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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2025

Paper 3

Since
$$
\nabla\cdot(\phi\nabla\psi-\psi\nabla\phi)=\phi\nabla^2\psi-\psi\nabla^2\phi,
$$
the divergence theorem proves Green's second identity.

If $x_0=(x_0,y_0,z_0)$ and $x_0^*=(x_0,y_0,-z_0)$, the image solution is
$$
\psi(x)=-\frac1{4\pi}\left(\frac1{|x-x_0|}-\frac1{|x-x_0^*|}\right).
$$
It vanishes on the plane and has the required unit delta source.

Green's identity yields the Poisson-kernel <integral>
$$
\phi(x_0,y_0,z_0)=\frac1{2\pi}\iint_{x^2+y^2<1}
\frac{z_0\,dx\,dy}{[(x-x_0)^2+(y-y_0)^2+z_0^2]^{3/2}}.
$$
On the axis this becomes
$$
\boxed{\phi(0,0,z)=\int_0^1\frac{zr\,dr}{(r^2+z^2)^{3/2}}
=1-\frac z{\sqrt{1+z^2}}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 3 / 14D / Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 3 / 14D / Solution

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Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 3 / 14D / Solution

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