🚀 go-pugleaf

Dear Ayman,

you can see this as the transition from a finite difference to a
finit element formulation.

For example, inside a triangular element, you can interpolate the values by 
a plane formula:

\begin{equation} \label{eq:phi1}
\phi(x,y) 
\,=\,  
k_1 + k_2\, x + k_3\, y . 
\end{equation}

Using the edge coordinates 
$(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$
you get for the edge values
$\phi_1$, $\phi_2$ et $\phi_3$

\begin{eqnarray} \label{eq:phi2}
\phi_1 &\,=\,&  k_1 + k_2\, x_1 + k_3\, y_1 \cr 
\phi_2 &\,=\,&  k_1 + k_2\, x_2 + k_3\, y_2 \cr 
\phi_3 &\,=\,&  k_1 + k_2\, x_3 + k_3\, y_3  
\end{eqnarray}

Resolve \ref{eq:phi2} for $k_1$, $k_2$ et $k_3$
and substitute in \ref{eq:phi1}, you get
the interpolation at interior points:
\begin{equation}
\phi(x,y) 
\,=\,  
N_1(x,y)\, \phi_1 +N_2(x,y)\, \phi_2 +N_3(x,y)\, \phi_3  
\end{equation}
avec
\begin{eqnarray}
	N_1(x,y)&\,=\,&\frac{y_3 x_2 - x_3 y_2 + x\,(y_2 - y_3) + y\,(x_3 - x_2)}{2\Delta} \cr
	N_2(x,y)&\,=\,&\frac{x_3 y_1 - y_3 x_1 + x\,(y_3 - y_1) + y\,(x_1 - x_3)}{2\Delta} \cr
	N_3(x,y)&\,=\,&\frac{y_2 x_1 - x_2 y_1 + x\,(y_1 - y_2) + y\,(x_2 - x_1)}{2\Delta} 
\end{eqnarray}

There, the numerator represents the determinante,

\begin{eqnarray}
2\Delta 
&\,=\,&
\left\vert
\matrix{ 1 & x_1 & y_1 \cr
	 1 & x_2 & y_2 \cr
	 1 & x_3 & y_3 }
\right\vert \cr 
& &\cr
&\,=\,&
x_2 y_3 + x_3 y_1 + x_1 y_2 - x_2 y_1 - x_3 y_2 - x_1 y_3
\end{eqnarray}

which equals the double surface of the triangle.


Hope you are helped with this.

Salut

Bardo 

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