Exact Solutions to the Focusing Discrete Nonlinear Schrödinger Equation

Transcription

1 Exact Solutions to the Focusing Discrete Nonlinear Schrödinger Equation Francesco Demontis (based on a joint work with C. van der Mee) Università degli Studi di Cagliari Dipartimento di Matematica e Informatica The Seventh Imacs International Conference on Nonlinear Evolution Equation and Wave Phenomena. Athens, April 04-07, 2011 Supported by RAS under grant PO Sardegna , L.R. 7/2007

2 Contents a. Introduction to the IDNLS equation b. IDNLS equation and inverse scattering transform d. Explicit solutions of the IDNLS equation e. Examples slide 2 di 33

3 Introduction In this talk we consider the system of the Integrable Discrete Nonlinear Schrödinger (IDNLS) equations d i dτ u n = u n+1 2u n + u n 1 u n+1 w n u n u n w n u n 1, d i dτ w n = w n+1 2w n + w n 1 w n+1 u n w n w n u n w n 1, where n is an integer labeling position and u n and w n are N M and M N matrix functions depending on time τ R. When w n = u n we have the so-called focusing case: d i dτ u n = u n+1 2u n + u n 1 + u n+1 u nu n + u n u nu n 1. Our aim is to obtain the exact solutions for the IDNLS system (and for focusing equation) by using the Inverse Scattering Transform and the Marchenko method (i.e. formulating the inverse scattering problem in term of Marchenko equations). slide 3 di 33

4 Introduction Because the IDNLS can be seen as a finite difference approximation of the matrix NLS, it has the same applications of the matrix NLS, i.e. electromagnetic wave propagation in nonlinear media; surface waves on deep waters; signal propagation in optical fibers. slide 4 di 33

5 Introduction Important dates on the IDNLS: Ablowitz and Ladik ( ) were the first authors to show that IDNLS can be solved via the IST. In these works they only considered the scalar case. Gerdjikov and Ivanov ( ) developed the direct and inverse scattering theory of the discrete Zakharov-Shabat system in the matrix case. Tsuchida, Ujino and Wadati (2002) and Ablowitz, Prinari,Trubatch (2004) applied the IST to the matrix IDNLS. It is worthwhile to remark that also other techniques can be applied to solve the IDNLS. For example, the Hirota method can be used to derive some special solutions (in fact, D. Cai, A.P. Bishop, and N. Grønbech-Jensen did it to obtain the breather solutions to the IDNLS). slide 5 di 33

6 Introduction Representing the kernels of the Marchenko equation in a separated form by using two triplets of constant matrices, we explicitly solve the Marchenko equations by separation of variables. In particular, we get the following advantages: We obtain a compact solution formula in terms of two triplets of matrices. These solutions can be unzipped as a compound of polynomial, exponential and trigonometric functions by using Mathematica or Matlab. Choosing the triplets in an appropriate way, we can see that our formula contains the N-soliton and breather solutions as special cases (but also the multipole solutions). Our procedure also allows us to overcome the assumption (made in order to obtain appropriate symmetry properties of its scattering data) u n w n = w n u n = c n I N, n Z, (1) where {1 c n } n= is a sequence of nonzero scalars. In fact, we ll discuss one example of solution of IDNLS which does not satisfy (1). slide 6 di 33

7 Discrete Zakharov-Shabat System In order to solve the initial-value problem for the IDNLS by using the inverse scattering transform (IST) method, it is necessary to build the direct and inverse scattering for the following so-called matrix discrete Zakharov-Shabat (Z-S) system ( ) zin u v n+1 = n w n z 1 v I n, M where z is the (complex) spectral parameter and I N u n w n is assumed nonsingular for each n Z (which is always true in the focusing case) and the potentials {u n } n= and {w n } n= satisfy the l 1 -condition n= and denotes any matrix norm. { u n + w n } < +, slide 7 di 33

8 Direct and Inverse Scattering of the Zakharov-Shabat System We recall that THE DIRECT SCATTERING consists of: Determine the reflection coefficients, the bound states [poles of t r (λ)], and norming constants from the potentials u n, w n. THE INVERSE SCATTERING consists of: Reconstruct the potentials u n, w n from one reflection coefficient, the bound states, and the norming constants. slide 8 di 33

