$$
\newcommand{\uex}{{u_{\small\mbox{e}}}}
\newcommand{\half}{\frac{1}{2}}
\newcommand{\tp}{\thinspace .}
\newcommand{\Oof}[1]{\mathcal{O}(#1)}
$$
Linearization via a geometric mean approximation
- \( f(u')=bu'|u'| \) leads to a quadratic equation for \( u^{n+1} \)
- Instead of solving the quadratic equation, we use a geometric mean
approximation
In general, the geometric mean approximation reads
$$ (w^2)^n \approx w^{n-\half}w^{n+\half}\tp$$
For \( |u'|u' \) at \( t_n \):
$$ [u'|u'|]^n \approx u'(t_n+{\half})|u'(t_n-{\half})|\tp$$
For \( u' \) at \( t_{n\pm 1/2} \) we use centered difference:
$$
u'(t_{n+1/2})\approx [D_t u]^{n+\half},\quad u'(t_{n-1/2})\approx [D_t u]^{n-\half}
$$
