Error Analysis in L_2 \times H^1


50. Error Analysis in \(L_2 \times H^1\)#

The variational formulation is understood as mixed formulation either in \(H(\opdiv) \times L_2\), or in \([L_2]^2 \times H^1\). Formally, both formulations are equivalent.

Next, we play the following game:

  • Use the finite elements \(RT_k \times P^k\) of the \(H\opdiv) \times L_2\) formulation

  • Mimic the \(L_2 \times H^1\) setting for the error analysis

We use the norms on \(\Sigma_h\) and \(V_h\):

\[\begin{eqnarray*} \| \sigma_h \|_{\Sigma_h}^2 & = & \| \sigma_h \|_{L_2}^2 \\ \| v_h \|_{V_h}^2 & = & \sum_T \| \nabla v_h \|_{L_2(T)}^2 + \sum_E \frac{1}{h} \| [v_h ] \|_{L_2(E)}^2 \end{eqnarray*}\]

The jump-terms over edges compensate the missing \(H^1\)-continuity.

The infinit-dimensional LBB - condition for the \(L_2-H^1\) setting is trivial:

\[ \sup_{\sigma \in [L_2]^2} \frac{\int \sigma \nabla u}{\| \sigma \|_{L_2}} \geq \| \nabla u \|_{L_2} \]

Just take the candidate \(\sigma = \nabla u\).

This construction we can mimic for the \(RT_k-P^k\) elements. We give the proof for the lowest order elements:

Given an \(u_h \in P^0\). Choose a discrete candidate \(\sigma_h\) such that

\[ \sigma_h \cdot n = \frac{1}{h} [u_h] \]


\[ \frac{\sum_T \int_T \opdiv \sigma_h u_h}{ \| \sigma_h \|_{L_2} } = \frac{\sum_T \int_{\partial T} \sigma_h\cdot n \; u_h}{ \| \sigma_h \|_{L_2} } = \frac{\sum_E \int_E \frac{1}{h} [u_h]^2}{ \| \sigma_h \|_{L_2} } \approx \sum_E \frac{1}{h} \| [u_h] \|_E^2 \]

We used that \(\| \sigma_h \|_{L_2(\Omega)}^2 \approx \sum_E h \| \sigma_h \cdot n \|_{L_2(E)}^2 \approx \sum_E \frac{1}{h} \| [u_h] \|_{L_2(E)}\), what is proven by scaling arguments.

The rest of the error analyis follows the lines of the \(H(\opdiv) \times L_2\) analysis.