 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ib/paper-2/13a/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

By separation of variables, the axisymmetric normal modes satisfying regularity at

r = 0

and the fixed-edge condition at

r = 1

are

J_{0} (γ_{k} r) cos (γ_{k} t), J_{0} (γ_{k} r) sin (γ_{k} t),

where

J_{0} (γ_{k}) = 0

. The zero initial displacement removes all cosine terms, so

z (r, t) = \sum_{k = 1}^{\infty} C_{k} J_{0} (γ_{k} r) sin (γ_{k} t) .

Differentiating at

t = 0

gives the weighted Fourier-Bessel expansion

U 1_{{r < b}} = \sum_{k = 1}^{\infty} γ_{k} C_{k} J_{0} (γ_{k} r) .

Multiply by

r J_{0} (γ_{ℓ} r)

and integrate from zero to one. The orthogonality and norm from part (i) give

\frac{1}{2} γ_{ℓ} C_{ℓ} J_{0}^{'} (γ_{ℓ})^{2} = U \int_{0}^{b} r J_{0} (γ_{ℓ} r) d r .

Using

\frac{d}{d r} [r J_{1} (γ r)] = γ r J_{0} (γ r), J_{1} (x) = - J_{0}^{'} (x),

the remaining integral is

\int_{0}^{b} r J_{0} (γ_{ℓ} r) d r = - \frac{b}{γ _{ℓ}} J_{0}^{'} (γ_{ℓ} b) .

Hence

C_{k} = - \frac{2 b U J _{0}^{'} ( γ _{k} b )}{γ _{k}^{2} J _{0}^{'} ( γ _{k} ) ^{2}},

and therefore

z (r, t) = k = 1 \sum \infty C_{k} J_{0} (γ_{k} r) sin (γ_{k} t) .

 New to topics? Read the docs here!