Symbolic Computation of Conserved Densities
Generalized Symmetries, and Recursion Operators for
Nonlinear Evolution and Lattice Equations
Willy Hereman
Department of Mathematical and Computer Sciences
Colorado School of Mines
Golden, Colorado, USA
http://www.mines.edu/fs home/whereman/
whereman@mines.edu
Applied Mathematics Seminar
Department of Applied Mathematics
University of Colorado—Boulder
Thursday, February 28, 2002
Collaborators: Ünal Göktaş (WRI), Mark Hickman
Jan Sanders, & Jing-Ping Wang
Doug Baldwin, Jack Sayers, Holly Eklund, Paul Adams
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
Part I: Purpose, Motivation, Definitions, Software demo
• Definitions: dilation symmetry, conserved densities (fluxes), generalized symmetries, recursion operator (PDEs and DDEs)
• Prototypical examples
• Computer Demo
• Strategy
• Analogy between PDEs and DDEs
Part II: Algorithms for Partial Differential Equations (PDEs)
• Conserved densities for PDEs: Algorithm
• Extension to PDEs with transcendental nonlinearities
• Generalized symmetries of PDEs
• Algorithm for Recursion Operators of PDEs
Part III: Algorithms for Differential-difference Equations
(DDEs)
• Conserved densities and algorithm for DDEs
• Generalized symmetries for DDEs
• Recursion operators for DDEs
• Example: Ablowitz-Ladik lattice
• Application: Discretization of combined KdV-mKdV equation
2
Part IV: Software, Future Work, Publications
• Scope and Limitations of Algorithms
• Mathematica Software
• Conclusions & Future Research
• Publications
3
Part I Purpose, Motivation, Strategy, Demo
• Purpose
Design and implement algorithms to compute polynomial conservation laws, generalized symmetries, and recursion operators for nonlinear systems of PDEs and differential-difference equations (DDEs).
• Motivation
– Conservation laws describe the conservation of fundamental physical quantities (linear momentum, energy, etc.).
Compare with constants of motion (linear momentum, energy)
in mechanics.
– Conservation laws provide a method to study quantitative and
qualitative properties of equations and their solutions,
e.g. Hamiltonian structures.
– Conserved densities can be used to test numerical integrators.
– For PDEs and DDEs, the existence of a sufficiently large (in
principal infinite) number of conservation laws or symmetries
assures complete integrability.
– Conserved densities and symmetries aid in finding the recursion
operator (which guarantees the existence of infinitely many symmetries).
4
Definitions and Examples for 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
∂u
ut =
, uix = i .
∂t
∂x
F is polynomial in u, ux, u2x, ..., umx.
PDEs of higher order in t should be recast as a first-order system.
• Prototypical Examples
The Korteweg-de Vries (KdV) equation:
ut + uux + u3x = 0.
Fifth-order evolution equations with constant parameters (α, β, γ):
ut + αu2ux + βuxu2x + γuu3x + u5x = 0.
Special case. The fifth-order Sawada-Kotera (SK) equation:
ut + 5u2ux + 5uxu2x + 5uu3x + u5x = 0.
The 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.
5
A vector nonlinear Schrödinger equation (Verheest, Deconinck, Meuris):
Bt + (|B|2B)x + (B0 · Bx)B0 + e × Bxx = 0,
written in components, B0 = (a, b, 0), B = (u, v, 0), e = (0, 0, 1) :
2
2
ut + u(u + v ) + βu + γv − vx
vt + v(u2 + v 2) + θu + δv + ux
x
x
= 0,
= 0,
β = a2, γ = θ = ab, and δ = b2.
• Dilation invariance of PDEs
The KdV equation
ut + uux + u3x = 0
has scaling symmetry
(t, x, u) → (λ−3t, λ−1x, λ2u).
The Boussinesq system
ut + vx = 0,
vt + ux − 3uux − αu3x = 0,
is not scaling invariant (ux and u3x are conflicting terms).
If one introduces 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).
6
• Computation of the dilation symmetry (Lie-point symmetry)
The weight w of a variable is by definition the number of xderivatives the variable corresponds to.
The rank of a monomial is its total weight in terms of x-derivatives.
Example 1: KdV Equation
Set w(Dx) = 1 or w(x) = −1 and require that all terms in
ut + uux + u3x = 0
have the same rank. Hence,
w(u) + w(Dt) = 2w(u) + 1 = w(u) + 3.
Solve the linear system
w(u) = 2, w(Dt) = 3, or w(t) = −3.
So,
(t, x, u) → (λ−3t, λ−1x, λ2u).
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
to get
w(u) = 2, w(v) = 3, w(β) = 2, w(Dt) = 2.
Hence,
(x, t, u, v, β) → (λ−1x, λ−2t, λ2u, λ3v, λ2β).
7
• Conservation Law for PDEs
Dtρ + DxJ = 0
on PDE,
conserved density ρ and flux J.
Both are polynomial in u, ux, u2x, u3x, ....
Z +∞
P = −∞ ρ dx = constant in time
if J vanishes at infinity.
Examples
Conserved densities and fluxes for the Korteweg-de Vries (KdV)
equation
ut + uux + u3x = 0
u2
Dt(u) + Dx( + u2x) = 0.
2
ρ1 = u,
2u3
Dt(u ) + Dx(
+ 2uu2x − ux2) = 0.
3
2
2
ρ2 = u ,
ρ3 = u3 − 3ux2,
3
Dt u −3ux
...
2
3
+Dx  u4 −6uux2 +3u2u2x +3u2x2 −6uxu3x = 0.
4


ρ6 = u6 − 60u3ux2 − 30ux4 + 108u2u2x2
720 3 648
216 2
+
u2x −
uu3x2 +
u4x .
7
7
7
The Boussinesq system (before setting β = 1) :
ρ1 = u,
ρ3 = uv,
ρ2 = v,
ρ4 = βu2 − u3 + v 2 + αux2.
8
• Generalized Symmetries of PDEs.
G(x, t, u, ux, u2x, ...)
with G = (G1, G2, ..., Gn) is a symmetry iff it leaves the PDE
invariant for the replacement u → u + G within order . i.e.
