the poisson–boltzmann theory and variational implicit...

The Poisson–Boltzmann Theory andVariational Implicit Solvation of Biomolecules

Bo LiDepartment of Mathematics and

NSF Center for Theoretical Biological Physics (CTBP)UC San Diego

FUNDING: NIH, NSF, and CTBP

PDE/Applied Math SeminarDepartment of Mathematics, UC Santa Barbara

March 5, 2013

Solvation

protein folding host-guest system

Molecular Modeling: Explicit vs. Implicit

Molecular Dynamics(MD) Simulations

Statistical Mechanics

P(X ,Y ) = P0e−U(X ,Y )/kBT

U(X ,Y ) = Uuu(X ) + Uvv(Y ) + Uuv(X ,Y )

P(X ) =

∫

P(X ,Y ) dY = P0e−W (X )/kBT

W (X ) : Potential of Mean Force

A Variational Implicit-Solvent Model (VISM)

Dzubiella, Swanson, & McCammon, PRL & JCP, 2006.

n

boundarydielectric

xQi

iΩw Ωm

ε =80ε =1

w

m

Γ

Free-energy functional

Gtotal[Γ] = P Vol (Ωm) + γ0

∫

Γ(1− 2τH) dS

+ ρw

∫

ΩwUvdW dV + Gele[Γ],

where UvdW(x) =N∑

i=1

U(i)LJ(|x − xi |).

The level-set method: Vn = −δΓGtotal[Γ]

δΓGtotal[Γ] = P + 2γ0(H − τK )− ρwUvdW + δΓGele[Γ]

This talk: the PDE aspect of Gele[Γ] and δΓGele[Γ].

JCP 2007 & 2009, JCTC 2009, 2012, & 2013, PRL 2009, J. Comput. Phys. 2010.

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

1. The Poisson–Boltzmann Equation (PBE)

∇ · ε∇ψ +M∑

j=1

qjc∞

j e−βqjψ = −f

Poisson’s equation:

Charge density:

Boltzmann distributions:

∇ · ε(x)∇ψ(x) = −ρ(x)ρ(x) = f (x) +

∑Mj=1 qjcj(x)

cj(x) = c∞j e−βqjψ(x)

ε: dielectric coefficient

f : Ω → R: given, fixed charge density

cj : Ω → R: concentration of jth ionic species

c∞j : constant bulk concentration of jth ionic species

qj = Zje: charge of jth ionic species

β: inverse thermal energy

PBE ∇ · ε∇ψ +M∑

j=1

qjc∞

j e−βqjψ = −f

The linearized PBE (the Debye–Huckel approximation)

∇ · ε∇ψ − εκ2ψ = −f

Here κ > 0 is the ionic strength or the inverse Debyescreening length:

κ2 =β∑M

i=1 q2j c

∞

j

ε

“Derivation”: use the Taylor expansion and

Electrostatic neutrality:∑M

j=1 qjc∞

j = 0

The sinh PBE for 1:1 salt (q2 = −q1 = q, c∞2 = c∞1 = c∞)

∇ · ε∇ψ − 2qc∞ sinh(βqψ) = −f

I [φ] =

∫

Ω

[

ε

2|∇φ|2 − f φ+ β−1

M∑

j=1

c∞j e−βqjφ]

dV

Theorem (Li, Cheng, & Zhang, SIAP 2011). The functionalI : H1

g (Ω) → R has a unique minimizer ψ ∈ H1(Ω) ∩ L∞(Ω) whichis 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

ψλ(x) =

− λ if ψ(x) < −λ,ψ(x) if |ψ(x)| ≤ λ,

λ if ψ(x) > λ.Then ψλ is also a minimizer. Uniqueness =⇒ ψ = ψλ.

Step 4. Routine calculations. Q.E.D.

Electrostatic free-energy functional

G [c] =

∫

Ω

1

2ρψ + β−1

M∑

j=1

cj[

ln(Λ3cj)− 1]

−M∑

j=1

µjcj

dV

ρ(x) = f (x) +∑M

j=1 qjcj(x)

∇ · ε∇ψ = −ρ and ψ = 0 on ∂Ω

Equilibrium conditions

(δG [c])j = qjψ+β−1 ln(Λ3cj)−µj = 0 ⇐⇒ Boltzmann distributions

Minimum electrostatic free-energy (note the sign!)

