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Consider the system $$ \dot{x}=y,\quad \dot{y}=-x+y^2. $$

Obviously, $(0,0)$ is an equilibrium. The linearisation matrix at zero has purely imaginary eigenvalues. So, at least we know that zero is no hyperbolic equilibrium.

In fact, it is a center.

Whats the exact condition to have a center equilibrium? One conditions seems to be that we have only pure imaginary eigenvalues.

Rhjg
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  • If the eigenvalues are complex with zero real parts (i.e. purely imaginary) then the fixed point is a center. – fosho Jan 31 '16 at 12:40
  • That is +i and -i? – Rhjg Jan 31 '16 at 12:41
  • Or any (real) multiple. – fosho Jan 31 '16 at 12:41
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    @Daniel How about say $x'=y+x^3$, $y'=−x+y^3$? Hint: $xx'+yy'=x^4+y^4$. – John B Jan 31 '16 at 13:08
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    You should check Artem 's answer in the more general question of @Did http://math.stackexchange.com/questions/1333918/periodic-orbits-of-even-perturbations-of-the-differential-system-x-y-y/1377123#1377123 You can also check Evgeny 's answer on my question http://math.stackexchange.com/questions/1577274/formal-proof-of-lyapunov-stability for another approach. – RTJ Jan 31 '16 at 17:22
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    @CTNT Wow :) Funny fact is that this system is reversible too, with the same symmetry $(x(t), y(t)) \mapsto (x(-t), -y(-t))$. – Evgeny Jan 31 '16 at 21:56
  • So center means a pair of purely imaginary eigenvalues and reversible? – Rhjg Jan 31 '16 at 22:01
  • Do you want a quite general answer or do you want to study this particular system? – Evgeny Jan 31 '16 at 22:08
  • Rather this particular one – Rhjg Jan 31 '16 at 22:11
  • In this case it's better to stick to reversibility of this system. In general if you have $C^{\infty}$ (infinitely differentiable) system, having all zero Lyapunov values (that are mentioned in Artem's answer) is necessary and sufficient condition for having center equilibrium (as far as I remember). However, if your system is conservative or has symmetries, you can use this to conclude that equilibrium with purely imaginary eigenvalues is a center without computing all Lyapunov values. – Evgeny Jan 31 '16 at 22:16
  • @Evgeny Indeed, reversibility seems to be present in all systems in the form of a simple oscillator perturbed by even powers of the respective state as in Did's example. – RTJ Jan 31 '16 at 23:38
  • @CTNT Well, Artem showed that it's not always true that such system posess center equilibrium at the origin. However, if we correct Did's hypothesis a bit (require $f(x)$ to be odd instead of even), then the reversibility also applies. – Evgeny Feb 01 '16 at 07:01
  • @Evgeny I was referring to the system $$\ddot{x}+\omega^2 x+g_1(x)+g_2(\dot{x})=0$$ Correct me if I am wrong but if $g_2(\dot{x})$ is even ( $g_2(-\dot{x}) =g_2(\dot{x})$) and for arbitrary $g_1$ the reversibility seems to hold. If $(x(t),y(t))$ is a trajectory then $(x(-t),-y(-t))$ also is. – RTJ Feb 01 '16 at 07:47
  • @CTNT You are absolutely correct, this system is reversible too. I think that I've misunderstood you, I thought about system from this question: http://math.stackexchange.com/questions/1333918/periodic-orbits-of-even-perturbations-of-the-differential-system-x-y-y – Evgeny Feb 01 '16 at 08:30

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This is not a direct answer to the question, but a description of the real trajectories. enter image description here

enter image description here

The limit case $c=-\frac{1}{2}$ is the point $(x=0\:,\:y=0)$ which is an isolated point. Losely, one can say that a trajectory reduced to an isolated point is at equilibrium.

If we displace slightly $(x,y)$ from $(0,0)$ to any position close to $(0,0)$ but not exactly on $(0,0)$ it comes on an almost circular trajectory and so, doesn't come back to the center. In this sense, the equilibrium point $(0,0)$ can be said "unstable".

Around $(0,0)$ (with $-\frac{1}{2}<c\ll 0$ ) the equation of trajectories is approximately : $$y^2=(c+\frac{1}{2}) +(2c+1)x+(2c)x^2+O(x^3)$$ $$\left( \sqrt{-2c} \:x+x_0\right)^2+y^2=r^2+O(x^3)$$ where $x_0=\frac{c+\frac{1}{2}}{\sqrt{-2c} }$ and $r=\sqrt{ \frac{2c-1}{4c} (c+\frac{1}{2}) }$

The trajectory is an ellipse with center $(x_0\:,\:0)$ , with semiminor axis=$r$ and semimajor axis=$\frac{r}{\sqrt{-2c} }$.

The more $c$ is close to $-\frac{1}{2}$ , the more the trajectory is close to circular, with center close to $(0,0)$ and radius $r\simeq 0$ .

JJacquelin
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  • Don't call center equilibrium "unstable" :) it is Lyapunov stable, but not asymptotically Lyapunov stable – Evgeny Feb 04 '16 at 22:22
  • @Evgeny : Right, this is much more precise than loosely "unstable" in quotation marks. Thank you for the clarification. – JJacquelin Feb 05 '16 at 06:11