Symbolic Computation of Lax Pairs of Nonlinear Systems of Partial Difference Equations using Multidimensional Consistency

Transcription

1 Symbolic Computation of Lax Pairs of Nonlinear Systems of Partial Difference Equations using Multidimensional Consistency Willy Hereman Department of Applied Mathematics and Statistics Colorado School of Mines Golden, Colorado, U.S.A. whereman/ IMCA, Lima, Peru Tuesday, November 28, 2017, 4:00p.m.

2 Collaborators Terry Bridgman (Ph.D. student, CSM) Reinout Quispel & Peter van der Kamp La Trobe University, Melbourne, Australia This presentation is made in TeXpower

3 Outline Origins of nonlinear P Es Classification of integrable nonlinear P Es in 2D Lax pairs of nonlinear PDEs & gauge equivalence Lax pair of nonlinear P Es & gauge equivalence Examples of Lax pairs of nonlinear P Es Algorithmic computation of Lax pairs (Nijhoff 2001, Bobenko & Suris 2001) Software demonstration Conclusions and future work Addendum: Additional examples

4 Origins of nonlinear P Es full discretizations of completely integrable PDEs (Ablowitz, Ladik, Taha) fully discretized bilinear equations (Hirota) direct linearization of completely integrable PDEs (Quispel, Nijhoff) superposition principle (Bianchi permutability) for auto-bäcklund transformations between solutions of a completely integrable PDE classification of multi-dimensionally consistent P Es (Adler, Bobenko, Suris)

5 Example: 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 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 )

6 Example: 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 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 Concept of Consistency Around the Cube Superposition of auto-bäcklund transformations between 4 solutions x, x 1, x 2, x 3 (3 parameters: p, q, k) x 23 x 123 x 2 x 12 q x 3 x 13 k p x x 1

8 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 P E 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)

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

10 Verification of Consistency Around the Cube Example: discrete pkdv equation 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 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 Solve for x 13 = x p2 k 2 x 1 x 3 Compute x 123 : x 13 x 123 = x 2 p2 k 2 x 12 x 23

11 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 2 x 3 Compute x 123 : x 23 x 123 = x 1 q2 k 2 x 12 x 13 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 Upon substitution of x 12, x 13, and x 23 : x 13 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!

12 Tetrahedron property 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 ) is independent of x. Connects x 123 to x 1, x 2 and x 3. x 23 x 123 x 2 x 3 x 12 q k x 13 x p x 1

13 Classification of 2D scalar nonlinear P Es Adler, Bobenko, Suris (ABS) 2003, 2007 Consider a family P Es: Q(x, x 1, x 2, x 12 ; p, q) = 0 Assumptions (ABS 2003): 1. Affine linear Q(x, x 1, x 2, x 12 ; p, q) = a 1 xx 1 x 2 x 12 + a 2 xx 1 x a 14 x 2 + a 15 x 12 + a Invariant under D 4 (symmetries of square) Q(x, x 1, x 2, x 12 ; p, q) = ɛq(x, x 2, x 1, x 12 ; q, p) ɛ, σ = ±1 3. Consistency around the cube = σq(x 1, x, x 12, x 2 ; p, q)

14 Result of the ABS Classification List H H1 (x x 12 )(x 1 x 2 ) + q p = 0 H2 (x x 12 )(x 1 x 2 )+(q p)(x+x 1 +x 2 +x 12 )+q 2 p 2 =0 H3 p(xx 1 + x 2 x 12 ) q(xx 2 + x 1 x 12 ) + δ(p 2 q 2 ) = 0

15 List A A1 p(x+x 2 )(x 1 +x 12 ) q(x+x 1 )(x 2 +x 12 ) δ 2 pq(p q)=0 A2 (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

16 List Q Q1 p(x x 2 )(x 1 x 12 ) q(x x 1 )(x 2 x 12 )+δ 2 pq(p q) = 0 Q2 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 Q3 (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

17 Q4 (mother) Hietarinta s Parametrization sn(α + β; k) (x 1 x 2 + xx 12 ) sn(α; k) (xx 1 + x 2 x 12 ) sn(β; k) (xx 2 + x 1 x 12 ) + sn(α; k) sn(β; k) sn(α + β; k)(1 + k 2 xx 1 x 2 x 12 )=0 where sn(α; k) is the Jacobi elliptic sine function with modulus k. Other parameterizations (Adler, Nijhoff, Viallet) are given in the literature.

