Numerical Inverse Scattering Transform for Solving the Nonlinear - PowerPoint PPT Presentation

Numerical Inverse Scattering Transform for Solving the Nonlinear Schr¨ odinger Equation A. Aric` o, G. Rodriguez, S. Seatzu Department of Mathematics and Computer Science University of Cagliari · Italy Conference on Applied Inverse Problems July 20-24, 2009 · Vienna · Austria

Outline Nonlinear Schr¨ odinger equation and Inverse Scattering Transform Nonlinear Schr¨ odinger equation (NLS) Numerical method Issues related to the application of the IST to the NLS Numerical solution to the inverse scattering problem Time evolution of the scattering data Solution to the Marchenko systems Numerical experiments One soliton Truncated one soliton Numerical solution to the cubic NLS - AIP09 · Vienna slide 2/10

� � � � � � � � � Nonlinear Schr¨ odinger equation (NLS) We are interested in the IVP for the cubic Schr¨ odinger equation (NLS)  i q t = q xx + 2 q | q | 2 , x ∈ R , t > 0 ,  q ( x ; 0) given x ∈ R ,  where q = q ( x ; t ); we follow the path of the Inverse Scattering Transform (IST): direct initial potential scattering initial scattering data q ( x ; 0) S ( α ; 0) time evolution NLS: of scattering data: i q t = q xx + 2 q | q | 2 i S t = ± 4 S αα evolved potential evolved scattering data q ( x ; t ) S ( α ; t ) inverse scattering Numerical solution to the cubic NLS - AIP09 · Vienna slide 3/10

� � � � � � � � � Nonlinear Schr¨ odinger equation (NLS) We are interested in the IVP for the cubic Schr¨ odinger equation (NLS)  i q t = q xx + 2 q | q | 2 , x ∈ R , t > 0 ,  q ( x ; 0) given x ∈ R ,  where q = q ( x ; t ); we follow the path of the Inverse Scattering Transform (IST): direct initial potential scattering initial scattering data q ( x ; 0) S ( α ; 0) “The IST solves exactly the IVP for the NLS; time evolution NLS: of scattering data: i q t = q xx + 2 q | q | 2 i S t = ± 4 S αα all the physical properties are then preserved.” evolved potential evolved scattering data q ( x ; t ) S ( α ; t ) inverse scattering Numerical solution to the cubic NLS - AIP09 · Vienna slide 3/10

Numerical method Here we validate the last two step of the procedure. A further work will be the implementation of the direct scattering, thanks to a result due to van der Mee. Proposed algorithm 1. assume that we know S ( λ ; 0); choose { t 1 , . . . , t k } 2. for each t i : ◮ choose a discretization step h and sample S ( λ ; 0) with step h : S h (0) ◮ apply a numerical method to S h (0) and compute S h ( t i ) ◮ apply a numerical method to S h ( t i ) and compute [twice] q (0; t i ) ◮ check if q (0; t i ) is accurate: YES compute q ( x k ; t i ) for all x k NO halve h and restart the computation for the same t i Numerical solution to the cubic NLS - AIP09 · Vienna slide 4/10

Issues related to the application of the IST to the NLS ◮ eigenvalue system + FT + nonlinear least square problem; ◮ linear evolution problem; ◮ two systems of two Marchenko integral equations Numerical solution to the cubic NLS - AIP09 · Vienna slide 5/10

Issues related to the application of the IST to the NLS ◮ eigenvalue system + FT + nonlinear least square problem; ◮ linear evolution problem; ◮ two systems of two Marchenko integral equations Scattering data ◮ “ reflection coefficients ” from the left and from the right: L ( λ ), R ( λ ); ◮ “ transmission coefficients ”: T ( λ ); ◮ “ bound states ” λ j ∈ C + , and their “ norming constants ” Γ lj , Γ rj . In the numerical experiments for the Inverse Scattering, we assume to know the initial scattering data and use them to compute the potential. Numerical solution to the cubic NLS - AIP09 · Vienna slide 5/10

� � � � � Issues related to the application of the IST to the NLS ◮ eigenvalue system + FT + nonlinear least square problem; ◮ linear evolution problem; ◮ two systems of two Marchenko integral equations Time evolution of the relevant scattering data L ( λ ) R ( λ ) λ j Γ lj Γ rj e − 4 i λ 2 t e +4 i λ 2 t e +4 i λ 2 t e − 4 i λ 2 t 1 L ( λ ; t ) R ( λ ; t ) λ j ( t ) Γ lj ( t ) Γ rj ( t ) The scattering data determines the Marchenko integral kernels, i.e. the kernels of the integral system related to the inverse scattering step. Numerical solution to the cubic NLS - AIP09 · Vienna slide 5/10

� � Issues related to the application of the IST to the NLS ◮ eigenvalue system + FT + nonlinear least square problem; ◮ linear evolution problem; ◮ two systems of two Marchenko integral equations Kernels of the Marchenko integral systems Marchenko kernel from the right Marchenko kernel from the left Ω l ( α ) = ˆ Ω r ( α ) = ˆ � Γ lj e i λ j α � Γ rj e i λ j α R ( α ) + L ( α ) + j j i (Ω l ) t = 4(Ω l ) αα i (Ω r ) t = − 4(Ω r ) αα Ω r ( α ; t ) = ˆ � Ω r ( α ; t ) = ˆ � Γ rj ( t )e i λ j α Γ rj ( t )e i λ j α R ( α ; t ) + L ( α ; t ) + j j Numerical solution to the cubic NLS - AIP09 · Vienna slide 5/10

