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Put $E_k=e^{i(kx+k^{3}t)}$, $I={\int }_L\phi (l)d\lambda (l)$ and $q=-I_x$. For the dressing operator $\mathcal{F}h=h+i{E}_k{\int }_{L}h(l)/(k+l)d\lambda (l)$, differentiating gives $\mathcal{F}(M\phi )=0$ and $\mathcal{F}(N\phi )=3ikM\phi$, where $M={\partial }_x^2-ik{\partial }_x+q$ and $N={\partial }_t+{\partial }_x^3+3q{\partial }_x$. Uniqueness of the linear integral equation therefore supplies a Lax pair. Its commutator identity is $[N,M]=-3q_{x}M+(q_t+q_{xxx}+6q{q}_x)$, which gives the KdV equation when the eigenfunction is not identically zero. Differentiation under the integral and the uniqueness assumption are essential hypotheses.

For the three corner-normalized Lax pair eigenfunctions, let $s={\mu }_3(0,0,k)$ and $S=e^{4i{k}^{3}T{\hat{\sigma }}_3}{\mu }_2(0,T,k{)}^{-1}$. Their constant connection matrices give the displayed global relation. If their second columns are $(b,a{)}^T$ and $(B,A{)}^T$, its upper-right entry is $Ab-Ba=e^{8i{k}^{3}T}{\mu }_{3,12}(0,T,k)$. The right side is analytic in the lower half-plane. In the linear limit this is the backward-sign Airy half-line global relation at Fourier parameter $2k$, coupling all three boundary traces rather than allowing three independent boundary conditions.

Evaluating the Lax pair connection formula at $(0,T)$ gives $S^{-1}s=e^{2i{k}^{2}T{\hat{\sigma }}_3}{\mu }_3(0,T,k)$. The upper off-diagonal entry is the displayed relation, where $c_T=[{\mu }_3(0,T){]}_{12}$ is analytic in the upper half-plane and is $O(1/k)$. A finite-time relation must retain this final-state term.

Combine columns of the three normalized Lax pair eigenfunctions in their common Volterra analyticity sectors for a half-line NLS Lax pair, dividing by the relevant determinants $a,d,d^{♯},a^{♯}$. The resulting matrix is normalized to the identity at infinity and has jumps on the real and imaginary axes. Connection matrices determine each jump explicitly. Zeros of those determinants require residue conditions or local contour deformations.

The Harry Dym equation is the integrable partial differential equation $p_t=p^3{p}_{xxx}$. It admits a Lax pair based on the second-order operator $L=-{\partial }_x{p}^2{\partial }_x$.

An integrable partial differential equation has enough exact structure, such as a Lax pair, infinitely many conserved quantities, or an inverse scattering transform, to reduce its evolution to auxiliary linear or finite-dimensional data.

A linear integral equation depends linearly on its unknown function. For example, $(I+K)f=g$ with $(Kf)(x)=\int K(x,y)f(y)dy$ has a trivial homogeneous null space whenever its inhomogeneous problem has a unique solution: otherwise a nonzero homogeneous solution could be added to that solution. This is the uniqueness step used in dressing constructions of a Lax pair.

For the Robin boundary condition $q_x(0,t)=cq(0,t)$ with real $c$, the time Lax pair has a diagonal-conjugation symmetry under $k\mapsto -k$. It implies $A(-k)=A(k)$ and $B(-k)=-h(k)B(k)$. 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 $c$.

Assume a regular invertible normalized Riemann-Hilbert problem solution $\mu =I+{\mu }_1/k+{\mu }_2/k^2+\cdots$ with jump tending to the identity and conjugated by $e^{-i(kx+2k^{2}t){\sigma }_3}$. Residuals of its two differential equations have no jump and vanish at infinity, so Liouville theorem gives a Lax pair. Coefficient comparison gives the displayed reconstruction and $A=2Q$; compatibility yields $i{Q}_t-Q_{xx}{\sigma }_3+2Q^3{\sigma }_3=0$. A jump matrix tending to zero instead is inconsistent with the normalization. Recovering the initial jump from a prescribed initial potential still requires direct scattering.

