This is a preprint of: “Meromorphic integrability of the Hamiltonian systems with homogeneous potentials of degree -4”, Jaume Llibre, Yuzhou Tian, Discrete Contin. Dyn. Syst. Ser. B, 1–12, 2021. DOI: [10.3934/dcdsb.2021228] Meromorphic integrability of the Hamiltonian systems with homogeneous potentials of degree −4 ∗ Jaume Llibrea and Yuzhou Tianb,∗ a Departament de Matemàtiques, Universitat Autònoma de Barcelona, Edifici C, 08193 Bellaterra, Barcelona, Catalonia, Spain E-mail: jllibre@mat.uab.cat b School of Mathematics (Zhuhai), Sun Yat-sen University, Zhuhai 519082, P.R. China E-mail: tianyzh3@mail2.sysu.edu.cn Abstract We characterize the meromorphic Liouville integrability of the Hamiltonian systems with Hamiltonian H = p21 + p22 /2+V (q1 , q2 ), being V (q1 , q2 ) a homogeneous  potential of degree −4, with the exception of the potential V8 = 1/ q14 + 6µq12 q22 + q24 when µ ∈ {−5/3, −2/3}. For this potential we only can prove that it has no polynomial first integral. 2010 MSC: Primary 37K10, Secondary 37J30, 37C10. Keywords: Hamiltonian system with 2-degrees of freedom, homogeneous potential of degree −4, meromorphic integrability, Darboux point. 1 Introduction and main results Hamiltonian systems play an important role in the theory of the dynamical systems due to the fact that these systems occur frequently in mathematical physics, particularly in mechanics, engineering and other fields. In order to describe global information on the Hamiltonian systems is good to find sufficient number of first integrals. The fact that a Hamiltonian system has some additional first integral independent with its Hamiltonian is a rare phenomenon which lead to a difficult problem, how to determine whether a given Hamiltonian system has additional independent first integrals. In this work we consider the Hamiltonian systems of two degrees of freedom ∂V q̇i = pi , ṗi = − , i = 1, 2, (1.1) ∂qi with the Hamiltonian 2 1X H= pi + V (q) , (1.2) 2 i=1 where V (q) = V (q1 , q2 ) ∈ C [q1 , q2 ] is a homogeneous polynomial potential of degree k ∈ Z. Let A = A (q, p) and B = B (q, p) be two functions with q = (q1 , q2 ) and p = (p1 , p2 ). Their Poisson bracket is defined as X2   ∂A ∂B ∂A ∂B {A, B} = − . (1.3) ∂qi ∂pi ∂pi ∂qi i=1 ∗ Corresponding author (Yuzhou Tian). The first author is partially supported by the Ministerio de Ciencia, Innovación y Universidades, Agencia Estatal de Investigación grants M TM2016-77278-P (FEDER), the Agència de Gestió d’Ajuts Universitaris i de Recerca grant 2017SGR1617, and the H2020 European Research Council grant MSCA-RISE-2017-777911. The second author is partially supported by the National Natural Science Foundation of China (No. 11971495 and No. 11801582), China Scholarship Council (No. 201906380022) and Natural Science Foundation of Guangdong Province (No. 2019A1515011239). 