Artificial boundary conditions for dispersive equations. Christophe Besse

Transcription

1 Artificial boundary conditions for dispersive equations by Christophe Besse Institut Mathématique de Toulouse, Université Toulouse 3, CNRS Groupe de travail MathOcéan Bordeaux INSTITUT de MATHEMATIQUES de TOULOUSE

2 Outline of the talk 1 Why Artifical Boundary Conditions? 2 Derivation of some TBC 3 Numerical approximation

3 Introduction Typical equations The Schrödinger Eq. in R d i tψ + ψ + V (x, t, ψ) ψ = 0, (x, t) R d [0; T ] (S) lim ψ(x, t) = 0, t [0; T ] x + ψ(x, 0) = ψ 0(x), x R d ψ(x, t): wave function, complex V can be real potential, V = V (x, t) C (R d R +, R) nonlinear functional, V = f(ψ) Ex: V = q ψ 2 a combination: V = V (x) + f(ψ) ψ 0 compact support in Ω

4 Introduction Typical equations The Korteweg-de-Vries Eq. in R tu 6u xu + xu 3 = 0, (x, t) R [0; T ] (K) lim u(x, t) = 0, t [0; T ] x + u(x, 0) = u 0(x), x R u(x, t): real function u 0 compact support in Ω

5 Domain trunction Problem : Mesh an unbounded domain (here in 1D) R d [0; T ]

6 Domain trunction Problem : Mesh an unbounded domain (here in 1D) R d [0; T ] Truncation R [0; T ] Ω T := ]x l, x r[ [0; T ] Introduction of a fictitious boundary Σ T := Σ [0; T ] with Σ := Ω = {x l, x r} BC on the boundaryσ T

7 Domain trunction Problem : Mesh an unbounded domain (here in 1D) R d [0; T ] Truncation R [0; T ] Ω T := ]x l, x r[ [0; T ] Introduction of a fictitious boundary Σ T := Σ [0; T ] with Σ := Ω = {x l, x r} BC on the boundaryσ T Expression of the boundary condtion with the help of the Dirichlet-to-Neumann map: nψ + Λ + ψ = 0, on Σ T.

8 Domain trunction Problem : Mesh an unbounded domain (here in 1D) R d [0; T ] Truncation R [0; T ] Ω T := ]x l, x r[ [0; T ] Introduction of a fictitious boundary Σ T := Σ [0; T ] with Σ := Ω = {x l, x r} BC on the boundaryσ T Expression of the boundary condtion with the help of the Dirichlet-to-Neumann map: nψ + Λ + ψ = 0, on Σ T.

9 What happen if we do not take BC into account Airy Equation tu + xu 3 = 0, on ( 15, 15) [0; T ], u(x, 0) = exp( x 2 ), x ( 15, 15), xu( 15, t) = 0, t > 0, u(15, t) = 0, xu(15, t) = 0, t > 0.

10 Outline of the talk 1 Why Artifical Boundary Conditions? 2 Derivation of some TBC 3 Numerical approximation

11 TBC for Wave Eq. Homogeneous Wave Equation in 1D t 2 ψ xψ 2 = 0, (x, t) R x [0; T ], (P) ψ(x, 0) = ψ 0(x), x R x, tψ(x, 0) = ψ 1(x), x R x. Hypothesis : supp(ψ 0,1) B(0, R). Solution Simply, with ψ(x, t) = 1 2 (ψ0(x + t) + ψ0(x t)) x+t x t ψ(x, t) = ϕ 1(x + t) + ϕ 2(x t), x 2ϕ 1(x) = ψ 0(x) + 2ϕ 2(x) = ψ 0(x) x ψ 1(y)dy, ψ 1(y)dy. ψ 1(y)dy.

