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Adjoin $\dot{p}=0$ and use an extended centre manifold with local coordinates $(x,p)$. Count $p$ as order $x^2$, as appropriate to a steady-state fold, and writeThe centre-manifold invariance equation differentiates these graphs along $\dot{x}=y$. The derivatives $y_{x}y$ and $z_{x}y$ first contribute at weighted order three. Comparing the coefficients of $p$ and $x^2$ in the right sides therefore givesSince $q+1\ne 0$, their solution isHence the reduced normal form isBoth the quadratic and unfolding coefficients are nonzero, so the steady-state bifurcation is a saddle-node, with equilibria on the $p>0$ side. The exact steady equations confirm this: $y=0$, $z=x^2$, and $p=z+z^3=x^2+x^6$, giving two nearby equilibria for $p>0$ and none for $p<0$.

For $q>-1$ the transverse modes are attracting; the negative-$x$ branch is locally attracting and the positive-$x$ branch has one unstable direction. For $q<-1$, one transverse eigenvalue is already positive, so neither full equilibrium is asymptotically stable, although the reduced fold is still present with reversed flow orientation. The parameter $q=-1$ requires a higher-dimensional degeneracy analysis and is outside the stated assumption.

For parameter-dependent centre manifold theorem, append $\dot{\mu }=0$. At $(x,y,\mu )=(0,0,0)$ the eigenvalues are $0,-1,0$: the $y$ direction is strongly stable and $(x,\mu )$ span the extended centre directions. An extended centre manifold for a parameter is an invariant graph of a function $y=h(x,\mu )$ tangent to $y=0$; nearby trajectories approach its reduced dynamics along stable fibers. The stable eigenvalue remains separated from zero, so local bifurcation and stability are governed by this reduction. This is the parameter-extension content of the requested stable/centre-manifold discussion.

The centre-manifold invariance equation isReflection symmetry permits an invariant graph of a function with $h(-x,\mu )=h(x,\mu )$ and $h(0,\mu )=0$. At quadratic order the left side vanishes and the right side is $-h-x^2$, giving $h=-x^2+O(|\mu |x^2+x^4)$. More explicitly its first mixed coefficient is $2\mu x^2$, found by equating the $\mu x^2$ terms. The reduced equation isThus the origin is locally asymptotically stable for $\mu <0$, becomes a saddle for $\mu >0$, and creates two stable equilibria $x=\pm \sqrt{\mu /2}+O({\mu }^{3/2})$, $y=-\mu /2+O({\mu }^2)$. This is a supercritical pitchfork bifurcation. The exact nearby nonzero branches provide a check: the equilibrium equations givewhich are real and nonzero for small positive $\mu$. The omitted branch $y=1+\sqrt{1+\mu }$ is not near the origin. At $\mu =0$ the leading $-2x^3$ also attracts, although at an algebraic rather than exponential rate.

Near $(1,0)$ put $u=x-1$, $v=y$, $ϵ=\mu -1$, with $\dot{ϵ}=0$. The extended system isThe extended centre manifold for a parameter has $u=h(v,ϵ)$, $h(0,0)=0$, $Dh(0,0)=0$. Its centre-manifold invariance equation is $h_v\dot{v}=\dot{u}$. The left side starts at degree three. At degree two the right side is $-2h_2-v^2$, so $h_2=-v^2/2$, with no $vϵ$ or ${ϵ}^2$ term. Substitution givesThe leading normal form of a dynamical system $\dot{v}=ϵv-v^2$ has branches $v=0$ and $v\simeq ϵ$, exchanging stability at $ϵ=0$: this is a transcritical bifurcation. In the physical quadrant the interior branch is present only for $ϵ>0$.

