An excitable system like we saw in Exercise 26.16 has a critical role to play in neurons, as it can filter out unwanted noise

(i.e., random variability arising from random events at the level of individual ion channels). You want a neuron to respond to proper stimuli but you don’t want it to respond to any stimulus, no matter how small. If it did that, it would always be responding to tiny stimuli that are irrelevant, and so it wouldn’t be able to discriminate between what’s important and what’s not.

In this question we study a very simple model of such noise filtering in the bistable equation.

a. We begin by constructing a series of spikes, using trig and exponential functions. Plot the function We’ve seen something very similar to this before, in Exercise 5.15.

S(t) = −2.6+1.7 sin 

t − 9 2



+1.1 sin(t−9)−2.3 sin(2(t−9)), and then plot N(t) = 0.01e S(t)

. Show that, as t increases from zero, N(t) first has a few small bumps and then one larger bump.

b. Use this function N(t) as a driving function for the bistable equation, by considering the differential equation dv dt = 100v(v − α)(1 − v) + N(t).
A typical interpretation of this equation would be that v is the voltage across the membrane of the neuron, while N(t) is some input current to the neuron. It’s this input current that drives the response of the neuron, and it comes from a number of sources; other neurons, for example, or random variations in the neuron’s ion channels. This isn’t actually a good model for the whole neuron response (because a neuron isn’t actually bistable), but it’s a reasonable model for how a neuron gets activated.
Choose α = 0.01 and calculate the solution. Does v respond to the first two smaller bumps?

c. Now make the differential equation more excitable by setting α = 0.001 (If this confuses you, you should first do question 26.16.) Does v now respond to the initial bumps?