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Symbolic Computation of Lax Pairs of
Integrable Nonlinear Difference Equationson Quad-graphs
Willy HeremanDepartment of Mathematical and Computer Sciences
Colorado School of Mines
Golden, Colorado, U.S.A.
http://www.mines.edu/fs home/whereman/
Workshop on Discrete Systems and Special Functions
Isaac Newton Institute for Mathematical Sciences
Cambridge, UK
Monday, June 29, 2009, 2:00p.m.
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Collaborators
Prof. Reinout Quispel
Dr. Peter van der Kamp
Ph.D. students: Dinh Tran and Omar Rojas
La Trobe University, Melbourne, Australia
Research supported in part by NSF
under Grant CCF-0830783
This presentation is made in TeXpower
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Outline
• What are nonlinear P∆Es?
• Classification of integrable nonlinear P∆Es in 2D
• Lax pair 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
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.
What are nonlinear P∆Es?
• Nonlinear maps with two (or more) lattice points!
Some origins:
I full discretizations of PDEs
I discrete dynamical systems in 2 dimensions
I superposition principle (Bianchi permutability)
for Backlund transformations between 3
solutions (2 parameters) of a completely
integrable PDE
• Example: discrete potential Korteweg-de Vries
(pKdV) equation
(un,m − un+1,m+1)(un+1,m − un,m+1)− p2 + q2 = 0
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• Notation:
u is dependent variable or field (scalar case)
n and m are lattice points
p and q are parameters
• For brevity,
(un,m, un+1,m, un,m+1, un+1,m+1) = (x, x1, x2, x12)
• Alternate notations (in the literature):
(un,m, un+1,m, un,m+1, un+1,m+1) = (u, u, u, ˆu)
(un,m, un+1,m, un,m+1, un+1,m+1) = (u00, u10, u01, u11)
• discrete pKdV equation:
(un,m − un+1,m+1)(un+1,m − un,m+1)− p2 + q2 = 0
−→ (x− x12)(x1 − x2)− p2 + q2 = 0
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(un,m − un+1,m+1)(un+1,m − un,m+1)− p2 + q2 = 0
(x− x12)(x1 − x2)− p2 + q2 = 0
x = vn,m x1 = vn+1,m
x12 = vn+1,m+1x2 = vn,m+1 p
p
q q
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Classification of 2D nonlinear integrable P∆Es
Adler, Bobenko, Suris (ABS) 2003, 2007
• Family of nonlinear P∆Es in two dimensions:
Q(x, x1, x2, x12; p, q) = 0
• Assumptions (ABS, 2003):1. Affine linear
Q(x, x1, x2, x12; p, q) = a1xx1x2x12 + a2xx1x2 +
. . .+ a14x2 + a15x12 + a16
2. Invariant under D4 (symmetries of square)
Q(x, x1, x2, x12; p, q) = εQ(x, x2, x1, x12; q, p)
= σQ(x1, x, x12, x2; p, q)
ε, σ = ±1
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3. Consistency around the cube
Superposition of Backlund transformations
between 4 solutions x, x1, x2, x3 (3 parameters:
p, q, k)
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x
x123
x12
x13
x23
x1
x2
x3
p
qk
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• Trick: Introduce a third lattice variable `
• View u as dependent on three lattice points:
n,m, `. So, x = un,m −→ x = un,m,`
• Moves in three directions:
n→ n+ 1 over distance p
m→ m+ 1 over distance q
`→ `+ 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
x123 = un+1,m+1,`+1
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x
x123
x12
x13
x23
x1
x2
x3
p
qk
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Verification of Consistency Around the Cube
Example: discrete pKdV equation
? Equation on front face of cube:
(x− x12)(x1 − x2)− p2 + q2 = 0
Solve for x12 = x− p2−q2x1−x2
Compute x123 : x12 −→ x123 = x3 − p2−q2x13−x23
? Equation on floor of cube:
(x− x13)(x1 − x3)− p2 + k2 = 0
Solve for x13 = x− p2−k2
x1−x3
Compute x123 : x13 −→ x123 = x2 − p2−k2
x12−x23
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? Equation on left face of cube:
(x− x23)(x3 − x2)− k2 + q2 = 0
Solve for x23 = x− q2−k2
x2−x3
Compute x123 : x23 −→ x123 = x1 − q2−k2
x12−x13
? Verify that all three coincide:
x123 =x1 −q2 − k2
x12 − x13=x2 −
p2 − k2
x12 − x23=x3 −
p2 − q2
x13 − x23
Upon substitution of x12, x13, and x23 :
x123 =p2x1(x2 − x3) + q2x2(x3 − x1) + k2x3(x1 − x2)
p2(x2 − x3) + q2(x3 − x1) + k2(x1 − x2)
Consistency around the cube is satisfied!
