with respect to the type of ﬁrst integrals On a classiﬁcation of polynomial differential operators ScienceDirect

(1)

ScienceDirect

J. Differential Equations 260 (2016) 1993–2025

www.elsevier.com/locate/jde

On a classification of polynomial differential operators with respect to the type of first integrals

Jinzhi Lei

ZhouPei-YuanCenterforAppliedMathematics,TsinghuaUniversity,Beijing,100084,PRChina

Received 4November2010;revised 21September2012 Availableonline 21October2015

Abstract

ThispapergivesaclassificationofpolynomialdifferentialoperatorsX =X₁(x₁,x₂)δ₁+X₂(x₁,x₂)δ₂ (δ_i=∂/∂x_i).TheclassificationisdefinedthroughanorderderivedfromX.LetX=Xybe theassociated differentialpolynomial,theorderisdefinedastheorderofadifferentialidealthatisanessentialextension of{X}.Themainresultshowstheordercanonlybefourpossiblevalues:0,1,2,3,or∞.Furthermore,when theorderisfinite,theessentialextension= {X,A},whereAisadifferentialpolynomialwithcoefficients obtainedthrougharationalsolutionofapartialdifferentialequationgivenexplicitlybycoefficientsofX. Whentheorder isinfinite,the extensionisidenticalwith{X}. Inaddition,if,andonlyif,theorder is0,1,or2,theassociatedpolynomialdifferentialequationhasLiouvillianfirstintegrals.Examplesand connectionswithGodbillon–Veysequencesarealsodiscussed.

MSC:34A05;34A34;12H05

Keywords:Polynomialdifferentialoperator;Classification;Polynomialdifferentialequation;Differentialalgebra;

Liouvillianfirstintegral;Godbillon–Veysequence

E-mailaddress:jzlei@tsinghua.edu.cn.

(2)

1. Introduction 1.1. Background

This paper studies polynomial differential operators of form

X =X1(x1, x2)δ1+X2(x1, x2)δ2, (1) where δi =∂/∂x_i (i=1, 2), and X1(x₁, x₂), X₂(x₁, x₂)are polynomials of x1and x2. We further assume that X1≡0 without loss of generality.¹ In this paper, we give a classification for all operators of form (1), according to which the solution of the first order partial differential equation

Xω=0 (2)

is discussed.

The operator (1)closely relates to the following polynomial differential equation dx₁

dt =X₁(x₁, x₂), dx₂

dt =X₂(x₁, x₂), (3) and any non-constant solutions of (2)give a first integral of (3). Therefore, our results also yield a classification of the polynomial systems (3)with respect to the function type of its first integral.

The current study was motivated by investigating integrating methods of a polynomial differential equation of form (3). First, we take a look at a simple situation of how we can integrate(3).

If the equation (3)has an integrating factor ηthat is a rational function, a first integral ωis then obtained by an integral of a rational function, and satisfies

δ₁ω−a=0, (4)

with a=ηX1a rational function. Therefore, there is a non-constant function ωthat satisfies both equations (2)and (4). In this case, the differential operator DAdefined by

DAω=δ₁ω−a (5) is compatible with X. In other words, let two differential polynomials Xand Abe defined as

X=Xy, A=DAy,

then the differential ideal {X, A}is a nontrivial extension of the ideal {X} (refer to detailed definitions below). On the other hand, for any such differential polynomial A, we can always solve δ1ωand δ2ωfrom the equation {X =0, DAω=0}and therefore lead to the integration of the equation(3). This simple situation suggests that to integrate the equation (3) for first integrals, we need to find a differential polynomial Aso that {X, A}is a nontrivial extension

1 IfX₁≡0,onecanexchangex₁andx₂(andalsothederivationsδ₁andδ₂)indiscussionsbelow.Ourclassification isanintrinsicpropertyoftheoperatorX,andisunchangedwhenweexchangex₁andx₂,andevenwhenweapplya lineartransformationin(x₁,x₂).SeeourdiscussionfollowingTheorem 24.

(3)

of the ideal{X}. Such a differential polynomial A, if exists, is not unique. However, the lowest order among these differential polynomials is uniquely determined by the differential operator X (called the order of X, to be detailed below), and therefore provides an index for a classification.

