Finite Difference Approximation of the Nernst-Planck Equation and von
Neumann Stability Analysis
Oliver K. Ernst
September 14, 2015
1
Nernst-Planck Equation
In the presence of an electric field, ions undergoing diffusion give rise to an ionic current J(~r, t)
zF
J(~r, t) = −D ∇c(~r, t) +
c(~r, t)∇φ(~r, t) ,
RT
(1)
where c(~r, t) is the ionic concentration at position ~r and time t, D is the diffusion constant (taken
here as a constant for a particular ionic species), z is the ion charge, T is the temperature, φ(~r, t) is the
potential, and F, R are the Faraday and Universal Gas constants, respectively.
Together with the continuity equation
∂c
= 0,
(2)
∂t
the Nernst-Planck equation describing the time evolution of the ionic concentration is obtained [1, 2]
∇J +
as
∂c
zF
2
2
=D ∇ c+
(∇c∇φ + c∇ φ) .
∂t
RT
(3)
We restrict ourselves to the one dimensional case in x, t, although generalization to higher dimensions
is straight forward. For ease of notation, define α = zF/RT , then
∂c
∂φ
J = −D
+ αc
,
∂x
∂x
2
∂c
∂ c
∂c ∂φ
∂2φ
=D
+
α(
+
c
)
.
∂t
∂x2
∂x ∂x
∂x2
2
(4)
(5)
Finite Difference Approximation
We seek to obtain a finite difference approximation to the differential equation 5. Begin by discretizing
space into i = 1, 2, . . . , n compartments of width ∆x = xi+1 − xi with x1 = 0 and xn = L, and time
into j = 0, 1, 2, . . . steps of duration ∆t = tj+1 − tj , with t0 = 0. Let cji denote the concentration in
compartment i at time step j, and φji the potential.
1
2.1
Interior points
Consider the flux of ions into compartment i of volume V from compartments i − 1 and i + 1 across a
cross sectional area A. The net change in the number of particles in compartment i during a single time
step j → j + 1 using a forward difference scheme is
where Ja→b
cj+1
− cji
i
V = Ji−1→i A − Ji→i+1 A,
∆t
denotes the flux from compartment a into b. By 4,
Ji−1→i = −D
Ji→i+1 = −D
(6)
cji − cji−1
φj − φji−1
+ αcji i
∆x
∆x
!
cji+1 − cji
φj − φji
+ αcji+1 i+1
∆x
∆x
!
,
(7)
.
(8)
Furthermore, note that V = ∆x∆y∆z and A = ∆y∆z. Collecting terms
!
!)
j
j
cji − cji−1
cji+1 − cji
φji+1 − φji
j φi − φi−1
j
−D
+ αci
+D
+ αci+1
∆x
∆x
∆x
∆x
h io
D n j
j
j
j
j
j
j
j
j
j
j
c
−
2c
+
c
+
α
c
φ
−
2φ
+
φ
+
c
−
c
φ
−
φ
=
i
i−1
i
i+1
i
i−1
i+1
i
i+1
i
(∆x)2 i+1
cj+1
− cji
A
i
=
∆t
V
(
∴ cj+1
= cji + ∆t
i
2.2
(9)
(10)
h io
D n j
j
j
j
j
j
j
j
j
j
j
c
−
2c
+
c
+
α
c
φ
−
2φ
+
φ
+
c
−
c
φ
−
φ
i
i−1
i
i+1
i
i−1
i+1
i
i+1
i
(∆x)2 i+1
(11)
Endpoints
The boundary conditions required depend on the desired application of the Nernst-Planck equation. We
consider here the case of no-flux endpoints, i.e. Neumann boundary conditions of the type
∂c1 (t)
∂cn (t)
=
= 0.
∂t
∂t
Starting with boundary conditions for compartment i = 1, analogous to 6,
where J1→2
cj+1
− cj1
1
V = −J1→2 A,
∆t
is given by 8 for i = 1. Using the finite difference approximation gives
cj+1
= cj1 + ∆t
1
o
2D n j
j
j
j
j
c
−
c
+
αc
φ
−
φ
.
1
2
2
1
(∆x)2 2
(12)
(13)
(14)
Similar considerations for the i = n endpoint give
cj+1
= cjn − ∆t
n
cj+1
− cjn
n
V = Jn−1→n A,
∆t
(15)
o
2D n j
j
j
j
j
c
−
c
+
αc
φ
−
φ
.
n
n
n−1
n−1
(∆x)2 n
(16)
2
3
von Neumann Stability Analysis
Consider the following problem: given an initial concentration distribution c0i for all i on a domain [0, L],
a known potential function φji for all i, j, and having set D, ∆x, α - how is the time step ∆t to be chosen
such that the finite difference approximation is stable, in the sense that the errors in subsequent time
steps decrease? Clearly ∆t must be chosen to be sufficiently small, but this comes at the cost of increased
simulation time. An upper bound on ∆t that results in a stable simulation is desirable.
