= Solution
At the double threshold, $L\psi_0=L\psi_1=0$. For $j\ge2$, $q_j^2\ge9\pi^2/4$, so $|K^*-q_j^2|\ge13\pi^2/8>3\pi^2/8$ and every remaining <eigenvalue> is strictly negative. There is a <spectral gap> separating this two-dimensional kernel from the stable modes. The <centre manifold theorem> therefore gives a two-dimensional state <centre manifold>, with
$$
\psi=A\psi_0+B\psi_1+h(A,B),\qquad h=O((|A|+|B|)^2),\qquad \langle h,\psi_j\rangle=0\quad(j=0,1).
$$
If the two detuning parameters are adjoined as variables with zero time derivatives, the extended <centre manifold> has total dimension four; its state fibers still have dimension two.
The only nontrivial spatial <symmetry> of the bounded interval is the <reflection> $x\mapsto-x$. Since $\psi_0$ is even and $\psi_1$ odd, its <group action> on the amplitudes is $(A,B)\mapsto(A,-B)$. Thus <equivariance> requires $\dot A$ to be even and $\dot B$ odd in $B$. There is no sign symmetry $\psi\mapsto-\psi$, because the quadratic nonlinearity breaks it. The general smooth form is $\dot A=F(A,B^2)$, $\dot B=B G(A,B^2)$, with the zero solution preserved. The <quadratic even-odd mode interaction> follows on truncation:
$$
\dot A=\lambda_1A+a_1A^2+a_2B^2+O(3),\qquad
\dot B=\lambda_2B+a_3AB+O(3).
$$
Writing $m=\mu-\mu^*$ and $\kappa=K-K^*$, the exact linear growth rates are $m-3\pi^2\kappa/4-\kappa^2$ and $m+3\pi^2\kappa/4-\kappa^2$. To first order in detuning,
$$
\boxed{\lambda_1=m-\frac{3\pi^2}{4}\kappa,\qquad \lambda_2=m+\frac{3\pi^2}{4}\kappa.}
$$
The <orthogonal projection> of the quadratic term onto the neutral <eigenfunctions> determines the leading coefficients. Both <eigenfunctions> have squared $L^2$ norm one, and parity kills the unwanted products. The quadratic correction $h$ contributes no term through $Lh$, since $L$ is <self-adjoint> and the projected modes lie in its kernel. Consequently
$$
a_1=\int_{-1}^1\psi_0^3\,dx,\qquad
a_2=\int_{-1}^1\psi_0\psi_1^2\,dx,\qquad
a_3=2\int_{-1}^1\psi_0\psi_1^2\,dx.
$$
These projections are also a concrete way of computing the <normal form>. In this normalization they give $a_1=8/(3\pi)$, $a_2=32/(15\pi)$ and $a_3=64/(15\pi)$, consistent with the stated coefficient ratios.
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