Classification of integrable discrete Klein-Gordon models
aa r X i v : . [ n li n . S I] F e b Classification of integrable discrete Klein-Gordon models
Ismagil T. Habibullin Ufa Institute of Mathematics, Russian Academy of Science,Chernyshevskii Str., 112, Ufa, 450077, RussiaElena V. Gudkova Department of Applied Mathematics and Mechanics,Ufa State Petroleum Technical University,Kosmonavtov str., 1, Ufa, 450062, Russia
Abstract
The Lie algebraic integrability test is applied to the problem of classification of inte-grable Klein-Gordon type equations on quad-graphs. The list of equations passing the testis presented containing several well-known integrable models. A new integrable exampleis found, its higher symmetry is presented.
Keywords: quad-graph equations, classification, characteristic vector fields, Lie ring, inte-grability conditions, higher symmetry.
We study the integrability problem for the quad–graph equation of the form u , = f ( u, u , ¯ u ) . (1)Such kind equations have a large variety of applications in physics, biology, architecture, etc.Here the unknown u = u ( m, n ) is a function of two discrete variables m, n . For the sakeof convenience we use the following notations: u k = u ( m + k, n ), ¯ u k = u ( m, n + k ), u , = u ( m + 1 , n + 1). Function f is supposed to be locally analytic, it depends essentially on allthree arguments. In other words equation (1) can be rewritten in any of the following forms u i,j = f i,j ( u, u i , ¯ u j ) , with i = ± , j = ± . (2)Nowadays various approaches are known for studying integrable discrete phenomena. Theproperty of consistency around a cube [1], has been proposed as the integrability criterionfor quadrilateral difference equations [2], [3], [4]. Symmetry approach to the classification ofintegrable systems is adopted to discrete case [5], [6], [7], [8], [9]. Another characteristic propertyof an integrable equation is the vanishing of its algebraic entropy [10]. Alternative methods areused in [11], [12], [13]. In this article we study discrete phenomena from some different pointof view.Years ago it was observed that characteristic Lie algebras, introduced in [14], in the case ofintegrable hyperbolic type PDE’s like sine-Gordon and Tzitzeica-Zhiber-Shabat equations havea very specific property. The dimensions of the linear spaces spanned by multiple commutatorsof the generators grow essentially slower than in generic case. In [15] the problem of rigorous e-mail: [email protected] e-mail: [email protected] u x,y = f ( u, u x ) . In the recent article [16] we introduced and successfully tested a classification scheme (calledalgebraic test) based on investigation of multiple commutators of characteristic vector fieldsdefined by equation (1). Now we use the algebraic test (an explanation is given in Section 2below) to classify discrete Klein-Gordon type equation on quad graph u , + u = g ( u + ¯ u ) . (3)The list of equations which partially passed the algebraic test is studied additionally by applyingthe symmetry test. It is remarkable that the final list contains in addition to the well-knownequations as the discrete potential KdV equation, the discrete Liuoville equation and a discreteanalogue of the sine-Gordon equation also a new integrable model (see below the discussion onCorollary of theorem 2 and theorem 3): u , = αu ¯ u − βu ( α + u ¯ u ) . (4)Remind that the problem of complete description of integrable Klein -Gordon equations u x,y = g ( u ) (5)has been solved years ago by A.V.Zhiber and A.B.Shabat [17]. The authors proved that theonly integrable equations in (5) arelinear equation;the Liouville equation u x,y = e u ;the sine-Gordon equation u x,y = sin u ;the Tzitzeica-Zhiber-Shabat equation u x,y = e u + e − u .The article is organized as follows. In section 2 we define characteristic vector-fields and thetest Lie ring, formulate the algebraic test, consider an illustrative example. The classificationproblem for the model of the form u , + u = g ( u + ¯ u ) passing the test is investigated insection 3. The result of classification is summarized in Theorems 3 and 5. Note that for any integers p and q