Medium amplitude limit cycles of some classes of generalized Liénard systems

The bifurcation of limit cycles by perturbing a planar system which has a continuous family of cycles, i.e. periodic orbits, has been an intensively studied phenomenon; see for instance [13, 16, 2] and references therein. The simplest planar system having a continuous family of cycles is the linear center, and a special family of its perturbations is given by the generalized polynomial Liénard systems:


Introduction and statement of the results
The bifurcation of limit cycles by perturbing a planar system which has a continuous family of cycles, i.e. periodic orbits, has been an intensively studied phenomenon; see for instance [13,16,2] and references therein.The simplest planar system having a continuous family of cycles is the linear center, and a special family of its perturbations is given by the generalized polynomial Liénard systems: where µ, ν ∈ N, g i (x) and F i (x) are polynomials for i ≥ 1, and ε is a small parameter.The classical and generalized Liénard systems appear very often in several branches of science and engineering, as biology, chemistry, mechanics, electronics, etc., see for instance [20] and references therein.In particular Liénard systems are frequent specially in physiological processes, see for instance [10].Further, some planar systems can be transformed into (generalized) Liénard systems, see for example [5,15].In addition, the generalized polynomial Liénard systems is one of the most considered families in the study of limit cycles, see [18].
We assume F µ (x) ≡ 0 and g ν (x) ≡ 0, then we define m = max For a small enough ε, let H µ ν (m, n) be the maximum number of limit cycles of (1 ε ) that bifurcate from cycles of the linear center (1 0 ), i.e. the maximum number of medium amplitude limit cycles which can bifurcate from (1 0 ) under the perturbation (1 ε ), in short Maximum number of medium amplitude limit cycles of (1 ε ) .
The main problem concerning H µ ν (m, n) is finding its exact value.
We are interested in ν ≥ 1 because when g i (x) ≡ 0 for all i ≥ 1 the maximum number of medium amplitude limit cycles of (1 ε ) is well-known.Indeed, if we denote by H µ 0 (m) the maximum number of medium amplitude limit cycles of (1 ε ) in such a case, then we know from [17] that , where [•] denotes the integer part function.Moreover, by following [7,Theorem 3.1] we can prove that An explicit proof of this statement is provided in [1, Section 3.2.2]because it is known that the cyclicity of a non-degenerated center (as in our case) coincide with the cyclicity of the open period annulus surrounding it.See for instance [9].Theorem 1 (below) is a generalization of this result, and it also improves the results of Section 4.3.2 in [1] which prove that H µ ν (m, n) = [(m − 1)/2] for some families of generalized Liénard systems.A rewiev about the results conserning small and medium amplitude limit cycles of (1 ε ) is given in [19], where is also proved that However, the exact values of H 1 1 (m, n), H 2 2 (m, n), and H 3 3 (m, n) were not reported there.In this paper we give the exact value of H µ ν (m, n) for two subfamilies of (1 ε ).More precisely, we consider the families: where µ 0 is the smallest integer with 1 ≤ µ 0 ≤ µ such that F µ0 (x) ≡ 0. We will give the exact values of Hµ ν (m, n) and Hµ ν (m, n) the maximum number of medium amplitude limit cycles of systems in GL1 and GL2, respectively.We note that if µ 0 = µ, then Hµ ν (m, n) = H µ ν (m, n).Our main result is the following.
Theorem 1.The following statements hold.

(a) The exact value of Hµ
. Moreover, for each s with 0 ≤ s ≤ m−1 2 there exist systems in GL1 having exactly s hyperbolic limit cycles.
1 there exist systems in GL2 having exactly s hyperbolic limit cycles.
The assumptions on g i (x) and F i (x) in definitions of GL1 and GL2, respectively, are necessary.Otherwise, we can construct systems (1 ε ) having more medium amplitude limit cycles, see Remark 1 in Section 3.