9 Direct Scattering: Jost Solutions Let us define the four Jost solutions φ n (z), φ n (z), ψ n (z), and ψ n (z) as those (N + M) N, (N + M) M, (N + M) M, and (N + M) N matrix solutions to the discrete Z-S system φ n (z) z n ( IN 0 MN ) ( ), φn (z) z n 0NM, n, I M ψ n (z) z n ( 0NM I M ), ψ n (z) z n ( IN 0 MN ), n +. slide 9 di 33

10 Jost Solutions We can consider the (N + M) (N + M) matrices ( φn (z) φn (z) ) (, ψn (z) ψ n (z) ). We have used two colors because z n ψ n (z) and z n φ n (z) are continuous in z 1 and are analytic in z > 1 (RED), while z n ψ n (z) and z n φ n (z) are continuous in z 1 and analytic in z < 1 (BLUE). We have ( φn (z) φn (z) ) = ( ψn (z) ψ n (z) ) T(z), ( ψn (z) ψ n (z) ) = ( φ n (z) φn (z) ) T(z), where I N u n w n (and hence I M w n u n ) is nonsingular for each n Z and ( ) ) a(z) b(z) ( c(z) d(z) T(z) = def, T(z) = def, b(z) ā(z) d(z) c(z) are called the transition coefficient matrices. slide 10 di 33

11 Scattering matrices Using the analiticity properties of the Jost solutions, we arrive at the following Riemann-Hilbert problem ( ψ n (z) φ n (z) ) = ( φ n (z) ψ n (z) ) JS(z)J, z = 1, ( φn (z) ψ n (z) ) = ( ψ n (z) φ n (z) ) J S(z)J, z = 1, where J = I N ( I M ) ( ) tr (z) l(z) ( t S(z) =, S(z) = S(z) 1 = l (z) ρ(z) t l (z) l(z) ) ρ(z). t r (z) are called the scattering matrices. Morever, if there are no spectral singularities ρ(z) = z s ˆρ(s), ρ(z) = z sˆ ρ(s), l(z) = s= s= z sˆ l(s), l(z) = z sˆl(s). s= s= slide 11 di 33

12 Inverse Scattering Problem In order to (re)-construct the potential we have the following procedure: 1 Given the scattering data {ρ(z), ζ k, C rj } and { ρ(z), ζ k, C rj }, we build the following kernels F r (j) = ˆρ(j) + k ζ (j+1) k C rk, Fr (j) = ˆ ρ(j) k ζ j 1 k C rk. slide 12 di 33

13 Inverse Scattering Problem In order to (re)-construct the potential we have the following procedure: 1 Given the scattering data {ρ(z), ζ k, C rj } and { ρ(z), ζ k, C rj }, we build the following kernels F r (j) = ˆρ(j) + k ζ (j+1) k C rk, Fr (j) = ˆ ρ(j) k ζ j 1 k C rk. 2 Using the function F r (j) and F r (j), we can consider the following Marchenko equation κ(n, n + 2σ + 1) = F r (2[n + σ] + 1) + κ(n, n + 2σ + 1) F r (2[n + σ + σ ] + 1) F r (2[n + σ + σ] + 1), σ =0 σ =1 slide 12 di 33

14 Inverse Scattering Problem In order to (re)-construct the potential we have the following procedure: 1 Given the scattering data {ρ(z), ζ k, C rj } and { ρ(z), ζ k, C rj }, we build the following kernels F r (j) = ˆρ(j) + k ζ (j+1) k C rk, Fr (j) = ˆ ρ(j) k ζ j 1 k C rk. 2 Using the function F r (j) and F r (j), we can consider the following Marchenko equation κ(n, n + 2σ + 1) = F r (2[n + σ] + 1) + κ(n, n + 2σ + 1) F r (2[n + σ + σ ] + 1) F r (2[n + σ + σ] + 1), σ =0 σ =1 3 The potential u n (w n can be derived by considering a similar Marchenko equation) are connected to the above equations by means of the following relationship u n = κ(n, n + 1). slide 12 di 33

