This is a preprint of: “Limit cycles in piecewise polynomial systems allowing a non-regular switch- ing boundary”, Tao Li, Jaume Llibre, Phys. D, vol. 419, 132855:1–13, 2021. DOI: [10.1016/j.physd.2021.132855] LIMIT CYCLES IN PIECEWISE POLYNOMIAL SYSTEMS ALLOWING A NON-REGULAR SWITCHING BOUNDARY TAO LI1 AND JAUME LLIBRE2 Abstract. Continuing the investigation for the piecewise polynomial perturbations of the linear center ẋ = −y, ẏ = x from [Physica D 371(2018), 28-47] for the case where the switching boundary is a straight line, in this paper we allow that the switching boundary is non-regular, i.e. we consider a switching boundary which separates the plane into two angular sectors with angles α ∈ (0, π] and 2π − α. Moreover, unlike the aforementioned work, we allow that the polynomial differential systems in the two sectors have different degrees. Depending on α and for arbitrary given degrees we provide an upper bound for the maximum number of limit cycles that bifurcate from the periodic annulus of the linear center using the averaging method up to order N . The reachability of the upper bound is also reached for the first two orders. On the other hand, we pay attention to the perturbation of the linear center inside this class of all piecewise polynomial Liénard systems and give some better upper bounds in comparison with the one obtained in the general piecewise polynomial perturbations. Again our results imply that the non-regular switching boundary (i.e. when α 6= π) the piecewise polynomial perturbations usually leads to more limit cycles than the regular case (i.e. when α = π) where the switching boundary is a straight line. 1. Introduction and statement of main results In the qualitative theory of smooth differential systems, a classical and challenging objec- tive is to determine the maximum number of the limit cycles bifurcating from the periodic annulus of the linear center ẋ = −y, ẏ = −x, when it is perturbed inside the family formed by all planar polynomial differential systems of the form N N ! X X i i (1) (ẋ, ẏ) = −y + ε fi (x, y), x + ε gi (x, y) , i=1 i=1 where |ε| > 0 sufficiently small, and fi and gi are real polynomials of degree n. This is essentially the weak Hilbert’s 16th problem, see [1, 14, 18]. It was proved in [14] that system (1) has at most [N (n − 1)/2] limit cycles bifurcating from the periodic annulus for |ε| > 0 sufficiently small, where as usual [·] denotes the integer part function. Since this upper bound obtained in [14] is not reached in general, up to now we still do not know what is the exact maximum number of limit cycles under the general polynomial perturbation (1) except some special families of perturbations, such as the Liénard family, i.e. gi (x, y) = 0 and fi (x, y) is independent of the variable y, for which it was proved in [12] that at most [(n − 1)/2] limit cycles bifurcate and this number is reached due to [20]. 2010 Mathematics Subject Classification. 34C29, 34C25, 34C05. Key words and phrases. Averaging method, Hilbert’s 16th problem, limit cycles, discontinuous piecewise polynomial systems, Liénard systems. 1 2 TAO LI AND JAUME LLIBRE Figure 1. Switching boundary Σα and angular sectors Σ± α. As the discontinuity turns out to be ubiquitous in the real world (see for instance the papers in mechanical engineering [8,17], in neural sciences [11,32] and in electronic circuits [3, 26], ...), discontinuous piecewise smooth differential systems have attracted many researchers in recent years. Let Σα be the union of the non-negative x-axis and the ray starting at O − and forming with the non-negative x-axis an angle α ∈ (0, π], Σ+ α and Σα be two angular sectors separated by Σα with angles α and 2π − α respectively, see Figure 1. It is worth mentioning that Σα is just the x-axis if α = π. In this paper we consider the discontinuous piecewise polynomial differential systems of the form N N ! X X i ± i ± (2) (ẋ, ẏ) = −y + ε fi (x, y), x + ε gi (x, y) if (x, y) ∈ Σ±α, i=1 i=1 where fi+ and gi+ (respectively fi− and gi− ) are real polynomials of degree n (respectively m). In this case we say that the degree of system (2) is (n, m). Throughout this paper we will restrict our attention to the case of n ≥ m ≥ 1. Following the Filippov convention [10] we know that all points in Σα \ {O} where the vector fields of two subsystems simultaneously point outward or inward Σα \ {O} form the sliding regions, and the complement of sliding regions in Σα \ {O}, excluding the tangency points of the vector fields and Σα , defines the crossing regions. A periodic orbit is called crossing periodic orbit if it intersects Σα only at the crossing regions. Additionally, we call it crossing limit cycle if this periodic orbit is isolated. Analogous to the polynomial differential system (1), it is natural to ask what is the maximum number, denoted by Mα,N (n, m), of the crossing limit cycles of system (2) bifurcating from the periodic annulus of the linear center ẋ = −y, ẏ = x for given α, N and degree (n, m). Some contributions have been made for the study of Mα,N (n, m) with n = m. In the case of α = π where the switching boundary Σπ is a straight line, i.e. the x-axis, Buzzi, Pessoa and Torregrosa [6] considered the perturbations inside the class of all piecewise linear differential systems and they proved Mπ,N (1, 1) = 1, 1, 2, 3, 3, 3, 3 up to order N = 1, 2, 3, 4, 5, 6, 7 respectively. The piecewise quadratic and cubic perturbations were studied by Llibre and Tang in [24] where it was proved that Mπ,N (2, 2) = 2, 3, 5, 6, 8 and Mπ,N (3, 3) = 3, 5, 8, 11, 13 up to order N = 1, 2, 3, 4, 5 respectively. For the general piecewise polynomial perturbations of degree (n, n) separated by a straight line, Buzzi, Lima and Torregrosa [5] proved that Mπ,1 (n, n) = n, Mπ,2 (n, n) = 2n − 1 and Mπ,N (n, n) ≤ N n − 1 for N ≥ 3. When α ∈ (0, π) the switching boundary Σα is non-regular, in this case Cardin 3 and Torregrosa [7] studied the piecewise linear perturbations providing that Mα,N (1, 1) = 1, 2, 2, 3, 4 up to order N = 1, 2, 3, 4, 5 and Mα,6 (1, 1) ≤ 5. Regarding Mα,N (n, m) with n > m, there exist few works to be done. If Mα,N (n, n) is known, it provides an upper bound of Mα,N (n, m) with n > m, but how to know if this upper bound is exact? Therefore it is worth for us to study Mα,N (n, m) allowing n > m, although some information could be provided by studying Mα,N (n, n). On the other hand, to our knowledge an upper bound of Mα,N (n, m) for α ∈ (0, π) is still not given, even in the case of m = n. The case of α = π and n = m has been studied in [5] as recalled in the last paragraph. Stimulated by these two aspects, we provide some upper bounds for Mα,N (n, m) with α ∈ (0, π] and n ≥ m using the averaging theory up to order N as they are stated in the following result. Theorem 1. For system (2) satisfying α ∈ (0, π] and n ≥ m ≥ 1, the following statements can be obtained using the averaging theory up to order N . (i) If α = π, then Mα,1 (n, m) = n, Mα,2 (n, m) = 2n − 1 and Mα,N (n, m) ≤ N n − 1 for N ≥ 3. (ii) If α ∈ (0, π), then Mα,1 (n, m) = n, Mα,2 (n, m) ≤ 2n and Mα,N (n, m) ≤ N n for N ≥ 3. Moreover the upper bound Mα,2 (n, m) is reached for α ∈ (0, π/2]. The result in statement (i) of Theorem 1 was also obtained in [5, Theorem 1.1] in the particular case m = n, so the result of this statement generalizes Theorem 1.1 of [5]. From Theorem 1 we see that the upper bound of Mα,N (n, m) for α ∈ (0, π) is the upper bound for α = π plus one, which emphasizes the importance of the shape of the switching boundary in the study of the crossing limit cycles of the piecewise differential systems. Besides we observe that the upper bound of Mα,N (n, m) is usually determined by the subsystem with the higher degree, because all numbers obtained in Theorem 1 are independent of m. Among discontinuous differential systems, one of the most studied classes is the one formed by the discontinuous piecewise Liénard systems, which is widely used to model or analyze many real problems, as for instance the mechanical engineering with dry frictions [8], the integrate-and-fire neurons [32], the discontinuous control in the buck electronic converter [3, 11], ... Of course, the study of the limit cycles for the discontinuous piecewise Liénard systems is also of fundamental importance and many researchers are devoted to the study of this subject, see the papers [5,7,19,23,27,30,31] for one switching boundary and [9,16,25,28] for multiple ones. In this paper we also study the maximum number of crossing limit cycles bifurcating from the periodic annulus of the linear center ẋ = −y, ẏ = x when the perturbations are restricted to the class of all piecewise polynomial Liénard systems of the form N ! X (3) (ẋ, ẏ) = −y + i ± ε fi (x), x if (x, y) ∈ Σ± α, i=1 where fi+ (respectively fi− ) are real polynomials of degree n (respectively m) with n ≥ m ≥ 1. We denote the maximum number of the crossing limit cycles of system (3) bifurcating from the periodic annulus by Lα,N (n, m). When α = π, i.e. the switching boundary Σα is a straight line, Buzzi, Lima and Torregrosa [5] proved that Lπ,1 (n, n) = [(n − 1)/2] and Lπ,2 (n, n) = n + [(n − 1)/2] for any given n, and 4 TAO LI AND JAUME LLIBRE Lπ,N (n, n) = n + [(n − 1)/2] for n = 1, 2, 3, 4 up to order N = 2, 3, 4, 5, 6. In particular, if fi± are even polynomials, they further proved that the origin O is a center for every ε, while if fi± are odd polynomials, Lπ,N (n, n) = (n−1)/2 up to any order N . When α ∈ (0, π), i.e. the switching boundary Σα is non-regular, the result on the maximum number of crossing limit cycles of system (3) bifurcating from the periodic annulus is much fewer. As far as we know, only the case where fi± are linear was considered by Cardin and Torregrosa in [7]. In that paper they proved that Lα,N (1, 1) = 1, 2, 2, 2, 2, 2 up to a study of order N = 1, 2, 3, 4, 5, 6. Allowing α ∈ (0, π] and n ≥ m ≥ 1 again, we give some upper bounds for Lα,N (n, m) using the averaging theory up to order N = 1, 2 in the next result. Theorem 2. For system (3) satisfying α ∈ (0, π] and n ≥ m ≥ 1, the following statements can be obtained using the averaging theory up to order N = 1, 2. (i) If α = π, then Lα,1 (n, m) = [(n − 1)/2] and Lα,2 (n, m) = n + [(m − 1)/2]. (ii) If α ∈ (0, π), then Lα,1 (n, m) = n and Lα,2 (n, m) ≤ max{2m − 1, n} (respectively ≤ max{2m − 2, n}) for m odd (respectively even). The result in statement (i) of Theorem 2 is just Theorem 1.2 of [5] for m = n, while for m < n we provide a better upper bound up to order 2. Again we see that the non-regular switching boundary α ∈ (0, π) increases the number of limit cycles in comparison with the regular one α = π as it was observed in some publications, as for instance in [7]. The paper is organized as follows. In section 2 we shortly review some main tools used in this paper, including the averaging method, Descartes Theorem and some technical results on integrals. It is worth mentioning that another method studying the limit cycle bifurcations from a periodic annulus is the Poincaré-Pontryagin-Melnikov method and it turns out to be equivalent to the averaging one for planar autonomous systems, see [4, 13]. In sections 3 and 4 we are devoted to the proof of Theorems 1 and 2, respectively. Some comments and future directions are summarized in Section 5. 2. Preliminaries The averaging theory is an important tool for studying the number of limit cycles of differential systems. The classical averaging theory [29] usually requires that the considered system is smooth. However this theory, with the efforts of many researchers, has been generalized for piecewise smooth differential systems in recent years, see [15, 21, 22, 33]. Following the work [15] in this section we introduce the averaging theory that we shall use in order to obtain information on the existence and number of crossing limit cycles for the discontinuous piecewise polynomial system (2). To apply the averaging theory we need to write system (2) in a convenient normal form. As usual this normal form can be obtained using the polar coordinates (r, θ), where x = r cos θ and y = r sin θ. In this case system (2) becomes N dr X i  = ε cos θfi± (r cos θ, r sin θ) + sin θgi± (r cos θ, r sin θ) , dt i=1 (4) N dθ 1X i  =1− ε sin θfi± (r cos θ, r sin θ) − cos θgi± (r cos θ, r sin θ) , dt r i=1 5 for θ ∈ Iα± , where Iα+ = [0, α] and Iα− = [α, 2π]. Then taking θ as the new independent variable system (4) writes N dr X i ± (5) = ε Fi (θ, r) + O(εN +1 ) if θ ∈ Iα± , dθ i=1 where the expression of Fi± (θ, r) will be determined later on. From [15] the averaged function Fi (r) : (0, +∞) → R of order i is yi+ (α, r) − yi− (α − 2π, r) (6) Fi (r) = , i! and the functions yi± : [0, 2π] × (0, +∞) → R for i = 1, 2, ..., N are defined recurrently by Z θ y1± (θ, r) = F1± (ϕ, r)dϕ, 0 (7) Z θ i X l ! X 1 L ± Y ± ± ± bj yi (θ, r) = i! Fi (ϕ, r) + ∂ Fi−l (ϕ, r) yj (ϕ, r) dϕ. 0 K j=1 l=1 Sl L Here ∂ denotes the derivative of order L with respect to r, Sl is the set of all l-tuples of non-negative integers (b1 , b2 , ..., bl ) satisfying b1 + 2b2 + · · · + lbl = l, L = b1 + b2 + · · · + bl and K = b1 !b2 !2!b2 · · · bl !l!bl . Moreover we are assuming that F0± = 0 in (7) for convenience. From [15] we get the following result on the averaging theory for the non-autonomous discontinuous piecewise smooth differential system (5). Theorem 3. Consider the