Minimization of Electrostatic Free Energy and the Poisson-Boltzmann Equation for Molecular Solvation with Implicit Solvent

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1 Minimization of Electrostatic Free Energy and the Poisson-Boltzmann Equation for Molecular Solvation with Implicit Solvent Bo Li April 15, 009 Abstract In an implicit-solvent description of the solvation of charged molecules solutes, the electrostatic free energy is a functional of concentrations of ions in the solvent. The charge density is determined by such concentrations together with the point charges of the solute atoms, and the electrostatic potential is determined by the Poisson equation with a variable dielectric coefficient. Such a free-energy functional is considered in this work for both the case of point ions and that of ions with a uniform finite size. It is proved for each case that there exists a unique set of equilibrium concentrations that minimize the free energy and that are given by the corresponding Boltzmann distributions through the equilibrium electrostatic potential. Such distributions are found to depend on the boundary data for the Poisson equation. Pointwise upper and lower bounds are obtained for the free-energy minimizing concentrations. Proofs are also given for the existence and uniqueness of the boundary-value problem of the resulting Poisson-Boltzmann equation that determines the equilibrium electrostatic potential. Finally, the equivalence of two different forms of such a boundary-value problem is proved. 000 Mathematics Subject Class: 35J, 35Q, 49S, 8D, 9C. Key words and phrases: implicit solvent, electrostatic free energy, ionic concentrations, electrostatic potentials, the Poisson-Boltzmann equation, variational methods, nonlinear elliptic interface problems. Department of Mathematics and the NSF Center for Theoretical Biological Physics, University of California, San Diego, 9500 Gilman Drive, Mail code: 011. La Jolla, CA , USA. bli@math.ucsd.edu. 1

2 1 Introduction It has long been realized that the electrostatic potential of a charged molecular system extremizes an electrostatic free-energy functional [3, 6, 1, 13, 15, 18, 0, 6, 8, 9]. In a simple setting, this functional is given by F[c 1,...,c M ;ψ] = { ε 8π ψ + ρψ + β 1 [ c j lnλ 3 c j 1 ] } µ j c j dx, where c 1,...,c M are ionic concentrations, ψ is an electrostatic potential, ε is the dielectric constant, ρ is the charge density defined to be a linear combination of the ionic concentrations, β is the inverse thermal energy, Λ is the thermal de Broglie wavelength, and µ j is the chemical potential of the jth ionic species. Throughout, we use the electrostatics CGS units. We also use log x to denote the natural logarithm of x > 0. Extremizing this functional with respect to the concentrations and the potential lead to the Boltzmann distribution of concentrations and the Poisson equation for the equilibrium potential, respectively [3,6,1,13,15,6,8]. Notice, however, that this free-energy functional is concave with respect to the electrostatic potential. Therefore, the extremizing concentrations and potential do not minimize this free-energy functional, rather they form an unstable saddle point of the system [1,6,1,13]. This flaw of theory is removed in the free-energy minimization approach that was proposed in [1,0]. The key point in this new approach is that the electrostatic free-energy functional depends solely on the ionic concentrations and the electrostatic potential is determined by such concentrations through the Poisson equation. In the recent article [5], this free-energy minimization approach was revisited and applied to the implicit-solvent or continuum-solvent description of solvation. The present work is a mathematical study of the free-energy minimization approach to the electrostatics applied to the solvation of molecules with an implicit-solvent. Such application introduces additional mathematical complications due to the presence of point charges in solutes and the dielectric boundaries. We consider both the case of point ions ions modeled as points without volumes and that of ions with a uniform finite size. The finite-size effect of ions is known to be important in continuum modeling of electrostatics in molecular systems. Our analysis shows particularly that the free-energy minimizing ionic concentrations are uniformly bounded from above and away from zero at each spatial point. This uniform boundeness, which is proved by somewhat tedious constructions, is a consequence of the property that the free-energy minimizing concentrations have a large entropy. We do not consider the more general case of ions with different sizes for which there seems no explicit Boltzmann distributions. Consider now the solvation of charged molecules with an implicit solvent [7]. We divide the entire region of the solvation system into the region of solute molecules m R 3 that is possibly multiply connected, the region of solvent such as salted water R 3, and the solute-solvent interface Γ = m ; cf. Figure 1. This interface Γ serves as the dielectric boundary. Assume the solutes consist of N atoms with the ith

3 one located at x i and carrying a charge Q i. Assume also there are M ionic species in the solvent with q j = ez j the buck charge of the jth ionic species, where e is the elementary charge and z j the valence of jth ionic species. Denote by c j = c j x the local concentration at x of the jth ionic species. Following the common assumption that the mobile ions in the solvent can not penetrate the dielectric boundary Γ, we define c j x = 0 for all x m and 1 j M. x 1. n Γ s x. N x. i m n Figure 1. The geometry of a solvation system with an implicit solvent. We consider two mean-field approximations of the electrostatic free energy of the solvation system as functionals of the local ionic concentrations c = c 1,...,c M in the solvent region. In the first one, point ions are assumed, and the related electrostatic free-energy functional is given by [5,1,19,0,6] F 0 [c] = 1 Q i ψ ψ vac x i + 1 M q j c j ψdx M + β 1 c j [ loga 3 c j 1 ] dx µ j c j dx. 1.1 In the second approximation, all ions are assumed to have a uniform linear size, and the related free-energy functional is given by [3,0] F a [c] = 1 Q i ψ ψ vac x i + 1 M q j c j ψdx M + β 1 j=0 c j [ log a 3 c j 1 ] dx µ j c j dx, 1. where the summation in the β 1 term starts from j = 0 and ] c 0 x = a [1 3 a 3 c j x x

4 In 1.1 and 1., ψ is the electrostatic potential of the solvation system, ψ vac x = Q i ε m x x i 1.4 defines the electrostatic potential generated by all the point charges Q i at x i in a medium with the dielectric constant ε m usually taken as that in the vacuum, a > 0 is a constant, and µ j is the constant chemical potential of the jth ionic species. The constant a > 0 represents in 1.1 a non-physical cut-off which is often chosen to be the thermal de Broglie wavelength and in 1. the uniform linear size of ions. The electrostatic potential ψ is determined by the Poisson equation ε Γ ψ = 4πρ in, 1.5 where ε Γ is the dielectric coefficient and ρ is the charge density, together with a boundary condition which is usually taken to be ψ = ψ 0 on, 1.6 where ψ 0 is a given function. The dielectric coefficient is defined to be ε Γ x = { εm if x m, ε s if x, 1.7 where ε m and ε s are the dielectric constants of the solutes and the solvent, respectively. The charge density is given by ρ = Q i δ xi + q j c j in, 1.8 where δ xi denotes the Dirac delta function centered at x i. The first two terms in 1.1 or 1. represent the internal electrostatic energy which are often written formally as the integral of ρψ/ over the entire region. Based on Born s definition [], the contribution to the electrostatic free energy due to the solute point charges is given as the first term in 1.1 or 1. though the reaction field ψ ψ vac. The β 1 term represents the ideal gas entropy. The term 1 M a3 c j in 1. is the concentration of solvent molecules. It describes the effect of finite size of ions. The last term in 1.1 or 1. accounts for a constant chemical potential in the system. The osmotic pressure from the mobile ions is dropped, since it is only an additive constant to the free-energy functional in the present setting. We remark that the use of notations F 0 and F a does not indicate that we can obtain the functional F 0 by simply setting a = 0 in F a. In this work, we prove the following results: 4

