Classification of a Subclass of Two-Dimensional Lattices via Characteristic Lie Rings

The main goal of the article is testing a new classification algorithm. To this end we apply it to a relevant problem of describing the integrable cases of a subclass of two-dimensional lattices.

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Дата:	2017
Автори:	Habibullin, I., Poptsova, M.
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Опубліковано:	Інститут математики НАН України 2017
Назва видання:	Symmetry, Integrability and Geometry: Methods and Applications
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Цитувати:	Classification of a Subclass of Two-Dimensional Lattices via Characteristic Lie Rings / I. Habibullin, M. Poptsova // Symmetry, Integrability and Geometry: Methods and Applications. — 2017. — Т. 13. — Бібліогр.: 29 назв. — англ.

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spelling	irk-123456789-1487602019-02-19T01:25:09Z Classification of a Subclass of Two-Dimensional Lattices via Characteristic Lie Rings Habibullin, I. Poptsova, M. The main goal of the article is testing a new classification algorithm. To this end we apply it to a relevant problem of describing the integrable cases of a subclass of two-dimensional lattices. 2017 Article Classification of a Subclass of Two-Dimensional Lattices via Characteristic Lie Rings / I. Habibullin, M. Poptsova // Symmetry, Integrability and Geometry: Methods and Applications. — 2017. — Т. 13. — Бібліогр.: 29 назв. — англ. 1815-0659 2010 Mathematics Subject Classification: 37K10; 37K30; 37D99 DOI:10.3842/SIGMA.2017.073 http://dspace.nbuv.gov.ua/handle/123456789/148760 en Symmetry, Integrability and Geometry: Methods and Applications Інститут математики НАН України
institution	Digital Library of Periodicals of National Academy of Sciences of Ukraine
collection	DSpace DC
language	English
description	The main goal of the article is testing a new classification algorithm. To this end we apply it to a relevant problem of describing the integrable cases of a subclass of two-dimensional lattices.
format	Article
author	Habibullin, I. Poptsova, M.
spellingShingle	Habibullin, I. Poptsova, M. Classification of a Subclass of Two-Dimensional Lattices via Characteristic Lie Rings Symmetry, Integrability and Geometry: Methods and Applications
author_facet	Habibullin, I. Poptsova, M.
author_sort	Habibullin, I.
title	Classification of a Subclass of Two-Dimensional Lattices via Characteristic Lie Rings
title_short	Classification of a Subclass of Two-Dimensional Lattices via Characteristic Lie Rings
title_full	Classification of a Subclass of Two-Dimensional Lattices via Characteristic Lie Rings
title_fullStr	Classification of a Subclass of Two-Dimensional Lattices via Characteristic Lie Rings
title_full_unstemmed	Classification of a Subclass of Two-Dimensional Lattices via Characteristic Lie Rings
title_sort	classification of a subclass of two-dimensional lattices via characteristic lie rings
publisher	Інститут математики НАН України
publishDate	2017
url	http://dspace.nbuv.gov.ua/handle/123456789/148760
citation_txt	Classification of a Subclass of Two-Dimensional Lattices via Characteristic Lie Rings / I. Habibullin, M. Poptsova // Symmetry, Integrability and Geometry: Methods and Applications. — 2017. — Т. 13. — Бібліогр.: 29 назв. — англ.
series	Symmetry, Integrability and Geometry: Methods and Applications
work_keys_str_mv	AT habibullini classificationofasubclassoftwodimensionallatticesviacharacteristiclierings AT poptsovam classificationofasubclassoftwodimensionallatticesviacharacteristiclierings
first_indexed	2025-07-12T20:10:56Z
last_indexed	2025-07-12T20:10:56Z
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fulltext	Symmetry, Integrability and Geometry: Methods and Applications SIGMA 13 (2017), 073, 26 pages Classification of a Subclass of Two-Dimensional Lattices via Characteristic Lie Rings Ismagil HABIBULLIN †‡ and Mariya POPTSOVA † † Ufa Institute of Mathematics, 112 Chernyshevsky Str., Ufa 450008, Russia E-mail: habibullinismagil@gmail.com E-mail: mnpoptsova@gmail.com ‡ Bashkir State University, 32 Validy Str., Ufa 450076, Russia Received March 30, 2017, in final form August 24, 2017; Published online September 07, 2017 https://doi.org/10.3842/SIGMA.2017.073 Abstract. The main goal of the article is testing a new classification algorithm. To this end we apply it to a relevant problem of describing the integrable cases of a subclass of two- dimensional lattices. By imposing the cut-off conditions u−1 = c0 and uN+1 = c1 we reduce the lattice un,xy = α(un+1, un, un−1)un,xun,y to a finite system of hyperbolic type PDE. Assuming that for each natural N the obtained system is integrable in the sense of Darboux we look for α. To detect the Darboux integrability of the hyperbolic type system we use an algebraic criterion of Darboux integrability which claims that the characteristic Lie rings of such a system must be of finite dimension. We prove that up to the point transformations only one lattice in the studied class passes the test. The lattice coincides with the earlier found Ferapontov–Shabat–Yamilov equation. The one-dimensional reduction x = y of this lattice passes also the symmetry integrability test. Key words: two-dimensional integrable lattice; cut-off boundary condition; open chain; Darboux integrable system; characteristic Lie ring 2010 Mathematics Subject Classification: 37K10; 37K30; 37D99 1 Introduction In the present article we study the classification problem for the following class of two-dimen- sional lattices un,xy = α(un+1, un, un−1)un,xun,y. (1.1) Here the sought function u = un(x, y) depends on real x, y and on integer n. Function α = α(un+1, un, un−1) is assumed to be analytical in a domain D ⊂ C3. We request also that the derivatives ∂α(un+1,un,un−1) ∂un+1 and ∂α(un+1,un,un−1) ∂un−1 do not vanish identically. Constraint un0 = c0 where c0 is a constant parameter defines a boundary condition which cuts off the lattice (1.1) into two independent semi-infinite lattices un,xy = α(un+1, un, un−1)un,xun,y, for n > n0 (n < n0), un0 = c0. (1.2) Any solutions of the lattice located on the semiaxis n > n0 does not depend on the solutions of that located on n < n0 and vice versa. Turning to the general case of the lattices recall that the boundary conditions (or cut-off constraints) having such a property are called degenerate. It is well known that the degenerate boundary conditions are admitted by any integrable nonlinear This paper is a contribution to the Special Issue on Symmetries and Integrability of Difference Equations. The full collection is available at http://www.emis.de/journals/SIGMA/SIDE12.html mailto:habibullinismagil@gmail.com mailto:mnpoptsova@gmail.com https://doi.org/10.3842/SIGMA.2017.073 http://www.emis.de/journals/SIGMA/SIDE12.html 2 I. Habibullin and M. Poptsova lattice. They are compatible with the whole hierarchy of the higher symmetries [1, 8]. In the literature they are met in the connection with the so-called open chains (see, for instance, [15]). Since the symmetry approach which is a powerful classification tool