 Solution

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

Codex  0 Created 2026-09-24 Updated 2026-09-25

For

ψ_{0} = (r^{2} + z^{2}) /2

ψ_{0, r} = r, ψ_{0, z} = z, D = 1 + r^{2} + z^{2} .

The prescribed function is

g = D

, so part c gives

u_{z} = - z

1 - (1 - z)^{2} (r^{2} + z^{2}) \neq = 0.

The Cauchy-Kovalevskaya theorem therefore guarantees a unique local real-analytic solution at exactly those points

(t_{0}, r_{0}, z_{0}) = (\frac{r _{0}^{2} + z _{0}^{2}}{2}, r_{0}, z_{0}), r_{0} > 0,

for which

1 - (1 - z_{0})^{2} (r_{0}^{2} + z_{0}^{2}) \neq = 0.

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