= Solution
To first order in the potential, $a=A=1+O(q^2)$, while the spatial and temporal <Volterra integral equations> give
$$
b(k)=-\int_0^\infty e^{2ikx}q_0(x)dx+O(q^3),\qquad B(k)=-\int_0^T e^{4ik^2s}\bigl[2k g_0(s)+ig_1(s)\bigr]ds+O(q^3).
$$
Thus the linear limit of the <global relation> is exactly
$$
\widehat q_0(-2k)-e^{4ik^2T}\widehat q(-2k,T)=\int_0^T e^{4ik^2s}\bigl[2k g_0(s)+ig_1(s)\bigr]ds,
$$
for the linear equation $iq_t+q_{xx}=0$. This ties the two boundary traces together, rather than allowing them to be independent data.
The count is made explicit by a temporal <Laplace transform>. For $\operatorname{Re}p>0$, the transformed equation is $\widetilde q_{xx}+ip\widetilde q=iq_0$. Put $\alpha=\sqrt{-ip}$ with positive real part. Of the two homogeneous solutions $e^{\pm\alpha x}$, only $e^{-\alpha x}$ decays at infinity. There is therefore exactly one free complex coefficient to be fixed at the endpoint. In particular, prescribed Dirichlet trace $\widetilde g_0$ gives the definite Neumann trace
$$
\widetilde g_1(p)=-\alpha\widetilde g_0(p)-i\int_0^\infty e^{-\alpha y}q_0(y)dy.
$$
This follows by differentiating the decaying Dirichlet Green kernel $-(e^{-\alpha|x-y|}-e^{-\alpha(x+y)})/(2\alpha)$ at $x=0$. Thus \b[one complex boundary condition is needed], such as Dirichlet, Neumann or a suitable Robin combination. Prescribing both traces arbitrarily would overdetermine the problem. In a <Sobolev space> with the required trace regularity, the cubic term is a locally Lipschitz perturbation, so the small-data perturbative problem retains the same boundary-condition count; corner compatibility still must be imposed for a classical solution.
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