 Solution

Undo the commutator gauge by putting ${\psi }_j={\mu }_j{e}^{-i\theta {\sigma }_3}$. Then ${\psi }_{j,x}=(-ik{\sigma }_3+Q){\psi }_j$ and ${\psi }_{j,t}=(-2i{k}^2{\sigma }_3+V){\psi }_j$, with $V=2kQ-i{Q}_x{\sigma }_3-i|q{|}^2{\sigma }_3$. The trace of each coefficient matrix is zero, so all normalized fundamental matrices have determinant one. Their ratios ${\psi }_2^{-1}{\psi }_j$ are independent of both $x$ and $t$, because differentiation cancels the common left-hand coefficient.

Call these constant matrices $s$ for $j=3$ and $S$ for $j=1$. Returning to the commutator gauge givesHere $e^{\alpha {\hat{\sigma }}_3}C=e^{\alpha {\sigma }_3}C{e}^{-\alpha {\sigma }_3}$. Evaluation at $(0,0)$ gives $s={\mu }_3(0,0,k)$, computed solely by the spatial Volterra integral equation for the initial potential. At $(0,T)$ the normalization of ${\mu }_1$ givesThus $S$ is computed by integrating the time Lax pair for ${\mu }_2(0,t,k)$ from zero to $T$. It uses both boundary traces $q(0,t)$ and $q_x(0,t)$ before their compatibility is enforced.