Characteristic Algebras of Fully Discrete Hyperbolic Type Equations
The notion of the characteristic Lie algebra of the discrete hyperbolic type equation is introduced. An effective algorithm to compute the algebra for the equation given is suggested. Examples and further applications are discussed.
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Symmetry, Integrability and Geometry: Methods and Applications Vol. 1 (2005), Paper 023, 9 pages
Characteristic Algebras of Fully Discrete
Hyperbolic Type Equations
Ismagil T. HABIBULLIN
Institute of Mathematics, Ufa Scientific Center, Russian Academy of Sciences,
112 Chernyshevski Str., Ufa, 450077 Russia
E-mail: ihabib@imat.rb.ru
Received August 04, 2005, in final form November 30, 2005; Published online December 02, 2005
Original article is available at http://www.emis.de/journals/SIGMA/2005/Paper023/
Abstract. The notion of the characteristic Lie algebra of the discrete hyperbolic type
equation is introduced. An effective algorithm to compute the algebra for the equation
given is suggested. Examples and further applications are discussed.
Key words: discrete equations; invariant; Lie algebra; exact solution; Liuoville type equation
2000 Mathematics Subject Classification: 37D99; 37K60; 39A12
1 Introduction
It is well known that the characteristic Lie algebra introduced by A.B. Shabat in 1980, plays
the crucial role in studying the hyperbolic type partial differential equations. For example, if
the characteristic algebra of the equation is of finite dimension, then the equation is solved in
quadratures, if the algebra is of finite growth then the equation is integrated by the inverse
scattering method. More details and references can be found in [1]. Recently it has been
observed by A.V. Zhiber that the characteristic algebra provides an effective tool for classifying
the nonlinear hyperbolic equations. Years ago the characteristic algebra has been used to classify
integrable systems of a special type [2]. However, the characteristic algebras have not yet been
used to study the discrete equations, despite the fact that the discrete equations had become
very popular the last decade (see, for instance, survey [3]).
In this paper we show that the characteristic algebra can be defined for any discrete equation
of the hyperbolic type and it inherits most of the important properties of its continuous counter-
part. However, it has essentially more complicated structure. The work was stimulated by [4],
where the discrete field theory is studied and the question is posed which types of algebraic
structures are associated with the finite field discrete 3D Toda chains.
2 Invariants and vector fields
Consider a discrete nonlinear equation of the form
t(u + 1, v + 1) = f
(
t(u, v), t(u + 1, v), t(u, v + 1)
)
, (1)
where t = t(u, v) is an unknown function depending on the integers u, v, and f is a smooth
function of all three arguments. The following notations are used to shorten formulae: tu =
t(u + 1, v), tv = t(u, v + 1), and tuv = t(u + 1, v + 1). By using these notations one can rewrite
the equation (1) as follows tuv = f(t, tu, tv). Actually, the equation (1) is a discrete analog of the
partial differential equations. Particularly, the class of equations (1) contains difference schemes
for the hyperbolic type PDEs on a quadrilateral grid.
mailto:ihabib@imat.rb.ru
http://www.emis.de/journals/SIGMA/2005/Paper023/
2 I.T. Habibullin
The notations above are commonly accepted, but not very convenient to indicate the iterated
shifts. Below we use also different ones. Introduce the shift operators D and D̄, which act as
follows Df(u, v) = f(u + 1, v) and D̄f(u, v) = f(u, v + 1). For the iterated shifts we introduce
the notations fj = Dj(f) and f̄j = D̄j(f), so that t(u+1, v) = t1, t(u, v+1) = t̄1, t(u+2, v) = t2,
t(u, v + 2) = t̄2 and so on.
The equation (1) is supposed to be hyperbolic. It means that it can be rewritten in any of
the forms: tu = g(t, tv, tuv), tv = r(t, tuv, tu), and t = s(tu, tuv, tv) with some smooth functions
g, r, and s.
A function F = F (v, t, t1, t̄1, . . . ), depending on v and a finite number of the dynamical
variables is called v-invariant, if it is a stationary “point” of the shift with respect to v so that
(see also [5])
D̄F (v, t, t1, t̄1, . . . ) = F (v, t, t1, t̄1, . . . ), (2)
and really the function F solves the equation F (v + 1, tv, f, f1, t̄2, . . . ) = F (v, t, t1, t̄1, . . . ). Exa-
mining carefully the last equation one can find that:
Lemma 1. The v-invariant does not depend on the variables in the set {t̄j}∞j=1.
