A fully discrete low-regularity integrator for the 1D periodic cubic nonlinear Schrödinger equation
aa r X i v : . [ m a t h . NA ] J a n A FULLY DISCRETE LOW-REGULARITY INTEGRATOR FOR THE1D PERIODIC CUBIC NONLINEAR SCHR ¨ODINGER EQUATION
BUYANG LI AND YIFEI WU
Abstract.
A fully discrete and fully explicit low-regularity integrator is constructed forthe one-dimensional periodic cubic nonlinear Schr¨odinger equation. The method can beimplemented by using fast Fourier transform with O ( N ln N ) operations at every time level,and is proved to have an L -norm error bound of O ( τ p ln(1 /τ ) + N − ) for H initial data,without requiring any CFL condition, where τ and N denote the temporal stepsize and thedegree of freedoms in the spatial discretisation, respectively.
1. Introduction
This article concerns the numerical solution of the cubic nonlinear Schr¨odinger (NLS)equation ( i∂ t u ( t, x ) + ∂ xx u ( t, x ) = λ | u ( t, x ) | u ( t, x ) for x ∈ T and t ∈ (0 , T ] ,u (0 , x ) = u ( x ) for x ∈ T , (1.1)on the one-dimensional torus T = ( − π, π ) with a nonsmooth initial value u ∈ H ( T ), where λ = − H s ( T ) for s ≥
0; see [2].The construction of numerical methods for the NLS equation and related dispersiveequations with nonsmooth initial data has attracted much attention recently since the pio-neering work of Ostermann & Schratz [17], who introduced a low-regularity exponential-typeintegrator that could have first-order convergence in H γ ( T d ) for initial data u ∈ H γ +1 ( T d )and γ > d , where d denotes the dimension of space. Before their work, the traditionalregularity assumption for the NLS equation for a time-stepping method to have first-orderconvergence in H γ ( T d ) is u ∈ H γ +2 ( T d ) for γ ≥ H γ ( T ) for initial data u ∈ H γ ( T ) and γ > (without losing any derivative). Ostermann, Rousset & Schratz furthermore weakened theregularity assumption of initial data to u ∈ H ( T ) in [15] and u ∈ H s ( T ) with s ∈ (0 , u ∈ H ( T ) these methodswere proved to have L -norm error bounds of O ( τ ) and O ( τ − ǫ ), respectively, for the one-dimensional NLS equation. A general framework of low-regularity integrators for nonlinear Mathematics Subject Classification.
Key words and phrases.
Nonlinear Schr¨odinger equation, numerical solution, first-order convergence, lowregularity, fast Fourier transform. parabolic, dispersive and hyperbolic equations was introduced in [7], where the conditionfor the numerical solution of the NLS equation to have first-order convergence in L ( T ) is u ∈ H ( T ).Besides the NLS equation, the techniques of twisted variable and harmonic analysistechniques were also used in the construction of low-regularity integrators for other dispersiveequations; see [9, 18, 20, 23, 24] and the references therein.As far as we know, the analysis of all the low-regularity integrators for the NLS equationare limited to semidiscretisation in time (the error from spatial discretisation is unknown fornonsmooth initial data), and the regularity condition for the time-stepping method to havefirst-order convergence is u ∈ H γ ( T ) for γ ≥ . We are only aware of a fully discreteLawson-type exponential integrator for the Korteweg–de Vries equation [18], with first-orderconvergence in L ( T ) in both time and space under a CFL condition τ = O ( h ) for solutionsin C (0 , T ; H ( T )).The objective of this article is to construct a fully discrete and fully explicit lower-regularity integrator that has first-order convergence (up to a logarithmic factor) in bothtime and space for H initial data. The temporal low-regularity integrator is constructedusing twisted variables and with different harmonic analysis techniques in approximatingthe low- and high-frequency parts of the functions in the exponential integral. The spatialdiscretisation is integrated in the temporal low-regularity integrator by repeatedly using fre-quency truncation and Fast Fourier transform (FFT) techniques in every nonlinear operation(i.e., computing the product of two functions). By using a (4 N + 1)-point FFT for everyproduct of two (2 N + 1)-term Fourier series in the numerical scheme and then truncating theobtained (4 N + 1)-term product series to (2 N + 1)-term again, we avoid generating trigono-metric interpolation errors from using FFT. As a result, the spatial discretisation error ofour method is purely due to frequency truncation and therefore can be analysed togetherwith the temporal discretisation error in the frequency domain by using harmonic analysistechniques.The rest of this article is organised as follows. The fully discrete low-regularity integratorand the main theorem on the convergence rates of the method are presented in section 2.Some technical tools of harmonic analysis are presented in section 3, which are used in section4 in the construction of the numerical method and analysis of the consistency error. Theerror bound of proposed fully discrete low-regularity integrator is proved in section 5 byutilizing the consistency error bounds obtained in section 4 and the stability of the method,as well as the H -regularity of fully discrete numerical solution. The latter is proved to bebounded uniformly with respect to the temporal stepsize and the number of Fourier termsin the spatial discretisation. Numerical results are presented in section 6 to support thetheoretical analysis in this article.
2. The numerical method and main theoretical result
It is known that the solution of the NLS equation satisfies the following two conservationlaws (see e.g., [4]):(1) Mass conservation:12 π Z T | u ( t, x ) | dx = 12 π Z T | u ( x ) | dx for t > . (2.1)(2) Momentum conservation:12 π Z T u ( t, x ) ∂ x ¯ u ( t, x ) dx = 12 π Z T u ∂ x ¯ u dx for t > . (2.2)These two conserved quantities will be approximated based on the initial data and utilizedin the construction of the numerical method. We denote by Π and Π =0 the zero-mode and nonzero-mode operators, respectively,defined by Π f = 12 π Z T f ( x ) dx and Π =0 ( f ) = X k ∈ Z ,k =0 e ikx ˆ f k . (2.3)Then the conserved mass and momentum are denoted by M = 12 π Z T | u ( x ) | dx = Π ( | u | ) and P = 12 π Z T u ∂ x u dx = Π (cid:0) u ∂ x u (cid:1) , (2.4)respectively.For any positive integer N , we denote by I N the (4 N + 1)-point trigonometric interpo-lation operator, which can be obtained through the discrete Fourier transform (see [5, 25]) I N f ( x ) = N X k = − N e ikx ˜ f k with ˜ f k = 14 N + 1 N X n = − N e − ikx n f ( x n ) (2.5)where x n = 2 πn N + 1 for n = − N, · · · , N. If the Fourier coefficient ˆ f k of the function f satisfies that ˆ f k = 0 for | k | > N , then I N f = f and therefore ˜ f k = ˆ f k in the formula (2.5). In this case, both f ( x n ) = N X k = − N e ikx n ˆ f k , n = − N, · · · , N, (2.6)and ˆ f k = N X n = − N e − ikx n f ( x n ) k = − N, · · · , N, (2.7)can be computed with cost O ( N ln N ) by using the fast Fourier transform (FFT); see [5].Let S N be the subspace of functions f ∈ L ( T ) such that ˆ f k = 0 for | k | > N . If w, v ∈ S N and their Fourier coefficients ˆ w k and ˆ v k , k = − N, · · · , N , are stored in the computer (withˆ w k = ˆ v k = 0 for N < | k | ≤ N ), then the values w ( x n ) and v ( x n ), n = − N, . . . , N , canbe computed exactly by using (2.6) and FFT. Since ( wv ) k = 0 for | k | > N , it follows that wv = I N ( wv ). If we denote by F k [ v ] the k th Fourier coefficient of the function v , then F k [ wv ] = N X n = − N e − ikx n w ( x n ) v ( x n ) , k = − N, . . . , N, which can also be computed exactly by using FFT. Therefore, if we denote by Π N : L ( T ) → L ( T ) the projection operator defined by F k [Π N f ] = ( ˆ f k for | k | ≤ N, | k | > N, then the cost of computing the Fourier coefficients of Π N ( wv ) ∈ S N from the Fourier coeffi-cients of w, v ∈ S N is O ( N ln N ).For any positive integer L , let t n = nτ , n = 0 , , . . . , L , be a partition of the time interval[0 , T ] with stepsize τ = T /L . The fully discrete low-regularity integrator for the NLS equation(1.1) to be constructed in this paper is: For given u nτ,N ∈ S N compute u n +1 τ,N ∈ S N by u n +1 τ,N = Ψ( u nτ,N ) for n = 0 , . . . , L − , with u τ,N = Π N I N u ∈ S N , (2.8) whereΨ( f ) := e iτ ( − λP N ∂ − x − λM N + ∂ x ) f + (1 − e − iλτM N )Π f − iλτ Π (cid:2) Π N ( | f | ) f (cid:3) + λ∂ − x Π N (cid:2) (e iτ∂ x f ) · ∂ − x Π N ( | e iτ∂ x f | ) (cid:3) − λ e iτ∂ x ∂ − x Π N (cid:2) f · ∂ − x Π N ( | f | ) (cid:3) − λ h ∂ − x Π N (cid:16) (e − iτ∂ x ¯ f ) e iτ∂ x Π N ( f ) (cid:17) − e iτ∂ x ∂ − x Π N (cid:16) ¯ f Π N ( f ) (cid:17)i − λ iτ∂ x ∂ − x Π N h ∂ x ¯ f (cid:16) e − iτ∂ x Π N (cid:2) (e iτ∂ x ∂ − x f ) (cid:3) − Π N (cid:2) ( ∂ − x f ) (cid:3)(cid:17)i − iλτ e iτ∂ x ∂ − x Π N (cid:0) ∂ x ¯ f Π N ( f ) (cid:1) + 2 iλτ Π f e iτ∂ x ∂ − x Π N (cid:0) ∂ x ¯ f f (cid:1) − iλτ (Π f ) e iτ∂ x Π =0 ¯ f for f ∈ S N , (2.9)and M N = Π ( | u τ,N | ) and P N = Π (cid:0) u τ,N ∂ x u τ,N (cid:1) (2.10)are the approximate mass and momentum, respectively. By using (2.5) with FFT, the initialvalue u τ,N = Π N I N u can be obtained with cost O ( N ln N ). Then, at every time level, themethod only requires computing several functions in the following forms: • e iτ ( − M N − P N ∂ − x + ∂ x ) f, e ± iτ∂ x f and ∂ − x f for some given function f ∈ S N , • Π N ( f g ) for some given functions f, g ∈ S N , where F k [e iτ ( − M N − P N ∂ − x + ∂ x ) f ] = ( e − M N iτ ˆ f for k = 0 , e iτ ( − M N − P N ( ik ) − − k ) ˆ f k for k = 0 . F k [e ± iτ∂ x f ] = e ∓ iτk ˆ f k and ∂ − x f = ( k = 0 , ( ik ) − ˆ f k for k = 0 . Hence, the computational cost is O ( N ln N ) at every time level.The main theoretical result of this paper is the following theorem. Theorem 2.1. If u ∈ H ( T ) then there exist positive constants τ , N and C such thatfor τ ≤ τ and N ≥ N the numerical solution given by (2.8) – (2.9) has the following errorbound: max ≤ n ≤ L k u ( t n , · ) − u nτ,N k L ≤ C (cid:0) τ p ln(1 /τ ) + N − (cid:1) , (2.11) where the constants τ , N and C depend only on T and k u k H . The rest of this paper is devoted to the construction of the method (2.8)–(2.9) and theproof of Theorem 2.1.
Remark 2.2.
The analysis in this article can be easily extended to proving higher-orderconvergence of the spatial discretisation method when the initial data is smoother. Namely,for u ∈ H s ( T ) with s >
1, the error bound of of the proposed method should becomemax ≤ n ≤ L k u ( t n , · ) − u nτ,N k L ≤ C (cid:0) τ + N − s (cid:1) . (2.12)The proof of this result (with smoother initial data) is easier than the proof of Theorem 2.1and therefore omitted. The convergence results in (2.11) and (2.12) are illustrated by thenumerical experiments in section 6 for s = 1 and s = 2, respectively.
3. Notation and technical tools
In this section we introduce the basic notation and technical lemmas to be used inanalysing the error of the numerical method to be constructed.
The inner product and norm on L ( T ) are denoted by( f, g ) = Z T f ( x ) g ( x ) dx and k f k L = p ( f, f ) , respectively . The norm on the Sobolev space H s ( T ), s ∈ R , is denoted by (cid:13)(cid:13) f (cid:13)(cid:13) H s = 2 π X k ∈ Z (1 + k ) s | ˆ f k | . For a function f : [0 , T ] × T → C we denote by k f k L p (0 ,T ; H s ) its space-time Sobolev norm,defined by k f k L p (0 ,T ; H s ) = (cid:18) Z T k f ( t ) k pH s dt (cid:19) p for p ∈ [1 , ∞ ) , ess sup t ∈ [0 ,T ] k f ( t ) k H s for p = ∞ . The Fourier coefficients of a function f on T are denoted by F k [ f ] or simply ˆ f k , definedby ˆ f k = 12 π Z T e − ikx f ( x ) dx for k ∈ Z . The Fourier inversion formula is given by f ( x ) = X k ∈ Z e ikx ˆ f k . The Fourier coefficients are known to have the following properties: k f k L = 2 π X k ∈ Z (cid:12)(cid:12) ˆ f k (cid:12)(cid:12) (Plancherel identity); F k [ f g ] = X k ∈ Z ˆ f k − k ˆ g k (Convolution) . For any function σ : Z → C such that | σ ( k ) | ≤ C σ (1 + | k | ) m for some constants C σ and m ≥
0, we denote by σ ( i − ∂ x ) : H s ( T ) → H s − m ( T ) the operator defined by σ ( i − ∂ x ) f = X k ∈ Z σ ( k ) ˆ f k e ikx . For abbreviation, we denote h k i = (1 + k ) and J s = h i − ∂ x i s , which imply that (cid:13)(cid:13) f (cid:13)(cid:13) H s = (cid:13)(cid:13) J s f (cid:13)(cid:13) L and d J s f k = h k i s ˆ f k . Moreover, we denote by ∂ − x : H s ( T ) → H s +1 ( T ), s ∈ R , the operator such that F k [ ∂ − x f ] = ( ( ik ) − ˆ f k , when k = 0 , , when k = 0 . (3.1)We denote by A . B or B & A the statement A ≤ CB for some constant C >
0. Thevalue of C may depend on T and k u k H , and may be different at different occurrences, butis always independent of τ , N and n . The notation A ∼ B means that A . B . A .We denote by O ( Y ) any quantity X such that X . Y . For any function σ : Z m +1 → C and w ∈ H ( T ) we denote by T m ( σ ; w ) the class of functions f ∈ L ( T ) such thatˆ f k . X k + ··· + k m = k | σ ( k, k , · · · , k m ) | | ˆ w k | · · · | ˆ w k m | ∀ f ∈ T m ( σ ; w ) . (3.2) If F = R t t f ( t ) dt for some function f ( t ) ∈ T m ( σ ; v ( t )), then we simply denote F ∈ Z t t T m ( σ ; v ( t )) dt. (3.3) We will use the following version of the Kato–Ponce inequalities, which was originallyproved in [12] and subsequently improved to cover the endpoint case in [3, 13].
Lemma 3.1 (The Kato–Ponce inequalities) . (i) If s > and f, g ∈ H s ( T ) then k f g k H s . k f k H s k g k H s . (ii) If s ≥ , s > , f ∈ H s + s ( T ) and g ∈ H s ( T ) , then k f g k H s . k f k H s + s k g k H s . In addition to Lemma 3.1 we also need the following results, which are consequences ofthe Kato–Ponce inequalities.
