Ismagil Habibullin, Natalya Zheltukhina, and Aslı Pekcan Citation: Journal of Mathematical Physics 50, 102710 (2009); View online: https://doi.org/10.1063/1.3251334
View Table of Contents: http://aip.scitation.org/toc/jmp/50/10
Complete list of Darboux integrable chains of the form
t
1x= t
x+ d
„t,t
1…
Ismagil Habibullin,1,a兲 Natalya Zheltukhina,2,b兲 and Aslı Pekcan2 1
Ufa Institute of Mathematics, Russian Academy of Science, Chernyshevskii Str., 112, Ufa 450077, Russia
2
Department of Mathematics, Faculty of Science, Bilkent University, 06800 Ankara, Turkey
共Received 28 July 2009; accepted 23 September 2009; published online 30 October 2009兲 We study differential-difference equation 共d/dx兲t共n+1,x兲= f共t共n,x兲,t共n + 1 , x兲,共d/dx兲t共n,x兲兲 with unknown t共n,x兲 depending on continuous and discrete variables x and n. Equation of such kind is called Darboux integrable, if there exist two functions F and I of a finite number of arguments x, 兵t共n+k,x兲其k=−⬁ ⬁, 兵共dk/dxk兲t共n,x兲其
k=1
⬁ , such that D
xF = 0 and DI = I, where Dxis the operator of total
differentiation with respect to x and D is the shift operator: Dp共n兲=p共n+1兲. Re-formulation of Darboux integrability in terms of finiteness of two characteristic Lie algebras gives an effective tool for classification of integrable equations. The com-plete list of Darboux integrable equations is given in the case when the function f is of the special form f共u,v,w兲=w+g共u,v兲. © 2009 American Institute of Physics. 关doi:10.1063/1.3251334兴
I. INTRODUCTION
In this paper we continue investigation of integrable semidiscrete chains of the form
d
dxt共n + 1,x兲 = f
冉
t共n,x兲,t共n + 1,x兲, ddxt共n,x兲
冊
, 共1兲started in our previous paper1共see also Refs.2–4兲. Here t=t共n,x兲 and t1= t共n+1,x兲 are unknown. Function f = f共t,t1, tx兲 is assumed to be locally analytic andf/txis not identically zero.
Nowa-days discrete phenomena are very popular due to their applications in physics, geometry, biology, etc.共see Refs.5–8and references therein兲.
Below we use subindex to indicate the shift of the discrete argument: tk= t共n+k,x兲, k苸Z, and
derivatives with respect to x: t关1兴= tx=共d/dx兲t共n,x兲, t关2兴= txx=共d2/dx2兲t共n,x兲, t关m兴=共dm/dxm兲t共n,x兲, m苸N. Introduce the set of dynamical variables containing 兵tk其k=−⬁ ⬁;兵t关m兴其m=1⬁ .
We denote through D and Dxthe shift operator and the operator of the total derivative with
respect to x correspondingly. For instance, Dh共n,x兲=h共n+1,x兲 and Dxh共n,x兲=共d/dx兲h共n,x兲.
Functions I and F, both depending on x and a finite number of dynamical variables, are called, respectively, n- and x-integrals of共1兲, if DI = I and DxF = 0共see also Ref.9兲. One can see that any n-integral I does not depend on variables tm, m苸Z\兵0其, and any x-integral F does not depend on
variables t关m兴, m苸N.
Chain 共1兲 is called Darboux integrable if it admits a nontrivial n-integral and a nontrivial
x-integral.
Note that all Darboux integrable chains of the form共1兲are reduced to the d’Alembert equation
w1x− wx= 0 by the following “differential” substitution w = F + I. Indeed, Dx共D−1兲w=共D−1兲DxF
+ Dx共D−1兲I=0. This implies that two arbitrary Darboux integrable chains of the form共1兲
a兲Electronic mail: habibullinismagil@gmail.com. b兲Electronic mail: natalya@fen.bilkent.edu.tr.
50, 102710-1
u1x= f共u,u1,ux兲, v1x= f˜共v,v1,vx兲
are connected with one another by the substitution
F共x,u,u1,u−1, . . .兲 + I共x,u,ux,uxx, . . .兲 = F˜共x,v,v1,v−1, . . .兲 + I˜共x,v,vx,vxx, . . .兲,
which is evidently split down into two relations,
F共x,u,u1,u−1, . . .兲 = F˜共x,v,v1,v−1, . . .兲 − h, I共x,u,ux,uxx, . . .兲 = I˜共x,v,vx,vxx, . . .兲 + h,
where h is some constant.
The idea of such kind integrability goes back to Laplace’s discovery of cascade method of integration of linear hyperbolic-type partial differential equation with variable coefficients made in 1773共see Ref.10兲. Roughly speaking the Laplace theorem claims that a linear hyperbolic partial
differential equation admits general solution in a closed form if and only if its sequence of Laplace invariants terminates at both ends共see Ref.11兲. More than hundred years later Darboux applied
the cascade method to the nonlinear case. He proved that a nonlinear hyperbolic equation is integrable 共Darboux integrable兲 if and only if the Laplace sequence of the linearized equation terminates at both ends. This result has been rediscovered very recently by Anderson and Kamran12 and Sokolov and Zhiber.13
An alternative approach was suggested by Shabat and Yamilov in 1981 共see Ref.14兲. They
assigned two Lie algebras, called characteristic Lie algebras, to each hyperbolic equation and proved that the equation is Darboux integrable if and only if both characteristic Lie algebras are of finite dimensions.
The purpose of the present article is to study characteristic Lie algebras of the chain 共1兲 introduced in our papers2–4and convince the reader that in the discrete case these algebras provide a very effective classification tool.
We denote through Lxand Lncharacteristic Lie algebras in x- and n-directions, respectively.
Remind the definition of Lx. Rewrite first the chain 共1兲 in the inverse form tx共n−1,x兲
= g共t共n,x兲,t共n−1,x兲,tx共n,x兲兲. It can be done 共at least locally兲 due to the requirement 共f/tx兲
共t,t1, tx兲⫽0. An x-integral F=F共x,t,t⫾1, t⫾2, . . .兲 solves the equation DxF = 0. Applying the chain
rule, one gets KF = 0, where
K = x+ tx t+ f t1 + g t−1 + f1 t2 + g−1 t−2 + ¯ . 共2兲
Since F does not depend on the variable tx, then XF = 0, where X =/tx. Therefore, any vector
field from the Lie algebra generated by K and X annulates F. This algebra is called the charac-teristic Lie algebra Lx of chain 共1兲 in x-direction. The notion of characteristic algebra is very
important. One can prove that chain共1兲admits a nontrivial x-integral if and only if its Lie algebra
Lxis of finite dimension. The proof of the next classification theorem from Ref.1is based on the
finiteness of the Lie algebra Lx.
