Proof of Thomson's theorem on electrostatics using variational calculus

Are you familiar with using Lagrange multiplier to find extrema of a system unter constraints? If not, check Wikipedia. In short in 2d, one constraint case, $f(x,y)$ has extrema when $nabla mathcal{L(x,y)}=0$, in which $mathcal{L} = f(x,y)-lambda cdot g(x,y)$ with constraint $g(x,y)=0$.

In this case $g = Q^{(n)}-intrho dV^{(n)}_{int}-intsigma dS^{(n)}$. This comes from charge conservation. The total charge of a conductor Q is equal to all the charges inside the conductor and on the surface, i.e. $Q^{(n)}=intrho dV^{(n)}_{int}+intsigma dS^{(n)}$. Later in this article the same methode is deployed with constraints as results of Poisson's equation.