On Sat, 27 Sep 2008, Derek Gaston wrote:
> Now, for C1 continuous elements (such as Clough-Toucher's,
> Hermite's, etc.) you should be able to get the value of the gradient
> at the nodes pretty easily: it should be in your solution vector.
> Obviously, I've never used these elements or I would know the answer
> to that... maybe Roy could fill us in.
Currently the C1 elements have degrees of freedom corresponding to x,
y partial derivatives at vertex nodes, but they don't have any degrees
of freedom which can give you the whole gradient at a second-order
node.
In general, the right way to get the well-defined gradient on a C1
node is the same way you'd get the cheapest O(h^p) gradient
approximation on a C0 node: start from an adjoining element, and do a
dof-weighted sum of its shape functions at that point.
>> One way to get an answer (John suggested this to me once) is to
>> compute the gradients at quadrature points and then do an L2
>> projection of that solution, and then just sample the projected
>> solution at the nodes.
>
> Yep... this is what's calle "Gradient Recovery". There are several
> methods for doing this...
Yup. For instance: if you do just a sum (weighted by element size?
angle? I forget) of the local gradients approaching a node on all
elements touching that node, you do get an O(h^{p+1}) approximation
that way. That's the first step to the popular Zienkiewicz-Zhu error
estimator, and it may be good enough (and cheap enough) for whatever
Nasser needs too.
---
Roy
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