Free diffusion with a one absorbing boundary and one "re-loading" boundary
If we are considering free diffusion of a particle which is heavily damped then the evolution of its position $x$ can be expressed by a <PERSON> equation:
$\dot{x} = A\cdot\eta (t)$
where $A$ is constant and $\eta$ is a gaussian noise function with unit variance:
$\left\langle \eta(t) \eta (t')\right\rangle = \delta (t - t')$
I know that the unrestricted diffusive kernel or <PERSON>'s function for this problem looks like:
$G(x,x';\Delta t) = \dfrac{1}{A\sqrt{2\pi\cdot\Delta t}} \text{exp} \left( -\dfrac{1}{2\cdot A^2 \cdot\Delta t}(x-x')^2\right)$
where $\Delta t = t - t'$ and $x'$ is the position of the particle at the initial time $t'$.
My problem is this. I want to impose two boundary conditions. The first is an exit boundary condition at $x_1$ which would correspond to an absorptive kernel $G_e$. I know how to do this straightforwardly enough: you simply subtract the same free diffusive kernel above and set $x \rightarrow 2x_1 -x$ :
$G_{e}(x,x';\Delta t) = G(x,x';\Delta t) - G(2x_1 -x,x';\Delta t) $
$\Rightarrow G_{e}(x,x';\Delta t)= \dfrac{1}{A\sqrt{2\pi\cdot\Delta t}} \left[\text{exp} \left( -\dfrac{1}{2\cdot A^2 \cdot\Delta t}(x-x')^2\right) - \text{exp} \left( -\dfrac{1}{2\cdot A^2 \cdot\Delta t}(2x_1 -x-x')^2\right)\right]$
The second boundary condition is more complicated. 402
When the particle reaches another point $x_2$ I want the particle to be "sent back" to where it started i.e. $x'$. In this way it is "reloaded" back to the initial point although obviously at a later time. I have no idea how to go about this or even what sort of things to search for. If it all possible I would like to add a fixed time delay between reaching $x_2$ and being "transported" back to $x'$ but if that's too tricky even the instantaneous transport would be useful.
Thanks for any help people can offer. 483