Need help - Robin condition for a 1d wave equation on the first quardant

Question

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?

Answer 1

1. 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, $$ u(x,t)= u_+(x+ct) + u_-(x-ct), \qquad x,t~>~0. \tag{A}$$

2. 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 $x=ct$.

3. 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).

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