# Math Insight

### Video: A simple spiking neuron model: sodium and potassium channels

This video is found in the pages

List of all videos

#### Transcript of video

So far, we've built a one-dimensional autonomous differential equation model that just incorporates the effects of sodium. In the model, \begin{align*} \diff{v}{t} = -v(v-a)(v-1), \end{align*} the voltage lives between zero and one. If it starts below the threshold $v=a$, the voltage decays to zero. If it somehow gets kicked above the threshold, such as via receiving enough input from other neurons, then the voltage rises to one and stays there.

We want to bring the voltage back down so that the neuron can be ready to listen to more input. To allow the voltage to turn around, we must add another state variable so that we have a two-dimensional differential equation. We'll add a variable that represents the state of potassium channels.

Since the concentration of potassium is higher inside the cell than outside, when the potassium channels open, potassium rushes out of the cell and the voltage inside goes down. Hodgkin and Huxley determined that the potassium channels share an important property with the sodium channels: potassium channels tend to open when the voltage inside the cell increases. In this case, since potassium channels bring the voltage down, the potassium channels act as a negative feedback. If the voltage increases, potassium channels will open, which will counteract the voltage change, lowering the voltage.

There's another way in which potassium channels differ from sodium channels. Sodium channels are fast, opening quickly in response to increased voltage. Potassium channels are a bit more sluggish. When the voltage increases, potassium channels take their time in opening, allowing the fast positive feedback of the sodium channels to cause the voltage to shoot up quickly. But then, after a delay -- those lazy potassium channels take a whole millisecond or so to wake up -- the potassium channels get their act together and open up. The result is that they can finally overpower the sodium channels and bring the voltage back down. (The potassium channels actually get some help from the sodium channels tiring out and closing back up, but we aren't modeling that.)

Let's see if we can add this effect of potassium channels to our model so that, by the combination of sodium and potassium, we can get this transient upward and downward excursion of voltage, that is a spike, in response to an input.

First, we need another state variable to represent the state of the potassium channels. We'll use the variable $w(t)$ to indicate how many potassium channels are open at time $t$. Let's not be precise about what the actual values of $w$ are; we could say it's proportional to the number of open potassium channels.

Next, we need to come up with a reasonable form for the change in $w$, i.e., $\diff{w}{t}$. $\diff{w}{t}$ should depend on voltage, so that $w$ will increase when $v$ gets large. For simplicity, let's make the equation for $w$ be a linear differential equation that depends on both $w$ and $v$. As a first stab, let's try an equation of the form \begin{align*} \diff{w}{t} = v-\gamma w \end{align*} where $\gamma$ is a positive parameter.

If we pretend for a moment that $v$ is a constant number, we can analyze this equation. It has one equilibrium: $w = v/\gamma$. Do you know how to show that this equilibrium is stable? (Remember $\gamma$ is positive, so the derivative of the right hand side with respect to $w$ will always be $-\gamma$, a negative number.)

What the equation $\diff{w}{t}=v-\gamma w$ does is make $w(t)$ evolve toward the number $v(t)/\gamma$. You can think of $v/\gamma$ as a target for $w$ that $w$ is chasing. Since $v$ is not really a fixed number, but instead evolves with time, this target will be a moving target for $w(t)$.

Another property of the potassium channels is that they are slow. To make $w(t)$ slow, we need $\diff{w}{t}$ to be small. How do we make it small? We can simply multiply the expression for $\diff{w}{t}$ by a small number. If we call that small number $\varepsilon$, then we can write the equation for $w$ as \begin{align*} \diff{w}{t} = \varepsilon (v-\gamma w). \end{align*} With this addition of $\varepsilon$, $w(t)$ still chases $v(t)/\gamma$, but it just chases it more slowly.

We've introduced two positive parameters. $\varepsilon$ controls the speed of $w(t)$, and $\gamma$ controls how strongly the potassium channels depend on voltage.

The model now captures how potassium channels respond to an increased voltage by slowly opening. To complete the negative feedback effect of the potassium channels, we have to include in our equations the influence of potassium on voltage. Remember, when the potassium channels open, they cause the voltage to go down. In other words, they have a negative effect on the change in voltage $\diff{v}{t}$. Consistent with our simple-minded manner in deriving these equations, let's just subtract off $w$ from the $\diff{v}{t}$ equation. The equation for $v$ becomes \begin{align*} \diff{v}{t} = -v(v-a)(v-1) -w. \end{align*}

Combining the $\diff{v}{t}$ equation with the $\diff{w}{t}$ equation, we now have a coupled system of two autonomous differential equations for our state variables $v$ and $w$. They are coupled because the equation for $v$ depends on $w$ and the equation for $w$ depend on $v$. These equations for the activity of a neuron are called the Fitzhugh-Nagumo equations, which are a simplified version of the Hodgkin-Huxley equations for a spiking neuron.

Our remaining task is to analyze this two-dimensional system of equations and see what kind of behavior we'll get from it. We can show how, for example, you can get a single spike of voltage from an input, where the voltage shoots up and then goes back down. But, we'll stop here for now.