Lecture 4: Diffusion and the Fokker-Planck equation Outline: intuitive treatment Diffusion as flow...

Post on 19-Jan-2016

213 views 0 download

Tags:

transcript

Lecture 4: Diffusion and the Fokker-Planck equation

Outline:

• intuitive treatment• Diffusion as flow down a concentration gradient• Drift current and Fokker-Planck equation

Lecture 4: Diffusion and the Fokker-Planck equation

Outline:

• intuitive treatment• Diffusion as flow down a concentration gradient• Drift current and Fokker-Planck equation

• examples:• No current: equilibrium, Einstein relation• Constant current, out of equilibrium:

Lecture 4: Diffusion and the Fokker-Planck equation

Outline:

• intuitive treatment• Diffusion as flow down a concentration gradient• Drift current and Fokker-Planck equation

• examples:• No current: equilibrium, Einstein relation• Constant current, out of equilibrium:

• Goldman-Hodgkin-Katz equation• Kramers escape over an energy barrier

Lecture 4: Diffusion and the Fokker-Planck equation

Outline:

• intuitive treatment• Diffusion as flow down a concentration gradient• Drift current and Fokker-Planck equation

• examples:• No current: equilibrium, Einstein relation• Constant current, out of equilibrium:

• Goldman-Hodgkin-Katz equation• Kramers escape over an energy barrier

• derivation from master equation

Diffusion Fick’s law:

J = −D∂P

∂x

Diffusion Fick’s law: cf Ohm’s law

J = −D∂P

∂x

I = −g∂V

∂x

Diffusion Fick’s law: cf Ohm’s law

J = −D∂P

∂x

I = −g∂V

∂x

conservation:

∂P

∂t= −

∂J

∂x

Diffusion Fick’s law: cf Ohm’s law

J = −D∂P

∂x

I = −g∂V

∂x

conservation:

∂P

∂t= −

∂J

∂x

∂P

∂t= D

∂2P

∂x 2

=>

Diffusion Fick’s law: cf Ohm’s law

J = −D∂P

∂x

I = −g∂V

∂x

conservation:

∂P

∂t= −

∂J

∂x

∂P

∂t= D

∂2P

∂x 2

=> diffusion equation

Diffusion Fick’s law: cf Ohm’s law

J = −D∂P

∂x

I = −g∂V

∂x

conservation:

∂P

∂t= −

∂J

∂x

∂P

∂t= D

∂2P

∂x 2

=> diffusion equation

initial condition

P(x | 0) = δ(x)

Diffusion Fick’s law: cf Ohm’s law

J = −D∂P

∂x

I = −g∂V

∂x

conservation:

∂P

∂t= −

∂J

∂x

∂P

∂t= D

∂2P

∂x 2

=> diffusion equation

initial condition

P(x | 0) = δ(x)

solution:

P(x | t) =1

4πDtexp −

x 2

4Dt

⎝ ⎜

⎠ ⎟

Diffusion Fick’s law: cf Ohm’s law

J = −D∂P

∂x

I = −g∂V

∂x

conservation:

∂P

∂t= −

∂J

∂x

∂P

∂t= D

∂2P

∂x 2

=> diffusion equation

initial condition

P(x | 0) = δ(x)

solution:

P(x | t) =1

4πDtexp −

x 2

4Dt

⎝ ⎜

⎠ ⎟

http://www.nbi.dk/~hertz/noisecourse/gaussspread.m

Drift current and Fokker-Planck equationDrift (convective) current:

Jdrift (x, t) = u(x)P(x, t)

Drift current and Fokker-Planck equation

Combining drift and diffusion: Fokker-Planck equation:

Drift (convective) current:

Jdrift (x, t) = u(x)P(x, t)

∂P

∂t= −

∂xJdrift + Jdiff( )

Drift current and Fokker-Planck equation

Combining drift and diffusion: Fokker-Planck equation:

Drift (convective) current:

Jdrift (x, t) = u(x)P(x, t)

∂P

∂t= −

∂xJdrift + Jdiff( )

= −∂

∂xu(x)P − D

∂P

∂x

⎡ ⎣ ⎢

⎤ ⎦ ⎥= −

∂xu(x)P( ) + D

∂ 2P

∂x 2

Drift current and Fokker-Planck equation

Combining drift and diffusion: Fokker-Planck equation:

Slightly more generally, D can depend on x:

Drift (convective) current:

Jdrift (x, t) = u(x)P(x, t)

∂P

∂t= −

∂xJdrift + Jdiff( )

= −∂

∂xu(x)P − D

∂P

∂x

⎡ ⎣ ⎢

⎤ ⎦ ⎥= −

∂xu(x)P( ) + D

∂ 2P

∂x 2

Jdiff (x, t) = −∂

∂xD(x)P(x, t)( )

Drift current and Fokker-Planck equation

Combining drift and diffusion: Fokker-Planck equation:

Slightly more generally, D can depend on x:

=>

Drift (convective) current:

Jdrift (x, t) = u(x)P(x, t)

∂P

∂t= −

∂xJdrift + Jdiff( )

= −∂

∂xu(x)P − D

∂P

∂x

⎡ ⎣ ⎢

⎤ ⎦ ⎥= −

∂xu(x)P( ) + D

∂ 2P

∂x 2

Jdiff (x, t) = −∂

∂xD(x)P(x, t)( )

∂P

∂t= −

∂xu(x)P( ) +

∂ 2

∂x 2D(x)P( )

Drift current and Fokker-Planck equation

Combining drift and diffusion: Fokker-Planck equation:

Slightly more generally, D can depend on x:

=>

Drift (convective) current:

Jdrift (x, t) = u(x)P(x, t)

∂P

∂t= −

∂xJdrift + Jdiff( )

= −∂

∂xu(x)P − D

∂P

∂x

⎡ ⎣ ⎢

⎤ ⎦ ⎥= −

∂xu(x)P( ) + D

∂ 2P

∂x 2

Jdiff (x, t) = −∂

∂xD(x)P(x, t)( )

