Hamiltonian formalism in a problem of 3-th waves hierarchy
arXiv:0704.1649v1  [hep-lat]  12 Apr 2007
Hamiltonian formalism in a problem of 3-th
waves hierarchy
A. N. Leznov
Abstract
By the method of discrete transformation equations of 3-th wave hier-
archy are constructed. We present in explicit form two Poisson structures,
which allow to construct Hamiltonian operator consequent application of
which leads to all equations of this hierarchy. For calculations it will be
necessary results of previous paper [1], which for convenience of the reader
we present in corresponding place of the text. The obtained formulae are
checked by independent calculations.
1 Introduction
All system equations of 3-th wave hierarchy are invariant with respect to two
mutually commutative discrete transformation of this problem [1],[3],[5],[6][1].
In this introduction we present the solution of the same problem in the case A1
algebra follow to the paper [2].
We repeat here briefly the most important punks of general construction
from [2].
The discrete invertible substitution (mapping) defined as
ũ = T (u, u′, ..., ur) ≡ T (u) (1)
u is s dimensional vector function; ur its derivatives of corresponding order with
respect to ”space” coordinates.
The property of invertibility means that (1) can be resolved and ”old” func-
tion u may expressed in terms of new one ũ and its derivatives.
Freshet derivative T ′(u) of (1) is s× s matrix operator defined as
T ′(u) = Tu + Tu′D + Tu′′D
2 + ... (2)
where Dm is operator of m-times differentiation with respect to space coordi-
nates.
∗Universidad Autonoma del Estado de Morelos, CCICAp,Cuernavaca, Mexico
http://arxiv.org/abs/0704.1649v1
Let us consider equation
Fn(T (u)) = T
′(u)Fn(u) (3)
where Fn(u) is s-component unknown vector function, each component of which
depend on u and its derivatives not more than n order. It is not difficult to
understand that evolution type equation
ut = Fn(u)
is invariant with respect substitution (1).
Two other equations and its solutions are important in what follows
T ′(u)J(u)(T ′(u))T = J(T (u)), T ′(u)H(u)(T ′(u))−1 = H(T (u)) (4)
where (T ′(u))T = T Tu −DT
u′ +D
2T Tu′′ + ... and J(u), H(u) are unknown s× s
matrix operators, the matrix elements of which are polynomial of some finite
order with respect to operator of differentiation (of its positive and negative
degrees).
JT (u) = −J(u) may be connected with the Poisson structure and equation
(4) means its invariance with respect to discrete transformation T .
The second equation (4) determine operator H(u), which after application
to arbitrary solution of (3) F (u) leads to new solution of the same system
F̃ (u) = H(u)F (u)
And thus we obtain reccurent procedure to construct solutions of (3) from few
simple ones.
If it is possible to find two different J1, J2 (Hamiltonian operators,Poisson
structures) then
H(u) = J2J
1 (5)
satisfy second equation (4).
In [2] are presented arguments that Hamiltonian operator it is the sense find
in a form
J(u) = Fn(u)D
−1Fn(u)
i (6)
where Fn some solution of (3) and Ai some s × s matrices constructed from u
and its derivatives.
The direct generalization of (7) which will be used below is the following
J(u) =
Fi(u)D
−1Fi(u)
i (7)
where first term in (6) is changed on sum of some number of different solutions
of (3).
2 Necessary facts from [1]
In [1] was constructed equations of 3-th waves hierarchy of zero, first and second
order for six unknown functions f±1.0, f
0.1, f
1.1. The form of these equations will
be essentially used in what follows.
2.1 Equations of the zero order
1.0 = ±bf
0.1 = ±cf
1.1 = ±af
1.1 (8)
where (b, c) arbitrary numerical parameters a = b+ c.
2.2 Equations of the first order
1.1 = θ1.1(f
′ + σ1.1f
1.0 = ν1.0(f
′ + σ1.0f
0.1 = ν0.1(f
′ + σ0.1f
0.1 = ν0.1(f
′ + σ0.1f
1.1 (9)
1.0 = ν1.0(f
′ + σ1.0f
0.1(f
1.1 = θ1.1(f
′ + σ1.1f
where νi,j , σij , θ11 are numerical parameters connected by condition
ν01 − ν10 = 2σ10, ν01 + ν10 = 2θ11, −2σ11 = σ10 = σ01 (10)
Thus solution is defined by two independent parameters ν01, ν10 as in the case
of zero degree solution of the previous subsection.
