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Math 19. Lecture 13Remarks about Australian PredatorsT. JudsonFall 20051 A ModelWhy there are no large mammalian predators in Australia? Letk(t) = kangaroos at time tp(t) = predators at time t.Derive the following system,dkdt= αk − βk2− γkpdpdt= −σp + λkp.This is a basic predator-prey system with logistic growth for the kangaroos.We have the following null clines.• The k null clines arek = 0p = (α − βk)/γ.• The p null clines arep = 0k = σ/λ.1We have two cases.• α/β < σ/ λ• α/β > σ/ λ2 Calculating DD =∂∂k(αk − βk2− γkp)∂∂p(αk − βk2− γkp)∂∂k(−σp + λkp)∂∂p(−σp + λkp)=α − 2βk − γp −γkλp −σ +λkCase 1. α/β < σ/λ At (α/β, 0),D =−α −γα/β0 −σ + λα/β.Since α > 0 we know thatdet(D) = ασ −λα2βif and only ifσλ>αβtr(D) = −α − σ +λαβ< −α − σ + λσλ= −α < 0In this case, k = α/β and p = 0 is stable, and we are modeling large mam-malian predators.Case 2. α/β > σ/ λ At k = σ/λ and p = α/γ − βσ/(λγ)D =−βσλ−γσλλα − βσγ0det(D) =σλ(λα − βσ)tr(D) = −βσλ< 0,2Since λα > βσ. From the previous argument, k = α/σ, p = 0 is unstable. Inthis case, p > 0 and k > 0 must be the large reptile case.Readings and References• C. Taubes. Modeling Differential Equations in Biology. Prentice Hall,Upper Saddle River, NJ, 2001. Chapter 12.• “The Case of the Missing Meat Eaters,” pp.


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HARVARD MATH 19 - Lecture 13

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