Gmin =

∫

Ω

−ε2|∇ψ|2 + f ψ − β−1

M∑

j=1

c∞j

(

e−βqjψ − 1)

dV

G [c] =

∫

Ω

1

2ρψ + β−1

M∑

j=1

cj[

ln(Λ3cj)− 1]

−M∑

j=1

µjcj

dV

Theorem (B.L. SIMA 2009).

G has a unique minimizer c = (c1, . . . , cM). Moreover,∃ θ1 > 0, θ2 > 0 : θ1 ≤ cj(x) ≤ θ2 a.e. x ∈ Ω, j = 1, . . . ,M.

Boltzmann distributions: cj(x) = c∞j e−βqjψ(x), j = 1, . . . ,M.

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 7→ s(log s + α) with α ∈ R;

the superlinearity of s 7→ s log s; and

a lemma (cf. next slide). Q.E.D.

G [c] =

∫

Ω

1

2ρψ + β−1

M∑

j=1

cj[

ln(Λ3cj)− 1]

−M∑

j=1

µjcj

dV

ρ(x) = f (x) +∑M

j=1 qjcj(x)

∇ · ε∇ψ = −ρ and ψ = 0 on ∂Ω

Lemma (B.L. SIMA 2009). Given c . There exists c satisfying: c is close to c in L1(Ω) ∩ H−1(Ω); G [c] ≤ G [c]; ∃ θ1 > 0, θ2 > 0 : θ1 ≤ cj(x) ≤ θ2 a.e. x ∈ Ω, j = 1, . . . ,M.

Proof. By construction using the fact that the entropic change isvery large for cj ≈ 0 and cj ≫ 1. Q.E.D.

O s

slns

2. A Size-Modified Mean-Field Theory

G [c] =

∫

Ω

1

2ρψ + β−1

M∑

j=0

cj[

ln(a3j cj)− 1]

−M∑

j=1

µjcj

dV

ρ(x) = f (x) +∑M

j=1 qjcj(x)

∇ · ε∇ψ = −ρ and ψ = 0 on ∂Ω

c0(x) = a−30

[

1−∑Mi=1 a

3i ci (x)

]

Theorem (B.L. Nonlinearity 2009). G has a unique minimizer(c1, . . . , cM) characterized by

Bounds: There exist θ1, θ2 ∈ (0, 1) such that

θ1 ≤ a3j cj(x) ≤ θ2 ∀x ∈ Ω ∀j = 0, 1, . . . ,M;

Equilibrium conditions (i.e.,δG [c] = 0)

aj3a0

−3 log(

a30c0)

−log(

a3j cj)

= β (qjψ − µj) ∀j = 1, . . . ,M,

which determine uniquely cj = cj(ψ) (j = 1, . . . ,M).

(

aj

a0

)3

log(

a30c0)

− log(

a3j cj)

= β (qjψ − µj) ∀j = 1, . . . ,M.

The general case: Implicit Boltzmann distributions

Set DM = u = (u1, . . . , uM) ∈ RM : uj > 0, j = 0, 1, . . . ,M

u0 = a−30

(

1−∑Mj=1 a

3j uj

)

fj(u) = a3j a−30 log

(

a30u0)

− log(

a3j uj

)

, j = 1, . . . ,M.

Lemma. The map f : DM → RM is C∞ and bijective.

Proof. It is clear that f is C∞.

f is injective. det∇f 6= 0, use det(I + v ⊗ w) = 1 + v · w .

f is surjective. Note: fj(u) = rj ⇐⇒ ∂jz = ∂z/∂uj = 0, where

z(u) =M∑

j=0

uj[

log(

a3j uj)

− 1]

+M∑

j=1

rjuj .

Construction: minDMz < min∂DM

z . So all ∂jz = 0. Q.E.D.

