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. By imposing the cut-off conditions $u_{-1}=c_0$ and $u_{N+1}=c_1$ we reduce the lattice $u_{n,xy}=\alpha(u_{n+1},u_n,u_{n-1})u_{n,x}u_{n,y}$ 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 $\alpha$. 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.


Introduction
In the present article we study the classification problem for the following class of two-dimentional lattices u n,xy = α(u n+1 , u n , u n−1 )u n,x u n,y . (1.1) Here the sought function u = u n (x, y) depends on real x, y and on integer n. Function α = α(u n+1 , u n , u n−1 ) is assumed to be analytical in a domain D ⊂ C 3 . We request also that the derivatives ∂α(u n+1 ,un,u n−1 ) ∂u n+1 and ∂α(u n+1 ,un,u n−1 ) ∂u n−1 do not vanish identically. Constraint u n 0 = c 0 where c 0 is a constant parameter defines a boundary condition which cuts off the lattice (1.1) into two independent semi-infinite lattices: u n,xy = α(u n+1 , u n , u n−1 )u n,x u n,y , for n > n 0 (n < n 0 ), (1.2) Any solutions of the lattice located on the semiaxis n > n 0 does not depend on the solutions of that located on n < n 0 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 lattice. They are compatible with the whole hierarchy of the higher symmetries [1], [6]. In the literature they are met in the connection with the so-called open chains (see, for instance, [12]). Since the symmetry approach which is a powerful classification tool in 1+1 dimensions (see, for instance, [2], [9], [10]) loses its efficiency in higher dimensions (an explanation can be found in [11]) it became clear years ago that it is necessary to look for alternative classification algorithms. 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 worth mentioning as an important step in this direction article [14] 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 [7]. It was observed that any finitely generated subring of the characteristic Lie ring for the integrable case is of finite dimension. The statement was approved for 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 [3], [4]. State-of-the-art for the subject and the references can be found in [13].
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 = N 1 and n = N 2 (take N 1 < N 2 ) reduces the lattice (1.1) into a finite system of hyperbolic type equations (open chain) u N 1 = c 1 , u n,xy = α(u n+1 , u n , u n−1 )u n,x u n,y , N 1 < n < N 2 , (1.3) Initiated by the article [7] , where a large class of two-dimensional lattices is discussed we use the following Definition 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 N 1 , N 2 . 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, u x , u y , . . . is a y-integral if it satisfies the condition D y I = 0, where D y is the operator of the total derivative with respect to the variable y and u is a vector (u 0 , u 1 , . . . u k ) with the coordinates u N 1 +1 , u N 1 +2 , . . . , u N 2 −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 u y , u yy , . . .. In what follows we are interested only on nontrivial y−integrals, i.e. integrals containing dependence on at least one dynamical variable u, u x , . . ..
We justify our definition 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 [N 1 , N 2 ] 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.

Characteristic Lie rings
Since the lattice (1.1) is invariant under the shift of the variable n we can without loosing of generality take N 1 = −1 and concentrate on the system u −1 = c 0 , u n,xy = α n u n,x u n,y , 0 ≤ n ≤ N, (2.1) u N +1 = c 1 .
Assume that system (2.1) is Darboux integrable and that I(u, u x , . . .) is its nontrivial integral. Let us evaluate D y I in the equation D y I = 0 and get due to the chain rule an equation 2) Since the coefficients of the equation Y I = 0 depend on u i,y while its solution I does not depend on them we have a system of several linear equations for one unknown I: The last equation together with (2.3) implies N i=0 u i,y Y i I = 0. Since the variables u i,y are independent the coefficients of this decomposition all vanish. Now we use the evident relation [X k , Y s ] = 0 valid for ∀k, s. The condition X i I = 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 Consider the set R 0 (y, N ) of all multiple commutators of the characteristic vector fields Y 0 , Y 1 , . . . , Y N . Denote through R(y, N ) the minimal ring containing R 0 (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.
We say that the ring R(y, N ) is of finite dimension if there exists a finite subset {Z 1 , Z 2 , . . . Z L } ⊂ 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 = λ 1 Z 1 + . . . + λ L Z L with the coefficients λ 1 , . . . λ L which might depend on a finite number of the dynamical variables; 2) relation λ 1 Z 1 + · · · λ L Z L = 0 implies that λ 1 = . . . = λ L = 0.
Let us formulate now an effective algebraic criterion (see, for instance [16], [17]) of solvability of the system (2.6). For the sake of convenience we introduce the following notation ad X (Z) := [X, Z]. Stress that in our further study the operator ad Dx plays the crucial role. Below we apply D x to smooth functions of the dynamical variables u, u x , u xx , . . .. As it was demonstrated above on this class of functions we have D y = Y . Therefore relation [D x , D y ] = 0 immediately gives [D x , Y ] = 0. Replace now Y due to (2.5) and get Since in (2.7) the variables {u i,y } N i=0 are linearly independent, the coefficients should vanish. Consequently we have From this formula we can easily obtain that ad Dx : R(y, N ) → R(y, N ). Describe the kernel of this map (see also [14] ).

