J. theor. Biol. (1995) 177, 299–304

0022–5193/95/230299+06 $12.00/0 7 1995 Academic Press Limited

Numerical Analysis of the Nernst-Planck-Poisson System

M K

Department of General Education, Tokyo Medical and Dental University, 2-8-30 Konodai,Ichikawa-shi, Chiba 272, Japan

Received on 24 February 1995, Accepted in revised form on 4 August 1995

The Nernst-Planck equation is used to describe membrane transport phenomena in biological systems,such as nerve excitation and cell transport of ions. Unfortunately, solving this equation is difficult andsome assumptions are required. There are two assumptions which are widely used: the ‘‘constant field’’and the ‘‘electroneutrality’’ assumptions. Neither assumption can be justified, however, once the Poissonequation is taken into account (Nernst-Planck-Poisson system), because the results obtained from theseassumptions do not satisfy the Poisson equation. It was shown that the two assumptions are limitingcases of a certain dimensionless parameter which is related to the ratio of the Debye length to membranethickness. For the cases far from both limits, however, the equations should be solved numerically. Thenumerical solution of the Nernst-Planck-Poisson equation was used to obtain the following results: (1)The result of the constant field assumption is not in good agreement with the numerical result andthe first order perturbation considerably improved the result. (2) The range of applicability of theconstant field assumption is larger than expected from the analysis based on the perturbation expansion.Using the solution for the case in which no approximation can be applied, transition between theconstant field and the electroneutrality is shown.

7 1995 Academic Press Limited

1. Introduction

The Nernst-Planck equation is used to describemembrane transport phenomena in biological sys-tems, such as nerve excitation and cell transport ofions (Goldman, 1943; Hodgkin & Katz, 1949;Hodgkin & Huxley, 1952; Chen, Barcilon &Eisenberg, 1992; Chen & Eisenberg, 1993; andreferences therein). The equation is insufficientbecause it includes only the effects of the gradients ofconcentrations and electric potential. From a point ofview of non-equilibrium thermodynamics, it alsoneglects coupling effects. It still has some advantagesand does not lose its usefulness when illustratingmembrane transport phenomena. Unfortunately,solving this equation is difficult and some assump-tions are required. There are two assumptions whichare widely used: the ‘‘constant field’’ and the‘‘electroneutrality’’ assumptions. Neither assumptioncan be justified once the Poisson equation is taken

into account (Nernst-Planck-Poisson system), be-cause the results obtained from these assumptionsdo not satisfy the Poisson equation. If the electricfield is constant, the charge density has to vanisheverywhere as a consequence of the Poisson equation,but this is not satisfied by the results obtained fromthe constant field assumption. On the other hand, ifthe electroneutrality is assumed, the electric fieldbecomes constant as a consequence of the Poissonequation, but also in this case this is not satisfied bythe results obtained from the assumption. It wasshown that the two assumptions are two limitingcases of a certain dimensionless parameter which isrelated to the ratio of the Debye length to membranethickness (MacGillivray, 1968; MacGillivray & Hare,1969). For cases far from both limits, however, theequations should be solved numerically.

When one uses the Nernst-Planck equation, theconstant field assumption is useful because thesolution has explicit analytical form. Knowledge of

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the applicability of this assumption is very important.Such an attempt was made based on the perturbationexpansion (MacGillivray & Hare, 1969) and the resultshould be checked by numerical calculation.

In this paper, using the numerical solution ofthe Nernst-Planck-Poisson equation, the validity ofthe approximation is studied with the emphasis onthe constant field assumption. The solution of theequation for cases where the two assumptions are notvalid is also studied.

In Section 2 formulation of the problem ispresented for a simple system with two univalentspecies in one dimension. In Section 3 the results arepresented and Section 4 contains the conclusion. Inthe Appendix the numerical method is explained insome detail.

2. Formulation of the Problem

For simplicity, let us consider the system of twounivalent species in one dimension. In this case theNernst-Planck equations are

J1=−D1FRT

c1dc

dx−D1

dc1

dx, (1)

J2=D2FRT

c2dc

dx−D2

dc2

dx, (2)

and the Poisson equation is

d2c

dx2=−1ee0

r, (3)

where c1 is the concentration of cation, c2 is theconcentration of anion and Ji , Di are the flux and thediffusion coefficient of the species i. R,T and F arethe gas constant, the absolute temperature and theFaraday constant, respectively. c is the electrostaticpotential, e0 is the permittivity of the free space, e isthe dielectric constant and r is the charge density.

Let us consider the membrane carrying a constantdistribution of fixed univalent cations of density N(MacGillivray & Hare, 1969).

