This is a preprint of: “Complete integrability of vector fields in RN ”, Jaume Llibre, Rafael Orlando Ramı́rez, Valentı́n Ramı́rez, J. Geom. Phys., vol. 168, 104308:1–25, 2021. DOI: [10.1016/j.geomphys.2021.104308] COMPLETE INTEGRABILITY OF VECTOR FIELDS IN RN JAUME LLIBRE1 , RAFAEL RAMÍREZ2 AND VALENTÍN RAMÍREZ3 Abstract. We give necessary and sufficient conditions for the complete inte- grability of first order N –dimensional differential systems. We propose a new method to determine in the Jacobi Theorem the last N −1 first integral for the complete integrability of an N –dimensional differential system with N − 2 independent first integrals and with a Jacobi multiplier. As an application we study the complete integrability of some 3–dimensional differential systems, more precisely the complete integrability of the asymmet- ric and symmetric May–Leonard differential systems. 1. Introduction For the N –dimensional nonlinear differential systems the existence of K < N − 1 independent first integrals means that these systems are partially integrable. The existence of N − 1 independent first integrals means that the system is completely integrable, i.e. for such systems the intersection of the N −1 hypersurfaces obtained fixing the N − 1 first integrals provide the trajectories of the differential system. We give necessary and sufficient conditions under which the differential system (1) ẋj = Xj (x1 , . . . , xN ), for j = 1, . . . , N, or its associated vector field ∂ ∂ ∂ X = X1 + X2 + . . . + XN , ∂x1 ∂x2 ∂xN is completely integrable. Here Xj : U −→ RN are C 1 functions defined in an open subset U ⊆ RN . Using these necessary and sufficient conditions we propose a new method to determine the last N − 1 first integral in the Jacobi Theorem for the complete integrability of the differential system (1) having N − 2 independent first integrals and a Jacobi multiplier. This paper is organized as follows. In section 2 we state some basic definitions and results. In section 3 we give our main results. In section 4 we prove our results. Finally in section 5 we apply the obtained results to the study the completely integrability of the asymmetric and symmetric May–Leonard differential systems and Clebsch differential systems. 2010 Mathematics Subject Classification. 34C05, 34C07. Key words and phrases. first integral, Jacobi multiplier, Nambu bracket, completely integrable ordinary differential equations, Jacobi Theorem. 1 2 J. LLIBRE, R. RAMÍREZ AND V. RAMÍREZ 2. Preliminary results and definitions 2.1. Complete integrable vector fields. For simplicity we shall assume that all the functions which appear below are of class C r for r ≥ 2 although most of the results remain valid under weaker hypotheses. Let U be an open subset of RN . We say that a non-locally constant function H : U −→ R is a first integral of the differential system (1) if H = H(x1 (t), . . . , xN (t)) is constant for all values of t for which the solution (x1 (t), . . . , xN (t)) is defined and contained in U . Clearly a C 1 function H is a first integral of system (1) if and only if ∂H ∂H ∂H Ḣ = X1 + X2 + . . . + XN ≡ 0 in U. ∂x1 ∂x2 ∂xN If Hr : Ur −→ R for r = 1, . . . , K are K first integrals of system (1), we say that they are independent in ŨK := U1 ∩ U2 . . . ∩ UK if their gradients are independent in all the points of ŨK except perhaps in a zero Lebesgue measure set. We say that system (1) is completely integrable in an open set ŨN −1 if it has N − 1 independent first integrals. In this case the orbits of system (1) are contained in the curves n o n o n o H1 = h1 ∩ H2 = h2 ∩ . . . ∩ HN −1 = hN −1 , when h1 , h2 . . . , hN −1 vary in R. Let J = J(x1 , . . . , xN ) be a non–negative function non–identically zero on an open subset U of RN . Then J is a Jacobi multiplier of the differential system (1) if Z Z J(x1 , . . . , xN )dx1 . . . dxN = J(x1 , . . . , xN )dx1 . . . dxN , Ω ϕt (Ω) where Ω is any open subset of U , ϕt is the flow defined by the differential system (1), and ϕt (Ω) is the image of the domain Ω under the flow ϕt . The following result of Whittaker [20] plays a main role for detecting a Jacobi multiplier. Theorem 1. Let J be a non–negative C 1 function non–identically zero defined on an open subset of RN . Then J is a Jacobi multiplier of the differential system (1) if and only if the divergence of the vector field JX is zero, i.e. ∂(JX1 ) ∂(JXN ) (2) div(JX ) := + ... + = X (J) + Jdiv(X ) = 0. ∂x1 ∂xN Note that if N = 2 then the definition of Jacobi multiplier coincides with the definition of integrating factor. The following result goes back to Jacobi, for a proof see Theorem 2.7 of [12]. Theorem 2 (Jacobi Theorem). Consider the differential system (1) and assume that it has a Jacobi multiplier J and N −2 independent first integrals H1 , H2 , . . . , HN −2 . Then the system admits an additional first integral independent of the previous ones given by Z ˜  J (3) HN −1 = X̃2 dx1 − X̃1 dx2 , ∆˜ COMPLETE INTEGRABILITY OF VECTOR FIELDS IN RN 3 where ˜ denotes quantities expressed in the variables (x1 , x2 , h1 , . . . , hN −2 ) with (4) Hj (x1 , . . . , xN ) = hj , for j = 1, . . . , N − 2 and ∂H1 ∂H1 ∂H1 ... ∂x3 ∂x4 ∂xN ∂H2 ∂H2 ∂H2 ∆= ... . ∂x3 ∂x4 ∂xN .. .. .. . . ... . ∂HN −2 ∂HN −2 ∂HN −2 ... ∂x3 ∂x4 ∂xN Then system (1) is completely integrable. 2.2. Nambu bracket. In the 1970s Nambu in [16] proposed a new approach to the classical dynamics based on an N − dimensional Nambu–Poisson manifold replacing the even dimensional Poisson manifold and replacing a single Hamiltonian H for N − 1 Hamiltonian H1 , . . . , HN −1 In the canonical Hamiltonian formulation the equations of motion (Hamilton equations) are defined via the Poisson bracket. In Nambu‘s formalism the Poisson bracket is replaced by the Nambu bracket. Nambu had originally considered the case N = 3. Although the Nambu formalism is a generalization of the Hamiltonian formal- ism its real applications are not as rich as the applications of this last one. In the monographs of Galliulin [9] used the Nambu formalism to study some inverse problems in ordinary differential equations. In this work there is also an extensive bibliography on the Nambu formalism. Let U be an open subset of RN . Let Hj = Hj (x1 , . . . , xN ) for j = 1, 2, . . . , N − 1 be independent functions defined in U. Given functions Hj for j = 1, . . . , N − 1 the Nambu vector field (see for instance [13]) associated to these functions is the N –dimensional vector field ∂H1 ∂H1 ∂H1 ... ∂x1 ∂x2 ∂xN ∂H2 ∂H2 ∂H2 ... ∂x1 ∂x2 ∂xN n o .. .. .. H1 , H2 , . . . , HN −1 , ∗ := . . ... . . ∂HN −1 ∂HN −1 ∂HN −1 ... ∂x1 ∂x2 ∂xN ∂ ∂ ∂ ... ∂x1 ∂x2 ∂xN n o When we apply a Nambu vector field to a function F , i.e. H1 , H2 , . . . , HN −1 , F , the obtained function is called a Nambu bracket, (see [13, 16]). For properties on the Nambu braket see [13]. In particular from the property (iv) of [13] we get that 4 J. LLIBRE, R. RAMÍREZ AND V. RAMÍREZ the Nambu vector field has zero divergence i.e, n o X N ∂ n o (5) div H1 , H2 , . . . , HN −1 , ∗ = H1 , H2 , . . . , HN −1 , xj ≡ 0. j=1 ∂xj In [13] we study the inverse