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DENSITIES AND FLUXES OF
DIFFERENTIAL-DIFFERENCE EQUATIONS
MARK S. HICKMAN
Department of Mathematics and Statistics
University of Canterbury
Private Bag 4800, Christchurch, New Zealand
Email address: M.Hickman@math.canterbury.ac.nz
WILLY A. HEREMAN
Department of Mathematical and Computer Sciences
Colorado School of Mines
Golden CO 80401–1887, USA
Email address: whereman@mines.edu
An algorithm is presented that uses direct methods to find conserved densities and
fluxes of differential-difference equations. The use of the code is illustrated with
the modified Volterra lattice. This algorithm has been implemented in Maple in
the form of a toolbox.
1. Differential-Difference Equations
Dating back to the work of Fermi, Pasta, and Ulam in the 1950’s [2]
differential-difference equations (DDEs) have been the focus of many nonlinear studies. A number of physically interesting problems can be modeled
with nonlinear DDEs, including particle vibrations in lattices, currents in
electrical networks, pulses in biological chains, etc. DDEs play important
roles in queuing problems and discretizations in solid state and quantum
physics. Last but not least, they are used in numerical simulations of nonlinear PDEs.
Consider a nonlinear (autonomous) DDE of the form
d
un = f (un− , un−+1 , . . . , un , . . . , un+m−1 , un+m )
dt
with
∂f
∂f
= 0,
∂ un− ∂ un+m
1
(1.1)
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where n is an arbitrary integer. In general, f is a vector-valued function
of a finite number of dynamical variables and each uk is a vector-valued
function of t.
The index n may lie in Z, or the uk may be periodic, uk = uk+N . The
integers and m measure the degree of non-locality in (1.1). If = m = 0
then the equation is local and reduces to a system of ordinary differential
equations.
The (up-)shift operator D is defined by
D uk = uk+1 .
Its inverse, called the down-shift operator, is given by D−1 uk = uk−1 .
Obviously, uk = Dk u0 . The actions of D and D−1 are extended to functions
by acting on their arguments. For example,
D g(up , up+1 , . . . , uq ) = g(D up , D up+1 , . . . , D uq ) = g(up+1 , up+2 , . . . uq+1 ).
In particular,
∂
∂
D
g(up , up+1 , . . . , uq ) =
g(up+1 , up+2 , . . . , uq+1 ).
∂ uk
∂ uk+1
Moreover, for equations of type (1.1), the shift operator commutes with the
time derivative; that is,
d
d
un =
(D un ) .
D
dt
dt
Thus, with the use of the shift operator, the entire system (1.1) which may
be an infinite set of ordinary differential equations is generated from a single
equation
d
u0 = f (u− , u−+1 , . . . , u0 , . . . , um−1 , um )
dt
(1.2)
with
∂f ∂f
= 0.
∂ u− ∂ um
Next, we define the (forward) difference operator, ∆ = D − I, by
∆ uk = (D − I) uk = uk+1 − uk ,
where I is the identity operator. The difference operator extends to functions by ∆ g = D g − g. This operator takes the role of a spatial derivative
on the shifted variables as many examples of DDEs arise from discretization
of a PDE in (1 + 1) variables [6].
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For any function g = g(up , up+1 , . . . , uq ), the total time derivative Dt g
is computed as


q
q
∂g d
∂ g k
uk = 
D
Dt g =
f (u− , u−+1 , . . . , um )
∂ uk d t
∂ uk
k=p
k=p
on solutions of (1.1). A simple calculation shows that the shift operator D
commutes with Dt , and so does D with ∆.
A function ρ = ρ(up , up+1 , . . . , uq ) is a (conserved) density of (1.2) if
there exists a function J = J(ur , ur+1 , . . . , us ), called the (associated) flux,
such that
Dt ρ + ∆ J = 0
(1.3)
is satisfied on the solutions of (1.2). Eq. (1.3) is called a local conservation
law.
