Let us consider $V(x) = 2-\sin(x) - \sin(\sqrt{2} x)$ on $x\in \mathbb{R}$ so that $V(x)>0$ everywhere. One can see that $$ \frac{C_1}{t^2} \leq \min_{|x|\leq t} V(x)\leq \frac{C_2}{t^2} $$ for all $t > 0$. Can we get an asymptotic expansion of $\min_{|x|\leq t} V(x)$ as an exact form $$ \min_{|x|\leq t} V(x) \sim \frac{c}{t^2} \qquad \text{as}\; t\to\infty? $$
$\begingroup$
$\endgroup$
3
-
$\begingroup$ Using $\sin(x) \geq 1 - \frac{((x \mod 2\pi) - \frac{\pi}2)^2}{2}$, this can be seen to be closely linked to the Diophantine approximation $|\sqrt2 n - m + \frac{\sqrt2 - 1}4|$. I believe $|\sqrt2 n - m + \beta| \leq \frac{1+\varepsilon}{\sqrt8 n}$ can be shown to have infinitely many solutions, like the general $|\alpha n - m + \beta| \leq \frac{1+\varepsilon}{\sqrt5 n}$ but excluding $\phi$, and there's obviously a lower bound for $\beta = 0$, but I'm not sure how to show a lower bound for $\beta = \frac{\sqrt2 - 1}4$. $\endgroup$– Daniel WeberCommented Jan 6 at 4:50
-
$\begingroup$ A bit more precisely, given such a solution $\sqrt2 - m + \frac{\sqrt2 - 1}4 = \varepsilon$, we can set $x = 2\pi n + \frac{\pi}2 + \delta$, and then we have $V(x) \approx \delta^2 + (\sqrt2 \delta + \varepsilon)^2$, which is minimized by $\delta = -\frac{\sqrt2}3\varepsilon$, for which $V(x) \approx \frac{\varepsilon^2}3$, and assuming $\varepsilon \sim \frac1{\sqrt8 n}$ this gives $V(x) \sim \frac{1}{24 n^2} \sim \frac{\pi^2}{6 x^2}$ $\endgroup$– Daniel WeberCommented Jan 6 at 5:13
-
$\begingroup$ the best diophantine approximations (of every irrational number) grow at least exponentially, thus this minimum takes the same value for too long $\endgroup$– Fedor PetrovCommented Jan 7 at 20:06
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
This cannot work, and in fact the Diophantine properties of $\sqrt{2}$ play no role here. Basically, the reason is that the condition imposes a kind of behavior on $(x,\sqrt{2}x)$ that is much too orderly for an ergodic orbit.
If we consider the flow $\phi_t(x,y)=(x+t,y+\alpha t)$, $\alpha\notin\mathbb Q$, on the torus $T=\mathbb R^2/\mathbb Z^2$, then we are interested in when the function $V:T\to\mathbb R$, $V(x,y)=2-\sin 2\pi x-\sin2\pi y$, becomes small. Clearly, $V$ has a unique minimum at $z_0=(1/4,1/4)$, and $V(z_0+z)\simeq c|z|^2$.
So the situation the OP was hoping for can now be paraphrased as follows (I allow myself some mild artistic license with the constants, which I don't feel like keeping track of at all times). The initial piece of the trajectory of $(0,0)$, for $0\le t\le T$, has entered the disk about $z_0$ of radius $(1+o(1))/T$, but no smaller disk.
We can also consider the Poincare section along $x=1/4$, and then we obtain a discrete system $y\mapsto y+\alpha$ on the one-dimensional torus. Clearly, the situation described in the previous paragraph is now absurd: Fix a large $M$ such that $|y_n-1/4|$ is minimal on $0\le n\le M$ for $n=M$. By assumption, this minimal value is $\simeq 1/M$. Let's say $y_M-1/4\simeq 1/M$ (rather than $-1/M$). Now by time $4M/3$, we must at some point have come closer to $1/4$. More precisely, we must have $y_N-1/4\simeq \pm 3/(4M)$, for some $M<N\le 4M/3$. Let's again assume we have the $+$ sign here.
But then this whole sequence of steps from $M$ to $N$ moves us almost exactly $1/(4M)$ units to the left and thus takes us much too close to $1/4$ when repeated three more times; the distance is now kept positive only thanks to the error terms, but it is supposed to be $\gtrsim 1/(4M)$.
We can delay this disaster if we really got to the other side of $1/4$, but since there are only two sides, this excuse works only once, and the two sequences of steps combined lead to the same problem as above if we keep changing sides.
answered Jan 6 at 23:15