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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2019

Paper 4

Taking $r\to0$ and using $\sinh(r\sqrt s)/r\to\sqrt s$ gives
$$
U(0,s)=\frac{u_0}{s}
-\frac{ka^2\sqrt s}
{s^2\{a\sqrt s\cosh(a\sqrt s)+(ka-1)\sinh(a\sqrt s)\}}.
$$
Set $x=a\sqrt s$. 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
$$
\frac{ka^2\sqrt s}{a\sqrt s\cosh(a\sqrt s)+(ka-1)\sinh(a\sqrt s)}
=1-\left(\frac{a^2}{6}+\frac{a}{3k}\right)s+O(s^2).
$$
The singular part at $s=0$ is therefore
$$
U(0,s)=-\frac1{s^2}
+\frac1s\left(u_0+\frac{a}{3k}+\frac{a^2}{6}\right)+O(1).
$$
Using <long-time asymptotics from a Laplace transform> and part (a), the centre temperature has late-time behaviour
$$
\boxed{u(0,t)\sim u_0-t+\frac{a}{3k}+\frac{a^2}{6}}.
$$

Solution

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

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

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