Set g = (g1, . . . , gM) = f −1 : RM → DM

Bj(φ) = gj (β(q1φ− µ1), . . . , β(qMφ− µM))

B0(φ) = a−30

[

1−∑Mj=1 a

3j Bj(φ)

]

Define B(φ) = −M∑

j=1

qj

∫ φ

0Bj(ξ) dξ ∀φ ∈ R

Assume∑M

j=1 qjBj(0) = 0 (electrostatic neutrality)

Lemma. The function B is strictly convex. Moreover,

B ′(φ) = −M∑

j=1

qjBj(φ)

> 0 if φ > 0,

= 0 if φ = 0,

< 0 if φ < 0,

and B(φ) > B(0) = 0 for all φ 6= 0. o ψ

B

Proof. Direct calculations using the Cauchy–Schwarz inequality toshow B ′′ > 0. Also, use the neutrality. Q.E.D.

Computational Validation

Zhou, Wang, & Li, PRE, 2011. Wen, Zhou, Xu, & Li, PRE, 2012.

5 10 15 20 25 300

5

10

15

20

25

Distance to the charged surface (A)

Concentratio

nofcounterio

n(M

)

z1=+1, R

1=3.0, N

1=100

z2=+2, R

2=2.5, N

2=100

z3=+3, R

3=3.5, N

3=100

0 5 10 150

5

10

15

20

25

30

35

40

Distance to the charged surface (A)R

adia

lpartic

ledensity

(M

)

z1=+1, R

1=3.0, N

1=100

z2=+2, R

2=2.5, N

2=100

z3=+3, R

3=3.5, N

3=100

The stratification of counterions near a highly charged surfacedetermined by valence-to-volume ratios.Left: Mean-field theory. Right: Monte Carlo simulations.

3. Dielectric Boundary Force (DBF)

Dielectric coefficient

εΓ =

εm in Ωm

εw in Ωw

Typical: εm = ε0 and εw = 80ε0.

Write G = Gele.

n

boundarydielectric

xQi

iΩw Ωm

ε =80ε =1

w

m

Γ

PBE ∇ · εΓ∇ψ − χwB′(ψ) = −f

PB free energy G [Γ] =

∫

Ω

[

−εΓ2|∇ψ|2 + f ψ − χwB(ψ)

]

dV

B(ψ) = β−1M∑

j=1

c∞j

(

e−βqjψ − 1)

Γ =⇒ εΓ =⇒ ψ = ψΓ =⇒ G [Γ] =⇒ Fn := −δΓG [Γ]

Let V ∈ C∞

c (R3,R3). Define x : [0,∞)× R3 → R

3 by

x = V (x) for t > 0,

x(0,X ) = X .

Then Tt(X ) := x(t,X ) ≈ X + tV (X ) if |t| ≪ 1. Define

δΓ,VG [Γ] = limt→0

G [Γt ]− G [Γ]

t=

∫

Γw(X )[V (X ) · n(X )] dSX

for some w : Γ → R by the Structure Theorem.

Shape derivative δΓG [Γ](X ) = w(X ) ∀X ∈ Γ

Theorem (Li, Cheng, & Zhang, SIAP 2011). Let n point from Ωmto Ωw and f ∈ H1(Ω). Let ψ be the unique solution to theboundary-value problem of PBE. Then

δΓG [Γ] =1

2

(

1

εm− 1

εw

)

|εΓ∇ψ · n|2

+1

2(εw − εm) |(I − n ⊗ n)∇ψ|2 + B(ψ).

Corollary. If εw > εm, then −δΓG [Γ] < 0.

n

boundarydielectric

xQi

iΩw Ωm

ε =80ε =1

w

m

Γ

B. Chu, Molecular Forces Based on the Baker Lectures of PeterJ. W. Debye, John Wiley & Sons, 1967:

“Under the combined influence of electric field generatedby solute charges and their polarization in the surroundingmedium which is electrostatic neutral, an additional potentialenergy emerges and drives the surrounding molecules to thesolutes.”

Proof of Theorem. Let V ∈ C∞

c (R3,R3) be local, Γ0 = Γ, and

G [Γt ] = G [Γt , ψt ] = maxφ∈H1

g (Ω)G [Γt , φ].

