Need help - Robin condition for a 1d wave equation on the first quardant Given $\alpha \neq 0$ and
$$u_{tt}-c^2u_{xx} =0 \quad x,t>0,$$
$$u(x,0) = f(x), \quad x\geq 0,$$
$$u_t(x,0)=g(x), \quad x\geq 0,$$
$$u_x(0,t)+\alpha u(0,t)=0, \quad t\geq 0.$$
I know it’s a Robin 1d wave equation that’s homogenous and on the first quarter, and that the solution will involve probably the extension of $f$ and $g$ and give them some attribute that’ll help us extract $u$ from The general known solution.  beyond that i’m clueless - how do i solve this? What are the conditions for which $f$ and $g$ are real? Is the solution singular?
 A: *

*As OP correctly states that the general solution to the wave equation in 1+1D spacetime is a sum of a left- and a right-mover,
$$ u(x,t)= u_+(x+ct) + u_-(x-ct), \qquad x,t~>~0. \tag{A}$$


*From the initial conditions (IC) we get
$$ f(x)~=~u_+(x)+u_-(x),  \qquad x~\geq~0,\tag{B} $$
and
$$ g(x)~=~cu^{\prime}_+(x)-cu^{\prime}_-(x),  \qquad x~\geq~0.\tag{C} $$
Let $$G(x) ~:=~ \int_0^x g + c_1,  \qquad x~\geq~0,\tag{D}$$
be an antiderivative to $g$. It contains an integration constant $c_1$. Then we may write the IC (C) as
$$ G(x)~=~cu_+(x)-cu_-(x),  \qquad x~\geq~0. \tag{E}$$
Hence the chiral solutions are
$$ u_{\pm}(x)~\stackrel{(B)+(E)}{=}~\frac{cf(x)\pm G(x)}{2c},  \qquad x~\geq~0. \tag{F}$$
Note that the integration constant $c_1$ is irrelevant for the solution (A) in the spacetime triangle between the initial time $t=0$ and the diagonal $x=ct$.


*We still don't know the right-mover $u_-(x)$ for negative argument $x<0$. From the Robin boundary condition (RBC), we get
$$ u^{\prime}_+(ct) + u^{\prime}_-(-ct) + \alpha\left\{u_+(ct) + u_-(-ct)\right\}~=~0, \qquad t~\geq~0,\tag{G}$$
or
$$ e^{-\alpha x}\frac{d}{dx}\left(e^{\alpha x} u_+(x)\right) -e^{\alpha x}\frac{d}{dx}\left(e^{-\alpha x} u_-(-x)\right)~=~0, \qquad x~\geq~0.\tag{H}$$
We can then integrate the right-mover:
$$ u_-(-x)~=~ e^{\alpha x} \left\{\int_0^x e^{-2\alpha x}\frac{d}{dx}\left(e^{\alpha x} u_+(x)\right) + c_2 \right\}, \qquad x~\geq~0,\tag{I}$$
where the left-mover $u_+$ is known from eq. (F).


*Continuity of the solution (A) at the diagonal $x=ct$ leads to continuity of the right-mover $u_-(x)$ at $x=0$:
$$ \frac{cf(0) -c_1}{2c}~=~u_-(0) ~=~c_2. \tag{J}$$