Theorem 1 is a generalization of Theorem 1.1 in [22], where the case µ = ν = 1 was considered.We note that in such a case H1 gives the exact value of H 1 1 (m, n).The proof of Theorem 1 is based on computing the maximum number of isolated zeros of the first non-vanishing Poincaré-Pontryagin-Melnikov function of the displacement function of (1 ε ), by taking into account the restrictions: g i (x) odd for 1 ≤ i ≤ ν and F i (x) even for µ 0 < i ≤ µ, respectively.
The paper is organized as follows.In Section 2 we recall the definition of the displacement function of (1 ε ), as well as the algorithm to compute the Poincaré-Pontryagin-Melnikov functions.Preliminary results that allow us to provide elementary proofs of the main result are given in Section 3. Finally, in Section 4 we will prove Theorem 1.

Poincaré-Pontryagin-Melnikov functions
The linear center (1 0 ) is the Hamiltonian system associated to the polynomial H = (x 2 + y 2 )/2; hence its cycles are the circles γ c = {H − c = 0} with c > 0. By using c as a parameter, the first return map of (1 ε ) can be expressed in terms of ε and c: P(ε, c).Therefore the corresponding displacement function L(ε, c) = P(ε, c) − c is analytic for small enough ε and can be written as the power series in ε where L i (c) with i ≥ 1 is the Poincaré-Pontryagin-Melnikov function of order i, which is defined for c ≥ 0. Let L k (c) with k ≥ 1 be the first non-vanishing coefficient in (2).The zeros of L k (c) are important in the study of medium amplitude limit cycles of (1 ε ) because of the Poincaré-Pontryagin-Andronov criterion: The maximum number of isolated zeros, counting multiplicities, of L k (c) is an upper bound for H µ ν (m, n).Furthermore each simple zero c 0 of L k (c) corresponds to one and only one limit cycle of (1 ε ) with ε small enough bifurcating from the cycle γ c0 .Now, we will recall the algorithm to compute the functions L i (c).System (1 ε ) can be written as where with ω i = (g i (x) + f i (x)y) dx and ω i ≡ 0 for i > max{µ, ν}.
As we know, L 1 (c) is given by the classical Poincaré-Pontryagin formula L 1 (c) = γc ω 1 .A construction to compute the second order Poincaré-Pontryagin-Melnikov function of a perturbed system of the form dH − εω 1 with ω 1 an arbitrary polynomial 1-form was given by Yakovenko [1995].After, Françoise [1996] gave the algorithm to know the Poincaré-Pontryagin-Melnikov function of any order of dH − εω 1 .Finally, Iliev [1999] gave the result for computing the higher order Poincaré-Pontryagin-Melnikov functions of a perturbed system of the form dH − εω 1 − ε 2 ω 2 − • • • = 0, where ω i for i ≥ 1 are arbitrary polynomial 1-forms.His result is the following. where The proof of this result easily follows from the Poincaré-Pontryagin formula, and the Ilyashenko-Gavrilov theorem ( [12], [8]): If γc ω = 0 for all c ≥ 0, then ω = dQ + qdH, where Q and q are polynomials, and by applying an induction argument.For a detailed proof, see for instance [11], [14].
On the other hand, we know from [11] that L k (c) has at most [k(max{n, m} − 1)/2] positive zeros, counting multiplicities.However, this result does not give the value of H µ ν (m, n) because the upper bound for k depending on µ, ν, m, and n is unknown.In addition, the number of isolated zeros of the first nonvanishing Poincaré-Pontryagin-Melnikov function does not provide the number of limit cycles of (1 ε ) with ε small enough as shows next example.
