Consider the linear PDE $$\frac{1}{2}yu_{xx}+ryu_{xy}+\frac{1}{2}yu_{yy}=0,$$ where $|r|<1$.
When reducing the PDE to its canonical form, I study \begin{align*} \phi_x,\phi_y = \frac{\frac{1}{2}r y\pm i\sqrt{\frac{1}{4}y^2-\frac{1}{4}r^2y^2}}{\frac{1}{2}y}=r \pm i\sqrt{1-r^2}. \end{align*} which is independent of $x$ and $y$.
Thus, $$\phi(x,y) = \left(r + i\sqrt{1-\rho^2}\right)x + \left(r - i\sqrt{1-\rho^2}\right)y$$
and \begin{align*} \xi&=r (x+y) \\ \eta&=\sqrt{1-r^2}(x-y). \end{align*}
However, the determinant of the Jacobian is zero, $$\xi_x\eta_y+\xi_y\eta_x=0,$$ and the inverse function theorem does not apply. Does this mean this (degenerate) PDE cannot be transformed and simplified?