Dt(u + G) = F(u + G)
must hold up to order on the solutions of PDE.
Consequently, G must satisfy the linearized equation
DtG = F0(u)[G]
∂
F(u + G)|=0
=
∂
m
∂F
X
(Dxi G)
=
∂uix
i=0
where m is the highest order and F0 is the Fréchet derivative of F
in the direction of G.
Here, u is replaced by u + G, and uix by uix + DixG.
Example
For the KdV equation, ut = 6uux + u3x, the first few generalized
symmetries are:
G(1) = ux,
G(2) = 6uux + u3x,
G(3) = 30u2ux + 20uxu2x + 10uu3x + u5x,
G(4) = 140u3ux + 70u3x + 280uuxu2x + 70u2u3x
+70u2xu3x + 42uxu4x + 14uu5x + u7x.
9
• Recursion Operators of PDEs.
A recursion operator R connects symmetries
G(j+s) = RG(j), j = 1, 2, ...
s is seed (s = 1 in simplest case).
For n-component systems, R is an n × n matrix.
Defining equation for R :
∂R
DtR + [R, F0(u)] =
+ R0[F] + R ◦ F0(u) − F0(u) ◦ R = 0,
∂t
where [ , ] means commutator, ◦ stands for composition, and
m ∂F
X
F0(u) = DF = (
) Dxi
i=0 ∂uix
m is highest order, Dx is differential operator and
Dxi = Dx ◦ Dx ◦ · · · ◦ Dx (i times).
R0[F] is the Fréchet derivative of R in direction of F :
n
∂R
X
(Dxi F)
R0[F] =
∂uix
i=0
Example
The recursion operator for the KdV equation:
2
−1
R = D2x + 2u + 2DxuD−1
x = Dx + 4u + 2ux Dx .
Dx is differentiation and D−1
x is integration operator.
For example,
Rux = (D2x + 2u + 2DxuD−1
x )ux = 6uux + u3x ,
R(6uux + u3x) = (D2x + 2u + 2DxuD−1
x )(6uux + u3x )
= 30u2ux + 20uxu2x + 10uu3x + u5x.
10
Definitions and Examples for DDEs (lattices)
• Nonlinear system of DDEs
(continuous in time, discretized in space)
u̇n = F(..., un−1, un, un+1, ...),
un and F are vector dynamical variables.
F is polynomial with constant coefficients (parameters).
No restrictions on the level of the shifts or the degree of nonlinearity.
• Example
One-dimensional Toda lattice
ÿn = exp (yn−1 − yn) − exp (yn − yn+1).
yn is the displacement from equilibrium of the nth particle with unit
mass under an exponentially decaying interaction force
between nearest neighbors.
Change of variables:
un = ẏn,
vn = exp (yn − yn+1)
yields
u̇n = vn−1 − vn,
v̇n = vn(un − un+1).
Toda system is completely integrable.
11
• Dilation Invariance of DDEs
The Toda lattice
u̇n = vn−1 − vn,
v̇n = vn(un − un+1).
is invariant under the scaling symmetry
(t, un, vn) → (λ−1t, λun, λ2vn).
Weights w(un), w(vn) are defined in terms of t-derivatives.
d
Using w( dt
) = 1, w(un±p) = w(un), w(vn±p) = w(vn)
w(un) + 1 = w(vn),
w(vn) + 1 = w(vn) + w(un).
Hence,
w(un) = 1, w(vn) = 2.
The rank of a monomial is its total weight in terms of t-derivatives.
• Conservation Law for DDEs:
ρ̇n = Jn − Jn+1
on DDE,
density ρn, flux Jn. Both polynomials in un and its shifts.
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.
The first two density-flux pairs (computed by hand):
ρ(1)
n = un ,
1 2
Jn(1) = vn−1, and ρ(2)
n = 2 un + vn ,
12
Jn(2) = unvn−1.
• Generalized Symmetries of DDEs
A vector function G(..., un−1, un, un+1, ...) is a symmetry iff the
infinitesimal transformation
un → un + G(..., un−1, un, un+1, ...)
leaves the DDE system invariant within order .
G must satisfy the linearized equation
p
∂
∂F
X
DtG = F (un)[G] = F(un + G)|=0 =
(DiG)
,
∂
∂un+i
i=−q
0
where F0 is the Fréchet derivative of F in direction of G.
D is up-shift operator, D−1 is down-shift operator,
and Di = D ◦ D ◦ · · · ◦ D (i times).
Here,
un → un + G(..., un−1, un, un+1, ...),
un±k → un±k + G|n→n±k .
• Example
Consider the Toda lattice
u̇n = vn−1 − vn,
v̇n = vn(un − un+1).
Higher-order symmetry of rank (3, 4):
G1 = vn(un + un+1) − vn−1(un−1 + un),
G2 = vn(u2n+1 − u2n) + vn(vn+1 − vn−1).
13
• Recursion Operators of DDEs.
A recursion operator R connects symmetries
G(j+s) = RG(j), j = 1, 2, ...,
s is seed. For r-component systems, R is an r × r matrix.
Defining equation for R :
∂R
+ R0[F] + R ◦ F0(un) − F0(un) ◦ R = 0,
∂t
where [ , ] means commutator, ◦ stands for composition, and
p
∂F
X
F0(un) =
(
) Di
i=−q ∂un+i
p, q are bounds of the shifts, D is up-shift operator and
Di = D ◦ D ◦ · · · ◦ D (i times).
DtR + [R, F0(un)] =
R0[F] is the Fréchet derivative of R in direction of F :
0
R [F] =
p
X
i=−q
(DiF)
∂R
∂un+i
Example 1
The Volterra equation
u̇n = un(un+1 − un−1),
has recursion operator
R = un(1 + D)(unD − D−1un)(D − 1)−1
1
un
= unD + unD−1 + un + un+1 + un(un+1 − un−1)(D − 1)−1
Note: ρn = ln(un) and Jn = −(un + un−1) are density-flux pair.
