Discretization of Stochastic Differential Systems With Singular Coefficients Part II

Transcription

1 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 212

2 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

3 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

4 On PDEs driven by divergence form operators Consider elliptic or parabolic PDEs driven by the strongly elliptic divergence form operator where L := 1 2 div(a(x) ), < λ ξ 2 (a(x)ξ, ξ) Λ ξ 2 < + for all x, ξ R d.

5 Techniques related to the generation of semigroups in H 1 (R d ): Variational formulations: Aronson, Stroock. Dirichlet form theory applied to forms of the type E(u, u) := 1 u(x) a(x) u(x)q(x)dx, 2 where q is a strictly positive density. Pseudo SDEs (Lyons Zheng decompositions: Fukushima, Roskoz, Slominski): For all function φ in Wp 1,loc (R d ), there exists a pair of local martingales (M φ, N φ ) respectively adapted with respect to the filtration generated by (X t, t T ) and the filtration generated by (X T t, t T ), such that φ(x t ) = φ(x )+ 1 2 M φ t N φ t and M φ t =... t t a φ φ(x θ )dθ and N φ t = a(x θ ) p(θ, x, X θ ) φ(x θ )dθ, p(θ, x, X θ ) t a φ φ(x T θ )dθ.

6 Techniques related to the generation of semigroups in H 1 (R d ): Variational formulations: Aronson, Stroock. Dirichlet form theory applied to forms of the type E(u, u) := 1 u(x) a(x) u(x)q(x)dx, 2 where q is a strictly positive density. Pseudo SDEs (Lyons Zheng decompositions: Fukushima, Roskoz, Slominski): For all function φ in Wp 1,loc (R d ), there exists a pair of local martingales (M φ, N φ ) respectively adapted with respect to the filtration generated by (X t, t T ) and the filtration generated by (X T t, t T ), such that φ(x t ) = φ(x )+ 1 2 M φ t N φ t and M φ t =... t t a φ φ(x θ )dθ and N φ t = a(x θ ) p(θ, x, X θ ) φ(x θ )dθ, p(θ, x, X θ ) t a φ φ(x T θ )dθ.

7 Techniques related to the generation of semigroups in H 1 (R d ): Variational formulations: Aronson, Stroock. Dirichlet form theory applied to forms of the type E(u, u) := 1 u(x) a(x) u(x)q(x)dx, 2 where q is a strictly positive density. Pseudo SDEs (Lyons Zheng decompositions: Fukushima, Roskoz, Slominski): For all function φ in Wp 1,loc (R d ), there exists a pair of local martingales (M φ, N φ ) respectively adapted with respect to the filtration generated by (X t, t T ) and the filtration generated by (X T t, t T ), such that φ(x t ) = φ(x )+ 1 2 M φ t N φ t and M φ t =... t t a φ φ(x θ )dθ and N φ t = a(x θ ) p(θ, x, X θ ) φ(x θ )dθ, p(θ, x, X θ ) t a φ φ(x T θ )dθ.

8 Techniques related to the generation of semigroups in H 1 (R d ): Variational formulations: Aronson, Stroock. Dirichlet form theory applied to forms of the type E(u, u) := 1 u(x) a(x) u(x)q(x)dx, 2 where q is a strictly positive density. Pseudo SDEs (Lyons Zheng decompositions: Fukushima, Roskoz, Slominski): For all function φ in Wp 1,loc (R d ), there exists a pair of local martingales (M φ, N φ ) respectively adapted with respect to the filtration generated by (X t, t T ) and the filtration generated by (X T t, t T ), such that φ(x t ) = φ(x )+ 1 2 M φ t N φ t and M φ t =... t t a φ φ(x θ )dθ and N φ t = a(x θ ) p(θ, x, X θ ) φ(x θ )dθ, p(θ, x, X θ ) t a φ φ(x T θ )dθ.

9 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 one should first compute the transition density p(t, x, y) of the Markov process, that is, the fundamental solution.

