The Poisson Boltzmann Theory and Variational Implicit Solvation of Biomolecules

Transcription

1 The Poisson Boltzmann Theory and Variational Implicit Solvation of Biomolecules Bo Li Department of Mathematics and NSF Center for Theoretical Biological Physics (CTBP) UC San Diego FUNDING: NIH, NSF, and CTBP PDE/Applied Math Seminar Department of Mathematics, UC Santa Barbara March 5, 2013

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3 Solvation protein folding host-guest system

4 Molecular Modeling: Explicit vs. Implicit Molecular Dynamics (MD) Simulations Statistical Mechanics P(X,Y) = P 0 e U(X,Y)/k BT U(X,Y) = U uu (X)+U vv (Y)+U uv (X,Y) P(X) = P(X,Y)dY = P 0 e W(X)/k BT W(X) : Potential of Mean Force

5 A Variational Implicit-Solvent Model (VISM) Dzubiella, Swanson, & McCammon, PRL & JCP, n dielectric boundary Ω w Ω m ε m=1 ε =80 w Γ x Q i i The level-set method: V n = δ Γ G total [Γ] Free-energy functional G total [Γ] = P Vol(Ω m )+γ 0 (1 2τH)dS Γ +ρ w U vdw dv +G ele [Γ], Ωw N where U vdw (x) = U (i) LJ ( x x i ). i=1 δ Γ G total [Γ] = P +2γ 0 (H τk) ρ w U vdw +δ Γ G ele [Γ] This talk: the PDE aspect of G ele [Γ] and δ Γ G ele [Γ].

6 JCP 2007 & 2009, JCTC 2009, 2012, & 2013, PRL 2009, J. Comput. Phys

7 OUTLINE 1. The Poisson Boltzmann Equation 2. A Size-Modified Mean-Field Theory 3. Dielectric Boundary Force 4. Motion of a Cylindrical Dielectric Boundary 5. Discussions

8 1. The Poisson Boltzmann Equation (PBE) M ε ψ + q j cj e βqjψ = f j=1 Poisson s equation: Charge density: Boltzmann distributions: ε(x) ψ(x) = ρ(x) ρ(x) = f(x)+ M j=1 q jc j (x) c j (x) = c j e βq jψ(x) ε: dielectric coefficient f : Ω R: given, fixed charge density c j : Ω R: concentration of jth ionic species cj : constant bulk concentration of jth ionic species q j = Z j e: charge of jth ionic species β: inverse thermal energy

9 M PBE ε ψ + q j cj e βqjψ = f j=1 The linearized PBE (the Debye Hückel approximation) ε ψ εκ 2 ψ = f Here κ > 0 is the ionic strength or the inverse Debye screening length: κ 2 = β M i=1 q2 j c j ε Derivation : use the Taylor expansion and Electrostatic neutrality: M j=1 q jcj = 0 The sinh PBE for 1:1 salt (q 2 = q 1 = q, c 2 = c 1 = c ) ε ψ 2qc sinh(βqψ) = f

10 [ ε I[φ] = Ω 2 φ 2 fφ+β 1 M j=1 ] cj e βq jφ dv Theorem (Li, Cheng, & Zhang, SIAP 2011). The functional I : H 1 g(ω) R has a unique minimizer ψ H 1 (Ω) L (Ω) which is also the unique solution to the PBE. Proof. Step 1. By shifting, one may assume f = g = 0. Step 2. Existence and uniqueness by the direct method. Step 3. Key: The L -bound. Let λ 1 and define λ if ψ(x) < λ, ψ λ (x) = ψ(x) if ψ(x) λ, λ if ψ(x) > λ. Then ψ λ is also a minimizer. Uniqueness = ψ = ψ λ. Step 4. Routine calculations. Q.E.D.

