A Classification of Integrable Hydrodynamic Type Chains Using the Haantjes Tensor

8/8/2019 A Classification of Integrable Hydrodynamic Type Chains Using the Haantjes Tensor

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Classification of integrable hydrodynamic

chains using the Haantjes tensor.

D.G. Marshall.

Department of Mathematical Sciences

Loughborough University

Loughborough, Leicestershire LE11 3TU

United Kingdom

1


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Keywords 2

Keywords

Integrability

Nijenhuis tensor

Haantjes tensor

Hydrodynamic chains

Benney Chain

Diagonalizability

Generating function of conservations laws


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Abstract 3

Abstract

The integrability of an m-component system of hydrodynamic type, ut = v(u)ux,

by the generalized hodograph method requires the diagonalizability of the m mmatrix v(u). The diagonalizability is known to be equivalent to the vanishing of

the corresponding Haantjes tensor. This idea is applied to hydrodynamic chains

infinite-component systems of hydrodynamic type for which the matrix v(u)is sufficiently sparse. For such sparse systems the Haantjes tensor is well-defined,

and the calculation of its components involves only a finite number of summations.

The calculation of the Haantjes tensor is done by using Mathematica to performsymbolic calculations. Certain conservative and Hamiltonian hydrodynamic chains

are classified by setting Haantjes tensor equal to zero and solving the resulting sys-

tem of equations. It is shown that the vanishing of the Haantjes tensor is a necessary

condition for a hydrodynamic chain to possess an infinity of semi-Hamiltonian hy-

drodynamic reductions, thus providing an easy-to-verify necessary condition for the

integrability of such sysyems. In the cases of the Hamiltonian hydrodynamic chains

we were able to first construct one extra conservation law and later a generating

function for conservation laws, thus establishing the integrability.


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Acknowledgements 4

Acknowledgements

I would like to express my gratitude to the following people for their assistance

during my work on this thesis:

Prof. E. V. Ferapontov and Dr K. R. Khusnutdinova, who gave me such superb and

enthusiastic supervision along with unwavering support and encouragement. Thank

you for all the time you have given me and all the patience you have shown me

throughout this work.

All the of staff of the Mathematical Sciences Department at Loughborough Univer-sity for all their help.

Loughborough University, particularly the Mathematical Sciences Department, for

financial support.

All my friends for their moral support and understanding.


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Contents 5

Contents

1 Introduction 7

2 Hydrodynamic type systems in 1+1 dimensions 15

2.1 Riemann invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.2 Commutativity Condition . . . . . . . . . . . . . . . . . . . . . . . . 22

2.3 Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.4 Semi-Hamiltonian property . . . . . . . . . . . . . . . . . . . . . . . . 26

2.5 Generalized hodograph method . . . . . . . . . . . . . . . . . . . . . 28

2.6 The Haantjes tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.7 Reciprocal Transformations . . . . . . . . . . . . . . . . . . . . . . . 34

3 Hydrodynamic Chains 38

3.1 The Benney chain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.2 Hydrodynamic chains and the Haantjes tensor . . . . . . . . . . . . . 43

3.3 Another chain considered by Benney . . . . . . . . . . . . . . . . . . 53

3.4 Hydrodynamic reductions and diagonalizability . . . . . . . . . . . . 54

4 Conservative chains 65

4.1 Egorov case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

4.2 Method of hydrodynamic reductions . . . . . . . . . . . . . . . . . . 67

4.3 Integration of the Egorov case . . . . . . . . . . . . . . . . . . . . . . 72

4.4 More general conservative chains . . . . . . . . . . . . . . . . . . . . 76

4.5 Integration of the general conservative chains . . . . . . . . . . . . . . 79

5 Kupershmidts brackets 89

5.1 Integration of the system of integrability conditions . . . . . . . . . . 91

5.2 The vanishing of the Haantjes tensor . . . . . . . . . . . . . . . . . . 93


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Contents 6

5.3 Conservation laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

5.4 Generating functions for conservation laws of Hamiltonian chains cor-

responding to Kupershmidts bracket . . . . . . . . . . . . . . . . . . 98

5.4.1 Generating functions in the general case . . . . . . . . . . . . 100

5.4.2 Generating functions in the case = 0 . . . . . . . . . . . . . 100

5.4.3 Generating functions in the case + 2 = 0 . . . . . . . . . . 101

6 Manin-Kupershmidts bracket 103

6.1 Integration of the system of integrability conditions . . . . . . . . . . 106

6.2 Conservation laws for Manin-Kupershmidts bracket . . . . . . . . . 1176.3 Generating functions for conservation laws of Hamiltonian chains with

Manin-Kupershmidts bracket . . . . . . . . . . . . . . . . . . . . . . 119

7 Conclusions and further work 125


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Introduction 7

1 Introduction

This thesis is a study in the area of integrable systems. It is devoted to a classification

of integrable hydrodynamic chains of the type that orginally appeared in the context

of fluid mechanics, in particular with reference to the Benney chain [3]. Integrable

systems have been actively investigated since the discovery of the Inverse Scattering

Transform of the KdV equation [21]. There have been significant developments for

1 + 1 dispersive equations, see [2], and 1 + 1 dimensional dispersionless equations

[9]. Classification of integrable systems has been an important theme of much of

the recent research.In this thesis we are concerned with the integrability of special infinite component

dispersionless systems known as hydrodynamic chains. Our approach towards clas-

sification of such systems is based on the use of the so-called Haantjes tensor [27].

Let us begin by defining the form of finite component hydrodynamic type systems

ut = v(u)ux (1)

where u = (u1, u2,...,un)t is an n-component column vector and v(u) is an n nmatrix. In what follows we consider the strictly hyperbolic case when the eigenval-

ues of the matrix v(u), also called the characteristic speeds of the system (1), are

real and distinct.

Some systems of the form (1) are diagonalizable, i.e. reducible to the Riemann

invariant form, more details follow in Sect. 2. The diagonalizability of (1) is a

necessary condition for integrability via the generalized hodograph method [61].

There exists an efficient tensor criterion of the diagonalizability which does not

require the computation of eigenvalues and eigenvectors of the matrix v(u). This is


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Introduction 8

useful because calculating the eigenvalues and eigenvectors can be troublesome in

practice. Let us first calculate the Nijenhuis tensor [27] of the matrix v(u) = vij,

Nijk = vpj upv

ik vpkupvij vip(ujvpk ukvpj ), (2)

and introduce the Haantjes tensor [27]

Hijk = Ni

prvpj v

rk Npjr vipvrk Nprkvipvrj + Npjk virvrp. (3)

For strictly hyperbolic systems, i.e. when the eigenvalues are positive and distinct,

the condition of diagonalizability is given by the following theorem:

Theorem 1 [27] A hydrodynamic type system with mutually distinct characteris-

tic speeds is diagonalizable if and only if the corresponding Haantjes tensor (3) is

identically zero.

The focus of this work is hydrodynamic chains, these are infinite component systems

of form (1). More formally a hydrodynamic chain is,

ut = V(u)ux

where V(u) is a square matrix of infinite dimension, while u = (u1, u2, . . . , )t is

an infinite-component vector. If the infinite matrix V(u) is sufficiently sparse then

both tensors (2) and (3) make sense. The formal definition of sufficiently sparse is

Definition 1 [20] An infinite matrix V(u) is said to belong to the class C (chain

class) if it satisfies the following two properties:

(a) each row of V(u) contains finitely many nonzero elements;

(b) each matrix element of V(u) depends on finitely many variables ui.


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Introduction 9

All the chains considered throughout the paper are of the class C. Below is the

Benney chain written in matrix form, the matrix is of class C, it has only two

nonzero elements in every row apart from the first where there is only one. The sole

element in the first row is constant while of the two elements on every other row

one is constant and the other is only a function of one variable. The Benney chain

has the form

A0

A1

A2

A3

t

+

0 1 0 0 0 0 .

A0

0 1 0 0 0 .2A1 0 0 1 0 0 .

3A2 0 0 0 1 0 .

. . . . . . .

A0

A1

A2

A3

x

= 0.

The Benney chain is a classical example of a hydrodynamic chain. This chain was

derived in [3] and was shown to consist of an infinite number of conservation laws

for the system of equations that describe fluid under the action of gravity. The

derivation of the Benney chain is shown in Sect. 3.1.

In Sect. 3 we propose that a hydrodynamic chain is said to be diagonalizable if all

components of the corresponding Haantjes tensor (3) are zero. The work in Sect.

3 goes on to show that the vanishing of the Haantjes tensor is a necessary (and in

some cases sufficient) condition for a hydrodynamic chain to possess an infinity

of finite-component diagonalizable hydrodynamic reductions. The main advantage

of our approach is that the classification does not require any extra objects such

as commuting flows, Hamiltonian structures, Lax pairs, etc. The vanishing of the

Haantjes tensor turns out to be an easy to calculate classification criterion. As an


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Introduction 10

example let us consider the chain

unt = un+1x + u

1unx + cnunu1x

where cn = const. It can be verified that the vanishing of the Haantjes tensor implies

the relation cn+2 = 2cn+1 cn. Since the form of this chain is relatively simple itis straightforward to calculate the Haanjtes tensor by hand though it is quicker to

using symbolic computations. Throughout this thesis the Mathematica[43] package

has been used, the three files contained in the Appendices can be downloaded from

http://www-staff.lboro.ac.uk/%7emakk/Marshall.html. The PDF file of this theis

is also available online at the fore mentioned address.