non-autonomous discontinuous piecewise smooth differential sys- tem (5). Suppose that i0 is the first positive integer such that Fi = 0 for 1 ≤ i ≤ i0 − 1 and Fi0 6= 0. If Fi0 (ρ) = 0 and Fi00 (ρ) 6= 0 for some ρ ∈ (0, +∞), then for |ε| > 0 sufficiently small there exists a 2π-periodic solution r(θ, ε) of system (5) such that r(0, ε) → ρ as ε → 0. Theorem 3 states that a simple positive zero of the first non-vanishing averaged function provides a crossing limit cycle of system (2) bifurcating from the periodic annulus of the linear center ẋ = −y, ẏ = x. In other words we can obtain some information on the existence and number on the crossing limit cycles of system (2) for |ε| > 0 sufficiently small via studying the number of simple positive zeros of the first non-vanishing averaged function. In order to determine the maximum number of positive roots of a real polynomial in one variable we shall use the following theorem, namely the Descartes Theorem, a proof of it can be found in [2]. Theorem 4. Consider the real polynomial p(x) = ai1 1 xi1 + ai2 xi2 + ... + air xir with 0 = i1 < i2 < ... < ir and r > 1. If aij aij+1 < 0, we say that aij and aij+1 have a variation of sign. If the number of variations of signs is r0 ∈ {0, 1, 2, ..., r − 1}, then the polynomial p(x) has at most r0 positive roots. Furthermore, we can choose the coefficients of the polynomial p(x) in such a way that p(x) has exactly r − 1 positive roots. Let Z θ Z θ Ck (θ) : = cosk ϕdϕ, Sk (θ) := sink ϕdϕ, 0 0 Z π Z θ Z ϕ ck,l : = cosk ϕ sinl ϕdϕ, Ck,l (θ) := cosk ϕ cosl φdφdϕ, 0 0 0 6 TAO LI AND JAUME LLIBRE for k, l ≥ 0. We summarize some technical results on these integrals that we will need in the proofs of our results. Lemma 5. The following propositions hold. (i) Ck (π) = Ck (2π)/2 = (k−1)!! k!! σk , where σk = π if k is even and σk = 0 if k is odd. (ii) Ck (α) 6= 0 and Ck (α − 2π) = Ck (α) − Ck (2π) 6= 0 for α ∈ (0, π). (iii) Sk (π) = (k−1)!! k!! ςk , where ςk = π if k is even and ςk = 2 if k is odd. (iv) kSk (θ) = − cos θ sink−1 θ + (k − 1)Sk−2 (θ). (v) ck,k = 0 if k is odd and ck,k = 2−k Sk (π) if k is even. (vi) ck,l = k−1 k−1 l+1 ck−2,l+2 = l+k ck−2,l . k+l k+l k+l −1 k−1 (vii) Ck,l (π) = 1−(−1) l−1 (−1) l(k+l) + l Ck,l−2 (π) = k(k+l) + k Ck−2,l (π) and Ck,l (π) = 1−(−1) l(k+l) + (l−1)((−1)k+l−2 −1) lk(k+l−2) + (l−1)(k−1) lk Ck−2,l−2 (π). (viii) If l is odd, then Ck,l (π) = 0 (respectively > 0) for k odd (respectively even) and Ck,l (2π) = 0. (ix) Ck,l (α − 2π) = Ck,l (α) if l is odd. (x) Ck,l (θ) = Ck (θ)Cl (θ) − Cl,k (θ). We neglect the proof of Lemma 5 because these propositions can be computed using some standard methods on integrals, such as the integration by parts method, and properties on trigonometric functions. We must mention that some of the results presented in Lemma 5 already appeared in [5]. 3. Proof of Theorem 1 This section is devoted to the proof of Theorem 1. To this end we next give four propo- sitions where the maximum number of simple positive zeros of the averaged functions asso- ciated to the considered systems is computed. Proposition 6. The averaged function of order N associated to the piecewise polynomial differential system (2) satisfying α ∈ (0, π] and n ≥ m ≥ 1 has at most (i) n (respectively N n − 1) simple positive zeros if α = π and N = 1 (respectively N ≥ 2), (ii) N n simple positive zeros if α ∈ (0, π) and N ≥ 1. Proof. Using the polar coordinates we transform system (2) into system (5) with i−1 X ± qi−l (θ, r) X qk±1 (θ, r)qk±2 (θ, r) Fi± (θ, r) = p± i (θ, r) + p± l (θ, r) + +··· r r2 l=0 k1 +k2 =i−l 1≤k1 ,k2 ≤i−l ! X qk±1 (θ, r)qk±2 (θ, r) · · · qk±i−l (θ, r) + ri−l k1 +k2 +···+ki−l =i−l 1≤k1 ,k2 ,...,ki−l ≤i−l for i = 1, 2, ..., N , where p± 0 (θ, r) ≡ 0 and p± ± ± i (θ, r) = cos θfi (r cos θ, r sin θ) + sin θgi (r cos θ, r sin θ), (8) qi± (θ, r) = sin θfi± (r cos θ, r sin θ) − cos θgi± (r cos θ, r sin θ). 7 Since fi+ (x, y) and gi+ (x, y) (respectively fi− (x, y) and gi− (x, y)) are polynomials of degree n − − (respectively m), p+ + i (θ, r) and qi (θ, r) (respectively pi (θ, r) and qi (θ, r)) are also polyno- mials of degree n (respectively m) in the variable r with the coefficients that are polynomials in sin θ and cos θ. Then p± (θ)q1± (θ)i−1 (9) Fei± (θ, r) := ri−2 Fi± (θ, r) − 1 r are polynomials of degree in − 1 and im − 1 in r because n, m ≥ 1, respectively. Here we are writing p± ± ± ± 1 (θ, 0) and q1 (θ, 0) as p1 (θ) and q1 (θ) to simplify the notation. Let hi (q1± (θ), q1± (0)) be the function defined recurrently by h1 (q1± (θ), q1± (0)) =q1± (θ) − q1± (0),  hi (q1± (θ), q1± (0)) =(i − 1)! q1± (θ)i − q1± (0)i (10) i−1 X Z l X Mi−l θ ± i−l−1 dq1± (ϕ) Y + i! q1 (ϕ) hj (q1± (ϕ), q1± (0))bj dϕ, K 0 dϕ j=1 l=1 Sl for i = 2, 3, ..., N , where (−1)L (L + i − l − 2)! Mi−l = , (i − l − 2)! and Sl , K, L, bj are defined below (7). We claim that hi (q1± (θ), q1± (0)) (11) ỹi± (θ, r) := ri−2 yi± (θ, r) − i = 1, 2, ..., N, r are polynomials of degree in − 1 and im − 1 in r respectively, where yi± (θ, r) is defined in (7). This claim will be proved by induction and we only deal with ỹi+ (θ, r), because a similar procedure can be applied to ỹi− (θ, r). To alleviate the notation, we drop the superscript + in the proof of this claim. In fact, it follows from (7) and (9) that Z θ Z θ Z θ y1 (θ, r) = F1 (ϕ, r)dϕ = r e F1 (ϕ, r)dϕ + p1 (ϕ)dϕ 0 0 0 Z θ =r Fe1 (ϕ, r)dϕ + q1 (θ) − q1 (0). 0 Rθ Thus ỹ1 (θ, r) = 0 Fe1 (ϕ, r)dϕ by the definitions of h1 and ỹ1 . Since Fe1 (θ, r) is a polynomial of degree n − 1 in r as it was stated below (9), so also ỹ1 (θ, r) is a polynomial of the same degree, i.e. the claim holds for i = 1. To complete the proof of the claim, we next only need to prove it for i = k, provided that it holds for i = 1, 2, ..., k − 1. According to (9), Fek−l (θ, r) is a polynomial of degree (k − l)n − 1 in r, and then a direct calculation yields Mk−l p1 (θ)q1 (θ)k−l−1 + rP(k−l)n−1 (θ, r) (12) ∂ L Fk−l (θ, r) = , rL+k−l−1 for 1 ≤ k − l ≤ k − 1. Here P(k−l)n−1 (θ, r) is a polynomial of degree (k − l)n − 1 in r and its specific expression is neglected because it is not necessary in the next proof. Moreover 8 TAO LI AND JAUME LLIBRE we have l Y l  Y b hj (q1 (θ), q1 (0)) + rỹj (θ, r) j yj (θ, r)bj = j=1 j=1 rj−1 (13) l Y 1 rPln−1 (θ, r) = hj (q1 (θ), q1 (0))bj + , rl−L j=1 rl−L for l ≤ k − 1, since we are assuming that the claim holds for i = 1, 2, ..., k − 1 and b1 + 2b2 + · · · + lbl = l, b1 + b2 + · · · + bl = L, where Pln−1 (θ, r) is a polynomial of degree ln − 1 in r and we neglect the specific expression again. Joining (12) and (13) we obtain Z θ k−1 XX 1 Yl ∂ L Fk−l (ϕ, r) yj (ϕ, r)bj dϕ 0 l=1 S K j=1 l (14) Z k−1 X X Yl rPkn−1 (θ, r) 1 Mk−l θ k−l−1 = + p 1 (ϕ)q 1 (ϕ) hj (q1 (ϕ), q1 (0))bj dϕ, rk−1 rk−1 K 0 j=1 l=1 Sl where Pkn−1 (ϕ, r) is a polynomial of degree kn − 1 in r. By (7), (9), (14) and since we are assuming that F0+ = 0 in (7), it follows that Z θ k−1 l ! XX 1 Y L bj yk (θ, r) =k! Fk (ϕ, r) + ∂ Fk−l (ϕ, r) yj (ϕ, r) dϕ 0 K j=1 l=1 Sl Z θ k! k!rPkn−1 (θ, r) = k−1 rFek (ϕ, r) + p1 (ϕ)q1 (ϕ)k−1 dϕ + r 0 rk−1 k−1 Z l k! X X Mk−l θ Y + k−1 p1 (ϕ)q1 (ϕ)k−l−1 hj (q1 (ϕ), q1 (0))bj dϕ r K 0 j=1 l=1 Sl Z θ k! k!P kn−1 (θ, r) hk 1 (θ), q1 (0)) (q = k−2 Fek (ϕ, r)dϕ + k−2 + , r 0 r rk−1 because dq1 (θ)/dθ = p1 (θ). Thus Z θ ỹk (θ, r) = k! Fek (ϕ, r)dϕ + k!Pkn−1 (θ, r) 0 from its definition given in (11). Since both Fek (ϕ, r) and Pkn−1 (θ, r) are polynomials of degree kn − 1 in r, we immediately get that ỹk (θ, r) is also a polynomial of degree kn − 1 in r. This ends the proof of the claim. From (6) and (11) the averaged function of order N associated to system (2) is + yN − (α, r) − yN (α − 2π, r) 1 FeN (r) (15) FN (r) = = , N! N ! rN −1 where FeN (r) := rỹ + (α, r) + hN (q + (α), q + (0)) − rỹ − (α − 2π, r) − hN (q − (α − 2π), q − (0)). N 1 1 N 1 1 + − Since ỹN (α, r) and ỹN (α − 2π, r) are polynomials of degree N n − 1 and N m − 1 respectively, FeN (r) is of degree N n due to n ≥ m. This concludes that FeN (r), or equivalently FN (r), has at most N n simple positive zeros, which directly gives statement (i) for N = 1 and statement (ii) for any N . 9 To obtain statement (i) for N ≥ 2 we shall prove (16) hN (q1+ (π), q1+ (0)) − hN (q1− (−π), q1− (0)) = 0, if N ≥ 2 provided that F1 = 0. In fact, by (15) F1 = 0 implies h1 (q1+ (π), q1+ (0)) − h1 (q1− (−π), q1− (0)) = 0. Besides it follows from (8) that q1+ (π) = g1+ (0, 0), q1− (−π) = g1− (0, 0), q1+ (0) = −g1+ (0, 0), q1− (0) = −g1− (0, 0). Thus together with the definition of h1 given in (10), we get g1+ (0, 0) = g1− (0, 0), which implies hN (g1+ (0, 0), −g1+ (0, 0)) − hN (g1− (0, 0), −g1− (0, 0)) = 0, if N ≥ 2, i.e. (16) holds. Consequently, FeN (r) = rỹN + (π, r) − rỹN− (−π, r) and the averaged function in (15) becomes + − 1 ỹN (π, r) − ỹN (−π, r) (17) FN (r) = N −2 , N! r + − if N ≥ 2. As we have proved that ỹN (θ, r) and ỹN (θ, r) are polynomials of degree N n − 1 and N m − 1 in r respectively, the averaged function in (17) has at most N n − 1 simple positive zeros because n ≥ m, i.e. statement (ii) holds for N ≥ 2.  For the piecewise polynomial differential system (2) Proposition 6 provides upper bounds for the maximum number of simple positive zeros of the corresponding averaged functions of order N . Next we study the realitzation of these upper bounds for N = 1, 2. Proposition 7. Consider the piecewise polynomial differential system  !  Xn Xn   −y + ε i i ai y , x + ε bi y if (x, y) ∈ Σ+ α, (18) (ẋ, ẏ) = i=0 i=0    (−y, x) if (x, y) ∈ Σ− . α For any given α ∈ (0, π] and n ≥ 1 there exists a choice of the parameters ai and bi such that the first order averaged function associated to (18) has exactly n simple positive zeros. Proof. Using the polar coordinates we write system (18) in the form (5) with n X n X F1+ (θ, r) = ai cos θ sini θri + bi sini+1 θri , F1− (θ, r) = 0. i=0 i=0 Then according to the definition in (6), the first order averaged function is Z π Xn F1 (r) = F1+ (θ, r)dθ = bi Si+1 (π)ri 0 i=0 if α = π. Thus F1 (r) is a complete polynomial of degree n because Si+1 (π) 6= 0, see (iii) of Lemma 5, which implies that we can choose bi in such a way that F1 (r) has exactly n simple positive zeros. If α ∈ (0, π), then F1 (r) becomes Xn   ai F1 (r) = sini+1 α + bi Si+1 (α) ri . i=0 i+1 In this case it is also a complete polynomial of degree n because sin α 6= 0 for α ∈ (0, π). Again we can choose ai in such that F1 (r) has exactly n simple positive zeros.  10 TAO LI AND JAUME LLIBRE Proposition 8. Consider the piecewise polynomial differential system  n >   X  −y+ε i i i−1 (ai x + bi y + ci xy )     i=0   if (x, y) ∈ Σ+  n n  π, (19) (ẋ, ẏ) =  X X    x+ε i di y + ε 2 ei y i     i=0 i=0   (−y, x) if (x, y) ∈ Σ− π, where n ≥ 1, (20) d0 = 0, ci = −i%i ai − idi , for i = 0, 1, ..., n and either %i = 1 if i is odd, or %i = 0 if i is even. For any given n there exists a choice of the parameters ai , bi , di and ei such that the second order averaged function associated to (19) has exactly 2n − 1 simple positive zeros. Proof. Using polar coordinates system (19) with (20) writes in the form (5) with n X n X n n 1X X F1+ (θ, r) = µi (θ)ri , F2+ (θ, r) = ei sini+1 θri + µi (θ)ri νi (θ)ri , (21) i=0 i=0 r i=0 i=0 F1− (θ, r) = 0, F2− (θ, r) = 0, where µi (θ) = ai (cosi+1 θ−i%i cos2 θ sini−1 θ)+bi cos θ sini θ+di (sini+1 θ−i cos2 θ sini−1 θ), (22) νi (θ) = ai (sin θ cosi θ − i%i cos θ sini θ) + bi sini+1 θ − di (i + 1) cos θ sini θ. From (i) and (iii) of Lemma 5 we have Z π cosi+1 θ−i%i cos2 θ sini−1 θdθ = 0 for i = 0, 1, 2, ..., n, 0 Z π sini+1 θ−i cos2 θ sini−1 θdθ = 0 for i = 1, 2, ..., n, 0 together with the definition in (6), we obtain that the first order averaged function vanishes. By (6) and (21) the second order averaged function is Z π Z θ Xn 2n + + + i 1X F2 (r) = F2 (θ, r)+∂F1 (θ, r) F1 (ϕ, r)dϕdθ = ei Si+1 (π)r + Vi ri , 0 0 i=0 r i=0 where X Z π Z θ (23) Vi = µi1 (θ)νi2 (θ) + i1 µi1 (θ) µi2 (ϕ)dϕdθ. i1 +i2 =i 0 0 0≤i1 ,i2 ≤n Rπ It is easy to obtain that V0 = (a0 + b0 )2 0 cos θ sin θdθ = 0 because d0 = 0. Hence F2 (r) becomes n X 2n X (24) F2 (r) = ei Si+1 (π)ri + Vi ri−1 , i=0 i=1 11 and it has at most 2n − 1 simple positive zeros. To obtain the realization of this number, we need more information on Vi . In particular, by (22) and (23) we can write Vi in the form X Vi = ξ1i (i1 , i2 )ai1 ai2 + ξ2i (i1 , i2 )bi1 bi2 + ξ3i (i1 , i2 )di1 di2 i1 +i2 =i 0≤i1 ≤i2 ≤n (25) X + ξ4i (i1 , i2 )ai1 bi2 + ξ5i (i1 , i2 )ai1 di2 + ξ6i (i1 , i2 )bi1 di2 , i1 +i2 =i 0≤i1 ,i2 ≤n and the following four statements will be proved, (a) ξ2i (i1 , i2 ) = ξ3i (i1 , i2 ) = ξ6i (i1 , i2 ) = 0 for any i1 and i2 ; (b) if i1 = i2 = i/2 ≥ 3 is odd, then ξ4i (i/2, i/2) < 0 and ξ4i−1 (i/2, i/2 − 1) < 0; (c) if i1 = i2 = i/2 ≥ 2 is even, then ξ5i (i/2, i/2) < 0 and ξ5i−1 (i/2, i/2 − 1) < 0; (d) if i is even, then ξ1i (i/2, i/2) = 0. Regarding statement (a), ξ2i (i1 , i2 ), ξ3i (i1 , i2 ) and ξ6i (i1 , i2 ) are the coefficients of bi1 bi2 , di1 di2 and bi1 di2 respectively. Then, according to (22) and (23), we get Z π Z θ i i i i i ξ2i (i1 , i2 ) = cos θ sin 2 θ sin 2 +1 θ + cos θ sin 2 θ cos ϕ sin 2 ϕdϕdθ = 0, 0 2 0 and Z π   i i i i+2 i ξ3i (i1 , i2 ) = sin 2 +1 θ − cos2 θ sin 2 −1 θ − cos θ sin θ 2 0 2 2   Z θ ! i i +1 i 2 i −1 i +1 i 2 i −1 + sin 2 θ − cos θ sin 2 θ sin 2 ϕ − cos ϕ sin 2 ϕdϕ dθ 2 2 0 2 Z π  Z θ ! i i +1 i 2 i −1 i +1 i 2 i −1 = sin 2 θ − cos θ sin 2 θ sin 2 ϕ − cos ϕ sin 2 ϕdϕ dθ 2 0 2 0 2 Z π  i i i i i =− sin 2 +1 θ − cos2 θ sin 2 −1 θ cos θ sin 2 θdθ = 0, 2 0 2 if i1 = i2 = i/2, while if i1 6= i2 then Z π Z θ ξ2i (i1 , i2 ) = cos θ sini1 θ sini2 +1 θ + i1 cos θ sini1 θ cos ϕ sini2 ϕdϕdθ 0 0 Z π Z θ + cos θ sini2 θ sini1 +1 θ + i2 cos θ sini2 θ cos ϕ sini1 ϕdϕdθ = 0, 0 0 12 TAO LI AND JAUME LLIBRE and Z π   ξ3i (i1 , i2 ) = sini1 +1 θ − i1 cos2 θ sini1 −1 θ −(i2 + 1) cos θ sini2 θ 0 Z !  