The classification presented in this study is given through an order of the operator X defined below. This order is essential for understanding integrating methods of the polynomial differential equation (3)in different classes, and also the classification of un-integrable systems.

Furthermore, for a given equation (3), the differential polynomial Ain defining the nontrivial extension {X, A}provides additional informations for the first integral, which are important for further investigations of the structure of integrating curves (or foliations) of the polynomial differential equation.

1.2. Preliminary definitions

Before stating the main results, we give some preliminary concepts from differential algebra.

For details, refer to[1]and [2].

Let K be the field of all rational functions of (x₁, x₂) with complex number coefficients, δ₁, δ₂ are two derivationsof K, and are commutable to each other. Then K together with the two derivations form a differential field, with Cas the constant field. For a differential indetermi- natey, there is a usual way to add yto the differential field K, by adding an infinite sequence of symbols

y, δ₁y, δ₂y, δ₁δ₂y,· · ·, δ₁ⁱ¹δⁱ₂²y,· · · (6) to K[1]. This procedure results in a differential ring, denoted as K{y}. Each element in K{y}is a polynomial of finite numbers of the symbols in (6), and therefore is a differential polynomial in y with coefficients in K.

We say an algebra ideal in K{y}to be a differential idealif a∈implies δia∈ (i= 1, 2). Let be any set of differential polynomials. The intersection of all differential ideals containing is called the differential ideal generated by , and is denoted as {}. A differential polynomial Ais in {}if, and only if, Ais a linear combination of differential polynomials in and of derivatives, of various orders, of such differential polynomials.

For the operator X given by (1), we have

X=Xy=X1δ1y+X2δ2y∈K{y}, (7) and {X}denotes the differential ideal generated by X.

Definition 1. Let

w₁=δ₁ⁱ¹δ₂ⁱ²y, w₂=δ^j₁¹δ₂^j²y,

be two derivatives of y, w1islowerthan w2(or w2ishigherthan w1) if either i1< j1, or i1=j1

and i2< j2. Any element in Kis lower than the indeterminate y.

The infinite sequence (6)can be organized from low to high (here w1< w₂means that w1is lower than w2) as

(4)

a < y < δ₂y < δ₂²y <· · ·

< δ1y < δ1δ2y < δ1δ₂²y <· · ·

< δ₁²y < δ²₁δ₂y < δ₁²δ²₂y <· · ·

<· · ·. (8)

Here ameans any element in K.

Definition 2. Let Abe a differential polynomial, if Acontains y (or its derivatives) effectively, by theleaderof A, we mean the highest of those derivatives of ywhich are involved in A.

Definition 3. Let A1, A₂be two differential polynomials, we say A2to be of higherrankthanA₁, if either

(1) A₂has higher leader than A1; or

(2) A₁and A2have the same leader, and the degree of A2(as a polynomial of the leader) in the leader exceeds that of A1.

A differential polynomial which effectively involves the intermediate y is of higher rank than one which does not. Two differential polynomials of which no difference in the rank as created above is said to be of the same rank.

The following fact is basic [2, p. 3]:

Proposition 4. Any non-empty subset of differential polynomials ⊆K{y}contains a differen- tial polynomial which is not higher than any other differential polynomials in the subset .

For a differential polynomial A ∈K{y}, we associate with Aa differential operatorDAon analytic functions A(), where is an open subset of C², such that

DAu=A|y=u, ∀u∈A(). (9) By S(DA), we denote the singularity set of DA, which contains all singular points (i.e., the poles) in the coefficients of the differential polynomial A. Because all coefficients of Aare rational functions, the singularity set DAis a closed subset in C². Thus, for any u ∈A(), DAuis well defined in the open subset \S(DA).

Definition 5. A differential ideal in K{y}is anextensionof {X}, or anextensionof X, if {X} ⊆. The extension is anontrivial extensionif there exists an open subset ⊂C²and a non-constant function ω∈A(), such that DAω=0 in \S(DA)for any A ∈. Otherwise, is atrivial extension.

It is obvious that {X}itself is a nontrivial extension of X, and K{y}is a trivial extension of X. Main results in this paper are to show that nontrivial extensions of X with the lowest order (to be defined below) are essential to provide the classification of X.

