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Department of Mathematics,Shanghai The Hamiltonian Structure and Algebro-geometric Solution of a 1 + 1-Dimensional Coupled Equations Xia Tiecheng and Pan Hongfei Page 2 of 46
Section One A Short Overview Section two Section three Lax pairs of the coupled 1 + 1-dimens Section four The Hamiltonian structures Section five The Dubrovin-type equations Section six Algebro-geometric solutions Section seven Open problem Page 3 of 46
1 Algebraic-geometry solutions(also known as quasiperiodic solutions or finite-band solution) are an important exact solutions of nonlinear evolution equations and is a kind of a natural extension of soliton solution. It will be associated with the several branch of mathematics, such as the theory of differential equation, operator spectrum theory, complex variables functions, algebra geometry. And also many other types of solutions, such as soliton solution, and elliptic periodic solutions can be obtained by the degradation of algebraic geometry solutions. The most striking aspects of integrable system theory is one of nonlinear differential equation with cross application of algebraic geometry. It is considered to be an perfect example of today s modern mathematical physics. Page 4 of 46
In 1967 inverse scattering method of the solution of KdV equation with initial value problem have been found by Gardner, Greene, Kruskal, Miura. Last century,70 s Zakharov, Shabat, Ablowitz, Newell generalized inverse scattering method to a big class of nonlinear evolution equations. But as a result of the long-term behavior of time, there is no effective method to solve the inverse spectrum problem of periodic coefficient linear operator, so the inverse scattering method cannot be directly applied to the periodic initial value problem. Page 5 of 46
In 80 s Novikov Dubrovin Mckean Its Matveev Krichiev etc. successfully solve the periodic initial value problem by using inverse scattering method based on the algebraic geometry method. Thus these caused cross development between the integrable differential equation and algebraic geometry. Because of the periodic potential Schrodinger operator of the spectrum has band structure and the spectrum of absolutely continuous component is divided by a finite number of belt, and the corresponding potential is known as with finite-potential. Page 6 of 46
Analysis on KdV equations with finite potential discovered by Mr Novikov Matveev, Lax, Marchenko, independently. All with finite potential can be described by the high order KdV equations of steady state, and then Matveev noticed this with finite potential can be converted into a two sheets Riemann surface with genus N K N : y 2 = 2N+1 m=1 (z E m ) by using Jacobi inverse problem to describe. Page 7 of 46
Finally, the potential function of continuous spectrum can be expressed as the function based on Riemann surface N u(x, t) = E 0 + (E 2j 1 + E 2j 2λ j ) j=1 2 2 xlnθ(ψ Q0 A Q0 (P ) + αq 0 (D µ (x, t))) Page 8 of 46
In the last century, 90 s Gesztesy and Holden etc. developed an effective polynomial recursive method and succeeded to extend algebraic geometry solutions from a single equation to the entire equation hierarchy. Up to now, this method has been extended to the categories of integrable equation, including sine- Gordon equation, Camassa - Holm equation, Thirring equations, KP equation, discrete Toda equation, Ablowitz - Ladik equation, etc. Page 9 of 46
