On the Classification of Darboux Integrable Chains

On the Classiﬁcati on of Darb o ux In tegrable Chains Ismagil Habibullin 1 Nataly a Z heltukhina Aslı P ek can Departmen t of Mathematics, F acult y of Science, Bilk en t Univ ersity , 06800, Ank ara, T urke y Abstract W e study diﬀeren tial-diﬀerence equation of the form t x ( n + 1) = f ( t ( n ) , t ( n + 1) , t x ( n )) with unknown t = t ( n, x ) dep en ding on x , n . The equation is called Darb oux in tegrable, if there exist fun ctions F (calle d an x -integ r al) and I (called an n -integ r al), b oth of a ﬁn ite n um b er of v ariables x , t ( n ), t ( n ± 1), t ( n ± 2), . . . , t x ( n ), t xx ( n ), . . . , such that D x F = 0 and D I = I , where D x is the oper ator of tota l diﬀerentia tion with r esp ect to x , and D is the shift op erato r: D p ( n ) = p ( n + 1). The Darb oux in tegrabilit y prop ert y is reform u lated in terms of charac teristic Lie algebras that giv es an eﬀectiv e tool for classiﬁcatio n of in tegrable equations. The complete list of equations of the form ab ov e adm itting nontrivia l x -in tegrals is give n in the case when the function f is of the sp ecial form f ( x, y , z ) = z + d ( x, y ). Keywor ds: semi-discrete chain, class iﬁcation, x -in tegr a l, n -in tegral, c ha racteristic Lie algebra, in tegrability conditions. 1 In tro duction In this pap er w e study integrable semi-discrete c hains of the follo wing form t x ( n + 1) = f ( t ( n ) , t ( n + 1) , t x ( n )) , (1) where the unkno wn t = t ( n, x ) is a function of tw o indep enden t v a riables: discrete n and contin uous x . Chain (1) can also b e in terpreted as an inﬁnite system of ordinary diﬀeren tial equations for the sequence of the v ariables { t ( n ) } ∞ n = −∞ . Here f = f ( t, t 1 , t x ) is assumed to b e lo cally analytic function of three v ar iables satisfying at least lo cally the condition ∂ f ∂ t x 6 = 0 . (2) F or the sak e of con v enience w e intro duce subindex denoting shifts t k = t ( n + k , x ) (k eep t 0 = t ) and deriv ativ es t x = ∂ ∂ x t ( n, x ) , t xx = ∂ 2 ∂ x 2 t ( n, x ), and so on. W e denote through D and D x the shift op erator and, corresp ondingly , the op erator of total deriv at iv e with resp ect to x . F or instance, D h ( n, x ) = h ( n + 1 , x ) and D x h ( n, x ) = ∂ ∂ x h ( n, x ). Set of all the v ariables { t k } ∞ k = −∞ ; { D m x t } ∞ m =1 constitutes the set of dynamical v a riables. Belo w we consider the dynamical v ariables as indep enden t ones. Since in the literature t he term ”in tegra ble” has v ar io us meanings let us sp ecify the meaning used in the article. In tro duce ﬁrst notions of n - and x -in tegrals [1]. F unctions I and F , b oth dep ending on x and a ﬁnite num b er of dynamical v ariables, are called resp ectiv ely n - and x -in tegrals of (1), if D I = I and D x F = 0. Deﬁnition . Chain (1) is called integrable (D arb oux inte g rable) if it admits a non trivial n -in tegra l and a non trivial x -integral. 1 e-mail: habibullin i@mail.rb.ru, (O n leav e from Ufa Institute of Mathematics, Russian Academy of Science, Chernyshevskii Str. , 112, Ufa, 4 5007 7, Russia ) 1 Darb oux in tegrability implies the so-called C-in tegrabilit y . Kno wing b oth in tegrals F and I a Cole-Hopf t yp e diﬀeren tia l substitution w = F + I reduces the equation (1) to the discrete vers ion of D’Alem b ert w a ve equation w 1 x − w x = 0. Indeed, ( D − 1) D x ( w ) = ( D − 1) D x F + D x ( D − 1) I = 0 . It is remark able that an in tegrable chain is reduced to a pair consisting of an ordinary diﬀeren tial equation a nd an ordinary diﬀerence equation. T o illustrate it note ﬁrst that a ny n -in tegral might dep end only o n x and x -deriv ativ es of the v ariable t : I = I ( x, t, t x , t xx , ... ) a nd similarly any x -integral dep ends only on x and the shifts: F = F ( x, t, t ± 1 , t ± 2 , ... ). Therefore eac h solution of the integrable c hain (1) satisﬁes t w o equations: I ( x, t, t x , t xx , ... ) = p ( x ) , F ( x, t, t ± 1 , t ± 2 , ... ) = q ( n ) with prop erly c hosen functions p ( x ) and q ( n ). No wada ys the discrete phenomena are studied inten siv ely due to their v arious applicatio ns in ph ysics. F or the discussions and references w e refer to the articles [1], [2], [3], [4], [5]. Chain (1) is very close to a w ell studied ob ject – the partial diﬀeren tial equation of the h yp erb olic t yp e u xy = f ( x, y , u , u x , u y ) . (3) The deﬁnition of integrabilit y fo r equation (3) was in tro duced b y G . Darb oux. The famous Liouville equation u xy = e u pro vides an illustrativ e example of the Darb oux in tegra ble equation. An eﬀectiv e criterion of integrabilit y of (3) w as disco v ered b y Darb oux himself: equation (3 ) is in tegra ble if and only if the Laplace sequence o f the linearized equation terminates a t b oth ends (see [6], [7], [8]). This criterion of integrabilit y w as used in [8], where the complete list o f a ll D arb oux integrable equations of form (3) is giv en. An alternativ e appro ac h to the classiﬁcation problem ba sed on the notion o f t he c ha r acteristic Lie algebra of h yp erbo lic type system s was in tro duced years a go in [9], [10]. In these articles an algebraic criterion of Darb oux in tegrability prop ert y has b een form ula ted. An impo rtan t classiﬁcation r esult w as o btained in [9] for the exponential system u i xy = exp ( a i 1 u 1 + a i 2 u 2 + ... + a in u n ) , i = 1 , 2 , ..., n. (4) It w as prov ed that system (4) is Darb oux integrable if a nd only if the matrix A = ( a ij ) is the Cartan matrix of a semi-simple Lie alg ebra. Prop erties of the characteristic Lie algebras of the h yp erb olic systems u i xy = c i j k u j u k , i, j, k = 1 , 2 , ..., n (5) ha ve b een studied in [1 1], [12]. Hyp erb olic systems of general form admitting in tegrals are studied in [13]. A promising idea of adopting the c haracteristic Lie algebras to t he problem of classiﬁcation of the h yp erb o lic systems whic h are in tegrated by means of the inv erse scattering tra nsfor ms metho d is discuss ed in [1 4]. The metho d of c ha r acteristic Lie algebras is closely connected with the symmetry approa c h [15] whic h is prov ed to b e a very eﬀectiv e to ol to classify in tegrable nonlinear equations of ev olutionar y t yp e [16], [17 ], [1 8 ], [19], [20] (see also the surv ey [3] and references therein). How ev er, the symmetry approac h meets ve ry serious diﬃculties when a pplied to h yp erb o lic t yp e mo dels. After the pap ers [21] and [22] it b ecame clear that this case needs alternativ e metho ds. In this article an algorithm of classiﬁcation of integrable discrete c hains o f the f o rm (1) is suggested based on the no tion of the characteris tic Lie a lgebra (see also [23], [2 4], [25]). In tro duce necessary deﬁnitions. Deﬁne v ector ﬁelds Y j = D − j ∂ ∂ t 1 D j , j ≥ 1 , (6) 2 and X j = ∂ ∂ t − j , j ≥ 1 . (7) The follow ing theorem (see [24]) deﬁnes the c haracteristic Lie algebra L n of (1). Theorem 1 Equation (1) ad m its a n ontrivial n -inte gr al if and on l y if the fol lowing two c onditions hold: 1) Line ar sp ac e sp anne d by the op er ators { Y j } ∞ 1 is of ﬁnite dimensi o n, denote this di m ension by N ; 2) Lie algebr a L n gener ate d by the op er ators Y 1 , Y 2 , ..., Y N , X 1 , X 2 , ..., X N is of ﬁnite d i m ension. We c al l L n the char acteristic Lie algebr a of (1) in the dir e ction o f n . T o intro duce the c haracteristic Lie algebra L x of (1) in the direction of x , consider v ector ﬁelds K 0 = ∂ ∂ x + t x ∂ ∂ t + f ∂ ∂ t 1 + g ∂ ∂ t − 1 + f 1 ∂ ∂ t 2 + g − 1 ∂ ∂ t − 2 + . . . (8) and X = ∂ ∂ t x . (9) Note that a n x -in tegral F solv es the equation K 0 F = 0. O ne can get this equation b y applying the c hain rule to t he equation D x F = 0, here the function g is deﬁned b y the equation ( 1) rewritten due to (2 ) as t x ( n − 1) = g ( t ( n ) , t ( n − 1 ) , t x ( n )). Since F do es not dep end on the v ariable t x one gets X F = 0. Therefore, any v ector ﬁeld f r o m the Lie algebra g enerated b y K 0 and X ann ulates F . This algebra is called the c haracteristic Lie algebra L x of the c hain ( 1 ) in the x -direction. The following result is essen tial, its pro of is a simple consequence of the famous Jacobi theorem (Jacobi theorem is discusse d, for instance, in [10]). Theorem 2 Equation (1) admits a n ontrivial x -inte gr al i f and only if its Lie algebr a L x is of ﬁnite dimension. In the presen t pap er w e restrict ourselv es to consideration of existence of x -in tegrals for a partic- ular kind of c hain (1), namely , w e study c hains o f the for m t 1 x = t x + d ( t, t 1 ) (10) admitting nontrivial x -in tegrals. The main result of the pap er, Theorem 3 b elow , is the complete list of c hains (10 ) admitting non trivial x -integrals. Theorem 3 Chain (10) admits a n ontrivial x -inte gr al if and onl y if d ( t, t 1 ) is o ne of the kind: (1) d ( t, t 1 ) = A ( t − t 1 ) , (2) d ( t, t 1 ) = c 0 ( t − t 1 ) t + c 2 ( t − t 1 ) 2 + c 3 ( t − t 1 ) , (3) d ( t, t 1 ) = A ( t − t 1 )e αt , (4) d ( t, t 1 ) = c 4 (e αt 1 − e αt ) + c 5 (e − αt 1 − e − αt ) , 3 wher e A = A ( t − t 1 ) is a function of τ = t − t 1 and c 0 , c 2 , c 3 , c 4 , c 5 ar e some c onstants with c 0 6 = 0 , c 4 6 = 0 , c 5 6 = 0 , and α is a nonzer o c onstant. Mor e over, x -inte gr als in e ach of the c ases ar e i) F = x + R τ du A ( u ) , if A ( u ) 6 = 0 , F = t 1 − t , if A ( u ) ≡ 0 , ii) F = 1 ( − c 2 − c 0 ) ln | ( − c 2 − c 0 ) τ 1 τ 2 + c 2 | + 1 c 2 ln | c 2 τ 1 τ − c 2 − c 0 | for c 2 ( c 2 + c 0 ) 6 = 0 , F = ln τ 1 − ln τ 2 + τ 1 τ for c 2 = 0 , F = τ 1 τ 2 − ln τ + ln τ 1 for c 2 = − c 0 , iii) F = R τ e − αu du A ( u ) − R τ 1 du A ( u ) , iv) F = (e αt − e αt 2 )(e αt 1 − e αt 3 ) (e αt − e αt 3 )(e αt 1 − e αt 2 ) . The n -in tegrals of c hain (10) can b e studied in a similar w a y by using Theorem 1, but this problem is out of the frame of the presen t ar t icle. The a rticle is organized as follows. In Section 2 , b y using the prop erly chosen sequence of m ultiple comm utators, a v ery ro ug h classiﬁcation result is obtained: function d ( t, t 1 ) for ch ain (10) admitting x -in tegrals is a quasi-p olynomial on t with co eﬃcien ts depending of τ = t − t 1 . Then it is observ ed that the exp onents α 0 = 0, α 1 , ..., α s in the expansion (24) cannot b e arbitrary . F or example, if the co eﬃcien t b efo re e α 0 t = 1 is not identically zero then the quasi-p olynomial d ( t, t 1 ) is really a p olynomial on t with co eﬃcien ts depending on τ . In Section 3 w e pro v e that the degree of this p olynomial is at most one. If d con tains a term of the form µ ( τ ) t j e α k t with α k 6 = 0 then j = 0 (Section 4). In Section 5 it is prov ed tha t if d contains terms with e α k t and e α j t ha ving nonzero exp onen ts then α k = − α j . This last case contains chains ha ving inﬁnite dimensional c haracteristic Lie algebras for whic h the sequence of multiple commutators grows very slo wly . They ar e studied in Sections 6-7. O ne can ﬁnd t he w ell kno wn semi-discre te v ersion of the sine-Gordon (SG) mo del among them. It is w orth men tioning that in Section 7 the c haracteristic Lie algebra L x for semi-discrete SG is completely describ ed. The la st Section 8 con tains the pro of of the main Theorem 3 and here the metho d of constructing of x -inte grals is also brieﬂy discussed. 