$$
\newcommand{\uex}{{u_{\small\mbox{e}}}}
\newcommand{\half}{\frac{1}{2}}
\newcommand{\tp}{\thinspace .}
\newcommand{\Oof}[1]{\mathcal{O}(#1)}
\newcommand{\x}{\boldsymbol{x}}
\newcommand{\dfc}{\alpha} % diffusion coefficient
\newcommand{\Ix}{\mathcal{I}_x}
\newcommand{\Iy}{\mathcal{I}_y}
\newcommand{\If}{\mathcal{I}_s} % for FEM
\newcommand{\Ifd}{{I_d}} % for FEM
\newcommand{\basphi}{\varphi}
\newcommand{\baspsi}{\psi}
\newcommand{\refphi}{\tilde\basphi}
\newcommand{\xno}[1]{x_{#1}}
\newcommand{\dX}{\, \mathrm{d}X}
\newcommand{\dx}{\, \mathrm{d}x}
\newcommand{\ds}{\, \mathrm{d}s}
$$
The equations in Newton's method
$$
F_i =
\int_0^L (\dfc(u)u^{\prime}\baspsi_i^{\prime} + au\baspsi_i -
f(u)\baspsi_i)\dx + C\baspsi_i(0)=0,\quad i\in\If
$$
Easy to evaluate right-hand side \( -F_i(u^{-}) \) by numerical integration:
$$
F_i =
\int_0^L (\dfc(u^{-})u^{\prime}\baspsi_i^{\prime} + au\baspsi_i -
f(u^{-})\baspsi_i)\dx + C\baspsi_i(0)=0
$$
(just known functions)
