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Introduction The one dimensional case The linear 3D Poisson-Boltzmann PDE in Molecular Dynamics The semi-linear 3D Poisson-Boltzmann PDE in Mol Discretization of Stochastic Differential Systems With Singular Coefficients Part II Denis Talay,


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Introduction The one dimensional case The linear 3D Poisson-Boltzmann PDE in Molecular Dynamics The semi-linear 3D Poisson-Boltzmann PDE in Mol

Discretization of Stochastic Differential Systems With Singular Coefficients Part II

Denis Talay, INRIA Sophia Antipolis joint works with Mireille Bossy, Nicolas Champagnat, Sylvain Maire, Miguel Martinez, Nicolas Perrin ICERM - Brown – November 2012

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Introduction The one dimensional case The linear 3D Poisson-Boltzmann PDE in Molecular Dynamics The semi-linear 3D Poisson-Boltzmann PDE in Mol

Outline

1 Introduction 2 The one dimensional case 3 The linear 3D Poisson-Boltzmann PDE in Molecular Dynamics 4 The semi-linear 3D Poisson-Boltzmann PDE in Molecular Dynamics

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Introduction The one dimensional case The linear 3D Poisson-Boltzmann PDE in Molecular Dynamics The semi-linear 3D Poisson-Boltzmann PDE in Mol

Outline

1 Introduction 2 The one dimensional case 3 The linear 3D Poisson-Boltzmann PDE in Molecular Dynamics 4 The semi-linear 3D Poisson-Boltzmann PDE in Molecular Dynamics

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Introduction The one dimensional case The linear 3D Poisson-Boltzmann PDE in Molecular Dynamics The semi-linear 3D Poisson-Boltzmann PDE in Mol

On PDEs driven by divergence form operators

Consider elliptic or parabolic PDEs driven by the strongly elliptic divergence form operator L := 1

2div(a(x)∇),

where 0 < λ|ξ|2 ≤ (a(x)ξ, ξ) ≤ Λ|ξ|2 < +∞ for all x, ξ ∈ Rd.

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Techniques related to the generation of semigroups in H 1(Rd):

  • Variational formulations: Aronson, Stroock.
  • Dirichlet form theory applied to forms of the type

E(u, u) := 1 2

  • ∇u(x) · a(x)∇u(x)q(x)dx,

where q is a strictly positive density.

  • ‘Pseudo SDEs’ (Lyons–Zheng decompositions: Fukushima,

Roskoz, Slominski): For all function φ in W 1,loc

p

(Rd), there exists a pair of local martingales (M φ, N φ) respectively adapted with respect to the filtration generated by (Xt, 0 ≤ t ≤ T) and the filtration generated by (XT−t, 0 ≤ t ≤ T), such that φ(Xt) = φ(X0)+1 2M φ

t +1

2N φ

t +1

2 t a(Xθ)∇p(θ, x, Xθ) p(θ, x, Xθ) ·∇φ(Xθ)dθ, and M φt = t a∇φ·∇φ(Xθ)dθ and N φt = t a∇φ·∇φ(XT−θ)dθ.

  • . . .
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Introduction The one dimensional case The linear 3D Poisson-Boltzmann PDE in Molecular Dynamics The semi-linear 3D Poisson-Boltzmann PDE in Mol

Techniques related to the generation of semigroups in H 1(Rd):

  • Variational formulations: Aronson, Stroock.
  • Dirichlet form theory applied to forms of the type

E(u, u) := 1 2

  • ∇u(x) · a(x)∇u(x)q(x)dx,

where q is a strictly positive density.

  • ‘Pseudo SDEs’ (Lyons–Zheng decompositions: Fukushima,

Roskoz, Slominski): For all function φ in W 1,loc

p

(Rd), there exists a pair of local martingales (M φ, N φ) respectively adapted with respect to the filtration generated by (Xt, 0 ≤ t ≤ T) and the filtration generated by (XT−t, 0 ≤ t ≤ T), such that φ(Xt) = φ(X0)+1 2M φ

t +1

2N φ

t +1

2 t a(Xθ)∇p(θ, x, Xθ) p(θ, x, Xθ) ·∇φ(Xθ)dθ, and M φt = t a∇φ·∇φ(Xθ)dθ and N φt = t a∇φ·∇φ(XT−θ)dθ.

