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.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-313/1/b/ii/solution