ionic size effects: generalized boltzmann distributions...
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Ionic Size Effects: Generalized Boltzmann Distributions,
Counterion Stratification, and Modified Debye Length
Bo Li ∗ Pei Liu † Zhenli Xu ‡ Shenggao Zhou §
July 18, 2013
Abstract
Near a charged surface, counterions of different valences and sizes cluster; and theirconcentration profiles stratify. At a distance from such a surface larger than the Debyelength, the electric field is screened by counterions. Recent studies by a variationalmean-field approach that includes ionic size effects and by Monte Carlo simulationsboth suggest that the counterion stratification is determined by the ionic valence-to-volume ratios. Central in the mean-field approach is a free-energy functional of ionicconcentrations in which the ionic size effects are included through the entropic effectof solvent molecules. The corresponding equilibrium conditions define the generalizedBoltzmann distributions relating the ionic concentrations to the electrostatic potential.This paper presents a detailed analysis and numerical calculations of such a free-energyfunctional to understand the dependence of the ionic charge density on the electrostaticpotential through the generalized Boltzmann distributions, the role of ionic valence-to-volume ratios in the counterion stratification, and the modification of Debye lengthdue to the effect of ionic sizes.
1 Introduction
Electrostatic interactions between macromolecules and mobile ions in an aqueous solventgenerate strong, long-ranged forces that play a key role in biological processes. In suchinteractions, ionic sizes or excluded volumes can affect many of the detailed chemical and
∗Department of Mathematics and NSF Center for Theoretical Biological Physics, University of California,San Diego, 9500 Gilman Drive, Mail code: 0112, La Jolla, CA 92093-0112, USA. Email: [email protected].
†Department of Mathematics and Institute of Natural Sciences, Shanghai Jiao Tong University, 800Dongchuan Rd., Shanghai, 200240, P. R. China. Email: [email protected].
‡Department of Mathematics, Institute of Natural Sciences, and Ministry of Education Key Laboratoryin Scientific and Engineering Computing, Shanghai Jiao Tong University, 800 Dongchuan Rd., Shanghai,200240, P. R. China. Email: [email protected].
§Department of Mathematics and NSF Center for Theoretical Biological Physics, University of Cal-ifornia, San Diego, 9500 Gilman Drive, Mail code: 0112, La Jolla, CA 92093-0112, USA. Email:[email protected].
1
physical properties of an underlying biological system. For instance, differences in ionicsizes can affect how mobile ions bind to nucleic acids [3, 4, 32]. The size of monovalentcations can influence the stability of RNA tertiary structures [21]. The ionic size effectis more profound in the ion channel selectivity, see, e.g., [16, 24]. Detailed Monte Carlosimulations and integral equations calculations also confirm some of these experimentallyobserved properties due to the non-uniformity of ionic sizes [17, 28, 34,36].
The classical Poisson–Boltzmann (PB) equation (PBE) is perhaps the most widely usedand efficient model of the electrostatics in ionic solutions, particularly in biomolecular sys-tems [2,10,12,14,27,30]. Despite of its many successful applications, this mean-field theoryis known to fail in capturing the ionic size effects and ion-ion correlations [15,17,28,35,36].Numerical calculations based on the classical PBE often predict unphysically high concen-trations of counterions near a charged surface [5, 38].
Recently, there has been a growing interest in studying the ionic size effects withinthe PB-like mean-field framework [5–7,9,11,18–20,22,23,25,26,31,33,38]. One of the mainapproaches is to include the additional entropic effect of solvent molecules in the free-energyfunctional of all local ionic concentrations, c1 = c1(x), . . . , cM = cM(x) (M being the totalnumber of ionic species), similar to that in the classical PB model. Given the volume viof an ion of the ith species (1 ≤ i ≤ M) and the volume v0 of a solvent molecule, one candefine the local solvent concentration c0 = c0(x) by
c0(x) = v−10
[
1−M∑
i=1
vici(x)
]
. (1.1)
The size-modified electrostatic free-energy functional is then given by [5, 20, 22,26]
F [c1, · · · , cM ] =
∫
[
1
2ρψ + β−1
M∑
i=0
ci [ln (vici)− 1]−M∑
i=1
µici
]
dV. (1.2)
Here, ρ =∑M
i=1 qici is the ionic charge density, where qi = zie, and zi is the valence of anion of the ithe species and e is the elementary charge, β = (kBT )
−1 with kB the Boltzmannconstant and T the temperature, and µi is the chemical potential of an ion of the ith species.The electrostatic potential ψ is determined by Poisson’s equation −∇ · ε∇ψ = ρ, togetherwith some boundary conditions, where ε is the dielectric coefficient.
One easily verifies that the equilibrium conditions δciF = 0 (i = 1, . . . ,M) are given by
viv0
ln (v0c0)− ln (vici) = β (qiψ − µi) , i = 1, . . . ,M, (1.3)
where ψ is the electrostatic potential corresponding to the equilibrium ionic concentrationsc1, . . . , cM . It is shown in [22] that this system of nonlinear algebraic equations has a uniquesolution ci = Bi(ψ) (i = 1, . . . ,M), defining the generalized Boltzmann distributions. Ifv0 = v1 = · · · = vM , then these distributions are [22, 23]
ci =c∞i e
−βqiψ
1 +∑M
j=1 vjc∞j (e−βqjψ − 1)
, i = 1, . . . ,M,
2
where
c∞i =v−1i eβµi
1 +∑M
j=1 eβµj
, i = 1, . . . ,M.
Note that, for each i, c∞i is the value of ci when ψ = 0. In general, if v0, v1, . . . , vM are notall the same, then explicit formulas for ci = Bi(ψ) (i = 1, . . . ,M) seem unavailable.
In studying the ionic size effects in biomolecular reactions, Lu and Zhou [26] minimizedthe free-energy functional (1.2) by solving for a steady-state solution of a size-modifiedPoisson–Nernst–Planck system of partial differential equations. In [38], Zhou et al. devel-oped a constrained optimization method for numerically minimizing the functional (1.2)with Poisson’s equation as a constraint. They found the remarkable stratification of ionicconcentrations near a highly charged surface. Moreover, they found that the ionic valence-to-volume ratio is an important parameter that determines the stratification. Specifically,the counterion with the largest valence-to-volume ratio has the highest ionic concentrationnear the surface, the counterion with the second largest such ratio has also the second high-est ionic concentration, and so on. Subsequently, Wen et al. [37] performed Monte Carlosimulations that confirm such a phenomenon.
In this work, we further study the size-modified, mean-field free-energy functional (1.2).We begin with a description of two different forms of the electrostatic free-energy functional,one with and the other without the constraint of total number of ions for each ionic species.We show the equivalence of these two forms; cf. Theorem 2.2. We also derive the conditionsfor the equilibrium concentrations; cf. (1.3).
We then focus on the equilibrium conditions (1.3) to analyze the generalized Boltzmanndistributions. In particular, we characterize the behavior of the equilibrium ionic chargedensity ρi =
∑Mi=1 qici as a function of ψ. Our main results of analysis are as follows:
(1) The concentration of ions with the largest and that with smallest valence-to-volumeratios monotonically decreases and monotonically increases, respectively, with respectto the electrostatic potential ψ. As ψ → −∞ (or ∞), the concentration of ions withthe largest (or smallest) valence-to-volume ratio approaches to the inverse of its cor-responding volume, while all other concentrations approach zero.
(2) The total ionic charge density ρi = ρi(ψ) is a monotonic function of the potential ψ.If all ions are cations (i.e., all zj > 0), then this density is always positive, ρi(∞) = 0,and ρi(−∞) exists and is finite. If all ions are anions (i.e., all zj < 0), then thisdensity is always negative, ρi(−∞) = 0, and ρi(∞) exists and is finite. If both cationsand anions exist, then both ρi(−∞) and ρi(∞) exist and are finite, and the densityreaches zero at a unique value of the potential ψ. All these are summarized in Figure 2in Subsection 3.2.
(3) A size-modified Debye length, λD, is derived with a weak potential limit. A formulaof of λD is given in (3.23) in Subsection 3.3.
Note that ionizable groups of biomolecules can dissociate in aqueous solution to produceanions or cations. So, studying an ionic system with only cations or only anions is of interest.In general, such a system can serve a model approximation system, as near a charged surfacethe concentration of coions is usually very low.
We finally consider a charged spherical molecule of radius R0 in an ionic solution that
3
occupies a region defined by R0 < |x| < R∞ for some R∞ > R0, and study the PBE, bothclassical and size-modified. In the case R∞ is finite, the detailed properties of the equilibriumelectrostatic potential are described through two parameters: one is the prescribed surfacecharge density σ at the boundary |x| = R0 and the other is an effective surface charge densityσ∞ at the boundary |x| = R∞. This effective density is set by the total charge neutrality.When σσ∞ < 0 the potential ψ is monotonic. Otherwise, it decreases then increases, or itincreases then decreases. These are summarized in Theorem 4.2 and illustrated in Figure 3 inSection 4. With the same setting of a charged spherical molecule, we numerically minimizethe functional (1.2) and confirm our analysis; cf. Figure 4 in Section 5. If R∞ = ∞, thenonly in the case that both cations and anions exist and they neutralize in the bulk does thePBE (classical or size-modified) permit a unique solution, the electrostatic potential, thatis radially symmetric and monotonic, and vanishes at infinity. We remark that consideringa finite R∞ is of certain interest as it is known that the effect of finite size of charged systemcan be significant, cf. e.g., [29], and as it is practically useful in numerical computations.
