The Cauchy problem for one-dimensional spiking neuron models

Abstract

I consider spiking neuron models defined by a one-dimensional differential equation and a reset—i.e., neuron models of the integrate-and-fire type. I address the question of the existence and uniqueness of a solution on ${\mathbb{R}}$ for a given initial condition. It turns out that the reset introduces a countable and ordered set of backward solutions for a given initial condition. I discuss the implications of these mathematical results in terms of neural coding and spike timing precision.

Appendices

Appendix A: Numerical calculation of the degree of an initial condition

In this appendix, I briefly explain how to calculate numerically the degree of an initial condition in the case of the numerical example of Sect. “Implications for neural coding.” For a given initial condition, we compute the backward (continuous) trajectory u(·) by numerically integrating the differential equation and use the following lemma:

Lemma 1

Suppose that the model is leaky, not silent and that f(x,t) > 0 if  x > x _m for some value x _m . Let u(·) be the continuous solution of the differential equation with initial condition (x ₀,t ₀) (x ₀ < 1).

• The degree of (x ₀,t ₀) is zero if and only if there is a time t < t ^* it such that u(t) = 1.

• The degree of (x ₀,t ₀) is positive if and only if there is a time t < t ^* such that u(t) = x _m and there is no s∈[t,t ₀] verifying u(s) = 1. (then u(s) < x _m for all s < t).

In the case of an integrate-and-fire model with synaptic conductances, x _m is the minimum of all reversal potentials. In the case of the numerical example of Sect. “Implications for neural coding,” x _m  = E _i . We follow the trajectory u(·) backward until it reaches 1 or x _m . If it reaches 1, the algorithm stops and the degree is zero. If it reaches x _m , we know that the degree is positive. The degree is at least 2 if and only if the backward trajectory can split at some time s∈[t,t ₀] such that (1,s) has positive degree, and the splitting point is unique by Theorem 2. It is such that u(s) = 0 and (1,s) has positive degree. If there is no such point, then the degree is 1 and the algorithm stops. If there is a splitting point s, then the degree of (x ₀,t ₀) is the degree of (1,s) plus one, and is defined recursively.

Proof

The first point of the lemma is simply Theorem 1. Suppose that the degree of (x ₀,t ₀) is positive. Then u(s) < 1 for all s < t ₀, and, according to prop 1, u(·) is not bounded on ${\mathbb{R}}^-.$ It follows that there is a t < t ₀ such that x(t) = x _m . Conversely, if there is a t < t ₀ such that x(t) = x _m , then u(s) < x _m for all s < t because of the hypothesis on f so that u(·) defines a valid spiking solution on ] − ∞,t ₀] if there is no s∈[t,t ₀] verifying u(s) = 1. Therefore the degree of (x ₀,t ₀) is positive.□

The computation of the degrees can be accelerated thanks to the following lemma:

Lemma 2

Let (x ₀,t ₀) be an initial condition with degree zero, and u(·) the corresponding continuous solution. Then any point (x,t) such that t > t ₀ and x > u(t) (above the graph of u(·)) has degree zero.

In particular, if one follows the (forward) continuous solution u(·) starting from an initial condition (1,t ₀) such that f(1,t ₀) < 0, then all points (1,t) such that t > t ₀ and u(t) < 1 have degree zero. With balanced synaptic input and short time constant, this condition is fulfilled most of the time and the remaining set is already rather small.

Proof

Consider the backward spiking solution x(·) starting from (x,t). Following the trajectories backward, since the difference x(·) − u(·) increases by 1 at every spike and never changes sign otherwise, it must remain positive. In particular, we will have x(s) > 1 when u(s) = 1, so that (x,t) has degree zero.□

Appendix B: Geometric distribution of degrees in a simplified model

In this appendix, I briefly show for a simplified integrate-and-fire model with synaptic conductances that the degrees of initial conditions (1,t) are geometrically distributed. In particular the probability of finding a point (1,t) with infinite degree is zero, which proves a weak version of the conjecture for this model. I consider the same model as in the numerical example of Sect. “Implications for neural coding,” except that g _e (·) and g _i (·) are (possibly rectified) white noise processes with positive mean (preferably in Stratonovich sense). With this assumption, spiking solutions are Markov processes. I also assume that E _i  = E _L  =  reset value (shunting inhibition). I normalize the potential so that the reset is zero and the threshold is one.

Consider a backward trajectory starting from (1,t ₀). If it hits 1 then the degree is 0. If it hits 0 at time s then it must be negative for t < s, so that its degree is at least 1 and the (possible) splitting point is at time s. Applying the same process from (1,s), one can see that the probabilities of the two outcomes are independent from the two outcomes for the first trajectory. Therefore, the distribution of the degrees is geometric.