Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 31E (Dynamical Systems)

For $a=b=0$, take the Logarithmic first integral of the Lotka-Volterra equationsAlong a trajectory,Thus $V$ is a first integral. The scalar function $h(s)=s-\log s-1$ is nonnegative for $s>0$, vanishes only at $s=1$, and tends to infinity as $s\downarrow 0$ or $s\to \infty$. Moreover,is positive definite, so $V$ is strictly convex. It follows that $(1,1)$ is its unique critical point and every level set $V=c>0$ is a compact smooth simple closed curve in the positive quadrant. The vector field is nonzero on such a level curve, so uniqueness of solutions makes each connected level curve one periodic orbit. The zero level consists of the fixed point $(1,1)$. Hence every trajectory in $\Lambda$ is either periodic or the fixed point $(1,1)$.

Keep the same function $V$. For arbitrary $a,b$ its orbital derivative isLet $\gamma$ be a nonconstant orbit of the unperturbed system with period $T$. Sinceintegration over one period givesConsequently,If a perturbed periodic orbit persisted and converged to $\gamma$ as $(a,b)\to (0,0)$, its net change in $V$ over one period would vanish. The averaged first-integral obstruction to persistence of a periodic orbit would therefore requireWhen $ab<0$, the two leading terms have the same nonzero sign and their magnitude is bounded below by $r\min (A,B)(|a|+|b|)$. Thus no unperturbed periodic orbit persists for sufficiently small $a,b$ with $ab<0$.

Use the Dulac theorem on the simply connected positive quadrant with $\varphi (x,y)=1/(xy)$. If $F=(\dot{x},\dot{y})$, thenand henceFor $ab<0$, both terms have the same strict sign throughout $\Lambda$. The Bendixson-Dulac criterion therefore gives no periodic orbit in $\Lambda$. The same computation also excludes the case $ab=0$ unless $a=b=0$.

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

Part d leaves only the region above $a=-1$ and to the right of $b=-1$ lying below the hyperbola $ab=1$. Part c removes the two opposite-sign quadrants, and its displayed divergence also removes either coordinate axis except at the origin. Hence periodic orbits might still occur only inIn the $a$-$b$ plane these are the portion of the first quadrant below $ab=1$, the open square $(-1,0)\times (-1,0)$, and the origin. The origin is not merely possible: part a gives its family of periodic orbits.