Exact Solution and Vortex Filament for the Hirota Equation

Transcription

1 Exact Solution and Vortex Filament for the Hirota Equation Francesco Demontis (joint work with G. Ortenzi and C. van der Mee) Università degli Studi di Cagliari Dipartimento di Matematica e Informatica Workshop on Nonlinear Waves and Integrable Systems in Sicily Taormina, June Research supported by GNFM under the grant: Progetto Giovani 2013

2 Introduction In 1973 Hirota introduced the equation iq t + 3iα q 2 q x + ρq xx + iσq xxx + δ q 2 q = 0, where q is a scalar function, (x, t) R 2, i is the imaginary unit, and α, ρ, σ, δ are real constants which satisfy αρ = σδ. In this talk we consider the same equation written in a different way: iq t α 2 [ qxx + 2 q 2 q ] + iα 3 [ qxxx + 6 q 2 q x ] = 0, where α = 2α 3, δ = 2α 2, ρ = α 2 and σ = α 3. BLU... Modified Korteweg de-vries equation (mkdv) Magenta...Nonlinear Schrödinger equation (NLS) slide 2 di 33

3 Introduction iq t α 2 [ qxx + 2 q 2 q ] + iα 3 [ qxxx + 6 q 2 q x ] = 0 The Hirota equation appears in the description of the vortex string motion for a three dimensional Euler incompressible fluid. b X=X(s,t) t n slide 3 di 33

4 Introduction The motion of an incompressible Euler fluid is described by: ( ) v ρ + rot v + grad p = 0. t div v = 0, where v = (u, v, w) is the velocity (u, v, w depend on x, y, z, t), p(= p(x, y, z, t)) is the pressure and the density ρ is constant. slide 4 di 33

5 Introduction A generic vortex is characterized by its components p, q, r which are related with the velocity by the relation 2ω = rotv, and the motion equation are given by: p t + p x u + p y v + p z w = u x p + u y q + u z r q t + q x v + q y w + q z u = v x q + v y r + v z p r t + r x w + r y u + r z v = w x r + w y p + w z q Helmholtz theorem The material particles which describe the vortex filament at the initial time t = 0 give the vortex filament in any time. Problem: How to describe the motion of the vortex filament? More precisely, how to reconstruct the velocity field from the vorticity field (Biot-Savart problem)? slide 5 di 33

6 Introduction iq t α 2 [ qxx + 2 q 2 q ] + iα 3 [ qxxx + 6 q 2 q x ] = 0 a. The classical approximation model for this problem is the Local Induction Approximation (LIA). It was developed by Da Rios (1906) and Betchov (1965). b. Hasimoto (1972) showed that the model obtained by using the LIA is equivalent to the NLS under the hypothesis of constant vorticity and null velocity inside the vortex core. In fact, he got that ( s ) q = κ exp i τds where q is a solution of the NLS equation. c. Fukumoto and Miyazaki (1991) generalized the Hasimoto s result obtaining a model (equivalent to the Hirota equation) which takes into account the contribution of (constant) axial velocity. In that case, the constant α 3 encodes the correction to the LIA due to the axial velocity along the vortex filament. 0 slide 6 di 33

7 Our Purposes 1) To find explicit solitonic solutions of the Hirota equation by using the Inverse Scattering Transform (IST). Using a suitable triplet of constant real matrices (A, B, C) where A, B, C have dimensions p p, p 1 and 1 p, we obtain this solution formula: with Q = u(x, t) = 2B e A x Γ 1 (x, t)e A x+iφ( ia )t C, 0 ds e A s C Ce As, N = Γ(x, t) = I p + e A x+iφ( ia )t Qe 2Ax iφ(ia)t Ne A x φ(z) = 4α 2 z 2 8α 3 z 3. 0 dr e Ar BB e A r, slide 7 di 33

8 Our Purposes 2) To describe the time evolution of a vortex filament associated with an exact solitonic solution of the Hirota equation through the Sym-Pohlmeyer formula. The Sym-Pohlmeyer formula gives the cartesian components of the curve (for t fixed) of the vortex filament associated with a specific solution of the Hirota equation in the following way: γ l (λ, x, t) λ=0 Ψ 1 (x, λ; t) λ Ψ(x, λ; t) λ=0 = i 3 x i (x, t)σ i where σ i are the Pauli s matrices and the Jost solutions Ψ(x, λ; t) satisfies the AKNS pair of the Hirota equation Y x = XY Y t = TY. i=1 slide 8 di 33