10 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 one should first compute the transition density p(t, x, y) of the Markov process, that is, the fundamental solution.

11 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 R d, consider the parabolic diffraction problem t u(t, x) 1 2 div(a(x) )u(t, x) = for all (t, x) (, T ] Rd, u(, x) = f (x) for all x R d, Compatibility transmission conditions along the interfaces surfaces. Suppose that 1 2 div(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.

12 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

13 We consider the one dimensional parabolic problem ( ) t u(t, x) 1 2 x (a(x) x u(t, x)) =, (t, x) (, T ] (R {}), u(t, +) = u(t, ), t [, T ], u(, x) = f (x), x R, a(+) x u(t, +) = a( ) x u(t, ), t [, T ]. ( ) Suppose λ >, Λ >, < λ a(x) = (σ(x)) 2 Λ < + for all x R. Suppose also that σ is of class C 3 b (R {}) and is left and right continuous at point. Suppose finally that the first derivative of the function σ has finite left and right limits at.

14 The key SDE with weighted local time The one dimensional case allows specific analytical and numerical tools: Portenko (1979), Le Gall (1985), Lejay-Martinez (23)... Consider the one-dimensional stochastic differential equation with local time dx t = σ(x t )db t + σ(x t )σ (X t )dt + σ2 (+) σ 2 ( ) 2σ 2 dl t (X ). (+) Here L t (X ) is the right-sided local time corresponding to the sign function defined as sgn(x) := 1 for x > and sgn(x) := 1 for x 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).

15 Probabilistic interpretation Theorem 1 Let the bounded function f be in the set { W 2 := g Cb 2 (R {}), g (i) L 2 (R) L 1 (R) for i = 1, 2, Then the function a(+)g (+) = a( )g ( )}. u(t, x) := E x f (X t ), (t, x) [, T ] R, is the unique function in C 1,2 b ([, T ] (R {})) and continuous on [, T ] R which satisfies the diffraction PDE.

16 Difficulties: One cannot apply Itô-Tanaka s formula to u(t, X t ) because u is time dependent. Astonishingly, proving that u(t, x) satisfies the transmission condition is not simple.

17 Difficulties: One cannot apply Itô-Tanaka s formula to u(t, X t ) because u is time dependent. Astonishingly, proving that u(t, x) satisfies the transmission condition is not simple.

18 Proof of Theorem 1 Key observation: for all function g of class Cb 2 (R {}) having a second derivative in the sense of the distributions which is a Radon measure and satisfying the transmission condition a(+)g (+) = a( )g ( ), the Itô Tanaka formula applied to g(x t ) leads to where t x R, t >, E x g(x t ) = g(x) + E x Lg(X s )ds Lg(x) := σ(x)σ (x) x g (x) a(x) 2 xx g(x)i x.

19 First step: smoothness and boundedness. Let σ + (x) be an arbitrary C 3 b (R) extension of the function σ(x)i x> which satisfies, for a + (x) := (σ + (x)) 2, < λ a + (x) Λ < + for all x R. Denote by (X t + ) the unique strong solution to dx t + = σ + (X t + )db t + σ + (X t + )(σ + ) (X t + )dt. Let τ (X ) be the first passage time of the process (X t ) at point : τ (X ) := inf{s > : X s = }. Notice that τ (X ) = τ (X + ). Let r x (s) be the density under P x of τ (X ) T. For all function φ such that E φ(x t ) is finite, for all x >, t t E x φ(x t ) = E x φ(x t + ) E φ(x s + )r x (t s)ds+ E φ(x s )r x (t s)ds. It remains to prove estimates on the derivatives of r x (s).

20 Second step: the differential part of the diffraction PDE In view of the preceding key observation, for all < t < T, < ɛ < T t and x in R, u(t + ɛ, x) u(t, x) = E x f (X t+ɛ ) E x f (X t ) = In addition, by the strong Markov property, t+ɛ u(t + ɛ, x) u(t, x) = E x u(t, X ɛ ) u(t, x). Then easy calculations lead to t u(t, x) = Lu(t, x). t E x Lf (X s )ds.

