From

γ = c / r + O (r^{- 2})

\partial_{r} γ = - \frac{c}{r ^{2}} + O (r^{- 3}) .

(1)

The first hypersurface equation gives

\partial_{r} β = \frac{c ^{2}}{2 r ^{3}} + O (r^{- 4}),

(2)

and hence

β = β_{0} (u, θ) - \frac{c ^{2}}{4 r ^{2}} + O (r^{- 3}) .

(3)

Asymptotic flatness in the standard Bondi frame sets the integration function

β_{0} = 0

With this choice, the leading bracket in the second hypersurface equation is

\frac{\partial _{θ} c + 2 c c o t θ}{r ^{2}} + O (r^{- 3}) .

(4)

Therefore

\partial_{r} (r^{4} \partial_{r} U) = 2 (\partial_{θ} c + 2 c cot θ) + O (r^{- 1}) .

(5)

Writing

D = \partial_{θ} c + 2 c cot θ

and integrating twice,

U = U_{0} (u, θ) - \frac{D}{r ^{2}} - \frac{U _{3} ( u , θ )}{3 r ^{3}} + O (r^{- 4}) .

(6)

Standard asymptotic flatness sets

U_{0} = 0

;

U_{3}

is free integration data at the next order.

Solved by gpt-5.6-sol high.

b

 0  0

Solution

 0  0

The azimuthal metric component is

g_{ϕϕ} = r^{2} e^{- 2 γ} sin^{2} θ .

(1)

Since

γ = c / r + O (r^{- 2})

\partial_{u} g_{ϕϕ} = - 2 r^{2} e^{- 2 γ} (\partial_{u} γ) sin^{2} θ = - 2 r (\partial_{u} c) sin^{2} θ + O (1) .

(2)

Thus the Bondi news function satisfies

\partial_{u} c = - \frac{1}{2} r \to \infty lim r^{- 1} (sin θ)^{- 2} \partial_{u} g_{ϕϕ} .

(3)

Solved by gpt-5.6-sol high.

c

 0  0

Solution

 0  0

Only the transverse metric components are needed. At fixed

(u, r, θ)

\partial_{ϕ} x = - y, \partial_{ϕ} y = x .

(1)

Contracting this vector with the

x, y

block of the metric makes all angular factors combine to

ρ^{2}

g_{ϕϕ} = κ^{2} ρ^{2} + κλ ρ^{2} E = r^{2} sin^{2} θ (1 + \frac{λ}{κ} E) .

(2)

Substitution into the preceding news formula gives

\partial_{u} c = - \frac{λ}{2 κ} r \to \infty lim (r \partial_{u} E) .

(3)

Therefore the requested constants are

p = - \frac{λ}{2 κ}, q = 1 .

(4)

Solved by gpt-5.6-sol high.

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

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