Use of Artificial Boundary Conditions for Schr odinger equation by - PowerPoint PPT Presentation

Use of Artificial Boundary Conditions for Schr¨ odinger equation by Christophe Besse Collaborators: X. Antoine, S. Descombes, P. Klein, J. Szeftel, F. Xing Institut de Math´ ematiques de Toulouse, Universit´ e Toulouse 3, CNRS Colloque Couplages Num´ eriques INSTITUT de MATHEMATIQUES de TOULOUSE

Outline of the talk 1 Introduction 2 Derivation of TBC for the linear Wave Eq. 3 Derivation of ABC for the 1D Schr¨ odinger Eq. 4 Numerical approximation of ABCs 5 Schwarz Waveform Relaxation method 2016/09 – Nice

Introduction Typical equations odinger Eq. in R d The Schr¨  ( x, t ) ∈ R d × [0; T ] i∂ t ψ + ∆ ψ + V ( x, t, ψ ) ψ = 0 ,     ( S ) | x |→ + ∞ ψ ( x, t ) = 0 , lim t ∈ [0; T ]    x ∈ R d  ψ ( x, 0) = ψ 0 ( x ) , ψ ( x, t ) : wave function, complex V can be → real potential, V = V ( x, t ) ∈ C ∞ ( R d × R + , R ) Ex: V = q | ψ | 2 → nonlinear functional, V = f ( ψ ) → a combination: V = V ( x ) + f ( ψ ) ψ 0 compact support in Ω 2016/09 – Nice

Introduction Typical equations The Wave Eq. in R  ∂ 2 t u − ∂ 2 x u = 0 , ( x, t ) ∈ R × [0; T ]     ( W ) | x |→ + ∞ u ( x, t ) = 0 , lim t ∈ [0; T ]     u ( x, 0) = u 0 ( x ) , ∂ t u ( x, 0) = u 1 ( x ) x ∈ R u ( x, t ) : real function, u 0 , u 1 compact support in Ω 2016/09 – Nice

Domain trunction Problem : Mesh an unbounded domain (here in 1D) R d × [0; T ] 2016/09 – Nice

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 boundary Σ T 2016/09 – Nice

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 boundary Σ T Expression of the boundary condtion with the help of the Dirichlet-to-Neumann map: ∂ n ψ + i Λ + ψ = 0 , on Σ T . 2016/09 – Nice

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 boundary Σ T Expression of the boundary condtion with the help of the Dirichlet-to-Neumann map: ∂ n ψ + i Λ + ψ = 0 , on Σ T . 2016/09 – Nice

What happen if we do not take BC into account 1 Potential V ( x ) = x 0.9 0.8 Initial datum: 0.7 Evolution of | ψ | 0.6 gaussian function | u(x,t) | w.r.t time 0.5 ψ 0 ( x ) = e − x 2 +10 ix 0.4 ∆ t = 0 . 2 0.3 0.2 Domain: Ω = [ − 5; 15] 0.1 0 −5 0 5 10 15 x Exact Solution 2 0.9 1.8 0.8 1.6 0.7 1.4 0.6 1.2 0.5 t 1 t 0.4 0.8 0.3 0.6 0.2 0.4 0.1 0.2 0 −5 0 5 10 15 x x 2016/09 – Nice

What happen if we do not take BC into account 1 Potential V ( x ) = x 0.9 0.8 Initial datum: 0.7 Evolution of | ψ | 0.6 gaussian function | u(x,t) | w.r.t time 0.5 ψ 0 ( x ) = e − x 2 +10 ix 0.4 ∆ t = 0 . 2 0.3 0.2 Domain: Ω = [ − 5; 15] 0.1 0 −5 0 5 10 15 x Approximated numerical solution 2 0.9 1.8 0.8 1.6 0.7 Homogeneous Dirichlet 1.4 Boundary Conditions: 0.6 1.2 0.5 ψ | Σ = 0 t 1 t 0.4 0.8 0.3 0.6 ⇒ Parasistic reflexion 0.2 0.4 0.1 0.2 0 −5 0 5 10 15 x x 2016/09 – Nice

