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I simulate the narrow escape problem in a reflective volume, $V$, with a permeable surface $A$, which is either circular or ellipsoid with a semimajor axis $a$. Diffusion coefficient is $D$ and time step is $\Delta{t}$. Thus $l=\sqrt{2D\Delta{t}}$.

The problem is defined as $a\ll V^{1/3}$, which makes sense due to its name.

This is a 2D illustration of what I am trying to do in 3D. enter image description here

Here $l=1$, which is much larger than I use in simulations as this is just for illustration purposes. 189 is the number of iterations it takes for the particle to leave $V$.

My questions are:

  • Is it important to have $l\ll a$?
  • If the problem was a simple Brownian Motion simulation, i.e., no such surface as $A$, is it important to have $l\ll V^{1/3}$?

Thanks in advance.

ck1987pd
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    You can get an idea of the answers to these questions by studying the expected exit time, which satisfies a deterministic PDE as a function of the starting point. Specifically if the process behaves locally in time as $\sqrt{D} B_t$ then you have $\frac{D}{2} \Delta u = -1$. There is a way to implement the reflecting boundary in this setting although I forget exactly how it works. Then back in the simulation vary $l$ for fixed other stuff and see how the average compares to what it should be. – Ian Mar 02 '22 at 19:53
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    That said, qualitatively speaking, if $l \not \ll a$ then you will poorly resolve the "near misses" and "near hits". If there's no barrier but you still don't have $l \ll V^{1/3}$ then at best you will poorly resolve the distribution of the exit time because the typical number of steps will not be large, so the number of "common" exit times will be small. – Ian Mar 02 '22 at 20:02
  • @Ian Please clarify this. Do you mean that on average, the exit time $n_1\Delta{t}_1$ be smaller than $n_2\Delta{t}_2$, where $n_i$ are the number of iterations and the time resolutions satisfy $\Delta{t}_1>\Delta{t}_2$ such that $l_1\not\gg a$ and $l_2 \gg a$? Or do you mean that $n_1<n_2$? – ck1987pd Mar 02 '22 at 20:14
  • If you mean the second point, I meant that there may be some nontrivial curve in the underlying continuous exit time distribution, with most of the probability mass confined to within $N \Delta t$ where $N$ is not that big (say 100 or so). Then you have effectively "unsmeared" the distribution to the point that you don't really understand what it really looks like. – Ian Mar 02 '22 at 20:16
  • @Ian Thank you. I know I am taking too much of your time, but one final question. Do you mean that for very large $l$, I might be missing some important information? Does this also imply that after a certain small $l$, if I make it even smaller I will not receive any more meaningful information for the extra computational time? – ck1987pd Mar 02 '22 at 20:24
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    Both of those statements are correct. The first one is rather obvious at least in the case where you haven't split the boundary, because in that case if $l$ is really huge then you will almost always hit the boundary in one step. It's the second case that's a bit less obvious, although if the second thing weren't true then numerics wouldn't be very useful. – Ian Mar 02 '22 at 20:30

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