 Solution

The first equilibrium equation gives $y=x$, and the third gives $z=x^2/b$. The second then becomes $x(r-1-x^2/b)=0$. Hence the equilibria areAt the origin, one eigenvalue is $-b$ and the remaining two solve ${\lambda }^2+(\sigma +1)\lambda +\sigma (1-r)=0$. Their discriminant is $(\sigma -1{)}^2+4\sigma r>0$. For $r<1$ both are negative, so the origin is linearly stable, with all eigenvalues real and negative. For $r>1$ one is positive and one negative, so it is a saddle equilibrium with one unstable and two stable directions. At $r=1$ its eigenvalues are $0,-(\sigma +1),-b$: linearization has a neutral direction and cannot alone prove nonlinear stability. Given the source's global attraction for $r<1$, there are no nontrivial limit cycles or persistent chaotic invariant dynamics in that regime; all initial states tend to the origin.

To establish the local pitchfork, one needs the simple eigenvalue to cross zero, the two symmetry-related branches to exist and acquire the expected local stability, and nondegeneracy of the nonlinear coefficient. The derivative of the crossing eigenvalue is $\sigma /(\sigma +1)>0$. At $C_{\pm }$, the characteristic polynomial isFor $r>1$ sufficiently close to one, all coefficients are positive and $(\sigma +b+1)b(\sigma +r)>2\sigma b(r-1)$, so the cubic Routh-Hurwitz criterion makes both new equilibria stable. Together with the symmetry $(x,y,z)\mapsto (-x,-y,z)$ and the square-root branches, this yields the supercritical pitchfork. A local centre-manifold coordinate proportional to $x$ has leading normal form $\dot{q}=\frac{\sigma }{\sigma +1}(r-1)q-\frac{\sigma }{b(\sigma +1)}{q}^3+\cdots$, explicitly displaying the nonzero stabilizing cubic coefficient. At $r=1$ that cubic also gives weak local nonlinear attraction along the centre direction.