General solution to wave equation of half-line with nonhomogeneous Neumann boundary Problem
Use the general solution to solve the signalling problem with homogeneous wave equation on the half line, homogeneous  IC and nonhomogeneous Neumann boundary conditions. Where c>0 is a constant, and h is continuous function. The solution u should be continuous.
$u_{tt} - c^2 u_{xx} = 0, \quad\quad \quad \quad 0 < x < \infty,\quad t>0$
$u(x,0)=0, u_t (x,0)=0 \quad (For \quad All \quad 0 < x < \infty)$
$u_x(0,t)=h(t) \quad \quad \quad \quad \quad(For \quad All \quad t > 0)$
My attempt
General solution:
$ u(x,t) = F(x-ct) + G(x+ct)$
Using B.C. :
$ u_x(0,t) = h(t) $
$F'(-ct) + G'(ct) = h(t)$
$F'(z) + G'(-z) = h(-z/c)$
$F(z) - G(-z) = \int_{0}^{z} h(-s/c)ds$
$F(x-ct) - G(ct-x) = \int_{0}^{x-ct} h(-s/c)ds$
$F(x-ct) = \int_{0}^{x-ct} h(-s/c)ds + G(ct-x)$
SO:
$u(x,t) = \int_{0}^{x-ct} h(-s/c)ds + G(ct-x) + G(x+ct)$
But we know that $G(ct-x), G(x+ct) =0$ from IC - I think.
So we get:
$u(x,t) = \int_{0}^{x-ct} h(-s/c)ds$
This solution satisfy BC: $ u_x(0,t) = h(t) $
$u_x(x,t) = h(\frac{ct-x}{c})$
$u_x(0,t) = h(t)$ <- Yes?
But it does not satisfy the IC?
Is the answer u(x,t) = 0?
Been at this some time - and the folks over at free math help forum weren't able to assist to get a final answer. Thought I would have better luck here.
 A: First of all, you should distinguish the spacetime regions $x\le ct$ (boundary layer) and $x>ct$. When $x> ct$, the influence of the boundary is not felt, and the solution is zero due to the initial condition being zero. From now on assume $x\le ct$. 
The formula 
$$u(x,t) = \int_{0}^{x-ct} h(-s/c)ds + G(ct-x) + G(x+ct)$$
is correct, although the integral is awkwardly written: the upper limit is negative. I would introduce $H(t)=\int_0^t h(s)\,ds$ and write 
$$
\int_{0}^{x-ct} h(-s/c)ds = -c \int_0^{t-x/c} h(\xi)\,d\xi = -c H(t-x/c)
$$
So far,
$$u(x,t) = -c H(t-x/c) + G(ct-x) + G(x+ct)\tag{1}$$

But we know that $G(ct-x), G(x+ct) =0$ from IC

Yes. Note that plugging $t=0$ here is problematic because we work in the regime $x\le ct$. It's better to plug $x=ct$, where (by continuity) $u=0$. So, 
$$ 0  =  G(0) + G(2x)$$
hence $G$ is identically zero. 
The final answer: 
$$
u(x,t) = \begin{cases} -cH(t-x/c),\quad &0\le x\le ct; \\ 0,\quad & x>ct\end{cases} 
$$

Verification: in the boundary layer $x< ct$ we have $u_x(x,t) = h(t-x/c)$, which at $x=0$ gives $h(t)$ as required. The initial conditions are also satisfied, and the function is continuous. 
