Bifurcations of limit cycles from cubic Hamiltonian systems with a center and a homoclinic saddle-loop

Publicacions Matem`atiques, Vol. 44 (2000), 205–235 BIFURCATIONS OF LIMIT CYCLES FROM CUBIC HAMILTONIAN SYSTEMS WITH A CENTER AND A HOMOCLINIC SADDLE-LOOP Yulin Zhao and Zhifen Zhang Abstract It is proved in this paper that the maximum number of limit cycles of system dx dt =y, dy dt =kx −(k+1)x2+x3+(α+βx +γx2)y is equal to two in the finite plane, where k>11+√33 4,0<||1, |α|+|β|+|γ|= 0. This is partial answer to the seventh question in [2], posed by Arnold. 1. Introduction Consider the Abelian integral I(h)=Γh Y(x, y)dx −X(x, y)dy, h ∈Σ,(1.1) where H(x, y), X(x, y) and Y(x, y) are real polynomial of xand y,Γ h is the compact component of H(x, y)=h, Σ is the maximal interval of existence of Γh. Finding the lowest upper bound for the number of zeros of I(h) is called the weakend Hilber-16th problem [1], which is closed related to determining the number of limit cycles of perturbed system        dx dt =∂H ∂y +X(x, y), dy dt =−∂H ∂x +Y (x, y), (1.2) where 0 <||1. This work was done in 1995–1998, when the first author was a Ph.D. student in Peking University. 206 Y. Zhao, Z. Zhang In particular, suppose H(x, y)=1 2y2+U(x)=h,(1.3) where U(x) is a real polynomial of xwith degree n. In this case, finding the number of zeros of I(h) is one of the ten problems in [2]. When n= 3, this problem was investigated by many authors (e.g. [7], [8], [10], [11]). When n= 4, some results were given by [5], [12], [13], [16], [17], but this case is far from complete solving. In this paper, we study the case n= 4 and the Hamiltonian vector field dH = 0 possesses one center and one homoclinic saddle-loop, which has the following normal form        dx dt =y, dy dt =kx −(k+1)x2+x3, (1.4) where k>2. The system (1.4) has the first integral H(x, y)=1 2y2−1 2kx2+1 3(k+1)x3−1 4x4=h,(1.5) and the phase portrait is shown in Figure 1.1. The closed ovals Γh are defined for Hamiltonian values h∈(−2k+1 12 ,0). H(x, y)=−2k+1 12 corresponds the center (1,0), Γ0={(x, y)|H(x, y)=0,0<x<x 1= 2(k+1)−√2(k−2)(2k−1) 3}corresponds the saddle point (0,0) and homoclinic loop. The critical point (k,0) is a saddle. y Γh O(k,0) x Figure 1.1 Bifurcations of limit cycles 207 Denote Ii(h)=Γh xiy dx, i =0,1,2,(1.6) I(h)=αI0(h)+βI1(h)+γI2(h),(1.7) where the ovals Γh,h∈(−2k+1 12 ,0), has negative (clockwise) orientation coinciding with the orientation of the vector field (1.4), α,βand γare arbitrary constants. The central result of this paper is the following theorem: Theorem 1.1. The maximum number of limit cycles of the perturbed system        dx dt =y, dy dt =kx −(k+1)x2+x3+(α+βx +γx2)y, (1.8) is equal to two in the finite plane, where