Liouvillian first integrals for a class of generalized Liénard polynomial differential systems

We study the existence of Liouvillian first integrals for the generalized Liénard polynomial differential systems of the form xʹ = y, yʹ = –g(x) – f(x)y, where f(x) = 3Q(x)Qʹ(x)P(x) + Q(x)2 Pʹ(x) and g(x) = Q(x)Qʹ(x)(Q(x)2 P(x)2 – 1) with P,Q ∈ ℂ[x]. This class of generalized Liénard polynomial differential systems has the invariant algebraic curve (y + Q(x)P(x))2 – Q(x)2 = 0 of hyperelliptic type.


Introduction and statement of the main result
One of the more classical and difficult problem in the qualitative theory of planar differential systems depending on parameters is to characterize the existence and non-existence of first integrals in function of the parameters of the system.
We consider the polynomial differential system (1) x ′ = y, y ′ = −g(x) − f (x)y, called the generalized Liénard polynomial differential system, where x and y are complex variables and the prime denotes derivative with respect to the time t, which can be real or complex. Such differential systems appear in several branches of the sciences, such as biology, chemistry, mechanics, electronics, etc, see for instance [6,17] and the references quoted there. For g(x) = x the Liénard differential system (1) is called the classical Liénard polynomial differential system. Let X = y ∂ ∂x − (g(x) + f (x)y) ∂ ∂y be the polynomial vector field associated to system (1). Let U be an open and dense set in C 2 . We say that the non-locally constant function H : U → C is a first integral of the polynomial vector field X on U , if H(x(t), y(t)) = constant for all values of t for which the solution (x(t), y(t)) of X is defined on U . Clearly H is a first integral of X on U if and only if XH = 0 on U .
A Liouvillian first integral is a first integral H which is a Liouvillian function, that is, roughly speaking which can be obtained "by quadratures" of elementary functions. For a precise definition see [16]. The study of the Liouvillian first integrals is a classical problem of the integrability theory of the differential equations which goes back to Liouville.
As far as we know the Liouvillian first integrals of some multi-parameter family of planar polynomial differential systems has only been completely classified for the planar Lotka-Volterra systems, see [1,7,12,13,14,15].
Note that when g(x) = x, system (1) is the well-known classical Lienard polynomial differential system whose Liouvillian first integrals were studied in [9]. Moreover, when 2 ≤ deg g ≤ deg f the Liouvillian first integrals of these systems were studied in [10], and when deg g = deg f + 1 the Liouvillian first integrals of these systems were studied in [11].
The case of f and g being general polynomials is out of the reach right now. The study of Liouvillian first integrals is based on particular on the search of what is called an invariant algebraic curve. Let h = h(x, y) ∈ C[x, y] \ C. As usual C[x, y] denotes the ring of all complex polynomials in the variables x and y. We say that h = 0 is an invariant algebraic curve of the vector field X if it satisfies for some polynomial K = K(x, y) ∈ C[x, y] called the cofactor of h = 0. Clearly h has degree at most m = max{deg f + 1, deg g} − 1. We also say that h is a Darboux polynomial of system (1). Note that a polynomial first integral is a Darboux polynomial with zero cofactor. The invariant algebraic curves are important because a sufficient number of them forces the existence of a first integral. This result is the basis of the Darboux theory of integrability, see for instance [4,5,8].
An exponential factor E of system (3) is a function of the form for some polynomial L = L(x, y) of degree at most m, called the cofactor of E. It is easy to check the following result for any generalized Liénard polynomial differential system (1). Proposition 1. System (1) has the exponential factors exp(x j ) with cofactors x j−1 y for j = 1, . . . , max{deg f, deg g − 1} and the ones of the form exp(u(x)) with u(x) a polynomial of degree at most max{deg f, deg g − 1}. Moreover, if deg g ≤ deg f then system (1) has the exponential factors exp( The main difficulty for studying the Liouvillian integrability of a polynomial differential system is the characterization of the invariant algebraic curves and also the characterization of the exponential factors of the polynomial differential system. For that reason we restrict our study of the Liovillian integrability of the generalized Liénard polynomial differential systems (1) to the following ones Our main result on the Liouvillian integrability of the class of generalized Liénard polynomial differential system (3) is the following. Theorem 2. The following statements hold for the generalized Liénard polynomial differential system (3).
(a) When deg Q = 0, i.e., Q(x) = κ ∈ C, then system (3) is Liouvillian integrable with the first integral H = y + κ 2 P (x); (b) When deg P = 0, i.e., P (x) = κ ∈ C, then system (3) is Liouvillian integrable with the first integral and The statements (a) and (b) can be checked directly from the definition of first integral. We will divide the proof of statement (c) of Theorem 2 into different sections. In section 3 we will prove Theorem 2(c.1), while the proof of Theorem 2(c.2) will be given in section 4.