(5)

Proposition 6. Let be an extension of {X}, then either = {X}, or contains a differential polynomial Aof lower rank than X.

Proof. We only need to show that if {X} , then contains a differential polynomial Aof lower rank than X. Let ₀=\{X}, then 0is a non-empty subset of . Proposition 4yields that there is a differential polynomial A ∈0that has the lowest rank.

We claim that Ahas lower rank than X. If otherwise, the leader of A, denoted as δⁱ₁δ₂^jy, is not lower than δ1y (the leader of X), i.e., i≥1. Hence, G =δ₁ⁱ⁻¹δ^j₂Xis an element contained in {X}and has the same leader δ₁ⁱδ^j₂y as in A, and the degree of δ₁ⁱδ₂^jy in Gis 1. Thus, there is a differential polynomial Q ∈K{y}so that

A1=A−QG

has lower rank than A. It is obvious A1∈0, but with lower rank than A, which is contradiction to our choice of A, and the proposition is proved. 2

Let be a nontrivial extension of {X}, and A ∈with the lowest rank. Proposition 6yields that the leader of Aeither has form δ₂^ry (r≥0), or equals δ1y. This fact leads to the order of given below.

Definition 7. Let be a nontrivial extension of {X}, and A ∈ with the lowest rank. If the leader of Ais δ₂^ry (r≥0), ris called theorderof , denoted by ord(). If otherwise, = {X} and the order of is infinity (ord() = ∞).

For a nontrivial extension , a differential polynomial with the lowest rank can only take one of the following forms:

• a polynomial of y, with at least one coefficient that is non-constant (ord() =0); or

• a differential polynomial of yeffectively involves derivatives (1 ≤ord() <∞); or

• the differential polynomial X, and therefore = {X}(ord() = ∞).

Definition 8. An essential extensionof X is a nontrivial extension of X that has order not higher than any other nontrivial extensions. By theorderof X, denoted as ord(X), we mean the order of an essential extension of X.

We note that for a given differential operator X, essential extensions of X may not be unique, but all essential extensions must have the same order. Thus, the order of X is well defined. Main results presented in the next subsection show that ord(X)provides a classification of polynomial differential operators.

1.3. Main results

Here we state the main results in this paper, with proofs given in the next section.

Theorem 9. Let the polynomial differential operator X be given by (1), with coefficients X₁, X₂∈K, then either 0 ≤ord(X) ≤3, or ord(X) = ∞. Furthermore, when 0≤ord(X)≤3,

(6)

we can always select an essential extension of X, such that = {X, A}, with A ∈K{y}given below

(1) if ord(X) =0, then

A=y−a (a∈K\R); (10)

(2) if ord(X) =1, then

A=(δ2y)ⁿ−a (n∈N, a∈K); (11) (3) if ord(X) =2, then

A=δ²₂y−aδ₂y (a∈K); (12) (4) if ord(X) =3, then

A=2(δ2y)(δ₂³y)−3(δ₂²y)²−a(δ₂y)² (a∈K). (13) From Theorem 9, when the order of a differential operator X is finite, an essential extension of X is given by = {X, A}, with A ∈K{y}given by (10)–(13). Discussions in [2, Chapter 2]

have proved that the system of equations

Xy=0, DAy=0 (14) has solution in some extension field of K. It is easy to see that this solution gives a first integral of the polynomial differential equation (3). The following result for the classification of (3)is straightforward from Theorem 9.

Theorem 10. Consider the polynomial differential equation (3), and let X be the corresponding differential operator given by (1); we have the following

(1) if ord(X) =0, then (3)has a first integral ω∈K;

(2) if ord(X) =1, then (3)has a first integral ω, such that (δ2ω)ⁿ∈K for some n ∈N;

(3) if ord(X) =2, then (3)has a first integral ω, such that δ²₂ω/δ2ω∈K; (4) if ord(X) =3, then (3)has a first integral ω, such that

2(δ2ω)(δ³₂ω)−3(δ₂²ω)² (δ₂ω)² ∈K;

(7)

(5) if ord(X) = ∞, then any first integral of (3) does not satisfy any differential equation of form

DAy=0 with A ∈K{y}\{X}.

