This is a preprint of: “Generalized analytic integrability of a class of polynomial differential systems in C 2 ”, Jaume Llibre, Yuzhou Tian, Acta Appl. Math., vol. 173(1), 1–17, 2021. DOI: [10.1007/s10440-021-00407-4] 1 GENERALIZED ANALYTIC INTEGRABILITY OF A CLASS OF POLYNOMIAL DIFFERENTIAL SYSTEMS IN C2 JAUME LLIBRE1 AND YUZHOU TIAN1,2 Abstract. This paper study the type of integrability of the differential sys- tems with separable variables ẋ = h (x) f (y), ẏ = g (y), where h, f and g are polynomials. We provide a criterion for the existence of generalized ana- lytic first integrals of such differential systems. Moreover we characterize the polynomial integrability of all such systems. In the particular case h (x) = (ax + b)m we provide necessary and sufficient conditions in order that this subclass of systems has a generalized analytic first integral. These results extend known results from [5] and [13]. Such differential systems of separable variables are important due to the fact that after a blow-up change of variables any planar quasi-homogeneous polynomial differential system can be transformed into a special differential system of separable variables ẋ = xf (y), ẏ = g (y), with f and g polynomials. 1. Introduction and the main results Planar polynomial differential systems play an important role in the qualitative theory of dynamical systems due to their many applications in physics, chemist, biology, economics, .... Nowadays the qualitative theory has a gained wide devel- opment for polynomial systems. For a planar differential system, the existence of a first integral determines completely its global dynamical behavior. So a natural problem arises: Given a polynomial differential system in R2 or C2 , how to de- cide if this system has a first integral? For general polynomial differential systems this problem is very difficult to solve. During the past three decades many mathe- maticians investigated the integrability of different classes of polynomial differential systems, such as Liénard systems [5, 13, 15, 16], Lotka-Volterra systems [7, 8, 10–12], and quasi-homogeneous polynomial systems [1–4, 6, 17], etc. Let C be the set of complex numbers and C[x] be the ring of all complex polyno- mials in the variable x. We consider the following complex polynomial differential systems (1) ẋ = h (x) f (y) , ẏ = g (y) , where h ∈ C [x] and f, g ∈ C [y] are coprime. The associated vector field of this system is ∂ ∂ (2) X = h (x) f (y) + g (y) . ∂x ∂y 1 Corresponding author: Yuzhou Tian. 2010 Mathematics Subject Classification. Primary 37C10, Secondary 34A34, 37J35. Key words and phrases. Polynomial systems, generalized analytic integrability, polynomial first integrals, residue. 1 2 JAUME LLIBRE AND YUZHOU TIAN The integer d = max {deg hf, deg g} is the degree of the vector field X . Let U be an open set of C2 . A non-locally constant function H : U → C is called a first integral of system (1) if it is constant along any solution curve of system (1) contained in U . If H (x, y) is differentiable, then H is a first integral of system (1) if and only if ∂H ∂H (3) X H = h (x) f (y) + g (y) =0 ∂x ∂y in U . System (1) has an analytic first integral if there exists a first integral H (x, y) which is an analytic function in the variables x and y. A function of the form Qk γ ϕ(y) = a i=1 (y − αi ) i is called a product function with αj , γj ∈ C and a ∈ C \ {0}. The polynomial function ϕ (y) is square-free if it can be written as ϕ (y) = Qk a i=1 (y − αi ) with a ∈ C \ {0}, αi 6= αj for i, j = 1, . . . , k and i 6= j. We say that system (1) has a generalized analytic first integral if there exists a first integral H (x, y) which is an analytic function in the variable