2 The ﬁ rst in tegrabili ty condi tion Deﬁne a class F of lo cally analytic functions eac h of whic h dep ends only on a ﬁnite n um b er of dynamical v ariables. In particular we a ssume that f ( t, t 1 , t x ) ∈ F . W e will consider v ector ﬁelds giv en a s inﬁnite formal series of the form Y = ∞ X −∞ y k ∂ ∂ t k (11) with co eﬃcien ts y k ∈ F . In t r o duce notions of linearly dep endent and indep endent sets of the vec t or ﬁelds (11). Denote through P N the pro jection op erat o r acting according to the rule P N ( Y ) = N X k = − N y k ∂ ∂ t k . (12) 4 First w e consider ﬁnite ve ctor ﬁelds as Z = N X k = − N z k ∂ ∂ t k . (13) W e say that a set o f ﬁnite v ector ﬁelds Z 1 , Z 2 , ..., Z m is linearly dep enden t in some op en r egio n U , if there is a set o f functions λ 1 , λ 2 , ..., λ m deﬁned on U suc h that the function | λ 1 | 2 + | λ 2 | 2 + ... + | λ m | 2 do es not v anish iden tically and the condition λ 1 Z 1 + λ 2 Z 2 + ... + λ m Z m = 0 (14) holds for eac h p oin t of region U . W e call a set of the v ector ﬁelds Y 1 , Y 2 , ..., Y m of the fo rm (11) linearly dependen t in the region U if for each natura l N the following set of ﬁnite v ector ﬁelds P N ( Y 1 ), P N ( Y 2 ), ..., P N ( Y m ) is linearly dep enden t in this region. Otherwise w e call the set Y 1 , Y 2 , ..., Y m linearly indep enden t in U . The follow ing prop osition is v ery useful, its pro of is almost eviden t. Prop osition . If a ve ctor ﬁeld Y is expr esse d as a line ar c ombin ation Y = λ 1 Y 1 + λ 2 Y 2 + ... + λ m Y m , (15) wher e the set o f ve ctor ﬁelds Y 1 , Y 2 , ..., Y m is line arly indep endent in U and the c o eﬃcients of al l the ve ctor ﬁelds Y , Y 1 , Y 2 , ..,. Y m b elonging to F ar e de ﬁne d in U then the c o eﬃcients λ 1 , λ 2 , ..., λ m ar e in F . Belo w we concen tr a te on the class of c hains of the form (10). F or this case the Lie algebra L x splits down into a direct sum of tw o subalgebras. Indeed, since f = t x + d and g = t x − d − 1 one gets f k = t x + d + P k j =1 d j and g − k = t x − P k +1 j =1 d − k , for k ≥ 1, where d = d ( t, t 1 ) and d j = d ( t j , t j +1 ). Due to this observ at io n the v ector ﬁeld K 0 can b e rewritten as K 0 = t x ˜ X + Y , with ˜ X = ∂ ∂ t + ∂ ∂ t 1 + ∂ ∂ t − 1 + ∂ ∂ t 2 + ∂ ∂ t − 2 + . . . (16) and Y = ∂ ∂ x + d ∂ ∂ t 1 − d − 1 ∂ ∂ t − 1 + ( d + d 1 ) ∂ ∂ t 2 − ( d − 1 + d − 2 ) ∂ ∂ t − 2 + . . . . Due to the relations [ X, ˜ X ] = 0 and [ X, Y ] = 0 w e hav e ˜ X = [ X , K 0 ] ∈ L x , hence Y ∈ L x . Therefore L x = { X } L L x 1 , where L x 1 is the Lie algebra generated b y the op erators ˜ X and Y . Lemma 1 If e quation (10) admits a nontrivial x -inte gr al then it a d m its a nontrivial x -inte gr al F such that ∂ F ∂ x = 0 . Pro of . Assume that a non trivial x -integral of (10) exists. Then the Lie alg ebra L x 1 is of ﬁnite dimension. One can c ho ose a basis of L x 1 in the form T 1 = ∂ ∂ x + ∞ X k = −∞ a 1 ,k ∂ ∂ t k , T j = ∞ X k = −∞ a j,k ∂ ∂ t k , 2 ≤ j ≤ N . 5 Th us, there exists an x -inte gral F dep ending on x , t , t 1 , . . . , t N − 1 satisfying the system of equations ∂ F ∂ x + N − 1 X k =0 a 1 ,k ∂ F ∂ t k = 0 , N − 1 X k =0 a j,k ∂ F ∂ t k = 0 , 2 ≤ j ≤ N . Due to the famous Jacobi Theorem [10] there is a c hange o f v ariables θ j = θ j ( t, t 1 , . . . , t N − 1 ) that reduces the system to the form ∂ F ∂ x + N − 1 X k =0 ˜ a 1 ,k ∂ F ∂ θ k = 0 , ∂ F ∂ θ k = 0 , 2 ≤ j ≤ N − 2 that is equiv alen t to ∂ F ∂ x + ˜ a 1 ,N − 1 ∂ F ∂ θ N − 1 = 0 for F = F ( x, θ N − 1 ). There are tw o p ossibilities: 1) ˜ a 1 ,N − 1 = 0 and 2) ˜ a 1 ,N − 1 6 = 0. In case 1), we at o nce ha v e ∂ F ∂ x = 0. In case 2), F = x + H ( θ N − 1 ) = x + H ( t, t 1 , . . . , t N − 1 ) for some function H . Eviden tly , F 1 = D F = x + H ( t 1 , t 2 , . . . , t N ) is also an x -integral, and F 1 − F is a nontrivial x -in tegral not dep ending on x .  Belo w w e lo ok for x -integrals F dep ending on dynamical v ariables t , t ± 1 , t ± 2 , . . . only (not dep ending on x ). In other w o r ds, w e study Lie a lg ebra generated b y v ector ﬁelds ˜ X and ˜ Y , where ˜ Y = d ∂ ∂ t 1 − d − 1 ∂ ∂ t − 1 + ( d + d 1 ) ∂ ∂ t 2 − ( d − 1 + d − 2 ) ∂ ∂ t − 2 + . . . . (17) One can pro v e that the linear op erat o r Z → D Z D − 1 deﬁnes an automor phism of the c haracteristic Lie algebra L x . This automorphism plays the crucial r o le in all of our furt her considerations. F urther w e refer to it as the shift automorphism. F or instance, direct calculations sho w that D ˜ X D − 1 = ˜ X , D ˜ Y D − 1 = − d ˜ X + ˜ Y . (18) Lemma 2 Supp ose that a ve ctor ﬁeld of the form Z = P a ( j ) ∂ ∂ t j with the c o eﬃcients a ( j ) = a ( j, t, t ± 1 , t ± 2 , ... ) dep ending on a ﬁnite numb er of the dynamic al variab l e s solves an e quation of the form D Z D − 1 = λZ . If f o r some j = j 0 we hav e a ( j 0 ) ≡ 0 then Z = 0 . Pro of . By applying the shift a utomorphism to the v ector ﬁeld Z one gets D Z D − 1 = P D ( a ( j )) ∂ ∂ t j +1 . No w, to complete the pro of, we compare the co eﬃcien ts of ∂ ∂ t j in the equation P D ( a ( j )) ∂ ∂ t j +1 = λ P a ( j ) ∂ ∂ t j .  Construct an inﬁnite sequence of multiple comm utators of the v ector ﬁelds ˜ X and ˜ Y ˜ Y 1 = [ ˜ X , ˜ Y ] , ˜ Y k = [ ˜ X , ˜ Y k − 1 ] for k ≥ 2 . (19) 6 Lemma 3 We have , D ˜ Y k D − 1 = − ˜ X k ( d ) ˜ X + ˜ Y k , k ≥ 1 . (20) Pro of . W e prov e the statemen t b y induction on k . Base of induction holds. Indeed, b y (1 8) and (19), w e hav e D ˜ Y 1 D − 1 = D [ ˜ X , ˜ Y ] D − 1 = [ D ˜ X D − 1 , D ˜ Y D − 1 ] = [ ˜ X , − d ˜ X + ˜ Y ] = − ˜ X ( d ) ˜ X + ˜ Y 1 . Assuming the equation (20) holds for k = n − 1, w e hav e D ˜ Y n D − 1 = [ D ˜ X D − 1 , D ˜ Y n − 1 D − 1 ] = [ ˜ X , − ˜ X n − 1 ( d ) ˜ X + ˜ Y n − 1 ] = − ˜ X n ( d ) ˜ X + ˜ Y n , that ﬁnishes the pro of of the Lemma.  Since v ector ﬁelds X , ˜ X and ˜ Y a r e linearly indep endent, then the dimension of Lie algebra L x is at least 3. By (2 0), case ˜ Y 1 = 0 corresp o nds to ˜ X ( d ) = 0, or d t + d t 1 = 0 that implies d = A ( t − t 1 ), where A ( τ ) is an arbitrary diﬀeren tiable function of one v ariable. Assume equation (10) admits a no n trivial x -integral and ˜ Y 1 6 = 0. Consider the sequence o f the v ector ﬁelds { ˜ Y 1 , ˜ Y 2 , ˜ Y 3 , . . . } . Since L x is of ﬁnite dimension, then there exists a natural n umber N suc h that ˜ Y N +1 = γ 1 ˜ Y 1 + γ 2 ˜ Y 2 + . . . + γ N ˜ Y N , N ≥ 1 , (21) and ˜ Y 1 , ˜ Y 2 , . . . , ˜ Y N are linearly indep enden t. Therefore, D ˜ Y N +1 D − 1 = D ( γ 1 ) D ˜ Y 1 D − 1 + D ( γ 2 ) D ˜ Y 2 D − 1 + . . . + D ( γ N ) D ˜ Y N D − 1 , N ≥ 1 . Due to Lemma 3 and (21) the last equation can b e rewritten as − ˜ X N +1 ( d ) ˜ X + γ 1 ˜ Y 1 + γ 2 ˜ Y 2 + . . . + γ N ˜ Y N = = D ( γ 1 )( − ˜ X ( d ) ˜ X + ˜ Y 1 ) + D ( γ 2 )( − ˜ X 2 ( d ) ˜ X + ˜ Y 2 ) + . . . + D ( γ N )( − ˜ X N ( d ) ˜ X + ˜ Y N ) . Comparing co eﬃcien ts b efo r e linearly independent v ector ﬁelds ˜ X , ˜ Y 1 , ˜ Y 2 , . . . , ˜ Y N , w e obtain the follo wing system of equations ˜ X N +1 ( d ) = D ( γ 1 ) ˜ X ( d ) + D ( γ 2 ) ˜ X 2 ( d ) + . . . + D ( γ N ) ˜ X N ( d ) , γ 1 = D ( γ 1 ) , γ 2 = D ( γ 2 ) , . . . , γ N = D ( γ N ) . Since the co eﬃcien ts of the v ector ﬁelds ˜ Y j dep end only o n the v aria bles t, t ± 1 , t ± 2 , ... the factors γ j migh t dep end only on these v ariables (see Prop osition a b ov e). Hence the system of equations implies that all co eﬃcien ts γ k , 1 ≤ k ≤ N , are constan ts, and d = d ( t, t 1 ) is a function that satisﬁes the follo wing diﬀeren tial equation ˜ X N +1 ( d ) = γ 1 ˜ X ( d ) + γ 2 ˜ X 2 ( d ) + . . . + γ N ˜ X N ( d ) , (22) where ˜ X ( d ) = d t + d t 1 . Using the substitution s = t and τ = t − t 1 , equation (22) can b e rewritten as ∂ N +1 d ∂ s N +1 = γ 1 ∂ d ∂ s + γ 2 ∂ 2 d ∂ s 2 + . . . + γ N ∂ N d ∂ s N , (23) that implies that d ( t, t 1 ) = X k m k − 1 X j =0 λ k ,j ( t − t 1 ) t j ! e α k t , (24) 7 for some functions λ k ,j ( t − t 1 ), where α k are ro ots of m ultiplicity m k for characteristic equation of (23). Let α 0 = 0, α 1 , . . . , α s b e the distinct ro ot s of the c haracteristic equation (22 ) . Equation (2 2) can b e rewritten as Λ( ˜ X ) d := ˜ X m 0 ( ˜ X − α 1 ) m 1 ( ˜ X − α 2 ) m 2 . . . ( ˜ X − α s ) m s d = 0 . (25) and m 0 + m 1 + . . . + m s = N + 1, m 0 ≥ 1. Initiated by the fo r mula (17) deﬁne a map h → Y h whic h assigns to an y function h = h ( t, t ± 1 , t ± 2 , ... ) a v ector ﬁeld Y h = h ∂ ∂ t 1 − h − 1 ∂ ∂ t − 1 + ( h + h 1 ) ∂ ∂ t 2 − ( h − 1 + h − 2 ) ∂ ∂ t − 2 + ... . F or an y p olynomial with constan t co eﬃcie n ts P ( λ ) = c 0 + c 1 λ + ... + c m λ m w e hav e a form ula P ( ad ˜ X ) ˜ Y = Y P ( ˜ X ) h , wh ere ad X Y = [ X, Y ] , (26) whic h establishes an isomorphism b et wee n the linear space V o f all solutions of equation (23) and the linear space ˜ V = span { ˜ Y , ˜ Y 1 , ..., ˜ Y N } of the corresp onding v ector ﬁelds. Represen t the function (24) as a sum d ( t, t 1 ) = P ( t, t 1 ) + Q ( t, t 1 ) of the p olynomial pa r t P ( t, t 1 ) = P m 0 − 1 j =0 λ 0 ,j ( t − t 1 ) t j and the ”exp onen tial” part Q ( t, t 1 ) = P s k =1  P m k − 1 j =0 λ k ,j ( t − t 1 ) t j  e α k t . Lemma 4 Assume e quation (1 0 ) admits a nontrivial x -inte gr al. Then one of the functions P ( t, t 1 ) and Q ( t, t 1 ) vanishes. Pro of . Assume in contrary that neither of the functions v anish. First w e sho w that in this case algebra L x con tains vector ﬁelds T 0 = Y A ( τ )e α k t and T 1 = Y B ( τ ) for some functions A ( τ ) and B ( τ ). Indeed, tak e T 0 := Λ 0 ( ad ˜ X ) ˜ Y = Y Λ 0 ( ˜ X ) d ∈ L x , where Λ 0 ( λ ) = Λ( λ ) λ − α k . Eviden tly the function ˜ A ( t, t 1 ) = Λ 0 ( ˜ X ) d solves t he equation ( ˜ X − α k ) ˜ A ( t, t 1 ) = Λ( ˜ X ) d = 0 whic h implies immediately that ˜ A ( t, t 1 ) = A ( τ )e α k t . In a similar w ay one sho ws that T 1 ∈ L x . Not e tha t due to our assumption the functions A ( τ ) and B ( τ ) cannot v anish iden tically . Consider an inﬁnite sequence of the v ector ﬁelds deﬁned as follo ws T 2 = [ T 0 , T 1 ] , T 3 = [ T 0 , T 2 ] , . . . , T n = [ T 0 , T n − 1 ] , n ≥ 3 . One can sho w that [ ˜ X , T 0 ] = α k T 0 , [ ˜ X , T 1 ] = 0 , [ ˜ X , T n ] = α k ( n − 1) T n , n ≥ 2 , D T 0 D − 1 = − A e α k t ˜ X + T 0 , DT 1 D − 1 = − B ˜ X + T 1 , D T n D − 1 = T n − ( n − 1)( n − 2) 2 α k A e α k t T n − 1 + b n ˜ X + n − 2 X k =0 a ( n ) k T k , n ≥ 2 . Since algebra L x is of ﬁnite dimension then there exists n umber N suc h that T N +1 = λ ˜ X + µ 0 T 0 + µ 1 T 1 + . . . + µ N T N , (27) and v ector ﬁelds ˜ X , T 0 , T 1 , . . . , T N are linearly indep enden t. W e ha ve , D T N +1 D − 1 = D ( λ ) ˜ X + D ( µ 0 ) {− A e α k t ˜ X + T 0 } + . . . + D ( µ N ) n T N − ( N − 1)( N − 2) 2 α k A e α k t T N − 1 + . . . o . 