Soliton solutions to the ABS list

to the ABS list Department of Physics, University of Turku, FIN-20014 Turku, Finland in collaboration with James Atkinson, Frank Nijhoff and Da-jun Zhang DIS-INI, February 2009

The setting CAC The setting Consistency Around a Cube We consider lattice maps defined on an elementary square: x n,m = x 00 = x x n+1,m = x 10 = x [1] = x x n,m+1 = x 01 = x [2] = x x n+1,m+1 = x 11 = x [12] = x x [2] x [12] x x [1] The four corner values are related by a multi-linear equation: Q(x, x [1], x [2], x [12] ; p, q) = 0 This allows propagation from initial data given on a staircase or on a corner.

The setting Consistency Around a Cube CAC - Consistency Around a Cube One definition of integrability for such lattices is that they should allow consistent extensions from 2D to 3D (and higher) Adjoin a third direction x n,m x n,m,k and construct a cube. x 101 x 001 x 011 x 111 x 100 x 000 x 010 x 110 Use the same map (with different parameters) on each side. Given values at black disks, we can compute values at open disks uniquely, but x 111 can be computed in 3 different ways, they must agree.

ABS classification CAC The setting Consistency Around a Cube It can be used as a method to classify integrable equation. [Adler, Bobenko and Suris, Commun. Math. Phys. 233 513 (2003)]. Two additional assumptions: symmetry (ε, σ = ±1): Q(x 000, x 100, x 010, x 110 ; p, q) =εq(x 000, x 010, x 100, x 110 ; q, p) =σq(x 100, x 000, x 110, x 010 ; p, q) tetrahedron property : x 111 does not depend on x 000.

The setting Consistency Around a Cube ABS results: List H: () (x ˆ x)( x ˆx) + q p = 0, () (x ˆ x)( x ˆx) + (q p)(x + x + ˆx + ˆ x) + q 2 p 2 = 0, () p(x x + ˆx ˆ x) q(x ˆx + x ˆ x) + δ(p 2 q 2 ) = 0. List A: (A1) p(x + ˆx)( x + ˆ x) q(x + x)(ˆx + ˆ x) δ 2 pq(p q) = 0, (A2) (q 2 p 2 )(x x ˆx ˆ x +1)+q(p 2 1)(x ˆx + x ˆ x) p(q 2 1)(x x +ˆx ˆ x) = 0.

The setting Consistency Around a Cube Main list: (Q1) p(x ˆx)( x ˆ x) q(x x)(ˆx ˆ x) + δ 2 pq(p q) = 0, (Q2) p(x ˆx)( x ˆ x) q(x x)(ˆx ˆ x) + pq(p q)(x + x + ˆx + ˆ x) pq(p q)(p 2 pq + q 2 ) = 0, (Q3) (q 2 p 2 )(x ˆ x + x ˆx) + q(p 2 1)(x x + ˆx ˆ x) p(q 2 1)(x ˆx + x ˆ x) δ 2 (p 2 q 2 )(p 2 1)(q 2 1)/(4pq) = 0, (Q4) (the root model from which others follow) a 0 x x ˆx ˆ x + a 1 (x x ˆx + x ˆx ˆ x + ˆx ˆ xx + ˆ xx x) + a 2 (x ˆ x + x ˆx)+ ā 2 (x x + ˆx ˆ x) + ã 2 (x ˆx + x ˆ x) + a 3 (x + x + ˆx + ˆ x) + a 4 = 0, where the a i depend on the lattice directions and are given in terms of Weierstrass elliptic functions. This was first derived by Adler as a superposition rule of BT s for the Krichever-Novikov equation. [Adler, Intl. Math. Res. Notices, # 1 (1998) 1-4]

The setting Consistency Around a Cube J.H., JNMP 12 Suppl 2, 223 (2005). a simpler Jacobi form for (Q4) of ABS: (h 1 f 2 h 2 f 1 )[(xx [1] x [12] x [2] + 1)f 1 f 2 (xx [12] + x [1] x [2] )] + (f 2 1 f 2 2 1)[(xx [1] + x [12] x [2] )f 1 (xx [2] + x [1] x [12] )f 2 ] = 0, hi 2 = fi 4 + δfi 2 + 1, parametrizable by Jacobi elliptic functions.

