Weak Convergence of Numerical Methods for Dynamical Systems and Optimal Control, and a relation with Large Deviations for Stochastic Equations

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1 Weak Convergence of Numerical Methods for Dynamical Systems and, and a relation with Large Deviations for Stochastic Equations Mattias Sandberg KTH CSC

2 Outline The error representation for weak convergence, with convergence result Large Deviations Example: Phase Transitions

3 Approximation of an evolution equation Want to approximate g ( x(t ) ) where x : [0, T ] X solves an equation x (t) = f ( x(t) ), x(t 0 ) = x 0, (1) with a solution to x (t) = f ( x(t) ), x(t 0 ) = x 0. (2)

4 Approximation of an evolution equation Want to approximate g ( x(t ) ) where x : [0, T ] X solves an equation x (t) = f ( x(t) ), x(t 0 ) = x 0, (1) with a solution to x (t) = f ( x(t) ), x(t 0 ) = x 0. (2) How do we compute the error g ( x(t ) ) g ( x(t ) )?

5 One way: (x x) = f (x) f ( x) = f (x) f (x) + f (x) f ( x), and Grönwall s Lemma.

6 One way: (x x) = f (x) f ( x) = f (x) f (x) + f (x) f ( x), and Grönwall s Lemma. Gives (too pessimistic) a priori bound.

7 An error representation Introduce the value functions u(x 0, t 0 ) := g ( x(t ) ), x solves (1), ũ(x 0, t 0 ) := g ( x(t ) ), x solves (2), By definition, u is constant along the solution path, i.e. 0 = d dt u(x(t), t) = u t + u x x = u t + u x f. Hence for every (x, t) we have u t (x, t) + u x (x, t) f (x) = 0.

8 An error representation g( x(t )) g(x(t )) = u( x(t ), T ) u(x 0, t 0 ) = = = T t 0 T t 0 T d u( x(t), t)dt dt ( ut ( x(t), t) + u x ( x(t), t) f ( x(t)) ) dt ( ut ( x(t), t) + u x ( x(t), t) f ( x(t)) ) dt t 0 }{{} =0 T + u x ( x(t), t) ( f ( x(t)) f ( x(t)))dt t 0

9 From Solving Ordinary Differential Equations by Hairer, Nørsett, and Wanner.

10 The function f could be an approximation of f or a numerical method: x n+1 = x n + t f ( x n ). Then g( x(t )) g(x(t )) = t n+1 u x ( x(t), t) ( f ( x n ) f ( x(t)))dt It is an error representation of the form weight n residual n, n which makes it possible to develop adaptive methods. t n

11 Examples of usage Adaptive ODE: A varitational principle for adaptive approximation of ordinary differential equations, K.-S. Moon, R. Tempone, A. Szepessy, G. Zouraris, Numerische Mathematik, (2003). Convergence rates for adaptive approximation of ordinary differential equations, K.-S. Moon, R. Tempone, A. Szepessy, G. Zouraris, Numerische Mathematik, (2003). Adaptive SDE: Adaptive weak approximation of stochastic differntial equations, A. Szepessy, R. Tempone G. Zouraris, Comm. Pure Appl. Math, (2001)

12 The optimal control problem Minimize when X solves T 0 h ( X (s), α(s) ) ds + g ( X (T ) ) X (s) = f ( X (s), α(s) ), X (0) = x 0. 0 < s < T and X (s) R d, α(s) B.

13 Value function ( T u(x, t) = inf h(x (s), α(s))ds + g(x (T )) ) α Xt =x t Hamilton-Jacobi equation: u t + H(u x, x) = 0, u(x, T ) = g(x), (3) where H(λ, x) = min a ( λ f (x, a) + h(x, a) ) (See e.g. L. Evans Partial Differential Equations.)

14 Optimal controls are characteristics Theorem Assume H C 1,1 loc (Rd R d ), f, g, h smooth etc., and (α, X ) optimal control and state variables for the starting position (x, t) R d [0, T ]. Then dual path λ : [t, T ] R d : X (s) = H λ ( λ(s), X (s) ), λ (s) = H x ( λ(s), X (s) ), λ(t ) = g ( X (T ) ). (See e.g. P. Cannarsa, C. Sinestrari, Semiconcave Functions, Hamilton-Jacobi Equations, and. ) This is the Pontryagin principle for the case when the Hamiltonian is differentiable.