  • . . .
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Introduction The one dimensional case The linear 3D Poisson-Boltzmann PDE in Molecular Dynamics The semi-linear 3D Poisson-Boltzmann PDE in Mol

Techniques related to the generation of semigroups in H 1(Rd):

  • Variational formulations: Aronson, Stroock.
  • Dirichlet form theory applied to forms of the type

E(u, u) := 1 2

  • ∇u(x) · a(x)∇u(x)q(x)dx,

where q is a strictly positive density.

  • ‘Pseudo SDEs’ (Lyons–Zheng decompositions: Fukushima,

Roskoz, Slominski): For all function φ in W 1,loc

p

(Rd), there exists a pair of local martingales (M φ, N φ) respectively adapted with respect to the filtration generated by (Xt, 0 ≤ t ≤ T) and the filtration generated by (XT−t, 0 ≤ t ≤ T), such that φ(Xt) = φ(X0)+1 2M φ

t +1

2N φ

t +1

2 t a(Xθ)∇p(θ, x, Xθ) p(θ, x, Xθ) ·∇φ(Xθ)dθ, and M φt = t a∇φ·∇φ(Xθ)dθ and N φt = t a∇φ·∇φ(XT−θ)dθ.

  • . . .
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Introduction The one dimensional case The linear 3D Poisson-Boltzmann PDE in Molecular Dynamics The semi-linear 3D Poisson-Boltzmann PDE in Mol

Techniques related to the generation of semigroups in H 1(Rd):

  • Variational formulations: Aronson, Stroock.
  • Dirichlet form theory applied to forms of the type

E(u, u) := 1 2

  • ∇u(x) · a(x)∇u(x)q(x)dx,

where q is a strictly positive density.

  • ‘Pseudo SDEs’ (Lyons–Zheng decompositions: Fukushima,

Roskoz, Slominski): For all function φ in W 1,loc

p

(Rd), there exists a pair of local martingales (M φ, N φ) respectively adapted with respect to the filtration generated by (Xt, 0 ≤ t ≤ T) and the filtration generated by (XT−t, 0 ≤ t ≤ T), such that φ(Xt) = φ(X0)+1 2M φ

t +1

2N φ

t +1

2 t a(Xθ)∇p(θ, x, Xθ) p(θ, x, Xθ) ·∇φ(Xθ)dθ, and M φt = t a∇φ·∇φ(Xθ)dθ and N φt = t a∇φ·∇φ(XT−θ)dθ.

  • . . .
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Remark.

  • Pardoux–Williams have exhibited a Lyons–Zheng decomposition

for Dirichlet forms with degenerate Neumann boundary conditions.

  • The Lyons–Zheng decompositions cannot lead to algorithms since
  • ne should first compute the transition density p(t, x, y) of the

Markov process, that is, the fundamental solution .

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Remark.

  • Pardoux–Williams have exhibited a Lyons–Zheng decomposition

for Dirichlet forms with degenerate Neumann boundary conditions.

  • The Lyons–Zheng decompositions cannot lead to algorithms since
  • ne should first compute the transition density p(t, x, y) of the

Markov process, that is, the fundamental solution .

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Parabolic diffraction problems

Given a finite time horizon T and a positive matrix-valued function a(x) which is smooth except at the interface surfaces between subdomains of Rd, consider the parabolic diffraction problem        ∂tu(t, x) − 1 2div(a(x)∇)u(t, x) = 0 for all (t, x) ∈ (0, T] × Rd, u(0, x) = f (x) for all x ∈ Rd, Compatibility transmission conditions along the interfaces surfaces. Suppose that 1

2div(a(x)∇) is a strongly elliptic operator. Existence

and uniqueness of continuous solutions with possibly discontinuous derivatives along the surfaces hold true: see, e.g. Ladyzenskaya et al. Motivations: Neurosciences (3D brains!), Molecular Dynamics, Geophysics, etc.