9 IST: a brief presentation { iq t α 2 [ qxx + 2 q 2 q ] + iα 3 [ qxxx + 6 q 2 q x ] = 0 q(x, 0) q(x, 0) direct scattering at t=0 Hirota solution {S(λ)} time evolution q(x, t) inverse scattering at t {S(λ, t)} Direct Scattering: Construct the scattering data when the initial data q(x, 0) is given. Inverse Scattering Data: Build the potential if the scattering data are known. Time Evolution of the Scattering Data: How to pass from S(λ) to S(λ, t)? We develop the direct and inverse scattering problems related to ψ x = X ψ while we use ψ t = T ψ to discover the time evolution of the scattering data. Problem: How to choose X and T? The IST has not been completely developed for the Hirota equation (at least at the best of our knowledge). slide 9 di 33

10 IST: Choice of X The NLS and mkdv equations are associated to the same spatial operator X (the so-called Zakharov-Shabat (ZS) system) which depends only on the hierarchy but not on the particular equation. Then, also the Hirota equation is associated to the ZS system: Y iσ 3 (λ, x) V (x)y (λ, x) = λy (λ, x), x where σ 3 = ( ) 1 0, V (x) = iq(x) = i 0 1 ( ) 0 q(x), r(x) 0 λ is the spectral parameter and q(x), r(x) are the potentials (in L 1 R). ZS system can also be written as: Y x = [ iλσ 3 + Q]Y, where ( ) 0 q Q(x) = iσ 3 V (x) = q = i {(Im q)σ (Re q)σ 2 }. slide 10 di 33

11 IST: Choice of T Let T 1 and T 2 be the matrices related to the time evolution matrix of the NLS (mkdv) equation: T (1) = 2iλ 2 σ 3 iσ 3 V 2 + 2λV + iσ 3 V x T (2) = 4iλ 3 σ 3 + 2iλσ 3 V 2 + 4λ 2 σ 3 V + 2iλV x σ 3 V xx 2σ 3 V 3. The zero-curvature conditions for the NLS and mkdv are, respectively, Defining T as Y t T (1) x Y t T (2) x = YT (1) T (1) Y = YT (2) T (2) Y T = α 2 T (1) + α 3 T (2), where α 2 and α 3 are real constants, and rescaling the time variable we arrive at the zero-curvature for the Hirota equation: Y τ T x = YT TY. slide 11 di 33

12 Direct scattering: Jost solutions Jost matrix solutions Ψ(λ, x) = ( ) ψ(λ, x) ψ(λ, x) = e iλσ 3x [I 2 + o(1)], x +, Φ(λ, x) = ( ) φ(λ, x) φ(λ, x) = e iλσ 3x [I 2 + o(1)], x. The Jost solutions have to satisfy the following Volterra equations Ψ(λ, x) = e iλσ3x + iσ 3 dy e iλσ3(y x) V (y)ψ(λ, y), Furthermore, x x Φ(λ, x) = e iλσ3x iσ 3 dy e iλσ3(x y) V (y)φ(λ, y). Ψ(x, λ)a r (λ) = Φ(x, λ), Φ(x, λ)a l (λ) = Ψ(x, λ), a l (λ) = a r (λ) 1 = a r (λ). RED denotes analyticity in C + while BLUE analyticity in C and the symbol denotes transpose conjugation. slide 12 di 33

13 Riemann-Hilbert problem: the scattering matrix F + (λ, x) = ( φ(λ, x) ψ(λ, x) ), F (λ, x) = ( ψ(λ, x) φ(λ, x) ). As a result, we arrive at the Riemann-Hilbert problems: F (λ, x) = F + (λ, x)σ 3 S(λ)σ 3, F + (λ, x) = F (λ, x)σ 3 S(λ)σ 3, λ R where ( ) ( ) T (λ) L(λ) T (λ) L(λ) S(λ) =, S(λ) = R(λ) T (λ) R(λ) T (λ) are called scattering matrices. The scattering matrices S(λ) and S(λ) are each the inverse of the other and satisfy, i.e. S(λ) = S(λ) 1 = σ 3 S (λ)σ 3, λ R. slide 13 di 33

