2

In the book neuronal dynamics chapter 8.4 (https://neuronaldynamics.epfl.ch/online/Ch8.S4.html), the authors derive the Fokker-Planck equation for a noisy neuron. Starting from equation 8.39:

$ p(u, t+\Delta t) = [1- \Delta t \sum_{k} v_k(t)] e^{\Delta t / \tau} p(e^{\Delta t / \tau} u, t) + \Delta t \sum_k v_k(t) [p(u-w_k, t) - p(u,t)] $

First, they do a Taylor approximation of $ \Delta t $ around 0. Doing it myself,

$ p(e^{\Delta t / \tau} u, t) \approx p(u, t) + u \Delta t \frac{\partial}{\partial u} p(u, t) $

$ e^{\Delta t / \tau} \approx 1 + \frac{\Delta t}{\tau} $

Ignoring all the terms with $ \Delta t^2 $, I reach the following result

$ \frac{\partial}{\partial t} p(u, t) = \frac{1}{\tau} p(u, t) + u \frac{\partial}{\partial u} p(u,t) - \sum_k v_k ( p(u-w_k, t) - p(u, t)) $

However, the book claims that there is a factor of $ \frac{1}{\tau} $ multiplying $u \frac{\partial}{\partial u} p(u,t)$. In equation 8.40,

$ \frac{\partial}{\partial t} p(u, t) = \frac{1}{\tau} p(u, t) + \frac{1}{\tau} u \frac{\partial}{\partial u} p(u,t) - \sum_k v_k ( p(u-w_k, t) - p(u, t)) $

More interestingly, this term somehow disappears when going from 8.40 to 8.41. The goal is to approximate the equation once more with Taylor expansion with $w_k$ going to 0, keeping the terms up to $w_k^2$. When I assume 8.40 to be true and use the expansion:

$p(u-w, t) = p(u, t) - w\frac{\partial}{\partial u} p(u,t) + \frac{1}{2} w^2 \frac{\partial^2}{\partial u^2} p(u,t)$

I reach the following solution:

$\tau \frac{\partial}{\partial t} p(u, t) \approx u \frac{\partial}{\partial u} p(u, t) - \frac{\partial}{\partial u} [-u + \tau \sum_k v_k(t) w_k] p(u, t) + \frac{1}{2}[\tau \sum_k v_k(t) w_k^2] \frac{\partial^2}{\partial u^2} p(u, t) $

The solution in the book (equation 8.41) omits the term $u \frac{\partial}{\partial u} p(u, t)$:

$\tau \frac{\partial}{\partial t} p(u, t) \approx - \frac{\partial}{\partial u} [-u + \tau \sum_k v_k(t) w_k] p(u, t) + \frac{1}{2}[\tau \sum_k v_k(t) w_k^2] \frac{\partial^2}{\partial u^2} p(u, t) $

I would appreciate any help showing where I'm being mistaken.

Ѕᴀᴀᴅ
  • 34,263
  • 1
    Using $e^{\Delta t / \tau} \approx 1 + \frac{\Delta t}{\tau}$, you have $p(e^{\Delta t / \tau}u,t) \approx p((1 + \frac{\Delta t}{\tau})u,t) \approx p(u,t) + u\frac{\Delta t}{\tau}\frac{\partial}{\partial u} p(u,t)$. – stange Mar 20 '24 at 15:36

0 Answers0