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I'm having trouble with the following problem:

Consider the following dynamical system (Duffing equation). $$\ddot{x}+x+\epsilon x^3 =0$$

a. Rewrite this second order system as a two dimensional system.

b. For $\epsilon>0$, show that the system has a single fixed point and it is a nonlinear centre. Sketch its phase portrait.

c. For $\epsilon<0$, show that trajectories near the origin are closed. Are trajectories far from the origin closed? Sketch the phase portrait.

I was able to do part a: $$\dot{x}=y$$ $$\dot{y}=-\epsilon x^3-x$$

I haven't looked at part c yet so for now I was hoping I could get some help with part b.

This is my working for part b:

$$\dot{x}=0 \Rightarrow y=0$$ $$\dot{y}=0 \Rightarrow -\epsilon x^3-x=0 \Rightarrow x=0, \pm \frac{1}{\sqrt\epsilon}$$

So the fixed points are $(0,0), (\pm \frac{1}{\sqrt\epsilon},0)$

Firstly, in the question it says I'm supposed to show that the system has a single fixed point but it looks like I've found three fixed points. Secondly, I don't know what they mean by "nonlinear" though and how I'd show the fixed point is a nonlinear centre.

I would appreciate it a lot if someone could help me out!

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b) For $ϵ>0$ there is indeed only one real solution to $$0=-x-ϵx^3=-x(1+ϵx^2).$$

c) Multiply with $\dot x$ and integrate to get the form of the energy of a conservative mechanical system with a 4th degree (bi-quadratic) polynomial as potential energy.

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  • $\begingroup$ Oh right of course complex solutions aren't considered here! Thank you for that! Could you tell me what they mean by nonlinear centre? $\endgroup$ – user450248 Jun 3 '17 at 10:24
  • $\begingroup$ No, not really as the linearization $\dot x=y$, $\dot y=-x$ is perfectly regular. Of course the exact solution is quantitatively different from the solution of the linearization, but it is qualitatively the same. It could simply refer to the fact that the ODE is non-linear. $\endgroup$ – LutzL Jun 3 '17 at 12:05
  • $\begingroup$ @LutzL "Nonlinear centre" is a shortcut for "an equilibrium of nonlinear system which has two imaginary eigenvalues and is surrounded by closed trajectories". So, yeah, you are right :) $\endgroup$ – Evgeny Jun 3 '17 at 15:28

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