$$
\newcommand{\uex}{{u_{\small\mbox{e}}}}
\newcommand{\half}{\frac{1}{2}}
\newcommand{\tp}{\thinspace .}
\newcommand{\Oof}[1]{\mathcal{O}(#1)}
\newcommand{\x}{\boldsymbol{x}}
\newcommand{\dfc}{\alpha} % diffusion coefficient
\newcommand{\Ix}{\mathcal{I}_x}
\newcommand{\Iy}{\mathcal{I}_y}
\newcommand{\If}{\mathcal{I}_s} % for FEM
\newcommand{\Ifd}{{I_d}} % for FEM
\newcommand{\basphi}{\varphi}
\newcommand{\baspsi}{\psi}
\newcommand{\refphi}{\tilde\basphi}
\newcommand{\xno}[1]{x_{#1}}
\newcommand{\dX}{\, \mathrm{d}X}
\newcommand{\dx}{\, \mathrm{d}x}
\newcommand{\ds}{\, \mathrm{d}s}
$$
Similarity with Picard iteration
Rewrite the PDE for \( \delta u \) using
\( u^{n,k} + \delta u =u^{n,k+1} \):
$$
\begin{align*}
& \frac{u^{n,k+1} - u^{n-1}}{\Delta t} =
\nabla\cdot (\dfc(u^{n,k})\nabla u^{n,k+1}) + f(u^{n,k})\\
&\qquad + \nabla\cdot (\dfc^{\prime}(u^{n,k})\delta u\nabla u^{n,k})
+ f^{\prime}(u^{n,k})\delta u
\end{align*}
$$
Note:
- The first line is the same PDE as arise in the Picard iteration
- The remaining terms arise from the differentiations in Newton's method