Consider the $d$-dimensional Coulomb "kernel" defined by: \begin{equation} x \in \mathbb{R}^{d} \mapsto g(x):=\left\{\begin{array}{ll} \log \frac{1}{|x|} & \text { if } d=2 \\ \frac{1}{|x|^{d-2}} & \text { if } d \geq 3 \end{array}\right. \end{equation} It is known that $g$ satisfies, in the sense of Swchartz distributions, the Poisson equation: \begin{equation} \Delta g = - c_d \delta_{0} \end{equation} where $c_d$ is some constant, $\delta_{0}$ the dirac mass at the origin and $\Delta$ the Laplace operator.

Sometimes the Coulomb kernel is "softened" because it has a singularity and one considers the "Plummer kernel" defined by: \begin{equation} p_{\epsilon}(x) = \frac{1}{\left(\sqrt{|x|^{2}+\epsilon^{2}}\right)^{d-2}} \end{equation} for $\epsilon >0$.

My question is: does this kernel also satisfy some Poisson like equation ? More generally are there some known PDE for which this kernel is the solution ?

  • 2
    $\begingroup$ Every function satisfies a Poisson equation with some RHS. For your function, the RHS is a smooth function which tends to $\delta$ when $\epsilon\to 0$. $\endgroup$ Jul 28, 2021 at 13:55
  • $\begingroup$ Thanks ! Yes indeed but I was wondering if we could characterize it more precisely (like in the case $\epsilon = 0$) $\endgroup$ Jul 28, 2021 at 17:33

1 Answer 1


The radial Laplacian in spherical coordinates is $$\Delta=\frac{d^2}{dr^2}+\frac{n-1}{r}\frac{d}{dr}.$$ By differentiating we obtain $$\Delta p_\epsilon=-\epsilon^2n(n-2)(r^2+\epsilon^2)^{-1-n/2}.$$ This RHS tends to $-c_n\delta_0$ when $\epsilon\to 0$.

  • $\begingroup$ Thank you very much ! $\endgroup$ Jul 29, 2021 at 7:00

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