Symbolic Computation of Lax Pairs of Systems of Partial Difference Equations Using Consistency Around the Cube

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1 Symbolic Computation of Lax Pairs of Systems of Partial Difference Equations Using Consistency Around the Cube Willy Hereman & Terry Bridgman Department of Applied Mathematics and Statistics Colorado School of Mines Golden, Colorado, U.S.A. home/whereman/ Computational Aspects of Moving Frames SIAM Conference on Applied Algebraic Geometry Fort Collins, Colorado Saturday, August 3, 2013, 5:30p.m.

2 Collaborators Reinout Quispel & Peter van der Kamp La Trobe University, Melbourne, Australia Research supported in part by NSF under Grant CCF This presentation is made in TeXpower

3 Outline What are nonlinear P Es? Lax pairs of nonlinear PDEs Lax pair of nonlinear P Es Algorithm to compute Lax pairs of P Es (Nijhoff 2001, Bobenko & Suris 2001) Additional examples Conclusions and future work

4 What are nonlinear P Es? Nonlinear maps with two (or more) lattice points Some origins: full discretizations of PDEs discrete dynamical systems in 2 dimensions superposition principle (Bianchi permutability) for Bäcklund transformations between 3 solutions (2 parameters) of a completely integrable PDE Example 1: discrete potential Korteweg-de Vries (pkdv) equation (u n,m u n+1,m+1 )(u n+1,m u n,m+1 ) p 2 + q 2 = 0

5 u is dependent variable or field (scalar case) n and m are lattice points p and q are parameters Notation: (u n,m, u n+1,m, u n,m+1, u n+1,m+1 ) = (x, x 1, x 2, x 12 ) Alternate notations (in the literature): (u n,m, u n+1,m, u n,m+1, u n+1,m+1 ) = (u, ũ, û, ˆũ) (u n,m, u n+1,m, u n,m+1, u n+1,m+1 ) = (u 00, u 10, u 01, u 11 ) Example 1: discrete pkdv equation (u n,m u n+1,m+1 )(u n+1,m u n,m+1 ) p 2 + q 2 = 0 Short: (x x 12 )(x 1 x 2 ) p 2 + q 2 = 0

6 (u n,m u n+1,m+1 )(u n+1,m u n,m+1 ) p 2 + q 2 = 0 Short: (x x 12 )(x 1 x 2 ) p 2 + q 2 = 0 x 2 = u n,m+1 p x 12 = u n+1,m+1 q q x = u n,m p x 1 = u n+1,m

7 Systems of P Es Example 2: Schwarzian-Boussinesq System z 1 y x 1 + x = 0 z 2 y x 2 + x = 0 z y 12 (y 1 y 2 ) y(p y 2 z 1 q y 1 z 2 ) = 0 Note: System has two single-edge equations and one full-face equation.

8 Concept of Consistency Around the Cube Superposition of Bäcklund transformations between 4 solutions x, x 1, x 2, x 3 (3 parameters: p, q, k) x 23 x 123 x 2 q k x 3 x 12 x 13 x p x 1

9 Introduce a third lattice variable l View u as dependent on three lattice points: n, m, l. So, x = u n,m x = u n,m,l Move in three directions: n n + 1 over distance p m m + 1 over distance q l l + 1 over distance k (spectral parameter) Require that the same lattice holds on the front, bottom, and left faces of the cube Require consistency for the computation of x 123 = u n+1,m+1,l+1 (3 ways same answer)

10 x 23 x 123 x 2 q k x 3 x 12 x 13 x p x 1

11 Peter D. Lax (1926-) Seminal paper: Integrals of nonlinear equations of evolution and solitary waves, Commun. Pure Appl. Math. 21 (1968)

12 Refresher: Lax Pairs of Nonlinear PDEs Historical example: Korteweg-de Vries equation u t + α uu x + u xxx = 0 α IR Key idea: Replace the nonlinear PDE with a compatible linear system (Lax pair): ( ) ψ xx αu λ ψ = 0 ψ t + 4ψ xxx + αuψ x αu xψ = 0 ψ is eigenfunction; λ is constant eigenvalue (λ t = 0) (isospectral)

13 Lax Pairs in Operator Form Replace a completely integrable nonlinear PDE by a pair of linear equations (called a Lax pair): Lψ = λψ and D t ψ = Mψ Require compatibility of both equations L t ψ + (LM ML)ψ = 0 Lax equation: L t + [L, M] = O with commutator [L, M] = LM ML. Furthermore, L t ψ = [D t, L]ψ = D t (Lψ) LD t ψ and = means evaluated on the PDE

14 Example: Lax operators for the KdV equation L = D 2 x αu I M = ( ) 4 D 3 x + αud x αu x I Note: L t ψ + [L, M]ψ = 1 6 α (u t + αuu x + u xxx ) ψ