A simple computation gives L 1 (c) = γc ω 1 = −2πc.Hence system (5 ε ) does not have limit cycles bifurcating from the cycles of the linear center.However, for any ε > 0 small enough the system (5 ε ) has a limit cycle bifurcating from the infinity; more precisely, if we consider (5 ε ) on the Poincaré sphere S 2 , then the limit cycle bifurcates from the equator of S 2 which is known as "the circle at infinity" or "points at infinity" of R 2 [21].Indeed, by using the function ; moreover, it is easy to see that for ε ∈ (0, 1) the curve {V ε (x, y) = 0} has an oval surrounding the origin (the unique singularity of (5 ε )).Hence, R 2 \ {V ε (x, y) = 0} has a 1-connected component U ε , then the generalized Bendixon-Dulac theorem [6] ensures that (5 ε ) has a hyperbolic limit cycle Γ ε in U ε for each ε ∈ (0, 1).Thus, Γ ε contains the oval of {V ε (x, y) = 0}.See Section 4 of [4] for more details.Finally, a straightforward computation shows that the circle x 2 + y 2 = 1/(2ε) 2 is contained in the bounded region limited by the oval of {V ε (x, y) = 0}.This implies that Γ ε bifurcates from the "the circle at infinity" of R 2 .
In next section we will give some properties on ω i which will allow us to simplify the computation of the Poincaré-Pontryagin-Melnikov functions

Preliminary results
For computing L k (c) for (1 ε ) we will use the following two elementary lemmas whose proof is omitted.Lemma 3. Let P be a polynomial in the ring R x 2 , H .We define deg 2 P to be the degree of P in R[x 2 , H].
(a) For i, j ≥ 0 there are homogeneous polynomials If i = 0, then q ij ≡ 0.
(b) For i, j ≥ 0 there are homogeneous polynomials (c) For i, j ≥ 0 we have Lemma 4. If ω, η ∈ A and q ∈ S where The next two results are straightforward consequences of these two previous lemmas.
We now will prove two lemmas which will be useful in the proof of Theorem 1.
Lemma 7. Suppose k ≥ 2. Then the following statements hold.
where Ω l is defined as in (4).
Proof.(a) We proceed by induction on k.If k = 2, then statement (a) is true because Ω 1 = ω 1 ∈ A. We now assume that the statement is true for k − 1, and we will prove it for k.
From the induction hypothesis it follows that ω l , Ω l ∈ A for 1 ≤ l ≤ k − 2. Thus, by Corollary 5, Ω l = dQ l + q l dH with q l ∈ S for all 1 ≤ l ≤ k − 2, and by using Lemma 4 we conclude that Ω k−1 := (b) By Corollary 5, Ω l = dQ l + q l dH with q l ∈ S for all 1 ≤ l ≤ k − 1, and In addition, by the statement (a), ω l ∈ A for 1 ≤ l ≤ k −1.Thus, Ω k := i+j=k q i ω j ∈ A because of Lemma 4, which implies that γc Ω k ≡ 0 by Corollary 5. Finally, from Theorem 2 we have Before announce next lemma, we note that each polynomial h(x) = m−1 r=0 a r x r of degree m − 1 can be written as h(x) = ĥ where Lemma 8. Let ω = (g(x) + f (x)y) dx, where f (x) = m−1 r=0 a r x r and g(x) = n s=0 b s x s . (a) γc ω = γc f x 2 ydx and its value is −πc and (c) γc (y q) ω ≡ 0 if and only if q ≡ 0 or ĝ x 2 ≡ 0.
Proof.(a) By statements (a) and (b) of Lemma 3, γc ω = γc f x 2 ydx, and the value of this integral follows from Corollary 6.
(b) If γc ω ≡ 0, then f x 2 ≡ 0 by (a).This implies that ω = g(x)dx + x f x 2 ydx and by (7) we have From Lemma 3.(b) we obtain x 2r+1 ydx = d yQ r + (yq r ) dH for some homogeneous polynomials Q r , q r ∈ R x 2 , H with deg 2 Q r = r + 1 and deg 2 q r = r, respectively.Hence . Moreover, a simple computation shows that As (yq) ω = qĝ x 2 ydx + qg x 2 xydx + q f x 2 xy 2 dx and q f x 2 xy 2 dx = q f x 2 x 2H − x 2 dx, it follows that (yq) ω = qĝ x 2 ydx + dQ 2 + q 2 dH because of statements (a) and (b) of Lemma 3. Hence we obtain γc (y q) ω = γc qĝ x 2 ydx.That is, By using expression (8) of q r , a straightforward computation, and Lemma 3(c) we obtain the formula given in the statement.Finally, statement (c) follows from the formula given in statement (b).
