I am working with a fractional Laplacian that is based on it's Fourier transform, namely

$$(-\Box)^{\alpha}f(t) := \int_{-\infty}^\infty d\omega e^{i\omega t} |\omega|^\alpha \int_{-\infty}^\infty d\tau e^{-i\omega\tau}f(\tau),$$ where f(t) is any compatible function or distribution and $\alpha \in \mathbb{R}$. The equation I want to solve is $$(-\Box)^{\alpha}f(t) = A f(t),$$ where $A$ is a constant. The above equation would be finding the eigenfunctions of this fractional Laplacian. If the above equation turns out to be easier for a restricted $\alpha$, then $0 < \alpha < 1$ and $-1 < \alpha < 0$ are of specific interest.

Potentially some useful answers: https://math.stackexchange.com/a/376093/867243 , https://math.stackexchange.com/a/4117943/867243 and https://math.stackexchange.com/a/2072823/867243 .
Where the idea of the last link would be that for those distributions that are homogeneous of degree $\alpha$, we could $|\omega|^\alpha f(\tau) = f(|\omega|\tau)$, which could maybe be useful.

  • $\begingroup$ One usually uses $(\Delta)^{1/2}$ or $|\nabla|$ rather than $\square$ for the fractional Laplacian ... About your question, it seems to me that if $g$ is the Fourier transform of $f$, you are just looking to the solution to $(|y|^\alpha-A)g(y) = 0$ so $g$ is supported on a circle of radius $A^{1/\alpha}$, and the problem being radial I suppose $g$ should be a uniform measure on this circle. And this might be linked to the gradient of the indicator function of the associated ball? $\endgroup$
    – LL 3.14
    Dec 21, 2022 at 0:27


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