In 1992, Singer has proved that the cases (1)–(3) in Theorem 10(also refer toTheorem 24 below) are the only cases to have Liouvillian integrals, i.e., the first integral obtained from rational functions using finite steps of exponentiation, integration, an algebraic functions [3](also refer to[4]). In the cases (4)–(5), however, there is no first integral of (3)that can be obtained from rational functions in finite steps operation given above (refer to[3]or [4]). From the proof of Lemma 15given below, when ord(X) =3, a first integral of (3)can be expressed through finite step operations from rational functions and a solution of the partial differential equation of form(35). This result provides further classification of polynomial differential equations that are not Liouvillian integrable.

In the rest of this paper, we first give the proof of Theorem 9in Section2, then show examples for each type of equations in Section3, and a discussion for connections between our results and Godbillon–Vey sequence in Section4.

2. Proof of the main result 2.1. Outline of the proof

Hereinafter, we denote δ₂ⁱy by y_i (y0=y). For any essential extension of X, let A ∈ with the lowest rank. From the above definitions, if ord(X) =r(<∞), then Ais a polynomial of y0, y1, · · ·, yr, with coefficients in K. Write

A=

m

a_my^m₀⁰y₁^m¹· · ·y_r^m^r, (15)

where m =(m₀, m₁, · · ·, m_r) ∈Z^r⁺¹, and am∈K. To prove Theorem 9, we only need to deter- mine all possible non-zero coefficients in A. Let

IA= {m∈Z^r⁺¹|a_m=0}. (16) We only need to specify the finite set IA. The process is outlined below.

Let m =(m₀, m₁, · · ·, m_r) ∈Z^r⁺¹; we define an operator i,j :Z^r⁺¹→Z^r⁺¹for 0 < i <

j≤rsuch that i,j(m) ∈Z^r⁺¹is given by

i,j(m)=m+ej−i−ej (17)

where

ek=( 0

↓ 0,· · ·,0,

k

↓

1,0,· · ·,0).

(8)

Fig. 1. Flow chart of the proof ofTheorem 9.

Therefore

−1

i,j(m)=m−ej−i+ej. (18)

For any m, n ∈Z^r⁺¹, we say m nif there exist 0 < i < j≤r, such that

i,j(m)=n.

The proof below is done by showing that if r=ord(X) <∞, then IAcan only be one of the following cases:

(1) r=0, and IA= {(1), (0)}; or (2) r=1, and IA= {(0, n), (0, 0)}; or (3) r=2, and IA= {(0, 0, 1), (0, 1, 0)}, with

(0,0,1)(0,1,0); or

(4) r=3, and IA= {(0, 1, 0, 1), (0, 0, 2, 0), (0, 2, 0, 0)}, with relations

(0,2,0,0) (0,1,1,0) (0,1,0,1)

(0,0,2,0)

≺

Here (0, 1, 1, 0)is an auxiliary index with a(0,1,1,0)=0.

The final proof is complete following fourteen preliminary lemmas as shown by the flow chart in Fig. 1.

2.2. Preliminary notations

Before proving Theorem 9, we introduce some notations as follows. Hereinafter, we note X₁≡0, and assume X1(x₁⁰, x₂⁰) =0.

(9)

Let

[δ2,X] =δ2X −Xδ2=(δ2X1)δ1+(δ2X2)δ2, b₀= −X₁(δ₂X2

X1

), b_i=X₁(δ₂bi−1

X1

)= −X₁(δⁱ₂⁺¹X2

X1

), i=1,2,· · ·

For F ∈K{y}, and {X}the differential ideal generated by X=Xy, we write

F∼R (19)

if R∈K{y}such that F −R∈ {X}.

Let m, n ∈Z^r⁺¹, the degreeof nis higher than that of m, denoted by n >m, if there exists 0 ≤k≤rsuch that nk> m_k and

n_i=m_i, i=k+1,· · ·, r.

It is easy to verify that the relation implies >, and for any m ∈Z^r⁺¹and 0 < i < j≤r,

−1

i,j(m)m i,j(m), (20)

and

−1

i,j(m) >m> _i,j(m). (21) In the following discussion, by m^∗ we always denote the element in IA with the highest degree, and always assume Am^∗=1 without loss of generality. This is possible as the coefficients in Aare rational functions in K.

For any m ∈Z^r⁺¹, define

P(m)= {p∈IA|pm} (22) and #(m) = |P(m)|, the number of elements in P(m).

