For $\psi_0=(r^2+z^2)/2$,
$$
\psi_{0,r}=r,
\qquad
\psi_{0,z}=z,
\qquad
D=1+r^2+z^2.
$$
The prescribed function is $g=\sqrt D$, so part c gives $u_z=-z$. Hence the <non-characteristic hypersurface>[non-characteristic condition] reduces to
$$
1-(1-z)^2(r^2+z^2)\ne0.
$$
The <Cauchy-Kovalevskaya theorem> therefore guarantees a unique local <real analytic function>[real-analytic] solution at exactly those points
$$
(t_0,r_0,z_0)
=\left(\frac{r_0^2+z_0^2}{2},r_0,z_0\right),
\qquad r_0>0,
$$
for which
$$
1-(1-z_0)^2(r_0^2+z_0^2)\ne0.
$$

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/e/solution