# Notes on Neurons

## Now V0 is the quasi-steady voltage across the membrane — the voltage that gives no net radial flux jq,r.   So the outward radial current flux is jq,r = (V - V0)gtot .   Now let v(x,t) = V(x,t) -V0 .   And let's define the axon's space constant as λ = (aκ/2gtot)1/2             (12.8a) and its time constant as τ = C/gtot.             (12.8b) With these definitions and assumptions, the cable equation becomes the linear cable equation λ2 (d2v/dx2) -τ (dv/dt) = v.             (12.9) Letting v(x,t) = exp(-t/τ) w(x,t), we find that (12.9) becomes (λ2/τ) (d2w/dx2) = dw/dx which is the diffusion equation we studied in Sec.(4.6.5).   Using its solutions, we have with  N a normalization constant v(x,t) = N exp(-t/τ) t-1/2 exp(-x2/(4tλ2/τ)             (12.10) which is the passive-spread solution.   With a = 0.5 mm, gtot = 5 m-2Ω-1, C =10-2F m-2, and κ = 3 Ω-1 m-1, we get λ 12 mm            (12.11a) and τ = 2 ms.             (12.11b)

Here is a cartoon that describes both voltage-gated sodium channels
and
(slightly better) voltage-gated potassium channels:

When the membrane is depolarized by more than about 10 mV,  the voltage-gated sodium channels open.  When enough of them are open on a given patch of the axon's membrane, the sodium conductance momentarily jumps from the value of Table 11.1, which was gNa+   = 0.04 gKto
gK+ = 0.05 gNa+ = 2gCl-          (12.14)

that is, the sodium conductance jumps from

gNa+   = 0.04 gKto gNa+   = 20 gK+

which is to say by a factor of 25 x 20 = 500.   Sodium ions come flooding in when the sodium channels are open.

The voltage-gated potassium channels are slower than the voltage-gated sodium channels to respond to changes in the voltage across the membrane.   They open only after this voltage V = V2 - V1 has reached its maximum of about +40 mV.

The action potentials of some muscle, egg, and endocrine cells depend upon voltage-gated calcium channels.

We seek a traveling-wave solution, that is, one of the form
V(x,t) = V(t - x/s)
where s is the speed of the wave.   The text has a tilde on the second V and uses a Greek theta for the speed, but I am not good enough at html to be so fancy.  Now the spatial derivative of this
traveling wave is
d
V(x,t)/dx = dV(t - x/s)/dx = (-1/s) dV(t - x/s)/dt      (12.145)
and its second
spatial derivative is
d2V(x,t)/dx=  (1/s2) d2V(t - x/s)/dt2   .     (12.146)

If we put this
traveling-wave solution into the cable equation (12.7b), then we find

## (aκ/(2s2))d2V/dt2 =  jq,r + C (dV/dt)            (12.15)

which is an ordinary, not a partial, differential equation.

Our simplified voltage-gating hypothesis is that the outward radial current  jq,r is

jq,r  = i ( V - ViN ) gi(V).               (12.17)

Now we need to model the sodium conductance somehow.  The text uses the abbreviation v = V - V0 where V0 is the resting potential and models gNa+ (v) by means of the quadratic formula
gNa+ (v) = g0NaB v2  .                       (12.19)

Nelson then finds an action potential that rises from 0 to a voltage he calls v2
But an action potential rises and then falls slightly below 0.  Also, the voltage-gated sodium channels stop the sodium current once the voltage v has risen to about 100 mV (V rises from -60 mV to + 40 mV).   So  Eq.(12.19)  is a very bad approximation.

