$$ \newcommand{\uex}{{u_{\small\mbox{e}}}} \newcommand{\half}{\frac{1}{2}} \newcommand{\tp}{\thinspace .} \newcommand{\Oof}[1]{\mathcal{O}(#1)} \newcommand{\x}{\boldsymbol{x}} \newcommand{\dfc}{\alpha} % diffusion coefficient \newcommand{\Ix}{\mathcal{I}_x} \newcommand{\Iy}{\mathcal{I}_y} \newcommand{\If}{\mathcal{I}_s} % for FEM \newcommand{\Ifd}{{I_d}} % for FEM \newcommand{\basphi}{\varphi} \newcommand{\baspsi}{\psi} \newcommand{\refphi}{\tilde\basphi} \newcommand{\xno}[1]{x_{#1}} \newcommand{\dX}{\, \mathrm{d}X} \newcommand{\dx}{\, \mathrm{d}x} \newcommand{\ds}{\, \mathrm{d}s} $$
Algorithm for Newton's method
$$ \underline{F(u^{-})}_{\mbox{vector}} + \underline{J(u^{-})}_{\mbox{matrix}} \cdot \underline{\delta u}_{\mbox{vector}} =0$$
Solution by a two-step procedure:
solve linear system \( J(u^{-})\delta u = -F(u^{-}) \) wrt \( \delta u \)
update \( u = u^{-} + \delta u \)
Relaxed update: $$ u = \omega(u^{-} +\delta u) + (1-\omega)u^{-} = u^{-} + \omega\delta u $$