1 The functions A and B are in involution if {A, B} = 0. A non-constant function I = I (q, p) is called a first integral of the Hamiltonian system (1.1) if I is in involution with the Hamiltonian function H, i.e. {H, I} = 0. Since the Poisson bracket is antisymmetric, it is obvious that H itself is always a first integral. The functions H and I are functionally independent if their gradients are linearly independent for all points of C4 except perhaps in a zero Lebesgue measure set. The Hamiltonian system (1.1) is completely or Liouville integrable if it has two functionally independent first integrals H and I. During the past four decades, there have had an extensive study on the integrability of the Hamiltonian systems (1.1), as it is shown in the papers [1, 2, 4-6, 8, 9, 14, 15, 19]. The funda- mental tools to investigate the integrability problem for system (1.1) are Painlevé test [7] and direct methods [10]. In [22, 23] Ziglin proved a relation between the integrability of Hamiltonian systems and the monodromy group of variational equations along a particular solution, and gave the necessary conditions of integrability for complex Hamiltonian systems. Yoshida proved some criteria for the nonexistence of an additional first integral in Hamiltonian systems with homoge- neous potentials, see [20, 21]. Using the differential Galois group Morales and Ramis [18] obtained necessary conditions for the existence of an additional meromorphic first integral of Hamiltonian systems (1.1) with homogeneous potentials. Recently Maciejewski and Przybylska [16] completely solved the meromorphic integrability problem of Hamiltonian systems (1.1) with homogeneous polynomial potentials of degree 3. Ma- ciejewski et al. [17] and Llibre et al. [12] characterized all integrable Hamiltonian systems (1.1) with homogeneous polynomial potentials of degree 4. The integrability of Hamiltonian systems (1.1) with homogeneous potentials of degrees 2, 1, 0, −1 or −2 were characterized in [13]. We note that Hamiltonian systems (1.1) with homogeneous potentials of degrees 2, 1, 0 or −1 has always a polynomial first integral independent with the Hamiltonian. The polynomial integrability of Hamiltonian systems (1.1) with homogeneous potentials of degree −3 have been classified in [11]. The objective of this paper is to study the integrability of the Hamiltonian system (1.1) with homogeneous potentials of degree −4, i.e. with 1 1 V = V (q1 , q2 ) = = , (1.4) aq14 + bq13 q2 + cq12 q22 + dq1 q23 + eq24 Q (q1 , q2 ) where Q (q1 , q2 ) 6≡ 0. Our main results are the following. Proposition 1. After a linear change of variables and a rescaling, the potential (1.4) can be written in one of the following forms: V0 = ± q14 ; V1 = 1 4q13 q2 ; V2 = ± 6p12 p2 ; V3 = ± 1 2 ; V4 = ± q 2 1 ; V5 = ± q2 1 ; 1 1 2 (q12 +q22 ) 2 (6q12 −q22 ) 2 (6q12 +q22 ) 1 V6 = q14 +6µq12 q22 −q24 ; V7 = − q4 +6µq12 q2 +q4 with µ > − 31 and µ 6= 13 ; V8 = 1 q14 +6µq12 q22 +q24 with µ 6= ± 13 . 1 1 2 2 Proposition 1 is proved in Section 4. Theorem 1. Hamiltonian system (1.1) with the potential: (a) V0 is completely integrable with the additional polynomial first integral p2 . (b) V3 is completely integrable with the additional polynomial first integral q2 p1 − q1 p2 . (c) Vi for i = 1, 2, 4, 5, 6, 7 does not admit an additional meromorphic first integral.   (d) V8 does not admit an additional meromorphic first integral if µ 6∈ − 53 , − 32 . If µ ∈ − 53 , − 23 then V8 does not admit an additional polynomial first integral. The proof of Theorem 1 is given in Section 5. 2 2 Darboux point The point d = (d1 , d2 ) ∈ C2 \ {0} is called a Darboux point of system (1.1) if it satisfies the gradient equation V 0 (d) = γd, (2.1) where V 0 (d) is the gradient of V (d) and γ ∈ C\ {0}. If d is a Darboux point, then the point d̃ = ωd, where ω ∈ C\ {0} satisfies V d̃ = ω k−2 γ d̃ = γ̃ d̃, and consequently d̃ is also a Darboux 0 point. Thus these Darboux points form an equivalence class. We can view a Darboux point d as a point in a projective space d = [d1 : d2 ] ∈ CP1 , for more details see [17]. Given a Darboux point d we consider the Hessian matrix V 00 (q) evaluated at the Darboux point d, that is,  2  ∂ V ∂2 V 2 (d) ∂q1 ∂q2 (d)  ∂q1  V 00 (d) =  . (2.2) ∂2 V ∂2 V ∂q2 ∂q1 (d) ∂q 2 (d) 2 Since the potential V is a homogeneous function of degree k, the number k − 1 is always an eigenvalue for any Darboux point d. The eigenvalues of the Hessian matrix V 00 (d) are denoted by {k − 1, λ}. We say that λ is the non-trivial eigenvalue. Consider the following potential 1 V (q1 , q2 ) = (2.3) a0 q1m + a1 q1m−1 q2 + · · · + am−1 q1 q2m−1 + am q2m with m ≥ 2. Let z = q2 /q1 . The potential (2.3) can be rewrite as 1 V = , q1m v (z) where v (z) = a0 + a1 z + · · · + am z m . In order to calculate the Darboux points and the non-trivial eigenvalue λ associated to potential (2.3), we define the polynomials  h (z) = mv (z) − zv 0 (z) , g (z) = 1 + z 2 v 0 (z) − mzv (z) , (2.4) where v 0 (z) denotes the derivative of v (z) with respect to z. The next result is due to Llibre et al., see Proposition 6 of [11]. Proposition 2. Assume that g 6≡ 0. The Darboux points d = [1 : z∗ ] associated with potential (2.3) are given by the zeros of g (z) = 0 for which h (z) 6= 0. Moreover the non-trivial eigenvalue of V 00 (d) is given by λ (z∗ ) = g 0 (z∗ ) /h (z∗ ) + 1. 3 Morales-Ramis theory The necessary condition for the complete meromorphic integrability of Hamiltonian systems (1.1) with homogeneous potentials V was given by Morales and Ramis, see page 100 of [18]. Theorem 2 (Morales-Ramis). Assume that the Hamiltonian system defined by the Hamiltonian (1.2) with a homogeneous potential V of degree k ∈ Z \ {0} is completely integrable with meromor- phic first integrals, then the non-trivial eigenvalues λ associated to its Darboux points must satisfy the conditions of Table 1, where p is an integer. 3 Table 1: The Morales-Ramis table. Degree Eigenvalue λ Degree Eigenvalue λ k 25 1 12 2 k p + p (p − 1) 2 −3 24 − 24 5 + 6p 2 arbitrary 3 − 241 1 + 24 (2 + 6p)2 2 −2 arbitrary 3 1 − 24 1 3 + 24 2 + 6p 49 1 10 2 1 1 6 2 −5 40 − 40 3 + 10p 3 − 24 + 24 5 + 6p 49 1 2 1 1 12 2 −5 40 − 40 (4 + 10p) 3 − 24 + 24 5 + 6p 9 1 4  2 2 −4 8 − 8 3 + 4p 4 − 18 + 18 43 + 4p 25 1 2 9 1 10 2 −3 24 − 24 (2 + 6p) 5 − 40 + 40 3 + 10p  2 −3 25 1 3 24 − 24 2 + 6p 5 9 − 40 + 1 40(4 + 10p)2 25 1 6 2 1 k−1  −3 24 − 24 5 + 6p k 2 k + p (p + 1) k 4 Homogeneous potentials of degree −4 To prove Proposition 1, we need the following result which was given in Theorem 2.6 of Cima and Llibre in [3]. Theorem 3. For