Convergence of operator splitting for the KdV equation

Transcription

1 Convergence of operator splitting for the KdV equation H. Holden Norwegian University of Science and Technology Trondheim, Norway Joint work with K.H. Karlsen, N. H. Risebro, and T. Tao Institute for Mathematics and Applications, July 2009

2 Operator splitting { du dt = C(u), t > 0 u(0) = u 0 Assume that C(u) =A(u)+B(u), we wish to find an approximation to u(t) by solving alternately. { dv dt = A(v), dv dt = B(v)

3 Introduce a time step t t n = n t, t n+1/2 = t n + t/2, n =0, 1, 2,... v(0) = u 0, dv ( ] dt =2A(v), t t n,t n+1/2, dv ( ] dt =2B(v), t t n+1/2,t n+1. Linear ODE s du dt = Cu = Au + Bu, A, B, C m m matrices u (t n )=e t nc u 0 v (t n )= [ e tb e ta] n u0

4 Convergence by the Lie-Trotter-Kato formula: e t n(a+b) = lim t 0 [ e tb e ta] t n / t Notation { u (t) =C(u), u(0) = u 0 u(t) =Φ C (t) u 0 v (t n )=[Φ B ( t) Φ A ( t)] n u 0 {Φ B ( 2(t tn+1/2 ) ) v ( t n+1/2 ), t [ tn+1/2,t n+1 ] v (t) = Φ A (2(t t n )) v (t n ), t [ t n,t n+1/2 ].

5 Formal Taylor series... v n+1/2 = v n + ta(v n )+ 1 2 t2 da(v n )A(v n )+O ( t 3) v n+1 = v n+1/2 + tb(v n+1/2 ) t2 db(v n+1/2 )B(v n+1/2 )+O ( t 3)

6 Formal Taylor series... v n+1/2 = v n + ta(v n )+ 1 2 t2 da(v n )A(v n )+O ( t 3) v n+1 = v n + t (A(v n )+B(v n )) t2[ da(v n+1/2 )A(v n+1/2 ) ] +2dB(v n )A(v n )+db(v n )B(v n ) + O ( t 3)

7 Φ B ( t)φ A ( t) v n = v n + t (A(v n )+B(v n )) t2[ da(v n+1/2 )A(v n+1/2 ) ] +2dB(v n )A(v n )+db(v n )B(v n ) + O ( t 3) Φ (A+B) ( t) v n = v n + t(a + B)(v n )+ 1 2 t2 d(a + B)(v n )(A + B)(v n ) + O ( t 3)

8 Φ B ( t)φ A ( t) v n = v n + t (A(v n )+B(v n )) t2[ da(v n+1/2 )A(v n+1/2 ) ] +2dB(v n )A(v n )+db(v n )B(v n ) + O ( t 3) Φ (A+B) ( t) v n = v n + t(a + B)(v n )+ 1 2 t2 d(a + B)(v n )(A + B)(v n ) + O ( t 3) Local error: Φ(A+B) ( t) v n Φ A ( t)φ B ( t) v n = 1 2 t2 [A, B](v n )+O ( t 3) = 1 2 t2 (dab dba)(v n )+O ( t 3).

9 Global error found by summing the local errors. there are O(1/ t) terms. Error = t/ t n=0 1 2 t2 [A, B] K t t t2 = O( t)

10 Global error found by summing the local errors. there are O(1/ t) terms. Error = t/ t n=0 1 2 t2 [A, B] K t t t2 = O( t) Strang splitting: v n+1 = [(Φ A ( t/2)φ B ( t/2)) (Φ B ( t/2)φ A ( t/2))] v n

11 Global error found by summing the local errors. there are O(1/ t) terms. Error = t/ t n=0 1 2 t2 [A, B] K t t t2 = O( t) Strang splitting: v n+1 = [(Φ A ( t/2)φ B ( t/2)) (Φ B ( t/2)φ A ( t/2))] v n Local error: 1 2 ( t/2)2 ([A, B]+[B, A]) + O ( t 3)

12 Global error found by summing the local errors. there are O(1/ t) terms. Error = t/ t n=0 1 2 t2 [A, B] K t t t2 = O( t) Strang splitting: v n+1 = [(Φ A ( t/2)φ B ( t/2)) (Φ B ( t/2)φ A ( t/2))] v n Local error: 1 2 ( t/2)2 ([A, B]+[B, A]) + O ( t 3) Global error: O ( t 2)

13 Bootstrap lemma: Let v : [0,T] X, where X is a normed space. Assume that t v(t) is continuous. If we can find a constant α such that v(0) α; if for some t>0, v(t) α, then v(t) α/2; then v(t) α/2 for all t [0,T].

14 Doubling the time variable v(t, τ) v(0, 0) = u 0, v t (t, t n )=B(v(t, t n )), t (t n,t n+1 ], v τ (t, τ) =A(v(t, τ)), (t, τ) [t n,t n+1 ] (t n,t n+1 ], τ A t AB A AB t t

15 The error function w(t) =v(t, t) u(t)

16 The error function w(t) =v(t, t) u(t) A(f + g) =A(f)+dA(f)[g]+ B(f + g) =B(f)+dB(f)[g] (1 α)d (2) A(f + αg)[g] 2 dα, (1 α)d (2) B(f + αg)[g] 2 dα.