the variable u p,q is expressed in terms of the variables { u i , ¯ u j } ∞ i,j = −∞ in a recurrent way. Hence the variables u i , ¯ u j are called dynamical variables.They considered as independent ones. For example, u , = f ( u , u , u , ) = f ( u , u , f ( u, u , ¯ u )) , and u − , − = f − , − (¯ u − , u − , − , ¯ u − ) = f − , − (¯ u − , f − , − ( u, u − , ¯ u − ) , ¯ u − ) , Introduce the shift operators D and ¯ D , shifting the first and respectively the secondinteger arguments: Dh ( m, n ) = h ( m + 1 , n ) , ¯ Dh ( m, n ) = h ( m, n + 1). Explain the ac-tion of the operators D and ¯ D on the functions of dynamical variables. For the function h = h ( u i , u i − ..., u i ′ , ¯ u j , ¯ u j +1 , ..., ¯ u j ′ ) we have h k = D k h = h ( u i + k , u i + k − ..., u i ′ + k , ¯ u k,j , ¯ u k,j +1 , ..., ¯ u k,j ′ )2nd similarly ¯ h k = ¯ D k h = h ( u i,k , u i,k − ..., u i ′ ,k , ¯ u k + j , ¯ u k + j +1 , ..., ¯ u k + j ′ ) . In these two formulas one has to replace the double shifts u α,β through dynamical variables asit is discussed above.Characteristic vector fields for the hyperbolic type PDE were introduced by E.Goursat[18] in 1899. They provide a very effective tool for classification of Liouville type integrablesystems. For instance in [18] a list (almost complete) of hyperbolic type PDE was found ad-mitting integrals in both directions by using a method based on the notion of characteristicvector fields. Interest to the subject renewed after the paper [14], where the exhaustive de-scription of exponential type systems is given admitting complete set of integrals. The conceptof characteristic vector fields was adopted to the quad graph equations in [19]. In the recentpapers [20], [21], [22] the characteristic vector fields were used for the purpose of classification ofsemi-discrete Liouville type equations. All of these studies concern with the Liouville type inte-grability. In [15] and [16] some new applications of these important notions are suggested. Letus remind some necessary definitions. At first suppose that equation (1) admits a non-trivial n -integral, i.e. there is a function I depending on a finite number of the dynamical variables I = I ( u − j , u − j +1 , ...u k ) satisfying the equation ¯ DI = I . In an enlarged form the last equationis I ( r − j +1 , r − j +2 , ..., r, ¯ u, f, f , ...f k − ) = I ( u − j , u − j +1 , ...u k ) , (6)where the function r = f − , ( u, u − , ¯ u ) is defined in (2). Since the right hand side of (6) doesnot depend on the variable ¯ u we find ∂∂ ¯ u ¯ DI = ∂∂ ¯ u ¯ I = 0. It implies the equation Y I = 0 wherethe operator Y is defined as follows Y := ¯ D − ∂∂ ¯ u ¯ D . Direct computations give (see [16], [19]) Y = ∂∂u + x ∂∂u + 1 x − ∂∂u − + xx ∂∂u + 1 x − x − ∂∂u − + ..., (7)where x = ¯ D − ( ∂f ( u,u , ¯ u ) ∂ ¯ u ) = − ∂f , − ( u,u , ¯ u − ) /∂u∂f , − ( u,u , ¯ u − ) /∂u . We call the operators X := ∂∂ ¯ u − and Y characteristic vector fields. Now it is evident that the map f ( u, u , ¯ u ) = ⇒ Y is correctlydefined for any f due to the formula (7).Denote through T the set of vector fields obtained by taking all possible multiple commuta-tors of the operators X and Y , and taking all linear combinations with coefficients dependingon a finite number of the dynamical variables ¯ u − , u , u ± , u ± , ... . Evidently the set T hasa structure of the Lie ring. We call it test ring of the equation (1) in the direction of n . In asimilar way one can define the test ring ¯ T in the direction of m .Notice that for Liouville type integrable equations of the form (1) both rings T and ¯ T areof finite dimension. Actually the test ring is a subset of the characteristic Lie ring [19]-[22].Denote through V j the linear space over the field of locally analytic functions spanned by X , Y and all multiple commutators of X and Y of order less than or equal to j such that: V = { X, Y } , V = { X, Y, [ X, Y ] } , . . . . Remind that such kind sequences of linear spaces have important applications in geometry (see [23], [24]). Introduce the function ∆( k ) = dim V k +1 − dim V k . The following conjecture isapproved by numerous examples.
