Symbolic Computation of Lax Pairs
of Two-Dimensional Nonlinear
Partial Difference Equations
Willy Hereman
Department of Mathematical and Computer Sciences
Colorado School of Mines
Golden, Colorado, U.S.A.
http://www.mines.edu/fs home/whereman/
Department of Mathematics
Catholic University of Leuven, Belgium
Tuesday, July 29, 2008
.
Collaborators
Prof. Reinout Quispel
Dr. Peter van der Kamp
Miss Dinh Tran
Mr. Omar Rojas
La Trobe University, Melbourne, Australia
This presentation is made in TeXpower
.
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: discrete potential KdV, modified KdV,
and sine-Gordon equations; the lattices from the
ABS classification (except Q4)
•
Algorithm (Nijhoff 2001, Bobenko & Suris 2001)
•
Software demonstration
•
Additional examples
•
Conclusions and future work
.
What are nonlinear P∆Es?
•
Nonlinear maps in two dimensions (2D)
•
Origin: full discretizations of PDEs in 2D
•
Example: discrete potential Korteweg-de Vries
(pKdV) equation (from vt + α2 vx2 + vxxx = 0)
(vn,m − vn+1,m+1 )(vn+1,m − vn,m+1 ) − p2 + q 2 = 0
•
Notation:
v is dependent variable or field (others use u or x)
n and m are lattice points (others use ` and m)
p and q are parameters (others use α and β)
.
•
For brevity, denote
(vn,m , vn+1,m , vn,m+1 , vn+1,m+1 ) = (x, x1 , x2 , x12 )
•
discrete pKdV equation:
(x − x12 )(x1 − x2 ) − p2 + q 2 = 0
x2 = vn,m+1
p
q
x = vn,m
x12 = vn+1,m+1
q
p
x1 = vn+1,m
.
•
Alternatives (in the literature):
ˆ
(vn,m , vn+1,m , vn,m+1 , vn+1,m+1 ) = (v, ṽ, v̂, ṽ)
or (x, u, v, y) or (v00 , v10 , v01 , v11 )
•
discrete pKdV equation:
(x − x12 )(x1 − x2 ) − p2 + q 2 = 0
•
Alternatives (in the literature):
ˆ
(v − ṽ)(ṽ
− v̂) − p2 + q 2 = 0
or
or
(x − y)(u − v) − p2 + q 2 = 0
(v00 − v11 )(v10 − v01 ) − p2 + q 2 = 0
.
Classification of 2D nonlinear integrable P∆Es
Adler, Bobenko, Suris (2003)
•
Family of nonlinear P∆Es in two dimensions:
Q(x, x1 , x2 , x12 ; p, q) = 0
•
Constraints:
1. Linear in each argument (affine linear):
Q(x, x1 , x2 , x12 ; p, q) = a1 xx1 x2 x12 + a2 xx1 x2 +
. . . + a14 x2 + a15 x12 + a16
2. Invariant under the square symmetries:
Q(x, x1 , x2 , x12 ; p, q) = Q(x, x2 , x1 , x12 ; q, p)
= σQ(x1 , x, x12 , x2 ; p, q)
, σ = ±1
.
3. Consistency around the cube:
x23
x123
x2
x3
q
x
x12
x13
k
p
x1
.
Verification of Consisteny Around the Cube
Example: discrete pKdV equation
? Equation on front face of cube:
(x − x12 )(x1 − x2 ) − p2 + q 2 = 0
Solve for x12 = x −
Compute x123 :
p2 −q 2
x1 −x2
x12 −→ x123 = x3 −
p2 −q 2
x13 −x23
? Equation on floor of cube:
(x − x13 )(x1 − x3 ) − p2 + k2 = 0
Solve for x13 = x −
Compute x123 :
p2 −k2
x1 −x3
x13 −→ x123 = x2 −
p2 −k2
x12 −x23
? Equation on left face of cube:
