# 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. Mar 31, 2019 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. Mar 31, 2019 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. Mar 31, 2019 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. Mar 31, 2019 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. Apr 1, 2019 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! Apr 4, 2019 at 6:15