XIV -th International Conference ”Geometry, Integrability and Quantization” June 8–13, 2012, ”St. Constantin and Elena”, Varna On Multicomponent Derivative Nonlinear Schr¨ odinger Equation Related to Symmetric Spaces Tihomir Valchev Institute for Nuclear Research and Nuclear Energy, Bulgarian Academy of Sciences, Sofia, Bulgaria 0-0
1. Introduction Derivative nonlinear Schr¨ odinger equation (DNLS) has the form: i q t + q xx + i( | q | 2 q ) x = 0 , where q ( x, t ) is a smooth complex-valued function. DNLS describes the propagation of circular polarized nonlinear Alfv´ en waves in plasma. DNLS is S -integrable [Kaup-Newell, 1977], i.e. it possesses a quadratic bundle Lax pair: i ∂ x + λQ ( x, t ) − λ 2 σ 3 , L ( λ ) := 3 A k ( x, t ) λ k − 2 λ 4 σ 3 , � A ( λ ) := i ∂ t + k =1 where λ ∈ C is a spectral parameter and � � � � 0 q ( x, t ) 1 0 Q ( x, t ) = , σ 3 = . q ∗ ( x, t ) 0 0 − 1 Purpose of the talk: Study of certain examples of multicomponent generazitations of DNLS related to Hermitian symmetric spaces. 0-1
+ 2. Preliminaries • Multicomponent DNLS equation related to A . III symmetric space Our main object of study is: 2i q T q ∗ � �� � i q t + q xx + x = 0 , q n + 1 where q : R 2 → C n is an infinitely smooth function. It is also assumed that q obeys zero boundary conditions, i.e. x →±∞ q ( x, t ) = 0 . lim 0-2
• Lax representation and connection with Hermitian symmetric spaces i ∂ x + λQ ( x, t ) − λ 2 J, L ( λ ) := 4 � λ k A k ( x, t ) . A ( λ ) := i ∂ t + k =1 All coefficients above are Hermitian traceless ( n + 1) × ( n + 1) matrices. Moreover, the following Z 2 reduction is imposed on the Lax pair: C L ( − λ ) C = L ( λ ) , C A ( − λ ) C = A ( λ ) , where C = diag (1 , − 1 . . . , − 1). Due to the form of C the potential Q has the block structure: q T ( x, t ) � � 0 Q ( x, t ) = . q ∗ ( x, t ) 0 while J is block diagonal. More particularly, we pick it up in the form J = diag ( n, − 1 , . . . , − 1). 0-3
The matrix C represents action of Cartan’s involutive automor- phism to define SU ( n +1) /S ( U (1) × U ( n )) symmetric space of the type A . III . It induces a Z 2 grading in sl ( n + 1) as follows sl ( n +1) = sl 0 ( n +1)+ sl 1 ( n +1) , [ sl σ ( n +1) , sl σ ′ ( n +1)] = sl σ + σ ′ ( n +1) , where sl σ ( n + 1) := { X ∈ sl ( n + 1) | C X C − 1 = ( − 1) σ X } . It is easy to see that Q as well as A 1 and A 3 belong to sl 1 ( n + 1) while J , A 2 and A 4 belong to sl 0 ( n + 1). The subspace sl 0 ( n + 1) coincides with the centralizer of J . 0-4
• Direct scattering problem In order to formulate the direct scattering theory one introduces auxilary linear problem: L ( λ ) ψ ( x, t, λ ) = i ∂ x ψ ( x, t, λ ) + λ ( Q ( x, t ) − λJ ) ψ ( x, t, λ ) = 0 . It is evident that det ψ = 1. Since [ L, A ] = 0 any fundamental solution satisfies as well 4 � λ k A k ψ = ψf ( λ ) , A ( λ ) ψ = i ∂ t ψ + k =1 where 4 � λ k A k ( x, t ) = − ( n + 1) λ 4 J. f ( λ ) = lim x →±∞ k =1 is called dispersion law. It is a fundamental property of any soliton equation. 0-5