The rest of this paper is organized as follows: In Section 2, we describe the electro-static free-energy functionals and derive the conditions of equilibrium concentrations. InSection 3, we present an analysis on the equilibrium conditions and on the stratification ofcounterions of multiple valences and sizes near a charged surface. In Section 4, we study acharged spherical molecule immersed in an ionic solution. In Section 5, we report numericalcalculations to confirm our analysis. Finally, in Section 6, we draw conclusions.
2 The Free-Energy Functional and Equilibrium Con-
ditions
We consider an ionic solution with M (M ≥ 1) ionic species, occupying a region Ω that is abounded, open, and connected subset of R3 with a smooth boundary Γ 6= ∅. The boundaryΓ is divided into two disjoint and smooth parts ΓD and ΓN (with D for Dirichlet and N forNeumann). We assume that a surface charge density is given on ΓN and the electrostaticpotential is specified on ΓD. Our set up covers the case that ΓN = ∅ and ΓD = Γ and thatΓD = ∅ and ΓN = Γ. It also covers the important case of biomolecular solvation with animplicit solvent: The region Ω is the solvent region and the part of the boundary ΓN is thesolute-solvent interface; cf. Figure 1. The surface charge density on ΓN effectively replacesthe density of partial charges carried by solute atoms.
For each i (1 ≤ i ≤M), we denote by ci = ci(x) the local ionic concentration of the ithionic species at position x ∈ Ω. As before, ρ =
∑Mi=1 qici is the ionic charge density. We
also denote by vi (1 ≤ i ≤M) the volume of an ion of the ith species and by v0 the volumeof a solvent molecule. The local concentration of the solvent, c0 = c0(x), is then defined by(1.1). For convenience we denote c = (c1, . . . , cM). We consider minimizing the mean-fieldelectrostatic free-energy functional
F [c] =
∫
Ω
1
2ρψ dV +
∫
ΓN
1
2σψ dS + β−1
M∑
i=0
∫
Ω
ci [ln(vici)− 1] dV −M∑
i=1
∫
Ω
µici dV, (2.1)
4
+
+ ++
++
+
+
+
+
+
+
+
+
+
+
++
+ +
+
+
+
+
-
-
-
-
-
-
-
--
-
-
-
-
-
-
-
ΓD
ΓN
Ω
Figure 1: A schematic view of a charged molecule immersed in an ionic solution. Themolecule is represented by the inner sphere filled with symbols +.
with
∇ · ε∇ψ = −ρ in Ω,
ψ = ψ∞ on ΓD,
ε∂ψ
∂n= σ on ΓN.
(2.2)
Here, the first two terms in (2.1) describe the electrostatic potential energy. The electrostaticpotential ψ is the unique solution of the boundary-value problem of Poisson’s equation (2.2),where ε = ε(x) (x ∈ Ω) is the dielectric coefficient assumed to be Lebesgue measurable andbounded above and below by positive constants, and n denotes the unit exterior normalat the boundary Γ. Both ψ∞ : ΓD → R and σ : ΓN → R are given bounded and smoothfunctions, representing the electrostatic potential at ΓD and the surface charge density atΓN, respectively. The third term in (2.1) has the form −TS with S being the entropy. Thegiven parameters µi (i = 1, . . . ,M) in the last term in (2.1) are the chemical potentials ofthe ions.
We define ψD : Ω → R and ψN : Ω → R by
∇ · ε∇ψD = 0 in Ω,
ψD = ψ∞ on ΓD,
ε∂ψD
∂n= 0 on ΓN,
and
∇ · ε∇ψN = 0 in Ω,
ψN = 0 on ΓD,
ε∂ψN
∂n= σ on ΓN,
respectively. For a given function f : Ω → R that is in the Sobolev space H−1(Ω) [1, 13],let us denote by Lf : Ω → R the unique weak solution to the following boundary-value
5
problem of Poisson’s equation
∇ · ε∇(Lf) = −f in Ω,
Lf = 0 on ΓD,
ε∂(Lf)
∂n= 0 on ΓN.
(The uniqueness is guaranteed if ΓD has a nonzero surface measure.) Notice that L is alinear and self-adjoint operator from the space H−1(Ω) to that consisting of H1(Ω)-functionsvanishing on ΓD. With these the solution ψ to the boundary-value problem (2.2) can bedecomposed as
ψ = ψ + ψD + ψN, (2.3)
where ψ = L(
∑Mi=1 qici
)
. By integration by parts, we have
∫
ΓN
σψ dS =
∫
ΓN
ε∂ψN
∂nψ dS =
∫
Γ
ε∂ψN
∂nψ dS
=
∫
Ω
ψ∇ · ε∇ψN dV +
∫
Ω
ε∇ψ · ∇ψN dV =
∫
Ω
ε∇ψ · ∇ψN dV
=
∫
Ω
(
−∇ · ε∇ψ)
ψN dV =
∫
Ω
(
M∑
i=1
qici
)
ψN dV.
Therefore the free-energy functional F [c] can now be rewritten as
F [c] =
∫
Ω
1
2
(
M∑
i=1
qici
)
L
(
M∑
j=1
qjcj
)
dV +M∑
i=1
∫
Ω
[
qi
(
1
2ψD + ψN
)
− µi
]
ci dV
+ β−1
M∑
i=0
∫
Ω
ci [ln(vici)− 1] dV +
∫
ΓN
1
2σ (ψN + ψD) dS.
By formal and routine calculations using the decomposition (2.3) and the fact that L isself-adjoint, we obtain the first variations [8, 22, 23, 26]
δciF [c] = qi
(
ψ −1
2ψD
)
− µi + β−1
[
ln(vici)−viv0
ln (v0c0)
]
, i = 1, . . . ,M, (2.4)
where ψ is determined by (2.2) with ρ =∑M
i=1 qici. The equilibrium conditions δciF [c] = 0(i = 1, . . . ,M) are then given by
viv0
ln(v0c0)− ln(vici) = β
[
qi
(
ψ −1
2ψD
)
− µi
]
, i = 1, . . . ,M. (2.5)
The following result can be proved by the argument used in [22]:
6
Theorem 2.1. (1) The functional F [c] is convex. Moreover, it has a unique minimizer inthe set of all concentrations c1, . . . , cM such that ci ≥ 0 (i = 1, . . . ,M) and
∑Mi=1 vici ≤
1 in Ω.(2) There exist θ1, θ2 ∈ R with 0 < θ1 < θ2 < 1 that bound uniformly from below and above,
respectively, the free-energy minimizing ionic concentrations and the correspondingconcentration of solvent molecules.
(3) The minimizer is characterized by the equilibrium conditions (2.5).
We now consider a different formulation: minimize the electrostatic free-energy func-tional
F [c] =
∫
Ω
1
2ρψ dV +
∫
ΓN
1
2σψ dS + β−1
M∑
i=0
∫
Ω
ci [ln(vici)− 1] dV, (2.6)
with
∇ · ε∇ψ = −ρ in Ω,
ψ = ψ∞ on ΓD,
ε∂ψ
∂n= σ on ΓN,
(2.7)
subject to
∫
Ω
ci dV = Ni, i = 1, . . . ,M, (2.8)
where again ρ =∑M
i=1 qici. Note that the functional F [c] defined in (2.1) has the term of
the chemical potentials µi. Here the functional F [c] does not have that term. On the otherhand, the minimization of F [c] is subject to the conservation of total number of ions ineach of the M ionic species, defined in (2.8) in which Ni (i = 1, . . . ,M) are given positivenumbers. The system (2.7) is the same as (2.2).
It again follows from standard arguments that the functional F [c] with (2.7) has aunique minimizer c = (c1, . . . , cM) among all c = (c1, . . . , cM) subject to (2.8). By a similarargument used in deriving (2.4), we have for any c = (c1, . . . , cM) that
δciF [c] = qi
(
ψ −1
2ψD
)
+ β−1
[
ln(vici)−viv0
ln (v0c0)
]
, i = 1, . . . ,M. (2.9)
For the minimizer c, we have
∫
Ω
(
δciF [c])
di dV = 0 ∀di :
∫
Ω
di dV = 0, i = 1, . . . ,M.
By Lemma 2.1 below, these lead to the existence of constants, still denoted by µi (i =1, . . . ,M), such that δciF [c] = µi (i = 1, . . . ,M). By (2.9), this is exactly (2.5).
We have in fact proved the following:
Theorem 2.2. The variational problem of minimizing the functional (2.1) with (2.2) isequivalent to that of minimizing the functional F [c] with (2.7) and (2.8).
7
We remark that, if one knows the chemical potentials µi (i = 1, . . . ,M), then the firstformulation, i.e., minimizing F [c], can be mathematically and computationally simpler thanthe second one that has the constraint of total numbers of ions. The question is how thechemical potentials can be estimated by experiment or other models. A similar practicalissue for the second formulation with the constraint (2.8) is that one needs to have a goodestimate of each Ni, the total number of ions of the ith species in the system.
