On a class of Darboux integrable semidiscrete equations

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R E S E A R C H

Open Access

On a class of Darboux-integrable

semidiscrete equations

Kostyantyn Zheltukhin

1*

_{, Natalya Zheltukhina}

2

_{and Ergun Bilen}

1

*_{Correspondence:}

zheltukh@metu.edu.tr 1_{Department of Mathematics,}

Middle East Technical University, Universiteler Mahallesi, Dumlupinar Bulvari 1, Ankara, 06800, Turkey Full list of author information is available at the end of the article

Abstract

We consider a classiﬁcation problem for Darboux-integrable hyperbolic semidiscrete equations. In particular, we obtain a complete description for a special class of equations admitting four-dimensional characteristicx-rings and two-dimensional characteristicn-rings. For all described equations, the correspondingx- and n-integrals are constructed.

Keywords: semidiscrete equations; Darboux integrability; characteristic rings

1 Introduction

Classification problems play an important role in the study of integrable equations. For classification of hyperbolic equations, it is convenient to define integrability in terms of characteristic rings. The notion of a characteristic ring was introduced by Shabat for in-tegrable hyperbolic equations of exponential type (see [, ]) and then used by Zhiber to study general integrable hyperbolic equations (see [–]). Later, Habibullin extended this notion to the case of semidiscrete and discrete equations (see [–]). For more details on characteristic rings, see survey paper [].

We consider semidiscrete hyperbolic equations that admit nontrivialx- andn-integrals,

so-called Darboux-integrable equations []. It was proved in [] that a semidiscrete hy-perbolic equation is Darboux integrable if and only if its characteristicx- andn-rings are

ﬁnite-dimensional. Description of all equations with characteristicx- andn-rings of

fi-nite dimensions is a very difficult classification problem. The majority of known

Darboux-integrable semidiscrete equations possessx- andn-rings of dimensions not exceeding ﬁve

(see [, , ]). Necessary and suﬃcient conditions for a characteristicx-ring to be four-dimensional were obtained in [] (also see [] for a characterization of ﬁve-four-dimensional characteristicx-rings). In [] the conditions for a two-dimensional characteristicn-ring were obtained. We use these conditions to explicitly derive integrable equations with

four-dimensional characteristicx-rings and two-dimensional characteristicn-rings.

Consider the equation

tx=f(x,t,t,tx), ()

where the functiont(n,x) depends on the discrete variablenand continuous variablex. We use the notationstx=_∂∂_xtandt=t(n+ ,x). It is also convenient to denotet[k]= ∂

k

∂xkt,

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k∈N, andtm=t(n+m,x),m∈Z. It was proved in [] that if equation () has a

four-dimensional characteristicx-ring, then the functionf has the form

f =A(x,t,t)M(x,t,tx) +B(x,t,t)tx+C(x,t,t). ()

In this work, we assume that the functionMdepends only ontxandf does not depend

onx, that is, we study equations of the form

tx=A(t,t)M(tx) +B(t,t)tx+C(t,t). ()

It turns out that we have to consider two cases off linear and nonlinear intx. The results

of our investigation are given in the following theorems.

Theorem  Let f be a linear function of tx.Equation()has a four-dimensional

charac-teristic x-ring and a two-dimensional characcharac-teristic n-ring if and only if

f = γ(t) γ(t)tx–

γ(t)

γ(t)σ(t) +σ(t),

where the functionsγ andσsatisfy either of the relations

γ(t)σ(t)=γ(t)B+Bγ(t)σ(t)

or

γ(t)σ(t)=γ(t)

B+B

γ(t)σ(t)

with arbitrary constants Band B.

Theorem  Let f be a nonlinear function of tx.Equation()has a four-dimensional

char-acteristic x-ring and a two-dimensional charchar-acteristic n-ring if and only if

f =cη(t)η(t)

tx

or f =ce

c(t+t)

tx+P

–P,

where c,c,and P are arbitrary constants,andηis an arbitrary function of one variable,

or

f =√B_{– }_t

x+Ptx+Q+Btx+



P(B– ),

where B,P,and Q are arbitrary constants.