Lemma 3.2. (i) If s > and f, g ∈ H s ( T ) then k J − ( J f g ) k H s . k f k H s k g k H s . (ii) If f, g ∈ H ( T ) then k J − ( J f g ) k L . min (cid:8) k f k L k g k H , k g k L k f k H (cid:9) . Proof. (i) The desired inequality is equivalent to k J s − ( J f g ) k L . k f k H s k g k H s . By theduality between L ( T ) and itself, it suffices to prove( J s − ( J f g ) , h ) . k f k H s k g k H s k h k L ∀ h ∈ L ( T ) , which is equivalent to X k X k + k = k h k i s − h k i ˆ f k ˆ g k ˆ h k . k f k H s k g k H s k h k L . Since the term corresponding to k = 0 satisfies X k + k =0 h k i ˆ f k ˆ g k ˆ h = X k h k i ˆ f k h− k i ˆ g − k ˆ h . k ( h k i ˆ f k ) k ∈ Z k l k ( h− k i ˆ g − k ) k ∈ Z k l | ˆ h | . k f k H k g k H k h k L . k f k H s k g k H s k h k L when s > , we only need to prove the following result: X k =0 X k + k = k | k | s − | k | ˆ f k ˆ g k ˆ h k . k f k H s k g k H s k h k L . To this end, we decompose the left-hand side of the inequality above into two parts, i.e., X k =0 X k + k = k | k | s − | k | ˆ f k ˆ g k ˆ h k . X k =0 X k + k = k | k |≤ | k | | k | s − | k || ˆ f k || ˆ g k || ˆ h k | + X k =0 X k + k = k | k | > | k | | k | s − | k || ˆ f k || ˆ g k || ˆ h k | . (3.4) The first term on the right-hand side of (3.4) can be estimated by using Plancherel’sidentity and Lemma 3.1 as follows: X k =0 X k + k = k | k |≤ | k | | k | − s | k || ˆ f k || ˆ g k || ˆ h k | . X k =0 X k + k = k | k |≤ | k | | k | s | ˆ f k || ˆ g k || ˆ h k | . ( J s ( ˜ f ˜ g ) , ˜ h ) . (cid:13)(cid:13) ˜ f ˜ g (cid:13)(cid:13) H s k ˜ h k L . k ˜ f k H s k ˜ g k H s k ˜ h k L , where ˜ f , ˜ g and ˜ h are functions with Fourier coefficients | ˆ f k | , | ˆ g k | and | ˆ h k | , respectively. Since k ˜ f k H s ∼ k f k H s , k ˜ g k H s ∼ k g k H s and k ˜ h k L ∼ k h k L , it follows that X k =0 X k + k = k | k |≤ | k | | k | − s | k || ˆ f k || ˆ g k || ˆ h k | . k f k H s k g k H s k h k L . In the second term on the right-hand side of (3.4), we have | k | ∼ | k | > | k | . For s > we have | k | s − | k | = | k | − s | k | s − | k | ≤ | k | − s | k | s ∼ | k | − s | k | s | k | s and therefore X k =0 X k + k = k | k | > | k | | k | s − | k || ˆ f k || ˆ g k || ˆ h k | . X k =0 X k + k = k | k | > | k | | k | − s | k | s | k | s | ˆ f k || ˆ g k || ˆ h k | . X k =0 F k [ J s ˜ f J s ˜ g ] | k | − s | ˆ h k | . max k |F k [ J s ˜ f J s ˜ g ] | X k =0 | k | − s | ˆ h k | . k J s ˜ f J s ˜ g k L k ( | k | − s ) = k ∈ Z k l k ( | ˆ h k | ) = k ∈ Z k l . k J s ˜ f k L k J s ˜ g k L k ˜ h k L . k f k H s k g k H s k h k L . This completes the proof of (i).(ii) Similarly as (i), it suffices to prove X k =0 X k + k = k | k | − | k | ˆ f k ˆ g k ˆ h k . min( k f k L k g k H , k f k H k g k L ) k h k L ∀ h ∈ L ( T ) . In view of the proof of (i), we can assume ˆ f k ≥
0, ˆ g k ≥ h k ≥ f , g and h by ˜ f , ˜ g and ˜ h , respectively, in the estimates below).Then X k =0 X k + k = k | k | − | k | ˆ f k ˆ g k ˆ h k . X k =0 X k + k = k | k |≤ | k | | k | − | k | ˆ f k ˆ g k ˆ h k + X k =0 X k + k = k | k | > | k | | k | − | k | ˆ f k ˆ g k ˆ h k . (3.5)The first term on the right-hand side of (3.5) can be estimated by using Plancherel’s identityand Lemma 3.1: X k =0 X k + k = k | k |≤ | k | | k | − | k | ˆ f k ˆ g k ˆ h k . X k =0 X k + k = k | k |≤ | k | ˆ f k ˆ g k ˆ h k . X k =0 F k [ f g ]ˆ h k . k ( F k [ f g ]) = k ∈ Z k l k (ˆ h k ) = k ∈ Z k l . (cid:13)(cid:13) f g (cid:13)(cid:13) L k h k L . min( k f k L k g k L ∞ , k f k L ∞ k g k L ) k h k L . min( k f k L k g k H , k f k H k g k L ) k h k L . In the second term on the right-hand side of (3.5) we have | k | ∼ | k | > k . On the one hand,we have X k =0 X k + k = k | k | > | k | | k | − | k | ˆ f k ˆ g k ˆ h k . X k =0 X k | k | − | k | ˆ f k − k ˆ g k ˆ h k . (cid:18) sup k X k | ˆ f k − k | (cid:19) (cid:18) X k | k | | ˆ g k | (cid:19) X k =0 | k | − ˆ h k . k f k L k g k H k h k L . On the other hand, we have X k =0 X k = k + k | k | > | k | | k | − | k | ˆ f k ˆ g k ˆ h k . X k =0 X k | k | − | k | ˆ f k ˆ g k − k ˆ h k . (cid:18) X k | k | | ˆ f k | (cid:19) (cid:18) sup k X k | ˆ g k − k | (cid:19) X k =0 | k | − ˆ h k . k f k H k g k L k h k L . This completes the proof of (ii). (cid:3)
4. Construction of the method through analysing consistency error
In this section we construct the numerical method based on twisted variables andDuhamel’s formula through analysing the consistency errors in approximating the expo-nential integrals using harmonic analysis techniques. For readers’ convenience, we presentthe derivation of the numerical method in subsection 4.1 and defer the technical estimatesto subsection 4.2.
As mentioned in the introduction section and the beginning of section 2, the NLS equation(1.1) has a unique solution u ∈ C ([0 , T ]; H ( T )) satisfying the Duhamel’s formula: u ( t n +1 ) = e iτ∂ x u ( t n ) − iλ Z τ e i ( t n +1 − ( t n + s )) ∂ x | u ( t n + s ) | u ( t n + s ) ds, (4.1)as well as the mass and momentum conservations (2.1)–(2.2). The norm k u k C ([0 ,T ]; H ( T )) isbounded by a constant depending on k u k H ; see [2].Let v ( t ) := e − it∂ x u ( t ) be the twisted variable. Then v ∈ C ([0 , T ]; H ( T )) satisfies k v k C ([0 ,T ]; H ( T )) = k u k C ([0 ,T ]; H ( T )) and the following conservation laws simiarly as u , i.e.,(1) Mass conservation:12 π Z T | v ( t, x ) | dx = 12 π Z T | u ( t, x ) | dx = M for t > . (4.2) (2) Momentum conservation:12 π Z T v ( t, x ) ∂ x ¯ v ( t, x ) dx = 12 π Z T u ( t, x ) ∂ x ¯ u ( t, x ) dx = P for t > . (4.3)Applying the operator e − it n +1 ∂ x to the identity (4.1), we obtain v ( t n +1 ) = v ( t n ) − iλ Z τ e − i ( t n + s ) ∂ x (cid:2) | e i ( t n + s ) ∂ x v ( t n + s ) | e i ( t n + s ) ∂ x v ( t n + s ) (cid:3) ds. (4.4)The Fourier coefficients of both sides of (4.4) should be equal, i.e.,ˆ v k ( t n +1 ) = ˆ v k ( t n ) − iλ Z τ X k + k + k = k e i ( t n + s ) φ ˆ¯ v k ( t n + s )ˆ v k ( t n + s )ˆ v k ( t n + s ) ds, (4.5)with a phase function φ = φ ( k, k , k , k ) = k + k − k − k . Replacing τ and s in (4.5) by s and σ , respectively, we haveˆ v k ( t n + s ) = ˆ v k ( t n ) − iλ Z s X k + k + k = k e i ( t n + σ ) φ ˆ¯ v k ( t n + s )ˆ v k ( t n + σ )ˆ v k ( t n + σ ) dσ. (4.6)In view of (4.6) and the definition of T m ( M ; v ) in (3.2), we have v ( t n + s ) − v ( t n ) ∈ Z s T (1; v ( t n + σ )) dσ. (4.7)As a result, (4.5) can be written asˆ v k ( t n +1 ) = ˆ v k ( t n ) − iλ X k + k + k = k ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) Z τ e i ( t n + s ) φ ds + ˆ R ,k , (4.8)withˆ R ,k = − iλ Z τ X k + k + k = k e i ( t n + s ) φ (cid:0) ˆ¯ v k ( t n + s )ˆ v k ( t n + s )ˆ v k ( t