2
$\begingroup$

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?

$\endgroup$

1 Answer 1

3
$\begingroup$

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 well-approximable 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.

$\endgroup$
6
  • $\begingroup$ Can you explain how to bound that $a = \theta(\log N)$? $\endgroup$
    – Sean
    Commented 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 well-known 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? $\endgroup$
    – Sean
    Commented 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
  • $\begingroup$ Thank you a lot! $\endgroup$
    – Sean
    Commented Mar 6, 2020 at 20:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.