Lemma 1. If the vector field
satisfies the condition [D x , Z] = 0 then Z = 0.

Method of the test sequences
We call a sequence of the operators W 0 , W 1 , W 2 , . . . in R(y, N ) a test sequence if the following condition is satisfied for ∀ k: The test sequence allows one to derive integrability conditions for the hyperbolic type system (2.1) (see [8], [16], [17]). 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 W 0 , . . . W k are linearly independent while the operator W k+1 is expressed through them as follows: Let us apply the operator ad Dx to both sides of (3.2). As a result we find By collecting the coefficients before the independent operators we obtain a system of the differential equations for the coefficients λ 0 , λ 1 , . . . λ k . The system is overdetermined since all of the coefficients λ j are functions of a finite number of the dynamical variables u, u x , . . .. The consistency conditions of this overdetermined system generate integrability conditions for the hyperbolic type system (2.1) ). For instance, collecting the coefficients before W k we find the first equation of the mentioned system: which is also overdetermined. Below we use two different samples of the test sequences in order to find the function α n .

The first test sequence
Define a sequence of the operators in R(y, N ) due to the recurrent formula In the case of the first two members of the sequence we have already deduced commutation relations (see (2.4) above) which are important for our further studies By using these two relations and applying the Jacobi identity we get immediately It can be proved by induction that (3.5) is really a test sequence. Moreover it is easily verified that for k ≥ 0 where the factors p k , q k are evaluated as follows Due to the assumption that R(y, N ) is of finite dimension only a finite subset of the sequence (3.5) is linearly independent. So there exists M such that: where the operators Y 0 , Y 1 , W 1 , . . . , W M −1 are linearly independent and the tail might contain a linear combination of the operators Y 0 , Y 1 , W 1 , . . . , W M −2 . At the moment we are not interested in that part in (3.10).
Proof . Assume that Since the operators Y 0 , Y 1 are of the form Y 0 = ∂ ∂u 0 + · · · , Y 1 = ∂ ∂u 1 + · · · while W 1 does not contain summands like ∂ ∂u 0 and ∂ ∂u 1 then the factors µ 1 , µ 0 vanish. If in addition λ 1 = 0 then we have W 1 = 0. Now by applying the operator ad Dx to both sides of this relation we get due to (3.7) an equation which yields two conditions: Y 0 (α 1 u 1,x ) = α 1,u 0 u 1,x = 0 and Y 1 (α 0 u 0,x ) = α 0,u 1 u 0,x = 0. Those equalities contradict our assumption that ∂α(u n+1 ,un,u n−1 ) It is easy to check that equation (3.4) for the case of the sequence (3.5) takes the following form We simplify the formula (3.12) due to the relations A simple analysis of the equation (3.12) gives that λ = λ(u 0 , u 1 ). Therefore (3.12) gives rise to the equation By comparing the coefficients before the independent variables u 0,x , u 1,x we deduce an overdetermined system of the differential equations for λ: Let us derive and investigate the consistency conditions of the system (3.14). We differentiate the first equation with respect to u −1 and find Now we introduce a new variable z due to the relation α 0,u −1 = − 1 2 e z and reduce (3.16) to the Liouville equation z uu −1 = e z for which we have the general solution where P (u) and Q(u −1 ) are arbitrary differentiable functions. Thus for α 0 we can obtain the following explicit expression where H(u, u 1 ) is to be determined. Now we can find λ from the second equation in (3.14): Let us specify H(u, u −1 ) by replacing in (3.14) α 0 and λ in virtue of (3.17) ), (3.18). As a result we obtain Summarizing the reasonings above we can conclude that where the functions of one variable P (u), P ′ (u) and the integer M are to be found. The next step requires some additional integrability conditions. In what follows we derive them by constructing another test sequence.