Conjecture (algebraic test) . Any integrable model of the form (1) satisfies the followingcondition: there is a sequence of natural numbers { t k } ∞ k =1 such that ∆( t k ) ≤ The authors thank E.Ferapontov for drawing their attention to this fact. T admits an automorphism, generated by the shift operator D , T ∋ Z Aut → DZD − ∈ T, (8)which plays crucial role in our further considerations. It is important that X and Y consideredas operators on the set of functions depending on the variables ¯ u − , u, u ± , u ± , ... satisfy thefollowing conjugation relations [16] DXD − = pX and DY D − = 1 x Y, (9)where p = D ( ∂f − , − ( u,u − , ¯ u − ) ∂ ¯ u − ) = ∂f , − ( u,u , ¯ u − ) /∂ ¯ u − . Indeed, specify the coefficients of theoperator DXD − = P a i ∂∂u i + p ∂∂ ¯ u − by applying it to the dynamical variables and find that a i = DXD − u i = 0 for any integer i . Moreover p = DXD − ¯ u − = DXf − , − ( u, u − , ¯ u − ) = D ( ∂f − , − ( u,u − , ¯ u − ) ∂ ¯ u − ). In a similar way one can prove the second formula. Really apply theoperator DY D − = P c i ∂∂u i + d ∂∂ ¯ u − to u i and find c j = D ( Y u j − ) = Y u j . Then evaluate d = DY D − ¯ u − = f − , − u + x − f − , − u − . Since u − , − = f − , − ( u, u − , ¯ u − ) one gets the equation u = f − , − ( f ( u, u , ¯ u ) , u , ¯ u ). Let us differentiate it with respect to ¯ u and find D ¯ D ( ∂f − , − ∂u ) ∂f∂ ¯ u + D ¯ D ( ∂f − , − ∂u − ) = 0 or the same ∂f − , − ∂u + D − ¯ D − ( ∂f∂ ¯ u ) ∂f − , − ∂u − = 0. Now due to the equation x = ¯ D − ( ∂f∂ ¯ u ) one concludes that d = 0. Lemma Suppose that Z = P ∞−∞ b j ∂∂u j ∈ T satisfies the following two conditions:1) DZD − = cZ for some function c ;2) b j ≡ for some fixed value of j = j .Then Z = 0 . Proof of the Lemma can be found in [16].
Example . As an illustrative example we consider the following discrete Liouville typeequation (see [25]) u , = 1 u ( u − u − . (10)Find first an explicit form of the characteristic vector field Y . Since f = u ( u − u −
1) and ∂f∂ ¯ u = u ( u −
1) then x = ¯ D − u ( u −
1) = 1¯ u − ( u , − − . (11)Express u , − through the dynamical variables due to equation (10). Apply the operator ¯ D − to both sides of (10) and find u = u − ( u , − − u − x = u u − , therefore Y = ∂∂u + u u − ∂∂u + u − − u ∂∂u − + u u ( u − u − ∂∂u + ..., (12)For this equation p = u − u = x , hence DXD − = pX , DY D − = pY . Evaluate D [ X, Y ] D − = p [ X, Y ] + pX ( p ) Y − pY ( p ) X , since X ( p ) = Y ( p ) = 0 we get D [ X, Y ] D − = p [ X, Y ], moreover[
X, Y ] = P ∞ j =1 a j ∂∂u j + a − j ∂∂u − j . Now due to the Lemma 1 one obtains [ X, Y ] = 0 . So dimensionof the ring T for equation (10) equals two and ∆( k ) = 0 for all k ≥ Equations of the form u , + u = g ( u + ¯ u ) In this section we apply the
Conjecture to the following Klein-Gordon type particular classof discrete model (1) u , + u = g ( u + ¯ u ) . (13)where the function g is to be determined. Classification scheme . Obviously equation (1) passing the algebraic test above shouldsatisfy one of the conditions:i) ∆(0) < ∆ max (0) = 1;ii) ∆(0) = ∆ max (0), ∆(1) < ∆ max (1) = 2;iii) ∆(0) = ∆ max (0), ∆(1) = ∆ max (1), ∆(2) < ∆ max (2) = 3;iv) ∆(0) = ∆ max (0), ∆(1) = ∆ max (1), ∆(2) = ∆ max (2) and ∆( k ) ≤ k > max ( k ) stands for the greatest value of ∆( k ) for equation (1) when f ( u, u , ¯ u ) rangesthe class of arbitrary functions.Remark that in the case of equation (13) investigation of the first