8 By comparing the co eﬃcien ts b efore T N in the last equation one gets µ N − N ( N − 1) 2 α k A ( τ )e α k t = D ( µ N ) . It follow s that µ N is a function o f v a riable t only . Also, by applying ad ˜ X to b oth sides of the equation (27), one gets N α k T N +1 = [ ˜ X , T N +1 ] = ˜ X ( λ ) ˜ X + ( ˜ X ( µ 0 ) + µ 0 α k ) T 0 + . . . + ( ˜ X ( µ N ) + µ N ( N − 1) α k ) T N . Again, b y comparing co eﬃcien ts b efore T N , w e hav e N α k µ N = ˜ X ( µ N ) + ( N − 1) α k µ N , i.e., ˜ X ( µ N ) = α k µ N . Therefore, µ N = A 1 e α k t , where A 1 is some nonzero constant, and th us A ( τ )e α k t = A 2 e α k t − A 2 e α k t 1 . Here A 2 is some constan t. W e hav e, T 0 = A 2 e α k t ˜ X − A 2 S 0 , where S 0 = ∞ X j = −∞ e α k t j ∂ ∂ t j . Also, [ ˜ X , S 0 ] = α k S 0 , D S 0 D − 1 = S 0 . Consider a new sequence of v ector ﬁelds P 1 = S 0 , P 2 = [ T 1 , S 0 ] , P 3 = [ T 1 , P 2 ] , P n = [ T 1 , P n − 1 ] , n ≥ 3 . One can sho w that [ ˜ X , P n ] = α k P n , DP n D − 1 = P n − α k ( n − 1) B P n − 1 + b n ˜ X + a n S 0 + n − 2 X j =2 a ( n ) j P j , n ≥ 2 . Since algebra L x is of ﬁnite dimension, then there exists nu m b er M suc h that P M +1 = λ ∗ ˜ X + µ ∗ 2 P 2 + . . . + µ ∗ M P M , (28) and ﬁelds ˜ X , P 2 , . . . , P M are linearly indep enden t. Th us, D P M +1 D − 1 = D ( λ ∗ ) ˜ X + D ( µ ∗ 2 ) { P 2 + . . . } + . . . + D ( µ ∗ M ) { P M − α k ( M − 1) B P M − 1 + . . . } . W e compare the co eﬃcien ts b efore P M in the last equation and get µ ∗ M − M α k B ( τ ) = D ( µ ∗ M ) , (29) that implies tha t µ ∗ M is a function of v ariable t only . Also,b y applying ad ˜ X to b o t h sides of (28), one gets α k P M +1 = [ ˜ X , P M +1 ] = ˜ X ( λ ∗ ) ˜ X + ( ˜ X ( µ ∗ 2 ) + α k µ ∗ 2 ) P 2 + . . . + ( ˜ X ( µ ∗ M ) + α k µ ∗ M ) P M . Again, we compare the co eﬃcien ts b efore P M and ha ve α k µ ∗ M ( t ) = ˜ X ( µ ∗ M ( t )) + α k µ ∗ M ( t ), that implies that µ ∗ M is a constan t. It follo ws then from (29) that B ( τ ) = 0. This con tra diction sho ws that our assumption that b oth functions are not iden t ically zero was wrong .  9 3 Multiple zero r o ot In this section w e assume that equation (10) admits a no n trivial x -in tegral and that α 0 = 0 is a ro ot of the characteristic p olynomial Λ( λ ). Then, due to Lemma 4, zero is the only ro ot and therefore Λ( λ ) = λ m +1 . It follo ws f rom the form ula (24) with m 0 = m + 1 that d ( t, t 1 ) = a ( τ ) t m + b ( τ ) t m − 1 + . . . , m = m 0 − 1 ≥ 0 . The case m = 0 corresp o nds to a v ery simple equation t 1 x = t x + A ( t − t 1 ) , whic h is easily solve d in quadrat ures, so w e concen trate on t he case m ≥ 1. F or this case the characteristic algebra L x con tains a v ector ﬁeld T = Y ˜ κ with ˜ κ = a ( τ ) t + 1 m b ( τ ) . Indeed, T = 1 m ! ad m − 1 ˜ X ˜ Y = Y ˜ κ . (30) In tro duce a sequ ence of m ultiple comm utators deﬁned as follo ws T 0 = ˜ X , T 1 = [ T , T 0 ] = Y − a ( τ ) , T k +1 = [ T , T k ] , k ≥ 0 , T k , 0 = [ T 0 , T k ] . Note that T 1 , 0 = 0. W e will see below that the linear space spanned b y this sequence is no t in v arian t under the action of the shift automorphism Z → D Z D − 1 in tro duced ab ov e. W e extend the sequence to pro vide the in v ariance prop ert y . W e deﬁne T α with the m ulti-index α . F or an y sequence α = k , 0 , i 1 , i 2 , . . . , i n − 1 , i n , where k is an y natural n um b er, i j ∈ { 0; 1 } , denote T α =     T 0 , T k , 0 ,i 1 ,...,i n − 1  , if i n = 0;  T , T k , 0 ,i 1 ,...,i n − 1  , if i n = 1; m ( α ) =    k , if α = k ; k , if α = k , 0; k + i 1 + . . . + i n , if α = k , 0 , i 1 , . . . , i n ; l ( α ) = k + n + 1 − m ( α ) . The multi-index α is c haracterized by t w o quantities m ( α ) and l ( α ) whic h allo w t o order pa r t ially the sequenc e { T α } . W e hav e, D T 0 D − 1 = T 0 , DT D − 1 = T − ˜ κ T 0 , DT 1 D − 1 = T 1 + aT 0 . One can pro ve b y induction on k that D T k D − 1 = T k + aT k − 1 − ˜ κ X m ( β )= k − 1 T β + X m ( β ) ≤ k − 2 η ( k , β ) T β . (31) In general, for an y α , D T α D − 1 = T α + X m ( β ) ≤ m ( α ) − 1 η ( α , β ) T β . (32) W e can choose a system P of linearly indep enden t v ector ﬁelds in the follow ing w a y . 1) T a nd T 0 are linearly indep enden t. W e tak e them into P . 2) W e che ck whether T , T 0 and T 1 are linearly indep enden t or not. If they are dep enden t then 10 P = { T , T 0 } and T 1 = µT + λT 0 for some functions µ and λ . 3) If T , T 0 , T 1 are linearly indep enden t then we c hec k whethe r T , T 0 , T 1 , T 2 are linearly indep enden t or not. If they are dependent, then P = { T , T 0 , T 1 } . 4) If T , T 0 , T 1 , T 2 are linearly independen t , w e add v ector ﬁelds T β , m ( β ) = 2, β ∈ I 2 , (actually , by deﬁnition I 2 is the collection of suc h β ) in suc h a wa y that J 2 := { T , T 0 , T 1 , T 2 , ∪ β ∈ I 2 T β } is a system of linearly indep enden t v ector ﬁelds a nd for an y T γ with m ( γ ) ≤ 2 w e hav e T γ = P T β ∈ J 2 µ ( γ , β ) T β . 5) W e c hec k whether T 3 ∪ J 2 is a linearly indep enden t system. If it is not, then P consists of all elemen ts from J 2 , and T 3 = P T β ∈ J 2 µ ( γ , β ) T β . If it is, then to the system T 3 ∪ J 2 w e add v ector ﬁelds T β , m ( β ) = 3, β ∈ I 3 , in suc h a wa y that J 3 := { T 3 , J 2 , ∪ β ∈ I 3 T β } is a system of linearly indep enden t v ector ﬁelds and fo r any T γ with m ( γ ) ≤ 3 w e hav e T γ = P T β ∈ J 3 µ ( γ , β ) T β . W e con tin ue the construction of the sy stem P . Since L x is of ﬁnite dimension, then there exists suc h a natural n umber N that (i) T k ∈ P , k ≤ N ; (ii) m ( β ) ≤ N for any T β ∈ P ; (iii) for an y T γ with m ( γ ) ≤ N we ha v e T γ = P T β ∈ P ,m ( β ) ≤ m ( γ ) µ ( γ , β ) T β and also T N +1 = µ ( N + 1 , N ) T N + X T β ∈ P ,m ( β ) ≤ N µ ( N + 1 , β ) T β . It follow s that (iv) for any v ector ﬁeld T α with m ( α ) = N , tha t do es not b elong to P , the co eﬃcien t µ ( α, N ) b efor e T N in the expansion T α = µ ( α , N ) T N + X T β ∈ P µ ( α, β ) T β (33) is constan t. Indeed, b y (32), D T α D − 1 = T α + X m ( β ) ≤ N − 1 η ( α , β ) T β = µ ( α , N ) T N + X T β ∈ P µ ( α, β ) T β + X m ( β ) ≤ N − 1 η ( α , β ) T β . F rom (33) w e hav e also D T α D − 1 = D ( µ ( α , N )) D T N D − 1 + X T β ∈ P D ( µ ( α , β )) D T β D − 1 = D ( µ ( α , N )) { T N + . . . } + X T β ∈ P D ( µ ( α , β )) { T β + . . . } . By comparing the co eﬃcien ts b efore T N in these t wo expre ssions f or D T α D − 1 , w e hav e µ ( α, N ) = D ( µ ( α , N )) , that implies that µ ( α , N ) is a constan t indeed. Lemma 5 We have , a ( τ ) = c 0 τ + c 1 , wher e c 0 and c 1 ar e so me c onstants. Pro of . Since T N +1 = µ ( N + 1 , N ) T N + X T β ∈ P µ ( N + 1 , β ) T β , 11 then D T N +1 D − 1 = D ( µ ( N + 1 , N )) { T N + . . . } + X T β ∈ P D ( µ ( N + 1 , β )) { T β + . . . } . On the other hand, D T N +1 D − 1 = T N +1 + aT N − ˜ κ X m ( β )= N T β + X m ( β ) ≤ N − 1 η ( N + 1 , β ) T β . W e compare the co eﬃcien ts b efore T N in the last t wo expre ssions. F o r N ≥ 0 the equation is µ ( N + 1 , N ) + a − ˜ κ X T β ∈ P ,m ( β )= N µ ( β , N ) = D ( µ ( N + 1 , N )) . (34) Denote b y c = − P T β ∈ P ,m ( β )= N µ ( β , N ) and b y µ N = µ ( N + 1 , N ). By prop ert y (iv), c is a constan t. It follo ws from (34) that µ N is a function of v ariables t and n only . Therefore, a ( τ ) + c  a ( τ ) t + 1 m b ( τ )  = µ N ( t 1 , n + 1) − µ N ( t, n ) . By diﬀeren tiating b oth sides of the equation with resp ect to t and then t 1 , w e hav e − a ′′ ( τ ) − c  a ′′ ( τ ) t + a ′ ( τ ) + 1 m b ′′ ( τ )  = 0 , that implies that a ′′ ( τ ) = 0, or the same, a ( τ ) = c 0 τ + c 1 for some constan ts c 0 and c 1 .  V ector ﬁelds T 1 and T in new v a riables are rewritten as T 1 = ∞ X j = −∞ a ( τ j ) ∂ ∂ τ j , (3 5) T = − ∞ X j = −∞ { a ( τ j ) t j + 1 m b ( τ j ) } ∂ ∂ τ j = − ∞ X j = −∞ { a ( τ j )( t + ρ j ) + 1 m b ( τ j ) } ∂ ∂ τ j = − tT 1 − ∞ X j = −∞ { a ( τ j ) ρ j + 1 m b ( τ j ) } ∂ ∂ τ j , (36) where ρ j =    − τ − τ 1 − . . . − τ j − 1 , if j ≥ 1 ; 0 , if j = 0; τ − 1 + τ − 2 + . . . + τ j , if j ≤ − 1 . The follow ing tw o lemmas are to b e useful. Lemma 6 If the Lie algebr a gener ate d by the ve ctor ﬁelds S 0 = ∞ P j = −∞ ∂ ∂ w j and P = ∞ P j = −∞ c ( w j ) ∂ ∂ w j is of ﬁnite dimen sion then c ( w ) is one of the forms (1) c ( w ) = c 2 + c 3 e λw + c 4 e − λw , λ 6 = 0 ; (2) c ( w ) = c 2 + c 3 w + c 4 w 2 , wher e c 2 , c 3 , c 4 ar e so me c onstants. 