each real fourth-order binary form Q (q1 , q2 ) there exists some σ ∈ GL (2, R) which transforms Q (q1 , q2 ) in one and only one of the following canonical forms: (I) Q = q14 + 6µq12 q22 + q24 with µ < − 31 ;  (II) Q = α q14 + 6µq12 q22 + q24 with µ > − 31 and µ 6= 13 ; (III) Q = q14 + 6µq12 q22 − q24 ;  (IV) Q = αq22 6q12 + q22 ;  (V) Q = αq22 6q12 − q22 ; 2 (VI) Q = α q12 + q22 ; (VII) Q = 6αp21 p22 ; (VIII) Q = 4q13 q2 ; (IX) Q = αq14 ; (X) Q = 0; where α = ±1 and GL (2, R) is the linear group. Proof of Proposition 1. It follows immediately from Theorem 3 and the fact that the linear changes of variables are canonical changes in the theory of Hamiltonian systems. 5 Proof of Theorem 1 Potentials V0 and V3 . It is immediate to check that the Hamiltonian system (1.1) with the potentials V0 and V3 have polynomial first integral I = p2 and I = q2 p1 − q1 p2 , respectively. So statements (a) and (b) hold. 4 By Theorem 2 we know that if potential (1.4) is integrable, then its eigenvalue λ must be one of Table 1. For convenience, we define the following sets (  2 ) 1 9 1 4 Z−4 = − + 4p : p ∈ Z , 8 8 3 2 Z−4 = {p − 2(p − 1)p : p ∈ Z} ,     3 1 5 Z−4 = − 4p(p + 1) : p ∈ Z . 2 4 √ Potential V1 . Using Proposition 2 we obtain the Darboux points z∗= ±1/ 3 and the corresponding non-trivial eigenvalue λ = −1. It is easy to check that −1 6∈ Z−4 1 ∪ Z2 ∪ Z3 −4 −4 . By Theorem 2 V1 is not integrable. Potential V2 . By Proposition 2 the potential V2 has the  1Darboux points z∗ = ±1. The corresponding non-trivial eigenvalue is λ = −1. Since −1 6∈ Z−4 ∪ Z−4 ∪ Z−4 2 3 , by Theorem 2 the Hamiltonian system (1.1) with the potential V2 is not integrable. √ Potential V4 . Analogously the potential V4 has the Darboux points z∗ = ± 3/2 with non- trivial eigenvalue λ = −5/3. It is not difficult to check that −5/3 6∈ Z−4 1 ∪ Z2 ∪ Z3 −4 −4 . By Theorem 2 V4 is not integrable. √ Potential V5 . The potential V5 has the Darboux  1 points2 z∗ =3 ± 6/2. The corresponding non-trivial eigenvalue is λ = −1/3. Since −1/3 6∈ Z−4 ∪ Z−4 ∪ Z−4 , the Hamiltonian system (1.1) with the potential V5 is not integrable.  Potential V6 . Using the notations of Proposition 2, we get g (z∗ ) = 4 3µ − 1 − (3µ + 1)z∗2 z∗ and h (z∗ ) = 12z∗2 µ + 4. If µ = −1/3, then this potential has Darboux points z∗ = 0 with non-trivial eigenvalue −4 . Therefore the potential V6 for µ = −1/3 is not integrable. λ = −1 6∈ Z−4 1 ∪ Z2 ∪ Z3 −4 If µ 6= −1/3, it has 3 Darboux points of the form [1 : z∗ ] r 3µ − 1 z∗,1 = 0 and z∗,2 = ± , 3µ + 1 √ and 2 Darboux points  of the form [0 : z] z = ± 3 2. The corresponding non-trivial  1 eigenvalues are ±3µ and 4/ 1 + 9µ2 −1. If potential V6 is integrable, then eigenvalues ±3µ ∈ Z−4 2 ∪ Z3 ∪ Z−4 −4 . j So we need to analyse nine possible cases, that is, 3µ ∈ Z−4 and −3µ ∈ Z−4 for j, k = 1, 2, 3. k j The integer of Table 1 corresponding to the case 3µ ∈ Z−4 (respectively −3µ ∈ Z−4k ) is denoted by pp0 (respectively p). All pthe possible values of integers p0 and p are shown in Table 2, where η = 3 (3 − 8µ) and ξ = 3 (3 + 8µ) satisfy η 2 + ξ 2 = 18. Table 2: Integers p0 and p. Eigenvalue 3µ Integer p0 Eigenvalue −3µ Integer p 1 1 1 1 3µ ∈ Z−4 12 (−4 ± 3η) −3µ ∈ Z−4 12 (−4 ± 3ξ) 2 1 2 1 3µ ∈ Z−4 (3 ± η) 4 −3µ ∈ Z−4 (3 ± ξ) 4 3 1 3 1 3µ ∈ Z−4 4 (−2 ± η) −3µ ∈ Z−4 4 (−2 ± ξ) Case 3µ ∈ Z−4 1 and −3µ ∈ Z 1 . Since p , p ∈ Z we have (3η, 3ξ) ∈ N × N. From the equation −4 0 η 2 + ξ 2 = 18, it follows that (η, ξ) = (3, 3). Thus µ = 0. Case 3µ ∈ Z−4 1 and −3µ ∈ Z 2 . Similarly (3η, ξ) ∈ N × N. Using the equation η 2 + ξ 2 = 18 −4 we obtain (η, ξ) = (3, 3). So µ = 0. The remaining cases can be analyzed in a similar way. We summarize all the possible values for η, ξ and µ in Table 3. Consequently if the potential  V6 is completely  1 integrable, then µ = 0. For µ = 0 the non-trivial eigenvalue 4/ 1 + 9µ2 − 1 becomes 3 6∈ Z−4 2 ∪ Z3 ∪ Z−4 −4 . So the Hamiltonian system (1.1) with the potential V6 is not integrable. 5 Table 3: The values of η, ξ and µ. Condition (η, ξ) µ (3η, 3ξ) ∈ N × N (3, 3) 0 (3η, ξ) ∈ N × N (3, 3) 0 (η, 3ξ) ∈ N × N (3, 3) 0 (η, ξ) ∈ N × N (3, 3) 0 Potential V7 . From Proposition 2 it follows that the Darboux points of the form [1 : z∗ ] are given by the zeros of g (z∗ ) = 4(3µ − 1)z∗ z∗2 − 1 for which h (z∗ ) = −4 3µz∗2 + 1 6= 0. We have that the Darboux points √ [1 : z∗ ] are z∗,1 = 0 and z∗,2 = ±1, and the Darboux points of the form [0 : z] are z = ± 2. The corresponding non-trivial eigenvalues are λ1 = 3µ and 3 λ2 = −1 + 4/ (1 + 3µ).  The necessary conditions for the complete meromorphic integrability of potential V7 are λ1,2 ∈ Z−4 1 ∪ Z2 ∪ Z3 −4 −4 . We need to consider nine possible cases, that is, j λ1 ∈ Z−4 and λ2 ∈ Z−4 for j, k = 1, 2, 3. For each case there exist two integers p0 and p such k j that λ1 ∈ Z−4 and λ2 ∈ Z−4k . All the possible values of the integers p and p are given in Table 4, 0 p p where η = 3 (3 − 8µ) and ζ = 17 − 32/ (1 + 3µ) satisfy √  η 2 ζ 2 − 17ζ 2 − 17η 2 + 33 = 0. Recall √  that µ > −1/3 and µ 6= 1/3. Thus η ∈ [0, 1) ∪ 1, 17 and ζ ∈ [0, 1) ∪ 1, 17 . Table 4: Integers p0 and p. Eigenvalue λ1 Integer p0 Eigenvalue λ2 Integer p 1 1 1 1 λ1 ∈ Z−4 12 (−4 ± 3η) λ2 ∈ Z−4 12 (−4 ± 3ζ) 2 1 2 1 λ1 ∈ Z−4 (3 ± η) 4 λ2 ∈ Z−4 (3 ± ζ) 4 3 1 3 1 λ1 ∈ Z−4 4 (−2 ± η) λ2 ∈ Z−4 4 (−2 ± ζ) For λ1,2 ∈ Z−41 , we have (3η, 3ζ) ∈ N × N due to the fact that p and p are integers. Combining √  √  0 with the conditions η ∈ [0, 1) ∪ 1, 17 , ζ ∈ [0, 1) ∪ 1, 17 and η 2 ζ 2 − 17ζ 2 − 17η 2 + 33 = 0, we conclude that η and ζ do not exist. The analysis  1 of other cases are similar.  1 We 2list all3 the possible values of η, ξ and ζ in Table 5. So λ1 6∈ Z−4 ∪ Z−4 ∪ Z−4 or λ2 6∈ Z−4 ∪ Z−4 ∪ Z−4 . 2 3 Therefore the Hamiltonian system (1.1) with the potential V7 is not integrable. In summary statement (c) of Theorem 1 is proved. Table 5: The values of η, ζ and µ. Condition (η, ζ) µ (3η, 3ζ) ∈ N × N ∅ ∅ (3η, ζ) ∈ N × N ∅ ∅ (η, 3ζ) ∈ N × N ∅ ∅ (η, ξ) ∈ N × N ∅ ∅ Potential V8 . Applying Proposition 2 the Darboux points of the form [1 : z∗ ] are given by the zeros of g (z∗ ) = 4(1 − 3µ)z∗ z∗2 − 1 for which h (z∗ ) = 4 3µz∗2 + 1 6= 0. The Darboux points of √ √ the form [1 : z∗ ] and [0 : z] are respectively z∗,1 = 0 and z∗,2 = ±1, and z = ±i 3 2 with i = −1. The corresponding non-trivial eigenvalues are