1D wave equation $u_{t t} - c^2 u_{x x} = 0$ with forcing at one point I want to solve the 1-dimensional wave equation
$$u_{t t} - c^2 u_{x x} = 0$$
for $u(x,t)$ in the domain $-\infty \leq x \leq \infty, t\geq 0$ with initial data
$$ u(x,0) = g(x) \hspace{4mm} , \hspace{4mm} u_t(x,0) = 0$$
and forcing at location $x=0$ by a prescribed value of
$$u(0,t) = h(t) \; .$$
The condition
$$ h(0) = g(0) $$
is satisfied. The situation is shown in the following figure.

As is well known, in the case of zero forcing, the solution is
$$ u = \frac{1}{2} [g(x+c t) + g(x-c t) ] \; . $$
How do I deal with the forcing? Can you show me a suitable approach or recommend literature that explains this? Thank you.
 A: In the case of the semi-infinite string with fixed end. You have ( these are some notes)
$$ \begin{align}\begin{cases} \frac{\partial^{2} u}{\partial t^{2}}  = c^{2}\frac{\partial^{2} u}{\partial x^{2}} & x > 0  , t > 0  \\ u(x,0) = f(x)   & x \geq 0 \\ \frac{\partial u}{\partial x}(x,0) = g(x) & x \geq 0  \\ u(0,t) = h(t)  & t \geq 0 \end{cases} \end{align} \tag{1}$$
Then d'Alembert's Formula is 
$$u(x,t) = \phi(x+ct) +\psi(x-ct) \tag{2} $$
you plug in $x=0$ to get $h(t)$ 
$$ u(0,t) = h(t) = \phi(ct) +\psi(-ct) \tag{3}$$
you then introduce some constant $\alpha = -ct$ 
$$ \psi(\alpha) = h(\frac{-\alpha}{c}) - \phi(-\alpha) \tag{4} $$
you then replace $\alpha$ with $x-ct$
$$\psi(x-ct) = h(t-\frac{x}{c}) - \phi(ct-x) \tag{5} $$
then you get for $ 0 \leq x \leq ct$
$$ u(x,t) = h(t-\frac{x}{c})  + \frac{1}{2}\big[f(x+ct) -f(ct-x) \big] + \frac{1}{2c} \int_{ct-x}^{x+ct} g(\tau) d \tau \tag{6} $$
and for $ x > ct$ we have
$$ u(x,t) = \frac{1}{2}\big[f(x+ct) -f(x-ct) \big] + \frac{1}{2c} \int_{x-ct}^{x+ct} g(\tau) d \tau \tag{7} $$
now popping some stuff in..
$$ \begin{align}\begin{cases} \frac{\partial^{2} u}{\partial t^{2}}  = c^{2}\frac{\partial^{2} u}{\partial x^{2}} & x > 0  , t > 0  \\ u(x,0) = g(x)   & x \geq 0 \\ \frac{\partial u}{\partial x}(x,0) = 0 & x \geq 0  \\ u(0,t) = h(t)  & t \geq 0 \end{cases} \end{align} \tag{8}$$
we get the following answer
$$ u(x,t) = \begin{align}\begin{cases} h(t-\frac{x}{c})  + \frac{1}{2}\big[g(x+ct) -g(ct-x) \big]   &   0 \leq x \leq ct  \\ \frac{1}{2}\big[g(x+ct) -g(x-ct) \big]    & x  > ct \end{cases} \end{align} \tag{9}$$
you can enforce your last condition fairly easily because it is $u(0,0)$
A: I found a general solution in the book "Handbook of linear partial differential equations for engineers and scientists, 2nd edition", section 6.1.2, for the domain $0 \leq x < \infty$. In my case, the solution is
$$u(x,t) = \frac{1}{2} [g(x+c t) + g(x-c t)] + \operatorname{H}\left( t-\frac{x}{c} \right) h\left( t - \frac{x}{c} \right) $$
where $\operatorname{H}\left( \cdot \right)$ is the Heaviside step function.
