Mathematical Modeling of Biological Neurons

Douglas Zhou & David CaiSongting Li & Yanyang Xiao

Math. Dept. and the Institute of Natural SciencesShanghai Jiao Tong University

June. 18-23, 2012

Outline:

Introduction(1) cerebral cortex and neuron (2) synapse and neuronal network

Hodgkin-Huxley model(1) experimental setup(2) mathematical modeling

Theoretical and numerical results(1) phase plane analysis(2) numerical simulation of dynamics(3) your project

Neuron and neuronal network

Cerebral Cortex• 1011 neurons and 1015 connections in human brain

• 104 cells and 1 km wiring per mm2 in the cortex• different shape, size and function

Area: ~2200cm2

Thickness: 2~3mm

NeuronDendrite(树突)“input device”

Soma(胞体)“central processing unit”

Axon(轴突)“output device”

Action Potentials(电脉冲)

Encoding

Decoding

Information encoding and decoding

Synapses

• Axons of Presynaptic neuron (sending signal)

• Synapse

• Dendrites of Postsynaptic neuron (receiving signal)

• A neuron may be connected to 102-104 other postsynaptic neurons• Some axons reach several cm’s away!• Chemical synapses: most common in vertebrate brain • Electrical synapses (gap junctions): direct electrical coupling (may be involved in ‘synchronization’ of neurons)

The connection between two neurons

Synaptic cleft

What happens at a Synapse?

Synaptic cleft

Presynaptic axon

Action potential arrives at a synapse 1. Calcium channels open leading to calcium influx

2. Synaptic vesicles fuse with cell membrane

3. Neurotransmittersreleased into synapticcleft

4. Neurotransmitters diffuse across synaptic cleft and binds to receptors on postsynaptic dendrite

5. Ion channels open

6. Inward or outward flux of ions changes membrane potential of postsynaptic neuron

Postsynaptic dendrite

RC circuit (warm up)

Cell Membrane

Membrane: 3 to 4 nm thick, essentially impermeable

Ionic Channels: Selectively permeable

)()()( outin tVtVtVm −=

modeling

The Membrane: Capacitance

Current I

C = Q/V (Q = CV)

1 Farad = 1 Coulomb/ 1 Volt

dQ/dt = I

I = C dV/dt

Outside

Inside

I

V

Voltage V

The Membrane: Resistance

R = V/I

1 Ohm = 1 Volt/ 1 Ampere

I = V/R

Outside

Inside

I

V

Voltage V

Current I

The Membrane: Capacitance and Resistance

I = C dV/dt + V/R

(C*R) dV/dt = -V + IR

Outside

Inside

I

V

Current I

Voltage V

The Membrane Potential Two types of Ions (sodium Na+ and potassium K+)

Reversal Potential: when opposing currents balance each other out.

Nernst Equation:

Na+: +50 mV, K+: -77 mVEquilibrium Potential: averaged by different ions (-70 mV).

voltage gradient

concentration gradient

intracellular

extracellular

[extra]log[intra]

E ∝

[Na+]2

[Na+]1

[K+]2

[K+]1

Mathematical modeling

Basic Concepts

Detailed Model of Neuron

Reduced Model of Neuron

Network Model

Neuron modeling

Experiments observations

Hodgkin-Huxley model

Alan L. Hodgkin1914-1998

Andrew F. Huxley1917-

1963 Nobel Prize

Squid giant axon

Hodgkin-Huxley Equations

: resistivity of axonρ: radius of the axona

: time, ( , ): axial current, ( , ): internal voltaget i x t V x t( , ): trans-membrane current per unit areaI x t: membrane capacitance per unit areaC

: leakage conductance per unit areaLg, : conductance per unit area of , Na Kg g Na K+ +

: equilibrium potential for leakage ccurrentLE, : equilibrium potential for , Na KE E Na K+ +

: distance along axonx

Hodgkin-Huxley Equations

Charge conservation:

( ) ( ) ( )2

12 1, , 2 ,

x

xi x t i x t a I x t dxπ− = − ⋅

2

1

2 0x

x

i aI dxx

π∂ + = ∂ Since and are arbitrary, we have:1x 2x

2 0i aIx

π∂ + =∂

1x 2x

Hodgkin-Huxley Equations

Ohm’s law for axial current:

Membrane current is the sum of capacitance and ionic current

( ) ( ) ( )L L Na Na K KVI C g V E G V E G V Et

∂= + − + − + −∂

1x 2x

( )2

, a Vi x tx

πρ

∂= −∂

x x x+ Δ

Hodgkin-Huxley Equations

( )( ) ( )1 1m

m

V

VS S

α

β

∗⎯⎯⎯→←⎯⎯⎯

Let the sodium channel consist of 4 independent subunits, but 3 of them are one type ( ) and 1 is another type ( ). Let eachtype of subunits have two states with voltage dependent transitions

