Example: Constant astigmatism equation

Equation

$\displaystyle z_{yy}+{\bigl (}1/z{\bigr )}_{xx}+2=0.$

[B-M]

Zero curvature representation

Let

$A=\left({\begin{array}{cc}\displaystyle {\frac {1}{8}}{\frac {(\lambda ^{2}+1)z_{x}}{\lambda z}}+{\frac {1}{8}}{\frac {(-\lambda ^{2}+1)z_{y}}{\lambda }}&\displaystyle {\frac {1}{4}}{\frac {(\lambda +1)^{2}{\sqrt {z}}}{\lambda }}\\[2ex]\displaystyle {\frac {1}{4}}{\frac {(\lambda -1)^{2}{\sqrt {z}}}{\lambda }}&\displaystyle -{\frac {1}{8}}{\frac {(\lambda ^{2}+1)z_{x}}{\lambda z}}-{\frac {1}{8}}{\frac {(-\lambda ^{2}+1)z_{y}}{\lambda }}\end{array}}\right),$

$B=\left({\begin{array}{cc}\displaystyle {\frac {1}{8}}{\frac {(-\lambda ^{2}+1)z_{x}}{\lambda z^{2}}}+{\frac {1}{8}}{\frac {(\lambda ^{2}+1)z_{y}}{\lambda z}}&\displaystyle {\frac {1}{4}}{\frac {-\lambda ^{2}+1}{\lambda {\sqrt {z}}}}\\[2ex]\displaystyle {\frac {1}{4}}{\frac {-\lambda ^{2}+1}{\lambda {\sqrt {z}}}}&\displaystyle -{\frac {1}{8}}{\frac {(-\lambda ^{2}+1)z_{x}}{\lambda z^{2}}}-{\frac {1}{8}}{\frac {(\lambda ^{2}+1)z_{y}}{\lambda z}}\end{array}}\right).$

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

Lax pair reformulation

$\psi _{xx}={\biggl (}-{\frac {1}{2}}{\frac {\mu ^{2}z_{xx}}{(\mu +1)(\mu -1)z}}+{\frac {1}{2}}{\frac {\mu z_{xy}}{(\mu +1)(\mu -1)}}+{\frac {1}{4}}{\frac {\mu ^{2}(3\mu ^{2}-2)z_{x}^{2}}{(\mu +1)^{2}(\mu -1)^{2}z^{2}}}-{\frac {1}{2}}{\frac {\mu ^{3}z_{x}z_{y}}{(\mu +1)^{2}(\mu -1)^{2}z}}+{\frac {1}{4}}{\frac {\mu ^{2}z_{y}^{2}}{(\mu +1)^{2}(\mu -1)^{2}}}+{\frac {\mu ^{2}z}{(\mu +1)^{2}(\mu -1)^{2}}}{\biggr )}\psi ,$

$\psi _{y}=-{\frac {\mu }{z}}\psi _{x}-{\frac {1}{2}}{\frac {\mu z_{x}}{z^{2}}}\psi .$

Here $\lambda =(1-\mu )/(1+\mu )$, the corresponding gauge matrix being

$\left({\begin{array}{cc}\displaystyle 2{\frac {\sqrt {\lambda {\sqrt {z}}}}{{\sqrt {z}}(\lambda +1)}}&0\\[2ex]\displaystyle {\frac {1}{4}}{\frac {(\lambda -1)^{2}{\sqrt {\lambda {\sqrt {z}}}}z_{x}}{\lambda z^{\frac {3}{2}}(\lambda +1)}}-{\frac {1}{4}}{\frac {{\sqrt {\lambda {\sqrt {z}}}}(\lambda -1)z_{y}}{\lambda {\sqrt {z}}}}&\displaystyle {\frac {1}{2}}{\frac {{\sqrt {z}}(\lambda +1)}{\sqrt {\lambda {\sqrt {z}}}}}\end{array}}\right).$

References

[B-M] H. Baran and M. Marvan, On integrability of Weingarten surfaces: a forgotten class, J. Phys. A: Math. Theor 42 (2009) 404007, arXiv:1002.0989