Consider the nonlinear dynamical system $(1)$ :

$x' = y(1 + x−y^2)$,

$y' = x(1 + y−x^2)$, where $(x,y)\in\mathbb{R}^2$.

(i) Determine the equilibrium points of $(1)$

(ii) Classify the equilibrium points found in part (i)

(iii) Suppose that the equations model an experimental situation such that $x≥0$, $y≥0$
($x$, $y$ could for example be related to the concentrations of chemical species in a chemical reaction). Do periodic orbits exist?

I got:

(i) $(0,0)$, $(0,1)$, $(0,-1)$, $(1,0)$, $(-1,0)$, $(\frac{1+\sqrt5}2,\frac{1+\sqrt5}2)$, $(\frac{1-\sqrt5}2,\frac{1-\sqrt5}2)$

(I am pretty sure these are correct and I haven't missed out any)

(ii) Saddle point, Unstable Spiral, cannot be classified, Unstable Spiral, cannot be classified, Saddle point, Stable Spiral respectively to the order I wrote the equilibrium points.

How do you do (iii)????

  • $\begingroup$ Look at the eigenvalues, det, and the discriminate of the Jacobian. You may want to commit this plot to memory. $\endgroup$
    – dustin
    Nov 24 '14 at 23:16
  • $\begingroup$ Yeah pretty sure. I typed in wolfram alpha the two equations set to zero and it gave me what I wrote down. Why do you ask? $\endgroup$
    – snowman
    Nov 24 '14 at 23:19
  • $\begingroup$ OK cool. But I still don't know how to do the third part because we haven't gone through jacobian. I have no idea what that is so we aren't really expected to use it. $\endgroup$
    – snowman
    Nov 24 '14 at 23:31
  • 1
    $\begingroup$ Have you learned any theorems for periodic orbits? For example, any of these: scholarpedia.org/article/Periodic_orbit . Lastly, did you do a phase portrait plot as it can be quite revealing to validate the analysis? $\endgroup$
    – Amzoti
    Nov 24 '14 at 23:34
  • $\begingroup$ Bendixons negative criterion is the only one tbh $\endgroup$
    – snowman
    Nov 24 '14 at 23:37

To repeat a comment about (iii), which does not seem to have been taken into consideration: for most initial conditions in the positive quadrant, the solution ends up out of the positive quadrant (the exceptional initial conditions are the fixed points $(1,0)$ and $(0,1)$ and the diagonal $x=y$). Hence this system is ill-suited to model the evolution of a pair of populations. Proof:

$\qquad\qquad\qquad\qquad$ enter image description here



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