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Tetrahedron property
x123 =p2x1(x2 − x3) + q2x2(x3 − x1) + k2x3(x1 − x2)
p2(x2 − x3) + q2(x3 − x1) + k2(x1 − x2)
is independent of x. Connects x123 to x1, x2 and x3
x
x123
x12
x13
x23
x1
x2
x3
p
qk
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Result of the ABS Classification
• List H
I (H1)
(x− x12)(x1 − x2) + q − p = 0
I (H2)
(x−x12)(x1−x2)+(q−p)(x+x1+x2+x12)+q2−p2 =0
I (H3)
p(xx1 + x2x12)− q(xx2 + xx12) + δ(p2 − q2) = 0
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• List A
I (A1)
p(x+x2)(x1+x12)−q(x+x1)(x2+x12)−δ2pq(p−q) = 0
I (A2)
(q2 − p2)(xx1x2x12 + 1) + q(p2 − 1)(xx2 + x1x12)
−p(q2 − 1)(xx1 + x2x12) = 0
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• List Q
I (Q1)
p(x−x2)(x1−x12)−q(x−x1)(x2−x12)+δ2pq(p−q) = 0
I (Q2)
p(x−x2)(x1−x12)−q(x−x1)(x2−x12)+pq(p−q)
(x+x1+x2+x12)−pq(p−q)(p2−pq+q2)=0
I (Q3)
(q2−p2)(xx12+x1x2)+q(p2−1)(xx1+x2x12)
−p(q2−1)(xx2+x1x12)−δ2
4pq(p2−q2)(p2−1)(q2−1)=0
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I (Q4) (mother)
a0xx1x2x12
+a1(xx1x2 + x1x2x12 + xx2x12 + xx1x12)
+a2(xx12 + x1x2) + a2(xx1 + x2x12)
+a2(xx2 + x1x12) + a3(x+ x1 + x2 + x12) + a4 = 0
ai depend on the lattice parameters and lie onthe Weierstraß elliptic curve y2 = 4x3 − g2x− g3
(Weierstraß elliptic functions)
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I (Q4) Hietarinta’s Parametrization
sn(α; k) (x1x2 + xx12) + sn(β; k) (xx1 + x2x12)
−sn(α− β; k) (xx2 + x1x12)
+sn(α; k) sn(β; k) sn(α− β; k)(1 + k2xx1x2x12)=0
where sn(α; k) is the Jacobi elliptic sinefunction with modulus k
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(Q4) expresses Bianchi’s permutability diagram
for Backlund transformations for the
Krichever-Novikov equation:
ut = u3x −3
2
(u2
2x − p(u)
ux
)
= ux {u;x}+3
2
p(u)
ux
where
{u;x} =
(u2x
ux
)x
−1
2
u22x
u2x
=u3x
ux−
3
2
u22x
u2x
is the Schwarzian derivative, p(u) is the cubic
polynomial for the Weierstraß curve (quartic
polynomial after a Mobius transformation)
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Krichever-Novikov (KN) Equation
Take ut = u3x + f(u, xx, u2x), require infinitely
many (local) conservation laws, discard all
equations that can be reduced to the KdV
equation via a Backlund transformation
u = F (v, vx, v2x, ..., vmx), then only the KN
equation is left with
p(u) = P(u; g2, g3) = 4u3 − g2u− g3.