Definition 2 A hydrodynamic chain from the class C is said to be integrable if,

for any m, it possesses infinitely many m-component semi-Hamiltonian reductions

parametrised by m arbitrary functions of a single variable.

The method of hydrodynamic reductions is illustrated in Sect. 3.4 by considering

the Benney chain.

In Sect. 3.4 the following theorem is proved:

Theorem 2 The vanishing of the Haantjes tensor H is a necessary condition for

the integrability of hydrodynamic chains from the class C.

If the spectrum of the infinite matrix V is simple, that is, for any there exists a

unique eigenvector such that V = , the following stronger result is obtained:

Theorem 3 In the simple spectrum case the vanishing of the Haantjes tensor H is

necessary and sufficient for the existence of two-component reductions parametrized

by two arbitrary functions of a single variable.


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Introduction 11

Theorem 3 provides an easy-to-verify necessary condition for testing the integrabil-

ity of hydrodynamic chains. For finite component systems of hydrodynamic type, it

is known that any diagonalizable semi-Hamiltonian (38) system possesses infinitely

many conservation laws and commuting flows of hydrodynamic type, and can be

solved by the generalized hodograph method [61]. At this point it must be empha-

sized that the vanishing of the Haantjes tensor is not sufficient for the integrability in

general: one can construct examples of diagonalizable chains which possess infinitely

many diagonal reductions none of which are semi-Hamiltonian (38) (see the example

in Sect. 3.4). To eliminate these cases let us recall that for finite-component systems

(1) there exists a tensor object which is responsible for the semi-Hamiltonian prop-

erty [48]. This is a (1, 3)-tensor Pskij (see Appendix 1 for explicit formulas in terms

of the matrix vij). Similarly to the Haantjes tensor H, the tensor P is well-defined

for hydrodynamic chains that are of class C. Thus the following conjecture is made

Conjecture 1 The vanishing of both tensors H and P is necessary and sufficient

for the integrability of hydrodynamic chains from the class C.

The necessary part of this conjecture, the claim that the integrability implies that

both H and P vanish identically, is proved in Sect. 3.4. The sufficiency is a much

more difficult property to show, and this is not yet established in general. We point

out that the vanishing of H (in fact, the vanishing of the first few components of

H), is already sufficiently restrictive and implies the integrability in many cases (e.g.

for conservative chains, Hamiltonian chains, etc).

In Sect. 4 the Haantjes tensor criterion is used to classify conservative chains.

Initially we restrict the flux of the first equation to be only u2, this is known as the

Egorov case. The chain is

u1t = u2x, u

2t = g(u

1, u2, u3)x, u3t = h(u

1, u2, u3, u4)x, ... . (4)


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Introduction 12

The chain (4) was investigated in [47], where a class of new integrable examples

was found, see also [6], based on the symmetry approach. These papers provide a

classification of conservative chains of the form (4). It turns out that the condi-

tions H1jk = 0 are already sufficiently restrictive and give an over-determined system

expressing all second order partial derivatives of h(u1, u2, u3, u4) in terms of g, see

(103). The consistency conditions of these equations lead to a system of equations

expressing all third order partial derivatives of g in terms of its lower order deriva-

tives, see (104). Computer algebra was used to calculate the Haantjes tensor and to

verify the involutivity by calculating the consistency conditions. It must be empha-

sized that the same system of equations for g was found in [47] using the symmetry

approach, as well as in [17] based on the method of hydrodynamic reductions. So,

for conservative chains (4) the condition of diagonalizability is equivalent to the

integrability. The requirement of the vanishing of other components Hijk , i 2,imposes no additional constraints on h and g: these conditions reconstruct the re-

maining equations of the chain (4). Furthermore, the conditions Hmjk = 0 specify

the right hand side of the mth

equation umt = ..., etc.

Also contained in Sect. 4 is the classification of conservative chains where the flux

of the first equation is f(u1, u2), i.e. chains of the type

u1t = f(u1, u2)x, u

2t = g(u

1, u2, u3)x, u3t = h(u

1, u2, u3, u4)x, ... . (5)

As in the Egorov case, the conditions H1jk = 0 lead to expressions for all second

order partial derivatives of h in terms of g and f. The consistency conditions of

these equations result in a system of equations expressing all third order partial

derivatives of g and f in terms of lower order derivatives.

Sect. 5 and Sect. 6 are devoted to the classification of Hamiltonian chains which


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Introduction 13

are of the form

ut =

B

d

dx+

d

dxBt

h

u, (6)

where u = (u1, u2, u3,...)t is an infinite-component column vector of the dependent

variables, and B ddx +d

dx Bt is known as the Hamiltonian operator. Chains of the form

(6) are known as Hamiltonian chains, h is called the Hamiltonian density. We are

going to look at the problem of the classification of the densities for two particular

Hamiltonian operators:

Kupershmidts bracket: Bij = ((i 1) + )ui+j1, h = h(u1, u2), see [34];

Manin-Kupershmidts bracket: Bij = (i 1)ui+j2, h = h(u1, u2, u3), see [33];

so that the corresponding Hamiltonian system is integrable. The Kupershmidts

bracket will be dealt with in Sect. 5, while the work on the Manin-Kupershmidts

bracket will be in Sect. 6.

According to the results of Tsarev [61], the vanishing of the Haantjes tensor is nec-

essary and sufficient for the integrability of finite-component Hamiltonian systems

of hydrodynamic type by the generalized hodograph method. Thus, we formulate

our main conjecture regarding Hamiltonian hydrodynamic chains:

Conjecture 2 [20] The vanishing of the Haantjes tensor is a necessary and suffi-

cient condition for the integrability of Hamiltonian hydrodynamic chains. In par-

ticular, it implies the existence of infinitely many Poisson commuting conservation

laws, and infinitely many hydrodynamic reductions.

The necessity part of this conjecture follows from Theorem 2, which is proved in

Sect 3.4, which gives the vanishing of the Haantjes tensor as a necessary condition


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Introduction 14

for the integrability of hydrodynamic chains (not necessarily Hamiltonian). The suf-

ficiency is more difficult to establish. The conjecture is supported by all examples

of integrable Hamiltonian chains known to us. Since components of the Haantjes

tensor can be calculated using computer algebra, this approach provides an effective

classification criterion. We are going to demonstrate that the conjecture is indeed

true for Hamiltonian chains of the type (6) with Kupershmidts bracket and Manin-

Kupershmidts bracket, respectively.

We also investigate the existence of higher order (than the hamiltonian, h) con-

servation laws for the two types of Hamiltonian systems in question; recall that

the existence of infinitely many conservation laws is one of the main features of

the integrability. We show that the requirement of the existence of one additional

conservation law of the form,

p(u1, u2, u3)t = q(u1, u2, u3, u4)x for Hamiltonian systems with Kupershmidtsbracket;

p(u1, u2, u3, u4)t = q(u1, u2, u3, u4, u5)x for Hamiltonian systems with Manin-Kupershmidts bracket;

leads to the same relations as requiring H1jk = 0 for the respective systems. Further-

more the system of equations obtained from setting H1jk = 0, imply the existence

of a generating function of conservation laws and, hence, an infinity of conservation

laws. This establishes the integrability of all examples constructed in Sect. 5 andSect. 6.

The main results of this thesis were published in [18], [20].


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Hydrodynamic type systems in 1+1 dimensions 15

2 Hydrodynamic type systems in 1+1 dimensions

An m-component system of hydrodynamic type has the form,

ut = v(u)ux, (7)

where v(u) is a mm matrix and u is a column vector with components u1, u2, u3, . . . ,um. The integrability of these systems by the generalized hodograph method re-

quires that the matrix v(u) is diagonalizable, or in other words possesses Riemann

invariants. This section contains information on hydrodynamic type systems in 1+1

dimension. We are interested only in systems of hyperbolic type, this means all

eigenvalues of matrix v in (7) must be real and distinct.

2.1 Riemann invariants

Consider (7), such a system possesses Riemann invariants if there exists a change of

variables

u1, . . . , un R1(u), . . . , Rn(u) (8)

so that (7) goes to

Rit = i(R)Rix. (9)

If (7) possesses Riemann invariants then it is said that the matrix v(u) is diagonal-

izable.

To find Riemann invariants of a 2x2 system, proceed as follows:


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1. Write the equations in the form (7) and find the eigenvalues of matrix v(u),

det v(u) iI = 0.

2. Require the gradients of the Riemann invariants are left eigenvectors of matrix

v corresponding to eigenvalues 1, 2.

grad Ri

v(u) iI = 0.

3. Solve the resulting two pairs of PDEs to obtain the Riemann invariants

R1, R2,

(R1

u1,

R1

u2)

v(u) 1I = 0,(

R2

u1,

R2

u2)

v(u) 2I = 0.

An example follows to further highlight the steps to calculate Riemann Invariants.

Example Calculating Riemann invariants for the equations of gas dynamics

t + ux + ux = 0, ut + uux + 2x = 0. (10)

Write (10) in matrix form

u

t

+

u

2 u

u

x

= 0. (11)


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Comparing (11) to (9) we see that

u1u2

=

u

, v

i

j = u

2 u

.

In step one we find the eigenvalues of vij,

1,2 = u (1)1/2. (12)

In step two we find the Riemann invariants by requiring,

R1 R

1u

u 1 2 u 1

= 0,

the same is done for 2, R2. In the third step, multiply out and solving we see that

R1 = u +21/2(1)/2

1

.

