Computer Algebra Recipes

158 CHAPTER 4. NONLINEAR ODE MODELS
producing the phase-plane picture shown in Figure 4.3. In the actual experiment,
the time interval was only about 50 hours. On reducing the time range
to be from t =0to50inthephaseportrait command line, Heather ¯nds that
the curve terminates in the vicinity of the maximum value for Nc. Inthiscase,
both species of yeast survived over the 50 hours of the experiment, with the
cerevisiae dominating over the kephir. However, assuming that enough nutrient
was available, she can see that ultimately only one species would survive over
250 hours and it isn't the cerevisiae, but rather the kephir. This is because the
interaction coe±cient ¯kc is much smaller than ¯ck. In this struggle for existence,
there would be only the victorious kephir and the vanquished cerevisiae.
To see the explicit time evolution of each yeast population, Heather uses the
phase-plane portrait command again, but now creates two plots, one with the
scene option scene=[t,N[k]], the other with scene=[t,N[c]].
> plot1:=phaseportrait([sys],[N[k],N[c]],t=0..300,
[[N[k](0)=0.5,N[c](0)=0.5]],stepsize=0.1,
scene=[t,N[k]],color=red,linecolor=blue):
> plot2:=phaseportrait([sys],[N[k],N[c]],t=0..300,
[[N[k](0)=0.5,N[c](0)=0.5]],stepsize=0.1,
scene=[t,N[c]],color=red,linecolor=red):
On superimposing the two plots with the display command, Figure 4.4 results.
> display(fplot1,plot2g,tickmarks=[3,3]);
8
6
N
4
2
0
100 200 300
t
Figure 4.4: Time evolution of the kephir (saturable curve) and the cerevisiae.
By clicking with the mouse on the plot, the crossover point for the two population
densities occurs at about 164 hours. With her assignment completed,
Heather is grateful that her sister was able to recommend a useful text. She
will have to treat her to a Starbucks latte the next time they get together.