Upper bounds for the number of zeroes for some Abelian Integrals

UPPER BOUNDS FOR THE NUMBER OF ZEROES FOR SOME ABELIAN INTEGRALS ARMENGOL GASULL, J. TOM´ AS L´ AZARO, AND JOAN TORREGROSA Abstract. Consider the vector field x′=−yG(x, y), y′=xG(x, y),where the set of critical points {G(x, y) = 0}is formed by Kstraight lines, not passing through the origin and parallel to one or two orthogonal directions. We perturb it with a general polynomial perturbation of degree nand study which is the maximum number of limit cycles that can bifurcate from the period annulus of the origin in terms of Kand n. Our approach is based on the explicit computation of the Abelian integral that controls the bifurcation and in a new result for bounding the number of zeroes of a certain family of real functions. When we apply our results for K≤4 we recover or improve some results obtained in several previous works. 1. Introduction The problem of determining the number of limit cycles bifurcating from the period annulus of a system (˙x=−yG(x, y) + ε P (x, y), ˙y=xG(x, y) + ε Q(x, y),(1) where P(x, y), Q(x, y) are polynomials of a given degree, G(x, y) satisfies G(0,0) 6= 0 and εis a small parameter, has been widely studied (see for instance [2, 4, 7, 8, 9, 10, 12, 13]). Among this type of systems we will be concerned with those having G(x, y) = K1 Y j=1 (x−aj) K2 Y ℓ=1 (y−bℓ),(2) where ajand bℓare real numbers with ai6= ajand bi6= bjfor i6=j. The unperturbed system (ε= 0) presents a centre at the origin and any line x= ajor y= bℓconstitutes an invariant set of singular points of the system. This invariant set is formed by parallel and/or orthogonal invariant lines. The aim of this work is to provide, for small values of ε, upper bounds for the number of limit cycles bifurcating from periodic orbits of the unperturbed system in the period 2010 Mathematics Subject Classification. Primary: 34C08. Secondary: 34C07, 34C23, 37C27, 41A50. Key words and phrases. Abelian integrals, Weak 16th Hilbert’s Problem, limit cycles, Chebyshev system, number of zeroes of real functions. The first and third authors are partially supported by the MICIIN/FEDER grant number MTM200803437 and by the Generalitat de Catalunya grant number 2009SGR410. The second author is partially supported by the MICIIN/FEDER grant number MTM2009-06973 and by the Generalitat de Catalunya grant number 2009SGR859. 1 2 A. GASULL, J. T. L ´ AZARO, AND J. TORREGROSA annulus D=(x, y)∈R2|0<px2+y2< ρ := min j,ℓ {|aj|,|bℓ|}. Several previous works handle this problem for different particular choices of small values of K1and K2.The following cases have been studied: one line in [10]; two parallel lines in [13]; two orthogonal lines in [2]; three lines, two of them parallel and one perpendicular in [5]; and four lines with a special configuration in [1]. Other related works are [8], with G(x, y) any quadratic polynomial; [12], one