We obtain R2 by the same method, ultimately,

R1 = u +21/2

12

1 , R2 = u 2

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1 . (13)

So we have equation (9) for i = 1, 2, thus

R1t + 1R1x = 0, R2t + 2R2x = 0. (14)

Note at this point that from (13) we have

R1 + R2

2= u, R1 R2 = 4

1

212

1 . (15)


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Substitute (15) into (12) to obtain

1 =R1 + R2

2

+( 1)(R1 R2)

4

, 2 =R1 + R2

2

+( 1)(R2 R1)

4

. (16)

If (16) is substituted into (14) then we have

R1t +

R1 + R2

2+

( 1)(R1 R2)4

R1x = 0,

R2t +

R1 + R2

2+

( 1)(R2 R1)4

R2x = 0. (17)

One can verify that (11) indeed goes to (17) under the change of variables

R1 + R2

2= u, R1 R2 = 4

1

212

1 .

Thus, R1 and R2 are Riemann invariants of (10).

Example [38] This is a much more involved example of determining the Riemann

invariants for a given system. Equations of Chromatography are

uix + ai(u)t = 0, i = 1, , n. (18)

Different models are obtained via the choice ofai(u), called isotherms of adsorption.

In this example we are going to obtain the Riemann invariants for the Langmuir

isotherm, thus,

ai(u) =iu

i

, = 1 +

ns=1

sus. (19)


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First, we calculate the eigenvalues of matrix A where

uix + Ai

jujt = 0, (20)

It was shown in [38] that the characteristic equation, det(A I) = 0, has the form,

det(A I) =

1 21u1 122

+n

s=2

2sun(1 1)

2(s s)

2 2

. . .

n n

=1

1

+

(1

1)

2

ns=1

2sun

s s +n

p=2

p p

=n

p=1

p p

1

ns=1

s s

ns=1

2su

s

s s

,

so,

=n

s=1

2sus

s s , (21)

since, det(A I) = 0. Thus, substituting into (19) we obtain,

1 +n

s=1

sus =

ns=1

2sus

s (22)

The eigenvalues, i, of A are the solutions of (22).

We now show that Riemann invariants are given by Ri = i. It is required (9) is

satisifed modulo Ri = i. Take R = in (22), calculate Rx, Rt in terms of ujt ,


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and then substitue to show Rx +RRt = 0.

Rx =

ssR

s R su

s

pupt p

2

su

st

,

Rt = R

ssustsR

p

up2p(pR)2

.

Substitue into (9) to obtain,

n

s=1

2sust

s R +n

p=1

pupt + R

ns=1

sust

s R = 0, (23)

one can see that coefficients at ukt cancel out. Thus, Rx +R

Rt = 0 modulo R = ,

as required.

Now, we show the diagonal form of (20) is

Rix + Ri

ns=1 R

s

ns=1 sRit = 0. (24)

Using the result that Ri = i in (22) we have

ns=1

2sus

s R = ,

write as a polynomial

n

s=1

(s R) 21u1n

s=1(s R) 2nun

ns=n

(s R) = 0.


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The coefficient of Rn is (1)n. The coefficient of R0 is

n

s=1

s

n

i=1

iui

n

s=1

s = 0,

so, the polynomial can be written in the from,

Rn + + (1)n

s s

= 0.

It can also be written in the form (R R1) (R Rn) = 0, so by Viets theorem,equating the free term gives

(1)nn

s=1

Rs = (1)n

s s

,

thus, =

ssRs

, so we have (24) as required.

Now it is claimed that the formulas that connect Riemann invariants Ri with vari-

ables ui are,

ui =1

i

s=q

iRs

1s=i

is

1 . (25)

Indeed, this can be shown to be true as follows. From (24) we know that

i =Ri

= Ri

n

s=1Rss

.

This implies that =n

s=1sRs . Next, take (21) and substitute (25) in to eliminate


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uk, obtaining an expression for , in terms of s and Rs:

=n

s=1

2sus

s Rk

=n

i=1

i s(iRs

1)

(i Rk

)

s=i(is 1)

=

s

sRs

ni1

ns=1 i Rs

(i Rk)n

s=i(i s)=

s

sRs

ni=1

s=k i Rs

s=i(i s)

=

s

sRs

.

So, (25) is consistent as it gives the correct expression for when substituted into

(21). Thus, it has been shown that the system (19) has Riemann invariants and the

diagonal form is given in (24).

2.2 Commutativity Condition

For two systems to commute means that they are consistent with each other. Take

two systems of type (7), ut = V(u)ux, uy = W(u)ux, for them to commute means

uty = uyt.

Theorem 4 Let us consider two systems of the form (9),

Rit = i(R)Rix, R

iy =

i(R)Rix, i = 1 . . . n , (26)

here t and y are the corresponding times. It is claimed that for these equations to

be consistent then the following condition must be true,

j i

(j i) =j

i

(j i) , i = j. (27)

Where j =

Rj .


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Proof For the equations to be consistent it is required that,

Rity = Riyt. (28)

So, let us calculate the necessary condition for (28) to be satisfied.

Rity = (iRix)y = j

iRjyRix +

iRixy

= jijRjxR

ix +

i(iRix)x = jijRjxR

ix +

ij iRjxR

ix +

iiRixx; (29)

Riyt = (iRix)t = j iRjt Rix + iRixt = j i(jRjx)Rix + i(iRix)x

= jijRjxR

ix +

ijiRjxR

ix +

iiRixx. (30)

From substituting (30) and (29) into (28) one obtains

jijRjxR

ix +

ijiRjxR

ix +

iiRixx = jijRjxR

ix +

ijiRjxR

ix +

iiRixx,

coefficients of Rixx cancel, leaving

ji(j i) = j i(j i).

Which gives (27). (27) shall be referred to as the commutativity condition [61]. Fur-

thermore, if (27) is satisified then the systems (26) are said to be commuting flows.

Example Let us calculate commuting flows for the system,


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R1t = R2R1x,

R2t = R1R2x,

in the form,

R1y = 1R1x, (31)

R2y = 2R2x. (32)

Substituting into (27) gives

1

R1 R2 =2

1

2 1 ,1

R2 R1 =1

2

1 2 . (33)

Look for 1, 2 in the form,

1

(R1

, R2

) = Af1 + Bf + Cg, 2

(R1

, R1

) = Dg2 + Ef + F g. (34)

A,B,C,D,E,Fare certain functions ofR1, R2 to be determined, while f = f(R1), g =

g(R2) are arbitrary functions.

Thus the general solution is, 1 = (R2 R1)f + f g and 2 = (R2 R1)g + f g,which can be verified by substituting into (33).

2.3 Conservation Laws

Consider a diagonal system of type (9). A conservation law is a relationship

[f(R)]t = [g(R)]x (35)


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which must hold identically by the virtue of (9). The quantities f(R) and g(R) are

called the conserved density and the flux, respectively. By substituting (9) into (35)

we see that

if iRix = igR

ix

so that if i = ig for any i. For consistency we need

jig = ijg.

From this condition we obtain

ijf =j

i

j i if +i

j

i j jf, i = j.

Example Consider the system

R1t = R2R1x, R

2t = R

1R2x, (36)

here 1 = R2, 2 = R1. The equation (36) for conserved densities takes the form

21f =1

R1 R2 1f +1

R2 R12f =1f 2fR1 R2 . (37)

Its general solution is

f =p(R1) q(R2)

R1 R2 ;

which can be verified by a straightfoward differentiation. Here p(R1) and q(R2) are

arbitrary functions of one variable.

Remark The condition (35) is equivalent to the 1-form f dx+gdt being closed. This


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can be seen by applying the differential d to dF = f dx + gdt,

d2F = fxd2x + ftdxdt + gxdtdx + gtd

2t

d2F = (ft gx)dxdt.

Note, d2F is identically zero, thus ft = gx.

2.4 Semi-Hamiltonian property

The system (9) is said to be semi-Hamiltonian [61] if

k

j

i

j i

= j

k

i

k i

(38)

for any i = j = k = i. Introducing aij = ji

ji one can rewrite the previous equation

in the simpler form

kaij = jaik. (39)

From equation (39) it can be shown that

kaij = aijajk aijaik + aikakj . (40)

Proposition [61]. If (38) is satisfied then commuting flows depend on n arbitrary

functions of one argument.

Proof Consider commuting flows of (9),

j i = aij(

j i), ki = aik(k i). (41)


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Work out jki and kj

i and use (40) and (41) to eliminate nm and manp,

kji = (aij ajk

aij aik + aikakj )(

j

i)

+aij(ajk (k j ) aik(k i)), (42)

j ki = (aikakj aikaij + aijajk )(k i)

+aik(akj (j k) aij(j i)). (43)

Now subtract (43) from (42) and collect coefficients of i, j, k. It is straightfoward

to determine that the coefficients of i, j, k are equal to 0. We know all mixed

partial derivatives from (41), we also now know that all the partial derivatives are

consistent identically in i, j , k. Thus commuting flows depend on n arbitrary

functions of one argument, the functions of integration. Indeed, 1 can be defined

arbitrarily on the R1- axis, since we know j1 j = 1, 2 can be defined arbitrarily

on the R2-axis, etc.

Proposition [61]. If (38) is satisfied then conserved densities depend on n arbitrary

functions of one argument.