A special case of solutions are Jost solutions defined as follows: x →±∞ ψ ± ( x, t, λ )e i λ 2 Jx = 1 lim 1 . The Jost solutions are defined only on the real and imaginery axes in the λ -plane (continuous spectrum of L ( λ )). The transition ma- trix ψ − ( x, t, λ ) = ψ + ( x, t, λ ) T ( t, λ ) is called scattering matrix. Its time evolution is given by: T ( t, λ ) = e i f ( λ ) t T (0 , λ )e − i f ( λ ) t . i ∂ t T + [ f ( λ ) , T ] = 0 ⇒ • Fundamental analytic solutions There exist two fundamental solu- tions χ + ( x, λ ) and χ − ( x, λ ) to be analytic in the upper and lower half-plane of the λ 2 -plane respectively. They can be constructed from Jost solutions through the formulae: χ ± ( x, λ ) = ψ − ( x, λ ) S ± ( λ ) = ψ + ( x, λ ) T ∓ ( λ ) D ± ( λ ) . 0-6
The matrices S ± ( λ ), T ± ( λ ) and D ± ( λ ) are involved in the gener- alized Gauss decomposition T ( λ ) = T ∓ ( λ ) D ± ( λ )( S ± ( λ )) − 1 . As a simple consequence of their construction we see that χ + ( x, λ ) = χ − ( x, λ ) G ( λ ) for some sewing function G ( λ ) = ( S − ( λ )) − 1 S + ( λ ). • Reduction conditions on the Jost solutions, the scattering matrix and fundamental analytic solutions � − 1 � − 1 = T ( λ ) , � ψ † ± ( x, λ ∗ ) T † ( λ ∗ ) � = ψ ± ( x, λ ) , C ψ ± ( x, − λ ) C = ψ ± ( x, λ ) , C T ( − λ ) C = T ( λ ) , ( χ + ) † ( x, λ ∗ ) [ χ − ( x, λ )] − 1 , C χ + ( x, − λ ) C = χ − ( x, λ ) . = 0-7
3. Dressing method and special solutions • Dressing method Concept of the dressing method: Q 0 → L 0 → ψ 0 → ψ 1 → Q 1 . Realization: let ψ 0 be a fundamental solution of L 0 ψ 0 = i ∂ x ψ 0 + λ ( Q 0 − λJ ) ψ 0 = 0 where � � 0 q 0 ( x ) Q 0 ( x ) = , J = diag ( n, − 1 , . . . , − 1) . q ∗ 0 ( x ) 0 for some vector q T 0 = ( q 1 0 , . . . , q n 0 ) assumed to be known. Now construct another function ψ 1 ( x, λ ) := g ( x, λ ) ψ 0 ( x, λ ) and assume it satisfies the linear system L 1 ψ 1 = i ∂ x ψ 1 + λ ( Q 1 − λJ ) ψ 1 = 0 0-8
for some potential � � 0 q 1 ( x ) Q 1 ( x ) := . q ∗ 1 ( x ) 0 to be found. Therefore the dressing factor g satisfies: i ∂ x g + λQ 1 g − λgQ 0 − λ 2 [ J, g ] = 0 . The Z 2 reductions imposed on the Lax pair implies that g is obliged to fulfill similar set of symmetry conditions: � − 1 g † ( x, λ ∗ ) � = g ( x, λ ) , C g ( x, − λ ) C = g ( x, λ ) . We pick up the dressing factor in the form: µ ( λ − µ ) + λ C B ( x ) C λB ( x ) g ( x, λ ) = 1 1 + µ ( λ + µ ) , Re µ k � = 0 , Im µ k � = 0 . 0-9