θ + i1 sini1 +1 θ − i1 cos2 θ sini1 −1 θ sini2 +1 ϕ − i2 cos2 ϕ sini2 −1 ϕdϕ dθ 0 Z π   + sini2 +1 θ − i2 cos2 θ sini2 −1 θ −(i1 + 1) cos θ sini1 θ 0 Z !  θ i2 +1 2 i2 −1 i1 +1 2 i1 −1 + i2 sin θ − i2 cos θ sin θ sin ϕ − i1 cos ϕ sin ϕdϕ dθ 0 Z π  = − i1 sini1 +1 θ − i1 cos2 θ sini1 −1 θ cos θ sini2 θdθ Z0 π  − i2 sini2 +1 θ − i2 cos2 θ sini2 −1 θ cos θ sini1 θdθ = 0. 0 Here we used cos2 θ = 1 − sin2 θ and (iv) of Lemma 5 in the computation of ξ3i (i1 , i2 ). Furthermore Z π  ξ6i (i1 , i2 ) = cos θ sini1 θ −(i2 + 1) cos θ sini2 θ 0 Z θ + i1 cos θ sin θ sini2 +1 ϕ−i2 cos2 ϕ sini2 −1 ϕdϕdθ i1 0 Z π  + sin i2 +1 θ − i2 cos2 θ sini2 −1 θ sini1 +1 θ 0 Z  θ + i2 sini2 +1 θ − i2 cos2 θ sini2 −1 θ cos ϕ sini1 ϕdϕdθ 0 Z π = − (i1 + i2 + 1) cos2 θ sini1 +i2 θdθ 0 Z i1 + i2 + 1 π i1 +i2 +2 + sin θ−i2 cos2 θ sini1 +i2 θdθ i1 + 1 0 Z Z (i1 + i2 + 1)(i1 + i2 + 2) π i1 +i2 +2 (i1 + i2 + 1)2 π i1 +i2 = sin θdθ− sin θdθ i1 + 1 0 i1 + 1 0 =0, where we used cos2 θ = 1 − sin2 θ and (iv) of Lemma 5 in the second equality, and the last equality is a straightway application of (iii) of Lemma 5. These computations conclude statement (a). Now we prove statement (b). From (25) we know that ξ4i (i/2, i/2) and ξ4i−1 (i/2, i/2 − 1) are the coefficients of a 2i b 2i and a 2i b 2i −1 respectively. Since i/2 is odd, we get % 2i = 1 from 13 its definition given in Proposition 8. Thus, using (22) and (23), for i/2 ≥ 3 we get Z π  i i i i ξ4i (i/2, i/2) = cos 2 +1 θ − cos2 θ sin 2 −1 θ sin 2 +1 θdθ 0 2   Z θ  i i i i i + cos 2 +1 θ − cos2 θ sin 2 −1 θ cos ϕ sin 2 ϕ dϕdθ 2 2 0 Z π   i i i i + cos θ sin 2 θ sin θ cos 2 θ − cos θ sin 2 θ dθ 0 2 Z θ ! i i i i i + cos θ sin 2 θ cos 2 +1 ϕ − cos2 ϕ sin 2 −1 ϕdϕ dθ 2 0 2 Z π i i i =2 cos 2 +1 θ sin 2 +1 θ − cos2 θ sini θdθ 0 2 Z Z π i π i i i i + cos θ sin 2 θdθ cos 2 +1 θ − cos2 θ sin 2 −1 θdθ 2 0 0 2 ! i π 2 !! i (i − 1)!! = i − < 0, 2 2 ( 2i + 1)!! 2 ( 2i + 1)! where we used the integration by parts method in the second equality and statements (iii)(v) of Lemma 5 are applied to the third one. In a similar way we have Z π  i−1 i +1 i 2 i −1 i ξ4 (i/2, i/2 − 1) = cos 2 θ − cos θ sin 2 θ sin 2 θdθ 0 2   Z θ  i i i i i + cos 2 +1 θ − cos2 θ sin 2 −1 θ cos ϕ sin 2 −1 ϕ dϕdθ 2 2 0 Z π   i −1 i i i + cos θ sin 2 θ sin θ cos θ − cos θ sin θ dθ 2 2 0 2 Z θ ! i i −1 i +1 i 2 i −1 + ( − 1) cos θ sin 2 θ cos 2 ϕ − cos ϕ sin 2 ϕdϕ dθ 2 0 2 Z π i i i =2 cos 2 +1 θ sin 2 θ − cos2 θ sini−1 θdθ 0 2 Z π  Z θ  i i i −1 i + cos 2 +1 θ − cos θ sin 2 2 θ cos ϕ sin 2 −1 ϕ dϕdθ 0 2 0 Z π 2 i i i =(2 + ) cos 2 θ sin 2 θ − cos θ sini−1 θdθ +1 2 i 0 2 !Z i i π 2 !!( − 1)!! i =(2 + ) 2 2 − cos2 θ sini−1 θdθ i (i − 2)!! 2 0  i Z π =(i + 1) 21− 2 − 1 cos2 θ sini−1 θdθ < 0, 0 for i/2 ≥ 3, where we used the integration by parts method again in the second equality and the fourth one is due to (vi) of Lemma 5. So statement (b) holds. We see that ξ5i (i/2, i/2) and ξ5i−1 (i/2, i/2 − 1) are the coefficients of a 2i d 2i and a 2i d 2i −1 respectively. Moreover, % 2i = 0 because we are assuming that i/2 is even in statement (c). 14 TAO LI AND JAUME LLIBRE Thus, using (22) and (23) again, for i/2 ≥ 2 we get Z π   i i+2 i ξ5i (i/2, i/2) = cos 2 +1 θ − cos θ sin θ dθ 2 0 2 Z θ ! i i +1 i +1 i 2 i −1 + cos 2 θ sin 2 ϕ − cos ϕ sin 2 ϕdϕ dθ 2 0 2 Z π  i +1 i 2 i −1 i + sin 2 θ − cos θ sin 2 θ sin θ cos 2 θdθ 0 2  Z θ i i +1 i 2 i −1 i + sin 2 θ − cos θ sin 2 θ cos 2 +1 ϕdϕdθ 2 2 Z π Z π0 i i i i = − (i + 1) +2 cos 2 θ sin 2 θdθ + cos 2 θ sin 2 +2 θdθ 0 0 Z Z π i π i i i i + +1 cos 2 θdθ sin 2 θ − cos2 θ sin 2 −1 θdθ +1 2 0 0 2 Z π Z π i i i i = − (i + 2) cos 2 +2 θ sin 2 θdθ + cos 2 θ sin 2 θdθ 0 0 Z i π i i =− cos 2 θ sin 2 θdθ < 0, 2 0 where weRused the integration by parts method in the second equality and the third equality π i is due to 0 cos 2 +1 θdθ = 0 for i/2 even, and the last equality is obtained by (vi) of Lemma 5. Applying these techniques and (iv) of Lemma 5 to ξ5i−1 (i/2, i/2 − 1) we get Z π   i i i ξ5i−1 (i/2, i/2 − 1) = cos 2 +1 θ − cos θ sin 2 −1 θ dθ 0 2 Z θ ! i i +1 i i 2 i −2 + cos 2 θ sin ϕ − ( − 1) cos ϕ sin 2 2 ϕdϕ dθ 2 0 2 Z π  i i i i + sin 2 θ − ( − 1) cos2 θ sin 2 −2 θ sin θ cos 2 θdθ 0 2  Z θ i i i 2 i −2 i + ( − 1) sin θ − ( − 1) cos θ sin 2 2 θ cos 2 +1 ϕdϕdθ 2 2 0 Z π Z π i i i i −1 = − (i − 1) +2 cos 2 θ sin 2 θdθ + cos 2 θ sin 2 +1 θdθ 0 0 Z π Z θ ! i +1 i i 2 i −2 + cos 2 θ sin ϕ − ( − 1) cos ϕ sin 2 2 ϕdϕ dθ 0 0 2 Z π Z π i i i i =−i cos 2 +2 θ sin 2 −1 θdθ + cos 2 θ sin 2 +1 θdθ 0 0 Z π i i = − (i + 1) cos 2 θ sin 2 +1 θdθ < 0. 0 In summary statement (c) is proved. 15 Finally, since ξ1i (i/2, i/2) is the coefficient of a 2i a 2i , statement (d) follows from the fol- lowing computation Z π   i i +1 i 2 i −1 i i i ξ1 (i/2, i/2) = cos 2 θ − % 2i cos θ sin 2 θ sin θ cos θ − % 2i cos θ sin θ 2 2 0 2 2   Z θ ! i i +1 i 2 i −1 i +1 i 2 i −1 + cos 2 θ− % 2i cos θ sin 2 θ cos 2 ϕ− % 2i cos ϕ sin 2 ϕ dθ 2 2 0 2 Z π   i +1 i 2 i −1 i i i = cos 2 θ − % 2i cos θ sin 2 θ sin θ cos θ − % 2i cos θ sin θ dθ 2 2 0 2 2 Z π 2 i i i i + cos 2 +1 θ− % 2i cos2 θ sin 2 −1 θdθ 4 0 2 Z π   i +1 i 2 i −1 i i i = cos 2 θ − % 2i cos θ sin 2 θ sin θ cos θ − % 2i cos θ sin θ dθ 2 2 0 2 2 Z π   2 i +1 i i +1 i +1 i i +1 =− cos 2 θ + % 2i sin 2 θd cos 2 θ + % 2i sin 2 θ i+2 0 2 2 Z π   i i i i i − % 2i sin 2 −1 θ sin θ cos 2 θ − % 2i cos θ sin 2 θ dθ 0 2 2 Z π i i i = − % 2i sin 2 θ cos 2 θdθ = 0. 