Without specific values for k, a, and b, all you can say is that the graph passes through (0, 0), rises to a maximum somewhere between x= 0 and x= a, goes down to (a, 0), then down to a minimum somewhere between a and b, the back up to (b, 0).
A population p(t) satisfies dp/dt = f(p), where f(p) = kp (1-p/a)(1-p/b)
The population lies in the interval 0<=p<=b
k,a,b are positive constants with b>a
Determine all equilibrium points.
For this part f(p)=0 so p=0, p=a, p=b
Then I have to sketch f as a function of p. I know by expanding f(p) that is a cubic equation but I don't have any data about k,a,b except that they are all >0.
I found f'(p) and solved it to be equal to zero to find the stationary points, i found an answer but was a complete mess. I don't know what we are expected to sketch exactly without any further information.
I will appreciate any help. Thanks in advance!
Thank you very much, I deduced that but how do I justify my sketch concerning the stationary points? What can I say to justify as much as possible my sketch?
I know it's a cubic equation and everyone knows that it goes like that, but I want a more appropriate mathematical justification if it is possible. Thanks again for the help!
And another two questions if it is possible. I have to
1) find for which values of p(0) does p(t)->a as t-> infinity
2) sketch p as a function of t for the following initial conditions: p(0)=0, p(0)=1/2*a, p(0) = a, p(0)=1/2*(a+b), p(0)=b
My question especially for 2) is how to sketch p without solving explicitly the differential equation, which I don't know how to solve since it is a third degree non separable differential equation. Thank you very much again! Appreciate it!
1) when p(0)=0 then dp/dt=0 and the solution is p=0.
y' - y (1-y)(1-y/2) = 0, y(0)=0 - Wolfram|Alpha
2)when 0<p(0)<a then p goes to a because dp/dt>0.
y' - y (1-y)(1-y/2) = 0, y(0)=0.5 - Wolfram|Alpha
3)when a<p(0)<b then p goes to a because dp/dt<0.
3)when p(0)>b then p goes to infinity because dp/dt>0.
You can try various initial conditions.