Inhomogeneous diffusion equation on the half plane Consider one-dimensional linear PDE of the form $$ {\rm u}_{xx}\left(x,t\right) -  {\rm u}_{t}\left(x,t\right) = {\rm f}\left(x,t\right) \quad \mbox{in the domain}\quad \left\{x > 0\,,\ t <  \infty\right\}$$
Boundary conditions: ${\rm u}\left(0,t\right)=0\,,\quad{\rm u}\left(x,0\right)=0\ \quad{\rm u}\left(x,t\right) = 0$ at  x→∞.
I tried so solve the equation by using Green function G(x,t):
$$ G_{xx}-G_{t}=\delta(x-x_0)\delta(t-t_0)$$
Where $$ G(x,0)=G(0,t)=0 \quad and \quad G→0 \quad at \quad x→∞$$
By using Laplace transform we get:
$$ g(x,p)=\int_0^\infty e^{-pt}G(x,t)dt\, $$
and we get a newODE of the form:
$$ d^2g/dx^2-pg=\delta(x-x_0)$$
I don't know how to solve this equation for x>0.
The Internet is full of solutions for this equation for the whole plane but not for half of the plane (x>0).
Please help me if you have any idea how to solve this equation.
Thank you!
 A: I will rewrite the equation for the Green's function as
$$\frac{d^2 g}{dx^2} - p g = \delta(x-x')$$
We may write the general solution of this equation as
$$g(x,x') = \begin{cases} A e^{\sqrt{p} x} + B e^{-\sqrt{p} x} & 0 < x < x' \\ C e^{\sqrt{p} x} + D e^{-\sqrt{p} x} & x \gt x' \end{cases} $$
Because $g$ is zero as $x \to \infty$, $C=0$.  Also, because $g=0$ at $x=0$, we have 
$$A+B=0$$  
We get two more equations from a continuity condition at $x=x'$, and a derivative jump condition (from integrating the equation about a small neighborhood about $x=x'$).  These translate into
$$A e^{\sqrt{p} x'} + B e^{-\sqrt{p} x'} = D e^{-\sqrt{p} x'}$$
$$A e^{\sqrt{p} x'} - (B-D) e^{-\sqrt{p} x'} = -\frac{1}{\sqrt{p}} $$
Solution of these three equations completely determines this Green's function.  You would then need to perform an inverse Laplace transform to get G(x,x',t), which is far from trivial.  If you need help with that, let me know.
A: Because of the homogeneous BCs it looks like the problem might be a decent candidate for the Laplace Transform. The arbitrary forcing function $f(x,t)$ is pesky though... It must stay general?
