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For periodic $U(x,t,\lambda )$ with period $L$, solve $W_x=UW$, $W(0)=I$. The monodromy matrix is $w=W(L)$. If $U,V$ satisfy the zero-curvature condition and are periodic, then $W_t=VW-WV(0)$, hence $w_t=[V(0),w]$. The matrix trace of each power of $w$ is a first integral.

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.

Differentiating the two equations for a fundamental matrix of solutions givesEquality for all independent initial vectors is equivalent to the zero-curvature conditionA single possibly zero solution vector alone would not imply a matrix identity; compatibility is understood for the complete linear system.

Write $q={\varphi }_{\rho }$, $s=\sin \varphi$, $c=\cos \varphi$ and use Pauli matrices:The matrix products givewhereas $V_{\rho }=-(iqc/(4\lambda )){\sigma }_2+(iqs/(4\lambda )){\sigma }_3$. Thus these last two components cancel, and the remaining curvature is $(i/2)({\varphi }_{\rho \tau }-\sin \varphi ){\sigma }_1$. Thereforethe Sine-Gordon equation in light-cone coordinates.

Compatibility of the auxiliary system, computed on a fundamental matrix of solutions, gives the zero-curvature conditionIndeed equality of the mixed derivatives gives $(U_t+UV-V_x-VU)\Psi =0$. A single possibly zero vector solution would not justify the matrix identity; compatibility means a full fundamental family.

Put $H=W_t-VW$. Using the zero-curvature condition gives $H_x=UH$. At $x=0$, $W_t=0$ and hence $H(0)=-V(0)$. Uniqueness of the initial-value problem gives $H=-WV(0)$, or $W_t=VW-WV(0)$. Periodicity implies $V(2\pi )=V(0)=v$, so the monodromy matrix $w=W(2\pi )$ obeys the Lax equationCyclicity of the trace yields $\frac{d}{dt}\mathrm{tr}(w^k)=k\mathrm{tr}(w^{k-1}[v,w])=0$ for every $k\ge 1$.

There is a genuine factor-of-two error in the printed sine-Gordon pair. Assign the second printed matrix to $U$ and the first to $V_{\mathrm{print}}$:Direct multiplication givesThus the printed matrices encode $u_{xt}=2\sin u$.

For the stated Sine-Gordon equation, explicitly repair the first prefactor by taking $V=V_{\mathrm{print}}/2$, that is $1/(4i\lambda )$. Its zero-curvature condition becomes $u_{xt}=\sin u$. For this corrected pair, $\mathrm{tr}(w(t,\lambda {)}^k)$ is conserved for every $\lambda \ne 0$. Since $U$ is entire in $\lambda$, so are $W$ and these traces. Their Taylor coefficients in $\lambda$ supply an infinite family of first integrals, by the identity theorem extending conservation through $\lambda =0$. At a fixed spectral parameter the higher traces of a $2\times 2$ matrix are related by the Cayley-Hamilton theorem; no independence of the resulting first integrals is claimed.

The system prescribes both first derivatives of the same vector $\psi$, so it is overdetermined: the two mixed derivatives must agree. Differentiating givesThus every solution satisfiesFor the system to possess arbitrary nontrivial local solution data, the coefficient matrix must vanish:This is the zero-curvature condition, or the compatibility condition for an overdetermined linear system.

Equality of the mixed derivatives of the auxiliary vector $\Psi$ givesThus the zero-curvature condition for this convention isSubstituting the displayed matrices and collecting powers of the spectral parameter $\lambda$ makes every diagonal entry and every $\lambda$-dependent term cancel. The remaining matrix isConsequently compatibility is equivalent toso the requested constant isThis is the AKNS Lax pair for the nonlinear Schrodinger equation.

The reductionis preserved by the two equations, which become complex conjugates and reduce to the Focusing nonlinear Schrodinger equationThe analogous reductiongives the Defocusing nonlinear Schrödinger equation

For a complex field $r$ and rapidly decreasing boundary conditions, defineTheir variational derivatives, after integration by parts, areBoth equations therefore have the Hamiltonian field equationor, including the conjugate equation,This is the Hamiltonian form of the cubic nonlinear Schrodinger equations.

Now putwhere $E$ and $f$ are real and $f$ is smooth and rapidly decreasing. For the focusing equation,Multiplication by $f^{\prime }$ and use of the decay at infinity gives the first integralA nonzero solution requires $E<0$. Writing $E=-{\kappa }^2$ with $\kappa >0$, separation of variables, or direct substitution, givesHenceis the Bright standing soliton of the focusing nonlinear Schrodinger equation.

For the defocusing equation the profile instead satisfieswith first integralIf a nonzero rapidly decreasing profile existed, $|f|$ would attain a positive maximum $A$. At that point $f^{\prime }=0$, so the identity forces $E=A^2>0$. But along either tail, where $0<|f|<A$, the right-hand side $f^2(f^2-E)$ is negative, which is impossible. Thus there is no nonzero solution of the prescribed form, as recorded by No rapidly decaying standing wave for the defocusing cubic nonlinear Schrodinger equation.

Equality of the mixed derivatives givesFor a fundamental matrix solution this is equivalent to the zero-curvature condition