15 Lax Pairs in Matrix Form (AKNS Scheme) Write ψ xx + ( 1 6 αu λ ) ψ = 0 ψ t + 4ψ xxx + αuψ x αu xψ = 0 as a first-order system. Introduce, Ψ = ψ, then, D x Ψ = XΨ with ψ x X = 0 1 λ 1 6 αu 0

16 Use φ t = ψ xt = ψ tx and eliminate ψ x, ψ xx, ψ xxx,... to get D t Ψ = TΨ with 1 T = 6 αu x 4λ 1 3 αu 4λ αλu α2 u αu 2x 1 6 αu x Compatibility of D x Ψ = X Ψ D t Ψ = T Ψ yields Lax equation (zero-curvature equation): D t X D x T + [X, T] = 0 with commutator [X, T] = XT TX

17 Equivalence under Gauge Transformations Lax pairs are equivalent under a gauge transformation: If (X, T) is a Lax pair then so is ( X, T) with X = GXG 1 + D x (G)G 1 T = GTG 1 + D t (G)G 1 G is arbitrary invertible matrix and Ψ = GΨ. Thus, X t T x + [ X, T] = 0

18 Example: For the KdV equation 0 1 X = λ 1 6 αu 0 and X = ik 1 6 αu 1 ik Here, X = GXG 1 and T = GTG 1 with where λ = k 2 G = i k 1 1 0

19 Reasons to Compute a Lax Pair Compatible linear system is the starting point for application of the IST and the Riemann-Hilbert method for boundary value problems Confirm the complete integrability of the PDE Zero-curvature representation of the PDE Compute conservation laws of the PDE Discover families of completely integrable PDEs Question: How to find a Lax pair of a completely integrable PDE? Answer: There is no completely systematic method

20 Lax Pair of Nonlinear P Es Replace the nonlinear P E by ψ 1 = L ψ (recall ψ 1 = ψ n+1,m ) ψ 2 = M ψ (recall ψ 2 = ψ n,m+1 ) For scalar P Es, L, M are 2 2 matrices; ψ = f Express compatibility: g ψ 12 = L 2 ψ 2 = L 2 M ψ ψ 12 = M 1 ψ 1 = M 1 L ψ Lax equation: L 2 M M 1 L = 0

21 Equivalence under Gauge Transformations Lax pairs are equivalent under a gauge transformation If (L, M) is a Lax pair then so is (L, M) with L = G 1 L G 1 M = G 2 M G 1 where G is non-singular matrix and φ = Gψ Proof: Trivial verification that (L 2 M M 1 L) φ = 0 (L 2 M M 1 L) ψ = 0

22 Example 1: Discrete pkdv Equation (x x 12 )(x 1 x 2 ) p 2 + q 2 = 0 Lax pair: L = tl c = t x p2 k 2 xx 1 1 x 1 M = sm c = s x q2 k 2 xx 2 1 x 2 with t = s = 1 or t = 1 DetLc = 1 k 2 p 2 and s = 1 DetMc = 1. Here, t 2 k 2 q 2 t s s 1 = 1.

23 Example 2: Schwarzian-Boussinesq System z 1 y x 1 + x = 0 z 2 y x 2 + x = 0 z y 12 (y 1 y 2 ) y(p y 2 z 1 q y 1 z 2 ) = 0 Lax pair: L = tl c = t y 0 yz 1 kyy 1 pyz 1 0 z z 0 1 y 1

24 y 0 yz 2 M = sm c = s kyy 2 qyz 2 0 z z 0 1 y 2 with t = s = 1 y, or t = 1 y 1 and s = 1 y 2, or t = 3 z y 2 y 1 z 1 and s = z 3 y 2 y 2 z 2. Here, t 2 t s s 1 = y 1 y 2.

25 Lax Pair Algorithm for Scalar P Es (Nijhoff 2001, Bobenko and Suris 2001) Applies to P Es that are consistent around the cube Example 1: Discrete pkdv Equation Step 1: Verify the consistency around the cube Equation on front face of cube: (x x 12 )(x 1 x 2 ) p 2 + q 2 = 0 Solve for x 12 = x p2 q 2 x 1 x 2 = x 2x xx 1 +p 2 q 2 x 2 x 1 Compute x 123 : x 12 x 123 = x 3 p2 q 2 x 13 x 23 Equation on floor of cube: (x x 13 )(x 1 x 3 ) p 2 + k 2 = 0

26 Solve for x 13 = x p2 k 2 x 1 x 3 = x 3x xx 1 +p 2 k 2 x 3 x 1 Compute x 123 : x 13 x 123 = x 2 p2 k 2 x 12 x 23 Equation on left face of cube: (x x 23 )(x 3 x 2 ) k 2 + q 2 = 0 Solve for x 23 = x q2 k 2 x 3 x 2 Compute x 123 : x 23 x 123 = x 1 q2 k 2 x 12 x 13 x 2 x 3 = x 3x xx 2 +q 2 k 2 Verify that all three coincide: x 123 =x 1 q2 k 2 x 12 x 13 =x 2 p2 k 2 x 12 x 23 =x 3 p2 q 2 x 13 x 23 Upon substitution of x 12, x 13, and x 23 : x 123 = p2 x 1 (x 2 x 3 ) + q 2 x 2 (x 3 x 1 ) + k 2 x 3 (x 1 x 2 ) p 2 (x 2 x 3 ) + q 2 (x 3 x 1 ) + k 2 (x 1 x 2 ) Consistency around the cube is satisfied!