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Lax Pair of Nonlinear PDEs
• Historical example: Korteweg-de Vries equation
ut + αuux + uxxx = 0
• Lax equation: Lt + [L,M] = 0 (on PDE)
with commutator [L,M] = LM−ML
• Lax operators:
L =∂2
∂x2+α
6u
M = −4∂3
∂x3−α
2
(u∂
∂x+
∂
∂xu
)+A(t)
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• Note: Ltψ + [L,M]ψ = α6
(ut + αuux + uxxx)ψ
• Linear problem
? Sturm-Liouville equation: Lψ = λψ
For the KdV equation
ψxx +
(α
6u− λ
)ψ = 0
? Time evolution of data: ψt = Mψ
? Eigenvalues of L are constant: λt = 0
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? Compatibility of Lψ = λψ and ψt = Mψ gives
Ltψ + Lψt = λψt
Ltψ + LMψ = λMψ
= Mλψ
= MLψ
Thus,Ltψ + (LM−ML)ψ = O
Lax equation:
Lt + [L,M] = 0 (on PDE)
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Lax Pair —- Matrix Formalism (AKNS)
• Express compatibility of
Ψx = X Ψ, Ψt = T Ψ
where Ψ =
ψφ
• Lax equation (Zero-curvature equation):
Xt −Tx + [X,T] = 0 (on PDE)
with commutator [X,T] = XT−TX
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• Korteweg-de Vries equation
Lax matrices:
X =
0 1
λ− α6u 0
T =
a(t) + α6ux −4λ− α
3u
−4λ2 + α3λu+ α2
18u2 + α
6u2x a(t)− α
6ux
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Peter D. Lax (1926-)
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Reasons to compute a Lax pair
• Replace nonlinear PDE by linear scattering
problem and apply the IST
• Describe the time evolution of the scattering data
• 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
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.
Lax Pair of Nonlinear P∆Es (ABS Lattices)
• Require that ψ1 = Lψ, ψ2 = Mψ
Here L and M are 2× 2 matrices and ψ =
fg
So, ψ1 = ψn→n+1, ψ2 = ψm→m+1
• Express compatibility:
ψ12 = L2ψ2 = L2Mψ
ψ12 = M1ψ1 = M1Lψ
Hence, L2Mψ −M1Lψ = 0• Lax equation:
L2M−M1L = 0 (on P∆E)
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• Example 1: Discrete pKdV equation
(x− x12)(x1 − x2)− p2 + q2 = 0
(H1) after p2 → p, q2 → q• Lax operators:
L = tLc = t
x p2 − k2 − xx1
1 −x1
M = sMc = s
x q2 − k2 − xx2
1 −x2
with t = s = 1or t = 1√
DetLc= 1√
k2−p2and s = 1√
DetMc= 1√
k2−q2
Note: t2t
ss1
= 1
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• Note: L2M−M1L =(
(x− x12)(x1 − x2)− p2 + q2)
N
with
L =
x p2 − k2 − xx1
1 −x1
M =
x q2 − k2 − xx2
1 −x2
N =
−1 x1 + x2
0 1
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• Example 2: Discrete modified KdV equation
p(xx2 − x1x12)− q(xx1 − x2x12) = 0
(H3) for δ = 0 and x→ −x or x12 → −x12• Lax operators:
L = t
−px kxx1
k −px1
M = s
−qx kxx2
k −qx2
with t = 1