x whose coefficients are product functions in the variable y. Let ∞ X n F (z) = an (z − z0 ) n=−∞ be a Laurent expansion at a point z0 . The coefficient a−1 = Res [F (z) , z0 ] is the residue of F (z) at z0 . Differential system (1) of separable variables has a lot of applications. For ex- ample Giné et al. in Lemma 2.2 of [7] proved that any planar quasi-homogeneous polynomial differential system can be transformed into a polynomial differential system (1) of the form (4) ẋ = xf (y) , ẏ = g (y) , with f (y) , g (y) ∈ C [y]. Hence the study of the type of integrability of the quasi- homogeneous polynomial systems can be reduced to study the type of integrability of their corresponding polynomial systems (4). Note that the polynomial differential systems (4) is a subclass of polynomial differential systems (1). In this paper we generalize some known facts for the systems (4) to systems (1), and provide other new results. First we present a necessary condition for the existence of generalized analytic first integrals of system (1). Theorem 1. Assume that α1 , · · · , αk are the different roots of the polynomial g (y). If system (1) has a generalized analytic first integral, then it must satisfy one of the following two conditions. (a) The polynomials h (x) and g (y) are square-free, and deg f < deg g. (b) The roots of the polynomial h (x) are not simple and the Res [f (y) /g (y) , αi ] = 0 for all i = 1, . . . , k. The following result is due to Llibre and Valls, see Theorem 1.1 of [14]. GENERALIZED ANALYTIC INTEGRABILITY 3 Theorem 2. Let h (x) = x. Then the polynomial differential system (1) has a generalized analytic first integral if and only if g (y) is square-free, and deg f < deg g. m In the next theorem we generalize Theorem 2 when h (x) = (ax + b) , where a ∈ C \ {0} and m ∈ N. As usual N denotes the set of positive integers. m Theorem 3. Let h (x) = (ax + b) with a ∈ C \ {0} and m ∈ N. Assume that α1 , · · · , αk are the different roots of the polynomial g (y). The following statements hold. (a) If m = 1, then system (1) has a generalized analytic first integral if and only if g (y) is square-free, and deg f < deg g. (b) If m ≥ 2, then system (1) has a generalized analytic first integral if and only if Res [f (y) /g (y) , αi ] = 0 for all i = 1, . . . , k. System (4) has a polynomial first integral if and only if g (y) is square-free, deg f < deg g and Res [f (y) /g (y) , αj ] ∈ Q− for j = 1, . . . , k, see statement (viii) of Lemma 2.4 of [6]. For the more general polynomial differential system (1) we provide necessary and sufficient conditions for its polynomial integrability in the following theorem. Theorem 4. Let α1 , · · · , αk be different roots of the polynomial g (y). System (1) has a polynomial first integral if and only if the two following conditions hold. (a) The polynomials h (x) = ax + b and g (y) is square-free, and deg f < deg g. (b) aRes [f (y) /g (y) , αj ] ∈ Q− for j = 1, . . . , k. This paper is organized as follows. We present some preliminary results in section 2. The proofs of Theorems 1, 3 and 4 are given in section 3. In section 4 we illustrate our results with some examples. 2. Preliminaries In this section we introduce some necessary lemmas for the proof of Theorems 1, 3 and 4. The following lemma can be found in many textbooks, as for instance in [9]. Lemma 5. Assume that F, G ∈ C [y] are coprime with deg F < deg G = w. Let p be the coefficient of the monomial y w of the polynomial G (y) and q the one of the monomial y w−1 of the polynomial F (y). (a) If y1 , y2 , . . . , ys are the distinct roots of G (y) with multiplicity n1 , n2 , . . . , ns , respectively, then s n F (y) X X i ti,j (5) = j , G (y) i=1 j=1 (y − yi ) where