Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 31E / d

An interior equilibrium point of a dynamical system must satisfyIf $ab\ne 1$, the unique candidate isIt belongs to $\Lambda$ precisely in either of the two casesAt this equilibrium the Jacobian matrix has determinantIn the first region the equilibrium has Poincare index $+1$, so index considerations alone do not rule out a periodic orbit. In the second region it is a saddle equilibrium of index $-1$. Everywhere else there is no interior equilibrium, except at $a=b=-1$, where the segment $x+y=1$ consists of equilibria.

Every simple closed curve in the convex set $\Lambda$ bounds a region contained in $\Lambda$. A periodic orbit has index $+1$, equal to the sum of the indices of the equilibria that it encloses. Thus a region with no equilibrium has total index zero, while the region with the single saddle has total index $-1$; both contradict the index of a planar periodic orbit. At $a=b=-1$, a periodic orbit cannot cross the equilibrium segment and must lie in one of its two complementary convex regions, where it encloses no equilibrium, giving the same contradiction. Therefore index considerations exclude periodic orbits except possibly when