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Forthe extended centre manifold for a parameter gives the local reduced equationswhere $y_0=\pm \sqrt{2}$. Thus the first two points undergo respectively supercritical and subcritical symmetry-forced pitchfork bifurcations, while each of the last two points undergoes a saddle-node bifurcation.

There are axis equilibria $E_{\pm }=(\pm \sqrt{\mu },0)$ when $\mu \ge 0$, and $E_a=(a,a^2-\mu )$ for every $\mu$. At an axis equilibrium $(x_{\ast },0)$ the eigenvalues of the Jacobian matrix are $2x_{\ast }$ and $a-x_{\ast }$. Thus a negative axis root is a stable node when $a<x_{\ast }$ and a saddle when $a>x_{\ast }$. A positive axis root is an unstable node when $a>x_{\ast }$ and a saddle when $a<x_{\ast }$. Equalities give a zero eigenvalue.

At $E_a$,It is a saddle for $\mu <a^2$. For $\mu >a^2$, it is stable if $a<0$, unstable if $a>0$, with real node eigenvalues for $a^2<\mu \le 2a^2$ and a focus for $\mu >2a^2$. At $\mu =2a^2$ the node-focus transition is not a bifurcation of hyperbolicity when $a\ne 0$. For $a=0$, $E_a$ has imaginary eigenvalues when $\mu >0$ and the linearized system is a centre. For $a\ne 0$, the candidate local bifurcation values are $\boxed{\mu =0and\mu =a^2}$. When $a=0$ these zero-eigenvalue events coincide at $\mu =0$, while the off-axis equilibrium also has zero-real-part imaginary eigenvalues for every $\mu >0$.

The centre manifold theorem reduces local dynamics near zero-real-part eigenvalues to an invariant manifold tangent to their eigenspace; transverse directions with nonzero real parts retain their stable or unstable character. For a varying parameter, adjoin $\dot{\mu }=0$ and use an extended centre manifold for a parameter.

Take $a=1$. At $\mu =0$ near the origin, $y=0$ is an exact extended centre manifold. Its reduced equation isThis is a saddle-node bifurcation: no nearby axis equilibria for $\mu <0$, one at zero, and two for $\mu >0$. The transverse eigenvalue is positive, so the negative root is a saddle and the positive root an unstable node for small positive $\mu$.

At $\mu =1$, put $u=x-1$, $v=y$, $\nu =\mu -1$. Then $\dot{u}=2u+u^2-v-\nu$, $\dot{v}=-uv$, $\dot{\nu }=0$. Write the extended centre manifold as $u=h(v,\nu )$. Its invariance equation is $h_v(-hv)=2h+h^2-v-\nu$. Matching degrees givesHence the reduced dynamics areThe exact equilibrium branches are $v=0$ and $v=-\nu$, crossing at zero and exchanging stability in the centre direction. This is a transcritical bifurcation; the transverse eigenvalue remains positive. For $\nu <0$ the axis branch is an unstable node and the off-axis branch a saddle; for $\nu >0$ their roles reverse locally.

When $a=0$, the two parameter values merge and the origin at $\mu =0$ has a double zero eigenvalue with nonzero nilpotent linear part. The centre manifold is then two-dimensional, so the preceding one-dimensional classifications cannot simply be reused. Also the off-axis branch has imaginary eigenvalues for every $\mu >0$, with identically zero trace: this is not a generic Hopf bifurcation obtained by varying $\mu$. These degeneracies already follow without computing a new normal form.

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.

Write $z=x+y$. When ${\mu }_1={\mu }_2=0$ the system gives $\dot{z}=y-z^3$ and $\dot{y}=-z^3$. Choosingmakes $V$ positive definite and radially unbounded, while its orbital derivative isThe set $\{\dot{V}=0\}$ is $z=0$. A trajectory remaining there must also have $\dot{z}=y=0$, so its largest invariant subset is the origin. The LaSalle invariance principle therefore proves that the origin is globally asymptotically stable. In the phase plane, $\dot{x}=0$ on $y=0$ and $\dot{y}=0$ on $x+y=0$; every nonstationary trajectory crosses the nested level curves of $V$ inward and tends to the origin.

