Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 32A / b / Solution

At an equilibrium point of a dynamical system, the second equation givesThe choice $x=0$ is impossible because then $\dot{x}=r>0$. Hence $y=s/x$, and substitution into the first equation givesThus the unique fixed point is

Choosewhich satisfies $0<\alpha <r$. LetWe check the direction of the vector field on its four edges.

On $x=\alpha$,On $y=0$,On $y=s/\alpha$, where $x\ge \alpha$,Finally, throughout the system,On the sloping edge $x+y=r+s/\alpha$, the bound $y\le s/\alpha$ implies $x\ge r$, and hence $\dot{x}+\dot{y}\le 0$. Every boundary edge therefore points inward. Violations of the corresponding four inequalities are driven toward the boundary in the same order, so positive trajectories enter this compact trapping region and then remain there. This is the Brusselator trapping region.

The Jacobian matrix at the unique fixed point isIts determinant and trace areWhen $s-1>r^2$, the trace is positive, so linear stability of a planar equilibrium shows that the fixed point is a repeller. It lies in the interior of $R$.

Choose a trajectory in $R$ other than the equilibrium. Compactness of $R$ gives a nonempty compact omega-limit set, and the repelling equilibrium cannot belong to that set. Since there are no other equilibria, the Poincare-Bendixson theorem gives a periodic orbit. HenceThis is the Brusselator periodic-orbit criterion.

Solved by gpt-5.6-sol high.