14
1
un
Example 2
The quadratic Volterra equation
u̇n = u2n(un+1 − un−1),
has recursion operator
R = u2nD + u2nD−1 + 2unun+1 + 2u2n(un+1 − un−1)(D − 1)−1
1
.
un
Example 3
The Toda lattice
u̇n = vn−1 − vn,
v̇n = vn(un − un+1),
has Hamiltonian operator

H =


−1
−1

D vn − vn D
−unvn + unD vn 

−vnDun + unvn −vnDvn + vnD−1vn,
and co-symplectic operator
S =
−1
1 0 (D − 1)
0
1
vn
+ 0
A recursion operator is given by
R = HS.
15
1
vn
−1
D(D − 1)
1 0 .
• Key Observation
Conserved densities, generalized symmetries, and recursion operators are invariant under the dilation (scaling) symmetry of the given
PDE or DDE.
• Overall 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)
* Fréchet derivative
* calculus with operators
Use analogy between continuous and semi-discrete cases
16
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 (1 + D)(un D − D−1 un )
(D − 1)−1 u1n
Prototypical Examples
17
Part II Algorithms for PDEs
• Algorithm for Conserved Densities of 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).
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 Dx4 .
u2 has weight 4, introduce Dx2 .
u3 has weight 6, no derivative needed.
18
Apply the Dx derivatives.
Remove total derivative terms (Dxupx) and highest derivative terms
as follows:
[u4x] → [ ]
[ux2, uu2x] → [ux2]
empty list.
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.
Compute
m ∂ρ
∂ρ X
Dtρ =
+
Dkxut
∂t k=0 ∂ukx
∂ρ
=
+ ρ0(u)[F ].
∂t
since ut = F. Here, Dtρ = 3c1u2ut + 2c2uxuxt.
Replace ut by −(uux + u3x) and uxt by −(uux + u3x)x.
Hence,
E = Dtρ = −3c1u2(uux + u3x) − 2c2ux(uux + u3x)x.
Apply the Euler operator (variational derivative)
m
∂
X
Lu = (−1)k Dkx
∂ukx
k=0
∂
∂
∂
∂
=
− Dx(
) + D2x(
) + · · · + (−1)mDm
(
).
x
∂u
∂ux
∂u2x
∂umx
19
to E of order m = 4. Result:
Lu(E) = −(3c1 + c2)ux3.
This non-integrable term must vanish.
So, c1 = − 13 c2. Set c2 = −3, then c1 = 1.
Hence,
ρ = u3 − 3ux2.
Integration of E = −DxJ yields
3
J = u4 − 6uux2 + 3u2u2x + 3u2x2 − 6uxu3x.
4
Note: Rank J = Rank ρ + Rank Dt − 1.
If integration by parts fails: 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) and match DxJ = −E.
20
Single Mixed-Derivative Equations (PDEs)
with Transcendental Nonlinearities
• Nonlinear evolution equation
uxt = F (u, ux, u2x, ..., umx),
scalar u(x, t) and F. But F is transcendental in u but polynomial
in ux, ..., umx.
• Examples:
The sine-Gordon (and sinh-Gordon) equations:
uxt = α sin(u) and ux,t = α sinh(u).
The Liouville equation:
uxt = α exp(u).
• Dilation Invariance
The sine-Gordon and Liouville equations have scaling symmetry
(t, x, u, α) → (λ−1t, λ−1x, u, λ2α).
Here, w(u) = 0, w(Dx) = 1, w(Dt) = 1, and w(α) = 2.
The sine-Gordon and Liouville equations are uniform of rank 2.
(if α is introduced).
Without α, the equations would be uniform of rank 0, and
w(Dt) = −1.
Use densities algorithm on variable ũ = ux instead of u.
21
• Conserved densities for sine-Gordon equation
Rank 2:
ρ1 = u x 2 ,
J1 = 2α cos(u)
Rank 4:
ρ2 = ux4 − 4u2x2,
J2 = 4α cos(u)ux2.
Rank 6:
ρ3 = ux6 − 20ux2u2x2 + 8u3x2,
J3 = 6α cos(u)ux4 + 16α sin(u)ux2u2x − 8α cos(u)u2x2.
Rank 8:
ρ4 = ux8 − 56ux4u2x2 −
112 4 224 2 2 64 2
u2x +
ux u3x − u4x ,
5
5
5
J4 = 8α cos(u)ux6 + 64α sin(u)ux4u2x + 32α cos(u)ux2u2x2
+
128
128
384
α sin(u)u2x3 −
α cos(u)ux3u3x −
α sin(u)uxu2xu3x
5
5
5
+
64
α cos(u)u3x2.
5
• Conserved densities for Liouville equation
Similar results as for sine-Gordon equation.
Conserved density with time and space dependent coefficients:
1
ρ = f 0(x)ux + f (x)ux2, J = −f (x) exp(u).
2
22
Systems of Evolution Equations (PDEs)
with Transcendental Nonlinearities
• Nonlinear system of evolution equations
ut = F(u, ux, u2x, ..., umx)
in a (single) space variable x and time t,
u = (u1, u2, ..., un),
F = (F1, F2, ..., Fn).
F is transcendental in u and polynomial in ux, ..., umx.
• Prototypical Example:
The sine-Gordon system:
ut = v,
vt = uxx + α sin(u).
• Dilation Invariance
The sine-Gordon system has scaling symmetry
(t, x, u, v, α) → (λ−1t, λ−1x, u, λv, λ2α).
Here, w(u) = 0, w(v) = 1, w(Dx) = 1, w(Dt) = 1, and w(α) = 2.
First equation has rank 1, the second has rank 2.
Without parameter α, the system would not be uniform in rank.
23
• Example: Density of rank 2 for the sine-Gordon system
ut = v,
vt = uxx + α sin(u).
Step 1: Compute the weights:
w(u) = 0, w(v) = 1, w(α) = 2.
Step 2: Construct the form of the density.
ρ = αh1(u) + h2(u)v 2 + h3(u)ux2 + h4(u)uxv ,
where hi(u) are unknown functions.
Step 3: Determine the functions hi.