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Introduction The one dimensional case The linear 3D Poisson-Boltzmann PDE in Molecular Dynamics The semi-linear 3D Poisson-Boltzmann PDE in Mol

Outline

1 Introduction 2 The one dimensional case 3 The linear 3D Poisson-Boltzmann PDE in Molecular Dynamics 4 The semi-linear 3D Poisson-Boltzmann PDE in Molecular Dynamics

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We consider the one dimensional parabolic problem (⋄)          ∂tu(t, x) − 1

2∂x(a(x)∂xu(t, x)) = 0, (t, x) ∈ (0, T] × (R − {0}),

u(t, 0+) = u(t, 0−), t ∈ [0, T], u(0, x) = f (x), x ∈ R, a(0+)∂xu(t, 0+) = a(0−)∂xu(t, 0−), t ∈ [0, T]. (⋆) Suppose ∃λ > 0, Λ > 0, 0 < λ ≤ a(x) = (σ(x))2 ≤ Λ < +∞ for all x ∈ R. Suppose also that σ is of class C3

b(R − {0}) and is left and right

continuous at point 0. Suppose finally that the first derivative of the function σ has finite left and right limits at 0.

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The key SDE with weighted local time

The one dimensional case allows specific analytical and numerical tools: Portenko (1979), Le Gall (1985), Lejay-Martinez (2003). . . Consider the one-dimensional stochastic differential equation with local time dXt = σ(Xt)dBt + σ(Xt)σ′

−(Xt)dt + σ2(0+) − σ2(0−)

2σ2(0+) dL0

t (X ).

Here L0

t (X ) is the right-sided local time corresponding to the sign

function defined as sgn(x) := 1 for x > 0 and sgn(x) := −1 for x ≤ 0 and σ′

− is the left derivative of σ.

Under mild hypotheses on σ this SDE has a unique weak solution which is a strong Markov process : see, e.g., Le Gall (1984).

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Probabilistic interpretation

Theorem 1 Let the bounded function f be in the set W2 :=

  • g ∈ C2

b(R − {0}), g(i) ∈ L2(R) ∩ L1(R) for i = 1, 2,

a(0+)g′(0+) = a(0−)g′(0−)} . Then the function u(t, x) := Exf (Xt), (t, x) ∈ [0, T] × R, is the unique function in C1,2

b ([0, T] × (R − {0})) and continuous on

[0, T] × R which satisfies the diffraction PDE.

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Difficulties:

  • One cannot apply Itˆ
  • -Tanaka’s formula to u(t, Xt) because u is

time dependent.

  • Astonishingly, proving that u(t, x) satisfies the transmission

condition is not simple.

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Difficulties:

  • One cannot apply Itˆ
  • -Tanaka’s formula to u(t, Xt) because u is

time dependent.

  • Astonishingly, proving that u(t, x) satisfies the transmission

condition is not simple.

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Proof of Theorem 1

Key observation: for all function g of class C2

b(R − {0}) having a

second derivative in the sense of the distributions which is a Radon measure and satisfying the transmission condition a(0+)g′(0+) = a(0−)g′(0−), the Itˆ

  • –Tanaka formula applied to g(Xt) leads to

∀x ∈ R, ∀t > 0, Exg(Xt) = g(x) + t ExLg(Xs)ds where Lg(x) := σ(x)σ′

−(x)∂xg−(x) + 1

2a(x)∂2

xxg(x)Ix=.

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First step: smoothness and boundedness. Let σ+(x) be an arbitrary C3

b(R) extension of the function σ(x)Ix>

which satisfies, for a+(x) := (σ+(x))2, 0 < λ ≤ a+(x) ≤ Λ < +∞ for all x ∈ R. Denote by (X +

t ) the unique strong solution to

dX +

t

= σ+(X +

t )dBt + σ+(X + t )(σ+)′(X + t )dt.

Let τ0(X ) be the first passage time of the process (Xt) at point 0: τ0(X ) := inf{s > 0 : Xs = 0}. Notice that τ0(X ) = τ0(X +). Let r x

0 (s) be the density under Px of

τ0(X ) ∧ T. For all function φ such that E|φ(Xt)| is finite, for all x > 0, Exφ(Xt) = Exφ(X +

t )−

t E0φ(X +

s )r x 0 (t−s)ds+

t E0φ(Xs)r x

0 (t−s)ds.

It remains to prove estimates on the derivatives of r x

0 (s).

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Second step: the differential part of the diffraction PDE In view of the preceding key observation, for all 0 < t < T, 0 < ǫ < T − t and x in R, u(t + ǫ, x) − u(t, x) = Exf (Xt+ǫ) − Exf (Xt) = t+ǫ

t

ExLf (Xs)ds. In addition, by the strong Markov property, u(t + ǫ, x) − u(t, x) = Exu(t, Xǫ) − u(t, x). Then easy calculations lead to ∂tu(t, x) = Lu(t, x).

