Exponential Runge Kutta methods for the Schrödinger equation

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1 Exponential Runge Kutta methods for the Schrödinger equation Guillaume Dujardin Department of Applied Mathematics and Theoretical Physics, Center for Mathematical Sciences, Wilberforce road, CAMBRIDGE CB3 WA phone: ( , fax: ( February 3, 29 Abstract We consider exponential Runge Kutta methods of collocation type, and use them to solve linear and semi-linear Schrödinger Cauchy problems on the d- dimensional torus We prove that in both cases (linear and non-linear and with suitable assumptions, s-stage methods are of order s and we give sufficient conditions to achieve orders s + 1 and s + 2 We show and explain the effects of resonant time steps that occur when solving linear Schrödinger problems on a finite time interval with such methods This work is inspired by [14], where exponential Runge Kutta methods of collocation type are applied to parabolic Cauchy problems We compare our results with those obtained for parabolic problems and provide numerical experiments for illustration Keywords Exponential integrators, Runge Kutta methods, Schrödinger equation 1 Introduction Exponential integrators are specific geometric integrators They have become very popular recently for the numerical integration of first order in time problems (see for example [15], where their multiple discoveries from the 6 s are recalled For example, some specific exponential integrators have been used for solving the non-linear Schrödinger equation (see for example [3] where the numerical analysis is made in one periodical space dimension 1

2 Exponential Runge Kutta methods are particular exponential integrators These methods have been derived and analysed for semi-linear parabolic Cauchy problems (see for example [14] for collocation methods and [13] for explicit methods This article deals with such exponential Runge Kutta methods applied to the linear and semi-linear Schrödinger equations, namely t u(t,x ι u(t,x = f(t,x (linear and t u(t,x ι u(t,x = f(t,u (semi-linear, considered as Cauchy problems in time (no space discretisation is made Of course, ι stands for the imaginary unit (ι 2 = 1 The analysis in these cases is different from the one performed for parabolic problems since the spectra of the linear operators are different We provide a numerical analysis of exponential Runge Kutta methods of collocation type applied to linear and semi-linear problems on the d-dimensional torus (d N We show that s-stage collocation methods are of order s for suitable Sobolev norms Moreover, we give algebraic sufficient conditions on the collocation points of an s-stage method to achieve orders s+1 and s+2 when solving Schrödinger Cauchy problems These results need additional assumptions, for example on the non-linear term of semi-linear problems These assumptions are fulfilled in many cases of interest, for example in the case of the cubic non-linear Schrödinger equation We also mention, quantify and explain the effects of resonant time steps that appear when solving linear Schrödinger problems on a finite time interval Many other results for the finite-time integration of highly oscillatory (discretisations of PDEs exist in the literature For second-order in time problems, like wave equations, we mention the mollified impulse method analysed in [8] and generalised in [16], the Gautschi-type methods analysed in [12] and the analysis carried out in [9] based on a one-step formulation This paper is concerned with the numerical integration of Hamiltonian PDEs over a finite time interval For results on numerical integration of Hamiltonian PDEs over long times, the reader may refer for example to [4, 5, 1] for results on the non-linear wave equation obtained via modulated Fourier expansions and to [7] for results on splitting methods applied to the linear Schrödinger equation with small potential Let us mention that many other geometric integrators exist to integrate numerically Schrödinger Cauchy problems For example, some of them take advantage of the conservation of phase space properties by considering multisymplectic formulations (see for example [2] This article is organised as follows: in Section 2 we deal with exponential Runge Kutta methods of collocation type applied to linear Schrödinger Cauchy problems We first introduce some notation for these methods applied to linear problems Then we look at order conditions for s-stage methods We show that these methods are of order at least s and give sufficient conditions to achieve orders s+1 and s+2 Moreover, we study the effect of resonant time steps on the 2

3 order constants In section 3, we look at exponential Runge Kutta methods of collocation type applied to semi-linear Schrödinger Cauchy problems We set up some notation for these methods applied to non-linear problems We prove, with suitable assumptions, that s-stage methods provide unique numerical solutions and are of order at least s Moreover, we give sufficient order conditions for s-stage methods to be of order s + 1 and s + 2 We conclude the article with numerical experiments and some comments 2 Linear problems 21 Notation We denote by T the one-dimensional torus R/(2πZ For all given positive integers d, we denote by T d the d-dimensional torus For x = (x 1,,x d R d and y = (y 1,,y d R d, we denote by xy the real number x 1 y 1 + +x d y d and by x 2 the non-negative real number (x x2 d 1/2 For k Z d, we denote by p k the function on T d defined for x T d by p k (x = e ιkx L 2 (T d (or simply L 2 denotes the set of (classes of complex functions f on T d such that T d f(x 2 dx < +, endowed with the norm f L 2 (T d = ( (2π d T d f(x 2 dx 1/2 An unbounded linear operator A acting on L 2 is said to be diagonal if for all k Z d, p k is in the domain of A and there exists a λ k C such that Ap k = λ k p k For example, the Laplace operator acting on L 2 (T d is diagonal since for all k Z d, p k = k 2 2 p k Another example is the identity operator on L 2 denoted by Id For all such A and all functions ϕ from C to itself, we denote by ϕ(a the unbounded diagonal linear operator acting on L 2 whose domain is the set of linear combinations of functions (p k k Z d defined for all k Z d by ϕ(ap k = ϕ(λ k p k (21 For all functions f L 2 and all k Z d, we denote by ˆf k the Fourier coefficient (2π d T f(xe ιkx dx For α R +, we denote by H α (T d (or simply H α the space of (classes of complex functions f L 2 (T d such that d k Z d \{} ˆf k 2 k 2α 2 < +, endowed with the norm f H α = ( ˆf 2 + k Z d \{} ˆf 1/2 k 2 k 2α 2 Note that L 2 = H with the same norm With these notations, if ϕ is a function from C to C bounded by M on ιr, then for all h > and α, ϕ(ιh is a bounded linear operator from H α to itself with 1 ϕ(ιh H α Hα M In 1 Of course, for all α and all linear operators A from H α to itself, we denote A H α H α = sup v H α v Av H α v H α 3

