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.

  • 3
    $\begingroup$ Calculation. But you don't have $\frac{\partial^2}{\partial x_i^2}P(x,\xi) = 0$, only the sum results in $0$. $\endgroup$ – Daniel Fischer Nov 15 '13 at 18:21
  • 1
    $\begingroup$ Well, welcome to the research level. Myabe you can get a easy answer here and now, but later in your research (if you pretend to do so), probably you will have to start a lot of exhausting calculations. $\endgroup$ – Tomás Nov 15 '13 at 18:23
  • $\begingroup$ Okay, then I start with trying to calculate $\frac{\partial^2}{\partial x_i^2}P(x,\xi)$ for an arbitrary $0\leq i\leq n$, looking, if I then can see that the sum is $0$. $\endgroup$ – math12 Nov 15 '13 at 18:24
  • $\begingroup$ By the way, why do you write $B_1(0)\setminus \{0\}$? The function is harmonic in the open unit ball; there is nothing special about its behavior at the center of the ball. $\endgroup$ – user103402 Nov 17 '13 at 4:10
  • $\begingroup$ Maybe it is a mistake on the worksheet and it is meant $B_1(0)\setminus\left\{\xi\right\}$. $\endgroup$ – math12 Nov 17 '13 at 10:13

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 $\|x-\xi\|^{ n+2}$: $$ \begin{split} \|x-\xi\|^{ n+2}\Delta(uv) &= -2n \|x-\xi\|^{2} + (1-\|x\|^2) 2n +4nx \cdot (x-\xi) \\ & = 0 \end{split} $$

  • $\begingroup$ I used to do things like $(\|x\|^2)^{n/2}$ for computing gradients, but after a while writing $\nabla \|x\| = x/\|x\|$ became easier than doing that. $\endgroup$ – user103402 Nov 16 '13 at 18:47

It looks like your derivatives are correct, and then you get $\Delta P = 0$ as you should, if you sum and group terms to annihilate each other.

Here's a trick to make such calculations more manageable: give (meaningless) short names to the building blocks. For example, if we name

$$\begin{align} N &= 1 - \lVert x\rVert^2 = \lVert \xi\rVert^2 - \lVert x\rVert^2,\\ S &= \lVert x-\xi\rVert^2, \end{align}$$

we get $P = N\cdot S^{-n/2}$, and

$$\begin{gather} \partial_i P = (\partial_i N)S^{-n/2} - \frac{n}{2} N(\partial_i S)S^{-(n/2+1)}\\ \partial_i^2 P = (\partial_i^2 N)S^{-n/2} -n(\partial_i N)(\partial_i S)S^{-(n/2+1)} - \frac{n}{2}N(\partial_i^2 S)S^{-(n/2+1)} + \frac{n(n+2)}{4}N(\partial_i S)^2S^{-(n/2+2)}\\ S^{n/2+1}\partial_i^2 P = (\partial_i^2 N)S - n(\partial_i N)(\partial_i S) - \frac{n}{2}N(\partial_i^2 S) + \frac{n(n+2)}{4}N(\partial_i S)^2 S^{-1}. \end{gather}$$

Now we look at the partial derivatives of $N$ and $S$ and the needed sums,

$$\partial_i N = -2x_i;\; \partial_i^2 N = -2;\; \partial_i S = 2(x_i-\xi_i);\; \partial_i^2 S = 2;$$

which yields

$$\begin{align} \sum_i \partial_i^2 N &= -2n\\ \sum_i (\partial_i N)(\partial_i S) &= -4\lVert x\rVert^2 +4\langle x,\xi\rangle\\ \sum_i \partial_i^2 S &= 2n\\ \sum_i (\partial_i S)^2 &= 4S \end{align}$$

and hence

$$\begin{align} S^{-(n/2+1)}\Delta P &= -2nS + 4n\lVert x\rVert^2 - 4n\langle x,\xi\rangle -n^2N + n(n+2)N\\ &= 2nN + 4n\lVert x\rVert^2 - 4n\langle x,\xi\rangle - 2nS\\ \frac{S^{-(n/2+1)}\Delta P}{2n} &= \lVert \xi\rVert^2 - \lVert x\rVert^2 + 2\lVert x\rVert^2 - 2\langle x,\xi\rangle - \lVert x-\xi\rVert^2\\ &= \lVert \xi-x\rVert^2 - \lVert x-\xi\rVert^2\\ &= 0. \end{align}$$

If we start from

$$P_{x_i x_i}=\underbrace{-2\lVert x-\xi\rVert^{-n}}_A + \underbrace{4x_in\lVert x-\xi\rVert^{-n-2}(x_i-\xi_i)}_B - \underbrace{n\lVert x-\xi\rVert^{-n-2}(1-\lVert x\rVert^2)}_C - \underbrace{n(x_i-\xi_i)^2(-n-2)\lVert x-\xi\rVert^{-n-4}(1-\lVert x\rVert^2)}_D,$$

summing $A$ produces $-2n\lVert x-\xi\rVert^{-n}$, and summing $B$ leads to $4n\lVert x-\xi\rVert^{-n-2}\langle x,x-\xi\rangle$. Summing $C$ yields, including the sign, $-n^2\lVert x-\xi\rVert^{-n-2}(1-\lVert x\rVert^2)$, and since $\sum_i (x_i-\xi_i)^2 = \lVert x-\xi\rVert^2$, $D$ produces $n(n+2)\lVert x-\xi\rVert^{-n-2}(1-\lVert x\rVert^2)$. Thus

$$\begin{align} C+D &= 2n\lVert x-\xi\rVert^{-n-2}(1-\lVert x\rVert^2)\\ A+B &= 2n\lVert x-\xi\rVert^{-n-2}\left(2\langle x,x-\xi\rangle - \langle x-\xi,x-\xi\rangle\right)\\ &= 2n\lVert x-\xi\rVert^{-n-2}\langle x+\xi,x-\xi\rangle\\ &= 2n\lVert x-\xi\rVert^{-n-2}(\lVert x\rVert^2 - \lVert\xi\rVert^2). \end{align}$$

Since $\lVert \xi\rVert^2 = 1$, we have $A+B+C+D = 0$.

  • $\begingroup$ I have one request. Could you please explain to me how from my second derivation there follows $\Delta P=0$? Maybe in chat? By the way: Meanwhile I got again another second derivative. :-) $\endgroup$ – math12 Nov 16 '13 at 18:18
  • $\begingroup$ Adding that to the answer, but I'm typing slow, so it'll take a moment. $\endgroup$ – Daniel Fischer Nov 16 '13 at 18:21
  • $\begingroup$ Oh, I thank you so much! One last question: What do you mean f.e. with $\langle x,x-\xi\rangle$? $\endgroup$ – math12 Nov 16 '13 at 18:39
  • $\begingroup$ The (standard) inner product, $\langle u,v\rangle = \sum_i u_iv_i$. $\endgroup$ – Daniel Fischer Nov 16 '13 at 18:40

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