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(This is the $n$-dimensional analogue of the 3D case: Show that $H(x) := |x|^{-1} u(x/|x|^2) $ is harmonic if $u$ is harmonic)

Suppose that $u$ is a harmonic function on $\mathbb{R}^n$.
Prove that the function $\displaystyle H(x):=\frac{1}{|x|^{n-2}}u\left(\frac{x}{|x|^2}\right)$ is harmonic on $\mathbb{R}^n\backslash\{0\}$.

I tried to compute $\Delta H$ directly by following the brute force method in the linked question, but it gets tedious very soon. Therefore, I am wondering if there is a more elegant way?

I am thinking to use the converse of mean value property. So far I have worked out (probably) that

  • Under the mapping $f: x\mapsto \frac{x}{|x|^2}$,
    a circle with radius $r$ centered at $x_0$ would be mapped to a circle with radius $R=\frac{2r}{|x_0|^2-r^2}$ centered at $y_0=\frac{x_0}{|x_0|^2-r^2}$.
  • $f = f^{-1}$
  • Its Jacobian is $|Jf|=|x|^{2n}$.

I am stuck after doing change of variables in the integral $\displaystyle\frac{1}{|B_R(y_0)|}\int_{B_R(y_0)} H(y)dy$ , since the terms do not magically cancel out as wished, and expressing $|B_R(y_0)|$ in terms of the corresponding $|B_r(x_0)|$ also yields a mess.

I really appreciate any help. Other methods (or more efficient brute force) are also greatly welcomed.

Kamal Saleh
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Vigej
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    Solution is here https://math.stackexchange.com/questions/4097778/u-is-harmonic-prove-that-vx-frac1xn-2-cdot-u-left-fracxx2/4098093#4098093 – Svyatoslav Apr 12 '21 at 05:33
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    @Svyatoslav thank you for adding your comment, I knew I saw it recently but could not find it again – Calvin Khor Apr 12 '21 at 05:35
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    @Svyatoslav Thank you very much, the solution is so systematic! – Vigej Apr 12 '21 at 06:44
  • Also, would anyone advise on how to handle this question now? Should I flag it as a duplicate, or edit it to specifically asking for a solution using mean value property instead? In any case, thank you everyone for the inputs so far! – Vigej Apr 12 '21 at 06:47
  • I think the question on using MVP is interesting. If you prefer it to be closed as a dupe, I can vote to close and then you (as the post author) will have a button to have it closed – Calvin Khor Apr 12 '21 at 07:02
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    After reading your post again, I remembered the "method of image charge". Some googling led me to the Wikipedia page for the Kelvin transform and then consequently this post from 7+ years ago. Even there the calculation is explicit. – Calvin Khor Apr 12 '21 at 07:09

2 Answers2

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Note that $|x|^{2-n}$ is harmonic, in the following calculations we will let $\phi(x):=|x|^{2-n}$ and use the Einstein summation rule to simplify our procedure.

Since $\bar{u}(x) = \phi(x) u\bigl(\frac{x}{|x|^2}\bigr)$, we calculate to get $$\bar{u}_{x_i}(x) = \phi_{x_i}(x) u\biggl(\frac{x}{|x|^2}\biggr) + \phi(x) u_{x_j} \biggl(\frac{x_j}{|x|^2}\biggr)_{x_i}.$$ Differentiate again with respect to $x_i$ to obtain \begin{align*} \bar{u}_{x_ix_i} & = \phi_{x_ix_i} u\biggl(\frac{x}{|x|^2}\biggr) + 2\phi_{x_i} u_{x_j}\biggl(\frac{x}{|x|^2}\biggr) \biggl(\frac{x_j}{|x|^2}\biggr)_{x_i} \\ & \quad + \phi(x) u_{x_jx_k}\biggl(\frac{x}{|x|^2}\biggr) \biggl(\frac{x_j}{|x|^2}\biggr)_{x_i} \biggl(\frac{x_k}{|x|^2}\biggr)_{x_i} + \phi(x) u_{x_j} \biggl(\frac{x}{|x|^2}\biggr) \biggl(\frac{x_j}{|x|^2}\biggr)_{x_ix_i}. \end{align*} So $$\Delta \bar{u}(x) = 0 + A + B + C.$$ Next we calculate $A$, $B$ and $C$ seperately. Since $$\biggl(\frac{x_j}{|x|^2}\biggr)_{x_i} = \frac{\delta_{ij} |x|^2 - 2x_ix_j}{|x|^4}.$$ It follows that $$x_i \biggl(\frac{x_j}{|x|^2}\biggr)_{x_i} = -\frac{x_j}{|x|^2},$$ and $$\biggl(\frac{x_j}{|x|^2}\biggr)_{x_i} \biggl(\frac{x_k}{|x|^2}\biggr)_{x_i} = \frac{\delta_{jk}}{|x|^4}.$$ (note: the above equality is just the same as $D_x\bar{x}(D_x\bar{x})^T = |\bar{x}|^4I$) Therefore, by combining $\phi_{x_i}(x) = (2-n)|x|^{-n}x_i$, we obtain that \begin{align*} A & = 2(2-n)|x|^{-n} x_i \biggl(\frac{x_j}{|x|^2}\biggr)_{x_i} u_{x_j}\biggl(\frac{x}{|x|^2}\biggr) \\ & = 2(2-n)|x|^{-n} \biggl(-\frac{x_j}{|x|^2}\biggr) u_{x_j}\biggl(\frac{x}{|x|^2}\biggr) \\ & = 2(n-2)|x|^{-n-2} x\cdot Du\biggl(\frac{x}{|x|^2}\biggr). \end{align*} $$B = \phi(x) u_{x_jx_k}\biggl(\frac{x}{|x|^2}\biggr) \frac{\delta_{jk}}{|x|^4} = |x|^{-n-2} \Delta u\biggl(\frac{x}{|x|^2}\biggr).$$ Finally we have to calculate $\Bigl(\frac{x_j}{|x|^2}\Bigr)_{x_ix_i}$ (be careful to this term!), \begin{align*} \biggl(\frac{x_j}{|x|^2}\biggr)_{x_ix_i} & = \biggl(\frac{\delta_{ij} |x|^2 - 2x_ix_j}{|x|^4}\biggr)_{x_i} \\ & = \frac{\bigl(2\delta_{ij}x_i - 2(\delta_{ij}+1)x_j\bigr) |x|^4 -(\delta_{ij}|x|^2 - 2x_ix_j)4|x|^2x_i}{|x|^8} \\ & = \frac{(-2n+4)x_j}{|x|^4}. \end{align*} So $$C = \phi(x) u_{x_j}\biggl(\frac{x}{|x|^2}\biggr) \frac{(-2n+4)x_j}{|x|^4} = (-2n+4)|x|^{-n-2} x\cdot Du\biggl(\frac{x}{|x|^2}\biggr).$$

