# IAF neurons singularity¶

This notebook describes how NEST handles the singularities appearing in the ODE’s of integrate-and-fire model neurons with alpha- or exponentially-shaped current, when the membrane and the synaptic time-constants are identical.

[1]:

import sympy as sp
sp.init_printing(use_latex=True)
from sympy.matrices import zeros
tau_m, tau_s, C, h = sp.symbols('tau_m, tau_s, C, h')


For alpha-shaped currents we have:

[2]:

A = sp.Matrix([[-1/tau_s,0,0],[1,-1/tau_s,0],[0,1/C,-1/tau_m]])


## Non-singular case ($$\tau_m\neq \tau_s$$)¶

The propagator is:

[3]:

PA = sp.simplify(sp.exp(A*h))
PA

[3]:

$\displaystyle \left[\begin{matrix}e^{- \frac{h}{\tau_{s}}} & 0 & 0\\h e^{- \frac{h}{\tau_{s}}} & e^{- \frac{h}{\tau_{s}}} & 0\\\frac{\tau_{m} \tau_{s} \left(- h \left(\tau_{m} - \tau_{s}\right) e^{\frac{h \left(\tau_{m} + \tau_{s}\right)}{\tau_{m} \tau_{s}}} + \tau_{m} \tau_{s} e^{\frac{2 h}{\tau_{s}}} - \tau_{m} \tau_{s} e^{\frac{h \left(\tau_{m} + \tau_{s}\right)}{\tau_{m} \tau_{s}}}\right) e^{- \frac{h \left(2 \tau_{m} + \tau_{s}\right)}{\tau_{m} \tau_{s}}}}{C \left(\tau_{m} - \tau_{s}\right)^{2}} & \frac{\tau_{m} \tau_{s} \left(- e^{\frac{h}{\tau_{m}}} + e^{\frac{h}{\tau_{s}}}\right) e^{- \frac{h \left(\tau_{m} + \tau_{s}\right)}{\tau_{m} \tau_{s}}}}{C \left(\tau_{m} - \tau_{s}\right)} & e^{- \frac{h}{\tau_{m}}}\end{matrix}\right]$

Note that the entry in the third line and the second column $$A_{32}$$ would also appear in the propagator matrix in case of an exponentially shaped current

## Singular case ($$\tau_m = \tau_s$$)¶

We have

[4]:

As = sp.Matrix([[-1/tau_m,0,0],[1,-1/tau_m,0],[0,1/C,-1/tau_m]])
As

[4]:

$\displaystyle \left[\begin{matrix}- \frac{1}{\tau_{m}} & 0 & 0\\1 & - \frac{1}{\tau_{m}} & 0\\0 & \frac{1}{C} & - \frac{1}{\tau_{m}}\end{matrix}\right]$

The propagator is

[5]:

PAs = sp.simplify(sp.exp(As*h))
PAs

[5]:

$\displaystyle \left[\begin{matrix}e^{- \frac{h}{\tau_{m}}} & 0 & 0\\h e^{- \frac{h}{\tau_{m}}} & e^{- \frac{h}{\tau_{m}}} & 0\\\frac{h^{2} e^{- \frac{h}{\tau_{m}}}}{2 C} & \frac{h e^{- \frac{h}{\tau_{m}}}}{C} & e^{- \frac{h}{\tau_{m}}}\end{matrix}\right]$

## Numeric stability of propagator elements¶

For the lines $$\tau_s\rightarrow\tau_m$$ the entry $$PA_{32}$$ becomes numerically unstable, since denominator and enumerator go to zero.

1. We show that $$PAs_{32}$$ is the limit of $$PA_{32}(\tau_s)$$ for $$\tau_s\rightarrow\tau_m$$.:

[6]:

PA_32 = PA.row(2).col(1)[0]
sp.limit(PA_32, tau_s, tau_m)

[6]:

$\displaystyle \frac{h e^{- \frac{h}{\tau_{m}}}}{C}$

2. The Taylor-series up to the second order of the function $$PA_{32}(\tau_s)$$ is:

[7]:

PA_32_series = PA_32.series(x=tau_s,x0=tau_m,n=2)
PA_32_series

[7]:

