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Why not just Lagrange polynomials basis functions

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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.

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  • $\begingroup$ Could you please tell me which theorem are we talking about in the last paragraph, thanks! $\endgroup$ Commented 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
    Commented May 2, 2019 at 5:06
  • $\begingroup$ I believe it is called "Sobolev embedding theorem". $\endgroup$
    – knl
    Commented Dec 3, 2023 at 19:47
  • $\begingroup$ The Sobolev emedding theorem has more to do with H^s regularity translating into C^s' regularity (and depends on the space dimension). Here this is probably some Green's formula motivated argument. What I mean is, say, you have a Laplacian in your strong form, it "behaves" (Riesz + BC if you want to be rigorous) against v as a gradient against the gradient of v. I'm also not sure it's generalizable like that. For instance for Maxwell's equations I believe you only need tangential continuity of E and normal of H (or the other way around). $\endgroup$
    – Sardine
    Commented Jan 17 at 16:00

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