Wave equation solution properties $u\in C^2$ is a solution of the one-dimensional wave equation $u_{tt}=u_{xx}$ with initial values $u(0,x)=f(x)$ and $u_t(0,x)=g(x)$
Now I a little bit confused with the following:
I define a function $v\in C^2(\Omega)$ where $\Omega=\{(a,b)\in\mathbb R^2 | a+b>0\}$ as $v(a,b)=u(a+b,a-b)$
Question 1: I do not see the relation that $u$ is a solution iff $v$ satifies $v_{ab}=0$
My thought was the following: I define $t=a+b, x=a-b$ then $v_{ab}=u_{tx}(t,x)=0$ and from this fact it should somehow be followed that $u_{tt}=u_{xx}$
Question 2: How can it be followed from above that every solution has the form $u(t,x)=F(t-x)+G(t+x)$. What exactly is $F$ and $G$ ?
 A: This is based on D'Almbert's solution to the wave equation:
$$u(x,t) = F(x+t) + G(x -t)$$
It follows by substitution into the original equation. You may also consider the transformation $\alpha = x+t$, $\beta=x-t$, so that $x=(\alpha + \beta)/2$, $t=(\alpha-\beta)/2$.  Then let
$$ u\left (\frac{\alpha+ \beta}{2},  \frac{\alpha- \beta}{2}\right) = v(\alpha, \beta) $$
$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial \alpha} \frac{\partial \alpha}{\partial x} +\frac{\partial v}{\partial \beta} \frac{\partial \beta}{\partial x}  =  \frac{\partial v}{\partial \alpha} + \frac{\partial v}{\partial \beta}$$
$$\frac{\partial u}{\partial t} = \frac{\partial v}{\partial \alpha} \frac{\partial \alpha}{\partial t} +\frac{\partial v}{\partial \beta} \frac{\partial \beta}{\partial t} = \frac{\partial v}{\partial \alpha} - \frac{\partial v}{\partial \beta}$$
By differentiating again in this matter, the wave equation may then be expressed as
$$\frac{\partial^2 v}{\partial \alpha^2} + 2 \frac{\partial^2 v}{\partial \alpha \partial \beta} + \frac{\partial^2 v}{\partial \beta^2} = \frac{\partial^2 v}{\partial \alpha^2} - 2 \frac{\partial^2 v}{\partial \alpha \partial \beta} + \frac{\partial^2 v}{\partial \beta^2} $$
From which follows
$$  v_{\alpha \beta} = 0$$
