On Magnus integrators for time-dependent Schrödinger equations

Transkript

1 On Magnus integrators for time-dependent Schrödinger equations Marlis Hochbruck, University of Düsseldorf, Germany Christian Lubich, University of Tübingen, Germany FoCM conference, August 22

2 Outline Time dependent Schrödinger equations Magnus integrators i dψ dt = H(t)ψ, ψ(t ) = ψ Error bounds for Magnus integrators of order 2 and 4 Sketch of general procedure for deriving these bounds Numerical experiment

3 Short course on Magnus integrators y = A(t)y, y() = y Magnus approach 54: determine Ω(t) such that y(t) = exp ( Ω(t) ) y solution of y = A(t)y (Ω(t) = ta for A(t) A) Differentiate y (t) = dexp Ω(t) ( Ω (t) ) y(t), where dexp Ω (B) = ϕ(ad Ω )(B) = k 1 (k + 1)! adk Ω(B), ϕ(z) = ez 1 z and ad Ω (B) = [Ω, B] = ΩB BΩ. Obtain Ω as solution of A(t) = dexp Ω(t) ( Ω (t) ), Ω() =

4 Magnus integrators II If Ω(t) < π, then dexp Ω(t) is invertible obtain dexp 1 ( ) Ω(t) A(t) = k β k k! adk Ω(t) (A(t)), β k kth Bernoulli number Ω (t) = A(t) 1 1 [Ω(t), A(t)] + [Ω(t), [Ω(t), A(t)]] integration and Picard iteration yields Magnus expansion Ω(t) = t A(τ)dτ t t [ [ t τ τ [ τ A(σ)dσ, A(τ)] dτ σ [ A(µ)dµ, A(σ)] dσ, A(τ)] dτ A(σ)dσ, [ τ A(µ)dµ, A(τ)]] dτ +...

5 Magnus integrators III Numerical methods (review: Iserles, Munthe-Kaas, Nørsett, Zanna, ) y n+1 = exp(ω n )y n Ω n Ω(h) Approximation involves truncating the Magnus expansion (after k terms) k = 1 : k = 2 : t Ω(t) = A(t n + τ)dτ t Ω(t) = A(t n + τ)dτ 1 2 t τ [ A(t n + σ)dσ, A(t n + τ)] dτ approximating integrals by replacing A(t) by interpolation polynomial Â(t) for quadrature nodes t n + c j h

6 Examples: k = 1, exponential midpoint rule Ω n = ha(t n + h/2). k = 2, 2-point Gauß quadrature rule: Ω n = h 3h 2 2 (A 1 + A 2 ) + 12 [A 2, A 1 ], A j = A(t n + c j h), c j nodes of Gauß quadrature rule k = 2, method by Blanes, Casas, Ros Ω n = h 6 ( A(tn ) + 4A(t n+1/2 ) + A(t n+1 ) ) h2 12 [A(t n), A(t n+1 )].

7 Implementation issues options for computing exp(ω)y: splitting methods Chebyshev approximation Lanczos process

8 Situation Magnus integrators efficient for problems like Schrödinger equations (Tal Ezer, Kosloff, 92; Blanes, Moan, ) error behavior well understood for A(t) moderate (Iserles, Nørsett, 99; Iserles, Munthe-Kaas, Nørsett, Zanna ) no results for h A(t) 1

9 Problems for large h A(t) dexp Ω need not be invertible Magnus expansion need not converge known results on order of Magnus integrators valid for h A(t) (Iserles, Nørsett, 99) constants involve A(t) obtained by studying remainder of Magnus series Practice: Magnus integrators work extremely well even for h A(t) π WHY?

10 General assumptions A(t) = ih(t) = i ( U + V (t) ) where U s.p.d., V (t) hermitian satisfying dm dt m V (t) Mm, m =, 1, 2,... but no bound on U! define D = U 1/2

11 typical situation: U = + I, V bounded multiplication operator continuous case: (Q cube, periodic boundary conditions) Dv 2 = v 2 dx + Q Q v 2 dx Dv is H 1 Sobolev norm Dy(t) is essentially the kinetic energy of the solution (bounded a priori) discrete case, minimal grid spacing x: U x 2, D x 1

12 Exponential midpoint rule Theorem (H., Lubich 2) If the solution y satisfies the finite energy condition y(t) H(t)y(t) K, then the error of exponential midpoint rule is bounded by y n y(t n ) Ch 2 t n max t t n Dy(t) where C = C(M m, K), m 2 Error bound for classical implicit midpoint rule: y n y(t n ) Ch 2 t n max t t n d3 dt 3 y(t)

