Phase-Field Variational Implicit Solvation

Phase-Field Variational Implicit Solvation Bo Li Department of Mathematics and Quantitative Biology Graduate Program UC San Diego KI-Net Conference: Mean-Field Modeling and Multiscale Methods for Complex Physical and Biological Systems UC Santa-Barbara October 31 November 3, 2016

Collaborators Modeling and Computation J. Andrew McCammon (UC San Diego) Joachim Dzubiella (Helmholtz Institute, Berlin) Jianwei Che (Parallel Computing Labs, San Diego) Zuojun Guo (Plexxikon, Inc., Berkeley) Li-Tien Cheng (UC San Diego) Shenggao Zhou (Soochow University) Yanxiang Zhao (George Washington University) Hui Sun (UC San Diego & Cal State, Long Beach) Jiayi Wen (Yahoo) Analysis Yanxiang Zhao (George Washington University) Yuan Liu (Fudan University) Shibin Dai (New Mexico State University) Jianfeng Lu (Duke University) Funding: NIH and NSF

Solvation Molecular Modeling: Explicit Solvent vs. Implicit Solvent

Dielectric Boundary Based Implicit-Solvent Models Fixed-surface models solvent excluded surface (SES) probing ball vdw surface solvent accessible surface (SAS) G total = G surf + G ele-pb/gb decoupling fitting parameters cavities curvature correction Hasted, Ritson, & Collie, JCP 1948.

OUTLINE 1. A Variational Model of Biomolecules 2. The Poisson Boltzmann Electrostatics 3. Convergence of Free Energy 4. Convergence of Force 5. Convergence of Interface

1. A Variational Model of Biomolecules Free-energy functional (Dzubiella, Swanson, & McCammon, 2006) F [Γ] = P 0 Vol ( p ) + γ 0 Area (Γ) + ρ 0 U vdw dx + F ele [Γ] w [ F ele [Γ] = ε ] Γ 2 ψ Γ 2 + ρψ Γ χ w B(ψ Γ ) dx U vdw (x) = N i=1 U(i) LJ ( x x i ) U (i) LJ (r) = 4ε i [(σ i /r) 12 (σ i /r) 6] ε Γ ψ Γ χ w B (ψ Γ ) = ρ ψ Γ = ψ on B(s) = k B T M j=1 c j in ( ) e q j s/(k B T ) 1 dielectric boundary w Note: min B = B(0) = 0, B > 0, and B(± ) =. n p Γ ε p=1 ε w=80 xi Q i

Free-energy functional F [Γ] = P 0 Vol ( p ) + γ 0 Area (Γ) + ρ 0 w U vdw dx + F ele [Γ] Boundary force F n : Γ R F n = δ Γ F [Γ] = P 0 2γ 0 H + ρ 0 U vdw δ Γ F ele [Γ] δ Γ F ele [Γ] = 1 2 ( 1 1 ) ε Γ ψ Γ n 2 ε p ε w + 1 2 (ε w ε p ) (I n n) ψ Γ 2 + B(ψ Γ ) In particular, δ Γ F ele [Γ] < 0, since ε w > ε p. (Li, Cheng, & Zhang, SIAP 2011) dielectric boundary w n p Γ ε p=1 ε w=80 xi Q i

pmf G (kb tot T) 40 20 0 20 40 60 80 PS Tight 100 PS Loose 120 PB Tight PB Loose 140 4 6 8 10 12 14 16 18 20 d (Å)

χ A Phase-field model of interface uɛ 1 0 The van der Waals Cahn Hilliard functional [ ξ E ξ [φ] = 2 φ 2 + 1 ] ξ W (φ) dx W (φ) = 18φ 2 (φ 1) 2 Why phase-field modeling? An alternative numerical method of implementation. Proof of existence of a minimizer for the sharp-interface free-energy functional. Connection to the Lum Chandler Weeks (LCW) theory of solvation (J. Phys. Chem. B 1999). Description of both bulk and interfacial fluctuations.