∂P

∂t= −

∂xu(x)P( ) +

∂ 2

∂x 2D(x)P( )

First term alone describes probability cloud moving with velocity u(x)Second term alone describes diffusively spreading probability cloud

Examples: constant drift velocityhttp://www.nbi.dk/~hertz/noisecourse/gaussspreadmove.m

Examples: constant drift velocity

u(x) = u0

P(x, t) =1

4πDtexp −

x − u0t( )2

4Dt

⎣ ⎢ ⎢

⎦ ⎥ ⎥

http://www.nbi.dk/~hertz/noisecourse/gaussspreadmove.m

Solution (with no boundaries):

Examples: constant drift velocity

u(x) = u0

P(x, t) =1

4πDtexp −

x − u0t( )2

4Dt

⎣ ⎢ ⎢

⎦ ⎥ ⎥

http://www.nbi.dk/~hertz/noisecourse/gaussspreadmove.m

Solution (with no boundaries):

Stationary case:

Examples: constant drift velocity

u(x) = u0

P(x, t) =1

4πDtexp −

x − u0t( )2

4Dt

⎣ ⎢ ⎢

⎦ ⎥ ⎥

http://www.nbi.dk/~hertz/noisecourse/gaussspreadmove.m

Solution (with no boundaries):

Stationary case:Gas of Brownian particles in gravitational field: u0 = μF = -μmg

Examples: constant drift velocity

u(x) = u0

P(x, t) =1

4πDtexp −

x − u0t( )2

4Dt

⎣ ⎢ ⎢

⎦ ⎥ ⎥

http://www.nbi.dk/~hertz/noisecourse/gaussspreadmove.m

Solution (with no boundaries):

Stationary case:Gas of Brownian particles in gravitational field: u0 = μF = -μmg

μ =mobility

Examples: constant drift velocity

u(x) = u0

P(x, t) =1

4πDtexp −

x − u0t( )2

4Dt

⎣ ⎢ ⎢

⎦ ⎥ ⎥

http://www.nbi.dk/~hertz/noisecourse/gaussspreadmove.m

Solution (with no boundaries):

Stationary case:Gas of Brownian particles in gravitational field: u0 = μF = -μmg

μ =mobilityBoundary conditions (bottom of container, stationarity):

P(x) = 0, x < 0;

J(x) = 0

Examples: constant drift velocity

u(x) = u0

P(x, t) =1

4πDtexp −

x − u0t( )2

4Dt

⎣ ⎢ ⎢

⎦ ⎥ ⎥

http://www.nbi.dk/~hertz/noisecourse/gaussspreadmove.m

Solution (with no boundaries):

Stationary case:Gas of Brownian particles in gravitational field: u0 = μF = -μmg

μ =mobilityBoundary conditions (bottom of container, stationarity):

P(x) = 0, x < 0;

J(x) = 0 drift and diffusion currents cancel

Einstein relation

FP equation:

μmgP(x) + DdP

dx= 0

Einstein relation

FP equation:

Solution:

μmgP(x) + DdP

dx= 0

P(x) =μmg

D

⎝ ⎜

⎠ ⎟exp −

μmgx

D

⎝ ⎜

⎠ ⎟

Einstein relation

FP equation:

Solution:

But from equilibrium stat mech we know

μmgP(x) + DdP

dx= 0

P(x) =μmg

D

⎝ ⎜

⎠ ⎟exp −

μmgx

D

⎝ ⎜

⎠ ⎟

P(x) =mg

T

⎝ ⎜

⎠ ⎟exp −

mgx

T

⎝ ⎜

⎠ ⎟

Einstein relation

FP equation:

Solution:

But from equilibrium stat mech we know

So D = μT

μmgP(x) + DdP

dx= 0

P(x) =μmg

D

⎝ ⎜

⎠ ⎟exp −

μmgx

D

⎝ ⎜

⎠ ⎟

P(x) =mg

T

⎝ ⎜

⎠ ⎟exp −

mgx

T

⎝ ⎜

⎠ ⎟

Einstein relation

FP equation:

Solution:

But from equilibrium stat mech we know

So D = μT Einstein relation

μmgP(x) + DdP

dx= 0

P(x) =μmg

D

⎝ ⎜

⎠ ⎟exp −

μmgx

D

⎝ ⎜

⎠ ⎟

P(x) =mg

T

⎝ ⎜

⎠ ⎟exp −

mgx

T

⎝ ⎜

⎠ ⎟

Constant current: Goldman-Hodgkin-Katz model of an (open) ion channel

Pumps maintain different inside and outside concentrations of ions

Constant current: Goldman-Hodgkin-Katz model of an (open) ion channel

Pumps maintain different inside and outside concentrations of ions Voltage diff (“membrane potential”) between inside and outside of cell

Constant current: Goldman-Hodgkin-Katz model of an (open) ion channel

Pumps maintain different inside and outside concentrations of ions Voltage diff (“membrane potential”) between inside and outside of cellCan vary membrane potential experimentally by adding external field

Constant current: Goldman-Hodgkin-Katz model of an (open) ion channel

Pumps maintain different inside and outside concentrations of ions Voltage diff (“membrane potential”) between inside and outside of cellCan vary membrane potential experimentally by adding external fieldQuestion: At a given Vm, what current flows through the channel?

Constant current: Goldman-Hodgkin-Katz model of an (open) ion channel

Pumps maintain different inside and outside concentrations of ions Voltage diff (“membrane potential”) between inside and outside of cellCan vary membrane potential experimentally by adding external fieldQuestion: At a given Vm, what current flows through the channel?

outside insidex

x=0 x=d

Constant current: Goldman-Hodgkin-Katz model of an (open) ion channel

Pumps maintain different inside and outside concentrations of ions Voltage diff (“membrane potential”) between inside and outside of cellCan vary membrane potential experimentally by adding external fieldQuestion: At a given Vm, what current flows through the channel?

outside insidex

V(x)Vm

Vout= 0

x=0 x=d

Constant current: Goldman-Hodgkin-Katz model of an (open) ion channel

Pumps maintain different inside and outside concentrations of ions Voltage diff (“membrane potential”) between inside and outside of cellCan vary membrane potential experimentally by adding external fieldQuestion: At a given Vm, what current flows through the channel?

outside insidex

V(x)Vm

Vout= 0

ρout

ρin

x=0 x=d

Constant current: Goldman-Hodgkin-Katz model of an (open) ion channel

Pumps maintain different inside and outside concentrations of ions Voltage diff (“membrane potential”) between inside and outside of cellCan vary membrane potential experimentally by adding external fieldQuestion: At a given Vm, what current flows through the channel?

outside insidex

V(x)Vm

Vout= 0

ρout

ρin

x=0 x=d?

