
\subsection{\bf Triangular Elements in Two Dimensions}

\begin{slide}

\begin{itemize}
\item First consider evaluating $F$ on a triangle with a linear
interpolation function (Figure~FEM~3)

\vspace{5pt}
\begin{center}
\centerline{\psfig{figure=cps615.fig3.ps,clip=true,width=4.75in}}
Figure: FEM 3
\end{center}
\vspace{5pt}

\item so that 
\begin{eqnarray*}
\varphi (\underline{r}) & = & \varphi_0+ \left( \varphi_1 - \varphi_0 \right)
\frac{\left( \underline{r} - \underline{r}_0 \right) \wedge \left(
\underline{r}_2 - \underline{r}_0 \right) \vert_z}{\left( \underline{r}_1 
- \underline{r}_0 \right) \wedge \left( \underline{r}_2 -
\underline{r}_0 \right) \vert_z} \\
&& \\
&& \phantom{\varphi_0} + \left( \varphi_2 - \varphi_0 \right)
\frac{\left( \underline{r} - \underline{r}_0 \right) \wedge \left(
\underline{r}_1 - \underline{r}_0 \right) \vert_z}{\left( \underline{r}_2 
- \underline{r}_0 \right) \wedge \left( \underline{r}_1 -
\underline{r}_0 \right) \vert_z}
\end{eqnarray*}

\item It can be seen that $\varphi$ is linear in $\underline{r}$ and
has value $\varphi_0$ at $\underline{r}_0$, $\varphi_1$ at
$\underline{r}_1$, and $\varphi_2$ at $\underline{r}_2$
\end{itemize}

\end{slide}