$\displaystyle \frac{h e^{- \frac{h}{\tau_{m}}}}{C} + \frac{h^{2} \left(- \tau_{m} + \tau_{s}\right) e^{- \frac{h}{\tau_{m}}}}{2 C \tau_{m}^{2}} + O\left(\left(- \tau_{m} + \tau_{s}\right)^{2}; \tau_{s}\rightarrow \tau_{m}\right)$

Therefore we have

$$T(PA_{32}(\tau_s,\tau_m))=PAs_{32}+PA_{32}^{lin}+O(2)$$ where $$PA_{32}^{lin}=h^2(-\tau_m + \tau_s)*exp(-h/\tau_m)/(2C\tau_m^2)$$

3. We define

$$dev:=|PA_{32}-PAs_{32}|$$

We also define $$PA_{32}^{real}$$ which is the correct value of P32 without misscalculation (instability).

In the following we assume $$0<|\tau_s-\tau_m|<0.1$$. We consider two different cases

a) When $$dev \geq 2|PA_{32}^{lin}|$$ we do not trust the numeric evaluation of $$PA_{32}$$, since it strongly deviates from the first order correction. In this case the error we make is

$$|PAs_{32}-PA_{32}^{real}|\approx |P_{32}^{lin}|$$

b) When $$dev \le |2PA_{32}^{lin}|$$ we trust the numeric evaluation of $$PA_{32}$$. In this case the maximal error occurs when $$dev\approx 2 PA_{32}^{lin}$$ due to numeric instabilities. The order of the error is again

$$|PAs_{32}-PA_{32}^{real}|\approx |P_{32}^{lin}|$$

The entry $$A_{31}$$ is numerically unstable, too and we treat it analogously.

## Tests and examples¶

We will now show that the stability criterion explained above leads to a reasonable behavior for $$\tau_s\rightarrow\tau_m$$

[8]:

import nest
import numpy as np
import pylab as pl


Neuron, simulation and plotting parameters

[9]:

taum = 10.
C_m = 250.
# array of distances between tau_m and tau_ex
epsilon_array = np.hstack(([0.],10.**(np.arange(-6.,1.,1.))))[::-1]
dt = 0.1
fig = pl.figure(1)
NUM_COLORS = len(epsilon_array)
cmap = pl.get_cmap('gist_ncar')
maxVs = []

<Figure size 432x288 with 0 Axes>


Loop through epsilon array

[10]:

for i,epsilon in enumerate(epsilon_array):
nest.ResetKernel() # reset simulation kernel
nest.resolution = dt

# Current based alpha neuron
neuron = nest.Create('iaf_psc_alpha')
neuron.set(C_m=C_m, tau_m=taum, t_ref=0., V_reset=-70., V_th=1e32,
tau_syn_ex=taum+epsilon, tau_syn_in=taum+epsilon, I_e=0.)

# create a spike generator
spikegenerator_ex = nest.Create('spike_generator')
spikegenerator_ex.spike_times = [50.]

# create a voltmeter
vm = nest.Create('voltmeter', params={'interval':dt})

## connect spike generator and voltmeter to the neuron
nest.Connect(spikegenerator_ex, neuron, 'all_to_all', {'weight':100.})
nest.Connect(vm, neuron)

# run simulation for 200ms
nest.Simulate(200.)

# read out recording time and voltage from voltmeter
times = vm.get('events','times')
voltage = vm.get('events', 'V_m')

# store maximum value of voltage trace in array
maxVs.append(np.max(voltage))

# plot voltage trace
if epsilon == 0.:
pl.plot(times,voltage,'--',color='black',label='singular')
else:
pl.plot(times,voltage,color = cmap(1.*i/NUM_COLORS),label=str(epsilon))

pl.legend()
pl.xlabel('time t (ms)')
pl.ylabel('voltage V (mV)')

[10]:

Text(0, 0.5, 'voltage V (mV)')


Show maximum values of voltage traces

[11]:

fig = pl.figure(2)
pl.semilogx(epsilon_array,maxVs,color='red',label='maxV')
#show singular solution as horizontal line
pl.semilogx(epsilon_array,np.ones(len(epsilon_array))*maxVs[-1],color='black',label='singular')
pl.xlabel('epsilon')
pl.ylabel('max(voltage V) (mV)')
pl.legend()

[11]:

<matplotlib.legend.Legend at 0x7f68764a5750>

[12]:

pl.show()


The maximum of the voltage traces show that the non-singular case nicely converges to the singular one and no numeric instabilities occur.