On Certain Aspects of the Theory of Polynomial Bundle Lax Operators Related to Symmetric Spaces. Tihomir Valchev

Transcription

1 0-0 XIII th International Conference Geometry, Integrability and Quantization June 3 8, 2011, St. Constantin and Elena, Varna On Certain Aspects of the Theory of Polynomial Bundle Lax Operators Related to Symmetric Spaces Tihomir Valchev Institute for Nuclear Research and Nuclear Energy, Bulgarian Academy of Sciences, Sofia, Bulgaria Joint work with: A. Mikhailov, V. Gerdjikov and G. Grahovski

2 Introduction Heisenberg ferromagnet equation (spin of a 1D ferromagnetic) S t = S S xx, S 2 = 1 or in a matrix notation is t = 1 2 [S, S xx], S 2 = 11, S = 3 S k σ k. k=1 Its Lax representation reads L(λ) := i x λs(x, t), A(λ) := i t + A 0 (x, t) + λa 1 (x, t) + λ 2 A 2 (x, t). All matrices above belong to su(2). Purpose of the talk: studying properties of 2-component system iu t +u xx + (uu x + vv x)u x + (uu x + vv x) x u = 0, iv t +v xx + (uu x + vv x)v x + (uu x + vv x) x v = 0,

3 0-2 to generalize the Heisenberg model. The functions u and v are infinitely smooth to satisfy lim u(x, t) = 0, lim x ± v(x, t) = 1. x ± Moreover, u and v obey the constraint u 2 + v 2 = 1.

4 Preliminaries Lax representation NEE [L(λ), A(λ)] = 0, where the Lax pair (polynomial bundle) is given by: L(λ) := i x + λl 1 (x, t), A(λ) := i t + λa 1 (x, t) + λ 2 A 2 (x, t), where L 1 = A 1 = 0 u v u 0 0 v a b a 0 0 b 0 0,, A 2 = 1/ u 2 2/3 u v 0 v u v 2 2/3 a = iu x + i(uu x + vvx)u b = iv x + i(uu x + vvx)v.,

5 0-4 The specific structure of the matrices is a result of a Z 2 Z 2 reduction on generic Lax operators L (λ ) = L(λ), A (λ ) = Ă(λ), CL( λ)c = L(λ), CA( λ)c = A(λ), where C = diag (1, 1, 1) and the operation is defined as follows L(λ)ψ(x, t, λ) := i x ψ(x, t, λ) λψ(x, t, λ)l 1 (x, t, λ). The matrix C represents Cartan s involutive automorphism involved in the definition of SU(3)/S(U(1) U(2)), that is it induces a Z 2 -grading in the Lie algebra sl(3, C) sl(3) = sl 0 (3) sl 1 (3), sl σ (3) = {X g; CXC = ( 1) σ X}. Thus any function X(x, t, λ) with values in sl(3) is presented as X 0 (x, t, λ) + X 1 (x, t, λ), X 0,1 (x, t, λ) sl 0,1 (3).

6 0-5 Scattering problem Fundamental solutions Introduce the auxiliary linear system L(λ)ψ(x, t, λ) = i x ψ(x, t, λ) + λl 1 (x, t)ψ(x, t, λ) = 0. Since L(λ) and A(λ) commute ψ also satisfies the equation A(λ)ψ(x, t, λ) = [i t +λa 1 (x, t)+λ 2 A 2 (x, t)]ψ(x, t, λ) = ψ(x, t, λ)f(λ) for some matrix-valued function f(λ). We choose f(λ) = lim x ± g 1 as [λa 1 (x, t) + λ 2 A 2 (x, t)]g as = λ 2 I, where g as = , I = diag (1/3, 2/3, 1/3).

7 0-6 Jost solutions and scattering matrix lim ψ ±(x, t, λ)e iλjx gas 1 = 11, x ± where J = diag(1, 0, 1) is the diagonal form of the asymptotic lim x ± L 1 (x, t). The transition matrix T(t, λ) = [ψ + (x, t, λ)] 1 ψ (x, t, λ) is called scattering matrix. It can be shown that T evolves with time according to i t T+[f(λ), T] = 0 T(t, λ) = e if(λ)t T(0, λ)e if(λ)t.

8 0-7 Construction of fundamental analytic solutions χ ± (x, λ) = ψ (x, λ)s ± = ψ + (x, λ)t (λ)d ± (λ), where T(λ) = T (λ)d ± (λ)(s ± (λ)) 1. Riemann-Hilbert problem χ + (x, λ) = χ (x, λ)g(λ). Reduction conditions on the Jost solutions, the scattering matrix and fundamental analytic solutions [ ψ ±(x, λ )] 1 = ψ ± (x, λ), [ T (λ ) ] 1 = T(λ), Cψ ± (x, λ)c = ψ ± (x, λ), CT( λ)c = T(λ), (χ + ) (x, λ ) = [χ (x, λ)] 1, Cχ + (x, λ)c = χ (x, λ).

