Find a unique $\mathcal{C}^2$ solution of $u_{tt} - c^2 u_{xx} = 0$ in $(0, \infty) \times (0, \infty)$ with $u_x(0,t) + \alpha u(0,t) = u_2(t)$ 
Given $\alpha \in \Bbb R$, and $u_2 \in \mathcal C^2(0,\infty)$, show that there is a unique $\mathcal C^2$ solution $u(x,t)$ of the Cauchy problem
$$\left\{\begin{align*}u_{tt}(x,t) - c^2 u_{xx}(x,t) = 0 & \quad \text{in } \Omega := (0,\infty) \times (0, \infty) \\ u(x,0) = u_0(x) = 0 & \quad \text{in } [0,\infty)\\ u_t(x,0) = u_1(x) = 0 & \quad \text{in } 
[0,\infty) \\ u_x(0,t) + \alpha u(0,t) = u_2(t) & \quad \text{in }(0, \infty),\end{align*}\right.$$
where $u_2$ satisfies $$u_2(0) = u_2'(0) = u_2''(0) = 0$$

My work. My first instinct is to define
$$\Omega_r := \{(x,t) \in \Omega: x > ct\}$$ and $$\Omega_l := \{(x,t) \in \Omega: x < ct\}$$
We can solve the wave-equation in $\Omega_r$ using d'Alembert's formula, i.e.,
$$u(x,t) = \frac{1}{2}\left(u_0(x+ ct) + u_0(x- ct) \right) + \frac{1}{2c}\int_{x-ct}^{x+ct} u_1(\xi) \, d\xi$$
to get $u(x,t) \equiv 0$ in $\Omega_r$. By continuity, $u(x,t) \equiv 0$ on $\{x = ct\}$ also. How should I extend this solution to $\Omega_l$? One of the many troubles is that I'm not sure how to use the last boundary condition, namely $u_x(0,t) + \alpha u(0,t) = u_2(t)$ for $0 \le t < \infty$.
 A: I think the most general approach is to use the solution $u(x) = F(x - ct) + G(x + ct)$ for $C^2$ functions $F,G$ representing a wave traveling to the right and left, respectively. The condition $u(x,0) = 0$ for $x \geq 0$ then tells us $G(x) = -F(x)$ for $x \geq 0$, so $u(x) = F(x - ct) - F(x + ct)$ (since $x + ct \geq 0$ on the relevant domain). Now the condition $u_t(x,0) = 0$ for $x \geq 0$ corresponds to
$$ 0 = -cF'(x) - cF'(x) = -2cF'(x) \implies F'(x) = 0 $$
and so $F(x)$ is some constant value $C$ for $x \geq 0$. To see what $F(x)$ looks like for $x < 0$, we look at the condition
$$u_2(t) = u_x(0,t) + \alpha u(0,t) = F'(-ct) - F'(ct) + \alpha[F(- ct) - F(ct)] $$
$\implies u_2(t) = F'(-ct) + \alpha F(-ct) - \alpha C $ since $ct \geq 0$. Now we essentially need to solve the linear differential equation $F'(x) + \alpha F(x) = \alpha C + u_2(-x/c)$ on the domain $x \leq 0$, $F(0) = C$. This is given using Duhamel's principle by
$$ F(x) = C +\int_0^xu_2\left(-\frac{y}{c}\right)e^{\alpha(y-x)}dy;  $$
therefore all together we get
$$F(x) = \begin{cases}
0 & \quad \text{if } \ x \geq ct \\
\int_0^{x-ct}u_2\left(-\frac{y}{c}\right)e^{\alpha(y-x+ct)}dy & \quad \text{if } \ x < ct
\end{cases} $$
