æ
Symbolic Integrability Tests
for Nonlinear Partial Differential and
Differential-Difference Equations
Willy Hereman
Department of Mathematical and Computer Sciences
Colorado School of Mines
Golden, Colorado, 80401-1887, U.S.A.
http://www.mines.edu/fs home/whereman/
whereman@mines.edu
Minisymposium, ICIAM 2003
July 7-11, Sydney, Australia
Collaborators: Ünal Göktaş, Mark Hickman
Jan Sanders, & Jing-Ping Wang
Students: Doug Baldwin, Jack & Ryan Sayers
Holly Eklund, Paul Adams, Adam Ringler
Research supported in part by NSF under Grant CCR-9901929
Talk on WEB (MSRI):
http://www.msri.org/publications/ln/msri/1998/scga/hereman/1
1
Outline
Introduction
Purpose
Strategy
Part I: Partial Differential Equations (PDEs)
• typical examples with dilation invariance
• algorithm for conserved densities (homotopy operator)
Analogy between PDEs and DDEs
Part II: Differential-difference Equations (DDEs)
• typical examples with dilation invariance
• algorithm for conserved densities (homotopy operator)
Mathematica software
Future Work
Conclusions
Publications
2
Introduction – Purpose – Strategy
• Alternatives
– Painlevé test
– IST, Lax pair, soliton solutions...
• Purpose
Design and implement algorithms to compute polynomial
conservation laws, generalized symmetries, and recursion operators
for nonlinear systems of PDEs and DDEs.
• Motivation
– Conservation laws describe the conservation of physical
quantities (linear momentum, energy).
– Conservation laws help in the study of quantitative and
qualitative properties of PDEs and DDEs and their solutions.
– Conserved densities can be used to test numerical integrators.
– The existence of a sufficiently large (in principal infinite)
number of conservation laws and generalized symmetries assures
complete integrability.
– Conserved densities and symmetries aid in finding the recursion
operator (guarantees the existence of ∞ many symmetries).
3
• Key Observation
Conserved densities, generalized symmetries, and recursion operators are invariant under the dilation (scaling) symmetry of the given
PDE or DDE.
• Strategy
Exploit dilation symmetry as much as possible.
Keep the computations as simple as possible.
Use linear algebra
* solve linear systems
* construct basis vectors (building blocks)
* use linear independence
* work in finite dimensional spaces
Use calculus and differential equations
* derivatives
* integrals (as little as possible)
* solve systems of linear ODEs
Use tools from variational calculus
* variational derivative (Euler operator)
* higher Euler operators
−1
* use homotopy operators to compute D−1
x and (D − I)
* Fréchet derivative
* calculus with operators
Use analogy between continuous and semi-discrete cases
4
Part I: Partial Differential Equations (PDEs)
• Nonlinear system of evolution equations
ut = F(u, ux, u2x, ..., umx)
in a (single) space variable x and time t. Notation:
u = (u1, u2, ..., un), F = (F1, F2, ..., Fn),
∂u
∂ iu
ut =
, uix =
.
∂t
∂xi
In practice: denote components of u by u, v, w, · · · .
F is polynomial or transcendental function in u, ux, u2x, ..., umx.
• Examples — Dilation Invariance
Example 1: KdV equation, ut + uux + u3x = 0, has scaling
symmetry
(t, x, u) → (λ−3t, λ−1x, λ2u).
Example 2: Boussinesq (wave) equation:
utt − u2x + 3uu2x + 3ux2 + αu4x = 0,
written as a first-order system (v auxiliary variable):
ut + vx = 0,
vt + ux − 3uux − αu3x = 0,
is not scaling invariant (ux and u3x are conflicting terms).
Introduce an auxiliary parameter β, then
ut + vx = 0,
vt + βux − 3uux − αu3x = 0,
has scaling symmetry:
(x, t, u, v, β) → (λ−1x, λ−2t, λ2u, λ3v, λ2β).
5
Example 3: Sine-Gordon system:
ut = v,
vt = uxx + α sin(u).
has scaling symmetry
(t, x, u, v, α) → (λ−1t, λ−1x, λ0u, λv, λ2α).
• Computation of the dilation symmetry
weight w of variable: # x-derivatives variable corresponds to. The
rank of a monomial is its total weight.
Example 1: KdV equation, ut + uux + u3x = 0.
Set w(Dx) = 1 or w(x) = −1. Require uniformity in rank.
w(u) + w(Dt) = 2w(u) + 1 = w(u) + 3.
Solve the linear system
w(u) = 2, w(Dt) = 3, or w(t) = −3.
(t, x, u) → (λ−3t, λ−1x, λ2u).
So,
Example 2: Boussinesq system
ut + vx = 0,
vt + βux − 3uux − αu3x = 0.
Solve
w(u) + w(Dt) = w(v) + 1,
w(v) + w(Dt) = w(β) + w(u) + 1 = 2w(u) + 1 = w(u) + 3.
So,
Hence,
w(u) = 2, w(v) = 3, w(β) = 2, w(Dt) = 2.