Then the charge density r becomes

r=c1−c2+N. (4)

The boundary conditions are fixed as follows (theboundaries are at x=0 and x=L)

ci (0)=C, ci (L)=C+DC,

c(0)=0, c(L)=Dc. (5)

Introducing the following dimensionless variables(MacGillivray, 1968; MacGillivray & Hare, 1969)

c=RTF

c , x=Lx̃, ci=Nc̃i , Ji=NDi

LJ i , (6)

eqns (1–3) become

J 1=−c̃1dcdx̃

−dc̃1

dx̃, (7)

J 2=c̃2dcdx̃

−dc̃2

dx̃, (8)

d2cdx̃2=−

NF2L2

ee0RT(c̃1−c̃2+1)=−

1a2 (c̃1−c̃2+1),

(9)

where

a2=ee0RTNF2L2 . (10)

This dimensionless parameter a2 can be related tothe ratio of membrane thickness to the Debyelength (MacGillivray, 1968; MacGillivray & Hare,1969).

The boundary conditions of dimensionlessvariables become

c̃i (0)=CN

0g,

c̃i (1)=C+DC

N0g+d,

c (0)=0,

c (1)=DcFRT

0b. (11)

If a steady state is assumed, J 1 and J 2 becomeconstants and eqns (7) and (8) can be integrated as

J 1=−1

g1

0

ec dx̃

(c̃1(1) ec (1)−c̃1(0) ec (0)), (12)

J 2=−1

g1

0

e−c dx̃

(c̃2(1) e−c (1)−c̃2(0) e−c (0)), (13)

The right hand side of the Poisson equation (9)becomes small for large a2, and the situationapproaches the constant field. On the other hand, forsmall a2, the left hand side becomes small and thesituation approaches electroneutrality. In fact, it wasshown that the electroneutrality and the constant fieldare the zeroth order approximation of the pertur-bation expansion about a2 (MacGillivray, 1968) and1/a2 (MacGillivray & Hare, 1969), respectively.

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F. 1. Dimensionless potential (1/a2=1). The solid line is thenumerical result. The dotted line is the zeroth-order (constant field)and the dot-dash line is the zeroth plus first-order term.

F. 3. Dimensionless fluxes versus 1/a2. The solid line is thenumerical result and the dot-dash line is the zeroth plus first-orderterm. (zeroth order term equals the value at 1/a2 is zero and doesnot depend on 1/a2).

The Poisson equation can be considered as a kindof constraint to the solution of the Nernst-Planckequation. Without the Poisson equation, one canarbitrarily give either electric field or concentration,either being obtained when the other is given. In thelimit of 1/a2=a, the Poisson equation gives theelectroneutrality and only the solution which satisfiesthis condition is allowed. In the limit 1/a2=0, thePoisson equation gives the constant field and, also inthis case, only the solution which satisfies thiscondition is allowed. The parameter 1/a2 can beconsidered to show the significance of these twoconditions; for small 1/a2 the constant field conditiongives stronger constraint than the electorneutralitywhile for large 1/a2 the situation is reversed.

3. Results

For all calculations, parameters are fixed to b=1,g=1, d=0.5.

Figures 1 and 2 show the results of the calcu-lation for 1/a2=1. The results of the first orderperturbation expansion of 1/a2 (MacGillivray & Hare,1969; the zeroth order is the constant fieldassumption) are also shown in the figures. Analyticalexpressions of zeroth and first order perturbation

expansion of 1/a2 are given in MacGillivray & Hare(1969).

J 1 J 2The values of J i are 0th −1.791 0.709

0th+1st −1.630 0.773numerical −1.665 0.762

In this case the result of the constant field as-sumption is not in good agreement with the numericalresult and the first order perturbation considerablyimproves the result.

Figure 3 shows the 1/a2 dependence of thedimensionless fluxes J 1 and J 2. From this figure onecan also see that the result of constant fieldapproximation is not good, even for small 1/a2, andthat the first order perturbation improves the resultand widens the range of applicability of thisassumption to a certain extent. The range ofapplicability of the constant field assumption is,however, larger than expected from the analysis basedon the perturbation expansion. For example, if oneallows a 10% error, the zeroth-order approximationis applicable to 1/a211.7 (estimate by first-orderperturbation is 1.1) and first-order perturbation is to1/a212.5.

F. 4. Dimensionless potential for various 1/a2. The solid lineis the constant field assumption. The long broken lines is for1/a2=2. The dotted lines is for 1/a2=5. The dot-dash line is for1/a2=10. The broken line is for 1/a2=20.

F. 2. Dimensionless concentrations (1/a2=1). The solid linesare the numerical results. The dotted lines are the zeroth-order(constant field) and the dot-dash lines are the zeroth plus first-orderterms. c̃1 is represented by circles and c̃2 by triangles.