Lemma 2.1. Let u ∈ L2(Ω) be such that∫
Ω
u v dV = 0 ∀v ∈ L2(Ω) :
∫
Ω
v dV = 0.
Then there exists a constant µ ∈ R such that u = µ almost everywhere in Ω.
Proof. Denote for any w ∈ L2(Ω)
w = −
∫
Ω
w dV =1
|Ω|
∫
Ω
w dV,
where |Ω| is the volume (i.e., the Lebesgue measure) of Ω. Note that the integral of w − wover Ω vanishes. For any v ∈ L2(Ω), we have by the assumption of the lemma that
∫
Ω
(u− u) v dV =
∫
Ω
(u− u) (v − v) dV =
∫
Ω
u (v − v) dV = 0.
Therefore u = u almost everywhere in Ω. Set µ = u. The Lemma is proved. Q.E.D.
3 Equilibrium Ionic Concentrations and Modified De-
bye Length
We consider the equilibrium conditions (2.5) that determine the equilibrium ionic concen-trations. Introducing a shifted potential φ = ψ − ψD/2, we can rewrite (2.5) as
viv0
ln(v0c0)− ln(vici) = β (qiφ− µi) , i = 1, . . . ,M. (3.1)
As c0 is given in (1.1), this is a system of nonlinear algebraic equations for the unknownsc1, . . . , cM . In [22], Li proved the following:
Theorem 3.1. (1) For each φ ∈ R the system (3.1) has a unique solution ci = Bi(φ) ∈(0, v−1
i ) (i = 1, . . . ,M). Moreover each Bi : R → (0, v−1i ) is a smooth function.
(2) Define V : R → R by
V (φ) = −M∑
i=1
∫ φ
0
qiBi(ξ) dξ + V0, (3.2)
where V0 is a constant. Then the function V : R → R satisfies V ′′ > 0 on R and isthus strictly convex.
8
The relations ci = Bi(φ) (i = 1, . . . ,M) define the generalized Boltzmann distributionsfor the equilibrium concentrations c1, . . . , cM . In general, explicit formulas of such distribu-tions seem to be unavailable. Notice that −V ′(φ) =
∑Mi=1 qiBi(φ) with φ = ψ−ψD/2 is the
ionic charge density. The proof of V ′′ > 0 in R given in [22] (cf. the proof of Lemma 5.2in [22]) applies to our present case; cf. the remark in Subsection 3.1. The electrostaticpotential ψ is now the unique solution to the boundary-value problem of the generalizedand implicit PBE
∇ · ε∇ψ − V ′
(
ψ −ψD2
)
= 0. (3.3)
The boundary conditions are the same as in (2.2). The existence and uniqueness of theequilibrium concentrations ci (i = 1, . . . ,M) lead to that of the solution to this boundary-value problem.
3.1 Equilibrium ionic concentrations
For convenience, we denote z0 = 0, q0 = 0, and µ0 = 0. We assume that all the M ionicspecies have different ionic valence-to-volume ratios. We assume there are l ≥ 0 species ofanions andm ≥ 0 species of cations. SoM = l+m. If l = 0 then all ions are cations. Ifm = 0then all ions are anions. We label all the M ionic species and the solvent concentration by−l, . . . ,−1, 0, 1, . . . ,m, and assume that
z−lv−l
< · · · <z−1
v−1
<z0v0
= 0 <z1v1< · · · <
zmvm. (3.4)
For convenience, we denote J = −l, . . . ,−1, 0, 1, . . . ,m. With our notation, the equilib-rium conditions (2.5) then become
viv0
ln(v0c0)− ln(vici) = β (qiφ− µi) ∀i ∈ J. (3.5)
Note that by Theorem 3.1 ci = Bi(φ) for any i ∈ J with i 6= 0. We thus have by (1.1) that
c0 = B0(φ) = v−10
[
1−∑
i∈J,i 6=0
viBi(φ)
]
. (3.6)
Theorem 3.2. Let ci = Bi(φ) (i ∈ J) be the equilibrium concentrations defined by (3.5)and (3.6). Assume (3.4) holds true.(1) Monotonicity of concentrations with the largest and smallest valence-to-volume ratios.
We have c′−l(φ) > 0 and c′m(φ) < 0 for all φ ∈ R.(2) Stratification of concentration profiles. If i ∈ −l, . . . ,m− 1 then
(vici)1/vi < (vi+1ci+1)
1/vi+1 if and only if φ <µi+1/vi+1 − µi/viqi+1/vi+1 − qi/vi
.
If i ∈ −l + 1, . . . ,m then
(vici)1/vi > (vi−1ci−1)
1/vi−1 if and only if φ >µi−1/vi−1 − µi/viqi−1/vi−1 − qi/vi
.
9
(3) Asymptotic behavior of concentrations. We have
v−lc−l → 1 and ci → 0 (i = −l + 1, . . . ,m) as φ→ ∞, (3.7)
vmcm → 1 and ci → 0 (i = −l, . . . ,m− 1) as φ→ −∞. (3.8)
Proof. We first prove that
(vici)1/vi = e−β(µj/vj−µi/vi)eβ(qj/vj−qi/vi)φ(vjcj)
1/vj ∀i, j ∈ J. (3.9)
In fact, it follows immediately from (3.5) that
(v0c0)1/v0 = e−βµi/vieβ(qi/vi)φ(vici)
1/vi ∀i ∈ J. (3.10)
This also implies(vici)
1/vi = eβµi/vie−β(qi/vi)φ(v0c0)1/v0 ∀i ∈ J. (3.11)
It follows from (3.10) and (3.11) that (3.9) holds true if one of i and j is 0. Now let i, j ∈ Jwith i 6= j, i 6= 0, and j 6= 0. We have by (3.5) that
1
viln (vici) +
βqiviφ−
βµivi
=1
vjln (vjcj) +
βqjvjφ−
βµjvj
.
Consequently,(vici)
1/vi = e−β(µj/vj−µi/vi)eβ(qj/vj−qi/vi)φ(vjcj)1/vj .
This, together with (3.10) and (3.11), implies (3.9).We now prove parts (1)–(3).(1) Denote γ = v0c0 and ηi = eβµi for each i ∈ J. It follows from the equilibrium
conditions (3.5) that
ci =ηiviγvi/v0e−βqiφ ∀i ∈ J. (3.12)
This and the definition of c0 (cf. (1.1)) imply
∑
i∈J
ηiγvi/v0e−βqiφ = 1. (3.13)
Taking the derivative of both sides of (3.13) with respective to φ, and using our notationz0 = 0 and µ0 = 0, we obtain
γ′(φ) =βv0γ
∑
i∈J ηiqiγvi/v0e−βqiφ
∑
i∈J ηiviγvi/v0e−βqiφ
.
Fix k ∈ J. Taking the derivative of both sides of (3.12) with i = k with respect to φ andusing the above expression of γ′(φ), we get
c′k(φ) = βηkγvk/v0e−βqkφ
∑
i∈J ηivi (qi/vi − qk/vk) γvi/v0e−βqiφ
∑
i∈J ηiviγvi/v0e−βqiφ
. (3.14)
10
This, the assumption (3.4), and the definition qi = zie (i ∈ J) imply that c′m(φ) < 0 andc′−l(φ) > 0.
(2) This follows directly from (3.9).(3) Since 0 ≤ v−lc−l ≤ 1, we have again by (3.9) with j = −l and (3.4) that for any
i ∈ J with i > −l
(vici)1/vi ≤ e−β(µ−l/v−l−µi/vi)eβ(q−l/v−l−qi/vi)φ → 0 as φ→ ∞.
Since∑
i∈J vici = 1, we thus also have v−lc−l → 1 as φ → ∞. This proves (3.7). The proofof (3.8) is similar. Q.E.D.
Remark. By (3.14) and the Cauchy–Schwarz inequality, we obtain through a series ofcalculations that
∑
k∈J
qkc′k(φ) =
β[
(∑
i∈J qivici)2
−(∑
i∈J q2i ci) (∑
i∈J v2i ci)
]
∑
i∈J v2i ci
≤ 0.
Since qi/vi (i ∈ J) are not all the same, this is a strict inequality. Thus the functionV : R → R, defined in (3.2) with −V ′(φ) being the total ionic charge density, is a strictlyconvex function of φ; cf. Theorem 3.1.
3.2 Equilibrium charge density
We now study the properties of the ionic charge density V ′(φ) = −∑
i∈J qiBi(φ). By (3.2),the function V is given now by
V (φ) = −∑
i∈J
∫ φ
0
qiBi(ξ) dξ + V0. (3.15)
We distinguish three cases:(1) l = 0, m =M , and J = 0, 1, . . . ,M;(2) l =M , m = 0, and J = −M, . . . ,−1, 0;(3) l ≥ 1, m ≥ 1, l +m =M , and J = −l, . . . ,−1, 0, 1, . . . ,m.
We denote f(−∞) = lims→−∞ f(s) and f(∞) = lims→∞ f(s) if the limits exist.
Theorem 3.3. Assume (3.4) holds true.(1) If l = 0 and m =M , and J = 0, 1, . . . ,M, then V ′(−∞) = −qM/vM < 0, V ′(∞) =
0, and V ′(φ) < 0 for all φ ∈ R. Moreover, V (−∞) = ∞, and V (∞) = 0 with asuitably chosen constant V0 in (3.2).