ti,1 = Res [F (y) /G (y) , yi ] and ti,ni 6= 0 for i = 1, . . . , s. 4 JAUME LLIBRE AND YUZHOU TIAN Qw (b) If G (y) is square-free that is G (y) = i=1 (y − yi ), then w F (y) X ti (6) = , G (y) i=1 y − yi where w X (7) ti = Res [F (y) /G (y) , yi ] f or i = 1, . . . , w and ti = q/p. i=1 The rational function F (y) /G (y) is a square-free rational function if it satisfies statement (b) of Lemma 5. Lemma 6. The function ϕ (y) is a product function if and only if ϕ0 (y) /ϕ (y) is a square-free rational function. Qk γ Proof. Necessity. Assume that ϕ (y) is the product function ϕ (y) = a i=1 (y − αi ) i . Then k X (8) ln ϕ (y) = ln a + γi ln (y − αi ) . i=1 Derivating equation (8) with respect to y, we get that k ϕ0 (y) X γi (9) = ϕ (y) i=1 y − αi is a square-free rational function. Hence necessity is proved. Sufficiency. Since ϕ0 (y) /ϕ (y) is a square-free rational function we have equation (9). Integrating equation (9) we get k Y γ ϕ (y) = a (y − αi ) i , i=1 where a is an integration constant. The proof of Lemma 6 is completed.  Pm−n Consider h (x) = i=0 hn+i xn+i with m ≥ n ≥ 1,hn+i ∈ C and hm hn 6= 0. If system (1) has a generalized analytic first integral H (x, y), then H (x, y) can be written as a power series in x of the form X (10) H (x, y) = aj (y) xj , j≥0 where the coefficients aj (y) are product functions in the variable y. From equation (3) we obtain m−n ! X X X XH = jf (y) aj (y) hn+i x n+i xj−1 + g (y) a0j (y) xj j≥0 i=0 j≥0 (11) m−n XX X = f (y) jhn+i aj (y) xn+i+j−1 + g (y) a0j (y) xj = 0. i=0 j≥0 j≥0 GENERALIZED ANALYTIC INTEGRABILITY 5 The equation m−n XX jhn+i aj (y) xn+i+j−1 i=0 j≥0 can be decomposed into sum of the following equations: X for i = m − n, jhm aj (y) xm+j−1 ; j≥0 X X for i = m − n − 1, jhm−1 aj (y) xm+j−2 = (j + 1) hm−1 aj+1 (y) xm+j−1 ; j≥0 j≥0 for i = m − n − 2, X X 1 X jhm−2 aj (y) xm+j−3 = (j + 2) hm−2 aj+2 (y) xm+j−1 + jhm−2 aj (y) xm+j−3 ; j≥0 j≥0 j=0 for i = m − n − 3, X X 2 X jhm−3 aj (y) xm+j−4 = (j + 3) hm−3 aj+3 (y) xm+j−1 + jhm−3 aj (y) xm+j−4 ; j≥0 j≥0 j=0 for i = m − n − 4, X X 3 X jhm−4 aj (y) xm+j−5 = (j + 4) hm−4 aj+4 (y) xm+j−1 + jhm−4 aj (y) xm+j−5 ; j≥0 j≥0 j=0 .. . for i = 2, X X jhn+2 aj (y) xn+j+1 = (m − n + j − 2) hn+2 am−n+j−2 (y) xm+j−1 + j≥0 j≥0 m−n−3 X jhn+2 aj (y) xn+j+1 ; j=0 for i = 1, X X jhn+1 aj (y) xn+j = (m − n + j − 1) hn+1 am−n+j−1 (y) xm+j−1 + j≥0 j≥0 m−n−2 X jhn+1 aj (y) xn+j ; j=0 for i = 0, X X m−n−1 X jhn aj (y) xn+j−1 = (m − n + j) hn am−n+j (y) xm+j−1 + jhn aj (y) xn+j−1 ; j≥0 j≥0 j=0 X X m−2 X and g (y) a0j (y) xj = g (y) a0m+j−1 (y) xm+j−1 + g (y) a0j (y) xj . j≥0 j≥0 j=0 6 JAUME LLIBRE AND YUZHOU TIAN Then equating the coefficients of xj in (11) we get the equations that aj (y) must satisfy: (12) a0j (y) = 0 for j = 0, . . . , n − 1; j−n+1 f (y) X a0j (y) = − ihj−i+1 ai (y) for j = n, . . . , m − 2; g (y) i=1 m−n f (y) X a0j (y) = − (j + i − m + 1) hm−i f (y) aj+i−m+1 (y) for j ≥ m − 1. g (y) i=0 Remark 7. Note that a0 (y) is a constant. In the following we can assume a0 (y) = 0, because a first integral does not depend on the sum of an additional constant. The solutions of equations (12) are characterized by the following two lemmas. Pm−n Lemma 8. Let h (x) = i=0 hn+i xn+i with hn+i ∈ C and hm hn 6= 0. Assume that n = 1 and that the differential polynomial system (1) has a generalized analytic first integral (10). Then the following statements hold. (a) There exist polynomials Fj (u) with deg Fj = j and Fj (0) = 0 such that   Z  f (y) aj (y) = Fj exp −h1 dy for all j ∈ N. g (y) (b) The polynomial g (y) is square-free and deg f < deg g. Proof. (a) When n = 1 equations (12) can be rewritten as j−1 f (y) f (y) X (13) a0j (y) = −jh1 aj (y) − ihj−i+1 ai (y) for j = 1, . . . , m − 