12 Pro of. In tro duce v ector ﬁelds S 1 = [ S 0 , P ] , S 2 = [ S 0 , S 1 ] , ..., S n = [ S 0 , S n − 1 ] , n ≥ 3 . Clearly , w e ha ve S n = ∞ X j = −∞ c ( n ) ( w j ) ∂ ∂ w j , n ≥ 1 . (37) Since all ve ctor ﬁelds S n are elemen ts of L x , and L x is o f ﬁnite dimension, then there exists a natural n umber N suc h that S N +1 = µ N S N + µ N − 1 S N − 1 + ... + µ 1 S 1 + µ 0 P + µS 0 , (38) and S 0 , P , S 1 , ..., S N are linearly indep enden t. (Note that we may assume S 0 and P are linearly indep enden t). Since D S 0 D − 1 = S 0 , D P D − 1 = P and D S n D − 1 = S n for a ny n ≥ 1, then it follo ws from (38) that S N +1 = D ( µ N ) S N + D ( µ N − 1 ) S N − 1 + ... + D ( µ 1 ) S 1 + D ( µ 0 ) P + D ( µ ) S 0 and together with (38), it implies that µ, µ 0 , µ 1 , ..., µ N are all constan ts. By comparing t he co eﬃcien ts befo r e ∂ ∂ w in (38) o ne gets, with the help of (37 ) , the following equalit y c ( N +1) ( w ) = µ N c ( N ) ( w ) + ... + µ 1 c ′ ( w ) + µ 0 c ( w ) + µ. Th us, c ( w ) is a solution of the nonhomogeneous linear diﬀerential equation with constan t coeﬃcien t whose c haracteristic p olynomial is Λ( λ ) = λ N +1 − µ N λ N − ... − µ 1 λ − µ 0 . Denote b y β 1 , β 2 , ..., β t c haracteristic ro ot s and b y m 1 , m 2 , ..., m t their multiplic it ies. There are the follo wing p ossibilities: (i) There exists a nonzero c haracteristic ro ot, say β 1 , and its m ultiplicit y m 1 ≥ 2 , (ii) There exists zero c hara cteristic ro ot, say β 1 , and m 1 ≥ 3, µ = 0 or m 1 ≥ 2, µ 6 = 0, (iii) There are t wo distinct c haracteristic ro ots, sa y β 1 and β 2 with β 1 6 = 0, β 2 = 0, (iv) There are t w o nonzero distinct c haracteristic ro ots, sa y β 1 and β 2 . In case (i), consider Λ 1 ( λ ) = Λ( λ ) λ − β 1 and Λ (2) 1 ( λ ) = Λ( λ ) ( λ − β 1 ) 2 . Then Λ 1 ( S 0 ) c ( w ) = α 1 e β 1 w + α 2 and Λ (2) 1 ( S 0 ) c ( w ) = ( α 3 w + α 4 )e β 1 w + α 5 , where α j , 1 ≤ j ≤ 5, ar e some constan ts with α 1 6 = 0, α 3 6 = 0. W e ha ve, Λ 1 ( ad S 0 ) P = ∞ X j = −∞ ( α 1 e β 1 w j + α 2 ) ∂ ∂ w j = α 1  ∞ X j = −∞ e β 1 w j ∂ ∂ w j  + α 2 S 0 = α 1 P 1 + α 2 S 0 , Λ (2) 1 ( ad S 0 ) P = ∞ X j = −∞ (( α 3 w j + α 4 ) e β 1 w j + α 5 ) ∂ ∂ w j = α 3  ∞ X j = −∞ w j e β 1 w j ∂ ∂ w j  + α 4 P 1 + α 5 S 0 = α 3 P 2 + α 4 P 1 + α 5 S 0 13 are elemen ts from L x and therefore v ector ﬁelds P 1 = P ∞ j = −∞ e β 1 w j ∂ ∂ w j and P 2 = P ∞ j = −∞ w j e β 1 w j ∂ ∂ w j b elong t o L x . Since P 1 and P 2 generate a n inﬁnite dimensional Lie algebra L x then case (i) f a ils to b e true. In case (ii), consider Λ (3) 1 ( λ ) = Λ( λ ) λ 3 and Λ (2) 1 ( λ ) = Λ( λ ) λ 2 , if µ = 0 , or Λ (3) 1 ( λ ) = Λ( λ ) λ 2 and Λ (2) 1 ( λ ) = Λ( λ ) λ , if µ 6 = 0 . W e hav e Λ (3) 1 ( S 0 ) c ( w ) = α 1 w 3 + α 2 w 2 + α 3 w + α 4 and Λ (2) 1 ( S 0 ) c ( w ) = α 5 w 2 + α 6 w + α 7 , where α j , 1 ≤ j ≤ 7, are some constan ts with α 1 6 = 0, α 5 6 = 0. Straightforw ard calculations show that v ector ﬁelds Λ (3) 1 ( ad S 0 ) P = ∞ X j = −∞ ( α 1 w 3 j + α 2 w 2 j + α 3 w j + α 4 ) ∂ ∂ w j and Λ (2) 1 ( ad S 0 ) P = ∞ X j = −∞ ( α 5 w 2 j + α 6 w j + α 7 ) ∂ ∂ w j generate an inﬁnite dimensional Lie algebra. It prov es that case (ii) fails to b e true. In case (iii), consider Λ 1 ( λ ) = Λ( λ ) λ − β 1 and Λ 2 ( λ ) = Λ( λ ) λ . W e hav e Λ 1 c ( w ) = α 1 e β 1 w + α 2 and Λ 2 c ( w ) = α 3 w + α 4 , if µ = 0 , or Λ 1 ( S 0 ) c ( w ) = α 1 e β 1 w + α 2 and Λ 2 ( S 0 ) c ( w ) = α 5 w 2 + α 6 w + α 7 , if µ 6 = 0 , where α j , 1 ≤ j ≤ 7 , are constan ts with α 1 6 = 0 , α 3 6 = 0 , α 5 6 = 0 . Since ve cto r ﬁelds Λ 1 ( ad S 0 ) P and Λ 2 ( ad S 0 ) P generate an inﬁnite dimensional Lie algebra, then case (iii) also fails to exist. In case (iv), consider Λ 1 ( λ ) = Λ( λ ) λ − β 1 and Λ 2 ( λ ) = Λ( λ ) λ − β 2 . W e ha ve, Λ 1 ( S 0 ) c ( w ) = α 1 e β 1 w + α 2 , Λ 2 ( S 0 ) c ( w ) = α 3 e β 2 w + α 4 , where α 1 6 = 0, α 2 , α 3 6 = 0, α 4 are some constan ts. Note tha t Λ 1 ( ad S 0 ) P = α 1  ∞ X j = −∞ e β 1 w j ∂ ∂ w j  + α 2 S 0 and Λ 2 ( ad S 0 ) P = α 3  ∞ X j = −∞ e β 2 w j ∂ ∂ w j  + α 4 S 0 , and v ector ﬁelds P ∞ j = −∞ e β 1 w j ∂ ∂ w j and P ∞ j = −∞ e β 2 w j ∂ ∂ w j generate a n inﬁnite dimensional Lie algebra if β 1 + β 2 6 = 0. It follow s fro m (i), (ii), (iii), (iv) that c ( w ) is one of the forms (1) c ( w ) = c 2 + c 3 e λw + c 4 e − λw , λ 6 = 0; (2) c ( w ) = c 2 + c 3 w + c 4 w 2 , where c 2 , c 3 , c 4 are some constan ts.  14 Lemma 7 If the Lie alg e br a gener ate d by the ve ctor ﬁelds S 0 = ∞ P j = −∞ ∂ ∂ w j , Q = ∞ P j = −∞ q ( w j ) ∂ ∂ w j and S 1 = ∞ P j = −∞ { ˜ ρ j + ˜ b ( w j ) } ∂ ∂ w j is of ﬁnite dimension then q ( w ) is a c onstant function. Pro of. It follows fro m Lemma 6 that (1) q ( w ) = c 2 + c 3 w + c 4 w 2 , or (2) q ( w ) = c 2 + c 3 e λw + c 4 e − λw , λ 6 = 0, where c 2 , c 3 , c 4 are some constan ts. Consider case (1). W e ha ve , [ S 0 , Q ] = c 3 ∞ X j = −∞ ∂ ∂ w j + 2 c 4 ∞ X j = −∞ w j ∂ ∂ w j = c 3 S 0 + 2 c 4 ∞ X j = −∞ w j ∂ ∂ w j . If c 4 6 = 0, then P ∞ j = −∞ w j ∂ ∂ w j ∈ L x and P ∞ j = −∞ w 2 j ∂ ∂ w j ∈ L x . If c 4 = 0, c 3 6 = 0, then P ∞ j = −∞ w j ∂ ∂ w j = 1 c 3 ( Q − c 2 S 0 ) ∈ L x . If c 3 = c 4 = 0, then q ( w ) = c 2 and there is nothing to pro ve. Assume c 2 4 + c 2 3 6 = 0. Denote by P = P ∞ j = −∞ w j ∂ ∂ w j . Construct the v ector ﬁelds P 1 = [ P , S 1 ] , P n = [ P , P n − 1 ] , n ≥ 2 . W e hav e, D S 0 D − 1 = S 0 , D S 1 D − 1 = S 1 − (e w − ˜ c ) S 0 , D P D − 1 = P , D P 1 D − 1 = P 1 + ( − w e w + e w − ˜ c ) S 0 , D P 2 D − 1 = P 2 + ( − w 2 e w + w e w − e w + ˜ c ) S 0 . In general, D P n D − 1 = P n + ( − w n e w + R n − 1 ( w )e w + c n ) S 0 , n ≥ 3 , where R n − 1 is a p o lynomial of degree n − 1, and c n is a constan t. Since L x is o f ﬁnite dimension, then there exists a natural n umber N suc h that P N +1 = µ N P N + ... + µ 1 P 1 + µ 0 S 0 , and S 0 , P 1 , ..., P N are linearly indep enden t. Th us D P N +1 D − 1 = D ( µ N ) D P N D − 1 + ... + D ( µ 1 ) D P 1 D − 1 + D ( µ 0 ) S 0 , or the same, µ N P N + ... + µ 1 P 1 + µ 0 S 0 + ( − w N +1 e w + R N ( w )e w + c N +1 ) S 0 = D ( µ N ) { P N + ( − w N e w + R N − 1 ( w )e w + c N ) S 0 } + ... + D ( µ 1 ) { P 1 + ( − w e w + e w − ˜ c ) S 0 } + D ( µ 0 ) S 0 . 15 By comparing the co eﬃcien ts b efore P N , ..., P 1 w e hav e µ N = D ( µ N ) , ... , µ 1 = D ( µ 1 ) , that implies µ N , ..., µ 1 are all constan ts. By comparing the co eﬃcien ts b efore S 0 w e hav e µ 0 − w N +1 e w + R N ( w )e w + c N +1 = µ N ( − w N e w + R N − 1 ( w )e w + c N ) + ... + µ 1 ( − w e w + e w − ˜ c ) + D ( µ 0 ) . The last equalit y show s that D ( µ 0 ) − µ 0 is a function of w only . Th us D ( µ 0 ) − µ 0 is a constan t, denote it b y d 0 . The last equalit y b ecomes a con tradictory one: w N +1 e w = R N ( w )e w + c N +1 − µ N ( − w N e w + R N − 1 ( w )e w + c N ) − ... − µ 1 ( − w e w + e w − ˜ c ) − d 0 . This con tradiction pro v es that c 2 3 + c 2 4 = 0, i.e. c 3 = c 4 = 0 in case (1). Therefore, q ( w ) = c 2 . Consider case (2). Since [ S 0 , Q ] = λc 3 ∞ X j = −∞ e λw j ∂ ∂ w j − λc 4 ∞ X j = −∞ e − λw j ∂ ∂ w j , [ S 0 , [ S 0 , Q ]] = λ 2 c 3 ∞ X j = −∞ e λw j ∂ ∂ w j + λ 2 c 4 ∞ X j = −∞ e − λw j ∂ ∂ w j , then ve ctor ﬁelds Q λ = c 3 P ∞ j = −∞ e λw j ∂ ∂ w j and Q − λ = c 4 P ∞ j = −∞ e − λw j ∂ ∂ w j b oth b elong to L x . W e ha ve , D Q λ D − 1 = Q λ , D Q − λ D − 1 = Q − λ . Assume c 3 6 = 0. Construct v ector ﬁelds Q 1 = [ Q λ , S 1 ] , Q n = [ Q λ , Q n − 1 ] , n ≥ 2 . Direct calculations sho w that D Q 1 D − 1 = Q 1 − c 3 e (1+ λ ) w S 0 + (e w − ˜ c ) λQ λ , D Q 2 D − 1 = Q 2 − c 2 3 (1 + λ )e (1+2 λ ) w S 0 + 2 λc 3 e (1+ λ ) w Q λ . It can b e pro ve d by induction on n that D Q n Q − 1 = Q n − p n S 0 + q n Q λ , n ≥ 2 , where p n = c n 3 (1 + λ )(1 + 2 λ ) ... (1 + ( n − 1) λ )e (1+ nλ ) w , q n = nc n − 1 3 λ (1 + λ ) ... (1 + ( n − 2) λ )e (1+( n − 1) λ ) w . Since L x is of ﬁnite dimension, there exists suc h a natural n umber N that Q N +1 = µ N Q N + ... + µ 1 Q 1 + µ λ Q λ + µ 0 S 0 , and S 0 , Q λ , Q 1 , ..., Q N are linearly indep enden t. Then D Q N +1 D − 1 = D ( µ N ) D Q N D − 1 + ... + D ( µ 0 ) D S 0 D − 1 , 16 or µ N Q N + ... + µ 1 Q 1 + µ λ Q λ + µ 0 S 0 − p N +1 S 0 + q N +1 Q λ = D ( µ N ) { Q N − p N S 0 + q N Q λ } + ... + D ( µ 1 ) { Q 1 − p 1 S 0 + q 1 Q λ } + D ( µ λ ) Q λ + D ( µ 0 ) S 0 . By comparing the co eﬃcie n ts b efo r e Q N , ..., Q 1 , w e hav e that µ k , 1 ≤ k ≤ N , are all constan ts. Comparing co eﬃcien ts b efore S 0 giv es µ 0 − p N +1 = − µ N p N − ... − µ 2 p 2 − µ 1 p 1 + D ( µ 0 ) . (39) Since p k , 1 ≤ k ≤ N + 1, dep end on w only , then D ( µ 0 ) − µ 0 is a function of w , and therefore D ( µ 0 ) − µ 0 is a constan t, denote it by d 0 . If λ 6 = − 1 r for all r ∈ N , then p k 6 = 0 for all k ∈ N , and equation (39) fails to b e true. Consider case when λ = − 1 r for some r ∈ N . Substitution u j = e − λw j transforms vec t o r ﬁelds − 1 λc 3 Q λ , − 1 λ S 1 , − 1 λ S 0 in to vec tor ﬁelds Q ∗ λ = ∞ X j = −∞ ∂ ∂ u j , S ∗ 1 = ∞ X j = −∞ { ˜ ρ ∗ j + ˜ b ∗ ( u j ) } u j ∂ ∂ u j , S ∗ 0 = ∞ X j = −∞ u j ∂ ∂ u j , where ˜ ρ ∗ j =            j − 1 P k =0 ( u r k − ˜ c ) , if j ≥ 1 ; 0 , if j = 0; − − 1 P k = j ( u r k − ˜ c ) , if j ≤ − 1 , , ˜ b ∗ ( u j ) = ˜ b ( r ln u j ) . First consider the case r = 1. W e hav e, T : = [ Q ∗ λ , S ∗ 1 ] = ∞ X j = −∞ { j u j + ˜ ρ ∗ j + ˜ b ∗ ( u j ) + u j ˜ b ∗ ′ ( u j ) } ∂ ∂ u j , K : = 1 2 [ Q ∗ λ , T ] = ∞ X j = −∞ { j + c ( u j ) } ∂ ∂ u j , where c ( u j ) = ˜ b ∗ ′ ( u j ) + 1 2 u j ˜ b ∗ ′′ ( u j ), T 1 = [ T , K ] = γ 1 ∞ X j = −∞ { j 2 + j g ( j ) 1 , 1 ( u j ) + g ( j ) 1 , 0 ( u, u 1 , ..., u j ) } ∂ ∂ u j , T 2 = [ T , T 1 ] = γ 2 ∞ X j = −∞ { j 3 + j 2 g ( j ) 2 , 2 ( u j ) + j g ( j ) 2 , 1 ( u, u 1 , ..., u j ) + g ( j ) 2 , 0 ( u, u 1 , ..., u j ) } ∂ ∂ u j , where γ 1 = − 3 2 and γ 2 6 = 0. 17 Construct v ector ﬁelds, T n = [ T , T n − 1 ], n ≥ 3. D irect calculations sho w t ha t T n = γ n ∞ X j =0 n j n +1 + j n g n,n ( u j ) + n − 1 X k =0 j k g n,k ( u, u 1 , ..., u j ) o ∂ ∂ u j + − 1 X j = −∞ a j ∂ ∂ u j , n ≥ 1 . Since { T n } ∞ n =1 is an inﬁnite sequence of linearly indep enden t v ector ﬁelds from L x , then case r = 1 fails to exist. Consider case r ≥ 2. W e hav e, ad Q ∗ λ S ∗ 1 = [ Q ∗ λ , S ∗ 1 ] = ∞ X j = −∞ n sg n ( j ) r  j − 1 X k =0 u r − 1 k  u j + ˜ ρ ∗ j + ˜ b ∗ ( u j ) + u j ˜ b ∗ ′ ( u j ) o ∂ ∂ u j , and ad r Q ∗ λ S ∗ 1 = ∞ X j = −∞ n r ! j u j + sg n ( j ) r ! j − 1 X k =0 u k + d ( u j ) o for some function d , ad r +1 Q ∗ λ S ∗ 1 = ∞ X j = −∞ n 2 r ! j + d ′ ( u j ) o ∂ ∂ u j . Note that ve cto r ﬁelds ad r Q ∗ λ S ∗ 1 and ad r +1 Q ∗ λ S ∗ 1 ha ve co eﬃcie n ts of the same kind as v ector ﬁelds T and K (from case r = 1) ha ve. It means that ad r Q ∗ λ S ∗ 1 and ad r +1 Q ∗ λ S ∗ 1 generate an inﬁnite dimensional Lie algebra. This con tradiction implies that case r ≥ 2 also fails to exist. Th us, c 3 = 0. By in t erchanging λ with − λ , w e obtain that c 4 = 0 also. Hence c 3 = c 4 = 0 and q ( w ) = c 2 .  W e already know that a ( τ ) = c 0 τ + c 1 . The next lemma sho ws that c 0 6 = 0. Lemma 8 c 0 is a nonzer o c onstant. Pro of . Assume contrary . Then a ( τ ) = c 1 and c 1 6 = 0, v ector ﬁelds (35) and (36) b ecome T 1 = c 1 ∞ X j = −∞ ∂ ∂ τ j = c 1 ˜ T 1 , and T = − tT 1 − c 1 ∞ X j = −∞ { ρ j + 1 mc 1 b ( τ j ) } ∂ ∂ τ j = − c 1 t ˜ T 1 − c 1 ˜ T , where ˜ T 1 = ∞ X j = −∞ ∂ ∂ τ j , ˜ T = ∞ X j = −∞ { ρ j + 1 mc 1 b ( τ j ) } ∂ ∂ τ j . Since [ ˜ T 1 , [ ˜ T 1 , ˜ T ]] = 1 mc 1 ∞ X j = −∞ b ′′ ( τ j ) ∂ ∂ τ j and ˜ T 1 b oth b elong to a ﬁnite dimensional L x , then, b y Lemma 6, 1) b ′′ ( τ ) = ˜ C 1 + ˜ C 2 e λτ + ˜ C 3 e − λτ or 2) b ′′ ( τ ) = ˜ C 1 + ˜ C 2 τ + ˜ C 3 τ 2 for some constan ts ˜ C 1 , ˜ C 2 , ˜ C 3 . 