Reversal potential

If there is no current, equilibrium

=> ρin/ρout=exp(-βV)

Reversal potential

If there is no current, equilibrium

=> ρin/ρout=exp(-βV)

This defines the reversal potential

at which J = 0.

Vr = T logρ out

ρ in

⎝ ⎜

⎠ ⎟

Reversal potential

If there is no current, equilibrium

=> ρin/ρout=exp(-βV)

This defines the reversal potential

at which J = 0.

For Ca++, ρout>> ρin => Vr >> 0€

Vr = T logρ out

ρ in

⎝ ⎜

⎠ ⎟

GHK model (2)

outside insidex

V(x)

Vout= 0

ρout

ρin

x=0 x=d?

Vm< 0: both diffusive current and drift current flow in

Vm

GHK model (2)

outside insidex

V(x)Vout= 0

ρout

ρin

x=0 x=d?

Vm< 0: both diffusive current and drift current flow inVm= 0: diffusive current flows in, no drift current

GHK model (2)

outside insidex

V(x)

Vm

Vout= 0

ρout

ρin

x=0 x=d?

Vm< 0: both diffusive current and drift current flow inVm= 0: diffusive current flows in, no drift currentVm> 0: diffusive current flows in, drift current flows out

GHK model (2)

outside insidex

V(x)

Vm

Vout= 0

ρout

ρin

x=0 x=d?

Vm< 0: both diffusive current and drift current flow inVm= 0: diffusive current flows in, no drift currentVm> 0: diffusive current flows in, drift current flows outAt Vm= Vr they cancel

GHK model (2)

outside insidex

V(x)

Vm

Vout= 0

ρout

ρin

x=0 x=d?

Vm< 0: both diffusive current and drift current flow inVm= 0: diffusive current flows in, no drift currentVm> 0: diffusive current flows in, drift current flows outAt Vm= Vr they cancel

Jdrift = μqEρ (x) = −μqdV

dxρ (x) = −

μqVm

dρ (x), Jdiff = −D

dx

Steady-state FP equation

dJ

dx=

d

dx−D

dx−

μqVm

⎝ ⎜

⎠ ⎟= 0

Steady-state FP equation

dJ

dx=

d

dx−D

dx−

μqVm

⎝ ⎜

⎠ ⎟= 0

J = −Ddρ

dx−

μqVm

dρ = const.

Steady-state FP equation

dJ

dx=

d

dx−D

dx−

μqVm

⎝ ⎜

⎠ ⎟= 0

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+

βqVm

⎝ ⎜

⎠ ⎟Use Einstein relation:

Steady-state FP equation

dJ

dx=

d

dx−D

dx−

μqVm

⎝ ⎜

⎠ ⎟= 0

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+

βqVm

⎝ ⎜

⎠ ⎟

−J

μT=

dx+ κρ, κ =

βqVm

d

Use Einstein relation:

Steady-state FP equation

dJ

dx=

d

dx−D

dx−

μqVm

⎝ ⎜

⎠ ⎟= 0

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+

βqVm

⎝ ⎜

⎠ ⎟

−J

μT=

dx+ κρ, κ =

βqVm

d

ρ (x) = −J

μTκ+ ρ(0) +

J

μTκ

⎝ ⎜

⎠ ⎟exp −κx( )

Use Einstein relation:

Solution:

Steady-state FP equation

dJ

dx=

d

dx−D

dx−

μqVm

⎝ ⎜

⎠ ⎟= 0

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+

βqVm

⎝ ⎜

⎠ ⎟

−J

μT=

dx+ κρ, κ =

βqVm

d

ρ (x) = −J

μTκ+ ρ(0) +

J

μTκ

⎝ ⎜

⎠ ⎟exp −κx( )

Use Einstein relation:

Solution:

We are given ρ(0) and ρ(d). Use this to solve for J:

Steady-state FP equation

dJ

dx=

d

dx−D

dx−

μqVm

⎝ ⎜

⎠ ⎟= 0

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+

βqVm

⎝ ⎜

⎠ ⎟

−J

μT=

dx+ κρ, κ =

βqVm

d

ρ (x) = −J

μTκ+ ρ(0) +

J

μTκ

⎝ ⎜

⎠ ⎟exp −κx( )

J

μTκ1− exp −κd( )( ) = ρ out exp −κd( ) − ρ (d),

Use Einstein relation:

Solution:

We are given ρ(0) and ρ(d). Use this to solve for J:

Steady-state FP equation

dJ

dx=

d

dx−D

dx−

μqVm

⎝ ⎜

⎠ ⎟= 0

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+

βqVm

⎝ ⎜

⎠ ⎟

−J

μT=

dx+ κρ, κ =

βqVm

d

ρ (x) = −J

μTκ+ ρ(0) +

J

μTκ

⎝ ⎜

⎠ ⎟exp −κx( )

J

μTκ1− exp −κd( )( ) = ρ out exp −κd( ) − ρ (d), ρ (d) = ρ in = ρ out exp −βqVr( )

Use Einstein relation:

Solution:

We are given ρ(0) and ρ(d). Use this to solve for J:

Steady-state FP equation

dJ

dx=

d

dx−D

dx−

μqVm

⎝ ⎜

⎠ ⎟= 0

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+

βqVm

⎝ ⎜

⎠ ⎟

−J

μT=

dx+ κρ, κ =

βqVm

d

ρ (x) = −J

μTκ+ ρ(0) +

J

μTκ

⎝ ⎜

⎠ ⎟exp −κx( )

