ANALYTIC INTEGRABILITY OF A CLASS OF PLANAR POLYNOMIAL DIFFERENTIAL SYSTEMS

. In this paper we ﬁnd necessary and suﬃcient conditions in order that the diﬀerential systems of the form ˙ x = xf ( y ), ˙ y = g ( y ), with f and g polynomials, have a ﬁrst integral which is analytic in the variable x and meromorphic in the variable y . We also characterize their analytic ﬁrst integrals in both variables x and y . These polynomial diﬀerential systems are important because after a con- venient change of variables they contain all quasi–homogeneous polynomial diﬀerential systems in R 2 .

1. Introduction and statement of the main results. Let C be the set of complex numbers and C[y] the ring of all polynomials in the variable y with coefficients in C. In this paper we consider the polynomial differential systems of the forṁ x = xf (y),ẏ = g(y), where f, g ∈ C[y] and are coprime. The dot denotes the derivative with respect to the independent variable t real or complex. We denote by X = (xf (y), g(y)) the polynomial vector field associated to system (1), and we say that the degree of the system is n = max{deg xf (y), deg g(y)}. For the sake of simplicity, we assume for the rest of the paper that system (1) is not linear, that is n > 1. We recall that given a planar polynomial differential system (1), we say that a function H : U ⊂ C 2 → C with U an open set, is a first integral of system (1)  in U. We call the integrability problem the problem of finding such a first integral and the functional class where it belongs. We say the system 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. We say that the system has a pseudo-meromorphic first integral if there exists a first integral H(x, y) which is an analytic function in the variable x and a meromorphic function in the variable y. The aim of this paper is to characterize the existence of first integrals of system (1) that can be described by functions that are analytic or pseudo-meromorphic.
Let α l for l = 1, . . . , k be the zeros of g. We say that g is square-free if g(y) = k l=1 (y − α l ) with α l = α j for l, j = 1, . . . , k and l = j. When g is square-free we define γ l = f (α l )/g (α l ) for l = 1, . . . , k. With this notation we introduce the main result of the paper. Theorem 1.1. System (1) has a pseudo-meromorphic first integral if and only if g(y) is square-free. Moreover, if γ l < 0 for all l = 1, . . . , k then the first integral is analytic, otherwise it is a pseudo-meromorphic function with poles on y = α l if γ l < 0.
The proof of Theorem 1.1 is given in section 2. Furthermore, the specific form of the first integral is given in the proof of Theorem 1.1. Example 1. Consider the differential systeṁ x = xy 3 ,ẏ = y + 1.
Note that g(y) = y + 1 is square-free, α 1 = −1 and γ 1 = −1 < 0. Example 2. onsider the differential systeṁ This system has the pseudo-meromorphic first integral Note that g(y) = y − 1 is square-free, α 1 = 1 and γ 1 = 1 > 0. System (1) is of separate variables and appears in many situations. In Lemma 2.2 of [1] it is proved that there exists a blow-up change of variables that transforms any quasi-homogeneous polynomial differential system into a differential system (1). However we point out that not all the planar polynomial differential systems (1) come from quasi-homogenous polynomial differential systems. We recall that a polynomial differential systeṁ is quasi-homogeneous if there exists s 1 , s 2 , d ∈ N (here N denotes the set of positive integers) such that for arbitrary α ∈ C, P (α s1 x, α s2 y) = α s1−1+d P (x, y), Q(α s1 x, α s2 y) = α s2−1+d Q(x, y).
From Theorem 3.1b) of [1] and Proposition 1 of [3] it follows the next result.
Note that Theorem 1.1 extends the result of Theorem 1.2 only valid for the quasihomogeneous polynomial differential systems. Recall that quasi-homogeneous polynomial differential systems can be written as a subclass of the polynomial differential systems (1) using the Lemma 2.2 of [1].
2. Proof of Theorem 1.1. Assume that system (1) has a pseudo-meromorphic first integral. Then it can be written as a power series in x in the form where a l (y) is a meromorphic function in the variable y. Then, it must satisfy Hence, a 0 (y) = 0 that is a 0 (y) = constant and for l ≥ 1, lf (y)a l (y) + g(y)a l (y) = 0 that is a l (y) a l (y) = −lf (y) g(y) .
If deg f ≥ deg g and we consider the division of −lf (y) by g(y) we can write lf (y) = q(y)g(y) + r(y), where r(y) cannot be zero taking into account that f and g are coprime and deg r < deg g. Hence equation (4) takes the form a l (y) a l (y) = −q(y) − r(y) g(y) .