We define a function C:Z^r⁺¹→Zby

C(m)= r j=1

j m_j, (23)

where m =(m₀, m₁, · · ·, m_r) ∈Z^r⁺¹. It is easy to verify that if m p, then C(m) > C(p). In particular,

C(m)−C( i,j(m))=i (0< i < j≤r). (24)

(10)

2.3. Preliminary lemmas

Now, we start the proof process. First, Lemma 11below is straightforward from the definition of nontrivial extension.

Lemma 11. Let A ∈K{y}\{X}, the differential ideal = {A, X}is a nontrivial extension of X if, and only if, the equation

Xy = 0

DAy = 0 (25)

has a non-constant solution in A(), with an open subset of C². The following result is a direct conclusion from Lemma 11:

Lemma 12. If there existsa∈K, non-constant, such that Xa=0, then let A=y−a,

the differential ideal = {X, A}is a nontrivial extension of X. Lemma 13. If there exists a∈K, a=0, such that

Xa=nb₀a, (26) where nis non-zero integer, let

A=(δ₂y)^|ⁿ^|−a^|ⁿ^|^/n, (27) then = {X, A}is a nontrivial extension of X.

Proof. From Lemma 11, we only need to show that there is a non-constant solution for the differential equation

X1δ1y+X2δ2y = 0

(δ₂y)^|ⁿ^|−a^|ⁿ^|^/n = 0. (28) Let

u=a^1/n, v= −X₂ X₁u, and taking account of (26), direct calculations show that

δ₁u= 1 X1

(b₀u−X₂δ₂u)=δ₂v.

(11)

Thus, the 1-form vdx1+udx₂is closed, and therefore the function of form

ω=

(x₁,x₂) (x₁⁰,x⁰₂)

vdx1+udx2

is well defined and analytic on a neighborhood of some (x₁⁰, x₂⁰) ∈C². Further, δ₁ω=v, δ₂ω=u.

It is easy to verify that ωsatisfies (28), and the lemma is proved. 2 Lemma 14. If there exists a∈Kthat satisfies

Xa=b₀a+b₁, (29) let

A=δ²₂y−aδ₂y, (30)

then = {X, A}is a nontrivial extension of X.

Proof. Let

b= −X2

X1

a+ b0

X1

.

From (29), we have

δ₁a= −X₂

X₁δ₂a−(δ₂X₂

X₁)a+δ₂b₀ X₁=δ₂b.

Thus, the 1-form bdx₁+adx₂ is closed, and there exists a function μ that is analytic on a neighborhood of some (x₁⁰, x₂⁰) ∈C², such that

δ₁μ=b, δ₂μ=a.

Let u =exp(μ), then uis a non-zero function, and

Xu=u(X1δ1μ+X2δ2μ)=u(X1b+X2a)=b0u.

Thus, following the proof of Lemma 13, let v= −X2

X1

u,

(12)

then vdx1+udx₂is a closed 1-form, and the function

ω=

(x1,x2) (x₁⁰,x₂⁰)

vdx1+udx2

is well defined in a neighborhood of (x₁⁰, x₂⁰), non-constant, and satisfies X₁δ₁ω+X₂δ₂ω=0, δ₂ω−u=0.

Therefore,

X1δ1ω+X2δ2ω=0, δ₂²ω−aδ2u=0.

Thus, the non-constant function ωsatisfies equations X₁δ₁y+X₂δ₂y=0

δ₂²y−aδ2y=0 (31)

and hence the lemma is concluded from Lemma 11. 2 Lemma 15. If there exists a∈Kthat satisfies

Xa=2b0a+b₂, (32) let

A=2(δ2y)(δ₂³y)−3(δ₂²y)²−a(δ2y)², (33) then = {X, A}is a nontrivial extension of X.

Proof. We only need to show that there is a function ωthat is analytic on an open subset of C², non-constant, and satisfies

X1δ1ω+X2δ2ω=0

2(δ2ω)(δ³₂ω)−3(δ₂²ω)²−a(δ₂ω)²=0. (34) We divide the proof into two steps. First, define

f (x₁, x₂, u)= −δ²₂X2

X₁−X2

X₁a−(δ₂X2

X₁)u−1 2(X2

X₁)u², g(x₁, x₂, u)=a+1

2u².

