Timeline for explicit computation of fractional Laplacian of a function
Current License: CC BY-SA 4.0
23 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 22, 2020 at 22:12 | comment | added | GabS | Yes, the characteristic function on the interval (1,2). | |
| Dec 22, 2020 at 21:06 | comment | added | Mateusz Kwaśnicki | Do you mean $\chi_{1,2}(x) = 1$ if $1 < |x| < 2$ and $\chi_{1,2}(x) = 0$ otherwise? The expression you gave looks OK for $x > 2$, for general $x$ you need absolute values of $x\pm1$ and $x\pm2$ and appropriate sign corrections. | |
| Dec 18, 2020 at 20:13 | comment | added | GabS | Means that $(x-2)$ is negative when $x\in (1,2)$, then what happens to $(-1)^{1-2s}(2-x)^{-2s}.$ | |
| Dec 18, 2020 at 14:50 | comment | added | GabS | @ Mateusz Kwaśnicki If I consider $u(x)=\chi_{1,2}(x)$ and calculate $(-\Delta)^s u=constant\times [(x+1)^{-2s}-(x+1)^{-2s} -(x-2)^{-2s}+ (x+2)^{-2s}]$. If $x\in (1, 2)$ the third term is a bit tricky. Am I wrong. | |
| Nov 23, 2020 at 19:53 | comment | added | Mateusz Kwaśnicki | If you look at the expression for $(-\Delta)^s u(x)$, then you will clearly see that these are quite different functions for $u = \chi_{(1,\infty)}$ and $u = \chi_{(1,2)}$. | |
| Nov 23, 2020 at 18:25 | comment | added | GabS | @ Mateusz Kwaśnicki Is the fractional Laplacian of the characteristic function $\chi_{(1, \infty)}$ and $\chi_{(1, 2)}$ is the same? Do the end point contribute of the calculation. | |
| Sep 11, 2020 at 14:04 | comment | added | GabS | @ Mateusz Kwaśnicki In your calculation, I think $|x|^2$ should be replaced by $|x|.$ | |
| Sep 8, 2020 at 18:09 | comment | added | Mateusz Kwaśnicki | @GabS: "Quickest" in the sense of numerical methods? No idea. Mathematica seems to evaluate it reasonably well, but relatively slow. WolframAlpha also is able to evaluate single values. Regarding general introduction to Meijer G-function, I do not have anything to add to what we wrote in the article. Prudnikov's book is perhaps a good start. | |
| Sep 8, 2020 at 16:36 | comment | added | GabS | @ Mateusz Kwaśnicki What is the quickest way to compute the Meijur $G$ function. You are using a Meijur function and hypergeometric series. Is there a good book on this topic. | |
| Sep 7, 2020 at 21:39 | comment | added | Mateusz Kwaśnicki | Simply write $u$ as a linear combination of $|x|^p \mathbb{1}_{|x|<a}$ or $|x|^p \mathbb{1}_{|x|>a}$ for different $p$ and $a$, and you are good to go, right? | |
| Sep 7, 2020 at 12:24 | comment | added | Mateusz Kwaśnicki | @GabS: But what are $f$ and $g$? | |
| Sep 5, 2020 at 6:17 | comment | added | GabS | Thank you very much for the information. | |
| Sep 4, 2020 at 9:45 | comment | added | Mateusz Kwaśnicki | I expanded my answer in response to your last comment. If everything remains unclear, feel free to ask. | |
| Sep 4, 2020 at 9:44 | history | edited | Mateusz Kwaśnicki | CC BY-SA 4.0 |
added 1860 characters in body
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| Sep 4, 2020 at 8:43 | comment | added | GabS | I do agree, but I think there is a problem. The function in your paper $u(x)= |x|^{2\rho}(|x|^{2}-1)_{+}^{\sigma}$ considering $V(x)=1.$ The function what I want is a bit more complicated $(|x|^{2(s-1/2)}-1)_{+}$ which need $\rho$ to be zero and $\sigma$ to be $s-1/2$ or may be I am making a mistake. | |
| Sep 4, 2020 at 4:59 | comment | added | Mateusz Kwaśnicki | If you are saying about $(-\Delta)^{1/2} [\log |x|]$, then this is indeed zero (observe $\Gamma(0)$ in the denominator in your expression). However, $(-\Delta)^{1/2} [\log |x| \mathbb{1}_{|x| > 1}]$ is not identically equal to zero. | |
| Sep 3, 2020 at 9:52 | comment | added | GabS | @ Mateusz Kwaśnicki The calculation is $(-\Delta)^s u(x)= 2^{2s} \frac{\Gamma(s) \Gamma((1/2)}{\Gamma(0)\Gamma ((1-2s)/2) } |x|^{-1}= 2^{2s} \frac{\Gamma(s) \Gamma((1/2)}{\Gamma (1-2s/2) } |x|^{-1}.$ So I am wrong. | |
| Sep 2, 2020 at 22:32 | comment | added | Mateusz Kwaśnicki | @GabS: No, it is definitely not zero, unless $s = 1$. Why do you expect it to be zero? | |
| Sep 2, 2020 at 12:07 | comment | added | GabS | @ Mateusz Kwaśnicki I believe in both the cases the answer is zero as in one dimension, these are fundamental solution away from zero. | |
| Aug 31, 2020 at 8:02 | comment | added | GabS | @ Mateusz Kwaśnicki Yes, I should have been precise. I would like to calculate $(-\Delta)^{1/2}$. Here $u$ look like the fundamental solution except in a neighbourhood of zero | |
| Aug 30, 2020 at 20:38 | comment | added | Mateusz Kwaśnicki | @GabS: I guess you can divide by $(2s-1)$ and pass to the limit as $s \to \tfrac12$, no? (The $s$ in the definition of $u$ need not be the same as the exponent $s$ in $(-\Delta)^s$, if this matters.) | |
| Aug 30, 2020 at 17:06 | vote | accept | GabS | ||
| Aug 30, 2020 at 0:47 | history | answered | Mateusz Kwaśnicki | CC BY-SA 4.0 |