Question

Ok, I have a rather large an in-depth problem that I will need help with. Additional material below.

Answer 1

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

\(\displaystyle\frac{dy}{dt}=r(1-\frac{y}{K})y\)

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

\(\displaystyle\frac{dy}{dt}=r(1-\frac{y}{K})y-H(y,t)\)

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.

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 |9E\) 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 \(H(y,t)\) the equation becomes

\(\displaystyle\frac{dy}{dt}=r(1-\frac{y}{k})y-Ey\)

Questions:

(a) Show that if \(E<r\), then there are two equilibrium points, \(y_1=0\) and \(y_2=K(1-E/r)>0\)

(b) Show that \(y=y_1\) is unstable and \(y=y_2\) 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 \(y_2\). 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 \(Y_m\)

Answer 2

Well your equilibrium points are going to be where the derivative vanishes, so simply set it to zero and solve. If you want to determine the stability of those points, find the second derivative and plug them in -- a stable point will have a positive second derivative while an unstable point has a negative second derivative.

Answer 3

The rest is straightforward multiplication and maximization.

Answer 4

So I want to set

\(\displaystyle \frac{dy}{dt}=r(1-\frac{y}{K})y-Ey\)

equal to zero? Then I would solve for \(y\) I assume?

Answer 5

Yup.

Answer 6

Very cool, gimme just a minute.

Answer 7

Ok, I have two as follows.

\(y=0\)
and
\(\displaystyle y=k-\frac{EK}{r}\)

How far off am I?

Answer 8

Well, I mean, the answers are actually given in the question .... those are right.

Answer 9

I know, but I don't know where the heck I went wrong...

Answer 10

You're not wrong, I don't understand the issue.

Answer 11

I am such a dolt.

Answer 12

For part (b), how would I prove this? Is there any particular way?

Answer 13

Do what I said -- calculate the second derivative and plug in your two values for y.

Answer 14

Find the second derivative of the overall diff.eq. or the second derivative of the two points that I calculated?

Answer 15

The second derivative of a point doesn't mean anything. You need to find the second derivative of y.

Answer 16

With respect to t, if we're being precise.

Answer 17

Well, because it is with respect to t, isn't is zero?

Answer 18

Absolutely not, how did you come up with that?

Answer 19

It would be a partial derivative right? Because there are no "t's" in the original derivative you would treat everything else like a constant. But evidently that is incorrect. What should I have done?

Answer 20

No, it would not be a partial derivative, it would be a total derivative. Perhaps it would be safer to write it in this way:

\[ \frac{dy}{dt} = r\left( 1 - \frac{y(t)}{k} \right)- E\cdot y(t)\]

Answer 21

Oops, I missed a y(t) after the parenthesis -- but you get the idea.

Answer 22

I have a very gnarly looking answer, I hope this is correct.

\(\displaystyle\frac{d^2y}{dt^2}=\frac{(EK-K-2\cdot y(t))\cdot(r\cdot y(t)(1-\frac{y(t)}{K})-E\cdot y(t))}{K}\)

Answer 23

You were typing for a while and then you stopped, this has me worried.

Answer 24

My apologies -- I made a typo. It's not with respect to t, it's with respect to y. The reason I stopped was because it takes awhile to type all this in to LaTeX and do the derivatives in my head at the same time.

Answer 25

Here:
\[ \frac{dy}{dt} = f(y) \]
\[f'(y) = ? \]

Answer 26

Lets see...
I get
\[ f'(y)= r\left ( 1- \frac{y}{k}\right) - \frac{r}{k}\cdot y - E\]
\[ = \left( r \left(1-\frac{y}{k}\right) - \frac{r}{k} - E \right)\]
\[ = \left( r - r\frac{y}{k} - \frac{r}{k} - E\right)\]

Answer 27

Oops. Messed up on that second line, should be
\[ f'(y) = r - 2ry/k - E \]
for the first solution, y = 0, that yields
\[ f'(0) = r-E > 0 \]
based on your assumption that E < r.

What do you get when you plug in the second solution?

Answer 28

Well, according to my likely incorrect work, it should be,

\(\displaystyle f(K-\frac{EK}{r})=r-\frac{2r(K-\frac{EK}{r})}{K}-E\)

Answer 29

Oopsies, should be \(\displaystyle f'(K-\frac{EK}{r})\)

Answer 30

That's right, but it can be simplified greatly.
\[ f'( K - EK/r ) = r - 2r + 2E - E = -r + E = - (r-E) \]

Answer 31

You deserve so many medals. Thank you so much!

Answer 32

No problem. The rest is literally just multiplying things together and maximizing (with respect to E, now).

Answer 33

Yep, I think I can handle that. I will post what I get for parts (c) & (d) after I figure them out! At which point you will likely tell me that they can be reduced further! :P

Answer 34

For part (c) I get,

\(Y(E)=E\cdot(K-\frac{EK}{r})\)

\(Y(E)=EK=-\frac{E^2K}{r}\)

Part (d)

I believe you find the derivative of the above function and then set equal to zero right?

\(\displaystyle \frac{dY(E)}{E}=-\frac{2EK}{r}\)

Which when set equal to zero and solved for \(E\) is equal to zero...

That doesn't seem right.

Answer 35

You have two many equals signs in the second line...

Answer 36

Typo, ignore the second equal sign in the second line. Sorry.

Answer 37

which changes your answer to (d).

Answer 38

\(\displaystyle Y(E)=EK-\frac{E^2K}{r}\)

And how would it change the answer? \(\displaystyle\frac{dY(E)}{dE}(EK)=0\)

Answer 39

So that part would cancel right?

Answer 40

what?
\[Y'(E) = K(1 - 2E/R) = 0\]

Answer 41

Excuse me,

It would be,

\(Y'(E)=\displaystyle K-\frac{2EK}{r}\)

Answer 42

Wait, you beat me to it :P

And you are correct, that would change the answer.

When set equal to zero, \(E=r/2\)

Correct?

Answer 43

Yep.

Answer 44

Lastly, what does that imply for your maximum sustainable yield?

Answer 45

Would we plug that back into the original equation? Sorry, I am a little slow today.