Fully analytical solution for the "Leaky Integrate and Fire" Neuron model

[madness]

Does anyone know if it is possible to develop a fully analytical solution for a leaky integrate and fire neuron driven by arbitrary time-varying current? Here's what I have so far (setting as many possible constants to 0 and 1):

The equations:

$\dot{V} = - V + I(t)$ and if $V(t) = 1$ then $\lim_{\epsilon\rightarrow 0} V(t + |\epsilon|) = 0$

Some further notation:

Let $\{ t^{(1)}, t^{(2)}, ...\} = \{ t: V(t) = 1\}$

My attempt at a solution:

We can solve when not spiking as $V(t) = V(t_0) e^{-t} + \int_{t_0}^t e^{-t'} I(t-t') dt'$.

So the solution in general for $t^{(i-1)} < t < t^{(i)}$ is $V(t) = \int_{t^{(i-1)}}^{t} e^{-t'} I(t-t') dt'$.

We can also write the general equation as $\dot{V} = - V + I(t) - \sum_i \delta(t-t^{(i)})$.

It looks almost possible to solve these equations for the spike times $t^{(i)}$ but I'm not sure whether it can be done. We can write $1 = \int_{t^{(i-1)}}^{t^{(i)}} e^{-t'} I(t-t') dt'$, but I don't think spike times can be found from this analytically for arbitrary $I(t)$. I guess what I'd really like is way to plug something for $t^{(i)}$ into the equation $\dot{V} = - V + I(t) - \sum_i \delta(t-t^{(i)})$ and make things a bit more explicit, even if it ends up being a self-consistent equation.

 

[jim mcnamara]

Not my area at all...

This search on Google scholar gave me 5 "good" hits:
model for a leaky integrate and fire neuron driven by arbitrary time-varying current

Spikes are mentioned in two of them.

 

[madness]

I haven't yet seen a fully analytical solution for abitrary input currents. I notice that we can write my equation as $\dot{V} = -V -\delta(V-1) + I(t)$. It's an extremely compact equation when written this way, but can one write down the general solution?

 

[aheight]

Does anyone know if it is possible to develop a fully analytical solution for a leaky integrate and fire neuron driven by arbitrary time-varying current?

This reference: Leaky integrate and fire model has solution for general $I(t)$:

$\displaystyle v(t)=v_r\exp\left(-\frac{t-t_0}{\tau_m}\right)+\frac{R}{\tau_m}\int_0^{t-t_0} \exp\left(- \frac{s}{\tau_m}\right)I(t-s)ds$

 

[madness]

That's the solution to the the passive equation without spiking, i.e., $\dot{V} = -V + I(t)$. The full equation is $\dot{V} = -V + I(t) - \delta(V-1)$ (for $V_r = 0$ and$\tau_m=1$, which I've assumed to make things simpler).

 

[aheight]

That's the solution to the the passive equation without spiking, i.e., $\dot{V} = -V + I(t)$. The full equation is $\dot{V} = -V + I(t) - \delta(V-1)$ (for $V_r = 0$ and$\tau_m=1$, which I've assumed to make things simpler).

Ok, then just a long-shot: The pendulum with a unit impulse $y''+ay'+by=\delta(t-1)$ is solved via Laplace Transforms. Perhaps you can tailor the method to your equation. Also, perhaps would be a good idea to move to differential equations sub-forum.

 

[madness]

Ok, then just a long-shot: The pendulum with a unit impulse $y''+ay'+by=\delta(t-1)$ is solved via Laplace Transforms. Perhaps you can tailor the method to your equation. Also, perhaps would be a good idea to move to differential equations sub-forum.

Thanks. As far as I can see, the fact that the thing inside the delta function has to be determined through the dynamics is what makes this harder in my case. I think you might be right about moving to differential equations - it was only after I started the thread that I understood how compactly the model can be written as an ODE.

 

[jim mcnamara]

If you want the thread moved, please report it (bottom most faint characters in any post), and ask to move it. No problem. Or start completely new thread where it fits better.