Question

can someone help me find/ draw the phase portraits of these two differential equations:

Answer 1

For an autonomous equation like this one, the first thing to do is determine where the derivative \(\dfrac{dN}{dt}\) disappears - these are your equilibrium points. This clearly happens when either of the following equations hold:
\[\begin{cases}
rN=0\\[1ex]
1-\dfrac{N}{K}=0\\[1ex]
\dfrac{N}{A}-1=0
\end{cases}\]

Answer 2

Easy enough: the equilibrium points occur for \(N=0,K,A\). Do you have any facts about \(K\) and \(A\)? I assume they're constants, but do you know anything about their relative values? Are they both positive/negative? Is one larger than the other? etc

Answer 3

@SithsAndGiggles heres a picture of the assignment if it make any difference. I have completed everything except #3

Answer 4

So is it \(\dfrac{N}{A}\) or \(\dfrac{A}{N}\)? Your images conflict, but I assume it's the first judging by this wiki page: https://en.wikipedia.org/wiki/Allee_effect#Mathematical_models

Answer 5

Anyway, yes, both \(A\) and \(K\) are constants, and we assume \(0<A<K\) as per the wiki page, which makes sense.

So... a phase portrait is a diagram intended to give some idea of the behavior of any solutions to the ODE. Based on the sign of the derivative \(\dfrac{dN}{dt}\) between consecutive equilibrium points, you can determine whether the population \(N\) is growing or dying off.

Take a simple example: consider the autonomous ODE
\[y'=(y-1)(y-2)^2\]which has equilibrium points \(y=1\) and \(y=2\). Pick three test points, say \(y=0,\dfrac{3}{2},3\), and plug them into the ODE. Notice:
\[y=0~~\implies~~y'<0\\
y=\dfrac{3}{2}~~\implies~~y'>0\\
y=3~~\implies~~y'>0\]You get the following phase portrait (in one dimension).

Answer 6

Its the first one, but the questions is to show that this equation has the same phase portrait as the equation from wiki

Answer 7

Here are some sample solutions for the example.