Example: Constant astigmatism equation

Equation

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

[B-M]

Zero curvature representation

Let

$A=\left(\begin{array}{cc}\frac{1}{8}\frac{({\lambda }^2+1)z_x}{\lambda z}+\frac{1}{8}\frac{(-{\lambda }^2+1)z_y}{\lambda }& \frac{1}{4}\frac{(\lambda +1{)}^2\sqrt{z}}{\lambda }\\ \frac{1}{4}\frac{(\lambda -1{)}^2\sqrt{z}}{\lambda }& -\frac{1}{8}\frac{({\lambda }^2+1)z_x}{\lambda z}-\frac{1}{8}\frac{(-{\lambda }^2+1)z_y}{\lambda }\end{array}\right),$

$B=\left(\begin{array}{cc}\frac{1}{8}\frac{(-{\lambda }^2+1)z_x}{\lambda z^2}+\frac{1}{8}\frac{({\lambda }^2+1)z_y}{\lambda z}& \frac{1}{4}\frac{-{\lambda }^2+1}{\lambda \sqrt{z}}\\ \frac{1}{4}\frac{-{\lambda }^2+1}{\lambda \sqrt{z}}& -\frac{1}{8}\frac{(-{\lambda }^2+1)z_x}{\lambda z^2}-\frac{1}{8}\frac{({\lambda }^2+1)z_y}{\lambda z}\end{array}\right).$

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

Lax pair reformulation

${\psi }_{xx}=(-\frac{1}{2}\frac{{\mu }^2{z}_{xx}}{(\mu +1)(\mu -1)z}+\frac{1}{2}\frac{\mu z_{xy}}{(\mu +1)(\mu -1)}+\frac{1}{4}\frac{{\mu }^2(3{\mu }^2-2)z_x^2}{(\mu +1{)}^2(\mu -1{)}^2{z}^2}-\frac{1}{2}\frac{{\mu }^3{z}_x{z}_y}{(\mu +1{)}^2(\mu -1{)}^{2}z}+\frac{1}{4}\frac{{\mu }^2{z}_y^2}{(\mu +1{)}^2(\mu -1{)}^2}+\frac{{\mu }^{2}z}{(\mu +1{)}^2(\mu -1{)}^2})\psi ,$

${\psi }_y=-\frac{\mu }{z}{\psi }_x-\frac{1}{2}\frac{\mu z_x}{z^2}\psi .$

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

$\left(\begin{array}{cc}2\frac{\sqrt{\lambda \sqrt{z}}}{\sqrt{z}(\lambda +1)}& 0\\ \frac{1}{4}\frac{(\lambda -1{)}^2\sqrt{\lambda \sqrt{z}}{z}_x}{\lambda z^{\frac{3}{2}}(\lambda +1)}-\frac{1}{4}\frac{\sqrt{\lambda \sqrt{z}}(\lambda -1)z_y}{\lambda \sqrt{z}}& \frac{1}{2}\frac{\sqrt{z}(\lambda +1)}{\sqrt{\lambda \sqrt{z}}}\end{array}\right).$

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