2, g (y) g (y) i=1 and (14) m−2 f (y) f (y) X a0j (y) = −jh1 aj (y) − (j + i − m + 1) hm−i f (y) aj+i−m+1 (y) , g (y) g (y) i=0 for j ≥ m − 1. For m = 2 we only need to study equation (14) that is f (y) f (y) (15) a0j (y) = −jh1 aj (y) − (j − 1) h2 aj−1 (y) , g (y) g (y) with j ≥ 1. When j = 1 the solution of equation (15) is  Z  f (y) (16) a1 (y) = C1 exp −h1 dy , g (y) where C1 is an integration constant. Obviously F1 (u) = C1 u. Hence for j = 1 statement (a) holds. GENERALIZED ANALYTIC INTEGRABILITY 7 Assume that there exists polynomial Fj (u) with deg Fj = j and Fj (0) = 0 such that   Z  f (y) aj (y) = Fj exp −h1 dy . g (y) By the induction hypothesis and equation (15) we have (17)   Z  f (y) f (y) f (y) a0j+1 (y) = − (j + 1) h1 aj+1 (y) − jh2 Fj exp −h1 dy . g (y) g (y) g (y) The solution of the linear differential equation (17) is (18)  Z  f (y) aj+1 (y) = exp − (j + 1) h1 dy Cj+1 g (y) Z  Z    Z   f (y) f (y) f (y) − jh2 exp (j + 1) h1 dy Fj exp −h1 dy dy . g (y) g (y) g (y)  R (y)  Let u = exp −h1 fg(y) dy . Then du f (y) (19) = −h1 dy. u g (y) So equation (18) can be written as  Z  j+1 jh2 1 (20) aj+1 (y) = Fj+1 (u) = u Cj+1 + F (u) du . j+2 j h1 u Using the induction hypothesis deg Fj = j and Fj (0) = 0, we get deg Fj+1 = j + 1 and Fj+1 (0) = 0. The induction is proved and statement (a) follows for m = 2. For m ≥ 3 we need to consider equations (13) and (14), that is j−1 f (y) f (y) X (21) a0j (y) = −jh1 aj (y) − ihj+1−i ai (y) for j = 1, . . . , m − 2, g (y) g (y) i=1 and m−2 f (y) f (y) X (22) a0j (y) = −jh1 aj (y) − (j + i − m + 1) hm−i aj+i−m+1 (y) , g (y) g (y) i=0 for j ≥ m − 1. For j = 1 equation (21) becomes a01 (y) f (y) (23) = −h1 . a1 (y) g (y) It is easy to get that  Z  f (y) a1 (y) = C1 exp −h1 dy , g (y) where C1 is an integration constant. Let F1 (u) = C1 u. So statement (a) holds for j = 1. 8 JAUME LLIBRE AND YUZHOU TIAN Assume that for j = 1, . . . , l there exist polynomials Fj (u) with deg Fj = j such that   Z  f (y) aj (y) = Fj exp −h1 dy . g (y) Next we consider j = l + 1. If l + 1 ≤ m − 2, then l f (y) f (y) X (24) a0l+1 (y) = −(l + 1)h1 al+1 (y) − ihl+2−i Fi (u) , g (y) g (y) i=1  R  f (y) with u = exp −h1 g(y) dy . The solution of the linear differential equation (24) is l Z ! X f (y) Fi (u) (25) al+1 (y) = ul+1 Cl+1 − ihl+2−i dy , i=1 g (y) ul+1  R  f (y) with u = exp −h1 g(y) dy . From equation (19) it follows that l Z ! X ihl+2−i Fi (u) l+1 al+1 (y) = Fl+1 (u) = u Cl+1 + du . i=1 h1 ul+2 By the induction hypothesis deg Fj = j and Fj (0) = 0 for j = 1, . . . , l, we obtain deg Fl+1 = l + 1 and Fl+1 (0) = 0. If l + 1 ≥ m − 1, then (26) m−2 f (y) f (y) X a0l+1 (y) = −(l + 1)h1 al+1 (y) − (l + i − m + 2) hm−i Fl+i−m+2 (u) , g (y) g (y) i=0  R (y)  with u = exp −h1 fg(y) dy . By the same arguments as above one can get that (27) m−2 Z ! X (l + i − m + 2) hm−i Fl+i−m+2 (u) l+1 al+1 (y) = Fl+1 (u) = u Cl+1 + du . i=0 h1 ul+1 Applying the induction hypothesis deg Fj = j and Fj (0) = 0 for j = 1, . . . , l, we have deg Fl+1 = l + 1 and Fl+1 (0) = 0. The proof of statement (a) is done. (b) Let aj (y) = constant = Cj for j ∈ N. From statement (a) we know that there exists a polynomial Fj (u) such that  Z  f (y) aj (y) = Fj (u) = Cj with u = exp −h1 dy and Fj (0) = 0. g (y) Thus Cj = 0. Since the first integral H (x, y) is a non-locally constant function, there exists a positive integer j0 such that aj0 (y) is not a constant and ai (y) = 0 for i = 1, . . . , j0 − 1. Using equations (13) and (14) we have f (y) a0j0 (y) = −j0 h1 aj (y) . g (y) 0 GENERALIZED ANALYTIC INTEGRABILITY 9 Consequently  Z  f (y) (28) aj0 (y) = Cj0 exp −j0 h1 , g (y) with constant Cj0 6= 0. From Lemma 6 we get that aj0 (y) if and only if f (y) /g (y) is a square-free rational function. So g (y) is square-free and deg f < deg g. This completes the proof of this lemma.  