2.3 Equations of the second order
1.1 = ν1.1(f
′′ + γ1.1f
1.0(f
′ + δ1.1f
0.1(f
′ + f+1.1R11,
where a ≡ (b+ c), Rij ≡ 2aijf
1.1 + bijf
1.0 + cijf
1.0 = ν1.0(f
′′ + γ1.0f
0.1(f
′ + δ1.0f
1.1(f
′ + f+1.0R10
0.1 = ν0.1(f
′′ + γ0.1f
1.0(f
′ + δ0.1f
1.1(f
′ + f+0.1R01,
− ˙f−0.1 = ν0.1(f
′′ + γ0.1f
1.0(f
′ + δ0.1f
1.1(f
′ + f−0.1R01,
− ˙f−1.0 = ν1.0(f
′′ + γ1.0f
0.1(f
′ + δ1.0f
1.1(f
′ + f−1.0R10,
− ˙f−1.1 = ν1.1(f
′′ + γ1.1f
1.0(f
′ + δ1.1f
0.1(f
′) + f−1.1R11.
All numerical parameters in (11) may be expressed in terms of only two ones
and are connected by relations
ν11 = a11, γ1.1 + δ1.1 = (b11 − c11), ν10 = 2b10
γ1.0 + δ1.0 = 2(c10 − b10), ν01 = 2c01, γ0.1 + δ0.1 = 2(c01 − b01)
a11 = −2c10 b11 = a10 c11 = −3c10 − b10
a10 = c10 + b10 b10 = b10 c10 = c10
a01 = −3c10 − b10 b01 = c10 c01 = −b10 − 4c10
 (12)
δ10 = 4c10, γ10 = −2(c10 + b10), 2γ11 = δ10 − γ10,
2δ11 = −γ10, δ01 = −δ10, γ01 = γ10 − δ10
2.4 Hamiltonian form of equations
As it was shown in (1) equations of the previous subsections may be considered
as Hamiltonian ones with following non zero Poisson breakets
{f+1.1, f
1.1} =
, {f+1.0, f
1.0} = 1, {f
0.1, f
0.1} = 1 (13)
This fact leads to existence of the first Poisson structure, inverse to which is the
following
0 0 0 0 0 −2
0 0 0 0 −1 0
0 0 0 −1 0 0
0 0 1 0 0 0
0 1 0 0 0 0
2 0 0 0 0 0
3 Hamiltonian operator of 3-th waves problem
Let us seek in connection the proposition (7) of introduction the second Poisson
structure in a form J2 = J
2 + J
2 , where J
2 contain terms with D
−1 and J
terms with non negative degree of D
2 = −F
−1(F 10 )
T − F 20D
−1(F 20 )
where F
0 are two different solutions of equations of the zero order (first sub-
section of the previous section).
0 0 0 − 1
0 0 f+1.1 0 D −
0 −f+1.1 0 D 0
1.0 0 D 0 −f
1.1 0
0.1 D 0 f
1.1 0 0
0.1 −
1.0 0 0 0
In the last expression we present finally result. Really it is necessary to write
anti symmetrical matrix with arbitrary coefficients, which will be found after
calculations described below.
Now let us consider how reccurent operator H (5) acts on some solution of
(3) F . At first F it is necessary multiply on J−11 with the result
1 F =
−2F−1.1
−F−1.0
−F−0.1
2F+1.1
and this column vector multiply on J2 = J
2 from (15). In two terms of J
it is necessary multiply vector line (F i0)
T on the last vector column with scalar
result
(F i0)
1 F = −2a
i(f+1.1F
1.1+f
1.1)−b
i(f+1.0F
1.0+f
1.0)−c
i(f+0.1F
0.1+f
Thus input of two first terms of Jn2 into ”new solution” will be
1.1 = −f
−1(2a2(f+1.1F
1.1+f
1.1)+(ab)(f
1.0+f
1.0)+(ac)(f
0.1+f
0.1))
1.0 = −f
−1(2(ba)(f+1.1F
1.1+f
1.1)+b
2(f+1.0F
1.0+f
1.0)+(bc)(f
0.1+f
0.1))
0.1 = −f
−1(2(ca)(f+1.1F
1.1+f
1.1)+(cb)(f
1.0+f
1.0)+c
2(f+0.1F
0.1+f
0.1))
and the same expressions with opposite sign for components with negative upper
indexes. a2 =
aiai, (ab) =
aibi and so on.
