Project 1 Harvesting a Renewable Resource
Suppose that the population y of a certain species of fish (e.g., tuna or halibut) in a given area of the ocean is described by the logistic equation
dy/dt = r(1 - y/K)y.
If the population is subjected to harvesting at a rate H(y, t) members per unit time, then the harvested population is modeled by the differential equation
dy/dt = r(1 - y/K)y - H(y, t).
Although it is desirable to utilize the fish as a food source, it is intuitively clear that if too many fish are caught, then the fish population may be reduced below a useful level and possibly even driven to extinction. The following problems explore some of the questions involved in formulating a rational strategy for managing the fishery.
Project 1 PROBLEMS
1. Constant Effort Harvesting. At a given level of effort, it is reasonable to assume that the rate at which fish are caught depends on the population y: the more fish there are, the easier it is to catch them. Thus we assume that the rate at which fish are caught is given by H(y, t) = Ey, where E is a positive constant, with units of 1/time, that measures the total effort made to harvest the given species of fish. With this choice for H(y, t), Eq. (1) becomes
dy/dt = r(1 - y/K)y - Ey.
This equation is known as the Schaefer model after the biologist M. B. Schaefer, who applied it to fish populations.
(a) Show that if E < r, then there are two equilibrium points, y1 = 0 and y2 = K(1 - E/r) > 0.
(b) Show that y = y1 is unstable and y = y2 is asymptotically stable.
(c) A sustainable yield Y of the fishery is a rate at which fish can be caught indefinitely. It is the product of the effort E and the asymptotically stable population y2. Find Y as a function of the effort E. The graph of this function is known as the yield-effort curve.
(d) Determine E so as to maximize Y and thereby find the maximum sustainable yield Ym.
2. Constant Yield Harvesting. In this problem, we assume that fish are caught at a constant rate h independent of the size of the fish population, that is, the harvesting rate H(y, t) = h. Then y satisfies
dy/dt = r(1 - y/K)y - h = f(y).
The assumption of a constant catch rate h may be reasonable when y is large but becomes less so when y is small.
(a) If h < rK/4, show that Eq. (ii) has two equilibrium points y1 and y2 with y1 < y2; determine these points.
(b) Show that y1 is unstable and y2 is asymptotically stable.
(c) From a plot of f(y) versus y, show that if the initial population y0 > y1, then y -> y2 as t -> infinity, but if y0 < y1, then y decreases as t increases. Note that y = 0 is not an equilibrium point, so if y0 < y1, then extinction will be reached in a finite time.
(d) If h > rK/4, show that y decreases to zero as t increases regardless of the value of y0.
(e) If h = rK/4, show that there is a single equilibrium point y = K/2 and that this point is semistable. Thus the maximum sustainable yield is hm = rK/4, corresponding to the equilibrium value y = K/2. Observe that hm has the same value as Ym in Problem 1(d). The fishery is considered to be overexploited if y is reduced to a level below K/2.