I encountered the following question in my studies: Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a Bohr almost periodic function such that $\inf_{\mathbb{R}} f = 0$ but $f(x) > 0$ for all $x\in \mathbb{R}$. An example is $$ f(x) = 2\sin(2\pi x)  \sin(2\pi \sqrt{2}x).$$ If $\eta(\cdot)$ is the solution to the following ODE $$ \dot{\eta}(s) = f(\eta(s)), \qquad \eta(0) = 0.$$ Is there any tools that allow us to say something about the limit $$ \lim_{s\rightarrow +\infty} \frac{\eta(s)}{s}$$ and if the limit exists (I guess, by numerical implementations) can we say anything about the rate of convergence of $\frac{\eta(s)}{s}$ to that limit?
1 Answer
For the funciton you gave, the limit is 0, but my proof below only gives convergence as $1/\log s$. If $f=2\sin(2\pi x)\sin(2\pi Lx)$ where $L$ is Louiville's constant (or some appropriately chosen irrational number wellapproximable by rationals), it's possible the convergence will be much faster if $f$ is close to $0$ frequently. Thus things depend delicately on approximation properties of ratios of the periods. Also, if we take $f=3\sin(2\pi x)\sin(2\pi\sqrt{2} x)\sin(2\pi\sqrt{3} x)$, the proof below fails, so general behavior is unclear.
Proof:
By the equidistribution theorem, we have that multiples of $\sqrt{2}$ mod $1$ are equidistributed in the interval $[0,1]$. Thus if we take the interval $[0,N]$, then letting $$A_\epsilon=\left\{x\in [0,N]\,\bigg\vert\,\epsilon/2<x\lfloor x\rfloor\frac{1}{4}<\epsilon, \epsilon/2<\frac{x}{\sqrt{2}}\left\lfloor \frac{x}{\sqrt{2}}\right\rfloor\frac{1}{4}<\epsilon\right\}$$ we have that the measure of $A_\epsilon$ is $A_\epsilon=\theta(\epsilon^2N)$ (here we are thinking of $\epsilon$ as fixed and taking $N\to\infty$. On the set $A$, the we have $f=\theta(\epsilon^2)$, so the time required to traverse $A_\epsilon$ is $\theta(N)$. Then, since the sets $A_{1/2^j}$ are disjoint, the total time to traverse the interval $[0,N]$ is at least $$ \sum_{j=1}^a A_{1/2^j}\ge aN $$ where we can take $a\to\infty$ as $N\to\infty$. Actually, I believe we can take $a=\theta(\log N)$, which would give convergence rate $1/\log s$, but that would require using quantitative bounds on how fast equidistribution happens and I haven't worked it our carefully.

$\begingroup$ Can you explain how to bound that $a = \theta(\log N)$? $\endgroup$– SeanCommented Nov 25, 2019 at 22:11

1$\begingroup$ @Sean The idea is that the argument about the size of $A_\epsilon$ should work for any $\epsilon\in [1/N^{1/4},1/10]$ (not trying to optimize the bounds here). We use wellknown facts about the approximability/continued fraction expansion of $\sqrt{2}$ to get that there are some natural numbers $a=\theta(N^{1/3})$ with $a\sqrt{2}b=\theta(N^{1/3})$. This gives that taking integers up to size $\theta(N)$, they are equidistributed on scale $N^{1/3}\ll\epsilon$. So we can take $\epsilon$ in the stated interval. $\endgroup$ Commented Nov 25, 2019 at 22:24

$\begingroup$ Why does that estimate relate to the limit? $\endgroup$– SeanCommented Nov 26, 2019 at 3:10

1$\begingroup$ @Sean this means that $a$ from my answer can be taken to be $\theta(\log N)$. Thus is takes $\Omega(N\log N)$ time to get to $N$. Letting $s=cN\log N$ for some small $c$, we have $\eta(s)<N$, so $\eta(s)/s<1/(c\log N)<\frac{2}{c}(1/\log s)$ $\endgroup$ Commented Nov 26, 2019 at 4:56