17 The error function w(t) =v(t, t) u(t) A(f + g) =A(f)+dA(f)[g]+ B(f + g) =B(f)+dB(f)[g] (1 α)d (2) A(f + αg)[g] 2 dα, (1 α)d (2) B(f + αg)[g] 2 dα. w t da(u)[w] db(u)[w] =v t + v τ u t da(u)[w] db(u)[w] = v t + A(v) (A + B)(u) da(u)[w] db(u)[w] = v t B(v)+ ( A(v) A(u) da(u)[w] ) + ( B(v) B(u) db(u)[w] ) = F (t) (1 α)d (2) A(u + αw)[w] 2 dα (1 α)d (2) B(u + αw)[w] 2 dα

18 The error function w(t) =v(t, t) u(t) A(f + g) =A(f)+dA(f)[g]+ B(f + g) =B(f)+dB(f)[g] (1 α)d (2) A(f + αg)[g] 2 dα, (1 α)d (2) B(f + αg)[g] 2 dα. w t da(u)[w] db(u)[w] =v t + v τ u t da(u)[w] db(u)[w] = v t + A(v) (A + B)(u) da(u)[w] db(u)[w] = v t B(v)+ ( A(v) A(u) da(u)[w] ) + ( B(v) B(u) db(u)[w] ) = F (t) (1 α)d (2) A(u + αw)[w] 2 dα (1 α)d (2) B(u + αw)[w] 2 dα

19 The error function w(t) =v(t, t) u(t) A(f + g) =A(f)+dA(f)[g]+ B(f + g) =B(f)+dB(f)[g] (1 α)d (2) A(f + αg)[g] 2 dα, (1 α)d (2) B(f + αg)[g] 2 dα. w t da(u)[w] db(u)[w] =v t + v τ u t da(u)[w] db(u)[w] = v t + A(v) (A + B)(u) da(u)[w] db(u)[w] = v t B(v)+ ( A(v) A(u) da(u)[w] ) + ( B(v) B(u) db(u)[w] ) = F (t) (1 α)d (2) A(u + αw)[w] 2 dα (1 α)d (2) B(u + αw)[w] 2 dα

20 The error function w(t) =v(t, t) u(t) A(f + g) =A(f)+dA(f)[g]+ B(f + g) =B(f)+dB(f)[g] (1 α)d (2) A(f + αg)[g] 2 dα, (1 α)d (2) B(f + αg)[g] 2 dα. w t da(u)[w] db(u)[w] =v t + v τ u t da(u)[w] db(u)[w] = v t + A(v) (A + B)(u) da(u)[w] db(u)[w] = v t B(v)+ ( A(v) A(u) da(u)[w] ) + ( B(v) B(u) db(u)[w] ) = F (t) (1 α)d (2) A(u + αw)[w] 2 dα (1 α)d (2) B(u + αw)[w] 2 dα

21 The error function w(t) =v(t, t) u(t) A(f + g) =A(f)+dA(f)[g]+ B(f + g) =B(f)+dB(f)[g] (1 α)d (2) A(f + αg)[g] 2 dα, (1 α)d (2) B(f + αg)[g] 2 dα. F (t, τ) =v t (t, τ) B(v(t, τ)) w t dc(u)[w] =F (t)+ w(0) = (1 α)d (2) C(u + αw)[w] 2 dα, t > 0

22 The error function w(t) =v(t, t) u(t) A(f + g) =A(f)+dA(f)[g]+ B(f + g) =B(f)+dB(f)[g] (1 α)d (2) A(f + αg)[g] 2 dα, (1 α)d (2) B(f + αg)[g] 2 dα. F (t, τ) =v t (t, τ) B(v(t, τ)) w t dc(u)[w] =F (t)+ 1 0 (1 α)d (2) C(u + αw)[w] 2 dα, t > 0 w(0) = 0 { F τ + da(v)[f ]=[A, B](v, v), (t, τ) [t n,t n+1 ], F (t, t n )=0 τ t AB A A AB t t

23 Ordinary differential equations { u t = C(u), t > 0, u(0) = u 0 R n C(u) =A(u)+B(u), d 2 C = constant.

24 Ordinary differential equations { u t = C(u), t > 0, u(0) = u 0 R n C(u) =A(u)+B(u), d 2 C = constant. There exists a unique bounded solution u(t) K for t [0,T].

25 Ordinary differential equations { u t = C(u), t > 0, u(0) = u 0 R n C(u) =A(u)+B(u), d 2 C = constant. There exists a unique bounded solution u(t) K for t [0,T]. The bootstrap argument assume v(t, τ) α.

26 Ordinary differential equations { u t = C(u), t > 0, u(0) = u 0 R n C(u) =A(u)+B(u), d 2 C = constant. There exists a unique bounded solution u(t) K for t [0,T]. The bootstrap argument assume v(t, τ) α. w t + C (u)w = F + κ 2 w2, t > 0, w(0) = 0 F τ A (v)f =[A, B](v), (t, τ) [t n,t n+1 ] (t n,t n+1 ], F(t, t n ) = 0

27 Ordinary differential equations { u t = C(u), t > 0, u(0) = u 0 R n C(u) =A(u)+B(u), d 2 C = constant. There exists a unique bounded solution u(t) K for t [0,T]. The bootstrap argument assume v(t, τ) α. w t + C (u)w = F + κ 2 w2, t > 0, w(0) = 0 F τ A (v)f =[A, B](v), (t, τ) [t n,t n+1 ] (t n,t n+1 ], F(t, t n ) = τ F 2 = A (v)f 2 +[A, B](v)F KαF 2 + Kα 2 F

