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At an equilibrium point, $x=-y^2$ and $x^2=a^{2}y$, so $y(y^3-a^2)=0$. With $d=|a{|}^{2/3}>0$, the two equilibrium points and their Jacobian matrix areAt $(-d^2,d)$, $\det J=-3a^2<0$. The trace-determinant stability criterion therefore gives an unstable saddle equilibrium. At the origin, the eigenvalues are $\lambda =\pm i|a|$: its linearization is a center equilibrium. The linearization stability theorem does not classify this nonhyperbolic equilibrium of the nonlinear system.

Nonlinear classification. In fact the answer depends on $|a|$:Here is an explicit derivation, rather than an inference from the imaginary eigenvalues. When $a^2=1$, set $L=1-x+y$. Near the origin $L>0$, and the first integralhasIts Hessian matrix at the origin is the identity. Small level curves are closed ovals on which the vector field is nonzero, proving the center equilibrium classification and Lyapunov stability.

For $a^2\ne 1$, a higher-order Lyapunov function calculation settles the issue. Put $k=|a|$, $U=x$, $V=ky$, $\tau =kt$, $p=k^{-1}$ and $q=k^{-2}$. ThenWrite $R^2=U^2+V^2$ and introduce the differential operatorsSuccessively cancel the terms of degrees three, four and five in the derivative of $R^2+H_3+H_4+H_5$, using these homogeneous polynomials:Direct differentiation verifies $\mathcal{L}{H}_3=-\mathcal{N}{R}^2$, $\mathcal{L}{H}_4=-\mathcal{N}{H}_3$ and $\mathcal{L}{H}_5=-\mathcal{N}{H}_4$. The remaining degree-six term has circular averageFor example, the only nonzero averages of degree-six monomials needed here are $⟨U^6⟩=⟨V^6⟩=5/16$ and $⟨U^4{V}^2⟩=⟨U^2{V}^4⟩=1/16$. Since $\mathcal{L}={\partial }_{\theta }$ in polar coordinates, the zero-average part can be cancelled by a degree-six homogeneous polynomial $H_6$ satisfyingIndeed its restriction to the unit circle is obtained by integrating a zero-average trigonometric polynomial; its Fourier modes have orders $2,4,6$, so multiplication by $R^6$ gives a polynomial. Consequently the positive Lyapunov function $W=R^2+H_3+H_4+H_5+H_6$ satisfiesFor $k>1$ this is strictly negative near the origin, giving asymptotic stability. For $k<1$ it is strictly positive, and $W_{\tau }\ge C{W}^3$ in a sufficiently small neighbourhood; every nonzero starting point must eventually leave that neighbourhood. Finally ${\theta }_{\tau }=1+O(R)>0$, so the nearby trajectories rotate: these are weak foci, with the stability stated above. Merely calling the nonlinear origin a center for every $a\ne 0$ would miss this distinction.

With $y=\dot{x}$, the feedback system isFor $p=0$, $(x,y,z)=(0,0,0)$ is an equilibrium. Its stability matrix and characteristic polynomial areThus there is always a zero eigenvalue, with eigenvector $(1,0,0)$, so it is a nonhyperbolic equilibrium. If $q\ne -1$, the other two eigenvalues are off the imaginary axis: for $q>-1$ their real parts are negative, while for $q<-1$ their product is negative. At $q=-1$, all three eigenvalues are zero, explaining the exclusion in the next part.

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.

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.