in the dimension 1 + 1 (see, for instance, [2, 11, 13]) loses its efficiency in higher dimensions (an explanation can be found in [14]) it became clear years ago that it is necessary to look for alternative classification algorithms. Since then different approaches to the integrable multidimensional models have been invented (see, for instance, [3, 5, 6, 12, 16, 17, 18, 24]). In 1994 A.B. Shabat posed a problem of creating a classification algorithm by combining the concepts of the degenerate boundary condition, open chain and the characteristic Lie algebra. It is worth mentioning as an important step in this direction the article [19] where the structure of the Lie algebra was described for the two-dimensional Toda lattice. Some progress toward creating the classification method was done in [9]. It was observed that any finitely generated subring of the characteristic Lie ring for the integrable case is of finite dimension. The statement was verified for a large class of the known integrable lattices. Our interest to the Shabat’s problem was stimulated by the success of the method of the hydrodynamic type reductions in the multidimensionality proposed in [5, 6]. State-of-the-art for the subject and the references can be found in [16]. In the present article the lattice (1.1) is used as a touchstone for the created algorithm. Our aim is to explain the core of the method and approve its efficiency by solving a relevant classification problem. Boundary condition of the form (1.2) imposed at two different integers n = N1 and n = N2 (take N1 < N2 − 1) reduces the lattice (1.1) into a finite system of hyperbolic type equations (open chain) uN1 = c1, un,xy = α(un+1, un, un−1)un,xun,y, N1 < n < N2, (1.3) uN2 = c2. Initiated by the article [9], where a large class of two-dimensional lattices is discussed we use the following Definition 1.1. We call the lattice (1.1) integrable if the hyperbolic type system (1.3) obtained from (1.1) by imposing degenerate boundary conditions is Darboux integrable for any choice of the integers N1, N2. Recall that a system (1.3) of the hyperbolic type partial differential equations is Darboux integrable if it admits the complete set of functionally independent integrals in both of x and y directions. Function I of a finite number of the dynamical variables u,ux,uy, . . . is a y-integral if it satisfies the condition DyI = 0, where Dy is the operator of the total derivative with respect to the variable y and u is a vector with the coordinates uN1+1, uN1+2, . . . , uN2−1 coinciding with the field variables. Since the system (1.3) is autonomous we can restrict ourselves by considering only autonomous nontrivial integrals. It can be verified that the y-integral does not depend on uy,uyy, . . .. In what follows we are interested only on nontrivial y-integrals, i.e., integrals containing dependence on at least one dynamical variable u,ux, . . .. Note that currently the Darboux integrable discrete and continuous models are intensively studied (see, [7, 9, 10, 21, 22, 25, 26, 27, 28, 29]). We justify Definition 1.1 by the following reasoning. The problem of finding general solution to the Darboux integrable system is reduced to a problem of solving a system of the ordinary differential equations. Usually these ODE are explicitly solved. On the other hand side any solution to the considered hyperbolic system (1.3) is easily prolonged outside the interval [N1, N2] and generates a solution of the corresponding lattice (1.1). Therefore in this case the lattice (1.1) has a large set of the explicit solutions and is definitely integrable. Classification of a Subclass of Two-Dimensional Lattices via Characteristic Lie Rings 3 Let us briefly discuss on the content of the article. In Section 2 we recall the necessary definitions and study the main properties of the characteristic Lie ring which is a basic implement in the theory of the Darboux integrable systems. The goal of Section 3 consists in deriving some differential equations on the unknown α (it is reasonable to call them integrability conditions) from the finite-dimensionality property of the characteristic Lie ring. To this end we used two test sequences. In Section 4 by summarizing the integrability conditions we found the final form of the searched function α. It is remarkable that two test sequences turned out to be enough to complete the classification. The classification result is formulated in Theorem 5.1 (see Section 5) which claims: any lattice (1.1) integrable in the sense of Definition 1.1 can be reduced by an appropriate point transformation v = p(u) to the following one, found earlier in [4] and, independently, in [20] vn,xy = vn,xvn,y ( 1 vn − vn−1 − 1 vn+1 − vn ) . (1.4) We obtained also a new result concerned to the lattice (1.4) by proving that for any choice of the integer N ≥ 0 the system of the hyperbolic type equations v−1 = c0, vn,xy = vn,xvn,y ( 1 vn − vn−1 − 1 vn+1 − vn ) , (1.5) vN+1 = c1, 0 ≤ n ≤ N admits a complete set of functionally independent x- and y-integrals for any constant parame- ters c0, c1, i.e., is Darboux integrable. This fact follows immediately from Theorem 5.2 proved in Appendix A, which states that the characteristic Lie rings in both characteristic directions x and y for the system (1.5) are of finite dimension. In the particular case when N = 1 for the corresponding system v0,xy = v0,xv0,y ( 1 v0 − c0 − 1 v1 − v0 ) , v1,xy = v1,xv1,y ( 1 v1 − v0 − 1 c1 − v1 ) we give the y- and x-integrals in an explicit form I1 = v0,xv1,x (v0 − c0)(v1 − v0)(c1 − v1) , I2 = v1,xx v1,x + v0,x(v1 − c0) (v0 − c0)(v1 − v0) + 2v1,x c1 − v1 , J1 = v0,yv1,y (v0 − c0)(v1 − v0)(c1 − v1) , J2 = v1,yy v1,y + v0,y(v1 − c0) (v0 − c0)(v1 − v0) + 2v1,y c1 − v1 . 2 Characteristic Lie rings Since the lattice (1.1) is invariant under the shift of the variable n we can without loss of generality take N1 = −1 and concentrate on the system u−1 = c0, un,xy = αnun,xun,y, 0 ≤ n ≤ N, (2.1) uN+1 = c1. Here αn = α(un−1, un, un+1). Assume that system (2.1) is Darboux integrable and that I(u,ux, . . .) is its nontrivial integral. Let us evaluate DyI in the equation DyI = 0 and get due to the chain rule an equation Y I = 0, where Y = N∑ i=0 ( ui,y ∂ ∂ui + fi ∂ ∂ui,x + fi,x ∂ ∂ui,xx + · · · ) . (2.2) 4 I. Habibullin and M. Poptsova Here fi = αiui,xui,y. Since the coefficients of the equation Y I = 0 depend on ui,y while its solution I does not depend on them we have a system of several linear equations for one unknown I Y I = 0, XjI = 0, j = 1, . . . , N, (2.3) with Xi = ∂ ∂ui,y . It follows from (2.3) that for ∀ i the operator Yi = [Xi, Y ] = XiY − Y Xi also annihilates I. Let us give the explicit form of the operator Yi Yi = ∂ ∂ui +Xi(fi) ∂ ∂ui,x +Xi(Difi) ∂ ∂ui,xx + · · · . Due to the relation Dk xfi = ui,yXi(D k xfi) we represent (2.2) as Y = N∑ i=0 ui,y ( ∂ ∂ui +Xi(fi) ∂ ∂ui,x +Xi(Dxfi) ∂ ∂ui,xx + · · · ) = N∑ i=0 ui,yYi. (2.4) The last equation together with (2.3) implies N∑ i=0 ui,yYiI = 0. Since the variables ui,y are in- dependent the coefficients of this decomposition all vanish. Now we use the evident relation [Xk, Ys] = 0 valid for ∀ k, s. The condition XiI = 0 is satisfied automatically. Thus we arrive at the statement: function I is a y-integral of the system (2.1) if and only if it solves the following system of equations YiI = 0 for i = 0, 1, . . . , N. (2.5) Consider the set R0(y,N) of all multiple commutators of the characteristic vector fields Y0, Y1, . . . , YN . Denote through R(y,N) the minimal ring containing R0(y,N). We refer to R(y,N) as the characteristic Lie ring of the system (2.1) in y-direction. In a similar way one can define the characteristic Lie ring in the direction of x. Thus we have a complete description of the set of the linear first order partial differential equations the y-integral should satisfy to. Now the task is to find a subset of the linearly independent equations such that all the other equations can be represented as linear combinations of those ones. We say that the ring R(y,N) is of finite dimension if there exists a finite subset {Z1, Z2, . . . , ZL} ⊂ R(y,N) which defines a basis in R(y,N) such that 1) every element Z ∈ R(y,N) is represented in the form Z = λ1Z1 + · · · + λLZL with the coefficients λ1, . . . , λL which might depend on a finite number of the dynamical variables, 2) relation λ1Z1 + · · ·+ λLZL = 0 implies that λ1 = · · · = λL = 0. Let us formulate now an effective algebraic criterion (see, for instance [27, 28]) of solvability of the system (2.5). Theorem 2.1. The system (2.1) is Darboux integrable if and only if both characteristic Lie rings R(x,N), R(y,N) are of finite dimension. Corollary 2.2. The system (2.5) has a nontrivial solution if and only if the ring R(y,N) is of finite dimension. For the sake of convenience we introduce the following notation adX(Z) := [X,Z]. We stress that in our further study the operator adDx plays a crucial role. Below we apply Dx to smooth functions of the dynamical variables u,ux,uxx, . . .. As it was demonstrated above on this class Classification of a Subclass of Two-Dimensional Lattices via Characteristic Lie Rings 5 of functions the operators Dy and Y coincide. Therefore relation [Dx, Dy] = 0 immediately gives [Dx, Y ] = 0. Replace now Y due to (2.4) and get [Dx, Y ] = N∑ i=0 ui,y(αiui,xYi + [Dx, Yi]) = 0. (2.6) Since in (2.6) the variables {ui,y}Ni=0 are linearly independent, the coefficients should vanish. Consequently we have [Dx, Yi] = −αiui,xYi. (2.7) From this formula we can easily obtain that adDx : R(y,N) → R(y,N). The following lemma describes the kernel of this map (see also [19]) Lemma 2.3. If the vector field Z = ∑ i z1,i ∂ ∂ui,x + z2,i ∂ ∂ui,xx + · · · satisfies the condition [Dx, Z] = 0 then Z = 0. 3 Method of the test sequences We call a sequence of the operators W0,W1,W2, . . . in R(y,N) a test sequence if the following condition is satisfied for ∀m [Dx,Wm] = m∑ j=0 wj,mWj . The test sequence allows one to derive integrability conditions for the hyperbolic type sys- tem (2.1) (see [10, 27, 28]). Indeed, let us assume that (2.1) is Darboux integrable. Then the ring R(y,N) is of finite dimension. Therefore there exists an integer k such that the operators W0, . . . ,Wk are linearly independent while the operator Wk+1 is expressed through them as follows Wk+1 = λkWk + · · ·+ λ0W0. (3.1) Let us apply the operator adDx to both sides of (3.1). As a result we find k∑ j=0 wj,k+1Wj + wk+1,k+1 k∑ j=0 λjWj = k∑ j=0 Dx(λj)Wj + λk k∑ j=0 wj,kWj + λk−1 k−1∑ j=0 wj,k−1Wj + · · ·+ λ0w0,0W0. By collecting the coefficients before the independent operators we obtain a system of the differential equations for the coefficients λ0, λ1, . . . , λk. The system is overdetermined since all of the coefficients λj are functions of a finite number of the dynamical variables u,ux, . . .. The consistency conditions of this overdetermined system generate integrability conditions for the hyperbolic type system (2.1). For instance, collecting the coefficients before Wk we find the first equation of the mentioned system Dx(λk) = λk(wk+1,k+1 − wk,k) + wk,k+1, (3.2) which is also overdetermined. Below we use two different samples of the test sequences in order to find the function αn. 6 I. Habibullin and M. Poptsova 3.1 The first test sequence Define a sequence of the operators in R(y,N) due to the recurrent formula Y0, Y1, W1 = [Y0, Y1], W2 = [Y0,W1], . . . , Wk+1 = [Y0,Wk], . . . . (3.3) In the case of the first two members of the sequence we have already deduced commutation relations (see (2.7) above) which are important for our further studies [Dx, Y0] = −α0u0,xY0, [Dx, Y1] = −α1u1,xY1. (3.4) By using these two relations and applying the Jacobi identity we get immediately [Dx,W1] = −(α0u0,x + α1u1,x)W1 − Y0(α1u1,x)Y1 + Y1(α0u0,x)Y0. (3.5) It can be proved by induction that (3.3) is really a test sequence. Moreover it is easily verified that for k ≥ 2 [Dx,Wk] = pkWk + qkWk−1 + · · · , where the factors pk, qk are evaluated as follows pk = −(α1u1,x + kα0u0,x), qk = k − k2 2 Y0(α0u0,x)− Y0(α1u1,x)k. Due to the assumption that R(y,N) is of finite dimension only a finite subset of the se- quence (3.3) is linearly independent. So there exists M such that WM = λWM−1 + · · · , (3.6) where the operators Y0, Y1,W1, . . . ,WM−1 are linearly independent and the tail might contain a linear combination of the operators Y0, Y1,W1, . . . ,WM−2. At the moment we are not interested in that part in (3.6). Lemma 3.1. The operators Y0, Y1, W1 are linearly independent. Proof. Assume that λ1W1 + µ1Y1 + µ0Y0 = 0. Since the operators Y0, Y1 are of the form Y0 = ∂ ∂u0 + · · · , Y1 = ∂ ∂u1 + · · · while W1 does not contain summands like ∂ ∂u0 and ∂ ∂u1 then the factors µ1, µ0 vanish. If in addition λ1 6= 0 then we have W1 = 0. Now by applying the operator adDx to both sides of this relation we get due to (3.5) an equation Y0(α1u1,x)Y1 − Y1(α0u0,x)Y0 = 0, which yields two conditions: Y0(α1u1,x) = α1,u0u1,x = 0 and Y1(α0u0,x) = α0,u1u0,x = 0. Those equalities contradict our assumption that ∂α(un+1,un,un−1) ∂un±1 6= 0. Lemma is proved. � Lemma 3.2. If the expansion (3.6) holds then α(u1, u0, u−1) = P ′(u0) P (u0) +Q(u−1) + 1 M − 1 Q′(u0) P (u1) +Q(u0) − c1(u0). Classification of a Subclass of Two-Dimensional Lattices via Characteristic Lie Rings 7 Proof. It is easy to check that equation (3.2) for the case of the sequence (3.3) takes the following form Dx(λ) = −α0u0,xλ− M(M − 1) 2 Y0(α0u0,x)−MY0(α1u1,x). (3.7) We simplify the formula (3.7) due to the relations Y0(α0u0,x) = ( ∂ ∂u0 + α0u0,x ∂ ∂u0,x ) α0u0,x = ( α0,u0 + α2 0 ) u0x, Y0(α1u1,x) = α1,u0u1,x. A simple analysis of the equation (3.7) gives that λ = λ(u0, u1). Therefore (3.7) gives rise to the equation λu0u0,x + λu1u1,x = − ( αλ+ M(M − 1) 2 ( α0,u0 + α2 0 )) u0,x −Mα1,u0u1,x. By comparing the coefficients before the independent variables u0,x, u1,x we deduce an overde- termined system of the differential equations for λ λu0 = −α0λ− M(M − 1) 2 ( α0,u0 + α2 0 ) , λu1 = −Mα1,u0 . (3.8) Let us derive and investigate the consistency conditions of the system (3.8). We differentiate the first equation with respect to u−1 and find λ = −M(M − 1) 2 α0,u0u−1 + 2α0α0,u−1 α0,u−1 . (3.9) Since λu−1 = 0 we have (logα0,u−1)u0u−1 + 2α0,u−1 = 0. (3.10) Now