Suppose $\{a\_n(t)\}\_{n \geq 0}$ is a collection of differentiable (or simply smooth) functions such that i) $0 \leq a\_n(0) \leq 1$ for all $n\in \mathbb N$ (ii) $a\_n(t) \approx 1 - \mu2^{-n}$ uniformly in $t$ for $n \gg 1$ (iii) $a'\_n = a^2\_{n+1} - a\_n.$ My goal is to show that $$a\_n(t) \xrightarrow{t \to \infty} \mathrm{e}^{-\mu 2^{-n}} \quad \text{for all $n \in \mathbb N$.}$$ Any help (or hints) will be greatly appreciated!




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Remark: The condition (ii) is a bit unclear, I should emphasize here that $a\_n(t) < 1$ for all (fixed) $n \in \mathbb N$ and for all $t \geq 0$, this is the reason that I did not write condition (ii) as
condition (ii') $a\_n(t) \approx 1$ uniformly in $t$ for $n \gg 1$.




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Remark: As discussed in the comments, I need to impose an initial datum ${a\_n(0)}\_{n \geq 0}$ such that $\sup\_{n \in \mathbb N} a\_n(0) = 1$ but there does not exists an $N \in \mathbb N$ for which $a\_N(0) = 1$. Also, one can savely assume that $a\_n(0) < a^2\_{n+1}(0)$ for all $n \in \mathbb N$.



Showing convergence of an infinite ODE system
Theorem 2 below offers a sufficient condition for convergence to the nontrivial equilibrium that you are referring.


The results are contingent on the uniqueness of solutions to this infinite-dimensional ODE -- I will include the proofs when possible, as needed, but, and unless I am missing something, they are quite simple.


**Question.** Let $\left(a\_n(t)\right)\_{n\in\mathbb{N}}$ be the solution to the infinite-dimensional ODE
$$a\_n'(t)=a\_{n+1}^2(t)-a\_n(t) \,\,\,\,(\star)$$
for $n\in\mathbb{N}$, with initial condition $\left(a\_n(0)\right)\_{n\in\mathbb{N}}$. For which initial conditions, the solution $\left(a\_n(t)\right)\_{n\in\mathbb{N}}$ converges to an equilibrium distinct from the origin and the all 1's equilibrium?


**Remark 1 [Domain]**. The cube $\left[0,1\right]^{\mathbb{N}}$ is invariant to this dynamics. That is, if we initialize the system within this infinite-dimensional cube, then the solutions will remain there. You cannot impose condition i) as in your question. But, if you assume $a\_n(0)\in\left[0,1\right]$ for all $n$, then $a\_n(t)\in\left[0,1\right]$ for all $n,t$.


**Remark 2 [Equilibria parametrization].** The equilibria is given by $\mathcal{E}=\left\{\left(a\_{n}\right)\_{n\in\mathbb{N}}\,:\,a\_{n+1}^2=a\_n\mbox{ for all }n\right\}$ which can be rewritten as $\mathcal{E}=\left\{\left(b^{2^{-(n-1)}}\right)\_{n\in\mathbb{N}}\,:\,\mbox{ for all }b\in\left[0,1\right]\right\}$. Using your parametrization, we can further write it as $\mathcal{E}=\left\{\left(e^{-\mu 2^{-(n-1)}}\right)\_{n\in\mathbb{N}}\,:\,\mbox{ for all }\mu\geq0\right\}\cup \left\{\mathbf{0}\right\}$. In other words, given any $\mu\geq 0$, then the sequence $\left(e^{-\mu 2^{-(n-1)}}\right)\_{n\in\mathbb{N}}$ is an equilibrium for $(\star)$. That is, this provides a one-dimensional parametrization for the equilibria of the system. In particular, let us define for simplicity
$${\sf eq}\_{\mu}\overset{\Delta}=\left(e^{-\mu 2^{-(n-1)}}\right)\_{n\in\mathbb{N}}$$
the equilibrium associated with the parameter $\mu\geq 0$.


**Remark 3 [Causal structure].** Observe that the evolution of $\left(a\_m(t)\right)\_{t\geq 0}$, for $m\geq N$, does not depend on the evolution of $\left(a\_n(t)\right)\_{t\geq 0}$ for any $n<N$. The state variable $a\_{n+1}$ impacts $a\_n$ but not the other way around. This implies the following: the *tail* of the initial condition is what determines the asymptotic behavior of this dynamical system. In other words, for any $N$, the sub-sequence of the initial condition $\left(a\_n(0)\right)\_{n\leq N}$ is irrelevant for the asymptotic behavior of the system.


The next result reveals an important property of this ODE.


**Lemma 1[Monotonicity].** If $a\_n(0)\leq \overline{a}\_n(0)$ for all $n\in\mathbb{N}$, then $a\_n(t)\leq \overline{a}\_n(t)$ for all $n\in\mathbb{N}$ and $t\geq 0$.


