PDE classification That's a simple question, but I'm confused.
What form must $G$ have for the differential equation $u_{tt}-u_{xx}=G(x,t,u)$ to be linear? Linear and homogeneous?
For the first question, I thought about $G(x,t,u)=a(x,t)u$, but I confused if $G(x,t,u)=a(x,t)u+b(x,t)$ is actually the right answer.
For the second question, I thought about $G(x,t,u)=a(x,t)u$, because it implies that $u_{tt}-u_{xx}-a(x,t)u=0. $
Am I right?
 A: I could simply say that your classification of the equation 
$$
u_{tt}-u_{xx}=G(x,t,u)\tag{1}\label{1}
$$
is the right one according to the customary lexicon, but your questions since are non trivial ones, deserve a comprehensive non trivial answer, which I postpone after the following definition: 
Definition 1. Let $V_1,V_2$ be two $\mathbb{K}$-vector spaces (for the sake of simplicity we can implicitly assume that $\mathbb{K}=\mathbb{C}$ or $\mathbb{R}$) and $\mathfrak{L}:V_1\to V_2$ a function between them: $\mathfrak{L}$ is said to be linear if and only if
$$
\mathfrak{L}(a\mathbf{u}+b\mathbf{v})=a\mathfrak{L}(\mathbf{u})+b\mathfrak{L}(\mathbf{v})\quad \forall \mathbf{u},\mathbf{v}\in V_1\text{ and }\forall a,b\in\mathbb{K}\tag{2}\label{2}
$$
Loosely speaking, engineers and physicists say that $\mathfrak{L}$ is linear if and only if it satisfies the superimposition of the effects, which is precisely what is stated by equation \eqref{2}. Said that, we can proceed to classify the PDE above and say that


*

*Strictly speaking, the right answer to the first question would be $G(x,t,u)=a(x,t)u$: this is the only case for which the "weighted sum" (with weights in $\mathbb{K}$) of two solutions to the original equation is again a solution. This point of view is for example the one of Enzo Tonti ([1], p. 1945), who defines as "affine operators" the ones for which $G(x,t,u)=a(x,t)u+b(x,t)$, and clearly states that they are nonlinear, implicitly not satisfying \eqref{2}. However, when the operator involved in equation \eqref{1} is linear according to \eqref{2}, except for the presence of a term which does not depend on its unknown solution, it is customary to speak of linear problem or linear equation, as it is done in linear algebra. In sum, we customarily say that equation \eqref{1} is linear if 
$$
G(x,t,u)=a(x,t)u+b(x,t)
$$

*According to the customary convention explained in the first part of the answer, we say that equation \eqref{1} is linear homogeneous if 
$$
G(x,t,u)=a(x,t)u
$$
[1] Enzo Tonti (1984), "Variational Formulation for Every Nonlinear Problem", International Journal of Engineering Science, Vol. 22, No. 11/12, pp. 1343-1371, Zbl 0558.49022.
