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The Forward Euler scheme: $$ \begin{equation} \lbrack D_t^+ u = -au \rbrack^n \tag{25} \tp \end{equation} $$ Definition of the truncation error \( R^n \): $$ \begin{equation} \lbrack D_t^+ \uex + a\uex = R \rbrack^n \tag{26} \tp \end{equation} $$ From (9)-(10): $$ [D_t^+ \uex]^n = \uex'(t_n) + \half\uex''(t_n)\Delta t + \Oof{\Delta t^2}\tp$$ Inserted in (26): $$ \uex'(t_n) + \half\uex''(t_n)\Delta t + \Oof{\Delta t^2} + a\uex(t_n) = R^n \tp $$ Note: \( \uex'(t_n) + a\uex^n = 0 \) since \( \uex \) solves the ODE. Then $$ \begin{equation} R^n = \half\uex''(t_n)\Delta t + \Oof{\Delta t^2} \tag{27} \tp \end{equation} $$
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