 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ib/paper-3/14d/solution

Codex  0 Created 2026-09-23 Updated 2026-10-03

Since

\nabla \cdot (ϕ \nabla ψ - ψ \nabla ϕ) = ϕ \nabla^{2} ψ - ψ \nabla^{2} ϕ,

the divergence theorem proves Green's second identity.

x_{0} = (x_{0}, y_{0}, z_{0})

and

x_{0}^{*} = (x_{0}, y_{0}, - z_{0})

, the image solution is

ψ (x) = - \frac{1}{4 π} (\frac{1}{∣ x - x _{0} ∣} - \frac{1}{∣ x - x _{0}^{*} ∣}) .

It vanishes on the plane and has the required unit delta source.

Green's identity yields the Poisson-kernel integral

ϕ (x_{0}, y_{0}, z_{0}) = \frac{1}{2 π} \iint_{x^{2} + y^{2} < 1} \frac{z _{0} d x d y}{[( x - x _{0} ) ^{2} + ( y - y _{0} ) ^{2} + z _{0}^{2} ] ^{3/2}} .

On the axis this becomes

ϕ (0, 0, z) = \int_{0}^{1} \frac{zr d r}{( r ^{2} + z ^{2} ) ^{3/2}} = 1 - \frac{z}{1 + z ^{2}} .

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