Pm−n Lemma 9. Let h (x) = i=0 hn+i x n+i with hn+i ∈ C and hm hn 6= 0. Assume that n ≥ 2 and that the polynomial differential system (1) has a generalized analytic first integral (10), and α1 , · · · , αk are the different roots of the polynomial g (y). The following statements hold. (a) There exist polynomials Fj (u) such that Z  f (y) aj (y) = Fj dy , g (y) for j ≥ n, and aj (y) are constants for j = 1, . . . , n − 1. (b) Then Res [f (y) /g (y) , αi ] = 0 for all i = 1, . . . , k. Proof. (a) From equations (12) we have that (29) a0j (y) = 0 for j = 1, . . . , n − 1; j−n+1 f (y) X (30) a0j (y) = − ihj−i+1 ai (y) for j = n, n + 1, . . . , m − 2; g (y) i=1 and m−n f (y) X (31) a0j (y) = − (j + i − m + 1) hm−i aj+i−m+1 (y) for j ≥ m − 1. g (y) i=0 Then aj (y) = constant = Cj for j = 1, . . . , n − 1. For j = n equation (30) can be written as f (y) f (y) a0n (y) = −hn a1 (y) = −hn C1 . g (y) g (y) We get Z f (y) an (y) = −hn C1 dy + Cn , g (y) where Cn is an integration constant. Let Fn (u) = −hn C1 u + Cn . Thus statement (a) holds for j = n. The constants aj (y) = Cj for j = 1, . . . , n − 1 can be regarded as polynomials of degree 0. Assume that for j = 1, . . . , l there exist polynomials Fj (u) such that Z  f (y) aj (y) = Fj dy . g (y) If n ≤ l + 1 ≤ m − 2, then function al+1 (y) satisfy l−n+2 Z  f (y) X f (y) (32) a0l+1 (y) = − ihl−i+2 Fi dy . g (y) i=1 g (y) 10 JAUME LLIBRE AND YUZHOU TIAN R Let u = (f (y) /g (y)) dy. Note that du = (f (y) /g (y)) dy. Therefore al+1 (y) = Fl+1 (u) l−n+2 X Z Z  f (y) f (y) = − ihl−i+2 Fi dy dy i=1 g (y) g (y) l−n+2 X Z = − ihl−i+2 Fi (u) du. i=1 If l + 1 ≥ m − 1, then m−n f (y) X a0l+1 (y) = − (l + i − m + 2) hm−i Fl+i−m+2 (u) . g (y) i=0 Using similar arguments we obtain m−n X Z al+1 (y) = Fl+1 (u) = − (l + i − m + 2) hm−i Fl+i−m+2 (u) du. i=0 Therefore statement (a) is proved. (b) Suppose that aj (y) = constant = Cj for all j ≥ n. Then the first integral H (x, y) is independent of the variable y. This implies that ẋ = 0, which is a contradiction. Therefore there exists a positive integer j0 ≥ n such that aj0 (y) is not a constant and ai (y) = constant = Ci for i = 1, . . . , j0 − 1. From equations (30) and (31) we obtain that f (y) (33) a0j0 (y) = C , g (y) Pj−n+1 Pm−n where C = − i=1 ihj−i+1 Ci or i=0 (j + i − m + 1) hm−i Cj+i−m+1 . Hence Z f (y) (34) aj0 (y) = C dy + Cj0 , g (y) with constant C 6= 0. Since aj0 (y) is a product function, by Lemma 6, we get that C f (y) R f (y) C g (y) g(y) dy + C j0 is a square-free rational function. This implies that Z f (y) (35) dy g(y) is a rational function. We know that there exist two polynomials p (y) , r (y) ∈ C [y] such that f (y) = p (y) g (y) + r (y) and deg r < deg g. The polynomial r (y) cannot be zero due to the fact that f (y) and g (y) are coprime. Consequently Z Z f (y) r (y) (36) dy = Q (y) + dy, g (y) g (y) GENERALIZED ANALYTIC INTEGRABILITY 11 with Q0 (y) = p (y). Assume that α1 , . . . , αk are the distinct roots of g (y) with mul- tiplicity n1 , n2 , . . . , nk , respectively. Using Lemma 5 r (y) /g (y) can be expressed as k n r (y) X X i ci,j (37) = j , g (y) i=1 j=1 (y − αi ) where ci,ni 6= 0 for i = 1, . . . , k. Thus Z k ! k X ni Z r (y) Y ci,1 X ci,j (38) dy = ln (y − αi ) + j dy. g (y) i=1 i=1 j=2 (y − αi ) Since equation (35) is a rational function and Q (y) (see equation (36)) is polyno- mial, equation (38) is also a rational function. Note that j ≥ 2 in equation (38). This implies that Z ci,j j dy (y − αi ) is a rational function. Then ci,1 must be 0, that is Res [f (y) /g (y) , αi ] = 0 for all i = 1, . . . , k. The proof is done.  3. Proofs of Theorems 