Motion of a Cylindrical Dielectric Boundary

Transcription

1 Motion of a Cylindrical Dielectric Boundary Bo Li Department of Mathematics and NSF Center for Theoretical Biological Physics UC San Diego Collaborators Li-Tien Cheng, Michael White, and Shenggao Zhou Funding: CSC, NIH, and NSF Institute of Applied Mathematics, Chinese Academy of Sciences September 26, 2012

2 Outline 1. Introduction 2. Free-Energy Functional and Motion Law 3. Derivation of Dielectric Boundary Force 4. Linear Stability Analysis 5. Numerical Results 6. Conclusions

3 1. Introduction

4 Water molecules inside/near proteins Berne, et al. Nature McCammon, et al. JCTC 2012.

5 Variational Implicit-Solvent Model (VISM) (Dzubiella, Swanson, & McCammon, 2006) Free-energy functional G[Γ] = P Vol(Ω m )+γ 0 +ρ w i Γ (1 2τH)dS Ω w U (i) LJ ( x x i )dv x +G ele [Γ] n dielectric boundary Ω w Ω m ε m=1 ε =80 w Γ x Q i i

6 Electrostatic free energy G ele [Γ] = ε M Γ 2 ψ 2 +ρ f ψ χ w β 1 Ω j=1 c j ( e βq jψ 1) dv Poisson s equation ε Γ ψ = ρ Charge density ρ = ρ f +χ w ρ i = N i=1 Q iδ xi +χ w M j=1 q jc j Boltzmann distributions: c j = c j e βq jψ Dielectric coefficient { εm in Ω m ε Γ = ε w in Ω w Dielectric boundary force F n = δ Γ G ele [Γ] n dielectric boundary Ω w Ω m ε m=1 ε =80 w Γ x Q i i

7 Definition of δ Γ G ele [Γ] : Shape derivatives 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. Denote T t (X) = x(t,x). Then T t (X) = X +tv(x)+o(t 2 ) for small t > 0. Define δ Γ,V G ele [Γ] = d G ele [Γ t ] G ele [Γ] dt G ele [Γ t ] = lim. t=0 t 0 t Structure Theorem. There exists w : Γ R such that δ Γ,V G ele [Γ] = Γ w(x)[v(x) n(x)]ds X V C c (R 3,R 3 ). Shape derivative δ Γ G ele [Γ](X) = w(x) X Γ

8 2. Free-Energy Functional and Motional Law

9 8R x Ω + Γ: r=u(z) O Ω_ L z y r R 8 Ω + Γ: r=u(z) O Ω_ L z

10 Free-energy functional 1 F[Γ] = γ 0 Area(Γ)+ Ω 2 ρψ ΓdV { εγ ψ Γ = ρ in Ω ρ = 0 on r = R γ 0 > 0 : a given constant ρ = ρ(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

11 Equivalent elliptic interface problem (ψ = ψ Γ ) ε ψ = ρ in Ω ε + ψ = ρ 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: f = f Ω+ f Ω.

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

13 Corollary. If ε > ε +, then the force F n = δ Γ E[Γ] < 0. n dielectric boundary Ω w Ω m ε m=1 ε =80 w Γ x Q i i R 8 r O Ω + Ω_ Γ: r=u(z) L z B. Chu, Molecular Forces, Based on the Baker Lectures of Peter J. W. Debye, John Wiley & Sons, 1967: 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.

14 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 = ρ 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]

15 3. Derivation of Dielectric Boundary Force

16 r R 8 Ω + Γ: r=u(z) O Ω_ L z Step 1. Consider cylindrically symmetric diffeomorphisms T t : Ω Ω (0 t 1) with T 0 = Id. Γ(t) = T t (Γ), Ω (t) = T t (Ω ), and Ω + (t) = T t (Ω + ). n : the unit normal at Γ(t) pointing from Ω (t) to Ω + (t). Γ(t) is the graph of r = u(z,t) with u(z,0) = u(z). The potential ψ = ψ(r,z,t) with ψ(r,z) = ψ(r,z,0) satisfy: { εγ(t) ψ = ρ in Ω ρ = 0 on r = R

17 Step 2. By Poisson s equation and integration by parts, 1 [ E[Γ(t)] = Ω 2 ρψdv = ε ] Γ(t) Ω 2 ψ 2 +ρψ dv [ L ] u(z,t) ( = πε ψ 2 r +ψz 2 ) r dr dz 0 0 [ L R πε + 0 u(z,t) ] ( ψ 2 r +ψz 2 ) r dr dz + ρψdv. Ω By the Leibniz formula, integration by parts, and Poisson s eq., d dt E[Γ(t)] = ε ψ ψ t dv ε + ψ ψ t dv Ω (t) Ω +(t) L +π εγ(t) ψ 2 uu t dz + ρψ t dv = Γ(t) 0 ε Γ(t) n ψ ψ t ds +π L 0 Ω εγ(t) ψ 2 uu t dz.

