Timeline for Equilibrium partitioning between two domains with different mobility
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| Nov 16 at 23:11 | comment | added | Syrocco | ... a discontinuity. As given by eq. 34 when $\alpha\neq1$ (Altough I quite disagree with their saying that the stationnary state is an equilibrium one, it's just an effective one, with an effective potential). Well, just to wrap it up, I agree with your solution, but I think that it most physically relevant case with a position dependent diffusivity or mobility, you would not end up with the equation you wrote and would, therefore, obtain a position dependent density | |
| Nov 16 at 23:10 | comment | added | Syrocco | no indeed! I agree with your solution the way you wrote it. Altough, writing it like that assume that your underlying FP equation is: $\partial_t p = \partial_x [D(x)\partial_xp]$ which is a very non natural interpretation of a diffusive process because it corresponds to an anti-ito interpretation of the underlying Langevin equation. If you were taking an Ito interpretation, your Fokker-plank equation would look something like: $\partial_t p = \partial_{xx}(D(x)p)$, and you would effectively have a dirac delta at $x=0$ in your equation due to the step function form of $D$, this would create... | |
| Nov 16 at 21:21 | comment | added | Javi | @Syrocco didn't read in detail, but they seem to derive the same expression as me (eq. 34 SM). I don't see how a distribution could be not continuous, yet with derivative equal to zero in each point... I think (for this very simple process, at least) the only possibility for such a non-continuous distribution is to have non-zero current (non-equilibrium stationary distributions). | |
| Nov 16 at 17:12 | comment | added | Javi | @Syrocco That is actually a good point! I'll read the reference and think about it. | |
| Nov 16 at 16:20 | comment | added | Syrocco | It does not have to be continuous at 0, cf this arxiv.org/abs/2403.11928 | |
| Nov 16 at 16:03 | history | answered | Javi | CC BY-SA 4.0 |