k>11+√33 4,0<||1, |α|+|β|+|γ|=0. Corollary 1.2. For k>11+√33 4, either I(h)vanishes identically or its lowest upper bound of the number of zeros is equal to two, which is partial answer to the seventh problem in [2]. The paper is organized as follows: In section 2, Picard-Fuchs equation satisfied by I0(h), I1(h) and I2(h) is derived and the expansions of I(h) near its endpoints are given, the latter results reveal the connection between the Abelian integrals I(h) and the limit cycles of system (1.8) which tend to the center (1,0) or homoclinic loop of (1.4) as →0. In section 3, instead of estimating the number of zeros of I(h), we will prove that I(h) has at most two zeros, i.e., I(h) has at most two inflection points in (−2k+1 12 ,0), which implies the lowest upper bound of the number of zeros of I(h) does not exceed three in the same interval. Using the fact ω(h)=I 1(h) I 0(h)satisfies a Riccati equation, we get ω(h)>0. Hence, the curve  Ω={(ω,ν)|ω=I 1(h) I 0(h),ν=I 2(h) I 0(h),h∈(−2k+1 12 ,0)}can be defined. It is readily seen that the intersection points of line α+βω + γν = 0 with  Ωinων-plane correspond the zeros of I(h), which shows that the convexity of  Ω determinates the number of the zeros of I(h). 208 Y. Zhao, Z. Zhang In section 4, we make precise connection between the intersection points of L:α+βP +γQ = 0 with the centroid curve Ω = {(P, Q)|P= I1(h) I0(h),Q=I2(h) I1(h)}on one hand and the zeros of Abelian integral I(h) on the other hand. Finally, the main results of this paper are proved in section 5. Some techniques in section 4 and section 5 are borrowed from [4]. Remark. Unfortunately, the techniques we use in this present paper do not fit for the case of 2 <k<11+√33 4. Therefore, throughout this paper, we suppose k>11+√33 4>4 unless the opposite is claimed. Some computation in this paper is done by the computer program “Mathematica”. 2. Picard-Fuchs equation and the asymptotic expansions of I(h) near its endpoints In this section we shall derive Picard-Fuchs equation satisfied by Ii(h) and describe the behaviours of I(h) near h= 0 and h=−2k+1 12 . Lemma 2.1. I0(h),I1(h)and I2(h)satisfy the following Picard-Fuchs equation (4hE+S)J=NJ,(2.1) which is equivalent to G(h)J=RJ,(2.2) where Eis an unit matrix of order 3,J= col(I0,I 1,I 2), and S=  01 3k(k+1) 1 3(−k2+k−1) 01 3k(k2−k+1) −1 3(k+ 1)(k−1)2 01 3k(k+ 1)(k−1)21 3(−k4+k3+k2+k−1) , N=  300 −1 3(k+1) 4 0 1 3(−k2+k−1) −2 3(k+1) 5  , R=  a00(h)a01(h)a02(h) a10(h)a11(h)a12(h) a20(h)a21(h)a22(h) , Bifurcations of limit cycles 209 G(h) = 192hh−−2k+1 12 h−k3(k−2) 12 , a00(h) = 144h2+4 3(−10k4+21k3−k2+21k−10)h −4 3k3(2k−1)(k−2), a01(h)=−8 3(k+ 1)(k2+5k+1)h+14 9k2(k+ 1)(2k−1)(k−2), a02(h) = 20(k2−k+1)h−5 3k2(k−2)(2k−1), a10(h)=−16(k+1)h2+4 3k(k+ 1)(2k−1)(k−2)h, a11(h) = 192h2+8 3(−7k4+6k3+8k2+6k−7)h, a12(h) = 20(k+ 1)(k−1)2h, a20(h) = 16(−k2+k−1)h2+4 3k2(k−2)(2k−1)h, a21(h)=−32(k+1)h2−8 3k(k+ 1)(7k2−13k+7)h, a22(h) = 240h2+20k(k2−k+1)h. Proof: It follows from (1.5) that ∂y ∂h =1 y (2.3) and y∂y ∂x =kx −(k+1)x2+x3.