What is the explicit solution of $(-\Delta)^s u = \chi_{B}$ in $\mathbb{R}^N$ with $0< s<1$ and $\chi$ is the characteristic function and $B$ is the unit ball around the origin? The answer should follow from the potential theory (answer follows by convolution theory) but I am not getting it in a simple form like the laplacian. Note that by known facts $u$ is continuous and radially symmetric.
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$\begingroup$ Is $B$ the unit ball around the origin? $\endgroup$– Ben McKayCommented Jul 11, 2017 at 15:46
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$\begingroup$ Yes $B$ is a unit ball in $\mathbb{R}^N.$ $\endgroup$– sadiazCommented Jul 11, 2017 at 19:23
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$\begingroup$ @SanjuDiaz: It seems unreasonably optimistic to me to expect explicit closed form formulae for the case of fractional powers. $\endgroup$– Alex M.Commented Jul 12, 2017 at 11:09
2 Answers
The operator $(-\Delta)^s$ in $\mathbb R^N$ is the Fourier multiplier $c_{s,N}\vert \xi\vert^{2s}$, so the Fourier transform of the fundamental solution $E_{s,N}$ should be homogeneous with degree $-2s$, so $E$ should be homogeneous with degree $2s-N$: it is indeed the case for $s=1$ where the fundamental solution of the Laplace equation is, up to a multiplicative constant, $\vert x\vert^{2-N}$ in dimension $\not=2$. Also that fundamental solution must be radial in the sense that for each vector field $X$ tangential to an Euclidean sphere with center $0$, $XE=0$. A good guess is thus $$ E_{s,N}(x)=\sigma_{s,N}\vert x\vert^{2s-N} $$ and a solution of your equation $$ u(x)=\sigma_{s,N}\int_B\vert x-y\vert^{2s-N} dy. $$
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$\begingroup$ I know this already (application of potential theory), but I am expecting a simplified form, like the laplacian case. I am expecting this as the solution is radial. $\endgroup$– sadiazCommented Jul 11, 2017 at 19:27
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$\begingroup$ @SanjuDiaz using answer of Basin I tired to calculate the result in Mma. For $N=2$ and $s=1/2$ the result seems to express via lots of elliptic functions of all kinds including incomplete. $\endgroup$– AndrewCommented Jul 11, 2017 at 20:41
The answer is $$ u(x) = (-\Delta)^{-\alpha/2} \chi_B(x) = 2^{-\alpha} G_{2,2}^{1,1}\biggl(\begin{matrix}1-(N-\alpha)/2 & 1+\alpha/2 \\ 0 & 1 - N/2\end{matrix} \, \bigg\vert \, |x|^2\biggr) , $$ where $G$ is the Meijer G-function; see Corollary 3(i) in this paper. This reduces to the hypergeometric function $_2F_1$; if I copied the expressions correctly from Mathematica, one has: $$ u(x) = \frac{\Gamma((N-\alpha)/2)}{2^\alpha \Gamma(1+\alpha/2) \Gamma(N/2)} {_2F_1}(-\alpha/2, (N-\alpha)/2; N/2; |x|^2) $$ for $x \in B$ and $$ u(x) = \frac{\Gamma((N-\alpha)/2)}{2^\alpha \Gamma(\alpha/2) \Gamma(N/2-\alpha)} |x|^{\alpha-N} {_2F_1}(1-\alpha/2, (N-\alpha)/2; 1+N/2; 1/|x|^2) $$ for $x \in \mathbb{R}^N \setminus B$. The calculation dates back to Blumenthal–Getoor–Ray (or a subset), I believe.
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$\begingroup$ @ Mateusz Kwaśnicki Can this hypergeometric function be simplified a bit further. Thank you. $\endgroup$– sadiazCommented Sep 29, 2020 at 5:22
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$\begingroup$ @sadiaz: I very much doubt, except possibly for special values of $\alpha$ and $N$. $\endgroup$ Commented Sep 29, 2020 at 11:50
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$\begingroup$ @sadiaz: Sorry, I could not parse your comment. Can you phrase it more carefully? $\endgroup$ Commented Oct 2, 2020 at 17:21
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$\begingroup$ The result is definitely continuous. I may have copied the expressions in terms of $_2F_1$ incorrectly, but the expression with the Meijer G-function seems correct. You can play with it with WolframAlpha if you like; here is a plot for $N = 3$ and $s = \alpha/2 = 1/5$. $\endgroup$ Commented Oct 3, 2020 at 20:11
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$\begingroup$ @ Mateusz Kwaśnicki Just a question. Will there be a change in the computation if I would have consider characteristic function of the set $|𝑥|<a$ in the Meijer function. $\endgroup$– sadiazCommented Oct 9, 2020 at 11:49