The surface area is
$$
\operatorname{Area}(\Sigma)
=\int_{\mathbb R^2}\frac{a^2}{(1+|z|^2)^2}\,dx\,dy
=2\pi a^2\int_0^\infty\frac{r\,dr}{(1+r^2)^2}
=\pi a^2.
$$
Integrating the <Taubes equation> over the compact surface eliminates the Laplacian and gives
$$
4\pi N=\int_\Sigma\Omega(1-e^u)d^2x
<\operatorname{Area}(\Sigma)
$$
for a nontrivial solution. For $N=2$, the <Bradlow bound> therefore requires
$$
\boxed{\pi a^2>8\pi,
\qquad a^2>8}.
$$
Equality is the dissolved-vortex limit with identically vanishing Higgs field and does not give the stipulated ordinary two-vortex solution.

Solved by gpt-5.6-sol high.

Solution

Paper 313 / 1 / b / iii / Solution

For a zero of order $N$ at the origin and no other zeros,
$$
u(r)=2N\log r+O(1)
\quad(r\to0),
\qquad
u(r)\to0
\quad(r\to\infty).
$$
Away from the zero, the flat <Taubes equation> is
$$
\Delta u=e^u-1.
$$
If $u$ had a positive interior maximum, then the <Hessian matrix>[second-derivative test] would give $\Delta u\leq0$ there, while $e^u-1>0$, a contradiction. The boundary values are $-\infty$ at the zero and $0$ at infinity, so the <maximum principle for subharmonic functions>[maximum principle] gives $u\leq0$. Hence
$$
\boxed{|\phi|=e^{u/2}\leq1}.
$$

Solved by gpt-5.6-sol high.

Paper 313 / 1 / b / ii / Solution

Away from zeros of $\phi$, write
$$
\phi=e^{u/2+i\chi},
\qquad
u=\log|\phi|^2.
$$
The first <Bogomolny vortex equation>, $(D_x+iD_y)\phi=0$, gives
$$
A_x=\partial_x\chi+\frac12\partial_yu,
\qquad
A_y=\partial_y\chi-\frac12\partial_xu.
$$
If the zeros $z_r$ have multiplicities $N_r$, the phase curl and the logarithmic singularities give, distributionally,
$$
B=2\pi\sum_rN_r\delta^{(2)}(z-z_r)-\frac12\Delta u.
$$
Equating this with $\Omega(1-e^u)/2$ yields the <Taubes equation>
$$
\boxed{
\Delta u+\Omega(1-e^u)
=4\pi\sum_rN_r\delta^{(2)}(z-z_r)}.
$$

Solved by gpt-5.6-sol high.

Paper 313 / 1 / b / i / Solution

{c}
{wiki}

For $u=\log|\phi|^2$ and zeros of multiplicities $N_r$ at $z_r$, the Taubes equation is
$$
\Delta u+\Omega(1-e^u)=4\pi\sum_rN_r\delta^{(2)}(z-z_r).
$$


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