The two differentiated linear integral equations have now supplied both equations of the Lax pair. To verify the nonlinear consequence explicitly, put $D={\partial }_x$ and take the operator commutator in the order $[N,M]=NM-MN$. The product rule givesFor example, $[D^3,q]=3q_x{D}^2+3q_{xx}D+q_{xxx}$ and $[3qD,D^2]=-6q_x{D}^2-3q_{xx}D$, which explain the cancellations. Acting on $\phi$ yieldsThe integral equation has the large-real-$k$ normalization $\phi /E_k=1+O(k^{-1})$ under the stated contour regularity, so its eigenfunction is not zero for every $k$ at a fixed $(x,t)$. Consequently $q_t+q_{xxx}+6q{q}_x=0$.

Thus a time-independent contour and measure determine $\phi$ through a linear spectral integral equation; $q=-{\partial }_x{\int }_L\phi d\lambda$ then reconstructs the nonlinear solution. This is the asserted linearisation. It does not assert that sums of reconstructed nonlinear solutions are again solutions, and it depends on the given uniqueness and differentiability hypotheses.

Put $\theta =kx-4k^{3}t$. Conjugation by $e^{i\theta {\sigma }_3}$ is the meaning of $e^{i\theta {\hat{\sigma }}_3}$. The normalized eigenfunctions have the Volterra integral equation representationChoose ${\gamma }_1$ from $(0,T)$ down the time boundary and then right, ${\gamma }_2$ from $(0,0)$ up the time boundary and then right, and ${\gamma }_3$ horizontally from infinity. Compatibility of the Lax pair makes the resulting construction path-independent.

Let $a=\mathrm{Im}k$ and $b=\mathrm{Im}{k}^3$. The nontrivial exponential in column one has modulus $e^{2a(x-x^{\prime })-8b(t-t^{\prime })}$; column two has its reciprocal. For ${\gamma }_1$, $x-x^{\prime }\ge 0$ and $t-t^{\prime }\le 0$, giving signs $a,b\le 0$ for column one and $a,b\ge 0$ for column two. For ${\gamma }_2$, both differences are nonnegative, giving $(a\le 0,b\ge 0)$ and $(a\ge 0,b\le 0)$. For ${\gamma }_3$, $x-x^{\prime }\le 0$ and the time difference is zero, giving upper and lower half-planes, respectively.

Define $D_1:(a>0,b>0)$, $D_2:(a>0,b<0)$, $D_3:(a<0,b>0)$ and $D_4:(a<0,b<0)$. The bounded analytic column domains are thereforeThese are the Volterra analyticity sectors for reverse-dispersion mKdV, shown in the right panel of the figure. $D_1$ is the pair of upper outer sectors, $D_2$ the upper middle sector, $D_3$ the lower middle sector and $D_4$ the lower outer pair. Their boundaries are the six rays $\arg k=m\pi /3$. The Volterra integral equation converges and differentiates analytically where these kernels are bounded, assuming the usual sufficiently integrable spatial decay and smoothness needed by these normalizations. If decay is interpreted as merely pointwise convergence to zero without an integrability condition, existence and the stated large-$k$ normalization need additional hypotheses. A finite-path eigenfunction may be entire for fixed coordinates, but boundedness for large $k$ holds in the displayed domains; entire continuation does not enlarge those bounded sectors.

For real $q$, write $\bar{f}(k)=\overline{f(\bar{k})}$. The reduction symmetry and unit determinants give the scattering data parametrizationsTo see the symmetry, conjugate the coefficient matrices by $C=\left(\begin{array}{cc}0& 1\\ \lambda & 0\end{array}\right)$ after taking complex conjugates at $\bar{k}$; $C{\sigma }_3{C}^{-1}=-{\sigma }_3$ and the real-$q$ reduction restore the original Lax pair. DefineIf ${\mu }_j^{(l)}$ denotes column $l$, bounded columns produce the following normalized matrices:Each divisor is exactly the determinant of the unscaled column pair, so $\det M_j=1$. The eigenfunctions and divisors tend to $I$ and one, respectively, giving $M\to I$ at infinity. The contour for the sectorial half-line mKdV Riemann-Hilbert reconstruction isIt is six rays, not just the real axis or a four-quadrant contour. For the displayed meromorphic construction assume the divisors have no zeros on the contour; contour zeros require an indented or otherwise regularized formulation.

