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If cardinal basis functions are used (i.e. $\psi_{ij}=1$ iff $i=j$, and 0 otherwise) in interior penalty methods for elliptic equations such as SIPG & NIPG, shoudn't the jump in basis functions when computing surface integrals be [$\psi$] = 0 ? If that is the case the second term (consistency term) is always zero, right? How about the penalty term that has $[\psi]$ ? I am sure I am misunderstanding something here. For the poisson equation $-\nabla. (a\nabla u)=0\ \text{in} \ \Omega$ with $u=0$

Using the symbol v for the basis functions $\psi$ ,

$$a_{\epsilon}(u,v)=l(v)$$ where $$a_{\epsilon}(u,v)=\sum_{K\in T_h}\int_K a\nabla u\cdot \nabla v-\sum_{e\in \Gamma_h\cup \Gamma_D}\int_e\{a\nabla u\cdot n_e\}[v]+ \sum_{e\in \Gamma_h\cup \Gamma_D}\int_e\epsilon\{a\nabla v\cdot n_e\}[u] + \sum_{e\in \Gamma_h\cup \Gamma_D}\frac{\sigma_0}{|e|}\int_e[u][v]$$

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  • $\begingroup$ What do you mean by cardinal basis functions - i.e. what do $\Psi, i,j$ refer to? Integrals? Points? $\endgroup$ – Jesse Chan Nov 21 '15 at 20:00
  • $\begingroup$ I am thinking of a nodal DG method with legendre-gauss-lobatto integration points (i,j are those nodes). $\psi(x)$ are the Lagrange basis functions where $\psi_{ii}=1$ and $\psi_{ij}=0$ for all $i !=j$ $\endgroup$ – danny Nov 21 '15 at 20:20
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    $\begingroup$ Ah. Not quite - $[v] = v^+ - v^-$ (or the other way around). When you test on only one element (denote it the $-$ side), $[v] = v^-$. $\endgroup$ – Jesse Chan Nov 21 '15 at 21:29
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    $\begingroup$ Put another way, they're different basis functions from the + and - sides of the face, so the corresponding solution (which is a linear combination of all basis functions) can have different values at the vertices depending on which element you look from. $\endgroup$ – Christian Clason Nov 22 '15 at 10:38
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    $\begingroup$ Yes, correct. That's why you need to average it. $\endgroup$ – Wolfgang Bangerth Nov 22 '15 at 15:31

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