Phase field φ : w = {φ 0}, p = {φ 1}, and Γ = {φ 1/2}; Dielectric coefficient ε = ε(φ) C 1 (R): ε(φ) = { εw if φ 0 ε p if φ 1 and ε (φ) > 0 if 0 < φ < 1. Phase-field free-energy functional F ξ [φ] = P 0 φ 2 dx + γ 0 [ ξ 2 φ 2 + 1 ] ξ W (φ) dx + ρ 0 (φ 1) 2 U vdw dx + F ele [φ] [ F ele [φ] = ε(φ) ] 2 ψ φ 2 + ρψ φ (φ 1) 2 B(ψ φ ) dx ε(φ) ψ φ (φ 1) 2 B (ψ φ ) = ρ ψ φ = ψ on in

2. The Poisson Boltzmann Electrostatics F ele [φ] = [ ε(φ) ] 2 ψ φ 2 + ρψ φ (φ 1) 2 B(ψ φ ) dx ε(φ) ψ φ (φ 1) 2 B (ψ φ ) = ρ in (1) ψ φ = ψ on (2) Define A = { u H 1 () : u = ψ on }, [ ] ε(φ) E φ [u] = 2 u 2 ρu + (φ 1) 2 B(u) dx. Theorem 1. For any φ L 4 (), there exists a unique weak solution ψ φ A to (1) and (2) with ψ φ L () C. Moreover, F ele [φ] = E φ [ψ φ ] = min u A E φ[u].

Proof. Direct methods in the calculus of variations and standard regularity theory. Key: the L -bound. [ ε Let u = argmin H 1 0 () I [ ], where I [v] = 2 v 2 + B(v)] dx. Let λ > 0 with B (λ) > 1 and B ( λ) < 1. Define λ if u(x) < λ, u λ (x) = u(x) if u(x) λ, λ if u(x) > λ. Then I [u λ ] I [u] and u λ u a.e.. By the convexity of B, 0 χ {u>λ} [B(u) B(λ)] dx + χ {u< λ} [B(u) B( λ)] dx χ {u>λ} B (λ)(u λ) dx + χ {u< λ} B ( λ)(u + λ) dx = χ { u >λ} u λ dx 0. This implies that u λ a.e.. Q.E.D.

Theorem 2. Assume sup k 1 φ k L 4 () <, φ k φ in L 1 (), ψ φk = argmin A E φk [ ] (k = 1, 2,... ), and ψ φ = argmin A E φ [ ]. Then ψ φk ψ φ in H 1 () and E φk [ψ φk ] E φ [ψ φ ]. Proof. It suffices to show that any subsequence of {ψ φk } has a further subsequence that converges to ψ φ in H 1 (). With bounds, {ψ φk } has a subsequence that converges to ˆψ weakly in H 1 (), strongly in L 2 (), and a.e. in. Then ˆψ = ψ φ. Further, use the weak formulations for ψ φk and ψ φ to show that ψ φk ψ φ in H 1 () and E φk [ψ φk ] E φ [ψ φ ]. Q.E.D.

Now consider the dielectric boundary force. [ F ele [φ] = ε(φ) ] 2 ψ φ 2 + ρψ φ (φ 1) 2 B(ψ φ ) dx ε(φ) ψ φ (φ 1) 2 B (ψ φ ) = ρ ψ φ = ψ on in Heuristics: use δφ as the test function in the weak form of PBE. [ δ φ F ele [φ]δφ = ε (φ) 2 δφ ψ φ 2 ε(φ) ψ φ δ φ ψ φ + ρδ φ ψ φ Hence, 2(φ 1)δφB(ψ φ ) (φ 1) 2 B (ψ φ )δ φ ψ φ ]dx [ ] = ε (φ) 2 ψ φ 2 2(φ 1)B(ψ φ ) δφ dx. δ φ F ele [φ] = ε (φ) 2 ψ φ 2 2(φ 1)B(ψ φ ).

Define for any φ L 4 () [ ] 1 f ele (φ) = 2 ε (φ) ψ φ 2 + 2(φ 1)B(ψ φ ) φ, [ ] ε(φ) T ele (φ) = ε(φ) ψ φ ψ φ 2 ψ φ 2 + (φ 1) 2 B(ψ φ ) I. If φ W 1, (), then we have by direct calculations that f ele (φ) = T ele (φ) + ρ ψ φ. If G is open, G, and G C 2 with unit outer normal ν, then we define [ f 0,ele [ G] = 1 ( 1 1 ) ε(χ G ) ψ χg ν 2 2 ε p ε w 1 ] 2 (ε w ε p ) (I ν ν) ψ χg 2 B(ψ χg ) ν.

