From $\gamma=c/r+O(r^{-2})$,
$$
\partial_r\gamma=-\frac c{r^2}+O(r^{-3}).
$$
The first hypersurface equation gives
$$
\partial_r\beta=\frac{c^2}{2r^3}+O(r^{-4}),
$$
and hence
$$
\boxed{\beta=\beta_0(u,\theta)-\frac{c^2}{4r^2}+O(r^{-3})}.
$$
Asymptotic flatness in the standard Bondi frame sets the integration function $\beta_0=0$.

With this choice, the leading bracket in the second hypersurface equation is
$$
\frac{\partial_\theta c+2c\cot\theta}{r^2}+O(r^{-3}).
$$
Therefore
$$
\partial_r(r^4\partial_rU)
=2(\partial_\theta c+2c\cot\theta)+O(r^{-1}).
$$
Writing $D=\partial_\theta c+2c\cot\theta$ and integrating twice,
$$
\boxed{U=U_0(u,\theta)-\frac{D}{r^2}
-\frac{U_3(u,\theta)}{3r^3}+O(r^{-4})}.
$$
Standard asymptotic flatness sets $U_0=0$; $U_3$ is free integration data at the next order.

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 357 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-357/2/a/solution