What is the maximum self-energy of an electrostatic distribution subject to the constraints that:
- the total charge is $1$; and
- the areal charge density anywhere is either $1$ or $0$.
How can I formulate this problem in calculus of variations?
I came up with this problem because I was calculating self-energies of a planar hexagon and noticed it is less than the energy of a round disk of same area, and physically it is reasonable that the disk requires more energy to put together. However I want to know if I can solve this with calculus of variations.
I’d try to use
$$E = \iint {\rho(x) \, \rho(y) \over \lvert x-y\rvert}\, dx\,dy + \int \bigl[ \lambda(x)\, \rho(x) \, (1-\rho(x)) +\mu \, \rho(x)\bigr]\, dx$$
Variation with $\rho$ gives
$$2 \int {\rho(y)\over \lvert x-y\rvert} \,dy + \mu + \lambda(x) \,\bigl[1 - 2 \,\rho(x)\bigr] = 0$$
Let $U(x)$ be the first term, then
$$\rho(x) = 0 \implies U(x) = -\mu - \lambda(x)$$ $$\rho(x) = 1 \implies U(x) = -\mu + \lambda(x)$$
This is where I am now.