Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 4 / 32A / b

Adding the two equations gives $\dot{x}+\dot{y}=b-x$, so a fixed point hasAt this point the Jacobian has determinant $a+b^2>0$ and traceThe fixed point is asymptotically stable exactly when $T<0$, orThe equality curve exists only for $a\le 1/8$ and isFor $0<a<1/8$ the fixed point is unstable between these two curves and stable outside them; for $a>1/8$ it is stable everywhere.

On $x=0$, $\dot{x}=ay\ge 0$; on $y=0$, $\dot{y}=b>0$. On $x=x^{\ast }$, the identity $\dot{x}+\dot{y}=b-x$ gives a negative outward component because $x^{\ast }>b$. On $y=b/a$, one has $\dot{y}=-x^{2}b/a\le 0$. Finally, the sloping edge has outward normal proportional to $(1,1)$, and thereThus the vector field points inward or is tangent on every edge, so trajectories cannot leave the closed polygon $D$.

The region $D$ is compact and positively invariant and contains the unique fixed point. If that fixed point is unstable, choose an initial point in $D$ which is not on its stable set. Its nonempty compact $\omega$-limit set cannot be the fixed point. The Poincare-Bendixson theorem therefore supplies a periodic orbit. Hence a sufficient parameter condition, and the one selected by the preceding analysis, is