n + s ) − ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) (cid:1) ds ∈ Z τ Z s T (1; v ( t n + σ )) dσds, where the last inclusion is based on the definition in (3.3). If R denotes the function withFourier coefficients ˆ R ,k , then the relation above implies that (according to Lemma 4.1 (i)of the next subsection) (cid:13)(cid:13) R (cid:13)(cid:13) H . τ k v k L ∞ t H x . (4.9)This term will be dropped in our numerical scheme.In the following, we approximate the second term on the right-hand side of (4.8) byexpressions that can be evaluated efficiently with FFT. To this end, we consider the threecases k = 0, | k | > N and 0 = | k | ≤ N , separately. Case k = 0. In this case, (4.8) reduces toˆ v ( t n +1 ) =ˆ v ( t n ) − iλ X k + k + k =0 ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) Z τ e i ( t n + s )( k − k − k ) ds + ˆ R , (4.10)=ˆ v ( t n ) − iλτ X k + k + k =0 e it n ( k − k − k ) ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) + ˆ R , + ˆ R , =ˆ v ( t n ) − iλτ Π (cid:0)(cid:12)(cid:12) e it n ∂ x v ( t n ) (cid:12)(cid:12) e it n ∂ x v ( t n ) (cid:1) + ˆ R , + ˆ R , =ˆ v ( t n ) − iλτ Π (cid:2) Π N (cid:0)(cid:12)(cid:12) e it n ∂ x v ( t n ) (cid:12)(cid:12) )e it n ∂ x v ( t n ) (cid:3) + ˆ R , + ˆ R , + ˆ R ∗ , , (4.11)withˆ R ,k = − iλ X k + k + k = k b ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) Z τ (e i ( t n + s )( k − k − k ) − e it n ( k − k − k ) ) ds, R ∗ = − iλτ (cid:2) (1 − Π N ) (cid:0)(cid:12)(cid:12) e it n ∂ x v ( t n ) (cid:12)(cid:12) (cid:1)(cid:3) e it n ∂ x v ( t n ) ∈ τ e it n ∂ x v ( t n ) T (1 >N ; v ( t n )) , where (1 >N ) k = 1 | k | >N = ( | k | ≤ N, | k | > N. (4.12)Since k + k + k = 0, it follows that there holds k − k − k = 2 k k and therefore Z τ (cid:16) e i ( t n + s )( k − k − k ) − e it n ( k − k − k ) (cid:17) ds = τ O ( k k ) . As a result, the function R (with Fourier coefficients ˆ R ,k ) satisfies that R ∈ τ T ( k k ; v ( t n ))in view of the definition in (3.2). According to Lemma 4.1 (i)–(ii) of the next subsection, R and R ∗ satisfy the following estimates: | ˆ R , | . τ k v k L ∞ t H x , (4.13) | ˆ R ∗ , | . kR ∗ k L . τ k e it n ∂ x v ( t n ) k L ∞ k (1 − Π N )( | e it n ∂ x v ( t n ) | ) k L . τ N − k v k L ∞ t H x . (4.14)The two terms ˆ R , and ˆ R ∗ , will be dropped in our numerical scheme. Case | k | > N . Let R be the function with Fourier coefficientsˆ R ,k = − | k | >N iλ X k + k + k = k ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) Z τ e i ( t n + s ) φ ds. Then R ∈ τ T (1 >N ; v ( t n )) . Lemma 4.1 (i) of the next subsection implies that (cid:13)(cid:13) R (cid:13)(cid:13) H s . τ N − s k v k L ∞ t H x for s ∈ [0 , . (4.15)This term will be dropped in the numerical scheme. Case
3: 0 = | k | ≤ N . By using the identity1 = ( k + k ) + ( k + k ) − k k and symmetry between k and k , we can decompose the second term on the right-hand sideof (4.8) into two parts, i.e.,ˆ v k ( t n +1 ) = ˆ v k ( t n ) − iλ X k + k + k = k ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) Z τ k + k k e i ( t n + s ) φ ds (4.16a)+ iλ X k + k + k = k ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) Z τ k k e i ( t n + s ) φ ds (4.16b)+ ˆ R ,k . (4.16c)We furthermore truncate (4.16a) to the frequency domain | k + k | ≤ N , i.e.,(4.16a) = ˆ v k ( t n ) − iλ X k + k + k = k | k + k |≤ N (cid:18) Z τ k + k k e i ( t n + s ) φ ds (cid:19) ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) + ˆ R ,k , (4.17) withˆ R ,k = − iλ X k + k + k = k | k + k | >N (cid:18) Z τ k + k k e i ( t n + s ) φ ds (cid:19) ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) for 0 = | k | ≤ N, . The corresponding function R with Fourier coefficients ˆ R ,k satisfies that R ∈ τ T (cid:18) k + k k = | k |≤ N | k + k | >N ; v ( t n ) (cid:19) . By Lemma 4.1 (iii) in the next subsection and symmetry between k and k , and we have (cid:13)(cid:13) R (cid:13)(cid:13) H s . τ N − s k v k L ∞ t H x for s ∈ [0 , . (4.18)Since k + k + k = k , it is straightforward to verify that φ = 2( k + k )( k + k ) . As aresult, if k + k = 0 then Z τ k + k k e i ( t n + s ) φ ds = 12 ik ( k + k ) (cid:0) e it n +1 φ − e it n φ (cid:1) ; (4.19)If k + k = 0 then φ = 0 and k = k , and therefore Z τ k + k k e i ( t n + s ) φ ds = τ (cid:16) k k + 1 (cid:17) . (4.20)Substituting the two relations (4.19)–(4.20) into (4.17), we obtain(4.16a) = ˆ v k ( t n ) − λ X k + k + k = k = | k + k |≤ N k ( k + k ) (cid:0) e it n +1 φ − e it n φ (cid:1) ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) − iλτ X k + k =0 (cid:16) k k + 1 (cid:17) ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) + ˆ R ,k . Then we apply the mass and momentum conservations in (4.2)–(4.3), which imply that − iλτ X k + k =0 (cid:16) k k + 1 (cid:17) ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) = − iλτ P ( ik ) − ˆ v k ( t n ) − iλτ M ˆ v k ( t n ) . Therefore,(4.16a) =ˆ v k ( t n ) − λ X k + k + k = k = | k + k |≤ N k ( k + k ) (cid:0) e it n +1 φ − e it n φ (cid:1) ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) − iλτ P ( ik ) − ˆ v k ( t n ) − iλτ M ˆ v k ( t n ) + ˆ R ,k = e − iλτP ( ik ) − − iλτM ˆ v k ( t n ) − λ X k + k + k = k = | k + k |≤ N k ( k + k ) (cid:0) e it n +1 φ − e it n φ (cid:1) ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n )+ ˆ R ,k + ˆ R ∗ ,k , (4.21)whereˆ R ∗ ,k = ((cid:0) − iλτ P ( ik ) − − iλτ M − e − iλτP ( ik ) − − iλτM (cid:1) ˆ v k ( t n ) for 0 = | k | ≤ N, . From this expression we see that the function R ∗ with Fourier coefficients ˆ R ∗ ,k satisfies that R ∗ ∈ τ T (1; v ( t n )) . (4.22) Note that for k + k + k = k the following equalities hold: φ ( k, k , k , k ) = 2 kk + 2 k k , (4.23a)2 kk = k + k − ( k + k ) , (4.23b)2 k k = ( k + k ) − k − k . (4.23c)By using these relations, we have e i ( t n + s ) φ = e it n φ e iskk e isk k = e it n φ [ e iskk + ( e isk k −