5 1 For each of the free-energy functionals 1.1 and 1., there admits a unique minimizer c 1,...,c M, which is also the unique equilibrium, in an admissible set of concentrations. Moreover, such concentrations and the corresponding equilibrium electrostatic potential ψ are related by the boundary-data dependent Boltzmann distributions c j e βq j[ψx ˆψ 0 x/] for point ions, c j x = c j e βq j[ψx ˆψ 0 x/] a 3 M c i e βq i[ψx ˆψ 0 x/] for finite-size ions, for a.e. x and 1 j M, where c j = a 3 e βµ j and ˆψ 0 H 1 is determined by ε Γ ˆψ 0 ηdx = 0 η H0, ˆψ 0 = ψ 0 on. The free-energy minimizing concentrations are shown to be uniformly bounded above and below away from zero. These results are summarized in Theorems.3.5 and Lemmas 3.4 and 3.5; The equilibrium electrostatic potential ψ is the unique solution to the boundary-data dependent Poisson-Boltzmann equation PBE [3, 4, 10, 11, 16, 17, 0, 31], together with the boundary condition 1.6, ε Γ ψ + 4πχ s M for the case of point ions, and ε Γ ψ+4πχ s M q j c j e βq jψ ˆψ N 0 / = 4π Q i δ xi in 1.11 q j c j e βq jψ ˆψ 0 / 1 + a 3 M c i e βq iψ ˆψ 0 / = 4π Q i δ xi in 1.1 for the case of finite-size ions, where χ s is the characteristic function of. These equations can be written together as ε Γ ψ 4πχ s B ψ ˆψ 0 = 4π Q i δ xi in, 1.13 where B is the derivative of the function B : R R defined by β 1 c j e βq jψ for point ions, Bψ = M β 1 a 3 log 1 + a 3 c j e βq jψ for finite-size ions

6 See Theorem.1; 3 The boundary-value problem of the PBE 1.13 and 1.6 is equivalent to the elliptic interface problem ε m ψ = 4π Q i δ xi in m, ε s ψ 4πB ψ ˆψ 0 = 0 in, ψ = ε Γ ψ n = 0 on Γ, ψ = ψ 0 on Here and below, we denote for any function u on, u m = u m, u s = u s,, and u = u s u m on Γ. See Theorem.. Two variations of the PBE 1.11 with ψ 0 = 0 are commonly used [8,15,9]. First, we have by the Taylor expansion and the electrostatic neutrality M c j = 0 that M q j c j e βqjψ βqjc j ψ, if ψ is small, leading to the linearized PBE [9] ε Γ ψ ε s κ χ s ψ = 4π Q i δ xi in, where κ = 4πβ M q j c j /ε s is the ionic strength or the inverse Debye-Hückel screening length. Clearly, all of our results for the nonlinear PBE 1.11 hold true for the linearized PBE. Second, for the common z : z type of salt such as NaCl in the solution, we have M =, c 1 = c, and q 1 = q = ze. The PBE 1.11 reduces to the following sinh PBE: ε Γ ψ 8πzec 1 χ s sinhβzeψ = 4π Q i δ xi in. In proving the existence of minimizers of the functionals F 0 and F a, we use de la Vallée Poussin s criterion [5] of the sequential compactness in L 1. The uniqueness of such minimizers follows basically from the convexity of these functionals. A crucial step in defining and deriving equilibriums of F 0 and F a is the construction of L -concentrations that are bounded below in by a positive constant and that have low free energies. Such constructions are made by increasing the entropy of ionic concentrations through their small perturbations. The effect of inhomogeneous Dirichlet boundary data to the 6

7 Boltzmann distributions and hence to the PBE can be useful to guide practical numerical computations. The equivalence of the two formulations is a common property for many physical problems. The interface formulation of the boundary-value problem of the PBE has been used for numerical computations using boundary integral method [ 4]. The finite-size effect is important in modeling electrostatics [3,0]. The rest of the paper is organized as follows: In Section, we state our main results; In Section 3, we provide some lemmas; In Section 4, we prove our theorems on the boundary-value problem of PBE; In Section 5, we prove our theorems on the free-energy minimization. Finally, in Appendix, we give the proof of two lemmas. Main results Throughout the rest of the paper, we make the following assumptions: A1. The set R 3 is non-empty, bounded, open, and connected. The sets m R 3 and R 3 are non-empty, bounded, and open, and satisfy that m and = \ m. The N points x 1,...,x N for some integer N 1 belong to m. Both and Γ are of C. The unit exterior normal at the boundary of is denoted by n, cf. Figure 1; A. M is an integer. All a > 0, β > 0, Q i R 1 i N, q j R and µ j R 1 j M, ε m > 0, and ε s > 0 are constants; A3. The functions ψ vac and ε Γ are defined in 1.4 and 1.7, respectively. The boundary data ψ 0 is the trace of a given function, also denoted by ψ 0, in W,. Boundary values are understood as traces. When no confusion arises, the capital letter C, with or without a subscript, denotes a positive constant that can depend on all m,,, Γ, ε m, ε s, a, β, N, M, x i and Q i 1 i N, q j and µ j 1 j M, and ψ 0. For any open set U R 3 that contains all x 1,...,x N, we denote H U 1 = {u W 1,1 U : u Uα H 1 U α α > 0}, where U α = U \ N Bx i,α and Bx i,α denotes the ball centered at x i with radius α. Definition.1. A function ψ H 1 is a weak solution to the boundary-value problem of the PBE 1.13 and 1.6, if ψ = ψ 0 on, χ s Bψ L, and ] [ε Γ ψ η + 4πχ s B ψ ˆψ 0 η dx = 4π Q i ηx i η Cc..1 We remark that if φ H 1 U for some bounded and smooth domain U R 3, then e φ and hence Bφ may not be in L 1 U. For example, let U = B0, 1 be the unit ball of R 3 and α 0, 1/. Define φx = x α for any x U. Then φ H 1 U and 7

8 that e φ L 1 U. Notice by 1.14 that χ s Bψ L is equivalent to χ s e βqjψ L or χ s e βq jψ ˆψ 0 / L j = 1,...,M, which in turn are equivalent to χ s B ψ ˆψ 0 / L. Theorem.1. There exists a unique weak solution ψ H 1 to the boundary-value problem of the PBE 1.13 and 1.6. Moreover, ψ C \ N Bx i,α for any α > 0 such that the closure of N Bx i,α is contained in m, and ψ C m \{x 1,...,x N }. Definition.. A function ψ : R is a weak solution of the interface problem 1.15, if the following are satisfied: ψ m H 1 m and m ε m ψ ηdx = 4π Q i ηx i η Cc m ;. ψ s H 1, χ s Bψ L, and [ε s ψ η + 4πχ s B ψ ˆψ ] 0 η dx = 0 η Cc ;.3 and the third and fourth equations in 1.15 hold true. Theorem.. A function ψ : R is a weak solution to the boundary-value problem 1.13 and 1.6, if and only if it is a weak solution to the boundary-value problem Let U R 3 be a non-empty, bounded, and open set. Let f L 1 U. Assume fξ dx U <..4 sup 0 ξ L U H 1 0 U ξ H 1 U Since L U H0U 1 is dense in H0U, 1 we can identify f as an element in H 1 U, the dual of H0U, 1 with f,ξ = fξdx ξ L U H0U, 1 U and we write f L 1 U H 1 U. The H 1 U-norm of f is given by.4. We define X = c X = { c = c 1,...,c M L 1, R M : c = 0 a.e. m and c j L 1 + q j c j H 1 8 c = c 1,...,c M X. } q j c j H 1,

