Example: sine-Gordon equation

Equation

$u_{xy}=\sin u.$

Zero curvature representation

Let

$A=\left(\begin{array}{cc}-\frac{1}{2}\mathrm{i}{u}_x& \lambda \\ \lambda & \frac{1}{2}\mathrm{i}{u}_x\end{array}\right),$

$B=\left(\begin{array}{cc}0& \frac{1}{4}\frac{{\mathrm{e}}^{-\mathrm{i}u}}{\lambda }\\ \frac{1}{4}\frac{{\mathrm{e}}^{\mathrm{i}u}}{\lambda }& 0\end{array}\right)$

Then the sine-Gordon equation is equivalent to $D_{y}A-D_{x}B+[A,B]=0.$

Lax pair reformulation

${\psi }_{xx}=(-\frac{1}{2}\mathrm{i}{u}_{xx}-\frac{1}{4}{u}_x^2+{\lambda }^2)\psi ,$

${\psi }_y=\frac{1}{4}\frac{{\psi }_x}{{\mathrm{e}}^{\mathrm{i}u}{\lambda }^2}+\frac{1}{8}\frac{\mathrm{i}{u}_x}{{\mathrm{e}}^{\mathrm{i}u}{\lambda }^2}\psi .$

is due to gauge transformation with respect to

$\left(\begin{array}{cc}\frac{1}{\sqrt{\lambda }}& 0\\ \frac{-\frac{1}{2}\mathrm{i}{u}_x}{\sqrt{\lambda }}& \sqrt{\lambda }\end{array}\right).$

Using another gauge matrix

$\left(\begin{array}{cr}1& -\mathrm{i}\\ 1& \mathrm{i}\end{array}\right)$

we obtain the real gauge equivalent transcription

$A=\left(\begin{array}{cc}\lambda & -\frac{1}{2}{u}_x\\ \frac{1}{2}{u}_x& -\lambda \end{array}\right),$

$B=\left(\begin{array}{cc}\frac{1}{4}\frac{\cos \left(u\right)}{\lambda }& \frac{1}{4}\frac{\sin \left(u\right)}{\lambda }\\ \frac{1}{4}\frac{\sin \left(u\right)}{\lambda }& -\frac{1}{4}\frac{\cos \left(u\right)}{\lambda }\end{array}\right).$

Lax pair reformulation

${\psi }_{xx}=(-\frac{1}{2}\frac{u_{xxx}}{u_x}+\frac{3}{4}\frac{u_{xx}^2}{u_x^2}-\frac{\lambda u_{xx}}{u_x}-\frac{1}{4}{u}_x^2+{\lambda }^2)\psi ,$

${\psi }_y=-\frac{1}{2}\frac{\sin \left(u\right)}{\lambda u_x}{\psi }_x+(-\frac{1}{4}\frac{\sin \left(u\right)u_{xx}}{\lambda u_x^2}+\frac{1}{4}\frac{\cos \left(u\right)}{\lambda })\psi .$

is due to gauge transformation with respect to

$\left(\begin{array}{cc}2\frac{1}{\sqrt{-2u_x}}& 0\\ -\frac{u_{xx}-2\lambda u_x}{\sqrt{-2u_x}{u}_x}& \frac{1}{2}\sqrt{-2u_x}\end{array}\right).$