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Suppose we have an element in the polar coordinate as $0\leq r\leq 1$, $0\leq\theta\leq \alpha$. What are the correct basis functions in this element? For this kind of mesh, there are a lot of references on finite difference discretization. But I can not find a fem solution.

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  • $\begingroup$ What do you mean by "correct basis functions"? There are an infinite number of bases that could be used to expand the solution in such an element. $\endgroup$
    – Dan
    Jan 5, 2012 at 9:59
  • $\begingroup$ I tried to use the function basis which is linear in $r, \theta$. But I found I can not put d.o.f. at the origin point $r=0$ and the two vertices $r=1,\theta=0,\alpha$. If $\varphi=ar+b\theta+c$, and $\varphi(0,\theta)=0$ for any $\theta$, then $b=c=0$. $\endgroup$
    – Hui Zhang
    Jan 5, 2012 at 10:02
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    $\begingroup$ You could consider using isoparametric curvilinear elements with the nodes distributed using the warp & blend procedure do to T. Warburton suitable for explicit (read efficient) construction of interpolation nodes on the simplex (caam.rice.edu/~timwar/preprints/JEMnodesV1.pdf). $\endgroup$ Jan 5, 2012 at 11:12
  • $\begingroup$ Thanks for answering and pointed out the reference. But by using isoparametric elements, do you mean the sector boundary will only be approximated in cartesian coordinates $(x,y)$ by polynomials? Is there any way to keep exactly the shape of the sector? $\endgroup$
    – Hui Zhang
    Jan 5, 2012 at 13:01
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    $\begingroup$ If you use rectangular elements, then this is really a tough question. If you have a regular $n$-gon in the middle then you might piecewise linear shape functions on that object still (with different gradients on the n-gon) such that continuity is preserved along element edges. Still $r=0$ might be a source of error, and an adaptive strategy is a must-do. - For triangle elements, none of this problems arise. $\endgroup$
    – shuhalo
    Jan 6, 2012 at 1:00

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