(2) If l =M , m = 0, and J = −M, . . . ,−1, 0, then V ′(−∞) = 0, V ′(∞) = −q−M/v−M >0, and V ′(φ) > 0 for all φ ∈ R. Moreover, V (−∞) = 0 with a suitably chosen constantV0 in (3.2), and V (∞) = ∞.
(3) If l ≥ 1, m ≥ 1, l + m = M , and J = −l, . . . ,−1, 0, 1, . . . ,m, then V ′(−∞) =−qm/vm < 0, V ′(∞) = −q−l/v−l > 0, and −qm/vm < V ′(φ) < −q−l/v−l for all φ ∈ R.Moreover, minφ∈R V (φ) = 0 with a suitably chosen constant V0 in (3.2), V (−∞) = ∞,and V (∞) = ∞.
11
In Figure 2, we show the schematic behavior of the convex function V for the threedifferent cases.
V ′
φ
−qM/vM
0
V
φ0
Case 1. 0 = z0/v0 < z1/v1 < · · · < zM/vM .
V ′
φ
−q−M/v−M
0
V
φ0
Case 2. z−M/v−M < · · · < z−1/v−1 < z0/v0 = 0.
V ′
φ
−q−l/v−l
0
−qm/vm
φ0
V
φ0 φ0
Case 3. z−l/v−l < · · · < z−1/v−1 < z0/v0 = 0 < z1/v1 < · · · < zm/vm.
Figure 2: Graphs of V ′ = V ′(φ) (left) and V = V (φ) (right) for the three different cases.The value −V ′(φ) is the ionic charge density at a (shifted) electrostatic potential φ.
12
Proof of Theorem 3.3. (1) In this case, we have V ′(−∞) = −qM/vM < 0 by (3.8) andV ′(∞) = 0 by (3.7). Since V ′′ > 0 in R by Theorem 3.1, V ′ is a strictly increasing function.Therefore, V ′(−∞) < 0 and V ′(∞) = 0 imply that V ′(φ) < 0 for all φ ∈ R. Fix φ0 ∈ R
so that V ′(φ0) < 0. Then for any φ < φ0 we have V (φ) = V (φ0) + V ′(ξ)(φ − φ0), whereξ ∈ (φ, φ0). Notice that V
′(ξ) < V ′(φ0) < 0. So, V (φ) → ∞ as φ→ −∞, i.e., V (−∞) = ∞.Finally, setting j = 0 in (3.9), we obtain that
(viBi(φ))1/vi = (vici)
1/vi ≤
(
max1≤i≤M
eβµi/vi)
e−β(qi/vi)φ.
Therefore, the integral of∑M
i=1 qiBi(φ) from 0 to φ > 0 is a bounded function of φ. Thus,by the definition of V (cf. (3.2)), V (φ) is bounded below. Since V ′ < 0 on R, V decreases.Therefore, V (∞) exists. Choosing the constant
V0 =∑
i∈J
∫ ∞
0
qiBi(φ) dφ
in (3.15), we then have V (∞) = 0.(2) This part can be proved similarly.(3) The fact that V ′(−∞) = −qm/vm < 0 and V ′(∞) = −q−l/v−l > 0 can be proved
by the same argument used for proving part (1). Since V ′′ > 0 in R, V ′(φ) is a strictlyincreasing function. Therefore, −qm/vm < V ′(φ) < −ql/vl for all φ ∈ R. Again by the sameargument used in part (1), we have V (−∞) = ∞ and V (∞) = ∞. It is clear now that thereexists a unique φ0 ∈ R such that V ′(φ0) = 0 and V (φ0) = minφ∈R V (φ). Setting in (3.15),
V0 =∑
i∈J
∫ φ0
0
qiBi(φ) dφ,
we have minφ∈R V (φ) = 0. Q.E.D.
3.3 Modified Debye length
We now linearize V ′(φ) at φ = 0 to obtain a modified Debye length. We first definec∞i = Bi(0) for any i ∈ J. Note that
v0c∞0 = 1−
∑
i∈J,i 6=0
vic∞i . (3.16)
By (3.10) with φ = 0 and ηi = eβµi , we also have that
(v0c∞0 )vi/v0 = η−1
i vic∞i , i ∈ J. (3.17)
We denote again γ = v0c0 which depends on φ. We have for |φ| ≪ 1 that
γ = τ0 + τ1φ+O(
φ2)
, (3.18)
13
where τ0 and τ1 are two constants to be determined. Using our notation, (1.1) becomes
v0c0 = 1−∑
i∈J,i 6=0
vici. (3.19)
This and (3.16) imply that
τ0 = v0c0(0) = 1−∑
i∈J,i 6=0
vic∞i = v0c
∞0 . (3.20)
Substituting (3.18) into (3.12) for a fixed i ∈ J with i 6= 0, we obtain by (3.20) and (3.17)that
ci =ηivi
(
τ0 + τ1φ+O(
φ2))vi/v0 (1− βqiφ+O
(
φ2))
=ηiviτvi/v00
(
1 +τ1τ0φ+O
(
φ2)
)vi/v0(
1− βqiφ+O(
φ2))
= c∞i
(
1 +viτ1v0τ0
φ+O(
φ2)
)
(
1− βqiφ+O(
φ2))
= c∞i + c∞i
(
viτ1v0τ0
− βqi
)
φ+O(
φ2)
. (3.21)
It now follows from (3.18), (3.19), (3.21), and (3.20) that
τ0 + τ1φ+O(
φ2)
= 1−∑
i∈J,i 6=0
vic∞i −
∑
i∈J,i 6=0
vic∞i
(
viτ1v20c
∞0
− βqi
)
φ+O(
φ2)
.
Comparing the leading-order terms, we obtain exactly (3.20). Comparing the O(φ)-termsand noting that z0 = 0, we obtain
τ1 =βv20c
∞0
∑
i∈J qivic∞i
∑
i∈J v2i c
∞i
.
This and (3.21) then lead to
ci = c∞i + βc∞i
(
vi∑
j∈J qjvjc∞j
∑
j∈J v2j c
∞j
− qi
)
φ+O(
φ2)
∀i ∈ J with i 6= 0.
Consequently,
V ′(φ) = −∑
i∈J
qici = −∑
i∈J
qic∞i − β
[
(∑
i∈J qivic∞i
)2
∑
i∈J v2i c
∞i
−∑
i∈J
q2i c∞i
]
φ+O(
φ2)
.
In the small |φ| approximation, we then obtain by (3.3) and the shift φ = ψ− ψD/2 thelinearized PBE with ionic size effects
∇ · ε∇ψ − ελ−2D ψ = −
∑
i∈J
qic∞i −
1
2ελ−2
D ψD, (3.22)
14
where λD > 0 is defined by
λ−2D =
β
ε
[
∑
i∈J
q2i c∞i −
(∑
i∈J qivic∞i
)2
∑
i∈J v2i c
∞i
]
. (3.23)
The fact that the right-hand side of the above equation is nonnegative follows from theCauchy–Schwarz inequality
(
∑
i∈J
qivic∞i
)2
≤
(
∑
i∈J
q2i c∞i
)(
∑
i∈J
v2i c∞i
)
.
In the case that ε is a constant, ψD = 0 on ΓD, and∑
i∈J qic∞i = 0 which is the charge
neutrality, the linearized equation (3.22) simplifies to the familiar equation
∆ψ − λ−2D ψ = 0.
We call λD the size-modified Debye length. Recall that the Debye length λD > 0 in theclassical Debye–Huckel theory is defined by
λ−2D =
β
ε
∑
i∈J
q2i c∞i . (3.24)
It is clear that λD > λD. This indicates that finite ionic sizes lead to a longer Debyescreening length.
4 A Charged Spherical Molecule: Basic Properties
In this section, we consider a charged spherical molecule, centered at the origin and withradius R0 > 0, immersed in a solvent that occupies the region defined by R0 < r < R∞ in thespherical coordinates for some R∞ > R0, where R∞ can be finite or infinite. Correspondingto the general setting described in the previous section, we have in this case that
Ω = x ∈ R3 : R0 < |x| < R∞,
ΓN = x ∈ R3 : |x| = R0,
ΓD =
x ∈ R3 : |x| = R∞ if R∞ <∞,
∞ if R∞ = ∞.
We consider the boundary-value problem of the PBE, classical and size-modified, for theelectrostatic potential ψ (cf. (2.2) and (3.3)):
ε∆ψ = V ′(ψ) in Ω,
ε∂ψ
∂n= σ on ΓN,
ψ = 0 on ΓD.
(4.1)
15
If ψ = ψ(r) (r = |x|) is radially symmetric, then this system is the same as
ε
(
ψ′′ +2
rψ′
)
= V ′(ψ) if R0 < r < R∞,
ψ′(R0) = −σ
ε,
ψ(R∞) = 0.