three particular cases i )- iii ) allows one to extract a very short list of equations expected to be integrable. The list isexhaustive because the case iv ) is never realized (see below Corollaries of Theorems 2 and 4).Introduce vector fields R = [ X, Y ], P = [ X, R ] , Q = [ Y, R ], R = [ X, Q ], W = [ Y, Q ], Z = [ X, P ]. Using these vector fields we can span in addition to V and V two more linearspaces: V = V + { P , Q } , V = V + { W, Z, R } . In order to evaluate ∆( k ) we will use the automorphism (8). At first evaluate the factors x and p in formula (9) for the case (13). We have x = p = g ′ ( g − ( u + ¯ u − )), where the function β = g − ( α ) is the inverse to the function α = g ( β ). Inversely, knowing x = x ( u + ¯ u − ) onecan recover g ( β ) by using the equation β = g − ( α ) = Z ( g − ( α )) ′ dα = Z dαg ′ ( g − ( α ) = Z dαx ( α ) . (14)Specify the action of the characteristic operators X and Y on the variable x . Evidently, Xx = x ′ , Y x = xx ′ . It is found by direct calculation that DR D − = R − x ′ x Y − x ′ X,DP D − = xP − x ′ R − rY − xrX, r = x ′′ − x ′ x ,DQ D − = 1 x Q + x ′ x R − x ′′ x Y − x ′′ X,DW D − = 1 x W + ( 2 x ′′ x − x ′ x ) R − x ′′′ x Y − x ′′′ X, (15) DZD − = x Z + ( x ′ − xx ′′ ) R − qY − xqX, q = xx ′′′ − x ′ x ′′ + x ′ x ,DR D − = R − x ′ x Q + x ′ P − x ′ x R − sY − xsX, s = x ′′′ − x ′ x ′′ x , G of all multiple commutators of X and Y . Lemma 2 . The coefficients of any operator in G are functions of a finite number of thedynamical variables x , x ± , x ± , ... . Proof . Here x ( α ) is a fixed function of the argument α = u + ¯ u − . Therefore one canwrite x ′ = φ ( x ) for some function φ . Then X ( x ) = − φ ( x ) and Y ( x ) = xφ ( x ) =: ψ ( x ). Byusing the conjugation relations DXD − = xX and DY D − = x Y one derives that X ( x j ) = φ j ( x, x , ...x j ) and Y ( x j ) = ψ j ( x, x , ...x j ). Similarly X ( x − j ) = φ − j ( x, x − , ...x − j ) and Y ( x − j ) = ψ − j ( x, x − , ...x − j ). Now evidently R = X ( x ) ∂∂u + X ( x − ) ∂∂u − + X ( xx ) ∂∂u + · · · satisfies thestatement of the lemma. Due to the formulas R ( x ) = X ( x ) x ′ and DR D − = R + x ′ x Y − x ′ X one gets R ( x j ) = φ j ( x, x , ...x j ). Obviously the proof can be completed by using induction. Theorem 2 . Suppose that an equation of the form (13) satisfies one of the conditions i ) − iii ) of the Classification scheme then function x = x ( α ) solves the following ordinarydifferential equation x ′ = ( x + 1) γ + xν (16) with the constant coefficients γ, ν . Proof of the theorem 2 . Begin with the case i ), suppose that ∆(0) = 0, then we have R = λX + µY . It is evident that R = X ( x ) ∂∂u + ... , X = ∂∂ ¯ u − and Y = ∂∂u + x ∂∂u + ... hence λ = µ = 0 and therefore R = 0. By applying the automorphism above to both sides of thelast equation one finds − x ′ x Y − x ′ X = 0 . Since X and Y are linearly independent we get equation x ′ = 0 which is a particular case of(16). Evidently its solution is x = c and due to (14) it can be found that β = g − ( α ) = c α + c .Thus our equation α = g ( β ) (see (13)) is linear u , + u = c ( u + ¯ u + c ). For this casedim T = 2 so that ∆( k ) = 0 for k ≥