If any v-invariant is found, then each solution of the equation (1) can be represented as
a solution of the following ordinary discrete equation F (t, tu, . . . , tj) = c(u), where c(u) is an
arbitrary function of u.
Due to the Lemma 1 the equation (2) can be rewritten as
F (v + 1, tv, f, f1, . . . ) = F (v, t, t1, t2, . . . ).
The left hand side of the equation contains tv, while the right hand side does not. Hence the
total derivative of D̄F with respect to tv vanishes. In other words, the operator X1 = D̄−1 d
dtv
D̄
annihilates the v-invariant F : X1F = 0. In a similar way one can check that any operator of the
form Xj = D̄−j d
dtv
D̄j , where j ≥ 1, satisfies the equation XjF = 0. Really, the right hand side
of the equation D̄jF (v, t, t1, t̄1, . . . ) = F (v, t, t1, t̄1, . . . ) (which immediately follows from (2))
does not depend on tv and it implies the equation XjF = 0. As a result, one gets an infinite set
of equations for the function F . For each j the operator Xj is a vector field of the form
X =
∞∑
j=0
x(j)
∂
∂tj
. (3)
Consider now the Lie algebra Lv of the vector fields generated by the operators Xj with the
usual commutator of the vector fields [Xi, Xj ] = XiXj − XjXi. We refer to this algebra as
characteristic algebra of the equation (1).
Remark 1. Note that above we started to consider the discrete equation of the form (1) con-
jecturing that it admits nontrivial v-invariant. The definition of the algebra was motivated by
the invariant. However the characteristic Lie algebra is still correctly defined for any equation
of the form (1) even if it does not admit any invariant.
3 Algebraic criterion of existence of the invariants
Theorem 1. The equation (1) admits a nontrivial v-invariant if and only if the algebra Lv is
of finite dimension.
Characteristic Algebras of Fully Discrete Hyperbolic Type Equations 3
Proof. Suppose that the equation (2) admits a non-constant solution. Then the following
system of equations
XjF (v, t, t1, t2, . . . , tN ) = 0, j ≥ 1 (4)
has a non-constant solution. It is possible only if the linear envelope of the vector-fields {Xj}∞j=1
is of finite dimension.
It is worth mentioning an appropriate property of the vector fields above. If for a fixed j the
operator Xj is linearly expressed through the operators X1, X2, . . . , Xj−1, then any operator Xk
is a linear combination of these operators. Really, we are given the expression Xj = a1X1 + · · ·+
aj−1Xj−1. Note that Xj+1 = D̄−1XjD̄, hence Xj+1 = D−1(a1)X1 + · · ·+ D−1(aj−1)Xj . Thus,
in this case the characteristic algebra is generated by the first j− 1 operators which are linearly
independent. Due to the classical Jacoby theorem the system (4) has a non-constant solution
only if dimension of the Lie algebra generated by the vector fields Xi is no greater than N .
Thus, one part of the theorem is proved.
Suppose now that the dimension of the characteristic algebra is finite and equals, say, N ,
show that in this case the equation (1) admits a v-invariant. Evidently, there exists a function
G(t, t1, . . . , tN ), which is not a constant and that XG = 0 for any X in Lv. Such a function
is not unique, but any other solution is expressed as h(G). Due to the construction the map
X → D̄−1XD̄ leaves the algebra unchanged, hence G1 = D̄G is also a solution of the same
system XG = 0 and therefore G1 = h(G). In other words, one gets a discrete first order
equation: D̄G = h(G), write its general solution in the following form: C = F (v,G) where C
does not depend on v. Evidently the function F found is just the v-invariant needed. �
4 Computation of the characteristic algebra
In this section the explicit forms of the operators {Xj} are given. We show that the operators
are vector fields and give a convenient way to compute the coefficients of the expansion (3).
