Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 57 / 2 / a

Add a fourth level usingwith the first two equations unchanged. The positive equilibrium must beThe first three entries are positive automatically. The fourth is positive exactly when $b_{2}r/a_1>d_3$. Thus the printed coexistence assertion requires adequate herbivore production to support the third and fourth levels; it is not true for every positive parameter set. The top predator supplies the extra mortality that was absent in the unbounded three-level case.

For a chain of $N$ levels, the same equations give $b_{i-1}{X}_{i-1}-d_i-a_i{X}_{i+1}=0$ in the interior, with the basal and top equations fixing alternating levels. The interaction matrix has zero diagonal and opposite-sign nearest-neighbor entries. Its determinant is nonzero for even $N$ and zero for odd $N$: expansion at the end gives $\det A_N=a_{N-1}{b}_{N-1}\det A_{N-2}$, with $\det A_0=1$, $\det A_1=0$. For even $N$ this gives a unique equilibrium candidate, but its entries still need to be positive. For odd $N$ a compatibility condition is necessary, just as for the three-level example. This is Lotka-Volterra food-chain parity. The parity result is specific to an exponential unregulated resource and nearest-neighbor trophic interactions.