(2.4) Obviously, (2.3) implies that I i(h)=Γh xi ydx.(2.5) Mutiplying (2.4) by yand integrating over Γhgive I3=−kI1+(k+1)I2.(2.6) 210 Y. Zhao, Z. Zhang Use (1.5) and (2.5) to get Ii(h)=Γh xiy2 ydx =Γh xi(2h+kx2−2 3(k+1)x3+1 2x4) ydx =2hI i+kI i+2 −2 3(k+1)I i+3 +1 2I i+4. (2.7) On the other hand, using (2.3), (2.4) and integrating by parts, we have Ii(h)=−1 i+1Γh xi+1 dy =−1 i+1Γh xi+1(kx −(k+1)x2+x3) ydx =−1 i+1(kI i+2 −(k+1)I i+3 +I i+4). (2.8) Eliminating I i+4 from (2.7) and (2.8) yields (i+3)Ii=4hI i+kI i+2 −1 3(k+1)I i+3.(2.9) This gives 3I0=4hI 0+kI 2−1 3(k+1)I 3,(2.10) 4I1=4hI 1+kI 3−1 3(k+1)I 4,(2.11) 5I2=4hI 2+kI 4−1 3(k+1)I 5.(2.12) Substituting (2.6) into (2.10), we obtain the first equation of (2.1). The formula (2.8) implies I i+4 =−(i+1)Ii(h)−kI i+2 +(k+1)I i+3.(2.13) Taking i= 0 in (2.13) and using (2.6), the formula (2.11) give the second equation of (2.1). Repeating the same arguments, we obtain the third equation. The lemma has been proved. Denote P(h)=I1(h) I0(h),Q(h)=I2(h) I0(h),(2.14) where h∈[−2k+1 12 ,0]. Bifurcations of limit cycles 211 Proposition 2.2. P(h),Q(h)are analytic function for h∈[−2k+1 12 ,0), and Ii−2k+1 12 =0,I i(h)>0,i=0,1,2,i) P−2k+1 12 =Q−2k+1 12 =1,P(h)>0,Q(h)>0,ii) P−2k+1 12 =−k−2 2(k−1)2,Q −2k+1 12 =−k−3 2(k−1)2,iii) P −2k+1 12 =(k−2)(−257 + 257k−110k2) 72(k−1)5, Q −2k+1 12 =651 −788k+ 467k2−110k3 72(k−1)5. iv) Proof: The results i) and ii) follows from Green’s formula. P(−2k+1 12 )= Q(−2k+1 12 ) = 1 imply that P(h)=1+o(h−−2k+1 12 ) 1+o(h−−2k+1 12 ),Q(h)=1+o(h−−2k+1 12 ) 1+o(h−−2k+1 12 ) (2.15) as h→−2k+1 12 . Noting Ii(h) is analytic at h=−2k+1 12 (see [15]) and I0(h)>0 for h∈(−2k+1 12 ,0), the formula (2.15) implies that P(h) and Q(h) are analytic functions for h∈[−2k+1 12 ,0). Using P=I 1I0−I 0I1 I2 0 ,Q =I 2I0−I 0I2 I2 0 and system (2.2) give GP=a10 +(a11 −a00)P+a12Q−a01P2−a02PQ, GQ=a20 +a21P+(a22 −a00)Q−a01PQ−a02Q2. (2.16) Differentiating (2.16) once (resp. twice) yields iii) (resp. iv)). It is well known that I(h) has the expansion near h=−2k+1 12 (see [15]) I(h)=b1h−−2k+1 12 +b2h−−2k+1 12 2 +···.(2.17) 212 Y. Zhao, Z. Zhang Theorem 2.3. i) b1=(α+β+γ)I 0−2k+1 12 , b2=−(k−2)β+(k−3)γ 2(k−1)2I 0−2k+1 12 if b1=0, b3=5(k−2)β 6(k−1)3(k−3)I 0−2k+1 12 if b1=b2=0. ii) If b1=0(resp. b1=b2=0), b2=0(resp. b3=0), then there exists one (resp. two) zero of I(h)tend to h=−2k+1 12 , i.e., system (1.8)has at most one (resp. two) limit cycle tend to (1,0). iii) The conditions b1=b2=b3=0hold if and only if I(h)≡0. Proof: (i) It follows from (1.7) and (2.14) that I(h)=I0(h)(α+βP(h)+γQ(h)),(2.18) which gives bm=1 m!   m  j=1 m jI(m−j) 0(h)[α+βP(h)+γQ(h)](j)  h=−2k+1 12 .