12 TBC for Wave Eq. t x L R R L supp(ϕ 1(x + t)) supp(ϕ 2(x t)) t > 0 and x > R, the supports are disjoint. x = L x = L ψ(x, t) = ϕ 1(x + t) ψ(x, t) = ϕ 2(x t) xψ = tψ xψ = tψ Then, on x = ±L, the Neumann datum is expressed in function of the Dirichlet one nψ + tψ = 0 where n denotes the outwardly unit normal vector to Ω = ( L, L).

13 TBC for Wave Eq. The problem (P) is thus transformed in (P app) (P app) where Γ = { L, L}. t 2 ψ a xψ 2 a = 0, (x, t) Ω [0; T ], ψ a (x, 0) = ψ 0(x), x Ω, tψ a (x, 0) = ψ 1(x), x Ω, nψ a + tψ a = 0, (x, t) Γ [0; T ], ψ Ω (x, t) = ψ a (x, t), (x, t) Ω [0; T ]. Ths BCs do not perturb the solution: we call them Remark: Transparent Boundary Conditions (TBC) 2 t 2 x = ( t x)( t + x).

14 TBC for Wave Eq. Other way of derivation: Laplace transform L (u)(x, ω) = û(x, ω) = with frequency ω = σ + iτ, σ > 0 0 u(x, t)e ωt dt L ( tu)(x, ω) = ωû(x, ω) u(x, 0), L ( t 2 u)(x, ω) = ω 2 û(x, ω) ωu(x, 0) tu(x, 0). 1 Transmission problem Transmission conditions: continuity of traces (u and xu) Interior Problem ( t 2 2 x )v = 0, x Ω, t > 0, xv = xw, x Γ, t > 0, v(x, 0) = ψ 0 (x), x Ω, tv(x, 0) = ψ 1 (x), x Ω, Exterior problem ( t x)w 2 = 0, x Ω, t > 0, w(x, t) = v(x, t), x = ±L, t > 0, w(x, 0) = 0, x Ω, tw(x, 0) = 0, x Ω.

15 TBC for Wave Eq. 2 Laplace tranform w.r.t time for x > L : ω 2 ŵ 2 xŵ = 0. Solution: ŵ(x, ω) = A + (ω)e ω x + A (ω)e ω x 3 How to select the outgoing wave: solution of finite energy A + = 0. ŵ(x, ω) = e ω (x L) L (w(l, ))(ω). Taking derivative + continuity of trace xŵ(x, ω) x=l = ω ˆv(x, ω) x=l 4 Inverse Laplace tranform xw(x, t) x=l = tv(x, t) x=l. 5 Thus, we obtain the TBC for v (continuity of trace) nv + tv = 0

16 The Airy equation tu + xu 3 = 0, on R [0; T ]. u(x, 0) = u 0(x), x R. 1 Splitting between interior and exterior problems left exterior problem right exterior problem interior problem t=0

17 The Airy equation tu + xu 3 = 0, on R [0; T ]. u(x, 0) = u 0(x), x R. 1 Splitting between interior and exterior problems left exterior problem right exterior problem interior problem t=1 The B.Cs are different at left and right boundaries

18 The Airy equation 2 Transmission problem Interior problem ( t + x)u 3 = 0, x l < x < x r, t > 0, u(x, 0) = u 0 (x), x l < x < x r, x 2 u(x l, t) = x 2 v(x l, t), t > 0 xu(x r, t) = xw(x r, t), t > 0, xu(x 2 r, t) = xw(x 2 r, t), t > 0. Left exterior problem ( t + x 3 )v = 0, x < x l, t > 0, v(x, 0) = 0, x < x l, v(x l, t) = u(l, t), t > 0, xv(x l, t) = xu(l, t), t > 0, v 0, x. Right exterior problem ( t + x 3 )w = 0, x > xr, t > 0, w(x, 0) = 0, x > x r, w(x, x r) = u(x, x r), t > 0, w 0, x.

