I am emailing a link of this to everyone on the class list every week. If you are not receiving these emails or want to have them sent to another email address feel free to email me at jippo@campus.ie and I will add you to the mailing list.

## 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, $\mu$ $x$ 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).