The paper is organized as follows. First, we give proofs of Theorems  and , and in the

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2 Proofs of Theorem 1 and Theorem 2 2.1 Preliminary results

In what follows, all calculations are done on the set of solutions of equation (), that is, we consider . . . ,t–,t,t, . . . andtx,txx,txxx, . . . as independent dynamical variables. The

derivatives of . . . ,t–,t,t, . . . and shifts oftx,txx,txxx, . . . are expressed in terms of the

dy-namical variables using ().

Let us formulate necessary and suﬃcient conditions so that equation () has a

charac-teristicx-ring of dimension four and a characteristicn-ring of dimension two. First, we

consider then-ring. The following theorem was proved in [].

Theorem  Equation()has a characteristic n-ring of dimension two if and only if

D

We remark that equality () implies that

∂

sinceftdoes not depend ont. We use this observation later.

For the characteristicx-ring, we have to consider two cases:ftxtx= , that is,f is a linear

function oftx, andftxtx= , that is,f is a nonlinear function oftx.

The following theorems were proved in [].

Theorem  Equation()with ftxtx= has a characteristic ring Lxof dimension four if and

where K is the vector ﬁeld

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In the same way as in equation (), we have

∂

∂tm=  and

∂

∂tm˜ = . ()

For convenience of the reader, let us give deﬁnitions of x- and n-integrals and of

Darboux-integrable semidiscrete equations.

Deﬁnition  A functionF(x,t,t, . . . ,tk) is called anx-integral of equation () if

DxF(x,t,t, . . . ,tk) = 

for all solutions of (). HereDxis the operator of total diﬀerentiation with respect tox:

Dxg(n,x) = (d/dx)g(n,x).

A functionG(x,t,tx, . . . ,t[m]) is called ann-integral of equation () if

DG(x,t,tx, . . . ,t[m]) =G(x,t,tx, . . . ,t[m])

for all solutions of ().

Equation () is called Darboux integrable if it admits a nontrivialx-integral and a

non-trivialn-integral.

2.2 Proof of Theorem 1

We assume thatf is a linear function oftx. Thus

f(t,t,tx) =A(t,t)tx+B(t,t), ()

and equation () becomes

tx=A(t,t)tx+B(t,t). ()

The proof of the Theorem  is based on the following lemmas.

Lemma  Let ftxtx= .Then the characteristic n-ring of equation()has dimension two

if and only if

f(t,t,tx) =

cγ(t) γ(t)tx+

c

γ(t)–

cγ(t)σ(t)

γ(t) +σ(t), ()

whereγ andσare functions of one variable,and c,care constants.

Proof It follows from condition () that

ft

ftx

=At

Atx+ Bt

A ()

does not depend ont. Hence At

A and Bt

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form

f(t,t,tx) =γ(t)ϕ(t)tx+l(t)ϕ(t) +σ(t). ()

Applying condition () tof given by (), we get

γ(t) γ(t)

γ(t)ϕ(t)tx+l(t)ϕ(t) +σ(t)

+l

₍_t₎

γ(t)

= –γ(t)ϕ(t)tx–l(t)ϕ(t) –σ(t). ()

By comparing the coeﬃcients oftxin () we get

γ(t) γ(t) +

ϕ(t) ϕ(t) = ,

so thatϕ(t) = c

γ(t), wherecis some constant. Substituting thisϕinto equation () and collecting the terms independent oftx, we get

γ(t)σ(t) +γ(t)σ(t) +l(t) = . ()

Solving (), we ﬁnd

l(t) = –γ(t)σ(t) +˜c, ()

wherecis some constant. Substitutingϕandlfound into equation (), we get equation

(). We can check that condition () is satisﬁed for function ().

Now we can rewrite equation () as

tx=

cγ(t) γ(t)

tx+

c

γ(t)

–cγ(t)σ(t) γ(t)

+σ(t), ()

whereγ andσare functions of one variable, andc,care constants. The equation can be

simpliﬁed by introducing the new variable

τ=L(t), ()

whereLsatisﬁesL(t) =γ(t). Equation () becomes

τx=cτx+c–cQ(τ) +Q(τ) ()

for some functionQof one variable. We can check that condition () is satisﬁed for the new

equation. Hence our change of variable does not aﬀect the dimension of the characteristic

n-ring.

