= Solution
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 to
$$
\boxed{\overline A_2A_3^2A_4^2,\qquad \overline A_1A_2^2A_3\overline A_4.}
$$
The 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> is
$$
\begin{aligned}
\dot A_1&=A_1G_1+e\overline A_2A_3^2A_4^2+f\overline A_1A_2^2A_3\overline A_4,\\
\dot A_2&=A_2G_2+e\overline A_1A_4^2A_3^2+f\overline A_2A_1^2A_4\overline A_3,\\
\dot A_3&=A_3G_3+eA_4A_1^2\overline A_2^{\,2}+f\overline A_3\overline A_4^{\,2}A_1A_2,\\
\dot A_4&=A_4G_4+eA_3A_2^2\overline A_1^{\,2}+f\overline A_4\overline A_3^{\,2}A_2A_1.
\end{aligned}
$$
Here $e,f$ 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,
$$
\begin{aligned}
0&=\rho+z^2(e e^{-i\chi_2}+f e^{-i\chi_1}),\\
0&=\rho+z^2(e e^{-i\chi_2}+f e^{ i\chi_1}),\\
0&=\rho+z^2(e e^{ i\chi_1}+f e^{ i\chi_2}),\\
0&=\rho+z^2(e e^{-i\chi_1}+f e^{ i\chi_2}).
\end{aligned}
$$
Taking <imaginary parts>, adding and subtracting pairs, yields
$$
e\sin\chi_2=f\sin\chi_1=e\sin\chi_1=f\sin\chi_2=0.
$$
Thus, unless $e=f=0$, each invariant <phase> is zero or $\pi$ modulo $2\pi$. Equality of the <real parts> further gives
$$
(e-f)(\cos\chi_1-\cos\chi_2)=0.
$$
This 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, are
$$
\boxed{\begin{array}{c|c}
(\chi_1,\chi_2)&\text{equation for }z>0\\\hline
(0,0)&\sigma+Sz+(e+f)z^2=0\\
(\pi,\pi)&\sigma+Sz-(e+f)z^2=0.
\end{array}}
$$
Representatives 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 $e=f\ne0$, the mixed pairs $(0,\pi)$ and $(\pi,0)$ are allowed and are equivalent under square <symmetry>. They form \b[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>.
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