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Third step: u(t, x) satisfies the transmission condition. In view of of the preceding first step, for all fixed t the second partial derivative w.r.t. x of u(t, x) is a Radon measure. Thus the Itˆ

  • -Tanaka

formula applied to u(t, Xs) for 0 ≤ s ≤ ǫ and fixed time t leads to (a(0+)∂xu(t, 0+) − a(0−)∂xu(t, 0−)) E0L0

ǫ(X )

= 2a(0+) t+ǫ

t

E0Lf (Xs)ds − ǫ E0Lu(t, Xs)ds

  • .

It then remains to prove lim inf

ǫց0

E0L0

ǫ(X )

ǫ = +∞.

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Last step: uniqueness. As, for all real number x, x ∨ 0 = 1

2(x + |x|) and x ∧ 0 = 1 2(x − |x|),

Itˆ

  • –Tanaka’s formula implies

d(Xt ∨ 0) = 1 2dXt + 1 2sgn(Xt)dXt + 1 2dL0

t (X )

= IXt>dXt + 1 2dL0

t (X ),

d(Xt ∧ 0) = 1 2dXt − 1 2sgn(Xt)dXt − 1 2dL0

t (X )

= IXt<dXt − a(0−) 2a(0+)dL0

t (X ).

Now, let U (t, x) be an arbitrary solution. For all fixed t in [0, T] the function U (t − s, x) is of class C1,2

b ([0, t] × R − {0}) and its partial

derivatives have left and right limits when x tends to 0. Thus we may apply the classical Itˆ

  • ’s formula (no need of Itˆ
  • -Tanaka’s formula!) to

this function and the semimartingales (Xs ∨ 0) and (Xs ∧ 0) and to use that, by hypothesis, U (t, x) satisfies the transmission condition.

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Smoothness properties in L1(R) of the transition semigroup of (Xt)

Theorem 2 The probability distribution of Xt under Px has a density qX (x, t, y) which satisfies: ∃C > 0, ∀x ∈ R, ∀t > 0, Leb-a-e. y ∈ R − {0}, qX (x, t, y) ≤ C √ t and ∃C > 0, ∀x ∈ R, ∀t ∈ (0, T], ∀f ∈ L1(R), |u(t, x)| = |Exf (Xt)| ≤ C √ t f 1.

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Suppose in addition that the function σ is of class C4

b(R − {0}) and

that its three first derivatives have finite left and right limits at 0. Set W4 :=

  • g ∈ C4

b(R − {0}), g(i) ∈ L2(R) ∩ L1(R) for i = 1, . . . , 4

a(0+)g′(0+) = a(0−)g′(0−) and a(0+)(Lg)′(0+) = a(0−)(Lg)′(0− Then, for all j = 0, 1, 2 and i = 1, . . . , 4 such that 2j + i ≤ 4, ∃C > 0, ∀x ∈ R, ∀t ∈ (0, T], ∀f ∈ W4, |∂j

t ∂i xu(t, x)| ≤ C

√ t f ′γ,1, where γ = 1 if 2j + i = 1 or 2, and γ = 3 if 2j + i = 3 or 4, and gℓ,p :=

  • i=0

∂i

xgp.

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Proof of Theorem 2

A key argument: we can closely follow a part of the proof of Aronson’s estimate (see, e.g., Bass and Stroock), starting with

  • bserving that, owing to the condition transmision satisfied by

fonctions in W2, integrating by parts leads to ∀φ ∈ C1

b(R),

  • φ(x) Lf (x) dx = −
  • φ′(x) a(x) f ′(x) dx;

similarly, as Ptf (x) satisfies the transmission condition, two successive integrations by parts lead to ∀t > 0, ∀φ ∈ W2,

  • φ(x) L(Ptf )(x) dx =
  • Lφ(x) (Ptf )(x) dx.
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A SDE with discontinuous coefficients without local time

Set β+ :=

2a(0−) a(0+)+a(0−),

β− :=

2a(0+) a(0+)+a(0−),

β(x) := x

  • β−Ix< + β+Ix>
  • ,

β−1(x) =

x β− Ix< + x β+ Ix>.

