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?
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 $ε^{-1/2}$.