(4.2)
Here, we assume both the dielectric coefficient ε > 0 and surface charge density σ areconstants. We can consider the inhomogeneous Dirichlet boundary condition ψ = ψ∞ onΓD for a constant ψ∞ by replacing ψ by ψ − ψ∞. We assume that V : R → R is smooth,V ′′ > 0 in R, and infs∈R V (s) = 0. We then have one and only one of the three casescovered in Theorem 3.3: V ′ > 0 in R, V ′ < 0 in R, or there exists a unique φ0 such thatV (s) > V (φ0) = 0 for any s 6= 0. In the last case, V ′ < 0 in (−∞, φ0), V
′(φ0) = 0, andV ′ > 0 in (φ0,∞). The term −V ′(ψ) is the ionic charge density that is related to M ionicspecies with valences and volumes satisfying (3.4).
For the classical PBE, we have V (φ) =∑
i∈J c∞i
(
e−βqiφ − 1)
, where c∞i (i ∈ J, i 6= 0) arebulk ionic concentrations. Under the assumption on the charge neutrality
∑
i∈J qic∞i = 0,
this V satisfies the properties in the third case in Theorem 3.3 with φ0 = 0. So our analysiscovers this case.
We first consider the case that R∞ <∞.
Theorem 4.1. Let ε > 0, σ ∈ R, and 0 < R0 < R∞ < ∞. Assume that V : R → R issmooth, V ′′ > 0 in R, and infs∈R V (s) = 0. Then the boundary-value problem (4.1) has aunique weak solution. Moreover it is radially symmetric and smooth on Ω.
Proof. We assume that σ 6= 0. The case that σ = 0 can be proved similarly. We denoteH = φ ∈ H1(R0, R∞) : φ(R∞) = 0 and define I : H → (−∞,∞] by
I[φ] = 4π
∫ R∞
R0
[ε
2φ′(r)2 + V (φ)
]
r2 dr − 4πR20σφ(R0) ∀φ ∈ H. (4.3)
If φ ∈ H, then we have with δ = ε/(4σ) > 0 that
|φ(R0)| =
∣
∣
∣
∣
∫ R∞
R0
φ′(r) dr
∣
∣
∣
∣
≤√
R∞ −R0
(∫ R∞
R0
|φ′(r) |2dr
)1/2
≤R∞ −R0
4δ+δ
∫ R∞
R0
|φ′(r) |2dr,
leading to
I[φ] ≥ πεR20
∫ R∞
R0
|φ′(r) |2dr −4πσ2R2
0(R∞ −R0)
ε∀φ ∈ H.
Notice that by Poincare’s inequality, I[φ] controls ‖φ‖2H1(R0,R∞), up to an additive constant.By standard arguments of direct methods in the calculus of variations and by the fact thatH1(R0, R∞) can be compactly embedded into C([R0, R∞]), we obtain the existence of aminimizer ψ ∈ H of the functional I : H → [0,∞]. Moreover, the minimum value of I
16
over H is finite. The uniqueness of this minimizer follows from the strict convexity of I.Clearly, the minimizer ψ = ψ(r) is a smooth solution to the corresponding Euler–Lagrangeequation, the first equation in (4.2). Moreover, it satisfies the natural boundary conditionψ′(R0) = −σ/ε. Since ψ ∈ H, ψ(R∞) = 0. If we also denote ψ(x) = ψ(|x|), then it is clearthat this function is a radially symmetric solution to (4.1), and is smooth on Ω.
To prove the uniqueness of solution to (4.1), we first note that the solution ψ minimizesthe functional
Q[φ] =
∫
Ω
[ε
2|∇φ|2 + V (φ)
]
dV −
∫
ΓN
σφ dS
over all φ ∈ H1(Ω) such that φ = 0 on ΓD. Indeed, for any η ∈ H1(Ω) with η = 0 on ΓD,we have by integration by parts, the convexity of V , and (4.1) that
Q[ψ + η]−Q[ψ] =
∫
Ω
[ε
2|∇η|2 + ε∇ψ · ∇η + V (ψ + η)− V (ψ)
]
dV −
∫
ΓN
ση dS
≥
∫
Ω
[ε∇ψ · ∇η + V ′(ψ)η] dV −
∫
ΓN
ση dS
=
∫
Ω
[−ε∆ψ + V ′(ψ)] η dV +
∫
ΓN
ε∂ψ
∂nη dS −
∫
ΓN
ση dS
= 0. (4.4)
Now if ψ is another solution to (4.1), which is not assumed to be radially symmetric, thenboth ψ and ψ are minimizers of the functional Q. Since Q is strictly convex, it has at mostone minimizer. Therefore, ψ = ψ. Q.E.D.
We now study the behavior of the equilibrium potential ψ = ψ(|x|) that is the solutionto the boundary-value problem (4.1). We denote by T (r) the total amount of ionic chargesin the region x ∈ Ω : R0 < |x| < r:
T (r) = −4π
∫ r
R0
V ′(ψ(s))s2 ds ∀r ∈ [R0, R∞]. (4.5)
Then T∞ := T (R∞) represents the total amount of ionic charges in Ω. If we integrateboth sides of Poisson’s equation in (4.1), apply integration by parts, and use the boundarycondition at r = R0, then we have
T∞ + 4πR20σ + 4πR2
∞εψ′(R∞) = 0. (4.6)
This is the charge neutrality of the system, as σ∞ := εψ′(R∞) can be viewed as the effectivesurface charge density on the part of the boundary ΓD = x ∈ R
3 : |x| = R∞ which isset due to the finiteness of the underlying domain. The surface charge density σ in theboundary condition and the effective surface charge density σ∞ = εψ′(R∞) are the keyparameters that determine the monotonicity of the electrostatic potential ψ.
Theorem 4.2. Let ε > 0, σ ∈ R, and 0 < R0 < R∞ < ∞. Assume that V : R → R
is smooth, V ′′ > 0 in R, and infs∈R V (s) = 0. Consider V ′ > 0 in R, or V ′ < 0 in R,or V ′(φ0) = 0 for a unique φ0 ∈ R. Let ψ = ψ(|x|) be the solution to the boundary-valueproblem (4.1).
17
(1) If σ > 0 and σ∞ > 0, then there exists a unique R+c ∈ (R0, R∞) such that ψ′ < 0 in
(R0, R+c ) and ψ′ > 0 in (R+
c , R∞). If σ < 0 and σ∞ < 0, then there exists a uniqueR−
c ∈ (R0, R∞) such that ψ′ > 0 in (R0, R−c ) and ψ
′ < 0 in (R−c , R∞).
(2) If σσ∞ ≤ 0 and at least one of σ and σ∞ is nonzero, then ψ′ > 0 in (R0, R∞) if σ < 0or σ∞ > 0, and ψ′ < 0 in (R0, R∞) if σ > 0 or σ∞ < 0.
(3) The case that σ = σ∞ = 0 can only occur when V ′(0) = 0. In this case ψ = 0identically in (R0, R∞).
rR∞R0
ψ
(a) σ > 0 and σ∞ < 0.
rR∞R0
ψ
(b) σ > 0 and σ∞ > 0.
rR∞R0
ψ
(c) σ < 0 and σ∞ > 0.
rR∞R0
ψ
(d) σ < 0 and σ∞ < 0.
Figure 3: The graph of electrostatic potential ψ.
The main results of Theorem 4.2 are illustrated in Figure 3. To prove the theorem, wefirst prove a lemma.
Lemma 4.1. Given the assumption as in Theorem 4.2 and A := r ∈ (R0, R∞) : ψ′(r) = 0.(1) If V ′ > 0 in R or V ′ < 0 in R, then either A = ∅ or A contains exactly one point.(2) If V ′(φ0) = 0 at some unique φ0 ∈ R, then either A = ∅, or A contains exactly one
point, or A = (R0, R∞) in which case we have additionally that φ0 = 0 and ψ = 0identically in (R0, R∞).
18
Proof. Suppose there exist r1, r2 ∈ R such that R0 ≤ r1 < r2 ≤ R∞ and ψ′(r1) = ψ′(r2) =0. Denote ω = x ∈ R
3 : r1 < |x| < r2. Then, by the same argument used in provingTheorem 4.1 (cf. (4.4)) and the strict convexity of V , the restriction of ψ onto ω is theunique minimizer among all φ ∈ H1(ω) of the functional
G[φ] =
∫
ω
[ε
2|∇φ|2 + V (φ)
]
dV. (4.7)
(1) Suppose V ′ > 0. Since infs∈R V (s) = 0, we have V (−∞) = 0. Thus, with φk = −k(k = 1, 2, . . . ), we have 0 ≤ G[ψ] ≤ G[φk] → 0 as k → ∞. Thus G[ψ] = 0. But this isimpossible: If the integral of |∇ψ|2 vanishes, then ψ is a constant. But V (s) > 0 for anys ∈ R. Thus the integral of V (ψ) over ω is positive, a contradiction to G[ψ] = 0. For thiscase and similarly the case that V ′ < 0, the set A can then contain at most one point.
