Symbolic Computation of Lax Pairs of Systems of Partial Difference Equations Using Consistency Around the Cube Willy Hereman Department of Applied Mathematics and Statistics Colorado School of Mines Golden, Colorado, U.S.A. http://www.mines.edu/fs home/whereman/ Applied Mathematics Mathematical Sciences Stellenbosch University, Stellenbosch, South Africa Thursday, April 24, 2014, 2:00p.m.
Collaborators Reinout Quispel & Peter van der Kamp La Trobe University, Melbourne, Australia Terry Bridgman (Ph.D. student, CSM) Research supported in part by NSF under Grant CCF-0830783 This presentation is made in TeXpower
Outline What are nonlinear P Es? Classification of integrable nonlinear P Es in 2D Lax pairs of nonlinear PDEs Lax pair of nonlinear P Es Examples of Lax pair of nonlinear P Es Algorithm (Nijhoff 2001, Bobenko & Suris 2001) Software demonstration Additional examples Conclusions and future work
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
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: 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
(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
Systems of P Es 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.
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 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)
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.
Peter D. Lax (1926-) Seminal paper: Integrals of nonlinear equations of evolution and solitary waves, Commun. Pure Appl. Math. 21 (1968) 467-490
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 + 1 6 αu λ ψ = 0 ψ t + 4ψ xxx + αuψ x + 1 2 αu xψ = 0 ψ is eigenfunction; λ is constant eigenvalue (λ t = 0) (isospectral)
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
Example: Lax operators for the KdV equation L = D 2 x + 1 6 αu I M = ( ) 4 D 3 x + αud x + 1 2 αu x I Note: L t ψ + [L, M]ψ = 1 6 α (u t + αuu x + u xxx ) ψ
Alternate Operator Formulations Define L = L λ I and M = M Dt Then, the Lax pair becomes Lψ = 0 and Mψ = 0 and the Lax equation becomes [ L, M] = O Challenge: Find linear commuting operators modulo the (nonlinear) PDE If S is an arbitrary invertible operator, then ˆL = SLS 1, ˆM = SMS 1, ˆDt = SD t S 1 satisfy ˆL t + [ ˆL, ˆM] = O
Lax Pairs in Matrix Form (AKNS Scheme) Express compatibility of D x Ψ = X Ψ where Ψ = ψ 1 ψ 2. ψ N D t Ψ = T Ψ, X and T are N N matrices Lax equation (zero-curvature equation): D t X D x T + [X, T] = 0 with commutator [X, T] = XT TX
Example: Lax pair for the KdV equation 0 1 X = λ 1 6 αu 0 T = 1 6 αu x 4λ 1 3 αu 4λ 2 + 1 3 αλu + 1 18 α2 u 2 + 1 6 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
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
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
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
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
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
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.
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
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.
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
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 = 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
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 = 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!
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!
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: 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 provide additional constraints during homogenization (Step 2).
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
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 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
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 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
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. 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 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
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
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.
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
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.
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 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 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
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 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
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 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
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 8: 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 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 F y. Here, t 2 t s s 1 = 1.
Software Demonstration
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.
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.
Thank You
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.
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.
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.
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.
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 )).
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
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 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
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 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 )
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 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
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 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 )
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 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
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?
Example 20: (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 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
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?
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