27

28 x 23 x 123 x 2 x 12 q x 3 x 13 k x p x 1

29 Step 2: Homogenization Numerator and denominator of x 13 = x 3x xx 1 +p 2 k 2 x 3 x 1 and x 23 = x 3x xx 2 +q 2 k 2 x 3 x 2 are linear in x 3. Substitute x 3 = f g x 13 = xf+(p2 k 2 xx 1 )g f x 1 g into x 13 to get On the other hand, x 3 = f g x 13 = f 1 g 1. Thus, x 13 = f 1 g 1 = xf+(p2 k 2 xx 1 )g f x 1 g. Hence, f 1 = t ( xf + (p 2 k 2 xx 1 )g ) and g 1 = t (f x 1 g)

30 In matrix form f 1 = t g 1 x p2 k 2 xx 1 1 x 1 Matches ψ 1 = L ψ with ψ = f g f g Similarly, from x 23 = f 2 g 2 = xf+(q2 k 2 xx 2 )g f x 2 g ψ 2 = f 2 = s g 2 x q2 k 2 xx 2 1 x 2 f = M ψ. g

31 Therefore, L = t L c = t x p2 k 2 xx 1 1 x 1 M = s M c = s x q2 k 2 xx 2 1 x 2

32 Step 3: Determine t and s Substitute L = t L c, M = s M c into L 2 M M 1 L = 0 t 2 s (L c ) 2 M c s 1 t (M c ) 1 L c = 0 Solve the equation from the (2-1)-element for t 2t s s 1 = f(x, x 1, x 2, p, q,... ) Find rational t and s. Apply determinant to get t 2 s det L c = t s 1 det (L c ) 2 det (M c ) 1 det M c Solution: t = 1 det Lc, s = 1 det Mc Always works but introduces roots!

33 The ratio t 2 t s s 1 is invariant under the change t a 1 a t, s a 2 a s, where a(x) is arbitrary. Proper choice of a(x) = Rational t and s. No roots needed!

34 Lax Pair Algorithm Systems of P Es (Nijhoff 2001, Bobenko and Suris 2001) Applies to systems of P Es that are consistent around the cube Example 2: Schwarzian-Boussinesq System z 1 y x 1 + x = 0 z 2 y x 2 + x = 0 z y 12 (y 1 y 2 ) y(p y 2 z 1 q y 1 z 2 ) = 0 Note: System has two single-edge equations and one full-face equation

35 Edge equations require augmentation of system with additional shifted, edge equations z 12 y 2 x 12 + x 2 = 0 z 12 y 1 x 12 + x 1 = 0 Edge equations will provide additional constraints during homogenization (Step 2).

36 Step 1: Verify the consistency around the cube System on the front face: z 1 y x 1 + x = 0 z 2 y x 2 + x = 0 z 12 y 2 x 12 + x 2 = 0 z 12 y 1 x 12 + x 1 = 0 zy 12 (y 1 y 2 ) y(py 2 z 1 qy 1 z 2 ) = 0

37 Solve for x 12, y 12, and z 12 : x 12 = x 2y 1 x 1 y 2 y 1 y 2 y 12 = (x 2 x 1 )(qy 1 z 2 py 2 z 1 ) z(y 1 y 2 )(z 1 z 2 ) z 12 = x 2 x 1 y 1 y 2 Compute x 123, y 123, and z 123 : x 123 = x 23y 13 x 13 y 23 y 13 y 23 y 123 = py 23y 3 z 13 qy 13 y 3 z 23 z 3 (y 13 y 23 ) z 123 = x 23 x 13 y 13 y 23

38 System on the bottom face: z 1 y x 1 + x = 0 z 3 y x 3 + x = 0 zy 13 (y 1 y 3 ) y(py 3 z 1 ky 1 z 3 ) = 0 Solve for x 13, y 13, and z 13 : x 13 = x 3y 1 x 1 y 3 y 1 y 3 z 13 y 3 x 13 + x 3 = 0 z 13 y 1 x 13 + x 1 = 0 y 13 = (x 2 x 1 )(ky 1 z 3 py 3 z 1 ) z(y 1 y 3 )(z 1 z 2 ) z 13 = x 3 x 1 y 1 y 3

