Past exam of the mathematics course of the University of Cambridge / 2009 / ib / Paper 3 / 15A (Methods)

Vary $g$ to $g+\epsilon \eta$ with $\eta (a)=\eta (b)=0$. The first variation of the functional isThe endpoint term vanishes. Since every smooth test variation is allowed, stationarity gives the Euler-Lagrange equation

The constant prefactor in the specified energy does not affect its stationary equation. Its integrand gives $f_{{\varphi }^{\prime }}=r^2{\varphi }^{\prime }$ and $f_{\varphi }=r^2\varphi /{\lambda }^2$, soTake $\lambda >0$ as a screening length; for a nonzero real parameter of either sign its absolute value gives the same equation. Setting $u=r\varphi$ yields $u^{\prime \prime }-u/{\lambda }^2=0$, henceFor $0<R_1<R_2$ and the prescribed endpoint values, the convenient hyperbolic form isIt directly satisfies both endpoint values. The energy is a positive quadratic functional, so this stationary solution is its unique minimizer for fixed endpoints: any zero-endpoint variation adds a positive quadratic energy, with the cross term zero by the equation.

In two dimensions, cylindrical area measure replaces the radial factor $r^2$ by $r$. Up to an irrelevant common normalization, the corresponding energy and its modified Helmholtz equation areThus the two-dimensional radial equation is ${\varphi }^{\prime \prime }+r^{-1}{\varphi }^{\prime }-{\lambda }^{-2}\varphi =0$, rather than the three-dimensional equation with coefficient $2/r$ on the first derivative.