Set also ˜ σ(x) := σ ◦ β−1(x)

  • β−Ix≤ + β+Ix>
  • ,

˜ b(x) := σ ◦ β−1(x)σ′

− ◦ β−1(x)

  • β−Ix≤ + β+Ix>
  • .

Adapt a calculation in Le Gall and apply Itˆ

  • –Tanaka’s formula to

β(Xt). The process Y := β(X ) satisfies the SDE with discontinuous coefficients : Yt = β(X0) + t ˜ σ(Ys)dBs + t ˜ b(Ys)ds.

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Our Euler type discretization scheme

We now present a Euler type scheme. For other methods: see

  • A. Lejay.

Let hn := T n and tn

k := k hn.

Approximate (Yt) by Y

n t = Y n tn

k + ˜

σ(Y

n tn

k )IY n tn k =(Bt − Btn k ) + ˜

b(Y

n tn

k )IY n tn k =(t − tn

k ).

Then set X

n t = β−1(Y n t ), 0 ≤ t ≤ T.

Remark. The Euler scheme (X t) converges weakly to (Xt) since (Y t) converges weakly to (Yt) (see Yan). However, as the coefficients ˜ b and ˜ σ are discontinuous, no classical convergence rate estimate applies.

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Convergence rate estimates

Theorem 3 Under the above hypotheses on the function σ, there exists a positive number C such that, for all initial condition f in W4, all parameter 0 < ǫ < 1

2, all n large enough, and all x0 in R,

  • Ex0f (XT) − Ex0f (X

n T)

  • ≤ Cf ′1,1h(1−ǫ)/2

n

+Cf ′1,1

  • hn+Cf ′3,1h1−ǫ

n

. Theorem 4 Let f : R → R be in the space W :=

  • g ∈ C4

b(R − {0}), g(i) ∈ L2(R) ∩ L1(R) for i = 1, . . . , 4,

  • .

There exists a positive number C (depending on f ) such that, for all 0 < ǫ < 1

2, all n large enough, and all x0 in R,

  • Ex0f (XT) − Ex0f (X

n T)

  • ≤ Ch1/2−ǫ

n

.

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A discretization error decomposition

The discretization error satisfies ǫT :=

  • E(f ◦ β−1(YT)) − E(f ◦ β−1(Y

n T))

  • =
  • n−1
  • k=0

(E(u(T − tn

k , β−1(Y n tn

k ))) − Eu(T − tn

k+1, β−1(Y n tn

k+1)))

  • ,

from which ǫT ≤

  • n−2
  • k=0

E

  • u(θn

k , β−1(Y n tn

k )) − u(θn

k+1, β−1(Y n tn

k ))

+u(θn

k+1, β−1(Y n tn

k )) − u(θn

k+1, β−1(Y n tn

k+1))

  • +
  • Eu(θn

1 , β−1(Y n tn

n−1)) − Eu(0, β−1(Y

n T))

  • =:
  • n−2
  • k=0

E{Tk−Sk}

  • + |ERn−1|.
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Methodology

We distinguish several cases.

  • When Y

n tn

k and Y

n tn

k+1 are simultaneously positive or negative, we

use a Taylor expansion of u(tn

k+1, ·) around (tn k , Y n tn

k ) and then

apply accurate estimates of the derivatives of u(t, x) for t in (0, T] and x in R − {0}.

  • We prove that Y

n tn

k and Y

n tn

k+1 have opposite signs with small

probability when |Y

n tn

k | is large enough.

  • When |Y

n tn

k | is small, we explicit the expansion of u(tn

k+1, ·)

around 0 and use the fact that u(t, x) solves the transmission problem, which allows us to cancel the lower order term in the expansion.

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Methodology

We distinguish several cases.

  • When Y

n tn

k and Y

n tn

k+1 are simultaneously positive or negative, we

use a Taylor expansion of u(tn

k+1, ·) around (tn k , Y n tn

k ) and then

apply accurate estimates of the derivatives of u(t, x) for t in (0, T] and x in R − {0}.

  • We prove that Y

n tn

k and Y

n tn

k+1 have opposite signs with small

probability when |Y

n tn

k | is large enough.

  • When |Y

n tn

k | is small, we explicit the expansion of u(tn

k+1, ·)

around 0 and use the fact that u(t, x) solves the transmission problem, which allows us to cancel the lower order term in the expansion.

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Methodology

We distinguish several cases.