39 Compute x 123, y 123, and z 123 : x 123 = x 23y 12 x 12 y 23 y 12 y 23 y 123 = py 2y 23 z 12 ky 12 y 2 z 23 z 2 (y 12 y 23 ) z 123 = x 23 x 12 y 12 y 23

40 System on the left face: z 3 y x 3 + x = 0 z 2 y x 2 + x = 0 z 23 y 2 x 23 + x 2 = 0 z 23 y 3 x 23 + x 3 = 0 zy 23 (y 3 y 2 ) y(py 2 z 3 qy 3 z 2 ) = 0 Solve for x 23, y 23, and z 23 : x 23 = x 3y 2 x 2 y 3 y 2 y 3 y 23 = (x 2 x 1 )(ky 2 z 3 qy 3 z 2 ) z(y 2 y 3 )(z 1 z 2 ) z 23 = x 3 x 2 y 2 y 3

41 Compute x 123, y 123, and z 123 : x 123 = x 13y 12 x 12 y 13 y 12 y 13 y 123 = qy 1y 13 z 12 ky 1 y 12 z 13 z 1 (y 12 y 13 ) z 123 = x 13 x 12 y 12 y 13 Substitute x 12, y 12, y 12, x 13, y 13, z 13, x 23, y 23, z 23 into the above to get

42 x 123 = (pz 1(x 3 y 2 x 2 y 3 )+ qz 2 (x 1 y 3 x 3 y 1 )+ kz 3 (x 2 y 1 x 1 y 2 ) (pz 1 (y 2 y 3 ) + qz 2 (y 3 y 1 ) + kz 3 (y 1 y 2 ) y 123 = pqy 3z 1 (x 2 x 1 )+kpy 2 z 1 (x 1 x 3 )+kqy 1 (x 3 z 2 x 1 z 2 +x 1 z 3 x 2 z 3 ) z 1 (pz 1 (y 2 y 3 ) + qz 2 (y 3 y 1 ) + kz 3 (y 1 y 2 )) z 123 = z(z 1 z 2 )(x 3 (y 1 y 2 ) + x 1 (y 2 y 3 )+ x 2 (y 3 y 1 )) (x 1 x 2 )(pz 1 (y 2 y 3 ) + qz 2 (y 3 y 1 ) + kz 3 (y 1 y 2 )) Answer is unique and independent of x and y. Consistency around the cube is satisfied!

43 Step 2: Homogenization Observed that x 3, y 3 and z 3 appear linearly in numerators and denominators of x 13 = y 1(x + yz 3 ) y 3 (x + yz 1 ) y 1 y 3 y 13 = pyy 3z 1 kyy 1 z 3 z(y 1 y 3 ) z 13 = y(z 3 z 1 ) y 1 y 3 x 23 = y 2(x + yz 3 ) y 3 (x + yz 2 ) y 2 y 3 y 23 = qyy 3z 2 kyy 2 z 3 z(y 2 y 3 ) z 23 = y(z 3 z 2 ) y 2 y 3

44 Substitute x 3 = h H, y 3 = g G, and z 3 = f F. Use constraints (from left face edges) z 1 y x 1 + x = 0, z 2 y x 2 + x = 0 z 3 y x 3 + x = 0 Solve for x 3 = z 3 y + x (since x 3 appears linearly). Thus, x 3 = fy+f x F, y 3 = g G, and z 3 = f F.

45 Substitute x 3, y 3, z 3 into x 13, y 13, z 13 : x 13 = F gx F Gxy 1 fgyy 1 + F gyz 1 F (g Gy 1 ) y 13 = y(fgky 1 F gpz 1 ) F (g Gy 1 )z z 13 = Gy(F z 1 f) F (g Gy 1 ) Require that numerators and denominators are linear in f, F, g, and G. That forces G = F. Hence, x 3 = fy+f x F, y 3 = g F, and z 3 = f F.

46 Compute Hence, x 3 = fy + F x F y 3 = g F y 13 = g 1 F 1 z 3 = f F z 13 = f 1 F 1 x 13 = f 1y 1 + F 1 x 1 F 1 x 13 = fyy 1 + g(x + yz 1 ) F xy 1 g F y 1 = f 1y 1 + F 1 x 1 F 1 y 13 = y(fky 1 gpz 1 ) (g F y 1 )z = g 1 F 1 z 13 = y( f + F z 1) g F y 1 = f 1 F 1

47 Note that x 13 = fyy 1 + g(x + yz 1 ) F xy 1 g F y 1 = f 1y 1 + F 1 x 1 F 1 is automaticlly satisfied as a result of the relation x 3 = z 3 y + x. Write in matrix form: f 1 y 0 yz 1 ψ 1 = g 1 = t kyy 1 pyz 1 0 z z 0 1 y 1 F 1 f g F = L ψ Repeat the same steps for x 23, y 23, z 23 to obtain

