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This Week

In lectures, we have finished off section 3.

Next Week

I have managed to secure a new venue for the Thursday tutorials. This tutorial will now take place in WGB G 09 rather than the Windle Building.

Problems

Summer 2010 Q.1

Summer 2009 Q. 2 (b)

Autumn 2009 Q. 2(b)

Summer 2009 Q. 2(a), 6(a)

Autumn 2009 Q. 2(a), 4

Supplementary Notes

Summer 2011 Q. 2(a)

The logistic family of mappings is given by

Q_\mu(x)=\mu x(1-x),

where 0\leq\mu\leq 4 and 0\leq x\leq 1.

(a) Motivate the use of the logistic equation as a model for population growth explaining the reasoning behind each of the three terms, \mux and (1-x)

Solution : We want an equation to model a population P_n under the following two assumptions:

  1. For ‘small’ populations the growth is approximately geometric P_{n+1}\approx \mu \,P_n for some positive constant \mu.
  2. There is a maximum population M such that if the population reaches M then all the resources are exhausted and extinction ensues;  i.e. P_{n+1}=0 if P_n=M.

Consider the following model:

P_{n+1}=\mu\,P_n- \frac{\mu}{M}P_n^2=\mu P_n\left(1-\frac{P_n}{M}\right).  (*)

  1. When P_n is small in comparison to M then P_n/M\approx 0 so that P_{n+1}\approx \mu\,P_n(1-0)=\mu\,P_n as required.
  2. If P_n=M then P_{n+1}=\mu\,M(1-M/M)=kM(1-1)=0 as required.

Hence (*) satisfies these conditions.

Now we let x=P/M be the proportion of the maximum population (i.e. P_n=Mx_n):

Mx_{n+1}=\mu\,Mx_n(1-x_n),

\Rightarrow x_{n+1}=\mu\,x_n(1-x_n).

x is the proportion of the maximum population, \mu is the growth rate and the (1-x_n) term is a consequence of the fact that if x_n=1 (i.e. P_n=M), then x_{n+1}=0 (i.e. extinction).

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