Sirish KamarajugaddaS1579822

6/27/13Project 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 equationdydt

=r (1− yk ) yIf 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

dydt

=r (1− yk ) y−H ( y ,t ) (1)

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.

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

dydt

=r (1− yk ) y−Ey (i)

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.

0=r (1− yk ) y−Ey0=ry− r y

2

k−Ey

0= y (r−ryk −E)

y 1=0

0=r (1− yk ) y−Ey

Ey=ry (1− yk )

E=r (1− yk )

Sirish KamarajugaddaS1579822

6/27/13

E=r− ryk

ryk

=r−E

ry=kr−Ek

y 2=k (1− Er )

(b) Show that y = y1 is unstable and y = y2 is asymptotically stable.

Stability of y1

f ( y )=r (1− yk ) y−Ey -use an arbitrary small value, x, for y

f ( y )=r (1− xk ) x−Ex -solve for equation using x as a small positive number

f ( y )=(r−E ) x− x2

k-for arbitrarily small positive value of x; r-E > 0

From this equation we can see that the solutions near 0 move away from our y1 value of 0. This means that y1= 0 is an unstable critical point.

Stability of y2F(y) > 0

y=k (1− er )− x -since f(y) for y2 is a quadratic equation, f(y) changes

signs at all roots, therefore:

y=k (1− er )+x -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 this function is known as the yield–effort curve.

Sustainable yield implies that at a specific rate fish can continue to be caught without effecting the overall population. Therefore we look to our asymptotically stable solution

of, 2=k (1−Er

) , and by multiplying, y2 x E, we get, y 2=k (1− Er )∗E, which is the

quadratic equation of our Yield-Effort Curve.

Yield-Effort Curve

Sirish KamarajugaddaS1579822

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 r2

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

dydt

=r (1− yk ) y−h= f ( y ) (ii)

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.

0=r (1− yk ) y−h -Our first step is to set our equation equal to 0

0=ry− r y2

k−h -Next we expand our the equation

0=kry−r y2−hk0=r y2−kry+hk

0=( rk ) y2−ry+h -Here we see our equation needs to be factored to find

a solution

0=r ±√r2−4( rk )(h)

2( rk)

Sirish KamarajugaddaS1579822

6/27/13

0=r ±√r2−4 rhk

2 rk

y 1=r−√r2−4 rhk

2 rk

-Once factored we see our 2 equilibrium

solutions

y 2=r+√r2−4 rhk

2 rk

When the value h=rK4

is entered into the equation to replace the value of h we see that the

value under the radical becomes 0. However if h < rK4

then our value under the radical stays

positive.

(b) Show that y1 is unstable and y2 is asymptotically stable.

In order to determine the stability of these critical points we have calculated we need to take a look at the original function and find out how the solutions react as the values approach these critical points.

f ( y )=r (1− yk ) y−hAs it can be seen from the graph,the values tend away from y1which means that it is asymptoticallyunstable which the values tend towards y2which shows that it is asymptotically stable.Y1

Y2

Sirish KamarajugaddaS1579822

6/27/13(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 extinction will be reached in a finite time.

The solution to this part can be shown by the f(y) vs. y graph or by the 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 the 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 the 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 h=rK4

we get a 0 under the radical sign, and when

solved out we get that k = 0. If h>rK4

, we see that as t increases, our answer will

quickly approach 0 because 4 rhk

under the radical will be larger than r2 resulting in

imaginary solutions.

(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:

y 1=r−√r2−4 rhk

2 rk

Sirish KamarajugaddaS1579822

6/27/13

y 1=r−√r2−4 r ( rk4 )

k2 rk

y 1= r−√r2−r22 rk

y 1= r2 rk

y 1= r k2 r

y 1= k2