11 Electrostatic free-energy functional 1 M [ G[c] = ρψ +β 1 c j ln(λ 3 c j ) 1 ] M µ j c j 2 dv Ω j=1 ρ(x) = f(x)+ M j=1 q jc j (x) ε ψ = ρ and ψ = 0 on Ω Equilibrium conditions (δg[c]) j = q j ψ+β 1 ln(λ 3 c j ) µ j = 0 Boltzmann distributions Minimum electrostatic free-energy (note the sign!) G min = ε M ( 1) 2 ψ 2 +fψ β 1 e βqjψ dv Ω j=1 c j j=1

12 G[c] = Ω 1 2 ρψ +β 1 Theorem (B.L. SIMA 2009). M [ c j ln(λ 3 c j ) 1 ] M µ j c j dv j=1 G has a unique minimizer c = (c 1,...,c M ). Moreover, θ 1 > 0, θ 2 > 0 : θ 1 c j (x) θ 2 a.e. x Ω, j = 1,...,M. Boltzmann distributions: c j (x) = c j e βq jψ(x), j = 1,...,M. j=1 The potential ψ is the unique solution to the PBE. Proof. By the direct method in the calculus of variations, using: the convexity of G; the lower boundedness of s s(logs +α) with α R; the superlinearity of s slogs; and a lemma (cf. next slide). Q.E.D.

13 G[c] = Ω 1 2 ρψ +β 1 M [ c j ln(λ 3 c j ) 1 ] M µ j c j dv j=1 ρ(x) = f(x)+ M j=1 q jc j (x) ε ψ = ρ and ψ = 0 on Ω Lemma (B.L. SIMA 2009). Given c. There exists ĉ satisfying: ĉ is close to c in L 1 (Ω) H 1 (Ω); G[ĉ] G[c]; θ 1 > 0, θ 2 > 0 : θ 1 ĉ j (x) θ 2 a.e. x Ω, j = 1,...,M. Proof. By construction using the fact that the entropic change is very large for c j 0 and c j 1. Q.E.D. slns j=1 O s

14 2. A Size-Modified Mean-Field Theory G[c] = Ω 1 2 ρψ +β 1 M [ c j ln(a 3 j c j ) 1 ] M µ j c j dv j=0 ρ(x) = f(x)+ M j=1 q jc j (x) ε ψ = ρ and ψ = 0 on Ω [ c 0 (x) = a0 3 1 ] M i=1 a3 i c i(x) Theorem (B.L. Nonlinearity 2009). G has a unique minimizer (c 1,...,c M ) characterized by Bounds: There exist θ 1,θ 2 (0,1) such that θ 1 a 3 jc j (x) θ 2 Equilibrium conditions (i.e.,δg[c] = 0) j=1 x Ω j = 0,1,...,M; a j 3 a 0 3 log ( a 3 0c 0 ) log ( a 3 j c j ) = β(qj ψ µ j ) j = 1,...,M, which determine uniquely c j = c j (ψ) (j = 1,...,M).

15 ( aj a 0 ) 3 log ( a 3 0c 0 ) log ( a 3 j c j ) = β(qj ψ µ j ) j = 1,...,M. The general case: Implicit Boltzmann distributions Set D M = {u ( = (u 1,...,u M ) R M : u j > 0, j = 0,1,...,M} u 0 = a0 3 1 ) M j=1 a3 j u j f j (u) = a 3 j a 3 0 log ( a 3 0 u 0 ) log ( a 3 j u j ), j = 1,...,M. Lemma. The map f : D M R M is C and bijective. Proof. It is clear that f is C. f is injective. det f 0, use det(i +v w) = 1+v w. f is surjective. Note: f j (u) = r j j z = z/ u j = 0, where z(u) = M [ ( ) ] M u j log a 3 j u j 1 + r j u j. j=0 Construction: min DM z < min DM z. So all j z = 0. Q.E.D. j=1

16 Set g = (g 1,...,g M ) = f 1 : R M D M B j (φ) = g j (β(q 1 φ µ 1 ),...,β(q M φ µ M )) [ B 0 (φ) = a0 3 1 ] M j=1 a3 j B j(φ) M φ Define B(φ) = q j B j (ξ)dξ φ R Assume j=1 0 M j=1 q jb j (0) = 0 (electrostatic neutrality) Lemma. The function B is strictly convex. Moreover, M > 0 if φ > 0, B (φ) = q j B j (φ) = 0 if φ = 0, j=1 < 0 if φ < 0, and B(φ) > B(0) = 0 for all φ 0. Proof. Direct calculations using the Cauchy Schwarz inequality to show B > 0. Also, use the neutrality. Q.E.D. B o ψ

17 Computational Validation Zhou, Wang, & Li, PRE, Wen, Zhou, Xu, & Li, PRE, Concentration of counterion (M) z 1 =+1, R 1 =3.0, N 1 =100 z 2 =+2, R 2 =2.5, N 2 =100 z 3 =+3, R 3 =3.5, N 3 = Distance to the charged surface (Å) Radial particle density (M) z 1 =+1, R 1 =3.0, N 1 =100 z 2 =+2, R 2 =2.5, N 2 =100 z 3 =+3, R 3 =3.5, N 3 = Distance to the charged surface (Å) The stratification of counterions near a highly charged surface determined by valence-to-volume ratios. Left: Mean-field theory. Right: Monte Carlo simulations.

