Past exam of the mathematics course of the University of Cambridge 2017 ii Paper 3 12B Solution Created 2026-09-24 Updated 2026-10-05
A fraction of adults survives each time step; is the adult mortality fraction. Adults produce larvae, which mature one step later with recruitment reduced by adult-dependent competition through . Eliminating larvae gives the second-order difference equationAt an equilibrium point, either orOnly nonnegative populations are admissible. At extinction the Jacobian matrix is and its characteristic polynomial is . Its positive root exceeds exactly when . For both roots have modulus less than ; at the roots are , so linearization is marginal. On the nonnegative state space extinction is still attracting at this equality: if two consecutive adult values are bounded by , the next is at most , with strict decrease away from zero; the maximum of two consecutive adult values strictly decreases after two steps whenever it is positive; continuity on its bounded state region then excludes a positive limiting maximum. Thus the extinction equilibrium point is unstable exactly when the positive equilibrium point exists, with equality understood through nonlinear rather than strict linear stability.
At the positive equilibrium point, the trace and determinant of the Jacobian matrix are and . The Jury stability criterion for requires , and . The strict linear stability analysis conditions areFor there is no upper bound on ; for the stable region is .
At the lower boundary , the positive equilibrium point merges with extinction, with a multiplier and slow recovery. At the upper boundary a multiplier reaches (the other is ), giving alternating adult/larval fluctuations. The nonlinear map confirms a supercritical period-doubling bifurcation, as follows. For , put and . A nonconstant period-two adult sequence satisfies and . Subtracting and adding these equations givesThus positive unequal values emerge for , with amplitude proportional to . The two larval phases are . If are the trace and determinant of the two-step Jacobian matrix, direct multiplication givesAt onset and , so the other two strict Jury stability criterion inequalities hold for sufficiently small positive . Hence just beyond the upper boundary, a stable period-two population cycle replaces the stable equilibrium. This is a local conclusion, not stability for arbitrarily large reproduction rates.
At the upper boundary itself, the positive equilibrium point is still locally asymptotically stable, despite its multiplier . To decide this equality case, apply the centre manifold theorem for a discrete dynamical system. Put and write the centre graph as with . If the reduced map is , its invariance equation is . Expanding the original recruitment map and this equation through cubic order givesConsequently . The stability at a nondegenerate flip bifurcation criterion shows algebraic attraction on the centre direction; the transverse multiplier has modulus less than one. Including this boundary, the complete local asymptotic-stability condition for a positive population is thereforeThe shaded figure shows the strict linear-stability region; its upper boundary adds this nonhyperbolic attracting case.
Demographic stochasticity near the lower boundary can lead to absorption at extinction; near the upper boundary it excites alternating fluctuations and can blur the deterministic period-doubling bifurcation. These qualitative predictions depend on the chosen stochastic recruitment and mortality rules.