9 Soliton solutions 3.1. Dressing method Concept of the dressing method Let ψ 0 be a fundamental solution to L 0 (λ)ψ 0 (x, λ) = i x ψ 0 (x, λ) + λl 1,0 (x)ψ 0 (x, λ) = 0 with some known potential L 1,0. We construct another function ψ = gψ 0 and suppose it satisfies the same linear problem L(λ)ψ(x, λ) = i x ψ(x, λ) + λl 1 (x)ψ(x, λ) = 0. with a different potential to be found. This implies i x g + λ(l 1 g gl 1,0 ) = 0.

10 0-9 Ansatz for the dressing factor We pick up a dressing factor in the form g(x, λ) = A(x) + B(x) λ µ CB(x)C λ + µ, that is compatible with the reduction conditions The inverse of g looks like Cg(x, λ)c = g(x, λ), [ g (x, λ ) ] 1 = g(x, λ). CAC = A g 1 (x, λ) = A (x) + B (x) λ µ CB (x)c λ + µ. Algebraic constraints From the identity gg 1 = 11 it follows ) B (A + B µ µ CB C µ + µ = 0.

11 0-10 At this point we introduce the factorization B = XF T. Hence we have AF = F T F µ µ X + F T CF µ + µ CX. Solving the equation above: F, A X. But F, A =?. Analysis of the differential constraint on g At λ = 0 it leads to x A 1 µ x(b + CBC) = 0 A = 1 µ (B + CBC) + A 0. It suffices to pick up A 0 = 11. On the other hand comparing the residues leads us to the coclusion that i x F T µf T L 1,0 = 0 F T (x) = F0 T ψ0 1 (x, λ = µ). After substituting A one obtains F = (a + bj 1 ) X.

12 0-11 where a := µ F T F µ(µ µ ), b := µ F T CF µ(µ + µ ). By inverting the linear system above we get X expressed by F: X = X 1 X 2 = (a + b) 1 F1 (a b) 1 F2. X 3 (a b) 1 F3 In order to obtain a relation between L 1 and L 1,0 one takes the limit λ in i λ xg + L 1 g gl 1,0 = 0 to deduce that L 1 = AL 1,0 A. This relation allows us to generate another solution starting from a known one by following the sequence L 1,0 ψ 0 F X, A L 1.

13 0-12 Recovering the time dependence In order to recover the time dependence one needs to compare the linear problems A(λ)ψ = [i t + λa 1 + λ 2 A 2 ]ψ = ψf, A 0 (λ)ψ = [i t + λa 1,0 + λ 2 A 2,0 ]ψ 0 = ψ 0 f. Therefore the dressing factor satisfies i t g + [λa 1 + λ 2 A 2 ]g g[λa 1,0 + λ 2 A 2,0 ] = 0. A more analysis shows that F 0 depends on time exponentially ( ) F0 T F0 T e if(µ)t = e iµ2 t 3 F0,1, e 2iµ2 t 3 F 0,2, e iµ2 t 3 F0,3.

14 Elementary solitons Let us choose as a seed solution L 1,0 = Then for ψ 0 one obtains ψ 0 (x, t, λ) = cos λx 0 i sinλx i sinλx 0 cos λx. Inserting ψ 0 into the expressions for F leads to F 0,1 cos µx if 0,3 sinµx F(x, t) = F 0,2 F 0,3 cos µx if 0,1 sinµx.

15 0-14 Particular cases (elementary solitons): 1. Consider F 0,2 = 0. In this case the solution is stationary, namely u(x) = 0, v(x) = exp where ω := Re µ > 0, γ := Im µ > 0 and [ 4i arctan γ cos(2ωx + δ ] 0), ω cosh(2γx + ξ 0 ) sinhξ 0 = sinδ 0 = 2 F 0,1 F 0,3 cos(arg F 0,1 arg F 0,3 ), ( F 0,1 2 + F 0,3 2 ) 2 4 F 0,1 F 0,3 2 cos 2 (arg F 0,1 arg F 0,3 ) 2 F 0,1 F 0,3 sin(arg F 0,1 arg F 0,3 ). ( F 0,1 2 + F 0,3 2 ) 2 4 F 0,1 F 0,3 2 cos 2 (arg F 0,1 arg F 0,3 )

16 F 0,1 = F 0,3. Now F is given by F(x, t) = F 0,1e iµx F 0,2 F 0,1 e iµx and after recovering the time dependence the soliton solution reads u(x, t) = 4iωγ [ 2ωe γ(x 2ωt+ϑ 0) + (ω iγ)e γ(x 2ωt+ϑ 0) ] (ω iγ) [ 2ωe γ(x 2ωt+ϑ 0) + (ω + iγ)e γ(x 2ωt+ϑ 0) ] 2, v(x, t) = 1 where exp[iωx + i(γ 2 ω 2 )t + iφ 0 ], 8ωγ 2 (ω iγ) [ 2ωe γ(x 2ωt+ϑ 0) + (ω + iγ)e γ(x 2ωt+ϑ 0) ] 2, ϑ 0 = 1 γ ln F 0,1 F 0,2, φ 0 = arg F 0,2 arg F 0,1.

