= Solution
Put $\Psi=\mu e^{-i(kx+2k^2t)\sigma}$. The <Lax pair> becomes $\Psi_x=U\Psi$, $\Psi_t=V\Psi$, with
$$
U=-ik\sigma+Q,\qquad V=-2ik^2\sigma+2kQ-i(Q^2+Q_x)\sigma.
$$
Equality 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 identities
$$
[Q,Q^2\sigma]=2Q^3\sigma,\qquad
[Q,Q_x\sigma]=(QQ_x+Q_xQ)\sigma
$$
cancel the differentiated $Q^2$ term, leaving $Q_t+iQ_{xx}\sigma-2iQ^3\sigma=0$. Multiplying by $i$ proves
$$
\boxed{iQ_t-Q_{xx}\sigma_3+2Q^3\sigma_3=0.}
$$
Once $S$ 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.
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