(x, t, u, v, β) → (λ−1x, λ−2t, λ2u, λ3v, λ2β).
6
Example 3: Sine-Gordon system
ut = v,
vt = uxx + α sin(u).
Solve
w(u) + w(Dt) = w(v),
w(v) + w(Dt) = w(u) + 2 = w(α) + w(u) = w(α) + 3w(u) = · · ·
So,
w(u) = 0, w(v) = w(Dt) = 1, and w(α) = 2.
(t, x, u, v, α) → (λ−1t, λ−1x, λ0u, λv, λ2α).
Hence,
• Conservation Law for PDEs
Dtρ + DxJ = 0
on PDE,
conserved density ρ and flux J.
P =
Z +∞
−∞
ρ dx = constant in time
if J vanishes at infinity.
• Examples of Conserved Densities:
Example 1: Korteweg-de Vries (KdV) equation
ut + uux + u3x = 0
ρ(1) = u,
2
ρ(2) = u ,
u2
Dt(u) + Dx( + u2x) = 0.
2
2u3
Dt(u ) + Dx(
+ 2uu2x − ux2) = 0.
3
2
ρ(3) = u3 − 3ux2,
7
Dt u3 −3ux2
...
3
+Dx  u4 −6uux2 +3u2u2x +3u2x2 −6uxu3x = 0.
4


ρ(6) = u6 − 60u3ux2 − 30ux4 + 108u2u2x2
720 3 648
216 2
+
u2x −
uu3x2 +
u4x .
7
7
7
Example 2: Boussinesq system (before setting β = 1) :
ut + vx = 0,
vt + ux − 3uux − αu3x = 0,
ρ(1) = u,
ρ(3) = uv,
ρ(2) = v,
ρ(4) = βu2 − u3 + v 2 + αux2.
Example 3: Sine-Gordon system:
ut = v,
vt = uxx + α sin(u).
ρ(1) = 2α cos(u) + v 2 + ux2,
ρ(2) = uxv,
J(1) = −2uxv
1
1
J(2) = −( v 2 + ux2 − α cos(u))
2
2
ρ(3) = 12 cos(u)vux + 2v 3ux + 2vux3 − 16vxu2x
ρ(4) = 2 cos2(u) − 2 sin2(u) + v 4 + 6v 2ux2 + ux4 + 4 cos(u)v 2
+20 cos(u)ux2 − 16vx2 − 16u2x2.
8
• Algorithm Conserved Densities Polynomial PDEs
(i) Determine weights (scaling properties) of variables and
auxiliary parameters.
(ii) Construct the form of the density (find monomial building blocks).
(iii) Determine the constant coefficients (parameters).
(iv) Compute the flux with the homotopy operator.
Example: Density of rank 6 for the KdV equation
ut + uux + u3x = 0
Step 1: Compute the weights (dilation symmetry).
Solve
w(u) + w(Dt) = 2w(u) + 1 = w(u) + 3.
Hence,
w(u) = 2,
w(Dt) = 3.
Step 2: Determine the form of the density.
List all possible powers of u, up to rank 6 :
[u, u2, u3].
Introduce x derivatives to ‘complete’ the rank.
u has weight 2, introduce D4x.
u2 has weight 4, introduce D2x.
u3 has weight 6, no derivative needed.
Apply the Dx derivatives.
Remove total derivative terms (Dxupx) and highest derivative terms:
9
[u4x] → [ ]
empty list.
[ux2, uu2x] → [ux2]
since uu2x = (uux)x − ux2.
[u3] → [u3].
Linearly combine the ‘building blocks’:
ρ = c 1 u3 + c 2 ux 2 .
Step 3: Determine the coefficients ci.
Use the defining equation
Dtρ + DxJ = 0 (on PDE),
Compute
E =
=
=
=
m ∂ρ
∂ρ X
∂ρ
Dtρ =
+
Dkxut =
+ ρ0(u)[F ]
∂t k=0 ∂ukx
∂t
2
3c1u ut + 2c2uxuxt
−3c1u2(uux + u3x) − 2c2ux(uux + u3x)x.
−(3c1u3ux + 3c1u2u3x + 2c2u3x + 2c2uuxu2x + 2c2uxu4x).
Apply the Euler operator (continuous variational derivative)
m
X
k ∂
(−D
)
L(0)
=
x
u
∂ukx
k=0
∂
∂
∂
∂
=
− Dx
+ D2x
+ · · · + (−1)mDm
.
x
∂u
∂ux
∂u2x
∂umx
to E of order m = 4. Result:
L(0)
u (E) = −6(3c1 + c2 )ux uxx ≡ 0
So, c1 = − 13 c2. Set c2 = −3, then c1 = 1.
Hence,
ρ = u3 − 3ux2.
10
Step 4: Compute the flux J.
– Method 1: Integrate by parts (simple cases)
Integration of DxJ = −E yields
3
J = u4 − 6uux2 + 3u2u2x + 3u2x2 − 6uxu3x.
4
– Method 2: Build the form of J (cumbersome)
Note: Rank J = Rank ρ + Rank Dt − 1.