Proof Consider

jif = aijif + aji j f,

kif = aikif + akikf, (44)

compute kjif and j kif, substitute (44) to eliminate all terms of the form

mnf. One obtains

kj if = kaijfi + aij (aikfi + akifk) + kaji fj + aji(ajk fj + akj fk),

jkif = j aikfi + aik(aijfi + aji fj ) + jakifk + aki(akj fk + ajk fj). (45)


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Set these last two equations equal to each other and collect coefficients of fi, fj, fk.

The coefficients offi immediately cancel to give 0, while the coefficients of fj , fk are

0 due to (40). So we know all mixed partial derivatives of all f, we also know that

all these partial derivatives are consistent. Thus conserved densities depend on n

arbitrary functions of one variable, indeed one can define f arbitrarily on any of the

coordinate lines.

2.5 Generalized hodograph method

The generalized hodograph method allows the general solution of a system of the

form (7) to be found. First we will demonstrate with a scalar example and then give

a more general outline of the generalized hodograph method. Consider the scalar

Hopf equation

Rt = RRx. (46)

It is known that the general solution of (46) is given by the implicit formula

(R) = x + Rt (47)

where f(R) is an arbitaray function. Indeed, if we differentiate (47) by x and t

respectively and solve for Rx and Rt :

Rx = 1 + Rxt Rx = 1 t ,

Rt = R + Rtt Rt = R t . (48)


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By substituting (48) into (46) we obtain the identity. Thus, (47) is the general

solution.

Theorem 5 [61] The general solution of the diagonal system

Rit = i(R)Rix (49)

is given by

i(R) = x + i(R)t. (50)

where i(R) satisifes (38) and i(R) are characteristic speeds of commuting flows:

ji

j i =j

i

j i i = j. (51)

Proof Differentiate (50) by x and t respectively,

j iRjx + iiRix = j

iRjxt + iiRixt,

j iRjt + i

iRit = jiRjt t + i

iRitt + i. (52)

By substituting (50) into (51) and rearranging one obtains

ji = j

it. (53)

Now substitute (53) into (52) to obtain

Rix =1

ii iit , Rit =

i

ii iit . (54)


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Thus by substituing (54) into (49) we obtain the identity so (50) is the general so-

lution. General solution means that (50) has the same amount of freedom as the

system (46) being solved. Considering the initial value problem, t = 0, of (49), all

Ri can be defined arbitrarily at t = 0, thus we have the freedom of n arbitrary

functions of one variable in the system being solved. In (51) all derivatives of the

solution, i, are defined apart from the ith one, thus i can be defined arbitrarily on

the Ri axis, thus there are n arbitrary functions of one variable.

2.6 The Haantjes tensor

The Haantjes tensor (3) is used to verify the diagonalizability of the system (7).

Calculating the Haantjes tensor using computer algebra is relatively straightforward

for a well defined matrix and thus is an efficient way to find out if the system is

diagonalizable whereas calculating the eigenvalues and eigenvectors, in some cases,

is extremely difficult. As defined previously the Nijenhuis tensor and the Haantjes

tensor are,

Nijk = vpj pv

ik vpkpvij vipj vpk + vipkvpj .

Hijk = Ni

prvpj v

rk Npjr vipvrk Nprkvipvrj + Npjk virvrp,

where s =

us . Both the Nijenhuis tensor and the Haantjes tensor are skew sym-

metric in lower indices, this means reversing the two lower indices changes the sign,

Hijk = Hikj . A direct corollary of this is Hijj = 0 for j = 1 n.

Theorem 6 The matrix v(u) from (7) can be diagonalized if and only if Hijk 0.


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Proof To prove this we first derive the necessary condition for the Haanjtes tensor

to be identically zero. Then we obtain the condition for v(u) to be diagonalizable

and see that they are equivalent. Evaluated on pairs of vector fields X, Y, the

Nijenhuis tensor and the Haantjes tensor can be written in the form,

N(X, Y) = [vX,vY] + v2[X, Y] v[X,vY] v[vX,Y], (55)

H(X, Y) = N(vX,vY) + v2N(X, Y) vN(X,vY) vN(vX,Y). (56)

where [Z, Y] = (Zi Yj

ui Yi Zj

ui)

uj. Let Xi, Xj be eigenvectors of v, thus,

vXi = iXi, (57)

using this upon direct calculation one obtains,

N(Xi, Xj) = [iXi,

j Xj ] + v2[Xi.Xj] v[Xi, jXj] v[iXi, Xj]

= ij [Xi, Xj ] + iji Xj jij Xi + v2[Xi, Xj] V j [Xi, Xj ] vji Xj

vi[Xi, Xj] vij Xi (58)

= (iji jji )Xj + (iij jij)Xi + (j i + v2 (i + j)v)[Xi, Xj].

Then using (58) calculating the Haantjes tensor and collecting terms at Xi, Xj, [Xi, Xj],

we obtain

H(Xi, Xj) = (

i

j

+ v

2

v(i

+

j

))

2

[Xi, Xj] + (

i

j

+ v

2

v(i

+

j

))(i j)ijXi + (i j)ji Xj

= (ij + v2 (i + j)v)2[Xi, Xj].


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The coefficients of Xi, Xj cancel identically due to the fact vXk = kXk. Next, let

the expansions of commutators [Xi, Xj] ben

k=1 Ckij Xk. Thus,

H(Xi, Xj) = (ij + v2 (i + j)v)2[Xi, Xj],

= (ij + v2 (i + j)v)2CkijXk,

=

(ij + v2 (i + j )v)2Ckij Xk,=

(ij + (k)2 (i + j)k)2CkijXk,

= (k i)2(k j)2Ckij Xk.

Hence H(Xi, Xj ) = 0 if and only if

Ckij = 0, i = j, i = k, k = j. (59)

The matrix v is diagonalizable if and only if each distribution < Xi, Xj > is inte-grable, i.e [Xi, Xj] span(Xi, Xj), which is exactly (59). For an explanation letus consider 3D space. Take X1(u

1, u2, u3), X2(u1, u2, u3), X3(u

1, u2, u3) as eigenvec-

tors. We are looking for a change of variables of type (8). The Riemann invariant

R3(u1, u2, u3) is defined by the equations:

X1R3 = 0, X2R

3 = 0,

the compatibility condition gives X2X1R3X1X2R3 = 0, which is precisely, [X1, X2]R3 =

0. Similarly, the Riemann invariant R2(u1, u2, u3) is defined by:

X1R2 = 0, X3R

2 = 0,


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the compatibility condition gives [X1, X3]R2 = 0 if C312 = 0. Therefore v possesses

Riemann Invariants. This idea can be extended to n dimensions.

Example Consider the Benney-like system

a

v

w

t

+

v a 0

0 v 1

q(a) 0 0

a

v

w

x

= 0. (60)

Is (60) diagonalizable and under what restrictions on q(a)? To answer this ques-

tion we calculate all components of the Haanjtes tensor for the system (60). First

calculate Nijk . Note that Ni

jk = Nikj .

N112 = 0, N113 = 1, N123 = 0, N221 = 0,

N223 = 1, N213 = 0, N331 = 0, N332 = 0, N312 = aq(a). (61)

Now we can calculate Hijk . Note that Hi

jk = Hikj

H123 = 0, H113 = 0, H

112 = aq(a), H213 = aq(a), H223 = 0,

H221 = avq(a), H312 = av2q(a), H332 = a2q(a), H331 = avq (a).

Thus, Hijk 0 if and only if q(a) = 0. So, q(a) must be a constant. Thus (60) isdiagonalizable if q(a) is constant.


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2.7 Reciprocal Transformations

Let B(u)dx + A(u)dt and N(u)dx + M(u)dt be two conservation laws of the system

(7), equivalently we can view them as 1-form closed by virtue of (7). One can

introduce new variables X and T defined by

dX = Adx + Bdt, dT = Mdx + Ndt

Lets apply the transformation to equation (7). First we need to calculate ux, ut in

terms of the new variables X, T:

ut = NuT + BuX , ux = MuT + AuX , (62)

substitue these into (7) and write in the same form we get,

uT = w(u)uX ,

where w(u) = (Av BI)(N I Mv)1. These transformations originate from gasdynamics.

Example Take the diagonal form of Chromatography equations (24). Introduce

the new independent variable, X,T, by the formulae

dT = n+1(Rs +

Rs

dt + R

s

s

dx ,dX = n+1

(Rs +

Rs

dt +

Rss

dx

s,


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and the new dependent variable R = Ri+

Ri+ . So, (24) becomes

RiX

n s

(Rs +

+ Ri

T

nRs +

s

+ Ri s

Rs

sRi

T

n+1s

Rs +

Rs

+RiX n+1

(Rs + )Rs

s

s

= 0.

After simplification,

( + Ri)

s

(Rs + )

RiX +

(Ri + )

s

(Rs + )

RiT

s

= 0.

Divide by

s(Rs + ) to get

( + Ri)RiX + (Ri + )

s

Rs

sRiT = 0.

Dividing by ( + Ri) gives

RiX + Ri

s

Rs

sRiT = 0.

Multiply both sides by (Ri+)2 to obtain

RiX + Ri

s

Rs

sRiT = 0.

Proposition The semi-Hamiltonian property (38) is preserved under reciprocal

transformations


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Proof Applying the reciprocal transformation,

dX = Adx + Bdt,

dT = Mdx + Ndt

to (9) we obtain

RiX =iA B

N iMRiT

Thus we need to show

i

satisfies (38) where,

i =

iA B

N iM

.

