Hopf bifurcation question for the system $\dot{x} = y + \mu x$ and $\dot{y} = -x + \mu y - x^2y$

Question

So I'm analyzing this dynamical system.

$\dot{x} = y + \mu x$

$\dot{y} = -x + \mu y - x^2y$

So I'm trying to determine the bifurcation category for this system. I know this is a Hopf bifurcation. However I don't know whether it is subcritical or supercricitcal. So I'm hoping you guys would help me.

So I pick arbitrary values for $\mu$. I have $\mu= -0.1,0, 0.1$

when $\mu = -0.1$, the Jacobian matrix is

$ J= \left[ {\begin{array}{cc} -0.1 & 1 \\ -1-2xy & -0.1-x^2 \\ \end{array} } \right] $

Evaluating the Jacobian at $J(0,0)$, we have:

$ J= \left[ {\begin{array}{cc} -0.1 & 1 \\ -1 & -0.1 \\ \end{array} } \right] $

I call the trace of a matrix $\tau$ and determinant of the matrix $\Delta$,

So for $J(0,0)$, $\tau = -0.2$ and $\Delta = 1.01$ , so we know this is a stable spiral. The fate of all orbits is all orbits will spiral toward $(0,0)$ for $\mu < 0$

So when $\mu = 0$

$J(0,0)$ = $ \left[ {\begin{array}{cc} 0 & 1 \\ -1 & 0 \\ \end{array} } \right] $

$\tau = 0$ and $\Delta = 1$, we call this a center in our class. All orbits will slowly approach the fixed point (0,0). The closer it gets to the fixed point the the slower the orbit will get.

And for the last case when $\mu = 0.1$

$J(0,0)$ = $ \left[ {\begin{array}{cc} 0.1 & 1 \\ -1 & 0.1 \\ \end{array} } \right] $

$\tau = 0.2$ and $\Delta = 1.01$

This is an unstable spiral. The fate of all orbits will approach the limit cycle around the f.p $(0,0)$

So how do you determine whether this is a supercritical Hopf or subcritical Hopf case?

Answer 1

The case whether you have super- or sub- (or degenerate) Hopf bifurcation can be decided by looking at the stability properties of your equilibrium at the critical parameter value. In you situation this is $\mu=0$. So you have the system $$ \dot x=y,\\ \dot y=-x-x^2y. $$ Take the Lyapunov function $$ V(x,y)=x^2+y^2 $$ and compute $$ \dot V=2xy-2yx-x^2y^2=-x^2y^2\leq 0. $$ The derivative is zero at $x=0$ and $y=0$ but note that if $(x,y)\neq (0,0)$ none of the points with $x=0$ and $y=0$ is invariant, therefore the only candidate for the omega-limit (and alpha-limit set as well) set is the origin, which is hence globally asymptotically stable.

Since it is asymptotically stable hence you are dealing with supercritical bifurcation with a birth of a stable limit cycle. Since your equilibrium is unstable for $\mu>0$ then for $\mu>0$ and small enough there exists a unique asymptotically stable limit cycle.

Answer 2

In the case of supercritical Hopf bif. the stable limit cycle appear. Like you wrote, for $\mu>0$ you have stable limit cycle, hence it is supercritical case. The most reliable way to decide supercritical vs. subcritical is to compute first Lyapunov coeffiecient. If it is negative, you have supercritical Hopf and vice versa. However the computation of Lyapunov coefficients is quite difficult in general.