$$
\newcommand{\uex}{{u_{\small\mbox{e}}}}
\newcommand{\Aex}{{A_{\small\mbox{e}}}}
\newcommand{\half}{\frac{1}{2}}
\newcommand{\Oof}[1]{\mathcal{O}(#1)}
$$
Computation of the truncation error
- The residual \( R^n \) is the truncation error.
- How does \( R^n \) vary with \( \Delta t \)?
Tool: Taylor expand \( \uex \) around the point where the ODE is sampled
(here \( t_n \))
$$ \uex(t_{n+1}) = \uex(t_n) + \uex'(t_n)\Delta t + \half\uex''(t_n)
\Delta t^2 + \cdots $$
Inserting this Taylor series in (31) gives
$$ R^n = \uex'(t_n) + \half\uex''(t_n)\Delta t + \ldots + a\uex(t_n)$$
Now, \( \uex \) solves the ODE \( \uex'=-a\uex \), and then
$$ R^n \approx \half\uex''(t_n)\Delta t$$
This is a mathematical expression for the truncation error.