we introduce a new variable z due to the relation α0,u−1 = −1 2e z and reduce (3.10) to the Liouville equation zu0u−1 = ez for which we have the general solution ez = 2P ′(u0)Q ′(u−1) (P (u0) +Q(u−1))2 , where P (u0) and Q(u−1) are arbitrary differentiable functions. Thus for α0 we can obtain the following explicit expression α0 = − 1 2 ∫ ezdu−1 = P ′(u0) P (u0) +Q(u−1) +H(u0, u1), (3.11) where H(u0, u1) is to be determined. Now we can find λ from the second equation in (3.8) λ = −M ∫ α1,u0du1 = −M Q′(u0) P (u1) +Q(u0) +Mc(u0). (3.12) Let us specify H(u0, u1) by replacing in (3.9) α0 and λ in virtue of (3.11), (3.12). As a result we obtain H(u0, u1) = 1 M − 1 Q′(u0) P (u1) +Q(u0) − 1 M − 1 c(u0)− 1 2 P ′′(u0) P ′(u0) . Summarizing the reasonings we can conclude that α(u1, u0, u−1) = P ′(u0) P (u0) +Q(u−1) + 1 M − 1 Q′(u0) P (u1) +Q(u0) − c1(u0), (3.13) where the functions of one variable P (u0), Q(u0), c1(u0) = 1 M−1c(u0)+ 1 2 P ′′(u0) P ′(u0) and the integerM are to be found. � 8 I. Habibullin and M. Poptsova The next step requires some additional integrability conditions. In what follows we derive them by constructing another test sequence. 3.2 The second test sequence Now we concentrate on a test sequence generated by the operators Y0, Y1, Y2 and their multiple commutators. It is more complicated than the previous sequence Z0 = Y0, Z1 = Y1, Z2 = Y2, Z3 = [Y1, Y0], Z4 = [Y2, Y1], Z5 = [Y2, Z3], Z6 = [Y1, Z3], Z7 = [Y1, Z4], Z8 = [Y1, Z5]. (3.14) The members Zm of the sequence for m > 8 are defined due to the recurrence Zm = [Y1, Zm−3]. Note that it is the simplest test sequence generated by the iterations of the map Z → [Y1, Z] which contains the operator [Y2, [Y1, Y0]] = Z5. Lemma 3.3. Operators Z0, Z1, . . . , Z5 constitute a linearly independent set. Proof. Firstly we note that the operators Z0, Z1, . . . , Z4 are linearly independent. It can be verified by using reasonings similar to those from the proof of Lemma 3.2. We prove the lemma by contradiction. Assume that Z5 = 4∑ j=0 λjZj . (3.15) Now we specify the action of the operator adDx on the operators Zi. For i = 0, 1, 2 it is obtained from the relation [Dx, Yi] = −αiui,xYi. Recall that αi = α(ui−1, ui, ui+1). For i = 3, 4, 5 we have [Dx, Z3] = −(a1 + a0)Z3 + · · · , [Dx, Z4] = −(a2 + a1)Z4 + · · · , [Dx, Z5] = −(a0 + a1 + a2)Z5 + Y0(a1)Z4 − Y2(a1)Z3 + · · · . Here ai = αiui,x. Let us apply the operator adDx to both sides of (3.15) and obtain −(a0 + a1 + a2)(λ4Z4 + λ3Z3 + · · · ) + Y0(a1)Z4 − Y2(a1)Z3 + · · · = λ4,xZ4 + λ3,xZ3 − λ4(a1 + a2)Z4 − λ3(a0 + a1)Z3 + · · · . (3.16) By comparing the coefficients before Z4 in (3.16) we obtain the following equation λ4,x = −α0u0,xλ4 − α1,u0u1,x. (3.17) A simple analysis of the equation (3.17) shows that λ = λ(u0, u1). Hence the equation (3.17) splits down into two equations λ4,u0 = −α0λ4 and λ4,u1 = −α1,u0 . The former shows that λ4 = 0. Indeed if λ4 6= 0 then we obtain an expression for α0: α0 = − (log λ4)u0 which shows that (α0)u−1 = 0. It contradicts the assumption that α(u1, u0, u−1) depends essentially on u1 and u−1, therefore λ4 = 0. Then (3.17) implies α1,u0 = 0 and it leads again to a contradiction. � Turn back to the sequence (3.14). For the further study it is necessary to specify the action of the operator adDx on the members of this sequence. It is convenient to divide the sequence into three subsequences and study them separately {Z3m}, {Z3m+1}, and {Z3m+2}. Classification of a Subclass of Two-Dimensional Lattices via Characteristic Lie Rings 9 Lemma 3.4. Action of the operator adDx on the sequence (3.14) is given by the following relations [Dx, Z3m] = −(α0u0,x +mα1u1,x)Z3m + ( m−m2 2 Y1(α1u1,x)−mY1(α0u0,x) ) Z3m−3 + · · · , [Dx, Z3m+1] = −(α2u2,x +mα1u1,x)Z3m+1 + ( m−m2 2 Y1(α1u1,x)−mY1(α2u2,x) ) Z3m−2 + · · · , [Dx, Z3m+2] = −(α0u0,x +mα1u1,x + α2u2,x)Z3m+2 + Y0(α1u1,x)Z3m+1 + Y2(α1u1,x)Z3m − (m− 1) (m 2 Y1(α1u1,x) + Y1(α0u0,x + α2u2,x) ) Z3m−1 + · · · . Lemma 3.4 is easily proved by induction. Since the proof is quite technical we omit it. Theorem 3.5. Assume that Z3k+2 is represented as a linear combination Z3k+2 = λkZ3k+1 + µkZ3k + νkZ3k−1 + · · · (3.18) of the previous members of the sequence (3.14) and neither of the operators Z3j+2 with j < k is a linear combination of Zs with s < 3j +2. Then the coefficient νk is a solution to the equation Dx(νk) = −α1u1,xνk − k(k − 1) 2 Y1(α1u1,x)− (k − 1)Y1(α0u0,x + α2u2,x). (3.19) Lemma 3.6. Suppose that all of the conditions of the theorem are satisfied. In addition assume that the operator Z3k (operator Z3k+1) is linearly expressed in terms of the operator Zi with i < 3k. Then in this decomposition the coefficient before Z3k−1 vanishes. Proof. Assume in contrary that λ 6= 0 in the formula Z3k = λZ3k−1 + · · · . (3.20) Let us apply adDx to (3.20). As a result we find due to Lemma 3.4 −(α0u0,x + kα1u1,x)λZ3k−1 + · · · = Dx(λ)Z3k−1 − λ(α0u0,x + (k − 1)α1u1,x + α2u2,x)Z3k−1 + · · · . (3.21) Collect the coefficients before Z3k−1 and obtain an equation the coefficient λ must satisfy to Dx(λ) = λ(α2u2,x − α1u1,x). Due to our assumption above λ does not vanish and hence Dx(log λ) = α2u2,x − α1u1,x. (3.22) Since λ depends on a finite number of the dynamical variables then due to equation (3.22) λ might depend only on u1 and u2. Therefore (3.21) yields (log λ)u1u1,x + (log λ)u2u2,x = α2u2,x − α1u1,x. The variables u1,x, u2,x are independent, so the last equation implies α1 = −(log λ)u1 , α2 = (log λ)u2 . Thus α1 = α1(u1, u2) depends only on u1, u2. It contradicts our assumption that α1 depends essentially on u0. The contradiction shows that assumption λ 6= 0 is not true. That completes the proof. � 10 I. Habibullin and M. Poptsova Now in order to prove Theorem 3.5 we apply the operator adDx to both sides of (3.18) and then simplify due to the relation from Lemma 3.4. Comparison of the coefficients before Z3k−1 implies equation (3.19). Let us find the explicit expressions for the coefficients of the equation (3.19) Y1(α0u0,x) = α0,u1u0,x, Y1(α2u2,x) = α2,u1u2,x, Y1(α1u1,x) = ( α1,u1 + α2 1 ) u1,x and substitute them into (3.19) Dx(νk) = −α1νku1,x − k(k − 1) 2 ( α1,u1 + α2 1 ) u1,x − (k − 1)(α0,u1ux + α2,u1u2,x). (3.23) A simple analysis of (3.23) convinces that νk might depend only on the variables u0, u1, u2. Therefore Dx(νk) = νk,u0u0,x + νk,u1u1,x + νk,u2u2,x. (3.24) From the equations (3.23), (3.24) we obtain a system of the equations for the coefficient νk νk,u0 = −(k − 1)α0,u1 , (3.25) νk,u1 = −α1νk − k(k − 1) 2 ( α1,u1 + α2 1 ) , (3.26) νk,u2 = −(k − 1)α2,u1 . (3.27) Substitute the preliminary expression for the function α given by the formula (3.13) into the equation (3.25) and get νk,u0 = k − 1 M − 1 P ′(u1)Q ′(u0) (P (u1) +Q(u0))2 . Integration of the latter with respect to u0 yields νk = − k − 1 M − 1 P ′(u1) P (u1) +Q(u0) +H(u1, u2). Since νk,u2 = Hu2 the equation (3.27) gives rise to the relation Hu2 = (k − 1) P ′(u2)Q ′(u1) (P (u2) +Q(u1))2 . Now by integration we obtain an explicit formula for H H = −(k − 1) ( Q′(u1) P (u2) +Q(u1) +A(u1) ) , which produces νk = −(k − 1) ( 1 M − 1 P ′(u1) P (u1) +Q(u0) + Q′(u1) P (u2) +Q(u1) +A(u1) ) . Classification