1

I have the following dynamical system:

$$\begin{eqnarray} \dot{\omega} & = & -\omega+J \\ \dot{J} & = & -\beta_{2} \\ \dot{\beta}_{1} & = & -\omega\beta_{2} \\ \dot{\beta}_{2} & = & J-\hat{\alpha}\omega(1-\beta_{1}) \end{eqnarray}$$

Where $\hat{\alpha}$ is a constant. Computing the equilibrium points I find that I have two: $(0,0,1,0)$ and $(J_{0},J_{0},1-\hat{\alpha}^{-1},0)$. Computing the Jacobian for the first point one obtains the following characteristic equation: $$\lambda(\lambda+1)(\lambda^{2}+1)=0$$ The second point has the characteristic equation: $$\lambda(\lambda^{3}+\lambda^{2}+(1-\hat{\alpha}J_{0}^{2})\lambda-\hat{\alpha}J_{0}^{2}+2)=0$$

So the first question I have is: Does the $J_{0}$ in the second equilibrium mean that I don't have an isolated equilibrium point?

What do the different eigenvalues mean? I get that I'm going to have complicated structure, and the phase plane will be a phase 4-space.

Any suggestions?

Matthew Hunt
  • 543
  • 2
    I think you've overlooked that $(0, 0, \beta_1^{(0)}, 0)$ is also a family of equilibrium solutions. And yeah, each of these equilibrium points are non-isolated hence there is the zero eigenvalue of Jacobi matrix. The implication is this way, not vice-versa: for example, saddle-node equilibrium has zero eigenvalue, but it is an isolated equilibrium. – Evgeny Mar 22 '18 at 00:53
  • So the zero eigenvalue in both indicates that the equlibirium points are non-isolated? – Matthew Hunt Mar 23 '18 at 09:26
  • I've specifically mentioned what implies what and gave examples, read this comment carefully please. – Evgeny Mar 23 '18 at 13:02
  • If you say so. You still haven't been that clear. Perhaps someone else can answer? – Matthew Hunt Mar 24 '18 at 18:07

0 Answers0