
\subsection{\bf Finite Element Equations}

\begin{slide}

\begin{itemize}
\item  We now have
\begin{eqnarray*}
F & = & \frac{1}{4} \varphi^T {\twelvebxti S} \varphi \\
& = & \frac{1}{4} \sum_{ij} \varphi_j S_{ji} \varphi_i 
\end{eqnarray*}

\item Let
\begin{eqnarray*}
\varphi_i & \rightarrow & \varphi_i + \eta_i \hspace{1em} \hbox{be
variation} \\
F & = & \frac{1}{4} \sum_{ij} \varphi_j S_{ji} \varphi_i \\
& + & \frac{1}{4} \left( \sum_{ij} \bigg( \eta_j S_{ji} \varphi_i
\bigg) + \bigg( \varphi_j S_{ji} \eta_i \bigg) \right)
\end{eqnarray*}

\item Thus $\delta F = \frac{1}{2} \sum_{ij} \eta_j\, S_{ji}\,
\varphi_i$

\item This $\delta F$ form is zero if and only if 
\[
\sum_i S_{ji}\, \varphi_i = 0
\]

\item This condition can be split into internal and boundary terms
\[
\underbrace{\Sigma_{(j\,{\rm not\/}\in\,\partial\Omega)} S_{ij} 
\varphi_j}_{\rm Unknown} + 
\underbrace{\Sigma_{(j\in\partial\Omega)} S_{ij} \left( \varphi = f_j
\right)}_{\rm Known\ at\ Boundary} = 0
\]
\end{itemize}

\end{slide}

