Sirish Kamarajugadda S1579822 6/27/13 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
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
(1)
Although it is desirable to utilize the fish as a food source, it is intuitively clear t hat 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. Problems
1. Constant Effort Harvesting. At a given level of effort, it is reasonable to assume that t he rate at which fish are caught depends on the population y: the more fish there are, t he easier it is to catch them. Thus we assume that the rate at which fish are caught is is given by H(y, t)= Ey, where E is a positive constant, with units of 1/time, t hat measures the total effort made t o harvest the given species of fish. With this choice for H(y, t), Eq. (1) becomes
(i)
This equation is known as the Schaefer model afte r the biologist M. B. Schaefer, who applied it to fish populations. (a) Show that if E < r, then there are two e quilibrium points, y1=0 and y2=K(1−E/r) > 0.
( )
Sirish Kamarajugadda S1579822 6/27/13 (b) Show that y = y1 is unstable and y = y2 is asymptotically stable. Stability of y1
-use an arbitrary small value, x, for y
-solve for equation using x as a small positive number
-for arbitrarily small positive value of x; r-E > 0
From this equation we can see that the solutions near 0 move away from o ur y1 value of 0. This means that y1= 0 is an unstable critical point. Stability of y2 F(y) > 0
-since f(y) for y2 is a quadratic e quation, f(y) changes signs at all roots, therefore: -this shows that for f(y) < 0 we have a change of sign making y2 a stable solution
(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 t his function is known as the yield –effort curve. Sustainable yield implies that at a specific rate fish c an continue to be caught without effecting the overall population. Therefore we look to our asymptotically stable solution of,
, and by multiplying, y2 x E, we get,
quadratic equation of our Yield-Effort Curve.
Yield-Effort Curve
, which is the
Sirish Kamarajugadda S1579822 6/27/13 (d) Determine E so as to maximize Y and thereby find the maximum sustainable yield Ym. From analyzing the graph it is easy to see that in order to maximize the value of y and find Ym, we need to take a look at the effort axis and see that at the values of 0 and r, we have a yield of 0. Since this is a quadratic graph the maximum yield can be found at the midpoint between 0 and r which in this case is
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
(ii)
The assumption of a constant catch rate h m ay 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.
-Our first step is to set our equation equal to 0 -Next we expand our the equation
-Here we see our equation needs to be factored to find a solution
-Once factored we see our 2 equilibrium solutions
When the value
is entered into the equation to replace the value of h we see that the
value under the radical becomes 0. However if h < positive.
then our value under the radical stays
Sirish Kamarajugadda S1579822 6/27/13 (b) Show that y1 is unstable and y2 is asymptotically stable. In order to determine the stability of these critical points we have c alculated we need to take a look at the original function and find out how the solutions react as the values approach these critical points.
As it can be seen from the graph, Y2 the values tend away from y1 which means that it is asymptotically unstable which the values tend towards y2 which shows that it is asymptotically stable. Y1
(c) From a plot of f(y) versus y, show that if the initial population y0 > y1, then y→ y2 as t→∞, but if y0 < y1, then y decreases as t increases. Note that y= 0 is not an equilibrium point, so if y0 < y1, then ex tinction will be reached in a finite time. The solution to this part can be shown by the f(y) vs. y graph or by t he t vs. y graph used in the previous part. As it can be seen from the various solution curves in the T vs. Y graph, if y0 was greater than y1 we could follow any of t he lines greater than y1 which are increasing and are asymptotic to the y2 solution as t approaches infinity. However, since we know that y1 is an unstable solution and not semi-stable, we know that if y0 was less than y1 the curves would tend away from t he y1 solution and would diverge resulting in extinction.
(d) If h > rK/4, show that y decreases to zero as t increases regardless of the value of y0.
As it was discussed in part a, if
we get a 0 under the radical sign, and when
solved out we get that k = 0. If
, we see that as t increases, our answer will
under the radical will be larger than r resulting in
quickly approach 0 because imaginary solutions.
2
Sirish Kamarajugadda S1579822 6/27/13 (e) If h = rK/4, show that there is a single equilibrium point y = K/2 and that this point is semistable (see Problem 7, Section 2.4). 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. Here we will take y1 and show that when h = rK/4, our equilibrium value will equal K/2:
√