Start with the operator X1. Directly by definition one gets X1F (t, t1, . . . ) = D̄−1 ∂
∂tv
F (tv, f ,
f1, . . . ). Computing the derivative by the chain rule one obtains
X1F (t, t1, . . . ) = D̄−1
(
∂
∂tv
+
∂f
∂tv
∂
∂f
+
∂f1
∂tv
∂
∂f1
+ · · ·
)
F (tv, f, f1, . . . ),
and finally
X1 =
∂
∂t
+ D̄−1
(
∂f
∂tv
)
∂
∂tu
+ D̄−1
(
∂f1
∂tv
)
∂
∂t2
+ · · ·+ D̄−1
(
∂fj
∂tv
)
∂
∂tj+1
+ · · · . (5)
So the operator X1 can be represented as X1 =
∞∑
i=0
xi
∂
∂ti
where the coefficients xj are found
as xj = D̄−1
(
∂fj−1
∂tv
)
for j > 0 and x0 = 1. Actually the coefficients can be computed by the
following more convenient formula
xj+1 = x1D(x1)D2(x1) · · ·Dj(x1), (6)
or xj+1 = xjD
j(x1). Really,
D̄xj+1 =
∂fj
∂tv
=
∂fj
∂fj−1
∂fj−1
∂fj−2
· · · ∂f
∂tv
= Dj
(
∂f
∂tv
)
· · ·D
(
∂f
∂tv
)
∂f
∂tv
= D̄
(
Dj(x1) · · ·D(x1)x1
)
.
4 I.T. Habibullin
To find X2, use the following formula
X2F = D̄−1X1D̄F (t, t1, . . . ) = D̄−1
(
∂
∂t
+ x1
∂
∂tu
+ x2
∂
∂t2
+ · · ·
)
F (tv, f, f1, . . . ).
After opening the parentheses and some transformation the right hand side of the last formula
gets the form
X2F (t, t1, . . . ) = D̄−1
(
X1(f)
∂
∂f
+ X1(f1)
∂
∂f1
+ X1(f2)
∂
∂f2
+ · · ·
)
F (tv, f, f1, . . . )
So the operator can be written as
X2 = D̄−1 (X1(f))
∂
∂t1
+ D̄−1 (X1(f1))
∂
∂t2
+ D̄−1 (X1(f2))
∂
∂t3
+ · · · . (7)
Continuing this way, one gets
Xj = D̄−1 (Xj−1(f))
∂
∂t1
+ D̄−1 (Xj−1(f1))
∂
∂t2
+ D̄−1 (Xj−1(f2))
∂
∂t3
+ · · · . (8)
One can derive an alternative way to compute the coefficients of the vector fields Xk above.
Represent the operators as follows
Xk =
∞∑
j=0
nkj
∂
∂tj
. (9)
We will show that the coefficients nkj of the operators satisfy the following linear equation
D̄nk+1,j+1 = Dj
(
∂f
∂t
)
nk,j + Dj
(
∂f
∂t1
)
nk,j+1 + Dj
(
∂f
∂tv
)
D̄nk+1,j , (10)
closely connected with the direct linearization of the initial nonlinear equation (1). In order to
derive this formula, apply the operator Xk to the iterated shift fj = f(tj , tj+1, fj−1) and use the
chain rule
Xk(fj) = Dj
(
∂f
∂t
)
D̄−1(Xk−1(fj−1)) + Dj
(
∂f
∂t1
)
D̄−1Xk−1(fj) + Dj
(
∂f
∂tv
)
Xk(fj−1).(11)
Comparison of the two representations (8) and (9) of the operator Xk yields Xk(fj)=D̄nk+1,j+1.
By replacing in (11) X with n one gets the formula (10) required.
The characteristic algebra is invariant under the map X → D−1XD. First prove the formula
D−1X1D = D−1(x1)X1. To this end use the coordinate representation of the operator X1 =
∞∑
i=0
xi
∂
∂ti
and the formula (6) for the coefficients. Then check that D−1X2D = ρX2. Really,
D−1X2D = D−1D̄−1X1D̄D = D̄−1D−1X1DD̄ = D̄−1
(
D−1(x1)
)
X2.
Obviously similar representations can be derived for any generator of the characteristic algebra.
The following statement turns out to be very useful for studying the characteristic algebra.
Lemma 2. Suppose that the vector field
X =
∞∑
j=0
xj
∂
∂tj
satisfies the equation
D(X) = X, (12)
then X ≡ 0.