Issues related to the application of the IST to the NLS ◮ eigenvalue system + FT + nonlinear least square problem; ◮ linear evolution problem; ◮ two systems of two Marchenko integral equations Marchenko integral systems: ( B ( R , R + ; t )) For a fixed value t � 0, and for any value of x ∈ R , solve: � + ∞  B l 1 ( x , γ ; t ) − Ω l (2 x + γ + β ; t ) B l 2 ( x , β ; t ) d β = 0    left 0 � + ∞ Ω l (2 x + γ + β ; t ) B l 1 ( x , β ; t ) d β + B l 2 ( x , γ ; t ) = − Ω l (2 x + γ ; t )    0 and/or � + ∞  B r 1 ( x , γ ; t ) + Ω r ( − 2 x + γ + β ; t ) B r 2 ( x , β ; t ) d β = 0 ,    right 0 � + ∞ − Ω r ( − 2 x + γ + β ; t ) B r 1 ( x , β ; t ) d β + B r 2 ( x , γ ; t ) = Ω r ( − 2 x + γ ; t ) ,    0 Numerical solution to the cubic NLS - AIP09 · Vienna slide 5/10

Issues related to the application of the IST to the NLS ◮ eigenvalue system + FT + nonlinear least square problem; ◮ linear evolution problem; ◮ two systems of two Marchenko integral equations The Marchenko system returns the solution: Potential  2 B l 2 ( x , 0 + ; t ) , x � 0 ,  q ( x ; t ) = − 2 B r 2 ( x , 0 + ; t ) , x � 0;  Density energy: � + ∞  B l 1 ( x , 0 + ; t ) = − 1 | q ( y ; t ) | 2 d y ,    2  x � x B r 1 ( x , 0 + ; t ) = − 1 | q ( y ; t ) | 2 d y ,    2  −∞ Numerical solution to the cubic NLS - AIP09 · Vienna slide 5/10

Time evolution of the scattering data ◮ The time evolution of each bound state term is straightforward: Γ lj e i λ j α → Γ lj e i λ j α → Γ lj e i λ 2 Γ lj e i λ 2 � j t i � e i λ j α , � j t i � e i λ j α . Numerical solution to the cubic NLS - AIP09 · Vienna slide 6/10

� � Time evolution of the scattering data ◮ The time evolution of each bound state term is straightforward: Γ lj e i λ j α → Γ lj e i λ j α → Γ lj e i λ 2 Γ lj e i λ 2 � j t i � e i λ j α , � j t i � e i λ j α . ◮ About the time evolution of the reflection coefficients L ( λ ), R ( λ ): L ( λ ) R ( λ ) e − 4 i λ 2 t i e 4 i λ 2 t i F F � ˆ � ˆ L ( λ ; t i ) R ( λ ; t i ) L ( α ; t i ) R ( α ; t i ) Numerical solution to the cubic NLS - AIP09 · Vienna slide 6/10

� � � � � � � � � � � � � � Time evolution of the scattering data ◮ The time evolution of each bound state term is straightforward: Γ lj e i λ j α → Γ lj e i λ j α → Γ lj e i λ 2 Γ lj e i λ 2 � j t i � e i λ j α , � j t i � e i λ j α . ◮ About the time evolution of the reflection coefficients L ( λ ), R ( λ ): F − 1 F − 1 ˆ ˆ L ( λ ) R ( λ ) L ( α ) R ( α ) e − 4 i λ 2 t i i ˆ L t = − 4ˆ e 4 i λ 2 t i i ˆ R t = 4ˆ L αα R αα F F � ˆ � ˆ L ( λ ; t i ) R ( λ ; t i ) L ( α ; t i ) R ( α ; t i ) Numerical solution to the cubic NLS - AIP09 · Vienna slide 6/10

� � � � � � � � � � � � � � Time evolution of the scattering data ◮ The time evolution of each bound state term is straightforward: Γ lj e i λ j α → Γ lj e i λ j α → Γ lj e i λ 2 Γ lj e i λ 2 � j t i � e i λ j α , � j t i � e i λ j α . ◮ About the time evolution of the reflection coefficients L ( λ ), R ( λ ): F − 1 F − 1 ˆ ˆ L ( λ ) R ( λ ) L ( α ) R ( α ) e − 4 i λ 2 t i i ˆ L t = − 4ˆ e 4 i λ 2 t i i ˆ R t = 4ˆ L αα R αα F F � ˆ � ˆ L ( λ ; t i ) R ( λ ; t i ) L ( α ; t i ) R ( α ; t i ) 1. [ α 0 , α N ] support of ˆ R ( α ), i.e. at time t = 0; 2. approximate ˆ R ( α ; t i ) on [ α 0 , α N ]: integrate (multiply by e 4 i λ 2 t i ); apply IFFT ; apply FFT ; 3. check if ˆ R ( α ; t i ) is negligible at the boundary; (4.) enlarge [ α 0 , α N ] and recompute ˆ R ( α ; t i ) with the same stepsize. Numerical solution to the cubic NLS - AIP09 · Vienna slide 6/10

Solution to the Marchenko systems ◮ one routine to solve both the Marchenko systems ◮ discretization of the “left” system by the Nystr¨ om method ◮ numerical method to solve the linear system Numerical solution to the cubic NLS - AIP09 · Vienna slide 7/10