(2.19) Therefore, the result i) follows from Proposition 2.2 and above equality. (ii) In a neighbourhood of (1,0), The Poincare map is P(h)=I(h)+o(), which yields ii). (iii) The conditions b1=b2=b3= 0 hold if and only if      α+β+γ=0, (k−2)β+(k−3)γ=0, β=0, which implies α=β=γ= 0. Hence, I(h)≡0. Bifurcations of limit cycles 213 Rewrite (1.5) in the form 1 2y2+Φ(x)=h,(2.20) where Φ(x)=−1 2kx2+1 3(k+1)x3−1 4x4satisfying Φ(x)(x−1) >0 for x∈(0,1) ∪(1,x 1).(2.21) For any x∈(0,1),there is an unique x∈(1,x 1), such that Φ(x)=Φ(x),0<x<1<x<x 1.(2.22) Therefore, we can define a function x=x(x) for 0 <x<1 satisfying (2.22). By (2.21) and (2.22), we have dx dx =Φ(x) Φ(x)<0.(2.23) Lemma 2.4. x1<x+x<2,xx<1. Proof: Let a=x+x, and b=xx.(2.24) The equality (2.22) implies that 1 2ka −1 3(k+1)a2+1 4a3+b1 3(k+1)−1 2a=0.(2.25) Taking a=2 3(k+ 1) into (2.25), we have −1 18 (2k−1)(k−2) = 0, which contradicts the assumption k>2. This shows a=2 3(k+ 1). Hence b=6ka −4(k+1)a2+3a3 6a−4(k+1) .(2.26) To find the maximal or minimal value of a(x), we consider the equation da(x) dx = 0, which is equivalent to Φ(x)+Φ (x)=0.(2.27) The relationship x2+x=a2−2b and (2.26) yield Φ(x)+Φ (x)=1 2a(a−2)(a−2k).(2.28) 220 Y. Zhao, Z. Zhang where ω+(h)=D−A−(D−A)2+4BC 2B,(3.10) ω−(h)=D−A+(D−A)2+4BC 2B (3.11) with ω−−2k+1 12 =−(k+ 1)(2k−7) 10k2−31k+31,ω −(0) = 0. Differentiating (3.9) once, we have (3.12) (ω−)−2k+1 12  =35(k−2)(k+ 1)(2k−1)(2k2−11k+ 11) (k−1)2(10k2−31k+ 31)2>0. Assume dω− dh =0ath=hand (ω−)(h)>0 for h∈(−2k+1 12 , h), which implies (ω−)(h)<0. Differentiate (3.9) twice to get (ω−)(h)=432(k2−k+ 1)(ω−−k+1 3)(ω−−(k+1)(7k2−13k+7) 9(k2−k+1) ) B(h)(k+ 1)(2k−1)(k−2)(ω−−D−A 2B).(3.13) By Lemma 3.5 and (3.11), we have ω−−D−A 2B<0, B(h)<0 and ω−(h)<0. Therefore, the formula (3.13) gives (ω−)(h)>0. This contradicts the assumption, which yields that the isocline ω=ω−(h)is monotonically increasing function for h∈(−2k+1 12 ,0). Since Ii(h) is analytic at h=−2k+1 12 , it follows from (3.4) that A−2k+1 12 I 0−2k+1 12 +B−2k+1 12 I 2−2k+1 12 =0, which implies ω−2k+1 12 =−(k+ 1)(2k−7) 10k2−31k+31.(3.14) Lemma 3.1 and (3.14) show that ω(h) is analytic for h∈[−2k+1 12 ,0). On the other hand, the formula (2.30) gives ω(0) = limh→0 I 1 I 0 =0.