= Solution
For asymptotic analysis it is useful to separate the spectral density from the observation time. For a fixed horizon $T\ge t$, $H(k,T)$ already has this property. If the boundary traces extend to all positive times with suitable growth, their one-sided <Laplace transforms> give the infinite-horizon version
$$
G_j^\infty(k)=\int_0^\infty e^{w(k)s}g_j(s)\,ds
=\mathcal Lg_j(-w(k)),\qquad
H_\infty=-Q_0(\nu)+(k^2-\nu^2)\mathcal Lg_0(-w)-i(k-\nu)\mathcal Lg_1(-w).
$$
For bounded traces these integrals converge in $D_+$; boundary values are taken by an Abel limit where needed. The finite-horizon formula permits this replacement: the tail from $t$ to infinity contributes an analytic, decaying integrand with zero integral around $\partial D_+$. \b[The resulting spectral functions are independent of $t$.]
They can be made particularly explicit on the two contour pieces. On the real axis and upper curve, respectively,
$$
\nu_{\mathbb R}(k)=-\frac k2-\frac i2\sqrt{3k^2+4},\qquad
\nu_C(k)=\overline k.
$$
The second identity follows because $k^3+k$ is real on $C$: the remaining roots are $\overline k$ and $-2\operatorname{Re}k$. Therefore rewrite the solution as
$$
q=\frac1{2\pi}\int_{\mathbb R}e^{ikx-i(k+k^3)t}[Q_0(k)+H_{\infty,\mathbb R}(k)]\,dk
+\frac1{2\pi}\int_Ce^{ikx-i(k+k^3)t}H_{\infty,C}(k)\,dk.
$$
This isolates all large-time dependence in the exponential, allowing the <stationary phase method> or the <method of steepest descent>. Along $x/t=\xi$, the phase is $i[k\xi-k-k^3]$, and its <stationary points> satisfy $3k^2=\xi-1$. Thus the real saddles coalesce at $\xi=1$; any poles arising from the boundary <Laplace transforms> also have to be retained during contour deformation.
The source specifies smooth boundary traces, not their large-time decay or growth. Accordingly a universal numerical long-time profile cannot be inferred. The infinite-horizon formula requires the stated additional transform hypothesis; for more general data the finite-horizon density $H(k,T)$ remains valid, or an appropriately shifted Laplace contour must be used. This qualification does not leave an unknown boundary trace in the representation.
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