= Solution
At a <fixed point>, the first equation gives $y=0$, the second then gives $x=0$, and the third gives $z=-\mu$. Thus there is exactly one <equilibrium> for every real $\mu$:
$$
\boxed{(x,y,z)=(0,0,-\mu).}
$$
The <Jacobian matrix> there is
$$
J=\begin{pmatrix}0&1&0\\-1&5\epsilon\mu&0\\0&0&-1\end{pmatrix},\qquad
\det(\lambda I-J)=(\lambda+1)(\lambda^2-5\epsilon\mu\lambda+1).
$$
One <eigenvalue> is $-1$. The other two have product one and sum $5\epsilon\mu$. For $\mu<0$ their real parts are negative, so the <equilibrium> is <asymptotically stable>; for $\mu>0$ both real parts are positive, leaving two unstable directions and one stable direction. At $\mu=0$ the <eigenvalues> are $-1,i,-i$, with
$$
\left.\frac{d}{d\mu}\operatorname{Re}\lambda_\pm\right|_{\mu=0}=\frac{5\epsilon}{2}>0.
$$
Therefore \b[the only local <bifurcation> of this <equilibrium> is a <Hopf bifurcation> at $\mu=0$]. A repeated real <eigenvalue> away from zero does not itself cause a <bifurcation>: the <equilibrium> remains <hyperbolic> there.
The cubic <centre manifold> calculation below gives the <Hopf normal form> radial coefficient $-9\epsilon/8<0$, so this is a <supercritical Hopf bifurcation>. Small attracting <periodic orbits> occur on the $\mu>0$ side. In particular, at $\mu=0$ the leading cubic radial drift is inward, although <linear stability> alone cannot decide that boundary case.
Back to article page