0. In a similar way one checks that condition ii ) leads to(16). Indeed suppose that ∆(0) = 1 and ∆(1) < P = νQ + ǫR . (17)Here due to Lemma 2 functions ν and ǫ might depend only on x , x ± , x ± . . . . Apply theautomorphism (8) to both sides of (17) then simplify due to equations (15): x ( νQ + ǫR ) − x ′ R − rY − xrX = D ( ν )( 1 x Q + x ′ x R − x ′′ x Y − x ′′ X ) + D ( ǫ )( R − x ′ x Y − x ′ X ) . Comparison of the coefficients before linearly independent operators gives rise to the conditions Q : xν = 1 x D ( ν ); R : − x ′ + xǫ = x ′ x D ( ν ) + D ( ǫ ); Y : − r = − x ′ x D ( ǫ ) + x ′′ x D ( ν ); X : − xr = − x ′ D ( ǫ ) − x ′′ D ( ν ) . Simple analysis of these equations implies that equation (17) holds if and only if the followingthree conditions valid: ν = 0, ǫ = const, x ′ = ǫ ( x − . Actually, under these conditionsthe last two equations above are satisfied automatically. In a similar way one can check that Q = νP − ǫR is equivalent to the same three conditions. Hence if ∆(1) < k ) = 0 for any natural k ≥
1. Thus in this case dim T = 3.6et us suppose now that ∆(0) = 1, ∆(1) = 2 and ∆(2) ≤ iii ). At first consider the case when Z is linearly expressed through the other vector fields insubspace V : Z = γR + δP + ǫQ + φR + ψW. (18)Apply the automorphism (8) to both sides of equation (18) and compare coefficients beforelinearly independent operators W : x ψ = D ( ψ ) 1 x ,R : x φ = D ( φ ) ,Q : x ǫ = D ( ǫ ) 1 x + x ′ x φ,P : x δ = D ( δ ) x + x ′ φ,R : x γ + x ′ − xx ′′ = D ( δ ) x ′ + D ( γ ) . Since x = x ( u + ¯ u − ) we have ψ = 0, φ = 0, ǫ = 0, δ = 0, γ = const . Comparison ofthe coefficients of X and Y gives one more equation xq = γx ′ . Finally we get two ordinarydifferential equations for x which are absolutely the same as in the case of the equation u , − u = g ( u − ¯ u ) (see [16]): x x ′′′ − xx ′ x ′′ + x ′ = γx ′ , ( x − γ + x ′ − xx ′′ = 0 . The compatibility condition of these equations is equivalent to equation (16). In this case wehave Z = γR . It is remarkable that x solves equation (16) if and only if W is linearly expressedthrough the other elements of V and then W = γR . And the last possibility is when R islinearly expressed through X , Y , R , P , Q , W , Z . In this case x solves the equation x ′ = 0.The proof of the theorem is completed.In order to find x = x ( α ) evaluate the integral H ( x ) := Z dx q ( x + 1) γ + xν = α − α . (19)For the case γ = 0 the answer is given by the formula H ( x ) = 1 √ γ ln(2 √ x + 1 + xb + 2 x + b ) , b = νγ . (20)Now find x by solving the equation H ( x ) = α − α . x ( α ) = 14 e √ γ ( α − α ) − ν γ − (1 − ν γ ) e −√ γ ( α − α ) . In order to get the corresponding quad-graph equation (13) integrate again: β = g − ( α ) = Z dαx ( α ) . (21)Integration gives β = 1 √ γ ln (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) e √ γ ( α − α ) − b − e √ γ ( α − α ) − b + 2 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + β . (22)7hen the equation searched is given by the formula (take √ γ = 1) a e u , + u +¯ u + u + a e u , + u + a e u +¯ u + a = 0 , where a = e − α − β , a = − e − α , a = (2 − b ) e − β , a = 2 + b . By setting u = ln v one reducesit to a quadrilinear form a v , v ¯ v v + a v , v + a v ¯ v + a = 0 , (23)with arbitrary constants a j .If γ = 0, then x = x ( α ) solves the equation x ′ = √ νx . Thus obviously x = c ( α − α ) andtherefore β = g − ( α ) = Z dαc ( α − α ) = dα − α − β , d = − /c. The corresponding equation (13) follows from ( α − α )( β − β ) = d. It is of the form( u , + u − α )( u + ¯ u − β ) = d. (24) Corollary of Theorem 2 . If for a nonlinear chain (13) one of the conditions i ) − iii ) issatisfied then the chain is of one of the form:1) u , + u = c ( u + ¯ u + c ) ;2) a u , u ¯ u u + a u , u + a u ¯ u + a = 0 ;3) ( u , + u − α )( u + ¯ u − β ) = d . Let us study in details the list obtained. It is remarkable that it contains a new integrableexample of equation (1).
