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frank0520

  • one year ago

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

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  1. frank0520
    • one year ago
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  2. frank0520
    • one year ago
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  3. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    @SithsAndGiggles heres a picture of the assignment if it make any difference. I have completed everything except #3

  6. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Here are some sample solutions for the example.|dw:1444188408781:dw|

  10. frank0520
    • one year ago
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    |dw:1444189228506:dw| would it look like this

  11. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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 \(N=2K\).

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