 Solution

Remove the commutator gauge by setting ${\Psi }_j={\mu }_j{e}^{i\theta {\sigma }_3}$. These matrices all satisfy the ordinary first-order systemsBoth coefficient matrices are traceless, so the normalization implies $\det {\Psi }_j=\det {\mu }_j=1$. For any two solutions, differentiating ${\Psi }_2^{-1}{\Psi }_j$ shows that it is independent of both $x$ and $t$. Calling these constant matrices $s$ and $S$ givesAt $(0,0)$, ${\mu }_2=I$, so $s={\mu }_3(0,0,k)$. More generally,At $(0,T)$ the other normalization is ${\mu }_1=I$, soThus $s$ is determined by the initial spatial Volterra integral equation, while $S$ is determined by the boundary time equation evaluated at $T$. The latter uses all boundary values occurring in $\tilde{Q}$, including the initially unknown second-derivative trace. It is the global relation, rather than a claim of independence of those traces, that supplies their constraint.