The concepts and mathematics knowledge used in algebraic geometry solutions: complex function (residue, Lagrange interpolation formula) Operator spectrum theory (gradual spectrum parameters, Baker - Akhiezer function) Algebra curve theory ( hyperelliptic curve with finite genus N, meromorphic function, in addition to the Abel maps and Riemann theta function) Integrable system (zero curvature, separation of variables, recursive method, polynomial Mateev - Its trace formula, Dubrovin equations) Page 10 of 46
Fritz Gesztesy Augest, 1957, Mathematician. Department of Mathematics, University of Missouri Columbia, USA Differential equations, Operator spectral theory, Integrable systems Works: Soliton Equations and Their Algebro-Geometric Solutions Vol I: (1+1)-Dimensional Continuous Models, Cambridge University Press, Cambridge 2003 Soliton Equations and Their Algebro-Geometric Solutions Volume II: (1+1)-Dimensional Discrete Models Cambridge University Press, Cambridge 2008 Solvable Models in Quantum Mechanics: American Mathematical Society, Providence, RI, USA, 2005 Page 11 of 46
2 In this talk, we will see how to construct the Hamiltonian structure and search for the algebro-geometric solution of the following coupled 1 + 1-dimensional soliton equations as an example: q t = 1 2 (r xx 3q 2 r x + q x r 2 + 2qrr x ), r t = 1 2 (q xx + 3r 2 r x r x q 2 2qrq x ).. (1.1) Page 12 of 46
3 The hierarchy and Lax pairs of the coupled 1 + 1-dimensional soliton equations Consider the spectral problem : ( ) λ q + r ψ x = Uψ, U = λ(q r) λ. (1.2) and the auxiliary problem: ( ψ tm = V (m) ψ, V (m) = V (m) 11 V (m) 12 V (m) 21 V (m) 11 ). (1.3) Page 13 of 46
Where V (m) 11 = V (m) 12 = V (m) 21 = m j=0 m j=0 m j=0 S (3) j λ m+1 j, S (2) j λ m j, S (1) j λ m+1 j. Page 14 of 46
Then the compatibility condition of Eq.(1.2) and Eq.(1.3) is U tm V x (m) +[U, V (m) ] = 0, which is equivalent to the hierarchy of nonlinear evolution equations q tm = 1 2 (S(2) mx + S mx), (1) In brief, r tm = 1 2 (S(2) mx S (1) mx). (q tm, r tm ) T = X m, m 0. (1.4) X m = 1 ( ) ( ) S m (2) 2 S m (1). Page 15 of 46
The first two nontrivial equations are q t0 = q x, r t0 = r x. (1.5) and q t1 = 1 2 (r xx 3q 2 r x + q x r 2 + 2qrr x ), r t1 = 1 2 (q xx + 3r 2 r x r x q 2 2qrq x ). (1.6) Page 16 of 46
4 The Hamiltonian structures Let ( ) V = V = V11 V 12 V 21 V 11 where V 11 = j 0 S (3) j λ j+1, V 12 = j 0 (2.1) S (2) j λ j, V 21 = j 0 S (1) j λ j+1. It is easy to calculate tr(v U λ ) = 2V 11 + (q r)v 12 = (2S (3) j + (q r)s (2) j 1 tr(v U q ) = λv 12 + V 21 = (S (2) j + S (1) j )λ j+1, j 0 tr(v U r ) = λv 12 + V 21 = j 0 ( S (2) j + S (1) j )λ j+1. (2.2) j 1 )λ j+1 + 2 Page 17 of 46
According to the trace identity[tu paper,1989,jmp.], we have ) (2V 11 +(q r)v 12 ) = (λ s λ λs ) ( δ δq δ δr ( ) λv12 + V 21 λv 12 + V 21 Page 18 of 46
Comparing the coefficients of λ j+1, we obtain ) ( (2S (3) j +(q r)s (2) j 1 ) = ( j+2+s) ( δ δq δ δr S (2) j 1 + S(1) j 1 S (2) j 1 + S(1) j 1 ), Page 19 of 46
we set j = 1, and then get s = 1, and ( δ δq δ δr ) H j = ( S (2) j 1 + S(1) j 1 S (2) j 1 + S(1) j 1 where H j = 2S(3) j +(q r)s (2) j 1 j+1. ) = ( 1 1 1 1 ) ( S (2) j 1 (2.3) S (1) j 1 ), Page 20 of 46
Thus the soliton equations (1.4) can be expressed the following form : ( ) qr = 1 ( ) ( ) ( S (2) δ ) m t m 2 S m (1) = J δq δ H m+1, m 0, δr (2.4) where J = 1 ( ) 0 2 0. In speciality, the Hamiltonian structure of equation (1.1) is(m = 1): H 2 = 1 8 (q+r)2 (q r) 2 qq x 2rr x + qr x 2 + 1 (q x r x )r x. 2 Page 21 of 46