multiple singular line; or [7] for Kisolated singular points. As it is standard in this type of problems, we rewrite our system (1) in the equivalent form on D,˙x=−y+ε P(x, y)/G(x, y), ˙y=x+ε Q(x, y)/G(x, y),(3) where G(x, y) is defined in (2). Let us denote by γr={(x, y)|x2+y2=r2},with 0 < r < ρ, any periodic orbit of the unperturbed system. It is well known that the associated return map, whose isolated zeroes give rise to limit cycles, is Π(r, ε) = r+I(r)ε+O(ε2) where I(r) is the Abelian integral I(r) = Zγr Q(x, y)dx −P(x, y)dy G(x, y).(4) This integral is the so-called (first-order) Poincar´e-Melnikov-Pontryagin function. The return map Π(r, ε) is analytic in rranging on any compact subset of (0, ρ) for εsmall enough. It is known (see for instance [3]) that, provided I(r) does not vanish exactly, the number of zeroes of Π(r, ε),for εsmall enough, is at most the number of zeroes of I(r) taking into account their multiplicity. The problem of estimating this number in terms of the involved degrees is commonly called the weakened Hilbert’s 16th Problem. The goal of this paper is to provide an upper bound for the number of zeroes of the Abelian integral associated to a system of the form (1), depending on the number of critical straight lines and the degree nof the perturbative polynomials P(x, y) and Q(x, y).We prove: Theorem 1.1. Consider a system of the form (1), (˙x=−yG(x, y) + ε P (x, y), ˙y=xG(x, y) + ε Q(x, y), where G(x, y) = K1 Y j=1 (x−aj) K2 Y ℓ=1 (y−bℓ), P(x, y), Q(x, y)are polynomials of degree n, ajand bℓare real numbers with ai6= ajand bi6= bjfor i6=j, ε is a small parameter and K1≥K2≥0.Moreover, when K2= 0, K1≥1and Q0 ℓ=1(y−bℓ) := 1. UPPER BOUNDS FOR THE NUMBER OF ZEROES 3 Let I(r)be its associated Abelian integral defined in (4). Then, the number of real zeroes of I(r)in (0,min{|aj|,|bℓ|}),counting their multiplicities, Z(I),satisfies Z(I)≤      e K1n+ 3 2+hn 2i, K2= 0, (e K1+e K2)n+ 2 2+L+n−1 2+L, K2≥1, where e K1= Card{|a1|,...,|aK1|} ≤ K1, e K2= Card{|a1|,...,|aK1|,|b1|,...,|bK2|}− e K1≤K2, L= Card{a2 j+ b2 ℓ, j = 1,...,K1, ℓ = 1,...,K2} ≤ K1K2 and [s]denotes the integer part of s. Note that the symmetric situation, K2≥K1,evolves in a completely similar way changing (x, y) by (y, x). Recall that using this approach we know that the total number of limit cycles (counting their multiplicities) of system (1) which bifurcate from its periodic orbits is bounded by the maximum number of isolated zeroes (counting their multiplicities) of I(r) for 0< r < ρ = min{|aj|,|bℓ|}. The proof of Theorem 1.1 is based on two steps: a first one where the corresponding Abelian integral (4) is explicitly computed and a second one where an upper estimate on the number of its zeroes is provided. This is done in Section 2 and 4, respectively. This second part is supported on the following result, proved in Section 3, that we believe it is interesting by itself. Theorem 1.2. Consider a function of the form F(x) = P0(x) + K X j=1 Pj(x)1 √x+cj ,(5) where Pj(x), j = 0,...,K, are real polynomials and cj, j = 1,...,K,are real constants. Then its number of real zeroes, taking into account their multiplicities, Z(F),satisfies Z(F)≤Kmax j=1,...,K deg(Pj)+ 1+ deg(P0).