$$
\newcommand{\uex}{{u_{\small\mbox{e}}}}
\newcommand{\half}{\frac{1}{2}}
\newcommand{\tp}{\thinspace .}
\newcommand{\Oof}[1]{\mathcal{O}(#1)}
$$
The schemes are not equivalent wrt the initial conditions
$$ u^{\prime}=v=0\quad\Rightarrow\quad v^0=0,$$
so
$$
\begin{align*}
v^1 &= v^0 - \Delta t\omega^2 u^0 = - \Delta t\omega^2 u^0\\
u^1 &= u^0 + \Delta t v^1 = u^0 - \Delta t\omega^2 u^0 !=
\underbrace{u^0 - \frac{1}{2}\Delta t\omega^2 u^0}_{\mbox{from }[D_tD_t u +\omega^2 u=0]^n\mbox{ and }[D_{2t}u=0]^0}
\end{align*}
$$
The exact discrete solution derived earlier does not fit the Euler-Cromer
scheme because of mismatch for \( u^1 \).