x1and s = 1
x2, or t = s = 1
x,
or t= 1√DetLc
= 1√(p2−k2)xx1
and s= 1√DetMc
= 1√(q2−k2)xx2
Note: t2t
ss1
= xx1
xx2= x1
x2
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• Example 3: Discrete Boussinesq equation
(Tongas and Nijhoff 2005)
z1 − xx1 + y = 0
z2 − xx2 + y = 0
(x2 − x1)(z − xx12 + y12)− p+ q = 0
• Lax operators:
L = t
−x1 1 0
−y1 0 1
p− k − xy1 + x1z −z u
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M = s
−x2 1 0
−y2 0 1
q − k − xy2 + x2z −z u
with t = s = 1, or t = 1
3√p−k and s = 13√q−k
Note: t2t
ss1
= 1
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• Example 4: System of pKdV equations
(Xenitidis and Mikhailov 2009)
(x− x12)(y1 − y2)− p2 + q2 = 0
(y − y12)(x1 − x2)− p2 + q2 = 0
• Lax operators:
L =
0 0 tx t(p2 − k2 − xy1)
0 0 t −ty1
Ty T (p2 − k2 − x1y) 0 0
T −Tx1 0 0
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M =
0 0 sx s(q2 − k2 − xy2)
0 0 s −sy2
Sy S(q2 − k2 − x2y) 0 0
S −Sx2 0 0
with t = s = T = S = 1,
or t T = 1√DetLc
= 1α−γ and s S = 1
β−γ
Note: t2T
Ss1
= 1 and T2
tsS1
= 1
or T2TSS1 = 1 with T = t T, S = s S
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• Example 5: Discrete system related to the NLS
equation (Xenitidis and Mikhailov 2009)
y1 − y2 − y((x1 − x2)y + α− β
)= 0
x1 − x2 + x12
((x1 − x2)y + α− β
)= 0
• Lax operators:
L = t
−1 x1
y γ − α− yx1
M = s
−1 x2
y γ − β − yx2
with t = s = 1, or t= 1√
DetLc= 1√
α−γ and s= 1√β−γ
Note: t2t
ss1
= 1
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Algorithm to Compute a Lax Pair
(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
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px
q
x2
x3
x23
x1
k
x12
x13
x123
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? Equation on front face of cube:
(x− x12)(x1 − x2)− p2 + q2 = 0
Solve for x12 = x− p2−q2x1−x2
Compute x123 = x3 − p2−q2x13−x23
? Equation on floor of cube:
(x− x13)(x1 − x3)− p2 + k2 = 0
Solve for x13 = x− p2−k2
x1−x3
Compute x123 = x2 − p2−k2
x12−x23
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? Equation on left face of cube:
(x− x23)(x3 − x2)− k2 + q2 = 0
Solve for x23 = x− q2−k2
x2−x3
Compute x123 = x1 − q2−k2
x12−x13
After substitution of x12, x13, and x23
x123 =p2x1(x2 − x3) + q2x2(x3 − x1) + k2x3(x1 − x2)
p2(x2 − x3) + q2(x3 − x1) + k2(x1 − x2)
unique and independent of x (tetrahedron property)
Consistency around the cube is satisfied!
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• Step 2: Homogenization
Numerator and denominator of
x13 = x3x−xx1+p2−k2
x3−x1and x23 = x3x−xx2+q2−k2
x3−x2
are linear in x3
Substitute x3 = fg−→ x13 = f1
g1, x23 = f2
g2.