  • When Y

n tn

k and Y

n tn

k+1 are simultaneously positive or negative, we

use a Taylor expansion of u(tn

k+1, ·) around (tn k , Y n tn

k ) and then

apply accurate estimates of the derivatives of u(t, x) for t in (0, T] and x in R − {0}.

  • We prove that Y

n tn

k and Y

n tn

k+1 have opposite signs with small

probability when |Y

n tn

k | is large enough.

  • When |Y

n tn

k | is small, we explicit the expansion of u(tn

k+1, ·)

around 0 and use the fact that u(t, x) solves the transmission problem, which allows us to cancel the lower order term in the expansion.

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A key estimates A discrete version of Krylov’s inequality: There exists C > 0 such that, for all ξ ∈ Rd and 0 < ε < 1/2, there exists h0 > 0 satisfying ∀ h ≤ h0, h

Nh

  • k=0

f (kh)P(Xph − ξ ≤ h1/2−ε) ≤ Ch1/2−ε, where Nh := ⌊T/h⌋ − 1.

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Outline

1 Introduction 2 The one dimensional case 3 The linear 3D Poisson-Boltzmann PDE in Molecular Dynamics 4 The semi-linear 3D Poisson-Boltzmann PDE in Molecular Dynamics

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The Poisson-Boltzmann PDE

The Poisson-Boltzmann (PB) PDE in Molecular Dynamics describes the electrostatic potential around a biomolecular assembly, and is used to compute global characteristics of the system such as

  • the solvatation free energy,
  • the electrostatic forces exerted by the solvent on the molecule.

The implicit solvent equation, which means that the solvent is considered as a continuum, reads −∇ · (ε(x)∇u(x)) + κ2(x)u(x) = f (x), x ∈ R3, where

  • ε(x) is the permittivity of the medium,
  • κ2(x) is called the ion accessibility parameter .
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Introduction The one dimensional case The linear 3D Poisson-Boltzmann PDE in Molecular Dynamics The semi-linear 3D Poisson-Boltzmann PDE in Mol

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The geometry of the problem

The atomic structure of the molecule modelled as

  • N atoms at positions x1, . . . , xN in Ωi with radii r1, . . . , rN and

charge qi,

  • Ωi = ∪N

i=1B(xi, ri).

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Other difficulties

  • The source term is singular

f :=

N

  • i=1

qiδxi. This difficulty can be removed by considering the solution G of εi∆G = f , that is, G(x) = 1 4πεint

N

  • l=1

qj |x − xl| ∀x ∈ Ωint. Then v := u − χG solves the ‘smoothened PB equation’ with a smooth source term g provided that χ has compact support in Ωi and χ ≡ 1 in the neighborhood of {x1, . . . , xN }.

  • The function κ is discontinuous. We must deal with it.
  • The operator has divergence form with discontinuous coefficient

ε. We must deal with it.

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Introduction The one dimensional case The linear 3D Poisson-Boltzmann PDE in Molecular Dynamics The semi-linear 3D Poisson-Boltzmann PDE in Mol

The general case

Assume that Γ is a smooth (C 3) manifold in Rd. Notation:

  • π(x) for the orthogonal projection of x on Γ,
  • n(y) as the outward normal to Γ for y ∈ Γ,
  • ρ(x) as the signed distance between x and Γ.

ρ(x) := (x − π(x)) · n(π(x)).

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A martingale problem

We say that (Px)x∈Rd on (C, B) solves the martingale problem (MP) for L if, for all x ∈ Rd, one has Px{w ∈ C : w(0) = x} = 1, and, for all ϕ satisfying ϕ ∈ C 0

b (Rd) ∩ C 2 b (Rd \ Γ),

ε∇ϕ · (n ◦ π) ∈ C 0

b (N),

  • ne has

M ϕ

t (w) := ϕ(w(t)) − ϕ(w(0)) −

t Lϕ(w(s))ds is a Px martingale. Remark: the test functions satisfy the transmission property εint∇intϕ(x) · n(x) = εext∇extϕ(x) · n(x).

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Our main result

Theorem The martingale problem for L is well-posed. In addition, there is weak existence and uniqueness in law for the SDE    dXt =

  • 2ε(Xt)dBt + εext − εint

2εext n(Xt)dL0

t (Y ),

Yt = ρ(Xt), and the probability law of X solves the martingale problem for L.