The von Neumann stability analysis (sometimes known as Fourier stability analysis) gives such a bound
for finite difference approximations using Fourier decomposition of the error. We follow [3] approximately
in notation and steps, extending the result to the Nernst-Planck equation.
Define the round-off error δij in compartment i at time step j as
δij = Cij − cji ,
cji
where
is the exact (true) solution to 11 and
arithmetic. Then δij also satisfies 11, i.e.
Cij
(17)
is the solution obtained using finite precision
h io
D n j
j
j
j
j
j
j
j
j
j
j
δ
−
2δ
+
δ
+
α
δ
φ
−
2φ
+
φ
+
δ
−
δ
φ
−
φ
.
i
i−1
i
i+1
i
i−1
i+1
i
i+1
i
(∆x)2 i+1
(18)
The desired quantity to bound is the amplification factor of the error, defined as
∴ δij+1 = δij + ∆t
δij+1
(19)
|G| ≤ 1.
(20)
G=
,
δij
such that the desired bound for the error to decrease in subsequent time steps is
To evaluate the amplification factor, expand the error in a Fourier series
δij =
n
X
Ap (tj )eikp xi ,
(21)
p=1
where Ap (t) are the amplitudes and kp = πp/L for n = L/∆x as before. Assume that the amplitude
of the errors grows or decays exponentially in time - each term of the series then has the form
δij = eatj eikp xi ,
(22)
for some constant a. Furthermore, the other terms in 18 are
δij+1 = ea(tj +∆t) eikp xi ,
(23)
j
δi+1
j
δi−1
,
(24)
= eatj eikp (xi −∆x) .
(25)
=e
atj ikp (xi +∆x)
e
Substituting these terms into 18 (it is sufficient to consider an arbitrary term rather than the whole
Fourier series), we obtain
h
io
D n ikp ∆x
j
j
j
j
j
−ikp ∆x
ikp ∆x
e
−
2
+
e
+
α
φ
−
2φ
+
φ
+
e
−
1
φ
−
φ
(26)
i+1
i
i−1
i+1
i
(∆x)2
h
i
D
j
j
j
j
j
2 kp ∆x
ikp ∆x
= 1 + ∆t
−4 sin
+ α φi+1 − 2φi + φi−1 + e
− 1 φi+1 − φi
.
(27)
(∆x)2
2
ea∆t = 1 + ∆t
3
Since the amplification factor evaluates to
G=
δij+1
δij
=
ea(tj +∆t) eikp xi
= ea∆t ,
eatj eikp xi
(28)
the condition for a stable finite difference analysis becomes
h
i
j
j
j
j
j
2 kp ∆x
ikp ∆x
≤ 1,
1 + ∆t D
−4 sin
+ α φi+1 − 2φi + φi−1 + e
− 1 φi+1 − φi
(∆x)2
2
h
i
2 kp ∆x
∆t D
≤ 2,
+ α φji+1 − 2φji + φji−1 + eikp ∆x − 1 φji+1 − φji
(∆x)2 −4 sin
2
h
i j
j
j
j
2 kp ∆x
ikp ∆x
≤ 2.
∆t D
−4
sin
+
α
−φ
+
φ
+
e
φ
−
φ
i
i−1
i+1
i
(∆x)2
2
(29)
(30)
(31)
To obtain a more precise expression for ∆t, consider proceeding by using the triangle inequality
|z1 + z2 | ≤ |z1 | + |z2 | for any z1 , z2 ∈ C,
then a more rigorous condition on ∆t is
D
j
ikp ∆x j
j j 2 kp ∆x ∆t
−4 sin
φi+1 − φi ≤ 2.