4 particular, we have for all α, e ιh H α H α = 1 Following [14], we set for all k N z C, ϕ k (z = 1 ( z k e z k 1 z p, (22 p! and ϕ k ( = 1 k! We also set ϕ = exp Note that for all k N, ϕ k is holomorphic on C, and is bounded on ιr For all unbounded linear diagonal operator A and all h >, we have k N, ϕ k+1 ( ha = 1 h k+1 h e (h τaτk dτ (23 k! In order to compute numerical solutions of the following Cauchy problem t u(t,x + Au(t,x = f(t,x (t,x [,T] T d u(,x = u (x x T d, (24 where T >, A = ι, u and f are given, we consider the following numerical methods, called exponential Runge Kutta methods of collocation type; we refer to [14] for a derivation of such methods for semi-linear problems based on the variation-of-constants formula: n {,,N}, t n u n+1 = nh = e ha u n + h b i ( haf(t n + c i h, (25 where h >, s N, (c 1,,c s [,1] s are given and N = T/h We assume that for all (i,j {1,,s} 2, c i c j if i j The operators b i ( ha are defined by i {1,,n}, b i ( ha = 1 h h e (h τa l i (τdτ, (26 where for all i {1,,s}, l i is the i-th Lagrange polynomial with respect to the points (c j h j {1,,s} : i {1,,s}, l i (τ = s,j i τ h c j c i c j (27 Note that for all i {1,,s}, b i span(ϕ,,ϕ s 1 We therefore derive that b i is holomorphic on C and is bounded on ιr In the case A =, the linear PDE (24 reduces to the collection of linear ODEs for x T d du (t,x = f(t,x dt t 4 [,T],

5 with initial conditions u(,x = u (x Moreover, the exponential Runge Kutta method (25 reduces to a classical Runge Kutta method called the underlying Runge Kutta method with coefficients (b i ( i {1,,s} since for all i {1,,s}, b i ( ha = b i (Id 22 Order of exponential Runge Kutta methods for linear problems 221 An s-stage method is of order s Our first result shows that an s-stage exponential Runge Kutta method of collocation type (25 applied to a linear Schrödinger problem (24 is of order at least s, provided that the right hand side of (24 is sufficiently smooth with respect to the time, when considered as an L 2 (T d -valued function This somehow natural result is very similar to the one obtained in the context of parabolic linear equations in [14] (see Theorem 1 However, note that in the case of the linear Schrödinger equation, the order constant C does not depend on T Theorem 21 An s-stage exponential Runge Kutta method of collocation type (25 applied to a linear problem (24 such that f C s ([,T],L 2 (T d is of global order s in the sense that there exists a positive constant C depending only on (c 1,,c s such that for all h > and all n {,,N}, tn u n u(t n L 2 C h s f (s (τ L 2 dτ (28 Proof Let us denote for all n {,,N}, e n = u n u(t n The variationof-constants formula for the exact solution u of problem (24 reads for h ],T] and t [,T h], u(t + h = e ha u(t + h e (h sa f(t + sds (29 Using Taylor expansions of f in (25 and (29 and the properties of the Lagrange collocation polynomials (27 involved in the functions (b i i {1,,s} (see (26, one has e n+1 = e ha e n + h h e (h τa τ e (h τa l i (τdτ ci h (τ σ s 1 f (s (t n + σdσdτ (s 1! (c i h σ s 1 f (s (t n + σdσ (21 (s 1! Recall that e ha L 2 L 2 = 1 Denoting δ n+1 = e n+1 e ha e n, we have that for some constant C depending only on c 1,,c s, tn+1 δ n+1 L 2 C h s f (s (τ L 2 dτ t n 5

6 This inequality and the relation e n = e pha δ n p, complete the proof Remark 22 Notice that for all r >, estimate (28 holds true with the L 2 (T d - norm replaced by the H r (T d -norm provided that f C s ([,T],H r (T d 222 Achieving order s + 1 If its underlying Runge Kutta method is of order s + 1, then the exponential Runge Kutta method of collocation type (25 applied to the linear Schrödinger problem (24 also has order s + 1, provided that we assume that the the righthand side of (24 has higher spatial regularity than just a standard L 2 one (see Theorem 24 for a precise statement This is a difference with the case of parabolic linear equations studied in [14] (see Theorem 2 Let us first recall (see [14], formula (12 that Lemma 23 If the underlying Runge Kutta method is of order s + 1, then c s ib i ( = 1 s + 1 (211 We are now able to state the first result of this section: Theorem 24 Assume r is given, the underlying Runge Kutta method is of order s + 1 and f C s+1 ([,T],L 2 (T d is such that f (s C 1 ([,T],H r+2 Then, the exponential Runge Kutta method of collocation type (25 applied to the linear problem (24 is of order s + 1 in the sense that there exists a positive constant C depending only on (c 1,,c s such that we have for all h > and all n {,,N}, u(t n u n H r C T h s+1( tn f (s ( H r+2 + f (s+1 (τ H r+2 dτ (212 Proof With the notation of the previous proof, using another Taylor series expansion of f in (21, we write with δ (1 δ n+1 = δ (1 n+1 + δ(2 n+1, ( h τ n+1 = e (h τa (τ σ s 1 dσdτ (s 1! h ci e (h τa h (c i h σ s 1 l i (τdτ dσ f (s (t n, (s 1! 6