28 Ordinary differential equations { u t = C(u), t > 0, u(0) = u 0 R n C(u) =A(u)+B(u), d 2 C = constant. There exists a unique bounded solution u(t) K for t [0,T]. The bootstrap argument assume v(t, τ) α. w t + C (u)w = F + κ 2 w2, t > 0, w(0) = 0 F τ A (v)f =[A, B](v), (t, τ) [t n,t n+1 ] (t n,t n+1 ], F(t, t n ) = 0 τ F Kα F + Kα2, F(t n, 0) = 0

29 Ordinary differential equations { u t = C(u), t > 0, u(0) = u 0 R n C(u) =A(u)+B(u), d 2 C = constant. There exists a unique bounded solution u(t) K for t [0,T]. The bootstrap argument assume v(t, τ) α. w t + C (u)w = F + κ 2 w2, t > 0, w(0) = 0 F τ A (v)f =[A, B](v), (t, τ) [t n,t n+1 ] (t n,t n+1 ], F(t, t n ) = 0 F (t, τ) K α t

30 Ordinary differential equations { u t = C(u), t > 0, u(0) = u 0 R n C(u) =A(u)+B(u), d 2 C = constant. There exists a unique bounded solution u(t) K for t [0,T]. The bootstrap argument assume v(t, τ) α. w t + C (u)w K α t + κ 2 w2, w(0) = 0. F (t, τ) K α t

31 Ordinary differential equations { u t = C(u), t > 0, u(0) = u 0 R n C(u) =A(u)+B(u), d 2 C = constant. There exists a unique bounded solution u(t) K for t [0,T]. The bootstrap argument 1 2 assume v(t, τ) α. d dt w2 = C (u)w 2 + K α tw + κ 2 w3 Kw 2 + K α t w +( u + v ) κ 2 w2 Kw 2 + K α t w + K α w 2.

32 Ordinary differential equations { u t = C(u), t > 0, u(0) = u 0 R n C(u) =A(u)+B(u), d 2 C = constant. There exists a unique bounded solution u(t) K for t [0,T]. The bootstrap argument assume v(t, τ) α. d dt w K α w + K α t, w(0) = 0. w(t) e K αt tk α t K α t

33 Ordinary differential equations { u t = C(u), t > 0, u(0) = u 0 R n C(u) =A(u)+B(u), d 2 C = constant. There exists a unique bounded solution u(t) K for t [0,T]. The bootstrap argument assume v(t, τ) α. d dt w K α w + K α t, w(0) = 0. w(t) e K αt tk α t K α t v(t, τ) u(t) + w(t) + v(t, t) v(t, τ) K + K α t.

34 Ordinary differential equations { u t = C(u), t > 0, u(0) = u 0 R n C(u) =A(u)+B(u), d 2 C = constant. There exists a unique bounded solution u(t) K for t [0,T]. The bootstrap argument assume v(t, τ) α. d dt w K α w + K α t, w(0) = 0. w(t) e K αt tk α t K α t v(t, τ) u(t) + w(t) + v(t, t) v(t, τ) K + K α t. Choose α such that α/4 >K, then t such that K α t<α/4

35 Ordinary differential equations { u t = C(u), t > 0, u(0) = u 0 R n C(u) =A(u)+B(u), d 2 C = constant. There exists a unique bounded solution u(t) K for t [0,T]. The bootstrap argument assume v(t, τ) α. v(t, τ) α 4 + α 4 α 2 d dt w K α w + K α t, w(0) = 0. w(t) e K αt tk α t K α t v(t, τ) u(t) + w(t) + v(t, t) v(t, τ) K + K α t. Choose α such that α/4 >K, then t such that K α t<α/4

36 Ordinary differential equations { u t = C(u), t > 0, u(0) = u 0 R n C(u) =A(u)+B(u), d 2 C = constant. There exists a unique bounded solution u(t) K for t [0,T]. The bootstrap argument assume v(t, τ) α. v(t, τ) α 4 + α 4 α 2 As a consequence v(t, t) u(t) K α/2 t

37 The logistic equation u t = u(u 1), u(0) = u 0 (0, 1) u(t) = u 0 u 0 + e t (1 u 0 ) A(u) =u 2 Φ A (t) u 0 = u 0 1 u 0 t B(u) = u Φ B (t) u 0 = u 0 e t v(t n,t n )= u 0 (1 e t ) (1 e t )e t n + u0 t(1 e t n ) v(t n,t n ) u(t n ) = u 0 2 (e t n 1) ( t) 1 (1 e t ) 1 (u 0 + e t n (( t) 1 (1 e t ) u 0 ))(u 0 + e t n (1 u0 ))

38 The logistic equation u t = u(u 1), u(0) = u 0 (0, 1) Error t v(t n,t n ) u(t n ) = u 0 2 (e t n 1) ( t) 1 (1 e t ) 1 (u 0 + e t n (( t) 1 (1 e t ) u 0 ))(u 0 + e t n (1 u0 ))

39 The KdV equation John Scott Russell, 1834 "I was observing the motion of a boat which was rapidly drawn along a narrow channel by a pair of horses, when the boat suddenly stopped not so the mass of water in the channel which it had put in motion; it accumulated round the prow of the vessel in a state of violent agitation, then suddenly leaving it behind, rolled forward with great velocity, assuming the form of a large solitary elevation, a rounded, smooth and well-defined heap of water, which continued its course along the channel apparently without change of form or diminution of speed. I followed it on horseback, and overtook it still rolling on at a rate of some eight or nine miles (14 km) an hour, preserving its original figure some thirty feet long and a foot to a foot and a half in height. Its height gradually diminished, and after a chase of one or two miles (3 km) I lost it in the windings of the channel. Such, in the month of August 1834, was my first chance interview with that singular and beautiful phenomenon which I have called the Wave of Translation". u t +6uu x + u xxx =0 Joseph Boussinesq, 1871, Lord Rayleigh, 1876, Korteweg and de Vries 1895 Zabusky and Kruskal, Numerical evidence of waves of translation. Gardner, Greene, Kruskal and Miura, Inverse scattering. Sjöberg, Existence via semi-discrete numerical method. Bona and Smith, Well-posedness of initial value problem. Christ, Colliander and Tao, More general well-posedness.