From x13 : f1g1
= xf+(p2−k2−xx1)gf−x1g
Hence, f1 = t(xf + (p2 − k2 − xx1)g
)and
g1 = t (f − x1g)
or, in matrix formf1
g1
= t
x p2 − k2 − xx1
1 −x1
fg
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Matches ψ1 = Lψ with ψ =
fg
Similarly, from x23 :f2
g2
= s
x q2 − k2 − xx2
1 −x2
fg
or ψ2 = Mψ. Therefore,
L = t Lc = t
x p2 − k2 − xx1
1 −x1
M = sMc = s
x q2 − k2 − xx2
1 −x2
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• Alternate method of deriving Lc and Mc
L = t Lc = t
− ∂Q∂x2
−Q∂2Q
∂x2∂x12
∂Q∂x12
x2=x12=0
= t
x p2 − k2 − xx1
1 −x1
where Q(x, x1, x2, x12; p, k) and t(x, x1; p, k)
• Apply x1 → x2 and p→ q
M = sMc = s
x q2 − k2 − xx2
1 −x2
where s = t(x, x2; q, k)
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• Step 3: Determine t and s
? Substitute L = tLc,M = sMc into L2M−M1L = 0
−→ t2s(Lc)2Mc − s1t(Mc)1Lc = 0
? Solve the equation from the (2-1)-element for
t2t
ss1
= f(x, x1, x2, p, q, . . . )
? If f factors as
f =F(x, x1, p, q, . . .)G(x, x1, p, q, . . .)
F(x, x2, q, p, . . .)G(x, x2, q, p, . . .)
then try t= 1F(x,x1,...)
or 1G(x,x1,...)
and
s= 1F(x,x2,...)
or 1G(x,x2,...)
−→ No square roots needed!
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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
t2
t
s
s1=
√detLc
det (Lc)2
√det (Mc)1
detMc
? A solution: t = 1√detLc
, s = 1√detMc
−→ Introduces square roots!
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How to avoid square roots?
Remedy: Apply a change of variables x = F (X)
t2
t
s
s1= f(F (X), F (X1), F (X2), p, q, . . . )
=F(X,X1, p, q, . . .)G(X,X1, p, q, . . .)
F(X,X2, q, p, . . .)G(X,X2, q, p, . . .)
Example 1: (Q2) lattice
t2
t
s
s1=
q((x− x1)2 − 2p2(x+ x1) + p4
)p((x− x2)2 − 2q2(x+ x2) + q4
)=
q((X +X1)2 − p2
) ((X −X1)2 − p2
)p((X +X2)2 − q2
) ((X −X2)2 − q2
)after setting x = F (X) = X2, hence, x1 = X1
2 andx2 = X2
2
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Example 2: (Q3) lattice
t2
t
s
s1=q(q2−1)
(4p2(x2+x2
1)−4p(1+p2)xx1+δ2(1−p2)2)
p(p2−1)(4q2(x2+x2
2)−4q(1+q2)xx2+δ2(1−q2)2)
Set x = F (X) = δ cosh(X) then
4p2(x2+x21)−4p(1+p2)xx1+δ2(1−p2)2
=δ2(p−eX+X1)(p−e−(X+X1))(p−eX−X1)(p−e−(X−X1))
= δ2(p−cosh(X +X1)+sinh(X +X1))
(p−cosh(X +X1)−sinh(X +X1))
(p−cosh(X −X1)+sinh(X −X1))
(p−cosh(X −X1)−sinh(X −X1))
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Equivalence under Gauge Transformations
Lax pairs are equivalent under a gauge
transformation
If L and M form a Lax pair then so do
L = G1LG−1 and M = G2MG−1
where G is non-singular diagonal matrix (or scalar
factor)
and φ = Gψ
Proof: Trivial verification that
