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Hope this isn't a duplicate.

I was trying to solve the following problem,

Consider the following non-linear system : $$ \dot x = y-a_0-a_1 x - a_2 x^2$$ $$\dot y = -x$$ , where $a_0,a_1,a_2 \in \Bbb R \text{ and } a_2 \ne 0$ . Then prove that it has no isolated periodic orbits.

My attempt :

I considered the associated Linear system at the origin i.e.

$$Df(0) = \begin{bmatrix} -a_1 & 1 \cr -1 & 0 \end{bmatrix}$$

And then considering the eigenvalues I ended up having$$\lambda = \frac{-a_1 \pm \sqrt{{a_1}^2 -4}}{2}$$ and 3 cases : (i) ${a_1}^2 > 4$ , (ii) ${a_1}^2=4$ i.e. $a_1 = \pm 2$ and (iii) ${a_1}^2 < 4$ .

$(ii) \implies$ it (the Linear system) is either a stable or unstable node.

$(iii) \implies$ it is a center .

$(i) \implies$ it is either a stable node or an unstable node or a saddle.

All the observations $(i),(ii),(iii)$ are the expected behaviour of the Linear system at the origin $Df(0)$ .

Am I on the right track ? What do I have to do next ?

The fact that the origin is the only critical point of the non-linear system, would it help me to argue?

Thanks in advance for help.

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    No, I don't think that you are on the right path. Linearizing the system at any equilibrium can give you at most information about the behavior near that equilibrium. Do you know Bendixson–Dulac theorem? – user539887 Apr 30 '18 at 12:19
  • @user539887 Would you kindly consider posting an answer –  Apr 30 '18 at 15:27
  • "Would you kindly consider posting an answer" Why not answer @user539887's question first? – Did Apr 30 '18 at 17:44
  • @user539887 yes, but if you kindly explain your points in a bit more details, it will be extremely helpful. –  May 04 '18 at 08:13
  • It appears to me that, under your general assumptions, it is very hard to find a Dulac function (see the material linked in my previous answer; I may be wrong, of course). On the other hand, the Bendixson-Dulac criterion excludes periodic orbits, while you are interested in the nonexistence of limit cycles. Hamiltonian systems can have periodic orbits but cannot have limit cycles. However, your system does not, in general, look like Hamiltonian. – user539887 May 04 '18 at 08:28
  • @user539887 While this system doesn't look like Hamiltonian, when $a_1 =0$ it is reversible: if $(\tilde x (t), \tilde y (t))$ is a solution of a system, then $(-\tilde x(-t), \tilde y(-t))$ is also a solution. This means that an equilibrium with purely imaginary eigenvalues is truly a center and there exists a continuum of periodic orbits around it. So, the formulation of the problem is pretty misleading and I agree that a clarification is needed. – Evgeny May 05 '18 at 09:25
  • @Evgeny I have solved the problem for $a_1=0$ and its pretty easy. My problem is when the question is more general say for arbitrary 2 degree polynomial! –  May 05 '18 at 10:30
  • @ThatIs The point is that you should clarify your question as much as possible if you want others to help you. You haven't said before that you already solved case $a_1 = 0$, so we can't telepathically know that you did it. – Evgeny May 05 '18 at 12:21
  • @Evgeny I'm trying to answer the question for a general 2 degree polynomial. What isn't clear about it? –  May 05 '18 at 12:46
  • @ThatIs What you have already did, that is what is unclear. – Evgeny May 05 '18 at 13:18

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