7 and, after integration by parts, h τ n+1 = e (h τa (τ σ s f (s+1 (t n + σdσdτ s! h ci e (h τa h (c i h σ s l i (τdτ f (s+1 (t n + σdσ (213 s! δ (2 Accordingly, we set e (1 n = e pha δ (1 n p and e (2 n = e pha δ (2 n p Recall that e ιh = 1 On one hand, for some constant C H r H r 1 > depending only on (c 1,,c s,we have δ (2 n+1 C H r (T d 1 h s+1 t n+1 t n f (s+1 (τ dτ H r (T d Therefore, we derive that tn e (2 n H r (T d C 1 h s+1 f (s+1 (τ H dτ (214 t r (T d On the other hand, since δ (1 n+1 = hs+1( ϕ s+1 ( ha 1 c s i s! b i( ha f (s (t n, an Abel summation yields e (1 n = h s+1 ψ s+1 ( ha ( ( e pha n 2 f (s ( p ( + e kha( f (s (t n p 2 f (s (t n p 1, k= where ψ s+1 (z = ϕ s+1 (z 1 s s! cs i b i(z Since the underlying Runge Kutta method is of order s + 1, Lemma 23 and the fact that ϕ s+1 ( = ((s + 1! 1 ensure that ψ s+1 ( = Moreover, ψ s+1 is a holomorphic function on C and is bounded on ιr We derive that there exists a holomorphic function ψ s+1 that is bounded on ιr such that z C, ψ s+1 (z = z ψ s+1 (z Hence, there exists a positive constant C 2 depending only on (c 1,,c s such that ψ s+1 ( ha C H r (T d H r (T d 2 We derive that e (1 n H r (T d C 2h s+1( ( ha e pha f (s ( H r (T d k= n 2 ( p + ha e kha t n p 1 f (s+1 (τdτ H t r (T d n p 2 For all k Z d and p {,,N 1}, p ιh k 2 2 e ιlh k 2 2 (p + 1h k 2 2 T k 2 2 (215 l= 7

8 Therefore, for all p {,,N 1} and u H r+2, p ha e lha u H r T u H r+2 (216 l= This allows us to bound the H r -norm of e (1 n Together with estimate (214, this ensures that inequality (212 holds true Note that the order constant in the previous result depends on T This is also the case when the exponential Runge Kutta method (25 is applied to a linear parabolic problem (see [14], Theorem 2 In the case of linear Schrödinger problems (24, if the underlying Runge Kutta method is of order s + 1, it is possible to get an order s + 1 constant that does not depend on T, with a restriction on the values of the time step The latter has to be non-resonant in the following sense: Lemma 25 Assume the following non-resonance condition on the time step h > : there exists γ > and ν > 1 such that N N, 1 e ιhn h γ N ν (217 Then we have for all n {,,N}, for all r and all u H r+2ν+2 (T d, ha e pha u H r (T d 2 γ u H r+2ν+2 (T d (218 Proof With the help of (217, we get for k Z d \ {}, ιh k 2 2 e ιph k 2 2 Therefore, estimate (218 holds true 2h k e ιh k 2 2 γ k 2ν+2 2 Note that the set of time steps h (,h that do not satisfy (217 has a Lebesgue measure in o(h as h tends to See for example [11], Chapter 1, Lemma 63 In addition to the restriction on the values of the time step, a higher spatial regularity of the right-hand side f of (24 is assumed to prove the following: Theorem 26 Assume that r, γ > and ν > 1 are given Assume that the underlying Runge Kutta method is of order s+1 and f C s+1 ([,T],L 2 (T d is such that f (s C 1 ([,T],H r+2ν+2 Then the exponential Runge Kutta method of collocation type (25 applied to the linear problem (24 is of order s + 1 for non-resonant time steps in the sense that there exists a positive constant C depending only on (c 1,,c s and γ such that if h > satisfies (217, then we have for all n {,,N}, u(t n u n H r C h s+1( tn f (s ( H r+2ν+2+ f (s+1 (τ H r+2ν+2 dτ (219 8

9 Proof With the help of Lemma 25, estimate (216 can be replaced with estimate (218 in the proof of Theorem 24 to get e (1 n H r 2 C t 2 γ hs+1( f (s ( H r+2ν+2 + f (s+1 (τ H r+2ν+2 dτ (22 In addition to (214, we derive that (219 holds true The effect of resonant time steps on the global error over the finite time interval [, T] is illustrated by numerical experiments in Section 4 In particular, one can see the difference between time steps satisfying (217 for which estimate (219 is sharp and those that do not satisfy (217 for which estimate (212 is sharp 223 Concerning order s + 2 As in the case of exponential Runge Kutta methods applied to linear parabolic problems studied in [14] (see Theorem 3, if its underlying Runge Kutta method is of order s + 2 and one has sufficient spatial regularity in the right-hand side of (24, then the exponential Runge Kutta method (25 applied to the linear problem (24 also has order s + 2 Before giving precise statements, let us recall the following: (see [14], Lemma 3 Lemma 27 If the underlying Runge Kutta method is of order s + 2, then c s+1 i b i ( = We are now able to prove the following: t 1 s + 2, (221 c s i b i ( = 1 (s + 1(s + 2 (222 Theorem 28 Assume that r is given, the underlying Runge Kutta method is of order s+2 and f C s+2 ([,T],L 2 (T d is such that f (s C 2 ([,T],H r+4 Then the exponential Runge Kutta method of collocation type (25 applied to the linear Schrödinger equation (24 is of order s + 2 in the sense that there exists a positive constant C depending only on (c 1,,c s such that for all h > and all n {,,N}, we have Proof Let δ (2 n+1 u(t n u n H r C T h s+2( f (s ( H r+4 + f (s+1 ( H r+2 tn tn + f (s+1 (τ H r+4 dτ + f (s+2 (τ H r+2 dτ be defined by (213 and write δ(2 n+1 = δ(2,1 n+1 + δ(2,2 δ (2,1 n+1 = hs+2( ϕ s+2 ( ha 1 (s + 1! 9 c s+1 i b i ( ha n+1 with f (s+1 (t n,

