# Show that requiring Electrostatic potential to be a stationary point of Electrostatic potential energy is equivalent to Laplace's equation.

Suppose we want to find the electrostatic potential $\phi$(r) inside of some volume $V$ with known boundary conditions. The physical field configuration should minimize the electrostatic potential energy function, which is defined as

$U[\phi] = \frac{\epsilon_0}{2} \int_v ( \nabla \phi)^2$d r

Show that requiring $\phi$ to be a stationary point of $U$ is equivalent to Laplace's equation, $\nabla^2\phi = 0$.

A hint I was given: take a small deformation of $\phi$, that is $\phi + \delta\phi$, such that $\delta\phi$ vanishes on the boundary of V. We must show that $\delta U = U[\phi + \delta\phi] - U[\phi] = 0$, to linear order in $\delta\phi$.

I have tried limits from first principles and first order taylor expansions. I always seem to just move in a loop however, constantly reaching no conclusion. If anyone is able to show me how this can be done, that would be fantastic.

Up to $\ds{\delta\phi^{2}}$, it becomes $\ds{2\nabla\cdot\pars{\delta\phi\nabla\phi} - 2\delta\phi\nabla\cdot\nabla\phi =2\nabla\cdot\pars{\delta\phi\,\nabla\phi} - 2\delta\phi\,\nabla^{2}\phi}$
Then, $$0=\delta{\rm U}=\int\bracks{2\nabla\cdot\pars{\delta\phi\,\nabla\phi} - 2\delta\phi\,\nabla^{2}\phi}\,\dd{\bf r} =2\int_{S}\delta\phi\,\nabla\phi\cdot\dd\vec{\rm S} -2\int\delta\phi\,\nabla^{2}\phi\ \dd{\bf r}$$
For 'arbitrary variations' $\ds{\delta\phi}$, it leads to $\ds{\color{#66f}{\large\nabla^{2}\phi = 0}}$ provided $\ds{\int_{S}\delta\phi\,\nabla\phi\cdot\dd\vec{\rm S}}$ vanishes out.