15 Hamilton-Jacobi vs. Pontryagin Hamilton-Jacobi Pontryagin Global min + - High dimension - + Idea: Use Pontryagin for numerical methods, and Hamilton-Jacobi for theoretical evaluation.

16 Why use our error representation?

17 Why use our error representation? Given continuous solution {X, λ}, is there a discrete solution {X n, λ n } N n=0 to s.t. (X n+1, λ n+1 ) = A t (X n, λ n ), X 0 = x 0, λ N = g (X N ). { sup λn λ(n t) + X n X (n t) } = O( t r ), n where r is the order of A t? In particular, this would imply λ 0 λ(0) small. Consider the continuous solution with initial position (x 0, λ(0) = λ 0 ), and vary λ(0) (linearization).

18 Linearization of ODE Let X and X = X + εv solve In the limit X = f (X ), X (0) = x 0, X = X + εv = f ( X ) f (X ) + εf (X )v, v(0) = v 0. v = f (X )v, v(0) = v 0. Hence the mapping x 0 X (T ) is space-filling. Apply this to the Hamiltonian system.

19 Linearization of Hamiltonian system X = X + εv x, λ = λ + εv λ. Then v x = H λx (X, λ)v x + H λλ (X, λ)v λ, v x (0) = v0 x = 0, v λ = H xx (X, λ)v x + H xλ (X, λ)v λ, v λ (0) = v0 λ. Hence v λ (T ) may be zero for any initial condition v0 λ.

20 Variational representation of Hamilton-Jacobi If u : R d [0, T ] R viscosity solution of Hamilton-Jacobi equation (3): u(x, t) = inf α { T L(α, X )dt + g ( X (T ) ) X (t) = α(t), X (t) = x }, t where { } { } L(x, α) = sup α λ+h(λ, x), and H(λ, x) = inf λ α+l(x, α). λ α (Legendre-type transform)

21 Representation as discrete minimization Symplectic Euler, type I: corresponds to minimization of X n+1 = X n + th λ (X n, λ n+1 ), λ n = λ n+1 + th x (X n, λ n+1 ), t N 1 n=0 where X n+1 = X n + tα n. L(X n, α n ) + g(x N ),

22 Representation as discrete minimization Symplectic Euler, type II: corresponds to minimization of X n+1 = X n + th λ (X n+1, λ n ), λ n = λ n+1 + th x (X n+1, λ n ), t N 1 n=0 L(X n+1, α n ) + g(x N ).

23 A problem: Nondifferentiable Hamiltonian f = α { 1, 1}, inf 1 0 X 2 dt H(λ, x) = min a ( λ f (x, a) + h(x, a) ) = λ + x 2 T t x

24 Symplectic Euler(I or II) + Regularized Hamiltonian = Symplectic Pontryagin(I or II). With h δ (λ, x) = H δ (λ, x) λ Hλ δ (λ, x). we can define an approximate value function ( ū(x, t n ) = min g(xn ) + h δ (λ m+1, X m ) t ), X n=x m n ( ū(x, t n ) = min g(xn ) + h δ (λ m, X m+1 ) t ), X n=x m n (SP(I)), (SP(II)).

25 Convergence of Symplectic Pontryagin (I and II) Theorem Assume that the Hamiltonian satisfies H(λ 1, x) H(λ 2, x) C λ 1 λ 2, H(λ, x 1 ) H(λ, x 2 ) C x 1 x 2 (1 + λ ), H(, x) is concave. Assume that the terminal cost g is continuously differentiable, and g(x) k(1 + x ). Both for Symplectic Pontryagin I and II there exists a solution { X n, λ n } with associated value ū, such that u ū = O(δ + t). (S. Szepessy, M2AN (2006), S. arxiv (2010))

26 Proof 1. Let u δ solve Then u u δ = O(δ). 2. Use that u δ (x, t) = inf α u δ t + H δ (u δ x, x) = 0, u δ (x, T ) = g(x). { T L(α, X )dt+g ( X (T ) ) X (t) = α(t), X (t) = x }, t where { L(x, α) = sup α λ+h δ (λ, x) }, and H δ { } (λ, x) = inf λ α+l(x, α) λ α (Legendre-type transform)