18 3. Dielectric Boundary Force (DBF) Dielectric coefficient { εm in Ω m ε Γ = ε w in Ω w Typical: ε m = ε 0 and ε w = 80ε 0. Write G = G ele. n dielectric boundary Ω w Ω m ε m=1 ε =80 PBE ε Γ ψ χ w B (ψ) = f [ PB free energy G[Γ] = ε ] Γ 2 ψ 2 +fψ χ w B(ψ) dv Ω B(ψ) = β 1 M j=1 c j ( ) e βqjψ 1 Γ = ε Γ = ψ = ψ Γ = G[Γ] = F n := δ Γ G[Γ] w Γ x Q i i

19 Let V C c (R 3,R 3 ). Define x : [0, ) R 3 R 3 by {ẋ = V(x) for t > 0, x(0,x) = X. Then T t (X) := x(t,x) X +tv(x) if t 1. Define G[Γ t ] G[Γ] δ Γ,V G[Γ] = lim = w(x)[v(x) n(x)]ds X t 0 t for some w : Γ R by the Structure Theorem. Shape derivative δ Γ G[Γ](X) = w(x) Γ X Γ

20 Theorem (Li, Cheng, & Zhang, SIAP 2011). Let n point from Ω m to Ω w and f H 1 (Ω). Let ψ be the unique solution to the boundary-value problem of PBE. Then δ Γ G[Γ] = 1 2 ( 1 ε m 1 ε w ) ε Γ ψ n (ε w ε m ) (I n n) ψ 2 +B(ψ). Corollary. If ε w > ε m, then δ Γ G[Γ] < 0. n dielectric boundary Ω w Ω m B. Chu, Molecular Forces Based on the Baker Lectures of Peter J. W. Debye, John Wiley & Sons, 1967: ε m=1 ε =80 Under the combined influence of electric field generated by solute charges and their polarization in the surrounding medium which is electrostatic neutral, an additional potential energy emerges and drives the surrounding molecules to the solutes. w Γ x Q i i

21 Proof of Theorem. Let V C c (R 3,R 3 ) be local, Γ 0 = Γ, and G[Γ t ] = G[Γ t,ψ t ] = max G[Γ t,φ]. φ Hg(Ω) 1 Note ψ 0 = ψ. Define z(t,φ) = G[Γ t,φ T 1 t ]. We have Step 1. Easy to verify that z(t,ψ 0 ) z(0,ψ 0 ) t Hence G[Γ t ] = max z(t,φ). φ Hg(Ω) 1 G[Γ t] G[Γ] t t z(ξ,ψ 0 ) G[Γ t] G[Γ] t z(t,ψ t T t ) z(0,ψ t T t ). t t z(η,ψ t T t ), ξ,η [0,t].

22 Step 2. Direct calculations lead to [ t z(t,φ) = ε Γ 2 A (t) φ φ+(( (fv)) T t )φj t Ω χ w B(φ)(( V) T t )J t ]dv. Replacing t by η and φ by ψ t T t, respectively, we obtain lim tz(η,ψ t T t ) = t z(0,ψ 0 ) t 0 and hence provided that δ Γ,V G[Γ] = t z(0,ψ 0 ), lim ψ t T t ψ 0 H t 0 1 (Ω) = 0.

23 Step 3. The limit follows from: lim ψ t T t ψ 0 H t 0 1 (Ω) = 0 Weak form of the Euler Lagrange equation for the maximization of z(t, ) by ψ t T t for t > 0 and by ψ 0 for t = 0, respectively; Subtract one from the other; Use the properties of T t (X) and the convexity of B. Step 4. We now have δ Γ,V G[Γ] = t z(0,ψ 0 ). Direct calculations complete the proof. Q.E.D.

