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# On a class of Darboux-integrable semidiscrete equations

*Advances in Difference Equations*
**volume 2017**, Article number: 182 (2017)

## 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 characteristic *x*-rings and two-dimensional characteristic *n*-rings. For all described equations, the corresponding *x*- and *n*-integrals are constructed.

## 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 integrable hyperbolic equations of exponential type (see [1, 2]) and then used by Zhiber to study general integrable hyperbolic equations (see [3–7]). Later, Habibullin extended this notion to the case of semidiscrete and discrete equations (see [8–16]). For more details on characteristic rings, see survey paper [17].

We consider semidiscrete hyperbolic equations that admit nontrivial *x*- and *n*-integrals, so-called Darboux-integrable equations [18]. It was proved in [9] that a semidiscrete hyperbolic equation is Darboux integrable if and only if its characteristic *x*- and *n*-rings are finite-dimensional. Description of all equations with characteristic *x*- and *n*-rings of finite dimensions is a very difficult classification problem. The majority of known Darboux-integrable semidiscrete equations possess *x*- and *n*-rings of dimensions not exceeding five (see [14, 16, 19]). Necessary and sufficient conditions for a characteristic *x*-ring to be four-dimensional were obtained in [20] (also see [21] for a characterization of five-dimensional characteristic *x*-rings). In [12] the conditions for a two-dimensional characteristic *n*-ring were obtained. We use these conditions to explicitly derive integrable equations with four-dimensional characteristic *x*-rings and two-dimensional characteristic *n*-rings.

Consider the equation

where the function \(t(n,x)\) depends on the discrete variable *n* and continuous variable *x*. We use the notations \(t_{x}=\frac{\partial}{\partial x} t\) and \(t_{1}=t(n+1,x)\). It is also convenient to denote \(t_{[k]}=\frac{\partial^{k} }{\partial x^{k}} t\), \(k\in{\mathbb {N}}\), and \(t_{m}=t(n+m,x)\), \(m\in{\mathbb {Z}}\). It was proved in [20] that if equation (1) has a four-dimensional characteristic *x*-ring, then the function *f* has the form

In this work, we assume that the function *M* depends only on \(t_{x}\) and *f* does not depend on *x*, that is, we study equations of the form

It turns out that we have to consider two cases of *f* linear and nonlinear in \(t_{x}\). The results of our investigation are given in the following theorems.

### Theorem 1

*Let*
*f*
*be a linear function of*
\(t_{x}\). *Equation* (3) *has a four*-*dimensional characteristic*
*x*-*ring and a two*-*dimensional characteristic*
*n*-*ring if and only if*

*where the functions*
*γ*
*and*
*σ*
*satisfy either of the relations*

*or*

*with arbitrary constants*
\(B_{1}\)
*and*
\(B_{2}\).

### Theorem 2

*Let*
*f*
*be a nonlinear function of*
\(t_{x}\). *Equation* (3) *has a four*-*dimensional characteristic*
*x*-*ring and a two*-*dimensional characteristic*
*n*-*ring if and only if*

*where*
\(c_{1}\), \(c_{2}\), *and*
*P*
*are arbitrary constants*, *and*
*η*
*is an arbitrary function of one variable*, *or*

*where*
*B*, *P*, *and*
*Q*
*are arbitrary constants*.

The paper is organized as follows. First, we give proofs of Theorems 1 and 2, and in the last section, we provide *x*- and *n*- integrals for equations found in Theorems 1 and 2.

## Proofs of Theorem 1 and Theorem 2

### Preliminary results

In what follows, all calculations are done on the set of solutions of equation (1), that is, we consider \(\dots, t_{-1}, t_{0}, t_{1}, \dots\) and \(t_{x}, t_{xx}, t_{xxx}, \dots\) as independent dynamical variables. The derivatives of \(\dots, t_{-1}, t_{0}, t_{1}, \dots\) and shifts of \(t_{x}, t_{xx}, t_{xxx}, \dots\) are expressed in terms of the dynamical variables using (1).