27 Proof 3. Lower bound for ū(x s, 0) u δ (x s, 0). Extend minimizer {X n } to a piecewise linear function X (t). ū(x s, 0) u δ (x s, 0) = t = t = t = N 1 n=0 N 1 n=0 N 1 tn+1 n=0 N 1 n=0 L(X n, α n ) + g(x N ) u δ (x s, 0) L(X n, α n ) + u δ ( X (T ), T ) u δ ( X (0), 0) T L(X n, α n ) + 0 d dt uδ ( X (t), t)dt t n ( u δ t ( X (t), t) + u δ x( X (t), t) α n + L( X (t), α n ) + L(X n, α n ) L( X (t), α n ) ) dt

28 Proof Now use H δ (x, λ) = min α R d { λ α + L(x, α) }, and the fact that u δ solves a Hamilton-Jacobi equation. Thus ū(x s, 0) u δ (x s, 0) N 1 n=0 The following relation holds: tn+1 t n ( L(Xn, α n ) L( X (t), α n ) ) dt. L(X n, α n ) = α n λ n+1 + H δ (X n, λ n+1 ). By the Symplectic Pontryagin relation we have λ n K. Let H δ (x, λ) = H δ (x, λ) ( max( λ K, 0) ) 2. { L(x, α) = sup α λ + H δ (x, λ) } λ R d

29 Proof Then L(X n, α n ) = L(X n, α n ). Suppose that L( X (t), α n ) = α n λ + H δ ( X (t), λ ). Then L(X n, α n ) L( X (t), α n ) = max λ ( αn λ + H δ (X n, λ) ) max λ ( αn λ + H δ ( X (t), λ) ) H δ (X n, λ ) H δ ( X (t), λ ) = H δ (X n, λ ) H δ ( X (t), λ ) K X (t) X n K t.

30 Proof 4. Lower bound for u δ (x s, 0) ū(x s, 0). By H δ (x 1, λ) H(x 2, λ) C x 1 x 2 (1 + λ ) (4) the following result holds: Lemma Assume that the Hamiltonian H satisfies equation (4) and is concave in its second argument. Let x 1, x 2, and α 1 be any elements in R d, and β any positive number. Then there exists an element α 2 in R d such that α 2 α 1 C x 2 x 1 + β, and L(x 2, α 2 ) L(x 1, α 1 ) + C x 2 x 1.

31 Proof Given a point X n at time t n there is a function α : [t n, t n+1 ] R d, such that α(t) α(t) C X n X (t) + β, and L( X n, α(t)) L(X (t), α(t)) + C X n X (t). Let X 0 = x s and α : [t 0, t 1 ] R d as above. Let α 0 = 1 t t1 t 0 αdt, X1 = X 0 + t α 0. With this X 1 we define α : [t 1, t 2 ] R d, and so on. Then X n X (t n ) = O( t).

32 Proof T 0 Therefore L(X (t), α(t))dt N 1 tn+1 n=0 t n N 1 = t n=0 ( L( Xn, α(t)) C X n X (t) ) dt L( X n, α n ) O( t). u δ (x s, 0) ū(x s, 0) t N 1 n=0 L( X N 1 n, α n ) + g(x (T )) n=0 L( X n, α n ) g( X N ) O( t) = O( t)

33 t Nondifferentiable Hamiltonian example contd δ=1/30 δ=1/ X

34 t Nondifferentiable Hamiltonian example contd. (SPI) x

35 Differential inclusions By letting F (x) = a A f (x, a), control problems may be viewed as differential inclusions: X (t) F ( X (t) ), X (0) = X 0,

36 Forward Euler Method ξ n+1 ξ n + tf (ξ n ), n = 0, 1,..., N 1, t = T N ξ 0 = x 0. Need concept of distance between sets in order to compare solution sets. Hausdorff distance: where H(C, D) = inf { λ 0 C D + λb and D C + λb }. C + D = { c + d c C and d D }, λc = { λc c C }.

