I have read a bit about stable and unstable manifolds on this website:
https://math.libretexts.org/Bookshelves/Differential_Equations/Book%3A_Ordinary_Differential_Equations_(Wiggins)/06%3A_Stable_and_Unstable_Manifolds_of_Equilibria?fbclid=IwAR2s9s7XB5TxehoQ0XfNA066iEYwIxvpt9kdup47t0t3AZy7PmndKsSKXU0
I have a question regarding example 6.13.
They say that "By inspection, we see that the y axis (i.e. x = 0) is the global stable manifold for the origin", but I don't quite understand how they can see that?
And can the global stable manifold be equal to the stable space?
Best Answer
It seems that the equation on that example has a typo and $\dot{y}$ should instead be equal to $-y + x^2$. Anyway, the stable manifold of the origin is, by definition, formed by the orbits that converge to the origin as time goes to infinity. Since the vector field has an extremely simple form on the $y$-axis, that is, $$ \dot{x} = 0 \ \text{and} \ \dot{y} = -y $$ we can solve it and know that for a point $(0,y_0)$ on the $y$-axis, the solution that passes by it at time $0$ is $(0,y_0e^{-t})$ which converges to $(0,0)$, this implies that the $y$-axis is contained in the stable manifold of the origin and is in fact all of the stable manifold of the origin, since we know it's a one dimensional disk.
For your second question, this example shows exactly that the global stable manifold can in fact be equal to the stable space, but that was a coincidence caused by the linear format of the vector field on the $y$-axis, they are generally not equal. In fact, for vector fields in smooth manifolds in general, the stable space lives in the tangent space of the manifold while the stable manifold lives in the manifold itself, so they are two different objects.