approach to the ordinary differential equations. The inverse problem as the problem of finding the more general differential system of first order satisfying a set of given properties was stated by Erugin [7] and developed by Galiullin and his followers (see for instance [10, 11, 15]). The new approach of the inverse problem which we proposed in [13] uses as an essential tool the Nambu bracket. We deduce new properties of this bracket which plays a very important role in the proof of all the results of this work and in its applications. In particular we prove the following properties of the Nambu bracket (see Proposition 1.2.2.) Proposition 3. We define Ω (f1 . . . , fN −1 , g1 . . . , gN , G) := −{f1 . . . , fN −1 , G}{g1 . . . , gN } N ! X + {f1 , . . . , fN −1 , gn }{g1 , . . . , gn−1 , G, gn+1 , . . . , gN } , n=1 and Fλ (f1 . . . , fN −1 , g1 , . . . , gN ) := −{f1 . . . , fN −1 , λ{g1 . . . , gN }} N ! X + {g1 , . . . , gn−1 , λ{f1 . . . , fN −1 , gn }, gn+1 , . . . , gN } , n=1 for arbitrary functions f1 , . . . , fN −1 , G, g1 , . . . , gN , λ. Then the Nambu bracket sat- isfies the identities: (viii) Ω (f1 . . . , fN −1 , g1 . . . , gN , G) = 0, and (ix) Fλ (f1 . . . , fN −1 , g1 , . . . , gN ) = 0. Note this identity is a generalization of the Filippov fundamental identity (see [8]) which is obtained when λ = 1. It is interesting to observe that the identity Ω (f1 . . . , fN −1 , g1 . . . , gN , G) = 0 is more basic, in the sense that for the identity Fλ (f1 . . . , fN −1 , g1 , . . . , gN ) = 0, the following relation holds ∂   XN Fλ (f1 , . . . , fN −1 , g1 , . . . , gN ) = λΩ(f1 , . . . , fN −1 , g1 , . . . , gN , xj ) . j=1 ∂xj Let Hj be for j = 1, . . . , N arbitrary C r independent functions with r ≥ 2 satisfying (6) {H1 , H2 , . . . , HN } = 6 0, in Ũ ⊆ U with U \ Ũ is a set of zero Lebesgue measure. In the solution of the inverse problem in ordinary differential equations play a fundamental roll the following COMPLETE INTEGRABILITY OF VECTOR FIELDS IN RN 5 vector field ∂H1 ∂H1 ∂H1 ... λ1 ∂x1 ∂x2 ∂xN ∂H2 ∂H2 ∂H2 ... λ2 ∂x1 ∂x2 ∂xN 1 .. .. .. .. Y= . . ... . . {H1 , H2 , . . . , HN } (7) ∂HN ∂HN ∂HN ... λN ∂x1 ∂x2 ∂xN ∂ ∂ ∂ ... 0 ∂x1 ∂x2 ∂xN n o XN H1 , H2 , . . . , Hj−1 , ∗ , Hj+1 . . . , HN = − λj , j=1 {H1 , . . . , HN } where λk = λk (x1 , . . . , xN ) are convenient functions satisfying (8) Y(Hk ) = −λk , for k = 1, . . . , N. In view of the identity (see identity (vi) of [13]) ∂f n o XN H1 , H2 , . . . , Hj−1 , xm , Hj+1 , . . . , HN (9) m=1 ∂xm n o = H1 , H2 , . . . , Hj−1 , f, Hj+1 , . . . , HN , we obtain that  n o N X X N ∂  H1 , H2 , . . . , Hj−1 , xm , Hj+1 , . . . , HN div(Y) = − λj  j=1 m=1 ∂xm {H 1 , H2 , . . . , HN } N n X o = − H1 , H2 , . . . , Hj−1 , λ̃j , Hj+1 , . . . , HN , j=1 λj where λ̃j = . {H1 , H2 , . . . , HN } The function g : U −→ R and the set {(x1 , . . . , xN ) ∈ U : g(x1 , . . . , xN ) = 0} are called the partial integral and the invariant hypersurface of a vector field Y, respectively, if Y(g)|g=0 = 0. The differential system generated by Y can be written as follows n o XN H1 , H2 , . . . , Hj−1 , xm , Hj+1 , . . . , HN (10) ẋm = Y(xm ) = − λj := Ym , j=1 {H1 , H2 , . . . , HN } for m = 1, . . . , N. Clearly that if Hk is a first integral of Y then λk = 0, and if Hk is a partial integral then λk |Hk =0 = 0, and if Hk = Jk is a Jacobi multiplier then λk = −Jk divX , see formula (2). As we shall see in the next two theorems (see Theorem 1.3,1 and 1.4.1 of [13]) by choosing properly the functions H1 , H2 , . . . , HN and λ1 , λ2 , . . . , λN we can obtain 6 J. LLIBRE, R. RAMÍREZ AND V. RAMÍREZ the most general autonomous differential system (1) in U ⊂ RN having the set of partial integrals Hj = gj for j = 1, 2, . . . , M1 , and the given set of first integrals Hk for k = M1 + 1, M1 + 2, . . . , M1 + M2 = M, with M = M1 + M2 ≤ N defined in U and such that (6) holds The first result characterizes the differential systems (1) having a given set of M = M1 partial integrals with M ≤ N. Theorem 4. Let Hj = gj = gj (x1 , . . . , xN ) for j = 1, 2, . . . , M with M ≤ N be a given set of independent functions defined in an open set U ⊂ RN . Then any differential system defined in U which admits the set of partial integrals gj for j = 1, 2, . . . , M can be written as M X {g1 , . . . , gk−1 , xj , gk+1 , . . . , gN } ẋj = Φk + {g1 , g2 , . . . , gN } k=1 N X {g1 , . . . , gk−1 , xj , gk+1 , . . . , gN } λk = Y(xj ), {g1 , g2 , . . . , gN } k=M +1 where HM +j = gM +j = gM +j (x1 , . . . , xN ) for j = 1, . . . , N −M, are arbitrary func- tions defined in U which we choose in such a way that the Jacobian {g1 , . . . , gN } = 6 0, in the set Ũ ⊆ U and the functions λj = Φj = Φj (x1 , . . . , xN ), for j = 1, 2, . . . , M and λM +k = λM +k (x1 , . . . , xN ) for k = 1, 2, . . . N − M are arbitrary functions such that Φj |gj =0 = 0, for j = 1, . . . , M. The second main result characterizes the differential systems (1) having a given set of M1 partial integrals and M2 first integrals with 1 ≤ M2 < N and M1 + M2 ≤ N. Theorem 5. Let Hl = gl = gl (x1 , . . . , xN ) for l = 1, 2, . . . , M1 and Hk = Hk (x1 , . . . , xN ) for k = 1, 2, . . . , M2 with M1 + M2 = M ≤ N be independent functions defined in the open set U. Then the most general differential systems in U which admits the partial integrals gl for j = 1, . . . , M1 and the first integrals Hk for k = 1, . . . , M2 are (11) M1 X {g1 , . . . , gk−1 , xj , gk+1 , . . . , gM1 , H1 , . . . , HM2 , gM +1 . . . gN } ẋj = Φk + {g1 , . . . gM1 , H1 , . . . , HM2 , gM +1 , . . . , gN } k=1 N X {g1 , . . . , gM1 , H1 , . . . , HM2 , gM +1 , . . . , gk−1 , xj , gk+1 , . . . , gN } λk , {g1 , . . . , gM1 , H1 , . . . , HM2 , gM +1 , . . . , gN } k=M +1 for j = 1, 2, . . . , N, where HM +j = gM +j for j = 1, . . . , N − M are arbitrary functions satisfying {g1 , . . . gM1 , H1 , . . . , HM2 , gM +1 , . . . , gN } = 6 0 in the set Ũ ⊆ U and the functions λj = Φj = Φj (x1 , . . . , xN ), for j = 1, 2, . . . , M1 and λM +k = λk (x1 , . . . , xN ) for k = M +1, 2, . . . N are arbitrary functions such that Φj |gj =0 = 0, for j = 1, . . . , M. COMPLETE INTEGRABILITY OF VECTOR FIELDS IN RN 7 Moreover if M1 = 0 and M2 = N − 2 then system (11) becomes (12) ẋj = Yj (xj ) {H1 , . . . , HN −2 , xj , gN } {H1 , . . . , HN −2 , gN −1 , xj } = −λN −1 − λN , {H1 , . . . , HN −2 , gN −1 , gN } {H1 , . . . , HN −2 , gN −1 , gN } for j = 1, 2, . . . , N. for j = 1, . . . , N, and if M1 = 0 and M2 = N − 1 then system (11) becomes {H1 , . . . , HN −1 , xj } (13) ẋj = −λN . {H1 , . . . , HN −1 , gN } 3. Statement of the main results The first result is related with the completely integrability of vector field Y see [13, 14, 18] Theorem 6. Differential system (4) is completely integrable if and only if (14) λk = −µ{F1 , . . . , FN −1 , Hk }, for k = 1, . . . , N where µ, F1 , . . . , FN −1 are convenient independent functions in U . Moreover if (14) holds then vector field Y becomes (15) Y = µ{F1 , ..., FN −1 , ∗}, i.e. it is completely integrable with independent first integrals F1 , . . . , FN −1 and the Jacobi multiplier J = 1/µ. The following classical result is well known: If a 2–dimensional differential system (16) ẋ1 = X1 (x1 , x2 ), ẋ2 = X2 (x1 , x2 ), has an integrating factor J, then it can be written as 1 ∂H1 1 ∂H1 (17) ẋ1 = − = X1 , ẋ2 = = X2 , J ∂x2 J ∂x1 being H1 is a first integral of system (16). The next theorem will extend this classical result to N –dimensional differential systems. Corollary 7. A differential system (1) is completely integrable first integrals Hj for j = 1, . . . , N − 1 if and only if it can be written as 1n o (18) ẋj = H1 , H2 , . . . , HN −1 , xj = Yj , for j = 1, . . . , N, J where J = J(x1 , x2 , . . . , xN ) is a Jacobi multiplier. We note that Corollary 7 is the natural extension of (17) because 1 1 ẋ1 = {H1 , x1 } = Y1 , ẋ2 = {H1 , x2 } = Y2 , J J where {H1 , x1 } and {H1 , x2 }, are Nambu brackets, which in this case coincides with the Poisson bracket. We observe that Corollary 7 is a simple consequence of Theorem 6. Another proof of Corollary 7 appeared in [19, 4]. 