Timeline for Showing convergence of an infinite ODE system
Current License: CC BY-SA 4.0
28 events
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| S Oct 19, 2022 at 17:01 | history | bounty ended | CommunityBot | ||
| S Oct 19, 2022 at 17:01 | history | notice removed | CommunityBot | ||
| Oct 12, 2022 at 4:34 | history | edited | Fei Cao | CC BY-SA 4.0 |
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| Oct 12, 2022 at 4:32 | vote | accept | Fei Cao | ||
| Oct 12, 2022 at 4:32 | vote | accept | Fei Cao | ||
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| Oct 12, 2022 at 0:38 | answer | added | Augusto Santos | timeline score: 3 | |
| S Oct 11, 2022 at 15:18 | history | bounty started | Fei Cao | ||
| S Oct 11, 2022 at 15:18 | history | notice added | Fei Cao | Draw attention | |
| Oct 9, 2022 at 16:02 | history | edited | Fei Cao | CC BY-SA 4.0 |
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| Oct 9, 2022 at 15:57 | comment | added | Fei Cao | @AnthonyQuas Dear Professor, I totally agree with your comment, I was not careful enough in the statement of the initial condition | |
| Oct 9, 2022 at 15:55 | comment | added | Fei Cao | @AugustoSantos Hello, yes I agree, the choices of initial datum matter. The assumption that $a_n(t) < 1$ for all fixed $n$ does not prevent the situation where $\sup_{n \in \mathbb N} a_n(t) = 1$ or $\sup_{n \in \mathbb N} a_n(0) = 1$ | |
| Oct 9, 2022 at 12:44 | comment | added | Augusto Santos | On the other hand, if $a_n(0)=1$ for all $n>N$ for some $N\in\mathbb{N}$, then we can easily prove that $a_n(t)\longrightarrow 1$ as $t\rightarrow \infty$ for all $n$. To grant convergence to the equilibrium that you are asking, you can neither have initial conditions bounded away from 1, i.e., $\sup_n a_n(0)<1$, nor those whose tail touches 1, i.e., $a_n(0)=1$ eventually. Which is an interesting problem. | |
| Oct 9, 2022 at 8:04 | comment | added | Augusto Santos | @FeiCao: Contingent on the uniqueness of solutions to this infinite-dimensional ODE, you can show that: i)[Monotonicity] $a_n(0)\leq \overline{a}_n(0)\Rightarrow a_n(t)\leq \overline{a}_n(t)$ for all $n$ and $t\geq 0$; ii) [invariance of the diagonal] $a_n(0)=a_m(0)\Rightarrow a_n(t)=a_m(t)$ for all $n,t$; iii) [dynamics on the diagonal] $a_n(0)=b\in[0,1]$ for all $n$ implies $a'_n(t)=a_n(t)^2-a_n(t)$; iv) Combining i) and iv), we have that for any sequence $(a_n(0))_{n\in\mathbb{N}}$ fulfilling $\sup_{n\in\mathbb{N}}a_n(0)<1$, then $a_n(t)\longrightarrow 0$ for all $n$. | |
| Oct 9, 2022 at 5:44 | comment | added | Anthony Quas | The bottom line is that this system has a one parameter family of obvious equilibria, namely for each $\mu>0$, $a_n(t)=\exp(-\mu 2^{-n})$ is an equilibrium point for the system. I think the question should be "Under which conditions on the sequence $(a_n(0))$ belonging to $(0,1)^{\mathbb N}$ does the system converge to one of the equilibrium points? Is it for all initial conditions? Or only some initial conditions? Given an initial condition, is there a way to decide which fixed point (if any) the orbit converges to?" | |