Let us formulate necessary and sufficient conditions so that equation (3) has a characteristic *x*-ring of dimension four and a characteristic *n*-ring of dimension two. First, we consider the *n*-ring. The following theorem was proved in [12].

### Theorem 3

*Equation* (1) *has a characteristic*
*n*-*ring of dimension two if and only if*

*where*
*D*
*is the shift operator*: \(Dg(n,x)=g(n+1,x)\).

We remark that equality (4) implies that

since \(f_{t_{1}}\) does not depend on \(t_{2}\). We use this observation later.

For the characteristic *x*-ring, we have to consider two cases: \(f_{t_{x}t_{x}}= 0\), that is, *f* is a linear function of \(t_{x}\), and \(f_{t_{x}t_{x}}\ne0\), that is, *f* is a nonlinear function of \(t_{x}\).

The following theorems were proved in [20].

### Theorem 4

*Equation* (1) *with*
\(f_{t_{x}t_{x}}= 0\)
*has a characteristic ring*
\(L_{x}\)
*of dimension four if and only if*

*where*
*K*
*is the vector field*

*and*
\(m= {\frac {-(f_{xt_{x}}+t_{x}f_{t_{x}t}+ff_{t_{x}t_{1}})+f_{t}+f_{t_{x}}f_{t_{1}}}{f_{t_{x}}}}\).

### Theorem 5

*Equation* (1) *with*
\(f_{t_{x}t_{x}}\ne0\)
*has a characteristic ring*
\(L_{x}\)
*of dimension four if and only if*

*and*

*where*
\(\tilde{m}= \frac {f_{xt_{x}}+t_{x}f_{t_{x}t}+ff_{t_{x}t_{1}}-f_{t}-f_{t_{x}}f_{t_{1}}}{f_{t_{x}t_{x}}}\).

In the same way as in equation (5), we have

For convenience of the reader, let us give definitions of *x*- and *n*-integrals and of Darboux-integrable semidiscrete equations.

### Definition 6

A function \(F(x,t,t_{1},\dots,t_{k})\) is called an *x*-integral of equation (1) if

for all solutions of (1). Here \(D_{x}\) is the operator of total differentiation with respect to *x*: \(D_{x}g(n,x)=(d/dx)g(n,x)\).

A function \(G(x,t,t_{x},\dots,t_{[m]})\) is called an *n*-integral of equation (1) if

for all solutions of (1).

Equation (1) is called Darboux integrable if it admits a nontrivial *x*-integral and a nontrivial *n*-integral.

### Proof of Theorem 1

We assume that *f* is a linear function of \(t_{x}\). Thus

and equation (1) becomes

The proof of the Theorem 1 is based on the following lemmas.

### Lemma 7

*Let*
\(f_{t_{x} t_{x}}=0\). *Then the characteristic*
*n*-*ring of equation* (11) *has dimension two if and only if*

*where*
*γ*
*and*
*σ*
*are functions of one variable*, *and*
\(c_{1}\), \(c_{2}\)
*are constants*.

### Proof

It follows from condition (4) that

does not depend on \(t_{1}\). Hence \(\frac{A_{t}}{A}\) and \(\frac{B_{t}}{A}\) do not depend on \(t_{1}\). So we can write \(A(t,t_{1}) = \gamma(t)\varphi(t_{1})\) and \(B(t,t_{1}) = l(t)\varphi(t_{1}) + \sigma(t_{1})\) for some functions *γ*, *φ*, and *σ*. The function *f* takes form

Applying condition (4) to *f* given by (14), we get

By comparing the coefficients of \(t_{x}\) in (15) we get

so that \(\varphi(t_{1}) = \frac{c_{1}}{\gamma(t_{1})}\), where \(c_{1}\) is some constant. Substituting this *φ* into equation (15) and collecting the terms independent of \(t_{x}\), we get

Solving (16), we find

where \(c_{2}\) is some constant. Substituting *φ* and *l* found into equation (14), we get equation (12). We can check that condition (4) is satisfied for function (12). □

Now we can rewrite equation (11) as

where *γ* and *σ* are functions of one variable, and \(c_{1}\), \(c_{2}\) are constants. The equation can be simplified by introducing the new variable

where *L* satisfies \(L'(t) = \gamma(t)\). Equation (18) becomes

for some function *Q* of one variable. We can check that condition (4) is satisfied for the new equation. Hence our change of variable does not affect the dimension of the characteristic *n*-ring.