Lemma. If φ W 1, and V Cc 1 (, R 3 ), then f ele (φ) V dx = [T ele (φ) : V ρ ψ φ V ] dx. If φ = χ G with G open, G, C 2, then [ f 0,ele [ G] V ds = Tele (χ G ) : V ρ ψ χg V ] dx. Proof. By direct calculations using the fact that ψ φ is the solution to the BVP of PBE. In particular, for φ = χ G, ε p ψ χg = ρ in G, ε w ψ χg + B (ψ) = ρ in G c, ψ χg G = ψ χg G c ε p ψ χg G ν = ε w ψ χg G c ν Moreover, notice that on G, on G. ψ χg = ( ψ χg ν)ν + (I ν ν) ψ χg on G. Q.E.D.

Theorem 3. If sup k 1 φ k L 4 () < and φ k φ a.e., then [ Tele (φ k ) : V ρ ψ φk V ] dx lim k = [ Tele (φ) : V ρ ψ φ V ] dx V C 1 c (, R 3 ). Proof. The key step is to prove ε(φ k ) ψ φk ψ φk dx = lim k Notice that, with ψ k = ψ φk and ψ = ψ φ, ε(φ) ψ φ ψ φ dx. ε(φ k ) ψ k ψ k = ε(φ k ) [( ψ k ψ) ( ψ k ψ) + ψ ( ψ k ψ) + ( ψ k ψ) ψ + ψ ψ]. The result follows from ψ φk ψ φ in H 1 (), φ k φ a.e. in, and the Lebesgue Dominated Convergence Theorem. Q.E.D.

Define 3. Convergence of Free Energy F 0 [φ] = P 0 A + γ 0 P (A) + ρ 0 \A U vdw dx + F ele [φ]. Theorem 4. Let ξ k 0. Then Γ-lim k F ξk = F 0 with respect to L 1 ()-convergence. (1) The liminf condition. If φ k φ in L 1 () then lim inf k F ξ k [φ k ] F 0 [φ]. (2) The recovering sequence. For any φ L 1 (φ), there exist φ k L 1 () (k = 1, 2,... ) such that φ k φ in L 1 () and lim sup F ξk [φ k ] F 0 [φ]. k Corollary. There exists G with P (G) < such that F 0 [χ G ] = min φ L 1 () F 0 [φ], which is finite.

Proof of Theorem 4. (1) Assume φ k φ a.e. in and {F ξk [φ k ]} converges. Then, [ ξk 2 φ k 2 + 1 ] W (φ k ) dx <. ξ k sup k 1 Hence, φ = χ G and φ k χ G in L 2 (). Note that η k (x) := φk (x) 0 2W (t) dt χg a.e., η k = 2W (φ k ) φ k, and sup k 1 η k W 1,1 () <. Hence, P (G) lim inf η k dx k = lim inf 2W (φk ) φ k dx k lim inf k [ ξk 2 φ k 2 + 1 ] W (φ k ) dx. ξ k

(2) Assume F 0 [φ] <. So, φ = χ G BV () with G. Step 1. Assume G = D with D open, D in C, D in C 2, and H 2 ( D ) = 0. There exist φ k H 1 () such that 0 φ k χ G in, φ k = 1 in G k := {x G : dist(x, G) } ξ k, φ k = 0 in \ G, φ k χ G strongly in L 1 () and a.e. in, [ ξk lim sup k 2 φ k 2 + 1 ] W (φ k ) dx P (G). ξ k There exist r 0 > 0 such that N i=1 B(x i, r 0 ) G k G for all k 1. {φ k } is a recovering sequence.

Step 2. Assume φ = χ G with 0 < G < and P (G) <. Choose σ k 0 such that B(σ k ) := N i=1 B(x i, σ k ), U 0 on B(σ k ), and 0 < G B(σ k ) < for all k 1. Set Ĝ k = G B(σ k ). Then Ĝk+1 Ĝk and χĝk χ G in L 1 (). So, lim sup F 0 [χĝk ] F 0 [χ G ]. k Fix k 1. There exist open D k,j R 3 (j = 1, 2,... ) satisfying: G k,j := D k,j B(σ k /2); D k,j is C, D k,j is C 2, and H 2 ( D k,j ) = 0; G k,j Ĝk 0; Then lim j F 0[χ Gk,j ] = F 0 [χĝk ], k = 1, 2,...

