Find Bifurcations Points

Question

Hello everbody !

How to find bifurcations points without bifurcation diagramm? (for 2D system) For example, if I want to solve this problem, how can I proceed?

${dz\over dt}= y- {1\over 3}z^3+(r-1)z$

${dy\over dt}=-z$

I found fixed point $(0,0)$ and Jacobian matrix $\begin{bmatrix}r-1 & 1\\-1 & 0\end{bmatrix}$ Characteristic polynomial: $(r-1)^2-4$ when $r=3$ => two equilibrium points: $A= (r-1)/2$

when $r>3$ => two fixed points : $B= {(r-1)-\sqrt{(r-1)-4}\over 2}$ or $B= {(r-1)+\sqrt{(r-1)-4}\over 2}$ Thank you!

Answer 1

A bifurcation point typically describes a point in parameter space (in this case, the values that $r$ can take) at which the stability, nature or existence of equilibrium points changes. Therefore you should consider the linear stability of the equilibrium point $(0,0)$ as you vary $r$. The value(s) of $r$ at which the stability or nature of the point changes will be the bifurcation point that you're looking for.

Note: by the 'nature' of an equilibrium point I'm referring to whether it is e.g. a saddle or a node, which depends on the eigenvalues of the Jacobian at the equilibrium point. A quick search will give you detailed information on the classification of equilibrium points in plane autonomous systems.

Edit: more complete solution:

The eigenvalues of the Jacobian are the roots of $$\lambda^2-\lambda(r-1)+1=0\,,$$ which are given by $$\lambda_{\pm} = \frac{(r-1) \pm \sqrt{(r-1)^2-4}}{2}\,.$$ The stability of the equilibrium point is determined by the sign of the real part of $\lambda_{\pm}$. When the square root is real (i.e. when $r\geq3$ or $r\leq-1$), we have that $\lambda_{\pm}<0$ for $r<1$, so the equilibrium point is a stable node, with the conditions satisfied when $r\leq-1$. When $r\geq3$ we have the $\lambda_{\pm}>0$, so we have an unstable node.

For $-1<r<3$ the square root is imaginary, so we now see spirals/centres. We get a centre precisely when $\Re(\lambda_{\pm})=0$, i.e. when $r=1$. This is an example of a linearly neutrally stable point, where linear stability analysis is unlikely to be a good indicator of phase plane behaviour. For $r<1$ we have a negative real part, so we have a stable spiral. For $r>1$ we have a positive real part, so we have an unstable spiral.