1) + ( e iskk − e isk k − , and therefore (4.16b) can be decomposed into the following three terms:(4.16b) = iλ X k + k + k = k | k + k |≤ N (cid:18) Z τ k k e it n φ e iskk ds (cid:19) ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) (4.24-1)+ iλ X k + k + k = k | k + k |≤ N (cid:18) Z τ k k e it n φ (cid:0) e isk k − (cid:1) ds (cid:19) ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) (4.24-2)+ b R ,k + b R ∗ ,k , (4.24-3)where b R ,k = iλ X k + k + k = k | k + k |≤ N (cid:18) Z τ k k e it n φ (cid:0) e iskk − (cid:1)(cid:0) e isk k − (cid:1) ds (cid:19) ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) , (4.25) b R ∗ ,k = iλ X k + k + k = k | k + k | >N (cid:18) Z τ k k e i ( t n + s ) φ ds (cid:19) ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) , (4.26)for 0 = | k | ≤ N , and b R ,k = b R ∗ ,k = 0 for k = 0 and | k | > N . Lemma 4.2 of the nextsubsection implies that (cid:13)(cid:13) R (cid:13)(cid:13) H s . τ k v k L ∞ t H x for s ∈ ( , , (4.27a) (cid:13)(cid:13) R (cid:13)(cid:13) L . τ √ ln τ − k v k L ∞ t H x . (4.27b)Obviously, R ∗ ∈ τ T ( σ ; v ( t n )) with some | σ ( k, k , k , k ) | ≤ | k | − | k | = | k |≤ N | k + k | >N . By Lemma 4.1 (iii) and symmetry, and we have that for any s ∈ [0 , (cid:13)(cid:13) R ∗ (cid:13)(cid:13) H s . τ N − s k v k L ∞ t H x . (4.28)Note that(4.24-1) = X k + k + k = k | k + k |≤ N λ k e it n φ (cid:0) e iτ ( k + k − ( k + k ) ) − (cid:1) ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) , (4.29)(4.24-2) = iλ X k + k + k = kk =0 ,k =0 | k + k |≤ N (cid:18) Z τ k k e it n φ (cid:0) e isk k − (cid:1) ds (cid:19) ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n )= λ X k + k + k = kk =0 ,k =0 | k + k |≤ N k kk k e it n φ (cid:0) e iτk k − (cid:1) ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) − iλτ X k + k + k = kk =0 ,k =0 | k + k |≤ N k k e it n φ ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n )= λ X k + k + k = kk =0 ,k =0 | k + k |≤ N k kk k e it n φ (cid:16) e iτ (cid:0) ( k + k ) − k − k (cid:1) − (cid:17) ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) − iλτ X k + k + k = k | k + k |≤ N k k e it n φ ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n )+ 2 iλτ X k + k = k | k |≤ N k k e it n φ ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v ( t n ) − iλτ e it n φ ˆ¯ v k ( t n )ˆ v ( t n )ˆ v ( t n ) . (4.30)Substituting (4.29)–(4.30) into (4.24), and then substituting (4.21) and (4.24) into(4.16), we obtainˆ v k ( t n +1 ) = e − iλτP ( ik ) − − iλτM ˆ v k ( t n )+ λ X k + k + k = k = | k + k |≤ N ik ( ik + ik ) (cid:0) e it n +1 φ − e it n φ (cid:1) ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) − λ X k + k + k = k | k + k |≤ N ik ) e it n φ (cid:0) e iτ ( k + k − ( k + k ) ) − (cid:1) ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) − λ X k + k + k = kk =0 ,k =0 | k + k |≤ N ik ik )( ik )( ik ) e it n φ (cid:0) e iτ (( k + k ) − k − k ) − (cid:1) ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) − iλτ X k + k + k = k | k + k |≤ N k k e it n φ ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n )+ 2 iλτ X k + k = k | k |≤ N k k e it n φ ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v ( t n ) − iλτ e it n φ ˆ¯ v k ( t n )ˆ v ( t n )ˆ v ( t n )+ ˆ R ,k + ˆ R ,k + ˆ R ∗ ,k + ˆ R ,k + ˆ R ∗ ,k for k = 0 and | k | ≤ N. (4.31)Then substituting (4.10) and (4.31) into the expression v ( t n +1 ) = P k ∈ Z ˆ v k ( t n +1 ) e ikx yields v ( t n +1 ) =Φ n ( v ( t n ); M, P ) + R + ˆ R , + ˆ R ∗ , + R + R + R ∗ + R + R ∗ , (4.32)where Φ n ( f ; M, P ) := e − iλτP ∂ − x − iλτM f + (1 − e − iλτM )Π f − iλτ Π (cid:2) Π N (cid:0)(cid:12)(cid:12) e it n ∂ x f (cid:12)(cid:12) (cid:1) e it n ∂ x f (cid:3) + λ e − it n +1 ∂ x ∂ − x Π N h(cid:0) e it n +1 ∂ x f (cid:1) · ∂ − x Π N (cid:0) | e it n +1 ∂ x f | (cid:1)i − λ e − it n ∂ x ∂ − x Π N h(cid:0) e it n ∂ x f (cid:1) · ∂ − x Π N (cid:0) | e it n ∂ x f | (cid:1)i − λ (cid:20) e − it n +1 ∂ x ∂ − x Π N (cid:16)(cid:0) e − it n +1 ∂ x ¯ f (cid:1) · e iτ∂ x Π N (cid:0) e it n ∂ x f (cid:1) (cid:17) − e − it n ∂ x ∂ − x Π N (cid:16) e − it n ∂ x ¯ f Π N (cid:0) e it n ∂ x f (cid:1) (cid:17)(cid:21) − λ (cid:20) e − it n ∂ x ∂ − x Π N (cid:16)(cid:0) e − it n ∂ x ∂ x ¯ f (cid:1) · e − iτ∂ x Π N (cid:0) e it n +1 ∂ x ∂ − x f (cid:1) (cid:17) − e − it n ∂ x ∂ − x Π N (cid:16)(cid:0) e − it n ∂ x ∂ x ¯ f (cid:1) · Π N (cid:0) e it n ∂ x ∂ − x f (cid:1) (cid:17)(cid:21) − iλτ e − it n ∂ x ∂ − x Π N (cid:16) e − it n ∂ x ∂ x ¯ f Π N (cid:0) e it n ∂ x f (cid:1) (cid:17) + 2 iλτ Π ( f )e − it n ∂ x ∂ − x (cid:16) e − it n ∂ x ∂ x ¯ f e it n ∂ x f (cid:17) − iλτ (Π f ) Π =0 (cid:0) e − it n ∂ x ¯ f (cid:1) . (4.33)The numerical scheme can be defined by dropping the defect terms R j and R ∗ j in (4.32)and replacing the numbers M and P by their approximations M N and P N defined in (2.10),respectively. Namely, for given v n ∈ S N compute v n +1 ∈ S N by v n +1 = Φ n ( v n ; M N , P N ) , n = 0 , . . . , L −
1; with v = u . (4.34)Then, replacing v n and v n +1 by e − it n ∂ x u n and e − it n +1 ∂ x u n in (4.34), we obtain the numericalscheme (2.8)–(2.9). In this subsection, we present two technical lemmas, which are used in estimating thedefect terms R j and R ∗ j in the previous subsection. Lemma 4.1.
For any given v ∈ H ( T ) and s ∈ [0 , , the following results hold. (i) Let m ≥ , N ∈ Z + . Then, for any f ∈ T m (1; v ) and any g ∈ T m (1 >N ; v ) , (cid:13)(cid:13) f (cid:13)(cid:13) H . k v k mH ; (cid:13)(cid:13) g (cid:13)(cid:13) H s . N − s k v k mH . (ii) For any f ∈ T ( k k ; v ) there holds | Π f | . k v k H . (iii) Let N ∈ Z + , N ≥ and f ∈ T ( σ ; v ) . If | σ ( k, k , k , k ) | . | k | − | k j | = | k |≤ N | k + k | >N , for some j ∈ { , , } , then k f k H s . N − s k v k H . Proof.
Without loss of generality, we can assume that ˆ v k j , j = 1 , · · · , m are positive for any t ∈ [0 , T ]. Otherwise we replace ˆ v k j by | ˆ v k j | as we did in the proof of Lemma 3.2.(i) By the definition of T m ( σ ; v ) in (3.2), f ∈ T m (1; v ) implies that | ˆ f k | . X k + ··· + k m = k ˆ v k · · · ˆ v k m ∼ F k [ v m ] . By Plancherel’s identity and Lemma 3.1 (i), we obtain that k f k H . k v m k H . k v k mH . For g ∈ T m (1 >N ; v ), we use the inequality k g k H s . N − s k g k H together with the inequalityabove, which implies that k g k H . k v k mH . This yields the desired inequality for g , i.e., k g k H s . N − s k v k mH . (ii) For any f ∈ T ( k k ; v ) we have that | Π f | . X k + k + k =0 ˆ v k ( t ) | k | ˆ v k ( t ) | k | ˆ v k ( t ) . X k + k ′ =0 ˆ v k ( t ) F k ′ [( |∇| v ( t )) ] . Z T v ( |∇| v ) dx . k v k L ∞ k|∇| v k L . k v k H . (iii) We only consider the case when j = 1, since the other cases can be treated in thesame way. Since the Fourier coefficients of J s f satisfies F k [ J s f ] = h k i s ˆ f k . X k + k + k = k | k + k | >N = | k |≤ N h k i − s h k i ˆ v k ( t )ˆ v k ( t )ˆ v k ( t ) . N − s X k + k + k = k | k + k | >N h k i − h k + k ih k i ˆ v k ( t )ˆ v k ( t )ˆ v k ( t ) . N − s F k [ J − ( vJ ( vJ v ))] , it follows from Lemma 3.2 (ii) that k J s f k L . N − s k vJ v k L k v k H . N − s k v k L ∞ k J v k L k v k H . N − s k v k H . This proves the desired results in Lemma 4.1. (cid:3)
Lemma 4.2. If v ∈ L ∞ (0 , T ; H ( T )) then (cid:13)(cid:13) R (cid:13)(cid:13) L . τ √ ln τ − k v k L ∞ (0 ,T ; H ) . (4.35) Moreover, for any s ∈ ( , , (cid:13)(cid:13) R (cid:13)(cid:13) H s . τ k v k L ∞ (0 ,T ; H ) . (4.36) Proof.