9 Clearly, X, X is a Banach space. Let α R and define S α : [0, R by S α 0 = 0 and S α u = uα+log u if u > 0. It is easy to see that S α is bounded below on [0, and strictly convex on 0,. Define { } V 0 = c 1,...,c M X : c j 0 a.e. and S 0 c j dx <, j = 1,...,M, { W 0 = c 1,...,c M V 0 : there exists p > 3 } such that c j L p,j = 1,...,M, V a = {c 1,...,c M V 0 : c 0 = a 3 1 a 3 c j 0 a.e. }. Clearly, all V 0, W 0, and V a are non-empty and convex. For any c = c 1,...,c M V 0, there exists a unique weak solution ψ = ψc of the boundary-value problem 1.5 and 1.6 with the charge density ρ given by 1.8; in particular, ψ ψ vac is harmonic in m, cf. Lemma 3.. We shall call ψ = ψc the electrostatic potential corresponding to c. Therefore, F 0 : V 0 R and F a : V a R are well defined. We use V,F to denote V 0, F 0 or W 0,F 0 or V a, F a. Definition.3. An element c = c 1,...,c M V is an equilibrium of F : V R, if there exist γ 1 > 0 and γ > 0 such that γ 1 c j x γ a.e. x, j = 1,...,M,.5 for the case of point ions, or there exists θ 0 0, 1 such that a 3 c j x θ 0 a.e. x, j = 0, 1,...,M,.6 for the case of finite-size ions; and δf[c]e := lim t 0 F[c + te] F[c] t = 0 e X L, R M. Definition.4. An element c V is a local minimizer of F : V R, if there exists ε > 0 such that F[d] F[c] for any d V with d c X < ε. Theorem.3. There exists a unique minimizer of F 0 : V 0 R. It is also the unique local minimizer of F 0 : V 0 R. It is an open question if the unique minimizer of F 0 : V 0 R is an equilibrium of F 0 : V 0 R as defined in Definition.3. The answer to this question would be yes if this minimizer were in W 0 or if min d V0 F 0 [d] = min d W0 F 0 [d], neither of which is clearly true. This is the reason we introduce the class of concentrations W 0. See the proof of Lemma 3.4 in Appendix. Theorem.4. 1 There exists a unique equilibrium c = c 1,...,c M of F 0 : W 0 R. It is also the unique global minimizer and the unique local minimizer of F 0 : W 0 R. 9

10 If ψ = ψc is the corresponding electrostatic potential, then the Boltzmann distributions 1.9 for point ions holds true and ψ is the unique weak solution to the corresponding boundary-value problem of PBE 1.11 and 1.6. Moreover, min d W 0 F 0 [d] = 1 Q i ψ ψ vac x i + ε s 8π ψ ˆψ 0 dx β 1 Γ 1 ψ 8π ˆφ 0 ε Γ n ψ ˆψ 0 ds c j e βq jψ ˆψ 0/ dx..7 Theorem.5. 1 There exists a unique equilibrium c = c 1,...,c M of F a : V a R. It is also the unique global minimizer and the unique local minimizer of F a : V a R. If ψ = ψc is the corresponding electrostatic potential, then the Boltzmann distributions 1.9 for finite-size ions holds true and ψ is the unique weak solution to the corresponding boundary-value problem of PBE 1.1 and 1.6. Moreover, min F a [d] = 1 d V a ε Γ 8π Q i ψ ψ vac x i + 3 Some lemmas ψ ˆψ 0 dx β 1 a 3 Γ 1 ψ 8π ˆφ 0 ε Γ n ψ ˆψ 0 ds [ M ] 1 + log 1 + a 3 c j e βq jψ ˆψ 0 / dx..8 The key point of our first lemma below is the existence and continuity across the interface Γ of the normal flux for a solution of an elliptic interface problem. In terms of electrostatics, this means that the electrostatic potential and the normal component of electrostatic displacement are continuous across dielectric boundaries. These seem to be known results. For completeness, we give a proof here. Lemma 3.1. Let U R 3 be an open set such that Γ U. Let g L 1 U H 1 U. Suppose u H 1 U satisfies ε Γ u ηdx = gηdx η Cc U. 3.1 U Then u = 0 on Γ. If in addition g L U, then ε Γ n u = 0 on Γ. U Proof. Fix an open ball B U such that Γ B. Let η Cc U with suppη B. Let n j with 1 j 3 be the j-th component of n, the unit normal at the Γ, pointing from to m. It follows from the fact that u H 1 and integration by parts that u j η dx = j uη dx B B 10

11 = j uη dx + j uη dx B m B s = u j η dx u m ηn j ds u j η dx + B m Γ B B s = u j η dx + u s u m ηn j ds, j = 1,, 3. B Γ B Γ B u s ηn j ds This and the arbitrariness of η imply u = 0 on Γ. To show the continuity of ε Γ u n across Γ, we fix an open set U 0 R 3 such that Γ U 0 U 0 U and that the boundary U 0 is C. By the fact that u H 1 U and g L U, and by 3.1, we have ε t u t U0 t L U 0 t, R 3 and ε t u t U0 t = g L U 0 t, t = m,s. Therefore, by Theorem 1. in [30], the trace of ε t u t U0 t ν H 1/ U 0 t exists, and also by 3.1, ε t u t ηdx = gηdx + ε t u t νηds η Cc U 0 t, 3. U 0 t U 0 t U 0 t where ν denotes the unit exterior normal of the boundary U 0 t which contains Γ and t = m,s. Notice that the normals ν at Γ from both sides U 0 m and U 0 are in opposite directions. These traces are determined independent of the choice of U 0. In fact, if Q 0 R 3 is another open set such that Γ Q 0 Q 0 U and the boundary Q 0 is C, then the traces ε m u m Q0 m ν H 1/ Q 0 m and ε s u s Q0 ν H 1/ Q 0 m exist, and 3. hold true for t = m,s when U 0 is replaced by Q 0. Consider now 3. with t = m. Choose any η CcU 1 0 Q 0 such that η = 0 on U 0 m \ Γ and on Q 0 m \ Γ. Extend η by η = 0 to outside U 0 Q 0. By 3. with t = m and the corresponding equation with U 0 replaced by Q 0, we obtain that ε m u m νηds = ε m u m νηds. U 0 m Q 0 m The arbitrariness of η then implies that the trace of ε m u m ν U0 m on Γ determined by U 0 is the same as that determined by Q 0. By the same argument, we see that the trace of ε s u s n U0 on Γ determined by U 0 is the same as that determined by Q 0. Now, by the fact that U 0 = U 0 m U 0 and U 0 m U 0 =, and by our convention for the direction of the unit normal n along Γ, we obtain from 3.1 and 3. that for any η Cc U with suppη U 0 Γ ε m u m n ε s 11 u s n ηds = 0.

12 The arbitrariness of η implies ε Γ n u = 0 on Γ. Q.E.D. Let L : H 1 H 1 be the linear operator defined as follows: for any ξ H 1, Lξ H0 1 is the unique function in H0 1 that satisfies 1 ε Γ Lξ v dx = ξv v H 1 4π It is easy to see that ξ,η = ξlη defines an inner product of H 1. Denote by the corresponding norm of H 1, i.e., ξ = ξ,ξ = ξlξ for any ξ H 1. One can verify that there exist C 1 = C 1,ε m,ε s > 0 and C = C ε m,ε s > 0 such that C 1 ξ ξ H 1 C ξ ξ H It follows from [1] with minor modifications that there exists a unique G H 1 such that G = 0 on and ε Γ G ηdx = 4π Q i ηx i η Cc. 3.5 Clearly, G ψ vac is harmonic in m and G W 1,p for any p [1, 3/. Notice that the function ˆψ 0 H 1 defined in 1.10 is harmonic in m. The next lemma gives a solution decomposition of the Poisson equation 1.5 with its right-hand side consisting of Dirac masses and a function in H 1 that represents the density of ionic charges. This decomposition is a mathematical formulation of the Born cycle []. Lemma 3.. Let f L 1 H 1 be such that f = 0 in m. Then ψ := G+ ˆψ 0 +Lf is the unique function in H 1 that satisfies ψ = ψ 0 on and ε Γ ψ ηdx = 4π Moreover, Lf and ψ ψ vac are harmonic in m, and Q i ηx i + 4π fηdx η Cc. 3.6 Q i Lf x i = Gfdx. 3.7 Proof. From the definition of G, ˆψ 0, and L cf. 3.5, 1.10, and 3.3, we easily verify that the function ψ is in H 1, ψ = ψ 0 on, and 3.6 holds true. If ψ H 1 satisfies ψ = 0 on and ε Γ ψ η dx = 0 for all η C c, then clearly ψ H 1 0 and in fact ψ = 0 a.e.. This proves the needed uniqueness. 1