Note that the main result in statement (c.1) is the uniqueness of h 1 and h 2 as irreducible Darboux polynomials because their existence follows from [18]. We remark that exp(x j ) are exponential factors for any generalized Liénard polynomial differential system (1).

Auxiliary notions and results
The following result is well-known. For a proof see, for instance, Proposition 8.4 in [5].
Lemma 3. Assume f ∈ C[x, y] and let f = f n 1 1 · · · f nr r be its factorization into irreducible factors over C[x, y]. Then for a polynomial differential system (1), f = 0 is an invariant algebraic curve with cofactor K f if and only if f i = 0 is an invariant algebraic curve for each i = 1, . . . , r with cofactor K f i . Moreover K f = n 1 K f 1 + · · · + n r K f r . Proposition 4. The following statements hold.
(a) If E = exp(u/v) is an exponential factor for the polynomial differential system (3) and v is not a constant polynomial, then v = 0 is an invariant algebraic curve. (b) Eventually E = exp(u) can be exponential factors coming from the multiplicity of the invariant straight line at infinity.
For a geometric meaning of exponential factors and a proof of Proposition 4 see [3]. The existence of exponential factors exp(u/v) is due to the fact that the multiplicity of the invariant algebraic curve v = 0 is larger than 1, for more details see again [3].
The following result given in [3] characterizes the algebraic multiplicity of an invariant algebraic curve using the number of exponential factors of system (3) associated with the invariant algebraic curve.
Proposition 5. Given an irreducible invariant algebraic curve v = 0 of degree k of system (3), it has algebraic multiplicity ℓ if and only if the vector field associated In view of Proposition 5 if we prove that e u/v is not an exponential factor with deg u ≤ deg v, there are no exponential factors associated to the invariant algebraic curve v = 0.
We say that a C 1 function V = V (x, y) is an integrating factor if it satisfies where div stands for the divergence of the vector field X.
In 1992 Singer [16] proved that a polynomial differential system has a Liouvillian first integral, if and only if it has an integrating factor of the form where U 1 and U 2 are rational functions which verify ∂U 1 /∂y = ∂U 2 /∂x. In 1999 Christopher [2] improved the results of Singer showing that there are integrating factors of the form where u, v and f i are polynomials and λ i ∈ C. From the Darboux theory of integrability (see [5,8,16]) we have the following result.
Theorem 6. The polynomial differential system (3) has a Liouvillian first integral if and only if system (3) has an integrating factor of the form (4), or equivalently there exist p invariant algebraic curves f i = 0 with cofactors K i for i = 1, . . . , p, q exponential factors E j = exp(u j /v j ) with cofactors L j for j = 1, . . . , q and λ j , µ j ∈ C not all zero such that 3. Proof of Theorem 2(c.1) The proof of Theorem 2(c.1) will be a direct consequence of some auxiliary results.
Proof. From the fact that system (3) has degree equal to deg g = 2 deg f + 1 = m + 1 ≥ 3, and that K is a polynomial of degree at most m we write K as where K j (x) has degree at most m − j. By assumptions h satisfies where f and g were given in (3). We write h(x, y) = ∑ l j=0 h j (x)y j . Without loss of generality we can assume that h l (x) ̸ = 0. Computing the coefficient of y l+m in (6) we get Therefore repeating this argument for y l+m−1 , . . . , y l+2 , we get that K j (x) = 0 for Since h l (x) must be a polynomial in x we have that K 1 (x) = 0. This completes the proof of the proposition.
Proof. By direct computations we get that h 1 and h 2 are irreducible Darboux polynomials of system (3). Now we shall prove that these are the only irreducible Darboux polynomials of system (3). Let h = h(x, y) be another irreducible Darboux polynomial of system (3) with cofactor K. In view of Proposition 7 we have that K = K(x). Then, with f and g as in (3). Now we introduce the variables (X, Y ) with Then in these variables system (3) becomes Let h =h(X, Y ). Then, if we denote byh =h(X) the restriction ofh to Y = 0 we get thath ̸ = 0 (otherwise h would not be irreducible). Note thath is a Darboux polynomial of system (8) restricted to Y = 0, that is, where K(X) is the cofactor ofh, equal to the cofactor of h. Solving this linear differential equation we deduce that Let r(X) = −Q(X)(P (X) − 1). Without loss of generality we can assume that K and r are coprime, otherwise we divide by their common factor. We claim that (11) deg K < deg r.