1, 3 and 4 The main purpose of this section is to prove Theorems 1, 3 and 4. Proof of Theorem 1. We claim that if h (x) simultaneously has simple roots and multiple roots, then system (1) has no generalized analytic first integral. Let β1 and β2 be a simple root and a root of multiplicity n of h (x) with n ≥ 2, respectively. Assume that system (1) has a generalized analytic first integral. Doing the change of variables (x, y, t) 7→ (x + β1 , y, t), system (1) becomes (39) ẋ = h̃ (x) f (y) , ẏ = g (y) , Pm i=1 h̃i x with h̃i ∈ C and h̃m h̃1 6= 0. Since system (1) has a i where h̃ (x) = generalized analytic first integral, system (39) also has a generalized analytic first integral. By Lemma 8 we get that g (y) is square-free and deg f < deg g. This means that Res [f (y) /g (y) , αi ] 6= 0 for all i = 1, . . . , k. Under the transformation (x, y, t) 7→ (x + β2 , y, t) system (1) changes to (40) ẋ = h̄ (x) f (y) , ẏ = g (y) , Pm where h̄ (x) = i=n h̄i xi with h̄i ∈ C and h̄m h̄n 6= 0. From Lemma 9 it follows that Res [f (y) /g (y) , αi ] = 0 for all i = 1, . . . , k. This is in contradiction with Res [f (y) /g (y) , αi ] 6= 0 for all i = 1, . . . , k. So the claim is proved. In summary, the polynomial h (x) is square-free or it has no simple roots. If h (x) is square-free, using Lemma 8, we obtain statement (a). If h (x) has no simple roots, by Lemma 9, statement (b) holds. This completes the proof of the theorem.  12 JAUME LLIBRE AND YUZHOU TIAN Proof of Theorem 3. Doing the change of variables (x, y, t) 7→ ((x − b) /a, y, t/a), system (1) becomes 1 (41) ẋ = xm f (y) , ẏ = g (y) . a (a) From Theorem 2 it follows that statement (a) holds. (b) Necessity. Using statement (a) of Theorem 1 the necessity is obvious. Sufficiency. It is sufficient to show that system (41) has a generalized analytic first integral. Assume that α1 , . . . , αk are the distinct roots of g (y) with multiplicity n1 , n2 , . . . , nk , respectively. There exist two polynomials p (y) , r (y) ∈ C [y] such that f (y) = p (y) g (y) + r (y) and deg r < deg g. The polynomial r (y) cannot be zero due to the fact that f (y) and g (y) are coprime. By Lemma 5 we have Xk X ni f (y) r (y) ci,j (42) = p (y) + = p (y) + j , g (y) g (y) i=1 j=1 (y − αi ) where ci,ni 6= 0 for i = 1, . . . , k. Since Res [f (y) /g (y) , αi ] = 0 that is ci,1 = 0 for all i = 1, . . . , k, we obtain Z Z ni Z k X X f (y) ci,j (43) P (y) := dy = p (y) dy + j dy. g (y) i=1 j=2 (y − αi ) Note that j ≥ 2 in equation (43). Thus P (y) is a rational function, that is, a product function. Now we show that (m − 1) xm−1 (44) H (x, y) = 1 + a (m − 1) xm−1 P (y) is a generalized analytic first integral of system (41). Doing simple computations we have 2 2 ∂H (m − 1) xm−2 ∂H a (m − 1) x2m−2 f (y) = 2 and =− 2 g (y) . ∂x (1 + a (m − 1) xm−1 P (y)) ∂y (1 + a (m − 1) x m−1 P (y)) Therefore ∂H 1 ∂H X H = xm f (y) + g (y) = 0. ∂x a ∂y Moreover H (x, y) can be written as a power series in x (m − 1) xm−1 X j H (x, y) = m−1 = (m − 1) xm−1 aj (1 − m) P j (y) x(m−1)j . 1 − a (1 − m) x P (y) j≥0 This completes the proof of the theorem.  Proof of Theorem 4. Necessity. We claim that h (x) is square-free. GENERALIZED ANALYTIC INTEGRABILITY 13 Let β1 , . . . , βl be different roots of the polynomial h (x). Suppose that β is an arbitrary root of the polynomial h (x) with multiplicity n. By changing the variables (x, y, t) 7→ (x + β, y, t), system (1) is equivalent to (45) ẋ = h̃ (x) f (y) , ẏ = g (y) , Pm where h̃ (x) = i=n h̃i xi with h̃i ∈ C and h̃m h̃n 6= 0. Note that h̃n = h(n) (β)/n!. Since system (1) has a polynomial first integral, system (45) also has a polynomial first integral, that is X (46) H (x, y) = aj (y) xj , j≥0 where aj (y) are polynomials. Obviously