Exploring transients in the passive cable

Starting from the cable equation (12) we can rescale time by the membrane time constant, $\tau_m$ and space by the space constant, $\lambda$ and shift the voltage so that V=0 is rest. With inputs, s(t,x) the full equation is:

$$\frac{\partial V}{\partial T} = -V(X,T) + \frac{\partial^2V}{\partial T^2} + s(t,x).$$

So far, we have only considered the steady state effects of inputs and the role of the cable in decrementing the steady state inputs. However, the temporal filtering of dendrites is also very important. Dendrites can produce effective delays and thus can have far reaching consequences for synchronization and other temporal coding aspects. It is possible to mathematically address this question for a variety of cable types. Since this is beyond what we have time to do, I will instead approach the question of dendritic transients computationally. Let L be the total electrotonic length of the cable and assume sealed ends. We will break the cable into 21 compartments labeled 0 to 20. We will explore the effects of a synaptic current of the form:

$$a (t/\tau) e^{-t/\tau}$$

where $\tau$ is the time constant of the synapse. Since this is a linear problem, the actual amplitude of the input doesn't matter and what we will be concerned with is the relative amplitudes, the time to peak of the inputs relative to the stimulus site, and the rate of decay of the inputs.

The ODE file that we will use is

# a 20 compartment mode with a synaptic input and sealed conditions
# [x0-.5,x0+.5] is position of input
s(t,x)=heav(x+.5-x0)*heav(x0+.5-x)*amp*t*exp(-t/tau)/tau
v0'=-v0+(v1-v0)/(dx^2)+s(t,0)
v[1..19]'=-v[j]+(v[j+1]-2*v[j]+v[j-1])/(dx^2) + s(t,[j])
v20'=-v20+(v19-v20)/(dx^2) + s(t,20)
dx=L/20
par tau=1,amp=5,x0=0,L=2
@ total=20,meth=cvode,toler=1e-5,atoler=1e-4,dt=.1,yhi=.2,ylo=0
@ xhi=20,yp=v0
done

which is a ``discretization'' of the continuous model. There are several parameters that are useful. The parameter x0 is the position of the input and corresponds to the compartment number,0-20. The parameter L is the electrotonic length and tau is the decay constant of the synaptic input. (Note that this is relative to the membrane time constant since the time scale is relative to the membrane time constant $\tau_m.$ Thus, tau=.1 means that the time constant of the synapse is ten times faster than the membrane time constant.)

HOMEWORK

We will call compartment 0 the soma. So that x0 is the distance of the input relative to the soma. Here are some things to look at:

1.

Sketch the potential at the soma for the synaptic input at compartments 0, 5, 10, and 20. Determine the following parameters, the time to peak, the amplitude of the peak, and the time to decay to half the peak voltage. How do the peak amplitudes depend on distance? How about the time to peak? Does the peak appear to decay slower or faster for more distant inputs?

2.

Set the total integration time to 50 and the length of the x-axis to be 50. Put the synapse at x0=5 and now look at the following three sets of parameters. Set tau=1,amp=5 and solve the equations. Repeat with tau=.1, amp=20. Repeat with tau=10, amp=2. Observe that the decay rate of the input for the two smaller values of tau is roughly the same. But the decay rate for the slower synapse is much greater. Explain why the two faster synapses decay at about the same rate. (Hint: For the faster synapses, what dominates the decay of the synapse or the membrane decay rate?)