18 In case 1), b ( τ ) = C 1 + C 2 e λτ + C 3 e − λτ + C 4 τ 2 + C 5 τ and [ ˜ T 1 , [ ˜ T 1 , ˜ T ]] − λ 2 ˜ T − 2 C 4 − λ 2 C 1 mc 1 ˜ T 1 = − λ 2 ∞ X j = −∞ n ρ j + C 4 τ 2 j + C 5 τ j mc 1 o ∂ ∂ τ j is an elemen t in L x . In case 2), b ( τ ) = C 1 + C 2 τ + C 3 τ 2 + C 4 τ 3 + C 5 τ 4 and ˜ T − C 1 mc 1 ˜ T 1 = ∞ X j = −∞ n ρ j + C 2 τ j + C 3 τ 2 j + C 4 τ 3 j + C 5 τ 4 j mc 1 o ∂ ∂ τ j , b elongs to L x . T o ﬁnish the pro of of the Lemma it is enough to sho w that v ector ﬁelds ˜ T 2 := ∞ X j = −∞ { ρ j + C 2 τ j + C 3 τ 2 j + C 4 τ 3 j + C 5 τ 4 j } ∂ ∂ τ j , and ˜ T 1 = ∞ X j = −∞ ∂ ∂ τ j pro duce an inﬁnite dimensional Lie algebra L x for a ny ﬁxed constan ts C 2 , C 3 , C 4 and C 5 . One can pro ve it by sho wing tha t L x con tains v ector ﬁelds ∞ P j = −∞ j k ∂ ∂ τ j , for all k = 1 , 2 , . . . . Note that [ ˜ T 1 , ˜ T 2 ] = ∞ X j = −∞ ( − j + C 2 + 2 C 3 τ j + 3 C 4 τ 2 j + 4 C 5 τ 3 j ) ∂ ∂ τ j . There are four cases: a) C 5 6 = 0 and b) C 5 = 0 , C 4 6 = 0, c) C 5 = C 4 = 0, C 3 6 = 0 and d) C 5 = C 4 = C 3 = 0. In case a), [ ˜ T 1 , [ ˜ T 1 , [ ˜ T 1 , ˜ T 2 ]]] − 6 C 4 ˜ T 1 = ∞ X j = −∞ 24 C 5 τ j ∂ ∂ τ j = 24 C 5 P 1 ∈ L x , P 1 = ∞ X j = −∞ τ j ∂ ∂ τ j , [ ˜ T 1 , [ ˜ T 1 , ˜ T 2 ]] = ∞ X j = −∞ { 2 C 3 + 6 C 4 τ j + 12 C 5 τ 2 j } ∂ ∂ τ j ∈ L x , and therefore, P 2 := ∞ X j = −∞ τ 2 j ∂ ∂ τ j ∈ L x , and ˜ T 3 := [ ˜ T 1 , ˜ T 2 ] − C 2 ˜ T 1 − 2 C 3 P 1 − 3 C 4 P 2 = ∞ X j = −∞ ( − j + 4 C 5 τ 3 j ) ∂ ∂ τ j ∈ L x . 19 W e hav e, J 1 := − 1 3 ([ ˜ T 3 , P 1 ] + 2 ˜ T 3 ) = ∞ X j = −∞ j ∂ ∂ τ j ∈ L x . No w, [ J 1 , [ J 1 , P 2 ]] = 1 2 ∞ X j = −∞ j 2 ∂ ∂ τ j ∈ L x . Assuming J k = ∞ P j = −∞ j k ∂ ∂ τ j ∈ L x w e hav e that J k +1 := 1 2 [ J 1 , [ J k , P 2 ]] = ∞ X j = −∞ j k +1 ∂ ∂ τ j ∈ L x . In case b) w e hav e P 1 := 1 6 C 4 { [ ˜ T 1 , [ ˜ T 1 , ˜ T 2 ]] − 2 C 3 ˜ T 1 } = ∞ X j = −∞ τ j ∂ ∂ τ j ∈ L x and ˜ T 3 = [ ˜ T 1 , ˜ T 2 ] − C 2 ˜ T 1 − 2 C 3 P 1 = ∞ X j = −∞ ( − j + 3 C 4 τ 2 j ) ∂ ∂ τ j ∈ L x . W e hav e, J 1 := − 1 2 ([ ˜ T 3 , P 1 ] + ˜ T 3 ) = ∞ X j = −∞ j ∂ ∂ τ j ∈ L x , and P 2 = 1 6 C 4 ( ˜ T 3 − [ ˜ T 3 , P 1 ]) = ∞ X j = −∞ τ 2 j ∂ ∂ τ j ∈ L x . As it w as sho wn in the pro of of case a), J 1 and P 2 pro duce an inﬁnite dimensional Lie algebra. In case c), ˜ T 3 = [ ˜ T 1 , ˜ T 2 ] − C 2 ˜ T 1 = ∞ X j = −∞ ( − j + 2 C 3 τ j ) ∂ ∂ τ j ∈ L x , ˜ T 4 = [ ˜ T 3 , ˜ T 2 ] = ∞ X j = −∞ ( j ( j − 1) 2 − j C 2 − 2 C 3 j τ j + 2 C 2 3 τ 2 j ) ∂ ∂ τ j ∈ L x . Also, ˜ T 5 = [ ˜ T 3 , ˜ T 4 ] = 2 C 3 ∞ X j = −∞  j ( j + 1) 2 + C 2 j − 2 C 3 j τ j + 2 C 2 3 τ 2 j  ∂ ∂ τ j ∈ L x . Since ˜ T 4 and ˜ T 5 b oth b elong to L x then either c )( i ) J 1 = ∞ X j = −∞ j ∂ ∂ τ j ∈ L x , ˜ T 6 = ∞ X j = −∞ ( j 2 2 − 2 C 3 j τ j + 2 C 2 3 τ 2 j ) ∂ ∂ τ j ∈ L x , 20 or c )( ii ) C 2 = − 1 2 , ˜ T 6 = ∞ X j = −∞ ( j 2 2 − 2 C 3 j τ j + 2 C 2 3 τ 2 j ) ∂ ∂ τ j ∈ L x . In case c) (i), P 1 = 1 4 C 2 3 { [ ˜ T 1 , ˜ T 6 ] + 2 C 3 J 1 } = ∞ X j = −∞ τ j ∂ ∂ τ j ∈ L x . Since [ P 1 , ˜ T 6 ] = ∞ X j = −∞ ( − j 2 2 + 2 C 2 3 τ 2 j ) ∂ ∂ τ j , and [ P 1 , [ P 1 , ˜ T 6 ]] = ∞ X j = −∞ ( j 2 2 + 2 C 2 3 τ 2 j ) ∂ ∂ τ j b oth b elong to L x then J 2 = ∞ X j = −∞ j 2 ∂ ∂ τ j ∈ L x , P 2 = ∞ X j = −∞ τ 2 j ∂ ∂ τ j ∈ L x , P 2 and J 1 generate an inﬁnite dimensional Lie alg ebra. In case c) (ii), ˜ T 1 = ∞ X j = −∞ ∂ ∂ τ j , ˜ T 2 = ∞ X j = −∞  C 3 τ 2 j − 1 2 τ j + ρ j  ∂ ∂ τ j . Note that the Lie algebra generated b y the vec to r ﬁelds ˜ T ∗ 2 = ˜ T 2 −  C 3 τ 2 − 1 2 τ  ˜ T 1 = d ( τ , τ 1 ) ∂ ∂ τ 1 − d ( τ − 1 , τ ) ∂ ∂ τ − 1 + ( d ( τ , τ 1 ) + d ( τ 1 , τ 2 )) ∂ ∂ τ 2 + . . . and ˜ T 1 = ∞ X j = −∞ ∂ ∂ τ j is inﬁnite dimensional. It can b e prov ed b y comparing this a lgebra with the inﬁnite dimensional c haracteristic Lie a lg ebra of the c hain t 1 x = t x + C 3 ( t 2 1 − t 2 ) − 1 2 ( t 1 + t ) . (40) Indeed, the Lie algebra L x 1 for (40) is generated b y the op erato rs (16) and (17) with d ( t, t 1 ) = C 3 ( t 2 1 − t 2 ) − 1 2 ( t 1 + t ). T o k eep standard notations w e put a ( τ ) = − 2 C 3 τ − 1 and b ( τ ) = C 3 τ 2 + 1 2 τ . Note t ha t since C 3 6 = 0 function a ( τ ) is not a constant. It follo ws from Theorem 3 pro v ed b elow that the c har acteristic Lie alg ebras L x (and therefore algebra L x 1 ) for equation (40) is of inﬁnite dimension. Th us, in case c) (ii) w e also ha ve an inﬁnite dimensional Lie algebra L x . In case d), ˜ T 2 = ∞ X j = −∞ ( − τ − τ 1 − . . . − τ j − 1 + C 2 τ j ) ∂ ∂ τ j ∈ L x . 21 Then J 1 = c 2 ˜ T 1 − [ ˜ T 1 , ˜ T 2 ] = ∞ X j = −∞ j ∂ ∂ τ j ∈ L x , and J 2 = − 2  [ J 1 , ˜ T 2 ] −  1 2 + C 2  J 1  = ∞ X j = −∞ j 2 ∂ ∂ τ j ∈ L x . Assuming that J k , 1 ≤ k ≤ n b elong to L x , b y considering [ J n , ˜ T 2 ] one may show that J n +1 = ∞ P j = −∞ j k +1 ∂ ∂ τ j ∈ L x . It implies L x is of inﬁnite dimension.  Let us in tro duce new v ariables w j = ln  τ j + c 1 c 0  . V ector ﬁelds T 1 and T in v ariables w j can be rewritten as T 1 = c 0 ∞ X j = −∞ ∂ ∂ w j = c 0 S 0 , T = − tc 0 S 0 + c 0 ∞ X j = −∞ { ˜ ρ j + ˜ b ( w j ) } ∂ ∂ w j = − c 0 tS 0 + c 0 S 1 , where S 0 = ∞ X j = −∞ ∂ ∂ w j , S 1 = ∞ X j = −∞ { ˜ ρ j + ˜ b ( w j ) } ∂ ∂ w j , ˜ ρ j =            j − 1 P k =0 (e w k − ˜ c ) , if j ≥ 1; 0 , if j = 0; − − 1 P k = j (e w k − ˜ c ) , if j ≤ − 1 , ˜ c = c 1 c 0 , ˜ b ( w j ) = − 1 m  b ( τ j ) c 0 τ j + c 1  . W e hav e D S 0 D − 1 = S 0 , D S 1 D − 1 = S 1 − (e w − ˜ c ) S 0 . These lemmas allo w one to prov e the follow ing Theorem. Theorem 4 If e quation t 1 x = t x + a ( τ ) t m + b ( τ ) t m − 1 + . . . , m ≥ 1 admits a nontrivia l x -inte gr al, then (1) a ( τ ) = c 0 τ , b ( τ ) = c 2 τ 2 + c 3 τ , wher e c 0 , c 2 , c 3 ar e so me c onstants. (2) m = 1 . Pro of . Consider the case (1). Deﬁne v ector ﬁeld Q = [ S 0 , [ S 0 , S 1 ]] − [ S 0 , S 1 ] = ∞ X j = −∞ ( ˜ b ′′ ( w j ) − ˜ b ′ ( w j )) ∂ ∂ w j . 22 By Lemma 7, ˜ b ′′ ( w ) − ˜ b ′ ( w ) = C fo r some constant C . Th us, ˜ b ( w ) = C 0 + C 1 e w + C 2 w for some constan ts C 1 , C 2 , C 0 . Consider v ector ﬁelds P = ( C 2 − C 0 ) S 0 + S 1 − [ S 0 , S 1 ] = ∞ X j = −∞ ( C 2 w j + ˜ cj ) ∂ ∂ w j , R = [ S 0 , [ S 0 , S 1 ]] = ∞ X j =1 (  j X k =1 e w k  + C 1 e w j ) ∂ ∂ w j + C 1 e w ∂ ∂ w − − 1 X j = −∞ (  − 1 X k = j e w k  + C 1 e w j ) ∂ ∂ w j , R 1 = [ P , R ] , R n +1 = [ P , R n ] , n ≥ 1 . Then R n = X j ≥ 0 { e w j ( C 1 C n 2 w n j + P n,j ) + r n,j ( w , w 1 , . . . , w j − 1 ) } ∂ ∂ w j + X j ≤− 1 { e w j (( C 1 − 1) C n 2 w n j + P n,j ) + r n,j ( w − 1 , w − 2 , . . . , w j +1 ) } ∂ ∂ w j , where P n,j = P n,j ( w j , j ) is a p olynomial of degree n − 1 whose co eﬃcien ts dep end on j , r n,j are the functions that do not dep end on w j . Since all v ector ﬁelds R n b elong to a ﬁnite dimensional Lie algebra L x then C 1 C 2 = ( C 1 − 1) C 2 = 0, or the same C 2 = 0. Therefore, ˜ b ( w ) = C 0 + C 1 e w . Since C 2 = 0, then P = ˜ c ∞ X j = −∞ j ∂ ∂ w j , R = ∞ X j =1 (  j X k =1 e w k  + C 1 e w j ) ∂ ∂ w j + C 1 e w ∂ ∂ w − − 1 X j = −∞ (  − 1 X k = j e w k  + C 1 e w j ) ∂ ∂ w j and R n = ˜ c n ∞ X j =1 { e w 1 + 2 n e w 2 + ( j − 1) n e w j − 1 + j n C 1 e w j } ∂ ∂ w j − ˜ c n − 1 X j = −∞ { ( − 1) n e w − 1 + ( − 2) n e w − 2 + ( j ) n e w j + j n C 1 e w j } ∂ ∂ w j . Again, ve ctor ﬁelds R n b elong t o a ﬁnite dimensional Lie alg ebra only if ˜ c = 0, or the same c 1 = 0. It implies that a ( τ ) = c 0 τ , b ( τ ) = c 2 τ 2 + c 3 τ . Consider the case (2). Assume con trary , that is m ≥ 2. Then the following v ector ﬁeld 1 m ! ad m − 2 ˜ X ( ˜ Y ) = Y 1 2 a ( τ ) t 2 + 1 m b ( τ ) t + 1 m ( m − 1) c ( τ ) = − ∞ X j = −∞ ( 1 2 a ( τ j ) t 2 j + 1 m b ( τ j ) t j + 1 m ( m − 1) c ( τ j )) ∂ ∂ τ j = − ∞ X j = −∞ ( 1 2 a ( τ j )( t + ρ j ) 2 + 1 m b ( τ j )( t + ρ j ) + 1 m ( m − 1) c ( τ j )) ∂ ∂ τ j − t 2 2 ∞ X j = −∞ a ( τ j ) ∂ ∂ τ j − t ∞ X j = −∞ { a ( τ j ) ρ j + 1 m b ( τ j ) } ∂ ∂ τ j − ∞ X j = −∞ { 1 2 a ( τ j ) ρ 2 j + 1 m b ( τ j ) + 1 m ( m − 1) c ( τ j ) } ∂ ∂ τ j 23 is in L x . In v ariables w j = ln τ j , 1 m ! ad m − 2 ˜ X ( ˜ Y ) = − t 2 2 c 0 S 0 + tc 0 S 1 − c 0 S 2 , where S 2 = ∞ X j = −∞ { 1 2 ˜ ρ 2 j − ˜ b ( w j ) ˜ ρ j + ˜ c ( w j ) } ∂ ∂ w j , ˜ c ( w j ) = c ( τ j ) m ( m − 1) τ j . The v ector ﬁelds S 0 and S 1 are as in Lemma 7. W e hav e, [ S 0 , S 2 ] = 2 S 2 + C 0 S 1 + P , P = ∞ X j = −∞ r ( w j ) ∂ ∂ w j , r ( w ) = ˜ c ′ ( w ) − 2 ˜ c ( w ) − C 0 ˜ b ( w ) . Construct the sequence S 3 = [ S 1 , S 2 ] , S n +1 = [ S 1 , S n ] , n ≥ 2 . One can pro ve b y induction on n that [ S 0 , S n ] = nS n + n − 1 X k =0 ν n,k S k , and D S n D − 1 = S n +  n ( n − 1) 2 − 1  e w S n − 1 + n − 2 X k =0 η ( n, k ) S k , n ≥ 3 . Since L x is of ﬁnite dimension then there exists a natural num b er N suc h tha t S N +1 = µ N S N + µ N − 1 S N − 1 + . . . + µ 0 S 0 . Then D S N +1 D − 1 = D ( µ N ) D S N D − 1 + D ( µ N − 1 ) D S N − 1 D − 1 + . . . + D ( µ 0 ) D S 0 D − 1 . On the other hand, D S N +1 D − 1 = S N +1 +  ( N + 1) N 2 − 1  e w S N + . . . . W e compare the co eﬃcien ts b efore S N and ha ve t w o equations. D ( µ N ) = µ N +  ( N + 1) N 2 − 1  e w , N ≥ 2 , and D ( µ 1 ) = µ 1 + e w , N = 1 . Both equation are con tradictory . Therefore, our assumption that m ≥ 2 w as wrong.  24 4 Nonzero ro ot Lemma 9 Assume e quation (10) admits a nontrivial x -inte gr al. The n the ch a r acteristic p olynom ial of the e quation (23) c an have only sim ple no n zer o r o ots. Pro of . Assume that m 1 ≥ 2 . Intro duce p olynomials Λ (2) α 1 ( λ ) = Λ( λ ) ( λ − α 1 ) 2 , Λ α 1 ( λ ) = Λ( λ ) ( λ − α 1 ) . Consider v ector ﬁelds S ∗ 0 = Λ (2) α 1 ( ad ˜ X ) Y d = Y A ( τ )e α 1 t S ∗ 1 = Λ α 1 ( ad ˜ X ) Y d = Y ( A ( τ ) t + B ( τ ))e α 1 t from the the Lie algebra L x . In v ariables τ j = t j − t j +1 , v ector ﬁelds S ∗ 0 and S ∗ 1 b ecome S ∗ 0 = − e α 1 t ∞ X j = −∞ A ( τ j )e α 1 ρ j ∂ ∂ τ j = − e α 1 t S 0 , S ∗ 1 = − t e α 1 t S 0 − e α 1 t ∞ X j = −∞ { A ( τ j ) ρ j + B ( τ j ) } e α 1 ρ j ∂ ∂ τ j = − t e α 1 t S 0 − e α 1 t S 1 , with S 0 = ∞ P j = −∞ A ( τ j )e α 1 ρ j ∂ ∂ τ j and S 1 = ∞ P j = −∞ { A ( τ j ) ρ j + B ( τ j ) } e α 1 ρ j ∂ ∂ τ j . Direct calculations sho w that D S 0 D − 1 = e α 1 τ S 0 , D S 1 D − 1 = e α 1 τ S 1 + τ e α 1 τ S 0 . Deﬁne the sequenc e S 2 = [ S 0 , S 1 ] , S n +1 = [ S 0 , S n ] , n ≥ 2 . One can easily sho w that D S 2 D − 1 = e 2 α 1 τ S 2 + α 1 e 2 α 1 τ A ( τ ) S 1 + e 2 α 1 τ ( A ( τ ) − α 1 B ( τ )) S 0 . It can b e pro ve d by induction on n that D S n D − 1 = e nα 1 τ S n + α 1 n ( n − 1) 2 e nα 1 τ A ( τ ) S n − 1 + n − 2 X k =0 γ ( n, k ) S k . Since the dimension of L x is ﬁnite and S 0 , S 1 , . . . are elemen ts of L x then there exists a natural n umber N suc h that S N +1 = µ N S N + µ N − 1 S N − 1 + . . . + µ 0 S 0 , and S 0 , S 1 , . . . , S N are linearly indep enden t. Therefore, D S N +1 D − 1 = D ( µ N ) D S N D − 1 + D ( µ N − 1 ) D S N − 1 D − 1 + . . . + D ( µ 0 ) D S 0 D − 1 . On the other hand, D S N +1 D − 1 = e ( N +1) α 1 τ S N +1 + α 1 ( N + 1) N 2 e ( N +1) α 1 τ A ( τ ) S N + N − 1 X k =0 γ ( N + 1 , k ) S k . 