Domain decomposition 2016/09 – Nice

Domain decomposition Global Domain Decomposition in (here 3) subdomains. Iterative procedure: Schwarz waveform relaxation method 2016/09 – Nice

Domain decomposition B j = ∂ n j + S Informations exchanges from domain 2 → domain 1  L u k +1 = 0 , ( t, x ) ∈ (0 , T ) × ( a 1 , b 1 ) ,  1   u k +1 (0 , x ) = ψ 0 ( x ) , x ∈ ( a 1 , b 1 ) , 1 ∂ n 1 u k +1 = 0 , x = a 1 ,  1   B 1 u k +1 = B 1 u k 2 , x = b 1 . 1 2016/09 – Nice

Domain decomposition B j = ∂ n j + S Informations exchanges from domain 1 → domain 2 and domain 3 → domain 2  L u k +1 = 0 , ( t, x ) ∈ (0 , T ) × ( a 2 , b 2 ) ,  2   u k +1 (0 , x ) = ψ 0 ( x ) , x ∈ ( a 2 , b 2 ) , 2 B 2 u k +1 = B 2 u k 1 , x = a 2 ,  2   B 2 u k +1 = B 2 u k 3 , x = b 2 . 2 2016/09 – Nice

Domain decomposition B j = ∂ n j + S Informations exchanges from domain 2 → domain 3  L u k +1 = 0 , ( t, x ) ∈ (0 , T ) × ( a 3 , b 3 ) ,  3   u k +1 (0 , x ) = ψ 0 ( x ) , x ∈ ( a 3 , b 3 ) , 3 B 3 u k +1 = B 3 u k 2 , x = a 3 ,  3   ∂ n 3 u k +1 = 0 , x = a 3 . 3 2016/09 – Nice

Domain decomposition Examples of operators S for Schr¨ odinger equation S = − ip , p real parameter (Halpern, Szeftel ’09-10) � S = − i∂ t − V ( a, b j ) , (Halpern, Szeftel ’09-10) S : Dirichlet-to-Neumann map i Λ + (X. Antoine, E. Lorin, A.D. Bandrauk & F. Xing, CB) 2016/09 – Nice

Outline of the talk 1 Introduction 2 Derivation of TBC for the linear Wave Eq. 3 Derivation of ABC for the 1D Schr¨ odinger Eq. 4 Numerical approximation of ABCs 5 Schwarz Waveform Relaxation method 2016/09 – Nice

TBC for Wave Eq. Homogeneous Wave Equation in 1D  ∂ 2 t ψ − ∂ 2 x ψ = 0 , ( x, t ) ∈ R x × [0; T ] ,  ( P ) ψ ( x, 0) = ψ 0 ( x ) , x ∈ R x ,  ∂ t ψ ( x, 0) = ψ 1 ( x ) , x ∈ R x . Hypothesis : supp ( ψ 0 , 1 ) ⊂ B (0 , R ) . Solution � x + t ψ ( x, t ) = 1 2( ψ 0 ( x + t ) + ψ 0 ( x − t )) + 1 ψ 1 ( y ) dy. 2 x − t Simply, ψ ( x, t ) = ϕ 1 ( x + t ) + ϕ 2 ( x − t ) , with � x 2 ϕ 1 ( x ) = ψ 0 ( x ) + ψ 1 ( y ) dy , � x −∞ 2 ϕ 2 ( x ) = ψ 0 ( x ) − ψ 1 ( y ) dy . −∞ 2016/09 – Nice

TBC for Wave Eq. t x − L − R R L supp( ϕ ( x + t )) supp( ψ ( 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 ) ∂ x ψ = ∂ t ψ ∂ x ψ = − ∂ t ψ Then, on x = ± L , the Neumann datum is expressed in function of the Dirichlet one ∂ n ψ + ∂ t ψ = 0 where n denotes the outwardly unit normal vector to Ω = ( − L, L ) . 2016/09 – Nice

