Symbolic Computation of Lax Pairs of Integrable Nonlinear Partial Difference Equations on Quad-Graphs

Symbolic Computation of Lax Pairs of Integrable Nonlinear Partial Difference Equations on Quad-Graphs Willy Hereman & Terry Bridgman Department of Mathematical and Computer Sciences Colorado School of Mines Golden, Colorado, U.S.A. http://www.mines.edu/fs home/whereman/ Symbolic Analysis @ FoCM 11 Budapest, Hungary Wednesday, July 13, 2011, 17:50

Collaborators Prof. Reinout Quispel Dr. Peter H. van der Kamp La Trobe University, Melbourne, Australia Ph.D. student at CSM: Terry Bridgman Research supported in part by NSF under Grant CCF-0830783 This presentation is made in TeXpower

Outline What are Lax pairs of nonlinear PDEs and DDEs What are nonlinear P Es? Classification of integrable nonlinear P Es in 2D Lax pair of nonlinear P Es Examples of Lax pair of nonlinear P Es (including systems) Algorithm (Nijhoff 2001, Bobenko & Suris 2001) Software demonstration Additional examples Conclusions and future work

What are Lax Pair for Nonlinear PDEs? Historical example: Korteweg-de Vries equation u t + αuu x + u xxx = 0 Key idea: Replace the nonlinear PDE with a compatible linear system: ( ) ψ xx + 1 6 αu λ ψ = 0 ψ t = 4ψ xxx αuψ x 1 2 αu xψ + a(t)ψ ψ is eigenfunction; λ is constant eigenvalue (λ x = λ t = 0).

Lax Operators Replace a completely integrable nonlinear PDE by Lψ = λψ ψ t = Mψ Require compatibility of both equations L t ψ + Lψ t = λψ t L t ψ + LMψ = λmψ = Mλψ = MLψ Hence, L t ψ + (LM ML)ψ = O

Lax Equation: L t + [L, M] = 0 with commutator [L, M] = LM ML, and where = means evaluated on the PDE. Example: Lax operators for the KdV equation L = 2 x 2 + α 6 ui M = 4 3 x 3 α 2 ( u x + ) x u + a(t)i Note: L t ψ + [L, M]ψ = α 6 (u t + αuu x + u xxx ) ψ

Lax Pair for PDEs Matrices (AKNS) Express compatibility of Ψ x = X Ψ where Ψ = ψ 1 ψ 2. ψ N Ψ t = T Ψ, X and T are N N matrices. Lax equation (zero-curvature equation): X t T x + [X, T] = 0 with commutator [X, T] = XT TX

Example: Lax matrices for KdV equation 0 1 X = λ 1 6 αu 0 T = a(t) + 1 6 αu x 4λ 1 3 αu 4λ 2 + 1 3 αλu + 1 18 α2 u 2 + 1 6 αu 2x a(t) 1 6 αu x Lax pairs are not unique. A second Lax pair: X = ik 1 6 αu 1 ik T= 4ik3 + 1 3 iαku 1 6 αu x 4k 2 + 1 3 αu ) 1 (2k 3 α 2 u 1 6 αu2 +iku x 1 2 u xx 4ik3 1 3 iαku+ 1 6 αu x

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 + G x G 1 T = GTG 1 + G t G 1 G is arbitrary invertible matrix and Ψ = GΨ. Thus, X t T x + [ X, T] = 0

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

Peter D. Lax (1926-)

Reasons to compute a Lax pair Compatible linear system is the starting point for application of the IST. Describe the time evolution of the scattering data Indicator for complete integrability of the PDE. Zero-curvature representation of the PDE. Some Lax pairs lead to conservation laws. 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.

Lax Pair of Nonlinear DDEs Operators Example: Kac-van Moerbeke (Volterra) lattice u n = u n (u n 1 u n+1 ) Key idea: Replace the nonlinear DDE with a compatible linear system: Formulation 1: L n ψ n = λψ n ψ n = M n ψ n Lax equation: L n + [L n, M n ] = 0 with commutator [L n, M n ] = L n M n M n L n

Example: Lax operator for KvM equation L n = 2 u n S + u n 1 S 1 and M n = u n+1 un S 2 1 un 1 un 2 S 2 4 S is the forward shift operator Sf n = f n+1 S k f n = f n+k S 1 f n = f n 1 S 1 is the backward shift operator

Formulation 2: L n ψ n = ψ n+1 ψ n = M n ψ n Lax equation: L n + L n Mn M n+1 Ln = 0 Conversion: Ln = 1 λ SL n and M n = M n original equation L n + [L n, M n ] = 0 target equation L n +S L n S 1 Mn M n Ln =0 L n +S L n S 1 Mn M n Ln =0 L n + [L n, M n ] = 0

