I am trying to understand the following claim (I came across it while reading a paper):
Consider the map (Standard/Arnold map)
$T_{k}:(x,y)\mapsto(x+y+kf(x), y+kf(x))$, with $x\in\mathbb{R}/2\pi\mathbb{Z}$, $y\in\mathbb{R}$, $f$ is a periodic, real analytic function. $k\in\mathbb{R}^{\geq0}$ is a parameter.
We now consider a curve $(x(\theta,k),y(\theta,k))$ such that
$T_{k}(x(\theta,k),y(\theta,k))=(x(\theta+\omega,k),y(\theta+\omega,k))$
with $\omega\in(0,2\pi)$ fixed.
Assume we have a real analytic, $2\pi$-periodic (in $\theta$) function $u(\theta,k)$ such that
$y(\theta,k)=u(\theta,k)+\omega-u(\theta-\omega,k)$
Then the claim is that one can obtain the finite difference equation for $u$:
$D^{2}u-kf(\theta+u)=0$
where $D$ denotes the operator $Du(\theta)=u(\theta+\omega/2)-u(\theta-\omega/2)$.
I saw the Wikipedia article on finite differences but there only the way is discussed, how to solve a given ODE using finite differences method. Cannot construct anything "backwards" for my case.
Thank you in advance!
In the following discussion, we drop the parameter $k$ when this does not lead to confusion.
In the paper, the authors want to find a KAM curve with rotation number $\omega$ which is defined by the parametric equations $$x=x(\theta)+u(\theta),\ 1+u_{\theta}\neq 0;\ y=y(\theta).$$ Note that $Du(\theta)=u(\theta+\omega/2)-u(\theta-\omega/2)$ so we have that $$D^2u(\theta)=u(\theta+\omega)-2u(\theta)+u(\theta-\omega).$$ Note that from $y(\theta)=u(\theta)+\omega-u(\theta-\omega)$ we have $$y(\theta+\omega)-y(\theta)=u(\theta+\omega)-2u(\theta)+u(\theta-\omega)=D^2u(\theta).$$ Now from the definition of $T_k$ we see that $y(\theta+\omega)=y(\theta)+kf(x)$ and so we have $$D^2u(\theta)=kf(x)=kf(u+\theta).$$