If we have a nonlinear first order ODE system,

$$x'(t) =f (x, y)$$

$$y'(t) =g (x, y)$$

and we approximate it to a linear system

$$x'(t) = ax + by$$

$$y'(t) = cx + dy$$

and we get for a critical point $\vec{x_0}$ pure imaginary eigenvalues, $\lambda=\pm qi$, it is said that this critical point will be a center of the linear system, but that it may be either a center or a focus of the nonlinear system.

How can we know if it is a center or a focus of the nonlinear system?

For instance, if we have the system

$$x'(t) = x + y$$

$$y'(t) = 2x -2x^2-y$$

There is a center of the linear approximation at the critical point $(3/2,-3/2)$. How could we determine if it is a center or a focus of the nonlinear system?

It is easy to get the first integral, but not to draw it (without computer).

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    $\begingroup$ There are some cases when it's pretty easy: for example, system has a first integral or a system is reversible. If these don't include your case, Lyapunov values will help you: see beautiful answers to this question. $\endgroup$ – Evgeny Jun 21 '18 at 20:26
  • $\begingroup$ Ok, thanks, I see... But I have added an example in with I'm not sure about how to use the first integral once we have found it $\endgroup$ – Quaerendo Jun 22 '18 at 10:15
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    $\begingroup$ Wait, so you already know that there is a first integral? In that case just check Hessian matrix at point $(3/2, -3/2)$ — if its eigenvalues are of the same sign, then you are guaranteed to have a center equilibrium. This is basically a consequence of Morse lemma then: all level sets around such critical points are homeomorphic to circles, you know that there are no equilibria on them, hence trajectories staying on level sets close to critical point are periodic. $\endgroup$ – Evgeny Jun 22 '18 at 11:49
  • $\begingroup$ Oh, I see, we can transform the system into first order equation that is exact. $\endgroup$ – Evgeny Jun 22 '18 at 11:58

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