(6) Here deg(0) = −1. In the forthcoming paper [6] the above result is extended to a wider family of functions, studying in particular its sharpness and its relation with the theory of Chebyshev systems. We only comment here that when deg(Pj) coincide for all j= 1,...,K it can be seen that the result is sharp. In Section 5 we apply Theorem 1.1 to some particular cases, already studied by other authors, all satisfying K1+K2≤4.More concretely, in that cases we show that our theorem either gives new proofs or improve the known results for the upper bounds for the number of zeroes of I(r). The main differences between our work and the previous ones are: 4 A. GASULL, J. T. L ´ AZARO, AND J. TORREGROSA - We manage to study the case of having an arbitrary number Kof lines of critical points for the unperturbed system. All previous results consider at most 4 lines with several relative positions. - We prove a result (Theorem 1.2) for bounding the number of zeroes of a special type of functions which are precisely the ones that appear in the final expression of the Abelian integral I(r).The previous works apply general methods like, squaring the equation to eliminate radicals, or the principle of the argument, extending the function to C.It can be seen that when there are more than two square roots in the expression (5), Theorem 1.2 is sharper than those general methods, see [6]. This is the reason for which we can improve some of the previously given upper bounds for Z(I). Some comments about the sharpness of the upper bounds provided by Theorem 1.1 are given at the end of Section 4. 2. Explicit computation of the Abelian integral The aim of this section is to obtain an explicit expression of the Abelian integral presented in the introduction. The first two lemmas deal with the cases of one and two perpendicular singular lines. They are already known, see [2, 10], but the proof that we present is shorter. The next two results extend them to the case of an arbitrary number of parallel or perpendicular lines. Lemma 2.1. Let abe a non-zero real number. For any 0< r < |a|and any polynomial Rn+1(x, y)of degree n+ 1,define Ia n+1(r) = Z2π 0 Rn+1(rcos θ, r sin θ) rcos θ−adθ. Then, for n≥0,one has Ia n+1(r) = S[(n−1)/2]+1(r2) √a2−r2+T[n/2](r2), for suitable polynomials Ss(ρ)and Ts(ρ)of degree s. Moreover Ia 0(r) = −2πR0/√a2−r2. Proof. It is easy to check that Z2π 0 1 rcos θ−adθ =−2π1 √a2−r2. Hence the expression for Ia 0(r) follows. Let us now deal with n≥0.We will proceed inductively. When n= 0 we have that Ia 1(r) = Z2π 0 a0,0+a0,1rcos θ+a1,0rsin θ rcos θ−adθ. UPPER BOUNDS FOR