Symbolic Computation of Lax Pairs of
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/
Minisymposium on Recent Advances in
Continuous and Discrete Integrable Systems
2009 SIAM Annual Meeting, Denver, Colorado
Monday, July 6, 2009, 12:00
Collaborators
Prof. Reinout Quispel
Dr. Peter van der Kamp
La Trobe University, Melbourne, Australia
Ph.D. student: Terry Bridgman
Colorado School of Mines, Golden, Colorado
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 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
.
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 Bäcklund 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 + 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,
(un,m , un+1,m , un,m+1 , un+1,m+1 ) = (x, x1 , x2 , x12 )
•
Alternate notations (in the literature):
ˆ
(un,m , un+1,m , un,m+1 , un+1,m+1 ) = (u, ũ, û, ũ)
(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 + q 2 = 0
−→ (x − x12 )(x1 − x2 ) − p2 + q 2 = 0
(un,m − un+1,m+1 )(un+1,m − un,m+1 ) − p2 + q 2 = 0
(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
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) = a1 xx1 x2 x12 + a2 xx1 x2 +
. . . + a14 x2 + a15 x12 + a16
2. Invariant under D4 (symmetries of square)
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
Superposition of Bäcklund transformations
between 4 solutions x, x1 , x2 , x3 (3 parameters:
p, q, k)
x23
x123
x2
x3
q
x
x12
x13
k
p
x1
•
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
x23
x123
x2
x3
q
x
x12
x13
k
p
x1
Verification of Consistency 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)
(x − x12 )(x1 − x2 ) + q − p = 0
I
(H2)
(x−x12 )(x1 −x2 )+(q−p)(x+x1 +x2 +x12 )+q 2 −p2 = 0
I
(H3)
p(xx1 + x2 x12 ) − q(xx2 + xx12 ) + δ(p2 − q 2 ) = 0
•
List A
I
(A1)
p(x+x2 )(x1 +x12 )−q(x+x1 )(x2 +x12 )−δ 2 pq(p−q) = 0
I
(A2)
(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)
p(x−x2 )(x1 −x12 )−q(x−x1 )(x2 −x12 )+δ 2 pq(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+q 2 ) = 0
I
(Q3)
(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) (mother)
Hietarinta’s Parametrization
sn(α; k) (x1 x2 + xx12 ) + sn(β; k) (xx1 + x2 x12 )
−sn(α − β; k) (xx2 + x1 x12 )
+sn(α; k) sn(β; k) sn(α − β; k)(1 + k2 xx1 x2 x12 ) = 0
where sn(α; k) is the Jacobi elliptic sine
function with modulus k
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 ψ + [L, M]ψ =
•
Linear problem
α
6
(ut + αuux + uxxx ) ψ
? 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,
Lt ψ + (LM − ML)ψ = O
Lax equation:
Lt + [L, M] = 0 (on PDE)
Lax Pair – Matrix Formalism (AKNS)
•
Express compatibility of
Ψx = X Ψ,
Ψt = T Ψ
 
ψ
where Ψ =   , X and T are 2 × 2 matrices
φ
•
Lax equation (Zero-curvature equation):
Xt − Tx + [X, T] = 0
(on PDE)
with commutator [X, T] = XT − TX
•
Korteweg-de Vries equation
Lax matrices:

0
1


X=
λ − α6 u 0

a(t) + α6 ux
T=
−4λ2 + α3 λu +
α2 2
u
18
−4λ − α3 u
+ α6 u2x a(t) − α6 ux


Peter D. Lax (1926-)
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 (ABS Lattices)
•
Require that ψ1 = Lψ,
ψ2 = Mψ
 
f
Here L and M are 2 × 2 matrices and ψ =  
g
So, ψ1 = ψn→n+1 , ψ2 = ψm→m+1
•
Express compatibility:
ψ12 = L2 ψ2 = L2 Mψ
ψ12 = M1 ψ1 = M1 Lψ
•
Hence, L2 Mψ − M1 Lψ = 0
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 = tLc = t 
1
−x1


x q 2 − k2 − xx2

M = sMc = s 
1
−x2
with t = s = 1
1
or t = √DetL
= √
c
Note:
t2 s
t s1
=1
1
k2 −p2
and s =
√ 1
DetMc
= √
1
k2 −q 2
•
Note: L2 M − M1 L = (x − x12 )(x1 − x2 ) −
with


x p2 − k2 − xx1

L=
1
−x1


x q 2 − k2 − xx2

M =
1
−x2


−1 x1 + x2

N =
0
1
p2
+
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
1
1
and
s
=
,
or
t
=
s
=
,
x1
x2
x
√ 1
√ 1
√ 212
=
and
s
=
DetLc
DetMc
(p −k )xx1
with t =
or t =
Note:
t2 s
t s1
=
x x1
x x2
=
x1
x2
=√
1
(q 2 −k2 )xx2
•
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:

−x1
1
0





L = t
−y1
0 1


p − k − xy1 + x1 z −z u


−x2
1 0




M = s
−y2
0 1


q − k − xy2 + x2 z −z u
with t = s = 1, or t =
Note:
t2 s
t s1
=1
1
√
3
p−k
and s =
1
√
3
q−k
•
Example 4: System of pKdV equations
(Xenitidis and Mikhailov 2009)
(x − x12 )(y1 − y2 ) − p2 + q 2 = 0
(y − y12 )(x1 − x2 ) − p2 + q 2 = 0
•
Lax operators:


2 − k 2 − xy )
0
0
tx
t(p
1 



0

0
t
−ty1

L=


2
2
T y T (p − k − x1 y) 0

0


T
−T x1
0
0

2 − k 2 − xy )
0
0
sx
s(q
2 



0

0
s
−sy2

M =


2
2

Sy S(q − k − x2 y) 0
0


S
−Sx2
0
0

with t = s = T = S = 1,
or t T =
Note:
or
√ 1
DetLc
t2 S
T s1
T2 S
T S1
=
1
α−k
= 1 and
and s S =
T2 s
t S1
1
β−k
=1
= 1 with T = t T, S = s S
•
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:


−1
x1

L = t
y k − α − yx1


−1
x2

M = s
y k − β − yx2
1
1
1
with t = s = 1, or t = √DetL
= √α−k
and s = √β−k
c
t2 s
Note: t s1 = 1
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
Use the equation on floor
(x − x13 )(x1 − x3 ) − p2 + k2 = 0
p2 −k2
x1 −x3
to compute x13 = x −
=
Use equation on left face
x3 x−xx1 +p2 −k2
x3 −x1
(x − x23 )(x3 − x2 ) − k2 + q 2 = 0
to compute x23 = x −
q 2 −k2
x2 −x3
=
x3 x−xx2 +q 2 −k2
x3 −x2
x123
x23
x12
x2
q
x13
x3
k
x
p
x1
•
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
 
f
Matches ψ1 = Lψ with ψ =  
g
Similarly, from x23 :

 
 
f2
x q 2 − k2 − xx2
f
  = s
 
g2
1
−x2
g
or ψ2 = Mψ. Therefore,


x p2 − k2 − xx1

L = t Lc = t 
1
−x1


x q 2 − k2 − xx2

M = s Mc = s 
1
−x2
•
Alternate method of deriving Lc and Mc


∂Q
− ∂x2
−Q


L = t Lc = t
2
∂ Q
∂x2 ∂x12
∂Q
∂x12
x2 =x12 =0


x p2 − k2 − xx1

= t
1
−x1
where Q(x, x1 , x2 , x12 ; p, k) and t(x, x1 ; p, k)
•
Apply x1 → x2 and p → q


x q 2 − k2 − xx2

M = s Mc = s 
1
−x2
where s = t(x, x2 ; q, k)
•
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: 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
4
q (x − x1 ) − 2p (x + x1 ) + p
t2 s
=
2
2
4
t s1
p (x − x2 ) − 2q (x + x2 ) + q
2
2
2
2
q (X + X1 ) − p
(X − X1 ) − p
=
2
2
2
2
p (X + X2 ) − q
(X − X2 ) − q
after setting x = F (X) = X 2 , hence, x1 = X1 2 and
x2 = X2 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
2
2
2
2
2
2
4q (x +x2 )−4q(1+q )xx2 +δ (1−q )
Set x = F (X) = δ cosh(X) then
4p2 (x2 +x21 )−4p(1+p2 )xx1 +δ 2 (1−p2 )2
= δ 2 (p−eX+X1 )(p−e−(X+X1 ) )(p−eX−X1 )(p−e−(X−X1 ) )
= δ 2 (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 non-singular 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 = x1 or t = x11 and s =
Note: tt2 ss1 = xx xx12 = xx12
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
1
1
with t = δp±(x−x
and
s
=
,
δq±(x−x2 )
1)
1
and
or t = q
k(p−k)((δp+x−x1 )(δp−x+x1 ))
1
s= q
k(q−k)((δq+x−x2 )(δq−x+x2 ))
Note:
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 )+pq(p−q)
(x+x1 +x2 +x12 )−pq(p−q)(p2 −pq+q 2 ) = 0
•
Lax operators:


(k−p)(kp−x1 )+kx




2
2
L = t
−p k(k−p)(k −kp+p −x−x1 )+xx1 


p
− (k−p)(kp−x)+kx1


(k−q)(kq−x2 )+kx




2
2
M = s
−q k(k−q)(k −kq+q −x−x2 )+xx2 


q
− (k−q)(kq−x)+kx2
•
with
t=
1
q
k(k−p)((x−x1 )2 −2p2 (x+x1 )+p4 )
and
s=
1
q
k(k−q)((x−x2 )2 −2q 2 (x+x2 )+q 4 )
Note:
t2 s
t s1
=
=
)2
2p2 (x
p4
q (x − x1 −
+ x1 ) +
p (x − x2 )2 − 2q 2 (x + x2 ) + q 4
2
2
2
2
p (X + X1 ) − p
(X − X1 ) − p
2
2
2
2
q (X + X2 ) − q
(X − X2 ) − q
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
= δ 2 (p−eX+X1 )(p−e−(X+X1 ) )(p−eX−X1 )(p−e−(X−X1 ) )
= δ 2 (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 )
•
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 (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:


(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.
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.
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:




qx2
1
p −kx1
−
kx 

L=
M = x
px1
x2
− xk
−
q
x
k
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.
•
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.
Thank You