This is the definition which we need for the proof of the theorem :
This is the definition of the $\mathscr C''$ – equivalent :
There is the theorem:
Suppose E is an open set in $R^3$, $u$ $\in$ $\mathscr C''(E)$, and $G$ is a vector field in $E$, of class $C''$ .
$(a)$ if $F$ $=$ $\nabla$$u$, then $\nabla$ $\times$ $F$ $=$ $0$ .
$(b)$ if $F$ $=$ $\nabla$ $\times$ $G$, then $\nabla$ $\cdot$ $F$ $=$ $0$.
Furthermore, if $E$ is $\mathscr C''$-equivalent to a convex set, then $(a)$ and $(b)$ have converses, in which we assume that $F$ is a vector field in $E$, of class $\mathscr C'$:
$(a')$ if $\nabla$ $\times$ $F$ $=$ $0$, then $F$ = $\nabla$$u$ for some $u$ $\in$ $\mathscr C''(E)$.
$(b')$ if $\nabla$ $\cdot$ $F$ $=$ $0$, then $F$ $=$ $\nabla$ $\times$ $G$ for some vector field $G$ in $E$, of class $\mathscr C''$
There is the proof :
If we compare the definitions of $\nabla$$u$, $\nabla$ $\times$ $F$, and $\nabla$ $\cdot$ $F$ with the differential forms $\lambda_F$ and $\omega_F$ given by ($124$) and ($125$), we obtain the following four statements:
$F$ $=$ $\nabla$$u$ if and only if $\lambda_F$ $=$ $du$. ($\star$).
$\nabla$ $\times$ $F$ $=$ $0$ if and only if $d\lambda_F$ $=$ $0$. ($\ast$)
$F$ $=$ $\nabla$ $\times$ $G$ if and only if $\omega_F$ $=$ $d\lambda_G$. ($\oplus$)
$\nabla$ $\cdot$ $F$ $=$ $0$ if and only if $d\omega_F$ $=$ $0$. ($\circ$)
I could n't understand these four statements ($\star$),($\ast$),($\oplus$),($\circ$).
Any help would be appreciated.
Best Answer
For the first one, $F=\nabla u$ iff $F_i=D_iu, i=1,2,3$ where $D_iu$ is the partial derivative of $u$ w.r.t. the $i$-th variable. Since $du=(D_1u)\,dx+(D_2u)\,dy+(D_3u)\,dz$, we have $\lambda_F=du$ is equivalent to $F_i=D_iu$ for all $i$ and hence equivalent to $F=\nabla u$.
For the second one, we have
\begin{align*}d\lambda_F=&dF_1\wedge dx+dF_2\wedge dy+dF_3\wedge dz\\ =&(D_1F_1\,dx+D_2F_1\,dy+D_3F_1\,dz)\wedge dx+(D_1F_2\,dx+D_2F_2\,dy+D_3F_2\,dz)\wedge dy\\&+(D_1F_3\,dx+D_2F_3\,dy+D_3F_3\,dz)\wedge dz \end{align*}
where $dx\wedge dx=0$ and $dx\wedge dy=-dy\wedge dx$. Hence the above is equal to (after reorganizing terms) $$d\lambda_F=(D_1F_2-D_2F_1)dx\wedge dy+(D_1F_3-D_3F_1)dx\wedge dz+(D_2F_3-D_3F_2)dy\wedge dz$$ Comparing this with $\nabla\times F$, we get the second statement.
For the third one, it is very similar to the second one.
For the last one, \begin{align*} d\omega_F&=dF_1\wedge dy\wedge dz+dF_2\wedge dz\wedge dx+dF_3\wedge dx\wedge dy\\ &=D_1F_1dx\wedge dy\wedge dz+D_2F_2dy\wedge dz\wedge dx+D_3F_3dz\wedge dx\wedge dy\\ &=(D_1F_1+D_2F_2+D_3F_3)dx\wedge dy\wedge dz \end{align*} Hence $d\omega_F=0$ iff $\nabla\cdot F=0$.