λ1 = 3µ and λ2 = −1 + 4/ (1 + 3µ). By the same reasons as above one can get Table 4. Next we will divide the proof into two cases: Case (i) µ > − 13 and µ 6= 13 . Case (ii) µ < − 13 . The Case (i) is the same as the potential V7 . Thus p statement (d) holdsp for Case (i). Now we consider Case (ii). Since µ < − 13 , η = 3 (3 − 8µ) and ζ = 17 − 32/ (1 + 3µ), we 6 √ √ get that η > 17, ζ > 17 and r 256 ζ= + 17. (5.1) η 2 − 17 Since p must be an integer we have Table 6. Let η̃ = 3η and ζ̃ = 3ζ. Then equation (5.1) becomes r 2304 ζ̃ = 3 2 + 17, (5.2) η̃ − 153 √ √   with η̃ > 3 17 and ζ̃ > 3 17. If (3η, 3ζ) ∈ N × N, then η̃, ζ̃ ∈ N × N with η̃ ≥ 13 and ζ̃ ≥ 13. Furthermore, 13 ≤ ζ̃ ≤ 38 due to the fact that equation (5.2) is decreasing with η̃. For each value of ζ̃ = 13, 14, . . . , 38, equation (5.2) provides a value of η̃. We get the first row of Table 6. For the second row of Table 6, equation (5.1) can be rewrite as r 2304 ζ= 2 + 17, (5.3) η̃ − 153 √ √ with η̃ > 3 17 and ζ > 17. If (η̃, ζ) ∈ N × N, then η̃ ≥ 13 and ζ ≥ 5. Since equation (5.3) is decreasing with η̃, we have 5 ≤ ζ ≤ 12. The second row of Table 6 holds. The remaining rows of Table 6 are obtained in a similar way. Table 6: The values of η, ζ and µ. Condition (η, ζ) µ (3η, 3ζ) ∈ N × N (5, 7) or (7, 5) − 32 or − 53 (3η, ζ) ∈ N × N (5, 7) or (7, 5) − 32 or − 53 (η, 3ζ) ∈ N × N (5, 7) or (7, 5) − 32 or − 53 (η, ξ) ∈ N × N (5, 7) or (7, 5) − 23 or − 53 For µ = −2/3 the non-trivial eigenvalues are λ1 = −2 and λ2 = −5. For µ = −5/3, the non-trivial eigenvalues are λ1 = −5 and λ2 = −2. Obviously −2 ∈ Z−4 2 (p = 2) and −5 ∈ Z 2 −4 (p = −1). For µ = −2/3 or −5/3, we cannot use Theorem 2 to decide whether or not the Hamiltonian system is meromorphically integrable. Next we show that V8 does not admit an additional polynomial first integral when µ ∈ {−5/3, −2/3}. Consider the following potential 1 V = , (5.4) q14 + 6µq12 q22 + q24 with µ ∈ {−5/3, −2/3}. The corresponding Hamiltonian system (1.1) is   4q1 q12 + 3µq22 4q2 3µq12 + q22 q̇1 = p1 , q̇2 = p2 , ṗ1 = 2 , ṗ2 = 4 2 . (5.5) q14 + 6µq12 q22 + q24 q1 + 6µq12 q22 + q24 2 After a rescaling of the time variable dt = q14 + 6µq12 q22 + q24 ds, system (5.5) becomes 2 2 q̇1 = p1 q14 + 6µq12 q22 + q24 , q̇2 = p2 q14 + 6µq12 q22 + q24 ,   (5.6) ṗ1 = 4q1 q12 + 3µq22 , ṗ2 = 4q2 3µq12 + q22 . Doing the transformation (q1 , q2 , p1 , p2 ) 7→ (q1 , q2 , p1 , P ) with P = q2 p1 − q1 p2 , system (5.6) writes 2 q2 p1 − P 4 2 q̇1 = p1 q14 + 6µq12 q22 + q24 , q̇2 = q1 + 6µq12 q22 + q24 , q1 (5.7) 2 2  ṗ1 = 4q1 q1 + 3µq2 , Ṗ = 4 (1 − 3µ) q1 q2 (q1 + q2 ) (q1 − q2 ) . 7 Suppose that system (5.6) has a polynomial first integral F (q1 , q2 , p1 , p2 ) ∈ C [q1 , q2 , p1 , p2 ]. In the new variables (q1 , q2 , p1 , P ), it can be written as   n X q2 p1 − P F̄ (q1 , q2 , p1 , P ) = F q1 , q2 , p1 , = fj (q2 , p1 , P ) q1j , (5.8) q1 j=−n where fj (q2 , p1 , P ) ∈ C [q2 , p1 , P ]. Since F̄ (q1 , q2 , p1 , P ) is a first integral of system (5.7), we have ∂ F̄ ∂ F̄ ∂ F̄ ∂ F̄ q̇1 + q̇2 + ṗ1 + Ṗ = 0. (5.9) ∂q1 ∂q2 ∂p1 ∂P For clarity we introduce the following differential operators acting on fj (q2 , p1 , P ) ∈ C [q2 , p1 , P ] : ∂fj A [fj ] := jp1 fj + (q2 p1 − P ) , ∂q2 ∂fj ∂fj B [fj ] := 3µq24 A [fj ] − (1 − 3µ) q2 + 3µ , ∂P ∂p1   2  4 ∂fj ∂fj C [fj ] := 18µ + 1 q1 A [fj ] + 2 (1 − 3µ) q2 + . ∂P ∂p1 Using the above notions equation (5.9) can be written as n  X  q1j+7 A + 12µq22 q1j+5 A + 2q1j+3 C + 4q22 q1j+1 B + q28 q1j−1 A [fj ] = 0, (5.10) j=−n with fj ∈ C [q2 , p1 , P ]. Moreover F̄ (q1 , q2 , p1 , P ) is a first integral of system (5.7) if and only if the coefficients of q1j−1 in equation (5.10) are A [fj−8 ] + 12µq22 A [fj−6 ] + 2C [fj−4 ] + 4q22 B [fj−2 ] + q28 A [fj ] = 0, (5.11) where j = −n, . . . , n + 8 and fj = 0 if j < −n or j > n. Therefore system (5.6) has a polynomial first integral if and only if there exist 2n + 1 polynomials fj such that equations (5.11) hold. The existence of such polynomials are given by the following two lemmas. Lemma 1. If F̄ (q1 , q2 , p1 , P ) is a first integral of system (5.7), then fj (q2 , p1 , P ) = 0 for j = 1, . . . , n. Proof. For j = n + 8 equation (5.11) becomes A [fn ] = 0. The solution of A [fn ] = 0 is fn = α (p1 , P ) / (P − q2 p1 )n , where α (p1 , P ) is a function in the variables p1 and P . So fn = 0 due to the fact that fn ∈ C [q2 , p1 , P ]. If n = 1 we are done. When n ≥ 2, taking j = n + 8, n + 7, we have A [fn ] = 0 and A [fn−1 ] = 0. By the same reason as above, fn = fn−1 = 0. If n = 2, then the lemma holds. When n ≥ 3, taking j = n + 8, n + 7, n + 6, we get, respectively, A [fn ] = 0, A [fn−1 ] = 0 and A [fn−2 ] + 12µq22 A [fn ] = 0. This implies that A [fn−2 ] = 0. Using similar arguments we obtain fn = fn−1 = fn−2 = 0. If n = 3, the proof is finished. Consider n ≥ 4. Substituting j = n + 8, n + 7, n + 6, n + 5 into equation (5.11), we get, respectively, A [fn ] = 0, A [fn−1 ] = 0, A [fn−2 ]+12µq22 A [fn ] = 0 and A [fn−3 ]+12µq22 A [fn−1 ] = 0. Using similar arguments to the case n ≥ 3, it is easy to prove that fn = fn−1 = fn−2 = fn−3 = 0. If n = 4, the lemma is proved. When n ≥ 5, we analyze respectively j = n+8, n+7, n+6, n+5, n+4 in equation (5.11). Using similar arguments as in the previous cases, one can get that fn = fn−1 = fn−2 = fn−3 = fn−4 = 0. If n = 5 the lemma is confirmed. Next we prove this lemma by induction when n ≥ 6. Assume that fn = fn−1 = · · · = fi+1 = 0, where i ≥ 1. Now we consider equation (5.11) for j = i + 8, that is, A [fi ] + 12µq22 A [fi+2 ] + 2C [fi+4 ] + 4q22 B [fi+6 ] + q28 A [fi+8 ] = 0. (5.12) 8 By the induction hypothesis we have A [fi+2 ] = A [fi+8 ] = C [fi+4 ] = B [fi+6 ] = 0. Thus equation (5.12) reduces to A [fi ] = 0. From the above analysis we know that the only polynomial solution of differential equation A [fi ] = 0 is fi = 0. This ends the proof. Lemma 2. If F̄ (q1 , q2 , p1 , P ) is a first integral of system (5.7), then fj (q2 , p1 , P ) = 0 for j = −n, −n + 1, . . . , −1, and f0 (q2 , p1 , P ) is a constant. Proof. We consider equation (5.11) for j = −n, −n+1, −n+2, that is, q28 A [f−n ] = 0, q28 A [f−n+1 ] = 0 and 4q22 B [f−n ] + q28 A [f−n+2 ] = 0. So A [f−n ] = A [f−n+1 ] = 0. The polynomial solutions of the differential equations A [f−n ] = 0 and A [f−n+1 ] = 0 are respectively