Combining all these results, we conclude that $$\Delta\bar{u}(x) = A + B + C = B = |x|^{-n-2} \Delta u\biggl(\frac{x}{|x|^2}\biggr).$$

Stephen
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I'm putting this in as an answer as it's too long for a comment.

This will definitely be computation-heavy, but here are some formulas that could help. Note that the bolded nabla is understood to be the general tensor gradient, also known as the covariant derivative.

The gradient has a chain rule for $\varphi:\mathbb{R}^n\to\mathbb{R}$, $\mathrm{v}:\mathbb{R}^n\to\mathbb{R}^n$ $$\nabla(\varphi\circ \mathrm{v})=(\nabla\varphi\circ\mathrm{v})~\boldsymbol\nabla\mathrm{v}$$ The divergence has one as well, for $\mathrm{w}:\mathbb{R}^n\to\mathbb{R}^n$: $$\nabla\boldsymbol{\cdot}(\mathrm{v}\circ\mathrm{w})=\operatorname{tr}\big((\boldsymbol\nabla\mathrm{v}\circ\mathrm{w})\boldsymbol\nabla\mathrm w\big)$$

K.defaoite
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  • Yours is only 49 chars too long. With minor changes, look: This will definitely be computation-heavy, but here are some formulas that could help. Note that the bolded nabla is understood to be the general tensor gradient, also known as the covariant derivative.

    The gradient has a chain rule for $\phi:\Bbb R^n\to\Bbb R$, ${v}:\Bbb R^n\to\Bbb R^n$ $$\nabla(\phi\circ {v})=(\nabla\phi\circ{v})~\boldsymbol\nabla{v}$$ The divergence has one as well, for ${w}:\Bbb R^n\to\Bbb R^n$: $$\nabla\boldsymbol{\cdot}({v}\circ{w})=\operatorname{tr}((\boldsymbol\nabla{v}\circ{w})\boldsymbol\nabla w)$$

     – Calvin Khor Apr 12 '21 at 01:14
  • No words removed; you could have left 2 comments anyway. `This will definitely be computation-heavy, but here are some formulas that could help. Note that the bolded nabla is understood to be the general tensor gradient, also known as the covariant derivative.

    The gradient has a chain rule for $\phi:\Bbb R^n\to\Bbb R$, ${v}:\Bbb R^n\to\Bbb R^n$ $$\nabla(\phi\circ {v})=(\nabla\phi\circ{v})~\boldsymbol\nabla{v}$$ The divergence has one as well, for ${w}:\Bbb R^n\to\Bbb R^n$: $$\nabla\boldsymbol{\cdot}({v}\circ{w})=\operatorname{tr}((\boldsymbol\nabla{v}\circ{w})\boldsymbol\nabla w)$$`

     – Calvin Khor Apr 12 '21 at 01:16
  • Thanks for the input I guess Calvin? – K.defaoite Apr 12 '21 at 03:33
  • It seems my comments are not appreciated; sorry. This Answer is just not "too long for a comment", and I feel unanswered questions that can reasonably expect full answers should not have non-answers as the only Answer. – Calvin Khor Apr 12 '21 at 04:35