of a Subclass of Two-Dimensional Lattices via Characteristic Lie Rings 11 Let us substitute the values of α and νk found into the equation (3.26). We get a huge equation −(k − 1) M − 1 ( P ′′(u1) P (u1) +Q(u0) − P ′2(u1) (P (u1) +Q(u0))2 ) − (k − 1) ( Q′′(u1) P (u2) +Q(u1) − Q′2(u1) (P (u2) +Q(u1))2 +A′(u1) ) = (k − 1) ( P ′(u1) P (u1) +Q(u0) + 1 M − 1 Q′(u1) P (u2) +Q(u1) − c1(u1) ) × ( 1 M − 1 P ′(u1) P (u1) +Q(u0) + Q′(u1) P (u2) +Q(u1) +A(u1) ) − k(k − 1) 2 ( P ′′(u1) P (u1) +Q(u0) + 1 M − 1 Q′′(u1) P (u2) +Q(u1) − 1 M − 1 Q′2(u1) (P (u2) +Q(u1))2 + 1 M − 1 2Q′(u1)P ′(u1) (P (u1) +Q(u0))(P (u2) +Q(u1)) + 1 (M − 1)2 Q′2(u1) (P (u2) +Q(u1))2 − c′1(u1)− 2c1(u1) ( P ′(u1) P (u1) +Q(u0) + 1 M − 1 Q′(u1) P (u2) +Q(u1) ) + c21(u1) ) . (3.28) Evidently due to our assumption ∂ ∂u1 α(u1, u0, u−1) 6= 0, ∂ ∂u−1 α(u1, u0, u−1) 6= 0 the func- tions P ′(u2) and Q ′(u0) do not vanish. Therefore the variables Q′2(u1) (P (u2) +Q(u1))2 , P ′2(u1) (P (u1) +Q(u0))2 , P ′(u1)Q ′(u1) (P (u1) +Q(u0))(P (u2) +Q(u1)) are independent. By gathering the coefficients before these variables in (3.28) we get a system of two equations( 1− 1 M − 1 )( 1− k 2(M − 1) ) = 0, 1 + 1 (M − 1)2 = k M − 1 . (3.29) There are two solutions to the system (3.29): M = 0, k = −2 and M = 2, k = 2. The former does not fit since k should be positive, so we have the only possibility M = 2, k = 2. This finishes the proof of Theorem 3.5. 4 Finding the functions P , Q and c1 In this section we specify the function α given by (3.13). For this aim we should consider expansions (3.6), (3.18) using the fact that M = 2, k = 2. Let us rewrite the expansion (3.6) in the complete form W2 = λW1 + σY1 + δY0. (4.1) Theorem 4.1. Expansion (4.1) holds if and only if the function α in (1.1) is of the following form α(un+1, un, un−1) = P ′(un) P (un) +Q(un−1) + Q′(un) P (un+1) +Q(un) − 1 2 ( logQ′(un)P ′(un) )′ , (4.2) where the functions P (un), Q(un) are connected with each other by the differential constraint −3Q′′2P ′2 − 2P ′′′P ′Q′2 + 3P ′′2Q′2 + 2P ′2Q′′′Q′ = 0. (4.3) 12 I. Habibullin and M. Poptsova Proof. Firstly by using relations (3.4), (3.5) and applying the Jacobi identity we get [Dx,W2] = −(2a0 + a1)W2 − Y0(a0 + 2a1)W1 + (2Y0Y1(a0)− Y1Y0(a0))Y0 − Y0Y0(a1)Y1. Evidently only one summand in (4.1) contains the term ∂ ∂u1 , namely σY1, and only one summand contains the term ∂ ∂u0 , namely δY0. Hence σ = 0, δ = 0 and we have W2 = λW1. Now by applying the operator adDx to both sides of this relation we obtain −(2a0 + a1)W2 − Y0(a0 + 2a1)W1 + (2Y0Y1(a0)− Y1Y0(a0))Y0 − Y0Y0(a1)Y1 = Dx(λ)W1 + λ(−(a0 + a1)W1 + Y1(a0)Y0 − Y0(a1)Y1). Collecting the coefficients before W2, W1, Y1, and Y0 we find the following system Dx(λ) = −a0λ− Y0(a0 + 2a1), (4.4) −Y0Y0(a1) = −λY0(a1), (4.5) 2Y0Y1(a0)− Y1Y0(a0) = λY1(a0). (4.6) Setting M = 2 in (3.7) we obtain equation (4.4). The overdetermined system (3.8) takes the form λu0 = −α0λ− ( α0,u0 + α2 0 ) , (4.7) λu1 = −2α1,u0 . Thus λ = −2 Q′(u0) P (u1) +Q(u0) + 2c(u0), (4.8) α(u1, u0, u−1) = P ′(u0) P (u0) +Q(u−1) + Q′(u0) P (u1) +Q(u0) − 1 2 P ′′(u0) P ′(u0) − c(u0). (4.9) We rewrite (4.5), (4.6) due to the relations Y0(a0) = ( ∂ ∂u0 + α0u0,x ∂ ∂u0,x + · · · ) (α0u0,x) = ( α0,u0 + α2 0 ) u0,x, Y0(a1) = ( ∂ ∂u0 + α0u0,x ∂ ∂u0,x + · · · ) (α1u1,x) = α1,u0u1,x, Y0(a0 + 2a1) = Y0(a0) + 2Y0(a1) = ( α0,u0 + α2 0 ) u0,x + 2α1,u0u1,x, Y0Y0(a1) = ( ∂ ∂u0 + α0u0,x ∂ ∂u0,x + · · · ) (α1,u0u1,x) = α1,u0u0u1,x, Y1(a0) = ( ∂ ∂u1 + α1u1,x ∂ ∂u1,x + · · · ) (α0u0,x) = α0,u1u0,x, Y0Y1(a0) = ( ∂ ∂u0 + α0u0,x ∂ ∂u0,x + · · · ) (α0,u1u0,x) = ( α0,u0u1 + α0α0,u1 ) ux, Y1Y0(a0) = ( ∂ ∂u1 + α1u1,x ∂ ∂u1,x + · · · )(( α0,u0 + α2 0 ) u0,x ) = ( α0,u0u1 + 2α0α0,u1 ) u0,x as follows α1,u0u0 = λα1,u0 , α0,u0u1 = λα0,u1 . (4.10) Classification of a Subclass of Two-Dimensional Lattices via Characteristic Lie Rings 13 We substitute (4.8), (4.9) into (4.10) and find that c(u0) = 1 2 Q′′(u0) Q′(u0) . So we find that functions (4.8), (4.9) are given by λ(R) = λ(u0, u1) = −2 Q′(u0) P (u1) +Q(u0) + Q′′(u0) Q′(u0) , (4.11) α(u1, u0, u−1) = P ′(u0) P (u0) +Q(u−1) + Q′(u0) P (u1) +Q(u0) − 1 2 P ′′(u0) P ′(u0) − 1 2 Q′′(u0) Q′(u0) . (4.12) Substituting (4.11), (4.12) into (4.7) we obtain that the functions P , Q must satisfy the equality −3Q′′2P ′2 − 2P ′′′P ′Q′2 + 3P ′′2Q′2 + 2P ′2Q′′′Q′ = 0. Thus we have proved that if the expansion (3.6) holds then it should be of the form W2 = λ(R)W1. Or the same [Y0, [Y1, Y0]] = λ(R)[Y1, Y0]. � Let us define a sequence of the operators in R(y,N) due to the following recurrent formula Y0, Y1, W̃1 = [Y1, Y0], W̃2 = [Y1,W1], . . . , W̃k+1 = [Y1, W̃k], . . . . It slightly differs from (3.3) and can be studied in a similar way. We can easily check that the conditions (4.2), (4.3) provide the representation W̃2 = λ(L)W̃1. Or the same [Y1, [Y1, Y0]] = λ(L)[Y1, Y0] (4.13) with the coefficient λ(L) = − 2P ′(u1) P (u1) +Q(u0) + P ′′(u1) P ′(u1) . Let us consider expansion (3.18) setting k = 2, Z8 = λZ7 + µZ6 + νZ5 + ρZ4 + κZ3 + σZ2 + δZ1 + ηZ0. (4.14) Theorem 4.2. Expansions (4.1), (4.14) hold if and only if the function α in (1.1) is of one of the forms α0 = α(u1, u0, u−1) = P ′(u0) P (u0) + c1P (u−1) + c2 + c1P ′(u0) P (u1) + c1P (u0) + c2 − P ′′(u0) P ′(u0) , (4.15) α0 = α(u1, u0, u−1) = c3r(u−1)r ′(u0) c3r(u0)r(u−1) + c4r(u−1)− c1 + c2r(u−1) + c1r ′(u0) r(u0) ( c3r(u1)r(u0) + c4r(u0)− c1 + c2r(u0) ) − r′′(u0)r(u0)− r′2(u0) r(u0)r′(u0) , (4.16) where P (u0) and r(u0) are arbitrary smooth functions, c1 6= 0, c3 6= 0, c2, and c4 are arbitrary constants. 14 I. Habibullin and M. Poptsova Proof. By taking k = 2 in the statement of Lemma 3.4 we get [Dx, Z6] = −(α0u0,x + 2α1u1,x)Z6 + · · · , (4.17) [Dx, Z7] = −(α2u2,x + 2α1u1,x)Z7 − (Y1(α1u1,x) + 2Y1(α2u2,x))Z4 + · · · , (4.18) [Dx, Z8] = −(α0u0,x + 2α1u1,x + α2u2,x)Z8 + Y0(α1u1,x)Z7 + Y2(α1u1,x)Z6 − ( Y1(α1u1,x) + Y1(α0u0,x + α2u2,x) ) Z5 + · · · . (4.19) Now we apply the operator adDx to both sides of (4.14) and then simplify due to the relations (4.17), (4.18), (4.19). Comparison of the coefficients before Z7 and Z6 implies λ = 0 and µ = 0. Thus formula (4.14) is simplified Z8 = νZ5 + ρZ4 + κZ3 + σZ2 + δZ1 + ηZ0. (4.20) In what follows we will use the following commutativity relations [Dx, Z8] = −(a2 + 2a1 + a0)Z8 + Y0(a1)Z7 − Y2(a1)Z6 − Y1(a2 + a1 + a0)Z5 + Y1Y0(a1)Z4 − Y1Y2(a1)Z3 + (Y1Y2Y0(a1) + Z5(a1))Z1, (4.21) [Dx, Z5] = −(a0 + a1 + a2)Z5 + Y0(a1)Z4 − Y2(a1)Z3 + Y2Y0(a1)Z1. (4.22) Let us apply adDx to (4.20) then simplify by using (4.21), (4.22), (4.20) and gather the coeffi- cients at Z5 −(a2 + 2a1 + a0)ν − Y1(a2 + a1 + a0) = Dx(ν)− (a2 + a1 + a0)ν or the same Dx(ν) = −a1ν − Y1(a2 + a1 + a0). (4.23) Equation (4.23) implies that ν depends on three variables ν = ν(u, u1, u2) and splits down into three equations as follows νu = −α0,u1 , (4.24) νu1 = −α1ν − α1,u1 − α2 1, (4.25) νu2 = −α2,u1 . (4.26) Substituting α defined by (4.12) into (4.24) and integrating with respect to u, we obtain ν = − P ′(u1) P (u1) +Q(u0) +H(u1, u2). (4.27) From equation (4.26) we find ν = − Q′(u1) P (u2) +Q(u1) +R(u0, u1). (4.28) Comparison of (4.27) and (4.28) yields − P ′(u1) P (u1) +Q(u0) +H(u1, u2) = − Q′(u1) P (u2) +Q(u1) +R(u, u1). Due to the fact that variables u0, u1, u2 are independent we obtain − P ′(u1) P (u1) +Q(u0) −R(u0, u1) = − Q′(u1) P (u2) +Q(u1) −H(u1, u2) = −A(u1). Classification of a Subclass of Two-Dimensional Lattices via Characteristic Lie Rings 15 Hence H(u1, u2) = − Q′(u1) P (u2) +Q(u1) +A(u1) and then ν = − P ′(u1) P (u1) +Q(u0) − Q′(u1) P (u2) +Q(u1) +A(u1). (4.29) Note that λ(R) defined by (4.11) satisfies the equation (4.7), i.e., λ(R),u = −α0λ(R) − α0,u0 − α2 0. Then λ(R)1,u1 = −α1λ(R)1 − α1,u1 − α2 1, (4.30) where λ(R)1 = Dn(λ(R)). Here Dn is a shift operator Dnuk = uk+1. Let us subtract (4.30) from (4.25)( ν − λ(R)1 ) u1 = −α1 ( ν − λ(R)1 ) . Substituting functions (4.11) and (4.29) into the last equation we arrive at the equality − P ′(u1)B(u1) P (u1) +Q(u0) − Q′(u1)B(u1) P (u2) +Q(u1) + 1 2 ( logQ′(u1)P ′(u1) )′( Q′(u1) P (u2) +Q(u1) − P ′(u1) P (u1) +Q(u0) +B(u1) ) = Q′′(u1) P (u2) +Q(u1) − P ′′(u1) P (u1) +Q(u0) +B′(u1), where B(u1) = A(u1)− Q′′(u1) Q′(u1) . This equality is satisfied only if the following conditions hold Q′′(u1) = −Q′(u1)B(u1) + 1 2 Q′(u1) ( logQ′(u1)P ′(u1) )′ , (4.31) P ′′(u1) = P ′(u1)B(u1) + 1 2 P ′(u1) ( logQ′(u1)P ′(u1) )′ , (4.32) B′(u1) = 1 2 B(u1) ( logQ′(u1)P ′(u1) )′ . (4.33) The equation (4.33) is satisfied if B(u1) = 0 or (logB(u1)) ′ = 1 2 (logQ′(u1)P ′(u1)) ′. (4.34) If B(u1) = 0 then Q(u1) = c1P (u1) + c2 and α0 = P ′(u0) P (u0) + c1P (u−1) + c2 + c1P ′(u0) P (u1) + c1P (u0) + c2 − P ′′(u0) P ′(u0) , λ(M) := ν = − P ′(u1) P (u1) + c1P (u0) + c2 − c1P ′(u1) P (u2) + c1P (u1) + c2 + Q′′(u1) Q′(u1) , (4.35) λ(R) = − 2c1P ′(u0) P (u1) + c1P (u0) + c2 + P ′′(u0) P ′(u0) , (4.36) λ(L) = − 2P ′(u1) P (u1) + c1P (u0) + c2 + P ′′(u1) P ′(u1) . (4.37) Here c1 6= 0. 16 I. Habibullin and M. Poptsova If B(u1) 6= 0 then from the system of equations (4.31), (4.32), and (4.34) we obtain that Q(u1) = − c1 r(u1) + c2, P (u1) = c3r(u1) + c4 and α0 = c3r(u−1)r ′(u0) c3r(u0)r(u−1) + c4r(u−1)− c1 + c2r(u−1) + c1r ′(u0) r(u0) ( c3r(u1)r(u0) + c4r(u0)− c1 + c2r(u0) ) − r′′(u0)r(u0)− r′2(u0) r(u0)r′(u0) , λ(M) := ν = − c3r(u0)r ′(u1) c3r(u1)r(u0) + c4r(u0)− c1 + c2r(u0) − c1r ′(u1) r(u1) ( c3r(u2)r(u1) + c4r(u1)− c1 + c2r(u1) ) + r′(u1) r(u1) − 2r′2(u1)− r′′(u1)r(u1) r(u1)r′(u1) , (4.38) λ(R) = − 2c1r ′(u0) r(u0) ( c3r(u1)r(u0) + c4r(u0)− c1 + c2r(u0) ) + r′′(u0)r(u0)− 2r′2(u0) r(u0)r′(u0) , (4.39) λ(L) = −2c3r(u0)r′(u1)( c3r(u1)r(u0) + c4r(u0)− c1 + c2r(u0) ) + r′′(u1) r′(u1) . (4.40) Now let us apply adDx to (4.20) using (4.21), (4.22), (2.7) and the facts that Z4 = Dn(Z3) and [Z1, Z4] = Dn[Z0, Z3] = −Dn(W2) = −Dn(λ(R))W1 = Dn(λ(R))Z4 and write down coefficients before Z4 −(a2 + 2a1 + a0)ρ+ Y1Y0(a1) + Y0(a1)Dn(λ(R)) = νY0(a1) +Dx(ρ)− (a1 + a2)ρ. Then Dx(ρ) = −(a1 + a0)ρ+ Y1Y0(a1) + Y0(a1)Dn(λ(R))− νY0(a1). (4.41) The equation (4.41) implies that ρ = ρ(u, u1, u2) and splits down into three equations as follows ρu2 = 0, −α0ρ = ρu0 , −α1ρ+ α1,u0u1 + α1α1,u0 + α1,u0Dn(λ(R))− να1,u0 = ρu1 . If α0, ν, and λ(R) are defined by the formulas (4.15), (4.35), and (4.36) or by the formulas (4.16), (4.38), and (4.39) correspondingly then ρ = 0 and the last equations are satisfied. Now let us apply adDx to (4.20) using (4.21), (4.22), (2.7), (4.13) and write down coefficients before Z3 −(a2 + 2a1 + a0)κ− Y1Y2(a1)− Y2(a1)λ(L) = −νY2(a1) +Dx(κ)− (a1 + a2)κ. Then Dx(κ) = −(a2 + a1)κ− Y1Y2(a1)− Y2(a1)λ(L) + νY2(a1). (4.42) The equations (4.42) implies that κ = κ(u1, u2) and splits down into two equations as follows κu2 = −α2κ, κu1 = −α1κ− α1,u2u1 − α1α1,u2 − α1,u2λ(L) + να1,u2 . If α0, ν, and λ(L) are defined by the formulas (4.15), (4.35), and (4.37) or by the formulas (4.16), (4.38), and (4.40) correspondingly then κ = 0 and the last equations are satisfied. Classification of a Subclass of Two-Dimensional Lattices via Characteristic Lie Rings 17 Apply adDx to (4.20) taking into account that ρ = κ = 0 and write down coefficients before operators Z2, Z1 and Z0 Dx(σ) = −(2a1 + a0)σ, Dx(δ) = −(a2 + a1 + a0)δ − Y1Y2Y0(a1) + λY2Y0(a1), Dx(η) = −(a2 + 2a1). From these equations we obtain that σ = δ = η = 0. Thus we have proved that if the expansions (4.1), (4.14) hold then (4.14) should be as follows Z8 = λ(M)Z5. Or the same [Y1, [Y2, [Y1, Y0]]] = λ(M)[Y2, [Y1, Y0]], (4.43) where λM defined by the formula (4.35) or (4.38) and α0, λ(R), and λ(L) are defined by the for- mulas (4.15), (4.36), and (4.37) or by the formulas (4.16), (4.39), and (4.40) correspondingly. � Corollary of Theorems 4.1 and 4.2: Corollary 4.3. In both cases Q(u1) = − c1 r(u1) +c2, P (u1) = c3r(u1)+c4 and Q(u1) = c1P (u1)+c2 the constraint (4.3) is satisfied identically. In a similar way we check that the same conditions (4.15), (4.16) provides the representations [Y0, [Y2, [Y1, Y0]]t] = λ(R)[Y2, [Y1, Y0]], [Y2, [Y2, [Y1, Y0]]] = Dn(λ(L)[Y2, [Y1, Y0]]. 5 Comments on the classif ication result In this section we briefly discuss the statements of Theorems 4.2 and 5.2 (see below) claiming that the lattice (1.1) is integrable in the sense of Definition 1.1 only for two choices of the function α given by (4.15) and (4.16). In both cases the lattice has a functional freedom which is removed by an appropriate point transformation. Therefore we have Theorem 5.1. Any lattice (1.1) integrable in the sense above is reduced by the point transfor- mation v = p(u) to the following lattice vn,xy = vn,xvn,y ( 1 vn − vn−1 − 1 vn+1 − vn ) . (5.1) Specify the point transformations1 applied to the lattices. Change of the variables w = P (u) reduces (4.15) to wn,xy = wn,xwn,y ( 1 wn + c1wn−1 + c2 + c1 wn+1 + c1wn + c2 ) . (5.2) The latter is connected with (5.1) by the change of the variables vn = (−c1)nwn− c2 1+c1 if c1 6= −1 and by vn = wn − c2n in the special case c1 = −1. 1We are glad to acknowledge that these transformations are found by R.I. Yamilov and R.N. Garifullin (private communication). 18 I. Habibullin and M. Poptsova Change of the variables v = r(u) reduces (4.16) to vn,xy = vn,xvn,y ( vn−1 vnvn−1 + βvn−1 − γ + vn+1 + β vnvn+1 + βvn − γ ) , (5.3) where we denote β = c2+c4 c3 , γ = c1/c3. Then change of the variables v = β ( 1 w+c ) , γ = β2(c2+c) reduces (5.3) to wn,xy = 1 wn + c c+1wn−1 + 1 c+1 + c c+1 wn+1 + c c+1wn + 1 c+1 . The latter coincides with (5.2) if c1 = c c+1 , c2 = 1 c+1 . Note that equation (5.1) coincides with the Ferapontov–Shabat–Yamilov equation found in [20] and [4]. Theorem 5.2. The characteristic Lie rings in x- and y-directions for the following system of hyperbolic type equations v−1 = c0, vn,xy = vn,xvn,y ( 1 vn − vn−1 − 1 vn+1 − vn ) , 0 ≤ n ≤ N, (5.4) vN+1 = c1, are of finite dimension. The proof of Theorem 5.2 can be found in Appendix A. Corollary 5.3. The system (5.4) is Darboux integrable. Remark 5.4. The following lattice (see [20]) qn,xy = qn,xqn,y ( f(qn+1 − qn)− f(qn − qn−1) ) , f ′ = f2 − b2 is reduced by the point transformation to (5.1). Namely if b 6= 0 then f(q) = b tan(b(q + c)) = −ib tanh(ib(q + c)), where c is the constant of integration, i is the imaginary unit. So we have the lattice qn,xy = qn,xqn,y(−ib) ( tanh(ib(qn+1 − qn + c))− tanh(ib(qn − qn−1 + c)) ) . The change of variables qn = − i bvn−nc reduces the last lattice to (5.1). If b = 0 then f(q) = 1 q+c . By the change of variables qn = vn − nc we obtain (5.1). 