48 ψ 2 = f 2 g 2 F 2 = t y 0 yz 2 kyy 2 qyz 2 0 z z 0 1 y 2 f g F = M ψ Therefore, y 0 yz 1 L = tl c = t kyy 1 pyz 1 0 z z 0 1 y 1 y 0 yz 2 M = sm c = s kyy 2 qyz 2 0 z z 0 1 y 2

49 Step 3: Determine t and s Substitute L = t L c, M = s M c into L 2 M M 1 L = 0 t 2 s (L c ) 2 M c s 1 t (M c ) 1 L c = 0 Solve the equation from the (2-1)-element: t 2t s s 1 = y 1 y 2. Thus, t = s = 1 y, or t = 1 y 1 and s = 1 y 2, or t = 3 z y 2 y 1 z 1 and s = 3 z y 2 y 2 z 2.

50 Additional Examples Example 3: Lattice due to Hietarinta (2011) x 1 z y 1 x = 0 x 2 z y 2 x = 0 z 12 y x 1 ( px1 qx ) 2 = 0 x z 1 z 2 Note: System has two single-edge equations and one full-face equation. Lax pair: L = t yz x k x kx px 1 z yzz 1 x x 1 z z 1 xz 1 z 0 zz 1

51 and M = s yz x k x kx qx 2 z yzz 2 x x 2 z z 2 xz 2, z 0 zz 2 where t = s = 1 z, or t = 1 z 1, s = 1 z 2, or t = 3 x x 1 z 2 z 1, s = 3 x x 2 z 2 z 2. Here, t 2 t s s 1 = z 1 z 2.

52 Example 4: Discrete Boussinesq System (Tongas and Nijhoff 2005) z 1 xx 1 + y = 0 z 2 xx 2 + y = 0 (x 2 x 1 )(z xx 12 + y 12 ) p + q = 0 Lax pair: L = t x y p k xy 1 + x 1 z z x

53 x M = s y q k xy 2 + x 2 z z x with t = s = 1, or t = 1 3 p k and s = 1 Note: x 3 = f F, y 3 = g F, and ψ = 3 q k. f F, and t 2 t g s s 1 = 1.

54 Example 5: System of pkdv Lattices (Xenitidis and Mikhailov 2009) (x x 12 )(y 1 y 2 ) p 2 + q 2 = 0 (y y 12 )(x 1 x 2 ) p 2 + q 2 = 0 Lax pair: 0 0 tx t(p 2 k 2 xy 1 ) 0 0 t ty L = 1 T y T (p 2 k 2 x 1 y) 0 0 T T x 1 0 0

55 M = 0 0 sx s(q 2 k 2 xy 2 ) 0 0 s sy 2 Sy S(q 2 k 2 x 2 y) 0 0 S Sx with t = s = T = S = 1, or t T = 1 DetLc = 1 p k and s S = 1 DetMc = 1 q k. Note: t 2 T or T 2 T S s 1 = 1 and T 2 t s S 1 = 1, S S 1 = 1, with T = t T, S = s S. Here, x 3 = f F, y 3 = g G, and ψ = [f F g G]T.

56 Example 6: Discrete NLS System (Xenitidis and Mikhailov 2009) y 1 y 2 y ( (x 1 x 2 )y + p q ) = 0 ( x 1 x 2 + x 12 (x1 x 2 )y + p q ) = 0 Lax pair: L = t 1 x 1 y k p yx 1 M = s 1 x 2 y k q yx 2

57 with t = s = 1, or t= 1 DetLc = 1 α k and s= 1 β k. Note: x 3 = f F, ψ = f F, and t 2 t s s 1 = 1.

58 Example 7: Schwarzian Boussinesq Lattice (Nijhoff 1999) z 1 y x 1 + x = 0 z 2 y x 2 + x = 0 z y 12 (y 1 y 2 ) y(p y 2 z 1 q y 1 z 2 ) = 0 Lax pair: L = t y 0 yz 1 pyz 1 0 z z 0 1 y 1 kyy 1

59 M = s y 0 yz 2 pyz 2 0 z z 0 1 y 2 kyy 2 with t = s = 1 y, or t = 1 y 1 and s = 1 y 2, or t = z 3 y 2 y 1 z 1 and s = z 3 y 2 y 2 z 2. Note: x 3 = fy+f x F, y 3 = g F, z 3 = f F, ψ = f g F, and t 2 t s s 1 = y 1 y 2.

