= Solution
Let $v$ be the voltage with $v(a)=1$, $v(b)=0$, and harmonic values at every other vertex. For any other admissible $f$, write $f=v+g$, where $g(a)=g(b)=0$. Its <discrete Dirichlet energy> expands as
$$
\mathcal E(f)=\mathcal E(v)+\mathcal E(g)
+\sum_{x,y}c(x,y)(v(x)-v(y))(g(x)-g(y)).
$$
Discrete summation by parts turns the cross term into
$$
2\sum_xg(x)\sum_yc(x,y)(v(x)-v(y))=0,
$$
because $v$ is harmonic in the interior and $g$ vanishes at the boundary. Thus $v$ minimizes the energy. Under a unit voltage drop, its energy equals the total current from $a$ to $b$, namely the <effective conductance> $1/R_{\mathrm{eff}}(a,b)$. Therefore the <Dirichlet principle> gives
$$
\boxed{
\frac1{R_{\mathrm{eff}}(a,b)}
=\inf_{f(a)=1,\,f(b)=0}
\frac12\sum_{x,y}(f(x)-f(y))^2c(x,y).}
$$
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