= Solution
Let
$$
D=\{k:\operatorname{Re}\omega(k)<0\}=\{k:\operatorname{Im}k^3<0\},\qquad D_\pm=D\cap\{\pm\operatorname{Im}k>0\}.
$$
The upper domain is the sector $\pi/3<\arg k<2\pi/3$; the lower domain consists of the sectors $\pi<\arg k<4\pi/3$ and $5\pi/3<\arg k<2\pi$. Orient their boundaries with the corresponding domain to the left. For $0<x<L$ and $0<t<T$, the <Fokas method> gives the \b[integral representation]
$$
\boxed{\begin{aligned}
q(x,t)={}&\frac1{2\pi}\int_{\mathbb R}e^{ikx-\omega(k)t}Q_0(k)dk\\
&-\frac1{2\pi}\int_{\partial D_+}e^{ikx-\omega(k)t}\mathcal F(k,T)dk\\
&-\frac1{2\pi}\int_{\partial D_-}e^{ik(x-L)-\omega(k)t}\mathcal G(k,T)dk.
\end{aligned}}
$$
Here $\mathcal F,\mathcal G$ are the combinations of all six <finite-time spectral boundary transforms> defined above. Initially, <Fourier inversion> of the <global relation> gives the first integral together with $-\int_{\mathbb R}e^{ikx-\omega t}\mathcal F/(2\pi)$ and $+\int_{\mathbb R}e^{ik(x-L)-\omega t}\mathcal G/(2\pi)$. For the left endpoint, deform through the upper complementary sectors, where the spatial exponential decays and
$$
e^{-\omega t}F_j(k,t)=\int_0^te^{-\omega(t-s)}f_j(s)ds
$$
is bounded. This replaces the real line by $\partial D_+$. For the right endpoint, deform in the lower half-plane; the real portions of $\partial D_-$ run from right to left. Thus its integral is the negative of the real-line integral, explaining the minus sign in the last displayed term. The arc contributions vanish by spatial exponential decay and <integration by parts> in the transforms. Oscillatory real-line integrals can equivalently be defined by a decaying cutoff and its limit.
Finally use the <finite-horizon extension of a boundary transform>. The difference between upper limits $t$ and $T$ contains $\int_t^Te^{\omega(s-t)}f_j(s)ds$, bounded in $D$, while the appropriate spatial factor decays inside $D_+$ or $D_-$. The <Cauchy integral theorem> makes this difference vanish. Keeping $T>t$ also avoids an artificial endpoint half-value in temporal inversion. The <integral representation> at this stage deliberately includes the unknown traces; the rotated <global relations> eliminate them in the next step.
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