THE NUMBER OF ZEROES 5 For this case we can write Ia 1(r) =Ia 0(r) + a0,1Z2π 0 rcos θ rcos θ−adθ +a1,0Z2π 0 rsin θ rcos θ−adθ = Ia 0(r) + a0,1Z2π 0 rcos θ−a + a rcos θ−adθ =Ia 0(r) + 2πa0,1+a0,1aZ2π 0 1 rcos θ−adθ = 2πa0,1−2π(a0,0+a0,1a) 1 √a2−r2. Now we will prove the expression for Ia n+1(r) by induction on the degree. By hypothesis of induction it is enough to prove the formula when Rn+1(x, y) is a homogeneous polynomial of degree n+ 1.If we write Z2π 0 n+1 P i=0 rn+1ai,n+1−isiniθcosn+1−iθ rcos θ−adθ =rn+1 n+1 X i=0 Z2π 0 ai,n+1−isiniθcosn+1−iθ rcos θ−adθ, using symmetry properties of the integrated functions, we know that all the integrals with an odd exponent in sin θare zero. Then we have rn+1 [(n+1)/2] X j=0 a2j,n+1−2jZ2π 0 sin2jθcosn+1−2jθ rcos θ−adθ = rn+1 [(n+1)/2] X j=0 a2j,n+1−2jZ2π 0 (1 −cos2θ)jcosn+1−2jθ rcos θ−adθ = [(n+1)/2] X j=0 a2j,n+1−2j j X k=0 r2j−2k(−1)jj kZ2π 0 rn+1+2k−2jcosn+1−2j+2kθ rcos θ−adθ. When k < j we can use again the induction hypotheses for each term of the sum and the statement is proved because when we multiply by the monomial (r2)j−k,the degrees, in r2,of the polynomials S(r2) and T(r2) are ([(n+ 2k−2j−1)/2] + 1) + (j−k) = [(n−1)/2] + 1 and [(n+ 2k−2j)/2] + (j−k) = [n/2],respectively. The unique case that remains to check is k=j. For it we can write Z2π 0 rn+1 cosn+1 θ rcos θ−adθ =Z2π 0 rncosnθ(rcos θ−a + a) rcos θ−adθ = Z2π 0 rncosnθdθ + a Z2π 0 rncosnθ rcos θ−adθ. We will treat separately the cases neven and nodd. When nis even the first term of the above sum is a polynomial of degree n/2 in r2 and for the second term, using the induction hypothesis, the polynomials S(r2) and T(r2) are of degree [((n−1) −1)/2] + 1 = (n−2)/2 + 1 = n/2 = [(n−1)/2] + 1 and [(n−1)/2] = (n−2)/2,respectively. The statement is proved, in this case, because we should add the monomial of degree n/2 corresponding to the first summand. Hence the degree of the polynomial T(r2) is n/2 = [n/2]. 6 A. GASULL, J. T. L ´ AZARO, AND J. TORREGROSA When nis odd the first term in the sum is zero. For the second term, from the induction hypothesis, the polynomials S(r2) and T(r2) are of degree [((n−1)−1)/2]+1 = (n−3)/2 + 1 = (n−1)/2 = [(n−1)/2] and [(n−1)/2] = [n/2],respectively. Note that the contribution of this term to the polynomial S(r2) has one degree less than expected but the total degree is [(n−1)/2] + 1 because it appears for the other terms. More concretely in the (sin θ)[(n+1)/2] term.  Lemma 2.2. Let aand bbe non-zero real numbers. For any 0< r < min(|a|,|b|)and any polynomial Rn+1(x, y)of degree n+ 1 consider Ia,b n+1(r) = Z2π 0 Rn+1(rcos θ, r sin θ) (rcos θ−a)(rsin θ−b) dθ. Then, for n≥0,we have that Ia,b n+1(r) = 1 a2+ b2−r2U[n/2]+1(r2) √a2−r2+V[n/2]+1(r2) √b2−r2+W[(n−1)/2](r2), for some given polynomials Us(ρ), Vs(ρ)and Ws(ρ)of degree s. Moreover Ia,b 0(r) = 2πR0 a2+ b2−r2a √a2−r2+b √b2−r2. Proof. Applying for instance residues formula, we