 Solution

The Lorenz equilibria satisfy $y=x$, $z=x^2/b$ and $x(r-1-z)=0$. ThusAt the origin the third eigenvalue is $-b$ and the other two solve ${\lambda }^2+(\sigma +1)\lambda +\sigma (1-r)=0$. Hence the origin is stable for $r<1$, has one zero eigenvalue at $r=1$, and becomes a saddle equilibrium with one unstable direction for $r>1$. The symmetry $(x,y,z)\mapsto (-x,-y,z)$ interchanges the new branches.

The nonzero-equilibrium characteristic polynomial isNear $r=1$ its Routh-Hurwitz inequalities are strict, so both new branches are stable. More explicitly, center-manifold projection at $r=1$ givesThe linear coefficient crosses positively and the cubic coefficient is negative. Therefore the equilibrium bifurcation is a supercritical pitchfork at $r=1$, not a saddle equilibrium-node.

For completeness, the later stability boundary satisfiesIf $\sigma \le b+1$, the nonzero branches remain linearly stable for all $r>1$. If $\sigma >b+1$, they lose stability atAt that point $P=(\lambda +\sigma +b+1)({\lambda }^2+{\omega }_H^2)$ andThis establishes a genuine transverse Hopf crossing. It is subcritical for these positive parameters, so it is not the supercritical Hopf normal form of a dynamical system from (c). One way to check its nonlinear sign is the cubic Hopf formula. At $C_{+}$ write $s=\sqrt{b(r_H-1)}$ andFor right and adjoint eigenvectors $Aq=i{\omega }_{H}q$, $A^{\ast }p=-i{\omega }_{H}p$, $⟨p,q⟩=1$, the cubic coefficient has the sign ofSubstitution and clearing positive factors reduces its sign to that of the following polynomial in $h=\sigma -b-1>0$:Every term is positive. The radial cubic is therefore positive, giving the unstable small cycle on the stable-equilibrium side. There is no saddle equilibrium-node among the equilibrium branches for $\sigma ,b,r>0$: the only zero-eigenvalue branch intersection is the symmetry-forced pitchfork. This does not rule out separate global bifurcations of periodic orbits.