Sustainable Harvesting

(Adapted from Chapter 10, Project 1)

In this project, we investigate the effects of harvesting a population of fish.  We want to know what level of harvesting leads to a sustainable yield.  That is, what restrictions should be placed on fishing to make sure the population does not eventually die out?

Logistic growth, no harvesting

When there is no fishing, a population of fish satisfies the differential equation:

(dN/dt)= 2N - 0.01(N^2)

where N is the number of fish at time t, in years.  We will rescale this function to make the slope fields look nicer.  Let y(t) be the number of fish in hundreds, at time t, in years.  Then we have

(dy/dt)= 2y - 0.01(y^2)

dYdt[y_] = 2y - y^2;
Plot[dYdt[y], {y, 0, 2.5}]

Notice that when y = 0 or 2, dy/dt = 0.  That means that when the population of fish is zero or two hundred, the population neither increases or decreases, i.e. it is in equilibrium.  An equillibrium solution is constant for all values of the independent variable.  The graph is a horizontal line.  Equilibrium solutions can be identified by setting the derivative of the function to zero.

Here is the slope field for dy/dt.

StreamPlot[{1,dYdt[y]}, {x, 0, 3}, {y, -0.5, 2.5}]

Use the slope field to sketch solutions showing N against t for various initial values.  What do you see?

Notice the equilibrium solutions when y = 0 and 2.

An equilibrium solution is stable if a small change in the initial conditions gives a solution which tends toward the equilibrium as the independent variable tends to positive infinity.

An equilibrium solution is unstable if a small change in the initial conditions gives a solution curve which veers away from the equilibrium as the independent variable tends to positive infinity.

Is the equilibrium solution y = 0 stable or unstable?  What does that mean in terms of the population of fish?  What about the equilibrium solution y = 2?

Harvesting

Fish are now removed by fishermen at a continuous rate of 75 fish per year.  Let P be the number of fish at time t with harvesting.  Then P satisfies the differential equation:

(dP/dt)= 2P - 0.01(P^2) - 75

(Why is this the correct differential equation?)

Rescaling as before, let z(t) be the number of fish in hundreds, after t years.

dzdt[z_] = 2z - z^2 - .75;
Plot[dzdt[z], {z, 0, 2}]

What are the equilibrium solutions?

Here is the slope field for dz/dt.

StreamPlot[{1,dzdt[y]}, {x, 0, 2}, {y, 0, 2}]

For what initial values of P does P increase?  At what value does  P level off?

For what initial values of P does P decrease?

Use the slope field to sketch graphs of P against t, with initial values P = 40, 50, 60, 150, 170.

What are are the equilibrium values of P ?  Are they stable?  Unstable?