Theorem 1.1: Chain
t1x= tx+ d共t,t1兲 共3兲
admits a nontrivial x-integral if and only if d共t,t1兲 is one of the following kinds: 共1兲 d共t,t1兲=A共t1− t兲,
共2兲 d共t,t1兲=c1共t1− t兲t+c2共t1− t兲2+ c3共t1− t兲, 共3兲 d共t,t1兲=A共t1− t兲e␣t,
共4兲 d共t,t1兲=c4共e␣t1− e␣t兲+c5共e−␣t1− e−␣t兲,
where A = A共t1− t兲 is an arbitrary function of one variable and␣⫽0, c1⫽0, c2, c3, c4⫽0,
c5⫽0 are arbitrary constants. Moreover, some nontrivial x-integrals in each of the cases are 共i兲 F = x +兰t1−t共du/A共u兲兲, if A共u兲⫽0, F=t
共ii兲 F = 1 共− c2+ c1兲 ln
冏
共− c2+ c1兲t2− t1 t3− t2 + c2冏
+ 1 c2 ln冏
c2t2− t1 t1− t − c2+ c1冏
for c2共c2− c1兲⫽0, F = ln冏
t2− t1 t3− t2冏
+t2− t1 t1− t for c2= 0, F =t2− t1 t3− t2 + ln冏
t2− t1 t1− t冏
for c2= c1. 共iii兲 F =冕
t1−t e−␣udu A共u兲 −冕
t2−t1 du A共u兲. 共iv兲 F =共e ␣t− e␣t2兲共e␣t1− e␣t3兲 共e␣t− e␣t3兲共e␣t1− e␣t2兲.In what follows we study semidiscrete chains共3兲admitting not only nontrivial x-integrals but also nontrivial n-integrals. First of all we will give an equivalent algebraic formulation of the
n-integral existence problem. Rewrite the equation DI = I defining n-integral in an enlarged form, I共x,t1, f, fx, . . .兲 = I共x,t,tx,txx, . . .兲. 共4兲
The left hand side contains the variable t1 while the right hand side does not. Hence we have
D−1共d/dt
1兲DI=0, i.e., the n-integral is in the kernel of the operator
Y1= D−1Y0D, where Y1= t+ D −1共Y 0f兲 tx + D−1Y0共fx兲 txx + D−1Y0共fxx兲 txxx +¯ 共5兲 and Y0= d dt1 . 共6兲
It can easily be shown that for any natural j the equation D−jY
0DjI = 0 holds. Direct calculations show that D−jY0Dj= Xj−1+ Yj, jⱖ 2, where Yj+1= D−1共Y jf兲 tx + D−1Y j共fx兲 txx + D−1Y j共fxx兲 txxx + ¯ , jⱖ 1, 共7兲
Xj=
t−j
, jⱖ 1. 共8兲
The following theorem defines the characteristic Lie algebra Ln of the chain共1兲.
Theorem 1.2:共Ref.3兲 Equation(1)admits a nontrivial n-integral if and only if the following two conditions hold.
共1兲 Linear space spanned by the operators 兵Yj其1⬁is of finite dimension, denote this dimension by
N.
共2兲 Lie algebra Ln generated by the operators Y1, Y2, . . . , YN, X1, X2, . . . , XN is of finite dimen-sion. We call Ln the characteristic Lie algebra of(1) in the direction of n.
We use x-integral classification Theorem 1.1 and n-integral existence Theorem 1.2 to obtain the complete list of Darboux integrable chains of the form共3兲. The statement of this main result of the present paper is given in the next theorem.
Theorem 1.3: Chain(3)admits nontrivial x- and n-integrals if and only if d共t,t1兲 is one of the
kind.
共1兲 d共t,t1兲=A共t1− t兲, where A共t1− t兲 is given in implicit form A共t1− t兲=共d/d兲P共兲, t1− t = P共兲,
with P共兲 being an arbitrary quasipolynomial, i.e., a function satisfying an ordinary differ-ential equation,
P共N+1兲=NP共N兲+ ¯ +1P
⬘
+0P,with constant coefficientsk, 0ⱕkⱕN.
共2兲 d共t,t1兲=C1共t12− t2兲+C2共t1− t兲. 共3兲 d共t,t1兲=
冑C
3e2␣t1+ C4e␣共t1+t兲+ C3e2␣t. 共4兲 d共t,t1兲=C5共e␣t1− e␣t兲+C6共e−␣t1− e−␣t兲,where␣⫽0, Ci, 1ⱕiⱕ6, are arbitrary constants. Moreover, some nontrivial x-integrals F and n-integrals I in each of the cases are the following.
共i兲 F = x −兰t1−tds/A共s兲, I=L共D
x兲tx, where L共Dx兲 is a differential operator which annihilates
共d/d兲P共兲 where Dx= 1. 共ii兲 F =兵共t3− t1兲共t2− t兲其/兵共t3− t2兲共t1− t兲其, I=tx− C1t2− C2t. 共iii兲 F =兰t1−te−␣sds/
冑C
3e2␣s+ C4e␣s+ C3−兰t2−t1ds/冑C
3e2␣s+ C4e␣s+ C3, I = 2txx−␣tx 2 −␣C3e2␣t. 共iv兲 F =兵共e␣t− e␣t2兲共e␣t1− e␣t3兲其/兵共e␣t− e␣t3兲共e␣t1− e␣t2兲其, I=tx− C5e␣t− C6e−␣t.
Equation of the form x= A共兲, where = t1− t, is integrated in quadratures. But to get the final answer one should evaluate the integral and then find the inverse function. The general solution is given in an explicit form,
t共n,x兲 = t共0,x兲 +
兺
j=0 n−1P共x + cj兲, 共9兲
where t共0,x兲 and cjare arbitrary functions of x and j, respectively, and A共兲= P
⬘
共兲, t1− t = P共兲. Actually we havex= P共兲x= P共兲, which implies x= 1, so that共n,x兲= P共x+cn兲. By solvingthe equation t共n+1,x兲−t共n,x兲= P共x+cn兲 one gets the answer above. Requirement for x= A共兲 to
be Darboux integrable induces condition on function P to satisfy a linear ordinary differential equation with constant coefficients.
The x-integrals in the cases共2兲 and 共4兲 given in Theorem 1.3 are written as cross ratios of four points t, t1, t2, t3and, respectively, points et, et1, et2, et3. Due to the well known theorem, given four
points z1, z2, z3, z4in the projective complex planeCP can be mapped to other given four points
z = R共w兲 ªa11w + a12 a21w + a22
, 共10兲
such that zj= R共wj兲, where j=1,2,3,4, if and only if z4− z2 z4− z3 z3− z1 z2− z1 =w4− w2 w4− w3 w3− w1 w2− w1 . 共11兲
Evidently function F for the case共2兲 关as well as for the case 共4兲兴 can immediately be found from the equation I = c共x兲 which is equivalent to Riccati equation tx= C1t2+ C2t + c共x兲. It is well known that cross ratio of four different solutions of the Riccati equation does not depend on x.
Studying the examples below we briefly discuss connection between discrete models and their continuum analogs. The case共3兲 with C3= 1 and␣= 1 leads in the continuum limit to the equation
uxy= eu
冑
uy2+ 1 共12兲found earlier in Ref.13. Indeed set t共n,x兲=u共y,x兲 and C4= −2 +⑀2, where y = n⑀. Then substitute
=⑀uy+ O共⑀2兲 as⑀→0 into the equation t1x− tx= et
冑e
2+ C4e+ 1 and evaluate the limit as⑀→0 to get 共12兲. It is remarkable that Eq.共12兲 has the same integral 共y-integral兲 I=2uxx− ux2− e2uas itsdiscrete counterpart.