Theorem 3 . The chain 2) in the Corollary of Theorem 2 admits a symmetry of the form u t = g ( u , u − , u, ¯ u , ¯ u − ) (25) if and only if the condition holds k := a a = ± . Under this condition the symmetry is of theform u t = λu ku + u − ku − u − + µ ¯ u k ¯ u + ¯ u − k ¯ u − ¯ u − . Chain 3) admits a symmetry of the form (25) if and only if α = β and then it is reduced tothe discrete potential KdV equation. Theorem can be proved by using technique developed in [5]. Actually it claims that thechain 2) with a = ± a passes the symmetry test. For the case k = 1 (or a = a ) the chain isa particular case of the Viallet equation (see [10]), this means that by a point transformationit can be reduced to one of the equations of ABS list [4]. Higher symmetries for this case arefound in [6]. For some special choice of the parameters chain 2) with k = − u , u ¯ u u −
1) + u , u − u ¯ u = 0 (26)found by R. Hernandez Heredero and C. Scimiterna. Recently in [5] it was proved that equation(26) admits higher symmetries. To the best of our knowledge general case of the equation 2)with k = − a = 0 and a = − a it is reduced to the well-known discrete Liouville equation e u , + u = e u +¯ u + 1 , having nontrivial integrals, i.e. functions solving the equations ¯ DF = F and D ¯ F = ¯ F , with F = e u − u + e u − u and ¯ F = e u − ¯ u + e ¯ u − ¯ u .The second part of the theorem claims that case 3) admits a symmetry of small orderonly if it coincides up to a point transformation with the discrete pkdv equation. The ques-tion is still open whether it admits any symmetry of a more complicated form, say u t = g ( u , u , u − , u, ¯ u , ¯ u − , ¯ u − ). There are some technical difficulties with applications of the sym-metry approach to discrete models (see discussion in [5]).Turn back to the set G consisting of X , Y and their all multiple commutators. Assign twointegers: order and degree to each element in G . Define the order of an element Z ∈ G asa number of its factors X and Y minus one. For instance ord [ X, Y ] = 1, ord [ X, [ X, Y ]] = 2and so on. Define the degree deg ( Z ) of Z as the exponent k in the expansion of the operatorobtained by applying the automorphism (8): DZD − = x k Z + ... , where the tail is a linearcombination of the elements with order less than ord ( Z ). Denote through G i,j a subset of G containing elements with the order i and the degree j . Let G i = S j G i,j be the union of all G i,j with one and the same i . Evidently the set G i,i − (as well as G i, − i +1 ) contains the only element Z i,i − = ad iX ( Y ) up to the factor − Z i, − i +1 = ad iY ( X ) upto the factor − ad is defined as follows ad X ( Y ) = [ X, Y ]. Theorem 4 . Suppose that Z k,k − (or Z k, − k +1 ) is in the basis of the linear space V k ⊃ G k forany integer k : ≤ k < N , but Z N,N − (respectively Z N, − N +1 ) is linearly expressed through theother operators in V N then function x = x ( u + ¯ u − ) solves an equation of the form x ′ = ǫ ( x − with the constant coefficient ǫ . The proof of the theorem is done exactly in the same way as the proof of Theorem 3 from[16], but presented in the article for the readers’ convenience.
