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The centre manifold theorem gives, near an equilibrium of a sufficiently smooth system, a local invariant manifold tangent to the generalized eigenspaces with zero-real-part eigenvalues, expressible as a graph of a function over them. Its reduced dynamics govern local equilibria and their stability in the presence of strictly stable transverse directions. The manifold need not be unique or analytic; finite Taylor jets can nevertheless be determined from invariance. Treating a parameter as a variable with $\dot{\mu }=0$ gives the extended centre manifold for a parameter.

At $r=1$ the linearized $(x,y)$ block has eigenvalues $0,-2$, and the $z$ direction has eigenvalue $-1$. Thus the origin is a nonhyperbolic equilibrium, with stable dimension two and non-extended centre dimension one. The extended centre dimension is two. The transformed equations areThe graph of a function satisfies $V(0,0)=Z(0,0)=0$ and $DV(0,0)=DZ(0,0)=0$, with invariance equations $V_u\dot{u}=\dot{v}$ and $Z_u\dot{u}=\dot{z}$. Symmetry permits $V$ odd and $Z$ even in $u$. Giving $u$ weight one and $\mu$ weight two, comparison yieldsHere $O_{\mathrm{w}}(j)$ means terms of weighted order at least $j$ under the stated scaling. If $a<1$, the pitchfork bifurcation is supercritical: stable branches $u\sim \pm \sqrt{\mu /(1-a)}$ appear for $\mu >0$ as the origin loses stability. If $a>1$, it is a subcritical pitchfork bifurcation: unstable branches occur for $\mu <0$, while the origin is stable there.

For $a=1$, assign $\mu$ weight four. The leading cubic terms in $\dot{u}$ cancel. The invariance equations now give $V=u^3/2+O_{\mathrm{w}}(5)$ and $Z=u^2+0u^4+O_{\mathrm{w}}(6)$: $Z_u\dot{u}$ first contributes at weight six and $v^2$ also first contributes there. Substitution then givesThe derivative along the nonzero branches is $-2\mu >0$, so they are unstable. The origin is stable for $\mu <0$, unstable for $\mu >0$, and nonlinearly unstable at $\mu =0$ because the leading term is $+u^5$. This is a degenerate subcritical pitchfork bifurcation.

After adding $ϵ>0$, the equilibrium equation isThe constant term breaks the reflection symmetry $x\mapsto -x$. Consequently each symmetry-forced pitchfork becomes an imperfect pitchfork bifurcation, while each crossing that produced a transcritical bifurcation is disconnected. The remaining generic turning points are saddle-node bifurcations. The lower row shows this unfolding for a representative small positive $ϵ$; the upper row gives the unperturbed diagrams.

For $a<0$, both pitchforks and both transcritical crossings are destroyed. For $a=0$, the simultaneous fifth-order meeting splits into separated folds. For $a>0$, both pitchforks become imperfect. Hence, under the allowed constant perturbation,This is the structural stability of a bifurcation distinction between the nongeneric pitchfork and transcritical diagrams and the generic saddle-node diagrams left by their unfoldings.

At a non-axis fixed point the determinant and trace areThese formulas determine the solid stable and dashed unstable branches in the bifurcation diagram. The pair $x_{\pm }$ is born in a saddle-node bifurcation at $\mu =-(\beta +1{)}^2/4$. For $\beta \ne 1$, it exchanges branches with $x=1$ at $\mu =-\beta$ and with $x=0$ at $\mu =0$, both transcritical bifurcations. If $0<\beta <1$, the saddle-node lies to the left of $x=1$; if $\beta >1$, it lies to the right. For $\beta =1$, the saddle-node and the collision at $x=1$ combine into a pitchfork bifurcation at $\mu =-1$.

At an equilibrium, the second equation gives $x=y=:s$. The first equation then givesHence the fixed-point branches in the $(a,x)$ plane areandThe three branches meet at $a=0$. The branch $x=a^2$ meets the positive square-root branch again when $a^2=\sqrt{a}$, so the second bifurcation value is

The Jacobian matrix at a general point isOn the branch $P_0=(a^2,a^2)$ it is triangular, with eigenvaluesThus $P_0$ is unstable for $a<0$, asymptotically stable for $0<a<1$, and unstable for $a>1$.

On a square-root branch write $s=\pm \sqrt{a}$. Thenwhose trace is $-1$ and determinant is $2s^2(s^3-1)$. The negative branch $s=-\sqrt{a}$ is therefore a saddle equilibrium for every $a>0$. The positive branch is a saddle for $0<a<1$ and asymptotically stable for $a>1$; close to $a=1$ it is a stable node.

To resolve the nonhyperbolic point at $a=0$, make the prescribed substitutionand append $\dot{a}=0$. The equations becomeThe centre subspace is $Y=X$. Seek the extended centre manifold for a parameter asThe centre-manifold invariance equation isTo second order, $\dot{X}=-aX+O(3)$, so comparison of coefficients givesand hence $A=C=0$, $B=1$. ThereforeSubstitution into the $X$ equation gives the reduced flowIts central branch $X=0$ is unstable for $a<0$ and stable for $a>0$, while the two nonzero branches $X\sim \pm \sqrt{a}$ for $a>0$ are unstable. Thus the bifurcation at zero is a subcritical pitchfork bifurcation with reversed normal-form parameter $\mu =-a$, exactly as recorded by the extended centre manifold of the 2023 Cambridge quadratic-product system.

The complete bifurcation diagram is therefore as follows. For $a<0$, only $P_0$ exists and is unstable. For $0<a<1$, $P_0$ is stable while both $P_{+}$ and $P_{-}$ are saddles. For $a>1$, $P_{+}$ is stable while $P_0$ and $P_{-}$ are saddles. At $a=1$, $P_0$ and $P_{+}$ cross and exchange stability, so the bifurcation is transcritical. This gives the bifurcation diagram of the 2023 Cambridge quadratic-product system.

Finally consider the phase plane near $(x,y)=(1,1)$ with $|a-1|\ll 1$. If $a<1$, the lower equilibrium $(a^2,a^2)$ is a stable node and the upper equilibrium $(\sqrt{a},\sqrt{a})$ is a saddle. If $a>1$, the lower equilibrium $(\sqrt{a},\sqrt{a})$ is the stable node and the upper equilibrium $(a^2,a^2)$ is the saddle. In each case the saddle has one-dimensional stable and unstable manifolds; one unstable separatrix runs toward the nearby stable node, while the other runs out of the local neighbourhood. The two equilibrium branches and their local invariant manifolds exchange roles as $a$ passes through one, which is the standard local phase portrait of a transcritical bifurcation.

The fixed point branches and their stability are as follows.

which is a negative-definite matrix, so this branch is stable.

Thus the $x$-versus-$\mu$ bifurcation diagram has the stable $x=0$ branch up to $\mu =1$, a stable branch $x=\sqrt{\mu -1}$ joining the bifurcation points $(1,0)$ and $(2,1)$, and the branch $x=\sqrt{\mu /2}$ changing from unstable to stable at $\mu =2$. The origin branch also lies at $x=0$ but remains unstable throughout.