I was asked the following vector calculus problem:
Let $D$ be the unit ball and let $S$ be the unit sphere in $\mathbb{R}^3$. Suppose that $F:\mathbb{R}^3\rightarrow \mathbb{R}^3$ is a $C^1$ vector field on some open neighborhood of $D$ which satisfies:
$(i) \nabla\times F=0$
$(ii) \nabla\cdot F=0$
$(iii)$ On $S$, $F$ is orthogonal to the radial vector.
Prove that $F=0$ on all of $D$.
Conditions $(i)$ and $(ii)$ imply that $F=\nabla g$ for some $g:\mathbb{R}^3\rightarrow \mathbb{R}$ where $g$ must be harmonic as well.
I know one solution (see end), however my initial instinct was to try to use the max/min property of harmonic functions, and I couldn't get it to work. Since the gradient is always orthogonal to the sphere, there must be a point on the sphere where it is $0$. (Hairy ball) If that was a local max or min in $\mathbb{R}^3$ we would be done, by taking a small neighborhood around it. If it is a saddle point this doesn't work. (We know that it must be a local max/min on $S$ since it is harmonic)
My question is: Is there any way to modify this approach, and solve the problem?
Thanks!
Other Solution: Here is one solution that first uses the fact that the radial vector is orthogonal, and then applies Gauss's Divergence theorem to the function $gF$. ($\nabla g=F$) That is $0=\iint_S (gF\cdot n)dS=\iiint_D \nabla\cdot (gF)dV=\iiint_D \|F\|^2dV,$ and since the integrand on the right hand side is non-negative, continuous and integrates to give zero, it must be zero.