My current research has lead to me solving a fractional Laplacian equation on $\mathbb{R}^d$. For Laplace's equation $-\Delta u = f$, I know we can solve it by an integral,

$$ u(x) = \int_{\mathbb{R}^d} \Phi(x-y) f(y) dy $$

where $\Phi$ is the fundamental solution to Laplace's equation. I desire a similar solution for the fractional Laplacian $(-\Delta)^{\alpha/2} u = f$, for $\alpha \in (0,2)$. Is there an integral representation of the solution similar to Laplace's equation? If it is helpful, I am specifically wanting to solve this equation for $\alpha = 1$.

Thank you for taking the time to read this post. Any insight into this is greatly appreciated!

  • 2
    $\begingroup$ You can work with a Fourier transform by noting that $u(k)=e^{\frac{\alpha}{2}\ln k^2}f(k)$. But I think that a proper definition is $-\Delta^\frac{\alpha}{2}u=f$. A proper treatment at $k^2=0$ is needed. $\endgroup$
    – Jon
    Mar 8, 2021 at 15:18

1 Answer 1


The fundamental solution of $(-\Delta )^{\frac \alpha 2} u = f$, $\alpha \in (0,2)$ is given by (up to a scaling constant) the Reisz potential of order $\alpha$: $$ \Phi_\alpha(x) = \frac 1 {\vert x \vert^{n-\alpha} } .$$

If you would like to read further, I would recommend looking at Fractional Thoughts by Nicola Garofalo and references therein. In particular, there is a section on the fundamental solution.

  • $\begingroup$ It might be important to notice that for the case $\alpha=1$ and $d=1$ this should be seen as a logarithm, also, there is a constant missing, but this is minor. $\endgroup$
    – Kernel
    Apr 26, 2021 at 7:17
  • 1
    $\begingroup$ @Kernal Yes, you're right there is a multiplicative constant which I've neglected and I also implicitly assumed $n>1$. Both of these issues are addressed in Fractional Thoughts, see Theorem 8.4 and Remark 8.7. $\endgroup$
    – JackT
    Apr 26, 2021 at 7:22

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