13 By the fact that f = 0 in m and the definition of L cf. 3.3, Lf is harmonic in m. The fact that ψ ψ vac is harmonic in m follows from 3.6 with η Cc so chosen that suppη m and ε m ψ vac ηdx = 4π Q i ηx i η Cc m. m It remains to prove 3.7. Denote ψ c = Lf H0. 1 Let α > 0 be sufficiently small and let B α = N Bx i,α. By the fact that G is harmonic in m \ B α and G ψ vac is harmonic in m, we obtain by a series of routine calculations that ε m G ψ c dx = ε m G ψ c dx + ε m G ψ c dx m m\b α B α = ε m Gψ c dx + m\b α G m = ε m ψ c m n ds + Γ G m = ε m ψ c m Γ n ds + + Bx i,α m\b α G ε m ψ c B α ν G ε m ψ c ds + Oα ν ds + Oα G ψ vac ε m ψ c B α ν ψ vac ε m ψ c ds + Oα ν N G m ε m ψ c m Γ n ds + 4π Q i ψ c x i as α 0, where ν is the exterior unit normal of m \B α and ν = n on Γ by our convention for the direction of n. Consequently, by the continuity of ε Γ G n across Γ cf. Lemma 3.1, the fact that G is harmonic in, and G = ψ c = 0 on, we obtain G s 4π Q i ψ c x i = ε s ψ c s Γ n ds + ε m G ψ c dx m = ε s Gψ c dx + ε s G ψ c dx + ε m G ψ c dx m = ε Γ G ψ c dx. 3.8 Since ψ c = Lf H0 1 is harmonic in m, we also have by the properties of G cf. [1] and integration by parts that ε m G ψ c dx = m ε m G ψ c dx + m\b α ε m G ψ c dx B α 13 ds

14 = ε m G ψ c dx + m\b α = Γ Γ ε m G m ψ c m n ds + m\b α ε m G ψ c B α ν ε m G ψ c ds + Oα ν ds + Oα ψ c m ε m G m ds as α n Let Ĝ H1 m be such that Ĝ = G on Γ = m. Replacing G in 3.9 by Ĝ and repeating the same calculations, we obtain ε m G ψ c dx = m ε m Ĝ ψ cdx. m 3.10 Define G : R by Gx = Ĝx if x m and by Gx = Gx if x. Clearly, G H 1 0. Since ψ c = Lf and f = 0 in m, we thus have by 3.8 and 3.10 that 4π Q i ψ c x i = ε m Ĝ ψ cdx + ε s G ψ c dx m = ε Γ G ψ c dx = 4π Gfdx = 4π Gfdx. This implies 3.7. Q.E.D. By Lemma 3., the potential ψ = ψc 1,...,c M corresponding to a set of concentrations c 1,...,c M is well defined with f = M q jc j and is given by M ψc 1,...,c M = G + ˆψ 0 + L q j c j Moreover, the functional F 0 : V 0 R and F a : V a R can be rewritten as F 0 [c] = 1 M M q j c j L q j c j dx + µ 0j c j dx M + β 1 S 1 c j dx + E 0, c = c 1,...,c M V 0, 3.1 F a [c] = 1 M M q j c j L q j c j dx + µ aj c j dx M + β 1 S 1 c j dx + E a c = c 1,...,c M V a, 3.13 j=0 14

15 respectively, where µ 0j x = q j Gx + 1 q j ˆψ 0 x + 3β 1 log a µ j x, j = 1,...,M, 3.14 E 0 = 1 Q i G + ˆψ 0 ψ vac x i, 3.15 µ aj x = q j Gx + 1 q j ˆψ 0 x µ j x, j = 1,...,M, 3.16 E a = 1 Q i G + ˆψ 0 ψ vac x i + 3β 1 a 3 log a, 3.17 where E denotes the Lebesgue measure of a Lebesgue measurable set E R 3. Lemma 3.3. Let D R 3 be a bounded and open set. Let α R. Let {u k } be a sequence of functions in L 1 D such that u k 0 a.e. D for each k 1 and that S α u k dx <. sup k 1 D Then there exists a subsequence {u kj } of {u k } such that {u kj } converges weakly in L 1 D to some u L 1 D with u 0 a.e. D and S α udx lim inf S α u k dx. k D Proof. Since S α : [0, R is bounded below, by passing to a subsequence if necessary, we may assume that the limit A := lim S α u k k D dx = lim inf S α u k dx 3.18 k exists and is finite. Since S α λ/λ + as λ +, {u k } is weakly sequentially compact in L 1 D by de la Vallée Poussin s criterion [5], Therefore, this sequence has a subsequence, not relabeled, that converges weakly in L 1 D to some u L 1 D. Clearly, u 0 a.e. D. Let ε > 0. By 3.18, there exists an integer K > 0 such that S α u k dx A + ε k > K D By Mazur s Theorem [7,3], there exist convex combinations v k of u K+1,...,u K+k for all k 1 such that v k u in L 1 D. Let v k = k λ k,ju K+j with λ k,j 0 for all j D D 15

16 and k, and k λ k,j = 1 for all k. Since S α : [0, R is convex, we have by Jensen s inequality and 3.19 that S α v k k λ k,j S α u K+j k λ k,j A + ε = A + ε k Since v k u in L 1 D, there exists a subsequence {v kj } of {v k } such that v k j x ux a.e. x D. Consequently, Since S α : [0, R is continuous and bounded below, we have by Fatou s Lemma and 3.0 that D S α uxdx = D lim S α v k j x dx lim inf j j concluding the proof by the arbitrariness of ε > 0. D Q.E.D. S α v k j x dx A + ε. The next two lemmas state some boundedness of concentrations that have low free energies. Their proofs are somewhat tedious, and are given in Appendix. Lemma 3.4. Let c = c 1,...,c M W 0 satisfy that c L, R M or there exists j {1,...,M} with {x : c j x < α} > 0 for all α > 0. Then for any ε > 0 there exist ĉ = ĉ 1,...,ĉ M W 0 that satisfies.5 with c replaced by ĉ, ĉ c X < ε, and F 0 [ĉ] < F 0 [c]. Lemma 3.5. Let c = c 1,...,c M V a and c 0 be defined by 1.3. Assume there exists j {0, 1,...,M} such that {x : a 3 c j x < α} > 0 for all α > 0. Let ε > 0. Then there exist ĉ = ĉ 1,...,ĉ M V a that satisfies.6 with c replaced by ĉ, ĉ c X < ε, and F a [ĉ] < F a [c]. 4 The Poisson-Boltzmann equation: Proof of Theorems.1 and. Proof of Theorem.1. It is easy to verify that the function B : R R defined in 1.14 is convex for both the case of point ions and that of finite-size ions. Let K := {u H 1 : u = ψ 0 on and χ s Bu L }. Clearly, K since ψ 0 K and K is convex since B : R R is convex. We show now that K is closed in H 1. Let u k K k = 1,,... and u k u in H 1 for some u H 1. Clearly, u = ψ 0 on. Up to a subsequence, not relabeled, u k x ux a. e. x. Since B : R R is convex and positive, we have d dv [Bv] = [B v] + BvB v > 0 v R. 16

