 Solution

Put $\Psi =\mu e^{-i(kx+2k^{2}t)\sigma }$. The Lax pair becomes ${\Psi }_x=U\Psi$, ${\Psi }_t=V\Psi$, withEquality of mixed derivatives, using invertibility, gives the zero-curvature condition $U_t-V_x+[U,V]=0$. The terms quadratic in $k$ cancel. The terms linear in $k$ cancel because $\sigma Q=-Q\sigma$, $\sigma Q_x=-Q_x\sigma$ and $Q^2$ commutes with $\sigma$. For the constant term, the identitiescancel the differentiated $Q^2$ term, leaving $Q_t+i{Q}_{xx}\sigma -2i{Q}^3\sigma =0$. Multiplying by $i$ provesOnce $S$ is known, its time dependence enters through the explicit exponential conjugation. The boundary relation and normalization form a linear problem for $\mu$, and $Q=i[\sigma ,{\mu }_1]$ reconstructs the nonlinear solution. This is the requested linearization. Its disadvantage as an initial-value method is that it does not itself determine $S$ from a prescribed $Q(x,0)$. The inverse scattering transform supplies that direct-scattering step, including any discrete spectral data and their pole conditions, before reconstruction. Solvability cannot be inferred merely from writing an arbitrary jump.