Note ψ0 = ψ. Define z(t, φ) = G [Γt , φ T−1t ]. We have

G [Γt ] = maxφ∈H1

g (Ω)z(t, φ).

Step 1. Easy to verify that

z(t, ψ0)− z(0, ψ0)

t≤ G [Γt ]− G [Γ]

t≤ z(t, ψt Tt)− z(0, ψt Tt)

t.

Hence

∂tz(ξ, ψ0) ≤G [Γt ]− G [Γ]

t≤ ∂tz(η, ψt Tt), ξ, η ∈ [0, t].

Step 2. Direct calculations lead to

∂tz(t, φ) =

∫

Ω

[

−εΓ2A′(t)∇φ · ∇φ+ ((∇ · (fV )) Tt)φJt

− χwB(φ)((∇ · V ) Tt)Jt

]

dV .

Replacing t by η and φ by ψt Tt , respectively, we obtain

limt→0

∂tz(η, ψt Tt) = ∂tz(0, ψ0)

and henceδΓ,VG [Γ] = ∂tz(0, ψ0),

provided thatlimt→0

‖ψt Tt − ψ0‖H1(Ω) = 0.

Step 3. The limit

limt→0

‖ψt Tt − ψ0‖H1(Ω) = 0

follows from:

Weak form of the Euler–Lagrange equation for themaximization of z(t, ·) by ψt Tt for t > 0 and by ψ0 fort = 0, respectively;

Subtract one from the other;

Use the properties of Tt(X ) and the convexity of B .

Step 4. We now have

δΓ,VG [Γ] = ∂tz(0, ψ0).

Direct calculations complete the proof. Q.E.D.

4. Motion of a Cylindrical Dielectric BoundaryCheng, Li, White, & Zhou, SIAP, 2013.

8R

Γ: r=u(z)

LO

x

y

zΩ_

Ω+

8R

Γ: r=u(z)

LOz

Ω_

Ω+

r

Free-energy functional

F [Γ] = γ0 Area (Γ) +

∫

Ω

1

2f ψΓ dV

∇ · εΓ∇ψΓ = −f in Ω

ψΓ = 0 on r = R∞

γ0 > 0 : a given constant f = f (r , z) : Ω → R: a given function, L-periodic in z dielectric coefficient

εΓ(x) =

ε− if x ∈ Ω−

ε+ if x ∈ Ω+

8R

Γ: r=u(z)

LOz

Ω_

Ω+

r

Equivalent elliptic interface problem (ψ = ψΓ)

ε−∆ψ = −f in Ω−

ε+∆ψ = −f in Ω+

JψK = 0 on Γ

JεΓ∂nψK = 0 on Γ

ψ = ψ(r , z) is L-periodic in z

ψ(R∞, z) = 0 ∀z ∈ R

Notation: JwK = w |Ω+ − w |Ω−.

Consider the electrostatic energy

E [Γ] =

∫

Ω

1

2f ψΓ dV

∇ · εΓ∇ψΓ = −f in Ω

ψΓ = 0 on r = R∞

Assume n points from Ω− to Ω+. Then

δΓE [Γ] =1

2

(

1

ε−− 1

ε+

)

|εΓ∇ψΓ · n|2 +1

2(ε+ − ε−) |(I − n ⊗ n)∇ψΓ|2

=1

2

(

1

ε−− 1

ε+

)

[εΓ (ψr − u′ψz)]2

1 + u′2+

1

2(ε+ − ε−)

(u′ψr + ψz)2

1 + u′2

Steepest descent: Vn = −δΓF [Γ]

ut = γ0

(

uzz

1 + u2z− 1

u

)

− 1

2

(

1

ε−− 1

ε+

)

[

εΓ(t) (ψr − uzψz)]2

√

1 + u2z

− 1

2(ε+ − ε−)

(uzψr + ψz)2

√

1 + u2z∀(z , t) ∈ (−∞,∞)× (0,T ]

u(z , t) is L-periodic in z for each t ∈ [0,T ]

u(z , 0) is given for all z ∈ (−∞,∞)

∇ · εΓ(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 ]

Linear Stability Analysis for the Case ε− > ε+

8R

Γ: r=u(z)

LO

x

y

zΩ_

Ω+

Water molecules deep in a protein.