We proceed by contradiction. Assume that and consider the Euclidean division of K and r. We have where ψ(X) cannot be zero taking into account that K and r are coprime and deg ψ < deg r. Hence equation (12) becomes Integrating this equation and taking into account (10) we have that wheres ′ (X) = s(X). Therefore, the first factor in (14) cannot cancel with the second factor of (14), and this gives a contradiction with the fact thath(X) is a polynomial. Hence, we conclude that deg K < deg r, which proves (11).
We say that the polynomial r(X) is square-free if r(X) = ∏ k l=1 (X − α l ) with α l ̸ = α l for l, j = 1, . . . , k and l ̸ = j. We claim that (15) the polynomial r must be square free.
We again proceed by contradiction. Using an affine transformation of the form X → X + α with α ∈ C if necessary, we can assume that X is a factor of the polynomial r with multiplicity µ > 1. Then we write it as r(X) = X µ s(X) with s(0) ̸ = 0. We know that K(0) ̸ = 0 since K and r are coprime. Now we develop K(X)/r(X) in simple fractions of X, that is where α 1 (X) is a polynomial with deg α 1 < deg s and c i ∈ C, for i = 1, 2, . . . , µ.
Equating both expressions we get that c µ = K(0)/s(0) ̸ = 0. Therefore equation (10) becomes where C ∈ C \ {0}. The first exponential cannot be simplified with any part of the second exponential. Thus, we get a contradiction with the fact thath must be a polynomial. Therefore we conclude that r must be square-free and (15) is proved. Hence we have Integrating (10) we get Sinceh must be a polynomial we must have that γ i ∈ N ∪ {0} for i = 1, . . . , k. Now we introduce the variables (X, Y ) with (17) X = x and Y = h 2 = y + Q(x)(P (x) + 1).
Then in these variables system (3) becomes Let h =ĥ(X, Y ). Then, if we denote by h * = h * (X) the restriction ofĥ to Y = 0 we get that h * ̸ = 0 (otherwise h would not be irreducible). Here h * is a Darboux polynomial of system (18) Solving this linear differential equation we deduce that Proceeding as we did forh, if we denote by r * (X) = −Q(X)(P (X) + 1), then we must have that r * is square free and that
Note that if we denote by h = h(x, y) a Darboux polynomial of system (3) with cofactor K = K(x) then for some polynomials h 1 , h 2 ∈ C[x, y]. Moreover, from (16) we obtain where the prime denotes derivative with respect to x, and from (20) we get That is (21) The rational function u(x) cannot be a constant, because the polynomial P (x) has degree at least one. Clearly from (21) the function u(x) cannot be neither a polynomial, nor a quotient of two polynomials. This is, we have a contradiction. This concludes the proof of the proposition. Proof. We introduce the variables (X, Y ) as in (7) and we get system (8). Let h =h(X, Y ) be a polynomial first integral. Then, if we denote byh =h(X) the restriction ofh to Y = 0, thenh satisfies (9) with K(X) = 0, i.e.
Since we can assume without loss of generality that h has no constant terms, we havec = 0 and thush = 0. Now, introducing the variables (X, Y ) as in (17) we get system (18). Then if we denote by h * = h * (X) the restriction ofĥ to Y = 0, then h * satisfies (20) with Since we can assume without loss of generality that h has no constant terms, we have c * = 0 and thus h * = 0.
In short any polynomial first integral h can be written as for some polynomials g 1 , g 2 ∈ C[x, y]. Hence, . In other words g 3 must be a Darboux polynomial of system (3) with cofactor K = 2Q ′ (x)Q(x)P (x). In view of Proposition 8 and Lemma 3 we must have . This is not possible because m 1 and m 2 must be positive integers, and this contradiction completes the proof of the proposition.

Proof of Theorem 2(c.2)
We divide the proof of Theorem 2 in different steps. Proof. Let h 1 = y+Q(x)(P (x)−1) and E = exp(u/h 1 ) with u and h 1 being coprime. Clearly, after simplifying by exp(u/h 1 ), we get that u satisfies where L is a polynomial of degree at most m. We introduce the change of variables of (7) and equation (22) becomes whereū =ū(X, Y ) = u(x, y) andL =L(X, Y ) = L(x, y). If we denote byũ the restriction ofū to Y = 0 we have thatũ ̸ = 0 (otherwiseū would be divisible by Y ). Evaluating (23) on Y = 0 we conclude that Thereforeũ must be a polynomial that satisfies (9) with K(X) = −Q ′ (X)(Q(X)P (X)+ 1). Note that proceeding as in the proof of Proposition 8 we get that deg K(X) must be less than the degree of Q(X)(P (X) − 1), which is not the case. Hence, system (3) has no exponential factors of the form exp(u/h 1 ) with u and h 1 being coprime. Let h 2 = y + Q(x)(P (x) + 1) and E = exp(u/h 2 ) with u and h 2 being coprime. After simplifying by u/h 2 , we get that u satisfies where L is a polynomial of degree at most m. We introduce the change of variables of (17) and equation (24) becomes whereū =ū(X, Y ) = u(x, y) andL =L(X, Y ) = L(x, y). If we denote byũ the restriction ofū to Y = 0 we have thatũ ̸ = 0 (otherwiseū would be divisible by Y ). Evaluating (25) on Y = 0 we conclude that −Q(X)(P (X) + 1) dũ dX + Q ′ (X)(Q(X)P (X) − 1)ũ = 0.
