Existence/uniqueness of wave equation solution over parametrized domain The problem statement:

For which values of $\alpha  \ne  \pm 1$ there exists a unique solution of the following initial-boundary value problem:
  $$\left\{ \matrix{
  {u_{tt}} = {u_{xx}}{\rm{               }}    \quad\quad \alpha t < x < \infty {\rm{  }}\quad,\quad{\rm{  }}t > 0 \hfill \cr 
  u\left( {x,0} \right) = 0\quad\quad{\rm{             }}{u_t}\left( {x,0} \right) = 1 \quad{\rm{        for     }}\quad0 < x < \infty \hfill \cr 
  u\left( {\alpha t,t} \right) = 0 \hfill \cr}  \right.$$
  Solve the problem for those values.

What I've tried:
I'll try to prove there's a unique solution in the case of $\alpha  < 1$
Since it's the wave equation the solution should be of the form
$$u\left( {x,t} \right) = \psi \left( {x + t} \right) + \varphi \left( {x - t} \right)$$
Next we partition our region into two parts:


Part I:$\quad x > t$
From ${u_t}\left( {x,0} \right) = 1$ we have (after integration) $\psi \left( x \right) = \varphi \left( x \right) + x + c\quad$ for $x>0$
and from $u\left( {x,0} \right) = 0$   we have  ${\psi \left( x \right) + \varphi \left( x \right) = 0}\quad$ for $x>0$.
Together, this amounts to 
$$\eqalign{
  & \varphi \left( x \right) =  - {1 \over 2}\left( {x + c} \right) =  - {1 \over 2}x - {1 \over 2}c  \cr 
  & \psi \left( x \right) = {1 \over 2}\left( {x + c} \right) = {1 \over 2}x + {1 \over 2}c  \cr 
  & u\left( {x,t} \right) = {1 \over 2}\left( {x + t} \right) - {1 \over 2}\left( {x - t} \right) = t \quad  \cr} $$

Part II:$\quad \alpha t < x < t$
$$\eqalign{
  & u\left( {x,t} \right) = \varphi \left( {x - t} \right) + \psi \left( {x + t} \right)  \cr 
  &   \cr 
  & 0 = \varphi \left( {\left[ {\alpha  - 1} \right]t} \right) + \psi \left( {\left[ {\alpha  + 1} \right]t} \right){\rm{ }}  \cr 
  & \varphi \left( {\left[ {\alpha  - 1} \right]t} \right) =  - \psi \left( {\left[ {\alpha  + 1} \right]t} \right)  \cr 
  & s = \left[ {\alpha  - 1} \right]t < 0  \cr 
  & t = {s \over {\alpha  - 1}} > 0  \cr 
  & \varphi \left( s \right) =  - \psi \left( {\underbrace {{{\alpha  + 1} \over {\alpha  - 1}}s}_{ > 0}} \right) =  - {1 \over 2}{{\alpha  + 1} \over {\alpha  - 1}}s - {c \over 2}  \cr 
  & u\left( {x,t} \right) = \varphi \left( {x - t} \right) + \psi \left( {x + t} \right)  \cr 
  & u\left( {x,t} \right) =  - {1 \over 2}{{\alpha  + 1} \over {\alpha  - 1}}\left( {x - t} \right) + {1 \over 2}\left( {x + t} \right) \cr} $$

In conclusion:
$$u\left( {x,t} \right) = \left\{ \matrix{
  {\rm{          }}t{\rm{                \quad  \quad     \quad  \quad          \quad  \quad                 }}x > t \hfill \cr 
   - {1 \over 2}{{\alpha  + 1} \over {\alpha  - 1}}\left( {x - t} \right) + {1 \over 2}\left( {x + t} \right){\rm{  \quad         }}\alpha t \le x < t \hfill \cr}  \right.$$
is a solution that satisfies the PDE, the initial conditions and the boundry condition.
 A: Let us introduce $p = u_t + u_x$ and $q = u_t - u_x$ (see e.g. (1) chap. 12-*). From the PDE, we deduce $p_t = p_x$ and $q_t = -q_x$, i.e. two linear advection equations with speed $\mp 1$ are obtained. The characteristic curves along which $p$, $q$ are transported are straight lines with slope $\mp 1$ in the $x$-$t$ plane. Thus, the solution deduced from the method of characteristics is
$p(x,t) = 1$ and $q(x,t) = 1$ for $x > t$. Integrating $u_t = \frac12(p+q)$ in time, we obtain
$$
u(x,t) = t
\qquad\text{for}\qquad
x > t \qquad\text{(I.)}
$$
We may distinguish several cases:


*

*If $\alpha\geq 1$, then we have simultaneously $u(\alpha t, t) = 0$ due to the boundary conditions and $u(\alpha t, t) = t$ due to the above result. This is impossible.

*If $-1<\alpha<1$, then by following the characteristics, we still have $p(x,t) = 1$ for $\alpha t \leq x \leq t$.
Upon differentiation w.r.t. time of $u$ along the boundary $x = \alpha t$, we have
\begin{aligned}
\frac{\text d}{\text d t} u(\alpha t, t) = 0 &= \alpha u_x(\alpha t, t) + u_t(\alpha t, t) \\
&= \tfrac{1+\alpha}2 p(\alpha t, t) + \tfrac{1-\alpha}2 q(\alpha t, t)  .
\end{aligned}
From the value of $p$, we deduce that $q(x,t) =\frac{\alpha+1}{\alpha-1}$ along the boundary $x = \alpha t$, and also for $\alpha t\leq x\leq t$ by following the characteristics. Integrating $u_x = \frac12(p-q)$ in space, we have
$$
u(x,t) = \frac{x - \alpha t}{1-\alpha}
\qquad\text{for}\qquad
\alpha t \leq x \leq t \qquad\text{(II.)}
$$
which is exactly the same solution as the one proposed in OP.

*If $\alpha \leq -1$, then the boundary $x=\alpha t$ cannot be reached by following the characteristic lines. The solution in II. cannot be found.

(1) R. Habermann, Applied Partial Differential Equations: with Fourier Series and Boundary Value Problems, 5th ed. Pearson Education Inc., 2013.