17 Thus, v [Bv] is convex. It then follows from Fatou s lemma, Jensen s inequality, and the H 1 -boundedness of {u k } that 1 [Bu] dx lim inf k s [ ] 1 1 [Bu k ] dx lim inf B u k dx <. k s This implies that u K. Therefore, K is closed in H 1. Since K is convex, it is also weakly closed in H 1. Define now J : K R by J[u] = [ ε Γ u + 4πχ s B u + G ˆψ 0 ] dx u K, where G and ˆψ 0 are defined in 3.5 and 1.10, respectively. Note that ψ 0 K and that J[ψ 0 ] <. By the Poincaré inequality, there exist constants C 3 > 0 and C 4 0 such that J[u] C 3 u H 1 C 4 for all u K. Thus, α := inf u K J[u] is finite. Let v k K k = 1,... be such that lim k J[v k ] = α. Then, {v k } is bounded in H 1 and hence it has a subsequence, not relabeled, that weakly converges to some v H 1. Since K is weakly closed, v K. Since the embedding H 1 L is compact, up to a further subsequence, again not relabeled, v k v a. e. in. Therefore, since B : R R is continuous and non-negative, Fatou s lemma implies lim inf k χ s B v k + G ˆψ 0 dx χ s B v + G ˆψ 0 Since u ε Γ u dx is convex and H 1 -continuous, it is sequentially weakly lower semicontinuous. Consequently, lim inf k J[v k ] J[v]. Thus, v is a minimizer of J : K R. Notice that χ s B v + G ˆψ 0 / L. Simple calculations of the first variation of J : K R at any η Cc leads to [ε Γ v η + 4πχ s B v + G ˆψ ] 0 η dx = 0 dx. η C c. The function ψ = v + G is thus a needed solution. We now prove the uniqueness. Let φ be another weak solution. Let ξ = ψ φ. Then, ξ H, 1 ξ = 0 on, and ] } {ε Γ ξ η + 4πχ s [B B ψ ˆψ 0 φ ˆψ 0 η dx = 0 η C c. 17

18 Choosing the test functions η Cc so that suppη m, we find that ξ is harmonic in m. This and the fact that ξ H 1 imply that ξ H0. 1 Thus, the above test functions η can be chosen from H0. 1 In particular, setting η = ξ and using the convexity of B : R R, we obtain that ξ = 0 and hence ψ = φ in H 1. Let σ > 0 be such that the closure of B σ := N Bx i,σ is contained in m. Clearly, the unique week solution ψ H 1 satisfies ε Γ ψ ηdx = gηdx η Cc \ B σ. 4.1 \B σ \B σ where g = 4πχ s B ψ ˆψ 0 / L \ B σ. Therefore, ψ C \ B σ by the standard regularity theory [14]. Since ε Γ = ε m in m and ε Γ = ε s in, ψ is harmonic in m \ B σ. Hence ψ C m \ B σ. Notice that B C R, thus we have ψ C by a standard bootstrapping argument. Q.E.D. Proof of Theorem.. Let ψ H 1 be a weak solution to the boundary-value problem 1.13 and 1.6. Clearly, ψ m H 1 m. For any η Cc m, we extend η to the entire by defining η = 0 outside m. Then, we obtain. from.1. Since all x i m i = 1,...,N, we have ψ s H 1. Since χ s Bψ L, it follows from 1.14 that χ s B ψ L. For any η Cc, we again extend η to by defining η = 0 outside. Then, we obtain.3 from.1. Finally, by Lemma 3.1, 4.1, and 1.6, the last two equations in 1.15 hold true. Hence, ψ is a weak solution to Now let ψ : R be a solution to the boundary-value problem We first show that ψ H. 1 Let σ > 0 be small enough so that the closure of B σ := N Bx i,σ is contained in m. Since ψ m H 1 m, ψ m H 1 m \B σ. Thus, the trace ψ m Γ L Γ, and is independent on the choice of σ. Similarly, ψ s H 1, and hence ψ s Γ L Γ. Fix j {1,, 3}. Define ξ j : \ B σ R by ξ j = j ψ m in m \ B σ and ξ j = j ψ s in. Clearly, ξ j L \ B σ. Let n j be the j-th component of the unit exterior normal n at Γ, pointing from to m. Then, for any η Cc \ B σ, we have ξ j η dx = j ψ m η dx + j ψ s η dx \B σ m\b σ s = ψ j η dx ψ m ηn j ds ψ j η dx + ψ s ηn j ds m\b σ Γ Γ = ψ j η dx + ψ n j η ds \B σ Γ = ψ j η dx, \B σ where in the last step we used the fact that ψ = 0 on Γ. Thus, ξ j = j ψ L \ B σ, and hence ψ H 1 by the arbitrariness of σ > 0. 18

19 Clearly, χ s B ψ L and ψ = ψ 0 on. It remains to show that.1 holds true. Let η Cc. Let V 1 and V be two open sets in R 3 such that V 1 and V are of C, x i V for i = 1,...,N, and Γ V 1 V 1 V V. Let ζ Cc be such that suppζ V and ζ = 1 on V 1. Then, 1 ζη m Cc m, 1 ζη s Cc, and 1 ζx i ηx i = ηx i, i = 1,...,N. We thus have by. and.3 that [ε Γ ψ η + 4πχ s B ψ ˆψ ] 0 η dx = [ε Γ ψ 1 ζη + 4πχ s B ψ ˆψ ] 0 1 ζη dx ] [ε Γ ψ ζη + 4πχ s B + V ψ ˆψ 0 = ε Γ ψ 1 ζηdx m + [ε Γ ψ 1 ζη + 4πχ s B ψ ˆψ ] 0 1 ζη dx + [ε Γ ψ ζη + 4πχ s B ψ ˆψ ] 0 ζη dx V = 4π Q i ηx i + [ε Γ ψ ζη + 4πχ s B ψ ˆψ ] 0 ζη dx. 4. We now show that the second term in 4. is zero. Notice that ψ V H 1 V and x i V 1 i N. Denoting V m = V m and V s = V, we have by. and.3 that ε m ψ ξ dx = 0 ξ Cc V m, V m ε s φ ξ dx = 4π V s Consequently, since χ s B ψ ˆψ 0 / V s B ψ ˆψ 0 ζη dx ξ dx ξ C c V s. L, we infer from the regularity theory of elliptic boundary-value problems [14] that ψ Vm H V m and ψ Vs H V s, and that ε m ψ = 0 a.e. V m, ε s ψ 4πB ψ ˆψ 0 = 0 a.e. V s. 19

20 Therefore, the trace of ε m ψ n and that of ε s ψ n on Γ both exist. Moreover, [ε Γ ψ ζη + 4πχ s B ψ ˆψ ] 0 ζη dx V = ε m ψ ζηdx + [ε s ψ ζη + 4πχ s B ψ ˆψ ] 0 ζη dx V m V s = ε m ψζη dx ε m ψ nζη ds V m Γ + [ ε s ψζη + 4πχ s B ψ ˆψ ] 0 ζη dx + ε s ψ nζη ds V s Γ = ε ψ n ζη ds = 0, Γ 4.3 where in the last step, we used the third equation of Now, since η Cc is arbitrary, we obtain.1 from 4. and 4.3. Q.E.D. 5 Minimization of the electrostatic free energy: Proof of Theorems.3,.4, and.5 Proof of Theorem.3. Let t = 1 + β max 1 j M µ 0j L, where µ 0j j = 1,...,M are defined in It follows from 3.1 and 3.4 that there exists C 5 > 0 such that M F 0 [c] C 5 q j c j H 1 M +β 1 S t c j dx+e 0 c = c 1,...,c M V 0, 5.1 where E 0 is defined in Let z = inf c V0 F 0 [c]. Since S t : [0, R is bounded below, z is finite. Let c k = c k 1,...,c k [ M V 0 k = 1,,... be such that lim k F ] 0 c k = z. It { } follows from 5.1 that S t c k j dx is bounded for each j = 1,...,M. Therefore, { } by Lemma 3.3, up to a subsequence that is not relabeled, converges weakly in L 1 to some c j L 1, and S t c j dx lim inf k s S t c k j c k j dx < j = 1,...,M. 5. 0