The result of multiplication J
1 F is determined by usual rules of multi-
plication matrix on vector
1 F =
(F+1.1)
′ + 1
(f+0.1F
1.0 − f
(F+1.0)
′ − f+1.1F
0.1 −
(F+0.1)
′ + f+1.1F
1.0 +
−(F−0.1)
′ − f−1.1F
1.0 −
−(F+1.0)
′ + f−1.1F
0.1 +
(F−1.1)
′ − 1
(f−0.1F
1.0 − f
Now let us take for F right hand side zero degree equations (subsection 1 from
previous section) F±1.1 = ±(ν1.0 + ν0.1)f
1.1, F
1.0 = ±ν1.0f
1.0, F
0.1 = ±ν0.1f
(b = ν1.0, c = ν0.1). In this case input from J
2 terms (16) equal to zero and
input from J
2 terms exactly coincides with right hand side of equations of the
first order (subsection 2 from previous section). Really calculations must be
done in a back direction: in definition of J
2 (15) it is necessary to use arbitrary
skew symmetrical matrix and after comparison result of calculations above with
first order equations obtain finally form J
2 (15).
Now we repeat the same trick with equations of the first degree. In this case
1 F =
−(ν1.0 + ν0.1)(f
′ − ν1.0−ν0.1
−ν1.0(f
′ + ν1.0−ν0.1
−ν0.1(f
′ + ν1.0−ν0.1
ν0.1(f
′ − ν1.0−ν0.1
ν1.0(f
′ − ν1.0−ν0.1
(ν1.0 + ν0.1)(f
′ + ν1.0−ν0.1
We present result of action J
1 F below
ν1.0 + ν0.1
(f+1.1)
3ν1.0 − ν0.1
(f+1.0)
0.1 +
ν1.0 − 3ν0.1
1.0)(f
ν1.0 − ν0.1
1.1(f
1.0 − f
0.1) (18)
The terms in the first line exactly coincide with terms with derivatives in equa-
tion of the second order for f+1.1 component. Indeed (ν1.0+ν0.1) = 2b10+2c01 =
−8c10 = 4a11 = 4ν11, b10 =
, c10 = −
ν1.0+ν0.1
, δ11 = b10 + c10 =
3ν1.0−ν0.1
and so on. The same situation takes place with respect all other components:
terms with derivatives coincide with calculated from J
1 F .
Terms without derivatives arise from terms of the second line of (18) and
from (16). In the last equations after substitution F from equation of the first
order (and in all other cases) under the sign D−1 arises sign D which lead to
unity and (16) plus terms of second line of (18) look as
1.1(2θ11a
1.1+[(ab)ν10+
ν1.0 − ν0.1
]f+1.0f
1.0+[(ac)ν01−
ν1.0 − ν0.1
]f+0.1f
1.0([2(ba)θ11+
ν1.0 − ν0.1
1.1+b
2ν10f
1.0+[(bc)ν01−
ν1.0 − ν0.1
]f+0.1f
0.1([2(ca)θ11−
ν1.0 − ν0.1
]f+1.1f
1.1+[(cb)ν10+
ν1.0 − ν0.1
]f+1.0f
1.0+c
All terms above with scalar products arise from (16). All others from the ”second
line” of (18). Now it is necessary compleat these expressions with terms f+ijRij of
the right hand side of equation of the second order (subsection 3 of the previous
section). This comparison leads to the following conclusion:
a2 = b2 = c2 =
, (ab) = (ac) = −(bc) =
And now recurrent operator H is defined uniquely. From this result it is clear
that all attempts to construct H using only one solution in the anzats for J2
lead to contradiction and it was the main difficult to the author.