obtain Z2π 0 dθ (rcos θ−a)(rsin θ−b) =2π a2+ b2−r2a √a2−r2+b √b2−r2. Therefore the expression for Ia,b 0(r) follows. To obtain the result for all nwe will proceed inductively. When n= 0 we have that Ia,b 1(r) = Z2π 0 a0,0+a0,1rcos θ+a1,0rsin θ (rcos θ−a)(rsin θ−b) dθ. For this case we can write Ia,b 1(r) =Ia,b 0(r) + a0,1Z2π 0 rcos θ−a + a (rcos θ−a)(rsin θ−b) dθ+ a1,0Z2π 0 rsin θ−b + b (rcos θ−a)(rsin θ−b)dθ =Ia,b 0(r) + a0,1Z2π 0 1 rsin θ−bdθ+ a1,0Z2π 0 1 rcos θ−adθ + (a0,1a + a1,0b) Z2π 0 1 (rcos θ−a)(rsin θ−b) dθ = a0,1−2π √b2−r2+a1,0−2π √a2−r2+a0,0+a0,1a + a1,0b a2+ b2−r22πa √b2−r2+2πb √a2−r2. This last expression satisfies the statement with U(r2) and V(r2) polynomials of degree [n/2] + 1 = [0/2] + 1 = 1 in r2and the polynomial W(r2) is identically zero. Now we will prove the expression for Ia,b n+1(r) by induction on the degree of Rn(x, y). As in the proof of Lemma 2.1, by induction hypothesis, it is enough to prove the result UPPER BOUNDS FOR THE NUMBER OF ZEROES 7 when Rn+1(x, y) is a homogeneous polynomial of degree n+ 1.For this case we can write Z2π 0 n+1 P i=0 rn+1ai,n+1−isiniθcosn+1−iθ (rcos θ−a)(rsin θ−b) dθ = n+1 X i=0 Z2π 0 rn+1ai,n+1−isiniθcosn+1−iθ (rcos θ−a)(rsin θ−b) dθ = [n 2] X j=0 Z2π 0 rn+1a2j+1,n+1−(2j+1) sin θ(1 −cos2θ)jcosn+1−(2j+1) θ (rcos θ−a)(rsin θ−b) dθ+ [n+1 2] X j=0 Z2π 0 rn+1a2j,n+1−2j(1 −cos2θ)jcosn+1−2jθ (rcos θ−a)(rsin θ−b) dθ = [n 2] X j=0 j X k=0 j k(−1)ka2j+1,n−2jr2(j−k)Z2π 0 rn−2(j−k)+1 sin θcosn−2(j−k)θ (rcos θ−a)(rsin θ−b) dθ+ [n+1 2] X j=0 j X k=0 j k(−1)ka2j,n+1−2jr2(j−k)Z2π 0 rn+1−2(j−k)cosn+1−2(j−k)θ (rcos θ−a)(rsin θ−b) dθ. When k < j the statement is proved using the induction hypothesis in each of the terms of the sum since, after the multiplication by the monomial (r2)j−k,we have that the polynomials U(r2), V (r2) and W(r2) have degree [(n−2(j−k))/2]+1+(j−k) = [n/2]+1, [(n−2(j−k))/2]+1+(j−k) = [n/2]+1 and [(n−2(j−k)−1)/2]+(j−k) = [(n−1)/2], respectively. The unique case that remains to check is k=jfor both integrals: Z2π 0 rn+1 sin θcosnθ (rcos θ−a)(rsin θ−b) dθ and Z2π 0 rn+1 cosn+1 θ (rcos θ−a)(rsin θ−b) dθ. For the first one we can write Z2π 0 rn+1 sin θcosnθ (rcos θ−a)(rsin θ−b) dθ =Z2π 0 rn(rsin θ−b + b) cosnθ (rcos θ−a)(rsin θ−b) dθ = Z2π 0 rncosnθ rcos θ−adθ + b Z2π 0 rncosnθ (rcos θ−a)(rsin θ−b) dθ. Thus we can take only the first term in the sum because for the second one we can apply once more the induction hypothesis. For this term, using the computation of Ia n(r),the corresponding polynomial W(r2),that is called Tin Lemma 2.1, has degree [(n−1)/2] and the polynomial U(r2) can be written as (a2+ b2−r2)S[(n−2)/2]+1(r2) and it has degree ([(n−2)/2] + 1) + 1 = [n/2] + 1. 