We will start by expanding

ijji

k

and using the fact that Bj = jAj, Nj =

jMj,

ij

j

ik

=

(N jM)(N iM)(ijA + (i j)Aj ) (iA B)(ij M + (i j )Mj)

(N iM)(j i)(AN MB)

=ij

N iM ij M

N iM (N iM)Aj (iA B)Mj

AN MBN jMN iM

=ij

N iM +Nj ijM iMj

N iM Nj iMjN iM

((N iM)Aj (iA B)Mj) (N j M)(AN MB)(N iM)

=ij

N iM+ ln(N

iM)j

(AN)j (BM)jAN BM

=

ij

j i + ln

N iMAN MB

j

k

.


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So, by symmetry we have

i

j

j

i

k

= ik

k

i

j

as required.


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Hydrodynamic Chains 38

3 Hydrodynamic Chains

A hydrodynamic chain is of the form (7),

ut = V(u)ux

where V(u) is a square matrix of infinite dimension, while u = (u1, u2, , )t isan infinite-component vector. We assume that V(u) belongs to class C as defined

in Def. 1 in the Introduction. A well known example of a hydrodynamic chain is

the Benney chain [3].

3.1 The Benney chain

The Benney chain is,

A0t + A1x = 0,

A1t + A2x + A

0A0x = 0,

A2t + A3x + 2A

1A0x = 0, (63)

A3t + A4x + 3A

2A0x = 0,

A4t + A5x + 4A

3A0x = 0,

. . .


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or, in matrix form,

A0

A1

A2

A3

t

+

0 1 0 0 0 0 .

A0 0 1 0 0 0 .

2A1 0 0 1 0 0 .

3A2 0 0 0 1 0 .

. . . . . . .

A0

A1

A2

A3

x

= 0.

Let us take a moment to look at how and where the (63) first appeared. This chain

was derived in [3] and was shown to consist of an infinite number of conservation

laws for the system of equations that describe long waves on shallow ideal fluid

under the action of gravity. These equations are,

ux + vy = 0, (64)

ut + uux + vuy + ghx = 0, (65)

v = 0, y = 0, (66)

ht + uhx v = 0, y = h. (67)

Here h(x,t, ) is the height of the water, u(x,y,t), v(x,y,t) are the horizontal and

vertical velocities of the water, g is gravity. The conservation laws of mass, momen-

tum and energy are all straightforward to derive. Equation (67) can be rewritten

as

ht +

x

ho

udy

= 0; (68)


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(64) is used to eliminate v. This is the mass conservation law. Integrate (65) w.r.t

y from 0 to h to obtain the momentum conservation law:

t

h

o

udy

+

x

h

o

u2dy +gh2

2

= 0; (69)

(64), (65), (67) have each been used. Finally multiply (65) by u and integrate w.r.t

y from 0 to h to obtain the energy conservation law:

t

1

2

ho

u2dy +gh2

2

+

x

1

2

ho

u3dy + gh

ho

udy

= 0. (70)

Note that the identity

x

r(x)s(x)

udy

=

r(x)s(x)

uxdy + u|y=rrx u|y=ssx

has been used. Benney went on to show that there are an infinte number of such

conservation laws as follows. Let us introduce the notation

An =

h0

un(x,y,t)dy,

so A0 = h, A1 =h0

u(x,y,t)dy, ,etc, then by multiplying (65) by un1 and thenintegrating from y = 0 to y = h. Using (64), (65), (66) we obtain an unclosed set of

equations for the moments:

Ant +

An+1x + nAn1

A0x = 0, n = 1, 2, . . . (71)

These equations are not conservative and nor is it in any way clear that they can

be brought into such a form. The existence of an infinity of conservation laws was


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demonstrated in [3, 32]. In our discussion we follow [22] which extends these ideas.

Instead of dealing directly with (71) introduce a generating function

= p +u1

p+

u2

p2+

u3

p3+ ... (72)

which was shown in [22] to satisfy, by virtue of (63), the relation

t px = ppt (p2/2 + u1)x

. (73)

Let us show that (72) does indeed satisfy (73) modulo (63). Using (73) and differ-

entiating we obtain,

t = pt +u1tp

u1pt

p2+

u2tp2

2u2pt

p3+ + u

kt

pk ku

kptpk+1

+ . . . . (74)

x = px +u1x

p u1px

p2+

u2x

p2 2u2px

p3+

+

ukx

pk kukpx

pk+1+ . . . . (75)

p = 1 u1

p2 2u

2

p3 3u

3

p4 ku

k

pk+1 . . . . (76)

Substituting (74), (75), (76) into (73) ,

t px = ppt (p2/2 + u1)x

,

pt +

uktpk

kukpt

pk+1

p

px +

ukxpk

kukpx

pk+1

=

1 ku

k

pk+1

pt (p + u1x)

,


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where k = 1, 2, 3, . . . , now collect t derivatives on the left hand side and x derivatives

on the right hand side,

pt +

uktpk

kukptpk+1

1 kuk

pk+1

pt = p

px +

ukxpk

kukpxpk+1

1 kuk

pk+1

px

,

after cancelation we are left with,

uktpk

=uk+1xpk

+kuku1xpk+1

, k = 1, 2, 3, . . .

upon equating powers of p we finally have,

ukt = uk+1x + (k 1)uk1u1x, k = 1, 2, 3, . . .

which is precisely (63) as expected. This relation provides an infinity of conserved

densities Hn(u) defined by the equation

pt = (p2/2 + u1)x

where one has to substitute the expression for p() obtained from (72): p = H1

H2

2 H3

3 .... Explicitly, one gets H1 = u1, H2 = u2, H3 = u3 + (u1)2, etc:

H1t = (H2)x,

H2t = (H3 H1

2 )x, (77)

H3t = (H4 H2H1)x,

H4t =

H5 H3H1 (H

2)2

2

x

.


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3.2 Hydrodynamic chains and the Haantjes tensor

Both the tensors (2) and (3) make perfect sense for infinite matrices which are of

class C as stated in the introduction. For matrices from the class C all contractions

in the expressions (2) and (3) reduce to finite summations so that each particular

component Hijk is a well-defined object which can be effectively computed. More-

over, for a fixed value of the upper index i there exist only finitely many non-zero

components Hijk .

We propose the following

Definition 3 [20] A hydrodynamic chain from the class C is said to be diagonaliz-

able if all components of the corresponding Haantjes tensor (3) are zero.

Although we have stated the above definition it must be noted that the Riemann

invariants of a hydrodynamic chain cannot be calculated because the chain is infinite.

Example In this example we illustrate how the Nijenhuis tensor and the Haantjes

tensor can be calculated for a general hydrodynamic chain of class C. First, calculate

the Nijenhuis tensor as this is required to calculate the Haantjes tensor. Then

calculate the Haanjtes tensor, in this example the upper index is fixed at 1, H1jk .

Consider the infinite system

u1t + u2x + v

11(u

1)u1x = 0,

u2

t + u3

x + v2

2(u1

, u2

)u2

x + v2

1(u1

, u2

)u1

x = 0, (78)

u3t + u4x + v

33(u

1, u2, u3)u3x + v32(u

1, u2, u3)u2x + v31(u

1, u2, u3)u1x = 0,

. . .


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Or, in matrix form,

u1

u2

u3

u4

t

+

v11 1 0 0 0 0 .

v21 v22 1 0 0 0 .

v31 v32 v

33 1 0 0 .

v41 v42 v

43 v

44 1 0 .

u1

u2

u3

u4

x

= 0.

(79)

Note that vi

j

= vi

j

(u1, . . . ui). It is easily seen that

vik =

vik(u1, . . . , ui) k < i + 1

1 k = i + 1

0 k > i + 1.

To calculate Nijk the index i will be considered fixed, this implies that we need only

consider j

i + 1, k

i + 1, 1

p

i + 1. Similarly when calculating Hijk with

i fixed, implies that we need only consider j i + 4, k i + 4, 1 p,r i + 3.In general if vik = 0 k > i + l, then to calculate Nijk we only need to consider

j i + l, k i + l, 1 p i + l. To calculate Hijk we only need to considerj i + 4l, k i + 4l, 1 p,r i + 3l.

Calculate Nijk , i 3, j , k 5, 1 p 4. The notation j = uj is used in the


45/174


following evaluation.

