Examples of Lagrange Interpolations

Several examples of a Lagrange interpolation are discussed at this section. At first the plane triangular domain is considered. The location of the interpolation points are shown at Fig. 7.5.

Figure 7.5: Colocation Points of a Plane Triangular Lagrange Interpolation
\begin{figure}\unitlength=1.000000pt
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...00){\circle{3.00}}
\put(090.00,160.00){\circle{3.00}}
\end{picture}
\end{figure}



The location of the interpolation points is sumarized at the table below.




\begin{displaymath}
x_{ip} \ = \
\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 ...
...quad ; \qquad \textrm{\footnotesize\((i=1,2\!\ ;\!\ p=1..3)\)}
\end{displaymath} (7.16)



The base vector \(z_{r}\); \((r=1..3)\) contains the monomial expressions of a complete first order polynomial as it has been described by Eq. (7.6).




\begin{displaymath}
z_{r} \ = \
\left[ \begin{array}{ccc}
1 & x_{1} & x_{2}
\e...
...ay} \right]
\qquad ; \qquad \textrm{\footnotesize\((r=1..3)\)}
\end{displaymath} (7.17)



These interpolation points \(x_{ip}\); \((i=1,2\!\ ;\!\ p=1..3)\) according to Eq. (7.16) are inserted into the base vector according to Eq. (7.17) to form the quadratic Vandermonde matrix \(z_{rp}\); \((r,p=1..3)\).




\begin{displaymath}
z_{rp} \ = \
\left[\begin{array}{ccc}
\ 1 \ & \ 1 \ & \ 1 \...
...y}\right]
\qquad ; \qquad \textrm{\footnotesize\((r,p=1..3)\)}
\end{displaymath}

The associated inverse matrix \(z_{pr}^{-1}\); \((r,p=1..3)\) is shown below.




\begin{displaymath}
z_{pr}^{-1} \ = \
\left[\begin{array}{rrr}
\ 0 \ & \ 1 \ & ...
...y}\right]
\qquad ; \qquad \textrm{\footnotesize\((r,p=1..3)\)}
\end{displaymath}



The three shape functions are finally obtained from Eq. (7.9).




\begin{displaymath}
h_{1}(x_{1},x_{2}) \ = \ x_{1} \qquad , \qquad
h_{2}(x_{1},x...
...{2} \qquad , \qquad
h_{3}(x_{1},x_{2}) \ = \ 1 - x_{1} - x_{2}
\end{displaymath}



The interpolation points \(x_{ip}\); \((i=1,2\!\ ;\!\ p=1..6)\), according to Eq. (7.16), may be inserted into the shape functions to check the conditions according to Eq. (7.13), Eq. (7.14) and Eq. (7.15).



The shape function \(h_{3}(x_{1},x_{2})\) of the linear Lagrange interpolation is sketched in Fig. 7.6. Similar figures of the shape functions are obtained at the corner points 1 and 2 of the triangular interpolation area according to Fig. 7.5.



Figure 7.6: Linear Shape Function over the Plane Triangle
\begin{figure}\begin{picture}(320.00,200.00)(0.00,0.00)
% Linear Shape Function ...
...6,1){048.00}}
\put(172.00,138.00){\line(6,1){024.00}}
\end{picture}
\end{figure}




In the next example a Lagrange interpolation is considered with four collocation points in a square two-dimensional domain. Their location is shown at Fig. 7.7. The location of the interpolation points is sumarized at the table below.




\begin{displaymath}
x_{ip} \ = \
\left[\begin{array}{rrrr}
-1 & 1 &\ \ 1 & -1 \...
...quad ; \qquad \textrm{\footnotesize\((i=1,2\!\ ;\!\ p=1..4)\)}
\end{displaymath} (7.18)

Figure 7.7: Colocation Points of a Square Plane Lagrange Interpolation
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...00){\circle{3.00}}
\put(210.00,160.00){\circle{3.00}}
\end{picture}
\end{figure}



The base vector \(z_{r}\); \((r=1..4)\) contains the monomial expressions of a complete first order polynomial as it has been described by Eq. (7.6). In addition the binomial term \((x_{1} x_{2})\) is provided.


\begin{displaymath}
z_{r} \ = \
\left[ \begin{array}{cccc}
1 & x_{1} & x_{2} &...
...ay} \right]
\qquad ; \qquad \textrm{\footnotesize\((r=1..4)\)}
\end{displaymath} (7.19)



These interpolation points \(x_{ip}\); \((i=1,2\!\ ;\!\ p=1..4)\) according to Eq. (7.18) are inserted into the base vector according to Eq. (7.19) to form the quadratic Vandermonde matrix \(z_{rp}\); \((r,p=1..4)\).