Solve the system of linear ODEs:
h2(u) − h3(u) = 0
0
h2 (u) = 0,
0
h3 (u) = 0,
0
h4 (u) = 0,
00
h2 (u) = 0,
00
h4 (u) = 0,
0
0
2h2 (u) − h3 (u) = 0,
00
00
2h2 (u) − h3 (u) = 0,
0
h1 (u) + 2 sin(u)h2(u) = 0,
00
0
h1 (u) + 2 sin(u)h2 (u) + 2 cos(u)h2(u) = 0,
Solution:
h1(u) = 2c1 cos(u) + c3,
h2(u) = h3(u) = c1,
h4(u) = c2.
24
After substitution in ρ
ρ = 2c1α cos(u) + αc3 + c1v 2 + c1ux2 + c2uxv .
After splitting
ρ0 = αc3,
ρ1 = 2α cos(u) + v 2 + ux2,
ρ2 = uxv.
• Conserved densities for sine-Gordon system
Rank 2 (two conserved densities):
ρ1 = 2α cos(u) + v 2 + ux2, J1 = −2uxv ,
1
1
ρ2 = uxv, J2 = −( v 2 + ux2 − α cos(u)).
2
2
Rank 4 (two new conserved densities) from
ρ̃3 = c1 [12α cos(u)vux + 2v 3ux + 2vux3 − 16vxu2x]
+c2 [α2(2 + 2 cos2(u) − 2 sin2(u)) + v 4 + 6v 2ux2
+ux4 + α(4 cos(u)v 2 + 20 cos(u)ux2) − 16vx2 − 16u2x2]
+c3 [−16αvux]
+c4 [−32α2 cos(u) + α(−16v 2 − 16ux2)]
+c5 [−16α2]
J3 is not shown (long).
After splitting and setting α = 1:
ρ3 = 12 cos(u)vux + 2v 3ux + 2vux3 − 16vxu2x,
ρ4 = 2 cos2(u) − 2 sin2(u) + v 4 + 6v 2ux2 + ux4 + 4 cos(u)v 2
+20 cos(u)ux2 − 16vx2 − 16u2x2.
25
• Algorithm for Generalized Symmetries of PDEs.
Consider the KdV equation, ut = 6uux + u3x, with w(u) = 2.
Step 1: Construct the form of the symmetry.
Compute the form of the symmetry with rank 7.
List all powers in u with rank 7 or less:
L = {1, u, u2, u3}.
For each monomial in L, introduce the needed x-derivatives, so that
each term exactly has rank 7. Thus,
Dx(u3) = 3u2ux, D3x(u2) = 6uxu2x + 2uu3x,
D5x(u) = u5x, D7x(1) = 0.
Gather the resulting (non-zero) terms
R = {u2ux, uxu2x, uu3x, u5x}.
The symmetry is a linear combination of these monomials:
G = c1 u2ux + c2 uxu2x + c3 uu3x + c4 u5x.
Step 2: Determine the unknown coefficients ci.
Compute DtG and use KdV to remove ut, utx, utxx, etc.
Compute the Fréchet derivative F 0(u)[G].
Equate the resulting expressions.
Group the terms:
(12c1 − 18c2)u2xu2x + (6c1 − 18c3)uu22x + (6c1 − 18c3)uuxu3x +
(3c2 − 60c4)u23x + (3c2 + 3c3 − 90c4)u2xu4x + (3c3 − 30c4)uxu5x ≡ 0.
26
Solve the linear system:
S = {12c1 − 18c2 = 0, 6c1 − 18c3 = 0, 3c2 − 60c4 = 0,
3c2 + 3c3 − 90c4 = 0, 3c3 − 30c4 = 0}.
c1
c2
c3
Solution: 30
= 20
= 10
= c4 .
Setting c4 = 1 one gets: c1 = 30, c2 = 20, c3 = 10.
Substitute the result into the symmetry:
G = 30u2ux + 20uxu2x + 10uu3x + u5x.
Note: ut = G is known as the Lax equation.
• x-t Dependent symmetries.
The KdV equation has also symmetries which explicitly depend on
x and t.
The same algorithm can be used provided the highest degree of x or
t is specified.
Compute the symmetry of rank 2, that is linear in x or t.
List all monomials in u, x and t of rank 2 or less:
L = {1, u, x, xu, t, tu, tu2}.
For each monomial in L, introduce enough x-derivatives, so that
each term exactly has rank 2. Thus,
Dx(xu) = u + xux, Dx(tu2) = 2tuux,
D2x(1) = D3x(x) = D5x(t) = 0.
Gather the non-zero resulting terms:
R = {u, xux, tuux, tu3x},
27
D3x(tu) = tu3x,
Build the linear combination
G = c1 u + c2 xux + c3 tuux + c4 tu3x.
Determine the coefficients c1 through c4:
2
1
G = u + xux + 6tuux + tu3x.
3
3
Two symmetries of KdV that explicitly depend on x and t :
G = 1 + 6tux, and G = 2u + xux + 3t(6uux + u3x),
of rank 0 and 2, respectively.
28
Recursion Operators for PDEs
• Key Observation
? The recursion operator of the KdV equation
2
−1
R = D2x + 2u + 2DxuD−1
x = Dx + 4u + 2ux Dx
has rank R = 2. Dx is differentiation and D−1
x is integration operator.
Indeed, compare the ranks of the symmetries
Rux = (D2x + 2u + 2DxuD−1
x )ux = 6uux + u3x ,
R(6uux + u3x) = (D2x + 2u + 2DxuD−1
x )(6uux + u3x )
= 30u2ux + 20uxu2x + 10uu3x + u5x.
? The terms in the recursion operator are monomials in Dx, Dx−1, u, ux, ...
? Recursion operators split naturally in R = R0 + R1.
R0 is a differential operator (no D−1
x terms).
R1 is an integral operator (with D−1
x terms).
? Application of R to a symmetry should not leave integrals.
For the KdV equation:
2
D−1
x (6uux + u3x ) = 3u + u2x is polynomial.
