## 6 Error Estimates: Semi-Discrete (SD) Problems
### Error Estimates: SD Problem in \(\Omega\)  
We define a Ritz-Galerkin type projection \(\mathcal{Q}_{h}:H(div,\Omega)\to\mathbf{W}^{E}_{0,h}(\Omega)\) by  
\[a(\mathcal{Q}_{h}E,\mathbf{v})=a(E,\mathbf{v})\qquad\forall\mathbf{v}\in \mathbf{W}^{E}_{0,h}(\Omega).\]  
Then using the estimates (see [46] for the details):  
\[[\|\xi\|]^{2}\leq-Ca(\xi,\xi)\leq C[\|\xi\|][\|\eta\|], \tag{50}\]  
and  
\[[\|\eta\|]\leq Ch^{\xi-1}\|E\|_{H^{\xi}(\Omega)},\qquad 1\leq\xi\leq 2. \tag{51}\]  
we can deduce that (see Lemma 5.4 in [46], \(H^{1}\)-estimate and interpolation),  
\[\|E-\mathcal{Q}_{h}E\|_{\Omega}+h^{\xi-1}[\|E-\mathcal{Q}_{h}E\|]\leq Ch^{\xi }\|E\|_{H^{\xi}(\Omega)},\quad 1\leq\xi\leq 2. \tag{52}\]  
Finally, we recall the Galerkin orthogonality:  
\[a(E-E_{h},\chi)=0,\qquad\forall\chi\in\mathbf{W}^{E}_{0,h}(\Omega). \tag{53}\]  
Our main result is the following theorem. Theorem 1: _Let \(E\) and \(E_{h}\) be the solutions for the continuous and semi-discrete, weakly formulated, problems (25) and (13), respectively. Assume that \(E(t),\partial_{t}E(t),\partial_{tt}^{2}E\in H^{\xi}(\Omega)\), \(1\leq\zeta\leq 2\). Then, there exists a constant \(C\), independent of \(x,t,h\), but dependant on \(\varepsilon\), such that for \(t\in[0,T]\), the error \(e(t)=E(\cdot,t)-E_{h}(\cdot,t)\) is estimated as_  
\[\|\partial_{t}^{\varepsilon}e(t)\|_{\Omega} \leq Ch^{\xi}\|\partial_{t}^{\varepsilon}E(t)\|_{H^{\xi}(\Omega)}+ Ch^{\xi}\xi^{1-\tau}\int_{0}^{t}\left\|\partial_{\sigma_{\sigma}}^{2\kappa}E \right\|_{H^{\xi}(\Omega)}ds,\quad r=0,1, \tag{54}\] \[\left[\|e(t)\|\right]_{\Omega} \leq Ch^{\xi-1}\|E(t)\|_{H^{\xi}(\Omega)}+Ch^{\xi-1}\int_{0}^{t} \left\|\partial_{\sigma_{\sigma}}^{2\kappa}E\right\|_{H^{\xi}(\Omega)}ds.\]Proof. : We shall prove the theorem for \(\zeta=2\). The case \(\zeta=1\) is a simpler version, and \(\zeta\in[1,2]\) follows using classical interpolation of the Sobolev spaces, see [47]. For \(\zeta=2\), we use a split of error via the Ritz projection:  
\[e:=E-E_{h}=(E-\mathcal{Q}_{h}E)+(\mathcal{Q}_{h}E-E_{h}):=\rho+\theta.\]  
To bound \(\theta\) we note that using the Galerkin orthogonality (53),  
\[ \begin{split}(\varepsilon\theta_{tt},\chi)+a(\theta,\chi)& =(\varepsilon\mathcal{Q}_{h}E_{tt},\chi)_{\Omega}+a(\mathcal{Q} _{h}E,\chi)_{\Omega}-(\varepsilon\partial_{tt}^{2}E_{h},\chi)_{\Omega}-a(E_{ h},\chi)_{\Omega}\\ &=(\varepsilon\mathcal{Q}_{h}E_{tt},\chi)_{\Omega}+a(\mathcal{Q} _{h}E,\chi)_{\Omega}-(\varepsilon\partial_{tt}^{2}E,\chi)_{\Omega}-a(E,\chi) _{\Omega}\\ &=-(\varepsilon\rho_{tt},\chi)_{\Omega},\qquad\qquad\forall\chi \in\mathbf{W}_{0,h}^{E}.\end{split} \]  
Then, for \(\chi=\theta_{t}\),  
\[(\varepsilon\theta_{tt},\theta_{t})_{\Omega}+a(\theta,\theta_{t})=-( \varepsilon\rho_{tt},\theta_{t})_{\Omega}.\]  
This yields (noting that \(\varepsilon\) is a function only in space)  
\[\frac{d}{dt}\Big{(}\varepsilon\|\theta_{t}\|_{\Omega}^{2}+a(\theta,\theta) \Big{)}\leq 2\|\varepsilon\rho_{tt}\|_{\Omega}\|\theta_{t}\|_{\Omega}.\]  
Note that, by the definition \(\theta(0)=\theta_{t}(0)=0\). Hence, integrating over \([0,t]\) we get  
\[ \begin{split}\varepsilon\|\theta_{t}(t)\|_{\Omega}^{2}+a(\theta (t),\theta(t))&\leq 2\int_{0}^{t}\|\varepsilon\rho_{ss}\|_{ \Omega}\|\theta_{s}\|_{\Omega}\,ds\\ &\leq 2\varepsilon\max_{s\in[0,t]}\|\theta_{s}\|_{\Omega}\int_{0}^{t} \|\rho_{ss}\|_{\Omega}\,ds\\ &\leq 2\Big{(}\int_{0}^{t}\|\rho_{ss}\|_{\Omega}\Big{)}^{2}+\frac{1}{ 2}\Big{(}\varepsilon\max_{s\in[0,t]}\|\theta_{s}\|_{\Omega}\Big{)}^{2}.\end{split}  \tag{55}\]  
Now, using the fact that \(a(\theta(t),\theta(t))\geq 0\), and since (55) holds for all \(t\in[0,T]\),  
\[\frac{1}{2}\varepsilon\Big{(}\max_{s\in[0,t]}\|\theta_{s}\|_{\Omega}\Big{)}^{ 2}\leq 2\Big{(}\int_{0}^{T}\|\rho_{ss}\|_{\Omega}ds\Big{)}^{2},\]  
which inserting into (55) and using (52) noting that \(\varepsilon\geq 1\) and \(h\sim\tau\) we get  
\[