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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2020

Paper 2

By <separation of variables>, the axisymmetric <normal modes> satisfying regularity at $r=0$ and the fixed-edge condition at $r=1$ are
$$
J_0(\gamma_kr)\cos(\gamma_kt),
\qquad
J_0(\gamma_kr)\sin(\gamma_kt),
$$
where $J_0(\gamma_k)=0$. The zero initial displacement removes all cosine terms, so
$$
z(r,t)=\sum_{k=1}^{\infty}
C_kJ_0(\gamma_kr)\sin(\gamma_kt).
$$
Differentiating at $t=0$ gives the weighted Fourier-Bessel expansion
$$
U\mathbf1_{\{r<b\}}
=\sum_{k=1}^{\infty}\gamma_kC_kJ_0(\gamma_kr).
$$
Multiply by $rJ_0(\gamma_\ell r)$ and integrate from zero to one. The orthogonality and norm from part (i) give
$$
\frac12\gamma_\ell C_\ell J_0'(\gamma_\ell)^2
=U\int_0^b rJ_0(\gamma_\ell r)\,dr.
$$
Using
$$
\frac{d}{dr}\bigl[rJ_1(\gamma r)\bigr]
=\gamma rJ_0(\gamma r),
\qquad
J_1(x)=-J_0'(x),
$$
the remaining integral is
$$
\int_0^b rJ_0(\gamma_\ell r)\,dr
=-\frac b{\gamma_\ell}J_0'(\gamma_\ell b).
$$
Hence
$$
\boxed{
C_k=-\frac{2bU\,J_0'(\gamma_kb)}
{\gamma_k^2J_0'(\gamma_k)^2}},
$$
and therefore
$$
\boxed{
z(r,t)=\sum_{k=1}^{\infty}
C_kJ_0(\gamma_kr)\sin(\gamma_kt)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ib / Paper 2 / 13A / ii / Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ib / Paper 2 / 13A / ii / Solution

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Past exam of the mathematics course of the University of Cambridge / 2020 / ib / Paper 2 / 13A / ii / Solution

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