Recall Hirota s direct method for constructing multisoliton solutions in the continuous case:

Recall Hirota s direct method for constructing multisoliton solutions in the continuous case: 1 find a dependent variable transformation into Hirota bilinear form

Recall Hirota s direct method for constructing multisoliton solutions in the continuous case: 1 find a dependent variable transformation into Hirota bilinear form 1 find a background or vacuum solution 2 find a 1-soliton-solution (1SS) 3 use this info to guess the transformation 2 construct the first few soliton solutions perturbatively 3 guess the general from (usually a determinant: Wronskian, Pfaffian etc) and prove it

What is a discrete Hirota bilinear form? The key property of an equation in Hirota s bilinear form is invariance under a gauge transformation or in the discrete case f i f i := e ax+bt f i. f j (n, m) f j (n, m) = An B m f j (n, m). We say an equation is in Hirota bilinear (HB) form if it can be written as c j f j (n + ν + j, m + µ + j ) g j (n + ν j, m + µ j ) = 0 j where the index sums ν + j + ν j = ν s, µ + j + µ j = µ s do not depend on j.

The background solution First problem in the perturbative approach: What is the background solution?

The background solution First problem in the perturbative approach: What is the background solution? Atkinson: Take the CAC cube and insist that the solution is a fixed point of the bar shift. The side -equations are then Q(u, ũ, u, ũ; p, r) = 0, Q(u, û, u, û; q, r) = 0. The equation is given by then the side-equations are (u ũ)(ũ û) (p q) = 0, (ũ u) 2 = r p, (û u) 2 = r q.

The side-equations then factorize as (ũ u a)(ũ u + a) = 0, (û u b)(û u + b) = 0,

The side-equations then factorize as (ũ u a)(ũ u + a) = 0, (û u b)(û u + b) = 0, Since the factor that vanishes may depend on n, m we actually have to solve ũ u = ( 1) θ a, û u = ( 1) χ b, where θ, χ Z may depend on n, m. From consistency θ, {n, 0},χ, {m, 0}. The set of possible background solution turns out to be an + bm + γ, 1 2 ( 1)n a + bm + γ, an + 1 2 ( 1)m b + γ, 1 2 ( 1)n a + 1 2 ( 1)m b + γ.

1SS CAC [Atkinson, JH and Nijhoff, JPhysA 41 (2008) 142001.] The BT generating 1SS for is (u ũ)(ũ ū) = p κ, (u ū)(ū û) = κ q. u is the background solution an + bm + γ, ū is the new 1SS, κ is the soliton parameter (the parameter in the bar-direction). We search for a new solution ū of the form ū = ū 0 + v, where ū 0 is the bar-shifted background solution ū 0 = an + bm + k + λ.

For v we then find (in the case of ) where ṽ = Ev v + F, v = Gv v + H, E = (a +k), F = (a k), G = (b +k), H = (b k), and k is related to κ by κ = r k 2.

For v we then find (in the case of ) where ṽ = Ev v + F, v = Gv v + H, E = (a +k), F = (a k), G = (b +k), H = (b k), and k is related to κ by κ = r k 2. Introducing v = f /g and Φ = (g, f ) T we can write the v-equations as matrix equations Φ(n +1, m) = N(n, m)φ(n, m), where N(n, m) = Λ Φ(n, m +1) = M(n, m)φ(n, m), ( ) ( ) E 0, M(n, m) = Λ G 0, 1 F 1 H In this case E, F, G, H are constants and we can choose Λ = Λ = 1.

Since the matrices N, M commute it is easy to find ( E n G m 0 Φ(n, m) = F n H m E n G m F n H m 2k ) Φ(0, 0).