The inverse of the dressing factor reads µ ∗ ( λ − µ ∗ ) + λ C B † ( x ) C λB † ( x ) [ g ( x, λ )] − 1 = 1 1 + µ ∗ ( λ + µ ∗ ) . There exists the following connection between Q 1 and Q 0 λQ 1 = − i ∂ x gg − 1 + λgQ 0 g − 1 + λ 2 [ J, g ] g − 1 . After dividing by λ and taking | λ | → ∞ we obtain Q 1 = AQ 0 A † + [ J, B − C B C ] A † , where 1 + 1 A = 1 µ ( B + C B C ) . 0-10
From the obvious identity gg − 1 = 1 1 it follows that the residue B satisfies: µB † µ C B † C � � B 1 + µ ∗ ( µ − µ ∗ ) + = 0 . 1 µ ∗ ( µ + µ ∗ ) B ( x, t ) is a degenerate matrix. Therefore we have B = XF T for some ( n + 1) × k rectangular matrices X ( x ) and F ( x ). Then the algebraic relation obtains the form � F T F ∗ µ − µ ∗ − F T C F ∗ F ∗ = µ ∗ � X. µ + µ ∗ C µ It can be solved easily to give � F T F ∗ � − 1 µ − µ ∗ − F T C F ∗ X = µ F ∗ . µ + µ ∗ C µ ∗ Thus we have expressed X through F . In order to find the latter we consider the differential equation for g . After calculating the residue at λ = µ we obtain i ∂ x F T − µF T ( Q 0 − µJ ) = 0 F T ( x ) = F T 0 [ ψ 0 ( x, µ )] − 1 . ⇒ 0-11
What remains is to recover the time evolution. For this to be done one must analyse some properties of the second Lax operator A ( λ ). Any fundamental solution of the bare linear problem also satisfies: λ k A (0) � i ∂ t ψ 0 + k ψ 0 = ψ 0 f ( λ ) k while the dressed fundamental solution solves λ k A (1) � i ∂ t ψ 1 + k ψ 1 = ψ 1 f ( λ ) . k As a result the dressing factor satisfies: 2 N 2 N λ k A (1) λ k A (0) � � i ∂ t g + k g − g = 0 . k k =1 k =1 Detailed analysis shows that 2 N i ∂ t F T − F T � µ k A k = 0 . k =1 0-12
Therefore we have i ∂ t F T 0 − F T 0 f ( µ ) = 0 . Thus we are able to propose a simple rule to derive the time de- pendence of potential, namely: F T 0 → F T 0 e − i f ( µ ) t . For the DNLS equation f ( λ ) = − ( n + 1) λ 4 J . 0-13
• Soliton solutions In the soliton sector Q 0 ≡ 0. Therefore we have: ψ 0 ( x, t, λ ) = e − i λ 2 Jx . We shall resrict ourselves with the case when the rank of B is 1. Then the column-vector F is given by e n i µ 2 x F 0 , 1 e − i µ 2 x F 0 , 2 F = . . . . e − i µ 2 x F 0 ,n +1 It proves to be convenient to adopt polar parametrization of the pole, i.e. µ = ρ exp(i ϕ ). Then the potential acquires the form: 0-14
n +1 ρ sin(2 ϕ )e − i σ l ( x ) e θ l ( x ) q j − 1 � ( x ) = ( Q 1 ) 1 j ( x ) = 2i( n + 1) p =2 e 2 θ p ( x ) × 1 e − 2i ϕ + � n +1 l =2 � � δ jl − 2i sin(2 ϕ )e θ j ( x )+ θ l ( x ) e i( δ j − δ l − 2 ϕ ) , e − 2i ϕ + � n +1 p =2 e 2 θ p ( x ) where ( n + 1) ρ 2 sin(2 ϕ ) x − ξ 0 ,p , θ p ( x ) = σ p ( x ) = ( n + 1) cos(2 ϕ ) x + δ 1 − δ p − ϕ. ξ 0 ,p = ln | F 0 , 1 /F 0 ,p | , δ 1 = arg F 0 , 1 , δ p = arg F 0 ,p . In order to recover the time dependence one uses the following rule: ξ 0 ,p − 2( n + 1) ρ 4 sin(4 ϕ ) t, ξ 0 ,p → δ 1 + 2 nρ 4 cos(4 ϕ ) t, δ 1 → δ p − 2 ρ 4 cos(4 ϕ ) t. δ p → 0-15
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