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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2024

Paper 3

For a radial <function> in three dimensions, the <radial Laplacian> gives
$$
\nabla^2\phi=\frac1{r^2}(r^2\phi')'.
$$
Here regularity at zero selects
$$
r^2\phi'(r)=\int_0^rs^2(s-1)e^s\,ds
=e^r(r^3-4r^2+8r-8)+8.
$$
A further integration gives
$$
\phi(r)=\frac{e^r(r^2-5r+8)-8}{r}+C,
$$
whose apparent singularity is removable, with <limit> $3+C$ at zero. Since the nonconstant term has value $4e-8$ at $r=1$, the <boundary condition> $\phi(1)=3$ gives $C=11-4e$. Therefore
$$
\boxed{\phi(r)=
\frac{e^r(r^2-5r+8)-8}{r}+11-4e}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 3 / 9B / c / i / Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 3 / 9B / c / i / Solution

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