# Construction of $C^1$/$H^2$-conforming finite element basis for triangular or tetrahedral mesh

In the paper Hierarchical Conforming Finite Element Methods for the Biharmonic Equation, P. Oswald claimed Clough-Tocher type elements has $C^1$-continuity while being a cubic polynomial on each triangle. He didn't give a set of explicit basis functions just the standard degrees of freedom on the quadrature points.

Similarly, in the book The Mathematical Theory of Finite Element Methods Chapter 3, the authors give us the construction of cubic Hermite finite elements, but they didn't mention the continuity of the cubic Hermite elements.

However, in the paper Differential complexes and numerical stability, Doulgas Arnold proposed that for $C^1$/$H^2$-conforming discrete space we should use the Hermite quintic(or rather Argyris) finite elements, which is very complicated to express explicitly.

So here are my questions:

(1) Is there any paper that comes up with an explicit formula for the $C^1$/$H^2$-conforming finite elements on triangular or tetrahedral mesh?

(2) Should piecewise cubic be the minimal degree of polynomials requirement for $C^1$-continuity?

The cubic Hermite elements have a continuous normal derivative but not full $C^1$ continuity. In particular, the normal derivatives may not match at the boundary of two elements, away from the vertices. If you want full $C^1$ continuity you will have to use the Argyris element or Hsieh-Clough-Tucker or something. I recommend the discussion in chapter 6 of Ciarlet's finite element book.

The degree of polynomial required for $C^1$ continuity will depend on your spatial dimension, but in 2D or 3D I don't think you can get away with less than cubic polynomials. You could consider some kind of nonconforming method which may allow a simpler finite element space.

• Err, if a function is continuous across the interface between two cells, and if the function on each cell is in $C^\infty$ as it must be if it's a polynomial, then how can the tangential derivative be discontinuous on a cell interface? Or did you mean that the tangential derivative can be discontinuous at the vertices, i.e. the end-points of each interface? – Wolfgang Bangerth May 3 '12 at 0:18
• You're absolutely right, I edited the answer. – Andrew T. Barker Jun 8 '12 at 14:51

I refer you to the book Splines on Triangulations. I cannot locate my copy at the moment to give you a better answer, but I recall a discussion/theorems on the polynomial order required for $C^1$ spaces. If I recall correctly, Lai proves that under certain conditions $p=3$ is OK, but $p=5$ is always sufficient.

Unfortunately, I also remember that Lai does not then show how to construct $C^1$ spaces, only prove they exist given a triangulation and a spline space. Once he has this proof, he solves his application with additional linear constraint equations to enforce the $C^1$ condition.

• welcome to scicomp Mr. Collier :) – Aron Ahmadia Apr 25 '12 at 6:25

You can refer to the following pages for a full listing of the basis functions for Argyris: FEMList.pdf Wikipedia entry (French)

Also, you can use the VT-ICAM ArgyrisPack that a colleague and I developed.