# Long time average of solution to ODE with almost periodic structure

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?

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 $$A_\epsilon=\left\{x\in [0,N]\,\bigg\vert\,\epsilon/2 we have that the measure of $$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.

• Can you explain how to bound that $a = \theta(\log N)$?
– Sean
Commented Nov 25, 2019 at 22:11
• @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. Commented Nov 25, 2019 at 22:24
• Why does that estimate relate to the limit?
– Sean
Commented Nov 26, 2019 at 3:10
• @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 $\eta(s)/s<1/(c\log N)<\frac{2}{c}(1/\log s)$ Commented Nov 26, 2019 at 4:56
• Thank you a lot!
– Sean
Commented Mar 6, 2020 at 20:08