Lemma For any integer k ≥ we have DZ k +1 ,k D − = x k Z k +1 ,k + c k x ′ x k − Z k,k − + · · · , (27) where c k ≥ (but c k > for k > and the tail contains a linear combination of the operatorswith the order less than k . Prove the Lemma 3 by induction. From the list of equations (15) one gets for Z = Z , and R = Z , the following representation DZ , D − = x Z , + ( x ′ − xx ′′ ) Z , − qY − xqX, q = xx ′′′ − x ′ x ′′ + x ′ x (28)showing that the statement is true for the case k = 3. Suppose now that DZ k,k − D − = x k − Z k,k − + c k − x ′ x k − Z k − ,k − + · · · and evaluate DZ k +1 ,k D − : DZ k +1 ,k D − = [ xX, x k − Z k,k − + c k − x ′ x k − Z k − ,k − + · · · ] = x k Z k +1 ,k + c k x ′ x k − Z k,k − + · · · where c k = c k − + ( k − > c k − ≥
0. The proof is completed.
Proof of the Theorem 4 . Suppose that Z N,N − = X ord ( Z ν )= N a ν Z ν + X ord ( Z µ )= N − b µ Z µ + · · · , (29)9here Z ν and Z µ range the basis of V N and the tail contains a linear combination of theoperators of less order. Apply the automorphism (8) to both sides of (29): x N − ( X ord ( Z ν )= N a ν Z ν + X ord ( Z µ )= N − b µ Z µ + · · · ) − x N − x ′ c N − Z N − ,N − + · · · == X ord ( Z ν )= N D ( a ν )( x k ν Z ν + · · · ) + X ord ( Z µ )= N − D ( b µ )( x k µ Z µ + · · · ) . Collect the coefficients before Z ν and get x N − a ν = x k ν D ( a ν ) , k ν = N − . (30)Due to Lemma 2 functions a ν and b µ depend on x and its shifts. Moreover it follows from(30) that a ν cannot depend on x , x ± , x ± , . . . at all. Therefore the only possibility is a ν = 0.Compare now the coefficients before Z N − ,N − and find: x N − b − c N − x N − x ′ = x N − D ( b ) , (31)where b is the coefficient of Z N − ,N − in the expansion (29). A simple analysis of equation (31)shows that b is constant. Thus (31) is equivalent to the equation x ′ = ǫ ( x −
1) with ǫ = b/c N − . Corollary of Theorem 4 . The case iv ) of the Classification scheme is never realized.
Proof . Suppose on contrary that such a case is realized. Then at least one of the vectorfields Z k,k − or Z k, − k +1 should be linearly expressed through other elements of V k , otherwise∆( k ) ≥
2. Therefore due to Theorem 4 we have x ′ = ǫ ( x −
1) which corresponds to the case ii )∆(0) = 1, ∆(1) < T = 3. The contradiction shows that our assumption is not true.The proof is completed.Let us summarize the results of reasonings above in the following theorem. Theorem 5 . Suppose that an equation of the form (13) passes the algebraic test then it isof one of the form given in Corollary of Theorem 2.
A classification scheme introduced recently in [16] is applied to quad-graph Klein-Gordon typeequation. The list of equations passed the test contains along with the well-known integrablemodels some new example which passes also the symmetry test.
Acknowledgments
The authors thank Prof. A.V.Zhiber and Dr. R.N.Garifullin for valuable advises. This work ispartially supported by Russian Foundation for Basic Research (RFBR) grants
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