60 Example 9: Toda modified Boussinesq System (Nijhoff 1992) y 12 (p q + x 2 x 1 ) (p 1)y 2 + (q 1)y 1 = 0 y 1 y 2 (p q z 2 + z 1 ) (p 1)yy 2 + (q 1)yy 1 = 0 y(p + q z x 12 )(p q + x 2 x 1 ) (p 2 + p + 1)y 1 + (q 2 + q + 1)y 2 = 0 Lax pair: 1+k+k k + p z 2 y L=t 0 p 1 (1 k)y p k x 1 k 2 y y 1 p 2 (y 1 y) ky(x 1 z)+yzx 1 y p(y 1+yx 1 +yz) y

61 1+k+k k + q z 2 y M =s 0 q 1 (1 k)y q k x 2 k 2 y y 2 q 2 (y 2 y) ky(x 2 z)+yzx 2 y q(y 2+yx 2 +yz) y with t = s = 1, or t = 3 y1 y and s = 3 y2 f Note: x 3 = f F, y 3 = g F, ψ = g F y. Here, t 2 t s s 1 = 1.

62 Conclusions and Future Work Mathematica code works for scalar P Es in 2D defined on quad-graphs (quadrilateral faces). Mathematica code has been extended to systems of P Es in 2D defined on quad-graphs. Code can be used to test (i) consistency around the cube and compute or test (ii) Lax pairs. Consistency around cube = P E has Lax pair. P E has Lax pair consistency around cube. Indeed, there are P Es with a Lax pair that are not consistent around the cube. Example: discrete sine-gordon equation.

63 Avoid the determinant method to avoid square roots! Factorization plays an essential role! Hard cases: Q4 equation (elliptic curves, Weierstraß functions) (Nijhoff 2001) and Q5 Future Work: Extension to more complicated systems of P Es. P Es in 3D: Lax pair will be expressed in terms of tensors. Consistency around a hypercube. Examples: discrete Kadomtsev-Petviashvili (KP) equations.

64 Thank You

65 Additional Examples Example 9: (H1) Equation (ABS classification) (x x 12 )(x 1 x 2 ) + q p = 0 Lax pair: L = t x p k xx 1 1 x 1 M = s x q k xx 2 1 x 2 with t = s = 1 or t = 1 k p and s = 1 k q Note: t 2 s t s 1 = 1.

66 Example 10: (H2) Equation (ABS 2003) (x x 12 )(x 1 x 2 )+(q p)(x+x 1 +x 2 +x 12 )+q 2 p 2 =0 Lax pair: L = t p k + x p2 k 2 + (p k)(x + x 1 ) xx 1 1 (p k + x 1 ) M = s with t = Note: t 2 t q k + x q2 k 2 + (q k)(x + x 2 ) xx 2 1 (q k + x 2 ) 1 and s = 1 2(k p)(p+x+x1 ) 2(k q)(q+x+x2 ) s s 1 = p+x+x 1 q+x+x 2.

67 Example 11: (H3) Equation (ABS 2003) p(xx 1 + x 2 x 12 ) q(xx 2 + x 1 x 12 ) + δ(p 2 q 2 ) = 0 Lax pair: L = t kx ( δ(p 2 k 2 ) ) + pxx 1 p kx 1 M = s kx ( δ(q 2 k 2 ) ) + qxx 2 q kx 2 with t = Note: t 2 t 1 and s = 1 (p 2 k 2 )(δp+xx 1 ) (q 2 k 2 )(δq+xx 2 ) s s 1 = δp+xx 1 δq+xx 2.

68 Example 12: (H3) Equation (δ = 0) (ABS 2003) p(xx 1 + x 2 x 12 ) q(xx 2 + x 1 x 12 ) = 0 Lax pair: L = t kx pxx 1 p kx 1 M = s kx qxx 2 q kx 2 with t = s = 1 x or t = 1 x 1 and s = 1 x 2 Note: t 2 s t s 1 = x x 1 x x 2 = x 1 x 2.

69 Example 13: (Q1) Equation (ABS 2003) p(x x 2 )(x 1 x 12 ) q(x x 1 )(x 2 x 12 )+δ 2 pq(p q) = 0 Lax pair: L = t (p k)x 1 + kx p ( δ 2 ) k(p k) + xx 1 p ( ) (p k)x + kx 1 M = s (q k)x 2 + kx q ( δ 2 ) k(q k) + xx 2 q ( ) (q k)x + kx 2 1 with t = δp±(x x 1 ) and s = 1 δq±(x x 2 ), 1 or t = k(p k)((δp+x x 1 )(δp x+x 1 )) and s = 1 k(q k)((δq+x x 2 )(δq x+x 2 )) Note: t 2 t s s 1 = q (δp+(x x 1 ))(δp (x x 1 )) p(δq+(x x 2 ))(δq (x x 2 )).

