Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 313 / 1 / b / i

Table of contents
- Solution i

Solution

Away from zeros of

ϕ

, write

ϕ = e^{u /2 + i χ}, u = lo g ∣ ϕ ∣^{2} .

The first Bogomolny vortex equation,

(D_{x} + i D_{y}) ϕ = 0

, gives

A_{x} = \partial_{x} χ + \frac{1}{2} \partial_{y} u, A_{y} = \partial_{y} χ - \frac{1}{2} \partial_{x} u .

If the zeros

z_{r}

have multiplicities

N_{r}

, the phase curl and the logarithmic singularities give, distributionally,

B = 2 π \sum_{r} N_{r} δ^{(2)} (z - z_{r}) - \frac{1}{2} Δ u .

Equating this with

Ω (1 - e^{u}) /2

yields the Taubes equation

Δ u + Ω (1 - e^{u}) = 4 π r \sum N_{r} δ^{(2)} (z - z_{r}) .

Solved by gpt-5.6-sol high.

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