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dN dt
= κ (C 0 − α N )N,
where N (t) is the bacterial density, C 0 is the available nutrient.
(a) Explain what the parameters represent, using the dimensions known. (b) Show that the equation can be rewritten as
dN dt
= r (1 −
do not nondimensionalize. (c) Let N (0) = N 0. Solve the new version of the equation.
dH dt
= rH H (1 −
a H P 1 + a H TH
dP dt
= rP P (1 −
k H
Take the following parameter values: rH = 0. 2 , K = 500, TH = 0. 5 , rP , k = 0.2 and a = 0. 001 , 0. 1 , 0 .3.
(a) When we find equilibrium points we set the time-derivative and solve for the dependent variables. Here, plot, in the P-H plane, the equations you arrive at when setting the time derivative equal to zero. Use the ”hold on” command in matlab to plot them on the same graph. One graph for each value of a. (b) Write a matlab function to solve these equations. Plot your results in the P-H plane. For each value of a test and state the effect of initial conditions (only try 3 or so). (c) Why is the a = 0.3 case different to what we’ve seen before (this will only be 1 point).