Proving that a given $u$ is a solution to $u_{tt} = c^2 u_{xx}.$
The question asks me to prove that $$u(x,t) = F(ct + x) + G(ct -x),$$ is a solution of the above equation on condition that we know that $F$ and $G$ are twice differentiable functions on $[0,L]$
My attempt:
I calculated $u_{tt} = c^2 (\partial^2F/\partial t^2 + \partial^2G/\partial t^2 )$, and $u_{xx} = (\partial^2F/\partial x^2 + \partial^2G/\partial x^2 )$, but I do not know what next should I do, the problem is that I have two different differentiations one with respect to $t$ and the other with respect to $x$ and I do not know how they can be cancelled with each other to satisfy the above equation, could anyone help me in this please?
Let $p = ct + x$ and $q = ct-x$. Then $F(p)$ and $G(q)$ are functions of a single variable.
By the multivariable chain rule we have
\begin{align} \frac{\partial F}{\partial x} &= \frac{dF}{dp}\frac{\partial p}{\partial x} = \frac{dF}{dp} \\ \frac{\partial F}{\partial t} &= \frac{dF}{dp}\frac{\partial p}{\partial t} = c\frac{dF}{dp} \end{align}
Do this a second time
\begin{align} \frac{\partial^2 F}{\partial x^2} &= \frac{d^2F}{dp^2} \\ \frac{\partial^2 F}{\partial t^2} &= c^2\frac{d^2F}{dp^2} \end{align}
which means $F$ satisfies the original PDE. The same logic goes for $G$.
By linearity, $u=F+G$ is also a solution.