8 J. LLIBRE, R. RAMÍREZ AND V. RAMÍREZ The proofs of Theorem 6 and Corollary 7 are given in section 4. From Corollary 7 it is immediate to prove the following result. Corollary 8. Assume that differential systems (1) has N − 2 independent first integrals H1 , H2 , . . . , HN −2 and a Jacobi multiplier J, then another independent first integral HN −1 can be obtained as a solution of the partial differential equation o {H1 , H2 , . . . , HN −2 , HN −1 , xj = JXj , for j = 1, . . . , N. Note that in order to compute the integral (3) by applying the Jacobi Theo- rem we need to solve the system of equations (3) for x3 , . . . , xN . In general to get xk (x1 , x2 , h1 , . . . , hN −2 ) for k = 3, . . . , N cap be a very difficult problem. Addition- ally, in general the explicit computation of the integral (3) is practically impossible to obtain. Corollary 8 provide a new method for computing a first integral HN −1 which avoid the necessity of the inversion of (3). In the next theorem we provide an extension of the classical result which goes back to Jacobi (see Theorem 2). Theorem 9. Consider the differential equations (1) with N − r independent first integrals H1 , . . . , HN −r and r −1 distinct Jacobi multipliers JN −r+1 , . . . , JN −1 such Jn that the functions Hn =: are not locally constants for n = N −r+1, . . . , N −2. JN −1 If the functions H1 , H2 , . . . , HN −2 are independent, then system (1) can be written as (19) {H1 , . . . , HN −r , log HN −r+1 , . . . , log HN −2 , xj , gN } ẋj = ν {H1 , . . . , HN −r , log HN −r+1 , . . . , log HN −2 , log JN −1 , gN } {H1 , . . . , HN −r , log HN −r+1 , . . . , log HN −2 , log JN −1 , xj } −λN = Yj , {H1 , . . . , HN −r , log HN −r+1 , . . . , log HN −2 , log JN −1 , gN } for j = 1, 2, . . . , N, where gN , ν and λN are functions which satisfy the first order partial differential equation {H1 , . . . , HN −r log HN −r+1 , . . . , log HN −2 , ν̃, gN } (20) +{H1 , . . . , HN −r log HN −r+1 , . . . , log HN −2 , JN −1 , λ̃N } = 0. where JN −1 ν ν̃ = , {H1 , . . . , HN −r , log HN −r+1 , . . . , log HN −2 , log JN −1 , gN } JN −1 λN λ̃N = . {H1 , . . . , HN −r , log HN −r+1 , . . . , log HN −2 , log JN −1 , gN } Moreover functions HN −r+1 , . . . , HN −2 are first integrals, independent with H1 , . . . , HN −r . and using the Jacobi multiplier JN −1 , the differential system (19) is completely in- tegrable. COMPLETE INTEGRABILITY OF VECTOR FIELDS IN RN 9 The additional first integral HN −1 can be determine as a solution of the first order partial differential equation {H1 , . . . , HN −r , log HN −r+1 , . . . , log HN −2 , HN −1 , xj } {H1 , . . . , HN −r , log HN −r+1 , . . . , log HN −2 , xj , gN } = ν̃ (21) {H1 , . . . , HN −r , log HN −r+1 , . . . , log HN −2 , log JN −1 , gN } {H1 , . . . , HN −r , log HN −r+1 , . . . , log HN −2 , log JN −1 , xj } −λ̃N . {H1 , . . . , HN −r , log HN −r+1 , . . . , log HN −2 , log JN −1 , gN } Corollary 10. Consider the differential equations (1) with N − r first integrals H1 , . . . , HN −r and r distinct Jacobi multipliers JN −r+1 , . . . , JN such that Hn =: Jn are not locally constant for n = N −r+1, . . . , N −1. If the functions H1 , H2 , . . . , HN −1 JN are independent, then system (1) can be written as {H1 , . . . , HN −r , log HN −r+1 , . . . , log HN −2 , log HN −1 , xj } ẋj = ν = Yj , {H1 , . . . , HN −r , log HN −r+1 , . . . , log HN −2 , log HN −1 , log JN } and it is completely integrable. Moreover if r = N then system (1) can be written as {log H1 , . . . , log HN −1 , xj } ẋj = ν = Yj , {log H1 , . . . , log HN −1 , log JN } 4. Proofs of the results Proof of Theorem 6. Assume that the vector field Y associated to differential sys- tem (4) is completely integrable, with the N −1 independent first integrals F1 , . . . , FN −1 , and consequently ∂F1 ∂F1 ... ∂x1 ∂xN .. .. . . .. .. = {F1 , . . . , FN −1 , xN } = 6 0, . . ∂FN −1 ∂FN −1 ... ∂x1 ∂xN XN ∂Fk in Ũ ⊆ U. Thus from the equations Y(Fk ) = Yj = 0 for k = 1, . . . , N − 1 j=1 ∂xj or, equivalent N X −1 ∂Fk ∂Fk Yj = −YN , j=1 ∂xj ∂xN 10 J. LLIBRE, R. RAMÍREZ AND V. RAMÍREZ for j = 1, . . . , N − 1. Solving this linear system in the unknown Y1 , . . . , YN −1 we get ∂F1 ∂F1 ∂F1 ∂F1 ∂F1 ... ... ∂x1 ∂xi−1 ∂xN ∂xi+1 ∂xN −1 ∂F2 ∂H2 ∂F2 ∂F2 ∂F2 ... ... YN ∂x1 ∂xi−1 ∂xN ∂xi+1 ∂xN −1 .. .. .. .. .. . ... . . . . ... ∂FN −1 ∂FN −1 ∂FN −1 ∂FN −1 ∂FN −1 ... ... ∂x1 ∂xi−1 ∂xN ∂xi+1 ∂xN −1 Yi = n o F1 , . . . , FN −1 , xN for i = 1, . . . , N − 1. Consequently ∂ ∂ ∂ n o Y = Y1 + Y2 + . . . + YN = µ F1 , F2 , . . . , FN −1 , ∗ , ∂x1 ∂x2 ∂xN −1 where µ = YN ({F1 , . . . , FN −1 , xN }) . So the “only if” part of the theorem follows. Now we shall prove the “if” part. We suppose that in system (4) we have that λl = −µ{F1 , ..., FN −1 , Hl }. Thus the vector field associated to differential system (4) takes the form N X {H1 , . . . , Hn−1 , xj , Hn+1 , . . . , HN } Y(xj ) = − λn n=1 {H1 , H2 , . . . , HN } N X {H1 , . . . , Hn−1 , xj , Hn+1 , . . . , HN } = µ {F1 , ..., FN −1 , Hn } . n=1 {H1 , H2 , . . . , HN } In view of the identity Ω (f1 . . . , fN −1 , g1 . . . , gN , G) = 0 with G = xj fj = Fj for j = 1, . . . , N − 1, and Hj = gj for j = 1, . . . , N we get that N X {H1 , . . . , Hn−1 , xj , Hn+1 , . . . , HN } Y(xj ) = µ {F1 , ..., FN −1 , Hn } n=1 {H1 , H2 , . . . , HN } {H1 , H2 , . . . , HN } = µ{F1 , ..., FN −1 , xj } = µ{F1 , ..., FN −1 , xj }. {H1 , H2 , . . . , HN } Thus Y = µ{F1 , ..., FN −1 , ∗}. Consequently Y(Fk ) = 0 for k = 1, . . . , N − 1 i.e. F1 , . . . , FN −1 are first integral. Hence the vector field Y is completely integrable. Finally we prove that 1/µ is a Jacobi multiplier. Indeed in view of relation (5) we get that Y1 Y2 YN   ∂( ) ∂( ) ∂( ) Y µ µ µ div == + + ... + = div ({F1 , . . . , FN −1 , ∗}) = 0. µ ∂x1 ∂x2 ∂xN Hence from Theorem 1 we obtain that 1/µ is a Jacobi multiplier. Thus the theorem is proved.  Proof of Corollary 7. It follows from Theorem 6. Indeed in this case from differen- tial system (13) which is the most general differential equations which have N − 1 COMPLETE INTEGRABILITY OF VECTOR FIELDS IN RN 11 independent first integral, we have that this system is completely integrable if and only if λN = −µ {H1 , . . . , HN −1 , gN } where J = 1/µ is a Jacobi multiplier.  