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I found a non-linear dynamical system which has a line of equilibrium points at $y=0$; when linearizing and evaluating at those points I find that Jacobian matrix is J=$\begin{bmatrix}0 &1\\0&0\end{bmatrix}$ on the line; what can I say about the kind of equilibrium?

Moo
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Mattew
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This indicates that there are no stable and no unstable dynamics around the critical point. You may want to look at the center manifold.

Riccardo Sven Risuleo
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  • If you look at the characteristic equation of the matrix you get $\lambda ^2 = 0$. So you have 2 poles at origin which would mean the system is unstable around operating point. Am I right ? – Brale Jul 05 '19 at 10:27
  • It's not asymptotically stable indeed. But you'd have to look at the center manifold to see if it is lyapunov stable. – Riccardo Sven Risuleo Jul 05 '19 at 10:31
  • So i can say that all those points are neither stable or instable, and to study stability of the origin i have to look at the center manifold? – Mattew Jul 05 '19 at 12:20
  • You cannot conclude anything about the points before looking at the center manifold. They may be either stable or unstable :) – Riccardo Sven Risuleo Jul 05 '19 at 16:36
  • Ok! So for example, in a system like this $\begin{cases}\dot{x}=(a-1)xy\ \dot{y}=-y^2\end{cases}$I study center manifold and i get that points along $y=0$ are unstable for every value of a; correct? Center manifold should be $y=(a-1)x$ and the respective motion $\dot{x}=(a-1)^2x^2$ ; right? – Mattew Jul 05 '19 at 17:51