(3.15) Bifurcations of limit cycles 221 Hence, ω(h) is the trajectory of (3.13) from J1to E1. Since (ω−)(h)>0, the graph of ω(h)=I 1 I 0 must stay in the region {(h, ω)|ω<ω −,h∈ (−2k+1 12 ,0)}, which implies ω(h)>0, see Figure 3.1. The inequality ii) follows from i), (3.14) and (3.15). ω J2 ω+(h) E2 h E1 ω−(h) J1 Figure 3.1 Corollary 3.7. i) If α+1 3(γk +βk +β)=0,γ=0, then I(h)has h=h∗=−(γk+βk+β)(k−2)(2k−1) 36γas the unique zero in (−2k+1 12 ,0). If α+1 3(γk +βk +β)=0, then h=h∗is not the zero of I(h). ii) If α+1 3(γk+βk+β)>0,−k−2 k−3β<γ<0, then I(h)has at most one zero in (−2k+1 12 ,0). iii) P(h)<0for h∈(−2k+1 12 ,0). Proof: (i) Lemma 3.2 yields (3.16) I(h)=(2k−1)(k−2)α−12γh (2k−1)(k−2) I 0 +36γh +(γk +βk +β)(2k−1)(k−2) (k+ 1)(2k−1)(k−2) I 1. 222 Y. Zhao, Z. Zhang If α+1 3(γk +βk +β) = 0, then I(h)=36γI 0(h)(ω(h)−k+1 3)(h−h∗) (k+ 1)(2k−1)(k−2) , which implies that I(h) has h=h∗as the unique zero. If α+1 3(γk + βk +β)= 0, then it follows from Lemma 3.1 and (3.16) that I(h∗)=α+1 3(γk +βk +β)I 0(h∗)=0. (ii) In the case of α+1 3(γk +βk +β)>0, −k−2 k−3β<γ<0, h=h∗is not a zero of I(h), and (3.16) is equivalent to I(h)= 36γ(h−h∗) (k+ 1)(2k−1)(k−2)I 0(h)q(h),(3.17) where q(h)=f(h)+ω(h) and f(h)=(k+ 1)[(2k−1)(k−2)α−12γh] 36γ(h−h∗),(3.18) which implies f(h)=−(k+ 1)(2k−1)(k−2)[α+1 3(γk +βk +β)] 36γ(h−h∗)2>0.(3.19) Therefore, by Proposition 3.6, we have q(h)=f(h)+ω(h)>0.(3.20) If −k+1 kβ≤γ<0, then h∗≥0. This and (3.20) imply that q(h) (i.e., I(h)) has at most one zero in h∈(−2k+1 12 ,0). On the other hand, if −k−2 k−3β<γ<−k+1 kβ<0, then h∗∈(−2k+1 12 ,0), α>−1 3(γk +βk +β). The inequality (3.19) gives f(h)<f(0) = (k+1)α γk +βk +β<0 for h∈(h∗,0). Hence, Proposition 3.6 yields q(h)<0 for h∗<h<0. It follows from (3.20) that q(h) has at most one zero in h∈(−2k+1 12 ,h ∗). We obtain ii) by using i), Lemma 3.1 and (3.17). Bifurcations of limit cycles 223 (iii) Consider the Abelian integral I(h)=αI0(h)+I1(h)=I0(h)(α+P(h)). If α>0, then I(h)>0. If α<0, then Lemma 3.1 and Proposition 3.6 show that I(h)=I 0(h)(α+ω(h)) <0, which implies the curve I(h) is concave for h∈(−2k+1 12 ,0). Therefore, noticing I(−2k+1 12 ) = 0 and I0(h)=0,α+P(h) has at most one zero for arbitrary constant α. This yields P(h) is monotonic for h∈(−2k+1 12 ,0). Suppose h=h1is the zero of I(h), the convexity of I(h) implies I(h1)= I0(h1)P(h1)≤0, i.e. P(h1)≤0. However, if P(h1) = 0, then I(h1)=I 0(h1)(α+P(h1))+2I 0(h1)P(h1)+I0(h1)P(h1) =I0(h1)P(h1)<0, which shows P(h1)<0, i.e., h=h1is the maximum point of P(h). This contradicts P(h1)≤0. The proof is finished. Proposition 3.8. ω(h)>0for h∈(−2k+1 12 ,0). Proof: We split the proof by several steps. 1) First, V(h, ω)=2D−2A−G −4Bω>0. It is readily seen V(h, 0) = −384 (2k−1)(k−2)[(k−2)(11k−1)+9]h +16 3(4k2−k+ 4)(k−1)2>0, Vh, −(2k−7)(k+1) 10k2−31k+31=−384(55k2−139k+ 139)h 10k2−31k+31 +16 3(10k2−31k+ 31)[k3(k−4)(40k2−58k+ 287)(3.21) + 582k3+ (211k2−414k) + 138] >0. Since V(h, ω) is linear function of ωand ω(h)>0, −(k+1)(2k−7) 10k2−31k+31 < ω<0 (see Proposition 3.6), it follows from (3.21) that V(h, ω)>0 for h∈(−2k+1 12 ,0). 