Loading [MathJax]/extensions/TeX/boldsymbol.js
\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.