Proof of Theorem 9. First by considering that ∂Jj ∂Jj ∂Jj div(Jj Y) = Y1 + Y2 + . . . + YN + Jj divY = Y(Jj ) + Jj divX = 0. ∂x1 ∂x2 ∂xN Hence Y(Jj ) = −Jj divY := −ν Jj = λj , for j = N − r + 1, . . . , N − 1. Now we shall construct the most general differential system (1) with N − r inde- pendent first integrals H1 , . . . , HN −r and r−1 Jacobi multipliers JN −r+1 , . . . , JN −1 . We use the vector field Y given in formula (4). Since Hj are first integrals, from (8) it follows that λj = 0 for j = 1, . . . , N − r and Hj = Jj , λj = −νJj for j = N − r + 1, . . . , N − 1, we get that vector field (7) becomes ∂H1 ∂H1 ∂H1 ... 0 ∂x1 ∂x2 ∂xN .. .. .. .. . . ... . . ∂HN −r ∂HN −r ∂HN −r ... 0 ∂x1 ∂x2 ∂xN ∂JN −r+1 ∂JN −r+1 ∂JN −r+1 ... −ν JN −r+1 ∂x1 ∂x2 ∂xN .. .. .. .. . . ... . . ∂JN −1 ∂JN −1 ∂JN −1 ... −ν JN −1 ∂x1 ∂x2 ∂xN ∂HN ∂HN ∂HN ... λN ∂x1 ∂x2 ∂xN ∂ ∂ ∂ ... 0 ∂x1 ∂x2 ∂xN Y= . {H1 , . . . , HN −r , JN −r+1 , . . . , JN −1 , HN } 12 J. LLIBRE, R. RAMÍREZ AND V. RAMÍREZ Multiplying the numerator and the denominator of the vector field Y by 1/(JN −r+1 . . . JN −1 ) we get that Y can be written as ∂H1 ∂H1 ∂H1 ... 0 ∂x1 ∂x2 ∂xN .. .. .. .. . . ... . . ∂HN −r ∂HN −r ∂HN −r ... 0 ∂x1 ∂x2 ∂xN ∂ log JN −r+1 ∂ log JN −r+1 ∂ log JN −r+1 ... −ν ∂x1 ∂x2 ∂xN (22) .. .. .. .. . . ... . . ∂ log JN −1 ∂ log JN −1 ∂ log JN −1 ... −ν ∂x1 ∂x2 ∂xN ∂HN ∂HN ∂HN ... λN ∂x1 ∂x2 ∂xN ∂ ∂ ∂ ... 0 ∂x1 ∂ x2 ∂xN Y= {H1 , . . . , HN −r , log JN −r+1 , . . . , log JN −1 , HN } Subtracting the file N −1 of the determinant of (22) to the files N −r +1, . . . , N −2 we obtain ∂H1 ∂H1 ∂H1 ... 0 ∂x1 ∂x2 ∂xN .. .. .. .. . . ... . . ∂HN −r ∂HN −r ∂HN −r ... 0 ∂x1 ∂x2 ∂xN ∂ log HN −r+1 ∂ log HN −r+1 ∂ log HN −r+1 ... 0 ∂x1 ∂x2 ∂xN .. .. .. .. . . ... . . (23) .. .. .. .. . . ... . . ∂ log HN −2 ∂ log HN −2 .. ∂ log HN −2 . 0 ∂x1 ∂x2 ∂xN ∂ log JN −1 ∂ log JN −1 ∂ log JN −1 ... −ν ∂x1 ∂x2 ∂xN ∂HN ∂HN ∂HN ... λN ∂x1 ∂x2 ∂xN ∂ ∂ ∂ ... 0 ∂x1 ∂ x2 ∂xN Y= {H1 , . . . , HN −r , log JN −r+1 , . . . , log JN −1 , HN } COMPLETE INTEGRABILITY OF VECTOR FIELDS IN RN 13 Jn where Hj = for j = N − r + 1, . . . , N − 2. Hence (19) follows. JN −1 This differential system has N − 2 first integrals Hj for j = 1, . . . , N − 2. Note that if Y is a vector field associated to the differential system (19), then it follows that Y(Hj ) = 0 for j = N − r + 1, . . . , N − 2, so these Hj ’s are first integrals. In fact it is well known that the quotient of two Jacobi multipliers is a first integral. Thus in view of Jacobi Theorem 2 we get that system (19) is completely integrable. The relation (20) holds. Indeed, the vector field (24) {H1 , . . . , HN −r log HN −r+1 , . . . , log HN −2 , xj , gN } JN −1 Y(xj ) == JN −1 ν {H1 , . . . , HN −r log HN −r+1 , . . . , log HN −2 , log JN −1 , gN } {H1 , . . . , HN −r log HN −r+1 , . . . , log HN −2 , log JN −1 , xj } −JN −1 λN , {H1 , . . . , HN −r , log HN −r+1 , . . . log HN −2 , log JN −1 , gN } has zero divergence, i.e. div(JN −1 Y) = 0. In view of identity (34) we get after some computations (20). The proof of relation (21) is obtained by considering that the constructed vector field is completely integrable with Jacobi multiplier JN −1 . Consequently there exists an additional first integral HN −1 . Thus the vector field Y can be written as (18) with J = JN −1 . i.e. JN −1 Y(xj ) = {H1 , H2 , . . . , HN −2 , HN −1 , xj } . Hence in view of (24) we obtain the relations (21). In short the theorem is proved.  Proof of Corollary 10. It follows from Theorem 9 taking gN = JN and λN = −ν JN , from (4) we obtain 14 J. LLIBRE, R. RAMÍREZ AND V. RAMÍREZ ∂H1 ∂H1 ∂H1 ... 0 ∂x1 ∂x2 ∂xN .. .. .. .. . . ... . . ∂HN −r ∂HN −r ∂HN −r ... 0 ∂x1 ∂x2 ∂xN ∂ log HN −r+1 ∂ log HN −r+1 ∂ log HN −r+1 ... 0 ∂x1 ∂x2 ∂xN .. .. .. .. . . ... . . .. .. .. .. . . ... . . ∂ log HN −2 ∂ log HN −2 ∂ log HN −2 ... 0 ∂x1 ∂x2 ∂xN ∂ log HN −1 ∂ log HN −1 ∂ log HN −1 ... 0 ∂x1 ∂x2 ∂xN ∂H log JN ∂ log JN ∂ log JN ... −ν ∂x1 ∂x2 ∂xN ∂ ∂ ∂ ... 0 ∂x1 ∂ x2 ∂xN Y= . {H1 , . . . , HN −r , log JN −r+1 , . . . , log HN −1 , log JN } {H1 , . . . , HN −r , log JN −r+1 , . . . , log HN −1 , ∗} Hence Y = ν where HN −1 = {H1 , . . . , HN −r , log JN −r+1 , . . . , log JN −1 , log JN } JN −1 /JN .  4.1. Construction of differential systems with given first integrals and Jacobi multiplers. We shall construct differential system (1) for N = 3 with a given first integral and Jacobi multipliers. First we study the case when the functions H1 , H2 and H3 and λ1 , λ2 and λ3 are |z| 1 |x| H1 = , H2 = , H3 = , |y||1 − x − y − z|−1+b3 |xyz(x + y + z − 1)| |y||1 − x − y − z|1−b2 λ1 = aκ(x, y, z) H1 , λ2 = bκ(x, y, z) H2 , λ3 = cκ(x, y, z) H3 , COMPLETE INTEGRABILITY OF VECTOR FIELDS IN RN 15 where κ = κ(x, y, z) is a convenient function and a, b, c are constant. The vector field Y given in (7) can be written as ∂ log H1 ∂ log H1 ∂ log H1 aκ ∂x ∂y ∂z ∂ log H2 ∂ log H2 ∂ log H2 bκ ∂x ∂y ∂z ∂ log H3 ∂ log H3 ∂ log H3 cκ ∂x ∂y ∂z ∂ ∂ ∂ 0 ∂x ∂y ∂z Y= . {log H1 , log H2 , log H3 } {log F1 , log F2 , ∗} If abc 6= 0 then Y = κ , where F1 and F2 are the first {log H1 , log H2 , log H3 } integrals such that F1 = H1c /H3a , F2 = H2c /H3b . Choosing properly the function κ we obtain, after some computations, that the differential equations generated by the vector field Y are  ẋ = x (a + b − 2c)(x − 1) + ((2 − b2 )(a + b) − (b2 + 2)c) y  + ((b2 + 1)(a − b) + (b3 − 4)(c − b)) z = Ỹ1 ,  ẏ = y (a + b + c)(y − 1) + (b2 (a + b) + (b3 + 1)c)x (25)  +((3 − b2 )a + (2 − b3 )(c + b)z = Ỹ2 ,  ż = z (−2a + b + c)(z − 1) + ((b2 + 2)(b − a) + (b3 − 3)(b − c))x  +(b3 (b + c) − (b3 + b2 + 1)a)y = Ỹ3 , which are the completely integrable Lotka–Volterra systems. Hence we get that (26) Ỹ(H1 ) − a H1 divY, Ỹ(H2 ) = −b H2 divỸ, Ỹ(H3 ) = −c H3 divỸ. ∂ ∂ ∂ where Ỹ = Ỹ1 + Ỹ2 + Ỹ3 and ∂x ∂y ∂z   divỸ = b (2b1 + b3 + 1)x + (4 + b3 − b2 )y + (7 − b2 − 2b3 )z − 3b . From (26) we get that if a = b = c = 1 then H1 , H2 and H3 are Jacobi multipliers. If a = c = 0 and b = 1 then H1 , H3 are first integrals and H2 is a Jacobi multiplier. Second we study differential system (19) for the case when N = 3 with the first integral H1 and the Jacobi multipliers J2 = H2 and J3 = H3 . We study differential system (12) for the case when N = 3 with first integral   x b3 −1 x b1 −1 H1 = log , z z 16 J. LLIBRE, R. RAMÍREZ AND V. RAMÍREZ and two Jacobi multiplier J2 , J3 . 