(x − x23 )(x3 − x2 ) − k2 + q 2 = 0
Solve for x23 = x −
Compute x123 :
q 2 −k2
x2 −x3
x23 −→ x123 = x1 −
q 2 −k2
x12 −x13
? Verify that all three coincide:
q 2 − k2
p2 − k 2
p2 − q 2
x123 = x1 −
= x2 −
= x3 −
x12 − x13
x12 − x23
x13 − x23
Upon substitution of x12 , x13 , and x23 :
x123
p2 x1 (x2 − x3 ) + q 2 x2 (x3 − x1 ) + k2 x3 (x1 − x2 )
=
p2 (x2 − x3 ) + q 2 (x3 − x1 ) + k2 (x1 − x2 )
Consistency around the cube is satisfied!
.
Tetrahedron property
x123
p2 x1 (x2 − x3 ) + q 2 x2 (x3 − x1 ) + k2 x3 (x1 − x2 )
=
p2 (x2 − x3 ) + q 2 (x3 − x1 ) + k2 (x1 − x2 )
is independent of x. Connects x123 to x1 , x2 and x3
x23
x123
x2
x3
q
x
x12
x13
k
p
x1
.
Result of the ABS Classification
•
List H
I
(H1) Equation
(x − x12 )(x1 − x2 ) + q − p = 0
I
(H2) Equation
(x−x12 )(x1 −x2 )+(q−p)(x+x1 +x2 +x12 )+q 2 −p2 = 0
I
(H3) Equation
p(xx1 + x2 x12 ) − q(xx2 + xx12 ) + δ(p2 − q 2 ) = 0
.
•
List A
I
(A1) Equation
p(x+x2 )(x1 +x12 )−q(x+x1 )(x2 +x12 )−δ 2 pq(p−q) = 0
I
(A2) Equation
(q 2 − p2 )(xx1 x2 x12 + 1) + q(p2 − 1)(xx2 + x1 x12 )
−p(q 2 − 1)(xx1 + x2 x12 ) = 0
.
•
List Q
I
(Q1) Equation
p(x−x2 )(x1 −x12 )−q(x−x1 )(x2 −x12 )+δ 2 pq(p−q) = 0
I
(Q2) Equation
p(x−x2 )(x1 −x12 )−q(x−x1 )(x2 −x12 )+δpq(p−q)
(x+x1 +x2 +x12 )−δ 2 pq(p−q)(p2 −pq+q 2 ) = 0
I
(Q3) Equation
(q 2 −p2 )(xx12 +x1 x2 )+q(p2 −1)(xx1 +x2 x12 )
2
δ
−p(q 2 −1)(xx2 +x1 x12 )−
(p2 −q 2 )(p2 −1)(q 2 −1)=0
4pq
I
.
(Q4) Equation (mother)
a0 xx1 x2 x12
+a1 (xx1 x2 + x1 x2 x12 + xx2 x12 + xx1 x12 )
+a2 (xx12 + x1 x2 ) + ā2 (xx1 + x2 x12 )
+ã2 (xx2 + x1 x12 ) + a3 (x + x1 + x2 + x12 ) + a4 = 0
the ai are given in terms of Weierstraß elliptic
functions, the ai depend on lattice parameters
.
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:
∂2
α
L=
+ u
2
∂x
6
∂3
α
M = −4 3 −
∂x
2
∂
∂
u
+
u + A(t)
∂x
∂x
.
•
Note: Lt φ + [L, M]φ =
•
Linear problem
α
6
(ut + αuux + uxxx ) φ
? Sturm-Liouville equation: Lψ = λψ
For the KdV equation
α
ψxx +
u−λ ψ =0
6
? Time evolution of data: ψt = Mψ
? Eigenvalues of L are constant: λt = 0
.
? Compatibility of Lψ = λψ and ψt = Mψ gives
Lt ψ + Lψt = λψt
Lt ψ + LMψ = λMψ
= Mλψ
= MLψ
Thus,
or
Lt ψ + (LM − ML)ψ = O
Lt + [L, M] = 0 (on PDE)
.
•
Eliminate the third order term in M :
Recall that Lψ = 0 −→ ψxx = (λ − α6 u)ψ
3
∂
α
∂
∂
M = −4 3 −
u
+
u + A(t)
∂x
2
∂x
∂x
α
α
∂
−→ M = A(t) + ux − 4λ + u
6
3
∂x
.
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
.
Lax Pair of Nonlinear P∆Es
•
Require that ψ1 = Lψ,
•
Compatibility:
ψ2 = Mψ
ψ12 = L2 ψ2 = L2 Mψ
ψ12 = M1 ψ1 = M1 Lψ
•
Hence, Lax equation: L2 M − M1 L = 0 (on P∆E)
•
.
Example 1: Discrete pKdV equation
(x − x12 )(x1 − x2 ) − p2 + q 2 = 0
•
H1 after p2 → p, q 2 → q
Lax operators:


x p2 − k2 − xx1

L = t
1
−x1


x q 2 − k2 − xx2

M = s
1
−x2
with t = s = 1 or t = √ 21 2 and s = √ 21 2
k −p
k −q
t2 s
Note: t s1 = 1
.
•
Note: L2 M − M1 L = (x − x12 )(x1 − x2 ) −
p2
+
with


x p2 − k2 − xx1

L=
1
−x1


x q 2 − k2 − xx2

M =
1
−x2


−1 x1 + x2
1


N = p
(p2 − k2 )(q 2 − k2 ) 0
1
q2
N
.
•
Example 2: Discrete modified KdV equation
p(xx2 − x1 x12 ) − q(xx1 − x2 x12 ) = 0
•
H3 for δ = 0 and x → −x or x12 → −x12
Lax operators:


−px kxx1

L = t
k
−px1


−qx kxx2

M = s
k
−qx2
1
and
x1
√ 212
(p −k )xx1
s=
t2 s
t s1
=
with t =
or t =
Note:
=
x x1
x x2
1
,
x2
or t = s =
and s = √
x1
x2
1
x
1
(q 2 −k2 )xx2
.
Algorithm to Compute a Lax Pair
(Nijhoff 2001, Bobenko & Suris 2001)
Applies to equations that are consistent on cube
Example: Discrete pKdV equation
•
Step 1: Verify the consistency around the cube
x123
x23
x12
x2
q
x13
x3
k
x
p
x1
.
? Equation on front face of cube:
(x − x12 )(x1 − x2 ) − p2 + q 2 = 0
Solve for x12 = x −
p2 −q 2
x1 −x2
Compute x123 = x3 −
p2 −q 2
x13 −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
? Equation on left face of cube:
(x − x23 )(x3 − x2 ) − k2 + q 2 = 0
Solve for x23 = x −
q 2 −k2
x2 −x3
Compute x123 = x1 −
q 2 −k2
x12 −x13
After substitution of x12 , x13 , and x23
x123
p2 x1 (x2 − x3 ) + q 2 x2 (x3 − x1 ) + k2 x3 (x1 − x2 )
=
p2 (x2 − x3 ) + q 2 (x3 − x1 ) + k2 (x1 − x2 )
unique and independent of x (tetrahedron property)
Consistency around the cube is satisfied!
.
•
Step 2: Homogenization
Numerator and denominator of
x13 =
x3 x−xx1 +p2 −k2
x3 −x1
and x23 =
x3 x−xx2 +q 2 −k2
x3 −x2
are linear in x3
Substitute x3 =
From x13 :
f1
g1
f
g
−→ x13 =
=
xf +(p2 −k2 −xx1 )g
f −x1 g
Hence, f1 = t xf +
g1 = t (f − x1 g)
(p2
−
k2
f1
,
g1
x23 =
f2
.
g2
− xx1 )g and
or, in matrix form
 

 
f1
x p2 − k2 − xx1
f
  = t
 
g1
1
−x1
g
.
Similarly, from x23 :
 