and g 1 (x) = 1 − x 2 does not satisfy the hypothesis in definition of GL1 because g 1 (x) is not an odd function.Here m = n = 2 and from Theorem 1.(a) it follows that H1 1 (2, 2) = 0; however, for ε small enough, this system has one medium amplitude limit cycle.Indeed, we need only to prove that the first non-vanishing coefficient of the displacement function ( 2), associated to this system, has a simple positive zero.For that we write system in the form (3 ε ) as dH − εω = 0 with ω = (1−x 2 −2xy)dx.By Lemma 8.(a), L 1 (c) ≡ 0, and by Theorem 2 and Lemma 8.(b), L 2 (c) = −πc(4−2c).
, and g 2 (x) = −5 + 25x 2 /6 does not satisfy the hypothesis in definition of GL2 because F 2 (x) is not an even function.In this case m = n = 3 and by Theorem 1.(b), H2 2 (3, 3) = 1; however, for ε small enough, the resulting system has two medium amplitude limit cycles.Indeed, following previous ideas, and using Theorem 2 and Lemma 8 it is easy to see that L 1 (c) ≡ 0, L 2 (c) ≡ 0, and

Proof of the Theorem 1
We can assume, after a linear change of variables if necessary, that m−1 r=0 a ir x r and g i (x) can be written as f i (x) = fi x 2 + x fi x 2 and g i (x) = ĝi x 2 + xg i x 2 , respectively, according to (6).
Proof of Theorem 1.(a) By hypothesis, g i (x) is odd for 1 ≤ i ≤ ν, which means that g i (x) = xg i x 2 for 1 ≤ i ≤ ν.Let L k (c) be the first non-vanishing Poincaré-Pontryagin-Melnikov function in (2).For proving the statement we will prove first that L k (c) has at most [(m − 1)/2] positive zeros, and then that for each s with 0 ≤ s ≤ [(m − 1)/2] we can choose systems in GL1 in such a way that L k (c) has exactly s simple positive zeros.
If k = 1, then the assertion is true.Indeed, we have Thus, as γc xg i x 2 dx ≡ 0 and γc f1 x 2 xydx ≡ 0 by Corollary 5, we obtain L 1 (c) = γc f1 x 2 ydx.From (7) we know that deg 2 f1 x 2 = [(m − 1)/2], which implies that L 1 (c) has at most [(m − 1)/2] positive zeros because of Corollary 6.In addition, since f1 x 2 has [(m − 1)/2] + 1 independent coefficients, for each s with 0 ≤ s ≤ [(m − 1)/2] we can choose suitable coefficients of f1 x 2 in such a way that L k (c) has exactly s simple positive zeros.Therefore, by applying the Poincaré-Pontryagin-Andronov criterion it follows that the corresponding system (1 ε ), which belongs to GL1, has exactly s hyperbolic limit cycles.In particular we have proved that Hµ ν (m, n) = [(m − 1)/2].Suppose then that k ≥ 2. If we prove that Ω l ∈ A for 1 ≤ l ≤ k − 1, then L k (c) = γc ω k by Lemma 7, and by applying the same idea as in previous paragraph the assertion follows.Therefore, it remains to show that Ω l ∈ A for 1 ≤ l ≤ k − 1.
We proceed by induction on k.If k = 2, then L 1 (c) ≡ 0, which implies that Hence the assertion is true for k = 2.We now assume that the assertion is true for k−2, and we will prove it for k−1.By induction hypothesis, Ω i ∈ A for 1 ≤ i ≤ k − 2, which implies that Ω i = dQ i + q i dH with q i ∈ S for 1 ≤ i ≤ k − 2 by Corollary 5. Furthermore, by Lemma 7, (b) Firstly we will show two properties concerning ω i and γc ω i which we will use along the proof of the statement.Then, we will split the proof into two cases: m odd and m even.