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So I have the system

\begin{align} x'&=y\\ y'&=-x-y\ln(x^2+4y^2) \end{align} To find the equilibrium points I need $x'=0$ and $y'=0$, thus I obtain

\begin{align} y&=0\\ -x-y\ln(x^2+4y^2)&=0 \end{align}

I don't see how to proceed here. If $y=0$ in the second equation we get $-x=0\Leftrightarrow x=0$ but if $x=0$ and $y=0$ the logarithm is undefined. So $(0,0)$ can't be an equilibrium point.

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    $\begingroup$ Maybe the second equation is meant to be interpreted as being extended to the origin by continuity? That is, $y'=-x-y\ln(x^2+4y^2)$ if $(x,y)\neq(0,0)$, but $y'=0$ if $(x,y)=(0,0)$? $\endgroup$ May 29, 2019 at 16:59
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    $\begingroup$ No, the question simply gives the system and asks to show that the system as a periodic solution. For this I need to find a positive invariant set that does not contain any equilibrium point and apply Pioncare-Benedixsons theorem. $\endgroup$
    – Parseval
    May 29, 2019 at 17:32
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    $\begingroup$ The question is poorly posed. $\endgroup$
    – copper.hat
    May 29, 2019 at 17:54
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    $\begingroup$ @copper.hat: the real question here is in Parseval's comment: show the system has a periodic orbit. And also, hello! $\endgroup$ May 29, 2019 at 17:56
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    $\begingroup$ @HansLundmark - You're absolutely right. The only thing I actually need is to find an invariant set that does not contian any equilibriums. If they exist outside of it does not change anything. $\endgroup$
    – Parseval
    May 29, 2019 at 21:08

2 Answers 2

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More likely than not, this is just a system that is defined on $\mathbb{R}^2/(0,0)$; that is, the origin is not part of the domain of the dynamical system.

However, that doesn't say anything about whether or not the neighborhood of the origin behaves as if the origin was a fixed point, and it is very likely that the singularity at the origin is removable.

To prove it, we can show using L'Hopital's rule that:

$$\lim_{y\rightarrow0} \frac{\ln(4y^2)}{1/y} = \lim_{y\rightarrow0} \frac{2/y}{-1/y^2} = \lim_{y\rightarrow0} -2y = 0$$

Therefore the singularity at the origin is removable, and you can analyze the system just as you would in the "normal" case as long as you take care to indicate that you are actually "tacking on" the $0$ value at the origin to define it over all of $\mathbb{R}^2$. In that situation, the origin is indeed a fixed point.

It is worthwhile to do this analysis, because as you can see below, the system possesses a stable limit cycle! You can formally prove it by constructing a trapping region and applying the Poincare-Bendixson theorem.

enter image description here

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  • $\begingroup$ The point of the compact set $A$ is my answer was to construct a 'trapping region'. Since $0 \notin A$, there is no need to extend the range of definition using continuity. $\endgroup$
    – copper.hat
    May 30, 2019 at 12:47
  • $\begingroup$ That’s correct, and your analysis above is correct—I thought to complement it by describing the “effectively” unstable nature of the origin, since the author had originally claimed that this was the central question (although comments made clear they were actually interested in finding a limit cycle). $\endgroup$ May 30, 2019 at 15:45
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Let $V((x,y)) = {1 \over 2} (x^2+y^2)$, and $\phi(t) = V((x(t),y(t)))$. Note that $\phi'(t) = -y^2 \ln(x^2+4y^2)$.

Let $A= \{ (x,y) | {1 \over 4} \le x^2+y^2 \le 4 \}$.

Note that $\phi'(t) \ge 0$ if $x(t)^2+y(t)^2= {1 \over 4}$.

Note that $\phi'(t) \le 0$ if $x(t)^2+y(t)^2= 4$.

$A$ contains no equilibrium points.

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