(2) In this case, V (φ0) = mins∈R V (s) = 0. Thus, the constant function φ = φ0 alsominimizes G over H1(ω). By the uniqueness of minimizer which follows from the strictconvexity of G, ψ = φ0 in [r1, r2] and [r1, r2] ⊆ A. Let s = inf A ≥ R0 and t = supA ≤ R∞.At any point in (s, t), ψ = φ0 and ψ′ vanishes. Moreover, ψ′ 6= 0 in (R0, s) if s > R0 and in(t, R∞) if t < R∞. We prove now s = R0 and t = R∞. Assume s > R0 and ψ
′ > 0 in (R0, s).Differentiating both sides of ∆ψ = V ′(ψ)/ε in R0 < r < t, we obtain
ψ′′′(r) +2
rψ′′(r) =
[
1
εV ′′(ψ(r)) +
2
r2
]
ψ′(r) ∀r ∈ (R0, t). (4.8)
Hence, setting u = ψ′, we have Lu := ∆u + c(r)u = 0 for R0 < r < t, where c(r) :=− [(1/ε)V ′′(ψ(r)) + 2/r2] < 0 and c(r) is bounded in (R0, s). Clearly u(s) = 0 ≤ u(r) forall r ∈ (R0, t). Hence u attains its minimum with the minimum value 0 at any interiorpoint x with |x| = s in x ∈ R
3 : R0 < |x| < t. By the Strong Maximum Principle (cf.Theorem 3.5 of [13]), u must be a constant in R0 < r < t. This contradicts the fact thatu > 0 in R0 < r < s and u = 0 in s < r < t. Hence ψ′ > 0 in (R0, s) is impossible.Similarly, ψ′ < 0 in (R0, s) is impossible. Therefore s = R0. Similarly, we have t = R∞.Finally A = (s, t) = (R0, R∞), and since ψ(R∞) = 0, ψ = φ0 = 0 in (R0, R∞). Q.E.D.
Proof of Theorem 4.2. (1) Assume σ > 0 and σ∞ > 0. Then ψ′(R0) < 0 and ψ′(R∞) > 0.Hence by Lemma 4.1 there exists a unique R+
c ∈ (R0, R∞) such that ψ′(R+c ) = 0. Clearly,
ψ′ < 0 in (R0, R+c ) and ψ
′ > 0 in (R+c , R∞). The case that σ < 0 and σ∞ < 0 can be proved
similarly.(2) Assume σ > 0 and σ∞ ≤ 0, i.e., ψ′(R0) < 0 and ψ′(R∞) ≤ 0. The other cases can
be treated similarly. It suffices to show that ψ′ does not vanish in (R0, R∞). Assume thereexisted r0 ∈ (R0, R∞) such that ψ′(r0) = 0. By Lemma 4.1, ψ′ does not vanish at otherpoints. Thus there are two cases. Case 1: ψ′ < 0 in (R0, r0) and ψ
′ < 0 in (r0, R∞). Case2: ψ′ < 0 in (R0, r0) and ψ
′ > 0 in (r0, R∞). We show that each case is impossible.Consider Case 1. If V ′ > 0 in R, then ∆ψ > 0 on x ∈ R
3 : r0 < |x| < R∞. Letx0 ∈ R
3 be such that |x0| = r0. Since ψ(|x0|) > ψ(|x|) for any x with |x| ∈ (r0, R∞).Lemma 3.4 in [13] then implies that ψ′(r0) 6= 0, a contradiction. If V ′ < 0, we obtainthe same contradiction ψ′(r0) 6= 0 by considering (R0, r0). Finally, assume V ′(φ0) = 0. If
19
ψ(r0) 6= φ0, then since ψ′(r0) = 0, ψ′′(r0) = V ′(φ(r0))/ε 6= 0. These imply that ψ′ changesits sign at r0 which is a contradiction in Case 1. Thus Case 1 is impossible.
Consider Case 2. In this case, we must have ψ′(R∞) = 0, since ψ′(R∞) > 0 would leadto ψ′(r1) = 0 for some r1 ∈ (r0, R∞) which is impossible. Moreover, ψ reaches its minimumat r0 and in particular ψ′′(r0) > 0. If V ′ < 0 then ∆ψ < 0 in Ω. The Strong MaximumPrinciple then implies that ψ must be a constant for r ∈ (R0, R∞). This is impossible asψ′(R0) < 0. If V ′ > 0, then ∆ψ > 0 and ψ(R∞) = 0 > ψ(r) for r ∈ (r0, R∞). By Lemma 3.4in [13], we must have ψ′(R∞) > 0, leading to a contradiction. Assume V ′(φ0) = 0 atsome unique φ0. We cannot have φ0 > ψ(r0) as this would imply that V ′(ψ(r0)) < 0 andhence ψ′′(r0) < 0 which contradicts ψ′′(r0) > 0. We cannot have φ0 ≤ ψ(r0) neither, sinceotherwise we would have that V ′(ψ) ≥ 0 and hence ∆ψ ≥ 0 on the region r0 < r < R∞.But ψ(R∞) = 0 > ψ(r) for any r ∈ (r0, R∞). Lemma 3.4 in [13] then implies ψ′(R∞) > 0,a contradiction. Thus Case 2 is also impossible.
(3) We have now ψ′(R0) = 0 and ψ′(R∞) = 0. As in the proof of Lemma 4.1 with r1 = R0
and r2 = R∞, ψ is a constant in [R0, R∞]. Since ψ(R∞) = 0, ψ = 0 identically in [R0, R∞].Thus V ′(0) = V ′(ψ) = ε∆ψ = 0. Hence φ0 = 0. Q.E.D.
We recall that for any r ∈ (R0, R∞), T (r) is the total amount of ionic charges in theregion x ∈ Ω : R0 < |x| < r; cf. (4.5). In the following corollary, we consider a specialand useful case but deliver more results:
Corollary 4.1. Let ε > 0, σ ∈ R, and 0 < R0 < R∞ < ∞. Assume that V : R → R issmooth, V ′′ > 0 in R, and V (s) > V (0) = 0 for any s ∈ R with s 6= 0. Let ψ = ψ(|x|) bethe solution to the boundary-value problem (4.1).(1) If σ = 0, then ψ = 0 and T = 0 identically.(2) If σ > 0, then ψ′ < 0, ψ′′ > 0, and T ′ < 0 in (R0, R∞).(3) If σ < 0, then ψ′ > 0, ψ′′ < 0, and T ′ > 0 in (R0, R∞).
Proof. (1) It is easy to verify that in this case ψ = 0 is the unique solution to (4.1).Moreover, T ′(r) = −V ′(ψ(r)) = −V ′(0) = 0 and T (R0) = 0. Hence T = 0 in (R0, R∞).
(2) Assume σ > 0. We then have σ∞ = εψ′(R∞) ≤ 0. Otherwise, σ∞ > 0. Thenby Theorem 4.2 there exists Rc ∈ (R0, R∞) such that ψ′ < 0 in (R0, Rc) and ψ′ > 0 in(Rc, R∞). We have ψ(Rc) < min(ψ(R0), ψ(R∞)) ≤ 0. Choose τ ∈ R such that ψ(Rc) < τ <min(ψ(R0), ψ(R∞)).Define ψ by ψ = ψ if ψ ≥ τ and ψ = τ if ψ < τ. Then I[ψ] < I[ψ], whereI[·] is defined in (4.3). This is impossible as ψ minimizes I[·] over H = φ ∈ H1(R0, R∞) :φ(R∞) = 0 that includes ψ. Now, since σ > 0 and σ∞ ≤ 0, we have ψ′ < 0 by Theorem 4.2.Since ψ(R∞) = 0, ψ > 0 in (R0, R∞). Hence V ′(ψ) > 0 in (R0, R∞). Consequently,
ψ′′(r) = −2
rψ′(r) +
1
εV ′(ψ(r)) > 0 in (R0, R∞).
Since V ′(ψ) > 0 in (R0, R∞), we also have T ′(r) < 0 for all r ∈ (R0, R∞).(3) This can be proved similarly. Q.E.D.
We now consider the case R∞ = ∞.
Theorem 4.3. Let ε > 0, σ ∈ R, and R∞ = ∞.
20
(1) Suppose V : R → R is smooth and V ′(0) 6= 0. Then there is no solution to theboundary-value problem (4.1).
(2) Suppose V : R → R is smooth, V ′′ > 0 on R, V (−∞) = ∞ and V (∞) = ∞, andV (0) = infs∈R V (s) = 0. Then there exists a unique solution to the boundary-valueproblem (4.1). Moreover, this solution ψ = ψ(|x|) is radially symmetric and smoothon Ω, ψ′ < 0 and ψ′′ > 0 in r = |x| if σ > 0, ψ′ > 0 and ψ′′ < 0 in r = |x| if σ > 0,and ψ = 0 for all r if σ = 0.
We remark that in general all charges in an ionic system are in a finite region. Thereforeit is natural to impose the far-field condition ψ(R∞) = 0. The properties on V assumedin part (2) of the theorem are exactly those covered in the third case in Theorem 3.3 butwith the minimum of V attained exactly at 0. The theorem states that the solution to (4.1)exists only when V ′(0) = 0. This is precisely the condition of charge neutrality in the bulk.Therefore, for the three cases of V stated in Theorem 3.3, only the third case with V ′(0) = 0permits a solution to (4.1).