+ |α| −φi + φi−1 + e
(∆x)2 2
(32)
(33)
Here, α contains the charge z of the ion and therefore may also be negative, but is always real. All
terms on the left are positive, and the condition on ∆t must be satisfied for all kp and φji (if the step ∆t
is fixed for the entire simulation; otherwise, it may be adjusted at each time step j). As a result, we need
to treat only the maximum terms possible. The maximum of sin2 is one, and the phase term does not
contribute to the magnitude, hence
D
j
j
j j
∆t
≤ 2,
(34)
4 + |α| max φi − φi−1 + φi+1 − φi i,j
(∆x)2
2(∆x)2
∆t ≤
D
−1
j
j
j j
4 + |α| max φi − φi−1 + φi+1 − φi .
i,j
(35)
Note the physical interpretation of the term on the right hand side,
φj − φji−1
∂φi
≈ i
,
∂t backward difference
∆x
φji+1 − φji
∂φi
,
≈
∆x
difference
∂t forward

 j
j
j j
∂φi 1  φi − φi−1 φi+1 − φi 
.
∴
+
≈ average 
2
∆x
∆x
∂t (36)
(37)
(38)
Finally, when no electric field is present, the potential terms vanish and we recover the stability
condition for the diffusion equation as expected,
∆t ≤
(∆x)2
.
2D
4
(39)
3.1
Constant Electric Field
Consider the case of an electric field which is constant in time and space and has magnitude E0 , such as
assumed in the derivation of the Goldman-Hodgkin-Katz (GHK) equation. By definition
φi − φi−1 φi+1 − φi ,
E0 = −
= −
(40)
∆x ∆x for all i, where we have dropped the superscript j to indicate that the potential time independent.
The maximum term in 35 becomes
1
j
j
j j
max φi − φi−1 + φi+1 − φi = 2E0 ,
∆x i,j
(41)
such that the bound is
∆x
∆t ≤
D
4
2
+ |α|E0
∆x
−1
.
(42)
Sample Mathematica Code
Mathematica (v.10.2) code is provided here.
Consider two example potentials φ: (1) a linear potential, and (2) an exponentially varying one.
The condition on ∆t for a linear potential is given by 42. Define the following time steps
∆t0.999
∆x
= 0.999 ×
D
∆t1.001
∆x
= 1.001 ×
D
2
+ |α|E0
∆x
−1
2
+ |α|E0
∆x
−1
,
(43)
,
(44)
such that ∆t0.999 is a stable time step and ∆t1.001 is an unstable step (see code example for sample
parameter values). For an initial concentration distribution in the shape of a sin function, Figure 1 shows
the initial and final concentration distributions for each of the two time steps. Despite only a 0.2%
difference in their magnitude, the time step ∆t1.001 clearly leads to an unstable simuatlion.
3.× 10-25
3.× 10-25
2.5 × 10-25
2.5 × 10-25
2.× 10-25
2.× 10-25
-25
�������� 1.5 × 10
1.5 × 10-25
1.× 10-25
1.× 10-25
5.× 10-26
5.× 10-26
0.5
1.0
1.5
2.0
0.5
1.0
1.5
2.0
dt0.999
dt1.001
initial dist.
Figure 1: Concentration distributions after 2000 steps with time steps as labeled, for a linear potential.
Similarly, consider the case of a potential in the shape of an exponential function, whose bound on
∆t is given by the more general 35. For
5
∆t0.99
2(∆x)2
= 0.99 ×
D
∆t1.01
2(∆x)2
= 1.01 ×
D
−1
4 + |α| max (|φi − φi−1 | + |φi+1 − φi |)
,
(45)
−1
4 + |α| max (|φi − φi−1 | + |φi+1 − φi |)
,
(46)
i
i
∆t0.99 is a stable time step and ∆t1.01 an unstable step. Figure 2 shows that again above the derived
bound, the simulation becomes unstable. Note that the bound is not exact, as the triangle inequality 32
provides a larger bound on ∆t than absolutely sufficient, although slightly smaller time step (e.g. ∆t1.0001 )
may also lead to stable simulations.
3.× 10-25
3.× 10-25
2.5 × 10-25
2.5 × 10-25
2.× 10-25
2.× 10-25
-25
��������� 1.5 × 10
1.5 × 10-25
1.× 10-25
1.× 10-25
5.× 10-26
5.× 10-26
0.5
1.0
1.5
2.0
0.5
1.0
1.5
2.0
dt0.99
dt1.01
initial dist.
Figure 2: Concentration distributions after 2000 steps with time steps as labeled, for an exponential
potential.
References
[1] Bertil Hille, Ion channels of excitable membranes, Sinauer, Sunderland, Mass, 2001.
[2] Christof Koch, Biophysics of computation information processing in single neurons, Oxford University Press, New York,
1999.
[3] Wikipedia, Von Neumann stability analysis — Wikipedia, The Free Encyclopedia (2015), https://en.wikipedia.org/
w/index.php?title=Von_Neumann_stability_analysis&oldid=674227751.
6