At an equilibrium point, $y=0$ andHence the origin always exists, while $x=\pm \sqrt{{\mu }_1}$ exist for ${\mu }_1>0$. At $(x_{\ast },0)$ the Jacobian matrix isA stationary bifurcation occurs when a real eigenvalue passes through zero, hence onA Hopf bifurcation requires zero trace and positive determinant. At the origin this givesand on either nonzero branch it givesThus the Hopf locus consists of the negative ${\mu }_1$-axis and the ray ${\mu }_2=3{\mu }_1$ in the first quadrant; the stationary locus is the ${\mu }_2$-axis.

Now set ${\mu }_2=-1$, write $\mu ={\mu }_1$, and append $\dot{\mu }=0$. At $(x,y,\mu )=(0,0,0)$ the extended centre manifold for a parameter is tangent to the $(x,\mu )$-plane, so put $y=h(x,\mu )$. Its centre-manifold invariance equation isUnder the ordering $\mu =O(x^2)$, solving through cubic order givesThis is the pitchfork bifurcation normal form: $x=0$ is stable for $\mu <0$ and unstable for $\mu >0$, while the two stable branches $x=\pm \sqrt{\mu }$ emerge for $\mu >0$. The bifurcation is therefore a supercritical pitchfork bifurcation.

A fixed point satisfiesThe branch with $y=0$ isEvery other fixed point can be parametrized by its $x$-coordinate:Equivalently,The $x_{-}$ pair exists for $0\le a\le 4$, and the $x_{+}$ pair for $3\le a\le 4$.

The Jacobian matrix isOn $P_0(a)$ its eigenvalues are $1$ and $a(3-a)$, so this branch has a nonhyperbolic equilibrium at $a=0$ and $a=3$. On a nonzero-$y$ branch,which vanishes at $x=2$, hence at $a=4$ and $y=\pm \sqrt{2}$. The three bifurcation values and locations are therefore

For the center manifold calculation, append $\dot{a}=0$ as in the extended centre manifold for a parameter.

At $(a,x,y)=(0,0,0)$ put $\mu =a$ and write the centre manifold as $x=h(y,\mu )$. Its centre-manifold invariance equation isIt gives $h=\mu -y^2+O(3)$, and henceThis is the pitchfork bifurcation normal form with stable nonzero centre branches for $\mu >0$: the bifurcation at $a=0$ is supercritical.

At $(a,x,y)=(3,3,0)$ put $u=x-3$ and $\mu =a-3$. ThenAgain $u=h(y,\mu )=\mu -y^2+O(3)$, soThe nonzero centre branches exist for $\mu >0$ and are unstable in the centre direction, whereas $y=0$ is centre-stable there. Thus the bifurcation at $a=3$ is a subcritical pitchfork bifurcation with reversed normal-form parameter.

Finally fix $y_0=\pm \sqrt{2}$ and putThe translated system isThe extended centre manifold hasand thereforeThis is the saddle-node bifurcation normal form. For $\mu <0$ there are two nearby fixed points, which coalesce and disappear at $\mu =0$. The complete reductions are recorded in the Extended centre-manifold reductions of the 2019 Cambridge reflection-symmetric system.

There are two bifurcations in $0\le k\le 1/2$, both at $k=0$.

Near $(0,0,0)$, append $\dot{k}=0$. The plane $y=0$ is invariant and tangent to the extended center subspace, so it is an extended center manifold. The reduced equation is exactlyIts leading terms $-kx-3x^2$ have two branches, $x=0$ and $x=-k/3+O(k^2)$, which cross and exchange their center-direction stability. Hence this is a transcritical bifurcation.

For the bifurcation at $(1,2,0)$, setThe extended system becomesSolving the centre-manifold invariance equation for $v=h(X,k)$ givesSubstitution into the $X$ equation yieldsFor $k>0$ this has two nearby equilibria $X=\pm \sqrt{k}+O(k)$, while for $k<0$ it has none. It is therefore a saddle-node bifurcation. These reductions are collected in bifurcations of the 2022 Cambridge quadratic-cubic system.

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.