Characteristic Algebras of Fully Discrete Hyperbolic Type Equations 5
Proof. It follows from (12) that
∞∑
j=0
D
(
xj(t̄1, t̄−1, t, t1, t2, . . . , tkj
)
) ∂
∂tj+1
=
∞∑
j=0
xj(t̄1, t̄−1, t, t1, t2, . . . , tkj
)
∂
∂tj
.
Comparison of the coefficients before the operators ∂
∂tj
in both sides of this equation yields:
x0 = 0, x1 = 0, . . . and so on. �
5 The commutativity property of the algebra
One of the unexpected properties of the characteristic Lie algebra is the commutativity of the
operators Xj . Consider first an auxiliary statement.
Lemma 3. The coefficients xki of the vector fields Xk, k ≥ 1, i ≥ 0 do not depend on the
variable tv.
In other words, the coefficients of the expansions
Xk =
∞∑
j=0
xkj
∂
∂tj
satisfy the equation d
dtv
xki = 0.
Proof. For the operator X1 one has x1,j+1 = D̄−1
(
∂fj
∂tv
)
. But the function fj does not depend
on t̄2 = D̄tv and on t̄3, t̄4, . . . as well. Hence the coefficients do not depend on t̄1 = tv. Similarly,
the functions X1(fj) do not depend on t̄2, so x2,j+1 = D̄−1 (X1(fj)) do not contain tv. One can
complete the proof by using induction with respect to k. �
Now return to the main statement of the section.
Theorem 2. For any positive integers i, j the equation holds [Xi, Xj ] = 0.
Proof. Remind that Xj = D̄−j d
dtv
D̄j , so that one can deduce
[Xi, Xj ] =
[
D̄−i d
dtv
D̄i, D̄−j d
dtv
D̄j
]
.
Suppose for the definiteness that k = j − i ≥ 1. Then
[Xi, Xj ] = D̄−i
[
d
dtv
, D̄i−j d
dtv
D̄j−i
]
D̄i = D̄−i
[
d
dtv
, Xk
]
D̄i = D̄−i
∞∑
j=0
d
dtv
(xkj)
∂
∂tj
D̄i.
But due to the Lemma 3 the last expression vanishes that proves the theorem. �
6 Characteristic algebra for the discrete Liouville equation
The well known Liouville equation ∂2v
∂x∂y = ev admits a discrete analog of the form [6]
tuv =
1
t
(tu − 1)(tv − 1), (13)
6 I.T. Habibullin
which can evidently be rewritten as
tu,−v =
1
t− 1
tut−v + 1.
Specify the coefficients of the expansions (5)–(7) representing the vector fields X1 and X2 for the
discrete Liouville equation (13). Find the coefficient x1 = D̄−1
(
∂f
∂tv
)
, remind that f(t, tu, tv) =
1
t (tu − 1)(tv − 1) and g(t−v, t, tu) = 1
t−1 tut−v + 1,
x1 = D̄−1
(
∂f
∂tv
)
= D̄−1
(
tu − 1
t
)
=
g − 1
t−v
=
tu
t− 1
.
Similarly
x2 = D̄−1
(
∂f1
∂f
∂f
∂tv
)
=
t2
t1 − 1
t1
t− 1
.
It can easily be proved by induction that xj =
j−1∏
k=0
tk+1
tk−1 for j ≥ 1, here t0 := t. Remind also that
x0 = 1. So that the vector field is
X1 =
∂
∂t
+
t1
t− 1
∂
∂t1
+
t1t2
(t− 1)(t1 − 1)
∂
∂t2
+
t1t2t3
(t− 1)(t1 − 1)(t2 − 1)
∂
∂t3
+ · · · ,
It is a more difficult problem to find the operator X2, it can be proved that
X2 =
(t− 1)t
(t−v − 1)t−v
(X1 −X−1),
where the operator X−1 is defined as follows X−1 = D̄ d
dt̄−1
D̄−1. For the coefficients x−,j of the
expansion X−1 =
∞∑
j=0
x−,j
∂
∂tj
one can deduce the formula x−,j =
j−1∏
k=0
tk+1−1
tk
, so that the operator
is represented as
X−1 =
∂
∂t
+
t1 − 1
t
∂
∂t1
+
(t1 − 1)(t2 − 1)
tt1
∂
∂t2
+
(t1 − 1)(t2 − 1)(t3 − 1)
tt1t2
∂
∂t3
+ · · · ,
X3 = D̄−1X2D̄ =
(t̄−1 − 1)t̄−1
(t̄−2 − 1)t̄−2
X2.