There exist H k = G k,jk such that for all k = 1, 2,... χ Hk χĝk L 1 () < 1/k and F 0 [χ Hk ] F 0 [χĝk ] < 1/k. Since χĝk χ G in L 1 (), we further have lim χ H k χ G L k 1 () = 0 and lim sup F 0 [χ Hk ] F 0 [χ G ]. k By Step 1, we can find for each k 1 a recovering sequence {φ k,l } l=1 for χ H k. Finally, a subsequence {φ k,lk } is a recovering sequence for χ G. Q.E.D.

Theorem 5. If φ k χ G a.e., and F ξk [φ k ] F 0 [χ G ] R, then lim φ 2 kdx = G, k [ ξk lim k 2 φ k 2 + 1 ] W (φ k ) dx = P (G), ξ k (φ k 1) 2 U vdw dx = U vdw dx, lim k \G lim F ele[φ k ] = F ele [χ G ]. k Proof. We have lim k a k = a and lim k b k = b, provided that lim (a k+b k ) = a+b, lim inf a k a 0 and lim inf b k b 0. k k k We have also that [ ξk lim inf k 2 φ k 2 + 1 ] W (φ k ) dx P (G), ξ k lim inf (φ k 1) 2 U vdw dx χ \G U vdw dx. Q.E.D. k {U vdw >0} {U vdw >0}

Phase-field forces 4. Convergence of Force f vol (φ) = 2P 0 φ φ, f ξ,sur (φ) = γ 0 [ ξ φ + 1 ] ξ W (φ) φ, f vdw (φ) = 2ρ 0 (φ 1)U vdw φ, in \ {x 1,..., x N } [ ] f ele (φ) = ε (φ) 2 ψ φ 2 2(φ 1)B(ψ φ ) φ. Phase-field stresses T vol (φ) = P 0 φ 2 I, {[ ξ T ξ,sur (φ) = γ 0 2 φ 2 + 1 ] } ξ W (φ) I ξ φ φ, T vdw (φ) = ρ 0 (φ 1) 2 U vdw I, T ele (φ) = ε(φ) ψ φ ψ φ [ ] ε(φ) 2 ψ φ 2 + (φ 1) 2 B(ψ φ ) I.

Lemma. We have for almost all points in that f vol (φ) = T vol (φ) if φ H 1 (), f ξ,sur (φ) = T ξ,sur (φ) if φ H 2 (), f vdw (φ) = T vdw (φ) ρ 0 (φ 1) 2 U vdw if φ H 1 (), f ele (φ) = T ele (φ) + ρ ψ φ if φ W 1, (). Moreover, we have for any V Cc 1 (, R 3 ) that f vol (φ) V dx = T vol (φ) : V dx if φ H 1 (), f ξ,sur (φ) V dx = T ξ,sur (φ) : V dx if φ H 2 (), [ f vdw (φ) V dx = TvdW (φ) : V + ρ 0 (φ 1) 2 U vdw V ] dx if {x 1,..., x N } supp (V ) = and φ H 1 (), f ele (φ) V dx = [T ele (φ) : V ρ ψ φ V ] dx if φ W 1, ().

Sharp-interface forces: Let G be open with G, G in C 2, x i G (i = 1,..., N), and ν the unit outer normal of G. Define f 0,vol [ G] = P 0 ν, f 0,sur [ G] = 2γ 0 Hν, f 0,vdW [ G] = ρ 0 U vdw ν, [ f 0,ele [ G] = 1 2 ( 1 ε p 1 ε w ) ε(χ G ) ψ χg ν 2 1 ] 2 (ε w ε p ) (I ν ν) ψ χg 2 B(ψ χg ) ν.

Lemma. We have for any V Cc 1 (, R 3 ) that f 0,vol [ G] V ds = T vol (χ G ) : V dx, G f 0,sur [ G] V ds = γ 0 (I ν ν) : V ds, G G f 0,vdW [ G] V ds = [T vdw (χ G ) : V G G +ρ 0 (1 χ G ) 2 U vdw V ] dx if {x 1,..., x N } supp (V ) =, f 0,ele [ G] V ds = [T ele (χ G ) : V ρ ψ χg V ] dx.