So let's try a more general expression for the sodium conductance:
gNa+ (v) = g0Naf(v) .                                           (A.1)
Then the outward radial current is from (12.17)

jq,r  = i ( V - ViN ) gi0(V)  +  ( V - VNa+N ) f(v)                 (A.2)
or by (12.2-3)
jq,r  = ( V - V0 ) gtot( V - V- (VNa+N - V0) ) f(v)         (A.3)

= v
gtot +  (v - H )  f(v       (A.4)
where H = VNa+N - V0 = 54 -(-66) mV = 120 mVSo the cable equation (12.15) now is
(aκ/(2s2)) d2V/dt2 =  v gtot +  (v - H f(v + C (dV/dt)                (A.5)
or with v(t) = V(t - x/s) - V0 more simply
(aκ/(2s2)) d2v/dt2 =  v gtot +  (v - H f(v + C (dv/dt) .               (A.6)
Now  v = V - V0 is between zero and about 100 mV, so the second term on the right-hand side of this equation is negative and represents the flow of sodium
thru the voltage-gated sodium channels into the axon.

Let's simplify (A.6).  First, we drop the first term
(aκ/(2s2)) d2v/dt2, which is reasonable when the  radius a of the axon is half a micron, as in mammals, rather than half a millimeter, as in giant squids.

Next, we take
the term  f(v) to be

f(v) =  f θ(v - 0.01) θ(v - 0.1)         (A.7)
in which the Heaviside function
θ(x) = 1 if x is positive and 0 otherwise and f is a constant, which we'll take to be
f = gNa+ = 20 gK+ = 20 x 25 x gtot/38.5 = 65 m-2 Ω-1  .     (A.8)
So we now have
0 =
v gtot +  (v - H ) f θ(v - v1 ) θ(v2 - v) + C (dv/dt) .               (A.9)

Now we imagine that at time t
= 0 the voltage v = v1 which we take to be just enough to open the voltage-gated sodium channels, roughly 10 mV.  While these channels are open,  Eq.(A.9) is
C (dv/dt) = f - ( gtot  f ) v .       (A.10)
We write this first-order, linear, inhomogeneous differential equation as
dv/[ f - ( gtot  f ) v ]  = dt/C    (A.11)
which we may integrate to
- (1/(
gtot + f )) ln {[ Hf - ( gtot + f ) v ]/[ Hf - ( gtot + f ) v1 ]} = t /(A.12)
or
[ Hf - ( gtot + f ) v ]/[ Hf - ( gtot + f ) v1 ] = e -(gtot + f ) t /C .    (A.12)
A little high-school algebra now gives
v(t) = Hf [ 1 - e -(gtot + f ) t /C ]/( gtot + f ) + v1 e -(gtot + f ) t /C .    (A.13)
This function starts at
v1 at t = 0 and then rises to Hf /( gtot + f ).
As
v(t) approaches Hf /( gtot + f ), we assume that the sodium channels close at t = t2 after which time the differential equation is simply
C
(dv/dt) =
- gtot v .       (A.14)
To integrate this simpler equation, we write it as
dv/v = - gtot dt / C .       (A.15)
Integrating both sides, we find
ln (
v/v2 ) = - gtot ( t - t2 )/ C .       (A.16)
Thus we find
v(t) = v2 e - gtot ( t - t2 )/ C  for t > t2 .       (A.17)
Here's a plot of v(t) as given by (A.13) followed by (A.17):

We may improve this model by including the voltage-gated potassium channels.  After the
voltage-gated sodium channels close and the voltage-gated potassium channels open, instead of (A.1), we have
gK+ (v) = g0K+  k(v) .                                           (A.18)
And instead of (A.4), we have
jq,r = v gtot +  (v - Kk(v)         (A.19)
in which K = VK+N - V0 = -75 -(-66) mV = -9 mV.   For simplicity, we will ignore the dependence of  k(v)  upon the voltage v and simply set k(v) = k a constant after the sodium channels close and while the potassium channels remain open.  Then in place of (A.9), we have
0 =  v (gtot + k)   - k K  + C (dv/dt)                    (A.20)
which we may integrate as we did
(A.9).  We get
v(t) k K [ 1 - e -(gtot + k ) (t-t0)/C ]/( gtot + k ) + v1 e -(gtot + k ) (t-t0) /C .    (A.21)
Here's a plot of the resulting action potential (A.13, 21, & 17):