Competition: surface energy vs. electrostatic energy.

Stability of an equilibrium dielectric boundary.

Step 1. Steady-state solutions: u = u0 and ψ = ψ0(r).

u0 =1

η0

(∫ u0

0sf (s) ds

)2

ψ0(r) =

− 1

ε−

∫ r

0

[

1

s

∫ s

0τ f (τ) dτ

]

ds + C2 if r < u0

− 1

ε+

∫ r

u0

[

1

s

∫ s

u0

τ f (τ) dτ

]

ds + C3 log r + C4 if r > u0

C2 − C3 log u0 − C4 =1

ε−

∫ u0

0

[

1

s

∫ s

0τ f (τ) dτ

]

ds

C3 = − 1

ε+

∫ u0

0sf (s) ds

C3 logR∞ + C4 =1

ε+

∫ R∞

u0

[

1

s

∫ s

u0

τ f (τ) dτ

]

ds

η = 2γ0

(

1

ε+− 1

ε−

)

−1

> 0 and ηR∞ <

(∫ R∞

0sf (s) ds

)2

Step 2. Linearization.

u = u(z , t, τ) = u0 + τu1(z , t) + · · · ,ψ = ψ(r , z , t, τ) = ψ0(r) + τψ1(r , z , t) + · · · ,u(z , 0, τ) = u0 + τu1(z , 0).

∂tu1 = γ0∂2zu1 +

[

γ0u20

−(

1

ε−− 1

ε+

)

ε2+ψ′

0(u+0 )ψ

′′

0(u+0 )

]

u1

−(

1

ε−− 1

ε+

)

ε2+ψ′

0(u+0 )∂rψ1(u

+0 , z , t) ∀z , t,

∆ψ1 = 0 if 0 < r < u0,

∆ψ1 = 0 if u0 < r < R∞,

ψ1(u+0 , z , t)− ψ1(u

−

0 , z , t) = −u1(z , t)[

ψ′

0(u+0 )− ψ′

0(u−

0 )]

∀z , t,ε−∂rψ1(u

−

0 , z , t) = ε+∂rψ1(u+0 , z , t) ∀z , t,

ψ1(R∞, z , t) = 0 ∀z , t.

Step 3. Dispersion relations.

Assume

u1(z , t) = Aeωte ikz with k = 2πk ′/L, k ′ ∈ Z,

ψ1(r , z , t) = u1(z , t)φk(r).

Then the dispersion relation ω = ω(k) is given by

ω = −γ0k2 +γ0u20

−(

1

ε−− 1

ε+

)

ε2+ψ′

0(u+0 )

[

ψ′′

0(u+0 ) + φ′k(u

+0 )

]

,

φ′′k(r) +1

rφ′k(r)− k2φk(r) = 0 if 0 < r < u0,

φ′′k(r) +1

rφ′k(r)− k2φk(r) = 0 if u0 < r < R∞,

φk(u+0 )− φk(u

−

0 ) = −[

ψ′

0(u+0 )− ψ′

0(u−

0 )]

,

ε−φ′

k(u−

0 ) = ε+φ′

k(u+0 ),

φk(R∞) = 0.

The modified Bessel differential equation

x2y ′′(x) + xy ′(x)− x2y(x) = 0.

The modified Bessel functions

I0(x) =∞∑

s=0

1

(s!)2

(x

2

)2sand K0(x) =

∫

∞

0

cos(xs)√1 + s2

ds.

φk(r) =

µε+[

ψ′

0(u+0 )− ψ′

0(u−

0 )]

I0(kr)[

K0(kR∞)I ′0(ku0)− I0(kR∞)K ′

0(ku0)]

if 0 < r < u0,

µε−[

ψ′

0(u+0 )− ψ′

0(u−

0 )]

I ′0(ku0)

[K0(kR∞)I0(kr)− I0(kR∞)K0(kr)] if u0 < r < R∞,

1

µ= ε−I1(ku0) [I0(kR∞)K0(ku0)− I0(ku0)K0(kR∞)]

+ ε+I0(ku0) [I1(ku0)K0(kR∞) + I0(kR∞)K1(ku0)] .