1  y  b b−1 1 y −1 z −2 , 3 J2 = , J3 = xyz|x + y + z − 1| z with λ2 = −ν J2 , λ3 = −ν J3 , where ν is a convenient function. After some computations we get that   {H1 , J2 , J3 } = 7 − 2b1 − b3 )x + (b1 + 2b3 + 1)y + (b1 − b3 + 4)z + 3 J22 J3 := ν1 J22 J3 , (b3 + 2b1 − 4) (2 − b1 − 2b3 ) {H1 , x, J3 } = J3 , {H1 , y, J3 } = J3 , yz xz   {H1 , J2 , x} = (2b1 + 3b3 − 5)x + (b1 + 2b3 − 3)y + (b1 + 3b3 − 4)z + 3 − b1 − 2b3 J22 ,   {H1 , J2 , y} = (3 − 2b1 − b3 )x + (5 − 2b3 − 3b1 )y + (4 − 3b1 − b3 )z + 2b1 + b3 − 3 J22 ,   {H1 , J2 , z} = (2b1 − b3 − 1)x + (b1 − 2b3 + 1)y + (b1 − b3 )z + b3 − b1 J22 . Hence the equation (12) for N = 3 becomes {H1 , x, J3 } {H1 , J2 , x} ν   ẋ = −λ2 − λ3 = {H1 , x, log J3 } + {H1 , log J2 , x} {H1 , J2 , J3 } {H1 , J2 , J3 } ν1 J2   ν J2 = H1 , log , x = Y(x), ν1 J2 J3 {H1 , y, J3 } {H1 , J2 , y} ν   ẏ = −λ2 − λ3 = {H1 , y, log J3 } + {H1 , log J2 , y} {H1 , J2 , J3 } {H1 , J2 , J3 } ν1 J2   ν J2 = H1 , log , y = Y(y), , ν1 J2 J3 {H1 , z, J3 } {H1 , J2 , z} ν   ż = −λ2 − λ3 = {H1 , z, log J3 } + {H1 , log J2 , z} {H1 , J2 , J3 } {H1 , J2 , J3 } ν1 J2   ν J2 = H1 , log , z = Y(z). ν1 J2 J3 Taking ν = ν1 we obtain the following differential system   1 J2 ẋ = x (1 − x − (b1 + b3 − 1)y − b1 z) = H1 , log , x = Y(x), J2 J3   1 J2 (27) ẏ = y (1 − y − (2 − b3 )z − b2 x) = H1 , log , y = Y(y), J2 J3   1 J2 ż = z (1 − z − (2 − b1 )x − b3 y) = H1 , log , z = Y (z). J2 J3 This differential system has the first integral H1 and two Jacobi multiplier J2 and J3 and consequently J2 /J3 is a first integral. It is easy to show that divY = −ν1 . We observe that the obtained differential system (27) is a particular case of the asymmetric May–Leonard system (see for instance [5]) COMPLETE INTEGRABILITY OF VECTOR FIELDS IN RN 17 5. New method for determining the additional first integral in the Jacobi Theorem 2 The problem on the determination the (N − 1)–th first integral in the Jacobi Theorem 2 was study in several papers. In particular for N = 3 the following result is given in [1]. Theorem 11. Suppose that the polynomial vector field ∂ ∂ ∂ Y = x(λ + ax + by + cz) + y(µ + dx + ey + f z) + z(ν + gx + hy + kz) ∂x ∂y ∂z has an analytic first integral F1 = xα y β z γ (1 + O(x, y, z)) with at least one of α, β, γ 6= 0, and a Jacobi multiplier J = xr y s z t (1 + O(x, y, z)) if the cross prod- uct of (r − i − 1, s − j − 1, t − k − 1) and (α, β, γ) is bounded away from zero for any integers i, j, k ≥ 0, then the system has a second analytic first integral of the form F2 = x1−r y 1−s z 1−t (1 + O(x, y, z)), and hence Y is completely integrable in a neighborhood of the origin. Jacobi (see Theorem 2) proposed a constructive method to build an extra first integral by knowing the multiplier J and N −2 first integrals for an N – dimensional differential system (1). In particular for N = 3 Theorem 2 writes: Theorem 12. Consider a three dimensional vector field (1) and assume that it admits a Jacobi multiplier and one first integral H1 (x1 , x2 , x3 ) = h1 . Then the system has the following extra first integral given by (see (3) for N = 3 Z ˜  J (28) H2 = X̃1 dx2 − X̃2 dx1 , ˜ ∆ where ˜ = ∂H1 (x1 , x2 , x3 ) ∆ , J˜ = J(x1 , x2 , x3 )|x3 =Ψ(x1 ,x2 ,h1 ) , ∂x3 x3 =Ψ(x1 ,x2 ,h1 ) X̃1 = X1 (x1 , x2 , x3 )|x3 =Ψ(x1 ,x2 ,h1 ) , X̃2 = X2 (x1 , x2 , x3 )|x3 =Ψ(x1 ,x2 ,h1 ) , and x3 = Ψ(x1 , x2 , h1 ) is a solution of the equation H1 = h1 . As we observe above the inversion of H1 = h1 , in terms of one of the variables is much more involved and the resulting extra first integral is given by the integral (28) which in general is very difficult to compute. Below we propose a new method for computing a first integral H2 which avoid the necessity of the inversion of H1 = h1 . Corollary 13. Assume that differential systems (1) with N = 3 has a first integral H1 and a Jacobi multiplier J, then a second first integral H2 can be obtained as a solution of the partial differential equation {H1 , H2 , xj } = JXj , for j = 1, 2, 3 which is equivalent to ∂H1 ∂H2 ∂H1 ∂H2 − = JX1 , ∂y ∂z ∂z ∂y ∂H1 ∂H2 ∂H1 ∂H2 (29) − = JX2 , ∂z ∂x ∂x ∂z ∂H1 ∂H2 ∂H1 ∂H2 − = JX3 , ∂x ∂y ∂y ∂x 18 J. LLIBRE, R. RAMÍREZ AND V. RAMÍREZ where x = x1 , y = x2 and z = x3 . Proof. The existence of the second first integral H2 follows from Jacobi Theorem 2. To obtain the function H2 we apply Theorem 7 with N = 3 from which we get equations (29).  Corollary 14. If ∂H1 ∂H1 ∂H1 (30) x +y +z = 0, ∂x ∂y ∂z then H2 is a solution of the differential equations   ∂H2 ∂H2 ∂H2 ∂H1 x +y +z = J(xX2 − yX1 ), ∂x ∂y ∂z ∂z (31) ∂H1 ∂H2 ∂H1 ∂H2 − = JX3 , ∂x ∂y ∂y ∂x J(xX2 − yX1 ) Moreover if := L is a homogenous function of degree one, then the ∂H1 ∂z solution of the first equation of (31) is (32) H2 = U (x/z, y/z)L(x, y, z). Proof. From the two first equations of (29) it follows that (33)     ∂H1 ∂H1 ∂H1 ∂H2 ∂H2 ∂H2 ∂H2 ∂H1 x +y +z − x +y +z = J(yX1 − xX2 ). ∂x ∂y ∂z ∂z ∂x ∂y ∂z ∂z Hence if H1 is a homogenous function of degree zero, i.e, (30) holds, then from equation (33) if follows the first equation of (31). If the function V is a homogenous function of degree one, i.e. ∂L ∂L ∂L x +y +z = L, ∂x ∂y ∂z then after some computations it follows that the solution of (33) can be written as (32).  Example 15. The particular case of Lotka-Volterra system a1 b4 (a2 − b2 ) ẋ = x(b3 + a1 x + a2 y + z = X1 , b2 (a1 − b1 ) ẏ = y(b3 + b1 x + b2 y + b4 z) = X2 , ż = b3 z = X3 is completely integrable with first integral H1 and Jacobi multiplier J such that H1 = xb2 (b1 −a1 ) y a1 (a2 −b2 ) z (a2 −b2 )(a1 −b1 ) Ga1 b2 −a2 b1 ,  1/(b2 (a1 −b1 ) J= y b1 b2 −a1 a2 ) z (a2 −b2 )(a1 −b1 ) Gb1 (a2 +b2 )−2a1 b2 , COMPLETE INTEGRABILITY OF VECTOR FIELDS IN RN 19 where G = b2 (a1 − b1 )x + (a2 − b2 )b2 y + b4 (a2 − b2 ) z = 0 is an additional invariant plane. It is easy to show that H1 and J are homogenous functions of degree zero and three respectively. In this case the first equation of (31) becomes   ∂H2 ∂H2 ∂H2 ∂H1 J xyzG (34) x +y +z = . ∂x ∂y ∂z ∂z a1 b4 (b2 − a2 )H2 By considering that J xyzG L := = L̃(x/z, y/z)G a1 b4 (b2 − a2 )H2 is a homogenous function of degree one we get that the solution of equation (34) is H = L̃(x/z, y/z)G + U (x/z, y/z). Inserting it into the second equation of (31) with X3 = b3 z we get the equation ∂H1 ∂H2 ∂H1 ∂H2 − = b3 z. ∂x ∂y ∂y ∂x After some computations we have that ∂H b2 a1 (a1 − b1 )x + a2 b2 (a1 − b1 )y + b4 (a2 − b2 )z ∂H (a2 − b2 )(b1 x + b2 y + b4 z)x − y ∂y b2 (a1 − b1 ) ∂x 1 + a2 b4 J xyzG = . a1 a1 b4 (b2 − a2 )H2 In view of relation ∂L b2 a1 (a1 − b1 )x + a2 b2 (a1 − b1 )y + b4 (a2 − b2 )z ∂L (a2 −b2 )(b1 x+b2 y+b4 z) − ≡ 0, ∂y b2 (a1 − b1 ) ∂x we obtain that the function U (x/z, y/z) satisfies the equation ∂U b2 a1 (a1 − b1 )x + a2 b2 (a1 − b1 )y + b4 (a2 − b2 )z ∂U (a2 − b2 )(b1 x + b2 y + b4 z) − ∂y b2 (a1 − b1 ) ∂x 1 + a2 b4 J xyzG = . a1 a1 b4 (b2 − a2 )H2 After the change ξ = x/z and η = y/z we obtain ∂ Ũ b2 a1 (a1 − b1 )ξ + a2 b2 (a1 − b1 )η + b4 (a2 − b2 ) ∂ Ū (a2 − b2 )(b1 ξ + b2 η + b4 )ξ − ∂η b2 (a1 − b1 ) ∂ξ 1 + a2 b4 = L̃(ξ, η)G(ξ, η, 1). a1 where Ũ = Ũ (ξ, η) := U |x=zξ, y=zη . 5.1. Applications of Corollary 13. Now we shall illustrated the applications of Corollary 13 in the determination of the second first integral in three particular cases of 3–dimensional Lotka-Volterra differential systems. More precisely, we study the existence of a second first integral H2 (i) of the integrable asymmetric May–Leonard model, (ii) of the integrable symmetric May–Leonard model. (iii) of some integrable cases for the special Lotka–Volterra systems studied in [1]. 