Lax Pair of Nonlinear DDEs Matrices Formulation 1: X n ψ n = λψ n T n ψ n = ψ n Lax equation (zero-curvature equation): Ẋ n ψ n + [X n, T n ] = 0 Formulation 2: X n ψ n = ψ n+1 T n ψ n = ψ n Lax equation (zero-curvature equation): X n + X n Tn T n+1 Xn = 0

Example: Lax matrices for KvM equation X n = 1 u n T n = λ 2 u n 1 λ 0 λu n 1 λ 2 u n 1 Lax pairs are not unique. A second Lax pair: X n = λ u n 1 0 T n = λ2 + u n λ λu n u n 1

Part I Scalar Nonlinear Lattices

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: 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

Notation: u is dependent variable or field (scalar case) n and m are lattice points p and q are parameters For brevity, (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 ) discrete pkdv equation: (u n,m u n+1,m+1 )(u n+1,m u n,m+1 ) p 2 + q 2 = 0 (x x 12 )(x 1 x 2 ) p 2 + q 2 = 0

(u n,m u n+1,m+1 )(u n+1,m u n,m+1 ) p 2 + q 2 = 0 (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

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

Introduce a third lattice variable l View u as dependent on three lattice points: n, m, l. So, x = u n,m is replaced by x = u n,m,l Moves 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 front, bottom, and left face of the cube Require consistency for the computation of x 123 = u n+1,m+1,l+1

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

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

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!

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

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 2 +... + a 14 x 2 + a 15 x 12 + a 16 2. 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)

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

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 12 + 1) + q(p 2 1)(xx 2 + x 1 x 12 ) p(q 2 1)(xx 1 + x 2 x 12 ) = 0

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

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 parametrizations (Adler, Nijhoff, Viallet) are given in the literature.

Lax Pair of Scalar Nonlinear P Es Replace the nonlinear P E by ψ 1 = Lψ (recall ψ 1 = ψ n+1,m ) ψ 2 = Mψ (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 Mψ ψ 12 = M 1 ψ 1 = M 1 Lψ Lax equation: L 2 M M 1 L = 0

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 LG 1 M = G 2 MG 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

Example 1: Discrete pkdv Equation (x x 12 )(x 1 x 2 ) p 2 + q 2 = 0 H1 after p 2 p, q 2 q Lax operators: L = tl c = t M = sm c = s with t = s = 1 or t = 1 and s = 1 DetMc = x p2 k 2 xx 1 1 x 1 x q2 k 2 xx 2 1 x 2 DetLc = 1 k 2 p 2 1. Here, t 2 k 2 q 2 t s s 1 = 1.

Note: L 2 M M 1 L = ( (x x 12 )(x 1 x 2 ) p 2 + q 2 ) N with L = M = x p2 k 2 xx 1 1 x 1 N = x q2 k 2 xx 2 1 x 2 1 x 1 + x 2 0 1

Example 2: Discrete modified KdV Equation p(xx 2 x 1 x 12 ) q(xx 1 x 2 x 12 ) = 0 H3 for δ = 0 and x x or x 12 x 12 Lax operators: with t = 1 x 1 t= 1 DetLc = L = t M = s px kxx 1 k px 1 qx kxx 2 k qx 2 and s = 1 x 2, or t = s = 1 x, or 1 and s= 1 (p 2 k 2 )xx 1 DetMc = 1 (q 2 k 2 )xx 2 Note: t 2 t s s 1 = x x 1 x x 2 = x 1 x 2.

Lax Pair Algorithm for Scalar P Es (Nijhoff 2001, Bobenko and Suris 2001) Applies to equations that are consistent on 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 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

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

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)

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

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

Step 3: Determine t and s Substitute L = tl c, M = sm 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,... ) If f factors as then try t= s= f = F(x, x 1, p, q,...)g(x, x 1, p, q,...) F(x, x 2, q, p,...)g(x, x 2, q, p,...) 1 F(x,x 1,...) or 1 G(x,x 1,...) and 1 F(x,x 2,...) or 1 G(x,x 2,...) No square roots needed!

Works for the following lattices: mkdv, H3 with δ = 0, Q1, Q3 with δ = 0, (α, β)-equation. Does not work for A1 and A2 equations! Needs further investigation! If f does not factor, apply determinant to get t 2 s det L c det (M c ) = 1 t s 1 det (L c ) 2 det M c A solution: t = 1 det Lc, s = 1 det Mc Always works but introduces square roots!