19 The Airy equation 3 Laplace transform w.r.t time tf + xf 3 = 0 ω ˆf + x 3 ˆf = 0 3 Solution: ˆf(x, ω) = c k (ω)e λk(ω)x with λ k (ω) = j k 1 3 ω, j = e 2iπ/3 k=1 Re λ 1(ω) < 0, Re λ 2(ω) > 0, Re λ 3(ω) > 0.

20 The Airy equation 4 Right exterior problem : c 2 = c 3 = 0 so ŵ(x, ω) = c 1e λ 1(ω)x. Using continuity, we have ŵ(x, ω) = L {u(x r, )}(ω)e λ 1(x x r). Taking first and second order derivatives and xŵ(x, ω) = λ 1L {u(x r, )}(ω)e λ 1(x x r) 2 xŵ(x, ω) = λ 2 1L {u(x r, )}(ω)e λ 1(x x r). Continuity of the first and second order derivatives xl {u(x r, )}(ω) = λ 1L {u(x r, )}(ω), 2 xl {u(x r, )}(ω) = λ 2 1L {u(x r, )}(ω).

21 The Airy equation 5 Inverse Laplace transform for right exterior problem u(x r, t) I 2/3 t u xx(x r, t) = 0, u x(x r, t) + I 1/3 t u xx(x r, t) = 0, where I α t with α is the fractional integral operator defined by I α t f(t) = 1 Γ(α) t 6 Left exterior problem : c 1 = 0 and thus, we have 0 (t τ) α 1 f(τ) dτ, t > 0. û(x l, ω) + 1 λ 1 û x(x l, ω) + 1 λ 2 1 û xx(x l, ω) = 0. 7 Inverse Laplace transform for left exterior problem u(x l, t) I 1/3 t u x(x l, t) + I 2/3 t u xx(x l, t) = 0.

22 The Airy equation The new IBVP is The Airy Eq. in Ω u t + u xxx = 0, (x, t) [x l, x r] (0, T ) u I 1/3 t u x + I 2/3 t u xx = 0, x = x l, t > 0, u I 2/3 t u xx = 0, x = x r, t > 0, u + I 1/3 t u xx = 0, x = x r, t > 0, u(x, 0) = u 0(x), x [x l, x r]

23 General Airy Eq. tu + U 1 xu + U 2 xu 3 = 0, on R [0; T ]. u(x, 0) = u 0(x), x R. We have to find the roots of the (depressed) cubic equation ω + U 1λ + U 2λ 3 = 0. We still have the separation property of the roots

24 General Airy Eq. The new IBVP is The General Airy Eq. in Ω u t + U 1u x + U 2u xxx = 0, (x, t) [x l, x r] (0, T ) ( ) ( u(t, x l ) U 2 L 1 u x(t, x l ) U 2 L 1 λ 1 (s) 2 s ( ) u(t, x r) L 1 1 λ 1 u (s) xx(t, x r) = 0, t > 0, 2 ( u x(t, x r) L 1 1 u(x, 0) = u 0(x), λ 1 (s) λ 1 (s) s ) u xx(t, x r) = 0, x = x r, t > 0, x [x l, x r]. ) u xx(t, x l ) = 0, t > 0, with λ k (ω) = j k 1 ζ(ω) 1 U U 2 j k 1 ζ(ω) and ζ(ω) = 1 ω ω /3 U 2 U 2 27 ( U1 U 2 ) 3 1/3.

25 The 1D Schrödinger Eq. TBC on {x l, x r} : Remark: nv + e iπ/4 1/2 t v = 0, on {x l, x r} [0; T ]. 2 x + i t = ( n + i i t)( n i i t) The Schrödinger Eq. in Ω i tψ + ψ = 0, (x, t) [x l, x r] [0, T ] (S) nψ + e iπ/4 1/2 t ψ = 0, on {x l, x r} [0, T ], ψ(x, 0) = ψ 0(x), x [x l, x r]