2 0 Here the third equality is obtained by a direct computation using (i)(iii) of Lemma 5, the last equality is obtain by joining (v) of Lemma 5 and the fact that % 2i = 0 if 2i is even. Under statements (a) and (d) the expression of Vi can be simplified as follows X X Vi = ξ1i (i1 , i2 )ai1 ai2 + ξ4i (i1 , i2 )ai1 bi2 + ξ5i (i1 , i2 )ai1 di2 . (26) i +i =i i +i =i 1 2 1 2 0≤i1 0, an−1 dn−2 < 0, an−2 bn−2 > 0, an−2 bn−3 < 0,..., en < 0, en−1 > 0, ..., e1 < 0, e0 > 0 in the next each step. Here the total number of steps is 2n − 1. From (26) and statement (b) we get F2 (r) = V2n r2n−1 with V2n = ξ42n (n, n)an bn < 0 when all parameters are zero except an and bn . In the first step we add the parameter bn−1 < 0, which satisfies an bn−1 < 0, and then from (26) and statements (b) the function F2 (r) is becomes (27) F2 (r) = V2n r2n−1 + V2n−1 r2n−2 , where V2n = ξ42n (n, n)an bn < 0 and V2n−1 = ξ42n−1 (n, n − 1)an bn−1 > 0. Thus one simple positive zero bifurcates from r = 0 by choosing bn−1 < 0 in such a way that |V2n−1 |  |V2n |. In the second step we further update F2 (r) adding the parameters an−1 and dn−1 satisfying an−1 dn−1 > 0. In this case (26) and statement (c) imply that (28) F2 (r) = V2n r2n−1 + V2n−1 r2n−2 + V2n−2 r2n−3 , with V2n−2 = ξ42n−2 (n − 1, n − 1)an−1 bn−1 + ξ52n−2 (n − 1, n − 1)an−1 dn−1 . Notice that V2n and V2n−1 in (28) may be different from the ones in (27) after adding the new parameters an−1 and dn−1 . We use the same notations only for the sake of simplification. Choosing an−1 and dn−1 such that V2n−2 < 0 and |V2n−2 |  |V2n−1 |, we get the second simple positive 16 TAO LI AND JAUME LLIBRE zero bifurcating from r = 0. Continue to add the parameters dn−2 with an−1 dn−2 < 0 and then F2 (r) = V2n r2n−1 + V2n−1 r2n−2 + V2n−2 r2n−3 + V2n−3 r2n−4 , where V2n−3 = ξ52n−3 (n − 1, n − 2)an−1 dn−2 > 0 by (26) and statement (c). Thus, in the third step the third simple positive zero occurs when |V2n−3 |  |V2n−2 |. Following the procedure, we finally obtain that 2n − 1 simple positive zeros bifurcate from r = 0. We note that it is reasonable to add the parameters ei for i = n, ..., 1, 0 from step n + 1 to 2n, because Si+1 (π) 6= 0 in (24). A similar procedure can be applied to the case where n is even, adding the parameters in the following order an dn > 0, an dn−1 < 0, an−1 bn−1 > 0, an−1 bn−2 < 0, an−2 dn−2 > 0, an−2 dn−3 < 0,..., en > 0, en−1 < 0,...,e1 < 0, e0 > 0 in the next 2n − 1 steps. This ends the proof of Proposition 8.  The procedure of choosing parameters in the proof of Proposition 8 relies on statements (a)-(d). As an example we give the explicit expression of F2 (r) for system (19) with n = 5, 25π 2 π F2 (r) = − a5 b5 r9 − (55a4 a5 +360a5 b4 +96a4 d5 )r8 − (40a5 b3 +15a3 b5 +12a4 d4 )r7 128 945 128 2 + (15a3 a4 − 55a2 a5 − 70a5 b2 − 24a3 b4 − 24a4 d3 − 16a2 d5 )r6 105 π − (10a5 b1 + 3a3 b3 + 4a4 d2 + 2a2 d4 − 5e5 )r5 16 2 − (3a2 a3 + 25a0 a5 + 25a5 b0 + 6a3 b2 + 12a4 d1 + 4a2 d3 − 8e4 )r4 15 π 2 1 − (3a3 b1 + 2a2 d2 − 3e3 )r3 − (3a0 a3 + 3a3 b0 + 2a2 d1 − 2e2 )r2 + e1 πr + 2e0 . 8 3 2 It is easy to see that the above F2 (r) satisfies all statements (a)-(d), and in each step we can add the parameters in the order a5 b5 , a5 b4 , a4 d4 , a4 d3 , e5 , e4 , e3 , e2 , e1 , e0 to produce 9 simple positive zeros. Proposition 9. Consider the piecewise polynomial differential system  n >   X  −y+ε i−1 (ai x y + bi y ) i     i=0   if (x, y) ∈ Σ+  n n  α, (29) (ẋ, ẏ) =  X X  i 2 i    x + ε ci x + ε di y     i=0 i=0  (−y − εb0 , x − εb0 ) if (x, y) ∈ Σ− α, where α ∈ (0, π/2], n ≥ 1, a0 = 0, ci = $i bi − ai with i = 0, 1, 2, ..., n, 2 sin α − cos α + 1 sini+1 α (30) $0 = , $i = i ≥ 1. cos α − 1 cosi+1 α − 1 For any given α and n there exists a choice of the parameters ai , bi and di such that the second order averaged function associated to (29) has exactly 2n simple positive zeros. 17 Proof. Writing system (29) in the form (5) and using the condition ci = $i bi − ai we obtain n X n X n n 1X X F1+ (θ, r) = ζi (θ)ri , F2+ (θ, r) = di sini+1 θri + ζi (θ)ri ηi (θ)ri , i=0 i=0 r i=0 i=0 1 2 F1− (θ, r) = −b0 (cos θ + sin θ), F2− (θ, r) = b (sin2 θ − cos2 θ), r 0 where ζi (θ) = bi ($i sin θ cosi θ + cos θ sini θ), ηi (θ) = ai cosi−1 θ + bi (sini+1 θ − $i cosi+1 θ). From (6) and (30) the first order averaged function is Xn Z α Z α−2π F1 (r) = ζi (θ)dθri + b0 cos θ + sin θdθ i=0 0 0 Xn $i (1 − cosi+1 α) + sini+1 α i = bi r + b0 (sin α − cos α + 1) = 0. i=0 i+1 Moreover the second order averaged function is Z α Z θ F2 (r) = F2+ (θ, r) + ∂F1+ (θ, r) F1+ (ϕ, r)dϕdθ 0 0 Z α−2π Z θ (31) − F2− (θ, r) + ∂F1− (θ, r) F1− (ϕ, r)dϕdθ 0 0 n X 2n 1X 1 = di Si+1 (α)ri + Wi ri + b20 sin α cos α, i=0 r i=0 r where X Z α Z θ Wi = ζi1 (θ)ηi2 (θ) + i1 ζi1 (θ) ζi2 (ϕ)dϕdθ i1 +i2 =i 0 0 0≤i1 ,i2 ≤n X = κi1 (i1 , i2 )bi1 ai2 + κi2 (i1 , i2 )bi1 bi2 , i1 +i2 =i 0≤i1 ,i2 ≤n and Z α κi1 (i1 , i2 ) = ($i1 sin θ cosi1 θ + cos θ sini1 θ) cosi2 −1 θdθ, 0 Z α κi2 (i1 , i2 ) = ($i1 sin θ cosi1 θ + cos θ sini1 θ)(sini2 +1 θ − $i2 cosi2 +1 θ) 0 Z θ i1 i1 + i1 ($i1 sin θ cos θ + cos θ sin θ) ($i2 sin θ cosi2 θ + cos θ sini2 θ)dθ. 0 Clearly rF2 (r) is a polynomial of degree 2n, implying that it has at most 2n simple positive zeros. To get the realization we need more information on the coefficient Wi . In particular we have the following statements for α ∈ (0, π/2] (e) κ02 (0, 0) + sin α cos α 6= 0; (f) κi1 (i1 , i2 ) 6= 0 when i1 ≥ 2 and i2 ≥ 2. 18 TAO LI AND JAUME LLIBRE Indeed, from the expression of κi2 (i1 , i2 ) we directly obtain Z α 1 − $02 0 κ2 (0, 0) = ($0 sin θ + cos θ)(sin θ − $0 cos θ)dθ = −$0 sin α cos α + sin2 α, 0 2 together with (30), which imply that   1 + $0 κ02 (0, 0) + sin α cos α = (1 − $0 ) sin α cos α + sin α = ($0 − 1) sin α 6= 0, 2 for α ∈ (0, π/2], i.e. statement (e) holds. Using the integration by parts method and (30) we get Z i2 − 1 α κi1 (i1 , i2 ) = (sini1 +1 θ − $i1 cosi1 +1 θ + $i1 ) cosi2 −2 θ sin θdθ i1 + 1 0 Z i2 − 1 α = ψi1 (θ)(cosi1 +1 θ − 1) cosi2 −2 θ sin θdθ, i1 + 1 0 where sini1 +1 θ sini1 +1 θ sini1 +1 α ψi1 (θ) = − $i = − . cosi1 +1 θ − 1 1 cosi1 +1 θ − 1 cosi1 +1 α − 1 Since dψi1 (θ) cosi1 θ − cos θ =(i1 + 1) sini1 θ , dθ (cosi1 +1 θ − 1)2 ψi1 (θ) is strictly decreasing in (0, π/2] when i1 ≥ 2, so that ψi1 (θ) > ψi1 (α) = 0 for θ ∈ (0, α) and α ∈ (0, π/2]. Consequently κi1 (i1 , i2 ) < 0 when i1 ≥ 2 and i2 ≥ 2, because the integrant is non-positive and is not identically zero, i.e. statement (f) holds. Consider the polynomial rF2 (r) where F2 (r) is given in (31). The constant term is W0 + b20 sin α cos α = b20 (κ02 (0, 0) + sin α cos α), because a0 = 0, and thus we can rewrite the polynomial as n+1 X 2n X rF2 (r) = b20 (κ02 (0, 0) + sin α cos α) + (di−1 Si (α) + Wi ) ri + Wi r i . i=1 n+2 Fixing b1 = b2 = · · · = bn = 1, we find that Wi with i = n + 2, n + 3, ..., 2n only contains the parameters ai−n , ai−n+1 , ..., an , and then Wi has one less parameter than Wi−1 . Moreover Wi with i = 1, 2, ..., n + 1 is independent of dj , j = 0, 1, 2, ..., n. Therefore joining statements (e), (f) and the fact that Si (α) > 0 for α ∈ (0, π/2], we get that all coefficients of rF2 (r) can be chosen arbitrarily, i.e. it is a complete polynomial of degree 2n. This concludes the proof of Proposition 9.  Finally, combining these propositions we can prove Theorem 1. Proof of Theorem 1. By the averaging theory statement (i) follows from statement (i) of Proposition 6, and Propositions 7 and 8. Statement (ii) follows from statement (ii) of Proposition 6, and Propositions 7 and 9.  