2006
DOI: 10.3934/dcds.2006.16.157
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The cyclicity of period annuli of some classes of reversible quadratic systems

et al.

Abstract: The cyclicity of period annuli of some classes of reversible and non-Hamiltonian quadratic systems under quadratic perturbations are studied. The argument principle method and the centroid curve method are combined to prove that the related Abelian integral has at most two zeros.

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Cited by 30 publications
(15 citation statements)
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“…Cases (r5) with b = 2 and (r10) were considered in [14] and [13], respectively. Cases (r5) for b = 2, 1 2 and (r6) with b ∈ ( 1 2 , 2) are studied in [2]. Case (r1) with b ∈ (−1, 1 3 ) and b ∈ ( 1 3 , 3) are considered in [22] and [15], respectively.…”
Section: The Generating Function In the Reversible Casementioning
confidence: 99%
See 1 more Smart Citation
“…Cases (r5) with b = 2 and (r10) were considered in [14] and [13], respectively. Cases (r5) for b = 2, 1 2 and (r6) with b ∈ ( 1 2 , 2) are studied in [2]. Case (r1) with b ∈ (−1, 1 3 ) and b ∈ ( 1 3 , 3) are considered in [22] and [15], respectively.…”
Section: The Generating Function In the Reversible Casementioning
confidence: 99%
“…Take real coordinates z = x + iy and consider a small quadratic perturbation as in (3.1). In the reversible case (2.4), the first integral and integrating factor are respectively [14] H(X, y) = X λ y 2…”
Section: The Generating Function In the Lotka-volterra Casementioning
confidence: 99%
“…There are some papers studying the limit cycles which bifurcate from the periodic orbits surrounding the centers of (2) with a = b, see for instance [4,18]. The study of the case a = b is more difficult because the first integral contains an exponential function.…”
Section: Introduction and Statement Of The Main Resultsmentioning
confidence: 99%
“…The bifurcation of limit cycles from system (1.2) under quadratic perturbations has been studied in recent years. Such as [1] for the isochronous centers; [17] for the unbounded heteroclinic loop; [6,11,16,18,12] for a = −3 with differentb ; [4] forā = −4 andā = 2 with 0 <b < 2; [5] forā = −1/2 with 0 <b < 2 and [15] forā = −3/2 with b ∈ (−∞, 0] ∪ {2} . However, to our knowledge, known results are very limited.…”
Section: Peng Sunmentioning
confidence: 99%