The following observation take place. Let us consider three elements of
Cartan subalgebra R3 =
h1+h2
, R2 = h2, R1 = h1 and the positive root system
of A2 algebra system X
3 = X
12, X
2 , X
1 . Let us define 3× 3 ”Cartan matrix”
KW by the condition
[Ri, X
j ] = K
a2 (ab) (ac)
(ba) b2 (bc)
(ca) (cb) c2
4 Hamiltonian formalism II
In calculations of the previous section results of [1] were used in the whole
measure. But the corresponding calculations were not simple and not strait
forward. Now having the explicit expression for J2 we are able to check that
equation it defined (4) is satisfied. Let us rewrite it notations of the previous
section
−T ′(f)F 10D
−1(F 10 )
T (T ′(f))T − T ′(f)F 20D
−1(F 20 )
T (T ′(f))T+
T ′(f)J
′(f))T = (Jn2 + J
2 )(T (f)) (20)
( we think that the same sign T for substitution and transposition will not lead
to mixing). We will do all calculations below with respect to T3 transformation
(explicit formulae for it and F ′3(f) reader can find in Appendix). We remind the
rule of multiplication of quadratical in derivatives operator on scalar function
(A+DB+DC2)R = AR+BRA′+CRT ′′+(BRA+2CRT ′)D+CDR2 (21)
Matrix elements of 3 first lines of F ′3(f) do not contain operator of differen-
tiation. Its fourth and fifth lines linear in D and its sixth one quadratical in D.
By definition (3) all terms without operator D lead to
F i0 . All others can be
simple calculated using (21) with the result
T ′3(f)F
F i0 +
1.0D − a
F i0 +
Γi (22)
where ∆i = ci − bi, ai = ci + bi, 03- three dimensional zero vector After substi-
tution (22) into (20) and cancelation equivalent terms in both sides we come to
the following equality have to be checked
Γi D−1
Γi D−1
Γi T )+
T ′3(f)J
3(f))
T = J
2 (T3(f)) (23)
The matrix of the first sum in the first line has different from zero only elements
of its last three columns. The second sum - only elements of its three last line.
And the last sum only elements of 3× 3 matrix in its left down corner.
At first let us calculate the first sum. Result is the following (we present
below only two first lines of this matrix operator)
F i0 D
03 0 0
D + 1
0.1D −
 (24)
(where 03 is three dimensional row vector)
(T ′3(f)J
3(f))
T )1l =
(T ′3(f))1m(J
2 )mk((T
3(f))
T )kl =
(T ′3(f))16(J
2 )6k(T
t)lk =
(because only one elements of the first line ((T ′3(f))16) of Frechet derivative
matrix and three elements of sixth line matrix J
2 are different from zero).
(f−1.1)
D(T ′3(f))
0.1(T
3(f))
1.0(T
3(f))
where (kD)t ≡ −Dkt. or the first line of the matrix T ′3(f)J
3(f))
T looks as
0.1 −
D − 1
Now let us calculate second line
(T ′3(f)J
3(f))
T )2l =
(T ′3(f))2m(J
2 )mk((T
3(f))
T )kl =
(T ′3(f))26(J
2 )6k(T
t)lk +
(T ′3(f))24(J
2 )4k(T
t)lk =
(because only two elements (T ′3(f))24, (T
3(f))26 of second line of Frechet deriva-
tive matrix are different from zero )
(f−1.1)
D(T ′3)
0.1(T
1.0(T
1.0(T
l1+D(T
1.1(T
Result of simple algebraical calculation lead to explicit form of second row
1.1 −
D + 3
0.1D +
1.0 +
[(f−0.1)
After summation (25) and (25) with corresponding lines of (24) we obtain
exactly first two lines of left hand side of (23).
By similar calculation it is possible to check that (23) is satisfied. Finally
expression for Jn2
27) and (15) solve the problem of the construction of the second Poisson struc-
ture for the equations of 3-th waves hierarchy.