70 Example 14: (Q1) Equation (δ = 0) (ABS 2003) p(x x 2 )(x 1 x 12 ) q(x x 1 )(x 2 x 12 ) = 0 which is the cross-ratio equation (x x 1 )(x 12 x 2 ) (x 1 x 12 )(x 2 x) = p q Lax pair: L = t (p k)x 1 + kx pxx 1 p ( ) (p k)x + kx 1 M = s (q k)x 2 + kx qxx 2 q ( ) (q k)x + kx 2

71 Here, t 2 t or t = s s 1 = q(x x 1) 2 p(x x 2. So, t = 1 ) 2 x x 1 and s = 1 x x 2 1 and s = 1. k(k p)(x x1 ) k(k q)(x x2 )

72 Example 15: (Q2) Equation (ABS 2003) p(x x 2 )(x 1 x 12 ) q(x x 1 )(x 2 x 12 )+pq(p q) (x+x 1 +x 2 +x 12 ) pq(p q)(p 2 pq+q 2 )=0 Lax pair: (k p)(kp x 1 )+kx L=t p ( k(k p)(k 2 kp+p 2 ) x x 1 )+xx 1 p ( ) (k p)(kp x)+kx 1 (k q)(kq x 2 )+kx M =s q ( k(k q)(k 2 kq+q 2 ) x x 2 )+xx 2 q ( ) (k q)(kq x)+kx 2

73 with t = 1 k(k p)((x x 1 ) 2 2p 2 (x+x 1 )+p 4 ) and s = 1 k(k q)((x x 2 ) 2 2q 2 (x+x 2 )+q 4 ) Note: t 2 t s = q ( (x x1 ) 2 2p 2 (x + x 1 ) + p 4) s 1 p ( (x x 2 ) 2 2q 2 (x + x 2 ) + q 4) = p ( (X + X 1 ) 2 p 2) ( (X X 1 ) 2 p 2) q ( (X + X 2 ) 2 q 2) ( (X X 2 ) 2 q 2) with x = X 2, and, consequently, x 1 = X 2 1, x 2 = X 2 2.

74 Example 16: (Q3) Equation (ABS 2003) (q 2 p 2 )(xx 12 +x 1 x 2 )+q(p 2 1)(xx 1 +x 2 x 12 ) p(q 2 1)(xx 2 +x 1 x 12 ) δ2 4pq (p2 q 2 )(p 2 1)(q 2 1)=0 Lax pair: 4kp ( p(k 2 1)x+(p 2 k 2 ) )x 1 L=t (p 2 1)(δ 2 k 2 δ 2 k 4 δ 2 p 2 +δ 2 k 2 p 2 4k 2 pxx 1 ) 4k 2 p(p 2 1) 4kp ( p(k 2 1)x 1 +(p 2 k 2 )x ) 4kq ( q(k 2 1)x+(q 2 k 2 ) )x 2 M=s (q 2 1)(δ 2 k 2 δ 2 k 4 δ 2 q 2 +δ 2 k 2 q 2 4k 2 qxx 2 ) 4k 2 q(q 2 1) 4kq ( q(k 2 1)x 2 +(q 2 k 2 )x )

75 with t= 1 2k p(k 2 1)(k 2 p 2 )(4p 2 (x 2 +x 2 1 ) 4p(1+p2 )xx 1 +δ 2 (1 p 2 ) 2 ) and s= 2k 1 q(k 2 1)(k 2 q 2 )(4q 2 (x 2 +x 2 2 ) 4q(1+q2 )xx 2 +δ 2 (1 q 2 ) 2 ).

76 Note: t 2 t s s 1 = q(q2 1) ( 4p 2 (x 2 +x 2 1 ) 4p(1+p2 )xx 1 +δ 2 (1 p 2 ) 2) p(p 2 1) ( 4q 2 (x 2 +x 2 2 ) 4q(1+q2 )xx 2 +δ 2 (1 q 2 ) 2) = q(q2 1) ( 4p 2 (x x 1 ) 2 4p(p 1) 2 xx 1 +δ 2 (1 p 2 ) 2) p(p 2 1) ( 4q 2 (x x 2 ) 2 4q(q 1) 2 xx 2 +δ 2 (1 q 2 ) 2) = q(q2 1) ( 4p 2 (x+x 1 ) 2 4p(p+1) 2 xx 1 +δ 2 (1 p 2 ) 2) p(p 2 1) ( 4q 2 (x+x 2 ) 2 4q(q+1) 2 xx 2 +δ 2 (1 q 2 ) 2)

77 where 4p 2 (x 2 +x 2 1) 4p(1+p 2 )xx 1 +δ 2 (1 p 2 ) 2 = δ 2 (p e X+X 1 )(p e (X+X1) )(p e X X 1 )(p e (X X1) ) = δ 2 (p cosh(x + X 1 )+sinh(x + X 1 )) (p cosh(x + X 1 ) sinh(x + X 1 )) (p cosh(x X 1 )+sinh(x X 1 )) (p cosh(x X 1 ) sinh(x X 1 )) with x = δ cosh(x), and, consequently, x 1 = δ cosh(x 1 ), x 2 = δ cosh(x 2 ).