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Lemma  Equation()has a four-dimensional characteristic x-ring if and only if

Q(τ) =Aτ+Aτ or Q(τ) =Aeατ+Ae–ατ ()

for some constants A,A,andα.

Proof Applying condition () to functionf=cτx+c–cQ(τ) +Q(τ), we get

cτx+c–cQ(τ) +Q(τ) Q(τ) –Q(τ)

cQ(τ) –Q(τ)

+Q

₍_τ

)(c–cQ(τ) +Q(τ)) Q(τ) –Q(τ)

–Q(τ) +Q(τ)

=cQ

₍_τ

) –Q(τ) Q(τ) –Q(τ)

τx+

Q(τ)(c–cQ(τ) +Q(τ)) Q(τ) –Q(τ)

+Q(τ) –Q(τ).

By comparing the coeﬃcients ofτxin this equality, we get

cD

cQ(τ) –Q(τ) Q(τ) –Q(τ)

=cQ

₍_τ

) –Q(τ) Q(τ) –Q(τ)

,

which implies that eitherc=  and cQ ₍_τ

)–Q(τ)

Q(τ)–Q(τ) is constant orcQ(τ) –Q(τ) = . In the

second case, we also getc= . Thus, equation () has the form

τx=τx+d(τ,τ). ()

Equations of this form were completely classiﬁed in [] (together with theirx– andn

-characteristic rings). It follows from [] thatQmust have the form given in the statement

of the lemma.

Returning to the original variabletin equation () withQgiven by equation (), we

get Theorem .

2.3 Proof of Theorem 2

We assume thatf is a nonlinear function oftx. Thus

f(t,t,tx) =A(t,t)M(tx) +B(t,t)tx+C(t,t), ()

and equation () becomes

tx=A(t,t)M(tx) +B(t,t)tx+C(t,t). ()

The proof of the Theorem  is based on the following lemmas.

Lemma  Let equation()have a characteristic n-ring of dimension two,and let M be a nonlinear function.Then the function M satisﬁes

M= –αM+αtx+α αM+αtx+α

, ()

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Proof If the dimension of the characteristicn-ring is two, then (ft

f_tx)t= . Hence, forf given by equation (), we have

AtM+txBt+Ct

AM+B

t

=(AttM+txBtt+Ctt)(AM+B) – (AtM+Bt)(AtM+txBt+Ct)

(AM+B) = .

This can be rewritten as

M(αM+αtx+α) = –(αM+αtx+α) ()

for some constantsαi,i= , , . . . , . Note that ifαM+αtx+α= , then eitherM=  or

αM+αtx+α= . In both cases, we get thatf is a linear function oftx. Hence we can

assume thatαM+αtx+α= , and we can write equality ().

The above lemma allows us to express the derivativeMin terms ofM. We can also

express the shiftDMin terms ofM. Indeed, as it was proved in [] (see Lemma ), if

equation () has a four-dimensional characteristicx-ring andftxtx= , then

Df = –H(t,t,t)tx+H(t,t,t)f+H(t,t,t) ()

for some functionsH,H, andH. Therefore,

D(AM+Btx+C) = –Htx+H(AM+Btx+C), ()

and

DM=Q(t,t,t)M+Q(t,t,t)tx+Q(t,t,t) ()

for some functionsQ,QandQ.

We use expressions () and () for the derivative and shift ofMin the next lemma.

Lemma  Let equation()have a characteristic n-ring of dimension two.Then M has either of the forms M= 

tx+P,or M=

t

x+Ptx+Q,or M=tx.

Proof Consider the vector ﬁeld X= _∂∂_t

x. We can easily check that DX =



f_txXD. Thus

DX(M) =_f

txX(DM). Using equation () forX(M) and equation () forDM, we get

–D

αM+αtx+α

αM+αtx+α

= 

AM+BX(QM+Qtx+Q).