Since the matrices N, M commute it is easy to find ( E n G m 0 Φ(n, m) = F n H m If we let ρ n,m = ( E F (ρ takes the role of e η ) E n G m F n H m 2k ) n ( ) G m ρ 0,0 = H ( a + k a k ) Φ(0, 0). ) n ( ) b + k m ρ 0,0, b k

Since the matrices N, M commute it is easy to find ( E n G m 0 Φ(n, m) = F n H m If we let ρ n,m = ( E F E n G m F n H m 2k ) n ( ) G m ρ 0,0 = H ( a + k a k (ρ takes the role of e η ) then we obtain v n,m = 2kρ n,m 1 + ρ n,m. Finally we obtain the 1SS for : u (1SS) n,m ) Φ(0, 0). ) n ( ) b + k m ρ 0,0, b k = (an + bm + λ) + k + 2kρ n,m 1 + ρ n,m.

Bilinearizing transformation In an explicit form the 1SS is u 1SS n,m = an + bm + λ + k(1 ρn,m) 1+ρ n,m

Bilinearizing transformation In an explicit form the 1SS is u 1SS n,m = an + bm + λ + k(1 ρn,m) 1+ρ n,m This suggests the dependent variable transformation u n,m = an + bm + λ g n,m f n,m.

Bilinearizing transformation In an explicit form the 1SS is u 1SS n,m = an + bm + λ + k(1 ρn,m) 1+ρ n,m This suggests the dependent variable transformation Then it is easy to show that (u ũ)(ũ û) p + q where u n,m = an + bm + λ g n,m f n,m. = [ H 1 + (a b)f f ][ + (a + b) f f ] /(f f f f ) + (a 2 b 2 ), H 1 ĝ f g f + (a b)( f f f f ) = 0, H 2 g f gf + (a + b)(f f f f ) = 0.

Casoratians CAC For given functions ϕ i (n, m, h) we define the column vectors ϕ(n, m, h) = (ϕ 1 (n, m, h), ϕ 2 (n, m, h),, ϕ N (n, m, h)) T,

Casoratians CAC For given functions ϕ i (n, m, h) we define the column vectors ϕ(n, m, h) = (ϕ 1 (n, m, h), ϕ 2 (n, m, h),, ϕ N (n, m, h)) T, and then compose the N N Casorati matrix from columns with different shifts h i, and then the determinant C n,m (ϕ; {h i }) = ϕ(n, m, h 1 ), ϕ(n, m, h 2 ),, ϕ(n, m, h N ). For example C 1 n,m(ϕ) := ϕ(n, m, 0), ϕ(n, m, 1),, ϕ(n, m, N 1) 0, 1,, N 1 N 1, C 2 n,m(ϕ) := ϕ(n, m, 0),, ϕ(n, m, N 2), ϕ(n, m, N) 0, 1,, N 2, N N 2, N,

Main result for CAC The equation (u ũ)(ũ û) (p q) = 0, is bilinearized with the substitution u n,m = an + bm + λ g n,m f n,m, p = r a 2, q = r b 2 and the resulting bilinear equations H 1 ĝ f g f + (a b)( f f f f ) = 0, H 2 g f gf + (a + b)(f f f f ) = 0. are solved by Casoratians f = N 1, g = N 2, N, with ψ given by ψ i (n, m, h; k i ) = ϱ + i k h i (a+k i ) n (b+k i ) m +ϱ i ( k i ) h (a k i ) n (b k i ) m.

The background solution/ The equation is (u ˆũ)(ũ û) + (q p)(u + ũ + û + ˆũ) + q 2 p 2 = 0. After reparametrization p = r a 2, q = r b 2, u = y 2 1 2 r the side equations lead to the background solution u (0ss) nm = [an + bm + γ] 2 1 2 r,

The 1SS/ CAC The 1SS is contructed as in the previous case using the cube. The result is where u 1SS n,m = U 2 n,m + 2kU n,m 1 ρ n,m 1 + ρ n,m + k 2, U n,m := an + bm + λ, ρ n,m = ( a + k a k ) n ( ) b + k m ρ 0,0, b k Note that in addition to the implicit dependency on the soliton parameter k through ρ, the 1SS also depends on k explicitly.

