\\
             \hbox{with } 
            {{\mathcal{A}}}_\tau = 
                \left(\begin{array}{cc}
                         i {{\mathcal{A}}}_{0,\tau}^{1/2} &  0
                    \\
                        0  & - i {{\mathcal{A}}}_{0,\tau}^{1/2}
                 \end{array}\right).
    \end{gathered}$$ Under this form, one easily sees that another energy for solutions of is given by $$\tilde E^{k+1/2}  =  \frac{1}{2}{\left\Verty^{k+1/2}_\tau\right\Vert}_{{{\mathcal{X}}}_0^2}^2 
        = {\left\Vert\frac{\varphi_\tau^{k+1}-\varphi_\tau^k}{\tau}\right\Vert}_{{{\mathcal{X}}}_0}^2 
        + {\left\Vert{{\mathcal{A}}}_{0, \tau}^{1/2}\left(\frac{\varphi_\tau^k+\varphi_\tau^{k+1}}{2}\right)  \right\Vert}_{{{\mathcal{X}}}_0}^2.$$ As one easily checks, it turns out that the energies $\tilde E^{k+1/2}$ and $E^{k+1/2}$ (defined in ) are equivalent when working within a filtered class $\mathfrak{C}(\delta/\tau)$ at scale $1/\tau$, thus making of no particular relevance to our purpose the fact that they do not coincide.

Now, one can easily show that if $y^{1/2} = \Phi$, where $\Phi$ is an eigenvector of ${{\mathcal{A}}}$ given by corresponding to the eigenvalue $i \mu$, then the solution $y_\tau$ of is given by $$\begin{gathered}
        \label{f-Newmark}
        y_\tau^{k+1/2} = \exp \left(i \frac{f(\mu\tau )}{\tau} k \tau \right) \Phi, \\
         \hbox{ with } f(\alpha) = 2 \arctan\left(\frac{\alpha}{2} \frac{1}{ \sqrt{1+ (\beta-1/4)\alpha^2 }}\right), \quad R = \infty.
    \end{gathered}$$

Hence the Newmark approximation scheme fits the assumptions of Theorem \[Thm-Main\].

Also note that the observation ${{\mathcal{B}}}_1 \varphi + {{\mathcal{B}}}_2 \varphi'$ for can be discretized in two different ways:

-   A natural discretization consists in taking $$\label{Formulation1-a}
                {{\mathcal{B}}}_1\left(\frac{\varphi_\tau^k + \varphi_\tau^{k+1}}{2}\right) + {{\mathcal{B}}}_2 \left(\frac{\varphi_\tau^{k+1} - \varphi_\tau^k}{\tau} \right)$$ for system , which 