# explicit computation of fractional Laplacian of a function

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 $$s\in (0, 1/2)$$?

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 $$0$$.

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.

• @GabS: I guess you can divide by $(2s-1)$ and pass to the limit as $s \to \tfrac12$, no? (The $s$ in the definition of $u$ need not be the same as the exponent $s$ in $(-\Delta)^s$, if this matters.) – Mateusz Kwaśnicki Aug 30 at 20:38
• @ Mateusz Kwaśnicki Yes, I should have been precise. I would like to calculate $(-\Delta)^{1/2}$. Here $u$ look like the fundamental solution except in a neighbourhood of zero – GabS Aug 31 at 8:02
• @ Mateusz Kwaśnicki I believe in both the cases the answer is zero as in one dimension, these are fundamental solution away from zero. – GabS Sep 2 at 12:07
• @GabS: No, it is definitely not zero, unless $s = 1$. Why do you expect it to be zero? – Mateusz Kwaśnicki Sep 2 at 22:32
• @ Mateusz Kwaśnicki The calculation is $(-\Delta)^s u(x)= 2^{2s} \frac{\Gamma(s) \Gamma((1/2)}{\Gamma(0)\Gamma ((1-2s)/2) } |x|^{-1}= 2^{2s} \frac{\Gamma(s) \Gamma((1/2)}{\Gamma (1-2s/2) } |x|^{-1}.$ So I am wrong. – GabS Sep 3 at 9:52