lecture 4: diffusion and the fokker-planck equation outline: intuitive treatment diffusion as flow...

Lecture 4: Diffusion and the Fokker-Planck equation

Outline:

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


Outline:


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


Outline:



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


Outline:



• 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


€

J = −D∂P

∂x

€

I = −g∂V

∂x

conservation:

€

∂P

∂t= −

∂J

∂x


€

J = −D∂P

∂x

€

I = −g∂V

∂x

conservation:

€

∂P

∂t= −

∂J

∂x

€

∂P

∂t= D

∂2P

∂x 2

=>


€

J = −D∂P

∂x

€

I = −g∂V

∂x

conservation:

€

∂P

∂t= −

∂J

∂x

€

∂P

∂t= D

∂2P

∂x 2

=> diffusion equation


€

J = −D∂P

∂x

€

I = −g∂V

∂x

conservation:

€

∂P

∂t= −

∂J

∂x

€

∂P

∂t= D

∂2P

∂x 2


initial condition

€

P(x | 0) = δ(x)


€

J = −D∂P

∂x

€

I = −g∂V

∂x

conservation:

€

∂P

∂t= −

∂J

∂x

€

∂P

∂t= D

∂2P

∂x 2


initial condition

€

P(x | 0) = δ(x)

solution:

€

P(x | t) =1

4πDtexp −

x 2

4Dt

⎛

⎝ ⎜

⎞

⎠ ⎟


€

J = −D∂P

∂x

€

I = −g∂V

∂x

conservation:

€

∂P

∂t= −

∂J

∂x

€

∂P

∂t= D

∂2P

∂x 2


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:

€


€

∂P

∂t= −

∂

∂xJdrift + Jdiff( )




€


€

∂P

∂t= −

∂


= −∂

∂xu(x)P − D

∂P

∂x

⎡ ⎣ ⎢

⎤ ⎦ ⎥= −

∂

∂xu(x)P( ) + D

∂ 2P

∂x 2



Slightly more generally, D can depend on x:


€


€

∂P

∂t= −

∂


= −∂

∂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 − D

∂P

∂x

⎡ ⎣ ⎢

⎤ ⎦ ⎥= −

∂

∂xu(x)P( ) + D

∂ 2P

∂x 2

€


∂xD(x)P(x, t)( )

€

∂P

∂t= −

∂

∂xu(x)P( ) +

∂ 2

∂x 2D(x)P( )




=>


€


€

∂P

∂t= −

∂


= −∂

∂xu(x)P − D

∂P

∂x

⎡ ⎣ ⎢

⎤ ⎦ ⎥= −

∂

∂xu(x)P( ) + D

∂ 2P

∂x 2

€


∂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):


€

u(x) = u0

P(x, t) =1

4πDtexp −

x − u0t( )2

4Dt

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥



Stationary case:


€

u(x) = u0

P(x, t) =1

4πDtexp −

x − u0t( )2

4Dt

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥



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


€

u(x) = u0

P(x, t) =1

4πDtexp −

x − u0t( )2

4Dt

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥




μ =mobility


€

u(x) = u0

P(x, t) =1

4πDtexp −

x − u0t( )2

4Dt

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥




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

€

P(x) = 0, x < 0;

J(x) = 0


€

u(x) = u0

P(x, t) =1

4πDtexp −

x − u0t( )2

4Dt

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥




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:


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:


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


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


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


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



outside insidex

V(x)Vm

Vout= 0

x=0 x=d



outside insidex

V(x)Vm

Vout= 0

ρout

ρin

x=0 x=d



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



This defines the reversal potential

at which J = 0.

€

Vr = T logρ out

ρ in

⎛

⎝ ⎜

⎞

⎠ ⎟

Reversal potential



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

dρ

dx

Steady-state FP equation

€

dJ

dx=

d

dx−D

dρ

dx−

μqVm

dρ

⎛

⎝ ⎜

⎞

⎠ ⎟= 0


€

dJ

dx=

d

dx−D

dρ

dx−

μqVm

dρ

⎛

⎝ ⎜

⎞

⎠ ⎟= 0

J = −Ddρ

dx−

μqVm

dρ = const.