15 Remarks Remark 1: The relation u n = κ(n, n + 1), holds under the assumption (1) (i.e. u n w n = w n u n = c n I N, n Z.) Remark 2: It is well known that the discrete spectrum of the discrete Z-S system is invariant under sign inversion. Then we have the transmission coefficients are even functions of z. the reflection coefficients are odd functions of z. the functions ˆρ(s), ˆ ρ(s) (as well as ˆl(s), ˆ l(s)) vanish if s is even. the Marchenko kernels F r (j) and F r (j) (as well as F l (j) and F l (j)) vanish if j is even. slide 13 di 33

16 Kernels in terms of the matrix triplets Taking into account the properties of the kernels F r (j) and F r (j), we assume the following representation for these kernels F r (j) = ˆρ(j) + [1 + ( 1) j+1 ]C r A (j+1) B r, F r (j) = ˆ ρ(j) + [1 + ( 1) j+1 ] C r Ā j 1 B r, where the triplets (A, B r, C r ), (Ā, B r, C r ) are in the following form: ( ) ( ) A 0 Br A =, B 0 A r =, C B r = ( ) C r C r, r (Ā ) ( ) 0 B Ā =, B 0 Ā r = r, C B r = ( ) C r C r. r Moreover, these triplets have to satisfy the following properties: A is nonsingular matrix having only eigenvalues of modulus larger than one. Ā is nonsingular matrix which have only eigenvalues of modulus less than one. slide 14 di 33

17 Explicit Solutions to the Marchenko Equations Substituting the two expressions of the kernels into the Marchenko equation, we get κ(n, n + 2σ + 1) = 2 C r Ā 2[n+σ] B r + 4 κ(n, n + 2σ + 1) C r A 2[n+σ +σ +1] B r C r Ā 2[n+σ +σ] B r. σ =0 σ =1 If we look for a solution in the following form κ(n, n + 2σ + 1) = H(n)Ā 2[n+σ+1] B r and define Q r = Ā 2σ B r C r A 2σ, N r = A 2σ B r C r Ā 2σ, σ=0 σ=0 we obtain H(n) = 2 C r Ā 2 [I 4Ā 2(n+1) Q r A 2(n+2) N r ] 1. slide 15 di 33

18 Solutions of IDNLS Since κ(n, n + 2σ + 1) = 2 C r Ā 2 [I 4Ā 2(n+1) Q r A 2(n+2) N r ] 1 Ā 2[n+σ+1] B r, and recalling that u n = κ(n, n + 1), we get Starting from the equation u n = 2 C r [Ā 2n 4Q r A 2(n+2) N r Ā 2 ] 1 Ā 2[n+σ+1] B r. K(n, n + 2σ + 1) = F r (2[n + σ] + 1) + K(n, n + 2σ + 1) F r (2[n + σ + σ ] + 1)F r (2[n + σ + σ] + 1), σ =0 σ =1 whose solution is related to IDNLS solution by w n = K(n, n + 1), and proceeding in the (essentially) same way we get w n = 2C r [A 2(n+1) 4N r Ā 2(n+1) Q r A 2 ] 1 B r. slide 16 di 33

19 The time dependence The final step to obtain the solutions of the IDNLS system consists of considering the time dependence of the scattering data. When the reflection coefficients vanish, the kernels must satisfy the following d i dτ F r (n; τ) = F r (n + 2; τ) 2F r (n; τ) + F r (n 2; τ), d i dτ F r (n; τ) = F r (n + 2; τ) 2 F r (n; τ) + F r (n 2; τ), and hence the kernels depending on time can be chosen as F r (j; τ) = 2C r A (j+1) e iτ(a A 1 ) 2 B r, F r (j; τ) = 2 C r e iτ(ā Ā 1 ) 2 Ā j 1 B r. slide 17 di 33

20 CRUCIAL OBSERVATION: Let us consider the kernels Making the following changes F r (j) = 2C r A (j+1) B r, F r (j) = 2 C r Ā j 1 B r. B r e iτ(a A 1 ) 2 B r, C r C r e iτ(ā Ā 1 ) 2, and leaving unchanged A, C r, B r, we get F r (j; τ) = 2C r A (j+1) e iτ(a A 1 ) 2 B r, F r (j; τ) = 2 C r e iτ(ā Ā 1 ) 2 Ā j 1 B r, which are the kernels depending on time. Under the substitutions above indicated we also have N r e iτ(a A 1 ) 2 N r e iτ(ā Ā 1 ) 2 while Q r remains unchanged. slide 18 di 33