25 By comparing the co eﬃcien ts b efore S N in the last t wo equations w e ha ve e ( N +1) α 1 τ µ N + α 1 ( N + 1) N 2 e ( N +1) α 1 τ A ( τ ) = D ( µ N )e N α 1 τ . It follow s at once that µ N is a constan t and then A ( τ ) = C (e − α 1 τ − 1) , C = 2 µ N α 1 N ( N + 1) . Let us construct a new inﬁnite sequence of ve ctor ﬁelds b elonging to L x , enume r a ted by a m ulti-index. T 0 := S 1 , T 1 := S 0 , T 2 = [ S 1 , T 1 ] , T n +1 = [ S 1 , T n ] , n ≥ 2 , T n, 0 = [ S 0 , T n ] , T n, 0 ,i 1 ,...,i n − 1 ,i n = [ S i n , T n, 0 ,i 1 ,...,i n − 1 ] , i j ∈ { 0; 1 } . Direct calculations sho w that D T 2 D − 1 = e 2 α 1 τ T 2 + e 2 α 1 τ ( α 1 B − A ) T 1 − α 1 e 2 α 1 τ AT 0 , D T 3 D − 1 = e 3 α 1 τ T 3 + e 3 α 1 τ (3 α 1 B − A + 3 α 1 τ A ) T 2 + τ e 3 α 1 τ T 2 , 0 + X m ( β ) < 2 ν (3 , β ) T β . Here and b elow w e use functions m = m ( β ) and l = l ( β ) deﬁned in Section 3. It can b e pro ve d b y induction on n that D T n D − 1 = e nα 1 τ T n +e nα 1 τ { c n B − A + c n τ A } T n − 1 + τ e nα 1 τ X m ( β )= n − 1 ,l ( β )=1 ν ∗ ( n, β ) T β + X m ( β ) ≤ n − 2 ν ( n, β ) T β , where c n = α 1 n ( n − 1) 2 , and ν ∗ ( n, β ) are constan ts for an y β with m ( β ) = n − 1 and l ( β ) = 1 . In general, for an y γ , D T γ D − 1 = e ( m ( γ )+ l ( γ )) α 1 τ T γ + X m ( β ) ≤ m ( γ ) − 1 ν ( γ , β ) T β . Among the v ector ﬁelds T β w e c ho ose a system P o f linearly indep enden t v ector ﬁelds in suc h a w ay that for some natural n umber N (i) T k ∈ P , k ≤ N , (ii) m ( β ) ≤ N for any T β ∈ P . (iii) for an y T γ with m ( γ ) ≤ N we ha v e T γ = P T β ∈ P ,m ( β ) ≤ m ( γ ) µ ( γ , β ) T β . Also T N +1 = µ ( N + 1 , N ) T N + P T β ∈ P µ ( N + 1 , β ) T β . (iv) for an y T γ / ∈ P with m ( γ ) = N and l ( γ ) = 1, w e ha ve µ ( γ , N ) = 0. Indeed, D T γ D − 1 = D ( µ ( γ , N )) D T N D − 1 + X T β ∈ P ,β 6 = N D ( µ ( γ , β )) D T β D − 1 . On the other hand, D T γ D − 1 = e ( m ( γ )+ l ( γ )) α 1 τ T γ + X m ( β ) ≤ N − 1 ν ( γ , β ) T β = e ( N +1) α 1 τ { µ ( γ , N ) T N + X T β ∈ P ,m ( β ) ≤ N ,β 6 = N µ ( γ , β ) T β } + X m ( β ) ≤ N − 1 ν ( γ , β ) T β . 26 By comparing the co eﬃcien ts b efore T N w e hav e e ( N +1) α 1 τ µ ( γ , N ) = D ( µ ( γ , N ) ) e N α 1 τ that pro ve s µ ( γ , N ) = 0 for an y γ with m ( γ ) = N and l ( γ ) = 1. W e ha ve , T N +1 = µ N T N + X T β ∈ P µ ( N + 1 , β ) T β , here µ N = µ ( N + 1 , N ). Then D T N +1 D − 1 = D ( µ N ) D T N D − 1 + X T β ∈ P D ( µ ( N + 1 , β )) D T β D − 1 . W e contin ue and hav e, e ( N +1) α 1 τ { µ N T N + X T β ∈ P µ ( N + 1 , β ) T β } + e ( N +1) α 1 τ { c N +1 B − A + c N +1 τ A } T N + τ e ( N +1) α 1 τ X m ( β )= N ,l ( β )=1 ν ∗ ( N + 1 , β ) T β + X m ( β ) ≤ N − 1 ν ( N + 1 , β ) T β = D ( µ N ) { e N α 1 τ T N + X m ( β ) ≤ N − 1 ν ( N , β ) T β } + X T β ∈ P D ( µ ( N + 1 , β )) { e ( m ( β )+ l ( β )) α 1 τ T β + X m ( r ) ≤ N − 1 ν ( β , r ) T r } . W e compare the co eﬃcien ts b efore T N and get e ( N +1) α 1 τ µ N + e ( N +1) α 1 τ { c N +1 B − A + c N +1 τ A } = e N α 1 τ D ( µ N ) . Note that, b y prop erty (iv), w e do not hav e term τ e ( N +1) α 1 τ in the left side of the last equalit y . Th us, using the express ion for A ( τ ) = C ( e − α 1 τ − 1) and the fact that µ N is a constan t, we hav e B ( τ ) = C 1 A + C 2 τ A = C 1 (e − α 1 τ − 1) + C 2 τ (e − α 1 τ − 1) , where C 1 = µ N C c N +1 + 1 c N +1 , C 2 = − 1 . W e intro duce new v ector ﬁelds ˜ S 0 = 1 C S 0 = (e − α 1 τ − 1) ∂ ∂ τ + . . . , ˜ S 1 = 1 C S 1 + C 1 C S 0 = τ (e − α 1 τ − 1) ∂ ∂ τ + . . . . ˜ S 2 = [ ˜ S 0 , ˜ S 1 ] , ˜ S n +1 = [ ˜ S 0 , ˜ S n ] , n ≥ 2 . W e hav e, D ˜ S 0 D − 1 = e α 1 τ ˜ S 0 , D ˜ S 1 D − 1 = e α 1 τ ˜ S 1 − τ e α 1 τ ˜ S 0 , D ˜ S n D − 1 = n X k =0 ˜ γ ( n, k ) ˜ S k , ˜ γ ( n, n ) = e nα 1 τ , where ˜ γ ( n, k ) are functions of τ only . Since all v ector ﬁelds ˜ S k b elong to a ﬁnite dimensional Lie algebra L x , then there exists suc h a natural num b er M that ˜ S M +1 = ˜ µ M ˜ S M + . . . + ˜ µ 0 ˜ S 0 , (41) 27 and ˜ S 0 , . . . , ˜ S M are linearly indep enden t. Then D ˜ S M +1 D − 1 = D ( ˜ µ M ) D ˜ S M D − 1 + . . . + D ( ˜ µ 0 ) D ˜ S 0 D − 1 , and ˜ γ ( M + 1 , M + 1) { ˜ µ M ˜ S M + . . . + ˜ µ 0 ˜ S 0 } + M X k =0 ˜ γ ( M + 1 , k ) ˜ S k = D ( ˜ µ N ) { ˜ γ ( M , M ) ˜ S M + . . . } + . . . . By comparing the co eﬃcien ts b efore ˜ S M , w e hav e e ( M +1) α 1 τ ˜ µ M + ˜ γ ( M + 1 , M ) = D ( ˜ µ M )e M α 1 τ that implies tha t ˜ µ M is a constant. In the same w a y , b y comparing the co eﬃcien ts b efore ˜ S M − 1 , and then b efore ˜ S M − 2 , and so on, one can sho w that all co eﬃcie nts ˜ µ k are constan ts. One can sho w b y induction on n that for n ≥ 2, ˜ S n = { α n − 2 1 ( − 1) n − 2 ( n − 2)!e − nα 1 τ + n − 1 X k =0 r ( n, k )e − α 1 k τ } ∂ ∂ τ + . . . , where r ( n, k ) are some constan ts. Return to equalit y (41) with constan t co eﬃcien t s ˜ µ k and compare the co eﬃcien ts b efore ∂ ∂ τ : α M − 1 1 ( − 1) M − 1 ( M − 1)!e − ( M +1) α 1 τ + M X k =0 r ( M + 1 , k )e − α 1 k τ = ˜ µ M  α M − 2 1 ( − 1) M − 2 ( M − 2)!e − M α 1 τ + M − 1 X k =0 r ( M , k )e − α 1 k τ  + . . . + ˜ µ 0 (e − α 1 τ − 1) . The last equality fails to b e true. It sho ws that our assumption that m ultiplicity m 1 of a nonzero ro ot α 1 can be 2 or more w as wrong.  If the c haracteristic p olynomial of (23) has only one nonzero ro ot α , then d ( t, t 1 ) = A ( t − t 1 ) e αt . In this case equation (10) admits a non trivial x -integral (see In tro duction, Theorem 3). In the next section w e consider a case when the c haracteristic polynomial of (23) has at least tw o nonzero roo ts. 5 Tw o nonzero r o ots Let α and β b e tw o nonzero ro ots. Consider the v ector ﬁelds S 0 = ∞ X j = −∞ A ( τ j )e αρ j ∂ ∂ τ j , S 1 = ∞ X j = −∞ B ( τ j )e β ρ j ∂ ∂ τ j from the Lie algebra L x , and construct a new sequence of v ector ﬁelds S 2 = [ S 0 , S 1 ] , S n +1 = [ S 0 , S n ] , n ≥ 1 . W e hav e, D S 0 D − 1 = e ατ S 0 , D S 1 D − 1 = e β τ S 1 , D S 2 D − 1 = e ( α + β ) τ S 2 + β A e ( α + β ) τ S 1 − αB e ( α + β ) τ S 0 . 28 In general, for an y n ≥ 3, D S n D − 1 = e (( n − 1) α + β ) τ { S n + ( c n α + d n β ) AS n − 1 + ( p n A ′ + q n A ) AS n − 2 + n − 2 X k =0 ν ( n, k ) S k } , where c n = ( n − 1)( n − 2) 2 , d n = n − 1 , p n +1 = n ( n − 1) 2  n − 2 3 α + β  , n ≥ 2 , q n +1 = n ( n − 2)( n − 1)(3 n − 1) 24 α 2 + ( n − 1) 2 n 2 αβ + n ( n − 1) 2 β 2 , n ≥ 2 . Let us consider a particular case when S 2 = µ 0 S 0 + µ 1 S 1 . (42) W e hav e, D S 2 D − 1 = D ( µ 0 )e ατ S 0 + D ( µ 1 )e β τ S 1 = e ( α + β ) τ S 2 + β A e ( α + β ) τ S 1 − αB e ( α + β ) τ S 0 = e ( α + β ) τ { µ 0 S 0 + µ 1 S 1 } + β A e ( α + β ) τ S 1 − αB e ( α + β ) τ S 0 . Comparing co eﬃcien ts b efore S 0 and S 1 pro duces the following t w o equations e ( α + β ) τ µ 0 − αB e ( α + β ) τ = D ( µ 0 )e ατ , e ( α + β ) τ µ 1 + β A e ( α + β ) τ = D ( µ 1 )e β τ . It follow s that µ 0 , µ 1 are constan ts and B ( τ ) = − µ 0 α (e − β τ − 1) , A ( τ ) = µ 1 β (e − ατ − 1) . And ﬁnally , comparing co eﬃcien ts b efore ∂ ∂ τ in equation (42) implies that α = − β . Let us return to the general case. Since L x is of ﬁnite dimension then there exists suc h n umber N tha t S 0 , S 1 , . . . , S N are linearly indep enden t and S N +1 = µ N S N + µ N − 1 S N − 1 + . . . + µ 0 S 0 . Then D S N +1 D − 1 = D ( µ N ) D S N D − 1 + D ( µ N − 1 ) D S N − 1 D − 1 + . . . + D ( µ 0 ) D S 0 D − 1 and therefore, e ( N α + β ) τ { ( µ N S N + µ N − 1 S N − 1 + . . . ) + A ( c N +1 α + d N +1 β ) S N + A ( p N +1 A ′ + q N +1 A ) S N +1 + . . . } = D ( µ N ) { e (( N − 1) α + β ) τ ( S N + A ( c N α + d N β ) S N − 1 + . . . ) } + D ( µ N − 1 ) { e (( N − 2) α + β ) τ S N − 1 + . . . } + . . . . By comparing the co eﬃcien ts b efore S N w e hav e e ( N α + β ) τ { µ N + A ( c N +1 α + d N +1 β ) } = D ( µ N )e (( N − 1) α + β ) τ It follow s that µ N is a constan t and then A ( c N +1 α + d N +1 β ) = µ N (e − ατ − 1) . 29 If c N +1 α + d N +1 β = N  N − 1 2 α + β  6 = 0, then A ( τ ) = C 1 (e − ατ − 1) for some constan t C 1 . If c N +1 α + d N +1 β = N  N − 1 2 α + β  = 0 ( in this case µ N = 0) w e compare co eﬃcien ts b efore S N − 1 and ha ve e ( N α + β ) τ { µ N − 1 + A ( p N +1 A ′ + q N +1 A ) } = D ( µ N − 1 )e (( N − 2) α + β ) τ . It follow s that µ N − 1 is a constan t and p N +1 AA ′ + q N +1 A 2 = µ N − 1 (e − 2 ατ − 1) . Note t ha t if c N +1 α + d N +1 β = N  N − 1 2 α + β  = 0 then p N +1 = − N ( N − 1)( N +1) 12 α 6 = 0 and q N +1 = − ( N − 1) N ( N +1) 24 α 2 6 = 0 for N ≥ 2. Therefore, 2 q N +1 p N +1 = α . Case N = 1 should b e studied separately ( S 2 = µ 1 S 1 + µ 0 S 0 ) and it w as already . Let us solv e the equation p N +1 AA ′ + q N +1 A 2 = µ N − 1 (e − 2 ατ − 1) . Denote b y y = A 2 . W e hav e, y ′ + αy = k 1 e − 2 ατ − k 1 for some constan t k 1 . It follo ws that A 2 ( τ ) = K 1 (e − 2 ατ + K 2 e − ατ + 1) for some constan ts K 1 and K 2 . Construct new sequence of v ector ﬁelds S ∗ 2 = [ S 1 , S 0 ] , S ∗ n +1 = [ S 1 , S ∗ n ] , n ≥ 2 . Note that S ∗ 2 = − S 2 . Since L x is of ﬁnite dimension then there exists n um b er M suc h that S 0 , S 1 , . . . , S ∗ M are linearly indep enden t and S ∗ M +1 = µ ∗ M S ∗ M + µ ∗ M − 1 S ∗ M − 1 + . . . + µ ∗ 0 S 0 . There are the follo wing p ossibilities . 1)  A ( τ ) = K 1 (e − ατ − 1) , B ( τ ) = K 3 (e − β τ − 1) , 2)  A ( τ ) = K 1 (e − ατ − 1) , B 2 ( τ ) = K 2 3 (e − 2 β τ + K 4 e − β τ + 1) , S ∗ M +1 = µ ∗ M S ∗ M + µ ∗ M − 1 S ∗ M − 1 + . . . + µ ∗ 0 S 0 , M − 1 2 β + α = 0 , 3)  B ( τ ) = K 3 (e − β τ − 1) , A 2 ( τ ) = K 2 1 (e − 2 ατ + K 2 e − ατ + 1) , S N +1 = µ N S N + µ N − 1 S N − 1 + . . . + µ 0 S 0 , N − 1 2 α + β = 0 , 4)  A 2 ( τ ) = K 2 1 (e − 2 ατ + K 2 e − ατ + 1) , S N +1 = µ N S N + µ N − 1 S N − 1 + . . . + µ 0 S 0 , N − 1 2 α + β = 0 , B 2 ( τ ) = K 2 3 (e − 2 β τ + K 4 e − β τ + 1) , S ∗ M +1 = µ ∗ M S ∗ M + µ ∗ M − 1 S ∗ M − 1 + . . . + µ ∗ 0 S 0 , M − 1 2 β + α = 0 , where K 1 , K 2 6 = − 2, K 3 , K 4 6 = − 2 are some constan ts, M , N ≥ 2. In case 1), v ector ﬁelds S 0 and S 1 generate an inﬁnite dimensional Lie algebra L x unless α + β = 0 . 