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 $H (div) \times L_{2}$ , or in $[L_{2}]^{2} \times H^{1}$ . Formally, both formulations are equivalent.

Next, we play the following game:

Use the finite elements $R T_{k} \times P^{k}$ of the $H div) \times L_{2}$ formulation
Mimic the $L_{2} \times H^{1}$ setting for the error analysis

We use the norms on $Σ_{h}$ and $V_{h}$ :

\begin{array}{rcl} ‖ σ_{h} ‖_{Σ_{h}}^{2} & = & ‖ σ_{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{array}

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_{σ \in [L_{2}]^{2}} \frac{\int σ \nabla u}{‖ σ ‖_{L_{2}}} \geq ‖ \nabla u ‖_{L_{2}}

Just take the candidate $σ = \nabla u$ .

This construction we can mimic for the $R T_{k} - P^{k}$ elements. We give the proof for the lowest order elements:

Given an $u_{h} \in P^{0}$ . Choose a discrete candidate $σ_{h}$ such that

σ_{h} \cdot n = \frac{1}{h} [u_{h}]

Thus

\frac{\sum_{T} \int_{T} div σ_{h} u_{h}}{‖ σ_{h} ‖_{L_{2}}} = \frac{\sum_{T} \int_{\partial T} σ_{h} \cdot n u_{h}}{‖ σ_{h} ‖_{L_{2}}} = \frac{\sum_{E} \int_{E} \frac{1}{h} [u_{h}]^{2}}{‖ σ_{h} ‖_{L_{2}}} \approx \sum_{E} \frac{1}{h} ‖ [u_{h}] ‖_{E}^{2}

We used that $‖ σ_{h} ‖_{L_{2} (Ω)}^{2} \approx \sum_{E} h ‖ σ_{h} \cdot n ‖_{L_{2} (E)}^{2} \approx \sum_{E} \frac{1}{h} ‖ [u_{h}] ‖_{L_{2} (E)}$ , what is proven by scaling arguments.

The rest of the error analyis follows the lines of the $H (div) \times L_{2}$ analysis.

Error Analysis in L_2 \times H^1

50. Error Analysis in L2×H1#

50. Error Analysis in $L_{2} \times H^{1}$ #