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

False. The constant function $u=1$ is a harmonic function and has zero normal derivative on every smooth boundary, but it is nonzero in $V$.

In fact, the first of Green's identities givesHence $\nabla u=0$; path connectedness makes $u$ constant throughout $V$. Thus this homogeneous Neumann boundary condition determines a harmonic function only up to an additive constant.

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 $\boxed{u=0}$.

False. Take $V=\{x^2+y^2+z^2<1\}$, the open unit ball, and $u(x,y,z)=x$. This is a nonzero harmonic function. On the unit sphere, the outward unit normal vector is $\mathbf{n}=(x,y,z)$, soThus the prescribed Robin boundary condition holds. Unlike the preceding case, Green's identities now give ${\int }_V|\nabla u{|}^2={\int }_{\partial V}{u}^2$, which permits nonzero solutions of the differential equation.

True. Two applications of Green's identities, or two integrations by parts, giveThe biharmonic equation makes the left side zero, while the two clamped boundary conditions $u={\partial }_{n}u=0$ annihilate the boundary integral. Consequentlyso $\Delta u=0$. Apply the first of Green's identities once more with $\varphi =\psi =u$: because $u=0$ on the boundary, it gives ${\int }_V|\nabla u{|}^2=0$. By path connectedness, $u$ is constant, and the zero boundary value implies $\boxed{u=0}$. This argument assumes the classical differentiability needed for the displayed integrations by parts.