The chain t1x− tx=共et1− et兲/2 goes to the equation uxy=
1 2e
uu
y in the continuum limit. Its n-integral I = tx−
1 2e
t
coincides with the corresponding y-integral of the continuum analog. The Darboux integrable chain t1x− tx= Ce共t1+t兲/2关it comes from the case 共3兲 for appropriate choice of the
parameters兴 being a discrete version of the Liouville equation uxy= eu, also has a common integral I = 2txx− tx
2
with its continuum limit equation. Note that the chain defines the Bäcklund transform for the Liouville equation.
Let us comment the list of the equations in Theorem 1.3. Case共1兲 is degenerate, it is reduced to a first order ordinary differential equation and easily integrated. Equation 共2兲 with C2= 0 is given in Ref. 9. Case 共3兲 for C4=⫾C3 is found in Ref. 15. To the best of our knowledge Eqs.
共2兲–共4兲are new except these two cases.
The article is organized as follows. In Sec. II general results related to the Lie algebra Lnof
Eq. 共1兲 are given. Section III is split into four subsections. Theorem 1.1 from Sec. I gives a complete list of Eq.共3兲admitting nontrivial x-integral. This list consists of four different types of equations共3兲. In each subsection of Sec. III one of these four different types from Theorem 1.1 is treated by imposing additional condition for an equation to possess nontrivial n-integrals. The conclusion is provided in Sec. IV.
II. GENERAL RESULTS
Define a class F of locally analytic functions each of which depends only on a finite number of dynamical variables. In particular, we assume that f共t,t1, tx兲苸F. We will consider vector fields
given as infinite formal series of the form
Y =
兺
k=0 ⬁ yk t关k兴 共13兲with coefficients yk苸F. Introduce notions of linearly dependent and independent sets of the vector
fields共13兲. Denote through PNthe projection operator acting according to the rule
PN共Y兲 =
兺
k=0 N yk t关k兴. 共14兲Z =
兺
k=0 N zk t关k兴. 共15兲We say that a set of finite vector fields Z1, Z2, . . . , Zmis linearly dependent in some open region U,
if there is a set of functions 1,2, . . . ,m苸F defined on U such that the function 兩1兩2+兩2兩2 +¯+兩m兩2 does not vanish identically and the condition
1Z1+2Z2+ ¯ + mZm= 0 共16兲
holds for each point of region U.
We call a set of the vector fields Z1, Z2, . . . , Zm of the form 共13兲 linearly dependent in the
region U if for each natural N the following set of finite vector fields PN共Z1兲, PN共Z2兲, ... , PN共Zm兲
is linearly dependent in this region. Otherwise we call the set Z1, Z2, . . . , Zmlinearly independent
in U.
Now we give some properties of the characteristic Lie algebra introduced in Theorem 1.2. The proof of the first two lemmas can be found in Ref.3. However, for the reader’s convenience we still give the proof of the second lemma.
Lemma 2.1: If for some integer N the operator YN+1is a linear combination of the operators Yiwith iⱕN: YN+1=␣1Y1+␣2Y2+ . . . +␣NYN, then for any integer j⬎N, we have a similar expres-sion Yj=1Y1+2Y2+¯+NYN.
Lemma 2.2: The following commutativity relations take place: 关Y0, X1兴=0, 关Y0, Y1兴=0, and 关X1, DX1D−1兴=0. Proof: We have 关Y0,X1兴 =
冋
d dt1 , d dt−1册
= 0, 关Y0,Y1兴 = D−1关DY0D−1,Y0兴D = D−1冋
d dt2, d dt1册
D = 0, 关X1,DX1D−1兴 = D关D−1X1D,X1兴D−1= D关X2,X1兴D−1= 0. 䊐 Note that Yk+1= D−1YkD, kⱖ 2, D−1Y1D = X1+ Y2. 共17兲 The next three statements turned out to be very useful for studying the characteristic Lie algebraLn.
Lemma 2.3: (Reference 1) If the Lie algebra generated by the vector fields S0=兺⬁j=−⬁/wjand S1=兺j=−⬁ ⬁c共wj兲/wj is of finite dimension then c共w兲 is one of the forms
共1兲 c共w兲=a1+ a2ew+ a3e−w,
共2兲 c共w兲=a1+ a2w + a3w2, where⫽0, a1, a2, and a3are some constants.
Lemma 2.4:
共1兲 Suppose that the vector field
Y =␣共0兲 t+␣共1兲 tx +␣共2兲 txx + ¯ ,
where␣x共0兲=0 solves the equation 关Dx, Y兴=兺k=−⬁ ⬁,k⫽0共k兲/tk, then Y =␣共0兲/t.
Y =␣共1兲 tx +␣共2兲 txx +␣共3兲 txxx +¯
solves the equation关Dx, Y兴=hY +兺k=−⬁ ⬁,k⫽0共k兲/tk, where h is a function of variables t, tx, txx, . . ., t⫾1, t⫾2, . . ., then Y = 0.
Lemma 2.5: For any mⱖ0, we have
关Dx,Ym兴 = −
兺
j=1 m D−j共Ym−j共f兲兲Yj−兺
k=1 ⬁ Ym共D−共k−1兲g兲 t−k −兺
k=1 ⬁ Ym共Dk−1f兲 tk . 共18兲 In particular, 关Dx,Y0兴 = −兺
k=1 ⬁ Y0共Dk−1f兲 tk , 共19兲 关Dx,Y1兴 = − D−1共Y0共f兲兲Y1−兺
k=1 ⬁ Y1共D−共k−1兲g兲 t−k −兺
k=1 ⬁ Y1共Dk−1f兲 tk . 共20兲Both Lemmas 2.4 and 2.5 easily can be derived from the following formula
关Dx,Y兴 = 共␣x共0兲 −␣共1兲兲 t−
兺
k=1 ⬁ Y共D−共k−1兲g兲 t−k −兺
k=1 ⬁ Y共Dk−1f兲 tk +兺
k=1 ⬁ 共␣x共k兲 −␣共k + 1兲兲 t关k兴. 共21兲 Suppose that Eq. 共1兲admits a nontrivial n-integral. Then, by Theorem 1.2, its characteristic Lie algebra Ln is of finite dimension. Linear space of the basic vector fields 兵Yk其1⬁ is also finite dimensional. We have the following theorem.Theorem 2.6: Dimension of span兵Yk其1⬁is finite and equal, say N if and only if the following
system of equations is consistent:
Dx共N兲 = N共AN,N− AN+1,N+1兲 − AN+1,N, Dx共N−1兲 = N−1共AN−1,N−1− AN+1,N+1兲 + NAN,N−1− AN+1,N−1, Dx共N−2兲 = N−2共AN−2,N−2− AN+1,N+1兲 + N−1AN−1,N−2+NAN,N−2− AN+1,N−2, ] Dx共2兲 = 2共A2,2− AN+1,N+1兲 + 3A3,2+ ¯ + NAN,2− AN+1,2, Dx共1兲 = 1共A1,1− AN+1,N+1兲 + 2A2,1+3A3,1+ ¯ + NAN,1− AN+1,1. 0 =1A1,0+2A2,0+3A3,0+ ¯ + NAN,0− AN+1,0. 共22兲 Here Ak,j= D−j共Y k−jf兲.