1S 2S

The channel is open if and only if all 4 subunits are in the state. Now consider a large population of channels, and let m be

the fraction of subunits in the state and h be the fractionof subunits in the state Then the conductance due to sodium channel is given by

S ∗

( )1S∗

3Na NaG g m h=

And m and h satisfy the kinetic differential equation as:

( )( ) ( )1 ,m mm V m V mt

α β∂ = − −∂

Sodium ion channels (Na+)

( )( ) ( )2 2h

h

V

VS S

α

β

∗⎯⎯⎯→←⎯⎯⎯

( )2S∗

( )( ) ( )1h hh V h V ht

α β∂ = − −∂

Hodgkin-Huxley Equations

Let the potassium channel consist of 4 identical and independentsubunits, each of which has two possible states ( and ) withvoltage dependent transitions

( )( )

n

n

V

VS S

α

β∗⎯⎯⎯→←⎯⎯⎯

S S ∗

The channel is open if and only if all 4 subunits are in the state. Now consider a large population of channels, and let n be

the fraction of subunits in the state . Then the conductance due to potassium channel is given by

S ∗

S ∗

4K KG g n=

And n satisfies the kinetic differential equation as:

( )( ) ( )1n nn V n V nt

α β∂ = − −∂

Potassium ion channels (K+)

( ) ( ) ( )2

3 422L L Na Na K K

V a VC g V E g m h V E g n V Et xρ

∂ ∂+ − + − + − =∂ ∂

( )( ) 1 ( )m mm V m V mt

α β∂ = − −∂

( )( ) 1 ( )h hh V h V ht

α β∂ = − −∂

( )( ) 1 ( )n nn V n V nt

α β∂ = − −∂

21 / , 0.0238 , 35.4 0.0354 mVC F cm a cm cm cmA

μ ρμ

= = = Ω =

2 2 20.3 / , 120 / , 36 /L Na KA A Ag cm g cm g cm

mV mV mVμ μ μ = = =

Hodgkin-Huxley Equations

54.387 , 50 , 77L Na KE mV E mV E mV= − = + = −Where

Hodgkin Huxley Equations

( )0.1 40401 exp

10

m

VV

α+

=+ − −

654exp18m

Vβ + = −

650.07exp20h

Vα + = −

1351 exp

10

h Vβ =

+ + −

( )0.01 55551 exp

10

n

VV

α+

=+ − −

650.125exp80n

Vβ + = −

1. System contains both time and space (PDE)2. The dimension is four (e.g. phase plane analysis)3. Functional forms comes from data fitting (complicated)4. Boundary conditions (geometrical structure)

, , : opening rate. , , : closing ratem h n m h nα α α β β β

Hodgkin Huxley Equations

( ) ( ) ( )2

3 422L L Na Na K K

V a VC g V E g m h V E g n V Et xρ

∂ ∂+ − + − + − =∂ ∂

( ) ( ) ( ) ( )3 4

0L L Na Na K KVC g V E g m h V E g n V E I tt

∂ + − + − + − =∂

The system becomes ordinary differential equations.

We can apply current to the membrane through the silver wire

Quantitative behavior of the Hodgkin-Huxley equations:1. Hodgkin and Huxley worked with the squid giant axon, an

axon which is so large that it is possible to insert a silver wire along its length.

2. The silver is an excellent conductor, this forces to be independent of x. Then m, h and n rapidly fall into like V and become x-independent also.

( ),V x t

Hodgkin-Huxley (HH) model

( ) ( ) ( ) ( )

( )

( )

( )

externalL Na

4K

3L KNa

1

1

1

m m

h h

n n

dVC G V G V G V I tdt

dm m

E

mdtdh h hdtdn n n

t

nm h

d

E E

α β

α β

α β

= − − − − − − + = − − = − − = − −

modeling

TV

NaE

KELE

~ 40 mv

~ -50 mv

~ -80 mv~ -70 mv

0 mv

( )( )

