basis_pm1_2D.tex

\begin{flushright} {\tiny {\color{gray} basis\_Pm1\_2D.tex}} \end{flushright}
%~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

On the reference element $\Omega=[-1,1]\times[-1,1]$ we have three nodes placed as follows:

\input{tikz/tikz_pm1_2D}

Let us assume that the function $f(r,s)$ is to be approximated on $[-1,1]\times[-1,1]$ by 
\[
f^h(r,s)=a+br+cs
\]
Note that this is a linear function, not a bilinear one (a direct consequence of this is the 
fact that this function cannot be continuous from one element to another). 
The function $f^h$ then must fulfill:
\begin{eqnarray}
f^h(r_1,s_1)&=&a \;\;\;\;\;\; =f_1    \nn\\
f^h(r_2,s_2)&=&a+\frac{b}{2}=f_2 \nn\\
f^h(r_3,s_3)&=&a+\frac{c}{2}=f_3 \nn
\end{eqnarray}
This leads to : 
\[
a=f_1
\quad
\quad
b=2(f_2-f_1)
\quad
\quad
c=2(f_3-f_1)
\]
Then
\[
f(r,s)=f_1 + 2(f_2-f_1) r + 2(f_3-f_1) s
\]
or, 
\[
f(r) = \sum_{i=1}^3 \bN_i(r,s) f_i
\]
with
\begin{mdframed}[backgroundcolor=blue!5]
\begin{eqnarray}
\bN_1(r) &=& 1-2(r+s)  \nonumber\\
\bN_2(r) &=& 2r   \nonumber\\
\bN_3(r) &=& 2s
\end{eqnarray}
\end{mdframed}

Note that we could also have placed the nodes at a different location: 

\input{tikz/tikz_pm1_2D_bis}

and we would then have
\begin{mdframed}[backgroundcolor=blue!5]
\begin{eqnarray}
\bN_1(r) &=& 1-r-s  \nonumber\\
\bN_2(r) &=& r   \nonumber\\
\bN_3(r) &=& s
\end{eqnarray}
\end{mdframed}
as obtained in Section~\ref{ss:p1}.