f−n = α−n (P − q2 p1 )n and f−n+1 = α−n+1 (P − q2 p1 )n−1 , (5.13) where α−n = α−n (p1 , P ) and α−n+1 = α−n+1 (p1 , P ) are polynomials in the variables p1 and P . From equations 4q22 B [f−n ] + q28 A [f−n+2 ] = 0 and (5.13), we obtain f−n+2 = (α−n+2 + β−n+2 ) (P − q2 p1 )n−2 , (5.14) where α−n+2 = α−n+2 (p1 , P ) is an integration constant, β−n+2 = 1 β̃ 15q25 −n+2 , and ∂α−n ∂α−n β̃−n+2 = 15nq2 α−n + 5(1 − 3µ) (3P − 4p1 q2 ) q2 + 9µ (5p1 q2 − 4P ) . (5.15) ∂P ∂p1 Since α−n is a polynomial, we claim that β−n+2 is also a polynomial. Assume that β−n+2 is not a polynomial. Then β−n+2 (P − q2 p1 )n−2 is a rational function with denominator 15q25 . Since f−n+2 must be a polynomial, we have that α−n+2 (P − q2 p1 )n−2 is also a rational function with denominator 15q25 . This contradicts the fact that α−n+2 is independent of the variable q2 . This means that β−n+2 is a polynomial if and only if β̃−n+2 is divisible by q25 . Thus β̃−n+2 = 0. Evaluating (5.15) on q2 = 0, we have ∂α−n /∂p1 = 0 due to the fact that 9µ 6= 0. After division by 5q2 in equation (5.15), we obtain ∂α−n 3nα−n + (1 − 3µ) (3P − 4p1 q2 ) = 0. (5.16) ∂P Substituting q2 = 3P/4p1 into (5.16), we get α−n = 0. Consequently f−n = 0 and f−n+2 = α−n+2 (P − q2 p1 )n−2 . For j = −n + 3 equation (5.11) becomes 4q22 B [f−n+1 ] + q28 A [f−n+3 ] = 0. In the same way as f−n+2 , we get f−n+3 = (α−n+3 + β−n+3 ) (P − q2 p1 )n−3 , (5.17) where α−n+3 = α−n+3 (p1 , P ) is an integration constant, β−n+3 = 1 β̃ 15q25 −n+3 , and ∂α−n+1 ∂α−n+1 β̃−n+3 = 15q2 α−n+1 (n − 1) + 5(1 − 3µ) (3P − 4p1 q2 ) q2 + 9µ (5p1 q2 − 4P ) . ∂P ∂p1 (5.18) By the same arguments as above one can get that α−n+1 = 0 and β−n+3 = 0. Therefore f−n+1 = 0 and f−n+3 = α−n+3 (P − q2 p1 )n−3 . Using exactly the same steps as in the previous cases, we can prove that f−n+3 = · · · = f−3 = 0, f−2 = α−2 (P − q2 p1 )2 and f−1 = α−1 (P − q2 p1 ), where α−2 = α−2 (p1 , P ) and α−1 = α−1 (p1 , P ) are polynomials in the variables p1 and P . From Lemma 1 we know that fj = 0 for j ≥ 1. For j = 8 equation (5.11) becomes A [f0 ] = 0, that is ∂f0 (q2 , p1 , P ) /∂q2 = 0. This implies that f0 does not depend on q2 . Finally we prove that f−1 = f−2 = 0 and f0 = constant. Consider equation (5.11) for j = 0, that is h i B [f−2 ] = B α−2 (P − q2 p1 )2 = 0. (5.19) 9 The general solution of the linear partial differential equation (5.19) is   1 p 1 q2 α−2 (p1 , P ) = ϕ P − p 1 q2 + , 9µ2 (P − p1 q2 )2 3µ where ϕ is a function in the variable P − p1 q2 + p3µ1 q2 . Since α−2 does not depend on q2 and 3µ 1 6= 0, we have α−2 = 0. Consequently f−2 = 0. Substituting j = 1 into (5.11) we have B [f−1 ] = B [α−1 (P − q2 p1 )] = 0. The general solution of this linear partial differential equation is   1 p 1 q2 α−1 (p1 , P ) = ψ P − p 1 q2 + . 3µ (P − p1 q2 ) 3µ By the same reasons as above we get α−1 = 0. From Lemma 1 and f−6 = f−4 = f−2 = 0, we obtain that ∂f0 ∂f0 (3µ − 1) q2 + 3µ = 0 for j = 2 in equation (5.11); ∂P ∂p1 (5.20) ∂f0 ∂f0 (1 − 3µ) q2 + = 0 for j = 4 in equation (5.11). ∂P ∂p1 System (5.20) is linear in the unknowns q2 ∂f0 /∂P and ∂f0 /∂p1 . 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