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ISSN 2074-1871 Уфимский математический журнал. Том 8. № 2 (2016). С. 114-120.
RECURSION OPERATOR FOR A SYSTEM WITH NON-RATIONAL LAX REPRESENTATION
K. ZHELTUKHIN
Abstract. We consider a hydrodynamic type system, waterbag model, that admits a dispersionless Lax representation with a logarithmic Lax function. Using the Lax representation, we construct a recursion operator of the system. We note that the constructed recursion operator is not compatible with the natural Hamiltonian representation of the system.
Keywords: recursion operator, hydrodynamic type systems, non-rational Lax representation.
Mathematics Subject Classification: 17B80, 37K10, 37K30, 70H06
1. Introduction
In the present paper we consider the so-called waterbag model [1],[2]. This hydrodynamic type system admits a dispersionless Lax representation with a logarithmic Lax function. Such systems have important applications in the topological field theories, see [3], [4] and the references therein. For a better understanding of such systems one needs to know a bi-Hamiltonian structure of a system and the corresponding recursion operator, see [5]-[7]. For the systems admitting dispersionless Lax representation the construction of bi-Hamiltonian structures and recursion operators is well understood in the case of a polynomial or rational Lax function [8]-[12]. The non-rational Lax functions present a much more difficult case. In the present paper we construct a recursion operator for the case of logarithmic Lax function. To our knowledge, in the literature, there are no other examples of recursion operators corresponding to a non-rational Lax function.
Let us give needed definitions. We introduce the algebra of Laurent series
( ~ }
A = < ^^ Uip1 : Ui are smooth functions decaying fast at infinity > , (1.1)
with the Poisson bracket given by
dp dx dx dp' (12)
Taking the Lax function
L = p - mIn(p - c1) + In(p - c2)+-----+ In(p - cm+1) (1.3)
К. ЖЕЛТУХИН, РЕКУРСИВНЫЙ ОПЕРАТОР ДЛЯ СИСТЕМ С ИРРАЦИОНАЛЬНЫМ ПРЕДСТАВЛЕНИЕМ ЛАКСА.
© К. ZHELTUKHIN. Поступила 08 декабря 2015 г.
and using the Gel'fand-Dikii construction [13], we can write the hierarchy of integrable equations
Ltn = {(Ln )>0,L} n = 1,2,.... (1.4)
The second equation of the hierarchy
Lt = {(L%a,L} (1.5)
leads to the waterbag model
ci = dx
j )2
(¿I 2
+ mc1 — c2 — • • • — ¿
,m+1
:í.6)
where j -operator
1, 2,..., (m +1). As we show, the above hierarchy admits the following recursion
n = Ad:
1
:1.7)
where the matrix A = (yij) has the entries
7ii
m+1 i i
cX — cJx
j=2
C1 - C>
Ikk = Cx — m—
oX — cX
+
c1 — ck
hk
m+1
z
_
c1 — ck
CX — cX ck — cj '
Ik1
Iki =
m
_ çk
c1 — ck
CX — c%x ck — ci
j=2,j=k
k = i, and k,i = 2, 3,... ,m +1.
We observe that the above system has an obvious Hamiltonian representation with the Hamil-tonian operator V = Jdx, where J is the matrix having one on the incidental diagonal and its other entries are zero.
In general, if a system has a bi-Hamiltonian representation with respect to a pair of Hamiltonian operators Vi and "Z^, one can construct a recursion operator TZ = V2V-1. Hence, one has V2 = TVi. For systems admitting dispersionless Lax representation one can generate the whole hierarchy of Hamiltonian operators T>n = Tin'Di [14]. It turns out that in our case, if we apply the recursion operator T to the Hamiltonian operator V, the resulting operator is not Hamiltonian. Thus, the recursion operator T and the Hamiltonian operator V are not compatible. For further studies, it is an interesting open question to find a bi-Hamiltonian representation of system (1.6).
The paper is organized as follows. In Section 2 we give a construction of the recursion operator of system (1.6) for general m. In Section 3 we give examples of system (1.6) and the corresponding recursion operator for m = 1, 2, 3.
