Timeline for Showing convergence of an infinite ODE system
Current License: CC BY-SA 4.0
12 events
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| Oct 19, 2022 at 17:01 | history | bounty ended | CommunityBot | ||
| Oct 13, 2022 at 11:36 | comment | added | Augusto Santos | Via monotonicity (in a version that extends Lemma 1), we would have convergence to some ${\sf eq}_{\mu}$ for any initial condition given by: $a_n(0)>e^{-\mu_1 2^{-(n-1)}}$ for any $n$ odd and $a_n(0)<e^{-\mu_2 2^{-(n-1)}}$ for any $n$ even. Solutions of initial conditions bounded this way would be squized to the equilibrium ${\sf eq}_{\mu}$ by the solution initialized as in my previous comment. This would extend Theorem 2. However, for initial conditions with the block format, it seems quite clear that the system cannot pick one equilibrium to converge to (unless uniqueness does not hold). | |
| Oct 13, 2022 at 11:21 | comment | added | Augusto Santos | @AnthonyQuas: Indeed, to avoid convergence to equilibria, I start believing that we need to initialize the system with something along your block example (with blocks scaling up to infinite). For example, let $\mu_1>\mu_2$. If we initialize with $a_n(0)= e^{-\mu_1 2^{-(n-1)}}$ if $n$ is odd, and $a_n(0)= e^{-\mu_2 2^{-(n-1)}}$ for $n$ even, then (I believe, but not sure) we have $a^2_{n+1}(t)>a_n(t)$ for all $t$ when $n$ is odd and $a^2_{n+1}(t)<a_n(t)$ for all $t$ when $n$ is even. Which would still grant convergence to ${\sf eq}_\mu$ for some $\mu_2<\mu<\mu_1$. | |
| Oct 13, 2022 at 4:48 | comment | added | Anthony Quas | @AugustoSantos: somehow I suspect that simple oscillation between $\mu$ and $\mu’$ may not be sufficient to avoid convergence to an equilibrium, but I have done no calculation whatsoever. On the other hand, I think easy soft arguments should work for block behavior. | |
| Oct 12, 2022 at 9:13 | comment | added | Augusto Santos | @FeiCao: Thank you. Of course, please, feel free to use whatever needed in your paper if it helps. | |
| Oct 12, 2022 at 9:07 | comment | added | Augusto Santos | @AnthonyQuas: Thank you. Incidently, while thinking about the stability of the equilibria, I was also thinking about possibilities of no-convergence to equilibrium (e.g., oscillatory behavior and chaoticity). I don't think we would need to go as far as resorting to this "bigger-blocks" initial conditions example -- It is already not clear to me, whether a simple oscillatory tail would not compromise convergence to equilibrium. The extra condition "$a_{n}(0)<a_{n+1}(0)^2$" prevents the oscilatory behavior (if everything is in place) yielding a well enough behaved tail (in view of Theorem 1). | |
| Oct 12, 2022 at 4:32 | vote | accept | Fei Cao | ||
| Oct 12, 2022 at 4:32 | vote | accept | Fei Cao | ||
| Oct 12, 2022 at 4:32 | |||||
| Oct 12, 2022 at 4:30 | comment | added | Fei Cao | Thank you for your wonderful answer! I have added more contexts in my OP (yes, the requirements on the initial datum can be assumed to be true as well). By the way, may I cite this answer (as I might want to use a simplified version of this in a short Appendix of a paper) ? | |
| Oct 12, 2022 at 2:09 | comment | added | Anthony Quas | Another comment: your answer shows, I think that if $2^n\log a_n(0)\to\mu$ then the system converges to the $\mu$ equilibrium. This is along the lines of the OP’s question. | |
| Oct 12, 2022 at 2:07 | comment | added | Anthony Quas | Nice! I haven’t yet written anything down, but I suspect that one can construct solutions that don’t converge to any equilibrium: let $\mu\ne \mu’$ and set $b_n=\exp(2^{-n}\mu)$ and $b_n’=\exp(2^{-n}\mu’)$. Then if $a_n(0)$ is set to be $b_n$ for a long block; then $b_n’$ for an even longer block; and then $b_n$ for a still longer block etc. then it seems as though (all finite portions of) the system would oscillate between behaving like the $\mu$ equilibrium and line the $\mu’$ equilibrium. I don’t think this is incompatible with anything you’ve written? | |
| Oct 12, 2022 at 0:38 | history | answered | Augusto Santos | CC BY-SA 4.0 |