$$ \newcommand{\uex}{{u_{\small\mbox{e}}}} \newcommand{\vex}{{v_{\small\mbox{e}}}} \newcommand{\half}{\frac{1}{2}} \newcommand{\tp}{\thinspace .} \newcommand{\Oof}[1]{\mathcal{O}(#1)} $$

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Carrying out the truncation error analysis

Using (15)-(16) and (5)-(6) we get $$ \begin{align*} \lbrack mD_tD_t \uex + \beta D_{2t} \uex\rbrack^n &= m\uex''(t_n) + \beta\uex'(t_n) + \\ &\quad \left(\frac{m}{12}\uex''''(t_n) + \frac{\beta}{6}\uex'''(t_n)\right)\Delta t^2 + \Oof{\Delta t^4} \end{align*} $$ The terms $$ m\uex''(t_n) + \beta\uex'(t_n) + \omega^2\uex(t_n) + s(\uex(t_n)) - F^n,$$ correspond to the ODE (= zero).

Result: accuracy of \( \Oof{\Delta t^2} \) since $$ \begin{equation} R^n = \left(\frac{m}{12}\uex''''(t_n) + \frac{\beta}{6}\uex'''(t_n)\right)\Delta t^2 + \Oof{\Delta t^4}, \tag{39} \end{equation} $$

Correction terms: complicated when the ODE has many terms...

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