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Consider the system of DE $$\frac{dx}{dt}=-axy+b$$ $$\frac{dy}{dt}=axy-cy$$

Where a,b,c are positive constants.

  1. Show that the system has a unique equilibrium point.

  2. Show that any solution in the system that starts close enough to it's equilibrium point tends finally to the equilibrium point when $t$ tends to infinity.

I did 1.

The equilibrium point is $\overline x=(\frac{c}{a},\frac{b}{c})$, and to prove uniqueness I assumed there was another equilibrium point $x^* $ and after calculation I got $x^*=\overline x.$

And I'm not quite sure what should I do to solve 2.

Can someone help me please?

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user441848
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  • "And I'm not quite sure what should I do to solve 2." ?? This is entirely standard stuff. Which references are you using? – Did Jan 06 '18 at 15:06
  • @Did standard stuff? Of course this is very easy for you to solve, not for me. References: my notebook – user441848 Jan 09 '18 at 00:02
  • The point is that any set of notes on the subject has very specific indications about how to solve 2., hence you coming here and basically saying that you have zero clue about even starting to solve it, is odd. As a kind of post hoc confirmation, which parts of the (competent) answer by Andrei) was not known to you before Andrei posted it? I am curious, especially because a lot of your questions on the site follow the same pattern of being straight applications of the relevant chapter in any set of notes on the topic... – Did Jan 09 '18 at 07:11
  • @Did First, you don't know how my notes are redacted, i.e. you can't assume that there are specific indications to solve 2. I knew the theory of Andrei's answer but I didn't know how to 'apply' it to solve 2. Second, why are you wasting your time finding patterns on my questions? – user441848 Jan 09 '18 at 22:33
  • If I translate what you just wrote into common language, I reach something similar to: "Everything in Andrei's answer was in my notes from the start but I was too lazy/distracted/whatever to read them so I posted this on the site and waited for a full answer to pop up, since this strategy had also worked for my N previous questions similarly with zero context". Correct? – Did Jan 09 '18 at 22:39
  • "don't know how can I prove you that I spend time on every question that I post on MSE" Very simple: post some real context. – Did Jan 09 '18 at 22:54

1 Answers1

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Look for example at this question. The second answer also contains a detailed link about the theory behind this. What it boils down to is to calculate the Jacobian $$J(x, y) = \begin{bmatrix} \dfrac{\partial x'}{\partial x} & \dfrac{\partial x'}{\partial y} \\ \dfrac{\partial y'}{\partial x} & \dfrac{\partial y'}{\partial y} \end{bmatrix} $$ You need to calculate the eigenvalues of this matrix. You can then show that the real part of the eigenvalues are both negative, so the equilibrium is stable.

In case I did not do ant mistakes, you get $$\lambda_{1.2}=\frac{-\frac{ab}{c}\pm\sqrt{\left(\frac{ab}{c}\right)^2-4ab}}{2}$$ If the quantity under the square root is positive, it will be less than $ab/c$, so both solutions are negative. If the quantity is negative, you get two complex solutions, but the real part is $-\frac{ab}{2c}\lt 0$, since $a,b,c\gt 0$.

Andrei
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  • but I think I'm not being asked to calculate the stability of the equilibrium point – user441848 Jan 03 '18 at 02:57
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    What does a stable equilibrium mean? It means that the system tries to reach that point at $t\rightarrow\infty$ – Andrei Jan 03 '18 at 02:59
  • if the equilibrium point $\overline x$ is asymptotically stable then $\lim_{t\to\infty}x(t)=\overline x.$ So the point should't be just stable but asymptotically stable – user441848 Jan 03 '18 at 03:10
  • You can get a stable solution that is not asymptotically stable only if the eigenvalues are purely imaginary. You cannot have that in your system. You can use the Jacobian to liniarize the equations around the equilibrium point. Just solve such a system and find that a stable but not asymptotically stable solution means an oscillation without damping around the equilibrium point. The imaginary part of $\lambda$ is the oscillatory part, the real part is the damping. As long as both $a$ and $b$ are greater than $0$, you always have a negative value for the real part, therefore damping. – Andrei Jan 03 '18 at 03:28