26 Outline of the talk 1 Why Artifical Boundary Conditions? 2 Derivation of some TBC 3 Numerical approximation

27 Standard scheme Crank Nicolson u n+1 u n + ( U 1 x + U 2 3 ) u n+1 + u n x = 0, on [x l, x r] [0; N] t 2 u 0 (x) = u 0(x), x [x l, x r], T.B.C., x {x l, x r}. If U 1 = 0, approximation of I α t by C. Zheng, X. Wen, and H. Han, (2008) β α,j = { ( t) α α(α + 1)Γ(α) I α t f(t m) m β α,m jf j, j=0 1, j = 0, (j + 1) α+1 + (j 1) α+1 2j α+1, j > 0. Followed path: Derivation of continuous TBC, then discretization.

28 Full discretization Collaboration with Matthias Ehrhardt (Wuppertal) and Ingrid Lacroix-Violet (Lille) Other way: Full discretization, then derivation of discrete TBC adapted to the numerical scheme. Discretization parameters (t n) 0 n N uniform discretization to [0, T ] s.t. t n = n t with t = T/N 0 = t 0 < t 1 < < t N 1 < t N = T. (x j) 0 j J uniform discretization to [a, b] s.t. x j = a + j x with x = (b a)/j u (n) j approximation to u(t n, x j) a = x 0 < x 1 < < x J 1 < x J = b.

29 Numerical scheme (R-CN) (R-CN): Case U 1 = 0 u (n+1) j u (n) j t Proprerties of the scheme (R-CN) inconditionally stable E R CN = O( x + t 2 ) 4 nodes scheme ( + U2 u (n+1) 2( x) 3 j+2 3u (n+1) j+1 + 3u (n+1) j ) u (n+1) j 1 ( + U2 u (n) 2( x) 3 j+2 3u(n) j+1 + 3u(n) j u (n) j 1 ) = 0.

30 TBCs for (R-CN) Definition of the Z transform û(z) = Z{(u n ) n}(z) = u k z k, z > R 1, k=0 Application of the Z transform to (R-CN) (û j = û j(z) = Z{(u (n) j ) n)}) ) û j+2 3û j+1 + (3 + 2( x)3 z 1 û j û j 1 = 0, 1 j J 2. U 2 t z + 1 Solution on the external domain û j = 3 k=1 where l = l(z) is solution to l 3 3l 2 + c k (z) l j k (z), j 1 ou j J 2, (3 + 2( x)3 U 2 t ) z 1 l 1 = 0. z + 1

31 TBCs for (R-CN) separation property of the discrete roots l 1(z) < 1, l 2(z) > 1, l 3(z) > 1, for all z, decay property = û j(z) = c 2(z) l j 2(z) + c 3(z) l j 3(z), j 1 û j(z) = c 1(z) l j 1(z), j J 2

32 TBCs for (R-CN) - Left B.C Scheme u (n+1) j û j satisfies u (n) j t = for j = 1 û j(z) = c 2(z) l j 2(z) + c 3(z) l j 3(z), j 1 ( + U2 u (n+1) 2( x) 3 j+2 3u (n+1) j+1 + 3u (n+1) j ) u (n+1) j 1 ( + U2 u (n) 2( x) 3 j+2 3u(n) j+1 + 3u(n) j u (n) j 1 = one need only one left BC û j+1(z) ( l 2(z) + l 3(z) ) û j(z) + l 2(z)l 3(z) û j 1(z) = 0. ) = 0. Z 1 {l 2(z)l 3(z)} d u (n) 0 Z 1 {l 2(z) + l 3(z))} d u (n) 1 + u (n) 2 = 0, n = 0, 1, 2..., where P d u (n) i = n k=0 P (k) u (n k) i.

