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

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Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 4 / 14D / c / Solution by

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

Taking

r \to 0

and using

sinh (r s) / r \to s

gives

U (0, s) = \frac{u _{0}}{s} - \frac{k a ^{2} s}{s ^{2} { a s c o s h ( a s ) + ( ka - 1 ) s i n h ( a s )}} .

x = a s

. The Taylor series of the denominator is

x cosh x + (ka - 1) sinh x = ka x + \frac{ka + 2}{6} x^{3} + O (x^{5}) .

Consequently

\frac{k a ^{2} s}{a s c o s h ( a s ) + ( ka - 1 ) s i n h ( a s )} = 1 - (\frac{a ^{2}}{6} + \frac{a}{3 k}) s + O (s^{2}) .

The singular part at

s = 0

is therefore

U (0, s) = - \frac{1}{s ^{2}} + \frac{1}{s} (u_{0} + \frac{a}{3 k} + \frac{a ^{2}}{6}) + O (1) .

Using long-time asymptotics from a Laplace transform and part (a), the centre temperature has late-time behaviour

u (0, t) \sim u_{0} - t + \frac{a}{3 k} + \frac{a ^{2}}{6} .

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