Why not just Lagrange polynomials basis functions


1 Answer 1


The weak form of the euler-bernoulli beam equations has second order weak derivatives. This means that the finite element space requires continuity in the 1st derivatives across each element boundary. Lagrange elements have continuity in the 0th derivative only (the values match at the boundary, but the 1st derivatives don’t). Hermite interpolants can be generalized to ensure continuity to any prescribed derivative order.

There is a theorem which states that for an nth order weak derivative in the weak form, you need (n-1)st order continuity in the interpolants between each element.

  • $\begingroup$ Could you please tell me which theorem are we talking about in the last paragraph, thanks! $\endgroup$ May 2, 2019 at 3:56
  • $\begingroup$ @Akash: it’s a property of the sobolev space. In order for a weak derivative of order n to be integrable, it has to have continuity of order (n-1). One of the main reasons why FEM is so popular is precisely because it “weakens” the restrictions on continuity of functions we can approximate with. $\endgroup$
    – Paul
    May 2, 2019 at 5:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.