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
y
=r 1−
y.
dt
K
If the population is subjected to harvesting at a rate of H(y, t) members per unit time,
then the harvested population is modeled by the differential equation
dy
y
=r 1−
y − H(y, t).
dt
K
If too many fish are caught, then the fish population may be driven to extinction. The
following problems explore some of the questions involved in formulating a rational
strategy for managing the fishery.
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. So assume the fish are caught at rate H(y, t) = Ey,
where E > 0 is constant with units 1/time which describes the amount of effort made
to harvest this species of fish. In this case, the ODE becomes
dy
y
=r 1−
y − Ey.
dt
K
This equation is known as the Schaefer model, after the biologist who applied it to fish
populations.
(1) Show that if a < r, then there are two equilibrium solutions, y1 = 0 and
y2 = K(1 − ar ) > 0. (See definitions below.)
(2) Show that y1 is asymptotically unstable and y2 is asymptotically stable. (See
definitions below.)
(3) A sustainable yield 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 = Y (E) as a function of effort, and graph it.
(4) Determine the value of E that maximizes Y and thereby find the maximum
sustainable yield Ymax .
Definition 1. y1 is an equilibrium solution of y 0 = f (y) iff f (y1 ) = 0 (i.e., if y is constant). An equilibrium solution is asymptotically stable iff f 0 (y1 ) < 0 and asymptotically
unstable iff f 0 (y1 ) > 0.
Intuitively, if the system is started in state y0 which is near a stable equilibrium
solution y1 , then the trajectory will tend to y1 , and if the system is started in state y0
which is near an unstable equilibrium solution y2 , then the trajectory will tend away
from y2 .
2
Harvesting a renewable resource
Constant yield harvesting. Now assume the fish are caught at a constant rate h
which is independent of the size of the fish population, so H(y, t) = h. Then the ODE
becomes
dy
y
=r 1−
y − h.
dt
K
The assumption of a constant catch rate may be reasonable when y is large (y >> h).
(5) If h < rK/4, show that this ODE has two equilibrium solutions y1 and y2 , with
y1 < y2 . Determine the values of these constant solutions.
(6) Show that y1 is unstable and y2 is stable.
(7) From a plot of f (y) as a function of y, show that if the initial population is
y0 > y1 , then y → y2 as t → ∞, and also show that if the initial population
is y0 < y1 , then y decreases as t → ∞. Note that y ≡ 0 is not an equilibrium
solution, so if y0 < y1 , then extinction occurs in finite time.
(8) If h > rK/4, then show that y decreases to 0 as t increases, regardless of the
value of y0 .
(9) If h = rk/4, show that there is a single equilibrium solution ym and determine
the value of this constant function. Show also that this solution is semistable,
i.e., that trajectories with initial values y0 > y1 will tend towards y1 , but
trajectories started even slightly below y1 will tend away from y1 .
(10) Compare the value of hm = rK/4 with the value Ymax that you found in Problem (4) and determine the critical level for this population (i.e., the population
is considered to be overexploited if it falls below this level).