The equations extended across $x=0$ are equivariant under $x\mapsto -x$, so near $(0,1)$ the reduced equation is odd in $x$. To verify its nondegenerate pitchfork bifurcation, set $\eta =\mu -1$ and $z=y-\mu$. ThenThe local centre manifold has $z=-x^2+O(|\eta |x^2+x^4)$, obtained by balancing the quadratic terms of its invariance equation. Consequently $\dot{x}=-2\eta x+x^3+O({\eta }^2|x|+|\eta ||x{|}^3+|x{|}^5)$: the cubic coefficient is nonzero. This is a subcritical pitchfork, with the two symmetry-related interior branches existing for $\eta >0$ in the extended plane; only the $x>0$ branch belongs to the physical quadrant.

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.

Adjoin $\dot{\mu }=0$ and put $w=z+\mu$. The extended centre manifold is a graph $w=H(x,y,\mu )$ tangent to $w=0$ at the origin. The equations becomeHere $ϵ>0$ is fixed. The centre-manifold invariance equation, obtained by the chain rule, isWrite $H=H_2+O({\rho }^3)$, where $\rho =|x|+|y|+|\mu |$ and $H_2$ is a homogeneous quadratic polynomial. To quadratic order the centre-manifold invariance equation isThe terms $\mu x,\mu y,{\mu }^2$ have zero coefficients: their equations are homogeneous, and the operator $1+y{\partial }_x-x{\partial }_y$ is invertible on these quadratic polynomials. For $H_2=a{x}^2+bxy+c{y}^2$, comparison of coefficients givesHence $a=3/5$, $b=-2/5$ and $c=2/5$, givingSubstitution into the $y$ equation givesThe constant in the big O notation may depend on fixed $ϵ$. Thus the equations printed in part (b) are the retained terms of the local Taylor series, rather than an exact identity for the full reduced vector field.

For the Hopf bifurcation classification in part (a), set $\mu =0$ and use $x=r\cos \theta$, $y=-r\sin \theta$. The radial derivative is $\dot{r}=(x\dot{x}+y\dot{y})/r$. Its cubic angular mean isHere $⟨\cdot ⟩$ denotes the mean over one revolution. Removing the oscillatory cubic terms by a near-identity transformation gives this same cubic radial coefficient in the Hopf normal form. The linear radial coefficient is $5ϵ\mu /2$.

Adjoin ${\mu }_{n+1}={\mu }_n$. At the origin the stable direction is the $x$ axis, with Floquet multiplier zero, while the $y$ and parameter directions have Floquet multiplier one. By the centre manifold theorem for a discrete dynamical system, write the extended centre manifold asOn it the scalar map is $Y=(1+\mu )y-\mu h(y,\mu )+y^3$. Its discrete centre-manifold invariance equation isPut $h=h_2+O({\rho }^3)$ with $\rho =|y|+|\mu |$. Since $Y=y+O({\rho }^2)$, the quadratic equation is simply $h_2(y,\mu )=\mu y$. ThusIn particular, setting the stable variable identically to zero would lose the leading parameter correction. The reduced map through total degree three isIn the natural pitchfork bifurcation scaling $\mu =O(y^2)$ this has leading form $y_{n+1}=y_n+\mu y_n+y_n^3$. Its nonzero fixed points satisfy $y^2\sim -\mu$, and their reduced fixed-point multipliers satisfy $1-2\mu +O({\mu }^2)>1$ for small $\mu <0$. The transverse Floquet multiplier remains inside the unit circle. Therefore the bifurcation at $\mu =0$ is a subcritical pitchfork bifurcation: the origin attracts for small negative $\mu$, each nonzero branch is a saddle fixed point of a map, and the origin is a saddle fixed point of a map for small positive $\mu$.

An exact check uses the full Jacobian matrix on either nonzero branch. Its trace is $T_{\ast }=1-2\mu +3{\mu }^2$ and its determinant is $D_{\ast }={\mu }^2$. Thuswhenever a nonzero branch exists. Also $p_{\ast }(0)={\mu }^2>0$, so one real Floquet multiplier lies in $(0,1)$ and the other in $(1,\infty )$. This proves that each nonzero branch is a saddle fixed point of a map throughout $\mu <0$ and $\mu >1$, in particular near the requested bifurcation.