Lecture 6: Dendrites and Axons

• Cable equation

• Morphoelectronic transform

• Multi-compartment models

• Action potential propagation

Refs: Dayan & Abbott, Ch 6, Gerstner & Kistler, sects 2.5-6; C Koch, Biophysics of Computation, Chs 2,6

Longitudinal resistance and resistivity

Longitudinal resistance and resistivity

2aLr

R LL Longitudinal resistance

Longitudinal resistance and resistivity

2aLr

R LL Longitudinal resistance

Longitudinal resistivity rL ~ 1-3 kmm2

Longitudinal resistance and resistivity

2aLr

R LL Longitudinal resistance

Longitudinal resistivity rL ~ 1-3 kmm2

x

V

r

aL

x

V

RV

RI

LLLL

211

Cable equation

Cable equation

rightLleftLemm x

Vra

xV

ra

iixatV

cxa

22

)(22

current balance:

Cable equation

rightLleftLemm x

Vra

xV

ra

iixatV

cxa

22

)(22

xxV

ra

xxV

ra

xV

ra

LrightLleftL

222

current balance:

on rhs:

Cable equation

rightLleftLemm x

Vra

xV

ra

iixatV

cxa

22

)(22

xxV

ra

xxV

ra

xV

ra

LrightLleftL

222

emL

m iixV

axart

Vc

2

21

current balance:

on rhs:

Cable equation:

Linear cable theory

m

restm r

VVi

Ohmic current:

Linear cable theory

m

restm r

VVi

rVVu

Ohmic current:

Measure V relative to rest:

Linear cable theory

m

restm r

VVi

rVVu

emL

m iru

xu

ra

tu

c

2

2

2

Ohmic current:

Measure V relative to rest:

Cable equation becomes

Linear cable theory

m

restm r

VVi

rVVu

emL

m iru

xu

ra

tu

c

2

2

2

mmmL

m crr

ar 2

Ohmic current:

Measure V relative to rest:

Cable equation becomes

Now define electrotonic length and membrane time constant:

Linear cable theory

m

restm r

VVi

rVVu

emL

m iru

xu

ra

tu

c

2

2

2

mmmL

m crr

ar 2 emm iru

xu

tu

2

22

Ohmic current:

Measure V relative to rest:

Cable equation becomes

Now define electrotonic length and membrane time constant:

Linear cable theory

m

restm r

VVi

rVVu

emL

m iru

xu

ra

tu

c

2

2

2

mmmL

m crr

ar 2 emm iru

xu

tu

2

22

Ohmic current:

Measure V relative to rest:

Cable equation becomes

Now define electrotonic length and membrane time constant:

Note: cable segment of length has longitudinal resistance = transverse resistance:

Linear cable theory

m

restm r

VVi

rVVu

emL

m iru

xu

ra

tu

c

2

2

2

mmmL

m crr

ar 2 emm iru

xu

tu

2

22

22 ar

ar

R Lm

Ohmic current:

Measure V relative to rest:

Cable equation becomes

Now define electrotonic length and membrane time constant:

Note: cable segment of length has longitudinal resistance = transverse resistance:

dimensionless units:

/ˆ/ˆ ttxx

extTextLT iriirri ˆˆ

dimensionless units:

/ˆ/ˆ ttxx

extTextLT iriirri ˆˆ

Removes , m from equation.

dimensionless units:

),(),(),(),(

2

2

txirtxux

txut

txuem

/ˆ/ˆ ttxx

extTextLT iriirri ˆˆ

Removes , m from equation.

Now remove the hats:

dimensionless units:

),(),(),(),(

2

2

txirtxux

txut

txuem

/ˆ/ˆ ttxx

extTextLT iriirri ˆˆ

Removes , m from equation.

Now remove the hats:

(t really means t/m, x really means x/)

Stationary solutions0

tuNo time dependence:

Stationary solutions0

tu

)(ˆ)()()(

2

2

xixirxux

xueem

No time dependence:

Static cable equation:

Stationary solutions0

tu

)(ˆ)()()(

2

2

xixirxux

xueem

No time dependence:

Static cable equation:

General solution where ie = 0:

Stationary solutions0

tu

)(ˆ)()()(

2

2

xixirxux

xueem

)exp()exp()( 21 xcxcxu

No time dependence:

Static cable equation:

General solution where ie = 0:

Stationary solutions0

tu

)(ˆ)()()(

2

2

xixirxux

xueem

)exp()exp()( 21 xcxcxu

)()(ˆ 0 xixie

No time dependence:

Static cable equation:

General solution where ie = 0:

Point injection:

Stationary solutions0

tu

)(ˆ)()()(

2

2

xixirxux

xueem

)exp()exp()( 21 xcxcxu

)()(ˆ 0 xixie

xixu e)( 021

No time dependence:

Static cable equation:

General solution where ie = 0:

Point injection:

Solution:

Stationary solutions0

tu

)(ˆ)()()(

2

2

xixirxux

xueem

)exp()exp()( 21 xcxcxu

)()(ˆ 0 xixie

xixu e)( 021

No time dependence:

Static cable equation:

General solution where ie = 0:

Point injection:

Solution:

Stationary solutions0

tu

)(ˆ)()()(

2

2

xixirxux

xueem

)exp()exp()( 21 xcxcxu

)()(ˆ 0 xixie

xixu e)( 021

)(ˆe)( 21 xixdxu e

xx

No time dependence:

Static cable equation:

General solution where ie = 0:

Point injection:

Solution:

Solution for general ie:

Boundary conditions at junctions

Boundary conditions at junctions

V continuous

Boundary conditions at junctions

V continuous

Sum of inward currents must be zero at junctionxV

a 2

Boundary conditions at junctions

0

xV

V continuous

Sum of inward currents must be zero at junction

closed end:

xV

a 2

Boundary conditions at junctions

0

xV

V continuous

Sum of inward currents must be zero at junction

closed end:

xV

a 2

open end: V = 0

Green’s functionResponse to delta-function current source (in space and time)

Green’s function

)()(),(),(),(

2

2

txtxux

txut

txu

Response to delta-function current source (in space and time)

Green’s function

)()(),(),(),(

2

2

txtxux

txut

txu

Response to delta-function current source (in space and time)

Spatial Fourier transform:

Green’s function

)()(),(),(),(

2

2

txtxux

txut

txu

)(),(),(),( 2 ttkutkuk

ttku

Response to delta-function current source (in space and time)

Spatial Fourier transform:

Green’s function

)()(),(),(),(

2

2

txtxux

txut

txu

)(),(),(),( 2 ttkutkuk

ttku

)(])1(exp[),( 2 ttktku

Response to delta-function current source (in space and time)

Spatial Fourier transform

Easy to solve:

Green’s function

)()(),(),(),(

2

2

txtxux

txut

txu

)(),(),(),( 2 ttkutkuk

ttku

)(])1(exp[),( 2 ttktku

Response to delta-function current source (in space and time)

Spatial Fourier transform

Easy to solve:

Invert the Fourier transform:

Green’s function

)()(),(),(),(

2

2

txtxux

txut

txu

)(),(),(),( 2 ttkutkuk

ttku

)(])1(exp[),( 2 ttktku

),(4

exp4

)(),(

2

txGt

xt

t

ttxu

Response to delta-function current source (in space and time)

Spatial Fourier transform

Easy to solve:

Invert the Fourier transform:

Green’s function

)()(),(),(),(

2

2

txtxux

txut

txu

)(),(),(),( 2 ttkutkuk

ttku

)(])1(exp[),( 2 ttktku

),(4

exp4

)(),(

2

txGt

xt

t

ttxu

),(ˆ),(),( txittxxGxdtdtxu e

t

Response to delta-function current source (in space and time)

Spatial Fourier transform

Easy to solve:

Invert the Fourier transform:

Solution for general ie(x,t) :

Pulse injection at x=0,t=0:

tx

tt

txu4

exp4

1),(

2

Pulse injection at x=0,t=0:

tx

tt

txu4

exp4

1),(

2

Pulse injection at x=0,t=0:

tx

tt

txu4

exp4

1),(

2

u vs t at various x: x vs tmax:

Pulse injection at x=0,t=0:

tx

tt

txu4

exp4

1),(

2

u vs t at various x: x vs tmax:

At what t does u peak?

Pulse injection at x=0,t=0:

tx

tt

txu4

exp4

1),(

2

0log4 2

12

t

tx

tdtd

u vs t at various x: x vs tmax:

At what t does u peak?

Pulse injection at x=0,t=0:

tx

tt

txu4

exp4

1),(

2

0log4 2

12

t

tx

tdtd

021

41 2

2

tt

x

u vs t at various x: x vs tmax:

At what t does u peak?

Pulse injection at x=0,t=0:

tx

tt

txu4

exp4

1),(

2

0log4 2

12

t

tx

tdtd

021

41 2

2

tt

x02)2( 22 xtt

u vs t at various x: x vs tmax:

At what t does u peak?

