# Numerical integration - parameter effects in Izhikevich Neuron model

Hy guys,

I am trying to examine the effects of the parameter $a$ in the Izhikevich neuron model that is represented by 2 first order differential equations solved via numerical integration.

Model: \begin{align*} v'&= 0.04v^2 + 5v + 140 - u + I \\ u'&= a(bv - u) \end{align*}

$v,u$ are variables representing membrane voltage and recovery variable of the membrane voltage; $a,b$ are constants defining the dynamics of the neuron model. Initial values are usually $u=-10$, $v=-65$. $I$ represents incoming current, that in a network is again a function of $v$ (and the connections among the neurons).

The equations are usually solved with Euler integration or alternatively Runge-Kutta.

Izhikevich claims on the above site that $a$ determines the decay of $u$, and therefore a larger value results in faster spiking. However clearly the effects of $a$ depend on the sign of $v$ and $u$.

I did some simulations in MATLAB plotting $a$ vs. $v$ and saw all sorts of relationships (cubic mostly), but never linear. The results were also highly dependent on the Euler step size.

I would be super grateful for any hits to solve this problem analytically.

Cheers.

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Okay, first of all, let's ignore $I$ for now. Let's also ignore the $v^2$ term and the constant $140$.

Then we have a simple linear differential equation, $(u,v)' = (u,v) M$, for some fixed 2x2 matrix M. This we can solve, it is $(u,v) = (u_0, v_0) e^{Mt}$.

Now we add the constant in, and we get \begin{align} (u,v) = ((u_0, v_0) + (0,140)M^{-1}) e^{Mt} - (0,140) M^{-1} \end{align}

Now the tricky bit; we have to add the $v^2$ term back in. I don't know how to do this off the top of my head, but what we can do is look to see when that term takes over. When that term takes over, we have $v' = 0.04v^2$, which looks like $v = \frac{1}{0.04(t_f-t)}$, that is, $v$ spikes at time $t_f$. If that term takes over when $v = v_1$ and $t=t_1$, then $v_1 = \frac{1}{0.04(t_f-t_1)}$ or $t_f = t_1 + 25/v_1$.

Notice that the assumption that the other term in the differential equation is unimportant is consistent: if $v$ grows as $1/(t_f-t)$, the other term implies a growth of $ln(t_f-t)$ which is slower.

So when does the $v^2$ term take over? When $0.04 v^2 > |140 + 5v - u|$ or $(u,v) diag(0,0.04) (u,v)^T > |140 + (-1,5). (u,v)|$. We can write this in terms of $M$ and $(u_0,v_0)$ but the basic upshot is the larger the eigenvalues of M, the faster this happens.

For generic $I$, I'm pretty sure this is impossible to solve analytically. However, you could probably do a semi-quantitative analysis like this without too much difficulty.

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right thanks that was helpful – ben Sep 3 '11 at 15:40