2nd order partial differential equations I have a question regarding based on text regarding 2nd-order partial differential equations.
Consider the quasilinear 2nd-order partial differential equation: 
$$-\text{div}(a(x,y,\nabla u)) + c(x,u,\nabla u) = g.$$
I have some questions regarding a paragraph describing $a(x,u,\frac{}{})$. The paragraph is as follows:
$p'' \in (1, + \infty)$ will denote the growth of the leading nonlinearity $a(x,u,\frac{}{})$ which essentially determines the setting and the other data qualification. Also, $a(x,u,\frac{}{})$ will be assumed to behave monotonically which is related to the adjective 'elliptic'. For the linear case $a(x,r,s) = \mathcal{A}s $, the montonicity and coercivity below implies the matrix $\mathcal{A}$ is positive definite, which is conventionally called "elliptic", contrary to $\mathcal{A}$ indefinte which is addressed as hyperbolic(resp parabolic). 
Questions:
1.I first want to confirm 'nonlinearity' here simply means that the mapping $a: \Omega \times \mathbb{R} \times \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ is a nonlinear mapping?
2.What is $\mathcal{A}s$ explicitly?
3.Why is $a(x,r,s) = \mathcal{A}s$ only if $a(x,r,s)$ is linear?
4.How would elliplicity be defined for the nonlinear case?
Thanks a lot for assistance. 
 A: 
'nonlinearity' here simply means that the mapping $a:\Omega \times \mathbb R\times \mathbb R^n \to \mathbb  R^n$ is a nonlinear mapping?

Close, but not quite. For example, the equation $\operatorname{div}((1+x^2)\nabla u) =0$ is linear. What matters is how the PDE involves  the unknown $u$, the dependence on $x$ need not be linear. Precisely, we need 
$$
 a(x,\alpha u+\beta v,\nabla (\alpha u+\beta v)) = 
\alpha a(x,u,\nabla u) +\beta  a(x,v,\nabla v)\tag{1}
$$
to hold. And similarly for $c$. That is, the PDE is linear if and only if 


*

*$(r,s)\mapsto a(x,r,s)$ is a linear map for every fixed $x$.

*$(r,s)\mapsto c(x,r,s)$ is a linear map for every fixed $x$.


How $a$ and $c$ depend  on $x$ is not important for linearity of the PDE. However, if they are  independent of $x$ we get a nice linear equation with constant coefficients. 

What is $\mathbb As$ explicitly?

$\mathbb A$ is a constant matrix; $s$ is a vector. Thus, $\mathbb As$ is the result of applying the matrix to a vector. 

Why is $a(x,r,s)=\mathbb As$ only if $a(x,r,s)$ is linear?

Every linear map is represented by a matrix. A nonlinear map cannot be represented in this way.  

How would ellipticity be defined for the nonlinear case?

However the author of your book chooses to. Definitions vary a lot for nonlinear, possibly degenerate, PDE. An important property that $a$ should have is monotonicity: $\langle( a(x, s)-a(x,s') ,s-s'\rangle$ should be positive. This allows one to at least hope for   the uniqueness of the Dirichlet boundary problem, if the lower order terms cooperate. But more assumptions are needed to say something about the PDE, and what they will be  depends on what you want to do. 
A: I agree with Moses in that the term 'linear' doesn't seem standard even when used for PDEs. Let me give my definition based on Post No bulls answer and the book by Evans 'Partial Differential Equations'. Any comments and corrections would be welcome. 
This is the definition of a PDE given in Evans book, it is a function of the form: 
$F(D^{k}u(x),D^{k-1}u(x),...,Du(x),u(x),x) = 0$ 
where $F: \mathbb{R}^{n^{k}} \times \mathbb{R}^{n^{k-1}} \times ...\times \mathbb{R}^{n} \times U \rightarrow \mathbb{R}$ is given and $u: U \rightarrow \mathbb{R}$ is the unknown. 
Then in Evans book a PDE is linear if it has the form $\sum_{|\alpha| \leq k}a_{\alpha}(x)D^{\alpha}u = f(x)$ for given functions $a_{\alpha},f$. 
So by Post Bulls explanation and Evans book's definition I would say in conclusion that a linear PDE is one that it a linear transformation(satisfies additivity and homogeneity of degree 1) for all independent variables of $F$  other than $x$.
That seems to be the most general notion of linearity for PDEs, there are obviously weaker forms of linearity such as quasilinearity and semi-linearity as well.
I welcome any comments or criticisms regarding this.
