Robin BC in the 1D wave equation

Let us simplify a bit the provided solution and extend the domain of integration from $(0,x)$ to $(-\infty,x)$ which is more appropriate. The final solution to the ODE induced by the boundary condition is thus $$g(\xi)=f(\xi)+\mathrm{e}^{-\alpha\xi}\Bigl(A-2\alpha\int_{-\infty}^{\xi} \mathrm{e}^{\alpha s}f(s)\mathrm{d}s\Bigr)$$ where $A$ is the constant of integration, which leads to $$u(x,t)=f(ct+x)+f(ct-x)+\mathrm{e}^{-\alpha(ct-x)}\Bigl(A-2\alpha\int_{-\infty}^{ct-x} \mathrm{e}^{\alpha s}f(s)\mathrm{d}s\Bigr)$$ We notice that for a vanishing incident wave $f=0$, the term $A\mathrm{e}^{-\alpha(ct-x)}$ still participates in the solution. Uniqueness of $u$ is not guaranteed. This invites us to have a look at the original problem when initial conditions are considered. The problem is now: $$ \begin{aligned} &\partial_2^2u(x,t)-c^2\partial_1^2u(x,t)=0\\ &\partial_1u(0,t)=\alpha u(0,t)\\ &u(x,0)=u_0(x)\quad\text{and}\quad \partial_2u(x,0)=v_0(x) \end{aligned} $$ where $u_0(x)$ and $v_0(x)$ are given. Basic developments [2] show that $$ \begin{aligned} 2f(x)&=u_0(x)+\frac{1}{c}\int_{-\infty}^x v_0(s)\mathrm{d}s-B\\ 2g(x)&=u_0(-x)-\frac{1}{c}\int_{-\infty}^{-x} v_0(s)\mathrm{d}s+B\\ \end{aligned} $$ where $B$ is a constant. Vanishing initial conditions, ie $u_0(x)=v_0(x)=0$, imply that $g$ is the constant function which in turn implies $A=0$. Since $A$ does not depend on the initial conditions, it always vanishes when the problem is read as an Initial Value Problem. Uniqueness of $u$ is now guaranteed.

Let us summarize:

• The term $A\mathrm{e}^{-\alpha(ct-x)}$ is the homogeneous solution to the Robin boundary condition of the 1D wave problem.
• This term vanishes ($A=0$) when the problem is read as an Initial Value Problem.
• Other boundary conditions (if the domain of interest has finite length) will most likely imply $A=0$.
• In the method of images, it is fair to assume that $A=0$ since $A\mathrm{e}^{-\alpha(ct-x)}$ is, in a way, a mathematical artifact, as explained above.
• In other words, this term is not observable.