Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 3 / 31A (Integrable Systems)

There is a literal inconsistency in the stated Riemann-Hilbert problem: a jump matrix $S(k)$ tending to zero is incompatible with ${\mu }_{\pm }\to I$. Along the real axis the conjugating exponentials are unitary, so the right side of the jump tends to zero while the left side tends to $I$. No normalized solution satisfies the printed decay condition. The intended condition is $S(k)\to I$, or replacement of $S$ in the jump by $I+S$ when $S$ denotes its decaying part.

For this corrected problem assume a regular, invertible, pole-free solution, differentiable in $x,t$, with the required expansion uniformly at infinity. These conditions matter: arbitrary jump data need not be solvable, and poles would contribute extra terms. Write $\sigma ={\sigma }_3$ and $\mu =I+{\mu }_1/k+{\mu }_2/k^2+\cdots$. Matching the nondecaying coefficients in the two differential expressions yieldsThe first expression makes $Q$ off-diagonal.

These matches really establish the differential equations, rather than just formal identities. Let $G=e^{-i(kx+2k^{2}t)\sigma }S(k)e^{i(kx+2k^{2}t)\sigma }$. It satisfies $G_x=-ik[\sigma ,G]$ and $G_t=-2i{k}^2[\sigma ,G]$. Differentiating ${\mu }_{+}={\mu }_{-}G$ consequently shows thathave matching boundary values. They therefore extend holomorphically across the contour. The chosen coefficients make both residuals $O(k^{-1})$ at infinity. Their absence of poles and Liouville theorem imply $R_x=R_t=0$, proving the required Lax pair under the corrected hypotheses. This is matrix NLS reconstruction from a normalized Riemann-Hilbert problem.

The $k^{-1}$ coefficient of the first equation is ${\mu }_{1x}+i[\sigma ,{\mu }_2]=Q{\mu }_1$. Substitution in the expression for $B$ gives $B=-2{\mu }_{1x}$. Decompose ${\mu }_1$ into diagonal and off-diagonal parts. Since $Q=i[\sigma ,{\mu }_1]$, its off-diagonal part is $iQ\sigma /2$. ThusThe commutator in the $k^{-1}$ equation has zero diagonal; multiplication of two off-diagonal two-by-two matrices is diagonal. Combining the two parts proves

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.