30 In case 2), w e make a substitution 1 − e ατ = e − αw . V ector ﬁelds S 0 and S 1 b ecome S 0 = K 1 ∂ ∂ w + . . . , S 1 = { K 2 3 ((1 − e − αw ) − 2 β α + K 4 (1 − e − αw ) − β α + 1) } 1 / 2 ∂ ∂ w + . . . = g ( w ) ∂ ∂ w + . . . . Note that if S ∗ M +1 = µ ∗ M S ∗ M + µ ∗ M − 1 S ∗ M − 1 + . . . + µ ∗ 0 S 0 , then all co eﬃcien ts µ ∗ k are constan ts. By comparing co eﬃcin ts befor e ∂ ∂ w in b oth sides of the last equation w e obtain that g ( w ) is a solution of linear diﬀerential equation with constan t co eﬃcien ts, that is g ( w ) = { K 2 3 ((1 − e − αw ) − 2 β α + K 4 (1 − e − αw ) − β α + 1) } 1 / 2 = X k R k ( w )e ν k w , (43) where R k ( w ) are some p olynomials. One can show that equalit y (4 3) holds o nly if B ( τ ) = K 3 (e ατ + 1 ). It can b e show n that in case 3) A ( τ ) = K 1 (e β τ + 1 ). In case 4) w e mak e substitution e ατ + K 1 2 + √ e 2 ατ + K 1 e ατ +1 = e αw . Then S 0 = K 1 ∂ ∂ w + . . . , S 1 = n K 2 3  1 2 e αw − K 1 2 +  K 2 1 8 − 1 2  e − αw  − 2 β α + K 4  1 2 e αw − K 1 2 +  K 2 1 8 − 1 2  e − αw  − β α + 1 o 1 / 2 ∂ ∂ w + . . . = g ( w ) ∂ ∂ w + . . . F or function g ( w ) to b e of the for m P k R k ( w )e ν k w , where R k ( w ) are p olynomials, function B ( τ ) has to b e of the form B ( τ ) = K 3 (e α + 1). Then, b y case 3), A ( τ ) = K 1 (e − ατ + 1). It has b een pro ved that in cases 1), 2), 3), 4) one has 1 ∗ )  A ( τ ) = K 1 (e − ατ − 1) , B ( τ ) = K 3 (e ατ − 1) , 2 ∗ )  A ( τ ) = K 1 (e − ατ − 1) , B ( τ ) = K 3 (e ατ + 1) , 3 ∗ )  A ( τ ) = K 1 (e − ατ + 1) , B ( τ ) = K 3 (e ατ − 1) , 4 ∗ )  A ( τ ) = K 1 (e − ατ + 1) , B ( τ ) = K 3 (e ατ + 1) . In case 1 ∗ ) function d ( t, t 1 ) in (10) has a fo rm d ( t, t 1 ) = c 4 ( e αt 1 − e αt ) + c 5 ( e − αt 1 − e − αt ), where c 4 and c 5 are some constan ts. Equation (10 ) with suc h f unction d ( t, t 1 ) admits a nontrivial x -in tegral (see In tro duction, Theorem 3 and § 8). In the next tw o sections w e sho w that Cases 3 ∗ ) and 4 ∗ ) b oth correspo nd to inﬁnite dimensional Lie algebra L x . Case 2 ∗ ) also pro duces an inﬁnite dimensional Lie algebra L x . It can b e prov ed in the same w a y as it is pro ved for case 3 ∗ ). 31 6 Characteris t ic Lie Algebra L x of the chain t 1 x = t x + A 1 (e αt 1 + e αt ) − A 2 (e − αt − e − αt 1 ) Since A ( τ ) = A 1 (e − ατ + 1) and B ( τ ) = A 2 (e ατ − 1) then A ( τ )e αt + k X j =1 A ( τ j )e αt j = A 1  e αt +  2 k − 1 X j =1 e αt j  + e αt k  , and B ( τ )e − αt + k X j =1 B ( τ j )e − αt j = A 2 (e − αt − e − αt k ) . W e hav e, 1 A 1 S 0 = (e αt + e αt 1 ) ∂ ∂ t 1 + ∞ X k =1  e αt +  2 k − 1 X j =1 e αt j  + e αt k  ∂ ∂ t k + ∞ X k =1  e αt +  2 k − 1 X j =1 e αt − j  + e αt − k  ∂ ∂ t − k , and 1 A 2 S 1 = e − ατ ˜ X − ∞ X k = −∞ e − αt k ∂ ∂ t k = e − ατ ˜ X − ˜ S 1 , where ˜ S 1 = ∞ X k = −∞ e − αt k ∂ ∂ t k . In v ariables w j = 1 α e αt j v ector ﬁelds ˜ S 1 and 1 A 1 S 0 can b e rewritten as ˜ S 1 = ∞ X k = −∞ ∂ ∂ w j , 1 A 1 S 0 = α 2 ∞ X k =1 { w k ( w + 2 k − 1 X j =1 w j ) + w 2 k } ∂ ∂ w k + α 2 ∞ X k =1 { w − k ( w + 2 k − 1 X j =1 w − j ) + w 2 − k } ∂ ∂ w − k . W e hav e T 1 = [ ˜ S 1 , [ ˜ S 1 , 1 α 2 A 1 S 0 ]] = 4 ∞ X k = −∞ k ∂ ∂ w k = 4 ˜ T 1 , ˜ T 1 = ∞ X k = −∞ k ∂ ∂ w k , T 2 = [ ˜ S 1 , [ ˜ T 1 , 1 α 2 A 1 S 0 ]] = 3 ∞ X k =1 { k 2 − k + 1 } ( ∂ ∂ w k + ∂ ∂ w − k ) = 3 ˜ T 2 − 3 ˜ T 1 + 3 ˜ S 1 , ˜ T 2 = ∞ X k = −∞ k 2 ∂ ∂ w k . Assume that ˜ T m = ∞ P k = −∞ k m ∂ ∂ w k , m = 1 , 2 . . . , n , are ve ctor ﬁelds from L x . Then T m +1 = [ ˜ S 1 , [ ˜ T m , 1 α 2 A 1 S 0 ]] = ∞ X k =1 { 2(1 + 2 m + 3 m + . . . + k m ) + 2 k m +1 − k m }  ∂ ∂ w k + ∂ ∂ w − k  = ∞ X k =1 n 2  k m +1 m + 1 + d m,m +1 k m + . . . + d 1 ,m +1 k + d 0 ,m +1  + 2 k m +1 − k m o ∂ ∂ w k + ∂ ∂ w − k  and therefore, ˜ T m +1 = ∞ P k = −∞ k m +1 ∂ ∂ w k ∈ L x . It sho ws tha t ˜ T n = ∞ P k = −∞ k n ∂ ∂ w k ∈ L x for a ll n = 1 , 2 , 3 , . . . , and L x is of inﬁnite dimension. 32 7 Characteris t ic Lie Algebra L x of the chain t 1 x = t x + A 1 (e αt 1 + e αt ) + A 2 (e − αt + e − αt 1 ) It w as observ ed in previous studies (see, for instance, [10]) that S-integrable mo dels ha ve the ch ar- acteristic L ie algebra of ﬁnite gro wth. The c hain studied in this section can easily b e reduced to the semi-discrete sine-G o rdon mo del t 1 x = t x + sin t + sin t 1 whic h belongs to the S-in tegrable class. It is remark able that its algebra L x is of ﬁnite grow th. Or, mor e exactly , the dimension of the linear space of multiple commu tators gro ws linearly with the multiplic it y . Belo w w e prov e that the linear space V n of all commutators of m ultiplicity ≤ n has a basis { P 1 , P 2 , P 3 , ...P 2 k ; Q 2 , Q 4 , ...Q 2 k } fo r n = 2 k and a basis { P 1 , P 2 , P 3 , ...P 2 k + 1 ; Q 2 , Q 4 , ...Q 2 k } for n = 2 k + 1, where the op erators P j and Q j are deﬁned consecutiv ely P 1 = [ S 0 , S 1 ] + αS 0 + αS 1 , Q 1 = P 1 , P 2 = [ S 1 , P 1 ] , Q 2 = [ S 0 , Q 1 ] , P 3 = [ S 0 , P 2 ] + αP 2 , Q 3 = [ S 1 , Q 2 ] − αQ 2 , P 2 n = [ S 1 , P 2 n − 1 ] , Q 2 n = [ S 0 , Q 2 n − 1 ] , P 2 n +1 = [ S 0 , P 2 n ] + α P 2 n , Q 2 n +1 = [ S 1 , Q 2 n ] − αQ 2 n , for n ≥ 1. D irect calculations sho w that D P 1 D − 1 = P 1 − 2 α ( S 0 + S 1 ) , D P 2 D − 1 = e − ατ ( P 2 + 2 αP 1 − 2 α 2 ( S 0 + S 1 )) , D P 3 D − 1 = P 3 + 2 αQ 2 − 2 αP 2 − 4 α 2 P 1 + 4 α 3 ( S 0 + S 1 ) , D P 4 D − 1 = e − ατ ( P 4 + 2 αQ 3 − 4 α 2 P 2 + 4 α 2 Q 2 − 4 α 3 P 1 + 4 α 4 ( S 0 + S 1 )) , D Q 2 D − 1 = e ατ ( Q 2 − 2 αP 1 + 2 α 2 ( S 0 + S 1 )) , D Q 3 D − 1 = Q 3 + 2 αQ 2 − 2 αP 2 − 4 α 2 P 1 + 4 α 3 ( S 0 + S 1 ) , D Q 4 D − 1 = e ατ ( Q 4 − 2 αP 3 + 2 α 2 ( P 2 − Q 2 ) + 4 α 3 P 1 − 4 α 4 ( S 0 + S 1 )) , P 3 = Q 3 , [ S 1 , P 2 ] = − α P 2 , [ S 0 , Q 2 ] = α Q 2 , [ S 1 , P 4 ] = − α P 4 , [ S 0 , Q 4 ] = αQ 4 . (44) The co eﬃcien t b efore ∂ ∂ τ in all v ector ﬁelds D P i D − 1 , D Q i D − 1 , 1 ≤ i ≤ 4 is zero. Lemma 10 F or n ≥ 1 we have, (1) D P 2 n +1 D − 1 + 2 α e ατ D P 2 n D − 1 = P 2 n +1 + 2 αQ 2 n , (2) e ατ D P 2 n +2 D − 1 − αD P 2 n +1 D − 1 = P 2 n +2 + αQ 2 n +1 , (3) D Q 2 n +1 D − 1 − 2 α e − ατ D Q 2 n D − 1 = Q 2 n +1 − 2 αP 2 n , (4) e − ατ D Q 2 n +2 D − 1 + αD Q 2 n +1 D − 1 = Q 2 n +2 − αP 2 n +1 , (5) P 2 n +1 = Q 2 n +1 , 33 (6) [ S 1 , P 2 n +2 ] = − αP 2 n +2 , (7) [ S 0 , Q 2 n +2 ] = α Q 2 n +2 . Mor e o ver, the c o eﬃcient b efor e ∂ ∂ τ in al l ve ctor ﬁelds D P k D − 1 , D Q k D − 1 is zer o. Pro of . W e pro v e the Lemma b y induction on n . It follow s from (44) that the base of induction holds f or n = 1. Assume (1) − (7) are true for all n , 1 ≤ n ≤ k . Let us prov e that (1) is true f or n = k + 1 . D P 2 n +3 D − 1 = D ([ S 0 , P 2 n +2 ] + α P 2 n +2 ) D − 1 = [e ατ S 0 , D P 2 n +2 D − 1 ] + αD P 2 n +2 D − 1 = [e ατ S 0 , α e − ατ D P 2 n +1 D − 1 + e − ατ P 2 n +2 + α e − ατ Q 2 n +1 ] + α D P 2 n +2 D − 1 = − α 2 (1 + e − ατ ) D P 2 n +1 D − 1 + α e − ατ [e ατ S 0 , D P 2 n +1 D − 1 ] − α ( 1 + e − ατ ) P 2 n +2 − α 2 (1 + e − ατ ) Q 2 n +1 + P 2 n +3 − αP 2 n +2 + αQ 2 n +2 + αD P 2 n +2 D − 1 = − α 2 (1 + e − ατ ) D P 2 n +1 D − 1 + α e − ατ D [ S 0 , Q 2 n +1 ] D − 1 − α (2 + e − ατ ) P 2 n +2 − α 2 (1 + e − ατ ) Q 2 n +1 + P 2 n +3 + αQ 2 n +2 + αD P 2 n +2 D − 1 = − α 2 (1 + e − ατ ) D P 2 n +1 D − 1 + αQ 2 n +2 − α 2 P 2 n +1 − α 2 D Q 2 n +1 D − 1 − α (2 + e − ατ ) P 2 n +2 − α 2 (1 + e − ατ ) Q 2 n +1 − 2 α 2 Q 2 n +1 − 2 αP 2 n +2 + P 2 n +3 = − 2 α 2 D P 2 n +1 D − 1 + 2 αQ 2 n +2 − 2 α 2 Q 2 n +1 − 2 αP 2 n +2 + P 2 n +3 = 2 α P 2 n +2 + 2 α 2 Q 2 n +1 − 2 α e ατ D P 2 n +2 D − 1 + 2 αQ 2 n +2 − 2 α 2 Q 2 n +1 − 2 αP 2 n +2 + P 2 n +3 = − 2 α e ατ D P 2 n +2 D − 1 + 2 αQ 2 n +2 + P 2 n +3 . The pro o f of (3) is the same as the proo f of (1). Let us sho w t hat (5) is true for n = k + 1. W e hav e, D P 2 n +3 D − 1 = − 2 α e ατ D P 2 n +2 D − 1 + 2 αQ 2 n +2 + P 2 n +3 = − 2 α ( α D P 2 n +1 D − 1 + P 2 n +2 + αQ 2 n +1 ) + 2 αQ 2 n +2 + P 2 n +3 , and D Q 2 n +3 D − 1 = 2 α e − ατ D Q 2 n +2 D − 1 − 2 αP 2 n +2 + Q 2 n +3 = 2 α ( − α D Q 2 n +1 D − 1 + Q 2 n +2 − αP 2 n +1 ) − 2 α P 2 n +2 + Q 2 n +3 . By (5), P 2 n +1 = Q 2 n +1 and therefore D ( P 2 n +3 − Q 2 n +3 ) D − 1 = − 2 α P 2 n +2 − 2 αQ 2 n +2 + 2 αQ 2 n +2 + 2 αP 2 n +2 = 0 . Hence, P 2 n +3 = Q 2 n +3 . Let us pro ve (2) is true for n = k + 1. W e hav e, e ατ D P 2 n +1 D − 1 = e ατ D [ S 1 , P 2 n +3 ] D − 1 = e ατ [e − ατ S 1 , D P 2 n +3 D − 1 ] = e ατ [e − ατ S 1 , − 2 α e ατ D P 2 n +2 D − 1 + 2 αQ 2 n +2 + P 2 n +3 ] = e ατ ( − 2 α 2 (1 + e ατ ) D P 2 n +2 D − 1 ) − 2 α e 2 ατ [e − ατ S 1 , D P 2 n +2 D − 1 ] + P 2 n +4 + 2 αQ 2 n +3 + 2 α 2 Q 2 n +2 = − 2 α 2 (e ατ + e 2 ατ ) D P 2 n +2 D − 1 + 2 α 2 e 2 ατ D P 2 n +2 D − 1 + P 2 n +4 + 2 αQ 2 n +3 + 2 α 2 Q 2 n +2 = − 2 α 2 e ατ D P 2 n +2 D − 1 + P 2 n +4 + 2 αQ 2 n +3 + 2 α 2 Q 2 n +2 = αD P 2 n +3 D − 1 − αP 2 n +3 − 2 α 2 Q 2 n +2 + P 2 n +4 + 2 αQ 2 n +3 + 2 α 2 Q 2 n +2 = αD P 2 n +3 D − 1 + αQ 2 n +3 + P 2 n +4 . 34 The pro of of (4) is similar to the pro of of (2). Let us pro ve that (6) is tr ue f o r n = k + 1. D [ S 1 , P 2 n +4 ] D − 1 = [e − ατ S 1 , α e − ατ D P 2 n +3 D − 1 + e − ατ P 2 n +4 + α e − ατ Q 2 n +3 ] = [e − ατ S 1 , α e − ατ ( − 2 α e ατ D P 2 n +2 D − 1 + P 2 n +3 + 2 αQ 2 n +2 ) + e − ατ P 2 n +4 + α e − ατ Q 2 n +3 ] = [e − ατ S 1 , − 2 α 2 D P 2 n +2 D − 1 + 2 α e − ατ P 2 n +3 + 2 α 2 e − ατ Q 2 n +2 + e − ατ P 2 n +4 ] = − 2 α 2 D [ S 1 , P 2 n +2 ] D − 1 − 2 α 2 e − 2 ατ (1 + e ατ ) P 2 n +3 − 2 α 3 e − 2 ατ (1 + e ατ ) Q 2 n +2 +2 α e − 2 ατ P 2 n +4 + 2 α 2 e − 2 ατ Q 2 n +3 + 2 α 3 e − 2 ατ Q 2 n +2 − α e − 2 ατ (1 + e ατ ) P 2 n +4 + e − 2 ατ [ S 1 , P 2 n +4 ] = 2 α 3 D P 2 n +2 D − 1 − 2 α 2 e − ατ P 2 n +3 + α (e − 2 ατ − e − ατ ) P 2 n +4 − 2 α 3 e − ατ Q 2 n +2 + e − 2 ατ [ S 1 , P 2 n +4 ] = α 2 e − ατ P 2 n +3 + 2 α 3 e − ατ Q 2 n +2 − α 2 e − ατ D P 2 n +3 D − 1 − 2 α 2 e − ατ P 2 n +3 + α (e − 2 ατ − e − ατ ) P 2 n +4 − 2 α 3 e − ατ Q 2 n +2 + e − 2 ατ [ S 1 , P 2 n +4 ] = − α 2 e − ατ P 2 n +3 + α (e − 2 ατ − e − ατ ) P 2 n +4 − αD P 2 n +4 D − 1 + α e − ατ P 2 n +4 + α 2 e − ατ Q 2 n +3 + e − 2 ατ [ S 1 , P 2 n +4 ] . Th us, D [ S 1 , P 2 n +4 ] D − 1 = e − 2 ατ [ S 1 , P 2 n +4 ] + α e − 2 ατ P 2 n +4 − αD P 2 n +4 D − 1 D ([ S 1 , P 2 n +4 ] + α P 2 n +4 ) D − 1 = e − 2 ατ ([ S 1 , P 2 n +4 ] + αP 2 n +4 ) . Hence, [ S 1 , P 2 n +4 ] = − α P 2 n +4 .  