Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 2 / 6B (Mathematical Biology)

The prey have per-capita growth ${\mu }_1$, loss through encounters with predators at rate ${\alpha }_{1}v$, and density-dependent competition $\delta u$. Predators suffer per-capita mortality ${\mu }_2$, gain from consuming prey at rate ${\alpha }_{2}u$, and undergo constant harvesting or removal $ϵ$. Constant harvesting makes the equations biologically valid only while the predator population remains nonnegative.

At an equilibrium with positive populations,Writing $A={\alpha }_2{\mu }_1-\delta {\mu }_2>0$ givesFor $0<ϵ<A^2/(4{\alpha }_1{\alpha }_2\delta )$, both roots are positive and give $u>{\mu }_2/{\alpha }_2>0$ via the second equilibrium equation. Thus there are two positive equilibria.

Using the equilibrium identities in the Jacobian givesThe smaller root satisfies $v_{-}{v}_{+}=\delta ϵ/({\alpha }_1{\alpha }_2)$ and $v_{-}<v_{+}$, so $v_{-}^2<\delta ϵ/({\alpha }_1{\alpha }_2)$. Hence its determinant is negative: it is always a saddle and unstable. For small $\delta ,ϵ$, this is the equilibrium with $u\sim {\mu }_1/\delta$ and $v\sim \delta ϵ/({\alpha }_2{\mu }_1)$, as required. The determinant argument is stronger than merely taking the small-parameter limit.