(L2M−M1L)φ = 0↔ (L2M−M1L)ψ = 0
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Software Demonstration
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Additional Examples
• Example 3: (H1) equation (ABS classification)
(x− x12)(x1 − x2) + q − p = 0
• Lax operators:
L = t
x p− k − xx1
1 −x1
M = s
x q − k − xx2
1 −x2
with t = s = 1 or t = 1√
k−p and s = 1√k−q
Note: t2t
ss1
= 1
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• Example 4: (H2) equation (ABS 2003)
(x−x12)(x1−x2)+(q−p)(x+x1+x2+x12)+q2−p2 =0
• Lax operators:
L = t
p− k + x p2 − k2 + (p− k)(x+ x1)− xx1
1 −(p− k + x1)
M = s
q − k + x q2 − k2 + (q − k)(x+ x2)− xx2
1 −(q − k + x2)
with t = 1√
2(k−p)(p+x+x1)and s = 1√
2(k−q)(q+x+x2)
Note: t2t
ss1
= p+x+x1
q+x+x2
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• Example 5: (H3) equation (ABS 2003)
p(xx1 + x2x12)− q(xx2 + xx12) + δ(p2 − q2) = 0
• Lax operators:
L = t
kx −(δ(p2 − k2) + pxx1
)p −kx1
M = s
kx −(δ(q2 − k2) + qxx2
)q −kx2
with t = 1√
(p2−k2)(δp+xx1)and s = 1√
(q2−k2)(δq+xx2)
Note: t2t
ss1
= δp+xx1
δq+xx2
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• Example 6: (H3) equation with δ = 0 (ABS 2003)
p(xx1 + x2x12)− q(xx2 + xx12) = 0
• Lax operators:
L = t
kx −pxx1
p −kx1
M = s
kx −qxx2
q −kx2
with t = s = 1
xor t = 1
x1and s = 1
x2
Note: t2t
ss1
= xx1
xx2= x1
x2
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• Example 7: (Q1) equation (ABS 2003)
p(x−x2)(x1−x12)−q(x−x1)(x2−x12)+δ2pq(p−q) = 0
• Lax operators:
L = t
(p− k)x1 + kx −p(δ2k(p− k) + xx1
)p −
((p− k)x+ kx1
)
M = s
(q − k)x2 + kx −q(δ2k(q − k) + xx2
)q −
((q − k)x+ kx2
)
with t = 1δp±(x−x1)
and s = 1δq±(x−x2)
,
or t = 1√k(p−k)((δp+x−x1)(δp−x+x1))
and
s = 1√k(q−k)((δq+x−x2)(δq−x+x2))
Note: t2t
ss1
=q(δp+(x−x1))(δp−(x−x1))p(δq+(x−x2))(δq−(x−x2))
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• Example 8: (Q1) equation with δ = 0 (ABS 2003)
p(x− x2)(x1 − x12)− q(x− x1)(x2 − x12) = 0
which is the cross-ratio equation(x− x1)(x12 − x2)
(x1 − x12)(x2 − x)=p
q• Lax operators:
L = t
(p− k)x1 + kx −pxx1
p −((p− k)x+ kx1
)
M = s
(q − k)x2 + kx −qxx2
q −((q − k)x+ kx2
)
Here, t2t
ss1
= q(x−x1)2
p(x−x2)2. So, t = 1
x−x1and s = 1
x−x2
or t = 1√k(k−p)(x−x1)
and s = 1√k(k−q)(x−x2)
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• Example 9: (Q2) equation (ABS 2003)
p(x−x2)(x1−x12)−q(x−x1)(x2−x12)+pq(p−q)
(x+x1+x2+x12)−pq(p−q)(p2−pq+q2)=0
• Lax operators:
L= t
(k−p)(kp−x1)+kx
−p(k(k−p)(k2−kp+p2−x−x1)+xx1
)p −
((k−p)(kp−x)+kx1
)
M=s
(k−q)(kq−x2)+kx
−q(k(k−q)(k2−kq+q2−x−x2)+xx2
)q −
((k−q)(kq−x)+kx2
)