6 Conclusion In [9] it was conjectured that any nonlinear integrable two-dimensional lattice of the form un,xy = g(un+1, un, un−1, un,x, un,y) (6.1) admits cut-off conditions reducing the lattice to a finite system of the hyperbolic type partial differential equations being integrable in the sense of Darboux when they are imposed at two points n = N1 and N2 chosen arbitrary. Classification of a Subclass of Two-Dimensional Lattices via Characteristic Lie Rings 19 In the present article we discussed the classification algorithm based on that conjecture. Actually we solved a problem of the complete description of the lattices (1.1) satisfying the suggested requirement. The lattice (1.1) is a particular case of the lattice (6.1) for which the mentioned cut-off condition is easily found: uN1 = c0, uN2 = c1. This circumstance essentially simplifies the situation. Nevertheless even in general when a priori the cut-off condition is also unknown the algorithm might be effective since the assumption on the existence of such boundary conditions puts severe restrictions on the characteristic operators. We show that the class of integrable lattices of the form (1.1) contains only one model up to the point transformations. This model coincides with the Ferapontov–Shabat–Yamilov equation. The one-dimensional reduction x = y of this lattice satisfies completely also the symmetry integrability conditions (see [23]). A Appendix The goal of the appendix is to prove Theorem 5.2. Let us introduce a special notation Yik,...,i0 for the multiple commutators. It is defined consecutively Yik,...,i0 = [Yik , Yik−1,...,i0 ]. (A.1) Number k is called the order of the operator (A.1). In order to prove Theorem 5.2 we show that the ring R(y,N) is of finite dimension. Actually we construct the basis in R(y,N) containing the operators {Yi}Ni=0, {Yi+1,i}N−1i=0 , {Yi+2,i+1,i}N−2i=0 , . . . , YN,N−1,...,0. (A.2) A.1 The base case of the mathematical induction In the previous section we have proved that [Y0, Y10] = λ(R)Y10, [Y1, Y10] = λ(L)Y10, (A.3) [Y0, Y210] = λ(R)Y210, [Y1, Y210] = λ(M)Y210, [Y2, Y210] = Dn(λ(L)Y210. (A.4) In what follows we will use the following relations which are easily verified [Dx, Y3210] = −(a3 + a2 + a1 + a0)Y3210 − Y3(a2)Y210 + Y0(a1)Y321, (A.5) [Dx, [Y0, Y3210]] = −(a3 + a2 + a1 + 2a0)[Y0, Y3210] − Y0(2a1 + a0)Y3210 − Y3(a2)[Y0, Y210] + Y0Y0(a1)Y321, (A.6) [Dx, [Y1, Y3210]] = −(a3 + a2 + 2a1 + a0)[Y1, Y3210]− Y1(a2 + a1 + a0)Y3210 − Y1Y3(a2)Y210 − Y3(a2)[Y1, Y210] + Y1Y0(a1)Y321 + Y0(a1)[Y1, Y321], (A.7) [Dx, [Y2, Y3210]] = −(a3 + 2a2 + a1 + a0)[Y2, Y3210]− Y2(a3 + a2 + a1)Y3210 − Y2Y3(a2)Y210 − Y3(a2)[Y2, Y210] + Y2Y0(a1)Y321 + Y0(a1)[Y2, Y321], (A.8) [Dx, [Y3, Y3210]] = −(2a3 + a2 + a1 + a0)[Y3, Y3210]− Y3(a3 + 2a2)Y3210 − Y3Y3(a2)Y210 + Y0(a1)[Y3, Y321]. (A.9) We prove the theorem by the mathematical induction. The base case consists in proving a lot of the formulas concerned to small order commutators up to order six. When constructing a linear expression for a given element in R(y,N) as a linear combination of those from (A.2) 20 I. Habibullin and M. Poptsova we always use Lemma 2.3. That is why we need in explicit expressions for [Dx, Yik,...,i0 ]. In the base case we prove a large set of the equalities. Since they all are proved by one and the same way we concentrate on one of them. Lemma A.1. We have [Y0, Y3210] = λ(R)Y3210. (A.10) Proof. By applying the operator adDx to Z = [Y0, Y3210]−λ(R)Y3210 and simplifying due to the equations (A.5), (A.6) we obtain [Dx, Z] = 0. Evidently Z satisfies the settings of Lemma 2.3. Due to this lemma we obtain Z = 0. Lemma A.1 is proved. � In what follows we need in the formulas [Y1, Y3210] = λ(M)Y3210, (A.11) [Y2, Y3210] = Dn ( λ(M) ) Y3210, (A.12) [Y3, Y3210] = D2 n ( λ(L) ) Y3210, (A.13) which are some versions of the formula (A.10) from Lemma A.1. Now using formulas [Dx, Y43210] = −(a4 + a3 + a2 + a1 + a0)Y43210 − Y4(a3)Y3210 + Y0(a1)Y4321, (A.14) [Dx, [Y0, Y43210]] = −(a4 + a3 + a2 + a1 + 2a0)[Y0, Y43210] − Y0(2a1 + a0)Y43210 − Y4(a3)[Y0, Y3210] + Y0Y0(a1)Y4321, (A.15) [Dx, [Y1, Y43210]] = −(a4 + a3 + a2 + 2a1 + a0)[Y1, Y43210] − Y4(a3)[Y1, Y3210]− Y1(a2 + a1 + a0)Y43210 − Y1Y4(a3)Y3210 + Y1Y0(a1)Y4321 + Y0(a1)[Y1, Y4321], (A.16) [Dx, [Y2, Y43210]] = −(a4 + a3 + a2 + a1 + a0)[Y2, Y43210] − Y2(a3 + a2 + a1)Y43210 − Y2Y4(a3)Y3210 − Y4(a3)[Y2, Y3210] + Y2Y0(a1)Y4321 + Y0(a1)[Y2, Y4321], (A.17) [Dx, [Y3, Y43210]] = −(a4 + 2a3 + a2 + a1 + a0)[Y3, Y43210] − Y3(a4 + a3 + a2)Y43210 − Y3Y4(a3)Y3210 − Y4(a3)[Y3, Y3210] + Y3Y0(a1)Y4321 + Y0(a1)[Y3, Y4321], (A.18) [Dx, [Y4, Y43210]] = −(2a4 + a3 + a2 + a1 + a0)[Y4, Y43210] − Y4(a4 + 2a3)Y43210 − Y4Y4(a3)Y3210 + Y0(a1)[Y4, Y4321] (A.19) by direct calculations we prove that [Y0, Y43210] = λ(R)Y43210, (A.20) [Y1, Y43210] = λ(M)Y43210, (A.21) [Y2, Y43210] = Dn ( λ(M) ) Y43210, (A.22) [Y3, Y43210] = D2 n ( λ(M) ) Y43210, (A.23) [Y4, Y43210] = D3 n ( λ(L) ) Y43210, (A.24) where λ(R), λ(M), and λ(L) are defined by the formulas (4.36), (4.35), and (4.37) or by the formulas (4.39), (4.38), and (4.40) correspondingly. Now, having explicit formulas for the small order commutators we are ready to work out an induction hypothesis. Classification of a Subclass of Two-Dimensional Lattices via Characteristic Lie Rings 21 A.2 Inductive step Theorem A.2. For n > 1 the multi-commutators satisfy the following formulas [Dx, Yn+1,n,...,0] = − ( n+1∑ i=0 ai ) Yn+1,n,...,0 − Yn+1(an)Yn,n−1,...,0 + Y0(a1)Yn+1,n,...,1, (A.25) [Dx, [Y0, Yn+1,n,...,0]] = −(an+1 + · · ·+ a1 + 2a0)[Y0, Yn+1,n,...,0]− Y0(a0 + 2a1)Yn+1,n,...,0 − Yn+1(an)[Y0, Yn,n−1,...,0] + Y0Y0(a1)Yn+1,n,...,1, (A.26) [Dx, [Yk, Yn+1,n,...,0]] = −(an+1 + · · ·+ 2ak + · · ·+ a0)[Yk, Yn+1,n,...,0] − Yk ( n+1∑ i=0 ai ) Yn+1,n,...,0 − YkYn+1(an)Yn,n−1,...,0 − Yn+1(an)[Yk, Yn,n−1,...,0] + YkY0(a1)Yn+1,n,...,1 + Y0(a1)[Yk, Yn+1,n,...,1], k = 1, 2, . . . , n, (A.27) [Dx, [Yn+1, Yn+1,n,...,0]] = −(2an+1 + an + · · ·+ a0)[Yn+1, Yn+1,n,...,0] − Yn+1(an+1 + 2an)Yn+1,n,...,0 − Yn+1Yn+1(an)Yn,n−1,...,0 + Y0(a1)[Yn+1, Yn+1,n,...,1]. (A.28) Proof by induction. For n = 2 and n = 3 formulas (A.25)–(A.28) are previously proved (see (A.5)–(A.9) and (A.14)–(A.19)). Assume that the multi-commutators satisfy the following formulas [Dx, Yn,...,0] = − ( n∑ i=0 ai ) Yn,...,0 − Yn(an−1)Yn−1,...,0 + Y0(a1)Yn,...,1, [Dx, [Y0, Yn,...,0]] = −(an + · · ·+ a1 + 2a0)[Y0, Yn,...,0] − Y0(a0 + 2a1)Yn,...,0 − Yn(an−1)[Y0, Yn−1,...,0] + Y0Y0(a1)Yn,...,1, [Dx, [Yk, Yn,...,0]] = −(an + · · ·+ 2ak + · · ·+ a0)[Yk, Yn,...,0]− Yk ( n∑ i=0 ai ) Yn,...,0 − YkYn(an−1)Yn−1,...,0 − Yn(an−1)[Yk, Yn−1,...,0] + YkY0(a1)Yn,...,1 + Y0(a1)[Yk, Yn,...,1], k = 1, 2, . . . , n− 1, [Dx, [Yn, Yn,...,0]] = −(2an + an−1 + · · ·+ a0)[Yn, Yn,...,0] − Yn(an + 2an−1)Yn,...,0 − YnYn(an−1)Yn−1,...,0 + Y0(a1)[Yn, Yn,...,1]. Then from these assumptions we deduce similar equations for n+ 1 [Dx, Yn+1,n,...,0] = [Dx, [Yn+1, Yn,n−1,...,0]] = [Yn+1, [Dx, Yn,n−1,...,0]]− [Yn,n−1,...,0, [Dx, Yn+1]] = [Yn+1,−(an + an−1 + · · ·+ a0)Yn,n−1,...,0 − Yn(an−1)Yn−1,n−2,...,0 + Y0(a1)Yn,n−1,...,1]− [Yn,n−1,...,0,−an+1Yn+1] = − ( n+1∑ i=0 ai ) Yn+1,n,...,0 − Yn+1(an + an−1 + · · ·+ a0)Yn,n−1,...,0Yn,n−1,...,0 − Yn+1Yn(an−1)Yn−1,n−2,...,0 − Yn(an−1)[Yn+1, Yn−1,n−2,...,0] + Yn+1Y0(a1)Yn,n−1,...,1 + Y0(a1)Yn+1,n,...,1 + Yn,n−1,...,0(an+1)Yn+1. (A.29) Note that Yi(aj) = 0 if \|i− j\| > 1, (A.30) 22 I. Habibullin and M. Poptsova i.e., if i 6= j, i 6= j ± 1, and [Ym, Yn,...,0] = 0 if m− n > 1 That is why the following terms in (A.29) are equal to zero Yn+1(an−1 + · · ·+ a0) = 0, Yn+1Yn(an−1) = 0, [Yn+1, Yn−1,n−2,...,0] = 0, Yn+1Y0(a1) = 0, Yn,n−1,...,0(an+1) = 0. Thus the equality (A.29) takes the form (A.25). Let us prove the formula (A.26), [Dx, [Y0, Yn+1,n,...,0]] = [Y0, [Dx, Yn+1,n,...,0]]− [Yn+1,n,...,0, [Dx, Y0]] = [ Y0,− ( n+1∑ i=0 ai ) Yn+1,n,...,0 − Yn+1(an)Yn,n−1,...,0 + Y0(a1)Yn+1,n,...,1 ] − [Yn+1,n,...,0,−a0Y0]. From this equality using property of linearity of the commutators and the equations (A.30) we obtain the formula (A.26). The formulas (A.27) and (A.28) are proved in a similar way. Theorem A.3. For m ≥ 1 the multi-commutators satisfy the following formulas [Y0, Ym+1,m,...,0] = λ(R)Ym+1,m,...,0, (A.31) [Yk, Ym+1,m,...,0] = Dk−1 n ( λ(M) ) Ym+1,m,...,0, k = 1, . . . ,m, (A.32) [Ym+1, Ym+1,m,...,0] = Dm n ( λ(L) ) Ym+1,m,...,0. (A.33) Proof by induction. For m = 1, 2, 3 formulas (A.31)–(A.33) are true (see (A.3), (A.4), (4.43), (A.10), (A.11), (A.12), (A.13), (A.20)–(A.24)). Assume that the multi-commutators satisfy the following formulas [Y0, Ym,m−1,...,0] = λ(R)Ym,m−1,...,0, (A.34) [Yk, Ym,m−1,...,0] = Dk−1 n ( λ(M) ) Ym,m−1,...,0, k = 1, . . . ,m− 1, (A.35) [Ym, Ym,m−1,...,0] = Dm−1 n ( λ(L) ) Ym,m−1,...,0. (A.36) Let us first prove the formula (A.31). The proof is rather tricky: we assume the expansion with undetermined coefficients [Y0, Ym+1,m,...,0] = λYm+1,m,...,0 + µYm+1,m,...,1 + νYm,m−1,...,0 + εYm+1,m,...,2 + ηYm,m−1,...,1 + ζYm−1,m−2,...,0 + · · · + θYm+1,m + · · ·+ ξY10 + σYm+1 + · · ·+ δY0, (A.37) and then evaluate the coefficients consecutively in the following way. We apply the opera- tor adDx to (A.37) and gather the coefficients before the linearly independent operators. For instance, by comparing the coefficients before the multi-commutator Ym+1,m,...,0 and then using formulas from Theorem 2.1 with n = m− 1 we find Dx(λ) = −a0λ− Y0(a0 + 2a1). The latter coincides with the equation (4.4) and, therefore, we can conclude that λ = λ(R). Classification of a Subclass of Two-Dimensional Lattices via Characteristic Lie Rings 23 Compare now the coefficients before Ym+1,m,...,1 to get an equation for determining µ. Note that by Theorem A.2 we have [Dx, Ym+1,m,...,1] = Dn[Dx, Ym,m−1,...,0] = \|Theorem A.2\| = Dn ( − ( m∑ i=0 ai ) Ym,m−1,...,0 − Ym(am−1)Ym−1,m−2,...,0 + Y0(a1)Ym,m−1,...,1 ) = − ( m+1∑ i=1 ai ) Ym+1,m,...,1 − Ym+1(am)Ym,m−1,...,1 + Y1(a2)Ym+1,m,...,2. (A.38) Due to the relation (A.38) the desired equation reduces to the form Dx(µ) = −2a0µ+ Y0Y0(a1)− λY0(a1). It is easily checked that the equation has the only solution µ = 0. Continuing this way we can prove that all of the other coefficients ν, ε, . . . , δ in (A.37) vanish. Now for the operator Z = [Y0, Ym+1,m,...,0]−λ(R)Ym+1,m,...,0 we have [Dx, Z] = 0. Due to Lemma 2.3 it implies Z = 0. That completes the proof of the formula (A.31). Now we prove the formula (A.32). To this end we assume that the equation holds [Yk, Ym+1,m,...,0] = λYm+1,m,...,0 + µYm+1,m,...,1 + νYm,m−1,...,0 + εYm+1,m,...,2 + ηYm,m−1,...,1 + ζYm−1,m−2,...,0 + · · · + θYm+1,m + · · ·+ ξY10 + σYm+1 + · · ·+ δY0 (A.39) with the coefficients to be determined. Let us apply adDx to (A.39) and write down the coefficients before the operator Ym+1,m,...,0 Dx(λ) = −akλ− Yk ( m+1∑ i=0 ai ) . Apply D −(k−1) n to the last equation Dx ( D−(k−1)n (λ) ) = −a1D−(k−1)n (λ)− Y1(a2 + a1 + a0). This equation coincides with the equation (4.23) then D −(k−1) n (λ) = λ(M) and λ = Dk−1 n ( λ(M) ) . Note that due to the formula (A.35) we have [Yk, Ym+1,m,...,1] = Dn[Yk−1, Ym,m−1,...,0] = Dn ( Dk−2 n (λ(M))Ym,m−1,...,0 ) = Dk−1 n ( λ(M) ) Ym+1,m,...,1. (A.40) Apply adDx to (A.39) using the formulas from Theorem A.2 and the formula (A.40) and write down the coefficients before the multi-commutator Ym+1,m,...,1 −(am+1 + · · ·+ 2ak + · · ·+ a0)µ+ YkY0(a1) + Y0(a1)D k−1 n (λ(M) = λY0(a1) +Dx(µ)− ( m+1∑ i=1 ) µ. 24 I. Habibullin and M. Poptsova Since λ = Dk−1 n (λ(M)) the latter can be brought to the form Dx(µ) = −(a0 + ak)µ+ YkY0(a1). (A.41) Evaluate the action of the product of the operators YkY0(a1) = ( ∂ ∂uk + αkuk,x ∂ ∂uk,x ) (α1,uu1,x) =  α1,uu1u1,x + α1α1,uu1,x, k = 1, α1,uu2u1,x, k = 2, 0, k > 2. Thus if k = 1 then the equality (A.41) takes the form Dx(µ) = −(α0ux + α1u1,x)µ+ (α1,uu1 + α1α1,u)u1,x. This equation implies that µ = µ(u, u1) and splits down into two equations as follows µu = −α0µ, µu1 = −α1µ+ α1,uu1 + α1α1,u. Then we can prove that µ = 0. If k = 2 then the equality (A.41) takes the form Dx(µ) = −(α0ux + α2u2,x)µ+ α1,uu2u1,x. This equation implies that µ = µ(u, u1, u2) and splits down into three equations as follows µu = −α0µ, µu1 = α1,uu2 , µu2 = −α2µ. And then again µ = 0. If k > 2 then the equality (A.41) takes the form Dx(µ) = −(α0ux + αkuk,x)µ. This equation implies that µ = µ(u, uk) and splits down into two equations as follows µu = −α0µ, µuk = −αkµ. Then µ = 0. In a similar way we can verify that all of the coefficients in (A.39) vanish except λ. Thus due to Lemma 2.3 formula (A.32) is correct. Now we check the formula (A.33). First we assume that the following decomposition takes place [Ym+1, Ym+1,m,...,0] = λYm+1,m,...,0 + µYm+1,m,...,1 + νYm,m−1,...,0 + εYm+1,m,...,2 + ηYm,m−1,...,1 + ζYm−1,m−2,...,0 + · · ·+ θYm+1,m + · · · + ξY10 + σYm+1 + · · ·+ δY0 (A.42) with undefined factors. Let us apply adDx to (A.42) and write down the coefficients before the multi-commutator Ym+1,m,...,0 Dx(λ) = −am+1λ− Ym+1(am+1 + 2am). Apply D−mn to this equation Dx ( D−mn (λ) ) = −a1D−mn (λ)− Y1(a1 + 2a0). This equation coincides with (4.4) then D−mn (λ) = λ(L) and λ = Dm n (λ(L)). Classification of a Subclass of Two-Dimensional Lattices via Characteristic Lie Rings 25 Let us apply adDx to (A.42) and write down the coefficients before the multi-commutator Ym+1,m,...,1 −(2am+1 + am + · · ·+ a0)µ+ Y0(a1)D m n (λ(L)) = λY0(a1) +Dx(µ)− ( m+1∑ i=1 ai ) µ. Note that λ = Dm n (λ(L)) then the last equality takes the form Dx(µ) = −(am+1 + a0)µ. Then µ = 0. Let us apply adDx to (A.42) and write down the coefficients before the multi-commutator Ym,m−1,...,0 Dx(ν) = −2am+1ν − Ym+1Ym+1(am) + λYm+1(am). (A.43) Note that λYm+1(am)− Ym+1Ym+1(am) = Dm n (λ(L))Ym+1(am)− Ym+1Ym+1(am) = Dm(λ(L))Y1(a0)− Y1Y1(a0)) = 0. Then the equation (A.43) takes the form Dx(ν) = −2am+1ν and we obtain that ν = 0. In a similar way we can prove the vanishing of the other coefficients in (A.42). Now by applying Lemma 2.3 it is easy to complete the proof of the formula (A.33). Theorem A.3 is proved. 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Classification of a Subclass of Two-Dimensional Lattices via Characteristic Lie Rings

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