TITLE: Distance of perturbed flow from the unperturbed stable manifold
QUESTION [0 upvotes]: Consider the following system
$ \dot x= v\\
\dot v= x - x^2(1+\varepsilon cost)$
Let $\phi_\varepsilon(x,v)=\psi_\varepsilon ^{2\pi}(x,v)$ the flow of system with initial condition $(x,v)$ at time $2\pi$. So we have a discrete map.
Defining the unstable manifold as
$$W_\varepsilon^{(unstable)}=\big\{  (x,v)\in R^2 : \phi^k_\varepsilon(x,v) \rightarrow 0 \,\,for \,\,k\rightarrow-\infty \big\}$$
In the same way one can define the stable manifold.
The origin is an equilibrium point for the system, even if $\varepsilon \neq 0 $.
Consider the unperturbed unstable manifold, that is $W_0^{(unstable)}$ (which coincide with stable manifold $ W_0^{(stable)}$).
Now my book says the following:
"By continuity of the ﬂow with respect to the parameter $\varepsilon$ we have $\phi^k_\varepsilon(x,v)$ close to the unperturbed unstable manifold for $0 ≤ k ≤ K$ , for some large $K$ , e.g., the distance is $O(\varepsilon^{1/2}) $ for $K \simeq \varepsilon^{−1/2}$ "
My question is:
Why the continuity can imply this property of distance of perturbed flow from the unperturbed manifold?
How can i see this?

REPLY [0 votes]: This seems to also use the smooth dependence on the perturbation parameter, just the continuous dependence is not sufficient for that claim.
Then the distance between perturbed and non-perturbed solutions grows in the first order like $εt$, so that at $t=2K\pi\sim ε^{-1/2}$ the distance is still small like $ε^{1/2}$. One could have also used $K\sim ε^{-2/3}$ to get a distance $\simε^{1/3}$ and so on.

Let $u$ be the $ε=0$ solution and $x$ the perturbation (vector) solution with the same initial condition. Then
$$
d(t)=|x(t)-u(t)|\le \int_0^t(|f(u(s))-f(x(s))|+ε|g(s,x(s))|)\,ds
$$
Now it should be intuitive that both solutions are bounded, $x$ at or close to the equilibrium is periodic, $u$ can not move away very fast. This means that $|g(t,x(t))|\le M$ and that there is a Lipschitz constant $L$ for $f$ so that in total
$$
d(t)\le\int_0^t(Ld(s)+εM)ds
$$
which is a typical Grönwall situation leading to
$$
d(t)\le εM\frac{e^{Lt}-1}{L}
$$
This is then an a-posteriori confirmation that $x$ can not explode away from $u$.
This now somewhat defeats the first argument, as now one needs $t\sim-\frac1{2L}\lnε$ which is much smaller than $ε^{-1/2}$.
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