37 Attainable sets: C n = { X (n t) X : [0, T ] R d solves X F (X ) }, D n = { ξ n {ξ i } N i=0 solves ξ n+1 ξ n + tf (ξ n ) }. Theorem Let F be a bounded and Lipschitz continuous (w.r.t. Hausdorff dist.) function from R d into the non-empty compact subsets of R d. Then max 0 n N H(C n, D n ) K t. (Result for F convex-valued by Dontchev and Farkhi (Computing, 1989), extended to non-convex-valued F by S. in SINUM (2008/2009), no. 1.)

38 Allen-Cahn as model problem for phase transitions ϕ t = δϕ xx δ 1 V (ϕ) + εη, ϕ(0, t) = ϕ(1, t) = 0. (5) Two stable equilibria: φ φ x

39 Goal: Find the probability of transition from ϕ + to ϕ in the finite time T. Use theory of large deviations. (Done by E, Ren, Vanden-Eijnden, Comm. Pure Appl. Math. 2004)

40 Large Deviations for ODE Consider first the SODE dx ε = b(x ε )dt + ε dw, X ε (0) = x 0, in the interval [0, T ]. Probability of rare events obtained using the action functional S(ξ) = 1 2 T 0 ξ (t) b ( ξ(t) ) 2 dt. Theorem Suppose D is a domain in R d such that the boundary of D coincides with the boundary of its closure. Then lim ε 0 ε log P{ X ε (T ) D } = inf { S(ξ) : ξ(0) = x 0, ξ(t ) D }. (Freidlin-Wentzell, Random Perturbations of Dynamical Systems) Hence, P { X ε (T ) D } exp( inf{s(ξ)}/ε).

41 Large Deviations for PDE The SODE case is extended to the present SPDE in e.g. W. Faris, G. Jona-Lasinio, J. Phys. A: Math. Gen. 15 (1982). Action functional I (ϕ) = 1 2 T ( ϕt δϕ xx + δ 1 V (ϕ) ) 2 dx dt.

42 Large Deviations for PDE Let S be an open subset of C([0, 1]). Let P ε be the probability that a solution ϕ to (5), with ϕ(0) = ϕ + satisfies ϕ(t ) S. Then and where I (S) lim inf ε 0 ε log P ε lim sup ε log P ε I ( S) ε 0 I (S) = inf I (ϕ). Hence, P ε exp( I (S)/ε).

43 Formulate as optimal control problem. Minimize v ϕ+,0(α), where and ϕ solves v ϕ0,t 0 (α) = T t 0 h ( α(t) ) dt + g ( ϕ(t ) ) h(α) = α 2 L 2 (0,1) /2, g(ϕ T ) = K ϕ T ϕ 2 L 2 (0,1) ϕ t = δϕ xx δ 1 V (ϕ) + α, ϕ(t 0 ) = ϕ 0. Introduce the value function and Hamiltonian u(ϕ 0, t 0 ) = inf { v ϕ0,t 0 (α) : α L 2( t 0, T ; L 2 (0, 1) )} H(λ, ϕ) = min = 1 α ( δλ x ϕ x δ 1 λv (ϕ) + λα + α2 2 )dx ( δλ x ϕ x δ 1 λv (ϕ) λ2 2 )dx

44 Spatial Discretization Introduce value function ū defined on FEM space V = {piecewise linear} ū( ϕ 0, t 0 ) = inf { T g( ϕ(t )) + h(ᾱ)ds : t 0 where ϕ C(t 0, T ; V ) solves ϕ(t 0 ) = ϕ 0, ᾱ L 2 (t 0, T ; V ) } ( ϕ t, v) = δ( ϕ x, v x ) + ( δ 1 V ( ϕ) + ᾱ, v), for all v V. Note however that T t 0 h(ᾱ)ds 1 2 T t 0 ϕ t δ ϕ xx + δ 1 V ( ϕ) 2 L 2 (0,1) ds. (The last integral is not well-defined since ϕ(, t) H 2 (0, 1)).

