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Prove the Poisson formula for a general ball $B_R(x_0)\subset\mathbb{R}^n$ $$ u(x)=\frac{1}{\sigma_n R}\int_{S_R(x_0)}\frac{R^2-\lVert x-x_0\rVert^2}{\lVert\xi-x\rVert^n}\varphi(\xi)\, d\sigma\text{ for }x\in B_R(x_0) $$ by starting from the Poisson formula of the unit ball $B_1(0)\subset\mathbb{R}^n$ $$ u(x)=\frac{1}{\sigma_n}\int_{S_1(0)}\frac{1-\lVert x\rVert^2}{\lVert \xi-x\rVert^n}\varphi(\xi)\, d\sigma\text{ for }x\in B_1(0). (*) $$

Edit:

I do not come along with this, because I do not exactly know what to do resp. what to start with.

My first idea is, to consider the coordinate transformation

$$ \psi\colon\mathbb{R}^n\to\mathbb{R}^n, (x_1,x_2,\ldots,x_n)\longmapsto (Rx_1+x_1^0,\ldots,Rx_n+x_n^0) $$ Now, I would simply put that in (*), i.e.

$$ u(\psi(x))=\frac{1}{\sigma_n}\int_{S_1(0)}\frac{1-\lVert\psi(x)\rVert^2}{\lVert\xi-\psi(x)\rVert^n}\varphi(\xi)\, d\sigma $$

Now I have to integrate by substitution I think.

How can I do so? Do I first have to write that integral in n-dim. ball coordinates?

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    $\begingroup$ Surely this is just a simple (linear) change of variable in the integral? $\endgroup$ – Old John Nov 14 '13 at 22:37
  • $\begingroup$ I do not know but the work sheet says that this is to show. $\endgroup$ – math12 Nov 14 '13 at 22:43
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    $\begingroup$ If the worksheet says you are to deduce the general result form the special case of the unit ball, then I think my previous hint shows you the way. $\endgroup$ – Old John Nov 14 '13 at 22:45
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You are substituting in the wrong place. $\varphi$ is defined on $S_R(x_0)$, so when using the Poisson integral for the unit ball, you must have $\varphi(\psi(\zeta))$ (with $\zeta\in S_1(0)$) in the integrand. And since that gives you a harmonic function in the unit ball, while you want a harmonic function in $B_R(x_0)$, the integral over the unit sphere gives you $u(\psi(y))$ for $y \in B_1(0)$, so

$$u(\psi(y)) = \frac{1}{\sigma_n} \int_{S_1(0)} \frac{1-\lVert y\rVert^2}{\lVert \zeta - y\rVert^n}\varphi(\psi(\zeta))\, d\sigma(\zeta).$$

Now writing $x = \psi(y)$ and $\xi = \psi(\zeta)$ gives you a relatively simple transformation to the desired form, since $\lVert \xi-x\rVert$ can be easily expressed using $\zeta-y$, and $\lVert x-x_0\rVert$ equally easily using $y$.

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  • $\begingroup$ I understand how the formula for $u(\psi(y))$ arises. But now I do not understand how to continue with your transformations $x=\psi(y)$ and $\xi=\psi(\zeta)$. $\endgroup$ – math12 Nov 15 '13 at 14:15
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    $\begingroup$ $\lVert \xi-x\rVert^n = R^n\lVert\zeta-y\rVert^n$, $R^2-\lVert x-x_0\rVert^2 = R^2(1-\lVert y\rVert^2)$, $d\sigma(\xi) = R^{n-1}d\sigma(\zeta)$. Just rewrite and count factors of $R$. $\endgroup$ – Daniel Fischer Nov 15 '13 at 14:23
  • $\begingroup$ And one more question, sorry: Why $\lVert y\rVert^2$ in the integrand?.. and not $\lVert\psi(y)\rVert^2$? $\endgroup$ – math12 Nov 15 '13 at 14:24
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    $\begingroup$ Yes. The integral is over an $n-1$-dimensional hypersurface, so we don't need the determinant of the $\mathbb{R}^n\to\mathbb{R}^n$ map, but the $n-1$-dimensional version. Whichever way that was done in your course (there are many). $\endgroup$ – Daniel Fischer Nov 15 '13 at 15:35
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    $\begingroup$ How exactly one applies the transformation formula depends on how you have defined $\int_{S_R(x_0)} H(\xi)\,d\sigma$. In some interpretation, you don't even use the transformation formula. If you view the integral as an integral over a rectangle $[0,2\pi]\times[0,\pi]^{n-2}\subset \mathbb{R}^{n-1}$ with a parametrisation of the sphere in spherical coordinates, you don't really use the transformation formula, you're just rewriting the integrands to see that both integrals are in fact the same. If you view the integral over the sphere as an integral with respect to an intrinsic measure on the $\endgroup$ – Daniel Fischer Nov 15 '13 at 16:29

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