Does anyone have a intuitive explanation of why Hermite polynomials have to be utilized as the shape functions in the FEM solution of the Euler Bernoulli Beam 4th order ODE?

I have been learning FEM on my own and can't figure out why any other 2nd order polynomial can't be utilized in the place of the cubic Hermite ones, especially because the weak form transfers two derivatives to the test functions...

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  • $\begingroup$ The presence of 2nd-order spatial derivative in the weak form requires that the approximation polynomials should be C1 continuous across the element boundaries. You can use uniform quadratic (or higher-order) b-splines instead of cubic Hermite polynomials. You can use any other type of polynomials as long as they are C1 continuous across the element boundaries. $\endgroup$ – Chenna K Jul 17 '19 at 16:02
  • $\begingroup$ Hi Chenna K, I have utilized the following shape functions (I "invented" them myself) w1 = (h^2 - x^2)/h^2 and w2 =(x/h)*(2 - x/h), where h is the element size. However, their second derivative are equal which makes the system impossible to solve. Is there a practical method for obtaining shape functions? $\endgroup$ – Marcus Jul 17 '19 at 16:35
  • $\begingroup$ The basis functions you chose are not valid. You can't choose the basis functions just like that. They need to possess characteristics such as the partition of unity and completeness. With quadratic polynomials, you need three basis functions, not two. Please refer to my PhD thesis for the details on solving PDEs with NURBS basis functions. $\endgroup$ – Chenna K Jul 17 '19 at 21:21
  • $\begingroup$ @ChennaK, you don't need partition of unity for the basis functions. It is common, but not mandatory. $\endgroup$ – nicoguaro Jul 17 '19 at 21:59
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    $\begingroup$ You can refer to the Introduction to FEM book by J N Reddy. $\endgroup$ – Chenna K Jul 22 '19 at 20:22

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