Let $s\in (0,1)$, how i can solve the equation: $$ (-\Delta)^su=1,\quad\text{in}\quad(-1,1)?$$ I have no idea, any help would be appreciated.
5
-
$\begingroup$ Is the equation on $\mathbb{R}$ or on $(-1,1)$ and then what are the boundary conditions? $\endgroup$– RaphaelB4Commented Oct 30, 2020 at 7:14
-
$\begingroup$ The equation is on $\mathbb{R}$ and i want that $u=0$ in $\mathbb{R}\setminus(-1,1)$. $\endgroup$– inocCommented Oct 30, 2020 at 7:27
-
2$\begingroup$ With the Fourier transform the equation is $ |k|^{2s}\hat{u}(k)=\frac{2\sin(k)}{k}$ $\endgroup$– RaphaelB4Commented Oct 30, 2020 at 8:17
-
$\begingroup$ Then i have to anti-transform the function $2\sin(k)/k|k|^{2s}$? How i can compute the Fourier anti-transform of this function? Can you give me the details please ? $\endgroup$– inocCommented Oct 30, 2020 at 8:43
-
$\begingroup$ @RaphaelB4 --- the inverse Fourier transform of the function $\hat{u}$ you wrote down is nonzero for $|x|>1$, while we seek a solution $u=0$ for $|x|>1$; the point is that we cannot assume that $\Delta^s u=0$ for $|x|>1$, we only know $\Delta^s u$ for $|x|<1$. $\endgroup$– Carlo BeenakkerCommented Oct 30, 2020 at 11:39
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
The solution to
$$(-\Delta)^s u(x)=1,\;\;-1<x<1,
$$
with $u(x)=0$ for $|x|\geq 1$ is
$$
u(x)=\frac{\sqrt{\pi}(1-x^2)^s }{4^{s}\Gamma(\tfrac{1}{2}+s)\Gamma(1+s)},$$
as follows from the integral definition of the fractional Laplacian.
$$\text{The $n$-dimensional generalization is} \qquad u(x)=\frac{\Gamma(n/2)(1-x^2)^s }{4^{s}\Gamma(\tfrac{n}{2}+s)\Gamma(1+s)}.$$
An early reference for this result is R. Getoor, First passage times for symmetric stable processes in space (1961) and a comment by Mateusz Kwaśnicki points to an earlier paper by Marcel Riesz.
The OP asks for a simple derivation. A calculation using only elementary calculus is in Some observations on the Green function for the ball in the fractional Laplace framework.
answered Oct 29, 2020 at 21:17
-
$\begingroup$ I am puzzled with the case where $s\le\frac12$, because one does not expect the solution to vanish at both ends. $\endgroup$ Commented Oct 29, 2020 at 21:36
-
1$\begingroup$ Noteworthy, for $s < \tfrac12$, the calculation essentially goes back to M. Riesz's 1938 paper, where it appeared in a rather disguised form. And I read once that the 1-D case was actually considered even earlier (by Weyl, perhaps?), but I fail to remember where I found that comment. $\endgroup$ Commented Oct 30, 2020 at 0:48
-
$\begingroup$ Is there a simple way for solve $(-\Delta)^su=1$? $\endgroup$– inocCommented Oct 30, 2020 at 7:31
-
1$\begingroup$ @inoc --- I have given a reference to a "simple" calculation, although it is still lengthy; I have not found a simple and quick derivation. $\endgroup$ Commented Oct 30, 2020 at 21:12