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

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*G. Bard Ermentrout*

*1/10/1998*