Here are explicit jump data without ambiguity about which side is used. DefineThe connection formulas give $M_j={\mu }_2{e}^{i\theta {\hat{\sigma }}_3}{C}_j$. Orient each ray outwards and let plus mean its left side. If that ray separates domains $D_i$ on the minus side and $D_j$ on the plus side, its jump isThis specifies every jump using only the spectral connection matrices; each jump has determinant one. Boundedness and consistency at the junction are inherited from the original eigenfunctions when the spectral data satisfy their global constraint.

If $a,\bar{a},d_{+}$ or $d_{-}$ has zeros in its corresponding domain, their simple zeros generate poles and require residue conditions. For example, at a simple zero $k_0$ of $a$ in $D_3$,At a simple zero of $d_{+}$ in $D_1$, the analogous coefficient for the first column is $\lambda \bar{b}{e}^{-2i\theta }/(A{d}_{+}^{\prime })$ times the second column; in $D_4$ it is $b{e}^{2i\theta }/(\bar{A}{d}_{-}^{\prime })$ for the second column. The $\bar{a}$ residue in $D_2$ is the conjugate first-column relation. These formulas assume the other displayed denominators are nonzero there; multiple or coincident zeros need their corresponding higher-order conditions. In a sufficiently small-data zero-free regime, the normalized Riemann-Hilbert problem has only the jumps. Finally the order-one term in ${\mu }_x-ik[{\sigma }_3,\mu ]=Q\mu$ givesThe column normalizations do not change this off-diagonal leading coefficient.

Let $F_{+}=(F+\ast F)/2$ and $F_{-}=(F-\ast F)/2$, using the oriented Euclidean metric. For a self-dual two-form $\omega$,Apply this componentwise in the Lie algebra. Since the three ${\omega }_i$ span the self-dual subspace, their vanishing wedges force $F_{+}=0$. Thus $\ast F=-F$. The gauge-theory Bianchi identity is $D_{A}F=0$, so $D_A\ast F=-D_{A}F=0$, the Euclidean Yang-Mills equations. This proves the implication locally; finite action would additionally be needed to call the solution an instanton.

Use the complex orientation for $w=x^1+i{x}^2$ and $z=x^3+i{x}^4$. The real and imaginary parts of $dw\wedge dz$, and $i(dw\wedge d\bar{w}+dz\wedge d\bar{z})$, span the self-dual two-forms. Accordingly the three gauge curvature equations areFor a spectral parameter $\lambda \in {\mathbb{CP}}^1$, consider the Lax pairTheir commutator isIt vanishes for every $\lambda$ exactly when all three coefficients vanish. The derivative directions ${\partial }_w-\lambda {\partial }_{\bar{z}}$ and ${\partial }_z+\lambda {\partial }_{\bar{w}}$ are mutually orthogonal null vectors for the complexified metric. Flatness on the corresponding two-planes makes the linear parallel-transport equations locally compatible. Conversely compatibility for an invertible fundamental matrix $\Psi$ forces the commutator to vanish; a single zero solution would not establish it. This is the Euclidean complex-coordinate anti-self-dual Lax pair.

For an Abelian U(1) connection, all commutators of the connection components vanish, and the gauge curvature equations therefore readLocally the first two equations say that $A^{1,0}$ is $\partial$-closed and $A^{0,1}$ is $\bar{\partial }$-closed. The Dolbeault-Poincaré lemma, and its complex conjugate, give scalar functions $u,v$ withSubstituting in the third equation givesIn a complex gauge transformation, use $g=e^{-u}$ and the derivative-plus-connection convention $A\mapsto A+g^{-1}dg$. ThenThis is the scalar Laplace equation, since for the stated metric $\Delta =2({\partial }_z{\partial }_{\bar{z}}+{\partial }_w{\partial }_{\bar{w}})$. The complex gauge transformation need not remain in $U(1)$.

A reduction using only a unitary gauge transformation is also available. Since $A$ is anti-Hermitian, $A_{\bar{w}}=-\overline{A_w}$ and $A_{\bar{z}}=-\overline{A_z}$, so the potential for the barred components can be chosen as $v=-\bar{u}$. Write $u=p+iq$ with real $p,q$. Then $A=\partial p-\bar{\partial }p+idq$, and $g=e^{-iq}$ removes the last term. With $f=-2p$ this givesIts gauge curvature is $\partial \bar{\partial }f$. Thus the Abelian anti-self-dual connections from harmonic scalar potentials reduction is local and does not require silently enlarging the real gauge group.

