482 Chapter 12. Nerve impulses[[Student version, January 17, 2003]]
T 2 Track 2
12.2.5′ The main qualitative feature of our formula for the velocity of an action potential (Equa-
tion 12.24 on page 463) is thatθ∝λaxon/τ;wecould have guessed such a result from dimensional
analysis. But how does the nonlinear cable equation select any velocity in the first place? To
answer the question, notice that Equation 12.23 has a familiar form. Interpretingyas a fictitious
“time” and ̄vas “position,” the equation resembles Newton’s law of motion for a particle sliding
with friction in a position-dependent force field. Such mathematical analogies allow us to apply
intuition from a familiar of system to a new one.
Write the right-hand side as−Qd ̄dvy−dd ̄Uv,whereU( ̄v)≡−s 4 v ̄^4 +1+ 3 sv ̄^3 −^12 v ̄^2. Then we can
think of our fictitious particle as sliding on an energy landscape defined byU.The landscape has
twopeaks and a valley in between.
The waveform we are seeking must go smoothly from ̄v=0aty→∞(resting potential, with
channels closed att→−∞)to ̄v=1aty→−∞(channels open att→+∞). In the language of
our particle analogy, we want a solution in which the particle starts out just below ̄v=1at large
negativey,rolls slowly off one of the two peaks of the potentialU,gains speed, then slows down
and approaches the other peak (at ̄v=0)wheny→∞.
Now ̄vmust pass through the value 1/2 at some intermediate valuey∗.Without loss of generality
wemay take this point to bey∗=0(any solution can be shifted inyto make another solution).
We now choose the “velocity” d ̄dvy
∣∣
∣
y∗
to be just large enough that the particle comes to rest at thetop of the ̄v=0peak. This value is unique: With a smaller push the particle would stall, then
slide back and end up at the bottom of the valley, aty=s−^1 ,whereas with a bigger push it would
run off to ̄v=−∞.
Wehave now used up all our freedom to choose constants of integration in our solution. Looking
at our solution for large negativey,itisvery unlikely that our solution will be perfectly at rest
right at the top of the other peak, at ̄v=1.The only way to arrange for this it to adjust some
parameter in the equation. The only available free parameter is the “friction” constantQ:Wemust
tuneQso that the solution does not over- or undershoot. Thus Equation 12.23 has a well-behaved
solution only for one particular value ofQ(namely the one quoted in Problem 12.6). The dashed
line in Figure 12.10 shows the result of attempting to find a solution by the above procedure with
adifferent value ofQ:Wecan at best make one asymptotic region satisfy its boundary condition,
but not both.