Build up form of J. Compute
m ∂J
∂J
X
DxJ =
+
u(k+1)x,
∂x k=0 ∂ukx
m is the order of J. Match DxJ = −E.
– Method 3: Use the homotopy operator (most powerful)
Higher Euler Operators:
L(i)
u =
 
∞
X
 
 
k=i
∂
k
(−Dx)k−i
.
i
∂ukx
Examples (scalar case, u = u1 = u):
∂
∂
∂
∂
− Dx
+ D2x
− D3x
+ ···
∂u
∂ux
∂u2x
∂u3x
∂
∂
2 ∂
3 ∂
L(1)
=
−
2D
+
3D
−
4D
+ ···
x
u
x
x
∂ux
∂u2x
∂u3x
∂u4x
∂
∂
2 ∂
3 ∂
L(2)
=
−
3D
+
6D
−
10D
+ ···
x
u
x
x
∂u2x
∂u3x
∂u4x
∂u5x
∂
∂
2 ∂
3 ∂
L(3)
=
−
4D
+
10D
−
20D
+ ···
x
u
x
x
∂u3x
∂u4x
∂u5x
∂u6x
L(0)
u =
11
The flux is
J(u) =
where
jr (u) =
Z 1 X
n
0 r=1
m−1
X
i=0
jr (u)[λu]
dλ
.
λ
Dxi (ur L(i+1)
ur (−E))
m is the order of E, and jr (u)[λu] means
u → λu, ux → λux, u2x → λu2x, etc.
Demonstration (scalar case, u = u1 = u, j1(u) = j(u)):
Compute J via the homotopy operator!
−E = 3u3ux + 3u2u3x − 6u3x − 6uuxu2x − 6uxu4x.
i
L(i+1)
(−E)
Dxi (uL(i+1)
(−E))
u
u
0 3u3 +24uu2x +18u4x +12u2x
3u4 +24u2u2x +18uu4x +12uu2x
1
−24uux −36u3x
−48uu2x −24u2u2x −36uxu3x −36uu4x
2
3u2 +24u2x
18uu2x +9u2u2x +24u22x +48uxu3x +24uu4x
3
−6ux
−18u22x −24uxu3x −6uu4x
Hence,
j(u) = 3u4 − 18uu2x − 12uxu3x + 9u2u2x + 6u22x.
Thus, the homotopy operator gives
dλ
0
λ
Z 1
= 0 (3λ3u4 − 18λ2uu2x − 12λuxu3x + 9λ2u2u2x + 6λu22x) dλ
3
= u4 − 6uux2 − 6uxu3x + 3u2u2x + 3u2x2.
4
J(u) =
Z 1
j(u)[λu]
12
Analogy PDEs and DDEs
Continuous Case (PDEs) Semi-discrete Case (DDEs)
System
ut = F(u, ux , u2x , ...)
u̇n=F(..., un−1 , un , un+1 , ...)
Conservation Law
Dt ρ + Dx J = 0
ρ̇n + Jn+1 − Jn = 0
Symmetry
Dt G = F0 (u)[G]
∂
= ∂
F(u + G)|=0
Dt G = F0 (un )[G]
∂
= ∂
F(un + G)|=0
Recursion Operator Dt R + [R, F0 (u)] = 0
Table 1:
Dt R + [R, F0 (un )] = 0
Conservation Laws and Symmetries
KdV Equation
Volterra Lattice
Equation
ut = 6uux + u3x
u̇n = un (un+1 − un−1 )
Densities
ρ = u, ρ = u2
ρ = u3 − 12 u2x
ρn = un , ρn = un ( 12 un + un+1 )
ρn = 13 u3n +un un+1 (un +un+1 +un+2 )
Symmetries
G = ux , G = 6uux + u3x
G = 30u2 ux + 20ux u2x
+10uu3x + u5x
G = un un+1 (un + un+1 + un+2 )
−un−1 un (un−2 + un−1 + un )
Recursion Operator R = D2x + 4u + 2ux D−1
x
Table 2:
R = un (I + D)(un D − D−1 un )
(D − I)−1 u1n
Prototypical Examples
13
Analogy PDEs and DDEs
Conservation laws for PDEs
Dtρ + DxJ = 0
density ρ, flux J. Compute E = Dtρ.
To guarantee the existence of J, apply the Euler operator
∂
∂ukx
k=0
∂
∂
∂
∂
=
− Dx(
) + D2x(
) + · · · + (−1)mDm
(
).
x
∂u
∂ux
∂u2x
∂umx
L(0)
u =
m
X
(−1)k Dkx
to E of order m. Dx is the differential operator.
If L(0)
u (E) = 0, then E is a total x-derivative (−Jx ).
If L(0)
u (E) 6= 0, the nonzero terms must vanish identically.
E must be in the kernel of L(0)
u operator, or equivalently, E must be
in the image of Dx operator.
Computation of flux J:
Apply the homotopy operator
J(u) =
Z 1 X
n
0 r=1
jr (u)[λu]
dλ
.