20 J. LLIBRE, R. RAMÍREZ AND V. RAMÍREZ 5.2. Integrability of the asymmetric May–Leonard model. Determination of the second first integral. A particular 3–dimensional Lotka–Volterra system is ẋ = x (1 − x − a1 y − b1 z) = X1 , (35) ẏ = y (1 − y − a2 z − b2 x) = X2 , ż = z (1 − z − a3 x − b3 y) = X3 . This system is known as the asymmetric May–Leonard model (see for instance [5]). This model describes the competitions between three species and depending on six non–negative parameters aj and bj for j = 1, 2, 3. The state space is the set  R3+ = (x, y, z) ∈ R3 : x ≥ 0 y ≥ 0and z ≥ 0, . We shall study the integrability of system (35) under the conditions that it has an additional invariant plane and the first integral H1 satisfying (30), consequently H1 = H1 (x/y, z/y). From (33) it follows that the additional first integral H2 is a solution of the first order partial differential system   ∂H2 ∂H2 ∂H2 ∂H1 x +y +z = −Jxy((b2 − 1)x + (1 − a1 )y + (a2 − b1 )z), ∂x ∂y ∂z ∂z ∂H1 ∂H2 ∂H1 ∂H2 − = JX3 , ∂x ∂y ∂y ∂x Proposition 16. Differential system (35) under the conditions a3 (a2 − 1) − a2 b3 + 1 a3 b1 − b1 b3 + b3 − 1 (36) b2 = , a1 = , 1 − b3 a3 − 1 i.e. differential system   a3 b1 − b1 b3 + b3 − 1 ẋ = x 1 − x − y − b1 z = X̃1 , a3 − 1   (37) a3 (a2 − 1) − a2 b3 + 1 ẏ = y 1 − y − a2 z − x = X̃2 , 1 − b3 ż = z (1 − z − a3 x − b3 y) = X̃3 . has the additional invariant plane g := g(x, y, z) = (1 − a3 )(a2 − 1)x + (1 − b1 )(b3 − 1)y + (a2 − 1)(b1 − 1)z = 0, with cofactor K = 1 − x − y − z. Moreover this differential system is completely integrable with first integral H1 and the Jacobi multiplier J given by  α1 α α  β β x y 2 z 3 y 2 z 3 −3 H1 = log , J= |g| , g g g g g where α1 = (a3 − 1)(b3 − 1)(a2 − 1), α2 = (1 − a3 )(b3 − 1)(b1 − 1), α3 = (1 − a2 )(b1 − 1)(a3 − b3 ), 1 − a 2 − b1 a3 (b1 − b3 − 1) − b1 b3 β2 = , β3 = , a2 − 1 (1 − b3 )(1 − a3 ) COMPLETE INTEGRABILITY OF VECTOR FIELDS IN RN 21 Moreover the first integral H2 becomes (38)  1+β2  1+β3  y 1+β2  z 1+β3 y z H2 = −λ x+Λ(y/x, z/x) = −λ |g|+Λ(y/x, z/x), |g| |g| x x where λ is a convenient constan and Λ = Λ(y/x, z/x) is a function which satisfies the first order partial differential equation 1+β2 1+β3 ∂Λ ∂Λ ξ η (39) T2 (1, ξ, η)) η − T3 (1, ξ, η) ξ = ∂η ∂ξ g(1, ξ, η) g(1, ξ, η) where ξ = y/x and η = z/x, and ∂H1 ∂H1 ∂H1 xg = T1 (x, y, z), yg = T2 (x, y, z), zg = T3 (x, y, z), ∂x ∂y ∂z Proof. After some computations we can check that X (g) = (1 − x − y − z)g, X (H1 ) = 0, div(JX ) = 0. This completes the proof of the first statement. The partial differential system (29) for system (37) becomes ∂H2 ∂H2 T3 y − T2 z = ω1 , ∂y ∂z ∂H2 ∂H2 (40) T1 z − T3 x = ω2 , ∂z ∂x ∂H2 ∂H2 T2 x − T1 y = ω3 , ∂x ∂y where ω1 = Jyz X̃1 , ω2 = Jxz X̃2 , ω3 = Jxy X̃3 . By considering that T1 + T2 + T3 = 0 from the two first equations of (40) we get ∂H2 ∂H2 ∂H2 yz X̃1 − λxz X̃2 x +y +z = − ∂x ∂y ∂z T3 1+β2 1+β3 (41) y z = −λxyzgJ = − x g g y 1+β2 z 1+β3 = −λg = F (y/x, z/x) x x x Hence we get that H2 = F (y/x, z/x) x + Λ(y/x, z/x). Inserting H2 into the third equation of (40) and by considering that ∂F (y/x, z/x) x ∂F (y/x, z/x) x (42) T1 y − T2 x − gJxyR = F (y/x, z/x) x ∂y ∂x we have that function Λ = Λ(y/x, z/x) is a solution of the equation ∂H2 ∂H2 T1 y − T2 x = xyzJg ∂y ∂x In view of the relations   ∂H2 ∂H2 ∂H2 ∂H2 ∂H2 y =ξ , x =− ξ +η ∂y ∂ξ ∂x ∂ξ ∂η 22 J. LLIBRE, R. RAMÍREZ AND V. RAMÍREZ we deduce that (42) becomes 1+β2 1+β3 ∂H2 ∂H2 ξ η (T1 + T2 )ξ + T2 η =x ∂ξ ∂η g(1, ξ, η) g(1, ξ, η) Hence by considering that T1 +T2 = −T3 and dividing by x after some computations we obtain (39). In short the proposition is proved.  Now we shall study integrability of another particular cases of differential system (35). Proposition 17. Differential system (35) under the conditions (43) a1 = 2 − b2 , a3 = 2 − b1 , a2 = 2 − b3 , i.e. differential equations (27) have the following five invariant planes with the corresponding cofactors: g1 = x, K1 = 1 − x − (2 − b2 )y − b1 z, g2 = y, K2 = 1 − y − (2 − b3 )z − b2 x, g3 = z, K3 = 1 − z − (2 − b1 )x − b3 y, g4 = 1 − x − y − z, K4 = −x − y − z, g5 = x + y + z, K5 = 1 − x − y − z. Proof. The fourth and firth invariant planes come from the relations d d (1 − x − y − z) = −(x+y+z)(1−x−y−z), (x + y + z) = (1−x−y−z)(x+y+z). dt dt This completes the proof.  Under the assumptions of Proposition 17 and Proposition Proposition 18. Differential system (27) is completely integrable. More precisely, system (27) has the first integral H1 and Jacobi multiplier J where  b3 −1  b1 −1  b2 −1 ! |x| |y| |z| H1 = log , |x + y + z| |x + y + z| |x + y + z| 1 J= . |xyz(x + y + z − 1)| A second first integral H2 is eΨ(y/x,z/x) (44) H2 = log , |x + y + z − 1| where Ψ(y/x, z/x) is a solution of the partial differential equation   ∂Ψ (2 − b1 − b3 )(1 + ξ + η) + (b1 + b2 + b3 − 3)(1 + η ξ ∂ξ (45)   ∂Ψ + (1 − b3 )(1 + ξ + η) + (b1 + b2 + b3 − 3)η η = −(1 + ξ + η), ∂η with ξ = x/y and η = z/y. COMPLETE INTEGRABILITY OF VECTOR FIELDS IN RN 23 If b1 + b2 + b3 = 3 then H1 and H2 becomes ! b −1 b −1 z 2 x 3 H1 = log = (b2 − 1)F1 + (b3 − 1)F2 , y y (46)  z 1−b2  x b3 −1  H2 = log y|x + y + z − 1|b3 −1 y|x + y + z − 1|1−b2 = (1 − b2 )F1 + (b3 − 1)F2 ,     |x| |z| where F1 = log , and F 2 = log .. |y||x + y + z − 1|1−b2 |y||x + y + z − 1|b3 −1 Moreover differential system (27) can be written as 1 1 1 (47) ẋ = {F1 , F2 , x} , ẏ = {F1 , F2 , y} , ż = {F1 , F2 , z} . J J J Proof. Partial differential system (29) can be rewritten as (40) with Tj = Tj (x, y, z), and ωj for j = 1, 2, 3 are given by T1 = (1 − b3 )(x + y + z) + (b1 + b2 + b3 − 3)x, T2 = (1 − b1 )(x + y + z) + (b1 + b2 + b3 − 3)y, T3 = (1 − b2 )(x + y + z) + (b1 + b2 + b3 − 3)z, (x + y + z)   ω1 = − (b1 − 1)z + (1 − b2 )y + x + y + z − 1 , (x + y + z − 1) x+y+z   ω2 = − (b2 − 1)x + (1 − b3 )z + x + y + z − 1 , (x + y + z − 1) x+y+z   ω3 = − (b3 − 1)y + (1 − b1 )x + x + y + z − 1 . (x + y + z − 1) x+y+z Hence by considering that T1 + T2 + T3 = 0, and ω1 − ω2 = − T3 , then x+y+z−1 from the two first equations and the first equation of (41) we get that ∂H2 ∂H2 ∂H2 x+y+z x +y +z =− . ∂x ∂y ∂z (x + y + z − 1) By considering the equation ∂H1 ∂H1 ∂H1 x +y +z = 0, ∂x ∂y ∂z we obtain that H2 satisfies the equation ∂H2 ∂H2 ∂H2 x+y+z x +y +z =− . ∂x ∂y ∂z x+y+z−1 A particular solution of this partial differential equation is − log(x + y + z − 1). Consequently H2 can be written as H2 = − log(x + y + z − 1) + Ψ(x, y, z), whereΨ = Ψ(x, y, z) is a solution of the equation ∂Ψ ∂Ψ ∂Ψ x +y +z = 0. ∂x ∂y ∂z 24 J. LLIBRE, R. RAMÍREZ AND V. RAMÍREZ Thus Ψ = Ψ(x/y, z/y). Hence H becomes (44). Inserting into the third equation of (40) and after some computations we obtain the equation ∂Ψ ∂Ψ T1 η + (T1 + T2 )ξ = −(x + y + z), ∂η ∂ξ where ξ = x/y and η = z/y. By considering that T1 + T2 = (x + y)(b1 + b2 + b3 − 3) + (2 − b1 − b3 )(x + y + z), then we get the equation   x ∂Ψ (2 − b1 − b3 )(x + y + z) + (b1 + b2 + b3 − 3)(x + y) y ∂ξ   z ∂Ψ + (1 − b3 )(x + y + z) + (b1 + b2 + b3 − 3)z = −(x + y + z), y ∂η and by dividing this equation by y and introducing the notations ξ = x/y and η = z/y after some computations we get equation (45). Now we assume that b1 + b2 + b3 = 3. From (45) we obtain that ∂Ψ ∂Ψ (b2 − 1))ξ − (b3 − 1)η = −1. ∂ξ ∂η   A particular solution of this equation is log η 1/(2−2b2 ) ξ 1/(2b3 −2) . Thus the general solution of the previous equation is   Ψ = log η 1/(2−2b2 ) ξ 1/(2b3 −2) + Θ(ξ 1−b3 η 1−b2 ), where Θ = (ξ 1−b3 η 1−b2 ) is an arbitrary function. Taking Θ = 0 and equation (44) becomes equation (46). The representation (46) and deduction of differential system (47) are easy to obtain. In short the proposition is proved.  