Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 78 / 3 / d / Solution

Seek an equilibrium in the invariant coordinate plane $z=0$ with $x,y\ne 0$. Put $U=x^2$ and $W=y^2$. The two nonzero-component equations areSubtracting gives $eU+cW=0$, so positive $U,W$ require $ce<0$. With $D=a(c-e)-ce$, the solutions areThey are actual additional equilibria when $D$ has the sign of $c$; the signs of $x,y$ are independently free, and cyclic permutation supplies the other two coordinate planes. For example, $a=\mu =c=1$, $e=-1$ gives $D=3$ and $x^2=y^2=1/3$. Thus additional two-coordinate equilibria do exist for suitable opposite-sign cross-couplings. The inequality $ce<0$ is necessary for this type, but without the positivity conditions on the displayed denominator it need not be sufficient for every coefficient choice.

The isotropy of a generic such point is just $⟨{\kappa }_z⟩$, whose fixed-point subspace is two-dimensional. The Equivariant branching lemma guarantees axial branches with one-dimensional fixed spaces; it does not rule out these lower-symmetry branches. Their existence therefore gives no contradiction.

At fixed $\mu >0$, letting $e$ cross zero with $c\ne 0$ sends $x^2\to \mu /a$, $y^2\to 0$: the new pair of opposite-$y$ branches meets an axis equilibrium. Its $y$ eigenvalue is $e\mu /a$, so this is a secondary symmetry-forced pitchfork bifurcation. Similarly, as $c$ crosses zero with $e\ne 0$, the branches meet the $y$ axis, whose $x$ eigenvalue is $-c\mu /a$. The cubic formula gives $y^2\sim -\mu e/(ac)$ near the first crossing, and $x^2\sim -\mu c/(ae)$ near the second. Reflections force the two signs of the newly nonzero coordinate, while cyclic symmetry gives the corresponding pitchforks on the other axes. The simultaneous $c=e=0$ case is degenerate and is not a generic single-parameter pitchfork.