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+\documentclass{scrartcl}
+\usepackage{amsmath}
+\begin{document}
+\noindent The Newmark scheme in its classical form with $\gamma = 1/2$
+and $\beta = 1/4$ reads
+\begin{align}
+  \label{eq:1} \dot \alpha_1
+  &= \dot \alpha_0 + \frac \tau 2 (\ddot \alpha_0 + \ddot \alpha_1 )\\
+  \label{eq:2} \alpha_1
+  &= \alpha_0 + \tau \dot \alpha_0 + \frac {\tau^2}4 ( \ddot \alpha_0 + \ddot \alpha_1 )\text.
+  \intertext{We can also write \eqref{eq:2} as}
+  \nonumber \ddot \alpha_1 + \ddot \alpha_0
+  &= \frac 4{\tau^2} ( \alpha_1 - \alpha_0 - \tau \dot \alpha_0)
+  \intertext{so that it yields}
+  \label{eq:3} \dot \alpha_1
+  &= \dot \alpha_0 + \frac 2\tau ( \alpha_1 - \alpha_0 - \tau \dot \alpha_0) = \frac 2\tau ( \alpha_1 - \alpha_0) - \dot \alpha_0
+\end{align}
+in conjunction with \eqref{eq:1}. The problem
+\begin{align*}
+  -\dot \alpha_1 \in \partial \psi(\alpha_1)
+\end{align*}
+then becomes
+\begin{align*}
+  \dot \alpha_0 -\frac 2\tau ( \alpha_1 - \alpha_0)
+  &\in \partial \psi(\alpha_1)\\
+  \psi(\beta) - \psi(\alpha_1)
+  &\ge (\dot \alpha_0 -\frac 2\tau ( \alpha_1 - \alpha_0), \beta - \alpha_1)
+  \quad \forall \beta\\
+  \frac 2\tau ( \alpha_1, \beta - \alpha_1) + \psi(\beta) - \psi(\alpha_1)
+  &\ge (\dot \alpha_0 + \frac 2\tau \alpha_0, \beta - \alpha_1)
+  \quad \forall \beta
+\end{align*}
+After which $\dot \alpha_1$ can be computed according to \eqref{eq:3}.
+\end{document}