Wave equation 1D inhomogeneous Laplace/Fourier Transforms vs Green's Function I am trying to solve the following 1D inhomogeneous wave equation. 
Forgive me if I a miss any rigorous mathematical concept.
$$ \frac{\partial^2 u}{\partial x^2} - \frac{1}{c^2}\frac{\partial^2 u}{\partial t^2} = \delta(x)f(t) $$
where $c$ is the wave velocity, $f(t)$ any force function and $\delta (x)$ is the dirac delta function. I don't know if I am doing right (and not how to apply also it) but I want that the energy vanishes at infinity $x^+_-$. The $\delta(x)$ is used to simulate a point source at $x=0$. 
I used the Fourier Transform $\mathcal{F}$ first to get to:
$$  \frac{\partial^2 U_{\omega}(\omega,x)}{\partial x^2} + \frac{\omega^2}{c^2} U_{\omega}(\omega,x) =  \delta(x)F(\omega)$$
$$  \frac{\partial^2 U_{\omega}(\omega,x)}{\partial x^2} + k^2 U_{\omega}(\omega,x) =  \delta(x)F(\omega)$$ (A)
with $ k^2 = \frac{\omega^2}{c^2} $ and the subscript $ U_{\omega}(\omega,x) = \mathcal{F_t}(u(t,x)) $ meaning Fourier Transformed from $t$.
That's the Helmholtz equation but let's continue applying now the Laplace $\mathcal{L}$ transform and applying the conditions $ U_{\omega}(\omega,0)  = 0 $ and $ \frac{dU_{\omega}(\omega,0)}{dx} = 0 $
$$  s^2U_{\omega s}(\omega,s) - sU_{\omega}(\omega,0)  - \frac{dU_{\omega}(\omega,0)}{dx}+ k^2 U_{\omega s}(\omega,s) = \mathcal{L}(\delta(x))F_{\omega} $$
$$  s^2U_{\omega s}(\omega,s) + k^2 U_{\omega s}(\omega,s) = F_{\omega} $$
$$  U_{\omega s}(\omega,s)\left(s^2 + k^2 \right) = F_{\omega} $$
$$  U_{\omega s}(\omega,s) = F_{\omega}\frac{1}{s^2 + k^2} $$
$$  U_{\omega}(\omega,x) = F_{\omega} \frac{\sin(kx)}{k} u(x) $$
with $u(x) $ as the Heaviside step function and $ U_{\omega s}(\omega,s) = \mathcal{L_x}(U_{\omega}(\omega,x)) $
the solution would than be 
$$ u(t,x) = \mathcal{F}^{-1}\{F_{\omega} \frac{\sin(kx)}{k} u(x)\} $$ (B)
If I use the Helmholtz approach from (A) with green's function I would get to :
$$ u(t,x) = \mathcal{F}^{-1}\{ G(x) \ast \left[ F_{\omega} \delta(x) \right] \} $$
$$ u(t,x) = \mathcal{F}^{-1}\{F_{\omega} \frac{j e^{jkx}}{2k}\} $$ (C)
Where $j = \sqrt{-1} $ imaginary unit and $ G(x) = \frac{j e^{jkx}}{2k} $ is the Green's function for the inhomogeneous Helmholtz equation.
Why the difference between (B) and (C), what am I missing?
Edit:
I used The boundary conditions are: 
$$ U_{\omega}(\omega,0) = \frac{dU_{\omega}(\omega,0)}{dx} = 0$$
I need the energy vanishes at infinity $x^+_-$ or $\lim_{|x| \to \infty} u(t,|x|) = 0 $ Don't know how to apply this though above.
 A: I think you think this question too complicated.
Similar to how to solve this PDE:
Let $\begin{cases}p=x+ct\\q=x-ct\end{cases}$ ,
Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial x}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial x}=\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}$
$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)=\dfrac{\partial}{\partial p}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)\dfrac{\partial p}{\partial x}+\dfrac{\partial}{\partial q}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)\dfrac{\partial q}{\partial x}=\dfrac{\partial^2u}{\partial p^2}+\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}=\dfrac{\partial^2u}{\partial p^2}+2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}$
$\dfrac{\partial u}{\partial t}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial t}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial t}=c\dfrac{\partial u}{\partial p}-c\dfrac{\partial u}{\partial q}$
$\dfrac{\partial^2u}{\partial t^2}=\dfrac{\partial}{\partial t}\left(c\dfrac{\partial u}{\partial p}-c\dfrac{\partial u}{\partial q}\right)=\dfrac{\partial}{\partial p}\left(c\dfrac{\partial u}{\partial p}-c\dfrac{\partial u}{\partial q}\right)\dfrac{\partial p}{\partial t}+\dfrac{\partial}{\partial q}\left(c\dfrac{\partial u}{\partial p}-c\dfrac{\partial u}{\partial q}\right)\dfrac{\partial q}{\partial t}=c^2\dfrac{\partial^2u}{\partial p^2}-c^2\dfrac{\partial^2u}{\partial pq}-c^2\dfrac{\partial^2u}{\partial pq}+c^2\dfrac{\partial^2u}{\partial q^2}=c^2\dfrac{\partial^2u}{\partial p^2}-2c^2\dfrac{\partial^2u}{\partial pq}+c^2\dfrac{\partial^2u}{\partial q^2}$
$\therefore\dfrac{\partial^2u}{\partial p^2}+2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}-\dfrac{1}{c^2}\left(c^2\dfrac{\partial^2u}{\partial p^2}-2c^2\dfrac{\partial^2u}{\partial pq}+c^2\dfrac{\partial^2u}{\partial q^2}\right)=\delta\left(\dfrac{p+q}{2}\right)f\left(\dfrac{p-q}{2c}\right)$
$4\dfrac{\partial^2u}{\partial pq}=\delta\left(\dfrac{p+q}{2}\right)f\left(\dfrac{p-q}{2c}\right)$
$\dfrac{\partial^2u}{\partial pq}=\dfrac{1}{4}\delta\left(\dfrac{p+q}{2}\right)f\left(\dfrac{p-q}{2c}\right)$
$u(p,q)=F(p)+G(q)+\dfrac{1}{4}\int_b^q\int_a^p\delta\left(\dfrac{r+s}{2}\right)f\left(\dfrac{r-s}{2c}\right)dr~ds$
$u(x,t)=F(x+ct)+G(x-ct)+\dfrac{1}{4}\int_b^{x-ct}\int_a^{x+ct}\delta\left(\dfrac{r+s}{2}\right)f\left(\dfrac{r-s}{2c}\right)dr~ds$