The second test sequence
Now we concentrate on a test sequence generated by the operators Y 0 , Y 1 , Y 2 and their multiple commutators. It is more complicated than the previous sequence: (3.21) The members Z m of the sequence for m > 8 are defined due to the recurrence Proof . Firstly we note that the operators Z 0 , Z 1 , . . . Z 4 are linearly independent. It can be verified by using resonings similar to those from the proof of Lemma 2. We prove the lemma by contradiction. Assume that Now we specify the action of the operator ad Dx on the operators Z i . For i = 0, 1, 2 it is obtained from the relation Let us apply the operator ad Dx to both sides of (3.22) and obtain By comparing the coefficients before Z 4 in (3.23) we obtain the following equation A simple analysis of the equation (3.24) shows that λ = λ(u 0 , u 1 ). Hence the equation (3.24) splits down into two equations λ 4,u 0 = −α 0 λ 4 and λ 4,u 1 = −α 1,u 0 . The former shows that λ 4 = 0. Indeed if λ 4 = 0 then we obtain an expression for α 0 : α 0 = − (log λ 4 ) u 0 which shows that (α 0 ) u −1 = 0. It contradicts the assumption that α(u 1 , u 0 , u −1 ) depends essentially on u 1 and u −1 , therefore λ 4 = 0. Then (3.24) implies α 1,u 0 = 0 and it leads again to a contradiction.
Turn back to the sequence (3.21). For the further study it is necessary to specify the action of the operator ad Dx on the members of this sequence. It is convenient to divide the sequence into three subsequences and study them separately {Z 3m }, {Z 3m+1 }, and {Z 3m+2 }.
Lemma 4. Action of the operator ad Dx on the sequence (3.21) is given by the following relations Lemma 4 is easily proved by induction. Since the proof is quite technical we omit it.
Theorem 2. Assume that Z 3k+2 is represented as a linear combination of the previous members of the sequence (3.21) and neither of the operators Z 3j+2 with j < k is a linear combination of Z s with s < 3j + 2. Then the coefficient ν k is a solution to the equation Lemma 5. Suppose that all of the conditions of the theorem are satisfied. In addition assume that the operator Z 3k (operator Z 3k+1 ) is linearly expressed in terms of the operator Z i with i < 3k. Then in this decomposition the coefficient before Z 3k−1 vanishes.
Proof . Assume in contrary that λ = 0 in the formula Let us apply ad Dx to (3.27). As a result we find due to Lemma 4: Collect the coefficients before Z 3k−1 and obtain an equation the coefficient λ must satisfy to Due to our assumption above λ = 0 and hence Since λ depends on a finite number of the dynamical variables then due to equation (3.29) λ might depend only on u 1 and u 2 . Therefore (3.28) yields The variables u 1,x , u 2,x are independent, so the last equation implies It contradicts our assumption that α 1 depends essentially on u, u 2 . The contradiction shows that assumption λ = 0 is not true. That completes the proof.
Now in order to prove the theorem we apply the operator ad Dx to both sides of (3.25) and then simplify due to the relation from the Lemma 4. Comparison of the coefficients before Z 3k−1 implies equation (3.26).
Let us find the explicit expressions for the coefficients of the equation (3.26): and substitute them into (3.26): A simple analysis of (3.30) convinces that ν k might depend only on the variables u 0 , u 1 , u 2 . Therefore From the equations (3.30), (3.31) we obtain a system of the equations for the coefficient ν k : Substitute the preliminary expression for the function α given by the formula (3.20) into the equation (3.32) and get Integration of the latter with respect to u yields Since ν k,u 2 = H u 2 the equation (3.34) gives rise to the relation Now by integrating we obtain an explicit formula for H: Let us substitute the values of α and ν k found into the equation (3.33). We get a huge equation (3.38) Evidently due to our assumption ∂ ∂u 1 α(u 1 , u, u −1 ) = 0, ∂ ∂u −1 α(u 1 , u, u −1 ) = 0 the functions P ′ (u 2 ) and Q ′ (u) do not vanish. Therefore the variables Q ′2 (u 1 ) (P (u 2 ) + Q(u 1 )) 2 , P ′2 (u 1 ) (P (u 1 ) + Q(u)) 2 , P ′ (u 1 )Q ′ (u 1 ) (P (u 1 ) + Q(u))(P (u 2 ) + Q(u 1 )) are independent. By gathering the coefficients before these variables in (3.38) we get a system of two equations In this section we specify the function α given by (3.20). For this aim we should consider expansions (3.10) ), (3.25) using the fact that M = 2, k = 2. Let us rewrite the expansion (3.10) in the complete form where the functions P (u n ), Q(u n ) are connected with each other by the differential constraint Proof . Firstly by using relations (3.6), (3.7) and applying the Jacoby identity we get Evidently only one summand in (4.1) contains the term ∂ ∂u 1 , namely σY 1 , and only one summand contains the term ∂ ∂u , namely σY 0 . Hence σ = 0, δ = 0 and we have Now by applying the operator ad Dx to both sides of this relation we obtain Collecting the coefficients before W 2 , W 1 , Y 1 , and Y 0 we find the following system: Setting M = 2 in (3.12) we obtain equation (4.5). The overdetermined system (3.14) takes the form: We rewrite (4.6), (4.7) due to the relations as follows: We substitute (4.10), (4.11) into (4.12) and find that c(u) = 1 2 Q ′′ (u) Q ′ (u) . So we find that functions (4.10) ), (4.11) are given by (4.14) The functions P , Q must satisfy the equality Thus we have proved that if the expansion (3.10) holds then it should be of the form Or the same That completes the proof of Theorem 1. Let us define a sequence of the operators in R(y, N ) due to the following recurrent formula It slightly differs from (3.5) and can be studied in a similar way. We can easily check that the conditions (4.2) ), (4.3) provide the representatioñ Or the same with the coefficient Let us consider expansion (3.25) setting k = 2.
Thus we have proved that if the expansions (4.1) ), (4.20) hold then (4.20) should be as follows (4.51) Or the same In a similar way we check that the same conditions (4.21), (4.22) provides the representations The proof of Theorem (5) can be found in Appendix.

