 Solution

The Bendixson-Dulac criterion says the following. Let $D\subseteq {\mathbb{R}}^2$ be a simply connected domain, let $F=(f,g)$ be continuously differentiable, and suppose there is a continuously differentiable function $B$ such thathas one sign throughout $D$ and does not vanish identically on any open subset. Then $\dot{z}=F(z)$ has no periodic orbit contained in $D$.

Indeed, if a periodic orbit $\Gamma$ existed, simple connectivity would put its interior $U$ inside $D$. The vector field $F$, and hence $BF$, is tangent to $\Gamma$, so its outward flux is zero. The divergence theorem would givecontradicting the sign hypothesis.

The Poincare-Bendixson theorem states that a nonempty compact omega-limit set of a continuously differentiable planar flow that contains no equilibrium point is a periodic orbit. More generally, when the omega-limit set contains only finitely many equilibria, it is either an equilibrium, a periodic orbit, or a union of equilibria and connecting trajectories.