We have an immediate corollary to Lemma 1.


**Corollary 1 [Invariant sub-regions].** Let $\mu\_1<\mu\_2$. If there exists $N\in\mathbb{N}$ so that $e^{-\mu\_2 2^{-(n-1)}}\leq a\_{n}(0)\leq e^{-\mu\_1 2^{-(n-1)}}$ for all $n\geq N$, then $e^{-\mu\_2 2^{-(n-1)}}\leq \lim\inf\_{t\rightarrow \infty}a\_{n}(t)\leq \lim\sup\_{t\rightarrow \infty}a\_{n}(t)\leq e^{-\mu\_1 2^{-(n-1)}}$ for all $n\in\mathbb{N}$.


As a consequence to Corollary 1, if the initial condition $\left(a\_n(0)\right)\_{n\in\mathbb{N}}$ is bounded as $e^{-\mu\_2 2^{-(n-1)}}\leq a\_{n}(0)\leq e^{-\mu\_1 2^{-(n-1)}}$ eventually, i.e., for $n\geq N$ for some $N$, then the dynamical system $\left(a\_n(t)\right)\_{n\in\mathbb{N}}$ cannot converge to ${\sf eq}\_{\mu}$ for any $\mu\notin \left[\mu\_1,\mu\_2\right]$.


**Theorem 1 [Monotonicity 2].** If $a\_n(0)<a\_{n+1}(0)^2$ for all $n$, then $a\_n(t)$ is increasing in $t$ for all $n$, i.e., $a\_{n}(t)>a\_{n}(t')$ for any $t>t'$ and for any $n$.


As a corollary to Theorem 1, and to the fact that $a\_n(t)\in\left[0,1\right]$ for all $t$, we have convergence to an equilibrium whenever the initial condition is given by $a\_n(0)<a\_{n+1}(0)^2$ for all $n$. In particular, via combining Corollary 1 and Theorem 1, we have the following sufficient condition for convergence to a nontrivial equilibrium.


**Theorem 2 [Sufficient condition].** If $a\_n(0)<a\_{n+1}(0)^2$ for all $n$, and further $e^{-\mu\_2 2^{-(n-1)}}<a\_n(0)<e^{-\mu\_1 2^{-(n-1)}}$ eventually for some $\mu\_1,\mu\_2\in\left(0,\infty\right)$, then there exists $\mu\in\left[\mu\_1,\mu\_2\right]$ so that $a\_n(t)\overset{t\rightarrow \infty}\longrightarrow e^{-\mu2^{-(n-1)}}$ for all $n$.


--------------------------- **Other results** --------------------------------------


**Lemma 2[Invariance of the diagonal].** If $a\_n(0)=b\in \left[0,1\right]$ for all $n$, then $a\_n(t)=a\_m(t)$ for all $n,m$. In particular, $a\_n'(t)=a\_n(t)^2-a\_n(t)$ for all $n$.


The proof is trivial, but relies on the uniqueness of solutions to the ODE $(\star)$.


**Theorem 3 [Origin's attraction].** If $\sup\_{n\in\mathbb{N}} a\_n(0)<1$, then, $a\_n(t)\overset{t\rightarrow \infty}\longrightarrow 0$ for all $n\in\mathbb{N}$.



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> *Proof.* Consider the sequence $\left(\overline{a}\_n(0)\right)\_n$ so that $\overline{a}\_n(0)=\sup\_{n\in\mathbb{N}}a\_n(0)=:c\in\left(0,1\right)$ for all $n$. Then, from Lemma 1, we have that $\overline{a}\_n(t)\geq a\_n(t)$ for all $n$ and $t\geq 0$. From Lemma 2, we have that $\overline{a}'\_n(t)=\overline{a}\_n(t)^2-\overline{a}\_n(t)$. Therefore, $\overline{a}\_n(t)\overset{t\rightarrow \infty}\longrightarrow 0$ for all $n$. Thus, $a\_n(t)\overset{t\rightarrow \infty}\longrightarrow 0$ from the boundedness and since $\left[0,1\right]^{\mathbb{N}}$ is invariant to the infinite-dimensional dynamical system $(\star)$.
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**Theorem 4 [1's attraction].** If $a\_n(0)=1$, for all $n>N$ for some $N\in\mathbb{N}$ large enough, then $a\_n(t)\overset{t\rightarrow \infty}\longrightarrow 1$ for all $n\in\mathbb{N}$.



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> *Proof.* It is trivial to check that $a\_n(t)=1$ for all $t\geq 0$ for all $n>N$. we have $a\_n'(t)=1-a\_n(t)$ for $n=N$ and hence $a\_n(t)\overset{t\rightarrow \infty}\longrightarrow 1$. From here, it is trivial to check that $a\_n(t)\longrightarrow 1$ for all $n$.
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