17 Generalized Fourier Transform Wronskian relations and squared solutions (χ ± ) 1 L 1 χ ± x= = dx ˆχ ± L 1,x χ ± Therefore one can write (χ ± ) 1 L 1 χ ±, E α = dx L 1,x, e ± α. The quantities e ± α(x, λ) = χ ± (x, λ)e α [χ ± (x, λ)] 1 introduced above are called squared solutions and is the Cartan-Killing form. X, Y := tr (XY ) Splitting of e ± α(x, λ) due to Z 2 -grading of the Lie algebra e α (x, λ) = H α (x, λ)+k α (x, λ), H α (x, λ) sl 0 (3), K α (x, λ) sl 1 (3).

18 0-17 In addition, H α (x, λ) and K α (x, λ) split into H α (x, λ) = H α (x, λ) + h α (x, λ)l 2 (x), H α, L 2 = 0, K α (x, λ) = K α (x, λ) + k α (x, λ)l 1 (x), K α, L 1 = 0, where L 2 := L 2 1 2/3 sl 0 (3). It can be proven that the squared solutions form a complete system in the space of smooth functions with values in sl(3)/ kerad L1. Hence they play a role quite similar to the exponential functions in the usual Fourier analysis by expanding all quantities involved in NEE one obtains a linearized version of the NEE. Recursion operators Consider the equation i x e α + λ[l 1, e α ] = 0. Due to the grading condition H α and K α are interrelated through i x H α + λ[l 1, K α ] = 0, i x K α + λ[l 1, H α ] = 0.

19 0-18 After we extract the terms proportional to L 1 we obtain L 2, x H α + x h α = 0 h α = h α, x L 2, x H α, L 1, x K α + x k α = 0 k α = k α, x L 1, x K α. On the other hand the orthogonal part reads: iπ x H α + ih α L 2,x = λ[l 1, K α ], iπ x K α + ik α L 1,x = λ[l 1, H α ]. After substituting h α and k α in the equations above we get Λ 1 K α = λh α k α,0 Λ 2 L 2, Λ 2 H α = λk α h α,0 Λ 1 L 1, where ( ) Λ 1 = iad 1 1 L 1 π x ( ) L 1,x 2 1 x x ( ), L 1, ( Λ 2 = iad 1 L 1 π x ( ) 3 ) 2 L 2,x x 1 x ( ), L 2.

20 0-19 Let us apply Λ 2 to the first relation and Λ 1 to the second one. The result reads: Λ ± 2 Λ± 1 K± α = λ 2 K ± α λh α,0 Λ ± 1 L 1 k α,0 Λ ± 2 Λ± 2 L 2, Λ ± 1 Λ± 2 H± α = λ 2 H ± α λk α,0 Λ ± 2 L 2 h α,0 Λ ± 1 Λ± 2 L 2. The constants h α,0 and k α,0 are determined by the asymptotic of the relevant squared solution for x (or for x ). More detailed analysis shows that the following equalities holds: Λ + 2 Λ+ 1 K± α(x, λ) = λ 2 K ± α(x, λ), Λ 2 Λ 1 K± ±α(x, λ) = λ 2 K ± ±α(x, λ) Λ + 1 Λ+ 2 H± α(x, λ) = λ 2 H ± α(x, λ), Λ 1 Λ 2 H± ±α(x, λ) = λ 2 H ± ±α(x, λ) The operator Λ ± introduced as Λ ± X := Λ ± 1 Λ± 2 X, X sl0 (3), Λ ± Y := Λ ± 2 Λ± 1 Y, Y sl1 (3). is the called recursion operator.

21 Integrable hierarchy Description of the integrable hierarchy in terms of the recursion operator Any member of the integrable hierarchy under consideration has a Lax pair in the form L(λ) = i x + λl 1 (x, t), A(λ) = N i t + λ k A k (x, t). k=1 As before the operators L and A are subject to the reductions CA 2q 1 C = A 2q 1 A 2q 1 sl 1 (3), CA 2q C = A 2q, A 2q sl 0 (3). The original Lax pair corresponds to the simplest nontrivial case (N = 2) of this general flow pair. The compatibility condition [L(λ), A(λ)] = 0 gives rise to the following set of recurrence relations:

22 0-21 [L 1, A N ] = 0, i x A N + [L 1, A N 1 ] = 0,... i x A k + [L 1, A k 1 ] = 0, k = 2,...,N 1,... x A 1 t L 1 = 0. It follows from the first relation that the highest order term is a polynomial of L 1 and hence we have two possibilities for A N : a) A N = f 2p L 2, for N = 2p, b) A N = f 2p+1 L 1, for N = 2p + 1. It suffices to restrict oursleves with the case when N = 2p. We shall split each element A k into two mutually orthogonal parts: A 2q 1 = A 2q 1 + f 2q 1 L 1, A 2q = A 2q + f 2q L 2.