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• with
t = 1√k(k−p)((x−x1)2−2p2(x+x1)+p4)
and
s = 1√k(k−q)((x−x2)2−2q2(x+x2)+q4)
Note:
t2
t
s
s1=
q((x− x1)2 − 2p2(x+ x1) + p4
)p((x− x2)2 − 2q2(x+ x2) + q4
)=
p((X +X1)2 − p2
) ((X −X1)2 − p2
)q((X +X2)2 − q2
) ((X −X2)2 − q2
)with x = X2, and, consequently, x1 = X2
1 , x2 = X22
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• Example 10: (Q3) equation (ABS 2003)
(q2−p2)(xx12+x1x2)+q(p2−1)(xx1+x2x12)
−p(q2−1)(xx2+x1x12)−δ2
4pq(p2−q2)(p2−1)(q2−1)=0
• Lax operators:
L= t
−4kp
(p(k2−1)x+(p2−k2)x1
)−(p2−1)(δk2−δ2k4−δ2p2+δ2k2p2−4k2pxx1)
−4k2p(p2−1) 4kp(p(k2−1)x1+(p2−k2)x
)
M=s
−4kq
(q(k2−1)x+(q2−k2)x2
)−(q2−1)(δk2−δ2k4−δ2q2+δ2k2q2−4k2qxx2)
−4k2q(q2−1) 4kq(q(k2−1)x2+(q2−k2)x
)
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• with
t= 1
2k√p(k2−1)(k2−p2)(4p2(x2+x2
1)−4p(1+p2)xx1+δ2(1−p2)2)
and
s= 1
2k√q(k2−1)(k2−q2)(4q2(x2+x2
2)−4q(1+q2)xx2+δ2(1−q2)2)
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Note:
t2
t
s
s1
=q(q2−1)
(4p2(x2+x2
1)−4p(1+p2)xx1+δ2(1−p2)2)
p(p2−1)(4q2(x2+x2
2)−4q(1+q2)xx2+δ2(1−q2)2)
=q(q2−1)
(4p2(x−x1)2−4p(p−1)2xx1+δ2(1−p2)2
)p(p2−1)
(4q2(x−x2)2−4q(q−1)2xx2+δ2(1−q2)2
)=q(q2−1)
(4p2(x+x1)2−4p(p+1)2xx1+δ2(1−p2)2
)p(p2−1)
(4q2(x+x2)2−4q(q+1)2xx2+δ2(1−q2)2
)
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where
4p2(x2+x21)−4p(1+p2)xx1+δ2(1−p2)2
= δ2(p−eX+X1)(p−e−(X+X1))(p−eX−X1)(p−e−(X−X1))
= δ2(p−cosh(X +X1)+sinh(X +X1))
(p−cosh(X +X1)−sinh(X +X1))
(p−cosh(X −X1)+sinh(X −X1))
(p−cosh(X −X1)−sinh(X −X1))
with x = δ cosh(X), and, consequently,
x1 = δ cosh(X1), x2 = δ cosh(X2)
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• Example 11: (Q3) equation with δ = 0 (ABS
2003)
(q2 − p2)(xx12 + x1x2) + q(p2 − 1)(xx1 + x2x12)
−p(q2 − 1)(xx2 + x1x12) = 0
• Lax operators:
L= t
(p2−k2)x1+p(k2−1)x −k(p2−1)xx1
(p2−1)k −((p2−k2)x+p(k2−1)x1
)
M=s
(q2−k2)x2+q(k2−1)x −k(q2−1)xx2
(q2−1)k −((q2−k2)x+q(k2−1)x2
)
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• with t = 1px−x1
and s = 1qx−x2
or t = 1px1−x and s = 1
qx2−x
or t = 1√(k2−1)(p2−k2)(px−x1)(px1−x)
and s = 1√(k2−1)(q2−k2)(qx−x2)(qx2−x)
Note: t2t
ss1
= (q2−1)(px−x1)(px1−x)(p2−1)(qx−x2)(qx2−x)
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• Example 12: (α, β)-equation (Quispel 1983)((p−α)x−(p+β)x1
) ((p−β)x2−(p+α)x12
)−((q−α)x−(q+β)x2
) ((q−β)x1−(q+α)x12
)=0
• Lax operators:
L= t
(p−α)(p−β)x+(k2−p2)x1 −(k−α)(k−β)xx1
(k+α)(k+β) −((p+α)(p+β)x1+(k2−p2)x
)
M=s
(q−α)(q−β)x+(k2−q2)x2 −(k−α)(k−β)xx2
(k+α)(k+β) −((q+α)(q+β)x2+(k2−q2)x
)
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• with t = 1(α−p)x+(β+p)x1)
