$$
\newcommand{\uex}{{u_{\small\mbox{e}}}}
\newcommand{\half}{\frac{1}{2}}
\newcommand{\tp}{\thinspace .}
\newcommand{\Oof}[1]{\mathcal{O}(#1)}
\newcommand{\x}{\boldsymbol{x}}
\newcommand{\dfc}{\alpha} % diffusion coefficient
\newcommand{\Ix}{\mathcal{I}_x}
\newcommand{\Iy}{\mathcal{I}_y}
\newcommand{\If}{\mathcal{I}_s} % for FEM
\newcommand{\Ifd}{{I_d}} % for FEM
\newcommand{\basphi}{\varphi}
\newcommand{\baspsi}{\psi}
\newcommand{\refphi}{\tilde\basphi}
\newcommand{\xno}[1]{x_{#1}}
\newcommand{\dX}{\, \mathrm{d}X}
\newcommand{\dx}{\, \mathrm{d}x}
\newcommand{\ds}{\, \mathrm{d}s}
$$
Example on a continuation method
$$ -\nabla\cdot\left( ||\nabla u||^q\nabla u\right) = f, $$
Pseudo-plastic fluids may be \( q=-0.8 \), which is a difficult problem for
Picard/Newton iteration.
$$ \Lambda\in [0,1]:\quad q=-\Lambda 0.8 $$
$$
-\nabla\cdot\left( ||\nabla u||^{-\Lambda 0.8}\nabla u\right) = f$$
Start with \( \Lambda = 0 \), increase in steps to \( \Lambda =1 \), use
previous solution as initial guess for Newton or Picard
