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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 fixed point zero has multiplier $F^{\prime }(0)=\mu$, so it is attracting for $|\mu |<1$ and repelling for $|\mu |>1$. Other real fixed points satisfyTheir multiplier is $M=1+bx+2a{x}^2$, and they are attracting exactly when $-2<bx+2a{x}^2<0$. At equality $M=\pm 1$ the linear test is inconclusive. For $b\ne 0$ two branches meet at $x_s=-b/(2a)$, ${\mu }_s=1+b^2/(4a)$, a generic saddle-node bifurcation. At $\mu =1$, the nonzero branch crossing zero gives a transcritical bifurcation when $b\ne 0$.

The three sketches use $(a,b)=(1,0),(1,1/2),(-1,-1/2)$ respectively; case (iii) therefore illustrates the subcritical-flip choice.

The fixed-point conditions areOn $I=0$ they give $(S,I)=(0,0)$ and the disease-free equilibrium $(1,0)$. An interior fixed point requires $I=(\beta -1)S$, so it exists only for $\beta >1$; substitution in the first equation gives the endemic equilibrium

The Jacobian matrix isAt $(1,0)$ its eigenvalues are $-1$ and $\beta -1$. Hence $(1,0)$ is asymptotically stable for $\beta <1$ and a saddle point for $\beta >1$. At the endemic equilibrium, the equivalent $(N,\theta )$ coordinates give eigenvalues $-1$ and $1-\beta$, so it is a stable node whenever it exists. The origin is unstable because every solution with $N(0)>0$ has $N(t)\to 1$.

For the phase portrait, the line $I=0$ is invariant and points toward $(1,0)$. Alsoso trajectories move toward the line $S+I=1$. If $\beta <1$, then $\dot{I}<0$ throughout the interior and every nonzero solution approaches $(1,0)$. If $\beta >1$, the additional infective nullcline is the ray $I=(\beta -1)S$; $I$ increases below this ray and decreases above it, and every trajectory with $I(0)>0$ approaches $(1/\beta ,1-1/\beta )$. At $\beta =1$ the disease-free and endemic branches exchange stability in a transcritical 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.

Suppose $a<0$. Near the pitchfork at $(\mu ,x)=(a,0)$, put $\lambda =\mu -a$. Keeping the lowest terms that distinguish the branches givesSince $-2a>0$, this is the pitchfork bifurcation normal form: the zero branch changes from stable to unstable and the stable branches $x=\pm \sqrt{\lambda }$ appear for $\lambda >0$.

Near $(\mu ,x)=(0,0)$, put $\lambda =\mu$. ThenHere $-a>0$, so for $\lambda >0$ the zero branch is stable and the two nearby nonzero branches are unstable. This confirms the parameter-reversed subcritical pitchfork bifurcation.

Finally fix either $x_0=\pm \sqrt{-2a}$ and putThe two vanishing factors have the leading expansionsWith the parameter-dependent coordinate $w=2x_{0}u-m$,This is the transcritical bifurcation normal form: the branches $w=0$ and $w=m$ cross and exchange stability.

PutThe equilibrium points of a dynamical system form the three branch familiesFor a simple equilibrium, fixed point stability for an autonomous differential equation is determined by the sign of $f_x$. On the three families,Thus solid branches in the figure are stable and dashed branches are unstable.

If $a<0$, the pair $x=\pm \sqrt{\mu -a}$ emerges stably from $x=0$ at $\mu =a$ in a pitchfork bifurcation. At $\mu =0$, the pair $x=\pm \sqrt{2\mu }$ meets $x=0$ with unstable nonzero branches; after reversing the parameter orientation this is a subcritical pitchfork bifurcation. At $\mu =-a$, the two positive branches cross and exchange stability, as do the two negative branches, giving two simultaneous transcritical bifurcations at $x=\pm \sqrt{-2a}$.

If $a=0$, every branch meets at $(\mu ,x)=(0,0)$. For $\mu <0$ only the stable trivial branch exists. For $\mu >0$, $x=0$ is stable, the inner branches $x=\pm \sqrt{\mu }$ are unstable, and the outer branches $x=\pm \sqrt{2\mu }$ are stable. This is a degenerate fifth-order bifurcation rather than a generic pitchfork.

If $a>0$, a supercritical pitchfork bifurcation occurs at $\mu =0$: the trivial branch loses stability and the stable branches $x=\pm \sqrt{2\mu }$ emerge. At $\mu =a$, the unstable branches $x=\pm \sqrt{\mu -a}$ meet the trivial branch, which regains stability; this is the parameter-reversed form of a subcritical pitchfork bifurcation.

These cases comprise the bifurcations of the 2020 Cambridge quintic system.

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, each product inmust vanish. This gives the two parameter-independent equilibriathe equilibria on $y=0$and the equilibria with both coordinates nonzerowhere matching signs are used.

The $y=0$ pair is born at $k=-9/4$, suggesting a saddle-node bifurcation. At $k=0$, one of these equilibria meets $(0,0)$, suggesting a transcritical bifurcation, while the two equilibria near $(1,2)$ are simultaneously born in a saddle-node. At $k=1$, the minus branch $(1-\sqrt{k},2-\sqrt{k})$ meets $(0,1)$, suggesting another transcritical bifurcation. At $k=4$, that branch meets the $y=0$ equilibrium $(-1,0)$, again suggesting a transcritical bifurcation. Thus the bifurcation parameter values are

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.

A transcritical bifurcation has normal formIts equilibrium branches are $x=0$ and $x=\mu$. Linearization gives derivative $\mu$ on the first and $-\mu$ on the second, so the branches cross at $(\mu ,x)=(0,0)$ and exchange stability: for $\mu <0$, $x=0$ is stable and $x=\mu$ unstable; for $\mu >0$ the labels reverse.

A small constant perturbationhas equilibria $x=(\mu \pm \sqrt{{\mu }^2+4\epsilon })/2$. For $\epsilon >0$ the two branches avoid the crossing; for $\epsilon <0$ there is a parameter interval with no equilibrium, bounded by two saddle-node points. Thus arbitrarily small perturbations change the bifurcation diagram, so the crossing is not structurally stable.

A bifurcation is structurally stable within a specified class of systems when every sufficiently small allowed perturbation preserves its local qualitative bifurcation diagram up to smooth changes of state and parameter coordinates. A generic saddle-node bifurcation is structurally stable in one-parameter families, whereas a pitchfork bifurcation requires a symmetry and a transcritical bifurcation requires intersecting equilibrium branches, so unrestricted perturbations destroy either one.