\begin{displaymath}
z_{rp} \ = \
\left[\begin{array}{rrrr}
\ 1 \ & \ 1 \ & \ \ ...
...y}\right]
\qquad ; \qquad \textrm{\footnotesize\((r,p=1..4)\)}
\end{displaymath}



The corresponding inverse matrix \(z_{pr}^{-1}\); \((r,p=1..4)\) is shown below.




\begin{displaymath}
z_{pr}^{-1} \ = \ \frac{1}{4}
\left[\begin{array}{rrrr}
\ 1 ...
...y}\right]
\qquad ; \qquad \textrm{\footnotesize\((r,p=1..4)\)}
\end{displaymath}

The four shape functions are finally obtained from Eq. (7.9).

\begin{eqnarray*}
h_{1}(x_{1},x_{2}) & = &
\frac{1}{4}\!\ \left( 1 - x_{1} - x_{...
...\frac{1}{4}\!\ \left( 1 - x_{1} + x_{2} - x_{1}\!\ x_{2} \right)
\end{eqnarray*}




The interpolation points \(x_{ip}\); \((i=1,2\!\ ;\!\ p=1..4)\), according to Eq. (7.18), may be inserted into the shape functions to check the conditions according to Eq. (7.13), Eq. (7.14) and Eq. (7.15). The shape functions \(h_{p}(x_{i})\); \((i=1,2\!\ ;\!\ p=1..4)\) are graphically displayed over the squared plane domain at Fig. 7.8.

Figure 7.8: Shape Functions of the Four Point Lagrange Interpolation in the Plane
\begin{figure}\unitlength=1.0pt
\begin{picture}(424.00,285.00)(0.00,0.00)
% Shap...
...\left( 1 - x_{1} \right) \left( 1 + x_{2} \right)\)}}
\end{picture}
\end{figure}

Now a parabolic Lagrange interpolation over the triangular domain is considered. The location of the interpolation points are shown at Fig. 7.9.

Figure 7.9: Colocation Points of a Lagrange Interpolation over a Triangular Area
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...00){\circle{3.00}}
\put(090.00,160.00){\circle{3.00}}
\end{picture}
\end{figure}



The base vector \(z_{r}\); \((r=1..6)\) contains the monomial expressions of a complete second order polynomial as it has been described by Eq. (7.6).




\begin{displaymath}
z_{r} \ = \
\left [\begin{array}{cccccc}
1 &
x_{1} &
x_{...
...ay} \right]
\qquad ; \qquad \textrm{\footnotesize\((r=1..6)\)}
\end{displaymath} (7.20)



The location of the interpolation points is sumerized at the table below and shown at Fig. 7.9.


\begin{displaymath}
x_{ip} \ = \
\left[\begin{array}{cccccc}
0 & 1 & 0 & 0.5 & ...
...quad ; \qquad \textrm{\footnotesize\((i=1,2\!\ ;\!\ p=1..6)\)}
\end{displaymath} (7.21)



These interpolation points \(x_{ip}\); \((i=1,2\!\ ;\!\ p=1..6)\) are inserted into the base vector according to Eq. (7.20) to form the quadratic Vandermonde matrix \(z_{rp}\); \((r,p=1..6)\).




\begin{displaymath}
z_{rp} \ = \
\left[\begin{array}{cccccc}
\ 1 \ & \ 1 \ & \ ...
...y}\right]
\qquad ; \qquad \textrm{\footnotesize\((r,p=1..6)\)}
\end{displaymath}

The associated inverse matrix \(z_{pr}^{-1}\); \((r,p=1..6)\) is shown below.




\begin{displaymath}
z_{pr}^{-1} \ = \
\left[\begin{array}{rrrrrr}
\ 1 & \ -3 & ...
...y}\right]
\qquad ; \qquad \textrm{\footnotesize\((r,p=1..6)\)}
\end{displaymath}



The six shape functions are finally obtained from Eq. (7.9).


$\displaystyle h_{1}(x_{1},x_{2})$ $\textstyle =$ $\displaystyle 1 - 3 \!\ (x_{1} + x_{2}) + 2 \!\ (x_{1} + x_{2})^{2}$ (7.22)
       
$\displaystyle h_{2}(x_{1},x_{2})$ $\textstyle =$ $\displaystyle x_{1} \!\ (2x_{1} - 1)$  
       
$\displaystyle h_{3}(x_{1},x_{2})$ $\textstyle =$ $\displaystyle x_{2} \!\ (2 x_{2} - 1)$  
       
$\displaystyle h_{4}(x_{1},x_{2})$ $\textstyle =$ $\displaystyle 4x_{1} \!\ (1 - x_{1} - x_{2})$  
       
$\displaystyle h_{5}(x_{1},x_{2})$ $\textstyle =$ $\displaystyle 4x_{1} \!\ x_{2}$  
       
$\displaystyle h_{6}(x_{1},x_{2})$ $\textstyle =$ $\displaystyle 4x_{2} \!\ (1 - x_{1} - x_{2})$ (7.23)