Use the conserved densities: ρ(1) = u, ρ(2) = u2, ρ(3) = u3 − 12 u2x
Dtρ(1) = Dtu = ut = −DxJ (1),
Dtρ(2) = Dtu2 = 2uut = −DxJ (2), and
1
0
Dtρ(3) = Dt(u3 − u2x) = ρ(3) (u)[ut] = (3u2 − uxDx)ut = −DxJ (3),
2
for polynomial J (i), i = 1, 2, 3.
−1
−1
2
So, application of D−1
x , or Dx u, or Dx (3u − ux Dx )
to 6uux + u3x leads to a polynomial result.
29
• Algorithm for Recursion Operators of PDEs.
Step 1: Determine the rank of the recursion operator.
Recall: symmetries for the KdV equation, ut = 6uux + u3x, are
G(1) = ux,
G(2) = 6uux + u3x,
G(3) = 30u2ux + 20uxu2x + 10uu3x + u5x.
Hence,
R = rank R = rank G(3) − rank G(2) = rank G(2) − rank G(1) = 2.
Step 2: Construct the form of the recursion operator.
(i) Determine the pieces of operator R0
List all permutations of type Dj uk of rank R, with j and k nonnegative integers.
L = {D2, u}.
(ii) Determine the pieces of operator R1
0
Combine the symmetries G(j) with D−1 and ρ(k) (u), so that every
term in
0
XX
R1 =
G(j)D−1ρ(k) (u)
j k
has rank R.
The indices j and k are taken so that
0
rank (G(j)) + rank (ρ(k) (u)) − 1 = R.
List such terms:
M = {uxD−1}.
30
(iii) Build the operator R
Linearly combine the term in
R = L M = {D2, u, uxD−1}.
[
to get
R = c1 D2 + c2 u + c3 uxD−1.
Step 3: Determine the unknown coefficients.
Require that
RG(k) = G(k+1),
k = 1, 2, 3, ...
Solve the linear system:
S = {c1−1 = 0, 18c1+c3−20 = 0, 6c1+c2−10 = 0, 2c2+c3−10 = 0},
Solution: c1 = 1, c2 = 4, and c3 = 2. So,
R = D2 + 4 u + 2 uxD−1.
Examples.
Sawada-Kotera (SK) equation:
ut = 5u2ux + 5uxu2x + 5uu3x + u5x.
Recursion operator:
R = D6 + 3uD4 − 3DuD3 + 11D2uD2 − 10D3uD + 5D4u
+ 12u2D2 − 19uDuD + 8uD2u + 8DuDu + 4u3
+ uxD−1(u2 − 2uxD) + G(2)D−1,
with G(2) = 5u2ux + 5uxu2x + 5uu3x + u5x.
31
The vector nonlinear Schrödinger system:
2
2
ut + u(u + v ) + βu + γv − vx
vt + v(u2 + v 2) + θu + δv + ux
x
x
= 0,
= 0.
Recursion operator:
2
−1
u θ + 2uv − D + 2uxD−1v 
 β − δ + 2u + 2ux D
.
R =
θ + 2uv + D + 2vxD−1u
2v 2 + 2vxD−1v


32
Analogy PDEs and DDEs
Conservation laws for PDEs
Dtρ + DxJ = 0
density ρ, flux J.
Compute E = Dtρ.
Use PDE to replace all t−derivatives: ut, utx, utxx, etc.
To avoid integration by parts, apply the continuous Euler operator
∂
∂uix
i=0
∂
∂
∂
∂
=
− Dx(
) + D2x(
) + · · · + (−1)mDm
(
).
x
∂u
∂ux
∂u2x
∂umx
Lu =
m
X
(−1)iDix
to E of order m.
Dx is the differential operator.
If Lu(E) = 0, then E is a total x-derivative (−Jx).
If Lu(E) 6= 0, the nonzero terms must vanish identically.
E must be in the kernel of Lu operator, or equivalently, E must be in
the image of Dx operator.
33
Conservation laws for DDEs
ρ̇n + Jn+1 − Jn = 0
density ρn, flux Jn.
Compute E = ρ̇n.
Use the DDE to remove all t−derivatives, u̇n u̇n±1, u̇n±2, etc.
To avoid pattern matching, apply the discrete Euler operator
p
X
∂
∂un+i
i=−q
∂
∂
∂
∂
=
+ D(
) + D2 (
) + · · · + Dq (
)
∂un
∂un−1
∂un−2
∂un−q
∂
∂
∂
+ D−1(
) + D−2(
) + · · · + D−p(
)
∂un+1
∂un+2
∂un+p
Lun =
D−i
to E with maximal shifts n − q, n + 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 Lun (E) = 0, then E matches −(Jn+1 − Jn).
If Lun (E) 6= 0, the nonzero terms must vanish identically.
34
Part III Algorithms for DDEs (lattices)
• Extra Tool: Equivalence Criterion
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.
Equivalence criterion:
Two monomials m1 and m2 are equivalent, m1 ≡ m2, if
m1 = m2 + [Mn − Mn+1]
for some polynomial Mn.
For 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).
For example, unun+2 is the main representative of the class
with elements un−1un+1, un+1un+3, etc.
Use lexicographical ordering to resolve conflicts.
For example, unvn+2 (not un−2vn) is the main representative of
the class with elements un−3vn−1, un+2vn+4, etc.
35
• Algorithm for Conserved Densities of DDEs.
Three-step algorithm to find conserved densities:
(i) Determine the weights.
(ii) Construct the form of density.
(iii) Determine the coefficients.
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(un) + w( ) = w(vn−1) = w(vn),
dt
d
w(vn) + w( ) = w(vn) + w(un) = w(vn) + w(un+1)
dt
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
d0 3
3
(u ) = un,
(unvn) = unvn,
dt0 n
dt0
d 2
(u ) = 2unvn−1 − 2unvn,
dt n
1
In general algorithm shifts are also needed: u3n , un un+1 un−1 , u2n un+1 , etc.
36
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}.
Replace members in the same equivalence class by their
main representatives .
For 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.
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 .
Apply the discrete Euler operator
p
∂
X
Lun =
D−i
∂un+i
i=−q
∂
∂
∂
∂
=
+ D(
) + D2 (
) + · · · + Dq (
)
∂un
∂un−1
∂un−2
∂un−q
∂
∂
∂
+ D−1(
) + D−2(
) + · · · + D−p(
)
∂un+1
∂un+2
∂un+p
37
to E with maximal shifts n − 1, n + 1.