5 The Dubrovin-type equations Let s consider the function det W which is a (2N + 2)th-order polynomial in λ with constant coefficients of the x- flow and t m -flow: det W = f 2 + gh = 2N+2 j=1 (λ λ j ) = R(λ), (3.1) Page 22 of 46
From (3.1) we see that: f λ=uk = R(u k ), f λ=vk = R(v k ). (3.2) Again we also obtain: g x λ=uk = (q+r)u kx h x λ=vk = (q r)v kx N j=1,j k N j=1,j k (u k u j ) = 2(q+r)f λ=uk, (v k v j ) = 2v k (q r)f λ=vk, Page 23 of 46
which together with (3.2) gives u kx = 2 R(u k ), 1 k N (3.3) N (u k u j ) j=1,j k v k,x = 2 R(v k ), 1 k N (3.4) N (v k v j ) j=1,j k Page 24 of 46
In a way similar to the above expression, by using (2.4)(m = 1, t 1 = t), we arrive at the evolution of {u k } and {v k } along the t m flow: Page 25 of 46
u k,t = 2f λ=u k V (1) (q+r) N j=1,j k 12 λ=u k (u k u j ) = 2 R(u k )[u k 1 2 (q+r)(q r)+1 2 ln(q+r)] = N j=1,j k (u k u j ) 2 R(u k )[u k ( N u j +α 1 )] N j=1,j k j=1 (u k u j ), (3.5) Page 26 of 46
v k,t = 2f λ=v k V (1) (q r)v k N j=1,j k 21 λ=v k (v k v j ) = 2 R(v k )[v k 1 2 (q+r)(q r) 1 2 ln(q r)] N (v k v j ) = j=1,j k 2 R(v k )[v k ( N v j +α 1 )] N j=1,j k j=1 (v k v j ). (3.6) Page 27 of 46
Therefore, if the (2N + 2) distinct parameters λ 1, λ 2,..., λ 2N+2 are given, and let u k (x, t) and v k (x, t) be distinct solutions of ordinary differential equations(3.3), (3.4), (3.5) and (3.6), then (q, r) determined by (A) is a solution of the coupled 1 + 1- dimensional equations (1.1). Page 28 of 46
6 Algebro-geometric solutions In this section, we will give the algebro-geometric solutions of the coupled 1 + 1-dimensional Equations (1.1). To this end, We first introduce the Riemann surface Γ of the hyperelliptic curve Γ : ζ 2 = R(λ), R(λ) = 2N+2 j=1 (λ λ j ), with genus N on Γ. On Γ there are two infinite points 1 and 2, which are not branch points of Γ. We equip Γ with a canonical basis of cycles: a 1, a 2,..., a N ; b 1, b 2,..., b N which are independent and have intersection numbers as follows: Page 29 of 46
a i a j = 0, b i b j = 0, a i b j = δ ij, i, j = 1, 2,..., N. We will choose the following set as our basis: ω l = λl 1 dλ R(λ), l = 1, 2,..., N, which are linearly independent from each other on Γ, and let A ij = ω i, B ij = ω i. a j b j It is possible to show that the matrices A = (A ij ) and B = (B ij ) are N N invertible matrices [?,?]. Page 30 of 46
Now we define the matrices C and τ by C = (C ij ) = A 1, τ = (τ ij ) = A 1 B. Then the matrix τ can be shown to symmetric (τ ij = τ ji ) and it has a positivedefinite imaginary part(im τ > 0).If we normalize ω j into the new basis ω j : ω j = Then we have: a j ω j = b j ω i = N C jl ω l, l = 1, 2,..., N. l=1 N C jl ω l = a j l=1 N C jl ω l = b j l=1 N C jl A li = δ ji l=1 N C jl B li = τ ji l=1 Page 31 of 46
Using B = (B jk ) g g is a symmetry matrix and ImB > 0(Positive definite) We can define theta function in the algebraic curve as follows: θ(ζ, B) = m Z g expπ 1(< Bm, m > +2 < ζ, m >), ζ C g For arbitrary m Z g, we have (1) θ( ζ, B) = θ(ζ, B); (2) θ(ζ + m, B) = θ(ζ, B) (3) θ(ζ + mb, B) = θ(ζ, B)e 1π(Bm,m) 2 1π(m,ζ) Page 32 of 46
As a series in the above definition, θ function of the compact set of C g is an uniform convergence and analytical function. Unit matrix (δ jk ) g g and symmetry matrix (B jk ) g g can be called Quasi periodic of θ function. Page 33 of 46
Now we introduce the Abel-Jacobi coordinates as follows: N p(uk ρ (1) (x,t)) N N uk λ l 1 dλ j (x, t) = ω j = C jl, k=1 p 0 k=1 l=1 λ(p 0 ) R(λ) (5.1) N p(vk ρ (2) (x,t)) N N vk (x,t) λ l 1 dλ j (x, t) = ω j = C jl, k=1 p 0 k=1 l=1 λ(p 0 ) R(λ) (5.2) where p(u k (x, t)) = (u k, R(u k )), p(v k (x, t)) = (v k, R(v k )), and λ(p 0 ) is the local coordinate of p 0. Page 34 of 46
