Timeline for helmholtz zero in R^3
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
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| May 10, 2010 at 17:55 | comment | added | Andrey Rekalo | @chris: you will probably get more attention if you ask for references in a separate question. | |
| May 10, 2010 at 17:50 | history | edited | Andrey Rekalo | CC BY-SA 2.5 |
added 620 characters in body
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| May 10, 2010 at 4:20 | comment | added | Harald Hanche-Olsen |
Typo in my previous comment: It should be $v(r)=k^{-1}u(x)\sin kr$.
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| May 10, 2010 at 4:09 | comment | added | chris | what are some of the best books for studying distributions from a very applied perspective? | |
| May 10, 2010 at 4:08 | vote | accept | chris | ||
| May 10, 2010 at 4:07 | comment | added | chris | it came up when we were researching for the same topic, something about distributions solving certain equations =) | |
| May 10, 2010 at 4:01 | comment | added | Harald Hanche-Olsen | This is clearly related to mathoverflow.net/questions/24039/maximum-decay-rate and for a similar reason: In the present case, if $\bar u(r)$ is the mean value over the sphere of radius $r$ then $v(r)=r\bar u(r)$ satisfies $v''+k^2v=0$. Also $v(0)=0$ and $v'(0)=u(x)$, so $v(r)=u(x)\sin kr$. | |
| May 10, 2010 at 2:19 | history | answered | Andrey Rekalo | CC BY-SA 2.5 |