Bilinearizing transformation/ The bilinearization is given by u NSS n,m = U 2 n,m 2U n,m g f + h + s, f where f = N 1, g = N 2, N, s = N 3, N 1, N, h = N 2, N + 1, with the same entries as before. These Casoratians satisfy h s = f ( N i=1 k i 2 ) and H 1 ĝ f g f + (a b)( f f f f ) = 0, H 2 g f gf + (a + b)(f f f f ) = 0, H 3 (a + b) f g + a f g + bf g + f h f h = 0, H 4 (a b)f g + a f ĝ b f g + f ĥ f h = 0, H 5 b( f g f ĝ) + f ĥ + f s gĝ = 0.

In terms of these bilinear equations, can be given as = 5 H i P i, i=1 where [ P 1 = 4(a + b) (Ũ Ũ a 2 + b 2 ) f f Ũ f g (a + b)f g ], [ P 2 = 4 (a b)(û Ũ a 2 + b 2 ) f f + (Ũ Ũ a 2 + b 2 ) f ĝ [ P 3 = 4 [ P 4 = 4 Ũ Ũ f g (a b)ũf g ], (a b)u f f + Û f ĝ Ũ f g (a + b)(ûf f f g) + Ũ( f g f g) P 5 = 4(a 2 b 2 ) f f, ] f ĥ + f h, ],

The background solution/ The equation is δ p(uũ + û ũ) q(uû + ũ ũ) δ(q 2 p 2 ) = 0. The side equation for T (x) = x read r(u 2 +ũ 2 ) 2puũ = δ(p 2 r 2 ), r(u 2 +û 2 ) 2quû = δ(q 2 r 2 ).

The background solution/ The equation is δ p(uũ + û ũ) q(uû + ũ ũ) δ(q 2 p 2 ) = 0. The side equation for T (x) = x read r(u 2 +ũ 2 ) 2puũ = δ(p 2 r 2 ), r(u 2 +û 2 ) 2quû = δ(q 2 r 2 ). The parametrization u nm = Ae ynm + Be ynm, p = r 1 + α2 2α, q = r 1 + β2 2β, then leads to the 0SS: u (0SS) = Aα n β m + Bα n β m, AB = rδ/4.

The 1SS/ CAC The algorithmic procedure given before leads to the 1SS u 1SS n,m = Aαn β m (1 + κ 2 ρ n,m ) + Bα n β m (1 + κ 2 ρ n,m ) 1 + ρ n,m, where κ is the soliton parameter and ρ is defined by ( ) S n ( ) T m ( α 2 κ 2 ) 1 n ( β 2 κ 2 ) 1 m ρ n,m = ρ 0,0 = Ω α 2 κ 2 β 2 κ 2 ρ 0,0, This suggests the reparametrization α 2 = a c a+c, β2 = b c b+c, κ2 = k c k+c, p2 = r 2 c 2 c 2 a 2, q 2 = r 2 c 2 c 2 b 2. with a new parameter c, after which ( ) a + k n ( ) b + k m ρ n,m = ρ 0,0. a k b k

Bilinearizing transformation/ The bilinearization is given by u NSS n,m = A α n β m f f + B α n β m f f, 4AB = rδ, where f = N 1 as before but its etries are given by ψ i (n, m, h) = ϱ + i (a+k i ) n (b+k i ) m (c+k i ) h +ϱ i (a k i ) n (b k i ) m (c k i ) h, and the bar-shift is in the h-index, associated with c.

There are (at least) two bilinearizations of, one in terms of another in terms of B 1 2cf f + (a c) f f (a + f = 0, c) f B 2 2cf f + (b c) f f (b + f = 0. c) f B 1 (b + c) f f + (a c)f f (a + b) f f = 0, B 2 (c b) f f (a + c)f + (a + f = 0, f b) f B 3 (c a)(b + f + (a + c)(b f + 2c(a b)f f = 0. c) f c) f These equations can be proved with the usual Laplace expansion techniques of the Casoratians.

Conclusions CAC We have considered the construction of multisoliton solution and bilinearization for the quadrilateral lattice equations in the ABS list. The technique presented in this talk has been applied for all but Q3,Q4. The structure of the soliton solution is similar to those of the Hirota-Miwa equation Explicit NSS has also been derived for Q3, using a connection to the NQC equation. (Atkinson, JH and Nijhoff, JPhysA 41 (2008) 142001.)