# On solving the fractional Laplacian

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!

• 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.
– Jon
Mar 8, 2021 at 15:18

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 $$\alpha$$: $$\Phi_\alpha(x) = \frac 1 {\vert x \vert^{n-\alpha} } .$$
• 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. Apr 26, 2021 at 7:17
• @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. Apr 26, 2021 at 7:22