$$
\newcommand{\uex}{{u_{\small\mbox{e}}}}
\newcommand{\vex}{{v_{\small\mbox{e}}}}
\newcommand{\half}{\frac{1}{2}}
\newcommand{\tp}{\thinspace .}
\newcommand{\Oof}[1]{\mathcal{O}(#1)}
$$
Exact solutions of the finite difference equations
How does the truncation error depend on \( \uex \) in finite differences?
- One-sided differences: \( \uex''\Delta t \) (lowest order)
- Centered differences: \( \uex'''\Delta t^2 \) (lowest order)
- Only harmonic and geometric mean involve \( \uex' \) or \( \uex \)
Consequence:
- \( \uex(t)=ct+d \) will very often give exact solution
of the discrete equations (\( R=0 \))!
- Ideal for verification
- Centered schemes allow quadratic \( \uex \)
Problem: harmonic and geometric mean (error depends on \( \uex' \) and \( \uex \))
