Partial derivatives-Why does this stand? In my notes there is the following:
$$u_{\xi \eta}=0 \Rightarrow \left\{\begin{matrix}
u_{\xi}=0 \Rightarrow u=g(\eta)\\ 
u_{\eta}=0 \Rightarrow u=f(\xi)
\end{matrix}\right.$$
I haven't understood why this stand... 
Isn't it as followed??
$$u_{\xi \eta}=0 \Rightarrow u_{\xi}=F(\xi) \text{ AND } u_{\xi \eta}=0 \Rightarrow u_{\eta}=G(\eta)$$
Why should $F(\xi)$ and $G(\eta)$ be equal to $0$??

EDIT:
I found this at the proof of the solution of the wave equation using the characteristics, which is the following:
$$u_{tt}-c^2u_{xx}=0$$
$\displaystyle{\frac{dx}{dt}=\frac{1}{a}(b \pm \sqrt{\Delta})=\pm c}$
$\left.\begin{matrix}
\xi=x-ct\\ 
\eta=x+ct
\end{matrix}\right\}$
$\displaystyle{\partial_x=\partial_{\xi}+\partial_{\eta}}$
$\displaystyle{\partial_t=-c \partial_{\xi}+c \partial_{\eta}}$
$\displaystyle{\partial_{xx}=\partial_{\xi \xi}+\partial_{\eta \eta}+2 \partial_{\xi \eta}}$
$\displaystyle{\partial_{tt}=c^2(\partial_{\xi \xi}+\partial_{\eta \eta}+2 \partial_{\xi \eta})}$
So $\displaystyle{(\partial_{tt}-c^2 \partial_{xx})u=0 \Rightarrow -4c^2 \partial_{\xi \eta}u=0 \Rightarrow}$
$$u_{\xi \eta}=0 \Rightarrow \left\{\begin{matrix}
u_{\xi}=0 \Rightarrow u=g(\eta)\\ 
u_{\eta}=0 \Rightarrow u=f(\xi)
\end{matrix}\right.$$
$$\Rightarrow u=f(\xi)+g(\eta) \Rightarrow u=f(x-ct)+g(x+ct)$$

You are correct. $u_\xi=F(\xi),\ u_\eta = G(\eta)$. Note that this implies (by integrating) that $u(\xi,\eta) = \widetilde F(\xi) + \widetilde G(\eta)$. You can verify this by trying something simple like $u(\xi,\eta) =\xi+\eta$