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Sketch of the proof

  • We construct a smooth local straightening ψ of Γ defined on a

neighborhood U of x, s.t. ψ1 = ρ and Zt := ψ(Xt) satisfies dZ 1

t =

  • 2ε(Xt)dBt + εext − εint

2εext dL0

t (Z 1) + (drift)(Zt)dt,

and there is no local time term in the SDE solved by Z 2

t , . . . , Z d t .

  • Girsanov’s formula allows one to remove the drift term, so that

Z 1

t solves a one-dimensional SDE.

  • Conditionnally to Z 1, Z 2, . . . , Z d solves a classical SDE with

time dependent coefficients.

  • Only weak existence.
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Lemma (Generalized Itˆ

  • -Meyer formula)

If X is a continuous semimartingale, Y := ρ(X ), and if φ is a test function for the MP for L, then φ(Xt) = φ(X0)+ t ∇intφ(Xs)·dXs+1 2

3

  • i,j=1

t ∂2u ∂xi∂xj (Xs)dX i, X js + 1 2 t g(Xs)dL0

s(Y ),

∀t ≥ 0 a.s., where g(x) := εint εext − 1

  • ∇intφ(π(x)) · n(π(x)).
  • The formula would be easily obtained from Itˆ
  • ’s and Itˆ
  • -Tanaka’s

formulas if the functions φ(x) − g(x)[ρ(x)]+ and g(x) were C2.

  • If (Xt) solves the preceding SDE, the local time terms cancel.
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Feynman-Kac formulas

Proposition (First Feynman-Kac representation) Let v be the solution of −∇ · (ε∇v) + κ2v = g, where g is a smooth

  • function. Then, for all x ∈ R3,

v(x) = Ex +∞ g(Xt) exp

t κ2(Xs)ds

  • dt
  • .

This representation does not allow one to develop an efficient numerical scheme because

  • One needs to precisely discretize X everywhere where g is

nonzero.

  • In general, the computation of g is costly.

Since X has (scaled) Brownian paths aaway from Γ, it is better to have formulas only involving informations on the entrance time and position in small neighborhoods of Γ.

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A second Feynman-Kac formula

Fix h > 0.

  • We define the stopping times

τk = inf{t ≥ τ ′

k−1 : ρ(Xt) = −h}

τ ′

k = inf{t ≥ τk : Xt ∈ Γ}

  • Since ∆(u − G) = 0 in Ωi, for all x s.t. ρ(x) ≤ −h,

u(x) = Ex[u(Xτ ′

1) − G(Xτ ′ 1)] + G(x).

  • For all x ∈ Ωe,

u(x) = Ex

  • u(Xτ1) exp

τ1 κ2(Xt)dt

  • .
  • Applying these two formulas recursively yields:
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Application

Theorem One has u(x) = Ex +∞

  • k=1
  • G(Xτk ) − G(Xτ ′

k )

  • exp

τk κ2(Xt)dt

  • .

Application: Analyze the convergence rates of (improved) Walk on Spheres algorithms introduced in this context by Mascagni and Simonov.

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Outline

1 Introduction 2 The one dimensional case 3 The linear 3D Poisson-Boltzmann PDE in Molecular Dynamics 4 The semi-linear 3D Poisson-Boltzmann PDE in Molecular Dynamics

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The semi-linear Poisson-Boltzmann PDE

The semi-linear PB equation reads −∇ · (ε(x)∇v(x)) + κ2(x) sinh(v(x)) = ˜ g(x), x ∈ R3.

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Interpretation in terms of Backward Stochastic Differential Equations

Consider the Backward Stochastic Differential Equation ∀T > 0, ∀0 ≤ t ≤ T, Y x

t = Y x T +

T

t

(g(X x

s ) − κ2(X x s ) sinh(Y x s ))ds

− T

t

Z x

s dBs.

Theorem (N. Perrin: Ph.D. thesis)

  • 1. There exists a unique solution (Y x, Z x) (in an appropriate space
  • f processes).
  • 2. There exists a unique weak solution to the smoothened PB

equation in the space M := {v ∈ H 1(R3) ; cosh(v)2 − 1 ∈ L2(R3)}. This solution belongs to C0

b(R3) C2(R3 − Γ). In addition,

v(x) = Y x

0 + χ(x) G(x).