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The problem is: Assuming that the parameters $a, b$ are real numbers and that $ab \neq0$, by transforming the system using polar coordinates, prove that the system $$x'=y+x(1-a^2x^2-b^2y^2)$$ $$y'=-x+y(1-a^2x^2-b^2y^2)$$ has at least one limit cycle in the phase plane.

My proof is: taking $x=r\cos\theta$ and $y=r\sin\theta$, we have $x^2+y^2=r^2$, i.e. $rr'=xx'+yy'$, so: $$r'=r(1-a^2r^2\cos^2\theta-b^2r^2\sin^2\theta)$$ $$\theta'=-1.$$ So, $\theta = \theta(0)-t$ and $$1-a^2r^2\cos^2\theta-b^2r^2\sin^2\theta=0 \implies \frac{x^2}{b^2}+\frac{y^2}{a^2}=\frac{1}{(ab)^2}.$$ That means the system has a periodic solution which is ellipse.

My question is how to prove that it is limit cycle? or the system has another solution which is limit cycle?

Any hints are welcome! Thank you!

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    $\begingroup$ Well, if you are suspicious that some $x(t)$ and $y(t)$ are solutions of the system -- just plug them in. I recommend you to plug and check, though the disappointment awaits you on this path (they aren't solutions, sorry). This problem is a classical example where Poincare-Bendixson theorem works really nice: you have to show that there is an annulus that contains no equlibria and vector field points inward on both its boundary components (or outward). Then you are done: no equilibria inside that annulus, it's forward (or backward) flow invariant, thus it must have at least one limit cycle $\endgroup$
    – Evgeny
    Jan 4, 2016 at 22:03
  • $\begingroup$ @Evgeny So, as I understand I can take the annulus with inner circle(radius $>0$) smaller than the ellipse described above, and outer circle greater than the ellipse. Hence, using the equation for $r'$ above, we conclude that the orbits which start inside of inner circle will enter the annulus. Similarly, the orbits which start outside the largest circle will also enter the annulus. Using Poincare-Bendixson (since the only equilibrium is zero which is not inside of the annulus) we conclude that there exists at least one limit cycle. Am I right? $\endgroup$
    – Kerr
    Jan 4, 2016 at 22:17
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    $\begingroup$ You are fantastically perfectly right :) btw, you can slightly weaken your reasoning: every trajectory that starts at the boundary of annulus, goes into it. To some extent, it doesn't really matter if every orbit that starts inside of inner circle will enter annulus. Only boundary is really important. That's the point of technique called "trapping region". $\endgroup$
    – Evgeny
    Jan 4, 2016 at 22:33
  • $\begingroup$ @Evgeny i understand now. Thank you very much! your explanation is excellent. $\endgroup$
    – Kerr
    Jan 4, 2016 at 22:36
  • $\begingroup$ My pleasure, I'm glad that it helped. $\endgroup$
    – Evgeny
    Jan 4, 2016 at 22:56

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