Writing a defocusing nonlinear Schrödinger field as turns it into a continuity equation for and a velocity equation for , including a quantum-pressure term.
In a common normalization, the defocusing case is . The sign distinguishes it from the focusing nonlinear Schrödinger equation.
For the Robin boundary condition with real , the time Lax pair has a diagonal-conjugation symmetry under . It implies and . Together with the half-line NLS global relation, this eliminates unknown boundary scattering from an equivalent Riemann-Hilbert problem determined by the initial data and .
Here and . The difference follows algebraically from the half-line NLS global relation and the Robin symmetry. Multiplication by the spatial-temporal jump factor gives , decaying in the second quadrant for . A triangular analytic change of the Riemann-Hilbert problem removes this finite-horizon defect while retaining the reconstruction and transferring all genuine pole information to the known determinant. No unproved large-time decay of the boundary values is needed.
The effective boundary coefficient is . All its ingredients come from initial scattering and the Robin boundary condition. Genuine zeros produce pole data; common zeros in numerator and denominator can instead be removable. It is incorrect to assert absence of boundary poles for every real Robin parameter.
Initial scattering gives and boundary-time scattering gives . The defocusing reduction and unit determinants give the displayed matrix forms. The initial functions are analytic in the upper half-plane, while the boundary functions are bounded in the first and third quadrants. Their finite-horizon compatibility is the half-line NLS global relation.
Evaluating the Lax pair connection formula at gives . The upper off-diagonal entry is the displayed relation, where is analytic in the upper half-plane and is . A finite-time relation must retain this final-state term.
For with real , the defocusing equation has first integralA nonzero smooth function tending to zero at both infinities cannot satisfy this identity: a nonzero extremum requires , but then the right side is negative for sufficiently small nonzero .
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