= Solution
Use negative-exponential spatial <Fourier transforms> and <finite-time spectral boundary transforms>:
$$
\begin{aligned}
Q_0(k)&=\int_0^Le^{-ikx}q_0(x)dx,&Q(k,t)&=\int_0^Le^{-ikx}q(x,t)dx,\\
F_j(k,t)&=\int_0^te^{\omega(k)s}\partial_x^jq(0,s)ds,&G_j(k,t)&=\int_0^te^{\omega(k)s}\partial_x^jq(L,s)ds\quad(j=0,1,2).
\end{aligned}
$$
Write $\mathcal F=k^2F_0-ikF_1-F_2$ and $\mathcal G=k^2G_0-ikG_1-G_2$. Integrate the <local relation> first over $0<x<L$ and then over time. The left and right endpoint fluxes have opposite signs, giving the <finite-interval Airy global relation>
$$
\boxed{Q_0(k)-e^{\omega(k)t}Q(k,t)=\mathcal F(k,t)-e^{-ikL}\mathcal G(k,t).}
$$
All these spatial and temporal <integral transforms> are <entire functions> of $k$ for sufficiently smooth data on the finite intervals. Put $\alpha=e^{2\pi i/3}$ and $k_j=\alpha^jk$, $j=0,1,2$. The <dispersion relation> obeys $\omega(k_j)=\omega(k)$; hence the same <global relation> holds at all three $k_j$. In particular, the missing traces satisfy the concrete <linear system>
$$
\begin{pmatrix}
-ik&-1&e^{-ikL}\\
-i\alpha k&-1&e^{-i\alpha kL}\\
-i\alpha^2k&-1&e^{-i\alpha^2kL}
\end{pmatrix}
\begin{pmatrix}F_1\\F_2\\G_2\end{pmatrix}
=\begin{pmatrix}R_0\\R_1\\R_2\end{pmatrix},
$$
where, with a common upper time limit $t$,
$$
R_j=Q_0(k_j)-e^{\omega(k)t}Q(k_j,t)-k_j^2F_0+e^{-ik_jL}(k_j^2G_0-ik_jG_1).
$$
The <finite-time spectral boundary transforms> $F_j,G_j$ are invariant under these rotations because their time kernel depends only on $k^3$. The coefficients multiplying them are not invariant; these differences make the three equations useful for eliminating the missing boundary traces.
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