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An equilibrium is nonhyperbolic when its Jacobian matrix has at least one eigenvalue on the imaginary axis. Linearization alone then need not determine the local dynamics, and a center manifold often supplies the relevant reduction.

The off-axis equilibria exist for $|a|\le 1$ and coalesce with an axial equilibrium at each endpoint. Thus the stationary bifurcations occur atFor the positive bifurcation, $a_0=y_0=1$. Putso thatWrite the extended center manifold as $Y=h(x,\mu )$. Its invariance equation isUnder the scaling $\mu =O(x^2)$, comparison of the terms of order $x^2$ givesSubstitution into the $x$ equation givesWith the reversed parameter $\nu =-2\mu =2(1-a)$, this is the pitchfork bifurcation normal form $\dot{x}=\nu x-x^3$. Therefore the positive stationary bifurcation is a supercritical pitchfork in the parameter $\nu =2(1-a)$, or equivalently the stable daughter equilibria exist on the $a<1$ side.

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.

Put $x=1+u$, $y=v$, and $\mu =-1+\nu$, adjoining $\dot{\nu }=0$. ThenThe zero eigenvalue has centre direction $v=-u$. Solving the center manifold invariance equation givesSubstitution yields the reduced equationIts equilibria are $u=0$ and $u=\pm \sqrt{\nu }+O(\nu )$ for $\nu >0$, exactly matching the three branches meeting at the pitchfork in part (b).

The transverse eigenvalues $-2$ and $-1$ are stable, so the center manifold reduction determines local asymptotic stability. If $a<1$, the leading center term has negative coefficient and $v$ flows toward zero from both sides. If $a>1$, it flows away. At $a=1$ the cubic term vanishes but the quintic coefficient isso the origin is again unstable. Therefore