Lun (E) = (3c1 − c2)u2nvn−1 + (c3 − 3c1)u2nvn
+(c3 − c2)vnvn+1 + (c2 − c3)unun+1vn + (c2 − c3)vn2
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 :
Use the equivalence criterion and main representatives to rearrange
E to match the pattern [Jn] − [Jn+1].
2
] − [unun+1vn + vn2 ].
E = [un−1unvn−1 + vn−1
Hence,
ρn = 31 u3n + un(vn−1 + vn),
2
Jn = un−1unvn−1 + vn−1
.
Analogously, conserved densities of rank ≤ 5:
ρ(1)
= un
n
2
1
ρ(2)
n = 2 un + v n
ρ(3)
=
n
3
1
u
n
3
+ un(vn−1 + vn)
ρ(4)
=
n
4
1
4 un
+ un2(vn−1 + vn) + unun+1vn + 12 vn2 + vnvn+1
ρ(5)
=
n
5
1
u
n
5
+ un3(vn−1 + vn) + unun+1vn(un + un+1)
+unvn−1(vn−2 + vn−1 + vn) + unvn(vn−1 + vn + vn+1).
38
• Algorithm for Generalized Symmetries of DDEs.
Consider the Toda system
u̇n = vn−1 − vn,
v̇n = vn(un − un+1).
with
w(un) = 1 and w(vn) = 2.
Compute the form of the symmetry of ranks (3, 4), i.e. the first
component of the symmetry has rank 3, the second rank 4.
Step 1: Construct the form of the symmetry.
List all monomials in un and vn of rank 3 or less:
L1 = {u3n, u2n, unvn, un, vn},
and of rank 4 or less:
L2 = {u4n, u3n, u2nvn, u2n, unvn, un, vn2 , vn}.
For each monomial in L1 and L2, introduce enough t-derivatives, so
that each term exactly has rank 3 and 4, respectively.
Using the DDEs, for the monomials in L1 :
d0 3
d0
3
(u ) = un,
(unvn) = unvn,
dt0 n
dt0
d 2
(u ) = 2unu̇n = 2unvn−1 − 2unvn,
dt n
d
(vn) = v̇n = unvn − un+1vn,
dt
d2
d
d
(u
)
=
(
u̇
)
=
(vn−1 − vn)
n
n
dt
dt
dt2
= un−1vn−1 − unvn−1 − unvn + un+1vn.
39
Gather the resulting terms:
R1 = {u3n, un−1vn−1, unvn−1, unvn, un+1vn}.
2
R2 = {u4n, u2n−1vn−1, un−1unvn−1, u2nvn−1, vn−2vn−1, vn−1
, u2nvn,
unun+1vn, u2n+1vn, vn−1vn, vn2 , vnvn+1}.
Linearly combine the monomials in R1 and R2
G1 = c1 u3n + c2 un−1vn−1 + c3 unvn−1 + c4 unvn + c5 un+1vn,
G2 = c6 u4n + c7 u2n−1vn−1 + c8 un−1unvn−1 + c9 u2nvn−1
2
+c10 vn−2vn−1 + c11 vn−1
+ c12 u2nvn + c13 unun+1vn
+c14 u2n+1vn + c15 vn−1vn + c16 vn2 + c17 vnvn+1.
Step 2: Determine the unknown coefficients.
Require that the symmetry condition DtG = F0(un)[G] holds.
Solution:
c1 = c6 = c7 = c8 = c9 = c10 = c11 = c13 = c16 = 0,
−c2 = −c3 = c4 = c5 = −c12 = c14 = −c15 = c17.
Therefore, with c17 = 1, the symmetry of rank (3, 4) is:
G1 = unvn − un−1vn−1 + un+1vn − unvn−1,
G2 = u2n+1vn − u2nvn + vnvn+1 − vn−1vn.
Analogously, the symmetry of rank (4, 5) reads
G1 = u2nvn + unun+1vn + u2n+1vn + vn2 + vnvn+1 − u2n−1vn−1
2
−un−1unvn−1 − u2nvn−1 − vn−2vn−1 − vn−1
,
G2 = un+1vn2 + 2un+1vnvn+1 + un+2vnvn+1 − u3nvn + u3n+1vn
−un−1vn−1vn − 2unvn−1vn − unvn2 .
40
• Example: The Ablowitz-Ladik DDE.
Consider the Ablowitz and Ladik discretization,
i u̇n = un+1 − 2un + un−1 + κu∗nun(un+1 + un−1),
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 the 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).
Since vn = u∗n, w(vn) = w(un).
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
w(un) + 1 = w(α) + w(un) = 2w(un) + w(vn) = 3w(un),
w(vn) + 1 = w(α) + w(vn) = 2w(vn) + w(un) = 3w(vn).
which yields
1
w(un) = w(vn) = ,
2
41
w(α) = 1.
Conserved densities (for α = 1, in original variables):
∗
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
The Ablowitz-Ladik lattice has infinitely many conserved densities.
Density we missed
∗
ρ(0)
n = ln(1 + un un ).
We cannot find the Hamiltonian (constant of motion):
H = −i
X
[u∗n(un−1 + un+1) − 2 ln(1 + unu∗n)],
since it has a logarithmic term.
42
• Application: Discretization of combined KdV-mKdV
equation.
Consider the integrable discretization
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]
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]
β 2
2
+ 2 [un+1(un+2 + un) − un−1(un−2 + un)] ,
Uniformity in rank requires
w(γ) = w(δ) = 2w(un),
w(α) = w(un),
w(β) = 0.
Then,
w(un) + 1 = 5w(un),
Hence,
w(un) = w(α) = 14 ,
w(γ) = w(δ) = 12 ,
43
w(β) = 0,
Conserved densities: Special cases
For the KdV case (β = 0) :
2
u̇n = −(γ + αh un)
+
α 2
2h [un+1
−
δ 1
( u
h3 2 n+2
u2n−1
− un+1 + un−1 − 12 un−2)
+ un(un+1 − un−1) + un+1un+2 − un−1un−2]
with γ = δ = 1 is a completely integrable discretization of the KdV
equation
ut + 6αuux + uxxx = 0.
