Timeline for Solutions to the eikonal equation
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Apr 24, 2012 at 0:05 | comment | added | Hans | @Robert Bryant: hmm, this is what I thought at the beginning, but I'm not convinced. It seems to me that the quadratic approximation is not enough to decide. Why should $L^+\geq L$ imply $\phi^+\geq \phi$ near $p$? (We really have $L^+\geq L$ in general and not $L^+ > L$, otherwise it would be clear to me...). Do I miss something? | |
| Apr 23, 2012 at 17:03 | comment | added | Robert Bryant | @Hans: Well, any smooth solution $\phi$ that vanishes at $p$ will have a critical point at $p$ and hence have a $p$-centered Taylor expansion $\phi = \tfrac12 L_{ij} x^ix^j + O(3)$ for a symmetric matrix $L$ that satisfies $g^{ij}(p):L_{ik}L_{jl}=h_{kl}$. One can now show that if $L^+ = (L^+_{ij})$ is the (unique) positive definite symmetric solution to this equation, then the corresponding $\phi^+$ (as constructed above) is unique near $p$, and, since $L^+\ge L$ for $L$ any other solution, the corresponding $\phi$ will (at least near $p$) be dominated by $\phi^+$. | |
| Apr 22, 2012 at 15:02 | comment | added | Hans | @Robert Bryant. Sorry, I was imprecise: I mean the first option you mentioned. | |
| Apr 22, 2012 at 14:05 | comment | added | Robert Bryant | @Hans: I'm not sure that I know what you mean by 'maximal'. Do you mean that this solution is greater than or equal to any other solution that vanishes at $p$, at least in a neighborhood of $p$, or do you just mean that there is no other solution that vanishes at $p$ that is greater than it near $p$? | |
| Apr 22, 2012 at 13:39 | comment | added | Hans | @Robert Bryant. Thanks for this answer. I was wondering if the solution you constructed, uniquely characterized among the smooth solutions by being $0$ in $p$ and nonnegative, could also be uniquely characterized among smooth solutions as the maximal solution. I'm pretty convinced this is true, but I don't see how to prove it. Does it follow somehow directly from your construction or are some other arguments needed? I would be very grateful if you or somebody else can comment on this! | |
| Dec 4, 2011 at 17:39 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Complete the proof by applying the Stable Manifold Thm
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| Nov 30, 2011 at 21:41 | vote | accept | Matthias Ludewig | ||
| Nov 30, 2011 at 16:40 | history | edited | Robert Bryant | CC BY-SA 3.0 |
added missing symbols
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| Nov 30, 2011 at 16:07 | history | answered | Robert Bryant | CC BY-SA 3.0 |