Solve $$\boxed{\frac{\partial^2 u}{\partial t^2}-c^2\frac{\partial^2u}{\partial x^2}=F(x,t),0<x<\infty} $$ with the boundary conditions $\frac{\partial u}{\partial x}=0$ at $x=0$ and $u\to 0$ as $x \to \infty$ and initial conditions $u(x,0)=0,\frac{\partial u(x,0)}{\partial t}=0$.
My attempt:
Applying a Green's function gives $$\frac{\partial^2 G}{\partial t^2}-c^2\frac{\partial^2 G}{\partial x^2}=\delta(x-\xi)(t-\tau)$$ and setting $x-\xi:=x, t-\tau:=t$ and taking Laplace transforms give $$s^2\overline{G}(s)-sG(0)-G'(0)-c^2\frac{\partial^2 \overline{G}}{\partial x^2}=\boxed{s^2\overline{G}(s)-c^2\frac{\partial^2 \overline{G}}{\partial x^2}=\delta(x)}$$ using the initial conditions.
For $x \neq 0 $ we have a standard ODE and I get $\overline{G}=Ae^{s/cx}+Be^{-s/cx} $. Since $u \to 0$, $A=0$ and $\boxed{\overline{G}=Be^{-s/cx}}$.
My problem is how do I find the constant $B$? For the 3D case, I have been shown to use the Gauss theorem to integrate a volume around $0$. I know the integral of the right hand side is 1 and I tried to integrate both sides to give $[s^2\overline{G}x]_0^\infty-c^2[\frac{\partial \overline{G}}{\partial x}]_0^\infty=1$ which substituting $Be^{-s/cx}$ into gave me $-s^2+s/cB=0$ but then $B$ depends on s.