# Hodgin-Huxley model for a single neuron - continuation

As continuation to This question I have posted. Another set of questions is given for the same model : This time it is also given that for $t<0$ the membrane potential is $u_0$, and at $t=0$ it is instantaneusly jumps to $u'$, and that $u_0,u',u_2,u_{ion}$ are maintained for $t \geq 0$. So, with this additional data I could actually compute $r(t)$ and $s(t)$ for $t \geq 0$ in the following way:

For $t \geq 0$, $u(t)=u'$ so $r_0(0)=r_0(u') \approx 1$ and $s_0(0)=s_0(u') \approx 0$. Also, $\tau_r(0)=\tau_r(u') = 1$ and $\tau_s(0)=\tau_2(u') = 15$.

So, we get: $r' = -\frac{r-1}{1}$ and $s'=-\frac{s-0}{15}$ $\Rightarrow$ $r' + r = 1$ and $s' + \frac{s}{15} = 0$.

These are a first order linear diff. equation of the form:

$y'(x)+p(x)y=q(x)$

So their solution is of the form:

$e^{-\int{p(x)dx}}(\int{q(x)^{\int{p(x)d(x)}}}+C)$ ($C$ is a constant)

So I get: $r(t)=C_1e^{-t}+1$ and $s(t)=C_2e^{-t/15}$.

I think that we have no way of knowing our initial conditions $r(t=0)$ and $s(t=0)$ (am I right?) So for simplisity let's assume that $C_1=C_2=1$ and we get:

$r(t)=e^{-t}+1$ and $s(t)=e^{-t/15}$.

So for example: $r(t=100) = e^{-100}+1 \approx 1$ and $s(t=100) = e^{-100/15} = 0.0013$ and $r(t=3) = e^{-3}+1 = 1.0498$ and $s(t=3) = e^{-3/15} = 0.8187$

So my question is: How come answers 4 and 7 below denoted as true? Actually there is a way to determine the initial conditions for this problem.

We assume that the gating variables $r$ and $s$ are at a steady state before the membrane potential jump from$u_0$ to $u'$.

If we look at the first graph(the steady state relation versus membrane potential) we can see that at the membrane potential $u_0$ that $r(t=0)\approx 0$ and $s(t=0)\approx 1$.

Then we make our jump in voltage from$u_0$ to $u'$ and now our $r(t)$ and $s(t)$ will change towards their new steady values $r'=1$ $s'=0$ with time constants given by the second graph.

Your equations $r(t) =C_1 e^{-t}+1$ and $s(t)=0+C_2 e^{-t/15}$ are correct, but using the information from the graphs we can se that $C_1= -1$ and $C_2= 1$.

From these equations we can see that $r(3)=1-e^{-3}\approx .9502$ and $s(3)= e^{\frac{-1}{5}}\approx .81$ (note that .81 is close enough to 1 for this particular problem). This means that the channel is mostly open as statement 4 asserts.

Now for statement 7 we have $t= 100$ and while $r(100)= 1-e^{-100}\approx1$, and $s(100)=e^{\frac{-100}{15}}\approx 0$. This means the product $r(100)^2* s(100)\approx 1^2 * 0 \approx 0$ thus the channel is mostly closed.

Hope this helps.

• All true. Just remember that a particular channel is pretty much either open or closed (as verified by single-channel electrical recordings), so 0.81 open means that 81% of channels are open at $t=3$.
– EdM
Jul 12 '15 at 15:52