Theorem 3. Dimension of the characteristic Lie algebra of the Liouville equation (13) equals
two.
Proof. It is more easy to deal with the operators Y+ = (t− 1)X1 and Y− = tX−1 rather than
the operators X1 and X−1. In order to prove the theorem it is enough to check the formula
[Y+, Y−]− Y+ + Y− = 0. (14)
Denote through X the left hand side of the equation (14) and compute D(X) to apply the
Lemma 2. It is shown straightforwardly that
D(Y+) =
t1 − 1
t1
(
Y+ − (t− 1)
∂
∂t
)
and D(Y−) =
t1
t1 − 1
(
Y− − t
∂
∂t
)
.
Characteristic Algebras of Fully Discrete Hyperbolic Type Equations 7
Compute now the shifted operator D([Y+, Y−]) = [D(Y+), D(Y−)] which can be represented as
follows
[D(Y+), D(Y−)] =
[
t1 − 1
t1
(
Y+ − (t− 1)
∂
∂t
)
,
t1
t1 − 1
(
Y− − t
∂
∂t
)]
= A1 + A2 + A3 + A4,
where
A1 =
[
t1 − 1
t1
Y+,
t1
t1 − 1
Y−
]
, A2 = −
[
(t1 − 1)(t− 1)
t1
∂
∂t
,
t1
t1 − 1
Y−
]
,
A3 = −
[
t1 − 1
t1
Y+,
tt1
t1 − 1
∂
∂t
]
, A4 =
[
(t1 − 1)(t− 1)
t1
∂
∂t
,
tt1
t1 − 1
∂
∂t
]
.
Direct computations give
A1 = [Y+, Y−]− 1
t1 − 1
Y−
1
t1Y+
, A2 =
1− t− t1
t1
∂
∂t
,
A3 =
1− t− t1
t1 − 1
∂
∂t
, A4 =
∂
∂t
.
Summarizing all computations above one gets the result
D[Y+, Y−] = [Y+, Y−]− 1
t1 − 1
Y− −
1
t1
Y+ +
(
1 +
t− 1
t1
+
t
t1 − 1
)
∂
∂t
,
which implies D([Y+, Y−]−Y+ +Y−) = [Y+, Y−]−Y+ +Y−. Now apply Lemma 2 to the function
X = [Y+, Y−]− Y+ + Y− to get X ≡ 0. �
7 How to find the invariants?
In this section two examples of the equations with the invariants are shown. Start with a simple
one.
Example 1. Consider a linear equation of the form
tuv = tu + tv − t + 1
so that f(t, tu, tv) = tu + tv − t + 1 and g(t−v, t, tu) = tu − t + t−v − 1. It is easy to see that the
algebra is of one dimension
X1 =
∂
∂t
+
∂
∂tu
+
∂
∂t2
+ · · · , X2 ≡ 0.
The first integral of the least order for the equation X1F = 0 can be taken as G0 = t1 − t.
Evidently it solves the equation D̄G0 = G0+1. Now, the invariant is to be taken as I = t1−t−v,
because D̄I = tuv − tv − v − 1 = tu − t − v = I. Any other v-invariant of the equation can be
represented as F = F (I,DI, D2I, . . . ,DkI).
Example 2. Return to the discrete Liouville equation discussed above. Find the intersection
of the kernels of the operators X1 and X−1, hence they constitute the basis of the characteristic
algebra of the Liouville equation (13). To this end first solve the equation X1F = 0 which is
reduced to the following infinite system of the ordinary differential equations:
dt
1
=
dt1
t1/(t− 1)
=
dt2
t1t2/(t− 1)(t1 − 1)
= · · · .
8 I.T. Habibullin
So that the invariants of the vector field X1 are I0 = t1
t−1 , I1 = D(I0), I2 = D(I1), . . . . Change
the variables in the vector fields by setting t̃ = t, t̃1 = I0, t̃2 = I1, . . . . Then one gets
X1 =
∂
∂t
and X−1 =
∂
∂t
− t̃1 + 1
t(t− 1)
∂
∂t̃1
− t̃2 + 1
t̃1t(t− 1)
∂
∂t̃2
− · · ·.