 
f2
x q 2 − k2 − xx2
f
  = s
 
g2
1
−x2
g
Therefore,


x p2 − k2 − xx1

L = t Lc = t 
1
−x1


x q 2 − k2 − xx2

M = s Mc = s 
1
−x2
.
•
Step 3: Determine t and s
? Substitute L = tLc , M = sMc into L2 M − M1 L = 0
−→ t2 s(Lc )2 Mc − s1 t(Mc )1 Lc = 0
? Solve the equation from the (2-1)-element for
t2 s
t s1
= f (x, x1 , x2 , p, q, . . . )
? If f factors as
F (x, x1 , p, q, . . .)G(x, x1 , p, q, . . .)
f =
F (x, x2 , q, p, . . .)G(x, x2 , q, p, . . .)
then try t = F (x,x1 1 ,...) or
s = F (x,x1 2 ,...) or
1
G(x,x1 ,...)
1
G(x,x2 ,...)
−→ No square roots needed!
and
.
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
s
s
det Lc
det (Mc )1
t2 s
=
t s1
det (Lc )2
det Mc
? A solution: t =
√ 1
,
det Lc
s=
−→ Introduces square roots!
√ 1
det Mc
.
How to avoid square roots?
Remedy 1: Work with the 3-leg form of the lattice
Remedy 2: Apply a change of variables x = F (X)
t2 s
t 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.
2
2
2
4
q (x − x1 ) − 2δp (x + x1 ) + δ p
t2 s
=
2
2
2
4
t s1
p (x − x2 ) − 2δq (x + x2 ) + δ q
2
2
2
2
q (X + X1 ) − δp
(X − X1 ) − δp
=
p (X + X2 )2 − δq 2 (X − X2 )2 − δq 2
after setting x = F (X) = X 2
.
Example 2: Q3 lattice.
q(q 2 −1)
t2 s
=
t s1 p(p2 −1)
4p2 (x2 +x21 )−4p(1+p2 )xx1 +δ 2 (1−p2 )2
2
4q 2 (x2 +x2 )−4q(1+q 2 )xx2 +δ 2 (1−q 2 )2
Set x = F (X) = δ cosh(X) then
4p2 (x2 +x21 )−4p(1+p2 )xx1 +δ 2 (1−p2 )2
= (p−eX+X1 )(p−e−(X+X1 ) )(p−eX−X1 )(p−e−(X−X1 ) )
= (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 ))
.
Equivalence under Gauge Transformations
Lax pairs are equivalent under a gauge
transformation
If L and M form a Lax pair then so do
L = G1 LG−1 and M = G2 MG−1
where G is diagonal matrix (or scalar factor)
and φ = Gψ
Proof: Trivial verification that
(L2 M − M1 L) φ = 0 ↔ (L2 M − M1 L) ψ = 0
.
Software Demonstration
.
•
Additional Examples
Example 3: H1 equation (ABS classification)
(x − x12 )(x1 − x2 ) + q − p = 0
•
Lax operators:


x p − k − xx1

L = t
1
−x1


x q − k − xx2

M = s
1
−x2
with t = s = 1 or t =
Note:
t2 s
t s1
=1
√1
k−p
and s =
√1
k−q
.
•
Example 4: H2 equation (ABS 2003)
(x−x12 )(x1 −x2 )+(q−p)(x+x1 +x2 +x12 )+q 2 −p2 = 0
•
Lax operators:


p − k + x p2 − k2 + (p − k)(x + x1 ) − xx1

L = t
1
−(p − k + x1 )


q − k + x q 2 − k2 + (q − k)(x + x2 ) − xx2

M = s
1
−(q − k + x2 )
1
2(k−p)(p+x+x1 )
p+x+x1
t2 s
=
t s1
q+x+x2
with t = √
Note:
and s = √
1
2(k−q)(q+x+x2 )
.
•
Example 5: H3 equation (ABS 2003)
p(xx1 + x2 x12 ) − q(xx2 + xx12 ) + δ(p2 − q 2 ) = 0
•
Lax operators:


kx − δ(p2 − k2 ) + pxx1

L = t
p
−kx1


kx − δ(q 2 − k2 ) + qxx2

M = s
q
−kx2
1
(p2 −k2 )(δp+xx1 )
δp+xx1
t2 s
=
t s1
δq+xx2
with t = √
Note:
and s = √
1
(q 2 −k2 )(δq+xx2 )
.
•
Example 6: H3 equation with δ = 0 (ABS 2003)
p(xx1 + x2 x12 ) − q(xx2 + xx12 ) = 0
•
Lax operators:


kx −pxx1

L = t
p
−kx1


kx −qxx2

M = s
q
−kx2
with t = s =
Note: tt2 ss1 =
1
1
or
t
=
x
x1
x x1
x1
=
x x2
x2
and s =
1
x2
•
•
Example 7: Q1 equation (ABS 2003)
p(x−x2 )(x1 −x12 )−q(x−x1 )(x2 −x12 )+δ 2 pq(p−q) = 0
Lax operators:


(p − k)x1 + kx −p δ 2 k(p − k) + xx1
L = t

p
− (p − k)x + kx1


(q − k)x2 + kx −q δ 2 k(q − k) + xx2
M = s

q
− (q − k)x + kx2
with t =
or t = q
s=
1
δp±(x−x1 )
and s =
1
1
,
δq±(x−x2 )
k(p−k)((δp+x−x1 )(δp−x+x1 ))
1
q
Note:
and
k(q−k)((δq+x−x2 )(δq−x+x2 ))
t2 s
t s1
=
q (δp+(x−x1 ))(δp−(x−x1 ))
p(δq+(x−x2 ))(δq−(x−x2 ))
•
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 )
p
=
(x1 − x12 )(x2 − x)
q
Lax operators:


(p − k)x1 + kx
−pxx1
L = t

p
− (p − k)x + kx1


(q − k)x2 + kx
−qxx2
M = s

q
− (q − k)x + kx2
Here,
t2 s
t s1
or t = √
=
q(x−x1 )2
.
p(x−x2 )2
1
k(k−p)(x−x1 )
1
So, t = x−x
and s =
1
1
and s = √
k(k−q)(x−x2 )
1
x−x2
•
Example 9: Q2 equation (ABS 2003)
p(x−x2 )(x1 −x12 )−q(x−x1 )(x2 −x12 )+δpq(p−q)
(x+x1 +x2 +x12 )−δ 2 pq(p−q)(p2 −pq+q 2 ) = 0
•
Lax operators:


(k−p)(δkp−x1 )+kx




2
2
L = t
−p δk(k−p)(δk −δkp+δp −x−x1 )+xx1 


p
− (k−p)(δkp−x)+kx1


(k−q)(δkq−x2 )+kx




2
2
M = s
−q δk(k−q)(δk −δkq+δq −x−x2 )+xx2 


q
− (k−q)(δkq−x)+kx2
.
•
with
t=
1
q
k(k−p)((x−x1 )2 −2δp2 (x+x1 )+δ 2 p4 )
and
s=
1
q
k(k−q)((x−x2 )2 −2δq 2 (x+x2 )+δ 2 q 4 )
Note:
t2 s
t s1
=
=
)2
2δp2 (x
δ 2 p4
q (x − x1 −
+ x1 ) +
p (x − x2 )2 − 2δq 2 (x + x2 ) + δ 2 q 4
2
2
2
2
p (X + X1 ) − δp
(X − X1 ) − δp
2
2
2
2
q (X + X2 ) − δq
(X − X2 ) − δq
with x = X 2 , and, consequently, x1 = X12 , x2 = X22
•
Example 10: Q3 equation (ABS 2003)
(q 2 −p2 )(xx12 +x1 x2 )+q(p2 −1)(xx1 +x2 x12 )
2
δ
−p(q 2 −1)(xx2 +x1 x12 )−
(p2 −q 2 )(p2 −1)(q 2 −1) = 0
4pq
•
Lax operators:


−4kp p(k2 −1)x+(p2 −k2 )x1




2
2
2
4
2
2
2
2
2
2
L = t
−(p −1)(δk −δ k −δ p +δ k p −4k pxx1 )


