$u$ is harmonic, prove that $v(x)=\frac{1}{|x|^{n-2}}\cdot u\left(\frac{x}{|x|^2}\right)$ is harmonic.

Question

$\Omega$ is open set in $\mathbb{R}^3, n\geq 3$. $u:\Omega\to\mathbb{R}$ is a harmonic function. Prove that $v(x)=\frac{1}{|x|^{n-2}}\cdot u\left(\frac{x}{|x|^2}\right)$ is harmonic.
I knew that $\Delta u=0$ since $u$ is a harmonic function, then I tried to directly calculate $\nabla v(x), \Delta v(x)$, but it became too complicated. I also thought of changing variables, let $y=\frac{x}{|x|^2}$, and $v(x)=|y|^{n-2}\cdot u(y)$, but I have no idea what to do next.

Answer 1

Let's introduce $\partial_i=\frac{\partial}{\partial x_i}$. Let also introduce vector $\vec a$ : $a_i=\frac{x^i}{|x|^2}$ and $\nabla_i=\frac{\partial}{\partial a_i}$.

Using summation over repeated indices $$|x|^2=x^2=\delta_{ij}x_ix_j=x_ix_i\,;\,\partial_i x_j =\delta_{ij}\,;\,\partial_ix_i=\delta{ii}=n\,; \,\partial_i\frac{x_j}{|x^2|}=\frac{\delta{ij}}{|x|^2}-\frac{2x_ix_j}{|x|^4}=\frac{x^2\delta_{ij}-2x_ix_j}{x^4}$$

$$\Delta\frac{1}{|x|^{n-2}}\cdot u(\vec a)=\partial_i\partial_i\Bigl(\frac{1}{x^{n-2}}\cdot u(\vec a)\Bigr)=\Delta\frac{1}{|x|^{n-2}}+2\partial_i\Bigl(\frac{1}{x^{n-2}}\Bigr)\partial_i\Bigl(u(\vec a)\Bigr)+ \frac{1}{x^{n-2}}\partial_i\partial_i\Bigl(u(\vec a)\Bigr)$$

$$\Delta\frac{1}{|x|^{n-2}}=0 \,\, \text {- this is a fundamental solution of Laplace equation in}\, R^n$$ (For example, Dirac delta function as a derivative of some function )

Second term

$$2\partial_i\Bigl(\frac{1}{x^{n-2}}\Bigr)\partial_i\Bigl(u(\vec a)\Bigr)=2(2-n)\frac{x_i}{x^{n}}\nabla_ju(\vec a)\partial_i\Bigl(\frac{x_j}{x^2}\Bigr)$$ $$=2(2-n)\frac{x_i}{x^{n}}\frac{x^2\delta_{ij}-2x_ix_j}{x^4}\nabla_ju(\vec a)=\boldsymbol{2(n-2)\frac{x_j}{x^{n+2}}\nabla_ju(\vec a)}$$

Third term (without multiplier $\frac{1}{|x|^{n-2}}$) $$\partial_i\partial_i\Bigl(u(\vec a)\Bigr)=\partial_i\Bigl(\nabla_ju(\vec a)\partial_i\Bigl(\frac{x_j}{x^2}\Bigr)\Bigr)=\Bigl(\nabla_k\nabla_ju(\vec a)\Bigr)\partial_i\Bigl(\frac{x_k}{x^2}\Bigr)\partial_i\Bigl(\frac{x_j}{x^2}\Bigr)+\nabla_ju(\vec a)\partial_i\partial_i\Bigl(\frac{x_j}{x^2}\Bigr)$$ Where $$\Bigl(\nabla_k\nabla_ju(\vec a)\Bigr)\partial_i\Bigl(\frac{x_k}{x^2}\Bigr)\partial_i\Bigl(\frac{x_j}{x^2}\Bigr)=\Bigl(\nabla_k\nabla_ju(\vec a)\Bigr)\frac{x^2\delta_{ik}-2x_ix_k}{x^4}\frac{x^2\delta_{ij}-2x_ix_j}{x^4}$$ $$=\Bigl(\nabla_k\nabla_ju(\vec a)\Bigr)\frac{x^4\delta_{kj}-4x^2x_kx_j+4x^2x_kx_j}{x^8}=\Bigl(\nabla_k\nabla_ju(\vec a)\Bigr)\frac{\delta_{kj}}{x^4}$$ But $$\nabla_k\nabla_ju(\vec a)\delta_{kj}=\Delta_a u(\vec a)=0$$ So, the third term (without multiplier $\frac{1}{|x|^{n-2}}\,$) is $$\nabla_ju(\vec a)\partial_i\partial_i\Bigl(\frac{x_j}{x^2}\Bigr)=\nabla_ju(\vec a)\partial_i\Bigl(\frac{\delta_{ij}}{x^2}-\frac{2x_ix_j}{x^4}\Bigr)=\nabla_ju(\vec a)\Bigl(\frac{-2x_i\delta_{ij}}{x^4}+\frac{8x_ix_ix_j}{x^6}-\frac{2\delta_{ii}x_j}{x^4}-\frac{2x_i\delta_{ij}}{x^4}\Bigr)$$ $$=\nabla_ju(\vec a)\Bigl(\frac{4x_j}{x^4}-\frac{2n x_j}{x^4}\Bigr)$$

Taking all together, the third term is $$\boldsymbol{-2(n-2)\frac{x_j}{x^{n+2}}\nabla_ju(\vec a)}$$ ...and it cancels the second term.

Thus,

$$\Delta \frac{1}{|x|^{n-2}}\cdot u\left(\frac{x}{|x|^2}\right)=0$$