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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 Langevin 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 Green'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. 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.

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1 Answer 1

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Edit: After posted I found that you were talking about re-loading. That is too weird a condition. You might want to double check of thie "re-loading" condition. I keep the post here in case you need a reflection boundary condition.


For bounce-back (reflection) boundary condition at $x_2$, the differential form boundary condition i s $$ - \frac{\partial N(x,t)}{\partial x } |_{x_2} = 0; $$ where $N(x,t)$ is the particle distribution function, a negative sign indicates the right side boundary condition (though it make no difference here).

In term of your diffusion kerner, the reflection boundary would be (assume the boundary at the right hand side $x_2 > x, x'$:

$$ G_r(x, x';t) = G(x, x'; t) + G(2 x_2 -x, x'). $$

This construction keeps the differentiation of $G_r(x, x';t)$ vanishes at $x_2$.

The reason of this boundary condition refers to an earlier article of S. Chandrasekhar, Stochastic Problems in Physics and Astronomy, Rev. Mod. Phys, 15 (1943).

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  • $\begingroup$ Have you any idea how to combine the reflecting boundary condition with the absorbing boundary condition? If I take the reflecting kernel you give above $G_r$ and then try to implement the absorbing boundary conditions using the absorbing kernel in my OP then I break the reflecting condition. Similar happens if I do it absorbing first then reflecting $\endgroup$
    – Belisarius
    Commented Feb 8, 2021 at 9:10
  • $\begingroup$ @Belisarius Yes. I do. It needs some scaling before mixed both boudary conditiosn. I will put it up later, when I access my PC. $\endgroup$
    – ytlu
    Commented Feb 9, 2021 at 2:36
  • $\begingroup$ Awesome thank you $\endgroup$
    – Belisarius
    Commented Feb 10, 2021 at 20:03
  • $\begingroup$ Sorry for rhis late correspodence. It is Lunar new year season in Taiwan, a family gathering time. For diffusion within a finite region, $x_1=0<x<L= x_2$, says the boundary condition at $0$ is absorbed, and reflective at $x=L$. The the image method ($x \to 2x_1- x$) will reander an infinite image points: $x \to 2x_1-x; \to 2x_2-x, 2x_2-2x_1-x; \to 2x_1- 2x_2-x, 4_x1-2x_2-x;....$. Fortunately, these serious converges very fast due to the Gaussian function. $\endgroup$
    – ytlu
    Commented Feb 14, 2021 at 9:57

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