= Solution
Take the linear limit and write $g_j(t)=\partial_x^jq(0,t)$. The spatial and temporal <Volterra integral equations> give, to first order,
$$
b(k)=-Q_0(2k),\qquad
B(k)=-\int_0^T e^{8ik^3s}[g_2(s)+2ikg_1(s)-4k^2g_0(s)]\,ds,
$$
while $a=A=1$ to first order and $c(k,T)=-\widehat q(2k,T)$. The minus sign in $b$ comes from the spatial path running from infinity to zero. For $B$, the inverse in $S=e^{4ik^3T\widehat\sigma_3}\mu_2(0,T)^{-1}$ supplies the minus sign; its conjugation cancels the final-time exponential in the temporal kernel. Substitution in the <global relation> therefore yields
$$
e^{8ik^3T}\widehat q(2k,T)
=Q_0(2k)+4k^2G_0-2ikG_1-G_2,\qquad
G_j=\int_0^T e^{8ik^3s}g_j(s)\,ds.
$$
This is the <backward-sign Airy half-line global relation>, because the linear limit is $q_t-q_{xxx}=0$, independently of $\lambda$.
Its constraint is especially transparent after a time <Laplace transform>. Let $\operatorname{Re}p>0$, choose the principal root $r=p^{1/3}$ with positive real part, and set $k=-ir/2$ in the lower transform half-plane. Then $8ik^3=-p$. Taking the infinite-horizon Laplace limit for a solution of sufficient exponential order gives
$$
\boxed{\widetilde g_2(p)=Q_0(-ir)-r\widetilde g_1(p)-r^2\widetilde g_0(p).}
$$
This proves that the second-derivative trace is fixed by the initial data and two boundary traces; it is not a third freely prescribable datum. Finite-horizon versions retain the terminal transform and express the same relation on $0<t<T$.
To see that two are necessary, rather than merely sufficient to eliminate a transform, the homogeneous transformed spatial equation is $\partial_x^3u-pu=0$. Its roots are $r$, $e^{2\pi i/3}r$, and $e^{4\pi i/3}r$. Since $-\pi/6<\arg r<\pi/6$, exactly two roots have negative real part. Decay at infinity retains two independent spatial modes. Their value/first-derivative boundary matrix has determinant equal to the difference of those two roots, which is nonzero. Thus prescribed $g_0,g_1$ fix both modes; one condition leaves a free mode, while three arbitrary traces overdetermine them.
\b[Two independent admissible boundary conditions are needed at $x=0$, for example $q(0,t)$ and $q_x(0,t)$.] This is the <two boundary traces for reverse-dispersion Airy flow> count and hence the perturbative count for the nonlinear <modified Korteweg-De Vries equation>. The argument determines the small-norm boundary count; it does not claim a global nonlinear existence theorem for arbitrary large boundary data or arbitrary incompatible choices of traces.
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