Let $D_1,D_2,D_3,D_4$ denote the open quadrants, counterclockwise from the first. Write $\theta =kx+2k^{2}t$. The Lax pair has the Volterra integral equation kernel $e^{-i[k(x-x^{\prime })+2k^2(t-t^{\prime })]{\hat{\sigma }}_3}$, obtained by integrating its closed matrix differential form from a normalization point.

In column one, the off-diagonal factor is $e^{2i[k(x-x^{\prime })+2k^2(t-t^{\prime })]}$, whose modulus is $e^{-2\mathrm{Im}[k(x-x^{\prime })+2k^2(t-t^{\prime })]}$. Column two has the reciprocal factor. For ${\mu }_2$, choose the path $(0,0)\to (0,t)\to (x,t)$, so both differences are nonnegative. Column one requires $\mathrm{Im}k\ge 0$ and $\mathrm{Im}{k}^2\ge 0$, giving $D_1$; column two requires the opposite inequalities, giving $D_4$.

For ${\mu }_1$, the time path starts at $T$ and runs backward, while the spatial path runs rightward. Its column-one conditions are $\mathrm{Im}k\ge 0$, $\mathrm{Im}{k}^2\le 0$, giving $D_2$; column two is bounded in $D_3$.

Finally, ${\mu }_3$ integrates spatially from infinity to $x$, where $x-x^{\prime }\le 0$. Column one is bounded in the lower half-plane and column two in the upper half-plane. Normalization at $(\infty ,T)$ is equivalent to normalization at $(\infty ,t)$ because the decaying potential vanishes on the vertical path at infinity. ThusThe Volterra integral equation Neumann series converges locally uniformly under the usual integrable smooth scattering assumptions, proving holomorphic dependence in the open domains and continuous boundary values. These are the Volterra analyticity sectors for a half-line NLS Lax pair; at $x=0$, absence of a spatial segment allows the larger time-only domains used for $A,B$.

Undo the commutator gauge by putting ${\psi }_j={\mu }_j{e}^{-i\theta {\sigma }_3}$. Then ${\psi }_{j,x}=(-ik{\sigma }_3+Q){\psi }_j$ and ${\psi }_{j,t}=(-2i{k}^2{\sigma }_3+V){\psi }_j$, with $V=2kQ-i{Q}_x{\sigma }_3-i|q{|}^2{\sigma }_3$. The trace of each coefficient matrix is zero, so all normalized fundamental matrices have determinant one. Their ratios ${\psi }_2^{-1}{\psi }_j$ are independent of both $x$ and $t$, because differentiation cancels the common left-hand coefficient.

Call these constant matrices $s$ for $j=3$ and $S$ for $j=1$. Returning to the commutator gauge givesHere $e^{\alpha {\hat{\sigma }}_3}C=e^{\alpha {\sigma }_3}C{e}^{-\alpha {\sigma }_3}$. Evaluation at $(0,0)$ gives $s={\mu }_3(0,0,k)$, computed solely by the spatial Volterra integral equation for the initial potential. At $(0,T)$ the normalization of ${\mu }_1$ givesThus $S$ is computed by integrating the time Lax pair for ${\mu }_2(0,t,k)$ from zero to $T$. It uses both boundary traces $q(0,t)$ and $q_x(0,t)$ before their compatibility is enforced.

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.

We derive the Robin spectral elimination while retaining the finite-time global relation. This distinguishes the actual ratio $B/A$ from the equivalent, time-independent boundary coefficient needed by the Riemann-Hilbert problem.

At $x=0$ the time Lax pair coefficient has off-diagonal entries $(2k+ic)q$ and $(2k-ic)\bar{q}$. LetDirect multiplication shows $V(0,t,k)=K(k)V(0,t,-k)K(k{)}^{-1}$. Because the commutator term depends on $k^2$ and $K$ commutes with ${\sigma }_3$, uniqueness of the time Volterra integral equation gives $S(k)=K(k)S(-k)K(k{)}^{-1}$. ThereforeThe exceptional parameter points of $K$ are treated by analytic continuation. These identities express linearizable Robin scattering for defocusing NLS.