λ
where jr (u) is computed with
jr (u) =
m−1
X
i=0
Dix(ur L(i+1)
ur (−E))
with higher Euler operators (continuous):
 
m k
X
 
k−i ∂
 (−Dx )
L(i)
=
.
u
∂ukx
k=i i
14
Conservation laws for DDEs
ρ̇n + Jn+1 − Jn = ρ̇n + ∆Jn = 0
density ρn, flux Jn.
Compute E = ρ̇n.
To guarantee existence of Jn, apply the discrete Euler operator
L(0)
un
p
X
∂
∂un+k
k=−q
∂
∂
∂
∂
=
+ D(
) + D2 (
) + · · · + Dq (
)
∂un
∂un−1
∂un−2
∂un−q
∂
∂
∂
) + D−2(
) + · · · + D−p(
)
+ D−1(
∂un+1
∂un+2
∂un+p
=
D−k
to E with maximal negative and positive shifts on u are q and p.
D is the up-shift operator, D−1 the down-shift operator.
Applied to a monomial m
D−1m = m|n→n−1 and Dm = m|n→n+1.
Note: D (up-shift operator) corresponds the differential operator Dx
due to the forward difference
∂J
Jn+1 − Jn
→
(∆x = 1)
∂x
∆x
If L(0)
un (E) = 0, then E matches −(Jn+1 − Jn ).
If L(0)
un (E) 6= 0, the nonzero terms must vanish identically.
15
In practice:
Compute Ẽ = Dq E (remove negative shifts) and apply
L(0)
un
∂ p+q
X
( D−k )
=
∂un k=0
∂
=
(I + D−1 + D−2 + · · · + D−(p+q))
∂un
Computation of flux J˜n
Apply the homotopy operator
J˜n(un) =
Z 1 X
m
0 r=1
j̃r,n(un)[λun]
dλ
.
λ
where j̃r,n(un) is computed with
j̃r,n(un) =
p+q−1
X
i=0
(D − I)i(ur,nL(i+1)
ur,n (−Ẽ))
with discrete higher Euler operators:
 
L(i)
un
∂ p+q
X k  −k
 D
).
=
( 
∂un k=i i
Down-shift J˜n by q steps: Jn = D−q J˜n.
16
Higher Euler Operators Side by Side
Continuous Case
∂
∂
∂
∂
+ D2x
− D3x
+ ···
− Dx
∂u
∂ux
∂u2x
∂u3x
∂
∂
2 ∂
3 ∂
L(1)
=
−
2D
+
3D
−
4D
+ ···
x
u
x
x
∂ux
∂u2x
∂u3x
∂u4x
∂
∂
2 ∂
3 ∂
L(2)
=
−
3D
+
6D
−
10D
+ ···
x
u
x
x
∂u2x
∂u3x
∂u4x
∂u5x
∂
∂
2 ∂
3 ∂
L(3)
−
4D
+
10D
−
20D
+ ···
=
x
u
x
x
∂u3x
∂u4x
∂u5x
∂u6x
L(0)
u =
Semi-discrete Case
∂
(I + D−1 + D−2 + D−3 + · · ·)
∂un
∂
L(1)
=
(D−1 + 2D−2 + 3D−3 + 4D−4 + · · ·)
un
∂un
∂
L(2)
=
(D−2 + 3D−3 + 6D−4 + 10D−5 + · · ·)
un
∂un
∂
L(3)
=
(D−3 + 4D−4 + 10D−5 + 20D−6 + · · ·)
un
∂un
L(0)
un =
17
Part II: Differential-difference Equations (DDEs)
• Nonlinear system of DDEs
(continuous in time, discretized in space)
u̇n = F(..., un−1, un, un+1, ...),
un = (u1,n, u2,n, · · · , um,n) and F = (F1, F2, · · · , Fm)
In practice: denote components of un by (un, vn, wn, · · ·).
F is polynomial with constant coefficients (parameters).
• Examples – Dilation Invariance
Example 1: Kac-van Moerbeke lattice
u̇n = un(un+1 − un−1).
is invariant under the scaling symmetry
(t, un) → (λ−1t, λun).
d
Using w( dt
) = 1 and w(un±p) = w(un),
w(un) + 1 = 2w(un).
Hence, w(un) = 1.
Example 2: One-dimensional Toda lattice
ÿn = exp (yn−1 − yn) − exp (yn − yn+1).
Change of variables:
un = ẏn,
vn = exp (yn − yn+1)
yields
u̇n = vn−1 − vn,
v̇n = vn(un − un+1).
18
System is invariant under the scaling symmetry
(t, un, vn) → (λ−1t, λun, λ2vn).
w(un) + 1 = w(vn),
w(vn) + 1 = w(vn) + w(un).
Hence,
w(un) = 1, w(vn) = 2.
Example 3: Ablowitz-Ladik lattice
i u̇n = un+1 − 2un + un−1 + κu∗nun(un+1 + un−1),
is an integrable discretization of the NLS equation:
iut + uxx + κu2u∗ = 0
u∗n is the complex conjugate of un.