19 4. Proof of Theorem 2 In this section we focus on the piecewise polynomial Liénard system (3) with N = 2 providing the proof of Theorem 2. In this section we take n X n X m X m X f1+ (x) = a+ i i x , f2+ (x) = b+ i i x , f1− (x) = a− i i x , f2− (x) = b− i i x . i=0 i=0 i=0 i=0 First we deal with the case of α = π where the switching boundary is a straight line. Proposition 10. Consider the piecewise polynomial Liénard system (3) with α = π and N = 2. For any given n ≥ m ≥ 1 the first (respectively second) order averaged function has at most [(n − 1)/2] (respectively n + [(m − 1)/2]) simple positive zeros. Moreover, these upper bounds are reached. Proof. For sake of convenience we rewrite f1± (x) as ± ± k X l X f1± (x) = a± 2j+1 x 2j+1 + a± 2j 2j x , j=0 j=0 where k + = [(n − 1)/2], k − = [(m − 1)/2], l+ = [n/2] and l− = [m/2]. Clearly, k + ≥ k − and l+ ≥ l− due to n ≥ m. Using polar coordinates system (3) becomes (5) with  F1± (θ, r) = cos θ R± (cos θ, r) + S ± (cos θ, r) , n X cos θ sin θ 2 F2+ (θ, r) = b+ i cos i+1 θri + R+ (cos θ, r) + S + (cos θ, r) , i=0 r m X cos θ sin θ 2 F2− (θ, r) = b− i cos i+1 θri + R− (cos θ, r) + S − (cos θ, r) , i=0 r where ± ± k X l X R± (cos θ, r) = a± 2j+1 cos 2j+1 θr2j+1 , S ± (cos θ, r) = a± 2j 2j 2j cos θr . j=0 j=0 By the definition in (6) and statement (i) of Lemma 5, the first order averaged function is + − k X k X F1 (r) = a+ 2j+1 C2j+2 (π)r 2j+1 + a− 2j+1 C2j+2 (π)r 2j+1 . j=0 j=0 Due to C2j+2 (π) 6= 0, all coefficients of F1 (r) can be chosen arbitrarily. By Theorem 4 and k + ≥ k − , eventually, F1 (r) has at most k + = [(n − 1)/2] simple positive zeros and there is a choice of the parameters a± i such that the maximum is reached. In order to compute the second order averaged function, we have to take F1 = 0, which is obviously equivalent to − (32) a+ 2j+1 = −a2j+1 for j = 0, 1, ..., k − , a+ 2j+1 = 0 for j = k − + 1, ..., k + , 20 TAO LI AND JAUME LLIBRE because k + > k − and C2j+2 (π) 6= 0. Notice that this condition implies R+ (cos θ, r) ≡ −R− (cos θ, r). We compute the following integrals, Z π Xn Z 1 π 2 F2+ (θ, r)dθ = Ci+1 (π)b+ i r i + cos θ sin θ R+ (cos θ, r)+S + (cos θ, r) dθ 0 i=0 r 0 Xn Z 1 1 2 = Ci+1 (π)b+ i r i + x R+ (x, r) + S + (x, r) dx i=0 r −1 Xn Z 4 1 = Ci+1 (π)b+ i r i + xR+ (x, r)S + (x, r)dx i=0 r 0 n X k+ X +l+ X 1 = Ci+1 (π)b+ i r i +4 a+ a+ r2j , i=0 j=0 j1 +j2 =j 2j + 3 2j1 +1 2j2 0≤j1 ≤k+ ,0≤j2 ≤l+ Z −π m X k− X +l− X 1 F2− (θ, r)dθ =− Ci+1 (π)b− i r i +4 a− a− r2j , 0 i=0 j=0 j1 +j2 =j 2j + 3 2j1 +1 2j2 0≤j1 ≤k− ,0≤j2 ≤l− and Z ±π Z θ Z ±π Z θ ∂F1± (θ, r) F1± (ϕ, r)dϕdθ = cos θ∂R± (cos θ, r) cos ϕR± (cos ϕ, r)dϕdθ 0 0 0 0 Z ±π Z θ + cos θ∂R± (cos θ, r) cos ϕS ± (cos ϕ, r)dϕdθ 0 0 Z ±π Z θ + cos θ∂S ± (cos θ, r) cos ϕR± (cos ϕ, r)dϕdθ 0 0 ± 2k X X = (2j1 +1)C2j1 +2,2j2 +2 (π)a± ± 2j1 +1 a2j2 +1 r 2j+1 j=0 j1 +j2 =j 0≤j1 ≤k± ,0≤j2 ≤k± k± X +l± X + (2j1 +1)C2j1 +2,2j2 +1 (π)a± ± 2j 2j1 +1 a2j2 r j=0 j1 +j2 =j 0≤j1 ≤k± ,0≤j2 ≤l± k± X +l± X + 2j2 C2j2 +1,2j1 +2 (π)a± ± 2j 2j1 +1 a2j2 r . j=0 j1 +j2 =j 0≤j1 ≤k± ,0≤j2 ≤l± The above computations together with the definition (6) and the condition (32), yield the second order averaged function n X k− X +l+ X F2 (r) = Ci+1 (π)b+ i i r + ϑj1 ,j2 a+ + 2j 2j1 +1 a2j2 r i=0 j=0 j1 +j2 =j 0≤j1 ≤k− ,0≤j2 ≤l+ m X k− X +l− X + Ci+1 (π)b− i r i + ϑj1 ,j2 a+ − 2j 2j1 +1 a2j2 r , i=0 j=0 j1 +j2 =j 0≤j1 ≤k− ,0≤j2 ≤l− 21 where 4 ϑj1 ,j2 = + (2j1 + 1)C2j1 +2,2j2 +1 (π) + 2j2 C2j2 +1,2j1 +2 (π). 2j + 3 It follows from (i) of Lemma 5 that Ci+1 (π) = 0 for i even and Ci+1 (π) 6= 0 for i odd, the first and third summations in F2 (r) contribute only to the odd terms r2j+1 , which are from j = 0 to [(n − 1)/2] because n ≥ m. Moreover the second and fourth ones contribute only to the even terms r2j , which are from j = 0 to k − +l+ because l+ ≥ l− . Then the total number of terms is at most [(n − 1)/2] + k − + l+ + 2 = n + [(m − 1)/2] + 1 using the definitions of k − and l+ . Therefore F2 (r) has at most n + [(m − 1)/2] simple positive zeros by Theorem 4. To get the realization of this upper bound, we prove ϑj1 ,j2 6= 0 if j2 ≥ j1 in what follows. In fact, by (i) and (x) of Lemma 5, ϑj1 ,j2 can be simplified to 4 ϑj1 ,j2 = + (2j1 + 1 − 2j2 )C2j1 +2,2j2 +1 (π). 2j + 3 If j2 = j1 , we have ϑj1 ,j1 > 0 directly, because C2j1 +2,2j1 +1 (π) > 0 from (viii) of Lemma 5. If j2 = j1 + 1, we have the recursion formula −4 2j1 + 1 ϑj1 ,j1 +1 = + ϑj −1,j1 , (2j1 + 3)(4j1 + 3)(4j1 + 1) 2j1 + 3 1 using property (vii) of Lemma 5. Hence, by induction we can prove that ϑj1 ,j1 +1 > 1/(2j1 + 3)2 > 0 for j1 ≥ 0, where ϑ0,1 = 2/9 > 1/9. If j2 = j1 + 2, we have the recursion formula 4(14j1 + 13) (2j1 + 4)(2j1 + 1) ϑj1 ,j1 +2 = + ϑj −1,j1 +1 , (2j1 + 5)(2j1 + 2)(4j1 + 5)(4j1 + 3) (2j1 + 5)(2j1 + 2) 1 using property (vii) of Lemma 5 again. By induction we can prove that ϑj1 ,j1 +2 < −1/(j1 + 3/2)2 < 0, where ϑ0,2 = −74/75 < −4/9. If j2 = j1 + k with k ≥ 2 now, we have −2(8j1 + 4k 2 + 3) (2j1 + 2k)(2k − 1) ϑj1 ,j1 +k = + ϑj ,j +k−1 . (2j1 + 2k + 1)(2k − 3)(4j1 + 2k + 1) (2j1 + 2k + 1)(2k − 3) 1 1 Since ϑj1 ,j1 +2 < 0 and we are assuming that k ≥ 2, we finally get ϑj1 ,j1 +k < 0 for k ≥ 2 by induction again. In conclusion we have that ϑj1 ,j2 6= 0 if j2 ≥ j1 . Let a−2j2 = 0 for j2 = 0, 1, ..., l − and b− i = 0 for i = 0, 1, ..., m. Joining the fact that Ci+1 (π) = 0 if i is even, the polynomial F2 (r) reduces to k X + k− X +l+ X (33) F2 (r) = C2j+2 (π)b+ 2j+1 r 2j+1 + ϑj1 ,j2 a+ + 2j 2j1 +1 a2j2 r . j=0 j=0 j1 +j2 =j 0≤j1 ≤k− ,0≤j2 ≤l+ To obtain n + [(m − 1)/2] simple positive zeros, we now choose the parameters a+ + 2j1 +1 , a2j2 and b+ 2j+1 following the next procedure. We start letting all parameters equal zero except a+ 0 = a+1 = 1, and then, in the next each step we add the rest of parameters one by one the order b+ + + + + + + + + + + 1 , a2 , b3 , a4 , ..., b2k+ −1 , a2k+ , b2k+ +1 , a2l+ , a3 , a5 , ..., a2k− +1 if n is even, while if n is + + odd, we need to interchange a2l+ and b2k+ +1 because l = k + + 1 (respectively k + ) for n + even (respectively odd). The sign of the added parameters must ensure that two adjacent terms have a variation of sign and it can be determined using the fact that C2j+2 (π) > 0 and ϑj1 ,j2 > 0 (respectively < 0) for j1 ≤ j2 ≤ j1 + 1 (respectively j2 ≥ j1 + 2). More precisely, in the first step we add b+ 1 < 0 and one simple positive zero occurs, because F2 (r) = C2 (π)b+ 1 r + ϑ 0,0 with ϑ 0,0 > 0 and C2 (π) > 0. The second one bifurcates from the infinity when we further add the parameter a+ 2 > 0 sufficiently small due to ϑ0,1 > 0. Then, 22 TAO LI AND JAUME LLIBRE due to C4 (π) > 0, in the third step we add b+ 3 3 < 0 in the way that the coefficient of r is far 2 less than the one of r , so the third simple positive zero bifurcates from infinity. Repeating the above procedure with the specific order, we can obtain n + [(m − 1)/2] simple positive zeros.  