Any shift of a density is trivially a density since
Dt Dk ρ + ∆ Dk J = Dk (Dt ρ + ∆ J) = 0,
with associated flux Dk J.
Constants of motion for (1.2) are easily obtained from a density and its
shifts. Indeed, for any density ρ with corresponding flux J, consider
Ω=
q
Dk ρ.
k=p
The total time derivative of Ω is
Dt Ω = −
q
∆Dk J = −
k=p
q
k+1
D
− Dk J = − Dq+1 − Dp J.
k=p
Applying appropriate boundary conditions (e.g. all uk → 0, as k → ±∞)
one gets the conservation law
∞
Dk ρ = lim Dq+1 J − lim Dp J = 0.
Dt
k=−∞
q→∞
p→−∞
For a periodic chain where uk = uk+N , after summing over a period, one
obtains
N
k
Dt
D ρ = DN +1 J − D0 J = J − J = 0.
k=0
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In either case, Ω is a constant of motion of (1.2) since Ω does not change
with time.
A function g = g(up , up+1 , . . . , uq ) is a total difference if there exists
another function h = h(up , up+1 , . . . , uq−1 ), such that g = ∆ h. A density
which is a total difference,
ρ = ∆F
(1.4)
(so that Dt ρ = ∆ Dt F and therefore J = −Dt F is an associated flux), is
called trivial. These densities lead to trivial conservation laws since
Ω=
q
Dk ∆ F = Dq+1 F − Dp F
k=p
holds identically (and not just on solutions of (1.2)).
Two densities ρ, ρ̃ are called equivalent if ρ − ρ̃ = ∆ F for some F.
Equivalent densities, denoted by ρ ∼ ρ̃, differ by a trivial density and yield
the same conservation law. Also, ρ ∼ Dk ρ, and (1.4) expresses that ρ ∼ 0.
The discrete Euler operator (variational derivative)
A necessary and sufficient condition for a function g to be a total difference
is that
E(g) = 0,
(1.5)
where E is the discrete Euler operator (variational derivative) [1,7] defined
by
E(g) =
q
D−k
k=p
∂
g .
∂ uk
Note that we can rewrite the Euler operator as


q
∂ 
D−k g  .
E(g) =
∂ u0
(1.6)
k=p
Also note that (1.5) implies
∂
∂
∂2 g
p
p
−p ∂ g
E(g) = D
. (1.7)
0=D
D
=
∂ uq−p
∂ uq−p
∂ up
∂ up ∂ uq
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2. The Algorithm for Computing Densities and Fluxes
Densities ρ and fluxes J are related by (1.3). In principle, we first need to
first solve E(Dt ρ) = 0 to find the density. This integrability condition for ρ
is a rather unusual “PDE” since it involves derivatives of both ρ and shifts
of ρ (and so involves ρ with different arguments). Surprisingly, perhaps,
the integrability condition is amenable to analysis. Next, to compute J =
−∆−1 (Dt ρ), we need to invert the operator ∆ = D − I. Working with the
formal inverse,
∆−1 = D−1 + D−2 + D−3 + . . . ,
is impractical, perhaps impossible.
We therefore present a simple algorithm which will yield both the density and the flux, and circumvent the above infinite formal series. The
algorithm does not require the densities or fluxes to be polynomial (see
[3,4,5] for an algorithm to find polynomial densities). The idea is to split
expressions into a total derivative term plus a term involving lower order
shifts. Using
I = (D − I + I) D−1 = ∆ D−1 + D−1 ,
(2.1)
any expression T can be split as follows,
T = ∆ D−1 T + D−1 T.
The first term will contribute to the flux whilst the second term has a
strictly lower shift than the original expression. These decompositions are
applied to terms that do not involve the lowest-order shifted variables. Once
all terms are “reduced” in this manner, left-over terms (all of which involve
the lowest-order shifted variable) yield the constraints for the undetermined
coefficients or unknown functions in the density.