The interpolation points \(x_{ip}\); \((i=1,2\!\ ;\!\ p=1..6)\), according to Eq. (7.21), may be inserted into the shape functions to check the conditions according to Eq. (7.13), Eq. (7.14) and Eq. (7.15). With respect to Eq. (7.21) the following expressions are obtained.



\begin{eqnarray*}
\sum_{p} x_{1p} \ h_{p} & = & h_{2}(x_{1},x_{2}) + \frac{1}{2}...
... \Big(
h_{5}(x_{1},x_{2}) + h_{6}(x_{1},x_{2}) \Big) \ = \ x_{2}
\end{eqnarray*}


The shape function \(h_{1}(x_{1},x_{2})\) of the parabolic Lagrange interpolation according to Eq. (7.22) is sketched in Fig. 7.10. Similar figures of the shape functions are obtained at the corner points 2 and 3 of the triangular interpolation area according to Fig. 7.9.

Figure 7.10: Corner Point Shape Function of the Parabolic Lagrange Interpolation
\begin{figure}\begin{picture}(320.00,200.00)(0.00,0.00)
% Corner Point Shape Fun...
...\qbezier(322.00, 63.90)(323.00, 63.70)(324.00, 63.60)
\end{picture}
\end{figure}




The shape function \(h_{6}(x_{1},x_{2})\) of the parabolic Lagrange interpolation according to Eq. (7.23) is sketched in Fig. 7.11. Similar figures of the shape functions are obtained at the midside points 4 and 5 of the triangular interpolation area according to Fig. 7.9.

Figure 7.11: Midpoint Shape Function of the Parabolic Lagrange Interpolation
\begin{figure}\begin{picture}(320.00,200.00)(0.00,0.00)
% Midpoint Shape Functio...
...\qbezier(322.00, 90.60)(323.00, 94.70)(324.00, 98.80)
\end{picture}
\end{figure}

Finally a three-dimensional Lagrange interpolation over the tetrahedral domain is considered. The location of the interpolation points is sumarized at the table below and shown at Fig. 7.12.




\begin{displaymath}
x_{ip} \ = \
\left[\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0...
...quad ; \qquad \textrm{\footnotesize\((i=1,3\!\ ;\!\ p=1..4)\)}
\end{displaymath} (7.24)

Figure 7.12: Colocation Points of a Tetrahedral Lagrange Interpolation
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\begin{picture}(424.00,240.00)(0.00,0.00)
%...
...00){\circle{3.00}}
\put(180.00,170.00){\circle{3.00}}
\end{picture}
\end{figure}



The base vector \(z_{r}\); \((r=1..3)\) contains the monomial expressions of a complete three-dimensional first order polynomial as it has been described by Eq. (7.7).




\begin{displaymath}
z_{r} \ = \
\left[ \begin{array}{cccc}
1 & x_{1} & x_{2} &...
...ay} \right]
\qquad ; \qquad \textrm{\footnotesize\((r=1..4)\)}
\end{displaymath} (7.25)



These interpolation points \(x_{ip}\); \((i=1,3\!\ ;\!\ p=1..4)\) according to Eq. (7.24) are inserted into the base vector according to Eq. (7.25) to form the quadratic Vandermonde matrix \(z_{rp}\); \((r,p=1..4)\).




\begin{displaymath}
z_{rp} \ = \
\left[\begin{array}{cccc}
\ 1 \ & \ 1 \ & \ 1 ...
...y}\right]
\qquad ; \qquad \textrm{\footnotesize\((r,p=1..4)\)}
\end{displaymath}

The corresponding inverse matrix \(z_{pr}^{-1}\); \((r,p=1..4)\) is shown below.




\begin{displaymath}
z_{pr}^{-1} \ = \
\left[\begin{array}{rrrr}
\ 0 \ & \ 1 \ &...
...y}\right]
\qquad ; \qquad \textrm{\footnotesize\((r,p=1..4)\)}
\end{displaymath}



The four shape functions are finally obtained from Eq. (7.9).


\begin{eqnarray*}
h_{1}(x_{1},x_{2},x_{3}) & = & x_{1} \\ \nonumber \\
h_{2}(x_...
...ber \\
h_{4}(x_{1},x_{2},x_{3}) & = & 1 - x_{1} - x_{2} - x_{3}
\end{eqnarray*}




The interpolation points \(x_{ip}\); \((i=1,2\!\ ;\!\ p=1..6)\), according to Eq. (7.24), may be inserted into the shape functions to check the conditions according to Eq. (7.13), Eq. (7.14) and Eq. (7.15).



The graphically representation of these shape functions would require a four-dimensional space and is threrfore not shown here.

Prof. Dr.-Ing. D. Maurer, Hochschule Landshut