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Appropriate predator-equilibrium

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edge
#1
Oct17-04, 11:55 PM
P: 14
I've been assigned a project but I'm having trouble on one isolated part. Could someone help me out a bit? Thanks.

Solve y'(t) = py(t)(1 - (y(t)/k) )

* 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 -> infinity. Note that only solutions satisfying y(0) >= 0 are relevant.

I've solved the equation (by seperation) though there might be some errors (please correct):

case 1: y<k

1/[1/(c*e^(rt)) + (1/k)]

case 2: y>k

(c*e^rt)/[c/(k*e^(rt)) - 1]

Now, I'm stuck on what is meant by the equilibria etc. For graphing with maple am I to assume any constant? Also, r and k are constant (reproduction and carrying capacity respectively) but not given. Am I to choose any values for these and graph solutions/direction fields for each possibility and record the general behavior? I don't have a solid understanding of the actual math and an even shakier knowledge of maple so it's pretty frustrating when putting the two together.Any suggestions and help would be greatly appreciated. Also, it's hinted that there is one "appropriate predator-equilibrium for the application under consideration." Basically, an equilibrium that can be used for y to solve:

x'(t) = rx(t)(1 - (x(t0?k) - px(t)y(t)

Any ideas?
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HallsofIvy
#2
Oct18-04, 07:53 AM
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Thanks
PF Gold
P: 39,348
An "equilibrium" solution is a constant solution.

In particular, if you have an equation that says dy/dt= f(y) (only y), then if y is an equilibrium solution it must satisfy f(y)= 0.

In this particular problem f(y)= py(t)(1 - (y(t)/k) ) so the equilibrium solutions are
the constant solutions y= 0 and y= k.

One good way of dealing with solutions NEAR an equilibrium solution is to "linearize" them. Your equation is

dy/dt= py(1-y/k)= py- py2/k. As long as y(t) is VERY close to 0, y2 will be even smaller and the linear equation closest to this is
dy/dt= py. What happens to that as t-> infinity?

To see what happens when y(t) is close to y(t)= k, Try making the substitution
v(t)= y(t)- k. Of course, dv/dt= dy/dt since dk/dt= 0 and y(t)= v(t)+ k. The equation becomes dv/dt= p(v+ k)(1-(v+k)/k)= p(v+k)(-v/k)= -pv2/k- pv/k.
Again, if y is VERY close to k, then v is VERY close to 0 and so the linear equation closest to this is dv/dt= -pv/k. What happens to v (and so y) as t goes to infinity?


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