78 Example 17: (Q3) Equation (δ) = 0 (ABS 2003) (q 2 p 2 )(xx 12 + x 1 x 2 ) + q(p 2 1)(xx 1 + x 2 x 12 ) p(q 2 1)(xx 2 + x 1 x 12 ) = 0 Lax pair: L=t (p2 k 2 )x 1 +p(k 2 1)x k(p 2 1)xx 1 (p 2 1)k ( (p 2 k 2 )x+p(k 2 ) 1)x 1 M =s (q2 k 2 )x 2 +q(k 2 1)x k(q 2 1)xx 2 (q 2 1)k ( (q 2 k 2 )x+q(k 2 ) 1)x 2

79 with t = 1 px x 1 and s = 1 qx x 2 or t = 1 px 1 x and s = 1 qx 2 x or t = 1 (k 2 1)(p 2 k 2 )(px x 1 )(px 1 x) and s = 1. (k 2 1)(q 2 k 2 )(qx x 2 )(qx 2 x) Note: t 2 t s s 1 = (q2 1)(px x 1 )(px 1 x) (p 2 1)(qx x 2 )(qx 2 x).

80 Example 18: (α, β)-equation (Quispel 1983) ( (p α)x (p+β)x1 ) ( (p β)x2 (p+α)x 12 ) ( (q α)x (q+β)x 2 ) ( (q β)x1 (q+α)x 12 ) =0 Lax pair: L=t (p α)(p β)x+(k2 p 2 )x 1 (k α)(k β)xx 1 (k+α)(k+β) ( (p+α)(p+β)x 1 +(k 2 p 2 )x ) M =s (q α)(q β)x+(k2 q 2 )x 2 (k α)(k β)xx 2 (k+α)(k+β) ( (q+α)(q+β)x 2 +(k 2 q 2 )x )

81 with t = 1 (α p)x+(β+p)x 1 ) and s = 1 (α q)x+(β+q)x 2 ) or t = or t = 1 (β p)x+(α+p)x 1 ) and s = 1 (β q)x+(α+q)x 2 ) 1 (p 2 k 2 )((β p)x+(α+p)x 1)((α p)x+(β+p)x 1) and s = Note: t 2 t 1 (q 2 k 2 )((β q)x+(α+q)x 2)((α q)x+(β+q)x 2) s s 1 = ( (β p)x+(α+p)x 1)((α p)x+(β+p)x 1) ((β q)x+(α+q)x 2)((α q)x+(β+q)x 2).

82 Example 19: (A1) Equation (ABS 2003) p(x+x 2 )(x 1 +x 12 ) q(x+x 1 )(x 2 +x 12 ) δ 2 pq(p q) = 0 (Q1) if x 1 x 1 and x 2 x 2 Lax pair: L = t (k p)x 1 + kx p ( δ 2 ) k(k p) + xx 1 p ( ) (k p)x + kx 1 M = s (k q)x 2 + kx q ( δ 2 ) k(k q) + xx 2 q ( ) (k q)x + kx 2

83 with t = 1 k(k p)((δp+x+x 1 )(δp x x 1 )) and s = 1 k(k q)((δq+x+x 2 )(δq x x 2 )) Note: t 2 t s s 1 = q (δp+(x+x 1 ))(δp (x+x 1 )) p(δq+(x+x 2 ))(δq (x+x 2 )). Question: Rational choice for t and s?

84 Example 20: (A2) Equation (ABS 2003) (q 2 p 2 )(xx 1 x 2 x ) + q(p 2 1)(xx 2 + x 1 x 12 ) p(q 2 1)(xx 1 + x 2 x 12 ) = 0 (Q3) with δ = 0 via Möbius transformation: x x, x 1 1 x 1, x 2 1 x 2, x 12 x 12, p p, q q Lax pair: k(p L=t 2 1)x ( p 2 k 2 +p(k 2 ) 1)xx 1 p(k 2 1)+(p 2 k 2 )xx 1 k(p 2 1)x 1 k(q M =s 2 1)x ( q 2 k 2 +q(k 2 ) 1)xx 2 q(k 2 1)+(q 2 k 2 )xx 2 k(q 2 1)x 2

85 with t = and s = 1 (k 2 1)(k 2 p 2 )(p xx 1 )(pxx 1 1) 1 (k 2 1)(k 2 q 2 )(q xx 2 )(qxx 2 1) Note: t 2 t s s 1 = (q2 1)(p xx 1 )(pxx 1 1) (p 2 1)(q xx 2 )(qxx 2 1). Question: Rational choice for t and s?

86 Example 21: Discrete sine-gordon Equation xx 1 x 2 x 12 pq(xx 12 x 1 x 2 ) 1 = 0 (H3) with δ = 0 via extended Möbius transformation: x x, x 1 x 1, x 2 1 x 2, x 12 1 x 12, p 1 p, q q Discrete sine-gordon equation is NOT consistent around the cube, but has a Lax pair! Lax pair: L = p kx 1 M = k x px 1 x qx 2 x 1 kx x 2 k q