Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics V / 2 / b / ii / Solution

True. The first of Green's identities and the Robin boundary condition ${\partial }_{n}u=-u$ giveBoth integrands are nonnegative, so $\nabla u=0$ in $V$ and $u=0$ on the boundary. By path connectedness, $u$ is a constant function, and its boundary value makes that constant zero. Therefore $\boxed{u=0}$.