, ,

, ,

m h n

m h n

V

V

α

β⎯⎯⎯⎯→←⎯⎯⎯⎯close open gating variables

, , m h n

Phase plane analysis

( ) ( ) ( ) ( )40

3L L Na Na K K

VC g V E g m h V E g n V Et

I t∂ + − + − + − =∂

( ) ( )( ) 1 ( ) 1 ( )( )m

mm

m V m V m V mV

mt τ

α β ∞=∂ = −∂

−−

( ) ( )( ) 1 ( ) 1 ( )( )h

hh

h V h V h V hV

ht τ

α β ∞=∂ = −∂

−−

( ) ( )( ) 1 ( ) 1 ( )( )n

nn

n V n V n V nV

nt τ

α β ∞=∂ = −∂

−−

Hodgkin Huxley Equations

Where( )( )

( ) ( )n

n n

Vn VV Vα

α β∞ =+

( )( ) ,( ) ( )

h

h h

Vh VV Vα

α β∞ =+

( )( ) ,( ) ( )

m

m m

Vm VV Vα

α β∞ =+

1( )( ) ( )n n

VV V

τα β∞ =

+1( ) ,

( ) ( )h h

VV V

τα β∞ =

+1( ) ,

( ) ( )m m

VV V

τα β∞ =

+

Hodgkin Huxley Equations, , , , , as functions of m h nm h n Vτ τ τ∞ ∞ ∞

1. As , , , , , , m n h h m nV α α β α β β↑ ↓↑ , and m n h ↑ ↓2. fast variable: , slow variables: , m h n (time scale separation)

Phase plane analysisIf we consider fast time scale (~1ms) dynamics and

h and n can be regarded as constants

we need only consider the dynamics of V and m

plot the curves along which V and m reaches steady states

0 ( )m m m Vt ∞

∂ = =∂

3 4

3 40 Na Na K K L L

Na K L

g m hE g n E g EV Vt g m h g n g

+ +∂ = =∂ + +

Notice that: 0K L NaE E E< < <

( )0 0I t =

0Vt

∂ =∂

0mt

∂ =∂

Depending on parameters and the values of h and n (here h and n are treated as parameters in the fast time scale analysis), the curves and may have one or three intersections. 0V

t∂ =∂ 0m

t∂ =∂

R: rest state (stable), T: threshold (unstable), E: excited state (stable)

Fast time scale dynamics: (m,V) planePhase plane analysis

Bistable!

Phase plane analysis

0Vt

∂ =∂

0mt

∂ =∂

Fast time scale dynamics: (m,V) plane

At T, there are two special trajectories (indicated by thick blacklines) that end up at T, all others turn away from T and end upat R or E. The special trajectories form a separatrix which divide the phase plane into two parts. Any initial condition to theleft of the separatrix will end at state R; any initial conditionto the right will end at state E

separatrix

Phase plane analysisSub-threshold and supra-threshold excitation:

Let the system be in state R prior to t=0, and letso that Q0 is the total charge per unit area delivered to the membrane by a current shock applied at t=0. This produces ajump in voltage Q0 /C. Since m cannot change instantaneously,the move in the phase plane is a horizontal shift △V. If this takesthe system past the separatrix (case in brown), the result is excitation, otherwise, the system returns to rest (case in magenta)

( ) ( )0 0I t Q tδ=

0Vt

∂ =∂

0mt

∂ =∂

Phase plane analysisSub-threshold and supra-threshold excitation:

Let the system be in state R prior to t=0, and letso that Q0 is the total charge per unit area delivered to the membrane by a current shock applied at t=0. This produces ajump in voltage Q0 /C. Since m cannot change instantaneously,the move in the phase plane is a horizontal shift △V. If this takesthe system past the separatrix (case in brown), the result is excitation, otherwise, the system returns to rest (case in magenta)

( ) ( )0 0I t Q tδ=

Phase plane analysisRecovery following excitation (effects of slow time scale process)

If h and n were really constants, the excited state would persist indefinitely (unless a negative stimulus were applied of sufficient magnitude to cross the separatrix again and return the system to rest). In fact, however, h and n change slowly, and the whole phase portrait evolves in such a way that the state E and T collide and mutually annihilate leaving only the state R, to which the system then returns.

Phase plane analysis

If h and n were really constants, the excited state would persist indefinitely (unless a negative stimulus were applied of sufficient magnitude to cross the separatrix again and return the system to rest). In fact, however, h and n change slowly, and the whole phase portrait evolves in such a way that the state E and T collide and mutually anihilate leaving only the state R, to which the system then returns.