21 Solutions of the IDNLS Then executing the substitutions B r e iτ(a A 1 ) 2 B r, C r C r e iτ(ā Ā 1 ) 2, N r e iτ(a A 1 ) 2 N r e iτ(ā Ā 1 ) 2, and leaving unchanged A, C r, B r, and Q r into the expression u n = 2 C r [Ā 2n 4Q r A 2(n+2) N r Ā 2 ] 1 Ā 2[n+σ+1] B r w n = 2C r [A 2(n+1) 4N r Ā 2(n+1) Q r A 2 ] 1 B r we get the solution to the IDNLS equation, i.e. u n (τ) = 2 C r [Ā 2n e iτ(ā Ā 1 ) 2 4Q r A 2(n+2) e iτ(a A 1 ) 2 N r Ā 2 ] 1 B r. w n (τ) = 2C r [A 2(n+1) e iτ(a A 1 ) 2 4N r e iτ(ā Ā 1 ) 2 Ā 2(n+1) Q r A 2 ] 1 B r slide 19 di 33

22 The focusing case Relating the matrix triplets in the following way: Ā = A 1, B r = A 1 C r, C r = B r A 1, the Marchenko kernels satisfy the relation F r (j) = F r (j), which characterize the focusing case. Introducing the quantity Q r = A 2σ C r C r A 2σ, N r = A 2σ B r B r A 2σ, σ=0 σ=0 we can written the general solutions before found as u n (τ) = 2B r [A 2(n+1) e iτ(a A 1 ) 2 +4Q r A 2(n+2) e iτ(a A 1 ) 2 N r A 2 ] 1 C r. Moreover, we can easily verify that w n (τ) = u n (τ), n Z. slide 20 di 33

23 Examples The easiest example is obtained choosing the triplet in the following way: A = (ae iθ ), B l = (1), C l = (c) with a > 1, and 0 c C. In fact, we easily compute u n (τ) = c a 2 a 2 c 2 }{{} modulus=1 e iτ[(a2 +a 2 ) cos 2θ 2] e ( 2inθ cosh 2n log(a)+τ(a 2 a 2 ) sin 2θ+log which is easily verified to satisfy the focusing IDNLS equation. ), 2 c a 2 a 2 If the matrix triplets have order equal or greater than two, it becomes impractical to unzip the solution formula because it take a lot of page!. For this reason we have applied Mathematica to plot the focusing IDNLS solutions. I m greatly indebted to Antonio Aricò for his assistance in developing the Mathematica code. slide 21 di 33

24 Soliton-soliton interaction Let us consider the focusing case, where A =, Br =, and Cr = 1 1 we obtain the so called soliton-soliton interaction (a) τ = (b) τ = (c) τ = (d) τ = 16 slide 22 di 33

25 Breather solutions Let us consider the focusing case, choosing A = Cr = 2 1 we obtain the breather solutions (a) τ = (c) τ = 0, (b) τ = , Br =, and (d) τ = 1 slide 23 di 33

26 Double poles soliton solutions 3 i 1 0 Let us consider the focusing case, where A =, Br =, and 0 3 i 1 Cr = 3 2. We are treating a double poles soliton solutions (a) τ = (c) τ = (b) τ = (d) τ = 20 slide 24 di 33

27 On the condition (1) Being the relationship between the solution of the Marchenko equation and the solutions of IDNLS derived under the assumption u n w n = w n u n = c n I N, n Z, the solutions found until now satisfy this condition. On the other hand, it is possible to formulate the inverse scattering problem in terms of more complicated modified Marchenko equation in a such a way that the relation between the solution of IDNLS and the solution of the modified Marchenko equation do not involve the condition (1): u n (τ) = 2C l [A 2n e iτ(a A 1 ) 2 4N l e iτ(ā Ā 1 ) 2 Ā 2n Q l A 2 ] 1 B l, w n (τ) = 2 C l [Ā 2(n+1) e iτ(ā Ā 1 ) 2 4Q l A 2(n 1) e iτ(a A 1 ) 2 N l Ā 2 ] 1 B l. slide 25 di 33