(13)

Then f and gare analytic at some point (x₁⁰, x₂⁰, u⁰) ∈C³. Now, we prove that there is a function u(x₁, x₂)that is analytic on a neighborhood of (x₁⁰, x₂⁰), and u(x₁⁰, x₂⁰) =u⁰, such that

δ1u =f (x1, x2, u),

δ2u =g(x1, x2, u) (35)

is satisfied in a neighborhood of (x₁⁰, x₂⁰). To this end, we need to show ( ∂

∂x₂+g(x₁, x₂, u)∂

∂u)f (x₁, x₂, u)=( ∂

∂x₁+f (x₁, x₂, u)∂

∂u)g(x₁, x₂, u) (36) and apply Lemma 28in Appendix A.

From (32), we have

δ1a= −δ₂³X₂

X₁−δ2(X₂

X₁a)−(δ2

X₂ X₁)a.

Thus, from (35), we have ( ∂

∂x2 +g(x₁, x₂, u)∂

∂u)f (x₁, x₂, u)

= −δ₂³X2

X1 −δ₂(X2

X1

a)−(δ₂²X2

X1

)u−1 2(δ₂X2

X1

)u²

−g(x₁, x₂, u)(δ₂X₂ X₁+X₂

X₁u)

= −δ₂³X₂

X₁ −δ2(X₂

X₁a)−(δ2

X₂

X₁)a−(δ₂²X₂

X₁)u−(X₂ X₁a)u

−(δ2

X₂ X₁)u²−1

2(X₂ X₁)u³,

= −δ₂³X₂

X₁ −δ₂(X₂

X₁a)−(δ₂X₂

X₁)a+uf (x₁, x₂, u)

=δ₁a+uf (x₁, x₂, u)

=( ∂

∂x₁+f (x₁, x₂, u)∂

∂u)g(x₁, x₂, u) and (36)is satisfied.

Now, assuming

u(x₁, x₂)=^∞

i=0

∞ j=0

u_i,j(x₁−x₁⁰)ⁱ(x₂−x₂⁰)^j (u_0,0=u⁰) (37)

and applying the Method of Majorants, we can obtain the coefficients ui,j by induction, and the power series (37)is convergent in a neighborhood of (x₁⁰, x₂⁰)(refer toLemma 28in Appendix A for detail). Thus, the function (37)gives an analytic solution of (35).

(14)

Next, we construct a solution ωof (34)from the above solution uof (35). Let v= −δ2

X₂ X₁ −X₂

X₁u.

It is easy to verify δ2v=δ₁u, and hence the 1-form vdx1+udx₂is closed. Let

μ=exp

⎡

⎢⎢

⎣

(x₁,x₂) (x₁⁰,x₂⁰)

vdx₁+udx₂

⎤

⎥⎥

⎦, (38)

then the function μis well defined, non-zero, and analytic on a neighborhood of (x₁⁰, x₂⁰)(here we note that X1(x₁⁰, x₂⁰) =0), and

Xμ=b0μ.

Following the proof of Lemma 13, there exists a non-constant function ω, analytic on a neighborhood of (x₁⁰, x₂⁰)(here we note that b0is analytic at (x₁⁰, x₂⁰)), such that

Xω=0, δ2ω=μ.

From (38)and (35), we have δ2μ =μu, and μδ²₂μ=μ

(δ2μ)u+μ(a+1 2u²)

=3

2(δ2μ)²+aμ². Taking account ofμ =δ2ω, we have

(δ₂ω)(δ₂³ω)−3

2(δ²₂ω)²−a(δ₂ω)²=0.

Thus, ωsatisfies (34)and the lemma is concluded. 2 Lemma 16. Let [δ₂, X]and yi be defined as previously, then (1) [δ2, X]=(^δ²_X^X¹

1 )X −b0δ2; (2) Xyj=δ2Xyj−1−(^δ²_X^X¹

1 )Xyj−1+b0yj. Proof. (1) is straightforward from

[δ2,X] =(δ2X1)δ1+(δ2X2)δ2

=δ₂X₁

X₁ (X1δ1+X2δ2)−X₂

X₁(δ2X1)δ2+(δ2X2) δ2

=δ₂X₁

X1 X +X₁X₁δ₂X₂−X₂δ₂X₁ X₁² δ₂

=δ2X1

X1 X −b₀δ₂.