Proof: Suppose that the dimension of span兵Yk其1⬁ is finite, say N, then, by Lemma 2.1,
YN+1=1Y1+2Y2+ ¯ + NYN. 共23兲
Take the commutator of both sides with Dxand get by using the main commutativity relation共18兲
the following equation:
−
兺
j=0 N+1
AN+1,jYj= Dx共1兲Y1+ Dx共2兲Y2+ ¯ + Dx共N兲YN
−
冉
1兺
j=0 1 A1,jYj+2兺
j=0 2 A2,jYj+ ¯ + N兺
j=0 N AN,jYj冊
.Now replace YN+1at the left hand side by共23兲and collect coefficients of the independent vector
fields to derive the system given in the theorem.
Suppose now that the system共22兲in the theorem has a solution. Let us prove that the vector field YN+1is expressed in the form共23兲. Let
Z = YN+1−1Y1−2Y2− ¯ − NYN. 共24兲 Let us find关Dx, Z兴. 关Dx,Z兴 = 关Dx,YN+1兴 − Dx共1兲Y1− ¯ − Dx共N兲YN−1关Dx,Y1兴 − 2关Dx,Y2兴 − ¯ − N关Dx,YN兴 = −
兺
j=0 N+1 AN+1,jYj− Dx共1兲Y1− ¯ − Dx共N兲YN −冉
1兺
j=0 1 A1,jYj+2兺
j=0 2 A2,jYj+ ¯ + N兺
j=0 N AN,jYj冊
+兺
k=−⬁,k⫽0 ⬁ 共k兲 tk .Replace now Dx共1兲, ... ,Dx共N兲 by means of the system共22兲. After some simplifications one gets
关Dx,Z兴 = − AN+1,N+1Z +
兺
k=−⬁,k⫽0 ⬁ 共k兲 tk . 共25兲 By Lemma 2.4 we get Z = 0. 䊐The proof of the next three results can be found in Ref.4.
Lemma 2.7: If the operator Y2= 0 then关X1, Y1兴=0.
The reverse statement to Lemma 2.7 is not true as the equation t1x= tx+ etshows共see Lemma
3.4 below兲.
Lemma 2.8: The operator Y2= 0 if and only if we have
ft+ D−1共ft1兲ftx= 0. 共26兲 Corollary 2.9: The dimension of the Lie algebra Lnassociated with n-integral is equal to 2 if
and only if共26兲holds, or the same Y2= 0. Now let us introduce vector fields
C1=关X1,Y1兴, Ck=关X1,Ck−1兴, k ⱖ 2. 共27兲
It is easy to see that
Cm= X1 m D−1共Y0共f兲兲 tx + X1mD−1共Y0Dx共f兲兲 txx + X1mD−1共Y0Dx 2共f兲兲 txxx + ¯ . 共28兲 Lemma 2.10: We have
关Dx,Cm兴 = − gtxX1 m D−1Y0共f兲X1− X1 m D−1Y0共f兲Y1−
兺
j=1 m A共m兲j Cj, 共29兲 where Aj共m兲= X1 m−j再
C共m, j − 1兲gt−1− C共m, j兲 gt gtx冎
, mⱖ 1, C共m,k兲 = m! k !共m − k兲!. In particular, 关Dx,C1兴 = − gtxX1D−1Y0共f兲X1− X1D−1Y0共f兲Y1−冉
gt−1− gt gtx冊
C1.Proof: We prove the lemma by induction on m. Note that for any vector field,
A =共0兲 t+共1兲 tx +共2兲 txx + ¯ ,
acting on the set of functions H depending on variables t−1, t, t关k兴, k苸N, formula共21兲becomes
关Dx,A兴 = − 共共0兲gt+共1兲gtx兲 t−1 +共x共0兲 −共1兲兲 t+共x共1兲 −共2兲兲 tx +共x共2兲 −共3兲兲 txx +共x共3兲 −共4兲兲 txxx + ¯ .
Applying the last formula with C1instead of A, we have
关Dx,C1兴 = − gtxX1D−1Y0共f兲X1− X1D−1Y0共f兲 t+
兺
k=1 ⬁ 兵DxX1D−1Y0Dx k−1共f兲 − X 1D−1Y0Dx k共f兲其 t关k兴. Since 关Y0,Dx兴G共t,t1,tx,txx,txxx, . . .兲 = ft1Gt1= ft1Y0G, i.e., Y0Dx= DxY0+ ft1Y0 and 关Dx,X1兴H共t−1,t,tx,txx,txxx, . . .兲 = − gt−1Ht−1= − gt−1X1H, then DxX1D−1Y0Dxk−1共f兲 − X1D−1Y0Dx k共f兲 = 兵D xX1D−1Y0− X1D−1Y0Dx其Dxk−1共f兲 = 兵DxX1D−1Y0 − X1D−1兵DxY0+ ft1Y0其其Dx k−1共f兲 =关Dx,X1兴D−1Y0Dxk−1共f兲X1共D−1Y0共f兲兲D−1Y0Dxk−1共f兲 − D−1共Y0共f兲兲X1D−1Y0Dx k−1共f兲 = − g t−1X1D−1Y0Dx k−1共f兲 − X1D−1共Y0共f兲兲D−1Y0Dx k−1共f兲 − D−1共Y 0共f兲兲X1D−1Y0Dx k−1共f兲. Therefore,关Dx,C1兴 = − gtxX1D−1Y0共f兲X1− X1D−1Y0共f兲 t−
兺
k=1 ⬁ X1共D−1Y0共f兲兲D−1Y0Dx k−1共f兲 t关k兴 − gt−1兺
k=1 ⬁ X1D−1Y0Dx k−1共f兲 t关k兴− D −1共Y 0共f兲兲兺
k=1 ⬁ X1D−1Y0Dx k−1共f兲 t关k兴= − gtxX1D −1Y 0共f兲X1 − X1D−1Y0共f兲Y1−共gt−1+ D−1共ft1兲兲C1that proves the base of mathematical induction. Assuming Eq.共29兲is true for m − 1, we have 关Dx,Cm兴 = 关Dx,关X1,Cm−1兴兴 = − 关X1,关Cm−1,Dx兴兴 − 关Cm−1,关Dx,X1兴兴 = 关X1,关Dx,Cm−1兴兴 + 关Cm−1,gt−1X1兴 =关X1,关Dx,Cm−1兴兴 + Cm−1共gt−1兲X1− gt−1Cm=
冋
X1,− gtxX1 m−1 D−1Y0共f兲X1− X1 m−1 D−1Y0共f兲Y1 −兺
j=1 m−1 A共m−1兲j Cj册
+ gt−1txX1m−1D−1Y0共f兲X1− gt−1Cm= − gt−1txX1m−1D−1Y0共f兲X1 − gtxX1 m D−1Y0共f兲X1− X1 m D−1Y0共f兲Y1− X1 m−1 D−1Y0共f兲C1−兺
j=1 m−1 X1共Aj共m−1兲兲Cj −兺
j=1 m−1 A共m−1兲j Cj+1+ gt−1txX1m−1D−1Y0共f兲X1− gt−1Cm= − gtxX1 m D−1Y 0共f兲X1− X1 m D−1Y 0共f兲Y1 −兵Am−1共m−1兲+ gt−1其Cm−兵X1m−1D−1Y0共f兲 + X1共A1共m−1兲兲其C1−兺
j=2 m−1 兵X1共A共m−1兲j 兲 + A共m−1兲j−1 其Cj = − gtxX1 m D−1Y0共f兲X1− X1 m D−1Y0共f兲Y1−兺
j=1 m A共m兲j Cj, where A1共m兲= X1m−1D−1Y0共f兲 + X1共A1共m−1兲兲 = X1m−1再
− gt gtx冎
+ X1X1m−2再
C共m − 1,0兲gt−1− C共m − 1,1兲 gt gtx冎
= X1m−1再
C共m,0兲gt−1− C共m,1兲 gt gtx冎
, A共m兲j = X1共Aj共m−1兲兲 + A共m−1兲j−1 = X1X1m−1−j再
C共m − 1, j − 1兲gt−1− C共m − 1, j兲 gt gtx冎
+ X1m−j再
C共m − 1, j − 2兲gt−1− C共m − 1, j − 1兲 gt gtx冎
= X1m−j再
C共m, j − 1兲gt−1− C共m, j兲 gt gtx冎
, Am m = Am−1共m−1兲+ gt−1=共m − 1兲gt−1− gt gtx + gt−1= mgt−1− gt gtxthat finishes the proof of the lemma. 䊏
Assume equation t1x= f共t,t1, tx兲 admits a nontrivial n-integral. Then we know that the
Consider case when the dimension of Lnis at least 3 and C1⫽0. Since linear space generated by vector fields C1, C2, C3, . . ., is of finite dimension, then there exists a natural number N, such that
CN+1=1C1+2C2+ ¯ +NCN,
and C1, C2, . . ., CNare linearly independent. By Lemma 2.10 we have
关Dx,CN+1兴 = − gtxA0共N+1兲X1− A0共N+1兲Y1− A1共N+1兲C1− ¯ − AN共N+1兲Cn− AN+1共N+1兲兵1C1+2C2+ ¯ +NCN其,
where A0共k兲= X1kD−1Y0共f兲. On the other hand,
关Dx,CN+1兴 = Dx共1兲C1+ Dx共2兲C2+ ¯ + Dx共N兲CN+1共− gtxA0共1兲X1− A0共1兲Y1− A1共1兲C1兲 +2共− gtxA0共2兲X1− A0共2兲Y1− A1共2兲C1− A2共2兲C2兲 + ¯ +N共− gtxA0共N兲X1− A0共N兲Y1 − A1共N兲C1− ¯ − AN共N兲CN兲.
