\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\pa}{\partial}
\newcommand{\dci}{\rlap{$\displaystyle\int\!\!\!\int_{B_{a,b,c}}$}
{\hspace{2.2pt}\bigcirc} \ \ }
\newcommand{\un}{\underline}
\begin{document}
\parindent=0pt


\begin{center}
\textbf{Vector Calculus}

\textit{\textbf{Grad, Div and Curl Identities}}
\end{center}

\textbf{Question}

It is given that $\phi$ and $\psi$ are scalar fields and $\un{F}$ and
$\un{G}$ are vector fields. They are all assumed to be smooth
functions. Prove the following identity

$$\nabla \times (\nabla \times \un{F}) = \nabla(\nabla \bullet \un{F})
- \nabla^2\un{F}$$


\textbf{Answer}

For $\nabla\times(\nabla\times\un{F})$, the first component is
$$\frac{\pa}{\pa y} \left ( \frac{\pa F_2}{\pa x} - \frac{\pa F_1}{\pa
y} \right ) - \frac{\pa}{\pa x} \left ( \frac{\pa F_1}{\pa z} -
\frac{\pa F_3}{\pa x} \right )$$
$$ = \frac{\pa^2F_2}{\pa y \pa x} - \frac{\pa^2 F_1}{\pa y^2} -
\frac{\pa^2F_1}{\pa z^2} + \frac{\pa^2F-3}{\pa z \pa x}.$$
For $\nabla(\nabla \bullet \un{F})$, the first component is
$$\frac{\pa}{\pa x} \nabla \bullet \un{F} = \frac{\pa^2 F_1}{\pa x^2}
+ \frac{\pa^2 F_2}{\pa x \pa y} + \frac{\pa^2 F_3}{\pa x \pa z}$$
For $-\nabla^2\un{F}$, the first component is
$$-\nabla^2F_1 = -\frac{\pa^2 F_1}{\pa x^2} - \frac{\pa^2 F_1}{\pa
y^2} - \frac{\pa^2 F_1}{\pa z^2}.$$

It can be seen that the first components of the identity agree, as do
the other components by symmetry. So
$$\nabla \times (\nabla \times \un{F}) = \nabla(\nabla \bullet \un{F})
- \nabla^2\un{F}$$


\end{document}