2
$\begingroup$

I read in a book about an $H^2$-conforming element on a rectangle, the Bogner-Fox-Schmit Rectangle element, and I was wondering if it has a three-dimensional extension to a cube. The degree of freedom of the BFS element is defined as $\frac{\partial \varphi}{\partial x}, \frac{\partial \varphi}{\partial y}, \frac{\partial^2 \varphi}{\partial x^2}, \frac{\partial^2 \varphi}{\partial y^2}, \frac{\partial^2 \varphi}{\partial x \partial y}$.

$\endgroup$
1
  • $\begingroup$ Can you add the bibliographic information of the book you are mentioning? $\endgroup$
    – nicoguaro
    Commented Nov 18 at 18:38

1 Answer 1

2
$\begingroup$

In this paper "MATLAB Implementation of C1 finite elements: Bogner-Fox-Schmit rectangle", they define it as products of Hermite polynomials in 2D, so you can try to generalize this to 3D. It should yield $4^3=64$ degreees of freedom per element. I suppose natural ones would be pointwise values ($8$), gradients ($3\times 8 = 24$), second mixed derivatives $\partial_{xy}, \partial_{xz}, \partial_{yz}$ ($3\times 8 = 24$), and triple mixed derivative $\partial_{xyz}$ ($8$).

You can generalize this to $n$ dimensions by multiplying $n$ Hermite polynomials. You have $2^n$ vertices of the $n$-dimensional hypercube. There are $4^n$ basis functions. Pointwise values $2^n$, first derivatives $n\times 2^n$, second mixed derivatives ${n \choose 2}\times 2^n$, third mixed derivatives ${n \choose 3} \times 2^n$, ..., $(n-1)$-st mixed derivatives ${n \choose n-1} \times 2^n$, and $n$-th mixed derivative $2^n$. Note that the mixed derivatives above are without coordinate repetition (e.g. $\partial_{xy}$ but not $\partial_{xx}$), that's why I consider combinations). It may be possible to prescribe $\partial_{xx}$ and friends, but it's not the natural boundary condition for a Hermite polynomial.

$\endgroup$
2
  • $\begingroup$ Are the base functions in the paper $C^1$? They only guarantee continuity in the cross-derivative. I am guessing that if you keep your domain as a rectangle it is the case but it will not hold for general quadrilaterals. $\endgroup$
    – nicoguaro
    Commented Nov 18 at 18:39
  • 1
    $\begingroup$ @nicoguaro If you look at the paper they assembled rectangles in a conforming fashion, so that you have no hanging nodes. Since the directional derivatives align in that case and are shared this should be $C^1$. It's like how you can get $C^1$ with cubic polynomials on triangles if the triangles are arranged nicely, but in the general setting you need quintic polynomials. I suppose that if you were to do something funny to the rectangles the above may not work. But that is the element Chandler asked for. $\endgroup$
    – lightxbulb
    Commented Nov 18 at 19:01

Your Answer

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

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