Timeline for Solutions to the eikonal equation
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Nov 30, 2011 at 21:41 | vote | accept | Matthias Ludewig | ||
| Nov 30, 2011 at 16:07 | answer | added | Robert Bryant | timeline score: 11 | |
| Nov 30, 2011 at 13:13 | comment | added | Robert Bryant | I think what you want to require is that $\phi$ be nonnegative near $p$ while $\phi(p)=0$. | |
| Nov 30, 2011 at 10:19 | comment | added | Willie Wong | Replacing $\varphi \mapsto \varphi + c$ gets you a new solution. By continuity of smooth functions, by possibly shrinking $U$ you can always translate any solution into one that is positive. So your "Edit" does not help at all with regards to the counterexamples given by Tom and Deane. | |
| Nov 30, 2011 at 10:11 | history | edited | Willie Wong |
edited tags
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| Nov 30, 2011 at 9:43 | history | edited | Matthias Ludewig | CC BY-SA 3.0 |
added 137 characters in body
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| Nov 30, 2011 at 5:03 | comment | added | Deane Yang | More generally, two functions $\phi$ and $\psi$ solve the equation if and only if $0 = \|\nabla\phi\|^2 - \|\nabla\psi\|^2 = \nabla(\phi - \psi)\cdot\nabla(\phi + \psi)$. This equation has many solutions, including $\phi = f(x) + g(y)$ and $\psi = f(x) - g(y)$, for any functions $f, g: R \rightarrow R$. | |
| Nov 30, 2011 at 1:13 | history | edited | David Roberts♦ | CC BY-SA 3.0 |
Corrected spelling error in title and question body
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| Nov 30, 2011 at 0:34 | comment | added | Tom Goodwillie | Um, not true as stated: in the plane with Euclidean metric, $x^2+y^2$ and $x^2-y^2$ have the same $||grad||^2$, which is $4(x^2+y^2)$. | |
| Nov 29, 2011 at 23:25 | comment | added | Alan Haynes | I don't know how helpful this will be, because it is outside my field of expertise. However I know someone who was a physicist and who did his PhD thesis on the eikonal equation. His thesis is available at cacr.caltech.edu/~sean/thesis.pdf . | |
| Nov 29, 2011 at 22:40 | history | asked | Matthias Ludewig | CC BY-SA 3.0 |