3
$\begingroup$

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.

$\endgroup$

1 Answer 1

7
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.