Example: sine-Gordon equation

Equation

$\displaystyle u_{xy}=\sin u.$

Zero curvature representation

Let

$A=\left({\begin{array}{cc}-{\frac {1}{2}}\mathrm {i} u_{x}&\lambda \\[1ex]\lambda &{\frac {1}{2}}\mathrm {i} u_{x}\end{array}}\right),$

$B=\left({\begin{array}{cc}0&\displaystyle {\frac {1}{4}}{\frac {\mathrm {e} ^{-\mathrm {i} u}}{\lambda }}\\[2ex]\displaystyle {\frac {1}{4}}{\frac {\mathrm {e} ^{\mathrm {i} u}}{\lambda }}&0\end{array}}\right)$

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

Lax pair reformulation

$\psi _{xx}={\biggl (}-{\frac {1}{2}}\mathrm {i} u_{xx}-{\frac {1}{4}}u_{x}^{2}+\lambda ^{2}{\biggr )}\psi ,$

$\psi _{y}={\frac {1}{4}}{\frac {\psi _{x}}{\mathrm {e} ^{\mathrm {i} u}\lambda ^{2}}}+{\frac {1}{8}}{\frac {\mathrm {i} u_{x}}{\mathrm {e} ^{\mathrm {i} u}\lambda ^{2}}}\psi .$

is due to gauge transformation with respect to

$\left({\begin{array}{cc}\displaystyle {\frac {1}{\sqrt {\lambda }}}&0\\[2ex]\displaystyle {\frac {-{\frac {1}{2}}\mathrm {i} u_{x}}{\sqrt {\lambda }}}&\displaystyle {\sqrt {\lambda }}\end{array}}\right).$

Using another gauge matrix

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

we obtain the real gauge equivalent transcription

$A=\left({\begin{array}{cc}\lambda &-{\frac {1}{2}}u_{x}\\[1ex]{\frac {1}{2}}u_{x}&-\lambda \end{array}}\right),$

$B=\left({\begin{array}{cc}\displaystyle {\frac {1}{4}}{\frac {\cos \left(u\right)}{\lambda }}&\displaystyle {\frac {1}{4}}{\frac {\sin \left(u\right)}{\lambda }}\\[2ex]\displaystyle {\frac {1}{4}}{\frac {\sin \left(u\right)}{\lambda }}&-\displaystyle {\frac {1}{4}}{\frac {\cos \left(u\right)}{\lambda }}\end{array}}\right).$

Lax pair reformulation

$\psi _{xx}={\biggl (}-{\frac {1}{2}}{\frac {u_{xxx}}{u_{x}}}+{\frac {3}{4}}{\frac {u_{xx}^{2}}{u_{x}^{2}}}-{\frac {\lambda u_{xx}}{u_{x}}}-{\frac {1}{4}}u_{x}^{2}+\lambda ^{2}{\biggr )}\psi ,$

$\psi _{y}=-{\frac {1}{2}}{\frac {\sin \left(u\right)}{\lambda u_{x}}}\psi _{x}+{\biggl (}-{\frac {1}{4}}{\frac {\sin \left(u\right)u_{xx}}{\lambda u_{x}^{2}}}+{\frac {1}{4}}{\frac {\cos \left(u\right)}{\lambda }}{\biggr )}\psi .$

is due to gauge transformation with respect to

$\left({\begin{array}{cc}\displaystyle 2{\frac {1}{\sqrt {-2u_{x}}}}&0\\[2ex]\displaystyle -{\frac {u_{xx}-2\lambda u_{x}}{{\sqrt {-2u_{x}}}u_{x}}}&\displaystyle {\frac {1}{2}}{\sqrt {-2u_{x}}}\end{array}}\right).$