In the next lemma, we give conditions for equation (20) to have a four-dimensional characteristic *x*-ring.

### Lemma 8

*Equation* (20) *has a four*-*dimensional characteristic*
*x*-*ring if and only if*

*for some constants*
\(A_{1}\), \(A_{2}\), *and α*.

### Proof

Applying condition (6) to function \(f=c_{1}\tau_{x} + c_{2} - c_{1}Q(\tau) + Q(\tau_{1})\), we get

By comparing the coefficients of \(\tau_{x}\) in this equality, we get

which implies that either \(c_{1}=1\) and \({\frac{c_{1}Q''(\tau _{1})-Q''(\tau)}{Q'(\tau_{1})-Q'(\tau)}}\) is constant or \(c_{1}Q''(\tau_{1})-Q''(\tau)=0\). In the second case, we also get \(c_{1}=1\). Thus, equation (20) has the form

Equations of this form were completely classified in [14] (together with their *x*− and *n*-characteristic rings). It follows from [14] that *Q* must have the form given in the statement of the lemma. □

Returning to the original variable *t* in equation (20) with *Q* given by equation (21), we get Theorem 1.

### Proof of Theorem 2

We assume that *f* is a nonlinear function of \(t_{x}\). Thus

and equation (1) becomes

The proof of the Theorem 2 is based on the following lemmas.

### Lemma 9

*Let equation* (24) *have a characteristic*
*n*-*ring of dimension two*, *and let*
*M*
*be a nonlinear function*. *Then the function*
*M*
*satisfies*

*where*
\(\alpha_{1}M + \alpha_{3}t_{x} + \alpha_{5} \ne0\).

### Proof

If the dimension of the characteristic *n*-ring is two, then \(( \frac{f_{t}}{f_{t_{x}}} )_{t_{1}} = 0\). Hence, for *f* given by equation (23), we have

This can be rewritten as

for some constants \(\alpha_{i}\), \(i=1,2,\dots, 5\). Note that if \(\alpha_{1}M + \alpha_{3}t_{x} + \alpha_{5} = 0\), then either \(M'=0\) or \(\alpha_{2}M + \alpha_{4}t_{x}+ \alpha_{6} = 0\). In both cases, we get that *f* is a linear function of \(t_{x}\). Hence we can assume that \(\alpha_{1}M + \alpha_{3}t_{x} + \alpha_{5} \neq0\), and we can write equality (25). □

The above lemma allows us to express the derivative \(M'\) in terms of *M*. We can also express the shift *DM* in terms of *M*. Indeed, as it was proved in [20] (see Lemma 12), if equation (1) has a four-dimensional characteristic *x*-ring and \(f_{t_{x}t_{x}} \neq0\), then

for some functions \(H_{1}\), \(H_{2}\), and \(H_{3}\). Therefore,

and

for some functions \(Q_{1}\), \(Q_{2}\) and \(Q_{3}\).

We use expressions (25) and (29) for the derivative and shift of *M* in the next lemma.

### Lemma 10

*Let equation* (24) *have a characteristic*
*n*-*ring of dimension two*. *Then*
*M*
*has either of the forms*
\(M = \frac{1}{t_{x} + P}\), *or*
\(M = \sqrt{t_{x}^{2} + Pt_{x} + Q}\), *or*
\(M = t_{x}^{2}\).

### Proof

Consider the vector field \(X= \frac{\partial }{\partial t_{x}}\). We can easily check that \(DX = \frac{1}{f_{t_{x}}}XD\). Thus \(DX(M) =\frac{1}{f_{t_{x}}}X(DM)\). Using equation (25) for \(X(M)\) and equation (29) for *DM*, we get

Using equation (25) and equation (29) once more, we get

where \(D\alpha_{k} =\tilde{ \alpha_{k}}\). Hence we can write

for some functions \(R_{k}\), \(k=1,2,\dots, 6\). Then, we find that

or

Since the function \(f=AM+Bt_{x}+C\) has a linear term \(Bt_{x}\) and a free term *C*, we can assume that *M* has the form given in the statement of the lemma. □

Now we consider each value of *M* obtained in the lemma, separately. We start with the simple case \(M =t_{x}^{2}\).

