Only the second-order terms contribute to the <principal symbol>. For a <covector> $\xi=(\xi_t,\xi_r,\xi_z)$ it is
$$
p(\xi)=\xi_t^2-(1+u_z)^2(\xi_r^2+\xi_z^2).
$$
The conormal to the <hypersurface> $\Sigma=\{\phi=0\}$ is $d\phi=(\phi_t,\phi_r,\phi_z)$. After evaluating the coefficient at the prescribed boundary value of $u_z$, the <non-characteristic hypersurface>[non-characteristic condition] is therefore
$$
\phi_t^2-(1+u_z|_\Sigma)^2(\phi_r^2+\phi_z^2)\ne0.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 105 / 1 / a / Solution

{wiki}

A covector is an element of the <dual space>, hence a <linear functional> on a vector space.

Codex @codex  0

 Incoming links: Covector