€

dJ

dx=

d

dx−D

dρ

dx−

μqVm

dρ

⎛

⎝ ⎜

⎞

⎠ ⎟= 0

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+

βqVm

dρ

⎛

⎝ ⎜

⎞

⎠ ⎟Use Einstein relation:


€

dJ

dx=

d

dx−D

dρ

dx−

μqVm

dρ

⎛

⎝ ⎜

⎞

⎠ ⎟= 0

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+

βqVm

dρ

⎛

⎝ ⎜

⎞

⎠ ⎟

−J

μT=

dρ

dx+ κρ, κ =

βqVm

d

Use Einstein relation:


€

dJ

dx=

d

dx−D

dρ

dx−

μqVm

dρ

⎛

⎝ ⎜

⎞

⎠ ⎟= 0

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+

βqVm

dρ

⎛

⎝ ⎜

⎞

⎠ ⎟

−J

μT=

dρ

dx+ κρ, κ =

βqVm

d

ρ (x) = −J

μTκ+ ρ(0) +

J

μTκ

⎛

⎝ ⎜

⎞

⎠ ⎟exp −κx( )


Solution:


€

dJ

dx=

d

dx−D

dρ

dx−

μqVm

dρ

⎛

⎝ ⎜

⎞

⎠ ⎟= 0

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+

βqVm

dρ

⎛

⎝ ⎜

⎞

⎠ ⎟

−J

μT=

dρ

dx+ κρ, κ =

βqVm

d

ρ (x) = −J

μTκ+ ρ(0) +

J

μTκ

⎛

⎝ ⎜

⎞

⎠ ⎟exp −κx( )


Solution:

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


€

dJ

dx=

d

dx−D

dρ

dx−

μqVm

dρ

⎛

⎝ ⎜

⎞

⎠ ⎟= 0

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+

βqVm

dρ

⎛

⎝ ⎜

⎞

⎠ ⎟

−J

μT=

dρ

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),


Solution:



€

dJ

dx=

d

dx−D

dρ

dx−

μqVm

dρ

⎛

⎝ ⎜

⎞

⎠ ⎟= 0

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+

βqVm

dρ

⎛

⎝ ⎜

⎞

⎠ ⎟

−J

μT=

dρ

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( )


Solution:



€

dJ

dx=

d

dx−D

dρ

dx−

μqVm

dρ

⎛

⎝ ⎜

⎞

⎠ ⎟= 0

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+

βqVm

dρ

⎛

⎝ ⎜

⎞

⎠ ⎟

−J

μT=

dρ

dx+ κρ, κ =

βqVm

d

ρ (x) = −J

μTκ+ ρ(0) +

J

μTκ

⎛

⎝ ⎜

⎞

⎠ ⎟exp −κx( )

€

J


J =μTκ ρ out exp −κd( ) − ρ (d)( )

1− exp −κd( )


Solution:



€

dJ

dx=

d

dx−D

dρ

dx−

μqVm

dρ

⎛

⎝ ⎜

⎞

⎠ ⎟= 0

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+

βqVm

dρ

⎛

⎝ ⎜

⎞

⎠ ⎟

−J

μT=

dρ

dx+ κρ, κ =

βqVm

d

ρ (x) = −J

μTκ+ ρ(0) +

J

μTκ

⎛

⎝ ⎜

⎞

⎠ ⎟exp −κx( )

€

J


J =μTκ ρ out exp −κd( ) − ρ (d)( )

1− exp −κd( )=

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

d 1− exp −βqVm( )( )


Solution:


GHK current, another wayStart from

€

J = −Ddρ

dx−

μqVm

dρ = const.