10 and h τ n+1 = e (h τa (τ σ s+1 f (s+2 (t n + σdσdτ (s + 1! h ci e (h τa h (c i h σ s+1 l i (τdτ f (s+2 (t n + σdσ (s + 1! δ (2,2 One deduces that there exists C > depending only on (c 1,,c s such that Therefore, δ (2,2 n+1 H r C h s+2 tn+1 t n e pha δ (2,2 n p H r C h s+2 f (s+2 (τ H r dτ tn p t n p 1 f (s+2 (τ H r dτ tn C h s+2 f (s+2 (τ H r dτ (223 t Since the underlying Runge Kutta method is of order s + 2, Lemma 27 ensures that s cs+1 i b i ( = 1 s+2 (see (221 Moreover, ϕ s+2( = ((s + 2! 1 Therefore, there exists a holomorphic function ζ s+2 on C such that 1 z C, c s+1 i b i (z ϕ s+2 (z = zζ s+2 (z (s + 1! Moreover, ζ s+2 is bounded on ιr Hence, we have and by an Abel summation, δ (2,1 n+1 = hahs+2 ζ s+2 ( haf (s+1 (t n, ( e pha δ (2,1 n p = hs+2 ζ s+2 ( ha ha ( e pha f (s+1 ( n 2 + ha ( p e kha( f (s+1 (t n p 2 f (s+1 (t n p 1 k= We derive with estimate (215 that there exists a positive constant C 1 depending only on (c 1,,c s such that e pha δ (2,1 n p H r C 1 T h s+2( f (s+1 ( H r+2 + t t f (s+2 (τ H r+2 dτ (224 To complete the proof, since the underlying Runge Kutta method is of order s+2, we get by Lemmas 23 and 27 that s cs i b i( = 1 s+1 and s cs i b i ( = (see relation (222 Therefore, there exists a holomorphic function 1 (s+1(s+2 κ s+2 on C such that ϕ s+1 (z 1 s! s cs i b i(z = z 2 κ s+2 (z Moreover, κ s+2 is 1

11 bounded on ιr As before, we get ( e pha δ (1 n p = hhs+1 κ s+2 ( ha ha ( e pha Af (s ( n 2 + ha ( p e kha A ( f (s (t n p 2 f (s (t n p 1 k= Therefore, using inequality (215, there exists a constant positive C 2 depending only on (c 1,,c 2 such that e pha δ (1 n p H r C 2 T h s+2( f (s ( H r+4 + t t f (s+1 (τ H r+4 dτ (225 The conclusion follows by adding (223, (224 and (225 together In the previous theorem, the order constant depends on T As in the previous section (see Theorem 26, if the underlying Runge Kutta method is of order s + 2, then the exponential Runge Kutta method (25 applied to the linear Schrödinger problem (24 is of order s + 2 with an order constant independent of T provided that the time step is non-resonant and the right-hand side f of (24 has higher spatial regularity A precise statement is the following: Theorem 29 Assume that r, γ > and ν > 1 are given as in Theorem 26 Assume that the underlying Runge Kutta method is of order s + 2 and that f C s+2 ([,T],L 2 (T d is such that f (s C 2 ([,T],H r+2ν+4 (T d Then the exponential Runge Kutta method of collocation type (25 applied to the linear Schrödinger equation (24 is of order s + 2 for non-resonant time steps in the sense that there exists a positive constant C depending only on (c 1,,c s and γ such that if h > satisfies (217, then we have for all n {,,N}, u(t n u n H r C h s+2( f (s ( H r+2ν+4 + f (s+1 ( H r+2ν+2 tn tn + f (s+1 (τ H r+2ν+4 dτ + f (s+2 (τ H r+2ν+2 dτ Proof With the help of Lemma 25, estimate (224 can be replaced by the following estimate e pha δ (2,1 n p H r C 1 γ hs+2( f (s+1 ( H r+2ν+2 + t t f (s+2 (τ H r+2ν+2 dτ, (226 11

12 where C 1 > depends only on (c 1,,c n The same lemma yields the existence of a positive constant C 2 depending only on (c 1,,c s such that e pha δ (1 n p H r C 2 γ hs+2( f (s ( H r+2ν+4 + t t f (s+1 (τ H r+2ν+4 dτ (227 This estimate replaces inequality (225 As in the proof of the previous theorem, the conclusion follows by adding (223, (226 and (227 together 3 Semi-linear problems 31 Notation We consider the following semi-linear Cauchy problem t u(t,x + Au(t,x = g(t,u(t,x (t,x [,T] T d u(,x = u (x x T d (31 where T >, A = ι, r, u H r (T d and g are given In the following and in many applications, for all t [,T], g(t comes from a non-linear function from C to C For example, for the so-called cubic non-linear Schrödinger equation, g(t,u = g(t(u = i u 2 u In order to solve numerically the problem (31, we consider the following numerical methods, also called exponential Runge Kutta methods of collocation type We also refer to [14] for a derivation of such methods for semi-linear problems based on the variation-of-constants formula We assume that s N, c 1,,c s [,1], h > are given and set N = T/h We also assume that if (i,j {1,,s} 2 are such that i j, then c i c j We denote for all n {,,N}, t n = nh For such n, if u n H r (T d is a given approximation of the exact solution u(t n of the problem (31, then we construct an approximation u n+1 of u(t n+1 by first solving the non-linear system consisting of the s following equations u n,i = e ciha u n + h a i,j ( hag(t n + c j h,u n,j, i {1,,s} (32 The s unknown quantities are u n,1,,u n,s H r (T d and the s 2 coefficients (a i,j ( ha (i,j {1,,s} 2 are defined by a i,j ( ha = 1 h ci h e (c ih τa l j (τdτ We recall that the (l j j {1,,s} are the Lagrange polynomials defined in (27 Then, we define u n+1 = e ha u n + h b i ( hag(t n + c i h,u n,i (33 12

13 Recall that the coefficients (b i i {1,,s} are defined in (26 Note that for all (i,j {1,,s} 2, a i,j ( ha span(ϕ (c i ha,,ϕ s 1 (c i ha and the coefficients in each linear combination depend only on the choice of the points (c 1,,c s Hence, each coefficient a i,j (z is bounded on ιr by a constant depending only on the choice of (c 1,,c s In the following, C denotes a bound of the functions (b i (z i {1,,s} and (a i,j (z (i,j {1,,s} 2 on ιr depending only on the choice of (c 1,,c s In the case A =, the non-linear Cauchy problem (31 reduces to the collection of non-linear ODEs for x T d du (t,x = g(t,u(t,x, dt t [,T], with initial values u(,x = u (x Moreover, the exponential Runge Kutta method of collocation type (32-(33 reduces to a classical Runge Kutta method whose coefficients are (a i,j ( (i,j {1,,s} 2 and (b i ( i {1,,s}, since in that case for all (i,j {1,,s} 2, a i,j ( ha = a i,j (Id and b i ( ha = b i (Id This method is called the underlying Runge Kutta method For t [, T], we denote f(t = g(t, u(t, the right-hand side of (31, where u denotes the exact solution of the problem The space H r (T d s is a Hilbert space when endowed with the norm ( 1/2 (u 1,,u s r,s = u i 2 H r (T d (34 For R, we denote by B(,R the set of u H r (T d s such that u r,s R 32 Existence and uniqueness of the numerical solution This section is devoted to the proof of existence and uniqueness of the numerical solution (u,,u N provided by the exponential Runge Kutta method (32- (33 of the semi-linear problem (31 for h sufficiently small and with some suitable assumptions on the exact solution and the right-hand side of (31 We will also give a first bound for the numerical error Our assumptions on the exact solution u of (31 and the non-linearity g are the following: Hypothesis 31 The function g satisfies: t [,T], g(t, = 13

