= Solution
Vary $g$ to $g+\varepsilon\eta$ with $\eta(a)=\eta(b)=0$. The <first variation> of the functional is
$$
\delta\mathcal E=\int_a^b(f_g\eta+f_{g'}\eta')\,dr
=\int_a^b\left(f_g-\frac d{dr}f_{g'}\right)\eta\,dr.
$$
The endpoint term vanishes. Since every smooth test variation is allowed, stationarity gives the <Euler-Lagrange equation>
$$
\boxed{\frac d{dr}\frac{\partial f}{\partial g'}-\frac{\partial f}{\partial g}=0.}
$$
The constant prefactor in the specified energy does not affect its stationary equation. Its integrand gives $f_{\phi'}=r^2\phi'$ and $f_\phi=r^2\phi/\lambda^2$, so
$$
(r^2\phi')'-\frac{r^2}{\lambda^2}\phi=0,
\qquad
\boxed{\frac1r(r\phi)''-\frac1{\lambda^2}\phi=0.}
$$
Take $\lambda>0$ as a screening length; for a nonzero real parameter of either sign its absolute value gives the same equation. Setting $u=r\phi$ yields $u''-u/\lambda^2=0$, hence
$$
\boxed{\phi(r)=\frac{Ae^{r/\lambda}+Be^{-r/\lambda}}r.}
$$
For $0<R_1<R_2$ and the prescribed endpoint values, the convenient hyperbolic form is
$$
\boxed{\phi(r)=\phi_1\frac{R_1}{r}
\frac{\sinh[(R_2-r)/\lambda]}{\sinh[(R_2-R_1)/\lambda]}.}
$$
It 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> are
$$
\mathcal E_2[\phi]=2\pi\int_{R_1}^{R_2}\left[\frac12\phi'^2+\frac{\phi^2}{2\lambda^2}\right]r\,dr,
\qquad
\boxed{\frac1r(r\phi')'-\frac1{\lambda^2}\phi=0.}
$$
Thus the two-dimensional radial equation is $\phi''+r^{-1}\phi'-\lambda^{-2}\phi=0$, rather than the three-dimensional equation with coefficient $2/r$ on the first <derivative>.
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