How to avoid square roots? Remedy: Apply a change of variables x = F (X) t 2 t s s 1 = f(f (X), F (X 1 ), F (X 2 ), p, q,... ) = F(X, X 1, p, q,...)g(x, X 1, p, q,...) F(X, X 2, q, p,...)g(x, X 2, q, p,...) Example 1: Q2 lattice 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) = q ( (X + X 1 ) 2 p 2) ( (X X 1 ) 2 p 2) p ( (X + X 2 ) 2 q 2) ( (X X 2 ) 2 q 2) setting x = F (X) = X 2. Hence, x 1 = X 1 2, x 2 = X 2 2.

Example 2: Q3 lattice ( t 2 s = q(q2 1) 4p 2 (x 2 +x 2 1 ) 4p(1+p2 )xx 1 +δ 2 (1 p 2 ) 2) t s 1 p(p 2 1) ( 4q 2 (x 2 +x 2 2 ) 4q(1+q2 )xx 2 +δ 2 (1 q 2 ) 2) Set x = F (X) = δ cosh(x) then 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 ))

Part II Systems of Nonlinear Lattices

Lax Pair Algorithm Systems of P Es (Nijhoff 2001, Bobenko and Suris 2001) Applies to systems of equations that are consistent on cube! Example: 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.

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 also provide additional constraints during homogenization (Step 2).

Step 1: Verify the consistency around the cube System of equations on 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

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

System of equations on 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

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

System of equations on 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

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

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. Tetrahedron property w.r.t. x, y only; not for z. Consistency around the cube is satisfied!

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

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.

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.

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

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 automatically 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 F 1 0 1 y 1 f g F = Lψ Repeat the same steps for x 23, y 23, z 23 to obtain

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 = Mψ 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

Step 3: Determine t and s Substitute L = tl c, M = sm 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 = z 3 y 2 y 1 z 1 and s = z 3 y 2 y 2 z 2.

Example 1: 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 operators: L = t x 1 1 0 y 1 0 1 p k xy 1 + x 1 z z x

x 2 1 0 M = s y 2 0 1 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.

Example 2: 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 operators: 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

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 2 0 0 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.

Example 3: 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 operators: L = t 1 x 1 y k p yx 1 M = s 1 x 2 y k q yx 2

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.

Example 4: 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 operators: L = t y 0 yz 1 pyz 1 0 z z 0 1 y 1 kyy 1

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.

Example 5: 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 operators: 1+k+k k + p z 2 y L=t 0 p 1 (1 k)y 1 1 0 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

1+k+k k + q z 2 y M =s 0 q 1 (1 k)y 2 1 0 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, and t 2 t F y. s s 1 = 1.

Software Demonstration

Additional Examples Example 3: H1 equation (ABS classification) (x x 12 )(x 1 x 2 ) + q p = 0 Lax operators: 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.

Example 4: 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 operators: 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.

Example 5: 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 operators: 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.

Example 6: H3 equation (δ = 0) (ABS 2003) p(xx 1 + x 2 x 12 ) q(xx 2 + x 1 x 12 ) = 0 Lax operators: 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.

Example 7: 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 operators: 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 )).

Example 8: 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 operators: 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

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 )

Example 9: 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 operators: (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

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.

Example 10: 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 operators: 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 )

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 ).

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)

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 ).

Example 11: 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 operators: 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

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).

Example 12: (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 operators: 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 )

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).

Example 13: 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 operators: 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

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 )). However, the choices t = cannot be used. 1 δp±(x+x 1 ) and s = 1 δq±(x+x 2 ) Needs further investigation!

Example 14: A2 equation (ABS 2003) (q 2 p 2 )(xx 1 x 2 x 12 + 1) + 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 operators: 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

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). However, the choices t = 1 p xx 1 and s = 1 q xx 2 or t = 1 pxx 1 1 and s = 1 qxx 2 1 cannot be used. Needs further investigation!

Example 15: 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 operators: L = p kx 1 M = k x px 1 x qx 2 x 1 kx x 2 k q

Example 16: Quotient-difference (QD) system x 2 y 2 xy 1 = 0 x 2 + y 12 x 1 y 1 = 0 QD system is defined on a stencil. QD system is NOT consistent around the cube, but has a Lax pair! Lax operators: L = y k y k x M = 1 k 0 k y k x + y 2

Conclusions and Future Work Mathematica code works for scalar P Es in 2D defined on quad-graphs (quadrilateral faces). Mathematica code is being 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. Examples: discrete sine-gordon equation and QD system.

Avoid the determinant method to avoid square roots! Factorization plays an essential role! Hard case: Q4 equation (elliptic curves, Weierstraß functions) (Nijhoff 2001). Software cannot handle Q4. Future Work: Extension to more complicated systems of P Es, not necessarily defined on quad-graphs. Holy Grail: Find Lax pairs for P Es in 3D: Lax pair will be expressed in terms of tensors. Consistency around a hypercube. Examples: discrete Kadomtsev-Petviashvili (KP) equations.

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