# How to see that this equilibrium is a center?

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.

• 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
• @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
• You should check Artem 's answer in the more general question of @Did math.stackexchange.com/questions/1333918/… You can also check Evgeny 's answer on my question math.stackexchange.com/questions/1577274/… for another approach. – RTJ Jan 31 '16 at 17:22
• @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

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

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 $r\simeq 0$ .

• Don't call center equilibrium "unstable" :) it is Lyapunov stable, but not asymptotically Lyapunov stable – Evgeny Feb 4 '16 at 22:22
• @Evgeny : Right, this is much more precise than loosely "unstable" in quotation marks. Thank you for the clarification. – JJacquelin Feb 5 '16 at 6:11