45 Theorem Let ϕ 0 V. Denote an optimal pair (control and state) for u(ϕ 0, t 0 ) by α : [t 0, T ] L 2 and ϕ : [t 0, T ] L 2 and an optimal pair for ū(ϕ 0, t 0 ) by ᾱ : [t 0, T ] V and ϕ : [t 0, T ] V. Then T t 0 ( p t (s) + H ( Pp ϕ(s), ϕ(s) )) ds ū(ϕ 0, t 0 ) u(ϕ 0, t 0 ) g ( Pϕ(T ) ) g ( ϕ(t ) ) + T t 0 ( H ( p # ϕ (s), Pϕ(s) ) H ( p # ϕ (s), ϕ(s) )) ds where p (s) = ( p ϕ(s), p t (s) ) L 2 (0, 1) R is any measurable function with values in D + u( ϕ(s), s), and p # (s) = ( p # ϕ (s), p # t (s) ) V R is any measurable function with values in D + ū(pϕ(s), s).

46 Proof. First inequality ū(ϕ 0, t 0 ) u(ϕ 0, t 0 ) =g ( ϕ(t ) ) + T t 0 h(ᾱ)dt u ( ϕ 0, t 0 ) =u ( ϕ(t ), T ) u ( ) T ϕ(t 0 ), t 0 + h(ᾱ)dt T = t 0 ( d dt u( ϕ(t), t ) + h ( ᾱ(t) )) dt Use that u is semiconcave (Cannarsa, Frankowska, Appl. Math. Optim. 26 (1992) p and 33 (1996) p. 1-33). The backward derivative satisfies almost everywhere that d dt u( ϕ(t), t ) p t (t) + ( p ϕ(t), ϕ t (t) ). ϕ t V (p ϕ, ϕ t ) = (Pp ϕ, ϕ t ). t 0

47 Hence the lower bound: d dt u( ϕ(t), t ) + h ( ᾱ(t) ) pt (t) + ( ϕ t (t), Ppϕ(t) ) + h ( ᾱ(t) ) = pt (t) δ ( ϕ x (t), (Ppϕ(t)) ) ( x δ 1 V ( ϕ(t) ) ), Pp (t) since H(λ, ϕ) = + ( ᾱ(t), Pp ϕ(t) ) ᾱ(t) 2 p t (t) + H ( Pp ϕ(t), ϕ(t) ), ( min δ(ϕ x, λ x ) δ 1( V (ϕ), λ ) α L 2 (0,1) Second inequality proved similarly. + (α, λ) α 2).

48 An element in D + u(ϕ 0, s) is given by ( λ(s), H ( λ(s), ϕ 0 ) ), where λ solves λ t = δλ xx δ 1 λv (ϕ), λ(t ) = 2K ( ϕ(t ) ϕ ).

49 Must estimate T t 0 ( H ( Pλ ϕ(s) (s), ϕ(s) ) H ( λ ϕ(s) (s), ϕ(s) )) ds. time T ϕ ϕ ϕ ϕ state

50 Must estimate T t 0 ( H ( Pλ ϕ(s) (s), ϕ(s) ) H ( λ ϕ(s) (s), ϕ(s) )) ds. time T ϕ ϕ ϕ ϕ state Theorem u(ϕ +, 0) ū(pϕ +, 0) = O( x 2 )

51 The Symplectic Euler Method ϕ n+1 = ϕ n + th λ ( λ n+1, ϕ n ), λ n = λ n+1 + th ϕ ( λ n+1, ϕ n ),

52 The Symplectic Euler Method ϕ n+1 = ϕ n + th λ ( λ n+1, ϕ n ), λ n = λ n+1 + th ϕ ( λ n+1, ϕ n ), Symplectic Euler solved in three steps

53 The Symplectic Euler Method ϕ n+1 = ϕ n + th λ ( λ n+1, ϕ n ), λ n = λ n+1 + th ϕ ( λ n+1, ϕ n ), Symplectic Euler solved in three steps Damped fixed point iterations in order to obtain good starting position for next step.

54 The Symplectic Euler Method ϕ n+1 = ϕ n + th λ ( λ n+1, ϕ n ), λ n = λ n+1 + th ϕ ( λ n+1, ϕ n ), Symplectic Euler solved in three steps Damped fixed point iterations in order to obtain good starting position for next step. Newton iterations.

55 The Symplectic Euler Method ϕ n+1 = ϕ n + th λ ( λ n+1, ϕ n ), λ n = λ n+1 + th ϕ ( λ n+1, ϕ n ), Symplectic Euler solved in three steps Damped fixed point iterations in order to obtain good starting position for next step. Newton iterations. Parameters K and δ and the grid size changed in subsequent Newton iterations.

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58 Thank you!