18 Systems of P Es Example: Schwarzian-Boussinesq System y x 1 z 1 + z = 0 y x 2 z 2 + z = 0 x y 12 (y 1 y 2 ) y (p x 1 y 2 q x 2 y 1 ) = 0 System has three dependent variable x, y, and z. Thus, x = (x, y, z). System has two single-edge equations and one full-face equation. System is consistent around the cube (CAC).

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

20 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)

21 Lax Pairs in Matrix Form (AKNS Scheme) Express compatibility of where Ψ = ψ ψ x D x Ψ = X Ψ D t Ψ = T Ψ Lax equation (zero-curvature equation): D t X D x T + [X, T] = 0 with commutator [X, T] = XT TX and where = means evaluated on the PDE

22 Example: Lax pair for the KdV equation 0 1 X = λ 1 6 αu 0 1 T = 6 αu x 4λ 1 3 αu 4λ αλu α2 u αu 2x 1 6 αu x Substitution into the Lax equation yields D t X D x T + [X, T] = 1 6 α 0 0 u t + αuu x + u 3x 0

23 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 Φ = GΨ where Φ goes with (X, T), i.e., D x Φ = X Φ and D t Φ = T Φ. Thus, D t X D x T + [X, T] = 0

24 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

25 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

26 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; x 3 = f F and ψ = F f For systems of P Es, L, M are N N matrices; x 3 = f F, y 3 = g F, z 3 = h, etc., ψ =[F f g h F...]T where T is transpose.

27 Express compatibility: ψ 12 = L 2 ψ 2 = L 2 M ψ ψ 12 = M 1 ψ 1 = M 1 L ψ L ψ 2 2 ψ12 M 1 M ψ L ψ 1 Lax equation: L 2 M M 1 L = 0

28 Equivalence under Gauge Transformations Lax pairs of the same size 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 an invertible matrix, φ = Gψ goes with (L, M) i.e., φ 1 = L φ, φ 2 = M φ. Proof: Trivial verification that (L 2 M M 1 L) φ = 0 (L 2 M M 1 L) ψ = 0

29 Examples of Lax Pairs of P Es Example 1: Discrete pkdv Equation (x x 12 )(x 1 x 2 ) p 2 + q 2 = 0 Lax pair: x 3 = f, ψ = [F f]t F L = tl c = t 1 x 1 x p 2 k 2 xx 1 M = sm c = s 1 x 2 x q 2 k 2 xx 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.

30 Example 2: Schwarzian-Boussinesq System y x 1 z 1 + z = 0 y x 2 z 2 + z = 0 x y 12 (y 1 y 2 ) y (p x 1 y 2 q x 2 y 1 ) = 0 Lax pair: ψ = [F g h] T L = tl c = t y kzy 1 x pyx 1 ky 1 x x 0 z 1 y 1

31 M = sm c = s y kzy 2 x qyx 2 ky 2 x x 0 z 2 y 2 with t = s = 1 y, or t = 1 y 1 and s = 1 y 2, or t = 3 x y1 2 y x and s = 3 1 x y 2 2 y x 2. Here, t 2 t s s 1 = y 1 y 2.

32 Lax Pair Algorithm for Scalar P Es (Nijhoff 2001, Bobenko and Suris 2001) Applies to P Es that are consistent around the cube Example: Discrete pkdv Equation Step 1: Verify the consistency around the cube Use the equation on floor (x x 13 )(x 1 x 3 ) p 2 + k 2 = 0 to compute x 13 = x p2 k 2 x 1 x 3 = x 3x xx 1 +p 2 k 2 x 3 x 1 Use the equation on left face (x x 23 )(x 3 x 2 ) k 2 + q 2 = 0 to compute x 23 = x q2 k 2 x 2 x 3 = x 3x xx 2 +q 2 k 2 x 3 x 2

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

34 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 F x 13 = xf+(p2 k 2 xx 1 )F f x 1 F into x 13 to get On the other hand, x 3 = f F x 13 = f 1 F 1. Thus, x 13 = f 1 F 1 = xf+(p2 k 2 xx 1 )F f x 1 F. Hence, F 1 = t (f x 1 F ) and f 1 = t ( xf + (p 2 k 2 xx 1 )F )

