The question

I consider the Laplacian $\Delta = \partial_1^2 + \partial_2^2 + \partial_3^2$ in $\mathbb{R}^3$. By the "standard" fundamental solution of the Laplacian, I mean the function

$$ \displaystyle E(x) = C|x|^{-1} $$

where $C$ is some normalization constant. I would like to know if one can construct a fundamental solution of the Laplacian that decays faster than any rational function at infinity. I have an idea how one could maybe construct this, but I'm not sure if my idea is correct as the idea of a suspicious fundamental solution seems highly suspicious to me. In the following, I sketch my idea:

An attempt at an answer

I already know that one can construct a rapidly decaying parametrix $F$ by multiplying $\hat E \sim |x|^{-2}$ with a cutoff function $\chi$ supported in the complement of the unit ball and then taking the inverse Fourier transform. Then, one has $$x^\alpha F(x) = \int D^\alpha(\chi\hat E)(\xi)e^{ix\xi}\text{d}\xi $$ and the right hand side is absolutely convergent for large $\alpha$. This proves the decay at infinity.

I want to modify this construction in order to turn the parametrix $F$ into a proper fundamental solution. To do this, I want to use a simplified version of the construction by Hörmander (see "Distribution" by Duistermaat and Kolk, Theorem 17.11). Letting $\eta$ be a unit vector, I set

$$ G(x) = (2\pi)^{-n} \int \frac{1- \chi(\xi)}{2\pi i}\int_{|z| = 100} \frac{e^{ix\cdot(\xi + z \eta)}}{|\xi + z\eta|^2}\frac{\text{d}z}{z}\text{d}\xi\,.$$

With Cauchy's integral theorem, one can show that $$\tilde E = F + G$$ is indeed a fundamental solution. The idea here is that we avoid the singularity at $0$ by taking a contour integral around it. It seems as if one can use the same trick that one used for $F$ to show the rapid decay of $G$. This suggest that one can indeed find a rapidly deacying fundamental solution.


1 Answer 1


No. You want the Fourier transform to satisfy $-|\xi|^2 \widehat{E}(\xi)=1$, so $\widehat{E}=-1/|\xi|^2 + \widehat{F}$, with $\textrm{supp}\:\widehat{F}=\{0\}$, but this says that $\widehat{F}$ is a linear combination of $\delta$ and its derivatives, so $F$ is a polynomial.

  • $\begingroup$ So this means that choosing any other fundamental solution than the standard one can only make the decay worse. Not what I hoped for but interesting. Thank you for your answer! $\endgroup$
    – ClemensB
    Feb 14, 2017 at 15:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .