Write
$$
D=1+\psi_r^2+\psi_z^2.
$$
Differentiating the prescribed identity $u(\psi(r,z),r,z)=0$ in its two <tangent vector>[tangential directions] gives
$$
u_r=-\psi_ru_t,
\qquad
u_z=-\psi_zu_t.
$$
The <unit normal> is $(1,-\psi_r,-\psi_z)/\sqrt D$, so the second item of <Cauchy data> becomes
$$
g=\partial_Nu
=\frac{u_t-\psi_ru_r-\psi_zu_z}{\sqrt D}
=\sqrt D\,u_t.
$$
Consequently
$$
u_t=\frac g{\sqrt D},
\qquad
u_r=-\frac{\psi_rg}{\sqrt D},
\qquad
u_z=-\frac{\psi_zg}{\sqrt D}.
$$
Substitution into the <principal symbol> from part a shows that the graph is <non-characteristic hypersurface>[non-characteristic] exactly where
$$
1-(1-\frac{\psi_zg}{\sqrt{1+\psi_r^2+\psi_z^2}})^2
(\psi_r^2+\psi_z^2)\ne0.
$$

Solved by gpt-5.6-sol high.


 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-105/1/c/solution