2. Evaluation of recursion operator Let us introduce new variables
c1 = u and vj 1 = c1 — c>, In terms of the new variables, system (1.6) becomes
j = 2, 3,... (m + 1)
ut = uux + vX + ...v. v\ = v1ux + (u — Vl)vX
m\„m
X
vm = vmUX + (u — Vm)V, System (2.2) admits a Lax representation
Lt = {(L%1,L}
(2.1)
(2.2)
(2.3)
1
c
X
with Lax function
L = p + u + + — J + — J + ••• + —
V p J V P J V P /
Thus, we have the whole hierarchy of the symmetries for the system (2.2) given by
Ltn = {(Ln)^i ,L} n =1, 2,'.'
Let us construct a recursion operator for the above hierarchy of the symmetries. We the recursion operator by direct analysis of the Lax representation. Let
Ln = anpn + an-ipn 1 + ... aip + + a-ip 1 + .
The next two lemmata provide some relations between coefficients of Ln and
Ltn = utn +
p + v1
+ ■ ■ ■ +
p + Vr
(2.4)
(2.5) construct
(2.6)
Lemme 2.1. For each k = 2, 3 .. .m and each n = 2, 3,... the identity
i=1
(2.7)
holds true.
Proof. Using (2.6) we can write the equation (2.5) as
utn +
p + V1
+ ■■■ +
p + vr
(nanpn +----+ 2a2p + 01) ux +
p + v1
+ ■■■ +
p + vr
- (an,xPn +-----+ a,2,xP2 + a,1,x) I 1 -
)
p(p + v1) p(p + vm)
Multiplying the above equation by (p + v1)(p + v2)... (p + vm) and then substituting p = -vk,
we obtain
That is,
^(-l)*-1 iai(vk y-1vk + J2(-iy-1ahxXvk y.
i=1
(n
i=1
i=1
(-iy-1at(vk )
x.
□
Lemme 2.2. For each n = 2, 3,..., the identity a0 = dx 1 utn holds true.
Proof. Lax equation (2.5) can be written as
Ltn = -{(Ln)^o,L} n =1, 2,...
Using (2.6) and collecting coefficients of zero power of p in the above equations we have utn ao,x.
□
The above lemmata allow us to express the coefficients of (L^+^p and (L>n+l1)x in terms of
coefficients of L>0 and Ltn.
1
r
v
V
t
t
n
n
1
1
r
r
V
V
V
V
t
t
x
x
It
It
1
V
V
k
V
t
n
k
V
t
n
Lemme 2.3. Let
Then
where r = 1, 2,.. .n. Let
n + 1
(4rai+1)) = bnpn-1 + ••• + b2p + bl
m r—1 m
br = ar—i + £ k)-ar—3 + k) —d—1 k=1 j=0
k=1
n + 1
(i-V^) = dnPn + ••• + d2p2 + d1p.
Then
m 1
dr = uxar + ^^ vk) — 1vkxar—j
+ ^(vk )—r—1vk dxx 1vk
k=1 j=0
k=1
where r = 1, 2,.. .n. Proof. We have
That is
n + 1
n + 1
=((
Lt+1)) =[(0nPn + ••• + ao) Ux +
(m
Ux + Y] k=1
P + Vk
-0
1
(2.8)
(2.9)
(2.10)
(2.11)
For each k = 1,.. .m, we expand -r as series in terms of p 1 at p = and multiply with
p + vk
(anpn + • • • + a0). Collecting coefficients at pk, k = 1,... m, in the above identity and using Lemma 2.1, we obtain formula (2.9). The formula (2.11) can be obtained in the same way. □
Using the above lemmata, we find a recursion operator for the hierarchy (2.5).
Lemme 2.4. The recursion operator for system (2.2) can be written as T = Cdx1, where C is an (m +1) x (m + 1) matrix. It is convenient to write matrix C as a sum of two matrices, C = (A + B). Matrix A = (a^) has the entries
an = Ux;
a1(j+1) = vx (v3 )—1
a(j+1)1 = <, au+1)u+1) = (ux- vx) a{i+1){j+1) = 0,
3 = 1, 2,... ,m; j = 1, 2,... ,m; j = 1, 2,... ,m; i = h hj = 1,2
Matrix B = (fcj) has the entries
, m;
ßn = 0; ßHj+1) = 0, ßu+1)1 = 0, j = 1, 2,...,-,
^1) = t 4 - VZ]f, ' = !• 2
k=1,k=j
, m;
Vlx - vlvx (VJ) 1
p(i+1)(j+1) =
V3 —
i = j, i,j = 1, 2,...,m.
