 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ib/paper-4/14d/b/solution

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

t

u (r, 0) = u_{0}

s U - u_{0} = \frac{1}{r} (r U)_{rr} .

Put

W = r U

and

q = s

. Then

W^{''} - s W = - u_{0} r,

U (r, s) = \frac{u _{0}}{s} + \frac{A ( s ) s i n h ( q r ) + B ( s ) c o s h ( q r )}{r} .

Finiteness at the centre forces

B (s) = 0

The transformed boundary condition is

\frac{1}{k} U_{r} (a, s) = \frac{u _{0}}{s} - U (a, s) - \frac{1}{s ^{2}} .

Since

U_{r} (a, s) = A (s) \frac{q a c o s h ( q a ) - s i n h ( q a )}{a ^{2}},

substitution and cancellation of the two

u_{0} / s

terms gives

A (s) [\frac{q a c o s h ( q a ) - s i n h ( q a )}{k a ^{2}} + \frac{s i n h ( q a )}{a}] = - \frac{1}{s ^{2}} .

Hence

A (s) = - \frac{k a ^{2}}{s ^{2} { q a c o s h ( q a ) + ( ka - 1 ) s i n h ( q a )}},

and the required explicit transform is

U (r, s) = \frac{u _{0}}{s} - \frac{k a ^{2} sinh ( r s )}{r s ^{2} { a s cosh ( a s ) + ( ka - 1 ) sinh ( a s )}} .

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