Consider the ODE $\dot x=F(x)$ with $F:U\to\mathbb R^2$ ($U\subset \mathbb R^2$ open) a $C^1$ function. Assume we have an equilibrium point $(x_0,y_0)$ ($F(x_0,y_0)=0$) that is an unstable focus (i.e., $(x_0,y_0)$ is an unstable focus in the linearized system). I am wondering if it's possible that $(x_0,y_0)\in\omega(x,y)$ for some $(x,y)\neq (x_0,y_0)$, where by $\omega(x,y)$ I mean the $\omega$-limit set.
Phrased differently: Is the stable manifold of $(x_0,y_0)$ equal to $\{(x_0,y_0)\}$?
My intuition (to the rephrased question) says yes: close to $(x_0,y_0)$ the solution behaves as a solution for the linearized system, and in the linear case we know the solution has to spiral outwards.
Is my intuition correct? If so, how to prove this?
Best Answer
According to the Hartman–Grobman linearization theorem, which applies since the system is of class $C^1$ and since unstable foci are hyperbolic equilibrium points (i.e., the linearized system has no eigenvalues on the imaginary axis), the system is topologically conjugate to its linearization in some neighbourhood of the equilibrium point. So your intuition is correct: the system does behave like its linearization near $(x_0,y_0)$.