Neuronal Activity

Abstract

We start off with the description of a single ganglion (a neuronal cell, nerve cell), that is stimulated by a given input. Experiments show basically that a certain minimal activation is necessary to provoke a reaction. This reaction has a form of one spike in case of a short input signal, or periodic spiking in case of a constant input. This behaviour is described by the Hodgkin-Huxley model, resp. by the simpler (but treatable) Fitzhugh-Nagumo model. In the next step, we consider the most simple network: the output signal is fed in again as input signal. If this feedback is negative, then we find periodic spiking. We aim to understand the origin of this observation. Last, we go to large networks. Of course, in general one is not able to state any theorem about large networks of neurons. However, we assume a certain topology (a two-dimensional lattice with interaction of nearest neighbours only) and consider a further abstraction: We derive Greenberg-Hastings automata, a special case of cellular automata. Using combinatorial methods, it is possible to prove some theorems about the behaviour. In particular, it is possible to derive a condition under which a network stays activated and can never go completely to the resting state.

Appendix: Neuronal Activity

6.1.1 1 Solutions

6.1 Let us assume that the delay in the Reichardt detector is τ, and that the inputs of the perceptons change only at time points i τ, $i \in \mathbb{N}$. The input at time i τ is given by I _i ⁽¹⁾ (perceptron 1) resp. I _i ⁽²⁾ (perceptron 2), $i \in \mathbb{N}$. Hence, the output of the detector can be written as

$$\displaystyle{O_{n} = I_{n}^{(1)}I_{ n-1}^{(2)} - I_{ n}^{(2)}I_{ n-1}^{(1)}.}$$

In order to get an idea about the purposes of the Reichardt detector we first consider identical inputs for both perceptrons, $I_{i}^{(1)} = I_{i}^{(2)} = I_{i}$. Then,

$$\displaystyle{O_{n} = I_{n}^{(1)}I_{ n-1}^{(2)} - I_{ n}^{(2)}I_{ n-1}^{(1)} = I_{ i}^{2} - I_{ i}^{2} = 0.}$$

Now we consider inputs that are different,

Table 2

That is, the detector produces an output, if an alternating pattern is fed into the detector, with a phase shift between the two perceptrons.

If a strip pattern is moved in front of the eye of the fly, this input is created. The Reichardt detector is a strip-detector.

6.2 In order to obtain an idea about the frequency of the integrate-and-fire model, we consider the ODE $\tau V ' = -V + I,$ V (0) = 0, for I constant, and estimate the time t until V (t) reaches a threshold, V (t) = V ₀,

$$\displaystyle{V _{0} = V (t) =\int _{ 0}^{t}Ie^{-(t-s)/\tau }\,ds = I\,\,\tau \,\,(1 - e^{-t/\tau }).}$$

If I ≫ 0, the time to the next event will be small; we approximate $e^{-t/\tau } \approx 1 - t/\tau$, and obtain t ≈ V ₀∕I. The frequency (as the inverse of the time interval) is proportional to I.

6.3 If r is small, then z is slow.

For determining the equilibria of the 2D model, we need to solve

$$\displaystyle\begin{array}{rcl} y - ax^{3} + bx^{2}& =& 0 {}\\ c - dx^{2} - y& =& 0 {}\\ \end{array}$$

Solving the second equation for y and inserting the result into the first equation yields the third order equation

$$\displaystyle{x^{3} + \frac{d - c} {a} x^{2} = \frac{c} {a}}$$

or by setting $p = \frac{d-c} {a}$ and $q = \frac{c} {a}$ we look for the roots of

$$\displaystyle{X(x) = x^{3} + px^{2} - q.}$$

We are only interested in the number of roots, not their exact value, so it helps to look at the critical points and their relative positions.

$$\displaystyle{\frac{dX} {dx} = 3x^{2} + 2px = 0 \Leftrightarrow x_{ 1} = 0,x_{2} = -2p/3}$$

with the corresponding values of the function

$$\displaystyle\begin{array}{rcl} X(x_{1})& =& -q {}\\ X(x_{2})& =& \frac{4} {27}p^{3} - q {}\\ \end{array}$$

We can use the second derivative to check for inflection points:

$$\displaystyle{\frac{d^{2}X} {dx^{2}} = 6x + 2p = 0,}$$

thus, the (unique) inflection point is located at $x = -p/3$ with the function value

$$\displaystyle{X(-p/3) = -\frac{4} {27}p^{3} - q.}$$

Hence, different cases are possible:

p < 0: :

The inflection point is located in the positive x area, with x between x ₁ and x ₂. Due to the general properties of X(x) as cubic function, x ₁ is a maximum and x ₂ is a minimum, and only one equilibrium point is possible (as the maximum is negative).

p = 0: :

x ₁ = x ₂ is a saddle point, with a negative function value, so also here, only one equilibrium is possible.

p > 0: :

Now, the inflection point is in the negative x area and $x_{2} = -2p/3$ is negative; x ₁ is a minimum and x ₂ a maximum. The number of equilibria depends on the sign of X(x ₂), with three possibilities: If it is negative, then again only one equilibrium is possible; if it is equal to zero, then we have the degenerated case of two equilibria (the curve is tangent to the x-axis in x ₂); if it is positive, then we have three equilibria, indeed. The condition for that can be formulated as

$$\displaystyle{X(x_{2}) > 0\qquad \Leftrightarrow \qquad \frac{4} {27}p^{3} > q.}$$

or, in terms of the original parameters d and b (assuming all parameters to be nonnegative)

$$\displaystyle{d > b.}$$