Now solve the equation dt̃2
dt̃1
= t̃2+1
t̃1(t̃1+1)
to find the common solution F0 =
(
t2
t1−1 + 1
) (
t−1
t1
+ 1
)
of both equations X1F = 0 and X−1F = 0. The invariant is
I(t, t1, t2) = F0(t, t1, t2) =
(
t2
t1 − 1
+ 1
) (
t− 1
t1
+ 1
)
(see also [5]). Evidently, the basis of common invariants can be chosen as follows: F0, D(F0),
D2(F0), . . . . It can be easily verified by the direct computation that F0 solves the equation
D̄F0 = F0.
Corollary 1. In the case of the Liouville equation each solution of the system of the equations
X1F = 0, X−1F = 0 gives the v-invariant.
Proof. Each solution of this system can be represented as F = F (F0, D(F0), D2(F ), . . . ), hence
D̄F
(
F0, D(F0), D2(F ), . . .
)
= F
(
D̄(F0), D̄(D(F0)), D̄(D2(F )), . . .
)
= F
(
F0, D(F0), D2(F ), . . .
)
. �
8 Discrete potentiated Korteweg–de Vries equation
Discrete equation of the form
(tuv − t + p + q)(tu − tv + q − p) = q2 − p2 (15)
is called the discrete potentiated KdV equation (DPKdV). The constant parameters p, q are
removed by a combination of the shift and scaling transformations of the form t(u, v) = ht̃(u, v)−
pv−qu, where h2 = q2−p2 if q2−p2 > 0 and h2 = −q2+p2 if q2−p2 < 0. After this transformation
the equation (15) takes one of the forms
(tuv − t)(tu − tv) = ±1.
The algebraic properties of the equation do not depend on the choice of the sign before the unity,
we will take the upper sign. Study the problem how to describe the characteristic Lie algebra
for the equation
(tuv − t)(tu − tv) = 1. (16)
Represent the equation (16) as tuv = f(t, tu, tv), where f(t, tu, tv) = t + 1
tu−tv
. Set x = ∂f
∂tv
=
1
(tv−tu)2
, then x = (f − t)2. Below we use the upper index to denote the shift with respect to v
and lower index to denote the shift with respect to u so that D̄−kDjx = xk
j . Then due to the
general rule the operator X1 is written as
X1 =
∂
∂t
+ x1 ∂
∂t1
+ x1x1
1
∂
∂t2
+ · · ·+ x1x1
1 · · ·x1
j
∂
∂tj+1
+ · · · .
Theorem 4. Characteristic Lie algebra of the DPKdV equation (16) is of infinite dimension.
Characteristic Algebras of Fully Discrete Hyperbolic Type Equations 9
Proof. In the case of the DPKdV equation one gets ∂f
∂t = 1, ∂f
∂t1
= −x, ∂f
∂tv
= x. The for-
mula (10) is specified
D̄nk+1,j+1 = nk,j − xjnk,j+1 + xjD̄nk+1,j .
It is more convenient to write it as
nk+1,j+1 = D̄−1nk,j − x1
jD̄
−1nk,j+1 + x1
jnk+1,j ,
remind that D−1xj = x1
j . It is clear now that the coefficients are polynomials of the finite
number of the dynamical variables in the set {xi
j}∞i,j=0. Order the variables in this set according
to the following rule: ord(xn
j ) > ord(xn
p ) if j > p and ord(xm
j ) > ord(xn
p ) if m > n.
Lemma 4. For any positive k, j the coefficient nk,j for the operator Xk in the DPKdV case is
represented as
nk,j = mk,jx
k
j−1 + rk,j
where mk,j and rk,j are polynomials of the variables with the orders less than the order of xk
j−1,
moreover, mk,j is a monomial.
Lemma can be easily proved by using the induction method. It allows to complete the proof
of the theorem. The vector fields {Xk}∞k=1 are all linearly independent. �
Due to the theorem the DPKdV equation do not admit any v-invariant. It is not surprising,
because the equation can be integrated by the inverse scattering method or, in other words, it is
S-integrable, it is well known in the case of partial differential equations that only C-integrable
equations (Liouville type) admit such kind objects called x- and y-integrals.