14 Inverse scattering: Marchenko equations Ψ(λ, x; t) = e iλσ3x + x dy α l (x, y; t)e iλσ3y, where α l (x, y; t) has to satisfy the Marchenko integral equations α l (x, y; t) + ω l (x + y; t) + Here ( K up ) (x, y; t) K up (x, y; t) α l (x, y; t) = K dn, (x, y; t) K dn (x, y; t) ω l (x; t) = ρ(x; t) + N j j=1 s=0 x 0 ρ(x; t) dz α l (x, z; t)ω l (z + y; y) = N j j=1 s=0 x s s! e xλ j [C l ] js (t) 0 x s s! e xλ j [C l ] js(t), where λ j are the bound states belonging to the upper (lower) half plane, and [C lr ] js are the norming constants associated with the bound states. slide 14 di 33

15 Inverse scattering: reconstruction of the potential In order to (re)-construct the potential we can follow the steps below: 1 Given the scattering data {R(λ, t), λ j, C js (t)}, we build the following function n N j 1 Ω l (y; t) = ρ y s (y; t) + c js;t s! eiλ j y. j=1 s=0 slide 15 di 33

16 Inverse scattering: reconstruction of the potential In order to (re)-construct the potential we can follow the steps below: 1 Given the scattering data {R(λ, t), λ j, C js (t)}, we build the following function n N j 1 Ω l (y; t) = ρ y s (y; t) + c js;t s! eiλ j y. j=1 s=0 2 Using the function Ω(y; t), we can consider the following Marchenko integral equation K up (x, y; t) Ω l (x+y; t) + where y > x. x dz x ds K up (x, z; t) Ω l (z+s; t) Ω l (s+y; t) = 0, slide 15 di 33

17 Inverse scattering: reconstruction of the potential In order to (re)-construct the potential we can follow the steps below: 1 Given the scattering data {R(λ, t), λ j, C js (t)}, we build the following function n N j 1 Ω l (y; t) = ρ y s (y; t) + c js;t s! eiλ j y. j=1 s=0 2 Using the function Ω(y; t), we can consider the following Marchenko integral equation K up (x, y; t) Ω l (x+y; t) + where y > x. x dz x ds K up (x, z; t) Ω l (z+s; t) Ω l (s+y; t) = 0, 3 The potential q(x, t) is connected to the above equation by means of the following relationship q(x; t) = 2K up (x, x; t). slide 15 di 33

18 Kernels of the Marchenko equations for solitonic solutions The soliton solutions are characterized by the condition R(λ) = 0. So, the kernel which appears in the Marchenko is given by Ω l (y; t) = N n j 1 j=1 s=0 c js (t) y s s! eiλ j y = Ce ya e iφ(ia)t B. The time evolution of the kernel Ω(y, t) is well-known for NLS and mkdv. Since the construction of the kernel is linear in the transmission and reflection coefficients, we obtain Ω l t iα 2 Ω l yy + 8α 3 Ω l yyy = 0. slide 16 di 33

19 Kernels of the Marchenko equations for solitonic solutions A solution (in the factorized form) of the PDE describing the time-evolution of the kernels is given by Ω(y; t) = Ce ya e iφ(ia)t B, where φ(z) = 4α 2 z 2 8α 3 z 3, where A is a p p real constant matrix, B, C are, respectively, p 1 and 1 p real constant matrices, the blue factor is the so-called time factor. Let us also assume that the eigenvalues of A have positive real parts and (A, B, C) is a minimal triplet, i.e., + r=1 ker CA r 1 = + r=1 ker B (A ) r 1 = {0}, Note that the triplet yielding a minimal realization is unique up to a similarity transformation (A, B, C) (EAE 1, EB, CE 1 ) for some unique matrix E. slide 17 di 33

20 Explicit solutions Sustituting Ω(y; t) = Ce ya e iφ(ia)t B in the Marchenko equation and looking for a solution in the form with easy calculation, we obtain where K up (x, y; t) = H(x, t)e A y+iφ( ia )t C, K up (x, y; t) = B e A x Γ 1 (x, t)e A y+iφ( ia )t C, q(x, t) = 2K up (x, x; t) = 2B e A x Γ 1 (x, t)e A x+iφ( ia )t C, Q = 0 ds e A s C Ce As, N = 0 dr e Ar BB e A r, Γ(x, t) = I p + e A x+iφ( ia )t Qe 2Ax iφ(ia)t Ne A x, and I p is the identity matrix of order p. slide 18 di 33