5.3. Determination of the second first integral H2 of the symmetric May– Leonard model. The differential system ẋ = x (1 − x − ay − bz) , (48) ẏ = y (1 − y − bx − az) , ż = z (1 − z − ax − by) , is well known as the symmetric May–Leonard model (see for instance [2]), where a and b are positive parameters. Clearly differential system (48) is obtained from (35) under the conditions bj = b and aj = a for j = 1, 2, 3. Proposition 19. Consider the symmetric May–Leonard model. (i) If a + b = 2 then this system has the first integral H1 and the Jacobi mul- tiplier J, where xyz 1 H1 = log , and J = . (x + y + z)3 |xyz(x + y + z − 1)| Moreover a second independent first integral is y z  H2 = (b − 1) log |x + y + z − 1| + Λ , , x x COMPLETE INTEGRABILITY OF VECTOR FIELDS IN RN 25 y z  where Λ = Λ , is a solution of the partial differential equation x x ∂ Λ̃ ∂ Λ̃ (49) −(2η − ξ − 1)ξ + (2ξ − η − 1)η = −(1 + ξ + η), ∂ξ ∂η where Λ̃ = Λ(ξ, η), ξ = y/x, and η = z/x. (ii) If a = b 6= 1 then y(x − z)   1−b 1 H1 = lg and J = |x|b+1 |z|2b−1 |y − x|b−2 |z − x|−b−1 , x(y − z) where H1 is a first integral and J is a Jacobi multiplier, moreover a second independent first integral is   1−b 1 y z  H2 = log |b − 1|(y − z)2 |x|b−3 |z|1−2b |y − x|2−b |z − x|b+1 +Λ , , x x where Λ is a solution of the partial differential equation ∂ Λ̃ ∂ Λ̃ (ξ − 1)ξ + η(η − 1) = −Φ1 (ξ, η), ∂ξ ∂η 2b−1 b+1 where Λ̃ = Λ(ξ, η), ξ = y/x, η = z/x and Φ1 (ξ, η) = η b−1 (η − 1) b−1 (ξ − 2−b 2 1) b−1 (ξ − η) . (iii) If a = b = 1 then y 1 H1 = log , and J = , x |xyz(x + y + z − 1)| where H1 is a first integral and J is a Jacobi multiplier, moreover a second z independent first integral is H2 = log . x Proof. For the case when a + b = 2 the symmetric May–Leonard systems has two Jacobi multiplier 1 1 J1 = , J2 = , |xyz(x + y + z − 1)| |(x + y + z)3 (x + y + z − 1)| J2 |xyz| hence H1 = log = log is a first integral. J1 |x + y + z|3 The second independent first integral can be obtained from equations (29) which, after some computations can be written as (40) with T1 = x + y + z − 3x, T2 = x + y + z − 3y, T3 = x + y + z − 3z, x+y+z ω1 = (1 − x − (2 − b)y − bz) , (x + y + z − 1) x+y+z ω2 = (1 − y − (2 − b)z − bx) , (x + y + z − 1) x+y+z ω3 = (1 − z − (2 − b)x − by) . (x + y + z − 1) By considering that T1 + T2 + T3 = 0 we get that the first of equation of system (41) becomes ∂H2 ∂H2 ∂H2 x+y+z x +y +z = (b − 1). ∂x ∂y ∂z H1 (x + y + z − 1) 26 J. LLIBRE, R. RAMÍREZ AND V. RAMÍREZ and considering the equation ∂H1 ∂H1 ∂H1 x +y +z = 0, ∂x ∂y ∂z we obtain that H2 satisfies the equation ∂H2 ∂H2 ∂H2 x+y+z x +y +z =− . ∂x ∂y ∂z x+y+z−1 The general solution of this equation is eΛ(x/y,z/y) H2 = (b − 1) log |x + y + z − 1| + Λ(x/y, z/y) = log( . |x + y + z − 1|1−b Inserting it into the second equation and introducing the notations ξ = x/y and η = z/y, we obtain ∂Λ ∂Λ (T1 + T2 )ξ + T1 η = −(x + y + z). ∂ξ ∂η Dividing this equation by y we deduce (49). So the proof of statement (i) is done. The proof of statement (ii) is the following. Symmetric May–Leonard system with a = b 6= 1 is completely integrable (see [2]). Moreover the Jacobi multiplier J and first integral H1 are   1−b 1 y(x − z) J = |x|b+1 |z|2b−1 |y − x|b−2 |z − x|−b−1 , H1 = log , x(y − z) respectively. The second independent first integral H2 = H can be obtained from equations (40) with T1 = y − z, T2 = z − x, T3 = y − x, ω1 = λ(x + b(y + z) − 1), ω2 = λ(y + b(x + z) − 1), ω3 = λ(z + b(y + x) − 1),   1−b 1 where λ = (y − z)2 xb−3 z 1−2b (y − x)2−b (z − x)b+1 . The differential system (41) in this case becomes ∂H2 ∂H2 ∂H2 x +y +z = (b − 1)λ, ∂x ∂y ∂z   ∂H2 ∂H2 ∂H2 ∂H2 ∂H2 (x − z)x + (y − z)y = −z x +y +z (50) ∂x ∂y ∂x ∂y ∂z ∂H2 ∂H2 ∂H2 +x2 + y2 + z2 ∂x ∂y ∂z = λ(b(x + y + z) − 1 − (b − 1)z). Hence in view of the first equation the second equation can be written as follows ∂H2 ∂H2 ∂H x2 + y2 + z2 = λ(b(x + y + z) − 1). ∂x ∂y ∂z By considering that the function λ = λ(x, y, z) can be rewritten as   y z b 1 1 1 1 λ = xΦ1 ( , ) = (xyz) b−1 Φ2 − , − , x x x y x z COMPLETE INTEGRABILITY OF VECTOR FIELDS IN RN 27 where   b+1 1/(b−1)  z y z z 2b−1  y 2−b  z y 2 Φ1 ( , ) = −1 −1 − , x x x x x x x 1    b+1  2−b !   1 1 1 1 1 1 1 1 b − 1 1 1 2 Φ2 − , − = − − − , x y x z x z x y z y after some computations the functions Φ1 and Φ2 satisfies the equations ∂Φ1 ∂Φ1 ∂Φ1 x +y +z = 0, ∂x ∂y ∂z ∂Φ2 ∂Φ2 ∂Φ2 x2 + y2 + z2 = 0. ∂x ∂y ∂z Consequently the function λ satisfies the following partial differential equations ∂λ ∂λ ∂λ x +y +z = λ, ∂x ∂y ∂z ∂λ ∂λ ∂λ b x2 + y2 + z2 = (x + y + z)λ. ∂x ∂y ∂z b−1 Thus the solution of the first equation of (50) is y z  H2 = (b − 1)λ(x, y, z) + Λ , , x x where Λ is a solution of the equation    ∂ ∂ ∂ y z y z  (51) x2 + y2 + z2 Λ , = −x Φ1 , = −λ(x, y, z). ∂x ∂y ∂z x x x x After the change x = x, y = xξ and z = xη and by considering that ∂Λ y ∂Λ z ∂Λ ∂Λ 1 ∂Λ ∂Λ 1 ∂Λ =− 2 − 2 , = , = , ∂x x ∂ξ x ∂η ∂y x ∂ξ ∂z x ∂η and dividing equation (51) by x we obtain, after some computations, that it becomes ∂ Λ̃ ∂ Λ̃ (ξ − 1)ξ + η(η − 1) = −Φ1 (ξ, η), ∂ξ ∂η 2b−1 b+1 2−b 2 where Λ̃ = Λ|y=xξ, z=xη and Φ1 (ξ, η) = η b−1 (η − 1) b−1 (ξ − 1) b−1 (ξ − η) . Hence statement (ii) is proved. Finally we prove statement (iii). Then the symmetric May–Leonard system with a = b = 1 is completely integrable. Moreover a Jacobi multiplier and a first integral are 1 y J= , H1 = log , |xyz(x + y + z − 1)| x respectively. The second independent first integral can be obtained from equations (29) which, after some computations, in this case becomes 1 ∂H2 1 1 ∂H2 1 1 ∂H2 1 ∂H2 1 = , = , − − = , y ∂z yz x ∂z xz x ∂y y ∂x yx 28 J. LLIBRE, R. RAMÍREZ AND V. RAMÍREZ where we have assumed that the absolute value of J is positive. If it negative the first integrals only changes the sign. The previous equations become ∂H2 1 ∂H2 ∂H2 = , x +y = −1. ∂z z ∂x ∂y The solution of these equations is s s z2 x z x x z H2 = log + Φ1 ( ) = log + log + Φ1 ( ) = log + Φ2 (H1 ), |xy| y x y y x x where Φj ( ) is an arbitrary function for j = 1, 2. Consequently a second indepen- y z dent first integral is log . In short the proposition is proved.  