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
Outline of the talk 1 Why Artifical Boundary Conditions? 2 Derivation of some TBC 3 Numerical approximation
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
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
Domain trunction Problem : Mesh an unbounded domain (here in 1D) R d [0; T ]
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
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.
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.
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.
Outline of the talk 1 Why Artifical Boundary Conditions? 2 Derivation of some TBC 3 Numerical approximation
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)) + 1 2 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.
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).
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).
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.
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
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
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
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.
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.
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).
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.
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]
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
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 1 1 2 U 2 j k 1 ζ(ω) and ζ(ω) = 1 ω ω + + 4 2 1/3 U 2 U 2 27 ( U1 U 2 ) 3 1/3.
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]
Outline of the talk 1 Why Artifical Boundary Conditions? 2 Derivation of some TBC 3 Numerical approximation
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.
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.
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.
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
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
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.
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...,
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...,
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.
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).
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
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
Example 1
Example 1
Example 1 - (R-CN) 10 1 10 1 10 2 10 2 rel.errl2 10 3 x rel.errtm 10 3 x 10 4 N=640 N=1280 N=2560 N=5120 N=10240 10 4 N=640 N=1280 N=2560 N=5120 N=10240 10 4 10 3 10 2 10 1 10 0 x 10 4 10 3 10 2 10 1 10 0 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
Example 1 - (C-CN) 10 1 10 1 10 2 10 2 10 3 10 3 rel.errl2 10 4 rel.errtm 10 4 10 5 x 2 10 5 x 2 N=640 10 6 N=1280 N=2560 N=5120 N=10240 10 7 10 4 10 3 10 2 10 1 10 0 x N=640 10 6 N=1280 N=2560 N=5120 N=10240 10 7 10 4 10 3 10 2 10 1 10 0 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
Example 1 - (R-CN) 10 1 10 1 10 0 10 0 10 1 10 1 rel.errl2 10 2 rel.errtm 10 2 10 3 t 2 10 3 t 2 10 4 10 4 10 4 10 3 10 2 10 1 t 10 4 10 3 10 2 10 1 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
Example 1 - (C-CN) 10 1 10 1 10 0 10 0 10 1 10 1 rel.errl2 10 2 rel.errtm 10 2 10 3 t 2 10 3 t 2 10 4 10 4 10 4 10 3 10 2 10 1 t 10 4 10 3 10 2 10 1 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
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 = 2 10 3, [a, b] = [0, 10], t = T/2560, x = 10/5000
Example 2
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