The voltage v(t) rests at 0 for the first 0.5 ms; then the sodium channels open and (A.13) raises v above 0.1 V in
0.5 ms; then the sodium channels close, and the potassium channels open, and (A.21) drops v drops, overshooting; then the potassium channels close, and (A.17) returns v to its resting value of 0.   The input values were: k = 50 m-2 Ω-1,  C = 0.01 F m-2, and K = - 9 mV.

Let's now return to the full Eq.(A.6) with the Heaviside representation (A.7) of the flow thru the voltage-gated sodium channels but without the potassium channels.   Before depolarization, the voltage v is lying at 0, and the channels are closed, so (A.6) is
(aκ/(2s2)) d2v/dt2 =  v gtot + C (dv/dt) .               (A.22)
This is a linear, homogeneous, second-order differential equation.
We may solve it by letting v = exp(pt).   Since
dv/dt = p v, and
d2v/dt2 = p2v,  we get

(aκ/(2s2))  p2v =  v gtot + C pv               (A.23)
which is a quadratic equation for p
(aκ/(2s2))  p2 - C p   - gtot =  0.               (A.24)

There are two solutions
p+ = s2 (  C + ( C2 + 2 gtot/s2 )1/2  ) /(aκ)    >  0    (A.25)
and
p- = s2 (  C - ( C2 + 2 gtot /s2 )1/2  ) /(aκ)    < 0 .    (A.26)
The general solution to Eq.(A.22) is then
v = c1 exp(
p+t ) + c2 exp( p-t )    .         (A.27)
There are three time zones here:  before the v-g sodium
channels open, while they are open,
and after they close.  Since the voltage v relaxes to 0 in the far past and in the far future, the solution
before the v-g sodium channels open must be
v(t) =
c1 exp( p+t )  for t < t1         (A.28)
and
v(t) = c2 exp( p-t )  for t > t2        (A.29)
after they close.

While they are open,  Eqs.(A.6&7) give
(aκ/(2s2)) d2v/dt2 =  v (gtot + f)   - H + C (dv/dt)               (A.30)
which is a linear, second-order, inhomogeneous DEQ.
It is easy to find a particular solution of the
inhomogeneous equation: the constant
v(t) = H f/(
gtot+ f )               (A.31)
will do.  To find the general solution of the homogeneous equation, we try v(t) = exp(qt).
Inserting this into
(A.26), we find that q must satisfy the quadratic equation

(aκ/(2s2)) q2 - Cq - (gtot + f)   =  0   .     (A.32)
The solutions are
q+ = s2 (  C + ( C2 + 2 aκ(gtot+ f ) /s2 )1/2  ) /(aκ)    >  0  (A.33)
and
q- = s2 (  C - ( C2 + 2 aκ(gtot + f )/s2 )1/2  ) /(aκ)    < 0 . (A.34)
So the general solution to the
inhomogeneous equation (A.26) is
v(t)  =
d1 exp( q+t ) + d2 exp( q-t )   +  H f/( gtot+ f )    for   t<  t < t2  .         (A.35)

The voltage-gated sodium channels open when v = v1
and  close when v = v2 .
So the solutions (A.24&25) must be
v(t) = v1 exp( p+(t - t1 ) )     for t < t1         (A.36)
and
v(t) = v2 exp( p-(t - t2 ) )       for t > t2      .  (A.37)

Let's allow a discontinuity in the time derivative
dv/dt of the voltage
at t =
t1
due to an applied stimulating voltage or to the sudden opening of the transmitter-gated sodium channels.  So we will write (A.31) as
v(t)  =
d1 exp( q+(t - t1 )) + d2 exp( q-(t - t1 ))   +  H f/( gtot+ f )    for   t<  t < t2           (A.38)
with coefficients that satisfy
d1 + d2   H f/( gtot+ f )  = v1   .     (A.39)
Here is a plot of (A.33-35) for s = 10 m/s, t1 = 0.0001 s, t2 = 0.0005 s, and d1 = 0.01 V:

This plot has two defects:  the time-derivative dv/dt is discontinuous at t2 = 0.0005 s and the voltage remains positive; there is no overshoot due to an efflux of potassium ions.  To try to remedy these defects, we model the voltage-gated potassium channels.