18 Step 3. Use the first interface condition ψ(u(z,t),z,t) = ψ(u(z,t)+,z,t). Take t and z : ψ t = u t ψ r and ψ z = u z ψ r. Then ψ t = ψ r u z ψ z 1+uz 2 = ψ n V n. At (r,θ,z), ψ = (ψ r cosθ,ψ r sinθ,ψ z ). At (ucosθ,usinθ,z) Γ(t), u t n = (cosθ,sinθ, u z) 1+u 2 z and V = (u t cosθ,u t sinθ,0), The first term is ε Γ(t) n ψ ψ t ds = Γ(t) = Γ(t) Γ(t) ε Γ(t) n ψ n ψ V n ds [ ε ( n ψ ) 2 ε+ ( n ψ +) 2 ] V n ds.

19 Recall Γ ξ(r,z)ds = 2π L ψ = n ψn+(i n n) ψ. 0 ξ(u(z), z)u(z) 1+u (z) 2 dz, The second term is π L 0 εγ(t) ψ 2 uu t dz = 1 2 = 1 2 = 1 2 Γ(t) Γ(t) Γ(t) εγ(t) ψ 2 V n ds ( ε+ ψ + 2 ε ψ 2) V n ds [ ε+ n ψ + 2 ε n ψ 2 +(ε + ε ) (I n n) ψ 2] V n ds d dt E[Γ(t)] = 1 [ ε n ψ 2 ε + n ψ Γ(t) +(ε + ε ) (I n n) ψ 2] V n ds

20 Use the second interface condition to get ε n ψ = ε + n ψ + = ε Γ(t) n ψ ε n ψ 2 ε + n ψ + 2 = Direct verification Finally, (I n n) ψ = u zψ r +ψ z 1+u 2 z d dt E[Γ(t)] = 1 2 Γ(t) ( 1 1 ) ε ε ε Γ(t) n ψ 2. + (u z cosθ,u z sinθ,1). { ( 1 1 ) [ εγ(t) (ψ r u z ψ z ) ] 2 ε ε + 1+uz 2 } +(ε + ε ) (u zψ r +ψ z ) 2 1+u 2 z V n ds.

21 4. Linear Stability Analysis

22 8R x Ω + Γ: r=u(z) O Ω_ L z y Assume ε > ε + : Water molecules deep in proteins; Competition: surface energy vs. electrostatic energy; Stability of an equilibrium dielectric boundary.

23 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 = ρ 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]

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

25 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.

26 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.

27 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 1 u0 2 1 ) ε 2 ε ε +ψ 0(u 0 + )[ ψ 0(u 0 + )+φ k (u+ 0 )]. +

28 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 unstable if and only if k > k c.

29 5. Numerical Results

30 Numerical methods Semi-implicit scheme u m+1 u m [ uzz m+1 = γ 0 t 1+(uz m ) 2 1 ] u m ( 1 1 ) [εγ m (ψr m uz m ψz m )] 2 ε ε (u m z ) 2 (ε + ε ) 2 (u m z ψ m r +ψ m z ) 2 1+(u m z ) 2. Cylindrical version of the CIM (Coupling Interface Method) for Poisson s equation; and for dielectric boundary forces. r i+2 r i+4 r i+3 r i+1 r i+2 r i q r i+1 r i q r i 1 r i 1 r i 2 r i 3 z j 1 z j z j+1 z j 1 z j z j+1

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

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

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

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

35 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

36 6. Conclusions

37 Summary A simple model for the competition between the surface and electrostatic energies. Derivation of the dielectric boundary force (DBF). The DBF always pushes toward charged solute molecules. Linear stability analysis for the case of ε > ε + : linearly unstable with k < k c, agreeing with MD simulations. Suggest that waters are unhappy when trapped deep in proteins. Developed numerical methods. Computations confirm the analysis.

38 Further Studies Numerical result: Near a steady state, the electrostatic energy increases as the dielectric boundary smoothens. True? If so, why? Well-posedness. Formation and development of singularities. Include the solute-solvent van der Waals interaction. Add noise to study the fluctuation of dielectric boundary.

39 Thank you!