For finite horizon $T$, the exact upper-half-plane ratio isConsequently the often used answer $B/A=b/a$ is not a literal finite-$T$ identity. Its effective time-independent replacement, in the two bounded time sectors, isThe second branch follows from the Robin symmetry; its argument $-k$ lies in the upper half-plane where the initial scattering functions are known. We now justify this replacement for the inverse problem without assuming that boundary values decay as $t\to \infty$.

Only $\Gamma =B^{♯}/(ad)$ is needed for the connection across the upper imaginary ray. The conjugate global relation at $-k$ and the Robin symmetry giveFor upper-half-plane $k$, the notation $b^{♯}(-k)=\overline{b(-\bar{k})}$ uses only values of $b$ in its legitimate upper-half-plane domain. PutThe effective boundary coefficient isThis Robin spectral determinant for defocusing NLS is explicitly computed from the initial data and $c$.

For completeness, the finite-horizon defect can be removed algebraically, not by an unproved limit. If $r=B^{♯}/A^{♯}$, then $\Gamma =r/[a(a-br)]$. Subtracting the same expression with $r_0=-h{b}^{♯}(-k)/a^{♯}(-k)$ givesIts jump multiplier isThe exponential decays in $D_2$ for $x>0$ and $t<T$, since $\mathrm{Im}k>0$ and $\mathrm{Im}{k}^2<0$. The coefficient is holomorphic away from the stated determinant zeros and is appropriately decaying at infinity. From the explicit matrices in part (c), $C_2=C_{1}L(\Gamma e^{2i\theta })$, where $L(z)=\left(\begin{array}{cc}1& 0\\ z& 1\end{array}\right)$. Hence replace the sector-two solution byThe reflected upper-triangular change is made in $D_3$. These analytic changes remove the finite-time defect, leave $M_1,M_4$ unchanged, and leave the reconstruction from the second column in $D_2$ unchanged. At apparent poles, the old $d$-dependent singularities cancel by the connection formula; genuine poles of the new determinant remain as explicit residue data. Equivalently, small circles can first isolate all zeros and the same algebra can be performed on the regularized contours. This is finite-horizon Robin spectral elimination.

Every modified jump is now explicitly known: keep $C_1,C_4$ from part (c) and replaceAll jumps are again $C_{-}^{-1}{C}_{+}$, now involving only $a,b,c$. If $\mathcal{D}$ has a simple zero $k_j$ giving a genuine pole, its coefficient iswith the reflected second-column condition at ${\bar{k}}_j$. Common numerator/denominator zeros are checked for removability, and zeros on the cross require contour indentation or local circle jumps. Thus no unknown boundary norming data remain. One must not assume away such zeros for every real $c$; attractive Robin parameters can support localized modes.

It follows that the equivalent normalized Riemann-Hilbert problem is specified entirely by $q_0$ and $c$, including its pole data. When two solutions exist, their quotient has no jump, its prescribed pole singularities cancel, and it tends to $I$ at infinity; Liouville theorem makes the quotient $I$, proving uniqueness. Smooth corner evolution additionally requires $q_0^{\prime }(0)=c{q}_0(0)$ and higher compatibility conditions at higher regularity. Without that compatibility, the same initial data are interpreted in the weak evolution sense, rather than as a smooth solution through the corner.

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.

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.

Let $E=e^{-ikx+k^{2}t}$. Expanding ${\partial }_t(Eq)+{\partial }_x(EX)$ shows that the required flux isIndeed $X_x-ikX=-q_{xx}-k^{2}q$, cancelling the $k^{2}q$ from the time derivative. Introduce a potential $\phi$ by ${\phi }_x=Eq$, ${\phi }_t=E(q_x+ikq)$. Equality of its mixed derivatives is precisely the conservation law. With $\phi =E\mu$, these equations become the linear auxiliary systemThis is a Lax pair in the affine-potential form used for linear evolution equations. If a homogeneous matrix pair is preferred, set $\Psi =(\mu ,1{)}^T$ and use ${\Psi }_x=\left(\begin{array}{cc}ik& q\\ 0& 0\end{array}\right)\Psi$, ${\Psi }_t=\left(\begin{array}{cc}-k^2& q_x+ikq\\ 0& 0\end{array}\right)\Psi$. Its zero-curvature condition has upper-right entry $q_t-q_{xx}$ and all other entries zero.