Timeline for One-sided version of the curve-shortening flow
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 26, 2023 at 15:13 | vote | accept | André Henriques | ||
| Jan 26, 2023 at 15:00 | answer | added | Joseph O'Rourke | timeline score: 4 | |
| Jan 26, 2023 at 14:40 | comment | added | RBega2 | The curves shortening flow is (in an appropriate gauge -- e.g. written as a normal graph) a strictly parabolic equation -- you flow is extremely degenerate when the curve is not convex. On a more concrete level the evolution of curvature in CSF is a semilinear heat equation while the evolution of curvautre in your example looks likes some sort of non-linear heat flow where the the thermal conductivity is supported on the part where the curvature is positive (i.e. the negative curved regions seem to act as perfect insulators) and so that part is static. | |
| Jan 26, 2023 at 13:34 | comment | added | André Henriques | @RBega2. This discussion is a bit moot given Joseph ORourke's answer, but I have to say that I'm not completely convinced by your claim that my 1-sided flow won't move the concave part of the graph of $\sin$. Yes, it won't move it to first order, but it might still move it... Consider the following related situation: apply the usual curve-shortening flow to a smooth Jordan curve that contains a straight line-segment. You might be tempted to say that this straight straight line-segment won't move. And yet, for any positive time, the deformed curve is real analytic. So it must have moved. | |
| Jan 26, 2023 at 13:28 | comment | added | André Henriques | @JosephO'Rourke. Thank you for your comment. You have answered my question (in the negative). If you want to post it as a (short) answer, I will accept it. | |
| Jan 26, 2023 at 1:44 | comment | added | RBega2 | Another pathological aspect of this flow is that in the example I wrote down the tangent at $(0,0)$ is constant in time and lies one the line $x=y$, so even if the positively curved part does converge to the line segment it can't do so in $C^1$. | |
| Jan 26, 2023 at 1:06 | comment | added | RBega2 | Your precise question is clearly not true if you consider the curve given by the graph of $y=sin(\pi x)$ as, for instance, the curvature with respect to the upward normal is negative on $[0,1]$ so the flow won't move this part. (It's a bit unclear what conventions you are using for the sign of the curvature -- for instance with the usual meanings of curvature and normal, the flow you write down is backwards parabolic when it is not degenerate). | |
| Jan 25, 2023 at 23:54 | comment | added | Joseph O'Rourke | Nice question. My guess is that simplicity (no crossings)---a feature of curve shortening---is not preserved. Consider a U-shape. The inside of the U has negative $\kappa$ and will remain fixed. Meanwhile the bottom of the U has positive $\kappa$ and will collide with the fixed portion. | |
| Jan 25, 2023 at 23:14 | history | asked | André Henriques | CC BY-SA 4.0 |