H (x, y) is a generalized analytic first integral. From the proof of Lemmas 8 and 9 we know that there exists a positive integer j0 such that aj0 (y) is not a constant, and ai (y) = constant = Ci for i = 1, . . . , j0 − 1. Assume that n ≥ 2. From the proof of statement (b) of Lemma 9, we get Z f (y) (47) aj0 (y) = C dy + Cj0 , g (y) with constant C 6= 0 (see equation (34)). By equation (38) aj0 (y) is not a polyno- mial. Thus β is a simple root of h (x), that is n = 1. Using Theorem 1 we obtain that the polynomials h (x) and g (y) are square-free, and deg f < deg g. Hence the claim is proved. From the proof of statement (b) of Lemma 8 we have  Z  f (y) (48) aj0 (y) = Cj0 exp −j0 h̃1 , g (y) with h̃1 = h0 (β) and the constant Cj0 6= 0 (see equation (28)). Applying Lemma 5 f (y) /g (y) can be expressed as k f (y) X µj (49) = , g (y) j=1 y − αj where µj = Res [f (y) /g (y) , αj ] for j = 1, . . . , k. Therefore k Y −j0 h̃1 µj (50) aj0 (y) = Cj0 (y − αi ) . j=1 Since aj0 (y) is a polynomial we have h̃1 µj ∈ Q− for all j = 1, . . . , k. Note that β is an arbitrary root of the polynomial h (x). Thus (51) h0 (βi ) · Res [f (y) /g (y) , αj ] ∈ Q− for i = 1, . . . , l and j = 1, . . . , k. This means that h0 (β1 ) /h0 (βi ) ∈ Q+ for i = 1, . . . , l. Assume that l ≥ 2. Using statement (b) of Lemma 5 1/h (x) can be written as X ti l 1 = , h (x) i=1 x − βi 14 JAUME LLIBRE AND YUZHOU TIAN with ti = Res [1/h (x) , βi ] = 1/h0 (βi ) 6= 0. From equation (7) we obtain l X l X 1 (52) ti = = 0. i=1 i=1 h0 (βi ) One can get l X h0 (β1 ) = 0, i=1 h0 (βi ) which is in contradiction with h0 (β1 ) /h0 (βi ) ∈ Q+ for i = 1, . . . , l. So l = 1, that is h (x) = ax + b with a ∈ C \ {0}. Then equation (51) becomes aRes [f (y) /g (y) , αj ] ∈ Q− for j = 1, . . . , k. This proves the necessity. Sufficiency. Let µj = Res [f (y) /g (y) , αj ] and consider     a1 Y k (53) e (x, y) = x + b H  −µ (y − αj ) j  . a j=1 Since aµj ∈ Q− there exists a positive integer N such that    aN  b N Y k e (x, y)  −µ aN (54) H (x, y) = H = x+ (y − αj ) j  a j=1 is a polynomial. Next we show that polynomial (54) is a first integral of system e (x, y) is a first integral of system (1). (1). In fact it is sufficient to prove that H Straightforward computations show that e ∂H e (x, y) H (55) = , ∂x ax + b and   e X k ∂H e (x, y)  µj . (56) = −H ∂y j=1 y − αj The polynomial g (y) is square-free with deg f < deg g. Using Lemma 5 we have k f (y) X µj (57) = . g (y) j=1 y − αj Equation (56) can be written as e ∂H e (x, y) f (y) . = −H ∂y g (y) Thus e e e = (ax + b) f (y) ∂ H + g (y) ∂ H = 0. XH ∂x ∂y That is He (x, y) is a first integral of system (1). This completes the proof of the theorem.  GENERALIZED ANALYTIC INTEGRABILITY 15 4. Examples In this section we present some applications of our results. Example 10. Consider the differential system 2 (58) ẋ = (x − 1) (x − 2) y, ẏ = y + 1. It has a first integral   x−2 1 H (x, y) = exp y + . (x − 1) (y + 1) x−2 By Theorem 1 system (58) has no generalized analytic first integral, because h (x) = 2 (x − 1) (x − 2) simultaneously has simple roots and multiple roots. Example 11. Consider the differential system m  (59) ẋ = (2x − 1) y 3 − 8y 2 + 29y − 26 , ẏ = (y − 3)3 (y + 1)2 , with m ∈ N and m ≥ 2. For this system we have f (y) = y 3 − 8y 2 + 29y − 26, g (y) = (y − 3)3 (y + 1)2 , α1 = −1 and α2 = 3. So Res [f (y) /g (y) , α1 ] = Res [f (y) /g (y) , α2 ] = 0. Applying Theorem 3 system (59) has the generalized analytic first integral (see equation (44)) 2 m−1 2(m − 1)(y − 1) (y − 3) (2x − 1) H (x, y) = 2 m−1 . (y + 1) (y − 3) − (m − 1) (2y 2 − 11y + 19) (2x − 1) Example 12. Consider the differential system  √ √   √  √  3 3 (60) ẋ = (5x − 1) 3 3 + 2 2 − 5y , ẏ = 6 y − 2 y − 3 . √  √  Using the notations of Theorem 4 we get that g (y) = 6 y − 2 y − 3 3 is square- √ √ √ √ free, α1 = 2, α2 = 3 3, h (x) = 5x − 1 and f (y) = 3 3 3 + 2 2 − 5y. For this system we have 5 5 5Res [f (y) /g (y) , α1 ] = − and 5Res [f (y) /g (y) , α2 ] = − . 