24 R 8 4. Motion of a Cylindrical Dielectric Boundary Cheng, Li, White, & Zhou, SIAP, R x Ω + Γ: r=u(z) O Ω_ L z y r Ω + Γ: r=u(z) O Ω_ L z

25 Free-energy functional 1 F[Γ] = γ 0 Area(Γ)+ Ω 2 fψ ΓdV { εγ ψ Γ = f in Ω ψ Γ = 0 on r = R γ 0 > 0 : a given constant f = f(r,z) : Ω R: a given function, L-periodic in z dielectric coefficient r ε Γ (x) = { ε if x Ω ε + if x Ω + R 8 Ω + Γ: r=u(z) O Ω_ L z

26 Equivalent elliptic interface problem (ψ = ψ Γ ) ε ψ = f in Ω ε + ψ = f in Ω + ψ = 0 on Γ ε Γ n ψ = 0 on Γ ψ = ψ(r,z) is L-periodic in z ψ(r,z) = 0 z R Notation: w = w Ω+ w Ω.

27 Consider the electrostatic energy 1 E[Γ] = Ω 2 fψ ΓdV { εγ ψ Γ = f in Ω ψ Γ = 0 on r = R Assume n points from Ω to Ω +. Then δ Γ E[Γ] = 1 ( ε ε + = 1 ( 1 1 ) [εγ (ψ r u ψ z )] 2 2 ε ε + ) ε Γ ψ Γ n (ε + ε ) (I n n) ψ Γ 2 1+u (ε + ε ) (u ψ r +ψ z ) 2 1+u 2

28 Steepest descent: V n = δ Γ F[Γ] u t = γ 0 ( uzz 1+u 2 z 1 u ) 1 ( 1 1 ) [ εγ(t) (ψ r u z ψ z ) ] 2 2 ε ε + 1+u 2 z 1 2 (ε + ε ) (u zψ r +ψ z ) 2 (z,t) (, ) (0,T] 1+u 2 z u(z,t) is L-periodic in z for each t [0,T] u(z,0) is given for all z (, ) ε Γ(t) ψ = f in Ω ψ(r,z,t) is L-periodic in z for each (r,t) [0,R ) [0,T] ψ(r,z,t) = 0 (z,t) (, ) [0,T]

29 Linear Stability Analysis for the Case ε > ε + 8R x Ω + Γ: r=u(z) O Ω_ L z y Water molecules deep in a protein. Competition: surface energy vs. electrostatic energy. Stability of an equilibrium dielectric boundary.

30 Step 1. Steady-state solutions: u = u 0 and ψ = ψ 0 (r). ) 2 u 0 = 1 ( u0 sf(s)ds η r [ 1 ε 0 s ψ 0 (r) = 1 r [ 1 ε + s s 0 s ] τf(τ)dτ τf(τ)dτ u 0 u 0 C 2 C 3 logu 0 C 4 = 1 ε 0 C 3 = 1 u0 sf(s)ds ε + 0 C 3 logr +C 4 = 1 R ε + u 0 ds +C 2 if r < u 0 ] ds +C 3 logr +C 4 if r > u 0 u0 [ 1 s ] τf(τ)dτ ds s 0 [ 1 s s u 0 τf(τ)dτ ] ds ( 1 η = 2γ 0 1 ) 1 ( R > 0 and ηr < sf(s)ds ε + ε 0 ) 2

31 Step 2. Linearization. u = u(z,t,τ) = u 0 +τu 1 (z,t)+, ψ = ψ(r,z,t,τ) = ψ 0 (r)+τψ 1 (r,z,t)+, u(z,0,τ) = u 0 +τu 1 (z,0). [ t u 1 = γ 0 zu 2 γ0 1 + u 2 0 ( 1 ( 1 1 ε ε + ψ 1 = 0 if 0 < r < u 0, ψ 1 = 0 if u 0 < r < R, ε 1 ε + ) ε 2 +ψ 0(u + 0 )ψ 0(u + 0 ) ]u 1 ) ε 2 +ψ 0(u 0 + ) rψ 1 (u 0 +,z,t) z,t, ψ 1 (u + 0,z,t) ψ 1(u 0,z,t) = u 1(z,t) [ ψ 0(u + 0 ) ψ 0(u 0 )] z,t, ε r ψ 1 (u0,z,t) = ε + r ψ 1 (u 0 +,z,t) z,t, ψ 1 (R,z,t) = 0 z,t.