28 On the condition (1): one example Let us consider the matrix triplet (A, B r, C r ) defined by where a > 1, β = that A = (a), B r = ( b 1 b N ), Cr = [ N c 1. c N, ] 1/2 [ j=1 b N ] 1/2 j 2 > 0, and γ = j=1 c j 2 > 0. We see u n (τ) = 1 β b 1. b N u n (τ) 1 γ ( ) c 1 cn, where u n (τ) is the (scalar) IDNLS solution of the type single-soliton which results by taking the matrix triplet (a, β, γ). slide 26 di 33

29 On the condition (1): one example Consequently, we calculate u n (τ)u n (τ) = 1 β 2 u n (τ) u n (τ) = 1 γ 2 and condition (1) is not satisfied. b 1. b N c 1. c N u n (τ) 2 ( b 1 b N ), u n (τ) 2 ( c 1 c N), slide 27 di 33

30 LITERATURE: I M.J. Ablowitz and J.F. Ladik, Nonlinear differential-difference equations, J. Math. Phys. 16, (1975). M.J. Ablowitz and J.F. Ladik, A nonlinear difference scheme and inverse scattering, Stud. Appl. Math. 55, (1976). M.J. Ablowitz and J.F. Ladik, Nonlinear differential-difference equations and nonlinearity, J. Math. Phys. 17, (1976). M.J. Ablowitz, B. Prinari, and A.D. Trubatch, Discrete and Continuous Nonlinear Schrödinger Systems, Cambridge University Press, Cambridge, D. Cai, A.P. Bishop, and N. Grønbech-Jensen, Perturbation theories of a discrete, integrable nonlinear Schrödinger equation, Phys. Rev. E 53, (1996). slide 28 di 33

31 LITERATURE: II V.S. Gerdjikov and M.I. Ivanov, Block discrete Zakharov-Shabat system. I. Generalized Fourier expansion, United Nuclear Research Institute Report E , Dubna, V.S. Gerdjikov and M.I. Ivanov, Block discrete Zakharov-Shabat system. I. Hamiltonian structures, United Nuclear Research Institute Report E , Dubna, V.S. Gerdjikov, M.I. Ivanov, and P.P. Kulish, Expansions of the squared solutions and difference evolution equations, J. Math. Phys. 25, (1984). A. Doliwa and P.M. Santini, Integrable dynamics of a discrete curve and the Ablowitz-Ladik hierarchy, J. Math. Phys. 36, (1995). slide 29 di 33

32 LITERATURE: III T. Tsuchida, Integrable discretizations of derivative nonlinear Schrödinger equations, J. Phys. A 35, (2002). T. Tsuchida, H. Ujino, and M. Wadati, Integrable semi-discretization of the coupled modified KdV equations, J. Math. Phys. 39, (1998). T. Tsuchida, H. Ujino, and M. Wadati, Integrable semi-discretization of the coupled nonlinear Schrödinger equations, J. Phys. A 32, (1999). slide 30 di 33

33 LITERATURE: IV T. Aktosun, F. Demontis, and C. van der Mee, Exact solutions to the sine-gordon equation, Journal of Mathematical Physics, 51, (2010). F. Demontis and C. van der Mee, Explicit solutions of the cubic matrix nonlinear Schrödinger Equation, Inverse Problems 24, (2008). C. Schiebold, Solutions of the sine-gordon equation coming in clusters, Revista Matemática Complutense 15, (2002). H. Dym, Linear Algebra in Action, Graduate Studies in Mathematics 78, Amer. Math. Soc., Providence, RI, slide 31 di 33

34 Acknowledgments Thank you!! slide 32 di 33

35 Q r, N r satisfy uniquely the following (matrix) equation Q r Ā 2 Q r A 2 = B r C r, N r A 2 N r Ā 2 = B r C r, Q r and N r are the unique solutions of the Stein equations Q r A 2 Q r A 2 = C r C r, N r A 2 N r A 2 = B r B r. slide 33 di 33