21 { M } Define c j = 0 on m for all j = 1,...,M. By 5.1, q jc k j is bounded in H 1. { M } Since H 1 is a Hilbert space, q jc k j has a subsequence, again not relabeled, that weakly converges to some F H 1. Let ξ L H0. 1 We have M M Fξ = lim q j c k k j ξdx = q j c j ξdx. Therefore, M q jc j H 1 and hence c = c 1,...,c M V 0. By 5. and the fact that the norm of a Banach space is sequentially weakly lower semicontinuous, we have z = lim inf k F 0 [ c k ] F 0 [c] z. This implies that c V 0 is a global minimizer of F 0 : V 0 R. Let d = d 1,...,d M V 0 be a local minimizer of F 0 : V 0 R. Then for λ 0, 1 close to 0, we have by the convexity of F 0 : V 0 R that F 0 [d] F 0 [λc + 1 λd] λf 0 [c] + 1 λf 0 [d], leading to F 0 [d] F 0 [c]. Thus, d is also a global minimizer of F 0 : V 0 R. Clearly, c+d/ V 0. Consequently, it follows from the definition of the norm and the Cauchy- Schwarz inequality with respect to the inner product ξ,η = ξlη ξ,η H 1 that [ ] c + d 0 F 0 min F 0 [e] e V 0 [ ] c + d = F 0 1 F 0[c] 1 F 0[d] = 1 q 8 j c j + d j 1 q 4 j c j 1 q 4 j d j M [ + β 1 cj + d j S 1 1 S 1 c j 1 ] S 1 d j dx M [ β 1 cj + d j S 1 1 S 1 c j 1 ] S 1 d j dx. This, together with the convexity of S 1 on [0,, implies that cj x + d j x S 1 = 1 S 1 c j x + 1 S 1 d j x j = 1,...,M, x \ ω s, for some ω s with ω s = 0. Let x \ ω s. Then it follows from the definition of S 1 : [0, R that c j x = 0 if and only if d j x = 0 for all j = 1,...,M. The strict 1

22 convexity of S 0 on 0, then implies that c = d a.e.. Hence, c = d in V 0. Q.E.D. Proof of Theorem.4. 1 Let c = c 1,...,c M W 0. We show that the following four statements are equivalent: i c is an equilibrium of F 0 : W 0 R; ii The property.5 holds true, and M q j L q j c j + µ 0j + β 1 log c j = 0 a.e., j = 1,...,M; 5.3 iii c is a global minimizer of F 0 : W 0 R; iv c is a local minimizer of F 0 : W 0 R. Assume i is true. Then.5 holds true by Definition.3. Let e = e 1,...,e M X L, R M. Notice that S 1u = log u for any u > 0. Thus, for each j {1,...,M} and each x, the Mean-Value Theorem implies the existence of θ j x [0, 1] such that S 1 c j x + te j x S 1 c j x = te j x logc j x + tθ j xe j x. Hence, by the Lebesgue Dominated Convergence Theorem, S 1c j + te j S 1c j lim dx = e j log c j dx, t 0 t j = 1,...,M. Therefore, it follows from Definition.3, the definition of the norm, 3.1, and 3.4 that 0 = δf 0 [c]e F 0 [c + te] F 0 [c] = lim t 0 t { 1 = lim t 0 t M M q j e j + q j e j L q j c j dx } + µ 0j e j dx + β 1 1 t [S 1c j + te j S 1 c j ] dx [ M = q j L q j c j + µ 0j + β 1 log c j ]e j dx e X L, R M. 5.4 This implies 5.3. Hence, ii is true. Assume ii is true. We show that iii is true. By Lemma 3.4, we need only to show that F 0 [c] F 0 [d] for any fixed d = d 1,...,d M W 0 that satisfies.5 with c replaced

23 by d. In fact, setting e = e 1,...,e M = d c X L, R M, we have by the convexity of S 1 : [0, R that S 1 d j S 1 c j d j c j S 1c j = e j log c j a.e.. Therefore, it follows from 3.1 and 5.3 that F 0 [d] F 0 [c] = 1 M M q j e j L q j e j dx + + = 0. M M q j e j L q j c j dx µ 0j e j dx + β 1 [S 1 d j S 1 c j ] dx [ M ] q j L q i c i + µ 0j + β 1 log c j e j dx Hence, F 0 [c] F 0 [d], and iii is true. Clearly, iii implies iv. Finally, assume iv is true. By Lemma 3.4,.5 holds true. For any e X L, R M, it is easy to see that δf 0 [c]e exists, cf Since F 0 [c + te] F 0 [c] for t small enough, we have δf 0 [c]e = 0. Therefore, c is an equilibrium of F 0 : W 0 R, and i is true. Let now ψ H 1 be the unique weak solution to the boundary-value problem of the PBE 1.11 and 1.6, cf. Theorem.1. Define c = c 1,...,c M : R by 1.9 for point ions and c j x = 0 for all x m and all j = 1,...,M. Clearly, c W 0. Moreover, by Theorem.1, ψ s C. This implies.5. It follows from 1.11, 1.6, 1.9 for point ions, and Lemma 3. with f = M q jc j that ψ is the electrostatic potential corresponding to c, i.e., ψ = G + ˆψ 0 + L M q jc j. This, together with the Boltzmann relations 1.9 for point ions and 3.14, implies 5.3. Hence, c is an equilibrium, and thus a local and global minimizer, of F 0 : W 0 R. The uniqueness of equilibria or local minimizers is equivalent to that of global minimizers, and can be proved by the same argument used in the proof of Theorem.3. It is clear that from our definition of c and ψ that we need only to prove.7. Since c is the unique minimizer of F 0 : W 0 R and ψ is the corresponding electrostatic potential determined by 3.11, we have by 1.1 and 1.9 for point ions that min F 0 [d] = F 0 [c] d W 0 = 1 Q i ψ ψ vac x i M q j c j ψdx

24 = 1 M + β 1 c j [ loga 3 c j 1 ] dx Q i ψ ψ vac x i 1 M β 1 µ j c j dx q j c j e βq jψ ˆψ 0/ ψ ˆψ 0 dx c j e βq jψ ˆψ 0/ dx. 5.5 Since ψ is the unique solution to the boundary-value problem of PBE 1.11 and 1.6, and since ˆψ 0 is harmonic in by 1.10, we have ε s ψ ˆψ 0 + 4π q j c j e βq jψ ˆψ 0 / = 0 a.e. s. Multiplying both sides of this equation by ψ ˆψ 0 and integrate the resulting terms over, we obtain by integration by parts and the fact that by Lemma 3.1 both ψ ˆψ 0 and ε Γ n ψ ˆψ 0 are continuous across Γ, ε s + 4π ψ ˆψ 0 dx + Γ q j c j e βq jψ ˆψ 0/ ε Γ ψ ˆψ 0 n ψ ˆψ 0 ds ψ ˆψ 0 dx = 0. This and 5.5 imply.7. Q.E.D. Proof of Theorem.5. 1 We first show that F a : V a R is a convex functional. Define T M = {u 1,...,u M R M : u j > 0 for j = 1,...,M, and M u j < 1} and ] hu = 1 u j [log 1 u j 1 u = u 1,...,u M T M. Clearly, T M is convex. We have ui u j hu = 1 M k=1 u k 1 for all 1 i,j M. Let Hu = ui u j h be the Hessian of h : T M R. Then, for any y = y 1,...,y M R M, we have y Huy = M k=1 y k /1 k=1 u k 0. Therefore, Hu is symmetric, semidefinite for any u T M. Hence, h : T M R is convex. Consequently, since V a is a convex subset of X and S 1 : [0, R is convex, we conclude by 3.13 that F a : V a R is convex. Let now c = c 1,...,c M V a. By the same argument used in the proof of Theorem.4, we obtain the equivalence of the following four statements: 4

