setting ${{\mathcal{X}}}= \mathcal{D}({{\mathcal{A}}}_0^{1/2})\times {{\mathcal{X}}}_0$, where $\mathcal{D}({{\mathcal{A}}}_0^{1/2})$ is endowed with the scalar product $\langle {{\mathcal{A}}}_0^{1/2}\cdot, {{\mathcal{A}}}_0^{1/2}\cdot \rangle_{{{\mathcal{X}}}_0} $ and $${{\mathcal{A}}}= \left(\begin{array}{cc} 0 & I \\ -{{\mathcal{A}}}_0 & 0 \end{array}\right).$$

But another way to write under the form is to set ${{\mathcal{X}}}= {{\mathcal{X}}}_0^2$, $$\label{NewmarkFormulation}
        y_1 = \varphi' + i {{\mathcal{A}}}_0^{1/2} \varphi, \quad y_2 = \varphi' - i {{\mathcal{A}}}_0^{1/2} \varphi, 
    \qquad {{\mathcal{A}}}= \left(\begin{array}{cc} i {{\mathcal{A}}}_0^{1/2} & 0 \\ 0 & - i {{\mathcal{A}}}_0^{1/2} \end{array}\right).$$ This formulation will be preferred to study since the eigenvectors are now given as $\Phi_j^+ = (\Psi_j, 0)$, $\Phi_j^- = (0, \Psi_j)$ where $\Psi_j$ are the eigenvectors of ${{\mathcal{A}}}_0$, and ${{\mathcal{A}}}\Phi_j^{\pm} = \pm i \sqrt{\lambda_j} \Phi_j^{\pm}$, with ${{\mathcal{A}}}_0 \Psi_j  =  {\lambda_j} \Psi_j$. (At this step, remember that ${{\mathcal{A}}}_0$ is assumed to be a self-adjoint positive definite operator, so its spectrum is given by a sequence of positive real numbers going to infinity.)

Now, setting $${{\mathcal{A}}}_{0, \tau} = \left(I+ \left(\frac{4\beta-1}{4} \right) \tau^2 {{\mathcal{A}}}_0\right)^{-1} {{\mathcal{A}}}_0,$$ considering $$\label{VariableNewmark}
        \left\{
        \begin{array}{ll}
        {\displaystyle}y_{1, \tau}^{k+1/2} = \frac{\varphi_\tau^{k+1}-\varphi_\tau^k}{\tau} + i {{\mathcal{A}}}_{0,\tau}^{1/2} \left( \frac{\varphi_\tau^k+\varphi_\tau^{k+1}}{2 }\right), 
        \smallskip
            \\
        {\displaystyle}y_{2, \tau}^{k+1/2} = \frac{\varphi_\tau^{k+1}-\varphi_\tau^k}{\tau} - i {{\mathcal{A}}}_{0,\tau}^{1/2} \left(
        \frac{\varphi_\tau^k+\varphi_\tau^{k+1}}{2}\right),
         \end{array}\right.$$ system can be written as $$\begin{gathered}
        \label{MidpointNewmark}
    {\displaystyle}\frac{y_\tau^{k+1/2} - y_\tau^{k-1/2}}{\tau} = {{\mathcal{A}}}_\tau \left(\frac{y_\tau^{k-1/2} + y_\tau^{k+1/2}}{2} \right), \quad k \in {\mathbb{Z}},
            