explicit computation of fractional Laplacian of a function

Question

For $x\in\mathbb R$ let $$ u(x)=\begin{cases} |x|^{2s-1}-1 &\mbox{if } |x|>1,\\ 0 & \mbox{otherwise}. \end{cases} $$

Is it possible to calculate explicitly the fractional Laplacian $(-\Delta)^{s} u(x)$ for a fixed $s\in (0, 1/2)$?

Answer 1

Yes, it is, as long as you are OK with special functions. See Corollary 3(ii) in my paper with B. Dyda and A. Kuznetsov titled Fractional Laplace operator and Meijer G-function, DOI:10.1007/s00365-016-9336-4. Here one needs to apply this result twice, with $d = 1$, $\alpha = 2 s$, $l = 0$, $\sigma = 0$ and $\rho$ equal to either $s-\tfrac12$ or $0$.

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Edit: To clarify, the result mentioned above implies that $$ (-\Delta)^s \bigl[|x|^{2\rho} (|x|^2-1)_+^\sigma\bigr] = 2^{2s} \Gamma(1+\sigma) G^{1,2}_{3,3}\biggl(\begin{array}{ccc}\tfrac12-s & 1+\rho+\sigma-s & -s \\ 0 & \rho - s & \tfrac12 \end{array} \; \bigg\vert \; |x|^2\biggr) . $$ If we set $\sigma = 0$, we find that $$ (-\Delta)^s \bigl[|x|^{2\rho} \mathbb{1}_{|x|>1}\bigr] = 2^{2s} G^{1,2}_{3,3}\biggl(\begin{array}{ccc}\tfrac12-s & 1+\rho-s & -s \\ 0 & \rho - s & \tfrac12 \end{array} \; \bigg\vert \; |x|^2\biggr) . $$

For $\rho = 0$, we get $$ (-\Delta)^s \bigl[\mathbb{1}_{|x|>1}\bigr] = 2^{2s} G^{1,2}_{3,3}\biggl(\begin{array}{ccc}\tfrac12-s & 1-s & -s \\ 0 & -s & \tfrac12 \end{array} \; \bigg\vert \; |x|^2\biggr) , $$ which simplifies to $$ \frac{\Gamma(2s) \sin(\pi s)}{\pi} ((|x|^2-1)^{-2s} - (|x|^2 + 1)^{-2s}) $$ when $|x| > 1$ (and to a similar expression when $|x| < 1$).

On the other hand, for $\rho = s-\tfrac12$ we find that $$ (-\Delta)^s \bigl[|x|^{2s-1} \mathbb{1}_{|x|>1}\bigr] = 2^{2s} G^{1,2}_{3,3}\biggl(\begin{array}{ccc}\tfrac12-s & \tfrac12 & -s \\ 0 & -\tfrac12 & \tfrac12 \end{array} \; \bigg\vert \; |x|^2\biggr) = 2^{2s} G^{1,1}_{2,2}\biggl(\begin{array}{ccc}\tfrac12-s & -s \\ 0 & -\tfrac12 \end{array} \; \bigg\vert \; |x|^2\biggr) , $$ which, for $|x| > 1$, is equal to $$ \frac{\Gamma(2s) \sin(\pi s)}{\pi |x|^2} ((|x|^2-1)^{-2s} + (|x|^2 + 1)^{-2s}) . $$

The result is found by subtracting the above two expressions: $$ (-\Delta)^s u(x) = \frac{\Gamma(2s) \sin(\pi s)}{\pi |x|^2} ((|x|^2 + 1)^{1-2s} - (|x|^2-1)^{1-2s}) $$ for $|x| > 1$ (unless I made an error when manipulating these terrible expressions).

Of course, this can be found by direct integration, too. The above method is more involved, but quite general. For example, it carries over to higher dimensions.