ω(k) = −γ0k2 +γ0u20

−(

1

ε−− 1

ε+

)

ε2+ψ′

0(u+0 )

[

ψ′′

0(u+0 ) + φ′k(u

+0 )

]

1 2 3 4 5k

-1.5

-1.0

-0.5

0.5

Ω2

Ω1

Ω

ω(k) = ω1(k) + ω2(k)

ω1(k) = −(

1

ε−− 1

ε+

)

ε2+ψ′

0(u+0 )

[

ψ′′

0(u+0 ) + φ′k(u

+0 )

]

ω2(k) = −γ0k2 +γ0u20

Conclusions: linearly stable if and only if k > kc .

Numerical results: The case ε− < ε+

0 π

2π 3π

22π

1.5

1.6

1.7

1.8

1.9

2

2.1

z

r

t = 0

t = 1

t = 32

t = 12

ε− = 2, ε+ = 80, and u(z , 0) = 2.025 + 0.1 sin(5z).

Numerical results: The case ε− > ε+

0 1 2 3 4 5 6

1.342

1.344

1.346

1.348

1.35

1.352

1.354

1.356

1.358

1.36

t = 0

t = 2

t = 4

t = 6

z

r

ε− = 80, ε+ = 2, u0 ≈ 1.35,, u(z , 0) = u0 + 10−3 sin z , k = 1 andω(k) > 0.

0 1 2 3 4 5 6

1.349

1.3495

1.35

1.3505

1.351

1.3515

t = 0

0 1 2 3 4 5 6

1.349

1.3495

1.35

1.3505

1.351

1.3515

t = 10

0 1 2 3 4 5 6

1.349

1.3495

1.35

1.3505

1.351

1.3515

t = 20

0 1 2 3 4 5 6

1.349

1.3495

1.35

1.3505

1.351

1.3515

t = 30

u(z , 0) = u0 + 10−3 sin(5z), k = 5 and ω(k) < 0.

0 2 4 6 8 10 12 14 16 18 20−2.5

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

x 10−3

Time

Cha

nge

in e

nerg

y

Electrostatic energy Surface energy Total energy

u(z , 0) = u0 + 10−3 sin(5z), k = 5 and ω(k) < 0.

0 2 4 60

1

2

3

Init ial inter face u(z ) = u 0 + ǫ

0 2 4 60

1

2

3

Init ial inter face u(z ) = u 0 − ǫ

t = 30

t = 27

t = 16u 0u 0 t = 16

t = 27

t = 25

z z

rr

u(z , 0) = u0 + 10−3 u(z , 0) = u0 − 10−3

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)

2n dS

F · the unit horizontal vector toward the center < 0.

PBE does not predict the like-charge attraction.

(2) The competition between the surface and electrostaticenergies, crucial in hydrophobic interactions.

Given f : R3 → R. In the large-εw limit modeling a perfectconducting solvent, one considers the energy functional

E [Ω] = Area (∂Ω) +

∫

Ω

∫

Ω

f (x)f (y)

|x − y | dxdy .

It is expected that E does not have a minimizer.

The motion of a boundary driven by the mean curvature andthe dielectric boundary force: Well-posedness? Singularityformation?

Fluctuation of a dielectric boundary.

(3) The Poisson–Nernst–Planck system.

∂ci∂t

−∇ · [Di (∇ci + βqici∇ψ)] = 0, i = 1, . . . ,M,

∇ · ε∇ψ = −M∑

i=1

qici .

Formally, Boltzmann distributions ci = c∞i e−βqiψ

(i = 1, . . . ,M) are steady-state solutions. Boundaryconditions?

Energy decay and bounds on concentration.

Consider the boundary conditions ci = 0 on ∂Ω for some i

and estimate the reaction rates

Ri =1

c∞i

∫

∂Ω

∂ci∂n

dS .

Thank you!