I wish to solve $u_{tt} = c^2u_{xx}$ for $0 < x < ∞, 0 ≤ t < ∞$
$u(x,0)=0$, $u_t(x,0)=V$,
$u_t(0,t)+au_x(0,t)=0$ where $V, a$ and $c$ are positive constants and $a>c$
So I know the general formula for the wave equation is $u(x,t)=\frac{1}{2}[\phi (x+ct) + \phi (x-ct)] + \frac{1}{2c}\int_{x-ct}^{x+ct} \psi(s)ds$, where $u(x,0)=\phi, u_t(x,0)=\psi$. Based on the info given, the formula simplifies to $$u(x,t) = \frac{1}{2c}\int_{x-ct}^{x+ct} \psi(s)ds \tag1 $$ Differentiating, we get $$\begin{align} u_x(x,t) &= \frac{1}{2c} (\psi(x+ct)-\psi(x-ct)) \\ u_t(x,t) &= \frac{1}{2c} (c\psi(x+ct)+c\psi(x-ct)) \end{align} \tag2$$ hence $$ au_x(0,t) + u_t(0,t) = \left(\frac12+\frac{a}{2c}\right) \psi(ct)+ \left(\frac12-\frac{a}{2c}\right)\psi(-ct) \tag3$$ Equating (3) to zero and using $\psi(ct)=V$ for all positive $t$, I get $\psi(-ct)=\frac{(1/2+a/2c)V}{a/2c-1/2}$ for all negative $t$. I must be missing something simple but but if this is correct, how do I substitute this back into Eqn1 to solve for u?