Recovery following excitation (effects of slow time scale process)

Phase plane analysisSlow time scale dynamics: (n,V) plane

As a preliminary step, we simply the situation by setting: 1h n= −In order to be consistent with HH equations, we also need:

( ) ( ) ( ) ( )1 and h hh V n V V Vτ τ∞ ∞+ = =Now we consider the slow time scale analysis, which is to assume that the fast scale m are always “at equilibrium” as n (and hence h) evolves, namely, ( )m m V∞=

( ) ( )( )( ) ( )3 40 1 0L L Na Na K KV g V E g m V n V E g n V Et ∞

∂ = − + − − + − =∂Record

( ) ( ) ( )( )( ) ( )3 4, 1L L Na Na K Kf n V g V E g m V n V E g n V E∞

= − + − − + −

Phase plane analysis

This is called the slow manifold. Note that there is an intervalof n between n1 and n2 , where there are three values of V corresponding to each value of n. These correspond to the three constant states R, T, E of the fast (m,V) phase plane. whenever the system is not on the slow manifold, it moves rapidly by means of changes in the fast variables. On the slow time scale, these rapid changes look like jumps. Since n can’t change fast, they appear as horizontal shift in the (n,V) phase plane. Combining with the fast time scale analysis, we know that all trajectories that start off the slow manifold jump horizontally to the branch of R or E.

Slow time scale dynamics: (n,V) plane

0Vt

∂ =∂

0nt

∂ =∂

Phase plane analysisSlow time scale dynamics: (n,V) plane

Once the system reaches the slow manifold, the system evolves according to ( )

( )n

n V nnt Vτ

∞ −∂ =∂

Where or depending on whether the system happens to be on the rest (R) branch or on the excited (E) branch of the slow manifold. ( and are the twostable solutions of and is defined for and is defined for , since , the domains overlap.

( )( )RV V n t= ( )( )EV V n t=

( )( )RV n t ( )( )EV n t( ), 0f n V = RV

EV1 1n n≤ ≤

20 n n≤ ≤ 1 2n n<

Phase plane analysisSlow time scale dynamics: (n,V) plane

The evolution along the slow manifold carries n towards . Thecurve intersects the R branch of the slow manifold, but not the E branch. Hence, on the R branch, the n dynamics carries the system towards a stable resting state, but on the E branch, the n dynamics drives the system towards and then past the value n=n2 , where the E branch terminates. The only possibility at that point is to jump to the R branch.

( )n V∞( )n n V∞=

0Vt

∂ =∂

0nt

∂ =∂

Phase plane analysisAction potential in slow time scale dynamics

The rest state of the neuron is the intersection of the curve: (the slow manifold) with the curve .

Let a neuron be at rest and then we apply a current pulse whichis sufficient large to step the voltage across the T branch of theslow manifold. Then the neuron jumps to the E manifold, whereit evolves to the top, jumps back to the R manifold, and then recoveries to its starting point.

( ), 0f n V = ( )n n V∞=

0Vt

∂ =∂

0nt

∂ =∂

Phase plane analysisAction potential in slow time scale dynamics

The rest state of the neuron is the intersection of the curve: (the slow manifold) with the curve .

Let a neuron be at rest and then we apply a current pulse whichis sufficient large to step the voltage across the T branch of theslow manifold. Then the neuron jumps to the E manifold, whereit evolves to the top, jumps back to the R manifold, and then recoveries to its starting point.

( ), 0f n V = ( )n n V∞=

Phase plane analysisRelative refractory period in dynamics

0Vt

∂ =∂

0nt

∂ =∂

Following an action potential, a second stimulus is given duringthe recovery phase (between point 4 and the returning point 0 in the diagram). Discuss the following issues: the threshold voltage that must be reached to achieve an

action potential. the size of the voltage step required to reach threshold the peak voltage achieved during the action potential the duration of the action potential

Phase plane analysisSpontaneous oscillations

Suppose the slow manifold is shifted so that it intersects the curve somewhere on the unstable T branch of theSlow manifold. Then the (n,V) phase plane looks like the following case, what is going to happen?

( )n n V∞=

0Vt

∂ =∂

0nt

∂ =∂

Phase plane analysisAnode-break excitation

From until , the trans-membrane potential is clamped at some value which is sufficiently negative that

(notice that n1 is the smallest value of n reached by the R branch of the slow manifold). At the voltage clamp is removed. What is going to happen? Would the result be thesame if a membrane at rest were suddenly stepped to the voltage ? Explain the difference between these two situations.

t = −∞ 0t =V∗

0t =

V∗

( ) 1n V n∞ ∗ <

0Vt

∂ =∂

0nt

∂ =∂

References

Hodgkin, A. L. & Huxley, A. F. A quantitative description of membrane current and its application to conduction and excitation in nerve. Journal of Physiology, 117, 500-544, (1952).

Koch, C. Biophysics of computation. Oxford: Oxford University Press. (1999).

Peskin, C. S. Mathematical aspects of neurophysiology. Lecture Notes. (2000).

Thank You!