35 In matrix form F 1 f 1 = t 1 x 1 x p 2 k 2 xx 1 Matches ψ 1 = L ψ with ψ = F f F f Similarly, from x 23 = f 2 F 2 = xf+(q2 k 2 xx 2 )F f x 2 F ψ 2 = F 2 f 2 = s 1 x 2 F = M ψ. x q 2 k 2 xx 2 f

36 Therefore, L = t L c = t 1 x 1 x p 2 k 2 xx 1 M = s M c = s 1 x 2 x q 2 k 2 xx 2

37 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!

38 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!

39 Algorithmic Computation of Lax Pairs for Systems of P Es Example 2: Schwarzian-Boussinesq System y x 1 z 1 + z = 0 y x 2 z 2 + z = 0 x y 12 (y 1 y 2 ) y (p x 1 y 2 q x 2 y 1 ) = 0 System has two single-edge equations and one full-face equation

40 Edge equations require augmentation of system with additional shifted, edge equations y 2 x 12 z 12 + z 2 = 0 y 1 x 12 z 12 + z 1 = 0 Edge equations will provide additional constraints during homogenization (Step 2). The way you handle edge equations leads to gauge-equivalent Lax pairs!

41 Step 1: Verify the consistency around the cube System on the front face: y x 1 z 1 + z = 0 y x 2 z 2 + z = 0 x y 12 (y 1 y 2 ) y (px 1 y 2 qx 2 y 1 ) = 0 Solve for x 12, y 12, and z 12 : x 12 = z 2 z 1 y 1 y 2 y 2 x 12 z 12 + z 2 = 0 y 1 x 12 z 12 + z 1 = 0 y 12 = y(px 1y 2 qx 2 y 1 ) x(y 1 y 2 ) z 12 = y 1z 2 y 2 z 1 y 1 y 2

42 Compute x 123, y 123, and z 123 : x 123 = z 23 z 13 y 13 y 23 y 123 = y 3(px 13 y 23 qx 23 y 13 ) x 3 (y 13 y 23 ) z 123 = y 13z 23 y 23 z 13 y 13 y 23

43 System on the bottom face: y x 1 z 1 + z = 0 y x 3 z 3 + z = 0 x y 13 (y 1 y 3 ) y (px 1 y 3 kx 3 y 1 ) = 0 Solve for x 13, y 13, and z 13 : x 13 = z 3 z 1 y 1 y 3 y 3 x 13 z 13 + z 3 = 0 y 1 x 13 z 13 + z 1 = 0 y 13 = y(px 1y 3 kx 3 y 1 ) x(y 1 y 3 ) z 13 = y 1z 3 y 3 z 1 y 1 y 3

44 Compute x 123, y 123, and z 123 : x 123 = z 23 z 12 y 12 y 23 y 123 = y 2(px 12 y 23 kx 23 y 12 ) x 2 (y 12 y 23 ) z 123 = y 12z 23 y 23 z 12 y 12 y 23

45 System on the left face: y x 3 z 3 + z = 0 y x 2 z 2 + z = 0 x y 23 (y 3 y 2 ) y (px 3 y 2 qx 2 y 3 ) = 0 Solve for x 23, y 23, and z 23 : x 23 = z 3 z 2 y 2 y 3 y 2 x 23 z 23 + z 2 = 0 y 1 x 23 z 23 + z 1 = 0 y 23 = y(qx 2y 3 kx 3 y 2 ) x(y 2 y 3 ) z 23 = y 2z 3 y 3 z 2 y 2 y 3

46 Compute x 123, y 123, and z 123 : x 123 = z 13 z 12 y 12 y 13 y 123 = y 1(qx 12 y 13 kx 13 y 12 ) x 1 (y 12 y 13 ) z 123 = y 12z 13 y 13 z 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

