 Solution

With $\sigma =1$, substitute $x=(u+v)/2$, $y=(u-v)/2$, $r=1+\mu$. The equations becomeThe extended centre variables are $u,\mu$. The centre manifold theorem gives smooth graph functions $V,Z$ with $V(0,0)=Z(0,0)=0$ and all first derivatives zero. Their invariance equations are $V_u\dot{u}=\dot{v}$ and $Z_u\dot{u}=\dot{z}$, after substitution of the graph.

Count $\mu$ as order $u^2$. Then $\dot{u}=O(u^3)$ on the graph, so the left side of the $Z$ equation contributes only at higher order. Its quadratic terms give $0=u^2/4-bZ$, hence $Z=u^2/(4b)+O(u^3)$. The cubic terms of the $V$ equation similarly give $0=-2V-\mu u/2+uZ/2$. ThusTerms purely in $\mu$ vanish because the origin remains an equilibrium for every parameter; these expansions describe the weighted scaling $\mu =O(u^2)$, not an unrestricted one-variable Taylor series. Substitution gives the reduced equationIts cubic coefficient is negative. For $\mu <0$ the origin is attracting in the centre direction; for $\mu >0$ it loses stability and two attracting branches emerge with $u\sim \pm 2\sqrt{b\mu }$. Therefore the bifurcation is a supercritical pitchfork. These branches agree with the exact Lorenz equilibria $x=y=\pm \sqrt{b\mu }$, $z=\mu$.