J

μTκ1− exp −κd( )( ) = ρ out exp −κd( ) − ρ (d), ρ (d) = ρ in = ρ out exp −βqVr( )

J =μTκ ρ out exp −κd( ) − ρ (d)( )

1− exp −κd( )

Use Einstein relation:

Solution:

We are given ρ(0) and ρ(d). Use this to solve for J:

Steady-state FP equation

dJ

dx=

d

dx−D

dx−

μqVm

⎝ ⎜

⎠ ⎟= 0

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+

βqVm

⎝ ⎜

⎠ ⎟

−J

μT=

dx+ κρ, κ =

βqVm

d

ρ (x) = −J

μTκ+ ρ(0) +

J

μTκ

⎝ ⎜

⎠ ⎟exp −κx( )

J

μTκ1− exp −κd( )( ) = ρ out exp −κd( ) − ρ (d), ρ (d) = ρ in = ρ out exp −βqVr( )

J =μTκ ρ out exp −κd( ) − ρ (d)( )

1− exp −κd( )=

μqVmρ out exp −βqVm( ) − exp −βqVr( )( )

d 1− exp −βqVm( )( )

Use Einstein relation:

Solution:

We are given ρ(0) and ρ(d). Use this to solve for J:

GHK current, another wayStart from

J = −Ddρ

dx−

μqVm

dρ = const.

GHK current, another wayStart from

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+ κρ

⎝ ⎜

⎠ ⎟

GHK current, another wayStart from

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+ κρ

⎝ ⎜

⎠ ⎟

J exp κx( ) = −μTdρ

dx+ κρ

⎝ ⎜

⎠ ⎟exp κx( )

GHK current, another wayStart from

Note

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+ κρ

⎝ ⎜

⎠ ⎟

J exp κx( ) = −μTdρ

dx+ κρ

⎝ ⎜

⎠ ⎟exp κx( ) = −μT

d

dxρ exp κx( )( )

GHK current, another wayStart from

Note

Integrate from 0 to d:

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+ κρ

⎝ ⎜

⎠ ⎟

J exp κx( ) = −μTdρ

dx+ κρ

⎝ ⎜

⎠ ⎟exp κx( ) = −μT

d

dxρ exp κx( )( )

GHK current, another wayStart from

Note

Integrate from 0 to d:

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+ κρ

⎝ ⎜

⎠ ⎟

J exp κx( ) = −μTdρ

dx+ κρ

⎝ ⎜

⎠ ⎟exp κx( ) = −μT

d

dxρ exp κx( )( )

J

κexp κd( ) −1( ) = −μT ρ in exp κd( ) − ρ out( )

GHK current, another wayStart from

Note

Integrate from 0 to d:

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+ κρ

⎝ ⎜

⎠ ⎟

J exp κx( ) = −μTdρ

dx+ κρ

⎝ ⎜

⎠ ⎟exp κx( ) = −μT

d

dxρ exp κx( )( )

J

κexp κd( ) −1( ) = −μT ρ in exp κd( ) − ρ out( )

J = −μTκρ in exp κd( ) − ρ out

exp κd( ) −1

GHK current, another wayStart from

Note

Integrate from 0 to d:

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+ κρ

⎝ ⎜

⎠ ⎟

J exp κx( ) = −μTdρ

dx+ κρ

⎝ ⎜

⎠ ⎟exp κx( ) = −μT

d

dxρ exp κx( )( )

J

κexp κd( ) −1( ) = −μT ρ in exp κd( ) − ρ out( )

J = −μTκρ in exp κd( ) − ρ out

exp κd( ) −1= −

μqVm ρ in exp βqVm( ) − ρ out( )

d exp βqVm( ) −1( )

GHK current, another wayStart from

Note

Integrate from 0 to d:

(as before)

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+ κρ

⎝ ⎜

⎠ ⎟

J exp κx( ) = −μTdρ

dx+ κρ

⎝ ⎜

⎠ ⎟exp κx( ) = −μT

d

dxρ exp κx( )( )

J

κexp κd( ) −1( ) = −μT ρ in exp κd( ) − ρ out( )

J = −μTκρ in exp κd( ) − ρ out

exp κd( ) −1= −

μqVm ρ in exp βqVm( ) − ρ out( )

d exp βqVm( ) −1( )

L =μqVmρ out exp −βqVm( ) − exp −βqVr( )( )

d 1− exp −βqVm( )( )

GHK current, another wayStart from

Note

Integrate from 0 to d:

(as before)

Note: J = 0 at Vm= Vr

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+ κρ

⎝ ⎜

⎠ ⎟

J exp κx( ) = −μTdρ

dx+ κρ

⎝ ⎜

⎠ ⎟exp κx( ) = −μT

d

dxρ exp κx( )( )

J

κexp κd( ) −1( ) = −μT ρ in exp κd( ) − ρ out( )

J = −μTκρ in exp κd( ) − ρ out

exp κd( ) −1= −

μqVm ρ in exp βqVm( ) − ρ out( )

d exp βqVm( ) −1( )

L =μqVmρ out exp −βqVm( ) − exp −βqVr( )( )

d 1− exp −βqVm( )( )

GHK current is nonlinear

(using z, Vr for Ca++)

V

J

GHK current is nonlinear

(using z, Vr for Ca++)

Vm → −∞ : qJ ≈ −μq2ρ out

Vm

d

V

J

GHK current is nonlinear

(using z, Vr for Ca++)

Vm → −∞ : qJ ≈ −μq2ρ out

Vm

d= σE, E = −Vm /d,

V

J

GHK current is nonlinear

(using z, Vr for Ca++)

Vm → −∞ : qJ ≈ −μq2ρ out

Vm

d= σE, E = −Vm /d, σ = μq2ρ out

V

J

GHK current is nonlinear

(using z, Vr for Ca++)

Vm → −∞ : qJ ≈ −μq2ρ out

Vm

d= σE, E = −Vm /d, σ = μq2ρ out

Vm → +∞ : qJ ≈ −μq2ρ out exp(−βqVr)Vm

d= σE,

V

J

GHK current is nonlinear

(using z, Vr for Ca++)

