Newton-Galerkin method, the most carefully guarded secret of finite element business. All examples I have seen so far feel like 'special tricks' and therefore I want to cover it once-and-for-all thoroughly, even just to remind myself.
Let us first consider searching an approximation to the zeros of a function $f : \mathbb{R} \rightarrow \mathbb{R}$ (i.e. standard Newton's method) because this is very analogous to what we do when solving zeros of functionals. Thus, let us solve
$$f(x)=0.$$ Suppose that we have a sufficiently good approximation $x_0$ to $x$. Then we attempt to compute the correct perturbation $\Delta x$ satisfying
$$x = x_0 + \Delta x.$$ Then using Taylor's expansion we get
$$0 = f(x) = f(x_0 + \Delta x) = f(x_0)+f'(x_0) \Delta x + \mathcal{O}(\Delta x^2).$$ Neglecting the higher order terms and solving for $\Delta x$ gives the approximate equation
$$\Delta x \approx - \frac{f(x_0)}{f'(x_0)}.$$ Then we have a new approximation to $x$:
$$x \approx x_0 - \frac{f(x_0)}{f'(x_0)},$$ which you might remember from high school (back then we used some stupid geometrical argument).
Similarly, let us consider a non-linear PDE for which the weak formulation is: find $u \in V$ such that
$$a(u;v)=l(v),\quad \forall v \in V,$$ where $a(u;v)$ is linear in $v$, non-linear in $u$, and $l(v)$ is linear. Here $V$ is some function space with the proper boundary conditions
Example. A weak form like this could, for example, arise from the strong formulation
$$-\nabla \cdot (u^3 \nabla u) = f,\quad \text{in $\Omega$,}$$ with the boundary condition $u|_{\partial \Omega}=0$. In this case
$$a(u;v)=\int_\Omega u^3 \nabla u \cdot \nabla v \,\mathrm{d}x$$ and the usual
$$l(v)=\int_\Omega f v\,\mathrm{d}x.$$
Anyways, we again assume that we have an approximation $u_0$ to $u$ and search for the correction $\delta u$ satisfying
$$u = u_0 + \delta u.$$ Then a similar Taylor's expansion as before gives
$$0 = a(u_0+\delta u;v) - l(v) = a(u_0;v)+D_u a(u_0;v) \delta u + \mathcal{O}(\delta u^2) - l(v),$$ where $D_u a(u_0;v) \delta u$ is the derivative of $a$ with respect to the first argument into the direction $\delta u$ evaluated at $(u_0,v)$. Rearranging and neglecting the HOT's gives
$$D_u a(u_0;v) \delta u \approx l(v)-a(u_0;v).$$ Note that the derivative $D_u a(u_0;v)$ is a linear operator acting on $\delta u$ and therefore in general we must discretize the previous equation before solving for $\delta u$. However, first we should compute an explicit representation for $D_u a(u_0;v) \delta u$ and this depends on how $a$ is defined. An example follows.
Example. Suppose
$$a(u;v)=\int_\Omega u^3 \nabla u \cdot \nabla v\,\mathrm{d}x$$ as in the previous example. Then by the definition of the directional derivative
\begin{align*}
D_u a(u_0;v) \delta u &= \frac{\partial a(u_0+t \delta u;v)}{\partial t}\Big|_{t=0}\\
&= \frac{\partial}{\partial t} \int_\Omega (u_0 + t \delta u)^3 \nabla(u+t \delta u) \cdot \nabla v\,\mathrm{d}x \Big|_{t=0} \\
&= \int_\Omega 3 u_0^2 \delta u \nabla u_0 \cdot \nabla v\,\mathrm{d}x + \int_\Omega u_0^3 \nabla \delta u \cdot \nabla v\,\mathrm{d}x.
\end{align*}
Now one step of the Newton-Galerkin iteration reads: find $\delta u \in V$ such that
$$D_u a(u_0;v) \delta u = l(v)-a(u_0;v)$$ holds for every $v \in V$.
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