$$
\newcommand{\uex}{{u_{\small\mbox{e}}}}
\newcommand{\half}{\frac{1}{2}}
\newcommand{\tp}{\thinspace .}
\newcommand{\Oof}[1]{\mathcal{O}(#1)}
\newcommand{\x}{\boldsymbol{x}}
\newcommand{\dfc}{\alpha} % diffusion coefficient
\newcommand{\Ix}{\mathcal{I}_x}
\newcommand{\Iy}{\mathcal{I}_y}
\newcommand{\If}{\mathcal{I}_s} % for FEM
\newcommand{\Ifd}{{I_d}} % for FEM
\newcommand{\basphi}{\varphi}
\newcommand{\baspsi}{\psi}
\newcommand{\refphi}{\tilde\basphi}
\newcommand{\xno}[1]{x_{#1}}
\newcommand{\dX}{\, \mathrm{d}X}
\newcommand{\dx}{\, \mathrm{d}x}
\newcommand{\ds}{\, \mathrm{d}s}
$$
Picard iteration defined from the variational form
$$
-(\dfc(u)u^{\prime})^{\prime} + au = f(u),\quad x\in (0,L),
\quad \dfc(u(0))u^{\prime}(0) = C,\ u(L)=D
$$
Variational form (\( v=\baspsi_i \)):
$$
F_i =
\int_0^L \dfc(u)u^{\prime}\baspsi_i^{\prime}\dx + \int_0^L au\baspsi_i\dx -
\int_0^L f(u)\baspsi_i\dx + C\baspsi_i(0) = 0
$$
Picard iteration: use "old value" \( u^{-} \) in \( \dfc(u) \) and \( f(u) \)
and integrate numerically:
$$
F_i = \int_0^L (\dfc(u^{-})u^{\prime}\baspsi_i^{\prime} + au\baspsi_i)\dx -
\int_0^L f(u^{-})\baspsi_i\dx + C\baspsi_i(0)
$$