25 i c is an equilibrium of F a : V a R; ii The property.6 holds true, and M q j L q i c i + µ aj + β 1 a 3 c j log 1 a 3 M c = 0 a.e., j = 1,...,M; i 5.6 iii c is a global minimizer of F a : V a R; iv c is a local minimizer of F a : V a R. Let ψ H 1 be the unique weak solution to the boundary-value problem of the PBE 1.1 and 1.6, cf. Theorem.1. Define c = c 1,...,c M : R by 1.9 for finite-size ions and c j x = 0 for all x m and all j = 1,...,M. Clearly, c V a. Moreover, by Theorem.1, ψ s C. This implies.6. By 1.9 for finite-size ions, we have M a 3 c j x = a 3 M c i e βq i ψ ˆψ 0 / This together with 1.9 for finite-size ions imply that c j 1 a 3 M c = c j e βq j ψ ˆψ 0 /, i j = 1,...,M. 5.7 It follows from 1.1, 1.6, 1.9 for finite-size ions, and Lemma 3. with f = M q jc j that ψ is the electrostatic potential corresponding to c, i.e., ψ = G + ˆψ 0 + L M q jc j. This, together with 5.7 and 3.16, implies 5.6. Hence, c is an equilibrium, and thus a local and global minimizer, of F a : V a R. The uniqueness of equilibria or local minimizers is equivalent to that of global minimizers, and can be proved by the same argument used in the proof of Theorem.3. We need only to prove.8. Since c is the unique minimizer of F a : V a R and ψ is the corresponding electrostatic potential determined by 3.11, we have by 1., 1.3, and 1.9 for finite-size ions that min F a [d] = F a [c] d V a = 1 = 1 Q i ψ ψ vac x i + 1 M + β 1 j=0 M q j c j ψdx c j [ loga 3 c j 1 ] dx Q i ψ ψ vac x i 1 5. µ j c j dx s q j c j e βq jψ ˆψ 0 / ψ ˆψ a 3 M c i e βq iψ ˆψ 0 / dx

26 β 1 a 3 [ 1 + log 1 + a 3 M ] c i e βq iψ ˆψ 0 / dx. 5.8 Since ψ is the unique solution to the boundary-value problem of PBE 1.1 and 1.6, and since ˆψ 0 is harmonic in by 1.10, we have ε s ψ ˆψ 0 + 4π q j c j e βq jψ ˆψ 0 / 1 + a 3 M c i e βq iψ ˆψ 0 / = 0 a.e.. Multiplying both sides of this equation by ψ ˆψ 0 and integrate the resulting terms over, we obtain by integration by parts and the fact that by Lemma 3.1 both ψ ˆψ 0 and ε Γ n ψ ˆψ 0 are continuous across Γ, ε s + 4π ψ ˆψ 0 dx + s q j c j e βq jψ ˆψ 0 / ψ ˆψ 0 Γ ε Γ ψ ˆψ 0 n ψ ˆψ 0 ds 1 + a 3 M c i e βq iψ ˆψ 0 / dx = 0. This and 5.8 imply.8. Q.E.D. Appendix We now prove Lemma 3.4 and Lemma 3.5 by constructing ionic concentrations that satisfy required conditions and that have lower free energies. The key idea here is based on the following observation: the function S α : [0, R, defined for any α R by S α 0 = 0 and S α u = uα + log u if u > 0, has a unique minimizer which is a positive number. Moreover, the magnitude S αu is very large if u is close to 0 or. Notice that S α represents the entropy of the system. Therefore, small changes of concentrations near zero or infinity can largely increase the corresponding entropy and hence decrease the free energy. Proof of Lemma 3.4. We first construct c W 0 that satisfies c j x γ a.e. x, j = 1,...,M, A.1 for some constant γ > 0, c c X < ε/, and F 0 [ c] F 0 [c] with a strict inequality if c L, R M. Let A > 0. Define c = c 1,..., c M : R by c j x = { cj x if c j x A 0 if c j x > A x, j = 1,...,M. A. 6

27 Clearly, c W 0 and A.1 holds true with γ = A. Moreover, M c j c j L 1 < ε/4 for A > 0 large enough. Denote τ j A = {x : c j x > A}, j = 1,...,M. Since c W 0, there exists p > 3/ such that each c j L p 1 j M. Thus, q j c j q j c j = q j χ τj Ac j 0 in L p as A. By the definition of L : H 1 H 1 0 and the regularity theory for elliptic problems M [14], we have L q s jχ τj Ac j W,p and M L q j χ τj Ac j C q j χ τj Ac j 0 as A. W,ps L ps Hence, by 3.4 and the embedding W,p L that M M q j c j q j c j C q j χ τj Ac j L q j χ τj Ac j dx H 1 L M C q j χ τj Ac j L q j χ τj Ac j 1 L L M C q j χ τj Ac j L q j χ τj Ac j p W,p C q j χ τj Ac j 0 as A. L p Therefore, c c X < ε if A > 0 is large enough. Notice that c j = c j χ τj Ac j for all j = 1,...,M. Thus, M 1 M q j c j L q j c j dx 1 M M q j c j L q j c j dx = 1 M M q j χ τj Ac j L q j c j + q j c j dx 1 q j d j A τ j A c j dx, A.3 7

28 where M d j A = L M q j c j + L q j c j. L s L s Since M L L q j c j C L M W L L q j c j C q j c j q j c j,p p p as A, we have max 1 j M d j A C as A > 0 large enough. For each fixed j {1,...,M} and x τ j A, we also have S 1 c j x S 1 c j x = S 1 c j x = c j x log c j x c j x log A. A.4 Therefore, it follows from 3.1, A.3, and A.4 that F 0 [ c] F 0 [c] 1 q j d j A + µ 0j L β 1 log A c j dx. τ j A If A > 0 is large enough, this is non-positive. If c L, R M then there exists j {1,...,M} such that τ j A > 0 for all A > 0. In this case, we have the strict inequality F 0 [ c] < F 0 [c]. We now construct ĉ W 0 that satisfies.5 with c replaced by ĉ, ĉ c X < ε/, and F 0 [ĉ] F 0 [ c] with a strict inequality if there exists j {1,...,M} such that {x : c j x < α} > 0 for all α > 0, all these implying that ĉ satisfies all the desired properties. If there exists γ 1 > 0 such that c j x γ 1 for a.e. x and j = 1,...,M, then ĉ = c with A γ 1 cf. A. satisfies all the desired properties with γ 1 = γ 1 and γ = γ. Assume otherwise there exists j 0 {1,...,M} such that {x : c j0 x < α} > 0 for all α > 0. This means that {x : c j0 x < α} > 0 for all α > 0. Define ρ j α = {x : c j x < α} α > 0, j = 1,...,M, I 0 = {j {1,...,M} : ρ j α > 0 α > 0}, I 1 = {1,...,M} \ I 0. Clearly, I 0. If I 1, then there exists α 1 > 0 such that c j x α 1 a.e. x, j I 1. Define for 0 < α < α 1 and 1 j M { cj x + αχ ρj αx if j I 0 ĉ j x = x. c j x if j I 1 8

29 Clearly, ĉ = ĉ 1,...,ĉ M W 0 and.5 holds true with c replaced by ĉ, γ 1 = α, and γ = γ + α. Moreover, M ĉ j c j L 1 < ε/4 if α > 0 is small enough. Furthermore, q j ĉ j q j c j = α q j χ ρj α α q j χ ρj α H 1 j I 0 H 1 j I 0 L α q j ρ j α 0 as α 0. A.5 j I 0 Hence, ĉ c X < ε/ if α > 0 is small enough. By the Mean-Value Theorem and the fact that S 1u = log u for any u > 0, [S 1 ĉ j S 1 c j ]dx = j I 0 ρ j α [S 1 ĉ j S 1 c j ] dx α log α j I 0 ρ j α. Consequently, it follows from 3.13, 3.4, A.5 that F 0 [ĉ] F 0 [ c] = 1 M q j c j + α M q j χ ρj α L q j c j + α q j χ ρj α dx j I 0 j I 0 1 M M q j c j L q j c j dx + α µ 0j dx j I 0 + β 1 j I 0 ρ j α [S 1 ĉ j S 1 c j ]dx α q j χ ρj α L q j χ ρj α dx j I 0 j I 0 M + α q j χ ρj α L q j c j dx j I 0 α ρ j α + α j I 0 µ 0j L ρ j α + α j I 0 β 1 logα ρ j α i I 0 q i M ρ j α + α L q j c j q j ρ j α j I 0 L s j I 0 + α j I 0 µ 0j L ρ j α + α j I 0 β 1 logα ρ j α = α j J 0 [ α j I 0 q j M + q j L q j c j L s 9