| Oct 9, 2022 at 1:54 | comment | added | Fei Cao | @AnthonyQuas Hello professor, the symbol $\approx$ in my original post is not interpreted rigorously like you said that $a_n - (1-\mathrm{e}^{-\mu2^{-n}}) \to 0$ is nothing different from $a_n - 1 \to 0$. I think the condition that $a_n(t) \in [0,1)$ is sufficient for my question to be "well-posed". Also, if you take $a_n = \mathrm{e}^{-\mu2^{-n}}$ then clearly it behaves like $1-\mu2^{-n}$ when $n \gg 1$. | |
| Oct 8, 2022 at 21:11 | comment | added | Anthony Quas | Fei: You have written three different versions of your condition (ii) in the question and the comments. The condition $a_n(t)\to 1-\mu 2^{-n}$ can never be satisfied. The version in your original question does not seem to be clearly formulated. If you don't have a clear question, we will not be able to answer it. | |
| Oct 8, 2022 at 20:03 | comment | added | Fei Cao | @AnthonyQuas Hello, like I said before, the ansatz that $a_n = \mathrm{e}^{-\mu2^{-n}}$ satisfies $a^2_{n+1} = a_n$ and it satisfies the condition (ii) as well because of the asymptotic $\mathrm{e}^{-x} \approx 1 - x$ for $0 < x \ll 1$. | |
| Oct 8, 2022 at 17:43 | comment | added | Anthony Quas | Hello Fei, I think if $a_n(t)\to 1-\mu 2^{-n}$ then $a_{n-1}(t)\to (1-\mu 2^{-n})^2\ne 1-\mu 2^{-(n-1)}$ so that your assumption cannot be satisfied. On the other hand, if $a_n(t)\to \exp(-\mu 2^{-n})$ then $a_{n-1}(t)\to \exp(-\mu 2^{-(n-1)})$. | |
| Oct 8, 2022 at 17:08 | comment | added | Fei Cao | Dear Professor, the condition that $a_n(t) \to 1 - \mu 2^{-n}$ as $ t \to \infty$ for all large enough $n$ is an assumption which can be of some help... | |
| Oct 8, 2022 at 17:02 | comment | added | Anthony Quas | Hello Fei, so this question is interesting to me, but I still don’t know exactly what you’re asking. One possible question would be: suppose $a_n(0)\in (0,1)$ for each $n$. Is it true that there exists a $\mu$ such that $a_n(t)\to \exp(-2^{-n}\mu)$. | |
| Oct 8, 2022 at 16:26 | history | edited | Fei Cao | CC BY-SA 4.0 |
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| Oct 8, 2022 at 16:23 | comment | added | Fei Cao | @AnthonyQuas Dear Professor, please see my edited question for further clarification. | |
| Oct 8, 2022 at 16:22 | comment | added | Fei Cao | @AugustoSantos Hello, you can check that the ansatz $a_n = \mathrm{e}^{-\mu 2^{-n}}$ indeed satisfies $a^2_{n+1} = a_n$ for all $n$ | |
| Oct 8, 2022 at 15:28 | comment | added | Augusto Santos | @FeiCao: something is a bit odd: the point $(e^{-\mu 2^{-1}}, e^{-\mu 2^{-2}},e^{-\mu 2^{-3}},...) \in \left[0,1\right]^{\infty}$ is not equilibrium of the system and thus, it cannot be the attractor of this dynamical system -- or am I missing something? The only equilibria that I could spot were the origin and $(1,1,1,...)$. It feels like the origin is the global attractor... | |
| Oct 8, 2022 at 5:56 | comment | added | Anthony Quas | Hello Fei. This doesn't sound like uniform convergence to me (which is what you wrote in the original question). Also, why is your condition different from "for every $t\ge 0$, $a_n(t)\to 1$ as $n\to\infty$"? | |
| Oct 8, 2022 at 4:03 | comment | added | Fei Cao | @AnthonyQuas Hello, condition (ii) enforces that for every $t \geq 0$, $|a_n(t) - (1-\mu2^{-n})|$ converges to $0$ when $n \to \infty$. | |
| Oct 8, 2022 at 3:34 | comment | added | Anthony Quas | Can you be more precise about your condition (ii)? | |
| Oct 7, 2022 at 20:27 | history | asked | Fei Cao | CC BY-SA 4.0 |