$c$ is constant, given
$u_{tt} = c^2u_{xx}, x > 0 , t >0 \\ u(x,0)=u_t(x,0)=0, \ for \ all \ x>0 \\ u(0,t)=f(t), t>0 \ f(0)=0$
I am asked to find the D'Alembert Solution for this problem. This is my attempt:
Let $ \xi = x+ct$ and $\eta = x-ct$. Then define $\Phi = \Phi(\xi,\eta) = u(x,t),$ where $x = \frac{\xi+\eta}{2}$ and $t = \frac{\xi-\eta}{2c}$.
I expressed the initial problem in terms of $\xi$ and $\eta$. (After a bunch of derivation)
$u_{\xi \eta}=0$
After integrating with respect to $\xi$ and $\eta$, I got:
$u(\xi ,\eta)= \alpha(\xi) + \beta(\eta)$.
Then using $u(x,0)=u_t(x,0)=0$,
I obtained $u(x,0)=\alpha(x) + \beta(x) =0$(1)
Using $u_t(x,0)=0$, I got $c\alpha'(x)-c\beta'(x)=0$, and so
$\alpha(x)-\beta(x)=\frac{K}{c}$ (2). Adding (1) and (2) side-by-side, I got $\alpha(x)=-\beta(x)=\frac{K}{2c}$.
However, I couldn't figure out how to move on. Can you help me with this? Thanks in advance.