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\begin{document}

{\bf Question}

$(*)$ To help determine a possible strategy for future whaling the
population is modelled mathematically.
\begin{enumerate}
\item
In the absence of any fishing occurring the population of humpback
whales, $Y(t)$, can be approximated as obeying the logistic
equation $$ {dY \over dt}=rY(1-Y/K)\;. $$ If $Y(0)=K/3$ find
$Y(t)$ and hence the time $\tau$ at which the population has
doubled.

\item
The model can be extended to take ``harvesting'' of the whales
into account. A simple model is to assume that the rate at which
whales are caught, called the yield, is proportional to the
population of whales. Specifically the yield is taken to be $EY$
(where $E$ is a constant determined by the amount resources
devoted to catching the whales) and the model is then $$ {dY \over
dt}=rY(1-Y/K)-EY $$ (This is known as the {\it Schaefer model}.)\\
Show that if $E<r$ there are two equilibrium points $Y=0$ and
$Y=K(1-E/r)$ and that the first of these is unstable and the
second stable.\\ From this solution find the yield (ie: $EY =$ the
rate at which whales will be caught) that will occur after a long
time (this is called the {\it sustainable yield}).  Find the value
of $E$ which gives the maximum sustainable yield (and hence the
level of whaling that will result in the maximum number of whales
being caught on a sustainable basis).\\ Comment on what might
occur if we take $E>r$?

\end{enumerate}




\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
$\ds \frac{dY}{dt}=rY\left(1-\frac{Y}{k}\right) \hspace{0.5in}
Y(0)=\frac{k}{3}$

solve by separation of variables
$\ds\int\frac{dY}{Y\left(1-\frac{Y}{k}\right)}=\int rdt$

Now by partial fractions
$\ds\int\frac{1}{Y}+\frac{\frac{1}{k}}{1-\frac{Y}{k}}dY=rt+A$

$\ln Y-\ln(k-Y)=rt+A$

use the initial data $\ds\,\,\, \ln
Y-\ln(k-Y)=rt+\ln\left(\frac{k}{3}\right)-\ln\left(\frac{2k}{3}\right)$

find $\ds\tau$ where $\ds Y(\tau)=\frac{2k}{3}$

$\ds \Rightarrow \ln\frac{2k}{3}-\ln\left(k-\frac{2k}{3}\right)=
r\tau+\ln\left(\frac{k}{3}\right)-\ln\left(\frac{2k}{3}\right)$

$\ds \Rightarrow \ln2=r\tau-\ln2 \Rightarrow \tau=\frac{2}{r}\ln2$

\item[b)]
plot solution curves of
$\ds\frac{dY}{dt}=rY\left(1-\frac{Y}{k}\right)-EY$

isoclines $\ds rY\left(1-\frac{Y}{k}\right)-EY=c,$

$\ds c=0 \Rightarrow Y=0 \rm{\ and\ }
Y=k\left(1-\frac{E}{r}\right)$

$\ds c$ is maximum when $\ds
Y=\frac{k}{2}\left(1-\frac{E}{r}\right)$

$\ds c>0, \,\,\,\, 0<Y<k\left(1-\frac{E}{r}\right)$

$\ds c<0 \left\{ \begin{array}{l}Y<0\\
Y>k\left(1-\frac{E}{r}\right)\end{array}\right.$

\begin{center}
\epsfig{file=114-5-1.eps, width=70mm}

Solution curves for $1-\frac{E}{r}>C$
\end{center}


Equilibrium at $\ds Y=0$ which is unstable, and at $\ds
Y=k\left(1-\frac{E}{r}\right)$ which is stable.

The yield is $\ds EY$ so at $\ds t\rightarrow\infty, \,\,\,\,
Y\rightarrow k\left(1-\frac{E}{r}\right)$

therefore yield $\ds \rightarrow Ek\left(1-\frac{E}{r}\right).$

The maximum yield occurs when $\ds \frac{d(YE)}{dE}=0$

$\ds \Rightarrow k\left(1-\frac{2E}{r}\right)=0 \Rightarrow
E=\frac{r}{2}$

i.e. catch fish at half the rate they are born.

If $\ds E>r$ then the solution curves are:

\begin{center}
\epsfig{file=114-5-2.eps, width=70mm}
\end{center}

Hence for physically relevant use of $\ds Y\geq0$ we have only one
equilibrium $\ds (Y=0)$ and it is stable.  If we harvest faster
than the fish are born $\ds (EY>rY)$ then we slowly lose the
population of fish.
\end{itemize}


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