For k + k + k = k and | k | ≥ | k | we claim that the following inequality holds: (cid:12)(cid:12)(cid:12)(cid:12) k k (cid:0) e iskk − (cid:1)(cid:0) e isk k − (cid:1)(cid:12)(cid:12)(cid:12)(cid:12) . τ | k | − α | k || k || k | α ∀ s ∈ [0 , τ ] , ∀ α ∈ [0 , . (4.37)In order to prove (4.37), we consider the following two cases: | k | ≥ | k | and | k | < | k | . Case | k | ≥ | k | . In this case, we use the following inequalities: (cid:12)(cid:12) e iskk − (cid:12)(cid:12) ≤ (cid:12)(cid:12) e isk k − (cid:12)(cid:12) ≤ τ | k || k | , it follows that (cid:12)(cid:12)(cid:12)(cid:12) k k (cid:0) e iskk − (cid:1)(cid:0) e isk k − (cid:1)(cid:12)(cid:12)(cid:12)(cid:12) ≤ τ | k | − | k || k || k | . τ | k | − α | k || k || k | α . Case | k | < | k | . In this case k + k + k = k and | k | ≥ | k | imply | k | ≤ | k | + | k | + | k | . | k | . We use the following inequalities: (cid:12)(cid:12) e iskk − (cid:12)(cid:12) ≤ τ | k || k | and (cid:12)(cid:12) e isk k − (cid:12)(cid:12) ≤ . Then we obtain (cid:12)(cid:12)(cid:12)(cid:12) k k (cid:0) e iskk − (cid:1)(cid:0) e isk k − (cid:1)(cid:12)(cid:12)(cid:12)(cid:12) ≤ τ | k | . Since | k | . | k | , it follows that | k | . | k || k | . | k || k | (cid:18) | k || k | (cid:19) α . This proves (4.37).By using the symmetry between k and k in the expression of | ˆ R ,k | in (4.25) andapplying (4.37) with α = 1 in the case | k | ≥ | k | and α = 0 in the case | k | < | k | , we obtainfor any k = 0, | ˆ R ,k | . τ X k + k + k = k | k |≥| k | , | k |≥| k | | k | − | k || k || k || ˆ¯ v k ( t n ) || ˆ v k ( t n ) || ˆ v k ( t n ) | + τ X k + k + k = k | k |≥| k | , | k | < | k | | k || k || ˆ¯ v k ( t n ) || ˆ v k ( t n ) || ˆ v k ( t n ) | . (4.38)Without loss of generality, we may assume that ˆ¯ v k ( t n ) , ˆ v k ( t n ) and ˆ v k ( t n ) are nonnegative.Otherwise we replace them by their absolute values as we did in the proof of Lemma 3.2.By the duality between L ( T ) and itself, it is sufficient to prove the following result toobtain (4.35): |hR , f i| . τ p ln( τ − ) k v k L ∞ (0 ,T ; H ) k f k L ∀ f ∈ L ( T ) . (4.39)From the definition below (4.26) we see that R , = 0. As a result, we have |hR , f i| . X k =0 | ˆ R ,k | | ˆ f k | . X | k | >τ − | ˆ R ,k | | ˆ f k | + X = | k |≤ τ − | ˆ R ,k | | ˆ f k | . (4.40)From the expression of R ,k in (4.25) we see that for | k | > τ − there holds | ˆ R ,k | ≤ τ X k + k + k = k | k | ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) . Hence, by the Cauchy–Schwartz inequality and Plancherel’s identity, we have X | k | >τ − | ˆ R ,k | | ˆ f k | ≤ τ X k X k + k + k = k | k | ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) | ˆ f k | = τ X k ,k X k | k − k − k | ˆ¯ v k − k − k ( t n )ˆ v k ( t n )ˆ v k ( t n ) | ˆ f k | . τ k ( ˆ f k ) k ∈ Z k l k ( k ˆ¯ v k ( t n )) k ∈ Z k l k (ˆ v k ( t n )) k ∈ Z k l k (ˆ v k ( t n )) k ∈ Z k l . τ k f k L k v k H , (4.41)where the last inequality uses the following result: k (ˆ v k ( t n )) k ∈ Z k l . k ( h k i − ) k ∈ Z k l k ( h k i ˆ v k ( t n )) k ∈ Z k l . k v k H . The second term in (4.40) can be estimated by using (4.38), i.e., X = | k |≤ τ − | ˆ R ,k | | ˆ f k | (4.42) . τ X = | k |≤ τ − X k + k + k = k | k |≥| k | , | k |≥| k | | k | − | k || k || k || ˆ f k | ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n )+ τ X = | k |≤ τ − X k + k + k = k | k |≥| k | , | k | < | k | | k || k || ˆ f k | ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) . τ X = | k |≤ τ − X | k |≤| k | X k | k | − | ˆ f k || k | ˆ¯ v k ( t n ) | k − k − k | ˆ v k − k − k ( t n ) | k | ˆ v k ( t n ) + τ X = | k |≤ τ − X | k | > | k | X k | ˆ f k || k | ˆ¯ v k ( t n ) | k − k − k | ˆ v k − k − k ( t n )ˆ v k ( t n ) . τ k ( k ˆ v k ( t n )) k ∈ Z k l k ( k ˆ v k ( t n )) k ∈ Z k l X = | k |≤ τ − | k | − | ˆ f k | X | k |≤| k | | k | ˆ v k ( t n )+ τ k ( k ˆ v k ( t n )) k ∈ Z k l k ( k ˆ v k ( t n )) k ∈ Z k l X = | k |≤ τ − | ˆ f k | X | k | > | k | ˆ v k ( t n ) . τ k v ( t n ) k H X = | k |≤ τ − | k | − | ˆ f k |k ( k ˆ v k ( t n )) k ∈ Z k l + τ k v ( t n ) k H X = | k |≤ τ − | ˆ f k |k ( h k i − ) | k | >k k l k ( h k i ˆ v k ( t n )) | k | >k k l . τ k v ( t n ) k H X = | k |≤ τ − | k | − | ˆ f k |k ( k ˆ v k ( t n )) k ∈ Z k l . τ k v ( t n ) k H k ( | k | − ) = | k |≤ τ − k l k ( ˆ f k ) | k |≤ τ − k l . τ k v ( t n ) k H p ln( τ − ) k f k L . (4.43)Substituting (4.41)–(4.43) into (4.40) yields (4.39), which implies the desired result in (4.35).It remains to prove (4.36). To this end, we use the following inequalities: | e iskk − | ≤ | e isk k − | . s || k | | k | , which imply that (cid:12)(cid:12)(cid:12)(cid:12) k k (cid:0) e iskk − (cid:1)(cid:0) e isk k − (cid:1)(cid:12)(cid:12)(cid:12)(cid:12) . τ | k | − | k || k | | k | ∀ s ∈ [0 , τ ] . By substituting this into the expression of ˆ R ,k in (4.25), and using Plancherel’s identity, weobtain (cid:13)(cid:13) R (cid:13)(cid:13) H s . τ (cid:13)(cid:13)(cid:13) |∇| − s (cid:16) |∇| ¯ v (cid:0) |∇| v (cid:1) (cid:17)(cid:13)(cid:13)(cid:13) L . Then using the Sobolev inequality, we get that for any s ∈ ( , (cid:13)(cid:13) R (cid:13)(cid:13) H s . τ (cid:13)(cid:13)(cid:13) |∇| ¯ v (cid:0) |∇| v (cid:1) (cid:13)(cid:13)(cid:13) L − s . τ (cid:13)(cid:13) |∇| ¯ v (cid:13)(cid:13) L (cid:13)(cid:13) |∇| v (cid:13)(cid:13) L − s . τ k v k H . This completes the proof of Lemma 4.2. (cid:3)
5. Proof of Theorem 2.1
The proof of Theorem 2.1 is divided into two parts. In subsection 5.1, we present anerror estimate for the numerical solution in H s ( T ) with s ∈ ( , H ( T ) uniformly with respect to τ and N . In subsection 5.2, we utilize the H -boundedness of the numerical solution to provethe desired error estimate in L ( T ). H ( T ) Lemma 5.1.
