How do you "linearize" a differential operator to get its symbol? Assuming I got some non-linear differential operator $D$, given by $(\partial_xu(x,t))^2-\partial_tu(x,t)=0$, what would its linearization (around a solution $\tilde u$) be? 
So far I thought of something like a Fréchet derivative, i.e. $Du = D\tilde u + A(u-\tilde u)+o(u-\tilde u)$, where $D\tilde u$ would vanish because $\tilde u$ is a solution. However, I don't see how this $A$ could be written as a "usual" DE so I can determine its symbol.
After some research, I guess this question could also be seen as a follow-up question to this comment.
 A: If you want to linearise a (partial) differential equation around a solution $\tilde{u}$, your goal is to investigate how the equation (or, to be more precise, its solutions) behave 'around' $\tilde{u}$. Therefore, the most straightforward thing to do is to substitute
\begin{equation}
 u = \tilde{u} + \epsilon v,
\end{equation}
where $\epsilon$ is some small number. We call the term $\epsilon v$ a perturbation of $\tilde{u}$.
When you perform this substitution in your PDE, you get
\begin{equation}
 (\partial_x \tilde{u})^2 + 2 \epsilon (\partial_x \tilde{u})(\partial_x v) + \epsilon^2 (\partial_x v)^2 - \partial_t \tilde{u} - \epsilon \partial_t v = 0,
\end{equation}
which we can rewrite as
\begin{equation}
 \left[(\partial_x \tilde{u})^2  - \partial_t \tilde{u}\right] + \epsilon\left[2 (\partial_x \tilde{u})(\partial_x v) -  \partial_t v\right] + \epsilon^2 (\partial_x v)^2 = 0
\end{equation}
Because $\tilde{u}$ is assumed to be a solution to the original PDE, the first term $\left[(\partial_x \tilde{u})^2  - \partial_t \tilde{u}\right]$ is zero by definition. Now, the 'linearisation' part of the procedure comes from the observation that if $\epsilon$ is very small (i.e. very close to zero), then $\epsilon^2$ is much smaller than $\epsilon$. Therefore, it seems a good approximation to neglect the term $\epsilon^2 (\partial_x v)^2$, because it is much smaller than the term $\epsilon\left[2 (\partial_x \tilde{u})(\partial_x v) -  \partial_t v\right]$. Actually, you can take $\epsilon$ as small as you want, or in other words, study the system as close to $\tilde{u}$ as you want. The 'linearisation' of the original PDE is therefore given by the PDE
\begin{equation}
2 (\partial_x \tilde{u})(\partial_x v) -  \partial_t v = 0.
\end{equation}
If we rewrite this in terms of a differential operator, we get
\begin{equation}
 A(\tilde{u}) v = 0,
\end{equation}
with the operator $A(\tilde{u})$ (which depends explicitly on $\tilde{u}$ !) given by
\begin{equation}
 A(\tilde{u}) = 2 (\partial_x \tilde{u})\,\partial_x - \partial_t.
\end{equation}
Note that $A$ is a linear operator, in the sense that it acts linearly on $v$. 
