6A (Mathematical Biology)

x_{n + 1} = g (x_{n}), g (x) = x (r + k e^{- λ x}) .

A positive equilibrium satisfies

1 = r + k e^{- λ x_{*}},

and hence

x_{*} = \frac{1}{λ} lo g (\frac{k}{1 - r}) .

It is strictly positive exactly when

k > 1 - r .

This is the nonzero equilibrium of the survival-augmented Ricker map.

b

Differentiate the map:

g^{'} (x) = r + k e^{- λ x} (1 - λ x) .

At the positive equilibrium,

k e^{- λ x_{*}} = 1 - r

, so

g^{'} (x_{*}) = 1 - (1 - r) λ x_{*} = 1 - (1 - r) lo g (\frac{k}{1 - r}) .

∣ g^{'} (x_{*}) ∣ < 1

. Existence already makes the logarithm positive, so the upper inequality is automatic. The lower inequality gives

(1 - r) lo g (\frac{k}{1 - r}) < 2.

1 - r < k < (1 - r) exp (\frac{2}{1 - r}) .

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