I am trying to understand how a LIF-Neuron (please take a look) works and how I come from this:

$$ I(t) = \frac{u(t)}{R} + C \frac{du}{dt} $$

by multiplying the equation by \$R\$ and call \$\tau_m = R\,C\$ the "leaky integrator":

$$ \tau_m \frac{du}{dt} = -u(t) + R\,I(t) $$

to this expression if we assume that \$u(t^{(1)}) = u_r = 0\$:

$$ u(t) = R\,I_0 \Bigg[1 - \text{exp}\Big\{- \frac{t - t^{(1)}}{\tau_m} \Big\} \Bigg] $$

It seems that I fail to do the integration part here. Could somebody help me to get over this?

The according circuit:

enter image description here


1 Answer 1


In the paper you reference ( there is the assumption of a constant current: $$I(t)=I_0$$ With this substitution, your problem reduces to a "first-order linear" differential equation: $$\tau_m\frac{du}{dt}=-u(t)+RI_0 \ \ \ \ \mbox{where}\ \ u(t^{(1)})=0.$$ One method would be to use an integrating factor. Put the equation into standard form: $$\frac{du}{dt}+\frac{u(t)}{\tau_m}=\frac{RI_0}{\tau_m}$$ The integrating factor is $$exp\Big\{\int\frac{dt}{\tau_m}\Big\}=e^{t/\tau_m}$$Multiply each term by the integrating factor: $$e^{t/\tau_m}\frac{du}{dt}+e^{t/\tau_m}\frac{u(t)}{\tau_m}=e^{t/\tau_m}\frac{RI_0}{\tau_m}$$ Notice that $$e^{t/\tau_m}\frac{du}{dt}+e^{t/\tau_m}\frac{u(t)}{\tau_m}=\frac{d}{dt}\Big[e^{t/\tau_m}\cdot u(t)\Big]$$ Thus, the original equation is reduced to $$\frac{d}{dt}\Big[e^{t/\tau_m}\cdot u(t)\Big]=e^{t/\tau_m}\frac{RI_0}{\tau_m}$$ $$\int\frac{d}{dt}\Big[e^{t/\tau_m}\cdot u(t)\Big]\ dt=\int e^{t/\tau_m}\frac{RI_0}{\tau_m}\ dt$$ $$e^{t/\tau_m}\cdot u(t)=RI_0\ e^{t/\tau_m}+C$$ From here the remaining matter is to plug in the given initial condition to recover C, and some algebra to put into the required form. The technique above can be found in most beginning textbooks on ordinary differential equations.

  • \$\begingroup\$ Thank you! That is a very good explanation! I have to say that I've never learned anything about differential equations. But I guess I'm going to learn that asap .. If it's not too much to ask: would you explain how you get the integrating factor? \$\endgroup\$ Apr 11, 2015 at 18:53
  • \$\begingroup\$ you can verify this substitution using the product rule. working backwards, perform the differentiation d/dt [~] on the RHS to get LHS. I tried to use an easier technique (separation of variables) but for some reason it didn't work... \$\endgroup\$ Apr 11, 2015 at 19:02
  • \$\begingroup\$ If found a good explanation for first order differential equations here. If you want you can add it to your answer. In any case thank you for your help! \$\endgroup\$ Apr 11, 2015 at 19:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.