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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.

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The problem you have is not with saddle points. By assumption $F|_S = \nabla (g|_S)$, and $g$ being a continuous function on $S$ must attain a max/min. –  Willie Wong Apr 27 '11 at 17:49
    
@Willie Wong: I edited to make it clearer. I meant meant local max/min in $\mathbb{R}^3$ as opposed to in $S$. We have a critical point, but maybe it is some sort of saddle. –  Eric Naslund Apr 27 '11 at 18:04
    
ah, then you should use the Strong Maximum Principle / Hopf's Lemma as Hans suggests. –  Willie Wong Apr 27 '11 at 18:41
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2 Answers

up vote 3 down vote accepted

Your second proof is a good one and entirely appropriate within vector calculus. The first attempt has a gap (I think).

A third proof relies on Hopf's Lemma (commonly taught in graduate level classes on partial differential equations) which implies here that if a function $u$ satisfying $\Delta u \le 0$ in $D$ attains a strict minimum at $z \in \partial D = S$, then the outer normal derivative at that point satisfies $\nu \cdot \nabla u(z) < 0$. Applying this with $g = u$, it follows that $g$ must attain its minimum in the interior of $D$ and hence must be constant.

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This is very interesting, thank you. Quick question: what is $\nu$? I assume its the vector going from $0$ to $z$. –  Eric Naslund Apr 27 '11 at 18:13
    
$\nu$ is the unit outward normal vector (to $S$). So yes, the vector going from $0$ to $z$. –  Willie Wong Apr 27 '11 at 18:42
    
A bit late, but I forgot to accept your answer! Thanks! –  Eric Naslund Jun 7 '11 at 16:34
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Eric, I am simply rephrasing your second proof to give you a more PDE style approach to this problem, let $\vec{F} = \nabla u$, the problem you gave was equivalent to solve the following boundary value problem: Find $u\in C^2(D)$ such that $$ \left\{\begin{eqnarray} -\Delta u &=& 0 \text{ in } D \\ \nabla u \cdot n &=& 0 \text{ on } \partial D \end{eqnarray} \right.$$ Now multiply both sides of the equation by a test function in some proper space(as large as possible), doing integration by parts to use the boundary condtion, and you will get the variational formulation of this problem: Find $u\in V = W^{1,2}(D)$ such that $$ \int_D \nabla u\cdot \nabla v = 0 \quad \forall v\in V $$ Notice there is no Dirichlet type boundary condition enforced for the test function $v$, which makes this problem yield non-unique solutions; In order to make this problem well-posed, instead in the space $V$, consider a new space $V_0 = \{v\in V | \int_D v dx = 0\} = V/R$ this variational problem is equivalent to the following functional minimization problem: If $u$ solves the original pde, then $u$ minimizes $$ J(u) = \min_{v\in V_0}\int_D |\nabla v|^2\,dx $$ Now assume $u$ is the solution, then $$ J(u) = \int_D |\nabla u|^2\,dx \leq \sup_{v\in V_0} \int_D \nabla u\cdot \nabla v = 0 $$ which proves for the solution $u\in V_0$: $\nabla u = 0$(other solutions are different with some specific $u$ from a constant by the definition of that quotient space, hence the gradient does not change)

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