Linearisation of a Quasilinear Elliptic PDE I have noticed that the word 'linearisation' can have different meanings in different places in the literature.
For example, if one has a second-order quasilinear elliptic PDE, what would be the linearisation of this PDE be and what would a solution be?  I assume it's not just discarding non-linear terms.
The equation (for $f$) is this one, where $g$ is a $3$-metric, $k$ denotes the corresponding second fundamental form.  $\phi$ is determined using another equation but for now, the form of $\phi$ could be assumed to obtain solutions.
$ \Bigg( g^{ij} - \frac{\phi^2 f^i f^j}{1 + \phi^2 |\nabla f|^2} \Bigg) \Bigg( \frac{\phi \nabla_{ij}f + \phi_i f_j + \phi_j f_i}{\sqrt{1 + \phi^2 |\nabla f|^2}} - k_{ij} \Bigg) =0,  $
$\phi_j$ denotes $\partial \phi/ \partial x_j$.  So essentially, what would one mean by a linearisation of this equation?
 A: A solution is a function which satisfies the PDE.
Given a solution $u_0$, the linearisation of aPDE about $u_0$ is obtained by substituting $u(x) = u_0(x)+\varepsilon v(x)$ then only keeping terms of order $\varepsilon$. This gives a linear PDE where the unknown function is now $v$. Note that "the linearisation of a PDE" doesn't make sense without specifying the solution about which you are linearising -  this is analogous to a Taylor series expansion which is always taken about a given point. Often, however, $u_0$ is obvious from context (it is usually an equilibrium solution for example) in which case authors will frequently just refer to the "linearisation about $u_0$" as simply "the linearisation".
Here is an example. Consider the equation $$(1+ \vert \nabla u \vert^2)\Delta u - \sum_{i,j=1}^n \frac{\partial u}{\partial x_i} \frac{\partial u}{\partial x_j} \frac{\partial^2u}{\partial x_ix_j}=0\qquad \text{in } B_1 \tag{$\ast$} $$ with $u=0$ on $\partial B_1$. (If you're curious this is the equation for prescribed zero mean curvature. I chose this because, to me, your equation looks of mean curvature type). An obvious solution to $(\ast)$ is $u_0(x) =0$ for all $x\in B_1$.
Now let us substitute $u(x) = u_0(x) + \varepsilon v(x)=\varepsilon v(x)$ into $(\ast)$. This gives $$0= \varepsilon(1+ \varepsilon^2\vert \nabla v \vert^2)\Delta v - \varepsilon^3\sum_{i,j=1}^n \frac{\partial v}{\partial x_i} \frac{\partial v}{\partial x_j} \frac{\partial^2v}{\partial x_ix_j}= \varepsilon\Delta v+ \varepsilon^3 \bigg (\vert \nabla v \vert^2\Delta v - \sum_{i,j=1}^n \frac{\partial v}{\partial x_i} \frac{\partial v}{\partial x_j} \frac{\partial^2v}{\partial x_ix_j} \bigg ). $$ From inspection, the order $\varepsilon$ equation is $$\Delta v=0 \qquad \text{in } B_1$$ with $v=0$ on $\partial B_1$. This is the linearisation of $(\ast)$  about the solution $u_0=0$. Tip: If it is not clear from inspection what the order $\varepsilon$ terms are then an effective strategy is to calculate $\frac{d}{d\varepsilon}\big \vert_{\varepsilon=0}$ for both sides.