Let u ∈ H ( T ) , and let u nτ,N , n = 0 , , . . . , L , be the numerical solution givenby (2.8) – (2.9) . Then there exist positive constants τ s and N s such that for τ ∈ (0 , τ s ] and N ≥ N s the following error bound holds: max ≤ n ≤ L k u ( t n , · ) − u nτ,N k H s . s τ + N − s ∀ s ∈ ( , , (5.1) where τ s and N s depend only on k u k H , T and s .Proof. Let v n = e − it n ∂ x u nτ,N . Then v n +1 = Φ n ( v n ; M N , P N ) as shown in (4.34). By usingthis identity we have v ( t n +1 ) − v n +1 = v ( t n +1 ) − Φ n ( v ( t n ); M, P ) + Φ n ( v ( t n ); M, P ) − Φ n ( v n ; M N , P N )=: L n + Φ n ( v ( t n ); M, P ) − Φ n ( v n ; M N , P N ) , (5.2)where L n = v ( t n +1 ) − Φ n ( v ( t n ); M, P ) = R + ˆ R , + ˆ R ∗ , + ˆ R , + R + R ∗ + R + R ∗ , which is shown in (4.32). From (4.9), (4.13), (4.15), (4.18), (4.27) and (4.28) we see that (cid:13)(cid:13) L n (cid:13)(cid:13) H s . τ + τ N − s ∀ s ∈ [0 , . (5.3)Note that the functional Φ n ( f ; M, P ) defined in (4.33) can be rewritten into the followingform:Φ n ( f ; M, P ) = f + (cid:0) e − iλτP ∂ − x − iλτM − iλτ P ∂ − x + 2 iλτ M (cid:1) f + (1 − e − iλτM )Π f − iλτ Π (cid:2) Π N (cid:0)(cid:12)(cid:12) e it n ∂ x f (cid:12)(cid:12) (cid:1) e it n ∂ x f (cid:3) − iλ X = | k |≤ N e ikx (cid:18) X k + k + k = k | k + k |≤ N Z τ k + k k e i ( t n + s ) φ ds (cid:19) ˆ¯ f k ˆ f k ˆ f k + iλ X = | k |≤ N e ikx X k + k + k = k | k + k |≤ N (cid:18) Z τ k k e it n φ e iskk ds (cid:19) ˆ¯ f k ˆ f k ˆ f k + iλ X = | k |≤ N e ikx X k + k + k = k | k + k |≤ N (cid:18) Z τ k k e it n φ (cid:0) e isk k − (cid:1) ds (cid:19) ˆ¯ f k ˆ f k ˆ f k . (5.4)For example, the third line of (5.4) comes from (4.21), which can be rewritten back into(4.17). This is how we obtain the third line in the expression above. The other terms areobtained similarly.From (5.4) we furthermore derive thatΦ n ( v ( t n ); M, P ) − Φ n ( v n ; M N , P N ) = v ( t n ) − v n + Φ n + Φ n + Φ n + Φ n + Φ n , (5.5)whereΦ n = (cid:0) e − iλτP ∂ − x − iλτM − iλτ P ∂ − x + 2 iλτ M + (1 − e − iλτM )Π (cid:1) v ( t n ) − (cid:0) e − iλτP N ∂ − x − iλτM N − iλτ P N ∂ − x + 2 iλτ M N + (1 − e − iλτM N )Π (cid:1) v n , Φ n = − iλτ Π (cid:16) | e it n ∂ x v ( t n ) | e it n ∂ x v ( t n ) − | e it n ∂ x v n | e it n ∂ x v n (cid:17) , Φ n = − iλ X = | k |≤ N e ikx X k + k + k = k | k + k |≤ N (cid:18) Z τ k + k k e i ( t n + s ) φ ds (cid:19)(cid:0) ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) − ˆ¯ v nk ˆ v nk ˆ v nk (cid:1) , Φ n = iλ X = | k |≤ N e ikx X k + k + k = k | k + k |≤ N (cid:18) Z τ k k e it n φ e iskk ds (cid:19)(cid:0) ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) − ˆ¯ v nk ˆ v nk ˆ v nk (cid:1) , Φ n = iλ X = | k |≤ N e ikx X k + k + k = k | k + k |≤ N (cid:18) Z τ k k e it n φ (cid:0) e isk k − (cid:1) ds (cid:19)(cid:0) ˆ¯ v k ( t n )ˆ v k ( t n )ˆ v k ( t n ) − ˆ¯ v nk ˆ v nk ˆ v nk (cid:1) . Note that P , M , P N and M N defined in (2.4) and (2.10) are all bounded numbers, withbounds depending on k u k H . In particular, | M − M N | = (cid:12)(cid:12)(cid:12)(cid:12) π Z T ( | u | − | u τ,N | ) dx (cid:12)(cid:12)(cid:12)(cid:12) . (cid:12)(cid:12)(cid:12)(cid:12) π Z T h ( u − u τ,N ) u + u τ,N ( u − u τ,N ) i dx (cid:12)(cid:12)(cid:12)(cid:12) . k u − u τ,N k L ( k u k L + k u τ,N k L ) . N − k u k H (5.6)and | P − P N | = (cid:12)(cid:12)(cid:12)(cid:12) π Z T ( u ∂ x u − u τ,N ∂ x u τ,N ) dx (cid:12)(cid:12)(cid:12)(cid:12) . (cid:12)(cid:12)(cid:12)(cid:12) π Z T h ( u − u τ,N ) ∂ x u + u τ,N ∂ x ( u − u τ,N ) i dx (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) π Z T h ( u − u τ,N ) ∂ x u − ∂ x u τ,N ( u − u τ,N ) i dx (cid:12)(cid:12)(cid:12)(cid:12) . k u − u τ,N k L ( k ∂ x u k L + k ∂ x u τ,N k L ) . N − k u k H . (5.7)From the expression of Φ n we see that its Fourier coefficients can be written as F k [Φ n ] = F ( M, P ; k )ˆ v k ( t n ) − F ( M N , P N ; k )ˆ v nk , with F ( M, P ; k ) := e − iλτP k − k =0 − iλτM − iλτ P k − k =0 + 2 iλτ M + (1 − e − iλτM )1 k =0 . By using Taylor’s expansion and mean value theorem, it is straightforward to verify that | F ( M, P ; k ) − F ( M N , P N ; k ) | . τ ( | P − P N | + | M − M N | ) . As a result, we have k Φ n k H s . k ( h k i s F k [Φ n ]) k ∈ Z k l . τ ( | P − P N | + | M − M N | ) k ( h k i s ˆ v k ( t n )) k ∈ Z k l + k ( h k i s (ˆ v k ( t n ) − ˆ v k )) k ∈ Z k l . τ ( | P − P N | + | M − M N | ) k v ( t n ) k H s + τ k v ( t n ) − v n k H s . τ N − k v k L ∞ (0 ,T ; H ) + τ k v ( t n ) − v n k H s , (5.8)where the last inequality follows from (5.6)–(5.7).Since Φ n is a constant, it is straightforward to show that (similarly as (5.6)) | Φ n | . τ (cid:0) k v n − v ( t n ) k L ( k e it n ∂ x v ( t n ) k L ∞ + k e it n ∂ x v n k L ∞ ) . τ (cid:0) k v n − v ( t n ) k L ( k v ( t n ) k H s + k v n k H s ) (this holds for s > ) . τ (cid:0) k v n − v ( t n ) k L ( k v ( t n ) k H s + k v n − v ( t n ) k H s ) . (5.9)Similarly, Φ n can be decomposed into several functions of the following form:Φ n = − i X = | k |≤ N e ikx X k + k + k = k | k + k |≤ N (cid:18) Z τ k + k k e i ( t n + s ) φ ds (cid:19) ˆ f ,k ˆ f ,k ˆ f ,k , where ˆ f j,k denotes the k th Fourier coefficient of the functions f j , and one of the threefunctions f j , j = 1 , ,
3, is v n − v ( t n ) or its conjugate; the other two of the three functions f j , j = 1 , ,
3, are either v n or v ( t n ) or their conjugates.We assume that ˆ f j,k , k ∈ Z are nonnegative; otherwise we consider functions with Fouriercoefficients | ˆ f j,k | as we did in the proof of Lemma 3.2 (ii). Then (cid:12)(cid:12)(cid:0) c Φ n (cid:1) k (cid:12)(cid:12) . τ X k + k + k = k | k + k || k | ˆ f ,k ˆ f ,k ˆ f ,k = F k [ τ J − ( f J ( f f ))] . As a result, by Plancherel’s identity and Lemma 3.2 (i), we have k Φ n k H s . k τ J − ( f J ( f f )) k H s . τ k f k H s k f f k H s (this requires s > ) . τ k f k H s k f k H s k f k H s . τ k v n − v ( t n ) k H s ( k v n k H s + k v ( t n ) k H s ) . τ k v n − v ( t n ) k H s ( k v n − v ( t n ) k H s + k v ( t n ) k H s ) . (5.10)Φ n and Φ n can be estimated similarly, i.e., k Φ n k H s + k Φ n k H s . τ k v n − v ( t n ) k H s ( k v n − v ( t n ) k H s + k v ( t n ) k H s ) . Hence, combining with the estimates of Φ nj , j = 1 , . . . ,
5, we have k Φ n ( v ( t n ); M, P ) − Φ n ( v n ; M N , P N ) k H s ≤ (1 + Cτ ) k v n − v ( t n ) k H s + Cτ k v n − v ( t n ) k H s + Cτ N − , which holds for any given s ∈ ( , k v ( t n +1 ) − v n +1 k H s ≤ C (cid:0) τ + τ N − s (cid:1) + (1 + Cτ ) k v n − v ( t n ) k H s + Cτ k v n − v ( t n ) k H s . By using the discrete Gronwall’s inequality with induction assumption on k v n − v ( t n ) k H s ≤ τ )max ≤ n ≤ L (cid:13)(cid:13) v ( t n ) − v n (cid:13)(cid:13) H s . τ + N − s . This proves the desired result in Lemma 5.1. (cid:3)
Lemma 5.1 implies that k v ( t n ) − v n k H s .