Linear independence of X1, Y1, C1, C2, . . ., CNallows us to compare coefficients before X1, Ck,
1ⱕkⱕN in the last two presentations for 关Dx, CN+1兴. We have
− A0共N+1兲= −1A0共1兲−2A0共2兲− ¯ −NA0共N兲, − A1共N+1兲−1A共N+1兲N+1 = −1A1共1兲−2A1共2兲− ¯ −NA1共N兲+ Dx共1兲, − Ak共N+1兲−kAN+1共N+1兲= −
再
兺
j=k N jAk共j兲冎
+ Dx共k兲, 2 ⱕ k ⱕ N − 3, − AN−2共N+1兲−N−2AN+1共N+1兲= −N−2AN−2共N−2兲−N−1AN−2共N−1兲−NAN−2共N兲 + Dx共N−2兲, − AN−1共N+1兲−N−1AN+1共N+1兲= −N−1AN−1共N−1兲−NAN−1共N兲 + Dx共N−1兲, − AN共N+1兲−NAN+1共N+1兲= −NAN共N兲+ Dx共N兲. 共30兲Thus we have proven the following theorem.
Theorem 2.11: Consistency of the system (30)is necessary for the existence of a nontrivial n-integral to the chain(1).
One can specify the system. Since
AN共N+1兲= X1
再
C共N + 1,N − 1兲gt−1− C共N + 1,N兲 gt gtx冎
=共N + 1兲N 2 gt−1t−1−共N + 1兲 gtt−1gtx− gtgtxt−1 gtx 2 , AN+1共N+1兲=再
C共N + 1,N兲gt−1− C共N + 1,N + 1兲 gt gtx冎
=共N + 1兲gt−1− gt gtx , AN共N兲=再
C共N,N − 1兲gt−1− C共N,N兲 gt gtx冎
= Ngt−1− gt gtx , the last equation of共30兲becomes再
共N + 1兲N 2 gt−1t−1−共N + 1兲 gtt−1gtx− gtgtxt−1 gtx 2冎
+N再
共N + 1兲gt−1− gt gtx冎
=N再
Ngt−1− gt gtx冎
− Dx共N兲,that can be rewritten as 共N + 1兲N 2 gt−1t−1−共N + 1兲 gtt−1gtx− gtgtxt−1 gt2x +Ngt−1= − Dx共N兲. 共31兲 If C1= 0 then, by Lemma 2.10, 0 = X1D−1Y0共f兲 = t−1 D−1共ft1兲. 共32兲
III. PROOF OF THEOREM 1.3 A. Case 1: t1x= tx+ A„t1− t…
Introduce= t1− t and rewrite the equation asx= A共兲. Study the question when this equation
admits a nontrivial n-integral or the same when the corresponding Lie algebra Ln is of finite
dimension. Since Y0f = A
⬘
共兲t1= DA共兲, Y0fx= A⬙
共兲A共兲 + A⬘
共兲A⬘
共兲 = DA共兲DA共兲, and Y0Dx k f =共DA共兲兲k+1, we can write Y1as Y1= t+兺
k=1 ⬁ D−1共DA共兲兲k Dx k t. 共33兲Now let us introduce new variables:+= t , = t1− t , −1= t − t−1, j= tj+1− tj. Since
t= + − + −1 , then the expression共33兲for Y1becomes
Y1= + − + −1 +
兺
k=1 ⬁ D−1共D A共兲兲k Dx k + . 共34兲One can ignore the term containing/since coefficients in the vector fields used below do not depend on. Multiply Y1by A共−1兲, A共−1兲Y1= A共−1兲 + + A共−1兲 −1 +
兺
k=1 ⬁ A共−1兲D−1共DA共兲兲k Dx k + . 共35兲 Introduce p共兲 = A共−1共兲兲, where d= d−1 A共−1兲 . 共36兲 Equation共35兲becomesA共−1兲Y1= p共兲 + + +
兺
k=1 ⬁ Dx k共p共兲兲 Dx k + . 共37兲Now instead of X1=/t−1, define
X ˜ 1= A共−1兲X1= − A共−1兲 −1 + A共−1兲 −2 . It is indeed with new variables
X ˜ 1= − + p共兲 p共−1兲 −1 . 共38兲
Note that关Dx, X˜1兴=Dx共p共兲/p共−1兲兲W1, where W1=/−1. Since关Dx, X1兴=−X1共g兲X1− X1共g−1兲X2, then关Dx, X˜1兴苸Ln. Therefore, we have two possibilities:
共i兲 Dx
共
p共兲 p共−1兲兲
= 0 or共ii兲 W1苸Ln.
First let us consider case共i兲. We have
Dx
冉
p共兲 p共−1兲冊
= p⬘
共兲p共−1兲 − p共兲p⬘
共−1兲 p2共 −1兲 = 0.Solving this differential equation we get p共兲=A共−1共兲兲=e. Since d/d−1= 1/A共−1兲, we have A共兲=+ c.
Now concentrate on case共ii兲. Since Dx共p共兲/p共−1兲兲W1苸Ln, then W1苸Ln and, due to共38兲, W =/苸Ln.
Lemma 3.1: If equationx= A共兲 admits a nontrivial n-integral then function p共兲, defined by (36), is a quasipolynomial.