5 Comments about multi-soliton solutions of the
systems of 3-th waves hierarchy
Let us find solution of the system equations of the second order (subsection 3 of
section 2) under additional condition f+i.j = 0. The system under consideration
looks as
− ˙f−0.1 = ν0.1(f
′′, − ˙f−1.0 = ν1.0(f
− ˙f−1.1 = ν1.1(f
′′ + γ1.1f
1.0(f
′ + δ1.1f
0.1(f
where ν11 =
ν10+ν01
, δ1.1 =
3ν10−ν01
, γ1.1 =
ν10−3ν01
Solution of two first linear equation are obvious
0.1 =
dµe−ν01tµ
2+µxq(µ), f−1.0 =
dλe−ν10tλ
2+λxp(λ)
Let us find partial solution on nonhomogineous third equation in a form
1.1 =
−(ν01µ
2+ν10λ
2)t+(µ+λ)x
r(µ, λ)
After substitution this anzats for f−1.1 and obtained above solutions for f
1.0, f
into third equation we come to an equality
[ν01µ
2 + ν10λ
ν10 + ν01
(µ+ λ)2]r(µ, λ) = (γ1.1µ+ δ1.1λ)q(µ)p(λ)
Quadratical multiplier of the left side is equal to 2(λ − µ)(γ1.1µ + δ1.1λ) and
finally we obtain
r(µ, λ) =
q(µ)p(λ)
1.1 =
dλe−(ν01µ
2+ν10λ
2)t+(µ+λ)x 1
q(µ)p(λ)
The last expression coincides (up to nonessential multiplier 1
) with solution in
the case of 3-th problem. But in resolving of equations of discrete transfor-
mation only differentiation with respect to space coordinate take place. Thus
all equations of discrete transformation in the case of 3-th wave problem and
in the case under consideration will be have the same solution except of time
dependent multiplier. And the form of multi soliton solution (up to this factor)
will be the same for all systems of 3-th waves hierarchy.
6 Outlook
The main result of the present paper the explicit expression for second Pois-
son structure (15) and (27), which allow to construct Hamiltonian reccurent
operator and obtain all equations of 3-th wave hierarchy in explicit form. From
the physical point of view these systems may be considered as three interacting
fields of nonlinear Schredinger hierarchy connected with A1 algebra. All equa-
tions depended on one arbitrary numerical parameter, which can be connected
with parameters of the particles of the fields describing by nonlinear Schredinger
hierarchy.
The discrete transformation for n-wave problem in the case of arbitrary
semisimple algebra was presented in [4] form and author have no doubts that
the problem of equations of n-wave hierarchy and its multi-soliton solution may
be resolved in explicit form.
The most riddle to the author remain question about the nature of the group
of discrete transformation. As it follows from its introduction [4] it has some
connection with the Weil group of the root space of semisimple algebra. Weil
group is discrete one but non commutative. Discrete transformation in the case
of the present paper is some reduction from group of discrete transformation
of four dimensional self-dual Yang-Mills equations [9]. And thus understanding
this situation in the case n-waves interaction will give possible guess to solution
of this problem in Yang-Mills case.
Aknowledgements
The author thanks CONACYT for financial support.
7 Appendix
We present here different from zero matrix elements of Frechet derivative for T3
discrete transformation - 6× 6 matrix operator (see Introduction). All calcula-
tions are done in connection with its definition (2) and explicit formulae for T3
discrete transformation presented below.
7.1 Discrete transformation T3
1.0= −
0.1= −2(f
′ − f−1.1f
1.0 −
(f−0.1)
(f−1.1)
1.0= −2(f
′ + f−1.1f
0.1 +
(f−1.0)
(f−1.1)
1.1= f
1.1+(ln f
(f−0.1)
0.1 − (f
(f+1.0f
1.0+f
0.1)+
7.2 Frechet derivative
We present below different from zero matrix elements of Frechet operator
F ′16= −
(f−1.1)
F ′24= −
F ′26=
(f−1.1)
F ′35=
F ′36= −
(f−1.1)
F ′42= −f
F ′44= −2D− Z +
(f−1.1)
F ′45= −
(f−0.1)
2f−1.1
F ′46= −f
1.0 +
2f−1.1
0.1(f
(f−1.1)
53= f
(f−1.0)
2f−1.1
55= −2D+ Z +
(f−1.1)
F ′56= f
0.1 −
2f−1.1
1.0(f
(f−1.1)
F ′61= (f
F ′62=
F ′63=
F ′64=
(f−1.0D−(f
1.0Z+
F ′65= −
(f−0.1D − (f
0.1Z +
F ′66= D
2 − 2
(f−1.1)
(f−1.1)
′(f−1.1)
+ 2f−1.1f
1.1 +
(f−1.0f
1.0 + f
0.1)−
where Z =
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