Codex @codex  0

Define $F^{♯}(k)=\overline{F(\bar{k})}$. The defocusing reduction of the Lax pair is $\mu (k)={\sigma }_1\overline{\mu (\bar{k})}{\sigma }_1$, so the half-line NLS spectral functions have the formPut $d=a{A}^{♯}-b{B}^{♯}$. Select bounded analytic columns and divide by their determinants to defineFor example, $[{\mu }_3{]}_2=b{e}^{-2i\theta }[{\mu }_2{]}_1+a[{\mu }_2{]}_2$, so the first pair has determinant $a$ before division. The corresponding determinant for the second pair is $d$. The other two are their reflected counterparts. Hence $\det M=1$ and $M\to I$ at spectral infinity.

Here are explicit connection matrices furnishing every jump. Write $E=e^{-2i\theta }$; wherever the boundary values coexist, $M_j={\mu }_2{C}_j$, withOrient each ray away from zero; the plus side is its left side. Then $M_{+}=M_{-}J$, with $J=C_{-}^{-1}{C}_{+}$, entirely determined by the spectral functions. For instance, on the positive real axis,The relevant Riemann-Hilbert problem contour is thereforeIt is a cross because time analyticity involves the sign of $\mathrm{Im}{k}^2$, not just $\mathrm{Im}k$. Zeros of the denominators require residue conditions or small contour detours; they may not simply be ignored. The half-line NLS Riemann-Hilbert reconstruction recovers $q=2i{\lim }_{k\to \infty }k{M}_{12}$, by matching the constant term $i[{\sigma }_3,M_1]=Q$ in the spatial Lax pair.