Theorem 6. Let G, P (G) <, F 0 [χ G ] is finite, φ k χ G a.e. in, and F ξk [φ k ] F 0 [χ G ]. Then, for any V Cc 1 (, R 3 ), lim T vol (φ k ) : V dx = T vol (χ G ) : V dx, k lim T ξk,sur(φ k ) : V dx = γ 0 (I ν ν) : V dh 2, k G [ (TvdW (φ k ) : V + ρ 0 (φ k 1) 2 U vdw V ] dx lim k = [ TvdW (χ G ) : V + ρ 0 (χ G 1) 2 U vdw V ] dx if {x 1,..., x N } supp (V ) =, lim [T ele (φ k ) : V ρ ψ φk V ] dx k = [T ele (χ G ) : V ρ ψ χg V ] dx. Proof. Each component of energy converges. Then, the key is the convergence of the surface energy. Q.E.D.

Force convergence for the van der Waals Cahn Hilliard functional. Theorem 7. Let R n be bounded and open. Let G, P (G) <, φ k χ G a.e. in, and [ ξk lim k 2 φ k 2 + 1 ] W (φ k ) dx = P (G). ξ k Then we have for any Ψ C c (, R n n ) that T ξk (φ k ) : Ψ dx = (I ν ν) : Ψ dh n 1. lim k If, in addition, all φ k W 2,2 (), G is open, and G is of C 2, then [ lim ξ k φ k + 1 ] W (φ k ) φ k V dx k ξ k = (n 1) Hν V ds V Cc 1 (, R n ). G G

Remark. Energy convergence is necessary. Proof. Asymptotic equipartition of energy: lim k lim k 2 ξk 2 φ W (φ k ) k dx = 0, ξ k k ξ 2 φ k 2 1 W (φ k ) ξ k dx = 0. Need only to prove that for any Ψ C c (, R n n ) T ξk (φ k ) : Ψ dx = (I ν ν) : Ψ dh n 1. lim k G

If suffices to prove that for any Ψ C c (, R n n ) ξ k φ k φ k : Ψ dx = ν ν : Ψ dh n 1. lim k G Assume this is true. Notice for any a R n that a 2 = a a : I. Then (I : Ψ)I C c (, R n n ). Hence, [ ξk lim k 2 φ k 2 + 1 ] W (φ k ) I : Ψ dx ξ k = lim ξ k φ k 2 I : Ψ dx k = lim ξ k φ k φ k : (I : Ψ)I dx k = ν ν : (I : Ψ)I dh n 1 G = I : Ψ dh n 1. G

Fix Ψ C c (, R n n ) and prove now ξ k φ k φ k : Ψ dx = lim k G ν ν : Ψ dh n 1. Let σ > 0. Recall that G = ( j=1 K j) Q, where K j s are disjoint compact sets, each being a subset of a C 1 -hypersurface S j, and Q G with G (Q) = 0. Moreover, j=1 Hn 1 (K j ) = H n 1 ( G) = G () = P (G) <. Choose J so that j=j+1 Hn 1 (K j ) < σ. Choose open U j such that K j U j U j (j = 1,..., J). For each j, we define d j : U j R to be the signed distance to S j, and zero-extend d j to \ U j. Choose ζ j Cc 1 () be such that 0 ζ j 1 on, ζ j = 1 in a neighborhood of K j, supp (ζ j ) U j, and ζ j d j C c (, R n ). Define ν J : R n by ν J = J j=1 ζ j d j. Note that ν j C c (, R n ), ν j 1 on, and ν j = ν on each K j (1 j J).

We rewrite ξ k φ k φ k as ( ξ k φ k φ k = ξk φ k + ) ξ k φ k ν J ξ k φ k ( 2W (φ k ) + ) ξ k φ k ν J ξ k φ k ξ k ν J 2W (φ k ) φ k. Finally, lim sup ξ k φ k φ k : Ψ dx ν ν : Ψ dh n 1 k G ( 4σ sup ) L ξ k φ k Ψ L 2 () () + 2σ Ψ L (). k 1 This completes the proof. Q.E.D.

5. Convergence of Interface Matched asymptotic analysis of the relaxation dynamics t φ = δ φ F ξ [φ] : t φ = 2P 0 φ + γ 0 [ξ φ 1 ] ξ W (φ) 2ρ 0 (φ 1)U vdw + 1 2 ε (φ) ψ φ 2 + 2(φ 1)B(ψ φ ), ε(φ) ψ φ (φ 1) 2 B (ψ φ ) = ρ, φ = 0 ψ = ψ on, on. Fast time scale. Instantaneous electrostatic relaxation. Perturb the unstable equilibrium φ 0 (x) = 1/2 of t φ = ξ φ 1 ξ W (φ) The most unstable mode is k c = 0. The fastest growth rate is ω(k c ) = O(1/ξ). So, the fast time scale is T = t/ξ.