Vm → −∞ : qJ ≈ −μq2ρ out

Vm

d= σE, E = −Vm /d, σ = μq2ρ out

Vm → +∞ : qJ ≈ −μq2ρ out exp(−βqVr)Vm

d= σE, σ = μq2ρ out exp(−βqVr) = μq2ρ in

V

J

GHK current is nonlinear

(using z, Vr for Ca++)

Vm → −∞ : qJ ≈ −μq2ρ out

Vm

d= σE, E = −Vm /d, σ = μq2ρ out

Vm → +∞ : qJ ≈ −μq2ρ out exp(−βqVr)Vm

d= σE, σ = μq2ρ out exp(−βqVr) = μq2ρ in

Vm ≈ Vr: qJ ≈μβq3Vr

d⋅

ρ outρ in

ρ out − ρ in

⋅(Vr −Vm )

V

J

Kramers escape

Rate of escape from a potential well due to thermal fluctuations

www.nbi.dk/hertz/noisecourse/demos/Pseq.matwww.nbi.dk/hertz/noisecourse/demos/runseq.m

V1(x)P1(x)

P2(x)

V2(x)

Kramers escape (2)

a b c

V(x)

Kramers escape (2)

a b c

V(x)

J

Kramers escape (2)

a b c

V(x)

Basic assumption: (V(b) – V(a))/T >> 1

J

Fokker-Planck equation

−∂P

∂t=

∂J

∂x=

∂xu(x)P −

∂xD(x)P( )

⎡ ⎣ ⎢

⎤ ⎦ ⎥Conservation (continuity):

Fokker-Planck equation

−∂P

∂t=

∂J

∂x=

∂xu(x)P −

∂xD(x)P( )

⎡ ⎣ ⎢

⎤ ⎦ ⎥

= −∂

∂xμ

∂V

∂xP + D

∂P

∂x

⎡ ⎣ ⎢

⎤ ⎦ ⎥

Conservation (continuity):

Fokker-Planck equation

−∂P

∂t=

∂J

∂x=

∂xu(x)P −

∂xD(x)P( )

⎡ ⎣ ⎢

⎤ ⎦ ⎥

= −∂

∂xμ

∂V

∂xP + D

∂P

∂x

⎡ ⎣ ⎢

⎤ ⎦ ⎥

= −D∂

∂x

∂(βV )

∂xP +

∂P

∂x

⎡ ⎣ ⎢

⎤ ⎦ ⎥

Conservation (continuity):

Use Einstein relation:

Fokker-Planck equation

−∂P

∂t=

∂J

∂x=

∂xu(x)P −

∂xD(x)P( )

⎡ ⎣ ⎢

⎤ ⎦ ⎥

= −∂

∂xμ

∂V

∂xP + D

∂P

∂x

⎡ ⎣ ⎢

⎤ ⎦ ⎥

= −D∂

∂x

∂(βV )

∂xP +

∂P

∂x

⎡ ⎣ ⎢

⎤ ⎦ ⎥

J = −Dexp −βV (x)( )∂

∂xexp βV (x)( )P[ ]

Conservation (continuity):

Use Einstein relation:

Current:

Fokker-Planck equation

−∂P

∂t=

∂J

∂x=

∂xu(x)P −

∂xD(x)P( )

⎡ ⎣ ⎢

⎤ ⎦ ⎥

= −∂

∂xμ

∂V

∂xP + D

∂P

∂x

⎡ ⎣ ⎢

⎤ ⎦ ⎥

= −D∂

∂x

∂(βV )

∂xP +

∂P

∂x

⎡ ⎣ ⎢

⎤ ⎦ ⎥

J = −Dexp −βV (x)( )∂

∂xexp βV (x)( )P[ ]

If equilibrium, J = 0,

Conservation (continuity):

Use Einstein relation:

Current:

Fokker-Planck equation

−∂P

∂t=

∂J

∂x=

∂xu(x)P −

∂xD(x)P( )

⎡ ⎣ ⎢

⎤ ⎦ ⎥

= −∂

∂xμ

∂V

∂xP + D

∂P

∂x

⎡ ⎣ ⎢

⎤ ⎦ ⎥

= −D∂

∂x

∂(βV )

∂xP +

∂P

∂x

⎡ ⎣ ⎢

⎤ ⎦ ⎥

J = −Dexp −βV (x)( )∂

∂xexp βV (x)( )P[ ]

If equilibrium, J = 0,

P(x)∝ exp −βV (x)( )

Conservation (continuity):

Use Einstein relation:

Current:

Fokker-Planck equation

−∂P

∂t=

∂J

∂x=

∂xu(x)P −

∂xD(x)P( )

⎡ ⎣ ⎢

⎤ ⎦ ⎥

= −∂

∂xμ

∂V

∂xP + D

∂P

∂x

⎡ ⎣ ⎢

⎤ ⎦ ⎥

= −D∂

∂x

∂(βV )

∂xP +

∂P

∂x

⎡ ⎣ ⎢

⎤ ⎦ ⎥

J = −Dexp −βV (x)( )∂

∂xexp βV (x)( )P[ ]

If equilibrium, J = 0,

Here: almost equilibrium, so use this P(x)

P(x)∝ exp −βV (x)( )

Conservation (continuity):

Use Einstein relation:

Current:

Calculating the current

−J

Dexp βV (x)( ) =

∂xexp βV (x)( )P(x)[ ] (J is constant)

Calculating the current

−J

Dexp βV (x)( ) =

∂xexp βV (x)( )P(x)[ ]

−J

Dexp βV (x)( )dx

a

c

∫ = exp βV (x)( )P(x)[ ]a

c

(J is constant)

integrate:

Calculating the current

−J

Dexp βV (x)( ) =

∂xexp βV (x)( )P(x)[ ]

−J

Dexp βV (x)( )dx

a

c

∫ = exp βV (x)( )P(x)[ ]a

c

≈ −exp βV (a)( )P(a)

(J is constant)

