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I am having a problem that was also mentioned at the accepted answer to this question by Wolfgang Bangerth. I need to calculate, as it was specified at the question at the link, F1 integral and for that I need to calculate u1 at gauss points of the new mesh but with a certain accuracy so I do not loose overall convergence of the scheme. I want to ask if anyone can give me a tip on how to do that? What I am currently doing is calculating "shape functions" at gauss points of the new mesh and multiplying u1 with those shape functions, but this makes my overall scheme loose convergence so I need to do it in a more accurate way.

Thanks in advance.

noname
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  • Projecting any function onto any nodal basis means evaluating the function at the respective points. If your function is a polynomial of certain degrees, and you project onto a polynomial basis of the same or higher degree, this is can be done without loss of information. Moreover, for a benign set of nodes, the basis transformation is well conditioned, so your whole task should be not a large problem. – davidhigh Mar 28 '21 at 19:28

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You will need to compute integrals like this: $$ F_i = \int_\Omega \phi_{2,i}(x) u_1(x) dx $$ which we break up as follows: $$ F_i = \sum_{K \in T_2} \int_K \phi_{2,i}(x) u_1(x) dx. $$

If the two meshes are unrelated, (one of) the issues is that the integrand $$ \int_K \phi_{2,i}(x) u_1(x) dx $$ is not a polynomial on $K$ and not smooth: $\phi_{2,i}$ is a polynomial because it is a shape function defined on the cells of $T_2$, but $u_1$ is a piecewise polynomial defined on the mesh $T_1$ but not on the cells of $T_2$ and so will, in general, have kinks on the cells $K$ of $T_2$. This means that Gauss quadrature will not be very accurate, pretty much regardless of the number of quadrature points you have. You will be better off with the iterated trapezoidal rule, for example.

Wolfgang Bangerth
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