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I'm attempting to solve a simple Dirichlet problem on the fractional Laplacian with boundary conditions:

$r^{+}(\nabla^s) v = f$

where $0 \leq s \leq 1/2$, $v$ is zero outside of $[0,1]$, $r^{+}$ restricts a function to $[0,1]$, and $f:[0,1] \rightarrow \mathbb{R}$ with $f(t) = t^{-s}$ or more generally, $f(t) = t^{k}$ for some fixed value $k$.

Most references I can find concern themselves with proving the regularity of solutions to the fractional Laplacian equation. Is there a simple way to solve this equation? I've looked in references such as: https://arxiv.org/pdf/1712.01196.pdf which purport to solve these equations, but I could not find a place in the text where a method for solving such equations is given.

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  • $\begingroup$ Not sure if I understand the question correctly; the Green's function for the fractional Laplacian in a ball (and in particular, for an integral) is known since late 50's, and in fact it goes back to Riesz's 1938 paper. I recently wrote a survey on the frational Laplacian (M. Kwaśnicki, Fractional Laplace Operator and its Properties, in: A. Kochubei, Y. Luchko, Handbook of Fractional Calculus with Applications. Volume 1: Basic Theory, De Gruyter Reference, De Gruyter, Berlin, 2019), see Theorem 3.4 there. I can copy-and-paste the formula here if this is what you are looking for. $\endgroup$ – Mateusz Kwaśnicki Mar 31 '19 at 9:07
  • $\begingroup$ Hi, the Green's function for the Fractional Laplacian is what I'm looking for. From what I know, Green's function for 1 dimensional ball is written in terms of a definite integral with the Poisson kernel. (Reference: arxiv.org/pdf/1502.06468.pdf, Definition 1.9, or web.ma.utexas.edu/mediawiki/index.php/…). A closed form without definite integrals would be useful for my application. Also, I can't find a copy of your survey online. Pasting the formula you mention would be great, especially if it also works for -1/2 < s < 0. $\endgroup$ – Timothy Chu Mar 31 '19 at 16:07
  • $\begingroup$ Well, there's no closed-form expression for the Green function, and one has to live with it. What I meant is given as Thm 3.1 in Claudia Bucur's paper you mentioned. Another expression involves the hypergeometric function $_2F_1$. For your particular $f$ there might be a simpler expression for the solution, as this is essentially the Mellin transform of the Green function. I once worked with the fractional Laplacian and Mellin transforms (here); I do not remember anything similar to your question, though. $\endgroup$ – Mateusz Kwaśnicki Mar 31 '19 at 16:56
  • $\begingroup$ One more comment: there is a huge literature on the one-dimensional case, that I do not know at all. In this case the fractional Laplacian is the composition of two one-sided fractional derivatives, and this often helps. $\endgroup$ – Mateusz Kwaśnicki Mar 31 '19 at 16:58
  • $\begingroup$ It just came to my mind that one can give an explicit solution in an integral form. I wrote up an answer below. I am in a rush, so please excuse me all typos and errors. $\endgroup$ – Mateusz Kwaśnicki Apr 1 '19 at 8:39
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Here is a solution in an integral form. I suppose it can be written in terms of hypergeometric functions (or maybe Meijer G-functions), but I did not attempt to do that. Once this is done, extension to general $\Re k > -1$ should follow by analytic continuation.

For $k \in \mathbb{C}$ let $$ f_k(x) = \begin{cases} x^k & \text{for $x > 0$,} \\ 0 & \text{otherwise.} \end{cases} $$ Let $L = (-\Delta)^{s/2} = |\nabla|^s$ denote the fractional Laplacian.

Lemma: If $-1 < \Re k < s$, we have $$ L f_k(x) = \begin{cases} a_k x^{k - s} & \text{for $x > 0$,} \\ b_k (-x)^{k - s} & \text{for $x < 0$,} \end{cases} $$ where $$ a_k = 2^{s-1} \frac{\Gamma(\tfrac{k+1}{2})\Gamma(\tfrac{k}{2})}{\Gamma(\tfrac{k+1-s}{2})\Gamma(-\tfrac{k}{2})} + 2^{s-1} \frac{\Gamma(\tfrac{k}{2}+1)\Gamma(\tfrac{k-1}{2})}{\Gamma(\tfrac{k-s}{2}+1)\Gamma(-\tfrac{k-1}{2})} $$ and $$ a_k = 2^{s-1} \frac{\Gamma(\tfrac{k+1}{2})\Gamma(\tfrac{k}{2})}{\Gamma(\tfrac{k+1-s}{2})\Gamma(-\tfrac{k}{2})} - 2^{s-1} \frac{\Gamma(\tfrac{k}{2}+1)\Gamma(\tfrac{k-1}{2})}{\Gamma(\tfrac{k-s}{2}+1)\Gamma(-\tfrac{k-1}{2})} \, . $$

Proof: In $\mathbb{R}^n$, it is known that $$ L[|x|^k] = 2^s \frac{\Gamma(\tfrac{k+n}{2})\Gamma(\tfrac{k}{2})}{\Gamma(\tfrac{k+n-s}{2})\Gamma(-\tfrac{k}{2})} \, |x|^{k - s} $$ and $$ L[|x|^{k - 1} x_1] = 2^s \frac{\Gamma(\tfrac{(k-1)+(n+2)}{2})\Gamma(\tfrac{k-1}{2})}{\Gamma(\tfrac{(k-1)+(n+2)-s}{2})\Gamma(-\tfrac{k-1}{2})} \, |x|^{(k - 1) - s} x_1 ; $$ the first identity is quite standard, the latter one is also likely well-known, and both follow, for example, from Theorem 1 in my paper Fractional Laplace operator and Meijer G-function with Bartłomiej Dyda and Alexey Kuznetsov, or Theorem 3.6 in my survey Fractional Laplace Operator and its Properties. Taking $n = 1$ and combining both identities, we get the desired result. $\square$

Corollary: Let $-1 < \Re k < s$, $$ v_k(x) = \frac{1}{a_{k+s}} \, f_{k+s}(x) - \frac{1}{a_{k+s} \Gamma(1 + \tfrac{s}{2}) |\Gamma(-\tfrac{s}{2})|} \int_1^\infty \frac{(x - x^2)^{s/2}}{(y^2 - y)^{s/2} (y - x)} \, f_{k+s}(y) dy $$ for $x \in (0, 1)$, and $v_k(x) = 0$ otherwise. Then $L v_k(x) = f_k(x)$ for $x \in (0, 1)$.

Proof: Note that $v_k$ is a difference of $f_{k+s} / a_{k+s}$ and an $L$-harmonic function in $(0, 1)$ (the integral term in the definition is just the $L$-harmonic reduction of $f_{k+s} / a_{k+s}$, that is, the integral of $f_{k+s} / a_{k+s}$ with respect to the Poisson kernel for $L$). Thus, $L v_k = L(f_{k+s}) / a_{k+s}) = f_k$ in $(0, 1)$ by our lemma. $\square$

The above works for any $s$ such that $\Re s > -1$, I suppose.

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  • $\begingroup$ Thanks for this in-depth solution! I'm slow to respond here since it will take me some time to verify all the details, but if this checks out I'll happily accept this answer! $\endgroup$ – Timothy Chu Apr 4 '19 at 6:15

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