Fast time scale. Assume T = t/ξ and φ(x, t) = φ 0 (x, T ) + ξφ 1 (x, T ) + ξ 2 φ 2 (x, T ) +, ψ(x, t) = ψ 0 (x, T ) + ξψ 1 (x, T ) + ξ 2 ψ 2 (x, T ) +, φ i = 0 (i 0), ψ 0 = ψ, and ψ i = 0 (i 1) on. Plug these into the governing equations and group terms with the same order of ξ to get T φ 0 = γ 0 W (φ 0 ) : x, φ 0 (x, T ) 0 or 1 exponentially as T, ε(φ 0 ) ψ 0 (φ 0 1) 2 B (ψ 0 ) = ρ. Conclusion: Quickly, the system is partitioned into two regions of different dielectric coefficient, and the electrostatics is relaxed.

Regular time scale. Assume φ(x, t) = φ 0 (x, t) + ξφ 1 (x, t) + ξ 2 φ 2 (x, t) +, ψ(x, t) = ψ 0 (x, t) + ξψ 1 (x, t) + ξ 2 ψ 2 (x, t) +, φ i = 0 (i 0), ψ 0 = ψ, and ψ i = 0 (i 1) on. Leading order equations W (φ 0 ) = 0, t φ 0 = 2P 0 φ 0 γ 0 W (φ 0 )φ 1 2ρ 0 (φ 0 1)U vdw + 1 2 ε (φ 0 ) ψ 0 2 + 2(φ 0 1)B(ψ 0 ), ε(φ 0 ) ψ 0 (φ 0 1) 2 B (ψ 0 ) = ρ. Consequences: (1) φ 0 = 0 or 1; (2) {φ 0 = 0} by the BC; and (3) {φ 0 = 1} by the property of U vdw and second equation. Conclusion: These expansions do not hold in the entire region.

Assumptions The entire region is the union of an outer and inner region. The outer region is the union of + ξ (t) := {φ(x, t) 1} and (t) := {φ(x, t) 0}. ξ The inner region is a thin layer of width O(ξ) that centers around Γ ξ (t) = {x : φ(x, t) = 1/2} enclosing + ξ (t). This interface is O(ξ)-close to a smooth interface Γ(t) that is independent of ξ. The inner and outer solutions match asymptotically.

Local coordinates for a point x in the inter region: x = Φ(y, t) + ξz(t)n(y, t). Φ(, t) : Q(t) Γ ξ (t) a smooth, principal-curvature parameterization, and Q(t) R 3 is an O(ξ)-neighborhood of Γ ξ (t) and Γ(t). z(t) = s(x, t)/ξ with s(x, t) = ±dist (x, Γ ξ (t)). n(y, t) = x s(x, t) is the unit normal to Γ ξ (t) at Φ(y, t). The normal velocity is v(y, t) = t Φ(y, t) n(y, t). The mean curvature is H(y, t) = y n(y, t)/2.

Assume the inner expansions φ(x, t) = φ 0 (z, y, t) + ξ φ 1 (z, y, t) + ξ 2 φ2 (z, y, t) +, ψ(x, t) = ψ 0 (z, y, t) + ξ ψ 1 (z, y, t) + ξ 2 ψ2 (z, y, t) +. Leading order equations: z ψ 0 = 0 and v z φ 0 = γ 0 [ zz 2 φ ] 0 W ( φ 0 ). Matching conditions: φ 0 ( ) = 1, φ 0 ( ) = 0, and z φ 0 (± ) = 0. Hence ( ) [ ] 2 1 v z φ 0 dz = γ0 2 ( z φ z= 0 ) 2 W ( φ 0 ) = 0. z= Therefore v = 0. Conclusion: To the leading order, the interface does not move at this time scale.

Slow time scale: τ = ξt. Outer expansions W (φ 0 ) = 0, Use the local coordinate ε(φ 0 ) ψ 0 (φ 0 1) 2 B (ψ 0 ) = ρ. x = Φ(y, τ) + ξz(τ)n(y, τ) to assume the local inner expansions φ(x, t) = φ 0 (z, y, τ) + ξ φ 1 (z, y, τ) + ξ 2 φ 2 (z, y, τ) +, ψ(x, t) = ψ 0 (z, y, τ) + ξ ψ 1 (z, y, τ) + ξ 2 ψ 2 (z, y, τ) +. Also, expand U vdw (x) = U vdw (Φ(y), τ) + O(ξ) = Ũ 0 (y, τ) + O(ξ).