### Lemma 11

*Equation* (24) *cannot have a four*-*dimensional characteristic*
*x*-*ring if*
\(M=t_{x}^{2}\).

### Proof

We can easily check that, for any

condition (7) is not satisfied. Hence equation (24) cannot have a four-dimensional characteristic *x*-ring if \(M=t_{x}^{2}\). □

Let us consider the case \(M = {\frac{1}{t_{x} + P}}\).

### Lemma 12

*Let*
\(M = {\frac{1}{t_{x} + P}}\), *and let equation* (24) *have a four*-*dimensional characteristic*
*x*-*ring and a two*-*dimensional characteristic*
*n*-*ring*. *Then equation* (24) *takes either of the forms*

### Proof

We have

Applying condition (7) to *f*, we get

From this equality we get

In this equality the coefficients of \((t_{x} + P)^{k}\), \(k =1,2,3\), must be zero. So we find

Since \(A(t, t_{1}) \neq0\) (otherwise \(f_{t_{x}t_{x}} = 0\)), we find \(B = 0\) and \(C = -P\). Thus we have

Using condition (4), we get

or

It follows that \(\frac {A_{t_{1}}(t_{1},t_{2})}{A(t_{1},t_{2})}\) does not depend on \(t_{2}\), so \(\frac{\partial^{2}}{\partial t_{2}\partial t_{1}}\ln A(t_{1},t_{2})=0\), and \(\frac {A_{t_{1}}(t,t_{1})}{A(t,t_{1})}\) does not depend on *t*, so \(\frac{\partial^{2}}{\partial t\partial t_{1}}\ln A(t,t_{1})=0\).

Hence we get \(A(t,t_{1}) = \varphi(t)\eta(t_{1})\) for some functions *φ* and *η*. Using equation (34), we obtain \(\frac{\varphi '(t_{1})}{\varphi(t_{1})} = \frac{\eta'(t_{1})}{\eta (t_{1})}\), which implies that \(\varphi(t_{1}) = c^{*}\eta(t_{1})\), where \(c^{*}\) is some constant. Hence we have

From condition (8) it follows that

does not depend on \(t_{1}\). So, either \(P = 0\) or \(\eta'(t_{1}) = c^{**}\eta(t_{1})\), which implies \(\eta (t_{1}) = e^{c^{**}t_{1}}\) with some constant \(c^{**}\). Thus we obtain equations (31). We can easily check that these equations have a two-dimensional characteristic *n*-ring and a four-dimensional characteristic *x*-ring. □

Let us consider the case \(M = \sqrt{t_{x}^{2} + Pt_{x} + Q}\).

### Lemma 13

*Let*
\(M = \sqrt{t_{x}^{2} + Pt_{x} + Q}\), *and let equation* (24) *have a four*-*dimensional characteristic*
*x*-*ring and a two*-*dimensional characteristic*
*n*-*ring*. *Then equation* (24) *takes the form*

*where*
*B*, *Q*, *P*
*are constants*, *and*
\(B^{2}\ne1\).

### Proof

We have

Applying condition (7), we find

or

Comparing the coefficients of \((\sqrt{t_{x}^{2} + Pt_{x} +Q} )^{i} (t_{x} )^{j}\) for \(i,j =0,1,2\), we get

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

Simplifying, we get

In the case \(P^{2}=4Q\), we have that \(M=\sqrt{t_{x}^{2}+Pt_{x}+Q}\) is a linear function of \(t_{x}\). Therefore we have to study only the case \(B^{2}=A^{2} + 1\). Thus we have

where \(B\ne1\). In the same way, we check that condition (4) in the form \((D \frac{f_{t}}{f_{t_{x}}} )^{2} - (f_{t_{1}})^{2}=0\) is satisfied for this function *f* if and only if

Hence we can write

for some function *φ*.