−4k2 p(p2 −1)
4kp p(k2 −1)x1 +(p2 −k2 )x


−4kq q(k2 −1)x+(q 2 −k2 )x2




2
2
2
4
2
2
2
2
2
2
M= s
−(q −1)(δk −δ k −δ q +δ k q −4k qxx2 )


−4k2 q(q 2 −1)
4kq q(k2 −1)x2 +(q 2 −k2 )x
.
•
with
t=
1
q
2k p(k2−1)(k2−p2 )(4p2 (x2 +x21 )−4p(1+p2 )xx1 +δ 2 (1−p2 )2 )
and
1
s=
2k
q
q(k2−1)(k2−q 2 )(4q 2 (x2 +x22 )−4q(1+q 2 )xx2 +δ 2 (1−q 2 )2 )
Note:
t2 s
t s1
2
2
2
2
2
2
2
2
q(q −1) 4p (x +x1 )−4p(1+p )xx1 +δ (1−p )
=
2
p(p2 −1) 4q 2 (x2 +x2 )−4q(1+q 2 )xx2 +δ 2 (1−q 2 )2
2
2
2
2
2
2
2
q(q −1) 4p (x−x1 ) −4p(p−1) xx1 +δ (1−p )
=
p(p2 −1) 4q 2 (x−x2 )2 −4q(q−1)2 xx2 +δ 2 (1−q 2 )2
2
2
2
2
2
2
2
q(q −1) 4p (x+x1 ) −4p(p+1) xx1 +δ (1−p )
=
2
2
2
2
2
2
2
p(p −1) 4q (x+x2 ) −4q(q+1) xx2 +δ (1−q )
where
4p2 (x2 +x21 )−4p(1+p2 )xx1 +δ 2 (1−p2 )2
= (p−eX+X1 )(p−e−(X+X1 ) )(p−eX−X1 )(p−e−(X−X1 ) )
= (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 = cosh(X), and, consequently,
x1 = cosh(X1 ), x2 = cosh(X2 )
.
•
Example 11: Q3 equation with δ = 0 (ABS 2003)
(q 2 − p2 )(xx12 + x1 x2 ) + q(p2 − 1)(xx1 + x2 x12 )
−p(q 2 − 1)(xx2 + x1 x12 ) = 0
•
Lax operators:


(p2 −k2 )x1 +p(k2 −1)x
−k(p2 −1)xx1

L = t
(p2 −1)k
− (p2 −k2 )x+p(k2 −1)x1


(q 2 −k2 )x2 +q(k2 −1)x
−k(q 2 −1)xx2

M = s
(q 2 −1)k
− (q 2 −k2 )x+q(k2 −1)x2
.
•
with t =
or t =
1
px−x1
1
px1 −x
or t = √
and s =
1
qx−x2
1
qx2 −x
1
(k2 −1)(p2 −k2 )(px−x1 )(px1 −x)
and s = √
Note:
and s =
1
(k2 −1)(q 2 −k2 )(qx−x2 )(qx2 −x)
t2 s
t s1
=
(q 2 −1)(px−x1 )(px1 −x)
(p2 −1)(qx−x2 )(qx2 −x)
.
•
Example 12: (α, β)-equation (Tran)
(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:


(p−α)(p−β)x+(k2−p2 )x1 −(k−α)(k−β)xx1

L = t
(k+α)(k+β) − (p+α)(p+β)x1 +(k2−p2 )x


(q−α)(q−β)x+(k2−q 2 )x2 −(k−α)(k−β)xx2

M = s
(k+α)(k+β) − (q+α)(q+β)x2 +(k2−q 2 )x
.
•
with t =
1
(α−p)x+(β+p)x1 )
or t =
1
(β−p)x+(α+p)x1 )
or t =
q
and s =
1
(α−q)x+(β+q)x2 )
1
(β−q)x+(α+q)x2 )
1
and s =
Note:
and s =
(p2 −k2 )((β−p)x+(α+p)x1 )((α−p)x+(β+p)x1 )
1
q
(q 2 −k2 )((β−q)x+(α+q)x2 )((α−q)x+(β+q)x2 )
t2 s
t s1
((β−p)x+(α+p)x1 )((α−p)x+(β+p)x1 )
= (β−q)x+(α+q)x (α−q)x+(β+q)x
(
2 )(
2)
•
Example 13: A1 equation (ABS 2003)
p(x+x2 )(x1 +x12 )−q(x+x1 )(x2 +x12 )−δ 2 pq(p−q) = 0
•
Q1 if x1 → −x1 and x2 → −x2
Lax operators:


(k − p)x1 + kx −p δ 2 k(k − p) + xx1
L = t

p
− (k − p)x + kx1


(k − q)x2 + kx −q δ 2 k(k − q) + xx2
M = s

q
− (k − q)x + kx2
.
•
with t =
s=
q
Note:
1
q
k(k−p)((δp+x+x1 )(δp−x−x1 ))
1
and
k(k−q)((δq+x+x2 )(δq−x−x2 ))
t2 s
t s1
=
q (δp+(x+x1 ))(δp−(x+x1 ))
p(δq+(x+x2 ))(δq−(x+x2 ))
However, the choices t =
1
δp±(x+x1 )
and s =
CANNOT be used.
This lattice needs further investigation!
1
δq±(x+x2 )
.
•
Example 14: A2 equation (ABS 2003)
(q 2 − p2 )(xx1 x2 x12 + 1) + q(p2 − 1)(xx2 + x1 x12 )
−p(q 2 − 1)(xx1 + x2 x12 ) = 0
Q3 with δ = 0 via Möbius transformation:
x → x, x1 →
•
1
, x2
x1
→
1
, x12
x2
→ x12 , p → p, q → q
Lax operators:


k(p2 −1)x
− p2 −k2 +p(k2 −1)xx1

L = t
p(k2 −1)+(p2 − k2 )xx1
−k(p2 −1)x1


k(q 2 −1)x
− q 2 −k2 +q(k2 −1)xx2

M = s
q(k2 −1)+(q 2 −k2 )xx2
−k(q 2 −1)x2
.
•
with t = √
and s = √
Note:
1
(k2 −1)(k2 −p2 )(p−xx1 )(pxx1 −1)
1
(k2 −1)(k2 −q 2 )(q−xx2 )(qxx2 −1)
t2 s
t s1
=
(q 2 −1)(p−xx1 )(pxx1 −1)
(p2 −1)(q−xx2 )(qxx2 −1)
However, the choices t =
or t =
1
pxx1 −1
and s =
1
p−xx1
and s =
1
q−xx2
1
qxx2 −1
CANNOT be used.
This lattice needs further investigation!
.
•
Example 15: Discrete sine-Gordon equation
(xx1 x2 x12 ) − pq(xx12 − x1 x2 ) − 1 = 0
H3 with δ = 0 via extended Möbius transformation:
x → x, x1 → x1 , x2 →
•
1
, x12
x2
→ − x112 , p → p, q →
1
q
Discrete sine-Gordon equation is NOT consistent
around the cube, but has a Lax pair!
Lax operators:


p −kx1

L=
px1
− xk
x

M =
qx2
 x
− xk2
1
− kx
q


.
•
Conclusions and Future Work
Mathematica code works for P∆Es in 2D defined
on quad-graphs (quadrilateral faces)
•
Code can be used to test (i) consistency around
the cube and (ii) Lax pairs
•
Avoid the determinant method to avoid square
roots! Factorization plays an essential role!
•
Work with equivalent representation of the lattice
(3-leg form)
•
Try to find the simplest Lax pair, in particular, for
the lattices (A1) and (A2)
.
•
Consistency around cube =⇒ P∆E has Lax pair
•
P∆E has Lax pair ; consistency around cube
Example: discrete sine-Gordon equation
•
Lax pair reveals itself from the map (by
eliminations)
•
Hard case: Q4 equation (elliptic curves,
Weierstraß functions) (Nijhoff, 2001)
•
P∆Es in 3D: Lax pair will consist of tensors.
Examples: discrete Kadomtsev-Petviashvili (KP)
equations (AKP, BKP, KP lattice)
.
Thank You