13 Fourth order Magnus methods assumptions commutator bounds: for a method of order p containing products of A(t n + c j h) with r terms we need [A(τ k ), [..., [A(τ 1 ), dm dt m V (τ )]]...]v K D k v (easy for spatially cont. case, discrete case: generalization of results of Jahnke, Lubich, ) m p, k + 1 rp. in our examples: p = 4, r = 2

14 Theorem (H., Lubich 2) If the commutator bounds hold for p = 4 and r = 2, then error of fourth order Magnus methods is bounded by y n y(t n ) Ch 4 t n max D 3 y(t) t t n for time steps h D c with C = C(M m, K, c), m 4. Remarks: explicit integrators require h D 2 c for stability, error bounds for implicit integrators require smallness of h D 2 for oscillatory problems

15 General procedure for deriving error bounds (complete proofs for exponential midpoint rule and fourth order methods) cannot use derivation, in particular not A = dexp Ω (Ω ) cannot use Taylor expansion of solution

16 1. Error resulting from truncating the Magnus expansion: Motivation: Magnus approach starts from y(t) = exp ( Ω(t) ) y, Ω skew truncation yields ỹ(t) = exp ( Ω(t) ) y, Ω skew differentiate: ỹ(t) is solution of perturbed problem ỹ (t) = Ã(t)ỹ(t) with Ã(t) = dexp Ω(t) ( Ω (t) ) with initial value ỹ() = y, Ã(t) skew hermitian t Lemma ỹ(t) y(t) E(τ)y(τ) dτ, E = Ã A

17 2. Remainder of dexp series recall: dexp Ω (B) = ϕ(ad Ω )(B) truncate ϕ appropriately ϕ(z) = ez 1 z = z (p 1)! zp p! zp 1 r p (z), obtain remainder of truncated dexp Ω series dexp Ω (B) = B [Ω, B] (p 1)! adp 2 Ω (B) + 1 ( ) p! r p(ad Ω ) ad p 1 Ω (B) we need a bound of type ( r p ) (ad Ω(t) ad p 1 Ω(t)( Ω (t) )) v Ch p D p 1 v, t h, requires h D c for p = 4, no restriction on h for p = 2

18 Bound on Ev = (Ã A)v exponential midpoint rule (p = 2): and Ω(t) = t A(t n + τ)dτ Ã(t) = dexp ( Ω Ω(t) (t) ) = A(t) + 1 ( ) 2 r ) ( Ω 2(ad Ω(t) ad Ω(t) (t)) =: A(t) + E 2 (t) Lemma For p = 2, 4: E p (t)y(t) Ch p D p 1 y(t), t h where for p = 4, h D c

19 3. Error resulting from approximating the integrals nth step: write Ω n = Ω(h) for truncated Magnus series quadrature: use Ω n instead of Ω n e.g., for the midpoint rule Ω n = h A(t n + τ)dτ ha(t n+1/2 ) =: Ω n since A (t) = V (t) M 2, we have Ω n Ω n 1 24 M 2h 3 Lemma ( Ωn Ω n ) v Ch p+1 D r 1 v (p = 2, r = 1, proof more difficult for p = 4, r = 2) Lemma (obvious for r = 1) exp( Ω n )v exp(ω n )v Ch p+1 D r 1 v

20 4. Putting all steps together define local error by Steps 1 3 ɛ n = y(t n+1 ) exp(ω n )y(t n ) = y(t n+1 ) exp ( Ωn ) y(tn ) + exp ( Ωn ) y(tn ) exp(ω n )y(t n ) = y(t n+1 ) ỹ(t n+1 ) + exp ( Ωn ) y(tn ) exp(ω n )y(t n ) ɛ n tn+1 t n Ch p+1 E p (τ)y(τ) dτ + Ch p+1 D r 1 y(t n ) max D p 1 y(t) t n t t n+1 error recursion for global error e n = y n y(t n ): e n+1 = exp(ω n )e n + ɛ n yields e n Ct n h p max D p 1 y(t) t t n

21 Numerical experiment i ψ t = 1 2 ψ + b(x, t)ψ, x = (x 1,..., x d ) R d, t > discretized such that h D 3.5 (N = 32,..., 248 Fourier modes)

22 Summary Known in practice: Magnus methods work well for time-dependent Schrödinger equations here: theoretical explanation for this behavior, in particular presented optimal-order error estimates for problems where h H(t) is large developed new mechanisms which lead to such bounds