N1jk =

1v11 + 1v

22 2v21 j = 2, k = 1

1v11 1v22 + 2v21 j = 1, k = 20 else

N2jk =

(v22 v11)1v22 + 1v21 + 1v32 v212v22 2v31 j = 2, k = 11v

33 + 2v

21 3v31 j = 3, k = 1

(v11 v22)1v22 1v21 1v32 + v212v22 + 2v31 j = 1, k = 22v

22 + 2v

33 3v32 j = 3, k = 2

1v33 2v21 + 3v31 j = 1, k = 32v22 2v33 + 3v32 j = 2, k = 30 else


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N3jk =

1(v31 + v

42) + (v

33 v11)1v32 + v321v22 + (v22 v33)2v31

2(v21v32) 2v41 + v323v31 v313v32 j = 2, k = 1

(v33 v

11)1v

33 + 1v

43 + 2v

31 v

212v

33 v

313v

33 3v

41 j = 3, k = 1

1v44 + 3v

31 4v41 j = 4, k = 1

(v11 v33)1v32 v321v22 1(v42 + v31) + 2(v21v32)+2(v

33 v22) + 2v41 + v313v32 v323v31 j = 1, k = 2

1v33 + 2v32 + (v33 v22)2v33 + 2v43 v323v33 3v42 j = 3, k = 22v

44 + 2v

32 4v42 j = 4, k = 2

(v11

v33)1v33

1v

43 + v

212v

33

2v

31 + v

313v

33 + 3v

41 j = 1, k = 3

1v33 + v

222v

33 2v32 v332v33 2v43 + v323v33 + 3v42 j = 2, k = 3

3(v44 + v

33) 4v43 j = 4, k = 3

1v44 3v31 + 4v41 j = 1, k = 42v44 3v32 + 4v42 j = 2, k = 43(v44 + v33) + 4v43 j = 3, k = 40 otherwise


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Now H1jk can be calculated,

H1jk =

v111v32

v31(2v

22

23v

32 + 2v

33) + v

32(22v

21

1v22 23v31 + 1v33) v222v31 + v332v31+v211v

32 1v31 v331v32 + 2v41 1v42 j = 1, k = 2

v111(v33 v22) + v22(1(v22 v33) 2v21 + 3v31)+

v21(22v33 3v32) + v33(2v21 3v31) + 1(v21 + v32 v43)

22v31 + 3v41 + v313v33 j = 1, k = 3

1v44 23v31 + 4v41 + 2v21 + 1v33 j = 1, k = 4

v32(22v21 + 1v22 + 23v31 1v33) + v31(2v22+2v

33 23v32) + v222v31 v332v31 v212v32 + 1v31 v111v32+

v331v32 2v41 + 1v42 j = 2, k = 1

v11(2v22 2v33 + 3v32) + v33(2v22 3v32)+v323v

33) + 1(v11 v22 + 2v33) + 2(2v21 v32 v43)

+v222v33 + 3(v

42 v31) j = 2, k = 3

2(v22 + v

33 v44) 23v32 + 4v42 j = 2, k = 4

v111(v22 v33) + v21(3v32 22v33) + v22(1(v33 v22) + 2v21

3v31) + v

33(3v

31 2v21) + 1(v43 v21 v32) + 22v31

3v41 v313v33 j = 3, k = 1


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H1jk =

v11(2(v22 + v

33) 3v32) + v33(2v22 + 3v32) 21v33+

1(v11 + v

22) + 3(v

31 v42) v222v33 + 2(v32 v43 2v21)

v323v

33 j = 3, k = 2

3(v44 v33) + 4v43 j = 3, k = 4

1v44 + 23v

31 4v41 2v21 1v33 j = 4, k = 1

2(v44

v22

v33) + 23v

32

4v

42 j = 4, k = 2

3(v44 + v

33) 4v43 j = 4, k = 3

0 otherwise.

It is a necessary condition that each component of H1jk is identically zero for the

matrix to be diagonalizable.

Example In this example we show how to calculate every component of the Haantjes

tensor for the Benney chain (63). As in the previous example the Nijenhuis tensor

is first calculated. Since the system under consideration is infinite, it means that in

theory the summation for Nijenhuis and Haantjes tensor may be infinite. However,

the matrix is of the class C so the summation is in fact finite, as is now shown. It

is clear that

vmm+1 = 1, vm1 = (m 1)Am2,


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all other vmn = 0. Now we calculate the Nijenhuis tensor, first consider Ni1k,

Ni1k = vp1pv

ik

vpkpv

i1

vip1v

pk + v

ipkv

p1 .

Let us consider each term separately for a moment, the first term:

vp1pvik =

(i 1)(i 2)Ai3 k = 10 otherwise

The second term:

vpkpvi1 =

(i 1)(i 2)Ai3 k = 1i 1 k = i0 otherwise

The third term:

vi

p1vpk =

1 i = k = 1

0 otherwise

The fourth term:

vipkvp1 =

k i = k

0 otherwise


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Combining all the terms we have

Ni1k =

1 i = k, i = 10 otherwise

=

0 0 0 0 0 0 . . .

0 1 0 0 0 0 .

0 0 1 0 0 0 .

. 1 .

. . .

. . .

. . .

.

Now we calculate the Nijenhuis tensor with j = 1, by the skew symmetry property,Nijk = Nikj , it is clear that,

Nijk =

1 j = 1, i = k, i = 11 k = 1, i = j, i = 10 otherwise

Next we must calculate the Haantjes tensor, first consider

Hi1k = Ni

prvp1v

rk Np1rvipvrk Nprkvipvr1 + Np1kvirvrp.

Work on each of the terms separately, the first term:

Niprvp1v

rk =

(i 1)Ai2 k = 20 otherwise


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The second term:

Np1rvipvrk =

iAi1 k = 1

1 k = i + 2

0 otherwise

The third term:

Nprk vi

pvr1 =

iAi1 k = 10 otherwise

The fourth term:

Np1kvirv

rp =

(i 1)Ai2 k = 21 k = i + 2

0 otherwise

Thus, Hi1k = 0, i,k. Now work on

Hi2k = (Ni

prvp2v

rk) (Np2rvipvrk) (Nprk vipvr2) + (Np2kvirvrp).

Again work on each of the terms separately, the first term:

Niprvp2v

rk =

(i 1)Ai2 k = 11 k = i + 1, i = 10 otherwise

The second term:

Np2rvi

pvrk =

1 i = 1, k = 20 otherwise


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The third term:

N

p

rk v

i

pv

r

2 =

1 k = i + 1

0 otherwise

The fourth term:

Np2kvirv

rp =

(i 1)Ai2 k = 10 otherwise

Thus, Hi2k

= 0,

i,k. Now work on

Hijk = (Ni

prvpj v

rk) (Npjr vipvrk) (Nprk vipvrj ) + (Npjk virvrp), j = 1, 2.

Again work on each of the terms separately, the first term:

Niprvpj v

rk =

1 i = j 1, k = 20 otherwise

The second term:

Npjr vi

pvrk =


The third term:

Nprk vi

pvr

j =



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The fourth term:

N

p

jk v

i

rv

r

p =


Thus, Hijk = 0, i ,k,j 3. We have previously shown that Hi1k = 0, Hi2k = 0.Thus, Hijk = 0, i,j,k.

3.3 Another chain considered by Benney

An extension to (63) is considered in [4],

Cnt

+ anCn+1

x+ bnCn1

C0x

= 0, n 0 (80)

such a system only posesses an infinity of conservation laws if

anbn+1n + 1 = constant. (81)

Calculating the Haanjtes tensor of system (80) and setting each component equal

to zero implies that

anbn+1 2an+1bn+2 + an+2bn+3 = 0. (82)


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This implies

anbn+1 = n + (83)

where , are constants. It is clear that (81) just (83) with = = 1, thus

requiring that the Haantjes tensor is zero has produced a slightly more general

result than obtained in [4]. Under the change of variables An a0 an1Cn, theBenney chain, (63), is transformed to (80) modulo (83).

3.4 Hydrodynamic reductions and diagonalizability

To illustrate the method of hydrodynamic reductions we consider the Benney chain

(63),

u1t = u2x,

u

2

t = u

3

x + u

1

u

1

x,

u3t = u4x + 2u

2u1x,

u4t = u5x + 3u

3u1x,

etc. Following the approach of [24, 25] let us seek solutions in the form ui =

ui(R1, . . . , Rm) where the Riemann invariants R1, . . . , Rm solve the diagonal system

(9),

Rit = i(R)Rix.


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Substituting this ansatz into the Benney equations and equating to zero coefficients

at Rix we arrive at the following relations:

iu2 = iiu, (84)

iu3 = ((i)2 u)iu, (85)

iu4 = ((i)3 ui 2u2)iu, (86)

iu5 = ((i)4 u(i)2 2u2i 3u3)iu, (87)

etc. Here u = u1, i = Ri, i = 1,...,m (no summation!) The consistency conditions

of the first three relations (84)(86) imply

iju =j

i

j i iu +i

j

i j ju,

jiiu + i

jju = 0,

ijiiu +

j ij ju + iuju = 0,

respectively. Solving these equations for ji we arrive at the Gibbons-Tsarev sys-

tem

j i =

j u

j i , iju = 2iuju

(i j)2 . (88)

It is a truly remarkable fact that all other consistency conditions (e.g., of the relation

(87), etc), are satisfied identically modulo (88). Moreover, the semi-Hamiltonian

property (38) is also automatically satisfied. Thus, the system (88) governs m-

component reductions of the Benney chain. Up to reparametrizations Ri fi(Ri)these reductions depend on m arbitrary functions of a single variable. Solutions

arising within this approach are known as multiple waves, or nonlinear interactions

of planar simple waves.


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The main result of this section is the proof of Theorem 2 formulated in the intro-

duction: The vanishing of the Haantjes tensor H is a necessary condition for the

integrability of hydrodynamic chains from the class C [20].

Two different proofs of this statement will be given.

Proof This proof is computational. Writing down the equations of the chain in the

form umt = Vm

n unx and substituting the ansatz u

i = ui(R1,...,Rm) we arrive at an

infinite set of relations

Vmn iun = iiu

m;

we point out that all summations here and below involve finitely many nonzero

terms. Applying the operator j , j = i, we obtain

Vmn,kiunju

k + Vmn ijun = j

iium + iiju

m. (89)

Interchanging the indices i and j and subtracting the results we arrive at the ex-

pression for ijum in the form

ijum =

ji

j i ium +

ij

i j jum +

Vmn,k Vmk,ni j iu

nj uk.

