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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.

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.

Let $r=Ra-R{a}_c$. The amplitude equation is $\dot{A}=A(r-A^2)$. For $r<0$, $A=0$ is the sole equilibrium and every solution tends monotonically to it. At $r=0$, the origin remains attracting but only algebraically. For $r>0$, the origin is unstable and the two equilibriaare stable: positive initial data tend to $A_{+}$, negative initial data tend to $A_{-}$, and $A(0)=0$ remains zero. A plot of $A(t)$ therefore shows a pitchfork bifurcation normal form at $Ra=R{a}_c$.

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 $\mu =-1$.