(P(c) very small)

integrate:

Calculating the current

−J

Dexp βV (x)( ) =

∂xexp βV (x)( )P(x)[ ]

−J

Dexp βV (x)( )dx

a

c

∫ = exp βV (x)( )P(x)[ ]a

c

≈ −exp βV (a)( )P(a)

⇒ J =Dexp βV (a)( )P(a)

exp βV (x)( )dxa

c

(J is constant)

(P(c) very small)

integrate:

Calculating the current

−J

Dexp βV (x)( ) =

∂xexp βV (x)( )P(x)[ ]

−J

Dexp βV (x)( )dx

a

c

∫ = exp βV (x)( )P(x)[ ]a

c

≈ −exp βV (a)( )P(a)

⇒ J =Dexp βV (a)( )P(a)

exp βV (x)( )dxa

c

If p is probability to be in the well, J = pr, where r = escape rate

(J is constant)

(P(c) very small)

integrate:

Calculating the current

−J

Dexp βV (x)( ) =

∂xexp βV (x)( )P(x)[ ]

−J

Dexp βV (x)( )dx

a

c

∫ = exp βV (x)( )P(x)[ ]a

c

≈ −exp βV (a)( )P(a)

⇒ J =Dexp βV (a)( )P(a)

exp βV (x)( )dxa

c

If p is probability to be in the well, J = pr, where r = escape rate

p = P(x)dxa−Δ

a +Δ

(J is constant)

(P(c) very small)

integrate:

Calculating the current

−J

Dexp βV (x)( ) =

∂xexp βV (x)( )P(x)[ ]

−J

Dexp βV (x)( )dx

a

c

∫ = exp βV (x)( )P(x)[ ]a

c

≈ −exp βV (a)( )P(a)

⇒ J =Dexp βV (a)( )P(a)

exp βV (x)( )dxa

c

If p is probability to be in the well, J = pr, where r = escape rate

p = P(x)dx = P(a)a−Δ

a +Δ

∫ exp β V (a) −V (x)( )[ ]dxa−Δ

a +Δ

(J is constant)

(P(c) very small)

integrate:

Calculating the current

−J

Dexp βV (x)( ) =

∂xexp βV (x)( )P(x)[ ]

−J

Dexp βV (x)( )dx

a

c

∫ = exp βV (x)( )P(x)[ ]a

c

≈ −exp βV (a)( )P(a)

⇒ J =Dexp βV (a)( )P(a)

exp βV (x)( )dxa

c

If p is probability to be in the well, J = pr, where r = escape rate

p = P(x)dx = P(a)a−Δ

a +Δ

∫ exp β V (a) −V (x)( )[ ]dxa−Δ

a +Δ

≈ P(a) exp − 12 β ′ ′ V (a)y 2

[ ]dy−∞

∫ = P(a)2π

β ′ ′ V (a)

⎝ ⎜

⎠ ⎟

12

(J is constant)

(P(c) very small)

integrate:

calculating escape rate

In integral integrand is peaked near x = b

exp βV (x)( )dxa

c

calculating escape rate

In integral integrand is peaked near x = b

exp βV (x)( )dxa

c

exp βV (x)( )dxa

c

∫ ≈ exp βV (b)( ) exp − 12 β ′ ′ V (b) x − b( )

2

( )−∞

∫ dx

calculating escape rate

In integral integrand is peaked near x = b

exp βV (x)( )dxa

c

exp βV (x)( )dxa

c

∫ ≈ exp βV (b)( ) exp − 12 β ′ ′ V (b) x − b( )

2

( )−∞

∫ dx

= exp βV (b)( )2π

β ′ ′ V (b)

⎝ ⎜

⎠ ⎟

12

calculating escape rate

In integral integrand is peaked near x = b

exp βV (x)( )dxa

c

exp βV (x)( )dxa

c

∫ ≈ exp βV (b)( ) exp − 12 β ′ ′ V (b) x − b( )

2

( )−∞

∫ dx

= exp βV (b)( )2π

β ′ ′ V (b)

⎝ ⎜

⎠ ⎟

12

r =J

p=

Dexp βV (a)( )P(a)

p exp βV (x)( )dxa

c

calculating escape rate

In integral integrand is peaked near x = b

exp βV (x)( )dxa

c

exp βV (x)( )dxa

c

∫ ≈ exp βV (b)( ) exp − 12 β ′ ′ V (b) x − b( )

2

( )−∞

∫ dx

= exp βV (b)( )2π

β ′ ′ V (b)

⎝ ⎜

⎠ ⎟

12

r =J

p=

Dexp βV (a)( )P(a)

p exp βV (x)( )dxa

c

=Dexp βV (a)( )P(a)

P(a)2π

β ′ ′ V (a)

⎝ ⎜

⎠ ⎟

12

exp βV (b)( )2π

β ′ ′ V (b)

⎝ ⎜

⎠ ⎟

12

calculating escape rate

In integral integrand is peaked near x = b

exp βV (x)( )dxa

c

exp βV (x)( )dxa

c

∫ ≈ exp βV (b)( ) exp − 12 β ′ ′ V (b) x − b( )

2

( )−∞

∫ dx

= exp βV (b)( )2π

β ′ ′ V (b)

⎝ ⎜

⎠ ⎟

12

r =J

p=

Dexp βV (a)( )P(a)

p exp βV (x)( )dxa

c

=Dexp βV (a)( )P(a)

P(a)2π

β ′ ′ V (a)

⎝ ⎜

⎠ ⎟

12

exp βV (b)( )2π

β ′ ′ V (b)

⎝ ⎜

⎠ ⎟

12

=Dβ

⎝ ⎜

⎠ ⎟ ′ ′ V (a) ′ ′ V (b)( )

12 exp −β V (b) −V (a)( )[ ]

calculating escape rate

In integral integrand is peaked near x = b

exp βV (x)( )dxa

c

exp βV (x)( )dxa

c

∫ ≈ exp βV (b)( ) exp − 12 β ′ ′ V (b) x − b( )

2

( )−∞

∫ dx

= exp βV (b)( )2π

β ′ ′ V (b)

