basis_p1_3D.tex

\begin{flushright} {\tiny {\color{gray} basis\_p1p\_3D.tex}} \end{flushright}
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The location of the four notes are:
\begin{verbatim}
(r_0,s_0) = (0,0,0)
(r_1,s_1) = (1,0,0)
(r_2,s_2) = (0,1,0)
(r_3,s_3) = (0,0,1)
\end{verbatim}

The basis polynomial is given by
\[
f(r,s,t)=c_0 + c_1 r + c_2 s + c_3 t
\]
and it needs to satisfy these four conditions:
\begin{eqnarray}
f_1 &=& f(r_1,s_1,t_1) = c_0 \\
f_2 &=& f(r_2,s_2,t_2) = c_0 + c_1\\
f_3 &=& f(r_3,s_3,t_3) = c_0 + c_2\\
f_4 &=& f(r_4,s_4,t_4) = c_0 + c_3
\end{eqnarray}
which yields:
\[
c_0=f_1
\quad
\quad
c_1=f_2-f_1
\quad
\quad
c_2=f_3-f_1
\quad
\quad
c_3=f_4-f_1
\]
Finally,
\begin{eqnarray}
f(r,s,t) 
&=& c_0 + c_1 r + c_2 s + c_3 t \nonumber\\
&=& f_1 + (f_2-f_1) r + (f_3-f_1) s + (f_4-f_1) t \nonumber\\
&=& f_1 (1-r-s-t) + f_2 r + f_3 s + f_4 t \nonumber\\
&=& \sum_i \bN_i(r,s,t) f_i \nonumber
\end{eqnarray}
Finally,
\begin{mdframed}[backgroundcolor=blue!5]
\begin{eqnarray}
\bN_1(r,s,t) &=& 1-r-s-t \nonumber\\
\bN_2(r,s,t) &=& r \nonumber\\
\bN_3(r,s,t) &=& s \nonumber\\
\bN_4(r,s,t) &=& t \nonumber
\end{eqnarray}
\end{mdframed}
Derivatives are trivial to obtain.