Does local stability of a unique equilibrium in a bounded region implies global stability?

Question

If I have a non-linear system of ODE's, $\dot{\mathbf{x}} = f(\mathbf{x})$, where $f(\mathbf{\cdot})$ is a smooth non-linear function. Additionally, I know that $\mathbf{x}$ is bounded and the system has only a unique steady state that is locally asymptotically stable, $\mathbf{x}^*$. Is it wrong to conclude that $\mathbf{x}^*$ is globally asymptotically stable? I feel like it is too naive.

Answer 1

There are plenty of counterexamples, many of them taken from "real life" problems, but let me build one just from scratch. In dimension two, take a system written in polar coordinates as $$ \begin{cases} \dot{r} = -r (r - 1)^2 \\ \dot{\theta} = 1. \end{cases} $$ The system has one, locally asymptotically stable equilibrium $(0,0)$ (in Cartesian coordinates), with attraction basin equal to $\{(x,y) : x^2 + y^2 < 1\}$, one periodic orbit $\{(x,y): x^2 + y^2 = 1\}$, unstable from inside and asymptotically stable from outside. All solutions are bounded in positive time. Moreover, the disk $\{(x,y): x^2 + y^2 \le 1\}$ is the global attractor.