Proof of Theorem 4.3. (1) Assume V ′(0) > 0. (The case that V ′(0) < 0 is similar.)Assume ψ is a solution to (4.1), which is not assumed to be radially symmetric. Clearly,ψ is smooth on Ω. Since ψ(∞) = 0 and V ′(0) > 0, there exist A ≥ R0 and α > 0 suchthat ε∆ψ = V ′(ψ) ≥ α for all x ∈ Ω with |x| ≥ A. In the spherical coordinates (r, θ, ϕ)(r ≥ 0, 0 ≤ θ ≤ π, 0 ≤ ϕ < 2π), ψ = ψ(r, θ, ϕ), and
∆ψ =1
r2∂
∂r
(
r2∂ψ
∂r
)
+1
r2 sin θ
∂
∂θ
(
sin θ∂ψ
∂θ
)
+1
r2 sin2 θ
∂2ψ
∂ϕ2≥ α ∀r ≥ A. (4.9)
Define
u(r) =1
4π
∫ 2π
0
∫ π
0
sin θψ(r, θ, ϕ) dθdϕ ∀r > R0.
Clearly, u(∞) = 0, since ψ → 0 as r → ∞. Multiplying both sides of the inequality in (4.9)by sin θ/(4π), and then integrating the resulting sides over θ ∈ [0, π] and ϕ ∈ [0, 2π), weobtain (1/r2) (r2u′(r))
′≥ α for all r ≥ A. This leads to
u′(r) ≥1
3αr +
[
A2u′(A)−1
3αA3
]
1
r2∀r ≥ A.
With one more time of integration, we get
u(r) ≥ u(A) +1
6α(
r2 − A2)
−
[
A2u′(A)−1
3αA3
](
1
r−
1
A
)
∀r ≥ A.
This contradicts u(∞) = 0. Part (1) is proved.(2) We first show the uniqueness. Suppose ψ1 and ψ2 are both solutions to (4.1). Clearly
they are smooth on Ω = x ∈ R3 : |x| ≥ R0. Let η = ψ1 − ψ2. This function attains its
maximum on Ω. Suppose P := maxx∈Ω η > 0. If there exists x0 ∈ Ω = x ∈ R3 : |x| > R0
such that η(x0) = P > 0, then there exists an open ball B(x0) ⊂ Ω centered at x0 such that
21
η > 0 on B(x0). Since ψ1 and ψ2 are bounded on Ω, ψ1(x), ψ2(x) ∈ K for some compactsubset of R for all x ∈ Ω. Setting γ0 = mins∈K V
′′(s) > 0, we obtain
ε∆η = V ′(ψ1)− V ′(ψ2) ≥ γ0η > 0 in B(x0). (4.10)
Since η(x0) = maxx∈B(x0) η > 0, the Strong Maximum Principle (cf. Theorem 3.5 of [13])then implies that η is a constant in B(x0). Consequently ∆η = 0 in B(x0), and (4.10) impliesη = 0 in B(x0), contradicting to η(x0) > 0. Thus there exists y0 ∈ ∂Ω with |y0| = R0 suchthat η(y0) = P > η(x) for all x ∈ Ω. But this and (4.10) imply ∂nη(y0) = 0 by Lemma 3.4in [13], contradicting ∂nη = 0 on ΓD = x ∈ R
3 : |x| = R0. Hence η ≤ 0 and ψ1 ≤ ψ2 in Ω.Similarly ψ2 ≤ ψ1 in Ω. Thus ψ1 = ψ2 in Ω.
We now prove the existence of solution to (4.1) that satisfies all the properties. If σ = 0then ψ = 0 in Ω is the desired solution. Assume σ > 0. (The case σ < 0 is similar.)Choose a sequence Rk ∈ R (k = 1, 2, . . . ) such that R0 < R1 < R2 < · · · < Rk < · · · andRk → ∞ as k → ∞. For each integer k ≥ 1, let ψk = ψk(|x)|) be the unique solution to theboundary-value problem
ε∆ψk = V ′(ψk) in Ωk = x ∈ R3 : R0 < |x| < Rk,
ε∂ψk∂n
= σ on ΓkN = x ∈ R3 : |x| = R0,
ψk = 0 on ΓkD = x ∈ R:|x| = Rk.
(4.11)
By Corollary 4.1, ψk > 0, ψ′k < 0, and ψ′′
k > 0 in (R0, Rk).Using the spherical coordinates, we have by the first equation in (4.11) that
ψ′′k(r) +
2
rψ′k(r) = V ′(ψk(r)) in (R0, Rk). (4.12)
Multiply both sides of this equation by ψ′k(r) and integrate the resulting sides over [R0, Rk].
Noting that ψ′kψ
′′k = (1/2)(ψ′2
k )′ and ψ′
kV′(ψk) = (d/dr)V (ψk), we then obtain by ψ′
k(R0) =−σ/ε, ψk(Rk) = 0 and V (0) = 0 that
1
2
[
(ψ′k(Rk))
2 −(σ
ε
)2]
+
∫ Rk
R0
2
r(ψ′
k(r))2dr = −V (ψk(R0)).
Thus, since V (s) > 0 for any s > 0, we have 0 ≤ V (ψk(r)) ≤ σ2/(2ε2) for all r ∈ [R0, Rk]and k ≥ 1. Since V is positive and strictly increasing in (0,∞) and V (∞) = ∞, and sinceψk(R0) ≥ ψk(r) for all r ∈ [R0, Rk], we have by setting φc > 0 with V (φc) = σ2/(2ε2) that0 ≤ ψk(r) ≤ φc for all r ∈ [R0, Rk] and all k ≥ 1. Moreover, since ψ′′
k > 0 in (R0, Rk),−σ/ε = ψ′
k(R0) ≤ ψ′k(r) ≤ ψ′
k(R∞) ≤ 0 for all r ∈ [R0, Rk] and all k ≥ 1. Define ψk = 0 in(Rk,∞) (k = 1, 2, . . . ). Then the sequence ψk
∞k=1 is bounded in W 1,∞(R0,∞).
There now exists a subsequence ψk,1∞k=1 of ψk
∞k=1 such that ψk,1 → ψ in H1(R0, R1)
as k → ∞ for some ψ ∈ H1(R0, R1). There exists a subsequence ψk,2∞k=1 of ψk,1
∞k=1 such
that ψk,2 → φ in H1(R0, R2) as k → ∞ for some φ ∈ H1(R0, R2) with φ = ψ on (R0, R1).Wedefine ψ = φ in (R1, R2) so that ψ ∈ H1(R0, R2). Repeating, we obtain ψ ∈ C([R0,∞)) with
22
ψ ∈ H1(R0, R) for any R > R0 and subsequences ψk,j∞k=1 of ψk
∞k=1 (j = 1, 2, . . . ), where
for each j ≥ 1, the sequence ψk,j+1∞k=1 is a subsequence of ψk,j
∞k=1.Moreover, ψk,j → ψ in
H1(R0, Rj) as k → ∞. Now ψk,k∞k=1 is a subsequence of ψk
∞k=1, and ψk,k
∞k=1 converges
to ψ in H1(R0, R), and hence also in C([R0, R]), for any R > R0.The limit ψ = ψ(|x|) is clearly radially symmetric. It is a (weak and hence strong)
solution to the first two equations in (4.1). We show now that limr→∞ ψ(r) = 0. For eachk ≥ 1, define
ηk(r) =
ψk(R0)
(
1
R0
−1
Rk
)−1(1
r−
1
Rk
)
if R0 ≤ r ≤ Rk,
0 if r > Rk.
One easily verifies that ηk(|x|) satisfies ε∆ηk = 0 in Ωk (cf. (4.11) for the definition of Ωk)and ηk = ψk on the boundary of Ωk. Since ε∆(ψk − ηk) ≥ 0 in Ωk, we thus have by theMaximum Principle that ψk ≤ ηk in Ωk and hence in Ω = x ∈ R
3 : |x| > R0. Denoteηk,k
∞k=1 the subsequence of ηk
∞k=1 that corresponds to ψk,k
∞k=1. Then we have
0 ≤ lim supr→∞
ψ(r) = lim supr→∞
limk→∞
ψk,k(r) ≤ lim supr→∞
limk→∞
ηk,k(r) = lim supr→∞
R0ψ(R0)
r= 0,
implying limr→∞ ψ(r) = 0.Since ψ′
k ≤ 0 on (R0, Rk) for all k ≥ 1 ψ′ ≤ 0 on (R0,∞). Let u(r) = ψ′(r). Then u(|x|)satisfies ∆u−c(r)u = 0 on x ∈ R
3 : |x| > R0, where c(r) = − [(1/ε)V ′′(ψ(r)) + 2/r2] < 0,cf. (4.8). Since u ≤ 0 in (R0,∞), the Strong Maximum Principle implies that u 6= 0in (R0,∞). Hence u = ψ′ < 0 in (R0,∞). Since ψ ≥ 0 and V (ψ) ≥ 0, we also haveψ′′(r) = −(2/r)ψ′ + V ′(ψ) > 0 for all r > R0. Q.E.D.
5 Numerical Results
We now present numerical results to illustrate our analysis. We consider minimizing nu-merically the free-energy functional (2.6) with (2.7) and (2.8). We choose
Ω = x ∈ R3 : R0 < |x| < R∞, ΓN = x ∈ R
3 : |x| = R0, ΓD = x ∈ R3 : |x| = R∞,
with R0, R∞ ∈ R such that R0 < R∞. We assume the dielectric coefficient ε and surfacecharge density σ are both constants. We also set ψD = 0 on ΓD. Note that our system isradially symmetric. We use the constrained optimization method developed in our previouswork [38] to solve numerically our optimization problem.