23 0-22 Substituting the splitting of A N 1 we have if 2p,x L 2 + if 2p L 2,x + [L 1, A 2p 1] = 0. After taking the Killing form., L 2 to separate the L 1 -commuting part and its orthogonal complement we deduce that f N = c N = const, A 2p 1 = ic 2p ad 1 L 1 L 2,x. Similarly, after extracting the L 1 -commuting part from the generic recurrence relations we determine for the coefficient f 2q 1 (resp. f 2q ) f 2q 1 = c 2q x f 2q = c 2q x (A ), L 2q 1 x 1, (A ), L 2q x 2, where c 2q 1 (resp. c 2q ) is a constant of integration. On the other hand from the orthogonal parts of generic recurrence relations one

24 0-23 can express A 2q 1 (resp. A 2q), namely : A 2q = Λ 1 ( A 2q+1 ) ic2q+1 ad 1 L 1 L 1,x, A 2q 1 = Λ 2 ( A 2q ) ic2q ad 1 L 1 L 2,x. The last recurrence relation yields to f 1 = c x (A 1 ) x, L 1, iad 1 L 1 t L 1 + Λ 1 A 1 ic 1 ad 1 L 1 L 1,x = 0. Finally for an arbitrary member of the integrable hierarchy we obtain a) t L 1 = b) t L 1 = p q=1 p q=1 c 2q (Λ 1 Λ 2 ) q 1 Λ 1 ad 1 L 1 L 2,x + c 2q (Λ 1 Λ 2 ) q 1 Λ 1 ad 1 L 1 L 2,x + p 1 q=0 p q=0 c 2q+1 (Λ 1 Λ 2 ) q ad 1 L 1 L 1,x, c 2q+1 (Λ 1 Λ 2 ) q ad 1 L 1 L 1,x.

25 0-24 The coefficients c k are involved in the dispersion law of the corresponding NEE: f(λ) = lim x ± g 1 as N λ k A k (x, t)g as. k=1 The dispersion law determines the evolution of scattering matrix T(t, λ) = e if(λ)t T(0, λ)e if(λ)t. It is not hard to check that the equalities below are valid a) f(λ) = b) f(λ) = p 1 q=0 c 2q+1 λ 2q+1 J + p c 2q+1 λ 2q+1 J + q=0 p c 2q λ 2q I, q=1 p c 2q λ 2q I, q=1 where J = diag(1, 0, 1), I = diag (1/3, 2/3, 1/3). The initial NEE can be derived from the above formulae in the simplest case N = 2 after inserting c 2 = 1 and c 1 = 0.

26 0-25 Integrable hierarchy in terms of the scattering data The following theorem holds true: Theorem 1 Let the Lax operator L be such that its potential satisfies the conditions: 1. L 1 (x) lim x ± L 1 is complex valued function of Schwartz type; 2. L 1 (x) is such that the principal minors of the scattering matrix have a finite number of simple zeroes which do not coincide; 3. The principal minors have no zeroes on the real axis of the complex λ-plane. Then the NEE are equivalent to each of the following set of linear evolution equations: i t S ± + [f(λ), S ± (λ)] = 0, t D + (λ) = 0, i t s ± k + [ f(λ ± k ), ] dλ ± s± k k = 0, = 0. d t

27 0-26 where S ± (λ) = exps ± (λ) and s + (λ) = s (λ) = 0 s+ α 1 s + α s + α s α s α 3 s α 2 0, s + k =, s k = 0 Res λ=λ + k s + α 1 Res λ=λ + k 0 0 Res 0 0 λ=λ + k 0 Res λ=λ k Res λ=λ k s + α 3 s + α s α s α 3 Res λ=λ k s α 2 0, The variables {s ± α(λ), λ R, s ± α;k, α +} n k=1 define a minimal set of scattering data.

28 0-27 Conclusions A 2-component generalization of HF equation has been studied. The direct scattering problem for the corresponding L operator has been developed in terms of Jost solutions, scattering matrix and fundamental analytic solutions. The soliton solutions have been constructed analytically. For that purpose we have used the dressing technique. The basic notions of the generalized Fourier intepretation of the inverse scattering method has been introduced. These are Wronskian relations, squared solutions and recursion operators. By using them we have described the integrable hierarchy of NEE, associated to L in terms of recursion operators and scattering data.