Integrating this equation we have
where C l is a constant of integration and Q (y) = q(y). Note that inserting a l (y) in the function H given in (3) we get Since we are assuming that H is a pseudo-meromorphic first integral, it must be a meromorphic function in the variable y, i.e., must be a meromorphic function in the variable y. Therefore, since the first factor of (6) is an analytic function, we must study the second factor in (6).
Assume that g is not square free. Using an affine transformation of the form z = y + α with α ∈ C if it is necessary, we can assume that z is a multiple of g, that is,g(z) = z m R(z), whereg(z) = g(z − α) with m > 1 an integer and R(0) = 0. Since f and g are coprime we also haver(0) = 0, wherer(z) = r(z − α). Now we developr(z)/g(z) in simple fractions of z, that is, where α(z) is a polynomial with deg α(z) < deg R(z), and c i ∈ C for i = 1, . . . , m.
Note that c m = 0. Therefore integrating this last expression we have Note that the first exponential factor cannot be simplified by any part of the second exponential factor. Moreover c m = 0 and we get a contradiction with the fact that the left hand side must be a meromorphic function in the variable y while exp(c m /((1 − m)z m−1 ) has an essential singularity at z = 0, and this it is not meromorphic in z. Therefore, we conclude that g(y) is square-free. Hence we write Then, and, consequently, Note that this expression is always a meromorphic function. If γ j > 0 for all j = 1, . . . , k then it is an analytic function in the variable y, otherwise it is meromorphic with poles on the α j such that γ j < 0. Hence a k (y) is an analytic function in y if γ j < 0 for j = 1, . . . , k, and it is meromorphic with poles on the α j with γ j > 0. Conversely, assume that g is square-free and that f (y) = q(y)g(y) + r(y). We will show that with γ i = r(α i )/g (α i ) for i = 1, . . . , k is a pseudo-meromorphic function, and it is analytic if all γ j < 0 for j = 1, . . . , k. Note that the function H defined in (8) is just the function H in (3) with only a term different from zero when l = 1, and thus satisfying (4) with l = 1 and (5) with Q(y) = q(y) dy, and with e r(z) g(z) dz as in (7). Now we show that indeed it is a first integral of system (1). We set φ(y) = (y − α 1 ) γ1 · · · (y − α k ) γ k . Note that 0 = xf (y) ∂H ∂x + g(y) ∂H ∂y = xf (y)e − q(y) dy φ(y) + xg(y)(−q(y)φ(y) − φ (y))e − q(y) dy = xe − q(y) dy f (y)φ(y) − g(y)q(y)φ(y) − g(y)φ (y) = xe − q(y) dy r(y)φ(y) − g(y)φ (y) .
To see that this last expression is identically zero it is equivalent to see that φ (y) φ(y) = r(y) g(y) .
Recalling the expression of φ(y) we have Taking common denominator and recalling that g(y) = c(y −α 1 )(y −α 2 ) · · · (y −α k ) we obtain φ (y) φ(y) = c g(y) Now substituting the values of γ i = r(α i )/g (α i ) and taking into account that The last expression in the sum, recalling that deg r < deg g, is the expression of the Lagrange polynomial which interpolates r(y) in the k points (α i , r(α i )), for i = 1, 2, . . . , k, see for more details [2]. Therefore this polynomial is r(y), and we conclude that the expression (9) is satisfied. This completes the proof of the theorem.