Using equation () and equation () once more, we get

–α˜(QM+Qtx+Q) +α˜(AM+Btx+C) +α˜

˜

α(QM+Qtx+Q) +α˜(AM+Btx+C) +α˜

=Q–Q

αM+αtx+α

αM+αtx+α

B–AαM+αtx+α

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whereDαk=α˜k. Hence we can write

RM– (Rtx+R)M+

Rt_x+Rtx+R

=  ()

for some functionsRk,k= , , . . . , . Then, we ﬁnd that

M=(Rtx+R)±

(Rtx+R)– R(Rtx+Rtx+R)

R ifR= 

or

M=Rt



x+Rtx+R

Rtx+R

ifR= .

Since the functionf =AM+Btx+Chas a linear termBtxand a free termC, we can assume

thatMhas the form given in the statement of the lemma.

Now we consider each value ofMobtained in the lemma, separately. We start with the

simple caseM=t_x.

Lemma  Equation()cannot have a four-dimensional characteristic x-ring if M=t

x.

Proof We can easily check that, for any

f =A(t,t)tx+B(t,t)tx+C(t,t),

condition () is not satisﬁed. Hence equation () cannot have a four-dimensional

char-acteristicx-ring ifM=t_x.

Let us consider the caseM=_t

x+P.

Lemma  Let M=_t

x+P,and let equation()have a four-dimensional characteristic

x-ring and a two-dimensional characteristic n-x-ring.Then equation()takes either of the forms

tx=

c∗η(t)η(t)

tx

or tx=

c∗ec∗∗(t+t)

tx+P

–P. ()

Proof We have

f(t,t,tx) =

A(t,t)

tx+P

+B(t,t)tx+C(t,t). ()

Applying condition () tof, we get

(tx+P)

B(tx+P)+ (C+P–BP)(tx+P) +A

=(tx+P)(B(tx+P)

₊_A₎

(B(tx+P)–A)

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From this equality we get

Using condition (), we get

At(t,t)A(t,t)

From condition () it follows that

˜

some constantc∗∗. Thus we obtain equations (). We can easily check that these

equa-tions have a two-dimensional characteristicn-ring and a four-dimensional characteristic

x-ring.

Let us consider the caseM=t

x+Ptx+Q.

Lemma  Let M=t

x+Ptx+Q,and let equation()have a four-dimensional

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Proof We have

f =A(t,t)

t

x+Ptx+Q+B(t,t)tx+C(t,t). ()

Applying condition (), we ﬁnd

–BP+ tx+ A

Comparing the coeﬃcients of (t

x+Ptx+Q)i(tx)jfori,j= , , , we get

We can check that these equalities are satisﬁed if and only if

C=PB–P and C+ CP=AP– Q– Q+ BQ.

satisﬁed for this functionf if and only if

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Hence we can write

Bt(t,t) =A(t,t)ϕ(t),Bt(t,t) =±A(t,t)ϕ(t) ()

for some functionϕ.

Using condition (), let us show thatBcan only be a constant function. We have

˜

Diﬀerentiating this equality with respect tot, we get

 =CBt((P±Q) + (Q–P)B)

Now we consider the caseP= . Then, using equation (), we have

μ=

Diﬀerentiating this equality with respect tot, we get

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B=±_φφ₍(_tt)

), thenμ=±

ϕ₍_t_{) –}_ϕ₍_t_{), and since}_μ

does not depend ont, we get thatφ

is a constant function, and henceBis a constant function. Using the equalityB₌_A_{+ ,}

we get the statement of the lemma.

The proof of Theorem  easily follows from the above lemmas.

3 Examples

The functionsf given in the Theorem  lead to the following examples.

Example  The equation

where functionsγ andσsatisfy the relation

The functionsf given in the Theorem  lead to the following examples.

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Example  The equation

In all examples, we can check thatF is an x-integral andI is ann-integral by direct

calculations.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors contributed equally to this work. All authors read and approved the ﬁnal manuscript.

Author details

1_{Department of Mathematics, Middle East Technical University, Universiteler Mahallesi, Dumlupinar Bulvari 1, Ankara,}

06800, Turkey.2_{Department of Mathematics, Bilkent University, Bilkent, Ankara, 06800, Turkey.}

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Received: 27 January 2017 Accepted: 13 June 2017 References

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