Timeline for First order PDE, singular at a point
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Dec 21, 2011 at 23:06 | vote | accept | Matthias Ludewig | ||
| Dec 21, 2011 at 15:18 | answer | added | Robert Bryant | timeline score: 4 | |
| Dec 21, 2011 at 14:57 | comment | added | Robert Bryant | @Willie: You do have to be careful because the linearization can just be $C^1$, and that won't help you understand the space of smooth functions Kofi wants to decribe. For example, the vector field $X = x\ \partial_x + (2y+x^2)\ \partial_y$ cannot be linearized by any $C^2$ change of coordinates at the origin. | |
| Dec 21, 2011 at 9:17 | comment | added | Willie Wong | On the flip side, if you know that your vector field $X$ is $C^\infty$, using a result of Guysinsky, Hasselblatt, and Rayskin, www.tufts.edu/as/math/Preprints/HasselblattGuysinskyRayskin.pdf you can assert that the linearization map is differentiable at the origin with derivative being the identity. So any rigidity conditions coming from the first derivatives can still be used. | |
| Dec 21, 2011 at 8:58 | comment | added | Willie Wong | Just a comment on your conjecture: if none of the $a_i$ vanish, then the first thing that comes to mind is the Hartman-Grobman theorem. Unfortunately that only guarantees Holder continuity of the coordinate transformation to "normal form", so the regularity argument cannot necessarily apply. But since all of your $a_i$ are signed, you may be able to say more. | |
| Dec 20, 2011 at 16:14 | history | edited | Matthias Ludewig | CC BY-SA 3.0 |
deleted 26 characters in body; added 19 characters in body
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| Dec 20, 2011 at 16:13 | comment | added | Matthias Ludewig | No, $\eta$ is not supposed to be an integral curve of the vector field $X$. It is a function $\eta \in C^{\infty}(U)$ ($U$ some neighborhood of $0$) that fulfills the differential equation that is centered above. Maybe I should rename variables... | |
| Dec 20, 2011 at 16:04 | comment | added | Jaap Eldering | I'm assuming that $\eta$ denotes a solution curve of $X$, but correct me if I understand wrongly. In that case, the ODE is $\dot{\eta}_i(t) = a_i\,\eta_i(t)$ which is linear and the solutions corresponding to eigenvalues $a_i$ are $\eta(t) = e^{a_i\,t}\,e_i$ where $e_i$ denotes a coordinate basis vector. | |
| Dec 20, 2011 at 16:00 | history | edited | Matthias Ludewig | CC BY-SA 3.0 |
added 69 characters in body; added 1 characters in body
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| Dec 20, 2011 at 15:58 | comment | added | Matthias Ludewig | It seems like you missed the $x^i$ in $a_i x^i \partial_i$. The Operator $\partial_X$ is singular at $0$, meaning, it vanishes there and is of order $0$, as opposed to the neighborhood where it is of order $1$. | |
| Dec 20, 2011 at 15:54 | comment | added | Jaap Eldering | The vector field $X$ describes a linear ODE, with diagonal matrix consisting of the $a_i$ in your choice of coordinates. Thus, the eigenvalues are exactly the $a_i$. No singularities involved. | |
| Dec 20, 2011 at 14:47 | history | asked | Matthias Ludewig | CC BY-SA 3.0 |