Substituting this back into (89) we arrive at a simple relation

j iiu

m + ij ju

m =Nmnkiu

njuk

i j

where N is the Nijenhuis tensor of V. This can be rewritten in the invariant form

jiiu + i

jju =N(iu, j u)

i j (90)

which implies the following four relations:


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(i)2jiiu + (

j)2ijj u =

V2N(iu, ju)

i

j

, (91)

(i)2jiiu +

ijijju =

V N(V iu, ju)

i j , (92)

ij j iiu + (

j )2ijju =

V N(iu, V ju)

i j , (93)

i

j

ji

iu + i

j

ij

j u =

N(V iu, V ju)

i j . (94)

The first relation can be obtained by applying the operator V2 to (90) and using

V iu = iiu. Similarly by applying V

i, V j, ij respectively, we can obtain

(92),(93),(94). Thus combining all four relations (91), (92), (93), (94),

V2N(iu, ju) V N(V iu, ju) V N(iu, V ju) + N(V iu, V ju) = 0. (95)

Relation (95) can be rewritten in the form H(iu, ju) = 0 where H is the Haantjes

tensor, indeed, a coordinate-free form of the relation (3) is

H(X, Y) = V2N(X, Y) V N(V X , Y ) V N(X , V Y ) + N(V X , V Y )

where X, Y are arbitrary vector fields. Keeping in mind that

iu and ju are eigenvectors of the matrix V corresponding to the eigenvaluesi and j;

i and j can take arbitrary values;

we conclude that H(X, Y) = 0 for any two formal eigenvectors of the matrix V.


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By formal eigenvectors we mean vectors X defined by AX = X. So, for the Benney

chain (63) we get, in matrix form, the relations,

0 1 0 0 0

u1 0 1 0 0

u2 0 0 1 0

u3 0 0 0 1

. . . . .

x1

x2

x3

.

=

x1

x2

x3

.

.

Assuming that eigenvectors of V span the space of dependent variables u (this is

true for all examples discussed in this paper), we obtain H = 0. In more detail,

let X() = (1(), 2(), 3(),...)t be a formal eigenvector of the matrix V corre-

sponding to the eigenvalue . Let us assume that these eigenvectors span the space

of dependent variables, that is, that there exist no non-trivial relations of the form

cii() = 0 with finitely many nonzero -independent coefficients ci. In other words,

i() are linearly independent as polynomials in .

The condition H(X(), X()) = 0, written in components, takes the form

Hijk j()k() = 0; recall that, since the matrix V belongs to the class C, these sums

contain finitely many terms for any fixed value of the upper index i. Taking into

account the linear independence of j() and k(), one readily arrives at Hijk = 0.


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As an illustration, let V be the matrix corresponding to the Benney chain (63).

V X = X,

0 1 0 0 0 0 .

u1 0 1 0 0 0 .

u2 0 0 1 0 0 .

u3 0 0 0 1 0 .

. . . . . . .

x1

x2

x3

x4

.

=

x1

x2

x3

x4

.

.

Then setting x1 = 1 we obtain the equations

x2 = x1,

x3 + u1x1 = x2,

x4 + u2x1 = x3,

. . .

Then, writing each xi in terms of

x1 = 1,

x2 = ,

x3 = 2 u1,

x4 = 3 u1 u2,

. . .

Thus we arrive at X() = (1, , 2u1, 3u1 2u2,...)t, so that each i() is apolynomial in of the degree i 1. These eigenvectors span the space of dependentvariables since polynomials i() are manifestly linearly independent.


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Notice that we have proved a more general result, namely, that the existence of

sufficiently many two-component reductions already implies the vanishing of the

Haantjes tensor H. Indeed, nothing changes in the proof if we set i = 1, j = 2

in the formula (90). As demonstrated in the Theorem 5 below, one can further

strengthen the result by reversing the above proof under the additional assumption

of the simplicity of the spectrum of the matrix V. This will require the relations

ijju =

N(V iu, ju) V N(iu, j u)(i j)2 , j

iiu =N(V j u, iu) V N(ju, iu)

(i j)2 ,

which can be obtained by applying V to both sides of (90) and solving for ij

j u

and jiiu.

Now, the second proof is given.

Proof 2 Our first remark is that for the chains from the class C one needs to know

only finitely many rows of the matrix V(u) to calculate each particular component

of the Haantjes tensor. Let us fix the values of indices i,j,k and denote by C(i,j,k)

the maximal number of rows needed to calculate Hijk (counting from the first row).

We need to show that Hijk = 0. Let us consider an m-component diagonal re-

duction ui(R1, . . . , Rm), i = 1, 2,.... Choosing the first m variables u1, . . . , um as

independent, we can represent the reduction explicitly as

um+1 = um+1(u1, . . . , um), um+2 = um+2(u1, . . . , um), . . . ,

etc. Substituting these expressions into the first m equations of the chain we ob-

tain an m-component system Sm for u1, . . . , um, while the remaining equations will

be satisfied identically (by the definition of a reduction). Notice that the Haantjes

tensor of the reduced system Sm is identically zero since the reduction is diagonal-


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izable. Let us now choose the number m sufficiently large so that the first C(i,j,k)

equations of the chain do not contain variables um+1, um+2, . . . (one can always do

so since any equation of the chain depends on finitely many us, and m can be ar-

bitrarily large). Then the first C(i,j,k) equations of the reduced system Sm will

be identicalto the first C(i,j,k) equations of the original infinite chain. Hence, the

corresponding components Hijk for the reduced system and for the infinite chain will

also coincide. This proves that all components of the Haantjes tensor of the chain

must be zero.

A straightforward modification of the second proof allows one to show that the ex-

istence of an infinity of semi-Hamiltonian reductions implies the vanishing of the

tensor P. This establishes the necessity of the conjecture formulated in the Intro-

duction.

We emphasize that the condition of diagonalizability alone is not sufficient for the

integrability in general. This can be seen as follows.

Example Let us consider the chain

u1t = u2x + p(u

1)u1x,

u2t = u3x + p(u

1)u2x + u1u1x,

u3t = u4x + p(u

1)u3x + 2u2u1x,

u4t = u5x + p(u

1)u4x + 3u3u1x,

etc, which is obtained from the Benney chain (63) ut = V(u)ux by the transfor-

mation V V + p(u1)E where E is an infinite identity matrix and p is a functionof u1. One can verify that the corresponding Haantjes tensor is zero (which is not

at all surprising since the addition of a multiple of the identity does not effect the


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diagonalizability). A simple calculation shows that hydrodynamic reductions of this

chain are governed by exactly the same equations as in the Benney case, the only

difference is that now the Riemann invariants Ri solve the equations

Rit = (i(R) + p(u1))Rix.

The semi-Hamiltonian property is satisfied if and only if p = 0. Thus, we have con-

structed examples which possess infinitely many diagonal hydrodynamic reductions

none of which are semi-Hamiltonian (if p = 0).

We now prove Theorem 3 from the Introduction: in the simple spectrum case the

vanishing of the Haantjes tensor H is necessary and sufficient for the existence

of two-component reductions parametrized by two arbitrary functions of a single

argument.

Proof The necessity part is contained in the first proof of the previous theorem. To

establish the sufficiency one has to show that the vanishing of the Haantjes tensor

implies the solvability of the equations

V 1u = 11u, V 2u =

22u, (96)

12um =

21

2 1 1um +

12

1 22um +

Vmn,k Vmk,n1 2 1u

n2uk (97)

and

122u =

N(V 1u, 2u)V N(1u, 2u)(12)2 ,

211u =

N(V 2u, 1u)V N(2u, 1u)(12)2 ,

(98)

which govern two-component reductions. Our first observation is that the vanishing


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of the Haantjes tensor implies that the vectors

N(V 1u, 2u)

V N(1u, 2u), N(V 2u, 1u)

V N(2u, 1u) (99)

are the eigenvectors of V with the eigenvalues 2 and 1, respectively. Let us show

that N(V 1u, 2u) V N(1u, 2u) is indeed the eigenvector of eigenvalue 2.

V X = X (100)

V (N(V 1u, 2u) V N(1u, 2u)) = 2 (N(V 1u, 2u) V N(1u, 2u))

V N(V 1u, 2u) V2N(1u, 2u) = 2N(V 1u, 2u) 2N(1u, 2u)V N(V 1u, 2u) V2N(1u, 2u) = N(V 1u, V 2u) V N(1u, V 2u)

(100) is satisfied modulo Haantjes tensor identically zero. The same process can

be carried out for the other eigenvector in question. By the assumption of the

simplicity of the spectrum, that is the eigenvectors are distinct, it is known that

(99) is proportional to 2u and 1u. Thus, equations (98) reduce to a pair of first

order PDEs for 1 and 2,

12 =

k1(1 2)2 , 2

1 =k2

(1 2)2 , (101)

here k1 and k2 are the corresponding coefficients of proportionality. The relations

(96) allow one to reconstruct all components u2, u3,... of the infinite vector u in terms

of its first component u1

. Finally, the equations (97), which are the consistencyconditions of (96), reduce to a single second order PDE for the first component u1

of the infinite vector u,

12u1 =

1

(1 2)3

k12u1 k21u1

+

V1n,k V1k,n1 2 1u

n2uk. (102)


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Thus, relations (96) (98) reduce to a pair of first order equations (101) plus one

second order PDE for u1. Up to reparametrizations R1 f1(R1), R2 f2(R2),their general solution depends on two arbitrary functions of a single variable.