Without loss of generality, we can assume that ρ = ρ(u0 , . . . , uq ). For
(1.2), we have
Dt ρ =
q
∂ρ k
D f (u− , . . . , um )
∂ uk
k=0
q
∂ρ
∂ρ
=
f (u− , . . . , um ) +
Dk f (u− , . . . , um ).
∂ u0
∂ uk
k=1
Applying (2.1) to the second term, with f = f (u− , . . . , um ), we obtain
q
∂ρ
∂ρ k
−1
−1
f + ∆D + D
D f
Dt ρ =
∂ u0
∂ uk
k=1
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q−1
q−1
∂ D−1 ρ k
∂ D−1 ρ k
∂ρ
f +∆
D f+
D f
∂ u0
∂ uk
∂ uk
k=0
k=0
q−1
q−1
∂ ∂ D−1 ρ k
∂ D−1 ρ k
−1
ρ+D ρ f +∆
D f+
D f
=
∂ u0
∂ uk
∂ uk
=
k=0
k=1
Next, we repeat this procedure by applying (2.1) to the last term. After a
further q − 2 applications, we get
∂ −1
−2
ρ+D ρ+D ρ f
Dt ρ =
∂ u0
q−2
q−1
q−2
∂ D−1 ρ
∂ D−2 ρ
∂ D−2 ρ k
k
D f+
D f +
Dk f
+∆
∂ uk
∂ uk
∂ uk
k=0
k=0
k=1
= ···


= E(ρ) f + ∆ 
q−j
q ∂ D−j ρ
j=1 k=0
∂ uk
Dk f 
by (1.6).
If E(ρ) f = 0 then ρ is a trivial density. For ρ to be a non-trivial density,
we require
E(ρ) f = ∆ h
(2.2)
for some h with h = 0. In this case the associated flux is
J = −h−
q−j
q ∂ D−j ρ
j=1 k=0
∂ uk
Dk f.
One could apply the discrete Euler operator to E(ρ) f to determine conditions such that (2.2) holds. Alternatively, one could repeat the above
strategy by splitting this expression into a part, A(0) , that does not depend
on the lowest shifted variable and the remaining terms. One then applies
(2.1) repeatedly to A(0) (removing the total difference terms generated by
(2.1) from A(0) ) until one obtains a term that involves both the lowest and
highest shifted variables. From (1.7), one obtains conditions for (2.2) to
hold. This procedure must be repeated until all the terms in E(ρ) f have
been recast as a total difference. We now illustrate this algorithm with a
simple example.
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The Modified Volterra Lattice
Consider the modified Volterra (mV) lattice [1]
d vn
= vn 2 (vn+1 − vn−1 ),
dt
or, equivalently,
d v0
= v0 2 (v1 − v−1 ).
(2.3)
dt
To keep the exposition simple, we search for densities of the form
ρ = ρ(v0 , v1 , v2 )
for the mV lattice (2.3), where f = v0 2 (v1 − v−1 ). The construction of densities involving a greater spread of dynamic variables can be accomplished
with the Maple code.
We write
Dt ρ = σ + ∆ K
with
σ = E(ρ) f
2
= v0 (v1 − v−1 )
∂ρ (v0 , v1 , v2 ) ∂ρ (v−1 , v0 , v1 ) ∂ρ (v−2 , v−1 , v0 )
+
+
∂v0
∂v0
∂v0
and
K=
q−j
q ∂ D−j ρ k
D f
∂ uk
j=1
k=0
∂ρ (v−1 , v0 , v1 )
∂ρ (v−1 , v0 , v1 )
+ v1 2 (v2 − v0 )
∂v0
∂v1
,
v
,
v
)
∂ρ
(v
−2
−1
0
+ v0 2 (v1 − v−1 )
.
