A supercritical planar cubic Hopf bifurcation

The origin is an equilibrium for all $\mu$, and its Jacobian is $A(\mu )=\left[\begin{array}{cc}\mu & -1\\ 1& \mu \end{array}\right]$ with eigenvalues $\mu \pm i$. Thus the critical pair is simple, $\beta (\mu )=\mu$, ${\beta }^{\mathrm{\prime }}(0)=1$, and there are no other eigenvalues.

At $\mu =0$, take $q=p=(1,-i{)}^T\mathrm{/}\sqrt{2}$. The quadratic derivative $B$ is zero. For the cubic term $-\mathrm{\parallel }(x,y){\mathrm{\parallel }}^2(x,y)$, its symmetric third derivative is $C(u,v,w)=-2((u\cdot v)w+(u\cdot w)v+(v\cdot w)u)$, using the complex-bilinear extension of the real dot product. Since $q\cdot q=0$ and $q\cdot \bar{q}=1$, $C(q,q,\bar{q})=-4q$, so $G_{21}=p^{\ast }C(q,q,\bar{q})=-4$.

Therefore ${\mathrm{\ell }}_1=-2$. The hypotheses of [F2] hold, and $\beta {\mathrm{\ell }}_1<0$ exactly when $\mu >0$; since there are no transverse modes, the small cycle on that side is orbitally stable.

In polar coordinates $R=\sqrt{x^2+y^2}$ the equations give $\dot{R}=\mu R-R^3$ and $\dot{\theta }=1$. For $\mu >0$ the nonzero radial equilibrium is $R=\sqrt{\mu }$, and its radial derivative is $\mu -3R^2=-2\mu <0$, verifying the stable cycle and period $2\pi$.