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Population Dynamics... again

alane1994

Active member
Oct 16, 2012
126
Hey MHB, I have this rather large and convoluted project that consists of two problems. I have finished the first problem that consisted of 4 parts. I am working through the second part right now. It has a preamble that is kinda important. I will post that as well as the problems that I am working on.

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

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

If the population harvested 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 desireable 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 questions involved in formulating a rational strategy for managing the fishery.

Problem 1)
This problem was based upon effort involved in harvesting. I can post this question along with my work for it if you guys are interested.

Problem 2)
This is the problem that I am currently working on.

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

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

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 \(h<rK/4\) shows that Eq.(ii) has two equilibrium points \(y_1\) and \(y_2\) with \(y_1<y_2\); determine these points.

My work,

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

\(\displaystyle ry(1-\frac{y}{K})-h=0\)

\(\displaystyle -\frac{ry^2}{K}+ry-h=0\)

\(\displaystyle y^2-yK+\frac{hK}{r}=0\)

\(\displaystyle y^2-yK=-\frac{hK}{r}\)

\(\displaystyle y^2-yK+\frac{K^2}{4}=\frac{K^2}{4}-\frac{hK}{r}\)

\(\displaystyle (y-\frac{K}{2})^2=\frac{K^2}{4}-\frac{hK}{r}\)

\(\displaystyle y-\frac{K}{2}=\sqrt{\frac{K^2}{4}-\frac{hK}{r}}\)

\(\displaystyle y=\frac{K}{2}\pm\sqrt{\frac{K^2}{4}-\frac{hK}{r}}\)


(b) Show that \(y_1\) is unstable and \(y_2\) is asymptotically stable.

This part is giving me some trouble. I think I would just calculate the second derivative of the original diff.eq and then you would plug in the two points found just above right?

Any help would be appreciated!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I agree with the equilibrium points you have found:

\(\displaystyle y_1=\frac{K}{2}-\sqrt{\frac{K^2}{4}-\frac{hK}{r}}\)

\(\displaystyle y_2=\frac{K}{2}+\sqrt{\frac{K^2}{4}-\frac{hK}{r}}\)

Now, regarding the stability of the equilibrium, consider that the phase plot will look something like:

phaseplot.jpg

Disregard the actual numbers on the axes, just consider that the plot of \(\displaystyle \frac{dy}{dt}\) is a function of $y$ and has two positive roots (the equilibrium points) and is parabolic, opening downward.

Can you tell by looking at the plot how \(\displaystyle \frac{dy}{dt}\) will behave near these equilibrium points?

For example, look at $y_1$, the smaller root. Below this root, in what direction are the solutions going? How about above this root? In both cases, are they being attracted to or repelled away from this point?