224 Y. Zhao, Z. Zhang 2) If h=h1satisfies ω(h1) = 0, then ω(h1)>0. Indeed, differentiate (3.6) twice to get (3.22) G(h1)ω(h1)=C(h1)+(D(h1)−A(h1))ω(h1)−Bω2(h1) +V(h1,ω(h1))ω(h1)−2B(h1)(ω)2(h1). By Lemma 3.5, Proposition 3.6 and step 1),we conclude that C(h1)+(D(h1)−A(h1))ω(h1)−B(h1)ω2(h1) =864(k2−k+1) (k+ 1)(2k−1)(k−2) ω(h1)−k+1 3 ×ω(h1)−(k+ 1)(7k2−13k+7) 9(k2−k+1) >0, V(h1,ω(h1))ω(h1)>0,−2B(h1)(ω)2(h1)>0.(3.23) Hence, the formulas (3.22) and (3.23) imply ω(h1)>0. 3) ω(−2k+1 12 )>0. To prove it, differentiating (3.6) twice, we get ω −2k+1 12 =35(k−2)(k+ 1)(2k−1) 72(k−1)5(10k2−31k+ 31)3g(k),(3.24) where g(k) = 2200k6−24924k5+ 129246k4−375481k3 + 604833k2−500511k+ 166837. This gives g(i)(4) >0, i=0,1,2,... ,6, which implies g(k)= 6  i=0 g(i)(4) i!(k−4)i>0,k∈(4,+∞).(3.25) Hence, the result ω(−2k+1 12 )>0 follows from (3.24) and (3.25). 4) Finally, we prove ω(h)>0. By step 3), starting from h=−2k+1 12 ,ifh=h1is the first point satisfying ω(h1) = 0, then ω(h1)≤0, which contradicts the result proved in step 2). This implies that ω(h) has no zero. Therefore, ω(h)>0. Theorem 3.9. I(h)has at most three zeros (counted with their multiplicities) inside the interval (−2k+1 12 ,0). Bifurcations of limit cycles 225 Proof: This theorem is proved by several parts. 1) We are going to prove that I(h) has at most two zeros (counted with their multiplicities), i.e., I(h) has at most two inflection points. Since I(−2k+1 12 ) = 0, this result implies that the maximum number of zeros of I(h) is at most three on the interval (−2k+1 12 ,0). It has been proved in Proposition 3.6 that ω(h)>0. Therefore, we can take ωas a new parameter and consider the curve ν=ν(h(ω)), defined by  Ω=(ω,ν)|ω=ω(h),ν=ν(h)=I 2 I 0 ,h∈−2k+1 12 ,0 (3.26) where h=h(ω) is the inverse function of ω=ω(h). It is easy to get that I(h)=I 0(h)(α+βω(h)+γν(h)), I(h)=I 0(h)(βω+γν)ifI(h)=0, I(4)(h)=I 0(h)(βω +γν)ifI(h)=I(h)=0, which implies that  Ω has the following properties: i) The intersection points of the lines l:α+βω +γν = 0 with the curve  Ωinων-plane correspond to the zeros of I(h). ii) I(h0)=I(h0) = 0 hold if and only if lis tangent to the  Ωat the point (ω(h0),ν(h0)). iii) If (νω−νγ)|h=h0= 0, then I(h0)=I(h0)=I(4)(h0)=0 hold if and only if α=β=γ= 0, i.e, I(h)≡0. Lemma 3.2 gives ν(h)=−12h (2k−1)(k−2) +36h (k+ 1)(2k−1)(k−2) +k k+1ω, which yields νω−ων=12 (k+ 1)(k−2)(2k−1)[6(ω)2+(k+1−3ω)ω].(3.27) It follows from Proposition 3.6, Proposition 3.8 and (3.27) that d2ν dω2=νω−ων (ω)3>0. This implies that  Ω is convex in ων-plane. Therefore, the maximum number of intersection points of the line l:α+βω +γν = 0 with  Ωis at most two. By the properties i)–iii) of  Ω, I(h) has at most two zeros (counted with their multiplicities). 