47 x 123 = x(x 1 x 2 ) ( y 1 (z 2 z 3 ) + y 2 (z 3 z 1 )+ y 3 (z 1 z 2 ) ) (z 1 z 2 ) ( px 1 (y 3 y 2 ) + qx 2 (y 1 y 3 ) + kx 3 (y 2 y 1 ) ) y 123 = q(z 2 z 1 )(kx 3 y 1 px 1 y 3 )+ k(z 3 z 1 )(px 1 y 2 qx 2 y 1 ) x 1 ( px1 (y 3 y 2 )+ qx 2 (y 1 y 3 )+ kx 3 (y 2 y 1 ) ) z 123 = px 1(y 3 z 2 y 2 z 3 )+ qx 2 (y 1 z 3 y 3 z 1 )+ kx 3 (y 2 z 1 y 1 z 2 ) px 1 (y 3 y 2 )+ qx 2 (y 1 y 3 )+ kx 3 (y 2 y 1 ) Answer is unique and independent of x and y. Consistency around the cube is satisfied!

48 Step 2: Homogenization Observed that x 3, y 3 and z 3 appear linearly in numerators and denominators of x 13 = z 3 z 1 y 1 y 3 y 13 = y(px 1y 3 kx 3 y 1 ) x(y 1 y 3 ) z 13 = y 1z 3 y 3 z 1 y 1 y 3

49 Substitute x 3 = f F, y 3 = g G, and z 3 = h H. Use constraints (from left face edges) y x 1 z 1 + z = 0, y x 2 z 2 + z = 0 y x 3 z 3 + z = 0 Solve for x 3 = z 3 z y Thus, x 3 = f F = h zh yh, y 3 = g G, and z 3 = h H

50 Substitute x 3, y 3, z 3 into x 13, y 13, z 13 : x 13 = G(h z 1H) H(y 1 G g) y 13 = y(px 1gF ky 1 fg) F x(y 1 G g) z 13 = y 1hG z 1 gh H(y 1 G g) Require that numerators and denominators are linear in f, g, h, F, G, and H. That forces H = G = F. Hence, x 3 = h zf yf, y 3 = g F, and z 3 = h F.

51 Compute x 3 = h zf yf y 3 = g F y 13 = g 1 F 1 z 3 = h F z 13 = h 1 F 1 x 13 = h 1 z 1 F 1 y 1 F 1 Hence, x 13 = h z 1F y 1 F g = h 1 z 1 F 1 y 1 F 1 y 13 = ypx 1g ky 1 h + kzy 1 F x(y 1 F g) z 13 = y 1h z 1 g y 1 F g = h 1 F 1 = g 1 F 1

52 Note that x 13 = h z 1F y 1 F g = h 1 z 1 F 1 y 1 F 1 is automatically satisfied as a result of the relation x 3 = z 3 z y. Write in matrix form: F 1 y ψ 1 = g 1 = t kzy 1 x 0 z 1 y 1 h 1 pyx 1 ky 1 x x F g = L ψ h Repeat the same steps for x 23, y 23, z 23 to obtain

53 ψ 2 = F 2 g 2 h 2 = s y kzy 2 x qyx 2 ky 2 x x 0 z 2 y 2 F g = M ψ h Therefore, L = tl core = t M = sm core = s y kzy 1 x pyx 1 ky 1 x x 0 z 1 y 1 y kzy 2 x qyx 2 ky 2 x x 0 z 2 y 2

54 Step 3: Determine t and s Substitute L = t L core, M = s M core into L 2 M M 1 L = 0 t 2 s (L core ) 2 M core s 1 t (M core ) 1 L core = 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 (from determinant method) t = 3 s = 3 x y 2 2 y x 2. x y 2 1 y x 1 and

55 Summary: Lax Pair for Schwarzian-BSQ System Option a: Solving the edge equation for x 3 = z 3 z y yields F x 3 = h zf yf, y 3 = g F, and z 3 = h F, ψ a = Corresponding Lax matrix: L a = 1 y kzy 1 y x 0 z 1 y 1 pyx 1 ky 1 x x g h

56 Option b: Solving the edge equation for z 3 = x 3 y + z yields F x 3 = f F, y 3 = g F, and z zf +yf 3 =, ψ F b = f g Corresponding Lax matrix: L b = 1 y z z y 1 y 0 0 kyy 1 pyz 1 x 1 x

57 Gauge Equivalences between these Lax Matrices L b =G 1 L a G 1, with ψ b =Gψ a G = z 0 1 y y 0 1 0

58 Software Demonstration

59 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.