32 Step 3. Dispersion relations. Assume u 1 (z,t) = Ae ωt e ikz with k = 2πk /L, k Z, ψ 1 (r,z,t) = u 1 (z,t)φ k (r). Then the dispersion relation ω = ω(k) is given by ( 1 1 ε ε + ω = γ 0 k 2 + γ 0 u0 2 ) ε 2 +ψ 0(u + 0 )[ ψ 0(u + 0 )+φ k (u+ 0 )], φ k (r)+ 1 r φ k (r) k2 φ k (r) = 0 if 0 < r < u 0, φ k (r)+ 1 r φ k (r) k2 φ k (r) = 0 if u 0 < r < R, φ k (u + 0 ) φ k(u 0 ) = [ ψ 0(u + 0 ) ψ 0(u 0 )], ε φ k (u 0 ) = ε +φ k (u+ 0 ), φ k (R ) = 0.

33 The modified Bessel differential equation x 2 y (x)+xy (x) x 2 y(x) = 0. The modified Bessel functions 1 ( x ) 2s cos(xs) I 0 (x) = (s!) 2 and K 0 (x) = 2 ds. s=0 0 1+s 2 [ µε + ψ 0 (u 0 + ) ψ 0(u0 )] I 0 (kr) [ K0 (kr )I 0(ku 0 ) I 0 (kr )K 0(ku 0 ) ] if 0 < r < u 0, φ k (r) = [ µε ψ 0 (u 0 + ) ψ 0(u0 )] I 0(ku 0 ) [K 0 (kr )I 0 (kr) I 0 (kr )K 0 (kr)] if u 0 < r < R, 1 µ = ε I 1 (ku 0 )[I 0 (kr )K 0 (ku 0 ) I 0 (ku 0 )K 0 (kr )] +ε + I 0 (ku 0 )[I 1 (ku 0 )K 0 (kr )+I 0 (kr )K 1 (ku 0 )]. ω(k) = γ 0 k 2 + γ 0 u 2 0 ( 1 1 ) ε 2 ε ε +ψ 0(u 0 + )[ ψ 0(u 0 + )+φ k (u+ 0 )] +

34 k Ω Ω 1 Ω 2 ω(k) = ω 1 (k)+ω 2 (k) ( 1 ω 1 (k) = 1 ) ε 2 ε ε +ψ 0(u 0 + )[ ψ 0(u 0 + )+φ k (u+ 0 )] + ω 2 (k) = γ 0 k 2 + γ 0 u 2 0 Conclusions: linearly stable if and only if k > k c.

35 Numerical results: The case ε < ε + r t = t = t = t = π 2 π 3π 2 2π z ε = 2, ε + = 80, and u(z,0) = sin(5z).

36 Numerical results: The case ε > ε + r t = 0 t = 2 t = 4 t = z ε = 80, ε + = 2, u ,, u(z,0) = u sinz, k = 1 and ω(k) > 0.

37 t = t = t = t = u(z,0) = u sin(5z), k = 5 and ω(k) < 0.

38 x Change in energy Electrostatic energy Surface energy Total energy Time u(z,0) = u sin(5z), k = 5 and ω(k) < 0.

39 Initial interface u(z) = u 0 + ǫ r t = 30 3 r 3 Initial interface u(z) = u 0 ǫ t = u 0 t = 16 u 0 t = t = z t = z u(z,0) = u u(z,0) = u

40 5. Discussions (1) Validity of the PB theory. ψ = V (ψ) inside walls/outside balls ψ = const. on the walls ψ = const. on bdry of balls V > 0 and V (0) = 0 The electrostatic surface force is given by F = 1 2 (balls) ( nψ) 2 nds F the unit horizontal vector toward the center < 0. PBE does not predict the like-charge attraction.

41 (2) The competition between the surface and electrostatic energies, crucial in hydrophobic interactions. Given f : R 3 R. In the large-ε w limit modeling a perfect conducting solvent, one considers the energy functional f(x)f(y) E[Ω] = Area( Ω)+ dxdy. x y It is expected that E does not have a minimizer. The motion of a boundary driven by the mean curvature and the dielectric boundary force: Well-posedness? Singularity formation? Fluctuation of a dielectric boundary. Ω Ω

42 (3) The Poisson Nernst Planck system. c i t [D i( c i +βq i c i ψ)] = 0, i = 1,...,M, M ε ψ = q i c i. i=1 Formally, Boltzmann distributions c i = c i e βq iψ (i = 1,...,M) are steady-state solutions. Boundary conditions? Energy decay and bounds on concentration. Consider the boundary conditions c i = 0 on Ω for some i and estimate the reaction rates R i = 1 c i ci Ω n ds.

43 Thank you!