302

F. 5. Dimensionless concentrations for various 1/a2. The solidlines is the constant field assumption. The long broken line is for1/a2=2. The dotted line is for 1/a2=5. The dot-dash line is for1/a2=10. The broken line is for 1/a2=20. c̃1 is represented by circlesand c̃2 by triangles.

assumption is not consistent with the Poissonequation.

4. Conclusion

The numerical results are compared with the resultsof constant field assumption for 1/a2=1, and itwas shown that for this 1/a2 the results of the con-stant field assumption are not in good agreement withthe numerical results, although the first orderperturbation considerably improves the results.

The dimensionless fluxes are calculated for variousvalues of 1/a2 and this also shows that the first orderperturbation improves the results to a certain extent.The range of applicability of the constant fieldassumption is, however, larger than expected from theanalysis based on the perturbation expansion.

The potential and concentration is calculated forvarious 1/a2. As 1/a2 becomes large, a bend appearsin the potential and concentration near the boundary.This bend becomes sharper as 1/a2 becomes larger andwill give a discontinuity in the case 1/a2=a. Thisresult shows the transition between the constant fieldand the electroneutrality.

The charge density was calculated for various 1/a2

and it was shown that the significance of the constantfield and the electroneutrality changes according to1/a2. As 1/a2 becomes larger, the electroneutralitycondition becomes stronger and the charge densityapproaches zero.

I wish to thank Professor M. Wada for his interest in andencouragement of this study.

REFERENCES

A, M. & S, I. A. (1972). Handbook ofMathematical Functions, New York: Dover.

C, D. P., B, V. & E, R. S. (1992). Constantfields and constant gradients in open ionic channels. Biophys. J.61, 1372–1393.

C, D. & E, R. (1993). Charges, currents, and potentialsin ionic channels of one conformation. Biophys. J. 64,1405–1421.

G, D. E. (1943). Potential, impedance, and rectification inmembranes. J. Gen. Physiol. 27, 37–60.

H, A. L. & H, A. F. (1952). A quantitativedescription of membrane current and its application toconduction and excitation in nerve. J. Physiol. 117, 500–544.

H, A. L. & K, B. (1949). The effect of sodium ions onthe electrical activity of the giant axon of the squid. J. Physiol.108, 37–77.

MG, A. D. (1968). Nernst-Planck equations and theelectroneutrality and Donnan equilibrium assumptions. J.Chem. Phys. 48, 2903–2907.

MG, A. D. & H, D. (1969). Applicability ofGoldman’s constant field assumption to biological systems. J.theor. Biol. 25, 113–126.

Figures 4 and 5 show the results of the calculationfor various 1/a2. These 1/a2 cannot be regarded assmall, and the results based on the constant fieldassumption cannot be expected to give goodestimations. In fact, one can see from these figuresthat the deviation from the constant field approxi-mation is large. These results are consideredas an example of the non-limiting cases and showthe transition between the constant field andthe electroneutrality. The bend of potential andconcentration near the boundary is a consequence ofthe electroneutrality. It becomes sharper as 1/a2

becomes large and when 1/a2=a (a2=0) it willbecome a discontinuity at the boudary (Donnanequilibrium).

Figure 6 shows the dimensionless charge densityc̃1−c̃2+1 for various 1/a2. This shows that as 1/a2

becomes larger, the electroneutrality condition be-comes more significant. For small 1/a2, charge densityis far from zero, however, as 1/a2 becomes larger thecharge density approaches zero (electroneutrality).From this figure one can also see that the chargedensity obtained from the constant field assumptiondoes not vanish and this means that the result of this

F. 6. Dimensionless charge density for various 1/a2. The solidline is the constant field assumption. The long dotted line is for1/a2=1. The long broken line is for 1/a2=2. The dotted line is for1/a2=5. The dot-dash line is for 1/a2=10. The broken line is for1/a2=20.

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APPENDIXThe eqns (7–9) are non linear and the three variablesc̃1, c̃2 and c are related to each other. As it is difficultto solve them directly, a different method is needed.

Here a kind of ‘‘self-consistent’’ calculation isapplied. The strategy is schematically shown below.

c1 and c2 are calculated by eqns (7) and (8) from c.Next (new) c is calculated by eqn (9) from these c1

and c2, and the convergence is checked. If it does notreach the convergence, c1 and c2 are calculated fromthis new c. This process is repeated until theconvergent (self-consistent) solution is obtained.

