Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 54 / 1

Let $q_j$ denote the four critical wavevectors, and write $A_j=R_j{e}^{i{\theta }_j}$. A spatial translation by $(s,t)$ acts on the amplitudes asThe two independent translation directions in phase space are $(2,2,1,1)$ and $(1,-1,2,-2)$. Their common orthogonal complement is spanned by $(2,-2,-1,1)$ and $(1,1,-2,-2)$. Hence translation invariants of Fourier-mode phases can be chosen asAway from zero amplitudes, a translation-invariant function of the phases is a function of these two invariant angles. The four magnitudes are themselves invariant. Thus an equivariant dynamical system on the eight-real-dimensional centre manifold descends, after quotienting translation symmetry, to four magnitude equations and two invariant-phase equations. The phase coordinates are singular where an amplitude vanishes; the original complex-amplitude equations remain smooth there.

The wavevector selection rule for equivariant monomials proves the cubic regularity. For a monomial ${\prod }_j{A}_j^{p_j}{\overline{A}}_j^{q_j}$ in the first equation, put $d_j=p_j-q_j$. Translation covariance requiresAt total degree $N$, additionally $\sum |d_j|\le N$ and $N-\sum |d_j|$ is even. Conversely, any integer vector satisfying these conditions yields a monomial, with any remaining even degree supplied by factors $|A_j{|}^2$. This makes the degree test exhaustive. For example, eliminate the first two entries:Checking the integer pairs $|d_3|+|d_4|\le N$, retaining only integral $d_1,d_2$ with the norm and parity conditions, givesAt degree three the only possibility is therefore $A_1|A_j{|}^2$, $j=1,\dots ,4$. Square symmetry gives the same conclusion for every other equation. All cubic terms are regular.

Two generators of the square dihedral group act as follows. Reflection in the $x$ axis gives $(A_1,A_2,A_3,A_4)\mapsto (A_2,A_1,A_4,A_3)$, while interchange of $x,y$ gives $(A_3,{\overline{A}}_4,A_1,{\overline{A}}_2)$. Rotation by $\pi$ conjugates every amplitude, forcing the coefficients in a steady normal form of a dynamical system to be real. With real unfolding parameter $\sigma$ and real coefficients $c_1,c_2,c_3,c_4$, defineThe most general cubic normal form is ${\dot{A}}_j=A_j{G}_j+O(|A{|}^5)$. There are no further coefficient identifications: the square-group stabilizer of one of these oblique wavevectors is trivial, so its three couplings to the other magnitudes are independent. At cubic order ${\dot{R}}_j=R_j{G}_j$ and all phases, including ${\chi }_1,{\chi }_2$, are constant.

The degree-five row of the wavevector selection rule for equivariant monomials in part (a) has two vectors other than $(1,0,0,0)$. They correspond exactly toThe vector $(1,0,0,0)$ gives only regular fifth-order terms $A_1|A_j{|}^2|A_k{|}^2$. Omitting those as required, and applying the same two square-symmetry generators, the truncated normal form of a dynamical system isHere $e,f$ are real because rotation by $\pi$ acts by complex conjugation. These are equations for steady patterns, so all four complex right-hand sides must vanish; constancy of the two invariant angles alone would only establish a relative equilibrium and could allow translation drift.

Take a nonzero equal-magnitude state, $|A_j|=R>0$, and set $z=R^2$, $S=c_1+c_2+c_3+c_4$, $\rho =\sigma +Sz$. Dividing each equation by its nonzero amplitude gives, in order,Taking imaginary parts, adding and subtracting pairs, yieldsThus, unless $e=f=0$, each invariant phase is zero or $\pi$ modulo $2\pi$. Equality of the real parts further givesThis identifies the equal-amplitude states of an eight-mode square pattern completely. For generic $e\ne f$, the two nonzero phase types, with their amplitude equations, areRepresentatives are respectively $(A_1,A_2,A_3,A_4)=R(1,1,1,1)$ and $R(-1,1,-1,1)$. Spatial translations change individual phases but leave the indicated phase pair fixed. Reflection acts on the invariant angles as $({\chi }_1,{\chi }_2)\mapsto (-{\chi }_1,{\chi }_2)$, and interchange of $x,y$ as $({\chi }_1,{\chi }_2)\mapsto (-{\chi }_2,-{\chi }_1)$. Thus the two rows are not equivalent under square symmetry and translations. Every positive root of the corresponding polynomial gives an amplitude branch of that type; if there is no positive root, that type is absent at those parameter values.

There are two coefficient degeneracies to include. When $e=f\ne 0$, the mixed pairs $(0,\pi )$ and $(\pi ,0)$ are allowed and are equivalent under square symmetry. They form one additional phase type, with amplitude equation $\boxed{\sigma +Sz=0}$; $iR(1,1,1,1)$ is a representative. When $e=f=0$, all invariant phases are unrestricted, and the same cubic amplitude equation gives a continuous family modulo translations and the finite square-group action. Even if $e+f=0$ but $e\ne f$, only the original two phase types remain: the equality of their amplitude equations does not remove the phase constraints. Finally, the all-zero state is always a separate steady state and has no defined polar phases.