1. Then, by using the triangle inequality andboundedness of the exact solution in H , we have k v n k H s . k v ( t n ) − v n k H s + k v ( t n ) k H s . . This result can be furthermore improved to the H norm, as shown in the following lemma. Lemma 5.2.
Let u ∈ H ( T ) , and let u nτ,N , n = 0 , , . . . , L , be the numerical solution givenby (2.8) – (2.9) . Then there exists a constant τ > such that for τ ∈ (0 , τ ] the followingestimate holds: max ≤ n ≤ L k u nτ,N k H . . (5.11) Proof.
Let v n = e − it n ∂ x u nτ,N . By using the expression of Φ n in (5.4), we immediately obtainthat k Φ n ( v n ; M N , P N ) k H ≤ k v n k H + Cτ k v n k H + Cτ k v n k H k v n k H s , (5.12)which holds for any fixed s ∈ ( , k v n k H s . k v n +1 k H ≤ k v n k H + Cτ k v n k H , (5.13)which implies max ≤ n ≤ L k v n k H . n . The desired result follows from therelation k v n k H = k u nτ,N k H . (cid:3) L ( T )From (4.9), (4.13), (4.15), (4.18), (4.27) and (4.28) we conclude that (cid:13)(cid:13) L n (cid:13)(cid:13) L ≤ C (cid:0) τ √ ln τ − + τ N − (cid:1) . (5.14)By choosing s = 0 in (5.8) and choosing a fixed s ∈ ( ,
1) in (5.9), we have k Φ n k L + k Φ n k L . τ N − + τ k v n − v ( t n ) k L . Instead of (5.10), we need to use the following estimate for Φ n : k Φ n k L . k τ J − ( f J ( f f )) k L . τ min( k f k H k f f k L , k f k L k f f k H ) . which is a consequence of Lemma 3.2 (ii). Recall that one of the three functions f j , j = 1 , , v n − v ( t n ) or its conjugate, and the other two functions are either v n or v ( t n ) (or theirconjugates). If f is v n − v ( t n ) or its conjugate, then we choose L norm on f ; otherwise wechoose L norm on f f . In either case we obtain k Φ n k L . τ k v n − v ( t n ) k L ( k v ( t n ) k H + k v n k H ) . τ k v n − v ( t n ) k L . The two terms Φ n and Φ n can be estimated similarly, i.e., k Φ n k L + k Φ n k L . τ k v n − v ( t n ) k L . Substituting the estimates of k Φ nj k L , j = 1 , . . . ,
5, into (5.5), we have k Φ n ( v ( t n ); M, P ) − Φ n ( v n ; M N , P N ) k L . τ N − + τ k v n − v ( t n ) k L . Then, substituting this into (5.2) and using estimate (5.14), we obtain k v ( t n +1 ) − v n +1 k L ≤ C (cid:0) τ √ ln τ − + τ N − (cid:1) + (1 + Cτ ) k v n − v ( t n ) k L . (5.15)Iterating this inequality yieldsmax ≤ n ≤ L k v ( t n ) − v n k L . k v ( t ) − v k L + τ √ ln τ − + N − . τ √ ln τ − + N − . This completes the proof of Theorem 2.1 in view of k v ( t n ) − v n k L = k u ( t n ) − u nτ,N k L . (cid:3)
6. Numerical experiments
In this section we present numerical experiments to support the theoretical analysispresented in Theorem 2.1. We consider the NLS equation (1.1) with λ = − u ( x ) = 110 X = k ∈ Z | k | − . − α e ikx , (6.1)which satisfies that u ∈ H α ( T ) and u / ∈ H α − . ( T ).We solve the problem by the proposed method (2.8)–(2.9) for α = 2 and α = 1, re-spectively, and present the time discretisation errors k u τ,N − u τ/ ,N k L in Tables 1–2 forseveral sufficiently large N . From the numerical results we can see that the error fromspatial discretisation is negligibly small in observing the temporal convergence rates, i.e.,almost first-order convergent as τ →
0. This is consistent with the theoretical result provedin Theorem 2.1.We present the spatial discretisation errors k u τ,N − u τ, N k L for α = 2 and α = 1in Tables 3–4 for several sufficiently small stepsize τ . From the numerical results we cansee that the error from temporal discretisation is negligibly small in observing the spatialconvergence rates, i.e., α th-order convergence for H α initial data. This is consistent withthe result proved in Theorem 2.1 and the comments in Remark 2.2. Table 1.
Temporal discretisation error k u τ,N − u τ/ ,N k L at T = 1with α = 2 in (6.1) (for H initial data). N = 2 N = 2 N = 2 τ = 2 − τ = 2 − τ = 2 − O ( τ . ) O ( τ . ) O ( τ . ) Table 2.
Temporal discretisation error k u τ,N − u τ/ ,N k L at T = 1with α = 1 in (6.1) (for H initial data). N = 2 N = 2 N = 2 τ = 2 − τ = 2 − τ = 2 − O ( τ . ) O ( τ . ) O ( τ . ) Table 3.
Spatial discretisation error k u τ,N − u τ, N k L at T = 1with α = 2 in (6.1) (for H initial data). τ = 2 − τ = 2 − τ = 2 − N = 16 2.430E–04 2.430E–03 2.430E–03 N = 32 6.237E–05 6.237E–05 6.237E–05 N = 64 1.574E–05 1.574E–05 1.574E–05convergence rate O ( N − . ) O ( N − . ) O ( N − . ) Table 4.
Spatial discretisation error k u τ,N − u τ, N k L at T = 1with α = 1 in (6.1) (for H initial data). τ = 2 − τ = 2 − τ = 2 − N = 16 5.056E–03 5.056E–03 5.056E–03 N = 32 2.559E–03 2.559E–03 2.559E–03 N = 64 1.283E–03 1.283E–03 1.283E–03convergence rate O ( N − . ) O ( N − . ) O ( N − . )
7. Conclusion
We have constructed a fast fully discrete low-regularity integrator for solving the NLSequation with nonsmooth initial data in one dimension. The method can be implementedby using FFT with O ( N ln N ) operations at every time level, and is proved to have an errorbound of O ( τ p ln(1 /τ ) + N − ) when the initial data is in H ( T ). For initial data in H s ( T )with s >
1, the numerical results show that the proposed method can have an error boundof O ( τ + N − s ). We expect that the techniques for constructing and analysing the spatialdiscretisation method in combination with the temporal low-regularity integrator may alsobe extended to other dispersive equations with nonsmooth data. References [1] C. Besse, B. Bid´egaray, and S. Descombes: Order estimates in time of splitting methods for the nonlinearSchr¨odinger equation.
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60 (2014), pp. 390–407.[22] Y. Wu and F. Yao: A first-order Fourier integrator for the nonlinear Schr¨odinger equation on T withoutloss of regularity. Preprint, arXiv:2010.02672[23] Y. Wu and X. Zhao: Optimal convergence of a first order low-regularity integrator for the KdV equation.arXiv:1910.07367, 2019.[24] Y. Wu and X. Zhao: Embedded exponential-type low-regularity integrators for KdV equation underrough data. arXiv:2008.07053, 2020.[25] Wikipedia: https://en.wikipedia.org/wiki/Discrete_Fourier_transform Buyang Li: Department of Applied Mathematics, The Hong Kong Polytechnic University,Hung Hom, Hong Kong.
Email address : [email protected] Yifei Wu: Center for Applied Mathematics, Tianjin University, 300072, Tianjin, P. R. China.
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