Pulse injection at x=0,t=0:

tx

tt

txu4

exp4

1),(

2

0log4 2

12

t

tx

tdtd

021

41 2

2

tt

x02)2( 22 xtt

2411

22x

t

u vs t at various x: x vs tmax:

At what t does u peak?

Pulse injection at x=0,t=0:

2

1)/(414

2max

m

x

m xxt

tx

tt

txu4

exp4

1),(

2

0log4 2

12

t

tx

tdtd

021

41 2

2

tt

x02)2( 22 xtt

2411

22x

t

u vs t at various x: x vs tmax:

At what t does u peak?

Restoring, m:

Compare with no-leak case:

tx

ttxu

4exp

4

1),(

2

Compare with no-leak case:

tx

ttxu

4exp

4

1),(

2

Just diffusion, no decay

Compare with no-leak case:

tx

ttxu

4exp

4

1),(

2

Just diffusion, no decay

Now when does it peak for a given x?

Compare with no-leak case:

tx

ttxu

4exp

4

1),(

2

0log4 2

12

t

tx

dtd

Just diffusion, no decay

Now when does it peak for a given x?

Compare with no-leak case:

tx

ttxu

4exp

4

1),(

2

0log4 2

12

t

tx

dtd

021

4 2

2

tt

x

Just diffusion, no decay

Now when does it peak for a given x?

Compare with no-leak case:

tx

ttxu

4exp

4

1),(

2

0log4 2

12

t

tx

dtd

021

4 2

2

tt

x 22 xt

Just diffusion, no decay

Now when does it peak for a given x?

Compare with no-leak case:

tx

ttxu

4exp

4

1),(

2

0log4 2

12

t

tx

dtd

021

4 2

2

tt

x 22 xt

2

2

max 2 x

t m

Just diffusion, no decay

Now when does it peak for a given x?

Restoring, m:

Finite cableMethod of images:

Finite cable

)],2(),2([),;,( 000000,0 ttxnLxGttxnLxGtxtxGn

L

Method of images:

Finite cable

)],2(),2([),;,( 000000,0 ttxnLxGttxnLxGtxtxGn

L

),(ˆ),;,(),( 0000,0

0

00 txitxtxGdxdttxu eL

t L

Method of images:

General solution:

Finite cable

)],2(),2([),;,( 000000,0 ttxnLxGttxnLxGtxtxGn

L

),(ˆ),;,(),( 0000,0

0

00 txitxtxGdxdttxu eL

t L

Method of images:

General solution:

Morphoelectronic transform

Frequency-dependent morphoelectronic transforms

Multi-compartment models

Discrete cable equations

)()( 11,11,

VVgVVg

A

Ii

dt

dVc e

mm

Discrete cable equations

)()( 11,11,

VVgVVg

A

Ii

dt

dVc e

mm

)/( 2aLrL Resistance between compartments:

Discrete cable equations

)()( 11,11,

VVgVVg

A

Ii

dt

dVc e

mm

)/( 2aLrL Resistance between compartments:

Current between compartments: )/()( 12 LrVVa L

Discrete cable equations

)()( 11,11,

VVgVVg

A

Ii

dt

dVc e

mm

)/( 2aLrL

)2)(/()( 12 aLLrVVa L

Resistance between compartments:

Current between compartments:

Current per unit area:

)/()( 12 LrVVa L

Discrete cable equations

)()( 11,11,

VVgVVg

A

Ii

dt

dVc e

mm

)/( 2aLrL

)2)(/()( 12 aLLrVVa L

Resistance between compartments:

Current between compartments:

Current per unit area:

)/()( 12 LrVVa L

21, 2 Lra

gL

Action potential propagation

extL

NaNaKKLLm IxV

ra

VVhmgVVngVVgtV

c

2

234

2)()()(

Action potential propagation

extL

NaNaKKLLm IxV

ra

VVhmgVVngVVgtV

c

2

234

2)()()(

hVhdtdh

VmVmdtdm

VnVndtdn

V hmn )()()()()()(

Action potential propagation

extL

NaNaKKLLm IxV

ra

VVhmgVVngVVgtV

c

2

234

2)()()(

hVhdtdh

VmVmdtdm

VnVndtdn

V hmn )()()()()()(

a “reaction-diffusion equation”

Action potential propagation

extL

NaNaKKLLm IxV

ra

VVhmgVVngVVgtV

c

2

234

2)()()(

hVhdtdh

VmVmdtdm

VnVndtdn

V hmn )()()()()()(

)(),( stxVtxV

a “reaction-diffusion equation”

Moving solutions: look for solution of the form

Action potential propagation

extL

NaNaKKLLm IxV

ra

VVhmgVVngVVgtV

c

2

234

2)()()(

hVhdtdh

VmVmdtdm

VnVndtdn

V hmn )()()()()()(

)(),( stxVtxV

a “reaction-diffusion equation”

