Parametric interpolation for a cubic function

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For a function y = x^3 - 2 x^2 -x + 4 \,, it is known the values for 3 points:
  • x_1 = 0, ~ x_2=1, ~  x_3= 2 \,
  • y_1 = 4, ~ y_2=2, ~  y_3= 2 \,
For a three nodes quadratic element, the interpolation of such function is:
x = \sum_{i=1}^3 N_i^{(e)}(\xi) x_i = \frac{1}{2} \xi (\xi -1) x_1 + (1- \xi^2) x_2 + \frac{1}{2} \xi (1+\xi) x_3


y = \sum_{i=1}^3 N_i^{(e)}(\xi) y_i = \frac{1}{2} \xi (\xi -1) y_1 + (1- \xi^2) y_2 + \frac{1}{2} \xi (1+\xi) y_3
For this quadratic approach, the resulting expression is:


x = (1- \xi^2) + \xi (1+\xi) = 1 + \xi \,
y = 2 \xi (\xi -1) + 2(1- \xi^2) + \xi (1+\xi) = \xi (\xi - 1) + 2 \,


and then:
y = f(x) = (x - 1)(x - 2) + 2 = x^2 - 3 x + 4 \,


The following picture shows the error, much bigger outside of the interval.


ParametricInterpolation.jpg


If the set of points is:
  • x_1 = 0, ~ x_2=\frac{2}{3}, ~  x_3= \frac{4}{3}, ~  x_4= 2\,
  • y_1 = 4, ~ y_2=\frac{74}{27}, ~  y_3= \frac{40}{27}, y_4= 2 \,
that means:
  • \xi_1 = -1, ~ \xi_2=-\frac{1}{3}, ~  \xi_3= \frac{1}{3}, \xi_4= 1 \,


For the four nodes cubic element, the interpolation of such function is:
x = \sum_{i=1}^4 N_i^{(e)}(\xi)x_i
y = \sum_{i=1}^4 N_i^{(e)}(\xi)y_i


Remember the expression to obtain each N_i^{(e)}(\xi):
N_i^{(e)}(\xi) = \prod_{j=1 (j \ne i)}^n \frac{\xi - \xi_j}{\xi_i - \xi_j}


  • 
N_1^{(e)}(\xi) = 
\frac{ (\xi+\frac{1}{3}) (\xi-\frac{1}{3}) (\xi-1) }{ (-1 +\frac{1}{3}) (-1-\frac{1}{3}) (-1-1) }
=
- \frac{9}{16} (\xi+\frac{1}{3}) (\xi-\frac{1}{3}) (\xi-1)
  • 
N_2^{(e)}(\xi) = 
\frac{ (\xi+1) (\xi-\frac{1}{3}) (\xi-1) }{ (-\frac{1}{3} + 1) (-\frac{1}{3}-\frac{1}{3}) (-\frac{1}{3}-1) }
=
\frac{27}{16} (\xi+1) (\xi-\frac{1}{3}) (\xi-1)
  • 
N_3^{(e)}(\xi) = 
\frac{ (\xi + 1) (\xi + \frac{1}{3}) (\xi-1) }{ (\frac{1}{3} + 1) (\frac{1}{3} + \frac{1}{3}) (\frac{1}{3}-1) }
=
- \frac{27}{16} (\xi + 1) (\xi + \frac{1}{3}) (\xi-1)
  • 
N_4^{(e)}(\xi) = 
\frac{ (\xi + 1) (\xi + \frac{1}{3}) (\xi - \frac{1}{3}) }{ (1 + 1) (1 + \frac{1}{3}) (1 - \frac{1}{3}) }
=
\frac{9}{16} (\xi + 1) (\xi + \frac{1}{3}) (\xi - \frac{1}{3})


therefore:

x =
 \frac{2}{3} \frac{27}{16} (\xi+1) (\xi-\frac{1}{3}) (\xi-1) 
- \frac{4}{3} \frac{27}{16} (\xi + 1) (\xi + \frac{1}{3}) (\xi-1) 
+ 2 \frac{9}{16} (\xi + 1) (\xi + \frac{1}{3}) (\xi - \frac{1}{3})
= (\xi + 1)



\begin{align}
y & = 
- 4 \frac{9}{16} (\xi+\frac{1}{3}) (\xi-\frac{1}{3}) (\xi-1)
+ \frac{74}{27} \frac{27}{16} (\xi+1) (\xi-\frac{1}{3}) (\xi-1) \\
& - \frac{40}{27} \frac{27}{16} (\xi + 1) (\xi + \frac{1}{3}) (\xi-1) 
+ 2 \frac{9}{16} (\xi + 1) (\xi + \frac{1}{3}) (\xi - \frac{1}{3}) \\
& = \xi^3 + \xi^2 - 2 \xi + 2 \\
& = (\xi + 1)^3 - 2 (\xi + 1)^2 - (\xi + 1) + 4
\end{align}


The resulting expression is exactly the same to the original one:
y(x) = x^3 - 2 x^2 -x + 4 \,
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