Now,
w(γ) = w(δ) = w(un), w(α) = 0.
Then,
w(un) + 1 = 3w(un).
So,
w(un) = w(γ) = w(δ) = 12 ,
From rank
3
2
and
5
2
w(α) = 0.
(after splitting):
ρ(1)
= un ,
n
ρ(2)
= un( 12 un + un+1),
n
ρ(3)
= un( 13 u2n + unun+1 + u2n+1 + α1 un+2 + un+1un+2)
n
ρ(4)
= un( 14 u3n + u2nun+1 + 32 unu2n+1 + u3n+1 + · · · + un+1un+2un+3)
n
ρ(5)
= un( 51 αu4n − 12 u3n − 2u2nun+1 + · · · + αun+1un+2un+3un+4)
n
44
For the mKdV case (α = 0) :
u̇n =
−(γ + βh2u2n) hδ3 ( 12 un+2 − un+1 + un−1 − 12 un−2)
β
2
2
+ 2h [un+1(un+2 + un) − un−1(un−2 + un)]
with γ = δ = 1 is a completely integrable discretization of the
modified KdV equation
ut + 6βu2ux + uxxx = 0.
Now,
w(γ) = w(δ) = 2w(un), w(β) = 0.
Then,
w(un) + 1 = 5w(un).
So,
w(un) = 14 ,
From rank
3
2
and
5
2
w(γ) = w(δ) = 12 ,
w(β) = 0.
(after splitting):
= unun+1,
ρ(1)
n
ρ(2)
= un( 12 unu2n+1 + β1 un+2 + u2n+1un+2)
n
ρ(3)
= un( 13 u2nu3n+1 + β1 unun+1un+2 + · · · + u2n+1u2n+2un+3)
n
ρ(4)
= un( 41 βu3nu4n+1 + u2nu2n+1un+2 +· · · + βu2n+1u2n+2u2n+3un+4)
n
45
Part IV Software, Future Work, Publications
• Scope and Limitations of Algorithms.
– Systems of PDEs and DDEs must be polynomial in dependent
variables. No explicitly dependencies on the independent variables. (Transcendental nonlinearities in progress).
– Currently, one space variable (continuous or discretized).
– Program only computes polynomial conservation laws and generalized symmetries (no recursion operators yet).
– Program computes conservation laws and symmetries that explicitly depend on the independent variables, if the highest degree
is specified.
– 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 evolution and 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.
– Complex dependent variables are allowed.
– PDEs and lattice equations must be of first-order in t.
46
• Conclusions and Future Research
– Implement the recursion operator algorithm for PDEs and DDEs.
– Generalization to (3+1)-dimensional case
(e.g. Kadomtsev-Petviashvili equation).
– Computation of first integrals of ODEs, constants of motion for
dynamical systems (e.g. Lorenz, Hénon-Heiles, Rossler systems).
– Improve software, compare with other packages.
– Add tools for parameter analysis (Gröbner basis, Ritt-Wu or
characteristic sets algorithms).
– Generalization towards broader classes of equations (e.g. uxt).
– Exploit other symmetries in the hope to find conserved densities.
of non-polynomial form
– Application: test model DDEs for integrability.
(study the integrable discretization of KdV-mKdV equation).
47
• Implementation in Mathematica – Software
– Ü. Göktaş and W. Hereman, The software package
InvariantsSymmetries.m and the related files are available at
http://www.mathsource.com/cgi-bin/msitem?0208-932.
MathSource is an electronic library of Mathematica material.
– Software: available via FTP, ftp site mines.edu
in
pub/papers/math cs dept/software/condens
pub/papers/math cs dept/software/diffdens
or via the Internet
URL: http://www.mines.edu/fs home/whereman/
48
• Publications
1). Ü. Göktaş and W. Hereman, Symbolic computation of
conserved densities for systems of nonlinear evolution equations,
J. Symbolic Computation, 24 (1997) 591–621.
2). 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.
3). Ü. Göktaş and W. Hereman, Computation of conserved densities
for nonlinear lattices, Physica D, 123 (1998) 425–436.
4). Ü. Göktaş and W. Hereman, Algorithmiccomputation of higherorder symmetries for nonlinear evolution and lattice equations,
Advances in Computational Mathematics 11 (1999), 55-80.
5). W. Hereman and Ü. Göktaş, Integrability Tests for Nonlinear
Evolution Equations. In: Computer Algebra Systems: A Practical Guide, Ed.: M. Wester, Wiley and Sons, New York (1999)
Chapter 12, pp. 211-232.
6). W. Hereman, Ü. Göktaş, M. Colagrosso, and A. Miller,
Algorithmic integrability tests for nonlinear differential and lattice equations, Computer Physics Communications 115 (1998)
428–446.
49
Application
A Class of Fifth-Order Evolution Equations
ut + αu2ux + βuxu2x + γuu3x + u5x = 0
where α, β, γ are nonzero parameters.
u ∼ Dx2 .
Special cases:
α
α
α
α
=
=
=
=
30
5
20
2
β = 20
β=5
β = 25
β=6
γ = 10
γ=5
γ = 10
γ=3
Lax.
Sawada − Kotera.
Kaup−Kupershmidt.
Ito.
What are the conditions for the parameters α, β and γ
so that the equation admits a density of fixed rank?
• Rank 2:
No condition
ρ = u.
• Rank 4:
Condition: β = 2γ
(Lax and Ito cases)
ρ = u2 .
50
• Rank 6:
Condition:
10α = −2β 2 + 7βγ − 3γ 2
(Lax, SK, and KK cases)
15
ux 2 .
(−2β + γ)
ρ = u3 +
• Rank 8:
1). β = 2γ (Lax and Ito cases)
ρ = u4 −
2). α = − 2β
2 −7βγ−4γ 2
6γ
6
uux2 + u2x2.