60 Avoid the determinant method to avoid square roots! Factorization plays an essential role! Future Work: Extension to more complicated systems of P Es.

61 Thank You

62 Additional Examples Example: Discrete pkdv Equation Lax pair: (x x 12 )(x 1 x 2 ) p 2 + q 2 = 0 L = tl core = t M = sm core = s x p2 k 2 xx 1 1 x 1 x q2 k 2 xx 2 1 x 2 with t = s = 1 or t = 1 DetLcore = 1 k 2 p 2 and s = 1 DetMcore = 1. k 2 q 2 Here, x 3 = f F, ψ = [f F ]T, and t 2 t s s 1 = 1.

63 Example: (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 Here, x 3 = f F, ψ = f F, and t 2 t s s 1 = 1.

64 Example: (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 = 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 ) Here, x 3 = f F, ψ = f F, and t 2 t s s 1 = p+x+x 1 q+x+x 2.

65 Example: (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 = 1 and s = 1 (p 2 k 2 )(δp+xx 1 ) (q 2 k 2 )(δq+xx 2 ) Here, x 3 = f F, ψ = f F, and t 2 t s s 1 = δp+xx 1 δq+xx 2.

66 Example: (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 Here, x 3 = f F, ψ = f F, and t 2 t s s 1 = x x 1 x x 2 = x 1 x 2.

67 Example: (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

68 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 )) Here x 3 = f F, ψ = f F, and t 2t 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?

69 Example: (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

70 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) Here, x 3 = f F, ψ = f F, and t 2t 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?

71 Example: (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 ))

72 Here, x 3 = f F, ψ = f F, and t 2t s s 1 = q (δp+(x x 1 ))(δp (x x 1 )) p(δq+(x x 2 ))(δq (x x 2 )).

73 Example: (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

74 t = 1 x x 1 and s = 1 x x 2 or t = 1 and s = 1. k(k p)(x x1 ) k(k q)(x x2 ) Here, x 3 = f F, ψ = f, and t 2 t F s s 1 = q(x x 1) 2 p(x x 2 ) 2.

75 Example: (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

76 with t = and s = 1 k(k p)((x x 1 ) 2 2p 2 (x+x 1 )+p 4 ) 1 k(k q)((x x 2 ) 2 2q 2 (x+x 2 )+q 4 ) Here, x 3 = f F, ψ = [f F ]T, and t 2 s = q ( (x x1 ) 2 2p 2 (x + x 1 ) + p 4) t 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.

77 Example: (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 )

78 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 ).

79 Here, x 3 = f F, ψ = f F, and 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)

80 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 ).

81 Example: (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

82 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) Here, x 3 = f F, ψ = f F, and t 2t s s 1 = (q2 1)(px x 1 )(px 1 x) (p 2 1)(qx x 2 )(qx 2 x).

83 Example: (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 )

84 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 = 1 (q 2 k 2 )((β q)x+(α+q)x 2)((α q)x+(β+q)x 2) Here, x 3 = f F, ψ = f F, and t 2t 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).

85 Example: 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

86 Example: Lattice due to Hietarinta (2011) z x 1 y 1 x = 0 z x 2 y 2 x = 0 z 12 y x 1 ( px1 qx ) 2 = 0 x z 1 z 2 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

87 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, x 3 = h G, y 3 = g G, z 3 = f G, ψ = t 2t s s 1 = z 1 z 2. f g, and G

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

89 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 Here, x 3 = f F, y 3 = g F, ψ = 3 q k. f F, and t 2 t g s s 1 = 1.

90 Example: 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

91 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. Here, x 3 = f F, y 3 = g G, ψ = [f F g G]T, and t 2 S T s 1 = 1 and T 2 s t S 1 = 1, or T 2 T S S 1 = 1, with T = t T, S = s S.

92 Example: 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

93 with t = s = 1, or t= 1 DetLc = 1 α k and s= 1 β k. Here, x 3 = f F, ψ = f F, and t 2 t s s 1 = 1.

94 Example: Schwarzian-Boussinesq Lattice (Nijhoff 1999) y z 1 x 1 + x = 0 y z 2 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

95 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. Here, 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.

96 Example: 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

97 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 Here, x 3 = f F, y 3 = g F, ψ = g, F y. and t 2 t s s 1 = 1.