33 TBCs for (R-CN) - right B.C scheme u (n+1) j u (n) j t û j verifies two relations = for j = J 2 û j(z) = c 1(z) l j 1(z), j J 2 ( + U2 u (n+1) 2( x) 3 j+2 3u (n+1) j+1 + 3u (n+1) j ) u (n+1) j 1 ( + U2 u (n) 2( x) 3 j+2 3u(n) j+1 + 3u(n) j u (n) j 1 = two BCs are needed û j+2(z) = l 1(z) 2 û j(z), et û j+1(z) = l 1(z) û j(z), u (n) J Z 1 {l 2 1(z)} d u (n) J 2 = 0, where P d u (n) i = n k=0 P (k) u (n k) i. ) = 0. u(n) J 1 Z 1 {l 1(z)} d u (n) J 2 = 0, n = 0, 1, 2...,

34 TBCs for (R-CN) - conclusion Numerical scheme for 1 j J 2 and n = 0, 1, 2... u (n+1) j u (n) j t ( + U2 u (n+1) 2( x) 3 j+2 3u (n+1) j+1 + 3u (n+1) j ) u (n+1) j 1 ( + U2 u (n) 2( x) 3 j+2 3u(n) j+1 + 3u(n) j u (n) j 1 ) = 0. Left BC Z 1 {l 2(z)l 3(z)} d u (n) 0 Z 1 {l 2(z) + l 3(z))} d u (n) 1 + u (n) 2 = 0, n = 0, 1, 2..., Right BC u (n) J Z 1 {l 2 1(z)} d u (n) J 2 = 0, u(n) J 1 Z 1 {l 1(z)} d u (n) J 2 = 0, n = 0, 1, 2...,

35 Scheme (C-CN) Numerical scheme (R-CN): Case U 1 0 u (n+1) j u (n) j t ( ) ( ) + U1 u (n+1) j+1 u (n+1) j 1 + U1 u (n) j+1 4 x 4 x u(n) j 1 ) ( + U2 u (n+1) 4( x) 3 j+2 2u (n+1) j+1 + 2u (n+1) j 1 + U2 4( x) 3 ( u (n) j+2 2u(n) j+1 + 2u(n) j 1 u(n) j 2 u (n+1) j 2 ) = 0. Properties inconditionally stable E C CN = O( x 2 + t 2 ) five nodes We still need two BCs on right boundary but also two on the left.

36 TBCs for (C-CN) Numerical scheme for 1 j J 2 and n = 0, 1, 2... u (n+1) j u (n) j t ( ) ( ) + U1 u (n+1) j+1 u (n+1) j 1 + U1 u (n) j+1 4 x 4 x u(n) j 1 ) ( + U2 u (n+1) 4( x) 3 j+2 2u (n+1) j+1 + 2u (n+1) j 1 + U2 4( x) 3 ( u (n) j+2 2u(n) j+1 + 2u(n) j 1 u(n) j 2 u (n+1) j 2 ) = 0. Left BC Right BC Z 1 {λ 1} d u (n) 0 Z 1 {λ 2} d u (n) 1 + u (n) 2 = 0 Z 1 {λ 2 1} d u (n) 0 Z 1 {λ 2 2} d u (n) 2 + 2Z 1 {λ 2} d u (n) 3 u (n) 4 = 0 Z 1 {ρ 1} d u (n) J 2 Z 1 {ρ 2} d u (n) J 1 + u(n) J = 0 Z 1 {ρ 2 1} d u (n) J 4 Z 1 {ρ 2 2} d u (n) J 2 + 2Z 1 {ρ 2} d u (n) J 1 u(n) J = 0 where λ 1 = l 3(z)l 4(z), λ 2 = l 3(z) + l 4(z), ρ 1 = l 1(z)l 2(z), ρ 2 = l 1(z) + l 2(z).