14 Hypothesis 32 The non-linearity g in the right-hand side of (31 is Lipschitzcontinuous in the sense that there exists L > and ρ > such that for all t [,T] and for all u,v H r (T d satisfying u H r (T d ρ and v H r (T d ρ, we have g(t,u g(t,v H r (T d L u v H r (T d (35 Hypothesis 33 Equation (31 admits an exact solution u : [,T] H r (T d that is sufficiently smooth In particular, there exists R > such that for all t [,T], u(t H r (T d R Hypothesis 34 The mapping f : [,T] H r (T d is sufficiently smooth In applications, Hypothesis 34 will often be a consequence of Hypothesis 33 and of the regularity of the function g Note that these assumptions are fulfilled in many cases of interest, for example in the case of the cubic non-linear Schrödinger equation, at least when r > d/2 and r d/2 / N This can be seen, for example, by adapting classical results of non-linear analysis (see for example [1], Chapter II, Proposition 22 Precise statements of the adaptations can be found in [6] Under these assumptions, if ρ is sufficiently big (see (37 and h is sufficiently small (see (36, then for all n {,,N} the non-linear system (32 has a unique solution (u n,1,,u n,s H r (T s such that (u n,1,,u n,s r,s 3R s Moreover, the approximation u n+1 defined by (33 is such that u n+1 H s (T d 2R To prove these assertions, we start with the following lemma: Lemma 35 Assume that h > is such that and h (3CLs 2 1, (36 ρ 3R s (37 Then for all h (,h, t [,T h] and y H r (T d such that y H r (T d 2R, the non-linear system v 1 v s = e c1ha y + h a 1,j ( hag(t + c j h,v j = e csha y + h a s,j ( hag(t + c j h,v j, admits a unique solution v = (v 1,,v s B(,3R s 14

15 Proof For h (,h, t [,T h], y H r (T d such that y 2R, H r (T d we define the function ( H r (T d s H r (T d s ( f h,t,y : (v i 1 i s e ciha y + h s a i,j( hag(t + c j h,v j 1 i s Hypothesis 32 and inequality (37 ensure that for all (v 1,,v s B(,3R s, we have f t,h,y (v 1,,v s (e c 1hA y,,e csha y r,s Chs 3/2 L(3R s Since (e c 1hA y,,e csha y r,s 2R s, the triangle inequality and (36 imply that f t,h,y (v 1,,v s r,s 3R s Therefore, f t,h,y (B(,3R s B(,3R s On the other hand, for v = (v 1,,v s B(,3R s and w = (w 1,,w s B(,3R s, by Hypothesis 32 we have Hence, using (36, one derives that f t,h,y (v f t,h,y (w r,s CLhs v w r,s f t,h,y (v f t,h,y (w r,s h h v w r,s Since hh 1 < 1, equation x = f t,h,y (x has exactly one solution in B(,3R s by the classical Picard fixed-point theorem We are now able to prove that if h satisfies another smallness condition, then, for all h (,h, the exponential Runge Kutta method of collocation type (32-(33 provides a unique numerical solution with for all n {,,N}, (u n,1,,u n,s r,s B(,3R s and u n H 2R Moreover, such an s- r (T d stage method is of order s when applied to the semi-linear Schrödinger problem (31 Theorem 36 Assume Hypotheses 31, 32, 33 and 34 are satisfied with (37 Set κ s = (s + 1( 1 s! + Cs (s 1! Assume that h > satisfies (36, and T 1 h s R (κ s e 2CsTL f (s (τ H r (T d dτ (38 Then, for all h (,h, the exponential Runge Kutta method (32-(33 provides a unique numerical solution such that for all n {,,N}, u n H s (T d 2R and (u n,1,,u n,s r,s 3R s Moreover, for all n {,,N}, u n u(t n H r (T d κ se 2CsLT h s tn f (s (τ H dτ (39 r (T d 15

16 We recall that for all n {,,N}, t n T Proof Let h (,h We prove the result by induction Assume n {,,N 1} is such that u n H 2R Lemma 35 ensures that the nonlinear system (32 has a unique solution (u n,1,,u n,s B(,3R s More- r (T d over, Hypotheses 33 and 34 ensure that, for the exact solution, we have u(t n+1 = e ha u(t n + h b i ( haf(t n + c i h + δ n+1, (31 with δ n+1 = h h e (h τa τ b i ( ha ci h (τ σ s 1 f (s (t n + σdσdτ (s 1! (c i h τ s 1 f (s (t n + τdτ, (s 1! using the quadrature rule properties Similarly, for all i {1,,s}, we have u(t n + c i h = e ciha u(t n + h a i,j ( haf(t n + c j h + n,i, (311 where n,i = h ci h τ e (c ih τa a i,j ( ha cj h (τ σ s 1 f (s (t n + σdσdτ (s 1! (c j h τ s 1 f (s (t n + τdτ (312 (s 1! Hence, if we denote e n = u n u(t n and e n,i = u n,i u(t n + c i h, then we have e n,i = e ciha e n + h a i,j ( ha ( g(t n + c j h,u n,j f(t n + c j h n,i (313 For all i {1,,s}, we have e ciha H r Hr = 1 and hence ( e n,i H r (T d e n H r (T d + hcl e n,j H + r (T d n,i H r (T d (314 Summing these s inequalities and using (36, we get e n,j H r (T d 2s e n H r (T d + 2 n,i H r (T d (315 On the other hand, subtracting (31 from (33 yields ( e n+1 = e ha e n + h b i ( ha g(t n + c i h,u n,i f(t n + c i h δ n+1 (316 16