9 Conclusion
The notion of the characteristic algebra for discrete equations is introduced. It is proved that the
equation is Darboux integrable if and only if its characteristic algebras in both directions are of
finite dimension. The notion can evidently be generalized to the systems of discrete hyperbolic
equations. It would be useful to compute the algebras for the periodically closed discrete Toda
equation, or for the finite field discrete Toda equations found in [4] and [8], corresponding in the
continuum limit to the simple Lie algebras of the classical series A and C.
Acknowledgments
The author thanks A.V. Zhiber for fruitful discussions. The work has been supported by the
grant RFBR # 05-01-00775.
[1] Leznov A.N., Savel’ev M.V., Group methods of integration of nonlinear dynamical systems, Moscow, Nauka,
1985 (in Russian).
[2] Shabat A.B., Yamilov R.I., Exponential systems of type I and the Cartan matrices, Preprint, Ufa, 1981.
[3] Zabrodin A.V., The Hirota equation and the Bethe ansatz, Teoret. Mat. Fiz., 1998, V.116, N 1, 54–100
(English transl.: Theoret. and Math. Phys., 1998, V.116, N 1, 782–819).
[4] Ward R.S., Discrete Toda field equations, Phys. Lett. A, 1995, V.199, 45–48.
[5] Adler V.E., Startsev S.Ya., On discrete analogues of the Liouville equation, Teoret. Mat. Fiz., 1999, V.121,
N 2, 271–284 (English transl.: Theoret. and Math. Phys., 1999, V.121, N 2, 1484–1495).
[6] Hirota R., The Bäcklund and inverse scattering transform of the K-dV equation with nonuniformities,
J. Phys. Soc. Japan, 1979, V.46, N 5, 1681–1682.
[7] Habibullin I.T., Characteristic algebras of the discrete hyperbolic equations, nlin.SI/0506027.
[8] Habibullin I.T., Discrete Toda field equations, nlin.SI/0503055.
1 Introduction
2 Invariants and vector fields
3 Algebraic criterion of existence of the invariants
4 Computation of the characteristic algebra
5 The commutativity property of the algebra
6 Characteristic algebra for the discrete Liouville equation
7 How to find the invariants?
8 Discrete potentiated Korteweg-de Vries equation
9 Conclusion
|
| id | nasplib_isofts_kiev_ua-123456789-209337 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1815-0659 |
| language | English |
| last_indexed | 2025-12-07T17:02:28Z |
| publishDate | 2005 |
| publisher | Інститут математики НАН України |
| record_format | dspace |
| spelling | Habibullin, I.T. 2025-11-19T12:21:05Z 2005 Characteristic Algebras of Fully Discrete Hyperbolic Type Equations / I.T. Habibullin // Symmetry, Integrability and Geometry: Methods and Applications. — 2005. — Т. 1. — Бібліогр.: 8 назв. — англ. 1815-0659 2000 Mathematics Subject Classification: 37D99; 37K60; 39A12 https://nasplib.isofts.kiev.ua/handle/123456789/209337 https://doi.org/10.3842/SIGMA.2005.023 The notion of the characteristic Lie algebra of the discrete hyperbolic type equation is introduced. An effective algorithm to compute the algebra for the equation given is suggested. Examples and further applications are discussed. The author thanks A.V. Zhiber for fruitful discussions. The work has been supported by the grant RFBR # 05-01-00775. en Інститут математики НАН України Symmetry, Integrability and Geometry: Methods and Applications Characteristic Algebras of Fully Discrete Hyperbolic Type Equations Article published earlier |
| spellingShingle | Characteristic Algebras of Fully Discrete Hyperbolic Type Equations Habibullin, I.T. |
| title | Characteristic Algebras of Fully Discrete Hyperbolic Type Equations |
| title_full | Characteristic Algebras of Fully Discrete Hyperbolic Type Equations |
| title_fullStr | Characteristic Algebras of Fully Discrete Hyperbolic Type Equations |
| title_full_unstemmed | Characteristic Algebras of Fully Discrete Hyperbolic Type Equations |
| title_short | Characteristic Algebras of Fully Discrete Hyperbolic Type Equations |
| title_sort | characteristic algebras of fully discrete hyperbolic type equations |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/209337 |
| work_keys_str_mv | AT habibullinit characteristicalgebrasoffullydiscretehyperbolictypeequations |