1
1
1
k
1
V
x
Proof. Using notations of Lemma 2.3 the Lax equation (2.5) can be written as
m vk
Utn
+ E -7+T =(n + 1)(bnPn-1 + ••• + hp + b1)[ux + V
m u
Vl
k=1
k=1
p + Vk
(2.12)
(n+1)( dnPn + •••+d2p2+di)[i -J2
m u
vK
k=1
p(p + Vk)
We multiply the above equation by (p + vl)(p + v2)... (p + vm) and substitute the expressions for h,di, i — 1, 2,...n, given in Lemma 2.3. Equating coefficients at pk, k — 1, 2,...m, we obtain a system of equations linear with respect to vkn+l, k = 1, 2,.. .m. Solving the system,
□
we obtain the recursion operator given above.
Remark 2.1. Let us a define vector V = (u, v1, v2,..., vm) and write system (2.2) as
V = K (V,VX). (2.13)
By straightforward calculations we check that the constructed above operator satisfies the criteria for recursion operators
K = DK K-K DK, (2.14)
where D^ is the Frechet derivative of K.
Returning back to the original variables c1,... cm+1, we obtain recursion operator (1.7).
3. Examples
some examples. We give examples in variables c1, c2,..., cm+1. Let us consider equation (1.6) with m = 1. The equation becomes d = c V + cL - c2
Let us consider Example 3.1.
2 2 + 1 _ 2
The above system admits the recursion operator
(
1 2
1 | °X °X
X + C1 _ C2
4 _ c2
Cx _4\
1 _C2
Cx _ Cx
1 _C2J
d.
Example 3.2. Let us consider equation (1.6) with m =2. The equation becomes
+ — C Cm + 2 CM C^ C,
c Crg \ 2 c^ c^. c,
C Cm \ 2 CM CM C,
The above system admits the recursion operator
c1 + —
X ' 1
Cx
c1 - c2
2 X x
c1 _ c2 r1 _ r3
2 x ^x
c1 — c3
q2 _ £>3
X \ X X
2 3
4 _ 4
c1 — c2
r1 - C'2 c2 — 2 ^-^
x -,1 _ n2
\
1 2 _ c2
c3 — c2
2
°x °x 2
_
c1 _ c3
2 3 °x °x
2 3
C1 — C
_ 1-3
1 3
\
3 2
d
1
2
t
2
2
t
3
t
3
2
C' — c~
3
x
Example 3.3. Let us consider equation (1.6) with m = 3. The equation becomes
c! = C1C1x + 3Cx — Cx — Cx — Cx
c2 _ (2 + 3Cx — Cx — Cx — Cx
<3 = ¿(i + 3Cx — Cx — Cx — Cx
c4 44 = CCx + 3cX — Cx — Cx — Cx
The above system admits the recursion operator
(
c1 -3 %-c1 —
c1 -3 %-c1 —
3 °x
c1 -
cX — cX
c1 — c2
r1 - C'2
2 n_x 4
Cx 3
f3 — r2 c3 — c2 cX — cX à- c2
c1 — c2
oX — cX c1 — c3
^X — cX
c2 — c3
r1 — r3
„3 _ 3 x 4 Cx 3 rJ_ c3
cX — cX
c4- CX
cX — cX ' CX — cX
cX — cX
c2 — cX
CX — CX
c3 — c4
Q4 - C1
_ 3_X X
d:
-1
— c})
+
£
3=2
c1 - CP
E
3=2,3=2 0
cx 4
c2 — ci
E
=2, =3
0
^X — CXy.
à — ci
E
=2, =4
cX — cX
c4 — cJ
d:
1
1
c
X
4
c
X
c
1
OX — ty
0
0
0
0
0
0
0
0
0
0
The author would like to thank Professor Maxim Pavlov for pointing out the question of constructing a recursion operator for the waterbag model and Professor Metin Gurses for fruitful discussions.
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Kostyantyn Zheltukhin
Department of Mathematics, Faculty of Science, Middle East Technical University, 06800 Ankara, Turkey E-mail: [email protected]
