Equivalence of two vector fields

Question

Let $V$ be a convex region in $\mathbb R^3$ whose boundary is a closed surface $S$ and let $\vec n $ be the unit outer normal to $S$. Let $F$ and $G$ be two continuously differentiable vector fields such that $\mathop{\rm curl} F=\mathop{\rm curl} G; \mathop{\rm div} F=\mathop{\rm div}G$ everywhere in $V$ and $G\cdot\vec n = F\cdot\vec n$ every where on $S$. Then is it true that $F=G$ everywhere in $V$?

I tried as follows: Let $H=F-G$ , then $\mathop{\rm curl} H=O$, so there is a scalar field $f$ such that $H=\nabla f$, and also $\mathop{\rm div} H=0$, so there is a vector field $L$ such that $H=\mathop{\rm curl} L$ but then I am stuck; please help. Thanks in advance.

Answer 1

• The result follows from Green's first identity $$ \int_V d^3 r (\phi \nabla^2 \psi+ \vec{\nabla}\psi \cdot \vec{\nabla} \phi) = \oint_S d^2 r\,\phi (\vec{\nabla} \psi) \cdot \vec{n}.$$ (this is the divergence theorem $\int_V d^3 \nabla \cdot \vec{A} = \oint_S d^2 r \vec{A}\cdot \vec{n}$ applied to $\vec{A} =\phi \vec{\nabla} \psi $)

• In order to apply it to your case, observe that because the fields $\vec{F}$ and $\vec{G}$ are rotation free, you can find a scalar field $\phi$ such that $\vec{F}- \vec{G} =\vec{\nabla} \phi$. Then set $\psi=\phi$ in Green's first identity.

The following statements hold:

(a) $\vec{\nabla} \phi = \vec{F} -\vec{G}$ by definition.

(b) $\nabla^2 \phi =0$ as $\vec{F}$ and $\vec{G}$ are divergence free.

(c) $(\vec{\nabla} \phi) \cdot \vec{n} =0$ on the surface $S$ (as $\vec F\cdot \vec{n} =0$ and $\vec G\cdot \vec{n} =0$).

• Thus, we have that $$ \int_V d^3 r (\phi\underbrace{ \nabla^2 \phi}_{=0}+ \vec{\nabla}\phi \cdot \vec{\nabla} \phi) = \oint_S d^2 r\,\phi \underbrace{(\vec{\nabla} \phi) \cdot \vec{n}}_{=0}.$$ Thus Green's first identity implies $$ 0= \int_V d^3 r \,\vec{\nabla}\phi \cdot \vec{\nabla} \phi= \int_V d^3 r\, |\vec{\nabla} \phi|^2 = \int_V d^3 r\,| \vec{F} -\vec{G}|^2$$ from which we conclude that $\vec{F} =\vec{G}$ on $V$.