# Solving Fractional Laplacian Equations with Boundary Condition

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.

• 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. – Mateusz Kwaśnicki Mar 31 '19 at 9:07
• 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. – Timothy Chu Mar 31 '19 at 16:07
• 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. – Mateusz Kwaśnicki Mar 31 '19 at 16:56
• 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. – Mateusz Kwaśnicki Mar 31 '19 at 16:58
• 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. – Mateusz Kwaśnicki Apr 1 '19 at 8:39

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.

• 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! – Timothy Chu Apr 4 '19 at 6:15