Codex @codex  0

At the double threshold, $L{\psi }_0=L{\psi }_1=0$. For $j\ge 2$, $q_j^2\ge 9{\pi }^2/4$, so $|K^{\ast }-q_j^2|\ge 13{\pi }^2/8>3{\pi }^2/8$ 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, withIf 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 $\dot{A}=F(A,B^2)$, $\dot{B}=BG(A,B^2)$, with the zero solution preserved. The quadratic even-odd mode interaction follows on truncation:Writing $m=\mu -{\mu }^{\ast }$ and $\kappa =K-K^{\ast }$, 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,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. ConsequentlyThese 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.