Treat un and vn = u∗n as independent variables and add the complex
conjugate equation. Set κ = 1 (scaling) and absorb i in scale on t :
u̇n = un+1 − 2un + un−1 + unvn(un+1 + un−1),
v̇n = −(vn+1 − 2vn + vn−1) − unvn(vn+1 + vn−1).
No uniformity in rank! Introduce an auxiliary parameter α with
weight.
u̇n = α(un+1 − 2un + un−1) + unvn(un+1 + un−1),
v̇n = −α(vn+1 − 2vn + vn−1) − unvn(vn+1 + vn−1).
Uniformity in rank leads to
d
w(un) + w( ) = w(α) + w(un) = 2w(un) + w(vn),
dt
d
w(vn) + w( ) = w(α) + w(vn) = 2w(vn) + w(un).
dt
19
d
For w( dt
) = 1,
w(un) + w(vn) = w(α) = 1.
So, one solution is
1
w(un) = w(vn) = , w(α) = 1.
2
System is invariant under the scaling symmetry
1
1
(t, un, vn, α) → (λ−1t, λ 2 un, λ 2 vn, λα).
d
NOTE: Alternatively, for w( dt
) = 0,
w(un) + w(vn) = 0,
w(α) = 0.
The second scale helps eliminate terms in candidate density ρ.
Example 4: Taha-Herbst lattice
2
u̇n = −(1 + αh un +
+
+
βh2u2n)
1 1
( u
h3 2 n+2
− un+1 + un−1 − 21 un−2)
2
α 2
−
u
[u
n−1 + un (un+1 − un−1 ) + un+1 un+2
n+1
2h
β
2
2
2h [un+1 (un+2 + un ) − un−1 (un−2 + un )] ,
− un−1un−2]
is an integrable discretization of a combined KdV-mKdV equation
ut + 6αuux + 6βu2ux + uxxx = 0.
Discretizations the KdV and mKdV equations are special cases.
Set h = 1 (scaling). No uniformity in rank!
Introduce auxiliary parameters γ and δ with weights.
u̇n =
−(γ + αun + βu2n) δ( 12 un+2 − un+1 + un−1 − 21 un−2)
+ α2 [u2n+1 − u2n−1 + un(un+1 − un−1) + un+1un+2 − un−1un−2]
+ β2 [u2n+1(un+2 + un) − u2n−1(un−2 + un)] ,
20
Then,
w(γ) = w(δ) = 2w(un),
w(α) = w(un),
w(β) = 0.
and
w(un) + 1 = 5w(un),
Hence,
w(un) = w(α) = 14 ,
w(γ) = w(δ) = 12 ,
w(β) = 0,
System is invariant under the scaling symmetry
1
1
1
1
(t, un, α, δ, γ) → (λ−1t, λ 4 un, λ 4 α, λ 2 δ, λ 2 γ).
21
• Conservation Law for DDEs:
ρ̇n + Jn+1 − Jn = ρ̇n + ∆Jn
on DDE,
density ρn, flux Jn.
d X
X
X
( ρn) = ρ̇n = (Jn − Jn+1)
n
n
dt n
if Jn is bounded for all n.
Subject to suitable boundary or periodicity conditions
X
n
ρn = constant.
• Examples of Conserved Densities:
Example 1: Toda lattice
u̇n = vn−1 − vn,
v̇n = vn(un − un+1).
ρ(0)
n = ln(vn )
Jn(0) = un
ρ(1)
n = un
Jn(1) = vn−1
1 2
ρ(2)
n = 2 un + vn
Jn(2) = unvn−1
22
Example 2: Ablowitz-Ladik lattice
u̇n = un+1 − 2un + un−1 + unvn(un+1 + un−1),
v̇n = −(vn+1 − 2vn + vn−1) − unvn(vn+1 + vn−1).
∗
ρ(1)
n = un un−1
∗
ρ(2)
n = un un+1
∗
∗
1 2 ∗2
ρ(3)
n = 2 un un−1 + un un+1 un−1 vn + un un−2
∗
∗
∗
1 2 ∗2
ρ(4)
n = 2 un un+1 + un un+1 un+1 un+2 + un un+2
∗
∗
∗
∗
∗
1 3 ∗3
ρ(5)
n = 3 un un−1 + un un+1 un−1 un (un un−1 + un+1 un + un+2 un+1 )
+unu∗n−1(unu∗n−2 +un+1u∗n−1)+unu∗n(un+1u∗n−2 +un+2u∗n−1)+ unu∗n−3
∗
∗
∗
∗
∗
1 3 ∗3
ρ(6)
n = 3 un un+1 +un un+1 un+1 un+2 (un un+1 +un+1 un+2 +un+2 un+3 )
+unu∗n+2(unu∗n+1 +un+1u∗n+2)+unu∗n+3(un+1u∗n+1 +un+2u∗n+2)+unu∗n+3
Example 3: Taha-Herbst lattice
2
u̇n = −(1 + αh un +
+
+
βh2u2n)
1 1
( u
h3 2 n+2
− un+1 + un−1 − 21 un−2)
2
α 2
un+2
2h [un+1 − un−1 + un (un+1 − un−1 ) + un+1
β
2
2
2h [un+1 (un+2 + un ) − un−1 (un−2 + un )] ,
− un−1un−2]
ρ(1)
= αun + βunun+1
n
ρ(2)
n
α2 2 α2
=
un + unun+1 − unun+1 + αun2un+1 + αunun+12
2β
β
1
+ βun2un+12 + unun+2 + αunun+1un+2 + βunun+12un+2.
