Timeline for Helmholtz decomposition allows incompressible flow with an irrotational component?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 5, 2017 at 16:41 | answer | added | Nosferatu | timeline score: 0 | |
| Feb 7, 2016 at 21:51 | vote | accept | wil3 | ||
| Feb 2, 2016 at 12:06 | comment | added | Emilio Pisanty | It sort of depends on exactly what you're willing to impose. All the partial derivatives of $\phi$ are also harmonic so that's easy. If you want them to be non-singular and harmonic throughout, then they achieve their maximum at the boundary. Whether you can go to something as strong as $\nabla\phi$ being constant isn't quite clear to me at the moment - but that's a good question for Mathematics - must a bounded harmonic function in three dimensions be constant? | |
| Feb 1, 2016 at 21:34 | comment | added | wil3 | @EmilioPisanty I was wondering if that might be the solution to my problem---let's say I want the velocity to be bounded at infinity, would that somehow force $\nabla \phi$ to be a constant? | |
| Feb 1, 2016 at 21:19 | comment | added | Emilio Pisanty | It sort of depends on what region you're working in and what happens at the boundary. Normally you would want $\phi$ to be bounded at infinity, and that is very restrictive for harmonic functions. | |
| Feb 1, 2016 at 21:15 | answer | added | Red Act | timeline score: 5 | |
| Feb 1, 2016 at 15:15 | history | edited | Red Act | CC BY-SA 3.0 |
Fix a missing phrase.
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| Feb 1, 2016 at 9:06 | comment | added | Bort | I feel like this would reduce $\nabla \Phi$ to be a constant, i.e. be identical to a choice of reference frame (which of course you are free to do, but has no physical significance) | |
| Feb 1, 2016 at 3:49 | history | asked | wil3 | CC BY-SA 3.0 |