37 Numerical procedure for inverse Z. A. Zisowsky,, Discrete transparent boundary conditions for systems of evolution equations, PhD, Technische Universität Berlin, 2003 Z of (u n) n : U(z) = u k z k, z > R. Inverse Z: u n = 1 2iπ u n = ρn 2π k=0 2π 0 S ρ U(z)z n 1 dz, ρ > R. U(ρe iϕ )e inϕ dϕ ρn N 1 U N k=0 z = ρe iϕ Recall the rule of discrete Fourier transform of ( f n ) N 1 n=0 F k = F{f n}(k) = where ω N = e i 2π N. U k = U(ρω k N ) = U(z k ) N 1 n=0 ( ρe i 2π N k) e i 2π N kn N 1 f nω nk N, f n = F 1 {F k }(n) = u n u N n = ρ n F 1 {U k }(n), 0 n < N. k=0 F k ω nk N

38 Example 1 C. Zheng, X. Wen, H. Han,, Numerical Solution to a Linearized KdV Equation on Unbounded Domain, Numer. Meth. Part. Diff. Eqs., 2008 Equation u t + u xxx = 0, x R, u(0, x) = e x2, x R, u 0, x. Exact solution E(t, x) = 3 1 ( ) x Ai 3t 3, 3t u exact(t, x) = E(t, x) e x2, where denotes the convolution product. Discretization parameters T = 4, [a, b] = [ 6, 6], t = 4/2560, x = 12/5000, r = 1.001

39 Example 1

40 Example 1

41 Example 1 - (R-CN) rel.errl x rel.errtm 10 3 x 10 4 N=640 N=1280 N=2560 N=5120 N= N=640 N=1280 N=2560 N=5120 N= x x Error functions rel.errt m = ( ( max e (n)), rel.errl2 = t 0<n<N where e (n) is the relative error l 2 at time t = n t: e (n) = u (n) exact u num (n) / 2 u (n) exact 2 ) 1/2 N (e (n) ) 2 n=1

42 Example 1 - (C-CN) rel.errl rel.errtm x x 2 N= N=1280 N=2560 N=5120 N= x N= N=1280 N=2560 N=5120 N= x Error functions rel.errt m = ( ( max e (n)), rel.errl2 = t 0<n<N where e (n) is the relative error l 2 at time t = n t: e (n) = u (n) exact u num (n) / 2 u (n) exact 2 ) 1/2 N (e (n) ) 2 n=1

43 Example 1 - (R-CN) rel.errl rel.errtm t t t t Error functions rel.errt m = ( ( max e (n)), rel.errl2 = t 0<n<N where e (n) is the relative error l 2 at time t = n t: e (n) = u (n) exact u num (n) / 2 u (n) exact 2 ) 1/2 N (e (n) ) 2 n=1

44 Example 1 - (C-CN) rel.errl rel.errtm t t t t Error functions rel.errt m = ( ( max e (n)), rel.errl2 = t 0<n<N where e (n) is the relative error l 2 at time t = n t: e (n) = u (n) exact u num (n) / 2 u (n) exact 2 ) 1/2 N (e (n) ) 2 n=1

45 Example 2 W.L. Briggs, T. Sarie,, Finite difference solutions of dispersive partial differential equations, Math. Comput. Simul., 1983 Equation u t + u x + u xxx = 0, x R, u(0, x) = exp( 8(x 5) 2 ) sin(50π/4), x R, Exacte solution û exact(t, ξ) = û 0 exp( (iξ iξ 3 )t) Discretization parameters T = , [a, b] = [0, 10], t = T/2560, x = 10/5000

46 Example 2

47 Conclusion TBC for continuous and discrete linear Korteweg de Vries equation modified KdV : Zheng (06) u t ± 6u 2 u x + u xxx = 0 Use of inverse scattering to get exact TBC. Example (Zheng) : solitary waves generated by an initial Gaussian profile Numerical simulation u 0(x) = exp ( 1.5x 2 of a modified KdV equation 333 ). Future: Fig. extension 5 Evolution of toa Gaussian nonlinear wave packet problem: consider U1u x as a nonlinear term. of the wave field. We see that the interaction of solitons is very elastic. After collision, two solitons remains their shapes and velocities, but their phases are