2
23
Algorithm Conserved Densities for DDEs (lattices)
• Tool: Up and Down Shift Operators
D−1 and D are the down-shift and up-shift operators.
For a monomial m :
D−1m = m|n→n−1,
and Dm = m|n→n+1.
Example
D−1un+2vn = un+1vn−1,
Dun−2vn−1 = un−1vn.
Compositions of D−1 and D define an equivalence relation.
All shifted monomials are equivalent.
Example
un−1vn+1 ≡ un+2vn+4 ≡ un−3vn−1.
• Tool: Equivalence Criterion
Two monomials m1 and m2 are equivalent, m1 ≡ m2, if
m1 = m2 + [Mn − Mn+1]
for some polynomial Mn.
Example: un−2un ≡ un−1un+1 since
un−2un = un−1un+1+[un−2un−un−1un+1] = un−1un+1+[Mn−Mn+1].
Main representative of an equivalence class is the monomial
with label n on u (or v).
Example: unun+2 is main representative of class
{un−1un+1, un+1un+3, · · ·}.
Use lexicographical ordering to resolve conflicts.
unvn+2 (not un−2vn) is the main representative of class
{un−3vn−1, un+2vn+4, · · ·}
24
• Algorithm
Three-step algorithm to find conserved densities:
(i)
(ii)
(iii)
(iv)
Determine the weights.
Construct the form of density.
Determine the coefficients.
Compute the flux with the discrete homotopy operator.
Example: Density of rank 3 of the Toda lattice,
u̇n = vn−1 − vn, v̇n = vn(un − un+1).
Step 1: Compute the weights.
Require uniformity in rank for each equation:
d
) = w(vn−1) = w(vn),
dt
d
w(vn) + w( ) = w(vn) + w(un) = w(vn) + w(un+1)
dt
w(un) + w(
d
Weights are shift invariant. Set w( dt
) = 1 and solve the linear
system: w(un) = w(un+1) = 1 and w(vn) = w(vn−1) = 2.
Step 2: Construct the form of the density.
List all monomials1 in un and vn of rank 3 or less:
G = {u3n, u2n, unvn, un, vn}.
For each monomial in G, introduce enough t-derivatives to obtain
weight 3. Use the DDE to remove u̇n and v̇n :
d0 3
3
0 (un ) = un ,
dt
1
d0
(unvn) = unvn,
dt0
In general algorithm shifts are also needed: u3n , un un+1 un−1 , u2n un+1 , etc.
25
d 2
(u ) = 2unvn−1 − 2unvn,
dt n
d
(vn) = unvn − un+1vn,
dt
d2
(un) = un−1vn−1 − unvn−1 − unvn + un+1vn.
dt2
Gather the resulting terms in a set
H = {u3n, unvn−1, unvn, un−1vn−1, un+1vn}.
Introduce main representatives.
Example: unvn−1 ≡ un+1vn are replaced by unvn−1.
Linearly combine the monomials in
I = {u3n, unvn−1, unvn}
to obtain
ρn = c1 u3n + c2 unvn−1 + c3 unvn.
Step 3: Determine the coefficients ci.
Require that ρ̇n + Jn+1 − Jn = 0 holds.
Compute ρ̇n. Use the DDE to remove u̇n and v̇n. Thus,
E = ρ̇n = (3c1 − c2)u2nvn−1 + (c3 − 3c1)u2nvn + (c3 − c2)vn−1vn
2
+c2un−1unvn−1 + c2vn−1
− c3unun+1vn − c3vn2 .
Shift E by q = 1 step up (remove negative shifts n − 1). Apply
L(0)
un
∂ p+q
∂
X
=
( D−k ) =
(I + D−1 + D−2 + · · ·)
∂un k=0
∂un
to Ẽ = DE.
26
The maximal shift p + q = 1 + 1 = 2 on un. Hence,
∂
L(0)
(I + D−1 + D−2)(Ẽ)
(
Ẽ)
=
un
∂un
= 2(3c1 − c2)unvn−1 + 2(c3 − 3c1)unvn
+(c2 − c3)un−1vn−1 + (c2 − c3)un+1vn ≡ 0
The maximal shift p + q = 0 + 1 = 1 on vn. Hence,
∂
(I + D−1)(Ẽ)
L(0)
(
Ẽ)
=
=
vn
∂vn
= (3c1 − c2)u2n+1 + (c3 − c2)vn+1 + (c2 − c3)unun+1
+2(c2 − c3)vn + (c3 − 3c1)u2n + (c3 − c2)vn−1 ≡ 0.