21 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 Itô-Tanaka formula applied to u(t, X s ) for s ɛ and fixed time t leads to (a(+) x u(t, +) a( ) x u(t, )) E L ɛ(x ) = 2a(+) It then remains to prove ( t+ɛ t lim inf ɛ E Lf (X s )ds E L ɛ(x ) ɛ = +. ɛ ) E Lu(t, X s )ds.

22 Last step: uniqueness. As, for all real number x, x = 1 2 (x + x ) and x = 1 2 (x x ), Itô Tanaka s formula implies d(x t ) = 1 2 dx t sgn(x t)dx t dl t (X ) = I Xt>dX t dl t (X ), d(x t ) = 1 2 dx t 1 2 sgn(x t)dx t 1 2 dl t (X ) = I Xt<dX t a( ) 2a(+) dl t (X ). Now, let U (t, x) be an arbitrary solution. For all fixed t in [, T ] the function U (t s, x) is of class C 1,2 b ([, t] R {}) and its partial derivatives have left and right limits when x tends to. Thus we may apply the classical Itô s formula (no need of Itô-Tanaka s formula!) to this function and the semimartingales (X s ) and (X s ) and to use that, by hypothesis, U (t, x) satisfies the transmission condition.

23 Smoothness properties in L 1 (R) of the transition semigroup of (X t ) Theorem 2 The probability distribution of X t under P x has a density q X (x, t, y) which satisfies: and C >, x R, t >, Leb-a-e. y R {}, q X (x, t, y) C t C >, x R, t (, T ], f L 1 (R), u(t, x) = E x f (X t ) C t f 1.

24 Suppose in addition that the function σ is of class Cb 4 (R {}) and that its three first derivatives have finite left and right limits at. Set { W 4 := g Cb 4 (R {}), g (i) L 2 (R) L 1 (R) for i = 1,..., 4 a(+)g (+) = a( )g ( ) and a(+)(lg) (+) = a( )(Lg) ( Then, for all j =, 1, 2 and i = 1,..., 4 such that 2j + i 4, C >, x R, t (, T ], f W 4, j t i x u(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 g l,p := l x i g p. i=

25 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 observing that, owing to the condition transmision satisfied by fonctions in W 2, integrating by parts leads to φ Cb 1 (R), φ(x) Lf (x) dx = φ (x) a(x) f (x) dx; similarly, as P t f (x) satisfies the transmission condition, two successive integrations by parts lead to t >, φ W 2, φ(x) L(P t f )(x) dx = Lφ(x) (P t f )(x) dx.

26 A SDE with discontinuous coefficients without local time Set Set also β + := β := 2a( ) a(+)+a( ), 2a(+) a(+)+a( ), β(x) := x ( β I x< + β + I x> ), β 1 (x) = x β I x< + x β + I x>. σ(x) := σ β 1 (x) ( β I x + β + I x> ), b(x) := σ β 1 (x)σ β 1 (x) ( β I x + β + I x> ). Adapt a calculation in Le Gall and apply Itô Tanaka s formula to β(x t ). The process Y := β(x ) satisfies the SDE with discontinuous coefficients : t t Y t = β(x ) + σ(y s )db s + b(y s )ds.

27 Our Euler type discretization scheme We now present a Euler type scheme. For other methods: see A. Lejay. Let Approximate (Y t ) by h n := T n and t n k := k h n. Y n t = Y n t n k + σ(y n t n k )I Y n t n k (B t B t n k ) + b(y n t n k )I Y n t n k (t t n k ). Then set X n t = β 1 (Y n t ), t T. Remark. The Euler scheme (X t ) converges weakly to (X t ) since (Y t ) converges weakly to (Y t ) (see Yan). However, as the coefficients b and σ are discontinuous, no classical convergence rate estimate applies.