8 A. GASULL, J. T. L ´ AZARO, AND J. TORREGROSA Now we will consider the last integral, that is Z2π 0 rn+1 cosn+1 θ (rcos θ−a)(rsin θ−b)dθ =Z2π 0 rncosnθ(rcos θ−a + a) (rcos θ−a)(rsin θ−b) dθ = Z2π 0 rncosnθ rsin θ−bdθ + a Z2π 0 rncosnθ (rcos θ−a)(rsin θ−b)dθ. Changing θby θ+π/2 we can use the expression for Ib n(r) computed in Lemma 2.1 to prove that, as in the previous computation, this integral satisfies the formula of the statement. The second term follows applying the induction hypothesis. Then the statement is proved.  Proposition 2.3. Let Kbe a natural number, let Pn(x, y)and Qn(x, y)be real polynomials of degree n, γr={(x, y)|x2+y2=r2},{a1,...,aK}different real numbers and I(r) = Zγr Qn(x, y)dx −Pn(x, y)dy K Q j=1 (x−aj) . Then, I(r) = e K X j=1 Sj [(n−1)/2]+1(r2) qea2 j−r2 +T[n/2](r2),(7) for suitable polynomials Sj s(ρ), Ts(ρ)of degree s, where e K= Card{|a1|,...,|aK|} and ea1,...,eae Kdenoting the different values of the set {|a1|,...,|aK|}. Proof. Parametrising γrusing polar coordinates, (x, y) = (rcos θ, r sin θ),we can write I(r) = Z2π 0 Rn+1(rcos θ, r sin θ) K Q j=1 (rcos θ−aj) dθ where Rn+1(rcos θ, r sin θ) = −rQn(rcos θ, r sin θ) sin θ−rPn(rcos θ, r sin θ) cos θ. Performing partial fraction decomposition and using Lemma 2.1, it turns out that I(r) = K X j=1 Z2π 0 1 K Q ℓ=1 ℓ6=j (aℓ−aj) Rn+1(rcos θ, r sin θ) rcos θ−aj dθ = K X j=1  Sj [(n−1)/2]+1(r2) qa2 j−r2 +Tj [n/2](r2) = K X j=1  Sj [(n−1)/2]+1(r2) qa2 j−r2 +T[n/2](r2), provided we define T[n/2](ρ) = PK j=1 Tj [n/2](ρ).Since this expression depends only on the absolute values |aj|we collect terms, consider new polynomials Sj s(ρ) and, at the end, get formula (7).  UPPER BOUNDS FOR THE NUMBER OF ZEROES 9 Proposition 2.4. Let K1, K2be natural numbers, Pn(x, y)and Qn(x, y)be real polynomials of degree n, γr={(x, y)|x2+y2=r2}and {a1,...,aK1,b1,...,bK2}real numbers satisfying that aj6= aℓand bj6= bℓfor j6=ℓ. Consider I(r) = Zγr Qn(x, y)dx −Pn(x, y)dy K1 Q j=1 (x−aj) K2 Q k=1 (y−bk) . Then, I(r) = K1 X j=1 K2 X k=1 Uj,k [n/2]+1(r2) a2 j+ b2 k−r2!1 qa2 j−r2 + K2 X k=1 K1 X j=1 Vj,k [n/2]+1(r2) a2 j+ b2 k−r2!1 pb2 k−r2+W[(n−1)/2](r2), for suitable polynomials Uj,k s(ρ), V j,k s(ρ)and Ws(ρ)of degree s. Proof. Parameterising γrusing polar coordinates, (x, y) = (rcos θ, r sin θ),we can write I(r) = Z2π 0 Rn+1(rcos θ, r sin θ) K1 Q j=1 (rcos θ−aj) K2 Q k=1 (rsin θ−bk) dθ, where Rn+1(rcos θ, r sin θ) = −rQn(rcos θ, r sin θ) sin θ−rPn(rcos θ, r sin θ) cos θ. Performing a partial fraction expansion and using Lemma 2.2 it follows that I(r) = K1 X j=1 K2 X k=1 Z2π 0 1 K1 Q ℓ=1 ℓ6=j (aℓ−aj) 1 K2 Q ℓ=1 ℓ6=k (bℓ−bk) Rn+1(rcos θ, r sin θ) (rcos θ−aj)(rsin θ−bk)dθ = K1 X j=1 K2 X k=1  1 a2 j+ b2 k−r2 Uj,k [n/2]+1(r2) qa2 j−r2 +Vj,k [n/2]+1(r2) pb2 k−r2 +Wj,k [(n−1)/2](r2) = K1 X j=1 K2 X k=1 Uj,k [n/2]+1(r2) a2 j+ b2 k−r2!1 qa2 j−r2 + K2 X k=1 K1 X j=1 Vj,k [n/2]+1(r2) a2 j+ b2 k−r2!1 pb2 k−r2+ K1 X j=1 K2 X k=1 Wj,k [(n−1)/2](r2), for a collection of polynomials Uj,k s(ρ), V j,k s(ρ), Wj,k s(ρ) of degree sfor all j= 1,...,K1 and k= 1,...,K2.Denoting W[(n−1)/2](r2) = K1 P j=1 K2 P k=1 Wj,k [(n−1)/2](r2),the claimed result follows. 