Comments on the classification result
In this section we briefly discuss the statements of the Theorems 4, 5 claiming that the lattice (1.1) is integrable in the sense of the Definition only for two choices of the function α given by (4.21) and (4.22). In both cases the lattice has a functional freedom which is removed by an appropriate point transformation. Therefore we have Theorem 6. Any lattice (1.1) integrable in the sense above is reduced by the point transformation v = p(u) to the following lattice: Specify the point transformations applied to the lattices. Change of the variables w = P (u) reduces (4.21) to w n,xy = w n,x w n,y 1 The latter is connected with (5.1) by the change of the variables v n = (−c 1 ) n w n − c 2 1+c 1 if c 1 = −1 and by v n = w n − c 2 n in the special case c 1 = −1.
Theorem 6 claims that the following system of the hyperbolic type equations admits a complete set of functionally independent x− and y−integrals for any natural N and arbitrary choice of the constant parameters c 0 , c 1 .

Remark 1.
The following lattice (see [15]) q n,xy = q n,x q n,y f (q n+1 − q n ) − f (q n − q n−1 ) , is reduced by the point transformation to (5.1). Namely if b = 0 then f (q) = b tg (b(q + c)) = −ib · th (ib(q + c)), where c is the constant of integration, i is the imaginary unit. So we have the lattice q n,xy = q n,x q n,y (−ib) th (ib(q n+1 − q n + c)) − th (ib(q n − q n−1 + c)) .
The change of variables q n = − i b v n −nc reduces the last lattice to (5.1). If b = 0 then f (q) = 1 q+c . By the change of variables q n = v n − nc we obtain (5.1).

A Appendix. The proof of Theorem 5
Let us introduce a special notation Y i k ,...,i 0 for the multiple commutators. It is defined consecutively Number k is called the order of the operator (A.1)). In order to prove the Theorem 5 we show that the ring R(y, N ) is of finite dimension. Actually we construct the basis in R(y, N ) containing the operators

A.1 The base case of the mathematical induction
In the previous section we have proved that In what follows we will use the following relations which are easily verified: 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) we always use Lemma 1. That is why we need in explicit expressions for [D x , Y i k ,...,i 0 ]. 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. In what follows we need in the formulas which are some versions of the formula (A.10) from the Lemma 6. Now using formulas by direct calculations we prove that where λ (R) , λ (M ) , and λ (L) are defined by the formulas (4.43) ), (4.42), and (4.44) or by the formulas (4.47), (4.46), and (4.48) correspondingly. Now, having explicit formulas for the small order commutators we are ready to work out an induction hypothesis.
This equation implies that µ = µ(u, u k ) and splits down into two equations as follows: Then µ = 0.

Conclusions
A problem of the integrable classification is studied for two-dimensional lattices (1.1). The classification method used in the article is based on the following observation. Any integrable lattice of the form u n,xy = g(u n+1 , u n , u n−1 , u n,x , u n,y ) (A.50) admits a special kind of the cut off constraints such that under these boundary conditions imposed at two different points N 1 and N 2 the lattice (A.50) is reduced to a Darboux integrable finite system of hyperbolic type equations. The lattice (1.1) is a particular case of (A.50) for which the mentioned degenerate boundary condition is easily found: u n 0 = c 0 . This circumstance drastically simplifies the situation. In general the degenerate BC is also unknown that causes an additional trouble. 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 reductions x = y of this lattice satisfy completely also the symmetry integrability test (see [18]).