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Poincaré-Bendixson Theorem states that let $\mathbf{F} : \mathbb R^2 \to \mathbb R^2$ be a $C^1$ vector field in $\mathbb R^2$ and consider the system $\mathbf{x'} = \mathbf{F(x)}$. Suppose $K$ is a set in $\mathbb R^2$ such that:

$(1)K$ is closed and bounded;

$(2)$ the system has no equilibrium point in $K$; and

$(3) K$ contains a forward trajectory of the system.

Then, the system has a non-trivial closed orbit in $K$.

I am wondered to know why it needs conditions $(1)$ and $(2)$. Any ideas?

Martin
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1 Answers1

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For 1) Clearly $\mathbb{R}^2$ is forward invariant, but need not contain a periodic orbit (think of $\dot x=1, \dot y=0$).

For 2) consider the flow $\dot x=-x, \dot y=-y$, and $K$ a ball around the origin.

If you are wondering about the closedness of $K$ in condition 1), you can take a flows as in 2) and consider a ball minus the origin. This is forwards invariant, not closed, and does not have a periodic orbit.

Thomas Rot
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