17 Hence, e n+1 H r (T d e n H r (T d + hcl e n,i H r (T d + δ n+1 H r (T d (317 Using (315, we thus get e n+1 H r (T d ( e n H r (T d + 2hCL s e n H r (T d + n,j H r (T d (1 + 2hCsL e n H r (T d + n,j H r (T d + δ n+1 H r (T d n (1 + 2hCsL p( n p,j H r (T d + δ n+1 p H r (T d n ( e 2NhCsL n p,j H r (T d + δ n+1 p H r (T d + δ n+1 H r (T d using also (36 Using relation (21 (which still holds for non-linear problems and relation (312, we get that for all j {1,,s} and all p {,,n}, ( 1 n p,j H r (T d, δ n+1 p H r (T d s! + Cs tn p+1 h s f (s (τ (s 1! H t r (T d dτ n p (318 We derive ( 1 e n+1 H r (T d e2nhcsl (s + 1 s! + Cs tn+1 h s f (s (τ (s 1! H r (T d dτ In view of (38, we derive that e n+1 H R Hence, by triangle inequality, r (T d we get u n+1 u(t H r (T d n+1 + e H r (T d n+1 H R + R and estimate r (T d (39 holds true t, 33 A sufficient condition for order s + 1 As in the linear case (see Theorem 24 and Theorem 26, the exponential Runge Kutta method of collocation type (32-(33 applied to the non-linear Schrödinger equation (31 is of order s + 1 provided that the underlying Runge Kutta method is of order s + 1 A precise result is stated in Theorem 38 Note that this result is very similar to the one obtained in [14] in the case of semi-linear parabolic problems (see [14], Theorem 5 Let us start with a discrete Gronwall Lemma: Lemma 37 Let X be a subset 2 of R + Assume that T > and a are given There exists a positive constant C such that for all non-negative functions 2 Namely, X will be (, h or the set of h (, h satisfying (217 17

18 b defined on X and all sequences (ε n n N of non-negative real numbers satisfying for all h X and n N with nh T, ε n ah k= ε k + b(h, we have ε n Cb(h Theorem 38 Under the hypotheses of Theorem 36, if f (s C 1 ([,T],H r+2 and the underlying Runge-Kutta method is of order s + 1, then there exists a positive constant C > depending only on (c 1,,c s, L, and T such that for all h (,h, we have for all n {,,N}, u n u(t n H r C h s+1 ( f (s ( H r+2 + T f (s+1 (τ H r+2 dτ + Proof By induction, relation (316 becomes e n = h ( k= T f (s (τ H r dτ e kha d n k 1 e kha δ n k, where d k = s b i( ha(g(t k + c i h,u k,i f(t k + c i h As in the proof of Theorem 24, there exists a constant C 1 > depending only on (c 1,,c s such that e kha δ n k H r C 1 h s+1( t f (s ( H r+2 + f (s+1 (τ H r+2 dτ k= (319 On the other hand, with Hypothesis 32, we have k= Then, using (315, we deduce k= k= e kha d n k 1 H r CL e kha d n k 1 H r 2sCL k= k= e k,i H r e k H r + 2CL k= k,i H r Estimate (318 provides the existence of a constant C 2 > depending only on (c 1,,c s and s such that Hence, k= tn k,i H r C 2 h s f (s (τ H r dτ k= ( T e n H r 2shCL e k H r + h s+1 2CLC 2 f (s (τ H r dτ +C 1 ( f (s ( H r+2 + T f (s+1 (τ H r+2 dτ 18

19 We conclude with the help of Lemma 37 with ε n = e n H r, a = 2sCL and b equal the term multiplied by h s+1 in the previous estimate Remark 39 Note that in this non-linear case, the order constant C appearing in Theorem 38 depends a priori exponentially on T This contrasts the linear case (see Theorem 24 and Theorem 26 Remark 31 We could write a counterpart of Theorem 38 for non-resonant time steps with suitable assumptions on f essentially by modifying inequality (319 However, we would still get an order constant depending on T (see Theorem 26 for the corresponding result for linear problems and the fact is that we did not manage to observe resonances for non-linear problems (see section 4 for numerical experiments 34 Achieving order s + 2 In order to give sufficient algebraic conditions on the coefficients of the underlying Runge Kutta method for the s-stage exponential Runge Kutta method of collocation type (32-(33 to be of order s+2, we reinforce our hypothesis on the smoothness of the non-linearity g The function g is assumed to be smooth in the sense of functions from R 2 to R 2 For convenience, we denote, for example, for v C = R 2, g u (t,v the R-linear mapping from C = R2 to C = R 2 corresponding to the first derivative of g(t,u as a function of u R 2 For v H r (T d, we also denote by g u (t,v the induced mapping between functions on Td defined for a function w on T d by x T d, g g (t,v(w(x = u u (t,v(x(w(x Our additional hypotheses on the non-linearity g are the following: Hypothesis 311 Assume that there exists positive constants L 1 and L 1 such that and t [,T], v H r (T d, g u (t,u(tv H r (T d L 1 v H r (T d, t [,T], v H r+2 (T d, g u (t,u(tv H r+2 (T d L 1 v H r+2 (T d Hypothesis 312 Assume that there exists a positive constant L 2 such that for all t [,T] and u,v,w H r (T d satisfying max{ u, v, w } 2R, H r H r H one has r g(t,v g(t,w g u (t,u(v w H r L 2 ( u v H r + u w H r v w H r 19