⎝ ⎜

⎠ ⎟

12

r =J

p=

Dexp βV (a)( )P(a)

p exp βV (x)( )dxa

c

=Dexp βV (a)( )P(a)

P(a)2π

β ′ ′ V (a)

⎝ ⎜

⎠ ⎟

12

exp βV (b)( )2π

β ′ ′ V (b)

⎝ ⎜

⎠ ⎟

12

=Dβ

⎝ ⎜

⎠ ⎟ ′ ′ V (a) ′ ′ V (b)( )

12 exp −β V (b) −V (a)( )[ ] =

μ

⎝ ⎜

⎠ ⎟ ′ ′ V (a) ′ ′ V (b)( )

12 exp −βEb( )

calculating escape rate

In integral integrand is peaked near x = b

exp βV (x)( )dxa

c

exp βV (x)( )dxa

c

∫ ≈ exp βV (b)( ) exp − 12 β ′ ′ V (b) x − b( )

2

( )−∞

∫ dx

= exp βV (b)( )2π

β ′ ′ V (b)

⎝ ⎜

⎠ ⎟

12

r =J

p=

Dexp βV (a)( )P(a)

p exp βV (x)( )dxa

c

=Dexp βV (a)( )P(a)

P(a)2π

β ′ ′ V (a)

⎝ ⎜

⎠ ⎟

12

exp βV (b)( )2π

β ′ ′ V (b)

⎝ ⎜

⎠ ⎟

12

=Dβ

⎝ ⎜

⎠ ⎟ ′ ′ V (a) ′ ′ V (b)( )

12 exp −β V (b) −V (a)( )[ ] =

μ

⎝ ⎜

⎠ ⎟ ′ ′ V (a) ′ ′ V (b)( )

12 exp −βEb( )________

More about drift current

Notice: If u(x) is not constant, the probability cloud can shrink or spread even if there is no diffusion

More about drift current

Notice: If u(x) is not constant, the probability cloud can shrink or spread even if there is no diffusion(like density of cars on a road where the speed limit varies)

More about drift current

Notice: If u(x) is not constant, the probability cloud can shrink or spread even if there is no diffusion(like density of cars on a road where the speed limit varies)

http://www.nbi.dk/~hertz/noisecourse/driftmovie.m

Demo: initial P: Gaussian centered at x = 2u(x) = .00015x

Derivation from master equation

∂P(x, t)

∂t= d ′ x w(x ← ′ x )P( ′ x , t) − w( ′ x ← x)P(x, t)[ ]∫ w(x ← ′ x ) ≡ r( ′ x ;x − ′ x ) :

Derivation from master equation

∂P(x, t)

∂t= d ′ x w(x ← ′ x )P( ′ x , t) − w( ′ x ← x)P(x, t)[ ]∫ w(x ← ′ x ) ≡ r( ′ x ;x − ′ x ) :

(1st argument of r: starting point; 2nd argument: step size)

Derivation from master equation

∂P(x, t)

∂t= d ′ x w(x ← ′ x )P( ′ x , t) − w( ′ x ← x)P(x, t)[ ]∫ w(x ← ′ x ) ≡ r( ′ x ;x − ′ x ) :

= d ′ x r( ′ x ;x − ′ x )P( ′ x , t) − r(x; ′ x − x)P(x, t)[ ]∫ ′ x = x − s :

(1st argument of r: starting point; 2nd argument: step size)

Derivation from master equation

∂P(x, t)

∂t= d ′ x w(x ← ′ x )P( ′ x , t) − w( ′ x ← x)P(x, t)[ ]∫ w(x ← ′ x ) ≡ r( ′ x ;x − ′ x ) :

= d ′ x r( ′ x ;x − ′ x )P( ′ x , t) − r(x; ′ x − x)P(x, t)[ ]∫ ′ x = x − s :

= ds r(x − s;s)P(x − s, t) − r(x;−s)P(x, t)[ ]∫

(1st argument of r: starting point; 2nd argument: step size)

Derivation from master equation

∂P(x, t)

∂t= d ′ x w(x ← ′ x )P( ′ x , t) − w( ′ x ← x)P(x, t)[ ]∫ w(x ← ′ x ) ≡ r( ′ x ;x − ′ x ) :

= d ′ x r( ′ x ;x − ′ x )P( ′ x , t) − r(x; ′ x − x)P(x, t)[ ]∫ ′ x = x − s :

= ds r(x − s;s)P(x − s, t) − r(x;−s)P(x, t)[ ]∫

Small steps assumption: r(x;s) falls rapidly to zero with increasing |s| on the scale on which it varies with x or the scale on which P varies with x.

(1st argument of r: starting point; 2nd argument: step size)

Derivation from master equation

∂P(x, t)

∂t= d ′ x w(x ← ′ x )P( ′ x , t) − w( ′ x ← x)P(x, t)[ ]∫ w(x ← ′ x ) ≡ r( ′ x ;x − ′ x ) :

= d ′ x r( ′ x ;x − ′ x )P( ′ x , t) − r(x; ′ x − x)P(x, t)[ ]∫ ′ x = x − s :

= ds r(x − s;s)P(x − s, t) − r(x;−s)P(x, t)[ ]∫

Small steps assumption: r(x;s) falls rapidly to zero with increasing |s| on the scale on which it varies with x or the scale on which P varies with x.

