Let $F:\mathbb{R}^n\to\mathbb{R}^n$ be a smooth map and $\bar{x}$ be a fixed point of $F$. Assume that the Jacobian of $F$ at $\bar{x}$ has only eigenvalues with magnitude strictly smaller or larger than $1$. When studying iterative processes of the form $x_{k+1} = F(x_k)$, the stable manifold theorem see e.g. Theorem 4.4 adapted from here, Theorem III.7, states in particular that around $\bar{x}$, there exists a neighborhood $U$ of $\bar{x}$ and a stable manifold $W^{s}$ such that:
- If $x\in W^{s}\cap U$ then $F(x)\in W^{s}$ (i.e. $W^{s}$ is locally invariant for $F$)
- If $\forall k\in\mathbb{N}$, $F^k(x)\in U$, then $x\in W^{s}$, where $F^k$ denotes the composition of $F$ with itself $k$-times.
My question concerns the analogous theorem for studying ODEs of the form $\frac{\mathrm{d}x}{\mathrm{d}t}(t) = F(x(t))$. We can get a similar statement around points $\bar{x}$ where $F(\bar{x})=0$ and the Jacobian has no eigenvalues with zero real part. In every version that I could find (e.g. Section 2.7), a statement equivalent to (1.) above is given (i.e. there exists a locally invariant stable manifold). However I could never find an equivalent version of point (2.) which would be:
- If a solution $x$ of the ODE is such that $\forall t\geq 0$, $x(t)\in U$, then $x(0)\in W^{s}$. That is, all the locally non-escaping trajectory are contained in $W^{s}$.
My question is, why is it not stated? Is it false, is it true but "obvious"? etc.