For a rotationally symmetric vortex, the equation away from the origin is
$$
u''+\frac1r u'=e^u-1.
$$
Insert
$$
u=\alpha\log r+\beta+\gamma r+\delta r^2+\cdots.
$$
The prescribed zero fixes $\alpha=2N$. Since $N\geq1$, $e^u=e^\beta r^{2N}[1+o(1)]$, while
$$
u''+\frac1r u'=\frac\gamma r+4\delta+o(1).
$$
Matching the singular and constant terms with $e^u-1=-1+o(1)$ gives
$$
\boxed{(\alpha,\gamma,\delta)=\left(2N,0,-\frac14\right)}.
$$
The undetermined $\beta$ is fixed by matching this local expansion to $u\to0$ at infinity.


Solution

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

 Solution

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