 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2020/paper-1/2/b/i/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2020 / Paper 1 / 2 / b / i / Solution by

Codex  0 2026-10-18

Substitute

U = R (r) F (θ)

into the axisymmetric Laplace equation and multiply by

r^{2} / (RF)

. The radial and angular terms must be separate constants:

\frac{1}{R} \frac{d}{d r} (r^{2} R^{'}) = - \frac{1}{F s i n θ} \frac{d}{d θ} (sin θ F^{'}) = λ .

Thus

r^{2} R^{''} + 2 r R^{'} - λ R = 0, \frac{1}{s i n θ} (sin θ F^{'})^{'} + λ F = 0.

With

x = cos θ

, the angular equation becomes

(1 - x^{2}) F_{xx} - 2 x F_{x} + λ F = 0.

Requiring boundedness on the entire polar interval, and using the supplied characterization of bounded Legendre polynomial solutions, gives

λ = n (n + 1)

and

F \propto P_{n} (cos θ)

for

n = 0, 1, 2, \dots

. Trying

R = r^{s}

gives

s (s + 1) = n (n + 1)

, hence

s = n

or

s = - n - 1

.

The globally angular-regular separated modes and their superposition are therefore

U_{n} = (A_{n} r^{n} + B_{n} r^{- n - 1}) P_{n} (cos θ),

U (r, θ) = n = 0 \sum \infty (A_{n} r^{n} + B_{n} r^{- n - 1}) P_{n} (cos θ) .

Coefficients and convergence are determined by the radial domain and boundary data. A domain containing the origin excludes negative radial powers for a regular solution. If the angular domain excludes a polar axis of the spherical coordinate system and singular angular behavior is allowed, additional Legendre-function branches are possible; the displayed family is the one regular over the complete sphere, as intended by the hint.

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