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

For the Equivariant branching lemma, assume a generic steady-state crossing $DF(0,\mu )=\mu I$ after parameter normalization, and nonzero leading nonlinear coefficient on each axial line. The lemma gives an equilibrium branch for each axial subgroup with a one-dimensional real fixed-point subspace of a group action.

For $H_{\mathrm{ax}}=⟨{\kappa }_y,{\kappa }_z⟩$, $\mathrm{Fix}(H_{\mathrm{ax}})=\{(r,0,0)\}$. Conjugation by $\rho$ supplies the other two coordinate axes. For $H_{\mathrm{diag}}=⟨\rho ⟩$, $\mathrm{Fix}(H_{\mathrm{diag}})=\{(r,r,r)\}$. Sign reflections supply all the equal-magnitude diagonals. These isotropy types are not conjugate: their subgroup orders are $4$ and $3$. Thus there are two distinct primary axial branch types: coordinate-axis states and equal-magnitude diagonal states.

The global sign reversal belongs to the group and makes the restricted scalar equations odd, so generically each has the form $\dot{r}=r(\mu +\gamma r^2+\cdots )$, with $\gamma \ne 0$. Its two nonzero solutions satisfy $r^2=-\mu /\gamma +\cdots$ on the appropriate parameter side. Symmetry-related solutions belong to the same branch type; their existence does not imply stability or that both types lie on the same side of the crossing.