and s = 1(α−q)x+(β+q)x2)
or t = 1(β−p)x+(α+p)x1)
and s = 1(β−q)x+(α+q)x2)
or t = 1√(p2−k2)((β−p)x+(α+p)x1)((α−p)x+(β+p)x1)
and s = 1√(q2−k2)((β−q)x+(α+q)x2)((α−q)x+(β+q)x2)
Note: t2t
ss1
=((β−p)x+(α+p)x1)((α−p)x+(β+p)x1)((β−q)x+(α+q)x2)((α−q)x+(β+q)x2)
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• Example 13: (A1) equation (ABS 2003)
p(x+x2)(x1+x12)−q(x+x1)(x2+x12)−δ2pq(p−q) = 0
(Q1) if x1 → −x1 and x2 → −x2
• Lax operators:
L = t
(k − p)x1 + kx −p(δ2k(k − p) + xx1
)p −
((k − p)x+ kx1
)
M = s
(k − q)x2 + kx −q(δ2k(k − q) + xx2
)q −
((k − q)x+ kx2
)
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• with t = 1√k(k−p)((δp+x+x1)(δp−x−x1))
and
s = 1√k(k−q)((δq+x+x2)(δq−x−x2))
Note: t2t
ss1
=q(δp+(x+x1))(δp−(x+x1))p(δq+(x+x2))(δq−(x+x2))
However, the choices t = 1δp±(x+x1)
and s = 1δq±(x+x2)
CANNOT be used.
Needs further investigation!
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• Example 14: (A2) equation (ABS 2003)
(q2 − p2)(xx1x2x12 + 1) + q(p2 − 1)(xx2 + x1x12)
−p(q2 − 1)(xx1 + x2x12) = 0
(Q3) with δ = 0 via Mobius transformation:
x→ x, x1 → 1x1, x2 → 1
x2, x12 → x12, p→ p, q → q
• Lax operators:
L= t
k(p2−1)x −(p2−k2+p(k2−1)xx1
)p(k2−1)+(p2 − k2)xx1 −k(p2−1)x1
M=s
k(q2−1)x −(q2−k2+q(k2−1)xx2
)q(k2−1)+(q2−k2)xx2 −k(q2−1)x2
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• with t = 1√(k2−1)(k2−p2)(p−xx1)(pxx1−1)
and s = 1√(k2−1)(k2−q2)(q−xx2)(qxx2−1)
Note: t2t
ss1
= (q2−1)(p−xx1)(pxx1−1)(p2−1)(q−xx2)(qxx2−1)
However, the choices t = 1p−xx1
and s = 1q−xx2
or t = 1pxx1−1
and s = 1qxx2−1
CANNOT be used.
Needs further investigation!
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• Example 15: Discrete sine-Gordon equation
xx1x2x12 − pq(xx12 − x1x2)− 1 = 0
(H3) with δ = 0 via extended Mobius
transformation:
x→ x, x1 → x1, x2 → 1x2, x12 → − 1
x12, p→ p, q → 1
q
Discrete sine-Gordon equation is NOT consistent
around the cube, but has a Lax pair!• Lax operators:
L =
p −kx1
−kx
px1
x
M =
qx2
x− 1kx
−x2
kq
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Conclusions and Future Work
• Mathematica code works for scalar P∆Es in 2D
defined on quad-graphs (quadrilateral faces).
• 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.
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• Avoid the determinant method to avoid square
roots! Factorization plays an essential role!
• Hard case: Q4 equation (elliptic curves,
Weierstraß functions) (Nijhoff, 2001).
• Future Work: Extension to 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.
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Thank You