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Predator vs. prey - trouble

by edge
Tags: predator, prey, trouble
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Oct11-04, 03:29 AM
P: 14
I am having a bit of trouble deciding how to tackle a project in my LDE class. Any advice on how to get started or what to look out for would be greatly appreciated. It's quite lengthy so I'll try to shorten it wherever possible.

Address the impact of a generalist predator on a population of preys. Assume that the evolution of the prey pop x and of the predator pop y can be modeled by a logistic law, and that the rate of prey killed is proportional to both, the size of the prey population and the size of the predator population:

(1){ x'(t) = r*x(t)*(1 - x(t)/K) - p*x(t)*y(t)
y'(t) = q*y(t)*(1 - y(t)/d)

1. Explain the biological meaning of the parameters r, K, p, q, d.
(in the following assume all 5 parameters are positive)

2. Solve the second equation of (1) and discuss qualitatively the solution diagram. Address in particular the question of equilibria and the behavior of solutions near these equilibria. Determine the behavior of solutions as t approaches infinity. Only solutions satisfying y(0) >= 0 are relevant.

3. There is one appropriate predator-equilibrium for the application under consideration. Use this equilibrium for y and solve the first equation of (1). Find a sharp condition which guarantees that the size of the prey population does not converge to 0 as te approaches infinity. Determine the limit in that case.

4. Using qualitative tools such as direction fields and uniqueness argue that the findings under 3. extend to the general case. "This requires great care."
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Oct11-04, 05:43 AM
P: 42


This means [tex](\frac{1}{y}+\frac{1}{d-y})\frac{dy}{dt}=q[/tex]

Integrate both sides to var t, you will get [tex]ln\frac{y}{d-y}=qt+C_0, C_0 = const[/tex]
[tex]\frac{y}{d-y}=C_1 e^q^t[/tex]

Take y out and substitute it into the equation of x and similarly try to solve that equation (x,t). Then Find lim of x when t goes to infinity, this is what you are being asked to do..

I don't know if I has made any mistake in this post but hehhe you check it again yourself.
Oct12-04, 02:19 AM
P: 14
Thank you very much! That helped =) I think this is simpler than I'm making it. Your help is greatly appreciated :)

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