Proof: Instead of Y1, X1, take the pair of the operators W =/ and
Z = A共−1兲Y1− W = p共兲 + + Dxp共兲 +x + Dx2共p共兲兲 +xx + ¯ . 共39兲
Construct a sequence of the operators
C1=关W,Z兴, C2=关W,C1兴, Ck=关W,Ck−1兴, k ⱖ 2. 共40兲
Since algebra Lnis of finite dimension then there exists number N, such that
CN+1=0Z +1C1+ ¯ +NCN, 共41兲
and vector fields Z, C1, . . . , CNare linearly independent.
Direct calculations show that关Dx, W兴=关Dx, Z兴=0. Therefore, we have 关Dx, Cj兴=0 for all j. It
follows from共41兲that
0 = Dx共0兲Z + Dx共1兲C1+ ¯ + Dx共N兲CN,
which implies Dx共j兲=0. Clearly j=j共兲 and Dx共j兲=j
⬘
共兲=0. Hence j is constant for all jⱖ0.Look at the coefficients of/+in共41兲and get
0p共兲 +1p
⬘
共兲 + ¯ +Np共N兲共兲 = p共N+1兲共兲. 共42兲p共兲 =
兺
j=1 sqj共兲ej. 共43兲
䊐
Lemma 3.2: Let p共兲 is an arbitrary quasipolynomial solving a differential equation of the
form 共42兲and which does not solve any equation of this form of less order. Then the equation
t1x= tx+ A共t1− t兲 with A found from the conditions,
A共−1兲 = p共兲,
−1=
冕
0 p共˜兲d ˜ admits a nontrivial n-integral.
Proof: Introduce L共Dx兲 = Dx N+1 −NDx N −N−1Dx N−1 − ¯ −1Dx−0.
Equation 共42兲 can be rewritten as L共Dx兲p共兲=0. However, L共Dx兲p共兲=L共Dx兲A共−1兲. Since
L共Dx兲t1x= L共Dx兲tx+ L共Dx兲A共兲 and L共Dx兲A共兲=0, we have L共Dx兲t1x= L共Dx兲tx. But L共Dx兲t1x = DL共Dx兲tx, therefore DL共Dx兲tx= L共Dx兲tx. Denote L共Dx兲tx= I so we have DI = I. Hence L共Dx兲txis an
n-integral. 䊐
Therefore the condition 共43兲is necessary and sufficient for our equation to have nontrivial
n-integral.
Example 1: Take p共兲=12e+12e−= cosh, then
A共−1兲 = cosh,
−1= sinh+ c,
or A共−1兲2−共−1− c兲2= 1 which gives A共−1兲=
冑
1 +共−1− c兲2. So t1x= tx+冑
1 +共t1− t − c兲2, where c is arbitrary constant, is Darboux integrable. Moreover, its general solution is given by t共n,x兲 = G共x兲+nc+兺k=0n−1sinh共x+ck兲, where G共x兲 is arbitrary function depending on x and ckare arbitraryconstants.
B. Case 2: t1x= tx+ c1„t1− t…t+c2„t1− t…2+ c3„t1− t…
Lemma 3.3: If equation t1x= tx+ d共t,t1兲=tx+ c1共t1− t兲t+c2共t1− t兲2+ c3共t1− t兲 admits a nontrivial
n-integral, then there exists a natural number k such that
kc1−共k + 1兲c2= 0. 共44兲
Proof: Introduce vector fields T1=关X1, Y1兴, Tm=关X1, Tm−1兴, mⱖ2. Direct calculations show
that 关Dx,T1兴 = 共− c1+ 2c2兲X1+共− c1+ 2c2兲Y1+共dt−1共t−1,t兲 − dt共t−1,t兲兲T1, 关Dx,Tm兴 = − Am−1共m兲Tm−1− Am共m兲Tm, 共45兲 where Aj共m兲= X1 m−j兵− C共m, j − 1兲d t−1共t−1,t兲 + C共m, j兲dt共t−1,t兲其, C共m,k兲 = m! k !共m − k兲!.
TM+1=1T1+2T2+ ¯ +MTM,
and T1, T2, . . ., TM are linearly independent. We have
关Dx,TM+1兴 = 关Dx,1T1+2T2+ ¯ +MTM兴,
that can be rewritten by共45兲in the following form: − A共M+1兲M TM− AM+1共M+1兲兵MTM+M−1TM−1+ . . . +1T1其
= Dx共1兲T1−1共c1− 2c2兲X1−1共c1− 2c2兲Y1−1A1共1兲T1+ Dx共2兲T2−2A1共2兲T1−2A2共2兲T2 + ¯ + Dx共M兲TM−MAM−1共M兲TM−1−MAM共M兲TM.
Compare coefficients before the operators. The coefficient before X1 and Y1 gives −1共c1− 2c2兲 = 0. In this case we have two choices: 1= 0 or c1− 2c2= 0. The second one gives 共44兲 with k = 1. If c1− 2c2⫽0, then1= 0. Using this, from the coefficient of T1we get −2A1共2兲= 0. Again, we have that either2= 0 or A1
共2兲= 0. If A 1
共2兲= 0 we stop, if not then
2= 0 and we continue to compare the coefficients. Using 1=2= 0, the coefficient before T2 gives −3A2共3兲= 0 which means 3 = 0 or A2共3兲= 0. Same as before: if A2共3兲= 0, we stop, if not then 3= 0 and we continue to the procedure.
If1=¯ =M= 0 then TM+1= 0 and关Dx, TM+1兴=0=−AM
共M+1兲T
M− AM+1共M+1兲TM+1= −AM M+1
TM. Since T1, . . ., TM are linearly independent then TM⫽0 and therefore AM
共M+1兲= 0. It follows A k−1 共k兲= 0 for some k = 1 , 2 , . . . , M + 1. Evaluate Ak−1共k兲, Ak−1共k兲 = − C共k,k − 2兲dt−1t−1共t−1,t兲 + C共k,k − 1兲dtt−1共t−1,t兲 = − k共k − 1兲共c2− c1兲 + k共c1− 2c2兲 = k兵kc1 −共k + 1兲c2其. 䊏 Let us rewrite the equation in case共2兲 as
x= c1t + c22+ c3, where= t1− t. We have two important relations.
共1兲 Y0f = Dxln H, where H = 1/⑀ 共+⑀兲1/⑀, = 1 , ⑀= c1 c2 − 1. 共46兲 共2兲 Y1f = Dxln RH−1, where H−1= D−1H, R = 共+⑀兲, when ⑀⫽ 0.
关The case⑀= 0, i.e c1= c2, is not realized due to Lemma 3.3. The case c2= 0, due to Lemma 3.3, leads to c1= 0, and the equation becomes t1x= tx+ c3共t1− t兲 with an n-integral I=tx− c3t.兴 These two relations allow us to simplify the basis operators Y0, Y1, X1. Really, we take
Y ˜
1= H−1Y1, Y˜0= HY0,
and get关Dx, Y˜0兴=0 and 关Dx, Y˜1兴=⌳Y˜0, where⌳=−共H−1/H兲Dxln共RH−1兲.