226 Y. Zhao, Z. Zhang 2) The multiplicity of zero of I(h) is at most three. If h=h0is the zero of multiplicity 3, then h=h0is an unique zero of I(h). Otherwise, suppose the multiplicity of h=h0is great than 3, i.e., I(h0)=I(h0)=I(h0)=I(h0). By step 1), I(h) has at most two zeros (counted with their multiplicities) in h∈(−2k+1 12 ,0), which implies I(4)(h0)= 0. Without loss of generality, assume I(4)(h)>0. Hence, I(h) is convex in the neighbourhood of h=h0. Noting I(−2k+1 12 )=0, there must exist one inflection point h=h1,h1∈(−2k+1 12 ,h 0), see Figure 3.2(a). Thus, I(h) has two zeros, one is simple and another is multiplicity two. This contradicts the conclusion proved in step 1). Suppose h=h0is the zero with multiplicity 3, i.e., I(h0)=I(h0)= I(h0)=0,I(h0)= 0. Without loss of generality, assume I(h0)>0. Hence, the graph of I(h) is convex for h>h 0and concave for h<h 0, |h−h0|0. Since I(−2k+1 12 )=0,I(h) has another inflection point h= h1between h=−2k+1 12 and h=h0, see Figure 3.2(b). By step 1), I(h) has no other inflection point except h=hi,i=0,1, which implies h=h0is an unique zero of I(h). 3) If h=h0is the zero of multiplicity two of I(h), then another zero h=h1(if there exists) must be simple. Obviously, h=h0satisfies I(h0)=I(h0)=0,I(h0)= 0. Without loss of generality, suppose I(h0)>0, i.e., h=h0is minimal point of I(h). Suppose h1>h 0. Then there must exist two inflection points between −2k+1 12 and h1. Hence, it follows from step 1) that h=h1must be simple zero of I(h). In the case of h1<h 0, we can get the result by the same arguments as above. Summing up above discussion, we get the theorem. I −2k+1 12 h1h0 h (a) (b) h I −2k+1 12 h1 h0 Figure 3.2 Bifurcations of limit cycles 227 4. The geometry of the centriod curve Definition 4.1. In PQ-plane, the curve Ω=(P,Q)|P=P(h),Q=Q(h),h∈−2k+1 12 ,0 (4.1) is called centroid curve. It has been proved in Corollary 3.7 that P(h)<0. Therefore, Pcan be taken as a new parameter and denote Ω as Q=Q(h(p)), where h(P) is the inverse function of P=P(h). The importance of concept of the centroid curve lies in the fact that its geometry contains the complete information of I(h) although the definition of Ω depends only on H(x, y)=h. From this section, denoted by Lsand by Lcthe tangents to Ω at (P(0),Q(0)) and (1,1), i.e., at the endpoints of Ω. Ldenotes the line α+ βP +γQ =0,|β|+|γ|=0. Using same arguments as [4], we have Theorem 4.2. i) For any h0∈(−2k+1 12 ,0), the equality I(h0)=0 holds if and only if the line Lpasses through the point (P(h0),Q(h0)). ii) The equalities I(h0)=I(h0)=0hold if and only if Lis tangent to the centroid curve Ωat the point (P(h0),Q(h0)). iii) If I(h0)=I(h0)=0, then I(h0)=0holds if and only if P(h0)Q(h0)−P(h0)Q(h0)=0, i.e., the curvature of Ωat (P(h0),Q(h0)) is zero. Proof: (i) Part i) of the statement follows from (2.19). (ii) The equation of the tangent line is Q(h0)P−P(h0)Q+Q(h0)P(h0)−Q(h0)P(h0)=0.