fundamental solution and verified the mean value over spheres centred at the origin. The mean value property over spheres not entered at the origin seems pretty brute force like computing the laplacian. Wondering if there is a better way.

I also tried to use the product rule $\Delta (fg) $ but did not get so far because $\Delta_x u\left(\frac{x}{|x|^2}\right)$ is kind of complicated.

I tried to interoperate what $v$ means. The map $x\mapsto \frac{x}{|x|^2}$ is the inversion about the unit sphere, and so the claim appears to be saying if we adjust the inversion up to a factor of $|x|^{2-n}$ (which is also harmonic) then the function $v$ obtained is also harmonic. How could this observation help?

shark
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    Interesting question. The only thing I can immediately observe is that if you define $$(\mathcal{I}u)(x) = |x|^{2-n}u\left(\frac{x}{|x|^2}\right),$$ then $$ (\mathcal{II}u)(x) = |x|^{2-n} (\mathcal{I}u)\left(\frac{x}{|x|^2}\right) = \underbrace{|x|^{2-n}\left|\frac{x}{|x|^2}\right|^{2-n}}{=1} u\left(\underbrace{\frac{x/|x|^2}{|x/|x|^2|^2}}{=x}\right)=u(x).$$This at least explains the factor $|x|^{2-n}$: you want an involution. – Ivo Terek Nov 09 '22 at 00:52
  • @IvoTerek That is a very neat observation. Also we know from elementary geometry that inversion about a circle causes lines not through the centre to become circles. So in this problem, does calculating derivatives in direction of standard bases amount to computing tangential derivatives? This is just a guess from intuition and I don't know how to make it precise. – shark Nov 09 '22 at 00:58
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    Not marking as a duplicate, but this question seems to have the same motivation (an elegant or conceptual argument). <> FWIW, this is the fourth question I've seen about the Kelvin transform in the past few weeks. – Andrew D. Hwang Nov 09 '22 at 02:29

1 Answers1

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Note that $x\mapsto \frac{x}{|x|^2}$ is an involution. So for any $m$ the map on functions

$$ u(x) \mapsto \frac{1}{|x|^m} u(\frac{x}{|x|^2})$$ is an involution ( see @Ivo Terek comment above). However, the Kelvin $\mathcal{K}$ has the extra property that it takes a constant function to a multiple of $\frac{1}{|x|^{n-2}}$, a harmonic function.

Note also that $\mathcal{K}$ commutes with orthogonal changes of coordinates, and behaves nicely with respect to central homotheties.

Let's see what $\mathcal{K}$ does to some known harmonic functions, say the potential of a point $u(x) = \frac{1}{|x-x_0|^{n-2}}$. When $x_0=0$ we get a constant ( $\mathcal{K}$ involution). From the above, it is enough to consider the case $x_0$ on the unit sphere, (in fact enough for $x_0 = (1,0, \ldots, 0)$). A simple calculation shows that $\mathcal K u = u$ ( one should check that). So $\mathcal{K}$ takes the potential of a point on the unit sphere to itself, and the potential of any other point to a multiple of the potential of the inverse of that point). Now the point potentials "linearly generate" the space of harmonic functions. We are done.

Note: this is more of a plea than a proof. Perhaps Kelvin realised how the transformation should work by looking at potentials, and then gave a formal proof. Calculations show that we have

$$\Delta \mathcal{K} = \frac{1}{|x|^4} \mathcal{K} \Delta $$

$\bf{Added:}$ The calculations for the point potential are simple. Let's start with an inversion map $\mathcal{I}$ of center $O$ and modulus $\rho$. Consider two points $x$, $y$, and their transforms $x'$, $y'$ under $\mathcal{I}$. The triangles $Oxy$, and $Oy'x'$ are similar, so we have $$\frac{x'y'}{yx} = \frac{Ox'}{Oy} = \frac{Ox'\cdot Ox}{Ox\cdot Oy}$$ that is $$|x'-y'| = \frac{\rho}{|x|\cdot |y|} \cdot |x-y|$$

(for us $\rho =1$). This is the only calculation we need to see how $\mathcal K$ acts on point potentials.

orangeskid
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