First we will restrict the set of the variables as follows: t1, t , t−1, tx, txx, . . . and change the
variables t+= t,
−1= t − t−1 keeping the other variables unchanged. Then some of the differentia-tions will change,
t= t++ −1 , t−1 = − −1 . So we have X1= −/−1= −Xˆ1and Y ˜ 1= H−1
冉
t++ −1冊
+兺
k=1 ⬁ H−1D−1共Y 0Dxk−1f兲 t关k兴.Since 关Dx, Xˆ1兴=Dx共ln R−1兲Xˆ1, one can introduce X˜1=共1/R−1兲Xˆ1 and get 关Dx, X˜1兴=0. Here R−1 = D−1R.
Introduce vector fields C2=关X˜1, Y˜1兴, C3=关X˜1, C2兴, Ck=关X˜1, Ck−1兴, kⱖ3. We have
关Dx,Cj+1兴 = X˜1
j共⌳兲Y˜
0, jⱖ 1.
Since the algebra Lnis of finite dimension then there is a number N, such that
CN+1=NCN+ ¯ +2C2+1˜Y1, 共47兲 where Y˜1, C1, C2, . . . are linearly independent.
Applying the commutator with Dxone gets Dx共j兲=0 for j=1, ... ,N and
共X˜1
N
−NX˜1N−1− ¯ −1兲⌳ = 0. 共48兲
All the operators in our sequence have coefficients depending on , −1, t. So do j
=j共,−1, t兲. But the relation Dxj共,−1, t兲=0 shows that j/t = 0, i.e., j=j共,−1兲. Since the minimal x-integral for an equation in case共2兲 depends on variables t, t1, t2, t3, the relation
Dx共j兲=0 implies thatjis constant for all j.
Introduce new variables t˜1, t˜, as
t˜1= t1, t˜ = t+, = ln
冢
−1 −1+ 1 ⑀共t1− t+兲冣
, or the same −1= ⑀冉
e 1 − e冊
. 共49兲 Then −1 = −1 , t+= t˜+ t+ , t1= t˜1+ t1 .In these new variables X˜1 takes the form
X ˜ 1= −1共−1+⑀兲 −1 −1 = ,
冉
dN dN−N dN−1 dN−1− ¯ −1冊
⌳ = 0, 共50兲 where ⌳ = −H−1 H 共xln R + Dxln H−1兲 = − H−1 H冉
f t + D −1f t1冊
= −H−1 H 共c1− 2c2兲共−−1兲. 共51兲Let us show that c1− 2c2= 0. Assume contrary. It follows from共50兲and共51兲that both functions
H−1 and−1H−1should solve the linear differential equation with constant coefficients,
冉
dNdN−N dN−1
dN−1− ¯ −1
冊
y共兲 = 0. Therefore, both functions H−1 and−1H−1must be quasipolynomials in .Due to共46兲and共49兲, we have
H−1= ⑀e共1 − e兲1/⑀−1 and −1H−1= 2 ⑀2e 2共1 − e兲1/⑀−2.
To be quasipolynomials init is necessary that⑀= 1/m for some natural mⱖ2. Rewrite our vector fields X˜1, Y˜1 in the new variables;
X ˜ 1= , Y ˜ 1= H−1 t˜+ H−1
冉
t++ −1冊
+ ¯ .Study the projection on the direction/.
The operators X˜1=/ and H−1共/t++/−1兲/ generate a finite dimensional Lie algebra over the field of constants. Due to Lemma 2.3 in this case the coefficient H−1/t should
be of one of the forms
c ˜1e␣˜+ c˜2e−␣˜+ c˜3 or ˜c12+ c˜ 2+ c˜3, 共52兲 but we have H−1
冉
t++ −1冊
=冉
1 +冉
1 ⑀− 1冊
e冊
共1 − e兲1/⑀,with 1/⑀= mⱖ2 and it is never of the form共52兲. This contradiction shows that c1− 2c2= 0. 䊐
C. Case 3: t1x= tx+ A„t1− t…e␣t
Introduce= t1− t and rewrite the equation asx= A共兲e␣t. Study the question when the
equa-tion admits a nontrivial n-integral or the same when the corresponding Lie algebra Lnis of finite
dimension.
Instead of the vector fields Y0=/t1and Y1=/t + D−1共f/t1兲共/tx兲+D−1共fx/t1兲共/txx兲
+¯, we will use the vector fields Y˜0= A共兲Y0 and Y˜1= A共−1兲Y1. They are more convenient since they satisfy more simple relations,
关Dx,Y˜0兴 = 0, 关Dx,Y˜1兴 = 1Y˜0
as operators acting on the enlarged set t1, t , t−1, t−2, . . .; tx, txx, txxx, . . .. Here the coefficient
1is
1=
A共−1兲
A共兲 共A
⬘
共兲 −␣A共兲 − A⬘
共−1兲e−␣−1兲e␣t.
Since the equation is represented asx= A共兲e␣tit is reasonable to introduce new variables as+ = t , −1= t − t−1, −2= t−1− t−2, such that t= + + −1 , t−1 = − −1 + −2 , t−2 = − −2 .
Instead of the operators X1=/t−1 and X2=/t−2 use new ones X˜1= A共−1兲e−␣−1/−1 and X˜2 = A共−2兲e−␣−2/−2. They satisfy relations关Dx, X˜2兴=0 and 关Dx, X˜1兴=X˜2. Here the coefficient is
=␣A共−1兲e−2␣−1+␣t.
Construct a sequence by taking X˜1, Y˜1, C2=关X˜1, Y˜1兴, C3=关X˜1, C2兴, Ck=关X˜1, Ck−1兴 for kⱖ3.
One can easily check that
关Dx,C2兴 = − Y˜1共兲X˜2+ X˜1共1兲Y˜0= b2X˜2+ X˜1共1兲Y˜0, 关Dx,C3兴 = X˜1
2共
1兲Y˜0−共C2+ X˜1Y˜1兲共兲X˜2= X˜1 2共
1兲Y˜0+ b3X˜2, and for any k共it can be proven by induction兲
关Dx,Ck兴 = X˜1
k−1共
1兲Y˜0+ bkX˜2.
Since the characteristic Lie algebra Lnis of finite dimension then there is a number N, such that CN+1=NCN+ ¯ +1Y˜1+0X˜1, 共53兲 where X˜1, Y˜1, C1, C2, . . . are linearly independent.
Commute both sides of共53兲with Dxand get X
˜
1
N共
1兲Y˜0+ bN+1X˜2= Dx共N兲CN+ ¯ + Dx共1兲Y˜1+ Dx共0兲X˜1+NX˜1N−1共1兲Y˜0+ ¯ +11˜Y0 +
再
兺
k=2 N
bkk
冎
X˜2.Collect the coefficients before the operators and get Dx共j兲=0 for j=0,1, ... ,N, and
共X˜1
N
−NX˜1N−1−N−1˜X1N−2− ¯ −1兲1= 0. 共54兲 Introduce new variables, −1as solutions of the following ordinary differential equations:
d−1
d = A共−1兲e
−␣−1, d−2
d−1= A共−2兲e
−␣−2. 共55兲
Thus our vector fields are rewritten as
X ˜ 1= , X˜2= −1 , Y˜0= A共兲 t1,
Y ˜ 1= e␣−1 + A共−1兲 + + Dx共A共−1兲兲 tx + ¯ .
By looking at the projection on/we get an algebra generated by / and e␣−1/
con-taining all possible commutators and all possible linear combinations with constant coefficients. Due to Lemma 2.3, we get that e␣−1can be only one of the forms共a兲 e␣−1= c
1e+ c2e−+ c3and 共b兲 e␣−1= c
12+ c2+ c3,
where, c1, c2, c3 are some constants.