(4.2) By (2.19), I(h0)=I(h0) = 0 is equivalent to α+βP(h0)+γQ(h0)=0, βP(h0)+γQ(h0)=0. (4.3) 228 Y. Zhao, Z. Zhang Solving (4.3) for αand β, we obtain that α=P(h0)Q(h0)−P(h0)Q(h0) P(h0)γ, β =−Q(h0) P(h0)γ. Hence, the equation L:α+βP +γQ = 0 is (4.2). (iii) The condition I(h0) = 0 when I(h0)=I(h0) = 0 is equivalent to βP(h0)+γQ(h0)=0. This and (4.3) imply the result. Theorem 4.3. i) The equation of Lcis −1−(k−3)P+(k−2)Q=0.(4.4) ii) The coefficient b1=0if and only if L passes through (1,1). iii) The conditions b1=b2=0hold if and only if L=Lc, where b1 and b2are defined as Theorem 2.3. Proof: (i) Part i) of the statement follows from Proposition 2.2. (ii) By Theorem 2.3, b1=I 0(−2k+1 12 )(α+β+γ), which implies ii). (iii) Theorem 2.3 shows that b1=b2= 0 if and only if α+β+γ=0, (k−2)β+(k−3)γ=0. Solving this system for αand β, we obtain α=−1 k−2γ,β=−k−3 k−2γ, which implies that the equation of Lis (4.4). Theorem 4.4. i) The equation of Lsis Q P=Q(0) P(0).(4.5) ii) The coefficient c0is zero if and only if Lpasses through (P(0),Q(0)). iii) The coefficient c0=c1=0is equivalent to L=Ls, where c0,c1is defined as (2.31). Proof: (i) By (2.30) and Lemma 3.1, limh→0I 0(h)=+∞, limh→0I 1(h)= I 1(0), limh→0I 2(h)=I 2(0), limh→0Ii(h)=Ii(0), i=0,1,2. Therefore, dQ dP h=0 =dQ dh dh dP h=0 =limh→0 I 2I0−I 0I2 I 1I0−I 0I1 = limh→0 I 2I0 I 0−I2 I 1I0 I 0−I1 =Q(0) P(0), which yields that the equation of Lsis (4.5). Bifurcations of limit cycles 229 (ii) By (2.31), the condition c0= 0 is equivalent to α+βP(0) + γQ(0) = 0, which implies ii). (iii) It follows from (2.31) that c0=c1= 0 if and only if α+βP(0) + γQ(0) = 0, α=0, which implies α=0,β=−Q(0) P(0) γ. Therefore, the equation of Lis (4.5). The result follows. Proposition 4.5. Lcs doesn’t intersect Ωfor h∈(−2k+1 12 ,0), where Lcs is the line passing through both (1,1) and (P(0),Q(0)). Proof: By the definition of Lcs and Theorem 4.3, Theorem 4.4, we have α+β+γ=0, αI0(0) + βI1(0) + γI2(0) = 0, which implies α=I1(0) −I2(0) I2(0) −I0(0)β, γ =I0(0) −I1(0) I2(0) −I0(0)β.(4.6) If β= 0, then γ= 0, which contradicts the assumption |β|+|γ|= 0. Without loss of generality, suppose β>0. The formula (4.6) and Lemma 2.7 give that γ<0 and α+1 3(γk +βk +β)= β 3[I2(0) −I1(0)] [(k−2)I2(0) −(k−3)I1(0) −I0(0)] >0, γ+k−2 k−3β=β (k−3)[I2(0) −I0(0)] [(k−2)I2(0) −(k−3)I1(0) −I0(0)] >0. Corollary 3.7 yields that I(h) has at most one inflection point. Since I(0) = I(−2k+1 12 )=I(−2k+1 12 ) = 0 (cf. Theorem 2.3), I(h) has no zero in (−2k+1 12 ,0). The result follows from Theorem 4.2.