20 Hypothesis 313 Assume that there exists positive constants L 3 and L 3 such that for all t [,T] and v,w H r+2 (T d, one has and 2 g u 2(t,u(t(v,w H r L 3 v H r w H r, 2 g t u (t,u(tv H r L 3 v H r, 2 g u 2 (t,u(t(v,w H r+2 L 3 v H r+2 w H r+2, 2 g t u (t,u(tv H r+2 L 3 v H r+2 Such hypotheses on g are coupled with (and may imply the following one on the exact solution u of problem (31 that reinforces Hypothesis 33: Hypothesis 314 Equation (31 has an exact solution u : [,T] H r+2 (T d that is sufficiently smooth In particular, u H r+2 (T d Notice that Hypotheses (311, (312 and (313 are satisfied in many cases of interest, for example for the cubic Schrödinger equation when r > d/2 with r d/2 N provided that u is sufficiently smooth (see [6] for details Under these hypotheses, we prove that the exponential Runge Kutta method (32-(33 applied to the semi-linear Schrödinger equation (31 is of order s + 2 provided that the underlying Runge Kutta method is of order s + 2 Theorem 315 Assume that r is given Assume that the underlying Runge Kutta method is of order s+2, and that the hypotheses of Theorem 36, Hypothesis 311, Hypothesis 312, Hypothesis 313 and Hypothesis 314 are fulfilled If f C s+2 ([,T],H r+4, then there exists a positive constant C > such that for all h (,h and all n {,,N}, one has u(t n u n H r Ch s+2 Remark 316 Of course, the constant C appearing in Theorem 315 also involves the final time T > and integrals of norms of the time-derivatives of f over [,T] Proof Let us set for n {,,N} and J n = g u (t n,u(t n, d n,i = g(t n + c i h,u n,i g(t n + c i h,u(t n + c i h J n e n,i As before, subtracting (31 from (33 yields e n+1 = e ha e n δ n+1 + h b i ( haj n e n,i + h 2 b i ( ha d n,i

21 By induction, we derive that for all n {,,N 1}, n e n+1 = e pha δ n+1 p (32 + h h + h b i ( ha b i ( ha b i ( ha n e pha ( J n p en p,i + n p,i (321 n e pha J n p n p,i (322 n e pha dn p,i (323 Since the underlying Runge Kutta method is of order s + 2, one deduces, as in the proof of Theorem 28 that there exists a positive constant C 1 such that for all h and n, n e pha δ n+1 p H r C 1 h s+2( f (s ( H r+4 + f (s+1 ( H r+2 T T + f (s+1 (τ H r+4 dτ + f (s+2 (τ H r+2 (324 To bound the error term (321, one uses (313 and then (315 to derive that e n p,i + n p,i H r e n p H r + ChL e n p,j H r (1 + 2Ch Ls e n p H r + 2CLh Using Hypothesis 311 and estimate (318, we derive that 3 n h b i ( ha e pha ( J n p,i en p,i + n p,i H r ( n Chs L 1 ((1 + 2Ch Ls e n p H r + 2CLh ChsL 1 (1 + 2Ch Ls n p,j H r n p,j H r n T e p H r + 2C 2 LL 1 κ s h s+2 f (s (τ H r dτ Hence, there exists positive constants C 2 and C 3 such that for all h and n, the H r -norm of the term (321 is bounded by C 2 h n e p H r + C 3h s+2 3 See the definition of κ s in Theorem 36 21

22 In order to bound the H r -norm of the error term (322, we set ψ i,s+1 (z = ϕ s+1 (c i zc s+1 i 1 a i,j (zc s s! j, and write to get h h h h + h b i ( ha b i ( n p,i = h s+1 ψ i,s+1 ( haf (s (t n p + n p,i, n e pha J n p n p,i = n e pha J n p h s+1 ψ i,s+1 (f (s (t n p (325 b i ( ha n e pha J n p h s+1( ψ i,s+1 ( ha ψ i,s+1 ( f (s (t n p (326 ( bi ( ha b i ( n e pha J n p h s+1 ψ i,s+1 (f (s (t n p (327 b i ( ha n e pha J n p n p,i (328 Since the underlying Runge Kutta method is of order at least s + 2, we derive that b i (ψ i,s+1 ( =, and therefore the term (325 reads n ( h s+2 e pha J n p b i (ψ i,s+1 ( f (s (t n p = On the other hand, there exists s holomorphic functions (Ψ i,s+1 i {1,,s} such that i {1,,s}, z C, ψ i,s+1 (z ψ i,s+1 ( = zψ i,s+1 (z Note that the functions (Ψ i,s+1 i {1,,s} are bounded on ιr Let M > be a bound for these functions For all i {i,, s}, an Abel summation yields n e pha J n p Ψ i,s+1 ( ha ( ha f (s (t n p = + ( n e pha J Ψ i,s+1 ( ha ( ha f (s ( ( p e qha t n p d [ g t n p 1 dt u (t,u(tψ i,s+1( ha ( ha ] f (s (t dt q= 22

23 On one hand, ( n e pha J Ψ i,s+1 ( ha ( ha f (s ( TL 1M f (s ( H r H r+2 On the other hand, since after differentiating and using Hypothesis 311 and 313, one has d [ g dt u (t,u(tψ i,s+1( ha ( ha f (t] (s H r hl 3 M f (s (t H r+2 + hl 3 M R f (s (t H r+2 + hl 1 M f (s+1 (t H r+2, where R denotes the maximum of u (t H r on [,T] (see Hypothesis 33, we deduce that there exists a positive constant C 4 such that for all h and n, the term (326 has an H r -norm bounded by C 4 h s+2 The same kind of calculations yields, denoting (θ i i {1,,s} the holomorphic functions such that i {1,,n}, z C, b i (z b i ( = zθ i (z = θ i (zz, ( bi ( ha b i ( n e pha J n p ψ i,s+1 (f (s (t n p = ( n ψ i,s+1 ( e pha θ i ( ha ( ha J f (s ( ( p + ψ i,s+1 ( e qha θ i ( ha ( ha t n p d [ g ] t n p 1 dt u (t,u(tf(s (t dt q= If we also denote by M a positive constant bounding the functions (θ i i {1,,s}, then e qha θ i ( ha ( ha t n p d [ g ] t n p 1 dt u (t,u(tf(s (t dt H r tn p Mh A d [ g ] Hr t n p 1 dt u (t,u(tf(s (t dt tn p Mh d [ g ] dt u (t,u(tf(s (t Hr+2 dt t n p 1 Hypotheses 311 and 313 ensure that d [ g ] dt u (t,u(tf(s (t H r+2 dt L 3 (1 + R tn p t n p 1 f (s (t H r+2 dt + L 1 tn p t n p 1 f (s+1 (t H r+2 dt, if R stands for a bound of u (t H r+2 on [,T](see Hypothesis 314 Hence, the H r -norm of term (327 is bounded by C 5 h s+2 for some positive constant C 5 23

