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.
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.