2 3 By Theorem 4 system (60) has the polynomial first integral  6 1  √ 15  √3 10 H (x, y) = x − y− 2 y− 3 . 5 Acknowledgements 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 Founda- tion of China (No. 11971495 and No. 11801582), China Scholarship Council (No. 201906380022) and Natural Science Foundation of Guangdong Province (No. 2019A1515011239). 16 JAUME LLIBRE AND YUZHOU TIAN References [1] A. Algaba, E. Freire, E. Gamero, C. Garcı́a, Monodromy, center-focus and integrability problems for quasi-homogeneous polynomial systems, Nonlinear Anal., 72 (2010), 1726– 1736. [2] A. Algaba, C. Garcı́a and J. Giné, Analytic integrability of some examples of degenerate planar vector fields, Acta Appl. Math. 141 (2016), 1–15. [3] A. Algaba, C. Garcı́a, M. Reyes, Rational integrability of two-dimensional quasi- homogeneous polynomial differential systems, Nonlinear Anal., 73 (2010), 1318–1327. [4] B. Garcı́a, J. Llibre, J. S. Pérez del Rı́o, Planar quasi-homogeneous polynomial differ- ential systems and their integrability, J. Differential Equations, 255 (2013), 3185–3204. [5] A. Gasull, J. Giné, Integrability of Liénard systems with a weak saddle, Z. Angew. Math. Phys., 68 (2017), Art. 13, pp. 7. [6] J. Giné, M. Grau, J. Llibre, Polynomial and rational first integrals for planar quasi- homogeneous polynomial differential systems, Discrete Contin. Dyn. Syst., 33 (2013), 4531– 4547. [7] J. Giné, V. G. Romanovski, Integrability conditions for Lotka-Volterra planar complex quintic systems, Nonlinear Anal. Real World Appl., 11 (2010), 2100–2105. [8] S. Gravel, P. Thibault, Integrability and linearizability of the Lotka-Volterra system with a saddle point with rational hyperbolicity ratio, J. Differential Equations, 184 (2002), 20–47. [9] S. Lang, Complex analysis, vol. 103 of Graduate Texts in Mathematics, Springer-Verlag, New York, fourthed., 1999. [10] C. Liu, G. Chen, C. Li, Integrability and linearizability of the Lotka-Volterra systems, J. Differential Equations, 198 (2004), 301–320. [11] C. Liu, Y. Li, The integrability of a class of cubic Lotka-Volterra systems, Nonlinear Anal. Real World Appl., 19 (2014), 67–74. [12] J. Llibre, R. Ramı́rez, V. Ramı́rez, Integrability of a class of N -dimensional Lotka-Volterra and Kolmogorov systems, J. Differential Equations, (2020), https://doi.org/10.1016/j.jde.2020.02.001. [13] J. Llibre, C. Valls, On the local analytic integrability at the singular point of a class of Liénard analytic differential systems, Proc. Amer. Math. Soc., 138 (2010), 253–261. [14] J. Llibre, C. Valls, Analytic integrability of a class of planar polynomial differential sys- tems, Discrete Contin. Dyn. Syst. Ser. B, 20 (2015), 2657–2661. [15] J. Llibre, C. Valls, Liouvillian first integrals for a class of generalized Liénard polynomial differential systems, Proc. Roy. Soc. Edinburgh Sect. A, 146 (2016), 1195–1210. [16] J. Llibre, C. Valls, On the analytic integrability of the Liénard analytic differential sys- tems, Discrete Contin. Dyn. Syst. Ser. B, 21 (2016), 557–573. [17] J. Llibre, X. Zhang, Polynomial first integrals for quasi-homogeneous polynomial differ- ential systems, Nonlinearity, 15 (2002), 1269–1280. 1 Departament de Matemátiques, Universitat Autónoma de Barcelona, 08193 Bel- laterra, Barcelona, Catalonia, Spain Email address: jllibre@mat.uab.cat 2 School of Mathematics (Zhuhai), Sun Yat-sen University, Zhuhai 519082, P.R. China Email address: tianyzh3@mail2.sysu.edu.cn