TBC for Wave Eq. The problem ( P ) is thus transformed in ( P app )  t ψ a − ∂ 2 x ψ a = 0 , ∂ 2 ( x, t ) ∈ Ω × [0; T ] ,    ψ a ( x, 0) = ψ 0 ( x ) , x ∈ Ω , ( P app ) ∂ t ψ a ( x, 0) = ψ 1 ( x ) , x ∈ Ω ,   ∂ n ψ a + ∂ t ψ a = 0 ,  ( x, t ) ∈ Γ × [0; T ] , where Γ = {− L, L } . ψ | Ω ( x, t ) = ψ a ( x, t ) , ( x, t ) ∈ Ω × [0; T ] . Ths BCs do not perturb the solution: we call them Transparent Boundary Conditions (TBC) Remark: ∂ 2 t − ∂ 2 x = ( ∂ t − ∂ x )( ∂ t + ∂ x ) . 2016/09 – Nice

TBC for Wave Eq. Other way of derivation: Laplace transform � ∞ u ( x, t ) e − ωt dt L ( u )( x, ω ) = ˆ u ( x, ω ) = 0 with frequency ω = σ + iτ , σ > 0 L ( ∂ t u )( x, ω ) = ω ˆ u ( x, ω ) − u ( x, 0) , L ( ∂ 2 ω 2 ˆ t u )( x, ω ) = u ( x, ω ) − ωu ( x, 0) − ∂ t u ( x, 0) . 1 Transmission problem Interior Problem Exterior problem ( ∂ 2 t − ∂ 2  ( ∂ t − ∂ 2  x ) v = 0 , x ∈ Ω , t > 0 , x ) w = 0 , x ∈ Ω , t > 0 ,   ∂ x v = ∂ x w, x ∈ Γ , t > 0 ,   w ( x, t ) = v ( x, t ) , x = ± L, t > 0 ,  v ( x, 0) = ψ 0 ( x ) , x ∈ Ω , w ( x, 0) = 0 , x ∈ Ω ,    ∂ t v ( x, 0) = ψ 1 ( x ) , x ∈ Ω ,  ∂ t w ( x, 0) = 0 , x ∈ Ω .  2016/09 – Nice

TBC for Wave Eq. 2 Laplace tranform w.r.t time for x > L : ω 2 ˆ w − ∂ 2 x ˆ w = 0 . w ( x, ω ) = A + ( ω ) e ω x + A − ( ω ) e − ω x Solution: ˆ → A + = 0 . 3 How to select the outgoing wave: solution of finite energy − w ( x, ω ) = e − ω ( x − L ) L ( w ( L, · ))( ω ) . ˆ Taking derivative ∂ x ˆ w ( x, ω ) | x = L = − ω ˆ w ( x, ω ) | x = L 4 Inverse Laplace tranform ∂ x w ( x, t ) | x = L = − ∂ t w ( x, t ) | x = L . 5 Thus, we obtain the TBC for v ∂ n v + ∂ t v = 0 2016/09 – Nice

Outline of the talk 1 Introduction 2 Derivation of TBC for the linear Wave Eq. 3 Derivation of ABC for the 1D Schr¨ odinger Eq. 4 Numerical approximation of ABCs 5 Schwarz Waveform Relaxation method 2016/09 – Nice

The 1D Eq. without potential 1 Splitting between interior and exterior problems Interior Problem Exterior problem ( i∂ t + ∂ 2 ( i∂ t + ∂ 2  x ) v = 0 , x ∈ Ω , t > 0 ,  x ) w = 0 , x ∈ Ω , t > 0 ,     ∂ x v = ∂ x w, x ∈ Σ , t > 0 ,  w ( x, t ) = v ( x, t ) , x = x l,r , t > 0 ,    v ( x, 0) = ψ 0 ( x ) , x ∈ Ω . | x |→ + ∞ w ( x, t ) = 0 , lim t > 0 ,       w ( x, 0) = 0 , x ∈ Ω . left exterior right exterior problem interior problem problem (x,t) v output: input: ��� ��� Neumann data Dirichlet data w x (x ,t) v (x ,t) L L ��� ��� x x R L 2 Laplace transform w.r.t time on Ω r i∂ t w + ∂ 2 w + ∂ 2 x w = 0 − → iω ˆ x ˆ w = 0 √− iω x + A − ( ω ) e − + √− iω x + w ( x, ω ) = A + ( ω ) e Solution: ˆ √· ) > 0 . with the determination of the square root s.t. Re ( + 2016/09 – Nice