The finite difference method is used to calculate c1,c2 and c from the differential eqns (7–9). To do this,the interval [0, 1] is divided into equal N+1 parts andthe functions are discretized

fk=f(kh), h=1

N+1(A.1)

Then the differentiation is substituted by the differ-ence. Using the five-point method (see, for example,Abramowitz & Stegun, 1972), the first derivative is

f'k=1

12h(fk−2−8fk−1+8fk+1−fk+2)+O(h4)

(k=2, . . . , N−1), (A.2)

and at the endpoints

f'1=−1

12h(3f0+10f1−18f2+6f3−f4)+O(h4)

(A.3)

f'N=−1

12h( fN−3−6fN−2+18fN−1

−10fN−3fN+1)+O(h4). (A.4)

The second derivative is

f 0k =1

12h2 (−fk−2+16fk−1−30fk+16fk+1−fk+2)

+O(h3)(k=2, . . . , N−1), (A.5)

and at the endpoints

f 01 =1

12h2 (11f0−20f1+6f2+4f3−f4)+O(h3)

(A.6)

f 0N=1

12h2 (−fN−3+4fN−2+6fN−1

−20fN+11fN+1)+O(h3). (A.7)

By this substitution, eqn (7) becomes

J 1=−c̃1,kc 'k−c̃1,k'

=−c̃1,k1

12h(c k−2−8c k−1+8c k+1−c k+2)

−1

12h(c̃1,k−2−8c̃1,k−1+8c̃1,k+1−c̃1,k+2) (A.8)

This can be written as

−c̃1,k−2+8c̃1,k−1−(c k−2−8c k−1+8c k+1−c k+2)

×c̃1,k−8c̃1,k+1+c̃1,k+2=12hJ 1

(k=2, . . . , N−1), (A.9)

and at the endpoints

(10+3c 0+10c 1−18c 2+6c 3−c 4)

×c̃1,1−18c̃1,2+6c̃1,3−c̃1,4=12hJ 1−3c̃1,0, (A.10)

c̃1,N−3−6c̃1,N−2+18c̃1,N−1−(10−c N−3

+6c N−2−18c N−1+10c N

+3c N+1)c̃1,N=12hJ 1+3c̃1,N+1. (A.11)

These are simultaneous equations and in the matrixform they are

Initial c 0004 Calculate c1, c2 0004 Calculate c 0004 Check 0004 Results

NG

OK

A1,1 −18 6 −1 0 0 . . . c̃1,1 12hJ 1−3c̃1,0

8 A2,2 −8 1 0 0 . . . c̃1,2 12hJ 1+c̃1,0

−1 8 A3,3 −8 1 0 . . . c̃1,3 12hJ 1* ... ... ... ... ... * = * (A.12)G

G

G

G

G

G

G

F

f

GG

G

G

G

G

G

J

j

GG

G

G

G

G

G

F

f

GG

G

G

G

G

G

J

j

GG

G

G

G

G

G

F

f

GG

G

G

G

G

G

J

j

0 0 −1 8 AN−2,N−2 −8 1 c̃1,N−2 12hJ 10 0 0 −1 8 AN−1,N−1 −8 c̃1,N−1 12hJ 1−c̃1,N+1

0 0 0 1 −6 18 AN,N c̃1,N 12hJ 1+3c̃1,N+1

304

where

A1,1=10+3c 0+10c 1−18c 2+6c 3−c 4, (A.13)

AN,N=−(10−c N−3+6c N−2−18c N−1+10c N+3c N+1) (A.14)

Ak,k=−(c k−2−8c k−1+8c k+1−c k+2) (k=2, . . . , N−1), (A.15)

and J 1 is calculated from eqn (12).The equations for c 2,k are obtained from (A.12) by changing the sign of c and replacing J 1 by J 2.

By a similar procedure, eqn (9) becomes

F JF J F J G G−20 6 4 −1 0 0 . . . c 1 −12h2

a2 (c̃1,1−c̃2,1+1)−11c 0

G G G G G GG G G G G G16 −30 16 −1 0 0 . . . c 2 −

12h2

a2 (c̃1,2−c̃2,2+1)+c 0

G G G G G GG G G G G G

−1 16 −30 16 −1 0 . . . c 3 −12h2

a2 (c̃1,3−c̃2,3+1)G G G G G GG G G G G G* ... ... ... ... * = *G G G G G G

0 −1 16 −30 16 −1 c N−2 −12h2

a2 (c̃1,N−2−c̃2,N−2+1)G G G G G GG G G G G GG G G G G G0 0 −1 16 −30 16 c N−1 −

12h2

a2 (c̃1,N−1−c̃2,N−1+1)+c N+1

G G G G G GG G G G G G

0 0 −1 4 6 −20 c N −12h2

a2 (c̃1,N−c̃2,N+1)−11c N+1f j f j f j(A.16)

The differential equations become simultaneous equations which can be solved relatively easily. Allthe calculations in Section 3 are performed with N=500.