Leading order equations 0 = z ψ0, 0 = 2 zz φ 0 W ( φ 0 ), v z φ 0 = 2P 0 φ 0 + γ 0 [ 2H φ 0 + 2 zz φ 1 W ( φ 0 ) φ 1 ] 2ρ 0 ( φ 0 1)Ũ0 + 1 2 ε ( φ 0 ) [ ] z ε( φ 0 ) z ψ 1 = 0. Matching conditions lim φ 0 (z, y, τ) = φ ± z ± 0 (x, τ), ψ 0 (z, y, τ) = ψ 0 ± (x, τ), lim z ± ψ 1 (z, y, τ) = ( ψ ± 1 + z ψ± 0 s) (x, τ) + o(1) We have then φ 0 (z, y, τ) = φ 0 (z) = 0.5 [1 + tanh(3z)], z ψ 1 (z, y, τ) = ψ 0 ± (x, τ) s(x, τ) + o(1) [ Γ(τ) ψ0 2 + ( z ψ1 ) 2] + 2( φ 0 1)B( ψ 0 ), as z ±. as z ±.

First and fourth equations imply ψ 0 = ψ+ 0 and ε w ψ 0 n = ε p ψ + 0 n on Γ(τ). Thus, ψ 0 is a solution to the BVP of PBE. Multiply the v equation by z ψ 0 and integrate against z to get vs = S := R 1 (z) φ 0(z)dz + [ ] 2 φ 0(z) dz = 1, R 2 (z) φ 0(z)dz + R 1 (z) := 2P 0 φ0 2ρ 0 ( φ 0 1)Ũ0 + 2( φ 0 1)B( ψ 0 ), [ R 2 (z) := γ 0 2H z φ 0 + zz 2 φ ] 1 W ( φ 0 ) φ 1, R 3 (z) := 1 [ 2 ε ( φ 0 ) Γ(τ) ψ 0 2 + ( z ψ 1 ) 2]. R 3 (z) φ 0(z)dz,

Integration by parts and matching conditions imply R 1 (z) φ 0(z) dz = P 0 + ρ 0 U vdw B(ψ 0 ), R 2 (z) φ 0(z)dz = 2γ 0 H. Using in addition the interface conditions for ψ 0 on Γ(τ) to obtain Finally, the normal velocity is R 3 (z) φ 0(z) dz = 1 2 (ε p ε w ) Γ(τ) ψ 0 2 + 1 ( 1 1 ) ε 2 ε w ε Γ(τ) ψ 0 n 2. p v = P 0 2γ 0 H ρ 0 U vdw + 1 2 + 1 2 (ε w ε p Γ(τ) ψ 0 2 + B(ψ 0 ). ( 1 1 ) ε ε p ε Γ(τ) ψ 0 n 2 w Exactly the same as in the sharp-interface model!

References [1] B. Li, X. Cheng, & Z. Zhang, Dielectric boundary force in molecular solvation with the Poisson Boltzmann free energy: A shape derivative approach, SIAM J. Applied Math., 71, 2093 2111, 2011. [2] B. Li & Y. Zhao, Variational implicit solvation with solute molecular mechanics: From diffuse-interface to sharp-interface models, SIAM J. Applied Math., 73, 1 23, 2013. [3] Y. Zhao, Y.-Y. Kwan, J. Che, B. Li, & J. A. McCammon, Phase-field approach to implicit solvation of biomolecules with Coulomb-field approximation, J. Chem. Phys., 139, 024111, 2013. [4] B. Li & Y. Liu, Diffused solute-solvent interface with Poisson Boltzmann electrostatics: Free-energy variation and sharp-interface limit, SIAM J. Applied Math, 75, 2072 2092, 2015. [5] H. Sun, J. Wen, Y. Zhao, B. Li, & J. A. McCammon, A self-consistent phase-field approach to implicit solvation of charged molecules with Poisson Boltzmann electrostatics, J. Chem. Phys., 143, 243110, 2015. [6] S. Dai, B. Li, & J. Liu, Convergence of phase-field free energy and boundary force for molecular solvation, 2016 (submitted).

Thank you!