Fractional Laplacian of $(a-x)_+^\alpha$ in $(0,1)$

Question

How can I compute the spectral fractional Laplacian of $(a-x)_+^\alpha$ on $\Omega = (0,1)$? Here the operator is defined as $$(-\Delta)^s u = c_{N,s} \int_0^\infty (e^{t\Delta_N}u(x) - u(x)) t^{-1 - s} dt ,$$ where $e^{t\Delta_N}u$ is the solution of the heat equation with homogeneous Neumann boundary conditions on $\Omega$.

Answer 1

The operator $(-∆)^s$ is the Fourier multiplier $\vert\xi\vert^{2s}$ and thus deserves to be denoted by $\vert D\vert^{2s}$. Now the distribution $u_\alpha(x)=x_+^\alpha$ is homogeneous with degree $\alpha$ and thus has a Fourier transform homogeneous with degree $-1-\alpha$. We have in fact $$ \hat{u}_\alpha(\xi)=c_\alpha (\xi-i0)^{-1-\alpha}. $$ You are interested in $v_\alpha(x)=u_\alpha(x-t)$ so that $$ \vert\xi\vert^{2s}\hat{v}_\alpha(\xi)=e^{-it \xi}c_\alpha (\xi-i0)^{-1-\alpha}\vert\xi\vert^{2s}, $$ and $ (\vert D\vert^{2s} v_\alpha)(x)=\int e^{i(x-t) \xi}c'_\alpha (\xi-i0)^{-1-\alpha}\vert\xi\vert^{2s}d\xi =(\vert D\vert^{2s} u_\alpha)(x-t). $ Translations commute with derivations, including fractional versions.

Answer 2

Let us denote the corresponding eigensystem $\{\lambda_n,\phi_n\}$. If $u(x)=\sum\limits_nu_n\phi_n(x)$, then we have $u(x,t)=\sum\limits_n\exp(-\lambda_n t)u_n\phi_n(x)$. Since $\int\limits^\infty_0(\exp(-\lambda_n t)-1)t^{-1-s}=-\lambda^s_n\frac{\Gamma(1-s)}{s}$ we obtain $$ \text{R.H.I.}=-C_{s,N}\frac{\Gamma(1-s)}{s}\sum\limits_n\lambda^s_nu_n\phi_n(x)$$