21 Admissible class It is natural to look for a larger class including triplets for which the explicit solution formula found above makes still sense, i.e., the integrals which define Q and N converge and Γ(x, t) is invertible for all (x, t) R 2. The triplet (A, B, C) of size p belongs to the admissible class A if: The triplet (A, B, C) corresponds to a minimal realization for Ω(y, t). None of the eigenvalues of A are purely imaginary and no two eigenvalues of A can occur symmetrically with respect to the imaginary axis in the complex plane. slide 19 di 33

22 Equivalent triplets We say that two triplets (A, B, C) and (Ã, B, C) are equivalent if they lead to the same potential u(x, t). PROBLEM: Starting from one triplet in the admissible class, is it possible to get an equivalent triplet such that the matrices A, B, C are real, give a minimal representation for the kernel Ω(y, t) and all the eigenvalues of A have positive real parts? (Ã, B, C) is in the admissible class How construct Equivalent Triplets? (A, B, C) minimal representation where Reλ j > 0 slide 20 di 33

23 Reflecting some eigenvalues of A We have the following: For any admissible triplet (Ã, B, C), there corresponds an equivalent admissible triplet (A, B, C) in such a way that all eigenvalues of A have positive real parts. The equivalent triplet (A, B, C) is built as follows: A 1 = Ã1, A 2 = Ã 2, B 1 = B 1 Ñ2Ñ 1 4 B 2, B 2 = Ñ 1 4 B 2, C 1 = C 1 C Q Q 3, C 2 = C 2 Q 1 4, where Ã1 and Ã2 contains, respectively, all the eigenvalues having positive real parts and all the eigenvalues having negative real parts. slide 21 di 33

24 Canonical triplets (real eigenvalue) A B 1 0 A 2 0 A =......, B = B 2., C = ( ) C 1 C 2 C m, 0 0 A m B m where in the case of a real positive eigenvalues ω j of A j the corresponding blocks are given by ω j ω j ω j 0 0 A j :=......, B.. j :=. 0, ω j ω j C j := ( c jnj c j2 c j1 ), A j having size n j n j, B j size n j 1, C j size 1 n j. slide 22 di 33

25 Canonical triplets (real eigenvalue) In the case of complex eigenvalues the corresponding blocks are given by Λ j I Λ j I Λ j A j :=......, B.. j :=. 0, Λ j I Λ j C j := ( γ jnj ɛ jnj... γ j1 ɛ j1 ), where γ js and ɛ js for s = 1,..., n j are real constants with (γjn 2 j + ɛ 2 jn j ) > 0, each column vector B j has 2n j components, each A j has size 2n j 2n j, and the 2 2 matrix Λ j is defined as ( ) αj β j Λ j :=. β j α j slide 23 di 33

26 Classification of classes of inequivalent triplets Block A j Solution behavior 1 1 real matrix 1-soliton solution with q(x, t) = f (x) 1 1 complex matrix 1-soliton solution with q(x, t) = f (x Vt) 2 2 matrix with 1-breather solution complex conjugate eigenvalues ( 2-particle bound state ) Jordan block of order s (2s) Multipole solution (multipole breather) slide 24 di 33

27 One important example: breather solution Choosing the triplet as follows ( ) 0 1 A =, B = 1 0 ( 1 1), C = ( 1 1 ), we get a solito solutions known as breather solutions. slide 25 di 33

28 One important example: breather solution with α 2 = 1, α 3 = 1 and κ = q k x k x k x k x k x k x k x k x k x slide 26 di 33

29 Vortex filament The cartesian components x i (x, t), (for i = 1, 2, 3) of the curve (for a fixed t) described by a vortex filament associated with a specific solutions of the Hirota equation [i.e., fixing the triplet (A, B, C)] can be found from γ l (λ, x, t) λ=0 Ψ 1 (x, λ; t) λ Ψ(x, λ; t) λ=0 = i 3 x i (x, t)σ i where σ i are the Pauli s matrices. Since the (matrix) Jost solution Ψ(x, λ; t) belongs to the unitary group SU(2) the components x i (x, t) can be uniquely determined. We recall that Ψ(λ, x; t) = def ( ψ (up) ψ (dn) ψ (up) ψ (dn) ) i=1 has to satisfy the ZS system. IDEA: Try to express the Jost solutions in terms of the matrix triplets. slide 27 di 33