As an example we give the explicit expression of F2 (r) in (33) for n = 9 and m = 5, π 2 3π 14 74 F2 (r) = 2a0 a1 + b1 r + ( a1 a2 + 2a0 a3 )r2 + b3 r3 + ( a2 a3 − a1 a4 + 2a0 a5 )r4 2 9 8 15 75 5π 58 26 1426 35π + b5 r5 + ( a3 a4 + a2 a5 − a1 a6 )r6 + b7 r 7 16 525 21 735 128 578 702 38914 63π +( a4 a5 − a3 a6 − a1 a8 )r8 + b9 r 9 945 1225 14175 256 1774 84806 381454 +( a5 a6 − a3 a8 )r10 − a5 a8 r12 , 24255 72765 945945 where we drop the superscript + to alleviate the notation. According to the last proof, we first let all parameters equal zero except a0 = a1 = 1 and 11 simple positive ze- ros can bifurcate from the infinity by perturbing the rest parameters following the order b1 , a2 , b3 , a4 , b5 , a6 , b7 , a8 , b9 , a3 , a5 . Now we focus on α ∈ (0, π) where the switching boundary is non-regular. Proposition 11. Consider the piecewise polynomial Liénard system (3) with N = 2. For any given α ∈ (0, π) and n ≥ m ≥ 1 the first order averaged function has at most n simple positive zeros, which are reached, and the second order averaged function has at most max{2m − 1, n} (respectively max{2m − 2, n}) simple positive zeros if m is odd (respectively even). Proof. In polar coordinates system (3) writes in the form (5) with n X m X F1+ (θ, r) = a+ i cos i+1 θri , F1− (θ, r) = a− i cos i+1 θri , i=0 i=0 n n !2 X 1 X F2+ (θ, r) = b+ i cos i+1 i θr + cos θ sin θ a+ i i cos θr i , i=0 r i=0 m m !2 X 1 X F2− (θ, r) = b− i cos i+1 i θr + cos θ sin θ a− i i cos θr i . i=0 r i=0 Thus the first order averaged function is n X m X F1 (r) = a+ j i Ci+1 (α)r − a− j i Ci+1 (α − 2π)r . i=0 i=0 Since α ∈ (0, π), we know Ci+1 (α) 6= 0 and Ci+1 (α − 2π) 6= 0 for any i ≥ 0 from (ii) of Lemma 5, so that all coefficients of F1 (r) are free, i.e. F1 (r) is a complete polynomial of degree n because n ≥ m. Therefore F1 (r) has at most n simple positive zeros and there is a choice of the parameters a± i such that the maximum is reached. 23 To get the second order averaged function, we have to take F1 = 0, which is equivalent to − a+ i Ci+1 (α) = ai Ci+1 (α − 2π) for i = 0, 1, ..., m, and (34) a+ i =0 for i = m + 1, ..., n. A direct computation yields the following integrals Z n Z n !2 α X 1 α X F2+ (θ, r)dθ = b+ i Ci+1 (α)r i + cos θ sin θ a+ i i cos θr i dθ 0 i=0 r 0 i=0 n Z n !2 X 1 cos α X = b+ i Ci+1 (α)r i − x a+ i i i x r dx i=0 r 1 i=0 n X 2n 1X X cosi+2 α − 1 + + i = b+ i i Ci+1 (α)r − ai1 ai2 r , i=0 r i=0 i1 +i2 =i i+2 0≤i1 ,i2 ≤n Z m X 2m α−2π 1X X cosi+2 α − 1 − − i F2− (θ, r)dθ = b− i i Ci+1 (α − 2π)r − ai1 ai2 r , 0 i=0 r i=0 i1 +i2 =i i+2 0≤i1 ,i2 ≤m and Z α Z θ 2n X X ∂F1+ (θ, r) F1+ (ϕ, r)dϕdθ = i1 Ci1 +1,i2 +1 (α)a+ + i−1 i1 ai2 r , 0 0 i=0 i1 +i2 =i 0≤i1 ,i2 ≤n Z α−2π Z θ 2m X X ∂F1− (θ, r) F1− (ϕ, r)dϕdθ = i1 Ci1 +1,i2 +1 (α − 2π)a− − i−1 i1 ai2 r . 0 0 i=0 i1 +i2 =i 0≤i1 ,i2 ≤m Combining the above computations and the condition (34), we get the second order averaged function n X m X 2m 1X X F2 (r) = b+ i i Ci+1 (α)r − b− i i Ci+1 (α − 2π)r + ωi1 ,i2 a+ + i i1 ai2 r , i=0 i=0 r i=0 i1 +i2 =i 0≤i1 ,i2 ≤m where cosi+2 α − 1 cosi+2 α − 1 Ci1 +1 (α)Ci2 +1 (α) ωi1 ,i2 = − + i1 Ci1 +1,i2 +1 (α) + i+2 i+2 Ci1 +1 (α − 2π)Ci2 +1 (α − 2π) Ci1 +1 (α)Ci2 +1 (α) − i1 Ci1 +1,i2 +1 (α − 2π) . Ci1 +1 (α − 2π)Ci2 +1 (α − 2π) Using (i) and (ii) of Lemma 5 we have Ci+1 (α) = Ci+1 (α − 2π) if i is even, and using (ix) of Lemma 5 we have Ci1 +1,i2 +1 (α) = Ci1 +1,i2 +1 (α − 2π) if i2 is even. So ωi1 ,i2 = 0 if both i1 ≥ 0 and i2 ≥ 0 are even. This implies that the constant term in the third summation of F2 (r) vanishes and the highest degree term r2m also vanishes when m is even. Consequently due to n ≥ m, F2 (r) is a polynomial of degree max{2m − 1, n} (respectively max{2m − 2, n}) when m is odd (respectively even), in other words F2 (r) has at most max{n, 2m − 1} (respectively max{n, 2m − 2}) simple positive zeros.  24 TAO LI AND JAUME LLIBRE Finally we can prove Theorem 2. Proof of Theorem 2. Using the averaging theory it follows statement (i) of Theorem 2 from Proposition 10, and statement (ii) from Proposition 11.  5. Conclusions In this paper we extended the weak Hilbert’16 problem to discontinuous piecewise poly- nomial differential systems with two zones separated by the switching boundary Σα , either a straight line when α = π or a non-regular one when α ∈ (0, π). Here we allow that the degree of each subsystem in the two zones be different. More precisely, we studied the maximum number of crossing limit cycles bifurcating from the periodic annulus of the linear center ẋ = −y, ẏ = −x when we perturb it inside this class of general piecewise polynomial differential systems. Depending on α = π or not, we provided upper bounds for the max- imum number using the averaging method up to any order. Besides we also restricted the perturbation to the family of piecewise polynomial Liénard systems and some better upper bounds were given. As observed by many researchers, our results emphasize the importance of the shape of the switching boundary in the investigation of limit cycles. We proved that all upper bounds obtained with the first order averaging method are reached. Regarding the upper bounds obtained with the second order averaging method, the main difficulty in the study of the realization, is to determine which terms of the cor- responding averaged functions are not identically zero and then to give a suitable choice of parameters. Overcoming these difficulties, for α = π we proved the realization whatever the perturbation is inside the general piecewise polynomial family or piecewise Liénard one, while for α ∈ (0, π), the realization was obtained only in the case of the general piece- wise polynomial perturbations with α ∈ (0, π/2]. Hence, what about the general piecewise polynomial perturbations with α ∈ (π/2, π) and the piecewise Liénard perturbations with α ∈ (0, π)? These need to be answered in the future. Acknowledgements The first author is partially suported by the grant CSC #201906240094 from the P.R. China. 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Gerstner, W. Piecewise-linear differential equations and integrate-and-fire neurons: Insights from two-dimensional membrane models, Phys. Rev. E 67 (2003), 021908. [33] L. Wei, X. Zhang, Averaging theory of arbitrary order for piecewise smooth differential systems and its application, J. Dyn. Diff. Equat. 30 (2018), 55–79. 1 Department of Mathematics, Sichuan University, 610064 Chengdu, Sichuan, P.R. China Email address: t.l.scumath@gmail.com 26 TAO LI AND JAUME LLIBRE 2 Departament de Matemàtiques, Universitat Autònoma de Barcelona, 08193 Bellaterra, Barcelona, Catalonia, Spain Email address: jllibre@mat.uab.cat