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

Use a basal resource $X_1$, herbivore $X_2$ and predator $X_3$, with positive encounter, conversion and mortality parameters:Only the resource has intrinsic positive growth; the consumers die without food. There is no resource carrying capacity, immigration or feeding on nonadjacent levels. These assumptions define the simplest food chain with Lotka-Volterra equations.

A positive equilibrium would require both $X_2=r/a_1$ and $X_2=d_3/b_2$. Hence three-level coexistence is possible only on the special balance $\kappa =b_{2}r/a_1-d_3=0$. The exact identityfollows by combining $b_{2}d\log X_1/dt$ and $a_{1}d\log X_3/dt$. It gives a useful classification rather than incorrectly postulating a generic coexistence fixed point.

If $\kappa <0$, the top predator cannot invade the underlying resource-herbivore cycle: its average per-capita growth there is $\kappa$. This extinction statement can be strengthened. The boundary equilibrium has $X_{1\ast }=d_2/b_1$, $X_{2\ast }=r/a_1$, $X_{3\ast }=0$. With weights $w_1=1$, $w_2=a_1/b_1$, $w_3=a_1{a}_2/(b_1{b}_2)$, defineDirect cancellation of paired trophic terms gives $\dot{V}=w_3\kappa X_3<0$. Its level sets bound the first two species above and away from zero and bound $X_3$. Integration and uniform continuity imply $X_3\to 0$; the limiting dynamics are the usual resource-herbivore closed orbits, not generally an attracting equilibrium.

If $\kappa =0$, there is a line of positive equilibria: $X_2=r/a_1$, $b_1{X}_1-a_2{X}_3=d_2$. Choosing any positive point on that line gives a conserved coercive version of $V$ with all three relative-entropy terms. Together with the product integral, it confines generic trajectories to closed curves, with neutral oscillations around the equilibrium line. All three can persist, but only under this finely balanced parameter condition.

If $\kappa >0$, the product identity grows without bound, so at least one of $X_1,X_3$ must become unbounded in the unregulated model. There is no bounded persistent three-species attractor. Finite-time blow-up is excluded by a weighted total-abundance bound whose encounter terms cancel and whose derivative is at most a constant times that abundance. The unbounded long-term behavior signals the missing biological resource regulation. The simplest odd chain does not generically permit bounded full coexistence; adding resource self-limitation changes this conclusion.

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.

Assume the equilibrium in part (a) is strictly positive and start with all four species positive. Choose weights recursively by $w_{i+1}=w_i{a}_i/b_i$ and defineThe weighted interaction matrix is skew-symmetric matrix: for every adjacent pair $w_i{a}_i=w_{i+1}{b}_i$. On differentiating $V$, the equilibrium removes the constant growth terms, while paired products cancel. Thus $\dot{V}=0$. Each summand is nonnegative and diverges as its positive argument approaches zero or infinity. A fixed level of $V$ therefore bounds every abundance above and away from zero. This proves genuine persistence of all four species, although their dynamics need not converge to a fixed point.

Integrate each per-capita equation over $[0,T]$ and divide by $T$. Since $\log X_i(T)$ is bounded, the left side tends to zero. Writing ${\overline{X}}_i(T)=T^{-1}{\int }_0^T{X}_i(t)dt$, the limiting balances givewith the explicit abundances listed in part (a). The limit exists because the invertible four-level interaction matrix solves the averaged linear equations, whose residuals tend to zero. Periodicity or an ergodic assumption is not needed.

The same proof works for any even nearest-neighbor chain with a feasible positive equilibrium. It does not make all even-chain parameters feasible. Odd chains generically lack the required positive balance, and in their exceptional balanced cases the averaged equations do not uniquely fix every abundance. Resource density dependence and more general feeding interactions alter both the conservation law and the simple parity classification. Coexistence and the stated mean abundances hold under positivity of the equilibrium, not as an unconditional rule for every food chain.

Not in general. A food web with Lotka-Volterra equations has per-capita balances ${\dot{X}}_i/X_i=r_i+{\sum }_j{A}_{ij}{X}_j$. If all species remain bounded above and away from zero, averaging requires $r+A\overline{X}=0$ with positive means. Many webs have no feasible positive solution. Even a feasible solution does not ensure stability or persistence when the weighted skew-symmetric matrix structure and its coercive first integral are absent.

A simple resource shared by two consumers already illustrates competitive exclusion principle. Suppose their per-capita equations are $\dot{Y}/Y=cX-d$ and $\dot{Z}/Z=eX-f$. ThenIf $d/c\ne f/e$, the ratio has exponential drift, incompatible with bounded persistence of both consumers. The lower break-even-resource consumer has the advantage; equal break-even requirements are a special neutral balance, not generic coexistence. More complicated feeding links can provide additional niches, but positive coefficients and a larger web do not by themselves guarantee that every species survives. One must test feasibility, invasion into the relevant boundary states and stability, with explicit assumptions about resource renewal and self-limitation.