= Solution
The spatial mean is conserved: averaging every term in the <partial differential equation> over the periodic cell gives $\partial_t\langle T\rangle=0$. Adding a constant to $T$ also preserves the equation. We therefore fix the mean, conveniently at zero. This qualification is essential: without it the constant <Fourier mode> contributes an additional neutral direction, so the centre subspace would have five real dimensions rather than four.
On a <Fourier mode> $e^{i(mx+ny)}$, the linearized <linear operator> has <eigenvalue>
$$
\lambda_{m,n}=\mu(m^2+n^2)-(m^2+n^2)^2.
$$
On the fixed-mean subspace, the first <eigenvalues> to reach zero are those with $m^2+n^2=1$. At $\mu=1$ these are the modes $e^{\pm ix},e^{\pm iy}$; every other nonconstant <Fourier mode> is strictly damped. The reality condition leaves two complex <amplitudes>, hence a \b[four-real-dimensional centre manifold on the fixed-mean subspace]. The trivial state loses <linear stability> as $\mu$ increases through one.
Use slow time $\tau=\varepsilon^2t$, and write $T=\varepsilon W+\varepsilon^3T_3+\cdots$, where
$$
W=A(\tau)e^{ix}+\overline A(\tau)e^{-ix}+B(\tau)e^{iy}+\overline B(\tau)e^{-iy}.
$$
There is no quadratic nonlinearity, so no forced second-order correction is needed. At order $\varepsilon^3$, with $\mu=1+\varepsilon^2\nu$, projection onto the critical <Fourier modes> gives
$$
W_\tau=-\nu\Delta W+(-\Delta-\Delta^2)T_3+\nabla\cdot(|\nabla W|^2\nabla W).
$$
The correction $T_3$ does not contribute to the critical projection because the linear operator vanishes on those modes. To find the nonlinear <coefficient> of $e^{ix}$, split the <divergence> term into
$$
\partial_x(W_x^3)+\partial_x(W_y^2W_x)+\partial_y(W_x^2W_y)+\partial_y(W_y^3).
$$
The first term contributes $-3|A|^2A$. In the second, only the zero-$y$ <Fourier coefficient> of $W_y^2$ contributes; it is $2|B|^2$, and differentiating $W_x$ once more gives $-2|B|^2A$. The third term has zero $y$-average, and the fourth has no $x$-dependent critical component. Interchanging $x,y$ gives the other <amplitude equation>. Thus the <conserved-mean convection amplitude equations> are
$$
\boxed{A_\tau=\nu A-3|A|^2A-2|B|^2A,\qquad B_\tau=\nu B-3|B|^2B-2|A|^2B,\quad\alpha=3,\ \beta=2.}
$$
To establish the selected <planform>, put $A=re^{i\theta}$, $B=se^{i\psi}$. The <phases> are constant and the magnitudes obey
$$
\dot r=r(\nu-3r^2-2s^2),\qquad \dot s=s(\nu-3s^2-2r^2).
$$
For $\nu>0$, the square state has $r=s=\sqrt{\nu/5}$. Its magnitude <Jacobian matrix> is $-(2\nu/5)\begin{pmatrix}3&2\\2&3\end{pmatrix}$, with <eigenvalues> $-2\nu$ and $-2\nu/5$. Its two neutral <phases> correspond to <translation symmetry>. The pure roll has $r^2=\nu/3,s=0$; an infinitesimal $B$ disturbance grows at rate $\nu-2\nu/3=\nu/3>0$. The other pure roll is equivalent, while the trivial state is unstable for $\nu>0$. Consequently \b[squares, with $|A|^2=|B|^2=\nu/5$, are the stable small-amplitude planform, modulo translations]. For $\nu<0$ the trivial state is stable.
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