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
dρ
dx
Steady-state FP equation
€
dJ
dx=
d
dx−D
dρ
dx−
μqVm
dρ
⎛
⎝ ⎜
⎞
⎠ ⎟= 0
Steady-state FP equation
€
dJ
dx=
d
dx−D
dρ
dx−
μqVm
dρ
⎛
⎝ ⎜
⎞
⎠ ⎟= 0
J = −Ddρ
dx−
μqVm
dρ = const.
Steady-state FP equation
€
dJ
dx=
d
dx−D
dρ
dx−
μqVm
dρ
⎛
⎝ ⎜
⎞
⎠ ⎟= 0
J = −Ddρ
dx−
μqVm
dρ = const.
= −μTdρ
dx+
βqVm
dρ
⎛
⎝ ⎜
⎞
⎠ ⎟Use Einstein relation:
Steady-state FP equation
€
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:
Steady-state FP equation
€
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( )
Use Einstein relation:
Solution:
Steady-state FP equation
€
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( )
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
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),
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
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( )
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
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( )
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
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( )
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
2π
⎛
⎝ ⎜
⎞
⎠ ⎟ ′ ′ 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
2π
⎛
⎝ ⎜
⎞
⎠ ⎟ ′ ′ V (a) ′ ′ V (b)( )
12 exp −β V (b) −V (a)( )[ ] =
μ
2π
⎛
⎝ ⎜
⎞
⎠ ⎟ ′ ′ 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
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
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