28 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 W 4, all parameter < ɛ < 1 2, all n large enough, and all x in R, E x f (X T ) E x f (X n T ) C f 1,1 h (1 ɛ)/2 +C f 1,1 hn +C f 3,1 h 1 ɛ n n. Theorem 4 Let f : R R be in the space W := { g C 4 b (R {}), g (i) L 2 (R) L 1 (R) for i = 1,..., 4, There exists a positive number C (depending on f ) such that, for all < ɛ < 1 2, all n large enough, and all x in R, E x f (X T ) E x f (X n T ) Ch 1/2 ɛ n. }.

29 A discretization error decomposition The discretization error satisfies ɛ T := E(f β 1 (Y T )) E(f β 1 (Y n T )) n 1 = (E(u(T tk n, β 1 (Y n t ))) Eu(T t n k k+1, β 1 (Y n n t ))), k+1 n k= from which n 2 { ɛ T E u(θk n, β 1 (Y n t )) k n u(θn k+1, β 1 (Y n t )) k n k= } +u(θk+1, n β 1 (Y n t )) k n u(θn k+1, β 1 (Y n t )) k+1 n + Eu(θ1 n, β 1 (Y n t )) Eu(, n 1 n β 1 (Y n T )) n 2 =: E{T k S k } + ER n 1. k=

30 Methodology We distinguish several cases. When Y n t and Y n k n t are simultaneously positive or negative, we k+1 n use a Taylor expansion of u(tk+1 n, ) around (t k n, Y n t ) and then k n apply accurate estimates of the derivatives of u(t, x) for t in (, T ] and x in R {}. We prove that Y n t and Y n k n t have opposite signs with small k+1 n probability when Y n t is large enough. k n When Y n t is small, we explicit the expansion of u(t n k n k+1, ) around and use the fact that u(t, x) solves the transmission problem, which allows us to cancel the lower order term in the expansion.

31 Methodology We distinguish several cases. When Y n t and Y n k n t are simultaneously positive or negative, we k+1 n use a Taylor expansion of u(tk+1 n, ) around (t k n, Y n t ) and then k n apply accurate estimates of the derivatives of u(t, x) for t in (, T ] and x in R {}. We prove that Y n t and Y n k n t have opposite signs with small k+1 n probability when Y n t is large enough. k n When Y n t is small, we explicit the expansion of u(t n k n k+1, ) around and use the fact that u(t, x) solves the transmission problem, which allows us to cancel the lower order term in the expansion.

32 Methodology We distinguish several cases. When Y n t and Y n k n t are simultaneously positive or negative, we k+1 n use a Taylor expansion of u(tk+1 n, ) around (t k n, Y n t ) and then k n apply accurate estimates of the derivatives of u(t, x) for t in (, T ] and x in R {}. We prove that Y n t and Y n k n t have opposite signs with small k+1 n probability when Y n t is large enough. k n When Y n t is small, we explicit the expansion of u(t n k n k+1, ) around and use the fact that u(t, x) solves the transmission problem, which allows us to cancel the lower order term in the expansion.

33 A key estimates A discrete version of Krylov s inequality: There exists C > such that, for all ξ R d and < ε < 1/2, there exists h > satisfying N h h h, h f (kh)p( X ph ξ h 1/2 ε ) Ch 1/2 ε, k= where N h := T /h 1.

34 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

35 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 where (ε(x) u(x)) + κ 2 (x)u(x) = f (x), x R 3, ε(x) is the permittivity of the medium, κ 2 (x) is called the ion accessibility parameter.

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37 The geometry of the problem The atomic structure of the molecule modelled as N atoms at positions x 1,..., x N in Ω i with radii r 1,..., r N and charge q i, Ω i = N i=1 B(x i, r i ).

38 Other difficulties The source term is singular f := N q i δ xi. i=1 This difficulty can be removed by considering the solution G of ε i G = f, that is, G(x) = 1 4πε int N l=1 q j x x l x Ω int. Then v := u χg solves the smoothened PB equation with a smooth source term g provided that χ has compact support in Ω i and χ 1 in the neighborhood of {x 1,..., x N }. The function κ is discontinuous. We must deal with it. The operator has divergence form with discontinuous coefficient ε. We must deal with it.