Solve the linear system
S = {3c1 − c2 = 0, c3 − 3c1 = 0, c2 − c3 = 0}.
The solution is 3c1 = c2 = c3. Choose c1 = 31 , and c2 = c3 = 1 :
Step 4: Compute the flux Jn.
– Method 1: Use equivalence criterion (simple cases)
Start from
2
E = ρ̇n = un−1unvn−1 + vn−1
− unun+1vn − vn2 .
Replace un−1unvn−1 by unun+1vn + [un−1unvn−1 − unun+1vn].
2
2
Replace vn−1
by vn2 + [vn−1
− vn2 ]. Thus
2
E = [un−1unvn−1 − unun+1vn] + [vn−1
− vn2 ]
Group the first and second terms in the square brackets to match
[Jn − Jn+1]. Hence,
2
E = [un−1unvn−1 + vn−1
] − [unun+1vn + vn2 ]
= [ Jn ] − [ Jn+1 ]
2
Jn = un−1unvn−1 + vn−1
.
27
– Method 2: Splitting method (elegant)
Repeatedly use the operator identity
I = (D − I + I) D−1 = ∆ D−1 + D−1.
Split
Dtρ = E = A(0) + A(1),
where A(0) does not depend on the lowest shift.
E = ∆ D−1A(0) + D−1A(0) + A(1).
? Finished if D−1A(0) + A(1) = 0. Then J = − D−1A(0).
? If D−1A(0) + A(1) 6= 0, split
D−1A(0) + A(1) = A(2) + A(3),
where A(2) does not depend on the lowest shift. Then,
E = ∆ D−1A(0) + ∆ D−1A(2) + D−1A(2) + A(3).
−1
(2)
? Finished if D A +A
(3)
−1
= 0. Then J = −D
A
(0)
+A
(2)
.
? If not, we repeat the process until it terminates.
Demonstration:
E =
=
=
=
=
A(0) + A(1)
2
[−unun+1vn − vn2 ] + [un−1unvn−1 + vn−1
]
∆ [D−1A(0)] + [D−1A(0) + A(1)]
2
2
2
∆[−un−1unvn−1 −vn−1
]+[−un−1unvn−1 −vn−1
+un−1unvn−1 +vn−1
]
2
∆[−un−1unvn−1 − vn−1
] + [ 0 ].
So,
J = − D−1A(0)
2
= un−1unvn−1 + vn−1
28
– Method 3: Use the homotopy operator (most powerful)
Discrete higher Euler operators:
 
p+q
∂
X
k  −k
 D
)
(
L(i)
=
un
∂un k=i i
Examples (scalar case, u1,n = un):
∂
(I + D−1 + D−2 + D−3 + · · ·)
∂un
∂
L(1)
=
(D−1 + 2D−2 + 3D−3 + 4D−4 + · · ·)
un
∂un
∂
(D−2 + 3D−3 + 6D−4 + 10D−5 + · · ·)
L(2)
=
un
∂un
∂
(D−3 + 4D−4 + 10D−5 + 20D−6 + · · ·)
L(3)
=
un
∂un
L(0)
un =
Similar formulas for L(i)
vn .
The flux is
J˜n =
Z 1
(j̃1,n(un)[λun] + j̃2,n(un)[λun])
0
dλ
λ
where,
j̃1,n(un) =
j̃2,n(un) =
p+q−1
X
(D − I)i(unL(i+1)
un (−Ẽ))
i=0
p+q−1
X
i=0
(D − I)i(vnL(i+1)
un (−Ẽ))
Note that p+q is the highest shift in Ẽ, and j̃r,n(un)[λun] means
un → λun, un+1 → λun+1, un+2 → λun+2, etc.
29
Demonstration (vector case, un = (un, vn)) :
Compute Jn via the homotopy operator!
Start from
2
−Ẽ = −DE = −unun+1vn − vn2 + un+1un+2vn+1 + vn+1
.
To find: flux Jn such that (D − I)Jn = −E.
Homotopy operator inverts the operator (D − I).
i
L(i+1)
(D − I)i(unL(i+1)
un (−Ẽ)
un (−Ẽ))
0 un−1vn−1 +un+1vn unun−1vn−1 +unun+1vn
1
un−1vn−1
un+1unvn −unun−1vn−1
i
(i+1)
i L(i+1)
vn (−Ẽ) (D − I) (vn Lvn (−Ẽ))
0 unun+1 +2vn
vnunun+1 +2vn2
Hence,
j̃1,n(un) = 2unun+1vn,
j̃2,n(un) = unun+1vn + 2vn2 .
Thus, the homotopy operator gives
J˜n =
=
Z 1
(j̃1,n(un)[λun] + j̃2,n(un)[λun])
0
dλ
λ
Z 1
2
2
(3λ
u
u
v
+
2λv
) dλ
n
n+1
n
n
0
= unun+1vn + vn2 .
Summary:
2
Jn = D−1Jn = un−1unvn−1 +vn−1
.
