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I encountered this example in Slotine,Lee:Nonlinear Control book. Consider the nonlinear system $$\dot{x_1} = g(x_2) + 4x_1x_2^2$$ $$\dot{x_2} = h(x_1) + 4x_1^2x_2$$ Is there a limit cycle on phase plane?
The solution calculates $\frac{\partial{f_1}}{\partial{x_1}} + \frac{\partial{f_2}}{\partial{x_2}} = 4(x_1^2 + x_2^2)$ and by Bendixson's theorem suggests than this value is strictly positive (except at the origin), so the system has no limit cycle.

Here is my question: What about the origin? Why origin in this case is not considered as a point which makes the application of the theorem useless? The origin is not even an equilibrium for the system. If it was, what difference would it make?

Excerpt from the book: Bensixson's theorem from Slotine, Li : Nonlinear Control

Zeta.Investigator
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1 Answers1

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The fact that the origin might be or not an equilibrium is irrelevant for the criterion in terms of the divergence.

John B
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  • yes. and the derivative is not strictly positive also. Why is it considered being nonzero? – Zeta.Investigator Jan 26 '16 at 22:57
  • I regret that I don't understand what you mean. Could you please clarify? – John B Jan 27 '16 at 13:11
  • Bendixson's condition requires the derivative being non-zero. 4(x1^2 + x2^2) is nonzero everywhere except at the origin. So we can't state anything about whether the system has limit cycle or not. Am I right? – Zeta.Investigator Jan 27 '16 at 14:14
  • I see. Indeed, it is as people usually formulate it. But you only need to look at the proof. Simply, if the integrand (the divergence) is positive everywhere except at one point, the integral will still be positive, and so the same argument applies. – John B Jan 27 '16 at 14:49
  • So "does not vanish" isn't quite a suitable term here? – Zeta.Investigator Jan 27 '16 at 14:52
  • Yes, I've read the proof which uses Stokes Theorem – Zeta.Investigator Jan 27 '16 at 14:53
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    It is as people usually formulate it (that it does not vanish and does not change sign). But it could simply be: is always nonnegative or always nonpositive, and vanishes at most at finitely many points (in fact it could vanish say at a smooth line, since it has zero measure). – John B Jan 27 '16 at 14:55