I am studying a flow in discrete time over a cylinder $ X= \mathbb{R} \times \mathbb{S}^1$ of the form:
$$\begin{cases}y_{n+1}= \frac{1}{2}y_n
\\ \theta_{n+1}= \theta_n + by_n + \frac{1}{2} \end{cases} $$
where the last identity is intended modulo 1, i.e. identifying $$\mathbb{S}^1= \mathbb{R}/\mathbb{Z}.$$
This system is very easy because I have that y contracts and I have unique invariant limit cycle in $y=0$ which is asymptotically stable and is an normally hyperbolic invariant manifold, since I can split the tangent space to the curve $y=0$ in $X$ as the direct sum of the tangent space in such curve "plus" the contractive direction.
Chosen smooth function $ f=f(\theta)\, \colon\, \mathbb{S}^1 \to \mathbb{R}$ and taken $\sigma $ small I want to perturbate the $y$ cooordinate such that I consider the system:
$$\begin{cases}y_{n+1}= \frac{1}{2}y_n + \sigma f(\theta_n)
\\ \theta_{n+1}= \theta_n + by_n + \frac{1}{2} \end{cases} $$
It is well known that a normally hyperbolic invariant manifold $M \subset X$ is preserved under small perturbation, in the sense that for $\sigma$ small enough there exists $M_{\sigma}$ sucht that it is normally hyperbolic and invariant for the perturbed system and is diffeomorphic to $M$. If such exists,
it is easy -I think- to see that trajectories converge exponentially to it, because:
$T_{M_{\sigma}}X= TM_{\sigma} \oplus M_{\sigma}^{s}$, where the latter is the perturbed stable direction and a trajectory starting in $(y,\theta) \in X$ has no other choice that getting to $M_{\sigma}$ by following the contractive direction (since I have no unstable manifold).
First question is : Is this argument true? Because I feel like I am missing something;
Second question:
The proof of persistence of invariant manifold relies on the fact that stable and unstable manifolds persist and I take the intersection. In such case I have no unstable manifold in the non perturbed system, hence I do not understand how this manifold "persists", in the sense explained above.
Basically the problem is, I perturb a system which has exponential converge to a set and I want to prove there still has exponential convergence to some set which is homeomorphic to the first set using persistence of normally hyperbolic invariant manifolds, but I am not understanding exactly everything. I would appreciate any sort of help.
Best Answer
Note that the map, even before considering $f$, need not be continuous, let alone differentiable.
But for the perturbation to also have a normally hyperbolic manifold you need the map to be differentiable and to also perturb in that class. You are right that you have a single limit cycle but for the rest you need differentiability (or at least to repeat the argument and see whether they can be generalized, which usually is not possible).
On your second question, there is no problem if there is only expansion or only contraction. In that case simply the perturbation has also only expansion or only perturbation, respectively.