ρn = 31 u3n +un(vn−1 +vn),
30
Analogously, conserved densities of rank ≤ 5:
ρ(1)
= un
n
2
1
ρ(2)
n = 2 un + v n
ρ(3)
=
n
3
1
3 un
+ un(vn−1 + vn)
ρ(4)
=
n
4
1
u
n
4
+ un2(vn−1 + vn) + unun+1vn + 12 vn2 + vnvn+1
ρ(5)
=
n
5
1
5 un
+ un3(vn−1 + vn) + unun+1vn(un + un+1)
+unvn−1(vn−2 + vn−1 + vn) + unvn(vn−1 + vn + vn+1).
31
Mathematica Software, Future Work, Publications
• Implementation in Mathematica – Software
* P.J. Adams and W. Hereman
TransPDEDensityFlux.m: Symbolic computation of conserved densities and fluxes for systems of partial differential equations with transcendental nonlinearities (2002).
* H. Eklund and W. Hereman
DDEDensityFlux.m: Symbolic computation of conserved densities and fluxes for nonlinear systems of differential-difference
equations (2002).
* Ü. Göktaş and W. Hereman
InvariantsSymmetries.m: A Mathematica integrability package for the computation of invariants and symmetries (1997).
Available from MathSource (Item: 0208-932, Applications/Mathematics)
via FTP: mathsource.wolfram.com or URL
http://www.mathsource.com/cgi-bin/MathSource/Applications/
* Ü. Göktaş and W. Hereman
CONDENS.M: A Mathematica program for the symbolic computation of conserved densities for systems of nonlinear evolution
equations (1996).
* Ü. Göktaş and W. Hereman
DIFFDENS.M: A Mathematica program for the symbolic
computation of conserved densities for systems of nonlinear differentialdifference equations (1997).
All codes are available via the Internet
URL: http://www.mines.edu/fs home/whereman/
and via anonymous FTP from mines.edu in directory
pub/papers/math cs dept/software/
32
• Scope and Limitations of Algorithms.
– Systems of DDEs must be polynomial in dependent variables.
– One discretized space variable (lattice point n)
– Program only computes polynomial conservation laws and generalized symmetries (no recursion operators yet). (Non-polynomial
densities in progress).
– Program does not compute conservation laws and symmetries
that explicitly depend on n.
– No limit on the number of equations in the system.
In practice: time and memory constraints.
– Input systems may have (nonzero) parameters.
Program computes the compatibility conditions for parameters
such that conservation laws and symmetries (of a given rank)
exist.
– Systems can also have parameters with (unknown) weight.
This allows one to test lattice equations of non-uniform rank.
– For systems where one or more of the weights is free,
the program prompts the user for info.
– Fractional weights and ranks are permitted.
– Lattice equations must be of first-order in t.
33
• Conclusions and Future Research
– Compute simple logarithmic and rational densities.
– Implement the recursion operator algorithm for DDEs.
– Improve software, compare with other strategies & packages.
– Add tools for parameter analysis (Gröbner basis, Ritt-Wu or
characteristic sets algorithms).
d
– Introduce multiple sets of weights based on w( dt
) = 0 and
d
) = 1.
w( dt
– Application: test model DDEs for integrability.
(study the integrable discretization of KdV-mKdV equation).
34
• Publications
1). P. J. Adams, Symbolic Computation of Conserved Densities
and Fluxes for Systems of Partial Differential Equations with
Transcendental Nonlinearities, MS Thesis, CSM, Dec. 2002.
2). H. Eklund, Symbolic Computation of Conserved Densities
and Fluxes for Nonlinear Systems of Differential-Difference
Equations, MS Thesis, Colorado School of Mines, Dec. 2002.
3). Ü. Göktaş and W. Hereman, Symbolic computation of
conserved densities for systems of nonlinear evolution equations,
J. Symb. Comput., 24 (1997) 591–621.
4). Ü. Göktaş, W. Hereman, and G. Erdmann, Computation of
conserved densities for systems of nonlinear differential-difference
equations, Phys. Lett. A, 236 (1997) 30–38.
5). Ü. Göktaş and W. Hereman, Computation of conserved densities
for nonlinear lattices, Physica D, 123 (1998) 425–436.
6). Ü. Göktaş and W. Hereman, Algorithmiccomputation of higherorder symmetries for nonlinear evolution and lattice equations,
Adv. in Comput. Math. 11 (1999), 55-80.
7). W. Hereman and Ü. Göktaş, Integrability Tests for Nonlinear
Evolution Equations. In: Computer Algebra Systems: A Practical Guide, Ed.: M. Wester, Wiley & Sons, New York (1999)
Chap. 12, pp. 211-232.
8). W. Hereman, Ü. Göktaş, M. Colagrosso, and A. Miller,
Algorithmic integrability tests for nonlinear differential and lattice equations, Comp. Phys. Comm. 115 (1998) 428–446.
9). M. Hickman and W. Hereman, Computation of Densities and
Fluxes of Nonlinear Differential-Difference Equations, Proc. Roy.
Soc. Lon. A (2003) in press.
35