## frank0520 one year ago can someone help me find/ draw the phase portraits of these two differential equations:

1. frank0520

2. frank0520

3. anonymous

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: $\begin{cases} rN=0\\[1ex] 1-\dfrac{N}{K}=0\\[1ex] \dfrac{N}{A}-1=0 \end{cases}$

4. anonymous

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

5. frank0520

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

6. anonymous

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

7. anonymous

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).|dw:1444188255191:dw|

8. frank0520

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

9. anonymous

Here are some sample solutions for the example.|dw:1444188408781:dw|

10. frank0520

|dw:1444189228506:dw| would it look like this

11. anonymous

You have three equilibrium points, and so there are four intervals you should be considering: (1) $$y>K$$, (2) $$A<y<K$$, (3) $$0<y<A$$, and (4) $$y<0$$.

12. anonymous

Sorry, replace $$y$$ with $$N$$ in that last comment. So let's say $$N<0$$, in particular $$N=-1$$. Then $\frac{dN}{dt}=r(-1)\left( -\frac{1}{A}-1\right)\left(1+\frac{1}{K}\right)$ It looks like $$r$$ is a positive constant, so keep that in mind. Since $$A$$ is a positive number, the term $$-\dfrac{1}{A}-1$$ is also negative. This multiplied by $$-1$$ gives a positive number. Multiplied by $$r$$, it's still positive. And since $$K>0$$, $$1+\dfrac{1}{K}$$ is too, and positive times positive is positive. Now let's say $$0<N<A$$. Let's take the point halfway between $$0$$ and $$A$$, i.e. $$N=\dfrac{A}{2}$$. Then $\frac{dN}{dt}=r\left(\frac{A}{2}\right)\left( \frac{A}{2A}-1\right)\left(1-\frac{A}{2K}\right)=r\left(\frac{A}{2}\right)\left( -\frac{1}{2}\right)\left(1-\frac{A}{2K}\right)$ Is this value positive or negative?

13. anonymous

You do the same sort of thing with the other intervals. Pick a convenient test point and check the sign of the derivative. For $$A<N<K$$, you can try $$N=\dfrac{K-A}{2}$$ (the midpoint between $$A$$ and $$K$$), and for $$N>K$$ I would use $$N=2K$$.