$$
\newcommand{\uex}{{u_{\small\mbox{e}}}}
\newcommand{\half}{\frac{1}{2}}
\newcommand{\tp}{\thinspace .}
\newcommand{\Oof}[1]{\mathcal{O}(#1)}
\newcommand{\x}{\boldsymbol{x}}
\newcommand{\dfc}{\alpha} % diffusion coefficient
\newcommand{\Ix}{\mathcal{I}_x}
\newcommand{\Iy}{\mathcal{I}_y}
\newcommand{\If}{\mathcal{I}_s} % for FEM
\newcommand{\Ifd}{{I_d}} % for FEM
\newcommand{\basphi}{\varphi}
\newcommand{\baspsi}{\psi}
\newcommand{\refphi}{\tilde\basphi}
\newcommand{\xno}[1]{x_{#1}}
\newcommand{\dX}{\, \mathrm{d}X}
\newcommand{\dx}{\, \mathrm{d}x}
\newcommand{\ds}{\, \mathrm{d}s}
$$
Picard iteration for Backward Euler scheme
A simple Picard iteration, not knowing anything about the nonlinear
structure of \( f \), must approximate \( f(u,t_n) \) by \( f(u^{-},t_n) \):
$$ \hat F(u) = u - \Delta t\, f(u^{-},t_n) - u^{(1)}$$
The iteration starts with \( u^{-}=u^{(1)} \) and proceeds with repeating
$$ u^* = \Delta t\, f(u^{-},t_n) + u^{(1)},\quad
u = \omega u^* + (1-\omega)u^{-},
\quad u^{-}\ \leftarrow\ u$$
until a stopping criterion is fulfilled.