(15)

(2) can be obtained by direct calculations below:

Xyj=Xδ2yi−1

=δ2Xy_j₋1− [δ2,X]y_j₋1

=δ₂Xy_j₋₁−(δ₂X₁

X1 X −b₀δ₂)y_j₋₁

=δ₂Xy_j₋₁−δ₂X₁

X₁ Xy_j₋₁+b₀δ₂y_j₋₁

=δ₂Xy_j₋₁−(δ₂X₁ X1

)Xy_j₋₁+b₀y_j. 2

Lemma 17. We have

Xy_j∼

j−1

i=0

c_i,jb_iy_j₋_i (j≥1) (39)

where c_i,j are positive integers, and c0,j=j. Proof. From Lemma 16, when j =1, we have

Xy₁=δ₂Xy₀−(δ₂X₁

X₁ )Xy₀+b₀y₁∼b₀y₁. Thus (39)holds for j=1 with c0,1=1.

Assume that (39)is valid for j=kwith positive integer coefficients ci,k, and c0,k=k, apply- ing Lemma 16, we have

Xy_k₊₁=δ₂Xy_k−(δ₂X₁ X1

)Xy_k+b₀y_k₊₁

∼δ₂(

k−1

i=0

c_i,kb_iy_k₋_i)−(δ2X1

X1

)(

k−1

i=0

c_i,kb_iy_k₋_i)+b₀y_k₊₁

=

k−1

i=0

c_i,k((δ₂b_i)y_k₋_i+b_iδ₂y_k₋_i)−

k−1

i=0

c_i,kδ₂X₁

X₁ b_iy_k₋_i+b₀y_k₊₁

=

k−1

i=0

ci,k

(δ2bi−δ₂X₁

X₁ bi)yk−i+biyk−i+1

+b0yk+1

=(c_0,k+1)b0y_k₊₁+

k−2

i=0

c_i,kX₁δ₂(b_i

X₁)+c_i₊_1,kb_i₊₁

y_k₋_i

+c_k₋_1,kX₁δ₂(b_k₋₁ X₁ )y₁

=(c0,k+1)b0yk+1+

k−2

i=0

(ci,k+ci+1,k)bi+1yk−i+ck−1,kbky1.

(16)

Thus, let

⎧⎪

⎨

⎪⎩

c_0,k₊₁=c_0,k+1=k+1, c_i,k₊1=c_i₋1,k+c_i,k, ck,k+1=ck−1,k,

(1≤i≤k−1),

which are positive integers, we have

Xy_k₊₁∼ k

i=0

c_i,k₊₁b_iy_k₊₁₋_i.

The lemma is proved by induction. 2 Lemma 18. We have

Xa_my^m∼(Xa_m+C(m)b0a_m)y^m+

r−1

i=1

r j=i+1

m_jc_i,jb_ia_my ^i,j^(m), (40)

where am∈K, y^m=y₀^m⁰y₁^m¹· · ·y^m_r^r, and ci,j is defined as in Lemma 17.

Proof. It is easy to have

Xa_my^m=(Xa_m)y^m+a_m r j=0

∂y^m

∂yj Xy_j.

From Lemma 17, we have

Xamy^m=(Xam)y^m+am

r j=0

m_jy^m⁻^e^jXy_i

∼(Xa_m)y^m+a_m r j=1

m_jy^m⁻^e^j(

j−1

i=0

c_i,jb_iy_j₋_i)

=(Xa_m)y^m+a_mb₀( r j=1

c_0,jm_j)y^m+a_m r j=1

j−1

i=1

m_jc_i,jb_iy^m⁺^e^j⁻ⁱ⁻^e^j

=(Xam+C(m) b0am)y^m+

r−1

i=1

r j=i+1

m_jc_i,jb_iamy ^i,j^(m),

and the lemma is concluded. 2

(17)

Lemma 19. Let be a nontrivial extension of X and = {X}, A ∈with the lowest rank and r=ord() (<∞). Let m^∗∈IA with the highest degree and assume that am^∗=1, then for any m ∈Z^r⁺¹, m <m^∗, we have

Xa_m=(C(m^∗)−C(m))b0a_m−

r−1

i=1

r j=i+1

(m_j+1)ci,jb_ia ₋1

i,j(m). (41)

Here am=0whenever m /∈IA. Proof. We can write

A=

m∈IA

amy^m=

m≤m∗

amy^m.