(1st argument of r: starting point; 2nd argument: step size)

x

s

Derivation from master equation (2)

∂P(x, t)

∂t= ds r(x − s;s)P(x − s, t) − r(x;−s)P(x, t)[ ]∫

expand:

Derivation from master equation (2)

∂P(x, t)

∂t= ds r(x − s;s)P(x − s, t) − r(x;−s)P(x, t)[ ]∫

= ds r(x;s)P(x, t) − s∂

∂xr(x,s)P(x, t)( ) + 1

2 s2 ∂ 2

∂x 2r(x,s)P(x, t)( ) +L − r(x;−s)P(x, t)

⎧ ⎨ ⎩

⎫ ⎬ ⎭

expand:

Derivation from master equation (2)

∂P(x, t)

∂t= ds r(x − s;s)P(x − s, t) − r(x;−s)P(x, t)[ ]∫

= ds r(x;s)P(x, t) − s∂

∂xr(x,s)P(x, t)( ) + 1

2 s2 ∂ 2

∂x 2r(x,s)P(x, t)( ) +L − r(x;−s)P(x, t)

⎧ ⎨ ⎩

⎫ ⎬ ⎭

= −∂

∂xsr(x,s)ds∫( )P(x, t)[ ] +

∂ 2

∂x 212 s2r(x,s)ds∫( )P(x, t)[ ] +L

expand:

Derivation from master equation (2)

∂P(x, t)

∂t= ds r(x − s;s)P(x − s, t) − r(x;−s)P(x, t)[ ]∫

= ds r(x;s)P(x, t) − s∂

∂xr(x,s)P(x, t)( ) + 1

2 s2 ∂ 2

∂x 2r(x,s)P(x, t)( ) +L − r(x;−s)P(x, t)

⎧ ⎨ ⎩

⎫ ⎬ ⎭

= −∂

∂xsr(x,s)ds∫( )P(x, t)[ ] +

∂ 2

∂x 212 s2r(x,s)ds∫( )P(x, t)[ ] +L

= −∂

∂xr1(x)P(x, t)( ) +

1

2

∂ 2

∂x 2r2(x)P(x, t)( ) +L

expand:

Derivation from master equation (2)

∂P(x, t)

∂t= ds r(x − s;s)P(x − s, t) − r(x;−s)P(x, t)[ ]∫

= ds r(x;s)P(x, t) − s∂

∂xr(x,s)P(x, t)( ) + 1

2 s2 ∂ 2

∂x 2r(x,s)P(x, t)( ) +L − r(x;−s)P(x, t)

⎧ ⎨ ⎩

⎫ ⎬ ⎭

= −∂

∂xsr(x,s)ds∫( )P(x, t)[ ] +

∂ 2

∂x 212 s2r(x,s)ds∫( )P(x, t)[ ] +L

= −∂

∂xr1(x)P(x, t)( ) +

1

2

∂ 2

∂x 2r2(x)P(x, t)( ) +L

expand:

Kramers-Moyal expansion

Derivation from master equation (2)

∂P(x, t)

∂t= ds r(x − s;s)P(x − s, t) − r(x;−s)P(x, t)[ ]∫

= ds r(x;s)P(x, t) − s∂

∂xr(x,s)P(x, t)( ) + 1

2 s2 ∂ 2

∂x 2r(x,s)P(x, t)( ) +L − r(x;−s)P(x, t)

⎧ ⎨ ⎩

⎫ ⎬ ⎭

= −∂

∂xsr(x,s)ds∫( )P(x, t)[ ] +

∂ 2

∂x 212 s2r(x,s)ds∫( )P(x, t)[ ] +L

= −∂

∂xr1(x)P(x, t)( ) +

1

2

∂ 2

∂x 2r2(x)P(x, t)( ) +L

expand:

Kramers-Moyal expansionFokker-Planck eqn if drop terms of order >2

Derivation from master equation (2)

∂P(x, t)

∂t= ds r(x − s;s)P(x − s, t) − r(x;−s)P(x, t)[ ]∫

= ds r(x;s)P(x, t) − s∂

∂xr(x,s)P(x, t)( ) + 1

2 s2 ∂ 2

∂x 2r(x,s)P(x, t)( ) +L − r(x;−s)P(x, t)

⎧ ⎨ ⎩

⎫ ⎬ ⎭

= −∂

∂xsr(x,s)ds∫( )P(x, t)[ ] +

∂ 2

∂x 212 s2r(x,s)ds∫( )P(x, t)[ ] +L

= −∂

∂xr1(x)P(x, t)( ) +

1

2

∂ 2

∂x 2r2(x)P(x, t)( ) +L

rn (x) = snr(x,s)ds∫

expand:

Kramers-Moyal expansionFokker-Planck eqn if drop terms of order >2

Derivation from master equation (2)

∂P(x, t)

∂t= ds r(x − s;s)P(x − s, t) − r(x;−s)P(x, t)[ ]∫

= ds r(x;s)P(x, t) − s∂

∂xr(x,s)P(x, t)( ) + 1

2 s2 ∂ 2

∂x 2r(x,s)P(x, t)( ) +L − r(x;−s)P(x, t)

⎧ ⎨ ⎩

⎫ ⎬ ⎭

= −∂

∂xsr(x,s)ds∫( )P(x, t)[ ] +

∂ 2

∂x 212 s2r(x,s)ds∫( )P(x, t)[ ] +L

= −∂

∂xr1(x)P(x, t)( ) +

1

2

∂ 2

∂x 2r2(x)P(x, t)( ) +L

rn (x) = snr(x,s)ds∫

expand:

rn(x)Δt = nth moment of distribution of step size in time Δt

Kramers-Moyal expansionFokker-Planck eqn if drop terms of order >2

Derivation from master equation (2)

∂P(x, t)

∂t= ds r(x − s;s)P(x − s, t) − r(x;−s)P(x, t)[ ]∫

= ds r(x;s)P(x, t) − s∂

∂xr(x,s)P(x, t)( ) + 1

2 s2 ∂ 2

∂x 2 r(x,s)P(x, t)( ) +L − r(x;−s)P(x, t) ⎧ ⎨ ⎩

⎫ ⎬ ⎭

= −∂

∂xsr(x,s)ds∫( )P(x, t)[ ] +

∂ 2

∂x 212 s2r(x,s)ds∫( )P(x, t)[ ] +L

= −∂

∂xr1(x)P(x, t)( ) +

1

2

∂ 2

∂x 2r2(x)P(x, t)( ) +L

rn (x) = snr(x,s)ds∫

r1(x) = u(x), r2(x) = 2D(x)

expand:

rn(x)Δt = nth moment of distribution of step size in time Δt

Kramers-Moyal expansionFokker-Planck eqn if drop terms of order >2