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This question is largely motivated by the famous paper "UNIFIED ANALYSIS OF DISCONTINUOUS GALERKIN METHODS FOR ELLIPTIC PROBLEMS"

The Bassi and Rebay biliniar DG form is given in page 11 in the paper mentioned above as: \begin{equation} B_h(u_h, v) = \int_{\Omega} (\nabla u_h \cdot \nabla v + r([[u_h]])r([[v]])) \,dx - \int_{\Gamma} (\{ \nabla u_h \} \cdot [[v]] + [[u_h]] \cdot \{ \nabla v \}) \,ds \end{equation} where the $\{ \cdot \}$ and $[[\cdot ]]$ are the standard average and jump operators.

The lift operator is defined as: \begin{equation} \int_{\Omega} r(\phi) \cdot \tau \,dx = - \int_{\Gamma} \phi \cdot \{ \tau \} \,ds \end{equation}

How can I calculate the volume contribution with the lift operator? My first approach would be to use the definition for the lift above to do something like this: \begin{equation} \int_{\Omega} r([[u_h]])r([[v]]) \,dx -> - \int_{\Gamma} [[u_h]] \{ r([[v]]) \} \,ds \end{equation} but then I have the average lift operator again.

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    $\begingroup$ It might help you to first understand the calculation of "local" lift contribution for each cell-edge. $\endgroup$
    – ConvexHull
    Jan 21 at 20:08
  • $\begingroup$ I am new to lifts so I need me some hand holding here. So, if I had something like $\int_{\Omega} r([[u]]) v \,dx$ then I will use the definition of the lift and do this $ -\int_{\Gamma} [[u]] \{v\} \,ds$ instead which is something I know how to calculate. What do I do next? Can I calculate the integral $\int_{\Omega} r([[u]]) v \,dx$ or maybe something simpler like $\int_{\Omega} r(u) v \,dx$ on the cell without going through the definition of the lift? Is this what your hinting at? How can I do this? $\endgroup$ Jan 21 at 20:54
  • $\begingroup$ It might help to pay less attention to the pure formulas. Suppose you have two elements sharing one edge. The solution across the edge is discontinuous. Now, you can calculate three different flux contributions. The first one is the conservative numerical flux, unique on the edge. The other two fluxes are the physical fluxes, which can be calculated directly by means of the polynomial solution in each element at the boundary. In order to calculate the lifting contribution in both volumes you simply substract the physical fluxes from the numerical flux and project the solution to the volumes. $\endgroup$
    – ConvexHull
    Jan 21 at 21:49

1 Answer 1

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The lift operator takes a function defined on the boundary of a cell and produces a function that lives in the interior of the cell. You will evaluate the bilinear form $B_h$ for basis functions $\varphi_i,\varphi_j$, and for each of these, you can compute the jump over the boundaries of the cell, $[[\varphi_i]]$ and then compute the lifted function $r([[\varphi_i]])$. The latter is a polynomial (look up which polynomial space the lifted function actually lives in), and you can evaluate it at quadrature points to compute the domain integral over the current cell.

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  • $\begingroup$ I have been looking step 82 in deal. If the descrete heessin is writen as $H_h(\phi) = D^2_h(\phi) - \sum_{e \subset \partial K} r_e\left[\left[\nabla_h \phi\right]\right] + \sum_{e \subset \partial K} b_e\left[\left[\phi\right]\right]$ , then is the expression $\alpha_h(v_h, u_h) = \int_\Omega H_h(v_h) : H_h(u_h) = \int_\Omega D^2_h(v_h) : D^2_h(u_h) - \int_\Omega r_e\left[\left[\nabla_h v_h\right]\right] r_e\left[\left[\nabla_h u_h\right]\right] + \int_\Omega b_e\left[\left[v_h\right]\right] b_e\left[\left[u_h\right]\right] \, dx$ correct? (maybe that negative sign is wrong) $\endgroup$ Jan 21 at 20:43
  • $\begingroup$ No. $(a-b)(c-d)$ is not equal to $ac-bd$, regardless of how you choose the sign :-) $\endgroup$ Jan 22 at 4:16
  • $\begingroup$ I am not sure if I understand the weak form correctly since it's not written explicitly. Is there anywhere in step 82 where the integral of the product of two lifts is calculated? $\endgroup$ Jan 23 at 9:31
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    $\begingroup$ @CuteCompute I didn't write step-82. You might want to look at the papers cited there. $\endgroup$ Jan 23 at 16:50

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