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<p>Dear Oliver,</p>
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<p>I understand. Thanks for your quick response and clear explanation!</p>
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<p>Best regards,</p>
<p>Genki</p>
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<div id="x_divRplyFwdMsg" dir="ltr"><font face="Calibri, sans-serif" color="#000000" style="font-size:11pt"><b>From:</b> Oliver Sander <oliver.sander@tu-dresden.de><br>
<b>Sent:</b> Thursday, July 22, 2021 9:41:46 PM<br>
<b>To:</b> dune@lists.dune-project.org; Unagami, Genki<br>
<b>Subject:</b> Re: [Dune] Question on the example problem from the tutorial</font>
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<div class="PlainText">Hi Genki,<br>
<br>
> Could you briefly explain the math behind the following lines? Why can you estimate the appropriate quadrature order by these calculations? Also, why are they slightly different from each other?<br>
<br>
> int order = 2 * (localFiniteElement.localBasis().order()-1); (P.18, "For DUNE ver. 2.7", 2020)<br>
<br>
the integrand is \nabla \phi_i \nabla \theta_j. Both shape functions \phi_i<br>
and \theta_j are polynomials of order [...].order(). Taking the gradient<br>
reduces the order by one. That makes [...].order()-1 for each of the two<br>
factors. And as you are integrating the product you have to add the orders.<br>
That explains the "2*".<br>
<br>
> int order = 2∗(dim∗localFiniteElement.localBasis().order()−1); (P.19 "For DUNE ver. 2.4", 2016)<br>
<br>
To my own surprise I don't even have that document anymore. Without digging<br>
in the git log I guess that the example was using a grid with quadrilateral<br>
elements back then. In that case the polynomial order of a Lagrange shape function<br>
is "dim*[...].order()".<br>
<br>
Best,<br>
Oliver<br>
<br>
> <br>
> <br>
> Best regards,<br>
> <br>
> Genki Unagami<br>
> <br>
> <br>
> _______________________________________________<br>
> Dune mailing list<br>
> Dune@lists.dune-project.org<br>
> <a href="https://lists.dune-project.org/mailman/listinfo/dune">https://lists.dune-project.org/mailman/listinfo/dune</a><br>
> <br>
<br>
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