30 Vortex filament and matrix triplets ψ (up) (λ, x; t) = e iλx + ψ (dn) (λ, x; t) = ψ (up) (λ, x; t) = x ψ (up) (λ, x; t) = e iλx + x x dyk (up) (x, y; t)e iλy, dyk (dn) (x, y; t)e iλy, dyk (up) (x, y; t)e iλy, x dyk (dn) (x, y; t)e iλy, where K (up), K (dn), K (up) and K (dn), are given by K (up) (x, y; t) = B e xa Γ 1 (x; t)e ya e iφ( ia )t C, K (dn) (x, y; t) = Ce xa P(x)Γ 1 (x; t)e ya e iφ( ia )t C, K (up) (x, y; t) = B e xa P(x; t)(γ 1 (x; t)) e ya e iφ(ia)t B, K (dn) (x, y; t) = Ce xa (Γ 1 (x; t)) e ya e iφ(ia)t B. slide 28 di 33

31 Vortex filament and matrix triplets γ l (λ, x, t) λ=0 Ψ 1 (x, λ; t) λ Ψ(x, λ; t) λ=0 = i 3 x i (x, t)σ i [ ] ψ (up) (λ, x; t) = e iλx 1 + ib e xa P(x)(Γ (x)) 1 (λi p ia) 1 e xa e iφ(ia)t B, [ ] ψ (dn) (λ, x; t) = e iλx ice xa (Γ (x)) 1 (λi p ia) 1 e xa e iφ(ia)t B, ψ (up) (λ, x; t) = e iλx [ ib e xa Γ 1 (x)(λi p + ia ) 1 e xa e iφ( ia )t C ], ψ (dn) (λ, x; t) = e iλx [ 1 ice xa P(x)Γ 1 (x)(λi p + ia ) 1 e xa e iφ( ia )t C ]. Furthermore, Ψ 1 (λ, x, t) = Ψ (λ, x, t) for λ R (we have to replace λ with λ in the right-hand side if λ C) i=1 slide 29 di 33

32 Open problems To try to obtain similar results with nonvanishing boundary conditions To plot the graphics of the vortex filament in both cases (vanishing and non vanishing cases) slide 30 di 33

33 Literature: I R. Betchov, On the curvature and torsion of an isolated vortex filament, J. Fluid. Mech. 22 (1965), L.S. Da Rios, Sul moto d un liquido indefinito con un filetto vorticoso (in Italian), Rend. Circ. Mat. Palermo (1906). H. Hasimoto, Soliton on a vortex filament, J. Fluid. Mech (1972). Y. Fukumoto, and T. Miyazaki, Three-dimensional distortions of a vortex filament with axial velocity, J. Fluid. Mech (1991) Y. Fukumoto, and T. Miyazaki, Three-dimensional distortions of a vortex filament with axial velocity, J. Fluid. Mech (1991) B.G. Konopelchenko, and G. Ortenzi, Quasi-classical approximation in vortex filament dynamics. Integrable systems, gradient catastrophe, and flutter, Studies in Pure and Applied Mathematics, 130, (2013). A. Sym, Geometric Unification of Solvable Nonlinearities, Lett. Nuovo Cim. 36 n (1983) K. Pohlmeyer, Integrable hamiltonian systems and interactions through quadratic constraints, Commun. Math. Phys. 46, (1976). slide 31 di 33

34 Literature: II R. Hirota, Exact envelope-soliton solutions of a nonlinear wave equation J. Math. Phys. 14, , M. J. Ablowitz and H. Segur, Solitons and the inverse scattering transform, SIAM, Philadelphia, M.J. Ablowitz, B. Prinari, and A.D. Trubatch, Discrete and Continuous Nonlinear Schrödinger Systems, Cambridge University Press, Cambridge, L.D. Faddeev, and L.A. Takhtajan, Hamiltonian Methods in the Theory of Solitons, Springer, Berlin and New York, T. Aktosun, T. Busse, F. Demontis, and C. van der Mee, Symmetries for exact solutions to the nonlinear Schrödinger equation, Journal of Physics A, 43, (2010). T. Aktosun, F. Demontis, and C. van der Mee, Exact solutions to the focusing nonlinear Schrödinger equation, Inverse Problems 23 (2007), slide 32 di 33

35 Acknowledgments Thank you!! slide 33 di 33