First, we use (A.18 & 19) to model the
voltage-gated potassium channels after the voltage-gated sodium channels have shut down.     So instead of (A.1), we have
gK+ (v) = g0K+  k(v) .                                           (A.40)
And instead of (A.4), we have
jq,r = v gtot +  (v - Kk(v)         (A.41)
in which K = VK+N - V0 = -75 -(-66) mV = -9 mV.   For simplicity, we will ignore the dependence of  k(v)  upon the voltage v and simply set k(v) = k a constant after the sodium channels close and while the potassium channels remain open.  Then in place of (A.30), we have
(aκ/(2s2)) d2v/dt2 =  v (gtot + k)   - k K  + C (dv/dt)     .               (A.42)
The general solution v(t) to this linear, inhomogeneous, second-order ordinary DEQ is the sum of a particular solution to the
inhomogeneous equation, such as
v = k K/(
gtot + k)                                                              (A.43)
and the general solution of the
homogeneous equation
(aκ/(2s2)) d2v/dt2 =  v (gtot + k)  + C (dv/dt)     .               (A.44)
The general solution to this linear,
homogeneous equation is
v(t) =  e1 exp(r+ t) + e2 exp(r- t)
(A.45)
in which
r+
s2 (  C + ( C2 + 2 aκ(gtot+ k ) /s2 )1/2  ) /(aκ)    >  0     (A.46)
and
r-s2 (  C - ( C2 + 2 aκ(gtot+ k ) /s2 )1/2  ) /(aκ)    <  0 .    (A.47)
So the
general solution v(t) to Eq.(A.44) is

v(t) = e1 exp(r+ t) + e2 exp(r- t)  +  k K/(gtot + k)  .     (A.48)

We now stitch together the solution v(t) = 0 while the neuron is resting, followed by the solution
(A.35) for 0.5 ms while the voltage-gated sodium channels are open (and the K ones closed), followed by the solution (A.48) for 0.5 ms while the voltage-gated potassium channels are open (and the Na ones closed), followed by the solution (A.29) for 1 ms after all the voltage-gated ion channels are closed.  We match the coefficients to keep the time-derivative dv/dt continuous except at t = 0.0005 s when the voltage v jumps due to the opening of the transmitter-gated sodium channels, which start the action potential.  It is interesting that for k = 50 m-2 Ω-1K = - 9 mV, a = 0.5 mm,  andκ= 3.0 m-1 Ω-1, the resulting solution v(t) behaves weirdly unless the speed s lies within the interval 67 ≤    77 m/s.

Here is a plot for speed s = 75 m/s:

The voltage v stays at zero until the initial depolarization opens the sodium channels at t = 0.5 ms.  The influx of Na+ ions increases v until around v = 0.1 V the sodium channels close at t = 1 ms, and the potassium channels open.  Then the voltage v drops below zero until the potassium channels close at 2 ms.  Then it slowly recovers to its resting value of zero, which the sodium-potassium pumps maintain.  In real life, the potassium channels are more sluggish.

In Eq.(A.42), the speed s enters only thru the combination
(aκ/(2s2)) and so s must be proportional to the square-root of aκ.  So if we reduce the radius a by a factor of 4, then the allowed speed s should drop by a factor of 2.    Here's a plot for a = 0.000125 and s = 37.5 m/s with all the other parameters unchanged: