Modelling of Nerve Impulses
Sophia Halassy
Department of Mathematics
McMaster University
Hamilton, Ontario, Canada
June 26, 2012
1
Abstract
This is a mathematical analysis of the theory that explains the propagation of a nerve impulse along
an axon. Beginning with the known Hodgkin-Huxley model, the general, non-linear one-dimensional
cable equation is used. Both the linear and non-linear cable equations are studied and analyzed. A
derivation of both equations is demonstrated and their properties are explained. XPPAUT and R
software techniques are used for analysis.
2
2.1
Introduction
Biological Neuron
The nervous system, composed of all nerve tissues in the body, has the important property of
integrating and controlling all of the body’s functions.
The neuron consists of three major parts; the cell body, the dendrites and the axon. It can be
electrically stimulated to transmit an electrical impulse. A synapse is where the nerve ending forms
a contact with the neuron, and where one neuron can send a message to another. The nerve endings
can also secrete chemical substances.
A nerve is a cluster of nerve fibers. Many bundles of nerve fibers are what make up a bodys
connective tissue. The nerve fibers themselves are long strings of cytoplasm (the living contents of
a cell surrounded by a cell membrane), which are an extension of the cell body of a neuron. The
messages that are transmitted by the nerve fibers are referred to as nerve impulses. Some fibers
are specialists, some performing unique functions such as transmitting chemical substances, while
others only transmit nerve impulses. Research on the nervous system has been in the works for a
very long time. For example, Nersnt investigated the physical properties behind the flow of ions
through membranes during the First World War. He then proposed that this flow was the essential
mechanism for nerve impulses.
1
Figure 1: [3] Figure demonstrating anatomy of a typical textbook nerve cell and displaying direction
of a nerve impulse.
The cell membrane has many important functions, including regulating the movement of chemicals
and ions inside and outside. It is primarily composed of two phospholipid layers which help to
maintain the internal environment of the cell. Proteins are one component of the membrane and have
ion channels which can transport ions across the membrane. This passage of ions induces currents
and ultimately affects the membrane potential. There is an uneven distribution of inorganic ions
across the cell membrane, notably the K + , N a+ , and Cl− ions. The distribution is explained by the
Donnan equilibrium. This phenomenon is directly attributed to the permeability of the membrane
in allowing the diffusion of ions. Essentially, it is how ions crossing the semi-permeable membrane
fail to evenly distribute themselves across both sides. In this case, the two sides of the membrane
have a unequal distribution of ions, and this causes what we refer to as an electrical potential. At
room temperature, the potassium and chlorine ions have a higher intracellular concentration than
the sodium ion which has a higher extracellular concentration. The ion channels can change their
permeability in response to membrane potentials and the binding of molecules to receptor sites on
the channels.
The nerve cell is more negatively charged on the inside compared to the outside of the cell. It is
possible to record a potential (in millivolts) across the membrane with the use of an intracellular
electrode. The resting potential of the neuron is usually found to be -65mV and this is the value
of the current that will be used when we introduce the Hodgkin-Huxley model. The neuron is an
excitable cell, meaning that it has the ability to transmit an action potential by responding to an
2
above-threshold stimulus.
Research has shown that nerve fibres have similar conduction properties to electrical wires. The
comparison is mostly the conduction of messages over long distances, and normally done over a
short period of time. The message is sent via constant-sized and constant speed pulses. Hodgkin
first proved this theory in 1937, when he discovered that nerve impulses are a series of traveling
electro-chemical stimulations on the cell membrane (which is made up of lipid and protein) covering
the nerve axon.
2.2
Modelling of Electrical Activity
In 1952, Hodgkin and Huxley made the first precise recording of electrical signals conducted by
neurons with the use of micro electrodes placed in a giant squid axon. They discovered that the
K + and N a+ (potassium ion and sodium ion, resp.) along with a leak current (associated with the
Cl− ions) were the main contributors to the electrical potential across the membrane and hence
determined the current inside the neuron. And so, the sum of the sodium, potassium, and leak
current give the ionic current of the cell. Though they did not know that ”channels” existed, they
assumed that these ions could flow through the membrane via tiny holes allowing their passage only
when a particular number of particles occupied certain positions.The first recordings of membrane
potential gave rise to a bulk of research by many scientists interested in studying action potentials
across cell membranes.
3
3.1
The Basic Model
The Hodgkin-Huxley Model
The Hodgkin-Huxley model properly describes the dynamics of the cell membrane’s action potential.
It serves as the base model for many other action potential systems. The equations are the following:
Cm
dV
= −g(V, m, n, h) + Iapplied
dt
(1)
where V denotes the membrane potential and V 0 = −65mV is the resting membrane potential;
Cm the membrane capacitance, with a usual value of Cm = 1µF/cm2 (where µF/cm2 stands
for microFarad per square centimetre); m the sodium activation gating variable; n the potassium
activation gating variable; h the sodium inactivation (since it becomes smaller when m becomes
larger) gating variable; Iapplied the external stimulus. The sodium current can be expressed, by
Ohm’s law:
gN a (V − VN a ) = gN¯a m3 h(V − VN a )
(2)
where VN a is the sodium reverse potential; gN a the conductance of sodium channel; gN¯a the
maximum conductance of sodium channel. Usually, VN a = 50mV and gN¯a = 120pS/cm2 (where
pS/cm2 stands for picoSiemen per square centimetre).
3
In a similar manner, the potassium current can be expressed as:
gK (V − VK ) = gK̄ n4 (V − VK )
(3)
where VK is the potassium reverse potential; gK the conductance of potassium channel; gK̄ the
maximum conductance of potassium channel. Usually, VK = −77mV and gK̄ = 36pS/cm2 .
The leakage current can also be expressed as:
gL̄ (V − VL )
(4)
with gL̄ = 0.3pS/cm2 and VL = −54.4mV .
It is important to note that the variables n, m, and h are the opening probabilities of the gate
channels and the exponents in (2) and (3) denote the number of gates.
The following is a list of all parameters used in the Hodgkin-Huxley model:
Table 1: The Hodgkin-Huxley Parameters
Parameters
Membrane Capacitance
Membrane Potential
Equilibrium Potential
Conductances
Gate Parameters
Symbol
Cm
V
VK
VN a
VL
gK̄
gN¯a
gL̄
n
m
h
Values
1
−65
−77
50
−54.4
36
120
0.3
Units
µF/cm2
mV
mV
mV
mV
pS/cm2
pS/cm2
pS/cm2
We can now describe the gate opening and closing rates, denoted by α and β resp. and are
generally functions of the membrane potential V . They are described as the following, displaying
how the gates have to be completely open or completely closed in order to conduct:
dn
= αn (1 − n) − βn n
dt
(5)
dm
= αm (1 − m) − βm m
dt
dh
= αh (1 − h) − βh h
dt
(6)
It is sometimes useful to understand what is occurring by referring to a circuit diagram:
4
(7)
Figure 2: [11]A circuit diagram demonstrating the behaviour. Here, each of the conductances
represents the average effect of the opening or closing of the gating channels for the various ions.
This diagram sufficiently explains what is occurring in an isolated section of a giant squid axon.
Hence, there is no axial resistance.
Hodgkin and Huxley varied the clamp voltage and, as a consequence, were able to determine
equilibrium values and time constants associated with the conductances.The equilibrium (or steady
state) solutions are:
αn
(αn + βn )
αm
mequ =
(αm + βm )
αh
hequ =
(αh + βh )
nequ =
5
(8)
(9)
(10)
Figure 3: The steady state functions representation where x = n, m, h.
The time it takes to approach the equilibrium value of the steady states are described by their
associated time constants,
τn =
1
αn + βn
(11)
τm =
1
αm + βm
(12)
τh =
1
αh + βh
(13)
6
Figure 4: The time constant functions representation where x = n, m, h.
The α and β opening and closing rates are empirically determined formulas in function of membrane potential, V , such that
αn (V ) =
0.01(V + 55)
1 − exp( −(V10+55) )
−(V + 65)
)
80
0.1(V + 40)
βn (V ) = 0.125exp(
αm (V ) =
1 − exp( −(V10+40) )
−(V + 65)
)
18
−(V + 65)
αh (V ) = 0.07exp(
)
20
1
βh (V ) =
1 + exp( −(V10+35) )
βm (V ) = 4exp(
7
(14)
(15)
(16)
(17)
(18)
(19)
3.1.1
The Nerve Impulse
When the applied current is above a certain threshold, it will generate an action potential before
settling to its equilibrium state, we refer to this as a nerve impulse. When plotting the membrane
potential V in function of time T :
Figure 5: The neuron excitability in response to an applied stimulus. A pulse of amplitude lower
than threshold fails to induce an action potential. However, above threshold elicits a response and
returns to rest after a refractory period.
Biologically, what is occurring is that as the sodium gates open and sodium ions rush into the cell,
the voltage increases and depolarizes the membrane. The sodium concentration is trying to reach its
Donnan equilibrium, which is approximately 58mV (referred to earlier). However, as the membrane
becomes depolarized, the potassium gates open and potassium exits the cell as they are repulsed
by the positive charge of the sodium ions inside the cell. At the same time, the sodium gates close.
This is where we see the action potential. The voltage decreases and the membrane hyper polarizes
as the potassium concentration tries to reach its Donnan equilibrium, which is around −93mV .
Eventually, the potassium gates close, then there is a refractory period and the voltage returns to
its resting potential of −65mV . Only one action potential can occur at a time, when the sodium
gated channels are either open or inactive, we refer to this as the ”absolute refractory period”.
8
4
Neuronal Cable Theory
Now that we somewhat have a clear understanding of what defines a nerve impulse and its characteristics, we will focus our attention on explaining the way the post-synaptic potentials propagate
the length of a dendrite toward a soma. In the 1940’s, even before Hodgkin had made the first
recording of a nerve impulse, he and other scientists had applied cable theory to a typical nerve
fibre.
We already know that some scientists have previously compared the propagation of a nerve impulse
through a neuron to an electric impulse through a cable. Furthermore, it is even more comparable
to a cable that is under water, since a nerve fibre is emerged in ionic fluid. There is a covering (
analogous to the myelin sheath of a neuron) and this insulation allows some sort of leakage. Though
there are many resemblances, there are also some distinctions. While normal electrical wire is made
of copper and allows for high conductance, a nerve cell has a salt solution that has a conductance
that is inferior to copper. Moreover, the insulation on a wire is much more efficient than that of a
neuron .
In the end, there is a continuous distribution of resistance and capacitance across a nerve cell
membrane. Furthermore, there is an axial resistance parallel to the membrane. There are also
variable resistances coming from the various ion-selective channels that are present. We have explained the mechanisms involved in the activation of the voltage-gated channels that leads to the
firing of a neuron. Now, since we’ve only so far demonstrated what occurs in a sectioned portion of
the membrane, we must broaden our model to understand what occurs as the impulse propagates
throughout the entire neuron. We will try and understand the diffusive and wave-like properties of
an action potential.
4.1
Linear Cable Equation
The classical linear cable equation is one-dimensional. We know that a nerve cell has a membrane
which has a negative voltage on the inside relative to the outside, and that it has the ability to store
and separate the charge. This membrane provides some insulation to the resistance of the intraand extra-cellular fluids. Due to this difference in the membrane and axial resistivity, the electrical
current inside of the neuron shows a tendency to flow parallel to the cylinder axis. As a result, we
can ignore the other two dimensions and only consider the propagation of the nerve impulse in the
x-direction. Note: The complete derivation of the equation is found in the Appendix.
The linear equation will then have a membrane voltage, denoted V , that is a function of time, t,
and the distance, x, along the short length of the neuron, hence V (x, t). We must consider some
key assumptions before we proceed:
• The membrane is independent of time, hence passive, and has a uniform distribution.
• The interior of the nerve cell has the same cross-sectional area along its length, and we can
calculate the Ohmic resistance of the intracellular fluid.
• We can ignore the resistivity of the extracellular fluid.
9
• Currents are the only form of inputs that we will consider.
Through a cylindrical section of the axon, there is a current that can either flow longitudinally
along the x-axis, or through the membrane. The longitudinally current, denoted Ii , experiences
a voltage drop when it encounters the resistance of the cytoplasm. As the current flows from the
left x-direction to the right, then we will say the current is positive, and when the current flows in
the opposite direction, it will have a negative value. The cytoplasmic resistance per unit length, is
expressed as ri and in units of Ω/cm (Ohm/centimetre). By Ohm’s Law,
1 ∂V
= −Ii
ri ∂x
(20)
The membrane is studded with various kinds of proteins, and one of these proteins are the ion
channels that allow currents to pass. And so, as the current propagates along the membrane, it can
cross the resting membrane channels, represented as a resistance rm (in Ω · cm) for unit length. It
can also encounter the membrane capacitance per unit length, cm (in F/cm). And so, if there is
no additional applied current, the charge per unit length, ∂Ii /∂x of the longitudinal current is the
density of the membrane current Im per unit length.
∂Ii
V
∂V
= −Im = −(
+ cm
)
∂x
rm
∂t
(21)
By combining both aforementioned equations, we get the linear cable equation:
1 ∂2V
∂V
V
= cm
+
2
ri ∂x
∂t
Rm
(22)
If we consider a 1µcm2 sectional area of a membrane, then we can calculate a specific capacitance
and a specific resistance of the membrane. This way, our calculations can be independent of the
the size of the neural compartment. We will denote the specific capacitance by Cm and the specific
resistance by Rm .
As before, we will use 1µcm2 as the value of the membrane capacitance Cm . And so, for a segment
of cylindrical membrane with a diameter d and length l, the actual capacitance, denoted CM , is
πdlCM . For the capacitance, Cm = lcm .
If the membrane channels are uniformly distributed over the small section of membrane, then
the conductance is directly proportional to the membrane area. As a direct consequence, the
membrane resistance is inversely proportional to the area, Rm = RM /(πdl) = rm /l (expressed in
units of Ω · cm2 ).
As a summary, here are the relationships
Cm = cm l = πdlCM
Rm =
rm
RM
=
l
πdl
10
(23)
(24)
We can also define the space constant, in cm, as
λaxon =
!
rm
ri
(25)
This is describing how far a current will propagate and spread inside of the axon. So the larger
the value of λaxon , the further the current will travel. This makes sense because the larger the
membrane resistance, the more current will remain in the cytosol to travel longitudinally along the
axon.
The membrane time constant is,
τm = rm cm = RM CM = Rm Cm
(26)
This is describing how long it takes a membrane potential to change in response to an injected
current. Hence, the larger the membrane capacitance Cm , the more current it will take to charge
and discharge a patch of membrane and the longer the process will take.
Now, we can write the linear cable equation as,
λ2axon
4.1.1
∂2V
∂V
− τm
=V
2
∂x
∂t
(27)
An Analytic Solution to the Linear Cable Equation
It is possible to determine an analytic solution to the linear cable equation by method of separation
of variables where we reduce the equation involving partial derivatives into one with ordinary
derivatives.
First, we will start by considering the following linear cable equation
∂2V
∂V
−
= V (X, T )
2
∂X
∂T
(28)
x
t
where X = λaxon
and T = τm
. Now, following the usual method of separation of variables, we must
consider a solution of the form
V (X, T ) = f (X)g(T )
(29)
Hence, if we substitute this into our linear cable equation:
Next, divide by f g,
or, rewritten as,
f "" g − f g " = f g
(30)
f "" g "
− =1
f
g
(31)
f ""
g"
= 1 + = −c2
f
g
(32)
11
where −c2 represents the separation constant. We can see that the only way for the function
g!
g
f !!
f
(a
function of f) can equal 1 + (a function of g) is if they are both equal to some constant, which we
have named −c2 . We square the constant to guarantee a positive value, and the negative in front of
it will give a negative constant. The negative constant will allow the solution, which is a function
of time, to not blow up. Furthermore, the squaring of the constant guarantees that the solution
won’t be a square root. We now have the two following sets of equations,
f "" = −c2 f
(33)
Hence, the second derivative of the function is the negative of itself times some constant, and
g " = −(1 + c2 )g
(34)
which implies that the first derivative of the function is itself times a constant.
From similar trends using separation of variables, it is obvious that the second equation is simply
the exponential function. We can now write our analytic solution as
V (X, T ) = (Asin(cX) + Bcos(cX))exp[−(1 + c2 )T ]
(35)
Here, we have three constants, A, B, and c, two boundaries, and one initial condition. We will
use sealed ends of the axon as our boundary conditions, hence
−1
(36)
−1
(37)
∂V
|X=0 = 0
λaxon ri ∂X
∂V
|X=L = 0
λaxon ri ∂X
We take the derivative with respect to X to get:
∂V (X, T )
= (cAcos(aX) + −cBsin(cX))exp[−(1 + c2 )T ]
∂X
(38)
In order for this to follow our boundary condition, then A = 0 at X = 0 (since sin(0)=0).
Therefore, if A = 0, then the derivative will also equal to 0 at X = L when sin(cL) = 0. This will
occur when cL = nπ, where n = 0, 1, 2, · · ·.
We can now rewrite our solution as
V (X, T ) =
∞
"
n=0
Bn cos(
nπX
nπ
)exp[−(1 + ( )2 )T ]
L
L
(39)
• We can determine the value of Bn from our initial condition V (X, 0).
• We must notice here that the voltage will decay in a number of ways. Most importantly, in
an isopotential cell, the voltage will only decay in response to the membrane’s time constant,
denoted τm .
12
• τm will decay to 1/e of its initial value in the isopotential cell.
• As a result, when n = 0, the exponential part of our analytic solution will be exp( −t
τ0 ), where
τ0 = τm = Rm Cm .
• When n > 0, then the exponential part is
exp[−(
2
(1 + ( nπ
L ) )t
)]
τ0
(40)
where τ0 = τm .
• We can therefore write the time constant as
2
[1 + ( nπ
1
L ) ]
=
τn
τ0
(41)
τ0
2
[1 + ( nπ
L ) ]
(42)
or,
τn =
We can finally write our final analytic solution to the linear cable equation as
V (x, t) =
∞
"
−t
(43)
Cn e τn
n=0
where Cn replaced the Bn cos(nπx/L) and our equation has returned to x and t instead of X and
T.
Note: The term Cn will depend on the initial conditions.
Here is an alternative approach. By defining:
t
W (x, t) ≡ e τn V (x, t)
(44)
Our linear cable equation now becomes
(λaxon )2 d2 W
dW
=
2
τn
dx
dt
(45)
where λaxon is the space constant τn is the time constant, as previously stated. This has the same
form as the heat equation, of which we know the solution (See P. Nelson, Biological Physics [9] for
details). It is now possible to adapt the result of the known heat equation to this one to get the
passive spread solution:
−t
V (x, t) = const. × e τn t
−1
2
−x2
e (4t(λaxon )2 /τn )
(46)
And since we know that the heat equation does not have any traveling wave solutions, it is given
that the above equation will not have any traveling wave solutions as well.
13
4.2
Non-Linear Cable Equation
Now that we have introduced and shown the derivation of the linear cable equation, we must note
that they completely ignore the presence of voltage-gated channels in the membrane, just as we saw
in the Hodgkin-Huxley model. Hence, potential will decrease in amplitude and spread out in time
as they propagate away from the source.
4.2.1
Deriving the Non-Linear Cable Equation
As an impulse propagates along an axon it does, in fact, display a wave pattern.This can be apparent
if we think of a chain that progressively shifts from a higher to lower groove.
Figure 6: Mechanical analog of the action potential. A heavy chain lies in a tilted channel , with
two troughs at heights differing by $h. (a) An isolated kink will move steadily to the left at a
constant speed v: successive chain elements are lifted from the upper trough, slide over the crest,
and fall into the lower trough. (b) A disturbance can create a pair of kinks if it is above threshold.
The two kinds then travel away from each other [9].
14
This analogy demonstrates that a nervous system could support a traveling wave of fixed speed and
definite waveform [9]. So we must ask ourselves: Is the linear cable equation reliable in explaining
the correct movement of an impulse when it is above threshold? It is then important to introduce
the correct model, which takes into account the way the action potential can access the energy
stored in the membrane by the selective ion pumps.
It was previously shown that we can represent the voltage-gates in the following way:
Iionic = IN a + IK + IL = gN¯a m3 h(V − VN a ) + gK̄ n4 (V − VK ) + gL̄ (V − VL )
(47)
We can rewrite this in a simpler way, known as the voltage-gating hypothesis:
Iionic =
"
i
(V − Vi )gi (V )
(48)
If we refer to the action potential introduced near the beginning of this paper (Section 3.1.1),
we know that it all begins with an initial sodium influx. We will start by concentrating on the
sodium ion and then expanding this to all the ion channels. We will suppose that the conductance
is dependent upon an instantaneous change in the potential v ≡ V −V 0 . Let’s consider the following
function,
0
2
gN a (v) = gN
(49)
a + Bv
0 represents the resting conductance/area. We can say that this resting potential along
where gN
a
with the other conductances are represented by gT0 otal , and B is some positive constant since conductance is always positive. This equation encompasses the idea that the membrane depolarizes as
the sodium gates open and sodium rushes into the cell.
Based on the earlier equation, we can now express our current through the membrane as the sum
of the sodium contribution and the Ohmic terms of the other ions:
Iionic =
"
i
(V − Vi )gi0 + (V − VN a )Bv 2 = vgT0 otal + (v − H)Bv 2
where H = VN a − V 0 is a constant.
15
(50)
Figure 7: The curve of current versus depolarization [9].
From this curve, we see that there are three different points where the membrane current is zero.
Hence, the last equation mentioned has three real roots. Let’s name them v = 0, v1 , and v2 . Further
work shows that
#
1
v1 , v2 = (H ∓ H 2 − 4gT0 otal /B)
(51)
2
Algebra shows that v1 v2 = gT0 otal /B and we can now write our non-linear cable equation,
(λaxon )2
d2 V
dV
v(v − v1 )(v − v2 )
−τ
=
2
dx
dt
v1 v2
(52)
where v ≡ V − V 0 . It is possible to see that unlike the linear cable equation, this equation does
not show the same form as the heat equation, and thus we cannot perform the same work to find
an analytical solution. However, we know that we can represent a travelling wave of speed s by a
function of one variable, say v̄(t), hence v(x, t) = v̄(t − ( xs )). Substitute this back into the previous
equation to get:
λaxon 2 d2 v̄
dv̄
v̄(v̄ − v1 )(v̄ − v2 )
(
)
−τ
=
(53)
s
dx2
dt
v1 v2
By letting ṽ ≡ v̄/v2 , y ≡ v2 /v1 , and Q ≡ τ s/λaxon , we get:
d2 ṽ
dṽ
= −Q + sṽ 3 − (1 + s)ṽ 2 + ṽ
2
dy
dy
(54)
This nonlinear equation does display a traveling wave whose explicit formula can be obtained [10].
The disturbance must be large enough for a wave to propagate. If it is below a certain threshold,
then we see the same behaviour displayed by the linear cable equation where there is no generated
action potential, only passive diffusion.
16
4.2.2
Deriving the Non-Linear Cable Equation Through the Hodgkin-Huxley Model
We can now refer to our original Hodgkin-Huxley model:
1 ∂2V
∂V
V
∂V
= cm
+
= cm
+ gN¯a m3 h(V − VN a ) + gK̄ n4 (V − VK ) + gL̄ (V − VL )
2
ri ∂x
∂t
Rm
∂t
(55)
where the non-linearities are included and
V
= Iionic = IN a + IK + IL
Rm
(56)
Let’s first assume that an action potential propagates through an axon like a wave with constant
velocity. This movement will depend on the distribution of the currents away from the initial
location of the action potential. Hence, an action potential has the ability to repeat itself along the
length of the axon.
So, we can obtain the non-linear cable equation by first concentrating on all of the terms that
depend on the axon’s radius, hence
1 ∂2V
∂V
= cm
+ gN¯a m3 h(V − VN a ) + gK̄ n4 (V − VK ) + gL̄ (V − VL )
2
ri ∂x
∂t
(57)
Here, the three right-hand side terms describe the membrane current in terms of current per length
of axon. Then, we can rewrite all of the membrane currents as current per area of membrane, and
substitute for the constants that varied with the radius of the cylinder:
πa2 ∂ 2 V
∂V
= 2πaCm
+ 2πa[ĝN¯a m3 h(V − VN a ) + ĝK̄ n4 (V − VK ) + ĝL̄ (V − VL )]
2
Ri ∂x
∂t
(58)
Now, we can isolate the terms with radius, a, on the left side of the equation. As before, we assume
that the segments are on a one-dimensional domain and we also take into account that the axon
contains non-ohmic active conductances. This balance of axial and longitudinal currents along every
point gives rise to the non-linear cable equation:
a ∂2V
∂V
= Cm
+ ĝN¯a m3 h(V − VN a ) + ĝK̄ n4 (V − VK ) + ĝL̄ (V − VL )
2
2Ri ∂x
∂t
(59)
We know that we can refer to a propagation of nerve impulse as a travelling wave, hence we can
write it as a function of one variable, length x,
V (x, t) = F (x − st)
(60)
where s is the velocity at which the action potential propagates. With this assumption and the use
of the chain rule, we get
∂2V
1 ∂2V
=
(61)
∂x2
s2 ∂t2
And the non-linear cable equation is now a differential equation that resembles:
a
d2 V
dV
=
+ U (V, t)
2
2
2Ri Ch dt
dt
(62)
where U (V, t) is are the Hodgkin-Huxley currents and do not depend on the radius of the axons.
17
4.3
Computational Analysis
While properly solving this equation analytically proves to be difficult, we can study appropriate
boundary conditions.
4.3.1
Boundary Condition
As always, it is important to look at the boundary condition when the two ends of the segment are
sealed (just as we did with the linear cable equation, see section 4.1.1: An Analytic Solution to the
Linear Cable Equation). So when the two ends, x0 and xL , are closed off, there is no net current
flow at either end.
∂V
∂V
(x0 , t) =
(xL , t) = 0
(63)
∂x
∂x
However, at time t = t0 ,the distribution of the membrane potential can be described by a
function, say σ(x).
∂V
(x, t0 ) = σ(x)
(64)
∂t
4.3.2
Branching
As a nerve impulse propagates from the dendrites into the axon, it experiences some forms of
branching. We could assign each branch of the neuron a specific interval. Set points xi (i =
0, 1, 2, 3, , L) such that x0 = 0 < x1 < x2 < x3 < · · · < xL . Hence between each point, we can
specify a certain spatial domain for a branch i between the interval [xi , xi+1 ).
Let’s assume that there exists only two types of branching.
• Non-terminal branch: connected to other branches at both ends.
• Terminal branch: connected to other branches at less than two ends.
In further detail, we know that a branch could be either an axon or dendrite segment (this would
be dependant upon which ions it is conducting in its spatial domain). We could also count the
soma as a branch, but it would only be a point. Therefore, all branches (except for the soma) has
a parent branch.
Future research could lead to modelling the branching of neurons and correct ordering of this
branching phenomenon. If the soma is represented as the point xL , it is possible to order the
branches as they branch away from this endpoint. All branches that are directly connected to the
soma could be assigned the value of 1, and the branches propagating away assigned a value of
n + 1 (n = 1, , maxn), where maxn is the furthest branch possible (these would be analogous to
the terminal branches). By correctly ordering these branches, it would enable us to assign new
boundary conditions depending on the branching instead of the sealed ends proposed earlier.
18
5
Conclusion
The nerve cell is excitable and has the ability to transmit an electrical potential along its body.
This nervous system signalling has been widely adapted by many organisms on this Earth. Its
importance is undeniable and through some basic physics, advanced mathematics, chemistry, and
biology, we are able to conceptually explain this phenomenon. This is clear evidence of how so many
fields of Science can come together to explain the basis of what is occurring in the human body.
By introducing the fundamentals of what constitutes a nerve impulse, it was possible to display
and successfully explain what constitutes an action potential. Furthermore, by concentrating on
this biological phenomenon, the Hodgkin-Huxley model proved to be quite useful in explaining
the propagating nerve impulse along a nerve cell. While the linear cable equation was useful in
explaining some of this behaviour, it failed to include some key parts of the action potential, such
as the ion gated channels that are present in our neurons. Hence, basing our knowledge on the
Hodgkin-Huxley model and what we know of basic physics, it was possible to include these nonlinearities into what we refer to as the non-linear cable equation.
It was demonstrated that each individual ion channel (N a+ and K + ) displays two main types of
voltage dependent behaviour. The first is a graded, diffusive response when exposed to a stimuli
that is below-threshold. However, if this impulse is above a certain threshold, there is a large,
fixed amplitude response. Therefore, the second behaviour is characterized by this above-threshold
stimulus that can take the form of a traveling wave with fixed shape and speed.
In the end, most of what generates a nerve impulse was introduced and explained. By studying
this aspect of the human body, scientists and mathematicians alike have made many advances in
the field of medicine. Through synapses and what we know of the nerve cell signalling, we can
introduce various drugs that can be used against a wide number of diseases such a cardiac failure
and epilepsy.
6
6.1
Appendix
The Complete Derivation of the Linear Cable Equation
First, we shall start by introducing a circuit diagram that represents a sort of conceptual model of
a small patch of neuron membrane. Note that the capacitor is in parallel with a resistor, which
represents the membrane current. Also note that the segments of membrane are all connected
together with another resistance which efficiently represents the intracellular environment.
19
Figure 8: A circuit diagram explaining the parameters and the trends observed in the linear cable
equation. All parameters and equations are explained in the body of this section [13].
We can now use this conceptual model to define a basic equation to model a neuron:
6.1.1
Voltage Parameters
Let’s start by defining the voltage parameters ,
• V2 -Electrostatic potential inside the cell
• V21 -Electrostatic potential inside at location x1
• V22 -Electrostatic potential inside at location x2
• V1 -Electrostatic potential outside the cell
• V11 -Electrostatic potential outside at location x1
• V12 -Electrostatic potential outside at location x2
• Er -The difference of the voltage inside versus the outside, hence the resting membrane potential. The voltage with no net current flow is:
Er = V 2 − V 1
20
(65)
• Vm - The membrane potential (V2 − V1 ).
We can thus write the equation for the potential V , which shows that it is zero at rest:
V = V 2 − V 1 − Er
6.1.2
(66)
Intracellular Parameters
• r2 - Axial resistance per unit length (in Ω/cm)
Note: The intracellular cytoplasm provides a resistance to current flow, the actual resistance will depend on the diameter of the dendrite, and will be independent of the length of
the dendrite.
• r2 $x- Axial resistance of a $x length, in cm, of dendrite (in Ω)
• r1 - Extracellular resistance per unit length of dendrite (in Ω/cm)
Note: The extracellular cytoplasm provides a resistance to the flow of current.
• r1 $x- Extracellular resistance of a $x length, in cm, of dendrite (in Ω)
• Ci - Capacitance per unit length of membrane (in F/cm)
• Ci $x- Capacitance of a section of a dendrite $x (in F )
• gi - Conductance per unit length of membrane (in S/cm)
• gi $x- Conductance of a membrane patch that has length $x (in S)
•
ri
$x -
Resistance of a section of a dendrite with length $x (in Ω)
Note: The conductance is the inverse of the resistance (and Ω is the inverse of S), hence
Resistance =
1
ri
=
gi $x
$x
(67)
• ri - Resistance of a unit length section of membrane (in Ω · cm)
Note: This is the inverse of gi .
6.2
The Derivation
We can now proceed with deriving the linear cable equation:
We know that Ohm’s Law states:
V = IR
(68)
Hence, we can say that for two different locations on a section on the cylinder of a neuron, x1
and x2 (with x2 > x1 )
$x = x2 − x1
(69)
$V = V2 − V1
21
(70)
Note: The current, denoted I, is positive with positive values of x. Also, outward current is positive
and inwards current is negative.
Now, if we consider a potation of circuit represented by current flowing through the resistance,
mentioned earlier as r2 $x, then by Ohm’s Law:
V21 − V22 = I1 r1 $x
(71)
−$V2 = I2 r2 $x
(72)
And, we can divide by $x, then take the limit as $x approaches 0
Similarly,
$V
= −I2 r2
$x
(73)
∂V2
= −I2 r2
∂x
(74)
∂V1
= I1 r1
∂x
(75)
For the next step, we must use Kirchoff’s Law which states that in any circuit, the sum of the
currents coming into a node equals the sum of currents flowing out. So, let’s define some current
parameters:
$I = I2 − I1
(76)
• Ii - Membrane current per unit length (A/cm, A for Amperes)
• Ii $x- Membrane current across a section of a dendrite that has length $x (in A)
If the currents are meeting at the node denoted x1 , then by Kirchoff’s Law, the current coming
into the node I1 must equal the currents leaving the node, so:
I1 = I2 + Ii $x
(77)
−$I = Ii $x
(78)
Furthermore,
Once again, we can divide by $x and take the limit as $x approaches zero,
$I
= −Ii
$x
(79)
∂I2
= −Ii
∂x
(80)
22
Similarly,
∂I1
= Ii
∂x
(81)
q = Cm ($V )
(82)
We must now use Ohm’s Law,
To explain the change of current with time, we can take the derivative of both sides with respect
to time:
dV
dq
= Cm (
)
(83)
dt
dt
Note: Cm does not depend on time
Hence,
dVm
dt
= V2 − V1 and the current is through the capacitance section of the circuit.
I$x|capacitance = Cm $x
where Vm
(84)
Divide by $x, we get the capacitive current,
Icapacitance
dVm
=
Cm
dt
(85)
And so, by Ohm’s Law(V = IR), we must now analyze the resistance section of the circuit.
Therefore, for V , we must consider the difference from V2 through the battery Er to V1 . Also, for
I and R, we must consider the number of terms from the circuit.
V2 − (V1 + Er ) = I$x|resistance
ri
$x
(86)
Now, replace V2 − V1 with Vm , then cancel the δx, and divide by ri
Iresistance =
V m Er
(Vm − Er )
−
=
ri
ri
rm
(87)
Put Icapacitance and Iresistance together,
I = Cm
dVm Vm − Er
+
dt
ri
(88)
Finally, let’s put the intracellular and extracellular equations for current flow together. Hence,
combine
∂V2
= −I2 r2
(89)
∂x
and
∂V1
= I1 r1
(90)
∂x
to get
∂Vm
= −I2 r2 + I1 r1
(91)
∂x
23
Next, take the derivatives with respect to x:
∂ 2 Vm
∂I2
∂I1
= −r2
+ r1
2
∂x
∂x
∂x
Substitute
∂I2
∂x
= −Ii and
∂I1
∂x
= Ii , and get
∂ 2 Vm
= r2 Ii + r1 Ii = Ii (r2 + r1 )
∂x2
Substitute Ii = Cm dVdtm +
(92)
Vm +Er
Vm
(93)
in place of Ii .
∂ 2 Vm
∂Vm Vm − Er
= (r2 + r1 )(Cm
+
)
∂x2
∂t
ri
(94)
And then rearrange the terms:
ri ∂ 2 V m
∂Vm
= ri Cm
+ (Vm − Er )
2
r2 + r1 ∂x
∂t
(95)
Now we can consider the voltage in terms of a displacement from an initial resting potential and
then substitute V = Vm − Er to get the linear cable equation:
λ2axon
∂2V
∂V
−τ
=V
2
∂x
∂t
(96)
where λaxon is the space constant and defined as:
λaxon =
!
rm
r 2 + r1
(97)
and where τ is the time constant and defined as:
τ = ri Cm
(98)
In terms of the length of the dendrite, we use ri , Cm , and r2 and have units Ω · cm, F/cm, and
Ω/cm, respectively. All of these values vary with the diameter, hence we can incorporate this by
doing the following:
First, define Ri as the resistance of a unit area of membrane (in Ω · cm2 ), where
or
Ri = ri πd
(99)
ri
(πd$x) = Ri
$x
(100)
where πd$x is the membrane area of a cylinder, Ci is the capacitance per unit area (in F/cm2 ),
defined as:
C
Ci =
(101)
πd
24
and R2 is the axial resistance through a cross sectional area per unit length (Ω · cm), defined by
R2 = r2 πr2
(102)
and r is the radius.
We can re-write the space and time constant as,
λaxon =
$
Ri d
4R2
τ = Ri Ci
6.3
(103)
(104)
Solution to problem 12.6 in Biological Physics, P. Nelson [9]
αy −1
Problem: (Analytical solution for simplified action potential) Show that
% thes function ṽ = (1+e )
solves Equation 12.24, if we take the parameter Q to be given by 2/s( 2 − 1). Hence derive the
speed of the action potential (Equation 12.25). α is another constant, which you are to find.
Solution: First, I changed ṽ to just v, for simplicity. Next, compute the first and second derivatives:
v(y) = (1 + eαy )−1
(105)
dv
= −αeαy (1 + eαy )−2
dy
(106)
d2 v
= 2α2 e2αy (1 + eαy )−3 − α2 eαy (1 + eαy )−2
dy 2
(107)
Next, the LHS of the equation now becomes:
Expanded, this becomes
2α2 e2αy
α2 eαy
−
(1 + eαy )3 (1 + eαy )2
(108)
eαy (−α2 ) + e2αy (α2 )
(1 + eαy )3
(109)
Now, looking at the RHS of the equation and expanding, we get:
= −Q(
−αeαy
s
(1 + s)
1
)+
−
+
αy
2
αy
3
αy
2
(1 + e )
(1 + e )
(1 + e )
(1 + eαy )
(110)
By performing some algebra and simplifying, we get
=
eαy (Qα − s + 1) + e2αy (Qα + 1)
(1 + eαy )3
25
(111)
Comparing factors on both sides, we get the following equalities:
α2 = −Qα + s − 1
(112)
α2 = Qα + 1
(113)
and
By setting both equations equal to each other and substituting Q =
s−2
=
2
!
2 s
( − 1)
s 2
%
2/s( 2s − 1), we get
(114)
to get s = 2.
The RHS of our equation now becomes,
=
(Qα + 1)(eαy + e2αy )
1 + eαy )2
(115)
Now, we find that Q = 0, hence α = ±1. Therefore, the LHS=RHS
−ey + e2y
−ey + e2y
=
(1 + ey )3
(1 + ey )3
(116)
And since both sides are equal, then it is obvious that v = (1 + eαy )−1 solves Equation 12.24.
6.4
The Hodgkin-Huxley Nerve Impulse Generation Code
# h h m o d e l f i x . ode
# C l a s s i c a l Hodgkin−Huxley model f o r Squid Giant Axon
# P o t e n t i a l s h i f t e d t o −65 mV
# S e t t i n g up i n i t i a l v a l u e s
i n i t v=−65, n = 0 . 3 1 8 , h = 0 . 5 9 6 , m=0.053
# S e t t i n g up p a r a m e t e r s
params cm=1
params gnabar =120 , gkbar =36 , g l =0.3
params vna =50 , vk=−77, v l =−54.4
params pon =20 , p o f f =30 , i p =2.4
# C r e a t i n g a s t e p f u n c t i o n ( head ( a r g ) ) f o r t h e a p p l i e d c u r r e n t , z e r o i f arg1 <0 and 1 #o t h e r w i s e .
i a p p l i e d = heav ( p o f f −t ) ∗ heav ( t−pon ) ∗ i p
# Gate o p e n i n g and c l o s i n g r a t e e q u a t i o n s
a lp ha n = 0 . 0 1 ∗ ( v + 5 5 ) / ( 1 − exp ( −(v + 5 5 ) / 1 0 ) )
betan = 0 . 1 2 5 ∗ exp ( −(v + 6 5 ) / 8 0 )
nequ = al ph an / ( al ph an+betan )
taun = 1 / ( a lp ha n+betan )
a lp ha h = 0 . 0 7 ∗ exp ( −(v +65)/20)
betah = 1 / ( 1 + exp ( −(v +35)/10) )
hequ = al ph ah / ( al ph ah+betah )
tauh = 1 / ( a lp ha h+betah )
26
alpham = 0 . 1 ∗ ( v +40)/( 1 − exp ( −(v +40)/10) )
betam = 4∗ exp ( −(v +65)/18)
mequ = alpham / ( alpham+betam )
taum = 1 / ( alpham+betam )
# S e t t i n g up p r o p e r c u r r e n t c o n d u c t a n c e e q u a t i o n s ( where n , h , m a r e p r o b a b i l i t y o f g a t e c o n d u c t a n c e )
i n a = gnabar ∗ mˆ3 ∗ h ∗ ( v−vna )
i k = gkbar ∗ nˆ4 ∗ ( v−vk )
i l = g l ∗ ( v−v l )
# The
v’ =
n’ =
h’ =
m’ =
#
@
@
@
@
4 ODEs :
( i a p p l i e d − i n a − i k − i l ) / cm
( nequ − n ) / taun
( hequ − h ) / tauh
( mequ − m) / taum
XPP i n f o r m a t i o n :
maxstor =12000
t o t a l =100 , dt = 0 . 5 , bounds =10000
method=g e a r , t o l e r = 0 . 0 1 , dtmin = 0 . 0 0 0 0 0 1 , dtmax=1
x l o =0 , x h i =100 , y l o =−80, y h i =35
done
6.5
The Steady State and Time Constant Functions Code
##S e t t i n g up my ODE where :
##y : t h e i n i t i a l v a l u e s f o r t h e ODE system
##parms : any p a r a m e t e r s used i n f u n c t h a t s h o u l d be m o d i f i a b l e
##time : t i m e s a t which e x p l i c i t e s t i m a t e s f o r y a r e d e s i r e d
without r e w r i t i n g the f u n c t i o n
p o t e n t i a l g r a d <−f u n c t i o n ( time , y , parms ) {
g<−with ( a s . l i s t ( c ( y , parms ) ) ,
{
I<− 0
i f ( time>=d e l & time<=d e l+dur )
I = amp / 1 0 0 0 ;
##The 4 g r a d i e n t s
c ( −(1/ c1 ) ∗ ( Gl ∗ (Vm−El)+Gk∗n ˆ 4 ∗ (Vm−Ek)+GNa∗h∗mˆ 3 ∗ (Vm−ENa) − I /A) ,
al p ha n (Vm)∗(1 −n)− betan (Vm) ∗ ( n ) ,
al p hah (Vm)∗(1 −h)− betah (Vm) ∗ ( h ) ,
alpham (Vm)∗(1 −m)−betam (Vm) ∗ (m)
})
l i s t (g)
}
#Checking e q u i l i b r i u m f o r n ( e q u a t i o n ( 6 ) )
alphan<−f u n c t i o n (Vm) { ( 0 . 0 1 ∗ (Vm+55))/(1 − exp ( −(Vm+ 5 5 ) / 1 0 ) ) }
betan<−f u n c t i o n (Vm) { 0 . 1 2 5 ∗ exp ( −(Vm+65)/80)}
al p ha n ( 0 ) / ( al ph an (0)+ betan ( 0 ) )
#Checking e q u i l i b r i u m f o r h ( e q u a t i o n ( 7 ) )
alphah<−f u n c t i o n (Vm) { 0 . 0 7 ∗ exp ( −(Vm+65)/20)}
betah<−f u n c t i o n (Vm){1/(1+ exp ( −(Vm+ 3 5 ) / 1 0 ) ) }
al p hah ( 0 ) / ( al ph ah (0)+ betah ( 0 ) )
#Checking e q u i l i b r i u m f o r m ( e q u a t i o n ( 8 ) )
27
alpham<−f u n c t i o n (Vm) { 0 . 1 ∗ (Vm+40)/(1− exp ( −(Vm+ 4 0 ) / 1 0 ) ) }
betam<−f u n c t i o n (Vm) { 4 ∗ exp ( −(Vm+65)/18)}
alpham ( 0 ) / ( alpham (0)+ betam ( 0 ) )
c u r v e ( alphan , from =−100, t o =100 , x l a b=”Vm” , y l a b =””)
c u r v e ( alphah , add=TRUE, c o l =2)
#This i s t h e f i g u r e f o r t h e e q u i l i b r i u m f u n c t i o n f o r t h e t h r e e v a r i a b l e s m, n , h
# i n t h e Hodgkin−Huxley model . The r e s t i n g p o t e n t i a l i s a t Vm=0.
# The a s y m p t o t i c v a l u e x ˆ ∗ (Vm) i s g i v e n by t h e t r a n s f o r m a t i o n x ˆ ∗ (Vm)= a l p h a x (Vm) / [ a l p h a x (Vm)+ b e t a x (Vm) ]
# Added a +65 s h i f t t o t h e c u r v e s
c u r v e ( al ph an ( x ) / ( a lp ha n ( x)+ betan ( x ) ) , from =−100, t o =100 , n=317 , y l a b=”x e q u ” , x l a b=”V” )
c u r v e ( al ph ah ( x ) / ( a lp ha h ( x)+ betah ( x ) ) , add=TRUE, c o l =2)
c u r v e ( alpham ( x ) / ( alpham ( x)+betam ( x ) ) , add=TRUE, c o l =4 , n=317)
l e g e n d ( ” b o t t o m r i g h t ” , c ( ” n ” , ” h ” , ”m” ) , l t y =1 , c o l=c ( 1 , 2 , 4 ) )
#This i s t h e f i g u r e f o r t h e time c o n s t a n t f o r t h e t h r e e v a r i a b l e s m, n , h i n t h e Hodgkin−Huxley model .
#The r e s t i n g p o t e n t i a l i s a t Vm=0.
#The time c o n s t a n t T(Vm) i s g i v e n by t h e t r a n s f o r m a t i o n T(Vm)=[ a l p h a x (Vm)+ b e t a x (Vm)]ˆ −1
#Added a +65 s h i f t t o t h e c u r v e s
c u r v e ( 1 / ( al ph an ( x)+ betan ( x ) ) , from =−100, t o =100 , y l i m=c ( 0 , 1 0 ) , n=317 , y l a b=”t a u x ” , x l a b=”V” )
c u r v e ( 1 / ( al ph ah ( x)+ betah ( x ) ) , add=TRUE, c o l =2)
c u r v e ( 1 / ( alpham ( x)+betam ( x ) ) , add=TRUE, c o l =4 , n=317)
l e g e n d ( ” t o p r i g h t ” , c ( ” n ” , ”h ” , ”m” ) , l t y =1 , c o l=c ( 1 , 2 , 4 ) )
params0<−c ( Gl = 0 . 3 , Gk=36 , GNa=120 , c1 =1 , El = −54.3 , Ek=−77, ENa=50 , A=1.26 e −5 , d e l =5 , dur =1 , amp=0.1)
p o t e n t i a l g r a d ( time =0 , y=c (Vm=0 , n=0 , h=0 , m=0) , parms=params0 )
l i b r a r y ( deSolve )
#y1 <− c (Vm=0 , n=0 ,h=0 ,m=0)
#y2 <− c (Vm=0 , n=al ph an ( 0 ) / ( al ph an (0)+ betan ( 0 ) ) , h=a lp ha h ( 0 ) / ( a lp ha h (0)+ betah ( 0 ) ) ,
# m=alpham ( 0 ) / ( alpham (0)+ betam ( 0 ) ) )
#y3 <− c (Vm=65 , n=a lp ha n ( 0 ) / ( al p han (0)+ betan ( 0 ) ) , h=al ph ah ( 0 ) / ( a lp ha h (0)+ betah ( 0 ) ) ,
# m=alpham ( 0 ) / ( alpham (0)+ betam ( 0 ) ) )
y4<− c (Vm=−65, n = 0 . 3 1 8 , h = 0 . 5 9 5 , m= 0 . 0 5 3 )
t v e c 1 <− s e q ( 0 , 3 0 , by =0.1)
HH1 <− ode ( y4 , t v e c 1 , p o t e n t i a l g r a d , params0 )
p l o t (HH1)
References
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28
[8] Nebylitsyn, V.D. 1972. Fundamental Properties of the Human Nervous System. Chapter 2: Structure of the Basic Properties
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