40 The KdV equation John Scott Russell, 1834 "I was observing the motion of a boat which was rapidly drawn along a narrow channel by a pair of horses, when the boat suddenly stopped not so the mass of water in the channel which it had put in motion; it accumulated round the prow of the vessel in a state of violent agitation, then suddenly leaving it behind, rolled forward with great velocity, assuming the form of a large solitary elevation, a rounded, smooth and well-defined heap of water, which continued its course along the channel apparently without change of form or diminution of speed. I followed it on horseback, and overtook it still rolling on at a rate of some eight or nine miles (14 km) an hour, preserving its original figure some thirty feet long and a foot to a foot and a half in height. Its height gradually diminished, and after a chase of one or two miles (3 km) I lost it in the windings of the channel. Such, in the month of August 1834, was my first chance interview with that singular and beautiful phenomenon which I have called the Wave of Translation". u t +6uu x + u xxx =0 Joseph Boussinesq, 1871, Lord Rayleigh, 1876, Korteweg and de Vries 1895 Zabusky and Kruskal, Numerical evidence of waves of translation. Gardner, Greene, Kruskal and Miura, Inverse scattering. Sjöberg, Existence via semi-discrete numerical method. Bona and Smith, Well-posedness of initial value problem. Christ, Colliander and Tao, More general well-posedness.

41 Solitons

42 Solitons One soliton u(x, t) = 2a 2 cosh 2 (a (x 4a 2 t))

43 Solitons One soliton u(x, t) = 2a 2 cosh 2 (a (x 4a 2 t)) N soliton solution ρ 1,0,..., ρ N,0, ρ n (t) =ρ n,0 e 8p3 n t, p 1,..., p N n =1,..., N c n,m (x, t) = ρn (t)ρ m (t) p n + p m e (p n+p m )x, u(x, t) = 2 2 x 2 [ log [det {I N + C(x, t)}] ]

44 Solitons One soliton u(x, t) = 2a 2 cosh 2 (a (x 4a 2 t)) N soliton solution ρ 1,0,..., ρ N,0, ρ n (t) =ρ n,0 e 8p3 n t, p 1,..., p N n =1,..., N c n,m (x, t) = Rational solutions ρn (t)ρ m (t) p n + p m e (p n+p m )x, u(x, t) = 2 2 x 2 [ log [det {I N + C(x, t)}] u(x, t) = 6x(x3 24t) 2 (x 3 2, u(x, t) = t) x 2 log ( x x 3 t 720t 2) ]

45 The KdV equation H s = u t = uu x u xxx, t > 0 { f } j f L 2,j=0,..., s u(x, 0) = u 0 (x) s (f, g) H s = j f j g dx j=0 s 3, there exists a unique solution such that u 0 H s C 1 u(,t) H s C 2, t [0,T]

46 The KdV equation H s = u t = uu x u xxx, t > 0 { f } j f L 2,j=0,..., s u(x, 0) = u 0 (x) s (f, g) H s = j f j g dx j=0 s 3, there exists a unique solution such that u 0 H s C 1 u(,t) H s C 2, t [0,T] Solitons

47 Burgers equation u t = uu x, x R, t > 0, u(x, 0) = u 0 (x)

48 Burgers equation u t = uu x, x R, t > 0, u(x, 0) = u 0 (x) characteristics d x(t) = u(x(t),t) dt

49 Burgers equation u t = uu x, x R, t > 0, u(x, 0) = u 0 (x) characteristics d d x(t) = u(x(t),t) dt dt u(x(t),t)=u dx t + u x dt = u t uu x =0

50 Burgers equation u t = uu x, x R, t > 0, u(x, 0) = u 0 (x) characteristics d d x(t) = u(x(t),t) dt dt u(x(t),t)=u dx t + u x dt = u t uu x =0 t u(x, T ) characteristics u(x, 0) x

51 Burgers equation u t = uu x, x R, t > 0, u(x, 0) = u 0 (x) characteristics d d x(t) = u(x(t),t) dt dt u(x(t),t)=u dx t + u x dt = u t uu x =0 weak solutions t u(x, T ) characteristics T 0 R = 1 2 uϕ t dxdt T 0 R u 2 ϕ x dxdt u(x, 0) x

52 Burgers equation u t = uu x, x R, t > 0, u(x, 0) = u 0 (x) characteristics d d x(t) = u(x(t),t) dt dt u(x(t),t)=u dx t + u x dt = u t uu x =0 weak solutions t u(x, T ) characteristics T 0 R = 1 2 uϕ t dxdt T 0 R u 2 ϕ x dxdt u(x, 0) Well posedness theory entropy solutions in L 1 B.V. x

