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There is an exercise in Evans' book (chapter 2, problem #11). Suppose $u:\mathbb{R}^n\to\mathbb{R}$ is harmonic. Show that $\overline{u}(x)=u(\overline{x})|\overline{x}|^{n-2}=u(x/|x|^2)|x|^{2-n}$ is harmonic, where $\overline{x}=x/|x|^2$.

I'm less interested in the result, but I like this problem because it seems to be good practice in vector calculus. I want to improve my understanding of operations in $\mathbb{R}^n$, as my research thus far has been in $\mathbb{R}$.

Anyways, I am rather confused about applying a Laplacian to this function. The hint is to show that $(D_x \overline{x})(D_x \overline{x})^T=|x|^4 I$, which I have completed.

But I want to understand the Laplacian in terms of matrices, products, transposes, traces, etc.

I know that $\Delta(fg)=g\Delta f+f\Delta g+2(Df)\cdot (Dg).$ If $f(x)=u(x/|x|^2)$ and $g(x)=|x|^{2-n}$, then this formula can be applied. But I am a bit lost on calculating $\Delta (u(x/|x|^2))$.

Any insight would be helpful. Thank you.

ProfOak
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  • I don't understand the question? You want an answer which is in terms of matrix multiplication? What happened when you used the brute force definitions of Laplacian in terms of partial derivatives (b/c then you can just use chain rule)? The expression should be "symmetric" in $x_1, \dots, x_n$, in the sense that after calculating the 2nd partial derivative of $u(x/|x|ˆ2)$ with respect to $x_1$, the result with respect to $x_2$ should be the same if you switch the roles of $x_1$ and $x_2$, ..., the result with respect to $x_n$ would be the same if switch roles of $x_1$ and $x_n$. – hasManyStupidQuestions Jul 31 '20 at 02:30
  • This answer to another question looks similar to what I said above, but admittedly I didn't read all or even most of it, so your mileage may vary: https://math.stackexchange.com/a/601785/606791 – hasManyStupidQuestions Jul 31 '20 at 02:32

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