x 5.4. Determination of the second first integral H2 in the Lotka–Volterra system which are integrable at the origin. Now we shall apply Proposition 13 to some particular Lotka–Volterra systems having a first integral and Jacobi multiplier given in the paper [1]. Other results on the integrability of 3–dimensional Lotka-Volterra systems can be found in [3]. Proposition 20. The Lotka–Volterra differential system ẋ = x(2 + ax), ẏ = y(−1 + dx + hy + kz), ż = z(1 + gx + hy + kz), d+a−g has the first integral H1 = z(2 + ax) a /(xy) and the Jacobi multiplier J = 5 2(g−2d)−a x 2 y 3 (2 + ax) 2a . Then it has the second first integral a+2d   (2 + ax) 2a 1 z (52) H2 = √ + T ( , x) . x y y z where the function T = T ( , x) is a solution of the partial differential equation y ∂T ∂T (53) (2 + (g − d)x)η + x(2 + ax) + (x d − 1)T (η, x) = h + kη, ∂η ∂x z with η = . y Proof. In this case the partial differential system (29) becomes a+2d ∂H2 ∂H2 (2 + ax) 2a y +z + √ = 0, ∂y ∂z y x a+2d ∂H2 ∂H (2 + ax) 2a x(2 + ax) + z(2 + (g − d)x) − (−1 + dx + hy + kz) √ = 0, ∂x ∂z y x a+2d ∂H2 ∂ H2 (2 + ax) 2a x(2 + ax) − y(2 + (g − d)x) − (1 + gx + hy + kz) √ = 0, ∂x ∂y y x COMPLETE INTEGRABILITY OF VECTOR FIELDS IN RN 29 a+2d (2 + ax) 2a The solution of the first equation is H2 given by the formula H2 = √ Λ(x, y), x where Λ(x, y) is a solution of the partial differential equation ∂Λ ∂Λ 1 y +z = . ∂y ∂z y Consequently Λ = 1/y + T (z/y, x). Hence a+2d     (2 + ax) 2a 1 1 H2 = √ + T (z/y, x) := λ(x) + T (z/y, x) . x y y Inserting this expression into the second equation and by considering that λ0 (x) = λ(x) z , and introducing the notation η = we obtain the differential equation x(2 + ax) y 1 ∂T ∂T (−1 + ax + hy + gz)λ(x) λ(x)(ax−1)( +T (z/y, x))+x(2+ax) +(2+(g−d)x)η = , y ∂x ∂η y thus after simplification we get differential equation (53).  Proposition 21. The Lotka–Volterra differential system ẋ = x(1 + gx + by + kz), ẏ = y(−2 + ey), ż = z(1 + gx + hy + kz), h−b has the first integral H1 = x (1 − ey/2) /z, and the Jacobi multiplier J = e √ −3 − 2b−4h+e z y (1 − ey/2) e . Then it has the second first integral  ey  2e  2h+e 2 1−  2 1 z H2 = √ − + T ( , y) , y z x where the function T = T (z/x, y) is a solution of the partial differential equation ∂T ∂T (54) (2 − ey)yη + (b − h)η 2 y = (1 + hy)ηT + ηk + g, ∂y ∂η z with η = . x Proof. The d1ifferential system (29) in this case becomes  ey  2e 2h+e ∂H ∂H 2 1 − (2 − ey) − (h − b)z = 2 (1 + gx + by + kz), 3 ∂y ∂z zy 2  ey  2e 2h+e ∂H ∂H 2 1 − x +z = 2 , 1 ∂x ∂z zy 2  ey  2e 2h+e ∂H ∂H 2 1 − (2 − ey) + (h − b)x = 2 (1 + gx + hy + kz), 3 ∂y ∂x zy 2 From the second equation and in a similar way to the determination of function (52) we get the function H2 . Inserting this function into the first and third equation and in analogous way to the proof of Proposition 20 we get that the function T satisfies equation (54).  30 J. LLIBRE, R. RAMÍREZ AND V. RAMÍREZ Proposition 22. The Lotka–Volterra differential system ẋ = x(1 + ax + by + 2kz), ẏ = y(−2 + ey − kz), ż = z(1 + kz), has the first integral H1 = yz 2 /(1 − ey/2 + kz) and the Jacobi multiplier J = b −2 −2 b (x/z) (1 − ey/2 + kz) e (1 + kz)1− e . Then the system has the second first inte- gral   eb   ey b 1 (55) H2 = z 1 − + kz (1 + kz)1− e − + T (y, z) , 2 x where T = T (y, z) is a solution of the partial differential equation ∂T ∂H (56) y(ey − kz − 2) + z(1 + kz) + (1 + by + 2kz)T = a. ∂y ∂z Proof. System (29) in this case becomes ∂H ∂H  ey  eb b y(ey − kz − 2) + z(1 + kz) = −z 1 − + kz (1 + kz)1− e ∂y ∂z 2   1 + ax + by + 2kz) (57) , x  ey  eb b ∂H z 1 − + kz (1 + kz)1− e = 2 . ∂x x2 The solution of the second equation is H2 given in (55). Inserting this function into the first equation of (57) and by considering that ∂λ ∂λ y(ey − kz − 2) + z(1 + kz) = −(1 + by + 2kz)λ, ∂y ∂z b b where λ = z (1 − ey/2 + kz) e (1 + kz)1− e , we obtain differential equation   ∂λ ∂λ 1 y(ey − kz − 2) + z(1 + kz) (− + T )+ ∂y ∂z x   ∂T ∂T 1 + ax + by + 2kz λ y(ey − kz − 2) + z(1 + kz) = λ. ∂y ∂z x Hence after some computations we get (56).  Remark 23. Propositions 16, 17, 18, 19, 20, 21 and 22 illustrate the possibilities of the method which we propose to determine the complementary first integral in the Jacobi Theorem. To determine the final expression for H2 it is necessary to solve a partial differential equation, which in general is a non trivial problem. 5.5. On the Clebsch vector fields. In physics and mathematics the vector field ∂ ∂ ∂ (58) X = X1 + X2 + X3 ∂x1 ∂x2 ∂x3 is called solenoidal if there exists a function J such that (2) holds. Any solenoidal ∂ ∂ vector field can be expressed as curl of some other vector field Y = A +B + ∂x ∂y COMPLETE INTEGRABILITY OF VECTOR FIELDS IN RN 31 ∂ C (see for instance [17]), i.e. ∂z (59) JX = curl(Y), or equivalently ∂B ∂C ∂C ∂A ∂A ∂B (60) JX1 = − , JX2 = − , JX3 = − , ∂z ∂y ∂x ∂z ∂y ∂x In physics and mathematics the Clebsch representation of the three-dimensional vector field Y is (61) Y = gradϕ + µgradψ, where the functions ϕ = ϕ(x, y, z), µ = µ(x, y, z) and ψ = ψ(x, y, z) are known as Clebsch potentials (see for instance [6]). Proposition 24. The 3–dimensional solenoidal vector field (58) satisfying (59) with µ and ψ independent functions is completely integrable if and only if Y is a Clebsch vector field. Proof. Clearly if (61) holds then JX = curl (Y) = gradµ ∧ gradψ, where a ∧ b is the cross product of the vectors a and b. Hence   ∂µ ∂µ ∂µ JX (µ) = J X1 + X2 + X3 = 0, ∂x ∂y ∂z   ∂ψ ∂ψ ∂ψ JX (ψ) = J X1 + X2 + X3 = 0, ∂x ∂y ∂z then µ and ψ are independent first integrals, i.e. differential system (58) for N = 3 is completely integrable. The reciprocity is obtained as follows. Assume that (58) is completely integrable with first integral H1 and H2 , i.e. admits the representation (18) for N = 3, and (29) holds. By considering that     ∂H1 ∂H2 ∂H1 ∂H2 ∂ ∂H1 ∂ ∂H1 − = H2 − H2 = JX1 , ∂y ∂z ∂z ∂y ∂z ∂y ∂y ∂z     ∂H1 ∂H2 ∂H1 ∂H2 ∂ ∂H1 ∂ ∂H1 − = H2 − H2 = JX2 , ∂z ∂x ∂x ∂z ∂x ∂z ∂z ∂x     ∂H1 ∂H2 ∂H1 ∂H2 ∂ ∂H1 ∂ ∂H1 − = H2 − H2 = JX3 . ∂x ∂y ∂y ∂x ∂y ∂x ∂x ∂y Consequently from (60) we get that the functions A, B and C are such that     1 1 ∂ Φ − H1 H2 ∂ Φ + H1 H2 ∂H2 2 ∂H1 2 A = H1 + = −H2 + , ∂x ∂x ∂x ∂x     1 1 ∂ Φ − H1 H2 ∂ Φ + H1 H2 ∂H2 2 ∂H1 2 B = H1 + = −H2 + , ∂y ∂y ∂y ∂y     1 1 ∂ Φ − H1 H2 ∂ Φ + H1 H2 ∂H2 2 ∂H1 2 C = H1 + = −H2 + , ∂x ∂x ∂x ∂x 32 J. LLIBRE, R. RAMÍREZ AND V. RAMÍREZ where H1 and H2 are independent first integrals, and Φ = Φ(x, y, z) is an arbitrary function. Hence the vector field Y becomes   1 Y = H2 grad H1 + grad Φ − H1 H2 2   1 = −H1 grad H2 + grad Φ + H1 H2 . 2 Thus the vector field Y is a Clebsh vector field with Clebsh potentials H2 (or -H1 ,) 1 1 H1 (or H2 ) and Φ− H1 H2 (or Φ+ H1 H2 ). In short the proposition is proved.  2 2 ∂ ∂ ∂ Example 25. The vector field X = x(y − z) + y(z − x) + z(x − y) is ∂x ∂y ∂z solenoidal. The vector field Y in this case is Y = xyzgrad(x + y + z) + gradϕ, where ϕ is an arbitrary C 1 function. It is easy to show that xyz and x + y + z are first integral of X . Acknowledgements The first author is supported by the Ministerio de Economı́a, Industria y Com- petitividad, Agencia Estatal de Investigación grant MTM2016-77278-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. References [1] W. Aziz and C. 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Whittaker, A treatise on the Analytic Dynamics of Particles and Rigid Bodies, Dover, New York, 1944. 1 Departament de Matemàtiques, Universitat Autònoma de Barcelona, 08193 Bel- laterra, Barcelona, Catalonia, Spain. Email address: jllibre@mat.uab.cat 2 Departament d’Enginyeria Informàtica i Matemàtiques, Universitat Rovira i Vir- gili, Avinguda dels Paı̈sos Catalans 26, 43007 Tarragona, Catalonia, Spain. Email address: rafaelorlando.ramirez@urv.cat 3 Universitat Autònoma de Barcelona, 08193 Bellaterra, Barcelona, Catalonia, Spain. Email address: valentin.ramirez@e-campus.uab.cat