53

54 The Airy equation

55 The Airy equation u t = u xxx, t > 0, x R, u(x, 0) = u 0 (x)

56 The Airy equation u t = u xxx, t > 0, x R, u(x, 0) = u 0 (x) ( ) x y u(x, t) =t 1/3 Ai u 0 (y) dy Ai(y) = 1 π 0 cos t 1/3 ( ) s ys ds

57 The Airy equation u t = u xxx, t > 0, x R, u(x, 0) = u 0 (x) ( ) x y u(x, t) =t 1/3 Ai u 0 (y) dy Ai(y) = 1 π 0 cos t 1/3 ( ) s ys ds

58 The Airy equation u t = u xxx, t > 0, x R, u(x, 0) = u 0 (x) ( ) x y u(x, t) =t 1/3 Ai u 0 (y) dy Ai(y) = 1 π 0 cos t 1/3 ( ) s ys ds d dt u L 2 = = = uu t dx uu xxx dx ( uu xx 1 ) 2 u2 x x dx =0 0.4

59 The Airy equation u t = u xxx, t > 0, x R, u(x, 0) = u 0 (x) ( ) x y u(x, t) =t 1/3 Ai u 0 (y) dy Ai(y) = 1 π 0 cos t 1/3 ( ) s ys ds d dt u L 2 = = = uu t dx uu xxx dx ( uu xx 1 ) 2 u2 x x dx =0 0.4 u(,t) H s = u 0 H s Well posedness theory in H s.

60 Solutions of u t = B(u) =uu x and u t = A(u) = u xxx live in different spaces...

61 Solutions of u t = B(u) =uu x and u t = A(u) = u xxx live in different spaces... Solutions to u t = A(u)+B(u) are in H s

62 Solutions of u t = B(u) =uu x and u t = A(u) = u xxx live in different spaces... Solutions to u t = A(u)+B(u) are in H s t u(x, T ) characteristics u(x, 0) x

63 Splitting for KdV Airy A(f) = f xxx, da(f)[g] = g xxx, d (2) A(f)[g, h] =0, Burgers B(f) =ff x, db(f)[g] =fg x + f x g, d (2) B(f)[g, h] =hg x + h x g, d (3) B(f)[g, h, k] =0. [A, B](f, f) = x(f x ) 2

64 Splitting for KdV Airy A(f) = f xxx, da(f)[g] = g xxx, d (2) A(f)[g, h] =0, Burgers B(f) =ff x, db(f)[g] =fg x + f x g, d (2) B(f)[g, h] =hg x + h x g, d (3) B(f)[g, h, k] =0. [A, B](f, f) = x(f x ) 2 error equation w t ( u x w + w x u w xxx ) = F + wwx, w(0) = 0

65 Splitting for KdV Airy A(f) = f xxx, da(f)[g] = g xxx, d (2) A(f)[g, h] =0, Burgers B(f) =ff x, db(f)[g] =fg x + f x g, d (2) B(f)[g, h] =hg x + h x g, d (3) B(f)[g, h, k] =0. [A, B](f, f) = x(f x ) 2 error equation w t ( u x w + w x u w xxx ) = F + wwx, w(0) = 0 forcing term { F = v t vv x F τ + F xxx = (v x ) 2, F (t n, τ) =0

66 The bootstrap assumption v(t, τ) H 2 α

67 The bootstrap assumption v(t, τ) H 2 α 1 2 d dt v(t, t n) 2 H 5 =(v, v t ) H 5 = = 5 j=0 j k=0 5 j=0 ( ) j k j v j (vv x ) dx j v k+1 v j k v dx.

68 The bootstrap assumption v(t, τ) H 2 α 1 2 d dt v(t, t n) 2 H 5 =(v, v t ) H 5 = = 5 j=0 j k=0 5 j=0 ( ) j k j v j (vv x ) dx j v k+1 v j k v dx. terms with j<5, or with j = 5 and k<5 j v k+1 v j k v dx max{j,k j} v L k+1 v L 2 min{j,j k} v L 2 K v 2 H s v H 2 K α v 2 H s.

69 The bootstrap assumption v(t, τ) H 2 α 1 2 d dt v(t, t n) 2 H 5 =(v, v t ) H 5 = = 5 j=0 j k=0 5 j=0 ( ) j k j v j (vv x ) dx j v k+1 v j k v dx. term with j = 5 and k =5 5 v 5+1 v v dx = 1 2 ( 5 v ) 2 v dx v L 5 v 2 L 2 K v H 2 v 2 H 5

70 The bootstrap assumption v(t, τ) H 2 α 1 2 d dt v(t, t n) 2 H 5 =(v, v t ) H 5 = = 5 j=0 j k=0 5 j=0 ( ) j k d dt v (t, t n) H 5 K α v (t, t n ) H 5 j v j (vv x ) dx j v k+1 v j k v dx.

71 The bootstrap assumption v(t, τ) H 2 α 1 2 d dt v(t, t n) 2 H 5 =(v, v t ) H 5 = = 5 j=0 j k=0 5 j=0 ( ) j k d dt v (t, t n) H 5 K α v (t, t n ) H 5 j v j (vv x ) dx j v k+1 v j k v dx. v(t, t n ) H 5 e K α t v(t n,t n ) H 5 e K αt n u 0 H 5 v(t, τ) H 5 K α

72 Estimating the force term v(t, τ) H 2 α v(t, τ) H 5 K α F τ + F xxx = (v x ) 2, F(t n, τ) = 0

73 Estimating the force term v(t, τ) H 2 α v(t, τ) H 5 K α F τ + F xxx = (v x ) 2, F(t n, τ) = τ F (t, τ) 2 H 2 = j=0 K F H 2 j F j+2 (v x ) 2) dx v 2 H x 4 K F H 2 v x 2 H 4 K F H 2 v 2 H 5 K α F H 2

