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This is a simple question whose answer was probably known to Poisson, but I was not able to find it by searching. I need explicit formulas for the Newtonian potential of the unit ball $\mathbb{B}^n$ and the unit sphere $\mathbb{S}^{n-1}$, with explicit constants. So I did the computation myself but I still need to check the constants. I thought MO would be a good place for this, since having the precise formulas here could be good as a reference. Feel free to close the question if it is too restricted in scope.

Let $E_2=\frac 1{2\pi}\log|x|$ and $E_n=\frac1{\omega_n(2-n)}|x|^{2-n}$ for $n\ge3$, where $\omega_n$ is the surface of the unit sphere.I denote with $1_{A}$ the characteristic function of the set $A$. Here is what I get for the unit ball ($n\ge3$): \begin{equation} E_{n}*1_{\mathbb{B}^{n}}(x)= \begin{cases} \frac{1}{2(2-n)}+\frac{1}{2n}|x|^{2} &\text{if $ |x|\le1 $,}\\ \frac{1}{n(2-n)}|x|^{2-n} &\text{if $ |x|\ge1 $} \end{cases} \end{equation} \begin{equation} E_{2}*1_{\mathbb{B}^{2}}(x)= \begin{cases} -\frac{1}{4}+\frac{1}{4}|x|^{2} &\text{if $ |x|\le1 $,}\\ \frac{1}{2}\log|x| &\text{if $ |x|\ge1 $.} \end{cases} \end{equation} and for the sphere ($\sigma_{\mathbb{S}^{n-1}}$ is the surface measure of the unit sphere): \begin{equation} E_{n}*\sigma_{\mathbb{S}^{n-1}}(x)= \frac{1}{2-n} \begin{cases} 1 &\text{if $ |x|\le1 $,}\\ |x|^{2-n} &\text{if $ |x|\ge1 $.} \end{cases} \end{equation} \begin{equation} E_{2}*\sigma_{\mathbb{S}^{1}}(x)=(\log|x|)^{+}. \end{equation} Are these formulas correct? of course, a reference where this calculation is done would be enough.

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  • $\begingroup$ I don't understand your equations; you have $E_n$ on the left-hand-side, which contains a factor $|x|^{2-n}$, and then on the right-hand-side you have a different dependence on $x$ for $|x|<1$. $\endgroup$
    – Carlo Beenakker
    Mar 26 at 15:06
  • $\begingroup$ On the left I do not have $E_n$ but the convolution of $E_n$ with the characteristic function of the unit ball, computed at point $x$ $\endgroup$
    – Piero D'Ancona
    Mar 26 at 15:13

1 Answer 1

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$\newcommand\si\sigma$Below is a verification of your results, with integration done in Mathematica.

Here I used the formulas
$$(E_n * \si_{aS^{n-1}})(x)=a(E_n * \si_{S^{n-1}})(x/a) \tag{1}\label{1} $$ for $n\ge3$ and real $a>0$, $$(E_n * \si_{aS^{n-1}})(x)=a(E_n * \si_{S^{n-1}})(x/a)+a\ln a \tag{2}\label{2} $$ for $n=2$ and real $a>0$, and $$(E_n * 1_{B^n})(x)=\int_0^1 da\,(E_n * \si_{aS^{n-1}})(x). \tag{3}\label{3} $$

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  • $\begingroup$ Great. What is the purpose of the additiona parameter? $\endgroup$
    – Piero D'Ancona
    Mar 26 at 16:20
  • $\begingroup$ @PieroD'Ancona : I am glad you liked the answer. But what do you mean by "the additiona parameter"? $\endgroup$
    – Iosif Pinelis
    Mar 26 at 16:21
  • $\begingroup$ I mean the number $a$ $\endgroup$
    – Piero D'Ancona
    Mar 26 at 16:44
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    $\begingroup$ @PieroD'Ancona : I used $a$ to get $(E_n * 1_{B^n})(x)$ simply by integrating $\,(E_n * \sigma_{aS^{n-1}})(x)$. $\endgroup$
    – Iosif Pinelis
    Mar 26 at 16:52
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    $\begingroup$ Thisis a shortcut to avoid the explicit integration. Assume $n \geq 3$. Then $u=E_n*\sigma$ is radial and harmonic for $|x|<1$ and for $|x| >1$. Moreover it tends to $0$ at infinity and it is continuous. Then $u(x)=c$ for $|x| \leq 1$ and $u(x)=c|x|^{2-n}$ for $|x| \geq 1$. To compute $c=1/(2-n)$ it is sufficient to compute $u(0)$ which is elementary. $\endgroup$
    – Giorgio Metafune
    Mar 26 at 18:52

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