The equation A共−1兲=共1/␣兲共d/d兲e␣−1 implies that in case 共a兲 we have A共−1兲=共/␣兲 ⫻共c1e− c2e−兲, or the same
A2共兲 =
2
␣2兵共e␣− c3兲2− 4c1c2其, 共56兲 and in case共b兲 we have A共−1兲=共1/␣兲共2c1+ c2兲, or the same,
A2共兲 =4c1
␣2e␣+
c22− 4c1c3
␣2 . 共57兲
In addition to the operators X˜1, X˜2, Y˜0, Y˜1 introduced above, we will use Y˜2 = A共−2兲D−1共Y1f兲tx+ A共−2兲D−1共Y1fx兲txx+¯ defined as Y˜2= A共−2兲Y2. It satisfies the commutativ-ity relation 关Dx,Y˜2兴 = Y˜1+Y˜0+X˜1, 共58兲 where = −A共−2兲 A共兲 D −1共Y 1f兲 = − A共−2兲 A共兲 兵共− A
⬘
共−1兲 +␣A共−1兲兲e −␣−1+ A⬘
共 −2兲e−␣−2−␣−1其e␣t, = −A共−2兲 A共−1兲D−1共Y1f兲 and = −e␣−1. 共59兲
Lemma 3.4:
共1兲 Equation t1x= tx+共/␣兲共e␣− c3兲e␣tadmits a nontrivial n-integral if and only if c3=⫾1. 共2兲 Equation t1x= tx+ c5e␣t, c5⫽0 does not admit a nontrivial n-integral.
Proof: In this case the equationx= A共兲e␣tis reduced by evident scaling of x and t to t1x= tx+ et or t1x= tx+ et1+et.
By induction on m one can easily see that for the equation t1x= tx+ et, the basic vector fields Ymare
Y1= t, Ym= et−共m−1兲 tx + et−共m−1兲共tx− et−共m−1兲兲 txx + ¯ .
Since these vector fields Ym, mⱖ1, are linearly independent then equation t1x= tx+ et does not
admit a nontrivial n-integral.
Y1= t+ e t tx + et共tx+ et兲 txx + ¯ , Ym=共 + 1兲et−共m−1兲 tx +共 + 1兲et−共m−1兲共tx+共1 − 兲et−共m−1兲兲 txx + ¯ .
One can see that vector fields Ym, mⱖ1, are linearly independent if ⫽ ⫾1. Therefore, if
⫽ ⫾1, equation t1x= tx+ et1+et does not admit a nontrivial n-integral. If =−1, the equation
becomes t1x= tx+ et1− et, and one of its n-integrals is I = tx− et. If =1, the equation becomes t1x = tx+ et1+ et, and one of its n-integrals is I = 2txx− tx2− e2t. 䊏 Lemma 3.5: Let equation t1x= tx+ A共t1− t兲e␣twith (a) A2共兲=共2/␣2兲兵共e␣− c3兲2− 4c1c2其 or (b)
A2共兲=共4c
1/␣2兲e␣+共c22− 4c1c3兲/␣2, admit a nontrivial n-integral. Then in case (a), we have,
A共t1− t兲=共/␣兲
冑
共e␣共t1−t兲− c3兲2− c3
2+ 1, where c
3is an arbitrary constant, and in case (b), we have
A共t1− t兲=ce共␣/2兲共t1−t兲, where c is an arbitrary constant.
In cases (a) and (b) the corresponding n-integrals are I =共␣/2兲tx2− txx+共␣/2兲e2␣t and I =
−共␣/2兲tx
2 + txx. Proof: Note that
Dx=, where = − A共−2兲
A共−1兲− e
␣−2.
This implies that the vector field,
R2= Y˜2−Y˜1, satisfies very simple and convenient relation,
关Dx,R2兴 =˜Y˜ 0+X˜1, ˜ = − A共−2兲 A共兲 D −1共Y 1f兲 −1, = e␣−1 A共−2兲 A共−1兲 D−1共Y1f兲. Study now the sequence
Rj+1=关Xˆ,Rj兴, j ⱖ 2, where Xˆ = X˜1+ e−␣−1X2. Direct calculations show that
关Dx,Rm兴 = Xˆ共m−2兲共˜兲Y˜ 0+ Xˆ共m−2兲共˜兲X˜1+ bmX˜2. 共60兲 Since X˜1, X˜2, Y˜0, R2 are linearly independent, then there exists a number Nⱖ2, such that
RN+1=NRN+N−1RN−1+ ¯2R2+1X˜1 and
关Dx,RN+1兴 = 关Dx,NRN+N−1RN−1+ ¯2R2+1X˜1兴. 共61兲 We use 关Dx, X˜1兴=␣A共−1兲e−2␣−1+␣tX˜2, 关Dx, X2兴=0, and 共60兲 to compare the coefficients before linearly independent vector fields Rkand Y˜0in共61兲. We have, Dx共k兲=0, k=2,3, ... ,N, and
Xˆ共N−1兲共˜兲 = NXˆ共N−2兲共˜兲 + ¯ + 2˜ . 共62兲 Under the change in variables
= z, −1= z−1− q共z兲, q共z兲 z = − e −␣−1, equation共62兲is reduced to 共DzN−1−NDN−2z − ¯ −2兲˜ = 0, 共63兲 where k=k共−1,−2兲=k共z,z−1兲. Since Dx共z−1兲=0, Dx共z兲=e␣t⫽0, and 0 = Dx共k兲
= Dz−1共k兲Dx共z−1兲+Dz共k兲Dx共z兲, then coefficientsk do not depend on variable z. Since, due to
共63兲, ˜ = −A共−2兲 A共兲 e −␣−1e␣t兵− A
⬘
共 −1兲 +␣A共−1兲 + A⬘
共−2兲e−␣−2其 + A共−2兲 A共兲 e ␣t兵A⬘
共兲 −␣A共兲 − A⬘
共−1兲e−␣−1其 + A共−1兲 A共兲 e ␣−2e␣t兵A⬘
共兲 −␣A共兲 − A⬘
共 −1兲e−␣−1其is a quasipolynomial in z =for anyand t, then共d/d兲共˜A共 兲e−␣t兲 is a quasipolynomial as well. Hence we have
共A
⬙
共兲 −␣A⬘
共兲兲兵A共−2兲 + A共−1兲e␣−2其is a quasipolynomial in z, which is possible only if A
⬙
共兲−␣A⬘
共兲=0 or A共−2兲+A共−1兲e␣−2is aquasipolynomial in z. In case 共a兲 we have
A
⬙
共兲 −␣A⬘
共兲 = −␣c4 e2␣ 共
冑
共e␣− c3兲2− c4兲3
, c4= 4c1c2, and in case共b兲 we have
A
⬙
共兲 −␣A⬘
共兲 = − 4c12␣−2e2␣冉
4c1 ␣2e␣+ c22− 4c1c3 ␣2冊
−3/2 .Therefore, A
⬙
共兲−␣A⬘
共兲=0 if c1c2= 0 in case 共a兲 and if c1= 0 in case 共b兲. Both these cases are considered in Lemma 3.4.It follows from dq/dz=−e−␣−1that, in case共a兲, if r=
冑
c3 2− 4c 1c2⫽0, then q共兲 = − 1 rln