I am struggling to come to an answer the satisfies myself with the current problem.
I have a Lipschitz vector field, restricted to an hyperbox $\mathcal{B}= [a_i,b_i] \times[a_2,b_2]\times\dots\times[a_N,b_N]$. Such a set $\mathcal{B}$ is forward invariant, therefore for every initial condition $\xi \in \mathcal{B}$ the corresponding forward orbit is contained in a compact set. For my particular system I can show that:
- For each $\xi$ the corresponding omega limit set $\omega(\xi)$ is made only of equilibria, i.e. every point in $\omega(\xi)$ is an equilibrium.
Is 1. enough to conclude that every solution converges to a certain, single equilibrium, even without assuming that the equilibria are isolated? If not, why not?
Best Answer
It isn't. For simplicity the following example will be in dimension $N = 2$ but it can be generalized to higher dimensions and we take the box $[-1,1]^2$. First take a vector field with an equilibrium in the center and all the other orbits converging in a spiraling way to the periodic orbit in the circle of radius $\frac{1}{2}$ that we will denote by $C$, for instance, the following vector field in polar coordinates that we denote by $X$ $$ \begin{cases} r' = \left( \frac{1}{4}r-r^3\right)\\ \theta' = 1 \end{cases} $$ Here orbits starting from $p$ with $\|p\| > \frac{1}{2}$ have their norm decreasing and converges to $C$ with a spinning motion. Now the key is using a bump function. Take $\varphi \colon \mathbb{R}^2 \to \mathbb{R}$ of class $\mathcal{C}^{\infty}$ such that $$ \varphi(p) \begin{cases} = 0, \ \text{if} \ \|p\| \leq \frac{1}{2} \ \text{or} \ \frac{3}{4} \leq \|p\|\\ >0, \text{if} \ \frac{1}{2} < \|p\| < \frac{3}{4} \end{cases} $$ and consider the field $Y = \varphi X$. Notice that outside the annulus $A = \{ p \in \mathbb{R}^2 : \frac{1}{2} < \|p\| < \frac{3}{4} \}$ all orbits are equilibriums, in particular the box is invariant. Also, since $Y$ is smooth in a compact set, it is Lipschitz. Now, all the orbits that start in the annulus $A$ converge to the circle $C$ in a spiraling (slower) way, so their omega limit set is the entire circle, but it is now full of equilibriums since the vector field on it is now $0$.