= Solution
The other roots of $w(\nu)=w(k)$ satisfy
$$
\nu^2+k\nu+k^2+1=0,\qquad
\nu_\pm=\frac{-k\pm\sqrt{-3k^2-4}}2.
$$
For $k\in D_+$, select the unique other root $\nu(k)$ with negative imaginary part. This is a <dispersion symmetry elimination of a boundary trace>, but only the root in the spatial-transform half-plane may be used. The root is <holomorphic> throughout $D_+$: the upper branch point $2i/\sqrt3$ lies above its boundary. Near the real axis one other root is above and one below; a root can cross the real axis only when $\operatorname{Re}w=0$, so the lower-root choice persists throughout $D_+$.
Evaluate the <global relation> at $\nu$, using the same time transforms because $w(\nu)=w(k)$:
$$
G_2=Q_0(\nu)+(1+\nu^2)G_0-i\nu G_1-e^{wT}\widehat q(\nu,T).
$$
Consequently
$$
F(k,T)=H(k,T)+e^{w(k)T}\widehat q(\nu(k),T),
$$
where the completely known spectral density is
$$
\boxed{H(k,T)=-Q_0(\nu(k))+(k^2-\nu(k)^2)G_0(k,T)-i(k-\nu(k))G_1(k,T).}
$$
The remaining term in the solution formula has integral
$$
\int_{\partial D_+}e^{ikx+w(k)(T-t)}\widehat q(\nu(k),T)\,dk=0.
$$
Its transformed argument stays in the lower half-plane, it is <holomorphic> inside $D_+$, and $e^{w(T-t)}$ does not grow there. Spatial exponential decay closes the truncated domain. Thus the formula in part (b), with $F$ replaced by $H$, uses only the prescribed initial value and the two prescribed boundary traces. It contains no unknown $g_2$ or final-time solution transform. Keeping the inadmissible upper root instead would invalidate this elimination.
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