From above, we get N N x ρ (1) u l 1 k u N kx j = C jl R(uk ) = k=1 which implies l=1 k=1 N l=1 C jl u l 1 k, N (u k u j ) j=1,j k x ρ (1) j = 2C jn = Ω (1) j, j = 1, 2,..., N. (5.3) With the help of the following equality: Page 35 of 46
N k=1 N j=1,j k u l 1 k (u k u j ) = δ ln, l = 1, 2,..., N. In a similar way, we obtain from (5.1), (5.2), (4.9), (4.10), (4.11) and (4.12): t ρ (1) j = 2C j,n 1 + 2α 1 C j,n = Ω (2) j, (5.4) x ρ (2) j = Ω (1) j, j = 1, 2,..., N. (5.5) t ρ (2) j = Ω (2) j, j = 1, 2,..., N. (5.6) Page 36 of 46
On the basis of these results, we obtain the following: ρ (1) j (x, t) = Ω (1) j x + Ω (2) j t + γ (1) j (5.7) ρ (2) j (x, t) = Ω (1) j x Ω (2) j t + γ (2) j (5.8) Where γ (i) (i = 1, 2) are constants, and γ (1) j = j N k=1 p(ũk (0,0)) p 0 ω j, γ (2) j = N k=1 p(ṽk (0,0)) ω j ρ (1) = (ρ ((1) 1, ρ (1) 2..., ρ (1) N )T, ρ (2) = (ρ ((2) 1, ρ (2) 2,..., ρ(2) N )T. Ω (m) = (Ω (m) 1, Ω (m) 2,..., Ω (m) )T, γ (m) = (γ (m) 1, γ (m) N p 0 2,..., γ (m) Page 37 of 46 N )T, m
Now we introduce the Abel map A(p): A(p) = p p 0 ω, ω = (ω 1, ω 2,..., ω N ) T, A( p k ) = n k A(p k ), k and Abel-Jacobi coordinates: N N p(uk ρ (1) ) = A( p(u k )) = ω, p 0 ρ (2) = A( k=1 N p(v k )) = k=1 k=1 N k=1 p(vk ) p 0 ω, According to the Riemann theorem[?,?], there exists a Riemann constant vector M C N such that the function: Page 38 of 46
F (m) (λ) = θ(a(p(λ)) ρ (m) M (m) ), m = 1, 2. has exactly N zeros at u 1, u 2,..., u N for m = 1 or v 1, v 2,..., v N for m = 2. To make the function single valued, the surface Γ is cut along all a k, b k to form a simple connected region, whose boundary is denoted by γ. By Refs.[1.2], the integrals I(Γ) = 1 2πi γ λd ln F (m), m = 1, 2 are constants independent of ρ (1) and ρ (2) with N I = I(Γ) = λω j a j j=1 Page 39 of 46
By the residue theorem, we have: N 2 u j = I Res λ= s λd ln F (1) (λ) (5.9) j=1 N v j = I j=1 s=1 2 Res λ= s λd ln F (2) (λ) (5.10) s=1 Here we need only compute the residues in (5.9) and (5.10). In a way similar to calculations in [?], we arrive at Page 40 of 46
Res λ= s λd ln F (m) (λ) = ( 1) s+m ln θ (m) Where s, m = 1, 2; s = 1, 2. (5.11) θ s (1) = θ(ω (1) x+ω (2) t+ξ s ), θ s (2) = θ( Ω (1) x Ω (2) t+η s ), ξ s and η s are constants. Thus from (5.9)-(5.11), we arrive at N j=1 u j = I ln θ(1) 1 θ (1) 2, N j=1 v j = I ln θ(2) 2 θ (2) 1. (5.12) Page 41 of 46
Substituting (5.12) into (4.4), then we get an algebrogeometric solution for the coupled 1 + 1-dimensional soliton equations(1.1): q = A(t) 2 exp( 1 2N+2 j=1 λ j + 2 1 I 2 ln θ(2) 2 + A(t) 2 exp( 1 2N+2 θ (2) 1 j=1 λ j + 2 1 I 2 ln θ(2) 1 θ (1) 2 1 exp( 2 ln θ(2) 2 + 1 exp( 2 ln θ(2) 2 Page 42 of 46 θ (2) 1 θ (2) 1
r = A(t) 2 exp( 1 2N+2 A(t) 2 exp( 1 2N+2 j=1 λ j + 2 1 I 2 ln θ(2) 1 θ (1) 2 j=1 λ j + 2 1 I 2 ln θ(2) 2 + 1 exp( 2 ln θ(2) 2 θ(1) 2 θ (2) 1 where A(t) is arbitrary complex functions about variable t. 1 exp( 2 ln θ 2 Page 43 of 46 θ (2) 1 θ(1) 1 θ 1
Arbitrary really 3 3 spectral problem, it is very difficult to calculate its algebraic geometry solutions. Page 44 of 46
[1].C. L. Siegel, Topics in Complex Function Theory, vol. 2, John Wiley and Sons, New York, NY,USA, 1971. [14] P.Griffiths and J.Harris, [2].Principles of Algebraic Geometry,Wiley Classics Library, John Wiley and Sons, New York, NY, USA, 1994. Page 45 of 46
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