€

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+ κρ

⎛

⎝ ⎜

⎞

⎠ ⎟


€

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+ κρ

⎛

⎝ ⎜

⎞

⎠ ⎟

J exp κx( ) = −μTdρ

dx+ κρ

⎛

⎝ ⎜

⎞

⎠ ⎟exp κx( )


Note

€

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+ κρ

⎛

⎝ ⎜

⎞

⎠ ⎟


dx+ κρ

⎛

⎝ ⎜

⎞

⎠ ⎟exp κx( ) = −μT

d

dxρ exp κx( )( )


Note

Integrate from 0 to d:

€

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+ κρ

⎛

⎝ ⎜

⎞

⎠ ⎟


dx+ κρ

⎛

⎝ ⎜

⎞

⎠ ⎟exp κx( ) = −μT

d

dxρ exp κx( )( )


Note


€

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+ κρ

⎛

⎝ ⎜

⎞

⎠ ⎟


dx+ κρ

⎛

⎝ ⎜

⎞

⎠ ⎟exp κx( ) = −μT

d

dxρ exp κx( )( )

€

J

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


Note


€

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+ κρ

⎛

⎝ ⎜

⎞

⎠ ⎟


dx+ κρ

⎛

⎝ ⎜

⎞

⎠ ⎟exp κx( ) = −μT

d

dxρ exp κx( )( )

€

J


J = −μTκρ in exp κd( ) − ρ out

exp κd( ) −1


Note


€

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+ κρ

⎛

⎝ ⎜

⎞

⎠ ⎟


dx+ κρ

⎛

⎝ ⎜

⎞

⎠ ⎟exp κx( ) = −μT

d

dxρ exp κx( )( )

€

J



exp κd( ) −1= −

μqVm ρ in exp βqVm( ) − ρ out( )

d exp βqVm( ) −1( )


Note


(as before)

€

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+ κρ

⎛

⎝ ⎜

⎞

⎠ ⎟


dx+ κρ

⎛

⎝ ⎜

⎞

⎠ ⎟exp κx( ) = −μT

d

dxρ exp κx( )( )

€

J



exp κd( ) −1= −



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



Note


(as before)

Note: J = 0 at Vm= Vr

€

J = −Ddρ

dx−

μqVm

dρ = const.

= −μTdρ

dx+ κρ

⎛

⎝ ⎜

⎞

⎠ ⎟


dx+ κρ

⎛

⎝ ⎜

⎞

⎠ ⎟exp κx( ) = −μT

d

dxρ exp κx( )( )

€

J



exp κd( ) −1= −



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


GHK current is nonlinear

(using z, Vr for Ca++)

V

J



€

Vm → −∞ : qJ ≈ −μq2ρ out

Vm

d

V

J



€


Vm

d= σE, E = −Vm /d,

V

J



€


Vm

d= σE, E = −Vm /d, σ = μq2ρ out

V

J



€


Vm


Vm → +∞ : qJ ≈ −μq2ρ out exp(−βqVr)Vm

d= σE,

V

J



€


Vm



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

V

J



€


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)

http://www.nbi.dk/hertz/course/demos/Pseq.mat

http://www.nbi.dk/hertz/course/demos/runseq.m

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):


€

−∂P

∂t=

∂J

∂x=

∂

∂xu(x)P −

∂

∂xD(x)P( )

⎡ ⎣ ⎢

⎤ ⎦ ⎥

= −∂

∂xμ

∂V

∂xP + D

∂P

∂x

⎡ ⎣ ⎢

⎤ ⎦ ⎥

Conservation (continuity):


€

−∂P

∂t=

∂J

∂x=

∂

∂xu(x)P −

∂

∂xD(x)P( )

⎡ ⎣ ⎢

⎤ ⎦ ⎥

= −∂

∂xμ

∂V

∂xP + D

∂P

∂x

⎡ ⎣ ⎢

⎤ ⎦ ⎥

= −D∂

∂x

∂(βV )

∂xP +

∂P

∂x

⎡ ⎣ ⎢

⎤ ⎦ ⎥




€

−∂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[ ]



Current:


€

−∂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)( )∂


If equilibrium, J = 0,



Current:


€

−∂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)( )∂



€

P(x)∝ exp −βV (x)( )



Current:


€

−∂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)( )∂



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

€

P(x)∝ exp −βV (x)( )



Current:

Calculating the current

€

−J

Dexp βV (x)( ) =

∂

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


€

−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:


€

−J

Dexp βV (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:


€

−J

Dexp βV (x)( ) =

∂


−J

Dexp βV (x)( )dx

a

c

∫ = exp βV (x)( )P(x)[ ]a

c


⇒ J =Dexp βV (a)( )P(a)

exp βV (x)( )dxa

c

∫

(J is constant)

(P(c) very small)

integrate:


€

−J

Dexp βV (x)( ) =

∂


−J

Dexp βV (x)( )dx

a

c

∫ = exp βV (x)( )P(x)[ ]a

c



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:


€

−J

Dexp βV (x)( ) =

∂


−J

Dexp βV (x)( )dx

a

c

∫ = exp βV (x)( )P(x)[ ]a

c



exp βV (x)( )dxa

c

∫


€

p = P(x)dxa−Δ

a +Δ

∫

(J is constant)

(P(c) very small)

integrate:


€

−J

Dexp βV (x)( ) =

∂


−J

Dexp βV (x)( )dx

a

c

∫ = exp βV (x)( )P(x)[ ]a

c



exp βV (x)( )dxa

c

∫


€

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:


€

−J

Dexp βV (x)( ) =

∂


−J

Dexp βV (x)( )dx

a

c

∫ = exp βV (x)( )P(x)[ ]a

c



exp βV (x)( )dxa

c

∫


€

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

∫



€

exp βV (x)( )dxa

c

∫

€

exp βV (x)( )dxa

c

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

2

( )−∞

∞

∫ dx



€

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



€

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

∫



€

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



€

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

∫


P(a)2π

β ′ ′ V (a)

⎛

⎝ ⎜

⎞

⎠ ⎟

12

exp βV (b)( )2π

β ′ ′ V (b)

⎛

⎝ ⎜

⎞

⎠ ⎟

12

=Dβ

2π

⎛

⎝ ⎜

⎞

⎠ ⎟ ′ ′ V (a) ′ ′ V (b)( )

12 exp −β V (b) −V (a)( )[ ]



€

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

∫


P(a)2π

β ′ ′ V (a)

⎛

⎝ ⎜

⎞

⎠ ⎟

12

exp βV (b)( )2π

β ′ ′ V (b)

⎛

⎝ ⎜

⎞

⎠ ⎟

12

=Dβ

2π

⎛

⎝ ⎜

⎞

⎠ ⎟ ′ ′ V (a) ′ ′ V (b)( )

12 exp −β V (b) −V (a)( )[ ] =

μ

2π

⎛

⎝ ⎜

⎞

⎠ ⎟ ′ ′ V (a) ′ ′ V (b)( )

12 exp −βEb( )



€

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

∫


P(a)2π

β ′ ′ V (a)

⎛

⎝ ⎜

⎞

⎠ ⎟

12

exp βV (b)( )2π

β ′ ′ V (b)

⎛

⎝ ⎜

⎞

⎠ ⎟

12

=Dβ

2π

⎛

⎝ ⎜

⎞

⎠ ⎟ ′ ′ V (a) ′ ′ V (b)( )

12 exp −β V (b) −V (a)( )[ ] =

μ

2π

⎛

⎝ ⎜

⎞

⎠ ⎟ ′ ′ 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


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)


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 ) :


€

∂P(x, t)


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


€

∂P(x, t)


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



€

∂P(x, t)



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



€

∂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.



€

∂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.


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:


€

∂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:


€

∂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:


€

∂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:


€

∂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


€

∂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


€

∂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:



€

∂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:

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



€

∂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


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

expand:

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


lecture 4: diffusion and the fokker-planck equation outline: intuitive treatment diffusion as flow...

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