At criticality choose the fundamental mode
$$
w_1=A\cos(ax)\sin(\pi z).
$$
Its quadratic self-interaction satisfies
$$
\frac12\partial_z^3(w_1^2)
=-A^2\pi^3[1+\cos(2ax)]\sin(2\pi z).
$$
The critical linear operator has eigenvalues $-64\pi^6$ on $\sin(2\pi z)$ and $-60(a^2+\pi^2)^3$ on $\cos(2ax)\sin(2\pi z)$. The slaved second-order correction is therefore
$$
w_2=A^2[d_0+d_1\cos(2ax)]\sin(2\pi z),
$$
where
$$
d_0=\frac1{64\pi^3},
\qquad
d_1=\frac{\pi^3}{60(a^2+\pi^2)^3}.
$$

Introduce a <slow time> $\tau$ and project the next-order equation onto the fundamental mode. Detuning $R-R_c$ contributes $a^2(R-R_c)A$, while the interaction of $w_1$ with $w_2$ has fundamental component
$$
-\frac14\pi^3(2d_0+d_1)A^3.
$$
The solvability condition is the <Landau amplitude equation>
$$
\boxed{
\frac{dA}{d\tau}
=a^2(R-R_c)A
-\frac14\pi^3(2d_0+d_1)A^3}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 331 / 2 / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-331/2/ii/solution