Show that the Poisson-kernel
$$
P(x,\xi):=\frac{1-\lVert x\rVert^2}{\lVert x-\xi\rVert^n}\text{ for }x\in B_1(0)\subset\mathbb{R}^n, \xi\in S_1(0)
$$
is harmonic as a function in $x$ on $B_1(0)\setminus\left\{0\right\}$.
On my recent worksheet, this task is rated with very much points. So I guess it is either very difficult or requires much calculation.
Am I right that I do have to show (most likely by a rather long calculation) that for any $1\leq i\leq n$
$$
\frac{\partial^2}{\partial x_i^2}P(x,\xi)=\frac{\partial^2}{\partial x_i^2}\left(\frac{1-\sum_{i=1}^{n}x_i^2}{(\sum_{i=1}^{n}(x_i-\xi_i)^2)^{\frac{n}{2}}}\right)=0?
$$
I ask, because I do not want to start this exhausting calculation if there is maybe another way or without having the affirmation that this is constructive.
For example a continuous function that fulfills the mean value property is harmonic. Maybe this is an alternative way here?
My result for the first derivative
Consider any $1\leq i\leq n$. Then my result for $P_{x_i}$ is
$$
P_{x_i}=\frac{-2x_i\lVert x-\xi\rVert^n-(1-\lVert x\rVert^2)\frac{n}{2}\lVert x-\xi\rVert^{n-2}(2x_i-2\xi_i)}{\lVert x-\xi\rVert^{2n}}.
$$
Here I used the quotient rule. Moreover, I used the chain rule to calculate
$$
\frac{\partial}{\partial x_i}(\lVert x-\xi\rVert^n)=\frac{1}{2}n\lVert x-\xi\rVert^{n-2}(2x_i-2\xi_i).
$$
Maybe you can say me if my calculation is correct to this point.
My final result
As the second derivative I get
$$
P_{x_i x_i}=-2\lVert x-\xi\rVert^{-n}+4x_in\lVert x-\xi\rVert^{-n-2}(x_i-\xi_i)-n\lVert x-\xi\rVert^{-n-2}(1-\lVert x\rVert^2)-n(x_i-\xi_i)^2(-n-2)\lVert x-\xi\rVert^{-n-4}(1-\lVert x\rVert^2)
$$
My question is if then
$$
\Delta P=\sum_{i=1}^{n}P_{x_i x_i}=0?
$$
Maybe you can say me if this is correct. Unfortunately I do not see how I can show with that result, that $\Delta P=0$. Maybe I am blind, maybe my result is wrong. I did it again and again and I always get this second derivative. Therefore I hope that you can help me finding the mistake or my error in reasoning.
I am aware of the fact that I probably won't get any help, because it is too much calculation, but maybe someone has pity with me and my effort.
Best Answer
Here is a vectorized approach (as in "look ma, no coordinates!"). Make sure you have the chain and product formulas written down in convenient form (they help with other calculations too): $$\nabla(\varphi(u)) = \varphi'(u) \nabla u\tag{1}$$ $$\Delta(u) = \operatorname{div} \nabla u \tag{2}$$ $$\operatorname{div} u \mathbf F = \nabla u\cdot \mathbf F + u \operatorname{div} \mathbf F\tag{3}$$ $$\Delta(uv) = u\Delta v+v\Delta u+2\nabla u\cdot \nabla v \tag{4}$$
Your function is $uv$ with $u=(1-\|x\|^2)$ and $v=\|x-\xi\|^{-n}$. We have $$\nabla u = -2 x,\quad \Delta u = -2n $$ Using (1): $$ \begin{split} \nabla v &= -n \|x-\xi\|^{-n-1}\nabla \|x-\xi\| \\ &= -n \|x-\xi\|^{-n-1}\frac{x-\xi}{\|x-\xi\|} \\ &= -n \|x-\xi\|^{-n-2}(x-\xi) \end{split} $$ Using (2) and then (3): $$ \begin{split} \Delta v &= -n \operatorname{div} ( \|x-\xi\|^{-n-2}(x-\xi)) \\ &= -n (-n-2) \|x-\xi\|^{-n-3}\frac{x-\xi}{\|x-\xi\|}\cdot (x-\xi) -n \|x-\xi\|^{-n-2} n \\ & =2n\|x-\xi\|^{-n-2} \end{split} $$ Finally, combine the results using (4). For convenience, I multiply the Laplacian by $\|x-\xi\|^{ n+2}$: $$ \begin{split} \|x-\xi\|^{ n+2}\Delta(uv) &= -2n \|x-\xi\|^{2} + (1-\|x\|^2) 2n +4nx \cdot (x-\xi) \\ & = 0 \end{split} $$