24 One can easily check by using Hypothesis 311 that there exists a positive constant C 6 such that for all h and n, the H r -norm of term (328 is bounded by C 6 h s+2 Hence, term (322 is bounded for all h and n by (C 4 + C 5 + C 6 h s+2 Eventually, one can estimate term (323 by writing, for i {1,,s}, h (,h, n {,,N 1}, and p {,,n}, using Hypothesis 312 d n p,i H r L 2 ( u(t n p u n p,i H r+ u(t n p u(t n p +c i h H r e n p,i H r Therefore, we get that h b i ( ha CL 2 h n n e pha H dn p,i r Note that we have using (315 and (318 ( e n p,i 2 2 e H r n p,i H r ( (s + 1 4s 2 e n p H r ( e n p,i 2 + Rh e H r n p,i H r n p,i 2 H r (s + 1 (4s 2 e n p 2H tn p+1 + 4sκ 2 sh 2s+1 r t n p f (s (τ 2 dτ H r tn p+1 (s + 1 (4s 2 R e n p H r + 4sκ 2 s h2s+1 f (s (τ 2 dτ H t r n p We derive that there exists a positive constant C 7 such that for all h and n, the H r -norm of term (323 is bounded by n C 7 h e p H r + C 7 h s+2 The conclusion follows using Lemma 37 4 Numerical experiments We provide two kinds of numerical experiments to illustrate our results Both of them have been performed with s = 2 collocation points The first method we choose is defined by c 1 = 1 2 and c 2 = 1 This method does not satisfy relation (211 Hence, its underlying Runge Kutta method exactly is of order 2 The other method we choose is defined by c 1 = 1 3 and c 2 = 1 This method satisfies relation (211 but not relation (221 Hence, its underlying Runge Kutta method exactly is of order 3 24

25 41 Linear problems Firstly, we consider a linear problem (24 with dimension d = 1 The functions u and f are chosen in such a way that the exact solution of the problem is ( 2 u(t,x = e ι t 2 sin(x We set the final time T = 2π We apply Methods 1 and 2 to this problem for different time steps h > such that T/h is an integer We plot in logarithmic scales the L 2 -norm of the final error e T as a function of h h on Figure 1 Calculations are carried out with Fast Fourier Transforms with 2 8 modes One can see that, in both cases, the error plot lies between two different straight lines with same slope The upper line is reached when T/h = 2π/h is close to the square of an integer, that is to say when the time step is resonant (for example, when T/h = 2π/ = 1 2, relation (217 does not hold In this case, inequality (212 of Theorem 24 holds and seems to be sharp On the other hand, for non-resonant time steps, estimate (219 of Theorem 26 holds and seems to be sharp Of course, estimate (212 also holds, but it does not seem to be sharp anymore 3 4 Method 1 Method 2 Final error (logarithmic scale Time step (logarithmic scale Figure 1: Final error as a function of the time step for a linear Schrödinger problem Method 1 (upper dotted line and 2 (lower starred line Logarithmic scales 25

26 42 Semi-linear problems Secondly, we consider a semi-linear problem (31 with dimension d = 1 The non-linearity is g(u = i u 2 u and the initial datum is u (x = 1+e2ix 2+sin(x We set the final time T = 2π We apply Methods 1 and 2 to this problem for different time steps h > such that T/h is an integer We plot in logarithmic scales the L 2 -norm of the final error e T as a function of h on Figure 2 Calculations are h carried out with Fast Fourier Transforms with 2 8 modes One can see that no resonance occurs for time steps between 1 3 and 1 1, even when T/h is the square of an integer, and that both methods have a numerical order consistent with Theorems 36 and 38: the upper straight line has a numerical slope close to 2, while the lower straight line has a numerical slope close to Method 1 Method 2 Final error (logarithmic scale Time step (logarithmic scale Figure 2: Final error as a function of the time step for the cubic Schrödinger equation Method 1 (upper dotted line, numerical slope and Method 2 (lower starred line, numerical slope Logarithmic scales 26

27 5 Conclusion This paper provides a numerical analysis of exponential Runge Kutta methods of collocation type applied to the linear and semi-linear Schrödinger equations on a d-dimensional torus These methods have been studied in [14] when applied to parabolic problems This paper shows how the results of [14] extend to the case of the Schrödinger equation and provides sufficient conditions to achieve orders s, s+1 and s+2 using an s-stage method (see Theorems 21, 24, 26, 28 and 29 for linear Schrödinger problems and Theorems 36, 38 and 315 for non-linear Schrödinger problems Essentially, when solving Schrödinger problems, the methods behave as they do when solving parabolic problems : under some suitable assumptions (on the right-hand side or on the non-linearity, an s-stage method is of order s and can achieve orders s + 1 and s + 2 However, the proofs of the results presented in this paper require some hypotheses (for example on the regularity of the exact solution of the problem, like Hypotheses 33 and 314 that seem more restrictive in the Schrödinger context than the corresponding ones in the parabolic context Moreover, this paper points out a major difference between parabolic problems and Schrödinger problems solved by such exponential Runge Kutta methods: it shows and explains the effect of resonant and non-resonant time steps over finite time intervals when solving linear Schrödinger problems (see Theorems 26 and 29 Acknowledgment The author would like to thank Philippe Chartier and Erwan Faou for their comments and ideas about this work References [1] A Alinhac, P Gérard, Opérateurs pseudo-différentiels et théorie de Nash- Moser, InterÉditions/Éditions du CNRS [2] H Berland, AL Islas, CM Schober, Conservation of phase space properties using exponential integrators on the cubic Schrödinger equation, Journal of Computational Physics, 255(1: (27 [3] H Berland, B Skaflestad, Solving the nonlinear Schrödinger equation using exponential integrators, Proceedings SIMS 25 46th Conference on Simulation and Modeling Trondheim, Norway [4] D Cohen, E Hairer, C Lubich, Long-time analysis of nonlinearly perturbed wave equations via modulated Fourier expansions, to appear in Arch Ration Mech Anal [5] D Cohen, E Hairer, C Lubich, Conservation of energy, momentum and actions in numerical discretizations of nonlinear wave equations, April 27, preprint 27

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