74 Estimating the force term v(t, τ) H 2 α v(t, τ) H 5 K α F τ + F xxx = (v x ) 2, F(t n, τ) = τ F (t, τ) 2 H 2 = j=0 K F H 2 j F j+2 (v x ) 2) dx v 2 H x 4 K F H 2 v x 2 H 4 K F H 2 v 2 H 5 K α F H 2 F H 2 K α t

75 Estimating the error v(t, τ) H 2 α w t ( u x w + w x u w xxx ) = F + wwx, w(0) = 0 F H 2 K α t

76 Estimating the error v(t, τ) H 2 α w t ( u x w + w x u w xxx ) = F + wwx, w(0) = 0 v(t, τ) H 5 K α F H 2 K α t

77 Estimating the error v(t, τ) H 2 α 1 2 w t ( u x w + w x u w xxx ) = F + wwx, w(0) = 0 d dt w H 2 = 2 j=0 j w j (uw x + u x w + ww x ) dx +(w, F ) H 2 v(t, τ) H 5 F H 2 K α K α t

78 Estimating the error v(t, τ) H 2 α 1 2 w t ( u x w + w x u w xxx ) = F + wwx, w(0) = 0 d dt w H 2 = 2 j=0 j w j (uw x + u x w + ww x ) dx +(w, F ) H 2 v(t, τ) H 5 F H 2 K α K α t terms with one or two derivatives on w j w k+1 w j k u dx w 2 j k H u 2 L j = {0, 1}, k j K w 2 H 2 u H 5 K α w 2 H 2

79 Estimating the error v(t, τ) H 2 α 1 2 w t ( u x w + w x u w xxx ) = F + wwx, w(0) = 0 d dt w H 2 = 2 j=0 j w j (uw x + u x w + ww x ) dx +(w, F ) H 2 terms with three derivatives on w 2 w 3 w u dx = 1 ( 2 w ) 2 u dx K u 2 H5 w 2 2 ww 3 w dx = 1 ( 2 w ) 2 w dx 2 K ( u H 5 + v H 5) w 2 H 2. H 2, v(t, τ) H 5 F H 2 K α K α t

80 Estimating the error v(t, τ) H 2 α 1 2 w t ( u x w + w x u w xxx ) = F + wwx, w(0) = 0 d dt w H 2 = 2 j=0 j w j (uw x + u x w + ww x ) dx +(w, F ) H 2 v(t, τ) H 5 F H 2 K α K α t 1 2 d dt w 2 H 2 K α w 2 H 2 + K α t w H 2

81 Estimating the error v(t, τ) H 2 α 1 2 w t ( u x w + w x u w xxx ) = F + wwx, w(0) = 0 d dt w H 2 = 2 j=0 j w j (uw x + u x w + ww x ) dx +(w, F ) H 2 v(t, τ) H 5 F H 2 K α K α t 1 2 d dt w 2 H K 2 α w 2 H + K 2 α t w H 2 d dt w H K 2 α w H 2 + K α t w H 2 K α t

82

83 Concluding the bootstrap...

84 Concluding the bootstrap... v(t, τ) H 2 α v(t, τ) H 5 K α F H 2 K α t w H 2 K α t

85 Concluding the bootstrap... v(t, τ) H 2 α v(t, τ) H 2 = v(t, t) H 2 + v(t, τ) H 2 v(t, t) H 2 }{{} =0 v(t, t) u(t) H 2 + u(t) H 2 = w H 2 + u H 2 K α t + K v(t, τ) H 5 K α F H 2 K α t w H 2 K α t

86 Concluding the bootstrap... v(t, τ) H 2 α v(t, τ) H 2 = v(t, t) H 2 + v(t, τ) H 2 v(t, t) H 2 }{{} =0 v(t, t) u(t) H 2 + u(t) H 2 = w H 2 + u H 2 K α t + K v(t, τ) H 5 K α F H 2 K α t w H 2 K α t 1. choose α/4 K 2. then choose t s.t. K α t<α/4

87 Concluding the bootstrap... v(t, τ) H 2 α v(t, τ) H 2 = v(t, t) H 2 + v(t, τ) H 2 v(t, t) H 2 }{{} =0 v(t, t) u(t) H 2 + u(t) H 2 = w H 2 + u H 2 K α t + K v(t, τ) H 5 K α F H 2 K α t w H 2 K α t v(t, τ) H 2 α 4 + α 4 = α 2

88 Concluding the bootstrap... v(t, τ) H 2 α v(t, τ) H 2 = v(t, t) H 2 + v(t, τ) H 2 v(t, t) H 2 }{{} =0 v(t, t) u(t) H 2 + u(t) H 2 = w H 2 + u H 2 K α t + K v(t, τ) H 5 K α F H 2 K α t w H 2 K α t v(t, τ) H 2 α 4 + α 4 = α 2 If u 0 H 5 then v(t, t) u(t) H 2 K α/2 t

89 Sjöberg s semi-discrete scheme { } l 2 = u i x i Z u 2 i < u i (t) u(i x, t), x small (u i,v i ) l 2 = x i u i v i, u i 2 l 2 =(u i,u i ) l 2 D ± u i = ± u i±1 u i x, D = 1 2 (D + + D ) u t = uu x u xxx d dt u i = 1 3 ( ui Du i + Du 2 i ) D D 2 +u i

