Solution of the Wave Equation, the not so simple direction I read that the solution of the one-dimensional wave equation
$$
 \frac{1}{c^2} \frac{\partial^2 u}{\partial t^2} = \frac{\partial^2 u}{\partial x^2}
$$
are exactly the function of the form
$$
 u(t,x) = f(x + ct) + g(x - ct)
$$
with $f$ and $g$ being twice differentiable. To show that every such function is a solution is simply, but what about the other direction, how can I show that if I have a solution $u$ then it must have the form as a sum of two functions f and g? In my textbooks I just find the simple direction, but not the other?
 A: Change independent variables from $(x,t)$ to $\xi=x+ct$ and $\eta=x-ct$. You will find that the equation transforms to $$\frac{\partial^2 u}{\partial\xi\partial\eta}=0.$$
Now integrate out the $\xi$ variable: $$\frac{\partial u}{\partial\eta}=\gamma(\eta)$$
for some function $\gamma$. ($\gamma(\eta)$ is the integration “constant”). Integrate with respect to $\eta$ and get $u=f(\xi)+g(\eta)$, where $g'=\gamma$ and $f(\xi)$ is another integration “constant”.
Edit: It appears that a more detailed argument is requested. In the first part, fix $\eta$ and consider the function $\xi\mapsto u_\eta(\xi,\eta)$. (I use subscript notation for partial derivatives here, and abuse the notation by still using the letter $u$ for a function of $\xi$ and $\eta$.) The equation states that the derivative of this function is zero, so it must be constant. But you might get a different constant for each choice of $\eta$, so the “constant” is really a function of $\eta$. I called it $\gamma(\eta)$.
Next, fix $\xi$ and consider the map $\eta\mapsto u(\xi,\eta)$ the equation we just proved says that the derivative of this is $\gamma(\eta)$. So the function itself is $g(\eta)$ plus some constant, where $g'=\gamma$. Again, the “constant” may depend on $\xi$, so we call it $f(\xi)$.
A: This is called d'Alembert's formula. Basically you transform the wave equation to $u_{\mu\eta}=0$ and solve this one instead.
If additionally initial conditions $u(x,0) = u_0(x)$ and $u_t(x,0) = u_1(x)$ are specified, one has
$$f(y) = \frac{1}{2}u_0(y) + \frac{1}{2c}\int_0^{y}u_1(\xi)\,d\xi\quad \text{and} \quad g(y)= \frac{1}{2}u_0(y) - \frac{1}{2c}\int_0^y u_1(\xi)\,d\xi.$$
