 Solution

Let a scalar fixed point have magnitude $r>0$. If $z=-r<0$, then $x$ and $y$ have the same sign; the two lifts are the fixed points $(r,1)$ and $(-r,-1)$. If $z=r>0$, their signs are opposite, and the lift is the period-two orbit $(r,-1)\leftrightarrow (-r,1)$. The scalar fixed-point multiplier isThe two-return fixed-point multiplier of the period-two lift is $m^2$, so in either case dynamical stability requires $m<1$. The negative fixed point represents one of two mirror-related single-lobe periodic orbits in the flow, whereas the positive fixed point represents a double-lobe periodic orbit.

For $\delta >1$, the nearby fixed point is $z=\mu +o(\mu )$ and its fixed-point multiplier tends to zero. Thus two stable single-lobe limit cycles for $\mu <0$ glue into one stable double-lobe limit cycle for $\mu >0$. Their periods diverge as the separatrix is approached; $\mu =0,z=0$ is not a finite-period limit cycle.

For $0<\delta <1$, the small branch instead satisfies $z\sim -\mathrm{sgn}(\mu )(|\mu |/A{)}^{1/\delta }$ and its fixed-point multiplier tends to infinity. Hence one unstable double-lobe limit cycle on $\mu <0$ splits into two unstable single-lobe limit cycles on $\mu >0$. Additional finite-size stable branches depend on the surrounding map; near the resonance $\delta =1$ with $A<1$ they are resolved by the saddle-node bifurcations in part (c).