∂v0
= v0 2 (v1 − v−1 )
The application of (1.7) directly to σ yields nothing since no term depends
on both the lowest shifted variable (v−2 ) and the highest shifted variable
(v2 ). We therefore split σ into a term which depends on v−2 and a term
that does not. We then apply (2.1) to the latter term.
Thus,
∂ρ (v−2 , v−1 , v0 )
∂v0
∂ρ (v0 , v1 , v2 ) ∂ρ (v−1 , v0 , v1 )
−1
2
+ (∆ + I) D
+
v0 (v1 − v−1 )
.
∂v0
∂v0
σ = v0 2 (v1 − v−1 )
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Next, we update σ and K by removing the total difference term from σ and
adding it to K. Thus,
∂ρ (v−2 , v−1 , v0 )
∂v0
∂ρ (v−1 , v0 , v1 ) ∂ρ (v−2 , v−1 , v0 )
2
+ v−1 (v0 − v−2 )
+
∂v−1
∂v−1
σ = v0 2 (v1 − v−1 )
and
∂ρ (v−1 , v0 , v1 ) ∂ρ (v−2 , v−1 , v0 )
+
∂v0
∂v0
∂ρ (v−1 , v0 , v1 )
+ v1 2 (v2 − v0 )
∂v1
∂ρ
(v−1 , v0 , v1 ) ∂ρ (v−2 , v−1 , v0 )
+ v−1 2 (v0 − v−2 )
+
.
∂v−1
∂v−1
K = v0 2 (v1 − v−1 )
Now we are ready to apply (1.7) to σ. Thus,
D2
∂2 σ
∂ 2 ρ (v0 , v1 , v2 )
∂ 2 ρ (v1 , v2 , v3 )
= v2 2
− v1 2
= 0.
∂ v−2 ∂ v1
∂v0 ∂v2
∂v1 ∂v3
(2.4)
A differentiation with respect to v0 yields
v2 2
∂ 3 ρ (v0 , v1 , v2 )
= 0.
∂v0 2 ∂v2
So,
ρ(v0 , v1 , v2 ) = ρ(1) (v0 , v1 ) − ρ(2) (v0 , v1 ) + ρ(2) (v1 , v2 ) + ρ(3) (v1 , v2 )v0
= ρ(1) (v0 , v1 ) + ∆ ρ(2) (v0 , v1 ) + ρ(3) (v1 , v2 )v0
for some unknown functions ρ(1) , ρ(2) and ρ(3) . The term ∆ ρ(2) (v0 , v1 ) leads
to a trivial density and can be ignored. Thus
ρ = ρ(1) (v0 , v1 ) + ρ(3) (v1 , v2 )v0 .
The integrability condition (2.4) is now
∂ (3)
∂ (3)
ρ (v1 , v2 ) − v1 2
ρ (v2 , v3 ) = 0.
∂v2
∂v3
A subsequent differentiation with respect to v3 gives
v2 2
− v1 2
∂ 2 (3)
ρ (v2 , v3 ) = 0.
∂v3 2
Consequently, ρ(3) (v2 , v3 ) = ρ(4) (v2 ) + ρ(5) (v2 )v3 . Equation (2.4) has become
v2 2 ρ(5) (v1 ) − v1 2 ρ(5) (v2 ) = 0
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from which we get
ρ(5) (v2 ) = c(1) v2 2
for some constant c(1) .
The density now has the form ρ = ρ(1) (v0 , v1 ) + (ρ(4) (v1 ) + c(1) v1 2 v2 )v0 .
The term ρ(4) (v1 )v0 can be absorbed into the term ρ(1) (v0 , v1 ). Thus
ρ(v0 , v1 , v2 ) = ρ(1) (v0 , v1 ) + c(1) v0 v1 2 v2 .
