Suppose $u$ were positive somewhere. Since it tends to zero at infinity and to $-\infty$ at its vortex zero, it would attain a positive interior maximum away from the origin. At such a maximum the <second-derivative test> gives $\Delta u\leq0$, whereas the smooth Taubes equation gives
$$
\Delta u=e^u-1>0.
$$
This contradiction with the <maximum principle for subharmonic functions> proves
$$
\boxed{u\leq0\quad\text{on }\mathbb R^2}.
$$
Consequently the Higgs magnitude of a vortex satisfies $|\phi|=e^{u/2}\leq1$ everywhere.

Solution

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

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At an interior local maximum of a twice differentiable function, its <Hessian matrix> is negative semidefinite; at a local minimum it is positive semidefinite. For a scalar function on Euclidean space, the Laplacian is therefore nonpositive at a local maximum and nonnegative at a local minimum.

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