 Solution

The order-$m$ principal symbol of the formal adjoint is $p_m^{\ast }=(-1{)}^m{p}_m$. Derivatives falling on coefficients contribute only lower-order terms. Thus the original plane remains a non-characteristic hypersurface.

For the curved face use the defining function ${\rho }_{\epsilon }=x+y^2-\epsilon$, whose conormal is $(1,2y)$. On the face with $x>0$, $|y|<\sqrt{\epsilon }$ and $0<x\le \epsilon$. Therefore $(x,y,1,2y)$ tends uniformly to $(0,0,1,0)$ as $\epsilon \downarrow 0$. Since $p_m^{\ast }(0,0;1,0)\ne 0$, continuity givesThe same argument supplies a common neighborhood where the transformed coefficient of the highest normal derivative is nonzero, permitting the real analytic normal form used below.