Hence,
σ = c(1) v−2 v−1 3 v0 (v0 − 2v−2 ) − v−2 v−1 2
∂ρ(1) (v−1 , v0 )
∂v−1
∂ρ(1) (v−2 , v−1 )
+ v−1 2 (v0 − v−2 )
∂v−1
∂ρ(1) (v−2 , v−1 )
+ (∆ + I) c(1) v−2 2 v−1 3 v0 + v−2 2 v−1
∂v−2
by (2.1). As before, we update σ by moving the total difference term into
K. Therefore,
σ = c(1) v−2 v−1 3 v0 (v0 − v−2 ) − v−2 v−1 2
+ v−1 2 (v0 − v−2 )
∂ρ(1) (v−1 , v0 )
∂v−1
∂ρ(1) (v−2 , v−1 )
∂ρ(1) (v−2 , v−1 )
+ v−2 2 v−1
∂v−1
∂v−2
and
K = c(1) v−1 v0 v−1 v0 2 v1 − v0 v1 2 v2 − v−2 v−1 2 v0 − v0 2 v1 2 + 3v−2 2 v−1 2
∂ (1)
∂ (1)
+ v0 2 (v−1 − v1 )
ρ (v−1 , v0 ) + v−2 2 v−1
ρ (v−2 , v−1 )
∂v0
∂v−2
(1)
∂ρ (v−1 , v0 ) ∂ρ(1) (v−2 , v−1 )
+
.
+ v−1 2 (v−2 − v0 )
∂v−1
∂v−1
Applying (1.7) to σ yields
∂ 2 ρ(1) (v1 , v2 ) ∂ 2 ρ(1) (v0 , v1 )
∂2 σ
D2
= v1 2 2c(1) v1 (v2 − v0 ) −
+
∂ v−2 ∂ v0
∂v1 ∂v2
∂v0 ∂v1
=0
which readily integrates to
ρ(1) (v0 , v1 ) = 12 c(1) v0 2 v1 2 + ρ(6) (v0 ) + c(2) v0 v1 .
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At this stage
σ = v−2 2 v−1
dρ(6) (v−2 )
dρ(6) (v−1 )
− v2 v−1 2
dv−2
dv−1
and
K = v−2 2 v−1
dρ(6) (v−2 )
dρ(6) (v−1 )
− v−1 2 v−2
+ c(1) v−1 v0 3 v1 2
dv−2
dv−1
dρ(6) (v−1 )
+ c(1) v−1 v0 2 v1 2 v2 + c(2) v−1 v0 2 v1 .
dv−1
One more application of (1.7) to σ yields
(6)
(6)
∂2 σ
(v0 )
(v1 )
d
d
2
2 dρ
2 dρ
D
=
v0
−
v1
= 0.
∂ v−2 ∂ v−1
dv0
dv0
dv1
dv1
This equation integrates to yield
1
ρ(6) (v0 ) = c(3) + c(4) log v0 .
v0
Thus the solution is given by
1
ρ = 12 c(1) (v0 2 v1 2 + 2v0 v1 2 v2 ) + c(2) v0 v1 + c(3) + c(4) log v0 .
v0
Consequently,
+ v−1 2 v0
σ = c(3) v−2 − c(3) v−1 = − c(3) ∆ v−2 ,
which can be absorbed into K to yield σ = 0. Finally, the associated flux is
J = −K
= −c(1) v−1 v0 2 v1 2 (v0 + v2 ) − c(2) v−1 v0 2 v1 + c(3) (v0 + v−1 ) − c(4) v−1 v0 .
Splitting the density ρ and flux J according to the independent constants
c(i) we obtain four non-trivial conversation laws.
3. Implementation
The strategy outlined in the above example has been implemented in Maple.
This implementation is in the form of a toolbox which provides code to
compute the reductions and generate the integrability conditions. The solution of the integrability conditions has not been automated. The code is
available from the author.
Acknowledgments
The first author wishes to thank the Department of Mathematical and
Computer Sciences of the Colorado School of Mines for its hospitality during
his sabbatical visit where this work was completed.
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References
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