Problem: Find a solution to $$v_{tt}-c^2v_{xx} = f(x,t)$$ $$v(x,0)=\phi(x)$$ $$v_t(x,0)=\psi(x)$$ $$v(0,t) = h(t)$$
on $0<x<\infty$
I know that the general solution to the wave function in the half-line with a source and non-homogeneous boundary conditions at $x=0$ is
$$v(x,t) = \frac{1}{2}[\phi(x+ct)+\phi(x-ct)] + \frac{1}{2c} \int_{x-ct}^{x+ct} \psi(s) \mathrm{d}s + \frac{1}{2c} \int \int_\Delta f(y,s) \mathrm{d}y \mathrm{d}s$$
for $x > ct$, and
$$v(x,t) = \frac{1}{2}[\phi(x+ct)+\phi(x-ct)] + \frac{1}{2c} \int_{ct-x}^{x+ct} \psi(s) \mathrm{d}s + h(t-x/c) + \frac{1}{2c} \int \int_D f(y,s) \mathrm{d}y \mathrm{d}s$$
for $x < ct$
(Correct me if I'm wrong though, please)
I know that the double integral over $\Delta$ is $\int_o^t \int_{x-c(t-s)}^{x+c(t-s)}$
And I know that the integral over $D$ is similar but it omits a chunk due to the boundary. I can't find the limits of this integral though, what would they be?
On this question: Wave with a source on the half-line
they state the area as $D = \{(y,s) : |x-c(t-s)| \leq y \leq x+c(t-s)\}$
I know that the right hand side inequality has to be the same as that for $\Delta$, but the absolute value on the left hand side confuses me.
Thank you