Moving solutions: look for solution of the form

Ordinary DE

Action potential propagation

extL

NaNaKKLLm IxV

ra

VVhmgVVngVVgtV

c

2

234

2)()()(

hVhdtdh

VmVmdtdm

VnVndtdn

V hmn )()()()()()(

)(),( stxVtxV

a “reaction-diffusion equation”

Moving solutions: look for solution of the form

extL

NaNaKKLLm Idt

Vdsr

aVVhmgVVngVVg

dtdV

c 2

2

234

2)()()(

Ordinary DE

Action potential propagation

extL

NaNaKKLLm IxV

ra

VVhmgVVngVVgtV

c

2

234

2)()()(

hVhdtdh

VmVmdtdm

VnVndtdn

V hmn )()()()()()(

)(),( stxVtxV

a “reaction-diffusion equation”

Moving solutions: look for solution of the form

extL

NaNaKKLLm Idt

Vdsr

aVVhmgVVngVVg

dtdV

c 2

2

234

2)()()(

Ordinary DE

HH solved iteratively for s (big success of their model)

Propagation speed

a/s2 must be independent of a

Propagation speed

a/s2 must be independent of a

as

Propagation speed

a/s2 must be independent of a

as

This is probably why the squid axon is so thick.

Multi-compartment model

Multicompartment calculation

Myelinated axonsNodes of Ranvier: active Na channels

Myelinated axonsNodes of Ranvier: active Na channels

Myelinated axons

Treat as multilayer capacitor each layer of thickness a:

Nodes of Ranvier: active Na channels

Myelinated axons

a

daLcC m

mm 2Treat as multilayer capacitor each layer of thickness a:

Nodes of Ranvier: active Na channels

Myelinated axons

a

daLcC m

mm 2Treat as multilayer capacitor each layer of thickness a:

Integrate up from a1 to a2 (inverse capacitances add)

Nodes of Ranvier: active Na channels

Myelinated axons

a

daLcC m

mm 2

Ldcaa

ada

LdcC mm

a

ammm 2)/log(

211 12

2

1

Treat as multilayer capacitor each layer of thickness a:

Integrate up from a1 to a2 (inverse capacitances add)

Nodes of Ranvier: active Na channels

Myelinated axons

a

daLcC m

mm 2

Ldcaa

ada

LdcC mm

a

ammm 2)/log(

211 12

2

1

Lx

V

r

a

t

VC

Lm

2

221

Treat as multilayer capacitor each layer of thickness a:

Integrate up from a1 to a2 (inverse capacitances add)

Negligible leakage between nodes: cable equation becomes

Nodes of Ranvier: active Na channels

Myelinated axons

a

daLcC m

mm 2

Ldcaa

ada

LdcC mm

a

ammm 2)/log(

211 12

2

1

Lx

V

r

a

t

VC

Lm

2

221

mLmLm drcaaa

rCLa

D2

)/log( 1221

21

Treat as multilayer capacitor each layer of thickness a:

Integrate up from a1 to a2 (inverse capacitances add)

Negligible leakage between nodes: cable equation becomes

Diffusion constant:

Nodes of Ranvier: active Na channels

How much myelinization? optimal a1/a2

Find the value of y = a1/a2 that maximizes D

How much myelinization? optimal a1/a2

21112 log0log20log yyyy

dyd

yy

Find the value of y = a1/a2 that maximizes D

How much myelinization? optimal a1/a2

21112 log0log20log yyyy

dyd

yy

Find the value of y = a1/a2 that maximizes D

221 6.0e/ aaa

How much myelinization? optimal a1/a2

21112 log0log20log yyyy

dyd

yy

Find the value of y = a1/a2 that maximizes D

221 6.0e/ aaa

Agrees with experiment

Speed of propagation

mLm dreca

D4

22

2

2

xV

DtV

Diffusion equation

with diffusion constant

Speed of propagation

mLm dreca

D4

22

2

2

xV

DtV

22

2

/,const axyyV

tV

Diffusion equation

with diffusion constant

Speed of propagation

mLm dreca

D4

22

2

2

xV

DtV

22

2

/,const axyyV

tV

Diffusion equation

with diffusion constant

Speed of propagation proportional to a2

Speed of propagation

mLm dreca

D4

22

2

2

xV

DtV

22

2

/,const axyyV

tV

Diffusion equation

with diffusion constant

Speed of propagation proportional to a2

(cf a1/2 for unmyelinated axon)