We set R0 = 10 A and R∞ = 80 A.We distribute uniformly some negative charges on the
sphere r = R0 with the surface charge density σ = −150/(4π2R20) e/A
2. We consider three
species of counterions K+, Ca2+, Al3+, and one species of coions Cl− in the solution thatoccupies Ω. So M = 4. We label these ionic species by 1, 2, 3, and 4, respectively. We labelthe solvent (water) by 0. The corresponding valences and ionic volumes are (z1, z2, z3, z4) =
(+1,+2,+3,−1) and (v1, v2, v3, v4) = (5.513, 4.753, 4.13, 6.373) A3, respectively. The volume
23
of a solvent (water) molecule is v0 = 2.753 A3. For these real ionic systems, it happens
that α4 < α0 = 0 < α1 < α2 < α3, where αk = zk/vk (0 ≤ k ≤ 4) with z0 = 0. We usea3 to approximate the volume of an ion of diameter a. The temperature is T = 300K.The dielectric coefficient is absorbed into the Bjerrum length lB = e2/(4πεkBT ). We setthe Bjerrum length to be lB = 7A. We consider two cases of the numbers of ions indifferent species. Case I: (N1, N2, N3, N4) = (60, 30, 20, 50). By simple calculations, we havethat in this case the effective surface charge density σ∞ > 0. Case II: (N1, N2, N3, N4) =(50, 50, 50, 50). In this case σ∞ < 0.
In Figure 4, we plot our computational results of the electrostatic potential ψ and theequilibrium ionic concentrations. For Case I, the potential ψ is monotonically increasing.This is exactly predicted by Theorem 4.2, since in this case, σ < 0 and σ∞ > 0. Noticethat the concentration of +3, which has the largest ratio of valence-to-volume among allthe species, monotonically deceases, while that of −1 monotonically increases. These arepredicted by Theorem 3.2. For Case II, σ < 0 and σ∞ > 0, and the behavior of thepotential ψ is again as predicted by Theorem 4.2. We observe the stratification of thecounterion concentrations in both cases.
6 Conclusions
We have analyzed a PB-like mean-filed model for electrostatic interactions of multivalentions with finite ionic sizes in an ionic solution near a charged surface. Our work continuesthat presented in [22] but gives more details on how equilibrium ionic concentrations dependon the electrostatic potential and how the important parameters, the valence-to-volumeratios of ions, determine the properties of ionic concentrations near the charged surface.
Here we discuss some of our main results. First, our analysis shows that for a strongpotential, the layering structure of counterion concentrations does depend on the valence-to-volume ratios. In general, the comparison of concentrations for different ionic species canbe subtle. Instead of comparing ci and cj, one may need to compare (vici)
1/vi and (vjcj)1/vj .
The difference of these quantities may only be observed for certain range of the potential.See part (2) of Theorem 3.2.
Second, we characterize the ionic charge density through the function V defined in (3.2);cf. also (3.15). We recall for the classical PB theory that the corresponding function V isdefined by
V ′(φ) = −∑
i∈J,i 6=0
qic∞i e
−βqiφ,
where c∞i (i ∈ J, i 6= 0) are positive constants. An important difference here is that, whenthe ionic size effects are included, the function V grows linearly at the −∞ and ∞, ratherthan exponentially as in the classical PB theory.
Third, when the ionic size effects are included, the Debye length is then modified. InTable 1 and Table 2, we compare the size-modified Debye length λD defined in (3.23) withthe classical Debye length λD defined in (3.24) for a few ionic systems. The differencebetween these two tables is that the diameters of ions listed in Table 1 do not, while those
24
10 20 30 40 50 60 70 80
−0.6
−0.5
−0.4
−0.3
−0.2
−0.1
0
0.1
Radial distance (A)
Electro
staticpotential(V
olt)
10 11 12 13 14 15 16 17 18 19 200
5
10
15
20
25
Radial distance (A)Concentrations(M
)
65 70 75 800
0.02
0.04
0.06
0.08
+1+2+3−1
10 20 30 40 50 60 70 80−0.5
−0.4
−0.3
−0.2
−0.1
0
0.1
Radial distance (A)
Electro
staticpotential(V
olt)
10 11 12 13 14 15 16 17 18 19 200
5
10
15
20
25
Radial distance (A)
Concentrations(M
)
65 70 75 800
0.02
0.04
0.06
0.08
+1+2+3−1
Figure 4: The electrostatic potential ψ (left) and the equilibrium ionic concentrations(right). Top. Case I: (N1, N2, N3, N4) = (60, 30, 20, 50), σ < 0, and σ∞ > 0. Bottom.Case II: (N1, N2, N3, N4) = (50, 50, 50, 50), σ < 0, and σ∞ < 0. The concentrations profilesare marked by the corresponding ionic valences.
25
in Table 2, include the size of a hydration shell. It is clear that the modification of the Debyelength due to the inclusion of the ionic size effect is not significant. This may suggest thata linearized system may not describe well the size effect. Clearly, further tests are neededon real ionic solutions and biomolecular systems.
Table 1: A comparison of the Debye length λD and the ionic size-modified Debye lengthλD for a few systems of ionic solutions. The diameter of an ion does not include that of awater molecule (hydration shell). Concentrations mean bulk concentrations.
Ionic systems Diameters of ions (A) Concentrations (M) λD (A) λD(A)Cl−, Na+ 3.62, 2.04 0.1, 0.1 9.715 9.726Cl−, K+ 3.62, 2.76 0.1, 0.1 9.715 9.720Cl−, Ca2+ 3.62, 2.00 0.2, 0.1 5.609 5.618Cl−, Na+, K+ 3.62, 2.04, 2.76 0.2, 0.1, 0.1 6.870 6.880Cl−, Na+, Ca2+ 3.62, 2.04, 2.00 0.3, 0.1, 0.1 4.858 4.870Cl−, Na+, K+, Ca2+ 3.62, 2.04, 2.76, 2.00 0.4, 0.1, 0.1, 0.1 4.345 4.358Cl−, K+, Ca2+, Al3+ 3.62, 2.76, 2.00, 1.35 0.6, 0.1, 0.1, 0.1 3.072 3.085
Table 2: A comparison of the Debye length λD and the ionic size-modified Debye lengthλD for a few systems of ionic solutions. The diameter of an ion includes that of a watermolecule (hydration shell). Concentrations mean bulk concentrations.
Ionic systems Diameters of ions (A) Concentrations (M) λD (A) λD(A)Cl−, Na+ 6.37, 4.79 0.1, 0.1 9.715 9.847Cl−, K+ 6.37, 5.51 0.1, 0.1 9.715 9.763Cl−, Ca2+ 6.37, 4.75 0.2, 0.1 5.609 5.701Cl−, Na+, K+ 6.37, 4.79, 5.51 0.2, 0.1, 0.1 6.870 6.970Cl−, Na+, Ca2+ 6.37, 4.79, 4.75 0.3, 0.1, 0.1 4.858 4.974Cl−, Na+, K+, Ca2+ 6.37, 4.79, 5.51, 4.75 0.4, 0.1, 0.1, 0.1 4.345 4.449Cl−, K+, Ca2+, Al3+ 6.37, 5.51, 4.75, 4.10 0.6, 0.1, 0.1, 0.1 3.072 3.172
Finally, for a spherical charged molecule immersed in an ionic solution occupying a finiteregion, the qualitative behavior of the electrostatic potential—either it is monotonic or itchanges the monotonicity only once in the entire range of the radial variable—is completelydetermined by the sign of two surface charge densities. One is the prescribed surface chargedensity and the other is the effective surface charge density. If the region of ionic solution isthe entire complement of the sphere in R
3, then the ionic charge neutrality is necessary forthe existence and uniqueness of the electrostatic potential that vanishes at infinity, governedby the PBE (classical or size-modified). Such a potential is always monotonic.
26
We conclude with two remarks on the mathematical model we have studied here. First,for the case of a uniform size for all ions and solvent molecules, the free-energy functional(2.1) can be derived from a lattice-gas model; cf. [20]. For a general and more interestingcase where the sizes are different, such a derivation seems not available. It is thereforeinteresting to give a rigorous derivation of such a free-energy functional using the notionof equilibrium statistical mechanics. Second, we have assumed mainly that the region Ω ofionic solution is bounded. The case that Ω has an infinite volume is a tricky situation todefine the functional F [c]; cf. (2.1) and (2.6). Usually, the concentration ci is taken to bethe bulk value c∞i (> 0) at infinity (or far away from the charged surface). Such a value canbe experimentally determined. If Ω is unbounded with an infinite volume, then the integralof ci over Ω will be just infinite if ci = c∞i at infinity. The PBE, however, can be defined onan unbounded domain as the electrostatic potential ψ can exist in the entire space.
Acknowledgment.
This work was supported by the National Science Foundation (NSF) through grant DMS-0811259 (B.L.), the NSF Center for Theoretical Biological Physics (CTBP) through grantPHY-0822283 (B.L.), the National Institutes of Health through the grant R01GM096188(B.L.), the Natural Science Foundation of China through grant NSFC-11101276 and NSFC-91130012 (P.L. and Z.X.), and the Alexander von Humboldt Foundation (Z.X.). The authorsthank the referee for helpful comments.
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