I'm just starting my first PDE class and ran into this small example that I am a little confused about.. I was hoping someone could explain it to me. Please note this example takes place before the book define the d'Alembert equation so I don't think I can use that. Here is a screen shot of the example.
INITIAL VALUE PROBLEM
The initial-value problem is to solve the wave equation $$\boxed{u_{tt} = c^2u_{xx}\quad\text{for } -\infty< x<+\infty}\tag 1$$ with the initial conditions $$\boxed{u(x,0) = \phi(x) \quad u_t(x,0) = \psi(x),}\tag 5$$ where $\phi$ and $\psi$ are arbitrary functions of $x$. There is one, and only one, solution of this problem. For instance, if $\phi(x) = \sin x$ and $\psi(x) = 0$, then $u(x,t) = \sin t\cos ct$.
Image.
The general solution of $\quad u_{tt}=c^2u_{xx}\quad $ is $\quad u(x,t)=f(x+ct)+g(x-ct)$
where $f$ and $g$ are any differentiable functions.
$u_t=cf'(x+ct)-cf'(x-ct)$
Conditions : $\begin{cases} u(x,0)=\phi(x)=f(x)+g(x) \\ u_t(x,0)=\psi(x)=cf'(x)-cg'(x) \end{cases}$
$\begin{cases} f'(x)+g'(x)=\phi'(x) \\ f'(x)-g'(x)=\frac{1}{c}\psi(x) \end{cases} \quad\to\quad \begin{cases} f'(x)=\frac{1}{2}\phi'(x)+\frac{1}{2c}\psi(x)\\ g'(x)=\frac{1}{2}\phi'(x)-\frac{1}{2c}\psi(x) \end{cases}$
$\begin{cases} f(x)=\frac{1}{2}\phi(x)+\frac{1}{2c}\int_0^x\psi(X)dX\\ g(x)=\frac{1}{2}\phi(x)-\frac{1}{2c}\int_0^x\psi(X)dX \end{cases}\qquad$ at $\quad t=0$
$\begin{cases} f(x+ct)=\frac{1}{2}\phi(x+ct)+\frac{1}{2c}\int_0^{x+ct}\psi(X)dX\\ g(x-ct)=\frac{1}{2}\phi(x-ct)-\frac{1}{2c}\int_0^{x-ct}\psi(X)dX \end{cases}$
$u(x,t)=\frac{1}{2}\left(\phi(x+ct)+\phi(x-ct)\right)+\frac{1}{2c}\left(\int_0^{x+ct}\psi(X)dX-\int_0^{x-ct}\psi(X)dX \right)$
$$u(x,t)=\frac{1}{2}\left(\phi(x+ct)+\phi(x-ct)\right)+\frac{1}{2c}\int_{x-ct}^{x+ct}\psi(X)dX $$