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 $\frac{\partial \varphi}{\partial x}, \frac{\partial \varphi}{\partial y}, \frac{\partial^2 \varphi}{\partial x^2}, \frac{\partial^2 \varphi}{\partial y^2}, \frac{\partial^2 \varphi}{\partial x \partial y}$.
1 Answer
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.
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$\begingroup$ 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
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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. $\endgroup$ Commented Nov 18 at 19:01