39 The general case Assume that Γ is a smooth (C 3 ) manifold in R d. 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)).

40 A martingale problem We say that (P x ) x R d on (C, B) solves the martingale problem (MP) for L if, for all x R d, one has and, for all ϕ satisfying one has M ϕ t (w) := ϕ(w(t)) ϕ(w()) P x {w C : w() = x} = 1, ϕ C b (R d ) C 2 b (R d \ Γ), ε ϕ (n π) C b (N ), t Lϕ(w(s))ds is a P x martingale. Remark: the test functions satisfy the transmission property ε int int ϕ(x) n(x) = ε ext ext ϕ(x) n(x).

41 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 dx t = 2ε(X t )db t + ε ext ε int n(x t )dl t (Y ), 2ε ext Y t = ρ(x t ), and the probability law of X solves the martingale problem for L.

42 Sketch of the proof We construct a smooth local straightening ψ of Γ defined on a neighborhood U of x, s.t. ψ 1 = ρ and Z t := ψ(x t ) satisfies dz 1 t = 2ε(X t )db t + ε ext ε int dl t (Z 1 ) + (drift)(z t )dt, 2ε ext and there is no local time term in the SDE solved by Zt 2,..., Zt d. 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.

43 Lemma (Generalized Itô-Meyer formula) If X is a continuous semimartingale, Y := ρ(x ), and if φ is a test function for the MP for L, then t φ(x t ) = φ(x )+ int φ(x s ) dx s t 3 i,j =1 t + 1 g(x s )dl 2 s(y ), ( ) εint where g(x) := 1 int φ(π(x)) n(π(x)). ε ext 2 u x i x j (X s )d X i, X j s t a.s., The formula would be easily obtained from Itô s and Itô-Tanaka s formulas if the functions φ(x) g(x)[ρ(x)] + and g(x) were C 2. If (X t ) solves the preceding SDE, the local time terms cancel.

44 Feynman-Kac formulas Proposition (First Feynman-Kac representation) Let v be the solution of (ε v) + κ 2 v = g, where g is a smooth function. Then, for all x R 3, [ + ( t ) ] v(x) = E x g(x t ) exp κ 2 (X s )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.

45 A second Feynman-Kac formula Fix h >. We define the stopping times τ k = inf{t τ k 1 : ρ(x t ) = h} τ k = inf{t τ k : X t Γ} Since (u G) = in Ω i, for all x s.t. ρ(x) h, u(x) = E x [u(x τ 1 ) G(X τ 1 )] + G(x). For all x Ω e, ( τ1 )] u(x) = E x [u(x τ1 ) exp κ 2 (X t )dt. Applying these two formulas recursively yields:

46 Application Theorem One has u(x) = E x [ + k=1 ( ( τk G(Xτk ) G(X τ )) exp κ 2 (X k t )dt) ]. Application: Analyze the convergence rates of (improved) Walk on Spheres algorithms introduced in this context by Mascagni and Simonov.

47 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

48 The semi-linear Poisson-Boltzmann PDE The semi-linear PB equation reads (ε(x) v(x)) + κ 2 (x) sinh(v(x)) = g(x), x R 3.

49 Interpretation in terms of Backward Stochastic Differential Equations Consider the Backward Stochastic Differential Equation T >, t T, Y x t = Y x T + T t T t Z x s db s. (g(x x s ) κ 2 (X x s ) sinh(y x s ))ds Theorem (N. Perrin: Ph.D. thesis) 1. There exists a unique solution (Y x, Z x ) (in an appropriate space of processes). 2. There exists a unique weak solution to the smoothened PB equation in the space M := {v H 1 (R 3 ) ; cosh(v) 2 1 L 2 (R 3 )}. This solution belongs to C b (R3 ) C 2 (R 3 Γ). In addition, v(x) = Y x + χ(x) G(x).