Note that proceeding as in the proof of Proposition 8 we get that the degree of (Q ′ (X)(Q(X)P (X) − 1)) must be less than the degree of Q(X)(P (X) + 1), which is not the case. Hence, system (3) has no exponential factors of the form exp(u/h 2 ) with u and h 2 being coprime.
In view of Proposition 5 and Lemma 10, system (3) has no exponential factors of the form E = exp(u/h n j ) with u ∈ C[x, y] coprime with h j for j = 1, 2 and n ≥ 1.
Then the unique possible exponential factors of system (3) are of the form e u with u ∈ C[x, y].
Lemma 11. If system (3) has a Liouvillian first integral, then it has an integrating factor of the form exp(u(x, y))h λ 1 1 h λ 2 2 where u ∈ C[x, y], λ 1 , λ 2 ∈ C and h 1 and h 2 are the Darboux polynomials of Theorem 2 (c.1). Moreover the cofactor of the exponential factor exp(u(x, y)) is a polynomial L = L(x).
Proof. Let L(x, y) be the cofactor of exp(u(x, y)). In order that system (3) has a Liouvillian first integral, by Theorems 6, Theorem 2 (c.1) and Lemma 10 we must We expand L in power series in the variable y as L(x, y) = ∑ m j=0 L j (x)y j . Computing the coefficients of y j with j > 0 in (26) we get that L j (x) = 0 for j = 1, . . . , n and thus L = L 0 (x). This concludes the proof.
Since we are looking for Liouvillian first integrals of system (3), in view of Lemma 11, we can restrict to study the exponential factors with cofactor L = L(x). Proof. Let E = exp(u) with u ∈ C[x, y] \ C and let L = L(x) = ∑ m k=0 β k x k be the cofactor associated to E with β k ∈ C. We write u = ∑ r j=0 u j (x)y j . Without loss of generality we can assume that u r (x) ̸ = 0. By the definition of exponential factor (2) we have with f and g as in (3). Then We write where l.o.t denotes the lower order terms in x. Now we consider two cases. Case 1: r ≥ 2. Computing in (28) the coefficient of y r+1 we get that u ′ r (x) = 0, that is, without loss of generality we can take u r (x) = 1. Now we claim that if we write u = u(x, y) = y r + r ∑ j=1 u r−j (x)y r−j , then for j = 1, . . . , r where A 1 = (3q + p)r, A 2 = q(2q + p)r + (3q + p) 2 r(r − 1), and for ℓ ≥ 2, Note that in view of (30) we have that A ℓ+1 > 0 for any ℓ = 0, . . . , r − 1.
We start the proof of the claim. For j = 1 computing the coefficient of y r in (28), we get that Integrating it we obtain which coincides with (29) for j = 1.
For j = 2 computing the coefficient of y r−1 in (28) we get that Integrating it we get which coincides with (29) for j = 2. Now we assume that (29) holds for j = 0, . . . , L with L < r and we will prove it for j = L + 1. Computing the terms in (28) with y r−L we get From (29) with j = r − 1 we obtain We recall that A r−1 > 0. Now computing the coefficient of y 0 in (28) we get Using (31) the degree of the polynomial in the left hand side of (32) is (r − 1)(2q + p) + 4q + 2p − 1 ≥ 6q + 3p − 1. Since the degree of the right-hand side is at most m = 4q + 2p − 2, we have a contradiction. Case 2: r ≤ 1. We write u = u(x, y) = u 0 (x) + u 1 (x)y. Computing the coefficient of y 2 in (28) we get u ′ 1 (x) = 0, and without loss of generality we can take u 1 (x) = 1. Furthermore the coefficient of y in (28) gives being β 0 a constant. Finally the coefficient of y 0 in (28) gives Since g(x) = Q(x)Q ′ (x)(Q(x) 2 P (x) 2 − 1) has degree m + 1, from (34) we get a contradiction. This concludes the proof of the proposition.
Proof of Theorem 2(c.2). The proof of Theorem 2 follows directly from Lemma 11 and Proposition 12.