α
α
(SK, KK and Ito cases)
45
ρ = u4 −
135
675
2
uux2 +
u
.
2x
2β + γ
(2β + γ)2
• Rank 10:
Condition:
β = 2γ
and
10α = 3γ 2
(Lax case)
ρ = u5 −
500
50 2 2 100
u ux + 2 uu2x2 − 3 u3x2.
γ
γ
7γ
51
What are the necessary conditions for the parameters α, β and γ
so that the equation admits ∞ many polynomial conservation laws?
3 2
• If α = 10
γ and β = 2γ then there is a sequence
(without gaps!) of conserved densities (Lax case).
• If α = 15 γ 2 and β = γ then there is a sequence
(with gaps!) of conserved densities (SK case).
• If α = 15 γ 2 and β = 52 γ then there is a sequence
(with gaps!) of conserved densities (KK case).
• If
2β 2 − 7βγ + 4γ 2
α=−
45
or
β = 2γ
then there is a conserved density of rank 8.
Combine both conditions: α =
2γ 2
9
and β = 2γ (Ito case).
SUMMARY: see tables (notice the gaps!)
52
Table 1:
Conserved Densities for the Sawada-Kotera and Lax 5th-order equations
Density
Sawada-Kotera equation
Lax equation
ρ1
u
u
ρ2
----
1 2
2u
ρ3
1 3
3u
− ux 2
1 3
3u
− 16 ux 2
ρ4
1 4
4u
− 49 uux 2 + 34 u2x 2
1 4
4u
− 12 uux 2 +
ρ5
----
1 5
5u
− u2 ux 2 + 15 uu2x 2 −
ρ6
1 6
6u
1 6
6u
− 53 u3 ux 2 −
−
25 3
2
4 u ux
+2u2x 3 −
ρ7
ρ8
1 7
7u
−
17
4
8 ux
21
2
8 uu3x
− 9u4 ux 2 −
+ 6u2 u2x 2
+ 83 u4x 2
54
4
5 uux
+
2
2
+ 648
35 ux u2x +
489
3
35 uu2x
2
− 288
35 u2x u3x +
81
2
35 uu4x
57 3
2
5 u u2x
−
−
5
u2x 3 −
+ 63
1 7
7u
261 2
2
35 u u3x
1 8
8u
+
1
2
70 u3x
+ 12 u2 u2x 2
1
2
252 u4x
− 52 u4 ux 2 − 56 uux 4 + u3 u2x 2
10
3
21 uu2x
5
− 42
u2x u3x 2 +
----
5
4
36 ux
1
2
14 uu3x
+ 12 ux 2 u2x 2 +
9
2
35 u5x
1
2
20 u2x
−
1
2
42 uu4x
− 72 u5 ux 2 −
3 2
2
14 u u3x
1
2
924 u5x
−
35 2
4
12 u ux
+ 74 u4 u2x 2
+ 72 uux 2 u2x 2 + 53 u2 u2x 3 +
53
7
4
24 u2x
+ 12 u3 u3x 2
− 14 ux 2 u3x 2 − 56 uu2x u3x 2 +
1 2
2
12 u u4x
7
+ 132
u2x u4x 2 −
1
2
3432 u6x
1
2
132 uu5x
+
Table 2:
Conserved Densities for the Kaup-Kuperschmidt and Ito 5th-order equations
Density
Kaup-Kuperschmidt equation
Ito equation
ρ1
u
u
ρ2
----
u2
2
ρ3
u3
3
− 81 ux 2
----
ρ4
u4
4
−
ρ5
----
ρ6
u6
6
−
9
2
16 uux
u7
7
−
35 3
2
16 u ux
31
4
256 ux
−
15
2
128 uu3x
27 4
2
8 u ux
2
− 171
640 u2x u3x +
+
+
− 49 uux 2 + 34 u2x 2
51 2
2
64 u u2x
----
3
2
512 u4x
369
4
320 uux
−
2
2
+ 2619
4480 ux u2x +
ρ8
u4
4
3
2
64 u2x
----
37
+ 256
u2x 3 −
ρ7
+
+
2211
3
2240 uu2x
27
2
560 uu4x
−
69 3
2
40 u u2x
−
----
477 2
2
1120 u u3x
9
2
4480 u5x
----
----
54
Table 3:
Conserved Densities for the Sawada-Kotera-Ito and Lax 7th-order equations
Density
Sawada-Kotera-Ito equation
Lax equation
ρ1
u
u
ρ2
----
u2
ρ3
−u3 + ux 2
−2u3 + ux 2
ρ4
3u4 − 9uux 2 + u2x 2
5u4 − 10uux 2 + u2x 2
ρ5
----
−14u5 + 70u2 ux 2 − 14uu2x 2 + u3x 2
ρ6
6
− 12
7 u +
150 3
2
7 u ux
+
3
2
− 16
21 u2x + uu3x −
ρ7
17
4
7 ux
48 2
2
7 u u2x
1
2
21 u4x
5u7 − 105u4 ux 2 − 42uux 4 +
+24ux 2 u2x 2 +
163
3
9 uu2x
2
2
− 32
9 u2x u3x + uu4x −
ρ8
−
−
133 3
2
3 u u2x
29 2
2
3 u u3x
− 73 u6 +
70 3
2
3 u ux
− 7u2 u2x 2
1
2
18 u4x
− 23 u7 +
35
4
9 uux
35 4
2
3 u ux
5
2
9 u2x u3x
3 8
2u
----
35
4
18 ux
3
2
− 10
9 u2x + uu3x −
− 73 ux 2 u2x 2 −
1
2
27 u5x
+
+
20
3
9 uu2x
− 19 uu4x 2 +
−
14 3
2
3 u u2x
+ u2 u3x 2
1
2
198 u5x
− 42u5 ux 2 − 35u2 ux 4 + 21u4 u2x 2
+42uux 2 u2x 2 + 20u2 u2x 3 + 72 u2x 4 − 6u3 u3x 2
−3ux 2 u3x 2 − 10uu2x u3x 2 + u2 u4x 2
7
+ 11
u2x u4x 2 −
55
1
2
11 uu5x
+
1
2
286 u6x