Timeline for Allocation of $\mathbb R^2$
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jun 15, 2023 at 13:51 | comment | added | Fawen90 | @IosifPinelis Many thx Iosif. Indeed, I just don't know how to check that the equilibrium points $x\in X$, i.e. $F(x)=0$, are unstable. For a generic $n$, it's hard to obtain the explicit expression of $x$, and so it's not trivial to compute the Jacobian matrix | |
| Jun 15, 2023 at 12:54 | comment | added | Iosif Pinelis | Previous comment continued: I guess one can try to use a dynamical systems approach to prove that the exceptional set $X$ is of measure $0$. | |
| Jun 15, 2023 at 12:54 | comment | added | Iosif Pinelis | I think your conjecture is true, even with the following additions: (i) the exceptional set (say $X$) of points $x$ is the finite union of algebraic curves; (ii) all points $x\in X$ are unstable; (iii) the movement starting at any $x\in X$ is either periodic or stationary and hence $\tau_x=\infty$ for $x\in X$; (iv) $\tau_x<\infty$ for $x\notin X$; (v) $\tau_x$ is continuous in $x\in\mathbb R^2$; in particular, $\tau_x\to\infty$ as $x$ goes to a point in $X$. However, I have no proof for any of these statements. | |
| Jun 15, 2023 at 5:11 | comment | added | Fawen90 | @IosifPinelis Thanks Iosif. I get what you mean. As $t\mapsto U(Y(t))$ is decreasing, then any point $x\neq x^*$ on $S$ should move along $Y_x(t)$ to either $z_1$ or$z_2$. I will think why my reasoning is wrong. Feel free to give an answer | |
| Jun 14, 2023 at 22:09 | comment | added | Iosif Pinelis | No, I did not say that we have monotonicity along $S$, and we do not have it. However, at each point $x\in S$, we have a strict decrease of $U$ either on the segment from $x$ to $z_1$ or on the segment from $x$ to $z_2$. | |
| Jun 14, 2023 at 21:07 | comment | added | Fawen90 | @IosifPinelis Why $U$ is strictly decreasing when moving on $S$? Indeed $U(x^*)=\max_{z\in S}U(z)$. How do you have the monotonicity along $S$? | |
| Jun 14, 2023 at 20:46 | comment | added | Iosif Pinelis | The strict concavity of the log implies for $n=2$ that the potential function $U$ is strictly concave on the straight line segment (say $S$) connecting $z_1$ and $z_2$. So, from any relatively interior point of $S$, you can move a bit towards $z_1$ or $z_2$ so that $U$ is strictly decreasing during such a movement. | |
| Jun 14, 2023 at 20:09 | comment | added | Fawen90 | @IosifPinelis I'm applying Hartman–Grobman theorem (see e.g. en.wikipedia.org/wiki/Stability_theory). I don' understand why the strict concavity plays a role here. Do you mind giving more details? | |
| Jun 14, 2023 at 20:03 | comment | added | Iosif Pinelis | The log function is strictly concave. So, for $n=2$, how can any point on the straight line segment connecting $z_1$ and $z_2$ be a stable equilibrium point? | |
| Jun 14, 2023 at 19:35 | history | answered | Fawen90 | CC BY-SA 4.0 |