30 ] + µ 0j L + β 1 logα ρ j α. Since I 0, this is strictly negative if α > 0 is small enough. Q.E.D. Proof of Lemma 3.5. We first construct c = c 1,..., c M V a such that a 3 c M 0 x = 1 a 3 c j x θ 1 a.e. x A.6 for some constant θ 1 0, 1, c c X < ε/, and F a [ c] F a [c] with a strict inequality if {x : a 3 c 0 x < α} > 0 for all α > 0. Denote for any α > 0 ω 0 α = {x : a 3 c 0 x < α}. If there exists a constant α 1 > 0 such that ω 0 α 1 = 0, i.e., a 3 c 0 x α 1 a.e., then c 1,..., c M = c 1,...,c M V a satisfies all the desired properties with θ 1 = α 1. Suppose ω 0 α > 0 for any α > 0. Let 0 < α < 1/4M. Let x ω 0 α. Then there exists some j = jx {1,...,M} such that a 3 c j x 1/M. In fact, if this were not true, then a 3 c i x < 1/M for all i = 1,...,M. Hence, a 3 c 0 x = 1 a 3 M c ix > 1/ > α. This would mean that x ω 0 α, a contradiction. Denoting H j α = { x ω 0 α : a 3 c j x 1 M }, j = 1,...,M, we thus have ω 0 α = M H j α. Since ω 0 α > 0, we have H j1 α > 0 for some j 1 1 j 1 M. If H j α\h j1 α = 0 for all j j 1, then we have ω 0 α = K 1 α H j1 α for some K 1 α ω 0 α with K 1 α = 0. Otherwise, H j α \ H j1 α > 0 for some j j 1. In case ω 0 α \ [H j1 α H j α] = 0, we have ω 0 α = K α H j1 α [H j α \ H j1 α] for some K α ω 0 α with K α = 0. By induction, we see that there exist m {1,...,M}, Km α ω 0 α with K m α = 0, and mutually disjoint sets K j1 α,...,k jm α ω 0 α such that K ji α H ji α and K ji α > 0 for i = 1,...,m, and ω 0 α = K m α [ m K ji α]. By relabeling, we may assume that j i = i for i = 1,...,m. Define now { cj x αa 3 χ Kj αx x, j = 1,...,m, c j x = c j x x, j = m + 1,...,M, [ ] 3 c 0 x = a 3 1 a 3 c j x x. A.7 30

31 It is easy to see that c 1,..., c M V a. Moreover, a 3 c 0 x = a 3 c 0 x + αχ ω0 αx α a.e. x, A.8 implying A.6 with θ 1 = α. Clearly, M c j c j L 1 αa 3 m K jα. Moreover, q j c j H m q j c j αa 3 q j χ Kj α 1 L m αa 3 qj K jα. Therefore, c c X < ε/, provided that α > 0 is small enough. If x K j α for some j with 1 j m, then c j x 1/Ma 3, and c j x 1/4Ma 3 since 0 < α < 1/4M. By the Mean-Value Theorem and the fact that S 1u = log u for any u > 0, there exists η j x with c j x η j x c j x such that S 1 [ c j x] S 1 [c j x] = [ c j x c j x] log η j x αa 3 log c j x αa 3 log 4Ma 3. By the same argument using A.8 and the definition of ω 0 α, we obtain S 1 [ c 0 x] S 1 [c 0 x] αa 3 logα a.e. x ω 0 α. Consequently, we have by 3.13, 3.4, and the embedding L H 1 that F a [ c] F a [c] = 1 M m M m q j c j αa 3 q j χ Kj α L q j c j αa 3 q j χ Kj α dx 1 M M m q j c j L q j c j dx αa 3 µ aj dx m + β 1 1 α a 6 j=0 αa 3 ω 0 α [S 1 c j S 1 c j ]dx m m q j χ Kj α L q j χ Kj α dx K j α m m m q j χ Kj α L q j c j dx + αa 3 µ aj L K j α + β 1 αa 3 logα ω 0 α + β 1 αa 3 log 4Ma 3 ω 0 α m Cα a 6 M L m q j χ Kj α + αa 3 L q j c j q j K j α L m + αa 3 µ aj L K j α + β 1 αa 3 log 8Ma 3 α m K j α 31

32 [ M = α Cαa 6 qj + a 3 q j L L q j c j + a 3 µ aj L + β 1 a 3 log 8Ma 3 α ] K j α, where C > 0 is a constant independent of α. Thus, F a [ c] F a [c] is non-positive for α > 0 sufficiently small. It is strictly negative, if ω 0 α = m K jα > 0 for all α > 0, i.e., if {x : a 3 c 0 x < α} > 0 for all α > 0. We now construct ĉ V a that satisfies.6 with c replaced by ĉ, ĉ c X < ε/, and F a [ĉ] F a [ c] with a strict inequality if there exists j {1,...,M} such that {x : a 3 c j x < α} > 0 for all α > 0, all these implying that ĉ satisfies all the desired properties. If there exists θ 0, 1 such that min 1 j M c j x θ for a.e. x and all j = 1,...,M, then ĉ = c with 0 < α < θ / cf. A.7 satisfies all the desired properties with θ 0 = minθ 1,θ /. Assume otherwise there exists j 0 {1,...,M} such that {x : c j0 x < α} > 0 for all α > 0. This means that {x : c j0 x < α} > 0 for all α > 0. Define σ j α = {x : a 3 c j x < α} α > 0, j = 1,...,M, J 0 = {j {1,...,M} : σ j α > 0 α > 0}, J 1 = {1,...,M} \ J 0. Clearly, J 0. If J 1, then there exists α > 0 such that a 3 c j x α a.e. x, j J 1. Define for 0 < α < min{α,θ 1 /M} and 1 j M Notice by A.6 that { cj x + αa 3 χ σj αx if j J 0 ĉ j x = x. c j x if j J 1 ] ĉ 0 x = a [1 3 a 3 ĉ j x x. a 3 ĉ 0 x = 1 a 3 ĉ j x = 1 a 3 c j x α χ σj α θ 1 αm > 0 a.e. x. j J 0 Thus, ĉ = ĉ 1,...,ĉ M V a. Clearly,.6 holds true for θ 0 = min{α,α,θ 1 αm}. Applying the same argument used above, we obtain that ĉ c X < ε/ for α > 0 small enough. 3

33 We have now by the Mean-Value Theorem that [S 1 ĉ j S 1 c j ]dx = [S 1 ĉ j S 1 c j ] dx j J σ j α 0 αa 3 log αa 3 σ j α. j J 0 Similarly, we have by.6 that [S 1 ĉ 0 S 1 c 0 ] dx αa 3 log a 3 θ 0 j J 0 σ j α. Consequently, we have by 3.13 and a similar argument that F a [ĉ] F a [ c] = 1 M q j c j + αa M 3 q j χ σj α L q j c j + αa 3 q j χ σj α dx j J 0 j J 0 1 M M q j c j L q j c j dx + αa 3 µ aj dx j J 0 + β 1 j J 0 σ j α Cα a 6 q j χ σj α j J 0 σ j α [S 1 ĉ j S 1 c j ]dx + β 1 [S 1 ĉ 0 S 1 c 0 ]dx L M + αa 3 q j c j q j σ j α L s j J 0 + αa 3 j J 0 µ aj L σ j α + β 1 αa 3 logα/θ 0 j J 0 σ j α α L [Cαa 6 qj + a 3 q j q j c j + a 3 µ aj L j J 0 ] + β 1 a 3 logα/θ 0 σ j α. Since J 0, this is strictly negative if α > 0 is sufficiently small. Q.E.D. Acknowledgments. This work was supported by the US National Science Foundation NSF through the grant DMS and DMS , by the NSF Center for Theoretical Biological Physics with the NSF grant PHY-0883, and by the US Department of Energy through the grant DE-FG0-05ER5707. The author thanks Dr. Jianwei Che, Dr. Joachim Dzubiella, and Dr. Benzhuo Lu for helpful discussions on the background of the underlying problem. 33