The second order linear PDE

$a(x,y)u_{xx}+ 2b(x,y)u_{xy}+c(x,y)u_{yy}= 0$

was transformed by introducing new independent variable $ξ=φ(x,y)$ and $η=ψ(x,y)$, transforming the PDE to the form

$A(ξ,η)u_{ξξ}+ 2B(ξ,η)u_{ξη}+C(ξ,η)u_{ηη}= 0$

with

$A=aphi_x^2 + 2bphi_xphi_y+cphi_y^2$

$B=aphi_xpsi_x+b(phi_xpsi_y+phi_ypsi_x)+cphi_ypsi_y$

$C=apsi_x^2+2bpsi_xpsi_y+cpsi_y^2$

Assume we are in the elliptic case.

Show that in this case we can obtain the situation $A=C$ and $B=0$ in case that $ψ$ and $φ$ are solutions of

$φ_x=frac{bψ_x+cφ_y}{sqrt{ac−b^2}},;;;φ_y=−frac{aψ_x+bφ_y}{sqrt{ac−b^2}}$

and $ψ$ is a solution of the Beltrami equation:

$left(frac{aψ_x+bφ_y}{sqrt{ac−b^2}}right)_x+ left(frac{bψ_x+cφ_y}{sqrt{ac−b^2}}right)_y= 0$.

I’ve tried to calculate $A-C=0$ and $B=0$ and set them even. But I’m stuck at

$frac{phi_x+ipsi_x}{phi_y+ipsi_y} = frac{-bpmsqrt{b^2-ac}}{a}$.

Can you guys please help me? 🙂