90 L 2 estimate u t = uu x u xxx

91 L 2 estimate u t = uu x u xxx 1 2 d dt u 2 L 2 =(u, uu x ) L 2 (u, u xxx ) L 2 =0

92 L 2 estimate u t = uu x u xxx 1 2 l 2 estimate d dt u = 1 3 (udu + Du2 ) D D 2 +u d dt u 2 L 2 =(u, uu x ) L 2 (u, u xxx ) L 2 =0

93 L 2 estimate u t = uu x u xxx 1 2 l 2 estimate d dt u = 1 3 (udu + Du2 ) D D 2 +u d dt u 2 L 2 =(u, uu x ) L 2 (u, u xxx ) L 2 =0 (v, D ± u) l 2 = (D v, u) l 2, (v, Du) l 2 = (Dv, u) l 2

94 L 2 estimate u t = uu x u xxx 1 2 l 2 estimate d dt u = 1 3 (udu + Du2 ) D D 2 +u d dt u 2 L 2 =(u, uu x ) L 2 (u, u xxx ) L 2 =0 (v, D ± u) l 2 = (D v, u) l 2, (v, Du) l 2 = (Dv, u) l 2 ( u, udu + Du 2 ) l 2 = ( u 2, Du ) l 2 + ( u, Du 2) l 2 =0

95 L 2 estimate u t = uu x u xxx 1 2 l 2 estimate d dt u = 1 3 (udu + Du2 ) D D 2 +u d dt u 2 L 2 =(u, uu x ) L 2 (u, u xxx ) L 2 =0 (v, D ± u) l 2 = (D v, u) l 2, (v, Du) l 2 = (Dv, u) l 2 ( u, udu + Du 2 ) l 2 = ( u 2, Du ) l 2 + ( u, Du 2) l 2 =0 ( u, D+ D 2 u ) l 2 = ( D 2 +D u, u ) l 2 = ( u, D D 2 +u ) l 2

96 L 2 estimate u t = uu x u xxx 1 2 l 2 estimate d dt u = 1 3 (udu + Du2 ) D D 2 +u d dt u 2 L 2 =(u, uu x ) L 2 (u, u xxx ) L 2 =0 (v, D ± u) l 2 = (D v, u) l 2, (v, Du) l 2 = (Dv, u) l 2 ( u, udu + Du 2 ) l 2 = ( u 2, Du ) l 2 + ( u, Du 2) l 2 =0 ( u, D+ D 2 u ) l 2 = ( D 2 +D u, u ) l 2 = ( u, D D 2 +u ) l 2 ( u, D+ D u 2 ) = 1 ( ( l u, D+ D 2 2 D 2 ) ) 2 +D u = x ( u, (D+ D ) 2 u ) 2 l 2 = x 2 D +D u 2 l 2 l 2

97 L 2 estimate u t = uu x u xxx 1 2 l 2 estimate d dt u = 1 3 (udu + Du2 ) D D 2 +u d dt u 2 L 2 =(u, uu x ) L 2 (u, u xxx ) L 2 =0 1 2 d dt u 2 l 2 = 1 3 ( u, udu + Du 2 ) l 2 ( u, D + D 2 +u ) l 2 = x 2 D +D u 2 l 2

98 Via interpolation D + D u l 2 ɛ D D 2 +u l 2 + c(ɛ) u l 2 one can show that du/dt l 2 K, D D 2 +u l 2 K for t<t. For t<t, via Arzela-Ascoli, the piecewise constant function u i (t) converges to u(x, t) in L 2. Furthermore u xxx L 2, and in L 2. u t = uu x u xxx

99 A fully discrete splitting method

100 A fully discrete splitting method gridfunctions u(i x, n t) u n i

101 A fully discrete splitting method gridfunctions u(i x, n t) u n i discrete derivatives D ± u n i = ± un i±1 un i x D = 1 2 (D + + D ) D t +u n i = un+1 i u n i t

102 A fully discrete splitting method gridfunctions u(i x, n t) u n i discrete derivatives D ± u n i = ± un i±1 un i x D = 1 2 (D + + D ) D t +u n i = un+1 i u n i t ū n i = 1 2 ( u n i 1 + u n i+1)

103 A fully discrete splitting method gridfunctions u(i x, n t) u n i discrete derivatives D ± u n i = ± un i±1 un i x D = 1 2 (D + + D ) D t +u n i = un+1 i u n i t ū n i = 1 2 ( u n i 1 + u n i+1) continuous equation u t = uu x u xxx

104 A fully discrete splitting method gridfunctions u(i x, n t) u n i discrete derivatives D ± u n i = ± un i±1 un i x D = 1 2 (D + + D ) D t +u n i = un+1 i u n i t ū n i = 1 2 ( u n i 1 + u n i+1) continuous equation u t = uu x u xxx discrete equation D t +u n i =ū n i Du n i D D 2 +u n+1 i + xd D + u n i

105 A fully discrete splitting method gridfunctions u(i x, n t) u n i discrete derivatives D ± u n i = ± un i±1 un i x D = 1 2 (D + + D ) D t +u n i = un+1 i u n i t ū n i = 1 2 ( u n i 1 + u n i+1) continuous equation u t = uu x u xxx discrete equation D t +u n i =ū n i Du n i D D 2 +u n+1 i + xd D + u n i