Pro of of (7) is similar to the pro of of (6). Corollary 1 We have, e − ατ D Q 2 n D − 1 + e ατ D P 2 n D − 1 = Q 2 n + P 2 n , D P 2 n +1 D − 1 = P 2 n +1 + n X k =1 ( µ (2 n +1) 2 k P 2 k + ν (2 n +1) 2 k Q 2 k ) + n − 1 X k =0 µ (2 n +1) 2 k + 1 P 2 k + 1 + µ (2 n +1) 0 S 0 + ν (2 n +1) 0 S 1 , D P 2 n D − 1 = e − ατ ( P 2 n + n − 1 X k =1 ( µ (2 n ) 2 k P 2 k + ν (2 n ) 2 k Q 2 k ) + n − 1 X k =0 µ (2 n ) 2 k + 1 P 2 k + 1 + µ (2 n ) 0 S 0 + ν (2 n ) 0 S 1 ) , D Q 2 n D − 1 = e ατ ( Q 2 n − n − 1 X k =1 ( µ (2 n ) 2 k P 2 k + ν (2 n ) 2 k Q 2 k ) − n − 1 X k =0 µ (2 n ) 2 k + 1 P 2 k + 1 − µ (2 n ) 0 S 0 − ν (2 n ) 0 S 1 ) . Mor e o ver, µ (2 n +1) 2 n = − 2 α , ν (2 n +1) 2 n = 2 α , µ (2 n ) 2 n − 1 = 2 α . Assume L x is of ﬁnite dimension. There are three p ossibilities: 1) S 0 , S 1 , P 1 , P 2 , Q 2 , P 3 , P 4 , Q 4 , ..., P 2 n − 1 are linearly indep enden t and S 0 , S 1 , P 1 , P 2 , Q 2 , P 3 , P 4 , Q 4 , ..., P 2 n − 1 , P 2 n are linearly dep enden t, 2) S 0 , S 1 , P 1 , P 2 , Q 2 , P 3 , P 4 , Q 4 , ..., P 2 n − 1 , P 2 n are linearly indep enden t and S 0 , S 1 , P 1 , P 2 , Q 2 , P 3 , P 4 , Q 4 , ..., P 2 n − 1 , P 2 n , Q 2 n are linearly dep enden t, 35 3) S 0 , S 1 , P 1 , P 2 , Q 2 , P 3 , P 4 , Q 4 , ..., P 2 n , Q 2 n are linearly indep enden t and S 0 , S 1 , P 1 , P 2 , Q 2 , P 3 , P 4 , Q 4 , ..., P 2 n , Q 2 n , P 2 n +1 are linearly dep enden t. In case 1), P 2 n = γ 2 n − 1 P 2 n − 1 + γ 2 n − 2 P 2 n − 2 + η 2 n − 2 Q 2 n − 2 + ... and D P 2 n D − 1 = D ( γ 2 n − 1 ) D P 2 n − 1 D − 1 + D ( γ 2 n − 2 ) D P 2 n − 2 D − 1 + D ( η 2 n − 2 ) D Q 2 n − 2 D − 1 + ... . (45) W e use Corollary 1 to compare the co eﬃcien ts b efore P 2 n − 1 in (45) a nd ha v e the contradictory equalit y , e − ατ ( γ 2 n − 1 + 2 α ) = D ( γ 2 n − 1 ) . It sho ws that case 1) is imp ossible to hav e. In case 2), Q 2 n = γ 2 n P 2 n + γ 2 n − 1 P 2 n − 1 + η 2 n − 2 Q 2 n − 2 + ... and D Q 2 n D − 1 = D ( γ 2 n ) D P 2 n D − 1 + D ( γ 2 n − 1 ) D P 2 n − 1 D − 1 + D ( η 2 n − 2 ) D Q 2 n − 2 D − 1 + ... . (46) W e use Corollary 1 to compare the co eﬃcien ts b efore P 2 n − 1 in (46) a nd ha v e the contradictory equation, e ατ ( γ 2 n − 1 − 2 α ) = D ( γ 2 n − 1 ) . It sho ws that case 2) is imp ossible to hav e. In case 3), P 2 n +1 = η 2 n Q 2 n + γ 2 n P 2 n + ... and D P 2 n +1 D − 1 = D ( η 2 n ) D Q 2 n D − 1 + D ( γ 2 n ) D P 2 n D − 1 + ... . (47) W e use Coro llary 1 to compare t he co eﬃcien ts b efore P 2 n in ( 47) and ha ve the contradictory equation, ( γ 2 n − 2 α ) = D ( γ 2 n )e − ατ . It sho ws that case 3) also fails to b e tr ue. Therefore, c har a cteristic Lie algebra L x is of inﬁnite dimension. 8 Finding x-in tegrals No w w e are ready to pro ve the main Theorem 3, formulated in Introduction. R eally , in the previous sections we prov ed that if c hain (10) admits a nontrivial x -in t egra l then it is one of t he forms ( 1) − (4). The list i ) − iv ) allo ws one to prov e t he inv erse statemen t: eac h of the equations from the list admits indeed a non trivial x -in tegral.  Let us explain brieﬂy how w e found the list i ) − iv ). Since for eac h equation (1) − (4) w e ha ve constructed the relat ed characteristic Lie algebra to ﬁnd x -in tegr a l F o ne has to solv e the corresp onding system of the ﬁrst order partial diﬀeren tial equations. Belo w we illustrate the metho d with the case (2), for whic h the basis o f the c haracteristic algebra L x is giv en by the v ector ﬁelds ˜ Y = ∂ x + Y a ( τ ) t + b ( τ ) , T 1 = Y − a ( τ ) , ˜ X = ∂ ∂ t + ∂ ∂ t 1 + ∂ ∂ t − 1 + ∂ ∂ t 2 + ∂ ∂ t − 2 + . . . , 36 where a ( τ ) = c 0 τ a nd b ( τ ) = c 2 τ 2 + c 3 τ . Note that x -integral F o f (2) should satisfy the equations ˜ Y F = 0, T 1 F = 0 and ˜ X F = 0. In tro duce new v ariables t , w , w ± 1 , . . . where w j = ln( τ j ) and τ j = t j − t j +1 . V ector ﬁelds ˜ X , T 1 and ˜ Y in new v aria bles are rewritten as ˜ X = ∂ ∂ t , T 1 = ∞ X j = −∞ c 0 ∂ ∂ w j , ˜ Y = ∂ ∂ x − t ∞ X j = −∞ c 0 ∂ ∂ w j + c 0 ∞ X j = −∞ { ˜ ρ j + ˜ b ( w j ) } ∂ ∂ w j = ∂ ∂ x − tT 1 + c 0 ∞ X j = −∞ { ˜ ρ j + ˜ b ( w j ) } ∂ ∂ w j , where ˜ ρ j =            j − 1 P k =0 e w k , if j ≥ 1; 0 , if j = 0; − − 1 P k = j e w k , if j ≤ − 1 , ˜ b ( w j ) = − 1 c 0 ( c 2 e w j + c 3 ) . Note that since w e hav e ˜ X F = 0, F do es not dep end o n t . Now let us consider the ve ctor ﬁeld ˜ Y + tT 1 = A = ∂ ∂ x + c 0 ∞ X j = −∞ { ˜ ρ j + ˜ b ( w j ) } ∂ ∂ w j . W e can write the ve cto r ﬁeld A explicitly as A = ∂ ∂ x + ∞ X j = −∞ n c 0 j − 1 X k =0 e w k  − c 2 e w j − c 3 o ∂ ∂ w j = ∂ ∂ x − c 3 c 0 T 1 + ∞ X j = −∞ n c 0 j − 1 X k =0 e w k  − c 2 e w j o ∂ ∂ w j . The comm utator [ T 1 , A ] give s [ T 1 , A ] = c 0 A − c 0 ∂ ∂ x + c 3 T 1 . Th us we ha v e three v ector ﬁelds A − ∂ ∂ x + c 3 c 0 T 1 := ˜ A = ∞ X j = −∞ n c 0 j − 1 X k =0 e w k  − c 2 e w j o ∂ ∂ w j , T 1 c 0 := ˜ T 1 = ∞ X j = −∞ ∂ ∂ w j , ˜ X 1 = ∂ ∂ x , that solv e ˜ AF = 0, ˜ T 1 F = 0, ˜ X 1 F = 0. Note tha t [ ˜ T 1 , ˜ A ] = ˜ A . Since ˜ X 1 F = 0, F do es not dep end o n x . Hence w e end up with t w o equations. By Jacobi theorem the system of equations has a no n trivial solution F ( w , w 1 , w 2 ) dep ending on three v ariables. Therefore we need ﬁrst t hree terms of ˜ A and ˜ T 1 ; ˜ A = − c 2 w ∂ ∂ w + ( c 0 e w − c 2 e w 1 ) ∂ ∂ w 1 + ( c 0 e w + c 0 e w 1 − c 2 e w 2 ) ∂ ∂ w 2 , ˜ T 1 = ∂ ∂ w + ∂ ∂ w 1 + ∂ ∂ w 2 . 37 No w w e again in tro duce new v ariables w = ǫ, w − w 1 = ǫ 1 , w 1 − w 2 = ǫ 2 . V ector ﬁelds ˜ A and ˜ T 1 in new v ar iables are rewritten as ˜ A = e ǫ n − c 2 ∂ ∂ ǫ + (( − c 2 − c 0 ) + c 2 e − ǫ 1 ) ∂ ∂ ǫ 1 + (( − c 2 − c 0 )e − ǫ 1 + c 2 e − ǫ 1 − ǫ 2 ) ∂ ∂ ǫ 2 o , ˜ T 1 = ∂ ∂ ǫ . T o ﬁnd t he x -integral ii) in Theorem 3 one has to solv e the equation n (( − c 2 − c 0 ) + c 2 e − ǫ 1 ) ∂ ∂ ǫ 1 + e − ǫ 1 (( − c 2 − c 0 ) + c 2 e − ǫ 2 ) ∂ ∂ ǫ 2 o F = 0 . 9 Conclus ion In this article the pr o blem of classiﬁcation of D arb oux in tegrable no nlinear semi-discrete c hains of hy p erb olic t yp e was studied. An approach based on the notion of characteristic Lie algebra w as pro p erly mo diﬁed and successfully used. W e gav e a complete list of hyperb olic t yp e c ha ins t 1 x = t x + d ( t, t 1 ) admitting non trivial x -in tegrals. W e demonstrated that the metho d of c ha r acteristic Lie algebras pro vides an eﬀectiv e to ol to classify in tegrable discrete c hains as we ll. The method did not get m uc h atten tion in the literature, to our knowledge there are only tw o studies (see [9] and [14]) where the c haracteristic Lie a lgebras are applied for solving the classiﬁcation problem for the partial diﬀerential equations and systems. Surprisingly ﬁrst of them was published in 1981 and the second one only t wen t y ﬁv e ye a rs later. Ac kno w ledgmen t s This w ork is partially supp or t ed b y the Scien tiﬁc and T ec hnological Researc h Council of T urk ey (T ¨ UB ˙ IT AK). O ne o f the authors (IH) thanks also Russian F oundation for Basic Researc h (R F BR) (gran t s # 06-01-9 2051KE-a and # 08- 0 1-00440 -a). References [1] V. E. Adler, S. Y a. Start sev, On discr ete analo gues of the Liouvil le e quation, T eoret. Mat. Fizik a, 121 , no. 2, 271-284 (1999), (English translation: Theoret. and Math. Ph ys. , 121 , no. 2, 1484-14 95, (1999)). [2] A.V. Z abro din, Hir ota diﬀer ential e quations (In Russian), T eor. Mat. Fiz., 113 , no. 2, 179-23 0 (1997), (English translation: Theoret. a nd Math. Ph ys. , 113 , no. 2, 1347- 1392 (1997)). [3] R. l. Y amilov, Symmetries as inte gr ability criteria for diﬀer ential diﬀer enc e e quations, J. Ph ys. A: Mat h. Gen. 39, 541-623 (2006). [4] F. W. Nijhoﬀ, H. W. Cap el, The di scr ete Kortewe g-de V rie s e quation , Acta Applicandae Mathematicae, 39 , 133- 158 (1 9 95). [5] B. Grammaticos, G. Karra, V. P apageorgio u, A. Ramani, Inte gr ability of di s cr ete-time sys- tems, Ch a otic d yna m ics, (P atras,199 1), NA TO Adv. Sci. Inst. Ser. B Ph ys. , 298 , 7 5-90, Plen um, New Y ork, (1992 ). [6] G. D arb oux, L e¸ cons sur la th ´ e orie g ´ e n ´ e r ale des surfac es et les applic ations ge ometriques d u c alcul in ﬁnitesimal, T.2. P aris: Gautier-Villa rs (1915). 38 [7] I. M. Anderson, N. K a mran, The variational bic omplex for hyp erb olic se c o nd-or der sc alar p artial d i ﬀ er ential e quations in the plane, Duk e Math. J. , 87 , no. 2, 265-319 (1997). [8] A. V. Zhib er, V. V. Sok o lo v, Exactly inte gr able hyp erb olic e quations of Liouvil le typ e , (In Russian) Usp ekhi Mat. Nauk 56, no. 1 (337), 63- 1 06 (2001), (English t r anslation: Russian Math. Surv eys, 56 , no. 1, 61-101 (2001)). [9] A. B. Shabat, R. I. Y a milov, Exp onential systems of typ e I and the C a rtan matric es , (In Russian), Preprin t, Bashkirian Branc h of Academ y of Science of the USSR, Ufa, (1981). [10] A. N. Lezno v, V. G. Smirno v, A. B. Shabat, Gr oup o f inner s ymm etries and inte gr ability c onditions for two-dimen sional dynamic al systems, T eoret. Mat. Fizik a, 51 , no. 1, 10 -21 (1982). [11] A. V. Zhib er, F. Kh. Mukmino v, Quadr atic systems, symm etries, cha r acteristic and c om- plete algebr as , Problems of Mathematical Ph ysics and Asymptotics of their So lut io ns, ed. L.A.Kaly akin, Ufa, Institute of Mathematics, RAN, 13-33 (1991). [12] A. A. Bormiso v, F. K h. Mukmino v, Symmetries of hyp erb olic systems of R i c c ati e quation typ e , (In Russian), T eoret. Mat. Fiz. 127 , no. 1, 47-62 (200 1 ), (English translation: Theoret. and Math. Ph ys. , 127 , no. 1, 446-459 (2001)). [13] V.V. Sok olov and S.Y a. Start sev, Symmetries of nonline ar hyp erb olic systems of the T o da chain typ e , Theoretical and Mathematical Ph ysics, 155 , no. 2, 802-811 (2008). [14] A. V. Zhib er, R. D. Murtazina, On the char acteristic Lie al g ebr as for the e quations u xy = f ( u, u x ), (In Russian), F undam. Prikl. Mat. , 12 , no. 7, 65-78 (2006). [15] N. Kh. Ibrag imov, A. B. Shabat, Evolution e quations with nontrivial Lie-B¨ acklund gr oup, F unktsional. Anal. i Prilozhen, 14 , no. 1, 25- 3 6 (1980). [16] A. V. Mikhailo v, A. B. Shabat, R. I. Y amilov , A symmetry appr o ach to the classiﬁc ation of nonline ar e quations. C omplete list of inte gr able systems , (In Russian), Usp ekhi Mat. Nauk, 42 , no. 4, 3-53 (1987). [17] R. I. Y amilov, D . Levi, Inte gr ability c onditions for $ n $ an d $ t $ dep endent dynamic al lattic e e quations , J. Nonlinear Math. Ph ys. , 11 , no. 1, 75- 101 (2004). [18] M. G ¨ urses, A. Kara su, Inte gr able KdV Systems: R e cursion Op er ators o f De gr e e F our, Ph ysics Letters A, 251 , 247-249 (1999) // arxiv : solv-in t/98 11013. [19] M. G ¨ urses, A. Karasu, R. T urha n, Nonautonomous Svi n olup ov Jor dan KdV Systems, Journal of Ph ysics A: Mathematical and General, 34 , 5705-5 7 11 (20 0 1) // arxiv : nlin.SI/0101031. [20] S. I. Svinolupov, On the analo gues of the Bur gers Equation , Ph ys. Lett. 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P ek can, On Some A lgebr aic Pr op erties of Semi- Discr ete Hyp erb olic T yp e Equations, T urkish Journal of Mathematics, 32, 1-17(2008) //arXiv:nlin/070306 5. 40

On the Classification of Darboux Integrable Chains

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