Away from a zero of the Higgs field, write
$$
\phi=e^{u/2+i\chi},
\qquad u=\log|\phi|^2.
$$
Separating real and imaginary parts of $(D_x+iD_y)\phi=0$ gives
$$
A_x=\partial_x\chi+\frac12\partial_yu,
\qquad
A_y=\partial_y\chi-\frac12\partial_xu.
$$
Thus $B=-\Delta u/2$ away from zeros. Combining this with $B=(1-e^u)/2$ gives
$$
\Delta u+1-e^u=0.
$$
If $\phi$ has a zero of multiplicity $N$ at the origin, its <vortex number> is the <winding number> $N$ and $u=2N\log r+O(1)$. Since $\Delta\log r=2\pi\delta^{(2)}(\mathbf x)$ as a <distributional identity>, the complete <Taubes equation> is
$$
\boxed{
\Delta u+1-e^u=4\pi N\delta^{(2)}(\mathbf x)}.
$$
The boundary conditions for a single $N$-vortex are
$$
\boxed{
u(r)=2N\log r+O(1)\quad(r\to0),
\qquad
u(r)\to0\quad(r\to\infty)}.
$$
The latter is the finite-energy condition $|\phi|\to1$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 313 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-313/2/a/solution