Hereinafter am=0 if m /∈IA. First, it is easy to have

XA=X1δ1A+X2δ2A∈.

On the other hand, from Lemma 18, we have XA=

m≤m^∗

Xa_my^m

∼

m≤m^∗

⎛

⎝(Xa_m+C(m)b0a_m)y^m+

r−1

i=1

r j=i+1

m_jc_i,jb_ia_my ^i,j^(m)

⎞

⎠

=

m≤m∗

⎛

⎝Xam+C(m)b0am+

r−1

i=1

r j=i+1

(mj+1)ci,jbia −1 i,j(m)

⎞

⎠y^m

Note that for any j > i, ⁻_i,j¹(m^∗) >m^∗, and thus ⁻_i,j¹(m^∗) /∈IA, i.e., a −1

i,j(m^∗)=0 for any j > i. Taking account of am^∗=1, we have Xa_m∗=0, and hence

XA∼C(m^∗)b₀y^m^∗

+

m<m^∗

⎛

⎝Xam+C(m)b0am+

r−1

i=1

r j=i+1

(mj+1)ci,jbia −1 i,j(m)

⎞

⎠y^m.

Therefore,

XA−C(m^∗)b₀A∼R=

m<m^∗

f_my^m, (42)

(18)

where the coefficients fmare

fm=Xam+(C(m)−C(m^∗))b0am+

r−1

i=1 r−1

j=i+1

(mj+1)ci,jbia −1

i,j(m). (43)

Now, we obtain a differential polynomial Rthat has lower rank than Aand is contained in the differential ideal . But Ais an element in with the lowest rank. Thus, we must have R≡0.

Therefore the coefficients (43)are zero, from which (41)is concluded. 2 Note that am∗=1 and ⁻_i,j¹(m^∗) /∈IA, the equation (41)is also valid for am∗. The equation (41)can be rewritten in another form below.

Lemma 20. In Lemma 19, for any m ≤m^∗, let k=#(m)and P(m) = {p1, · · ·, pk}, and assume

i_l,j_l(pl) =m(l=1, 2, · · ·, k), then the coefficients ap_l, amsatisfy Xam=(C(m^∗)−C(m))b0am−

#(m)

l=1

(mj_l +1)ci_l,j_lbi_lap_l. (44)

Lemma 21. Let be a nontrivial extension of X and = {X}, A ∈with the lowest rank and r=ord() >1. Let m^∗∈IA with the highest degree. Then for any m ∈IA, #(m) =0 if and only if C(m) =C(m^∗). Furthermore, if #(m) =0, then amis a constant.

Proof. First, we prove that if #(m) =0, then C(m) =C(m^∗).

If #(m) =0, Lemma 20yields

Xam=(C(m^∗)−C(m))b0am.

If otherwise C(m) =C(m^∗), then n =C(m^∗) −C(m)is a non-zero integer, and am=0 such that

Xa_m=nb₀a_m. From Lemma 13, let

A=(δ₂y)^|ⁿ^|−a_m^|ⁿ^|^/n,

then the differential ideal = {X, A}is a nontrivial extension of X and with order ≤1. This contradicts with the assumption that is an essential extension with order >1. Thus, we have concluded that C(m) =C(m^∗).

Next, we prove that if C(m) =C(m^∗), then #(m) =0.

If on the contrary, C(m) =C(m^∗)but #(m) >0, there exists m1∈P(m). From (24), we have C(m1) > C(m) =C(m^∗). Applying the previous part of the proof to m1, we have #(m1) >0.

Thus, we can repeat the above process, and obtain m2∈P(m1)such that C(m2) > C(m1) >

C(m^∗)and #(m2) >0. This procedure can be continued to obtain an infinite sequence {mk}^∞_k₌₁⊆ IAsuch that #(mk) >0 and C(mk+1) > C(mk) > C(m^∗). But IAis a finite set. Thus, we come to a contradiction, and therefore #(m) =0.