106 A fully discrete splitting method gridfunctions u(i x, n t) u n i discrete derivatives D ± u n i = ± un i±1 un i x D = 1 2 (D + + D ) D t +u n i = un+1 i u n i t ū n i = 1 2 ( u n i 1 + u n i+1) continuous equation u t = uu x u xxx discrete equation D t +u n i =ū n i Du n i D D 2 +u n+1 i + xd D + u n i numerical viscosity

107 Rewritten as a splitting method... two time indices u n,m D τ +u n,m = un,m+1 u n,m t D t +u n,m =ū n,m Du n,m + xd + D u n,m D τ +u n+1,m = D D 2 +u n+1,m+1

108 Rewritten as a splitting method... two time indices u n,m D τ +u n,m = un,m+1 u n,m t D t +u n,m =ū n,m Du n,m + xd + D u n,m D τ +u n+1,m = D D 2 +u n+1,m+1 One soliton

109 Rewritten as a splitting method... two time indices u n,m D τ +u n,m = un,m+1 u n,m t D t +u n,m =ū n,m Du n,m + xd + D u n,m D τ +u n+1,m = D D 2 +u n+1,m+1 Two solitons

110 Convergence of the difference scheme CFL-condition t x 3/2 u 0 L 2 ( 1+3 t x 3/2 u L 2 ) < 3 8 λ = t x

111 Convergence of the difference scheme CFL-condition t x 3/2 u 0 L 2 ( 1+3 t x 3/2 u L 2 ) < 3 8 λ = t x basic L 2 estimate ) ( u n 2L 2 D t + + x (λ D + D 2 u n+1 2L2 + D + D u n+1 2L ) Dun 2L2 0.

112 Convergence of the difference scheme CFL-condition t x 3/2 u 0 L 2 ( 1+3 t x 3/2 u L 2 ) < 3 8 λ = t x basic L 2 estimate ) ( u n 2L 2 D t + + x (λ D + D 2 u n+1 2L2 + D + D u n+1 2L ) Dun 2L2 0. estimating the time derivative v n = D t +u n 1

113 Convergence of the difference scheme CFL-condition t x 3/2 u 0 L 2 ( 1+3 t x 3/2 u L 2 ) < 3 8 λ = t x basic L 2 estimate ) ( u n 2L 2 D t + + x (λ D + D 2 u n+1 2L2 + D + D u n+1 2L ) Dun 2L2 0. estimating the time derivative D t + ( v n 2L 2 ) + x v n = D t +u n 1 ( λ D+ D v 2 n D+ D L 2 v n ) L 2 8λ Dvn 2 L 2 C max Du n v n 2 L 2.

114 Convergence of the difference scheme

115 Convergence of the difference scheme D t + ) ( ( v n 2L d 2 1 v n 2L + d 2 2 v n 3 L + v n 2 ) v n 1 2 L 2 L 2 d 1 and d 2 depend on the L 2 norm of u 0

116 Convergence of the difference scheme D t + ) ( ( v n 2L d 2 1 v n 2L + d 2 2 v n 3 L + v n 2 ) v n 1 2 L 2 L 2 d 1 and d 2 depend on the L 2 norm of u 0 v n L 2 is bounded by the solution of dα dt = d 1α +2d 2 α 3/2, α (0)) = a 0

117 Convergence of the difference scheme D t + ) ( ( v n 2L d 2 1 v n 2L + d 2 2 v n 3 L + v n 2 ) v n 1 2 L 2 L 2 d 1 and d 2 depend on the L 2 norm of u 0 v n L 2 is bounded by the solution of dα dt = d 1α +2d 2 α 3/2, α (0)) = a 0 this solution has a blow-up time T (a) = 2 ( log 1+ d ) 1 d 1 2ad 2

118 Convergence of the difference scheme D t + ) ( ( v n 2L d 2 1 v n 2L + d 2 2 v n 3 L + v n 2 ) v n 1 2 L 2 L 2 d 1 and d 2 depend on the L 2 norm of u 0 v n L 2 is bounded by the solution of dα dt = d 1α +2d 2 α 3/2, α (0)) = a 0 this solution has a blow-up time T (a) = 2 ( log 1+ d ) 1 d 1 2ad 2 This means that for t n <T /2 u n L 2 <C D t + u n L 2 <C D D 2 +u n L 2 <C

119 If t O ( x 3/2) then for t<t /2 the piecewise constant function u n i L 2. Furthermore u xxx L 2, and converges to u(x, t) in u t = uu x u xxx in L 2.

120 If t O ( x 3/2) then for t<t /2 the piecewise constant function u n i L 2. Furthermore u xxx L 2, and converges to u(x, t) in u t = uu x u xxx in L 2. Experimental error estimates for the single soliton

121 If t O ( x 3/2) then for t<t /2 the piecewise constant function u n i L 2. Furthermore u xxx L 2, and converges to u(x, t) in u t = uu x u xxx in L 2. Experimental error estimates for the single soliton x L 2 error

122 If t O ( x 3/2) then for t<t /2 the piecewise constant function u n i L 2. Furthermore u xxx L 2, and converges to u(x, t) in u t = uu x u xxx in L 2. Experimental error estimates for the single soliton

123 If t O ( x 3/2) then for t<t /2 the piecewise constant function u n i L 2. Furthermore u xxx L 2, and converges to u(x, t) in u t = uu x u xxx in L 2. Experimental error estimates for the single soliton

124 What s next?

125 What s next? Convergence of Strang splitting

126 What s next? Convergence of Strang splitting Splitting with spectral methods

127 What s next? Convergence of Strang splitting Splitting with spectral methods Other equations; BBM, Boussinesq, generalized KdV

128 Thank you