Using condition (8), let us show that *B* can only be a constant function. We have

where

Since *m̃* does not depend on \(t_{1}\), we have that \(\mu_{1}\), \(\mu _{2}\), and \(\mu_{3}\) also do not depend on \(t_{1}\). Using (41), we have

Since \(\mu_{2}\) does not depend on \(t_{1}\), either *φ* is a constant function or \(P=0\). Note that in both cases, we get \(\mu_{1}=0\) and \(\mu_{2}=0\). We start with the case where *ϕ* is some constant *C*. Using equation (41), we have

Differentiating this equality with respect to \(t_{1}\), we get

which gives \(B_{t_{1}}=0\) or \(((P^{2} \pm4Q)+(4Q-P^{2})B)=0\). Both equalities imply that *B* is a constant.

Now we consider the case \(P=0\). Then, using equation (41), we have

Differentiating this equality with respect to \(t_{1}\), we get

which implies that either \(B_{t_{1}}=0\) or \(B=\pm\frac{\varphi(t)}{\varphi (t_{1})}\). In both cases, we get that *B* is a constant function. Indeed, if \(B_{t_{1}}=0\), then \(B_{t}=0\) by equation (41), so *B* is a constant function, and if \(B=\pm\frac{\phi(t)}{\phi(t_{1})}\), then \(\mu _{3}=\pm\sqrt{\varphi^{2}(t)-\varphi^{2}(t_{1})}\), and since \(\mu_{3}\) does not depend on \(t_{1}\), we get that *ϕ* is a constant function, and hence *B* is a constant function. Using the equality \(B^{2}=A^{2}+1\), we get the statement of the lemma. □

The proof of Theorem 2 easily follows from the above lemmas.

## Examples

The functions *f* given in the Theorem 1 lead to the following examples.

### Example 14

The equation

where functions *γ* and *σ* satisfy the relation

has an *x*-integral \(F=\frac {(L(t_{3})-L(t_{1}))(L(t_{2})-L(t))}{(L(t_{3})-L(t_{2}))(L(t_{2})-L(t_{1}))} \), where \(L(t)=\int_{0}^{t} \gamma(\tau)\,d\tau\), and an *n*-integral \(I=\gamma (t)t_{x}-\sigma(t)\).

### Example 15

The equation

where functions *γ* and *σ* satisfy the relation

has an *x*-integral \(F=\frac {(e^{L(t)}-e^{L(t_{2})})(e^{L(t_{1})}-e^{L(t_{3})})}{(e^{L(t)}- e^{L(t_{3})})(e^{L(t_{1})}-e^{L(t_{2})})}\), where \(L(t)=\int_{0}^{t} \gamma(\tau)\,d\tau\), and an *n*-integral \(I=\gamma (t)t_{x}-\sigma(t)\).

The functions *f* given in the Theorem 2 lead to the following examples.

### Example 16

The equation

has an *x*-integral \(F=\int_{0}^{t_{3}}\eta^{-1} (\tau) \,d\tau- \int _{0}^{t_{1}}\eta^{-1} (\tau) \,d\tau\) and an *n*-integral \(I= {\frac{t_{x}}{c_{1}\eta(t)}+\frac{\eta(t)}{t_{x}}}\).

### Example 17

The equation

has an *x*-integral \(F=e^{-c_{2}t_{3}+c_{2}Px}-e^{-c_{2}t_{1}+c_{2}Px}\) and an *n*-integral \(I= {\frac{t_{x}+P}{c_{1}e^{c_{2}t}}+\frac {e^{c_{2}t}}{t_{x}+P}}\).

### Example 18

The equation

has an *x*-integral

and an *n*-integral

In all examples, we can check that *F* is an *x*-integral and *I* is an *n*-integral by direct calculations.

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Zheltukhin, K., Zheltukhina, N. & Bilen, E. On a class of Darboux-integrable semidiscrete equations.
*Adv Differ Equ* **2017, **182 (2017). https://doi.org/10.1186/s13662-017-1241-z

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### Keywords

- semidiscrete equations
- Darboux integrability
- characteristic rings