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.



Solving Fractional Laplacian Equations with Boundary Condition
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*](http://doi.org/10.1007/s00365-016-9336-4) with Bartłomiej Dyda and Alexey Kuznetsov, or Theorem 3.6 in my survey [*Fractional Laplace Operator and its Properties*](http://doi.org/10.1515/9783110571622-007). 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.