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Conservative chains 65

4 Conservative chains

In this section we classify chains of form (5),

u1t = f(u1, u2)x,

u2t = g(u1, u2, u3)x,

u3t = h(u1, u2, u3, u4)x,

. . . .

All known integrable hydrodynamic chains can be written in conservative form, forexample we showed that the Benney chain, (63), can be written in the conservative

form (77).

4.1 Egorov case

The Egorov case is when f(u1, u2) = u2 in (5) thus becoming (4). There are ten

nonzero components ofHijk

and the requirement that they vanish leads to all second

order partial derivatives of h(u1, u2, u3, u4) in terms of first order partial derivatives


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of h and g:

h11 =2g1g12 g2g11 + 2h1g13

g3,

h12 =g22g1 + g11 + g13h2 + g23h1

g3,

h22 =g2g22 + 2g12 + 2g23h2

g3,

h13 =g13(h3 g2) + g23g1 + g12g3 + g33h1

g3, (103)

h23 = g22 +g13 + h3g23 + h2g33

g3,

h33 = 2g23

g33(g2 2h3)g3

,

h14 =h4g13

g3, h24 =

h4g23g3

, h34 =h4g33

g3, h44 = 0.

The consistency conditions for the equations (103) lead to expressions for all third

order partial derivatives of the function g(u1, u2, u3) in terms of its lower order

derivatives:

g333 =2g233

g3, g133 =

2g13g33g3

, g233 =2g23g33

g3,

g113 =2g213

g3, g123 =

2g13g23g3

, g223 =2g223

g3,

g222 =2

g23

g2g

223 + g23(g3g22 + 2g13) g33(g2g22 + 2g12)

,

g122

=2

g23g

1g223

+ g13

(g3g22

+ g13

)

g33

(g1g22

+ g11

) , (104)g112 =

2

g23(g33(g2g11 2g1g12) g13(g2g13 2g3g12) g23(g3g11 2g1g13)) ,

g111 =2

g23

(g1 + g

22)g

213 + g

21g

223 + g

23(g

212 g11g22) g22g33g21

+g13g3(g11 + 2(g1g22 g2g12)) + 2g23(g2(g3g11 g1g13) g1g3g12)

g33((g1 + g22)g11 2g1g2g12)

.


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This systems general solution depends on ten integration constants, indeed, the

values of g and its partial derivatives up to the second order can be prescribed

arbitrarily at any fixed point u10, u2

0, u3

0. The system (104) was first derived in [47]

from the requirement that the chain (4) is embedded into a hierarchy of commuting

hydrodynamic chains of Egorovs type. Exactly the same equations for g were

obtained in [17] by applying the method of hydrodynamic reductions to the (2+1)-

dimensional PDE

utt = g(uxx, uxt, uxy) (105)

which is naturally associated with the chain (4); here the function g is the sameas in (4), (104). Thus, for hydrodynamic chains of the type (4) the condition of

diagonalizability is necessary and sufficient for the integrability. Let us summarize

the work referred to in [17].

4.2 Method of hydrodynamic reductions

First, introduce the notation uxx = a, uxt = b, uxy = c. Thus (105) becomes

utt = G(a,b,c). (106)

Now we write the PDE (106) in quasilinear form, that is in the form,

at = bx,

ay = cx,

by = ct, (107)

bt = G(a,b,c)x.

Look for hydrodynamic reductions in the form a = a(R1, , Rn), b = b(R1, , Rn),


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c = c(R1, , Rn), where the Riemann invariants satisfy the equations (26). We ar-rive at the following identities

ib = iia, (108)

ic = iia, (109)

(i)2 = Ga + Gbi + Gc

i. (110)

Now, differentiate (110) w.r.t j, to obtain,

2iij = aj

Gaa + jGab

+ ijGb + iaj

gab + jGbb + jGbc

+ ijGc +

aji

Gac + jGbc +

jGcc

, (111)

rearrange (111) to make ij the subject. Use (26) to eliminate i

j, and (110) to

eliminate i. Thus (111) becomes

ij =aj

i

j

[GaaGc + (i + j)GabGc +

ijGbbGc +

(Gac + iGbc)((

j)2 j Gb Ga)

((i)2 iGb Ga)(Gac + jGbc + GccGc

((j )2 jGb Ga))]. (112)

The compatibility conditions of (108), (109), that is (bi)j = (bj)i and (ci)j = (cj)i

respectively, imply that

ij a =

ijj i ia +

jii j ja. (113)

The consistency conditions of (112), (ij)k = (ik)j , are of the form

Xjaka.


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The coefficient X is some rational expression in i, j , k, the coefficients depend

on partial derivatives of G(a,b,c) up to the third order. To obtain the integrability

conditions only three component reductions need be considered, thus set i = 1, j =

2, k = 3. Requiring that X vanishes identically we obtain expressions for all third

order partial derivatives of G(a, b, c). The compatibility conditions of equations

(113), that is, j(ika) = k(ija), take the form

Y iajaka (114)

The coefficient Y is rational in i

, j

, k

. Upon equating Y to zero one obtains

the same conditions as when we required X is identically zero. These conditions are

equivalent to (104).

Returning to the the Haantjes tensor, it can be shown that the vanishing of other

components of the Haantjes tensor does not impose any additional constraints on the

derivatives ofg and h. Thus, writing the fourth equation of the chain (4) in the formu4t = s(u

1, u2, u3, u4, u5)x and setting H2

jk = 0, one obtains the expressions for all

second order partial derivatives ofs in terms ofh and g, which are analogous to (103).

The consistency conditions are satisfied identically modulo (103), (104). Similarly,

the condition H3jk = 0 specifies the right hand side of the fifth equation of the chain,

etc. Although we know no direct way to show that this recursive procedure works in

general, there exists an alternative direct approach to the reconstruction of a chain

from the function g(u1, u2, u3). To illustrate this procedure we consider a simple

example g = u3 12(u1)2 which automatically satisfies (104). The corresponding h,as specified by (103), is given by h = + u1 + u2 + u3 + u4 u1u2, which canbe reduced to a canonical form h = u4 u1u2 by redefining u4 appropriately (thistransformation freedom allows one to absorb arbitrary integration constants arising


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at each step of the construction). Thus, the first three equations of the chain are

u1t = u2x, u

2t = (u

3

1

2

(u1)2)x, u3t = (u

4

u1u2)x,..., (115)

etc. Equations (115) are nothing but the first three equations in the conservative

representation of the Benney chain (63). The recostruction of the remaining equa-

tions of the chain consists of three steps.

(i) One introduces the corresponding PDE (105). To find the required PDE we

proceed as follows, introduce the commuting chain:

u1y = u3x, u

2y = (116)

substitute (116) into (115) to eliminate u3x, thus obtaining two equations depending

only on u1, u2, that depend on independent variables x, t, y,

u1t = u2x,

u2t = u1y u1u1x. (117)

Make the substitution u1 = vx, u2 = vt, in (117), thus obtaining

vtt = vxy 12

v2xx, (118)

make the substitution u = vx to obtain the dispersionless KP equation,

utt = uxy 12

u2xx. (119)

It was demonstrated in [17] that the general PDE (105) is integrable by the method

of hydrodynamic reductions if and only if the function g satisfies the relations (104).


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(ii) One constructs a dispersionless Lax pair for the PDE (105), the procedure is as

follows. Look for functions R and Q such that,

t = R(uxx, uxy, uxt, uyy , uyt, x), (120)

y = Q(uxx, uxy, uxt, uyy , uyt, x) (121)

Differentiate both equations w.r.t x and then make the substitution x = p, thus

we have

pt = Rx, (122)

py = Qx (123)

the consistency conditions pty = pyt are satisfied identically modulo (105), or more

specifically in our example (119). The existence of such Lax pairs was established

in [47, 17] for any equation (105) provided g satisfies the compatibility conditions

(104). In our example it takes the form

pt = (1

2p2 + uxx)x, py = (

1

3p3 + uxxp + uxt)x;

(iii) One looks for p as an expansion in the auxiliary parameter ,

p = u1

u

2

2 u

3

3 ....; (124)

the substitution of this ansatz into the first equation of the Lax pair implies an

infinite hydrodynamic chain for the variables ui. The first three equations of this

chain identically coincide with (115). The substitution into the second equation of

the Lax pair produces a commuting chain (one has to set u1 = uxx, u2 = uxt). Both

chains possess infinitely many hydrodynamic reductions since this is the case for


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the generating equation (105). Thus, the Haantjes tensor will automatically vanish.

Notice that (124) does not work in general, and a more complicated dependance of

p on is required.

4.3 Integration of the Egorov case

Now let us return to the work in question, solving the system (104). To explicitly

calculate g(u1, u2, u3) we will follow [47]. The main observation is that the first six

equations in (104) imply that the function 1/g3 is linear, 1/g3 = +u1 + u2 +u3.

If

= 0 then, up to a linear change of variables, one can assume that 1/g3 = u

3.

Similarly, if = 0, = 0, one can set 1/g3 = u2. If = = 0, = 0 one has1/g3 = u

1. The last possibility is 1/g3 = 1. Thus, we have four cases to consider:

g = u3 +p(u1, u2), g =u3

u1+p(u1, u2), g =

u3

u2+p(u1, u2), g = ln u3 +p(u1, u2);

here the function p(u1, u2) can be recovered after the substitution into the remaining

four equations (104). In each of these cases the resulting equations for p(u1

, u2

)integrate explicitly, see [47], leading to

A Classification of Integrable Hydrodynamic Type Chains Using the Haantjes Tensor

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