Limit Cycles and Integrability in a Class of Systems with High-Order Nilpotent Critical Points

and Applied Analysis 3 Thus, the origin of system (13) is an element critical point. It could be investigated using the classical integral factor method. Now, we consider the following system: dx dt = y + A30x3n + A21x2ny + A12xny2 + A03y3, dy dt = −x2n−1 + xn−1 (B30x3n + B21x2ny + B12xny2 + B03y3) . (14) When n = 2k + 1, by those transformations, system (14) is changed into dx dt = −y − √2k + 1(A30x3 + A21 √2k + 1x 2y + A12 2k + 1xy + A03 (2k + 1)3/2) , dy dt = x − (B30x3 + B21 √2k + 1x 2y + B12 2k + 1xy + B03 (2k + 1)3/2y 3) , (15) where A30 = −2A2 − 2A3 − 3A4 + 2A2k + A4k 16(1 + 2k)3 , A21 = −9B1 + B2 + B3 + 3B1k 8(1 + 2k)5/2 , A12 = −2A2 − 2A3 + 9A4 + 2A2k − 3A4k 16(1 + 2k)2 , A03 = 3B1 + B2 + B3 − B1k 8(1 + 2k)3/2 , B30 = 3B1 − B2 + B3 − B1k 8(1 + 2k)5/2 , B21 = −2A2 + 2A3 − 45A4 + 2A2k + 15A4k 16(1 + 2k)2 , B12 = −9B1 − B2 + B3 + 3B1k 8(1 + 2k)3/2 , B03 = −2A2 + 2A3 + 15A4 + 2A2k − 5A4k 16 (1 + 2k) . (16) By transformation z = x + iy, w = x − iy, T = it, (17) system (15) is changed into dz dT = z + a30z3 + a21z2w + a12zw2 + a03w3, dw dT = −w − b30w3 − b21w2z − b12wz2 − b03z3, (18) where a30 = (3iA4 + 2B1) (k − 3) 16(1 + 2k)5/2 , a21 = −iA2 + B2 + iA2k 8(1 + 2k)5/2 , a12 = − i (A3 − iB3) 8(1 + 2k)5/2 , a03 = − iA4 (k − 3) 8(1 + 2k)5/2 , bij = aij. (19) After careful computation by using formula in (4), we have the following. Theorem 7. For system (18), the first 5 Lyapunov constants at the origin are given by λ1 = iA2 (k − 1) 4(1 + 2k)5/2 , λ2 = i (2A3B1 + 3A4B3) (k − 3) 64(1 + 2k)5 . (20) When A4B1 / = 0 λ3 = iA4 (3A4 − 2B1) (3A4 + 2B1) (k − 3) (−9 + 3k − 2p) (−3 + k + 6p) 8192(1 + 2k) , λ4 = iA4B2 (3A4 − 2B1) (3A4 + 2B1) (k − 3) 2 (−9 + 3k − 2p) 49152(1 + 2k) , λ5 = iA4B 2 1 (3A4 − 2B1) (3A4 + 2B1) (k − 3) 4 (−9 + 3k − 2p) 2654208(1 + 2k) . (21) When A4 = 0, B1 / = 0 λ2 = iA3B1 (k − 3) 32(1 + 2k)5 . (22) When A4 / = 0, B1 = 0 λ2 = 3iA4B3 (k − 3) 64(1 + 2k)5 , λ3 = 3iA4 (2A3 − 3A4 + A4k) (−2A3 − 27A4 + 9A4k) (k − 3) 8192(1 + 2k)15/2 , λ4 = − iA4B2 (−2A3 − 27A4 + 9A4k) (k − 3) 2 49152(1 + 2k)10 . (23) When A4 = B1 = 0 λ2 = λ3 = λ4 = ⋅ ⋅ ⋅ = 0. (24) In the above expression of λk, one has already let λ1 = λ2 = λ3 = λ4 = 0. 4 Abstract and Applied Analysis FromTheorem 7, we obtain the following assertion. Proposition 8. The first 5 Lyapunov constants at the origin of system (18) are zero if and only if one of the following conditions is satisfied: k = 3, A2 = 0, (25) A2 = 0, A3 = 3 (3k − 9) 2 A4, B3 = −2 (3k − 9) B1, (26) A2 = 0, 2A3B1 = −3A4B3, B1 = −3 2A4, (27) A2 = 0, 2A3B1 = −3A4B3, B1 = 3 2A4, (28) A2 = A3 = A4 = 0, (29) A2 = B1 = B2 = B3 = 0, A3 = −k − 3 2 A4, (30) k = 1, A3 = −9A4, B3 = 6B1, (31) k = 1, 2A3B1 = −3A4B3, B1 = −3 2A4, (32) k = 1, 2A3B1 = −3A4B3, B1 = 3 2A4, (33) k = 1, A3 = A4 = 0, (34) k = 1, B1 = B2 = B3 = 0, A3 = A4. (35) Furthermore, we have the following. Theorem 9. The origin of system (18) is a center if and only if the first 5 Lyapunov constants are zero; that is, one of the conditions in Proposition 8 is satisfied. Proof. When one of conditions (25), (27), (28), (29), (30), (32), (33), and (34) holds, according to Theorems 6, we get all μk = 0, k = 1, 2 . . .. When condition (26) holds, system (18) could be written as dx dt = −y + (k − 3)A4 2(1 + 2k)5/2 x 3 − B2 8(1 + 2k)5/2 x 2y + 3 (k − 3)A4 4(1 + 2k)5/2 xy 2 + 4B1k − B2 − 12B1 8(1 + 2k)5/2 y 3, dy dt = x + 4B1k + B2 − 12B1 8(1 + 2k)5/2 x 3 − 3 (k − 3)A4 2(1 + 2k)5/2 x 2y + B2 8(1 + 2k)5/2 xy 2 − (n − 3)A4 4(1 + 2k)5/2y 3. (36) System (36) has an analytic first integral H(x, y) = 2x2 + 1 2y + 4B1n + B2 − 12B1 32(1 + 2k)5/2 x 4 − (k − 3)A4 2(1 + 2k)5/2 x 3y + B2 16(1 + 2k)5/2 x 2y2 − k − 3 4(1 + 2k)5/2A4xy 3 − 4B1n − B2 − 12B1 32(1 + 2k)5/2 y 4. (37) When condition (31) holds, system (18) could be written as dx dt = −y + √3( A4 27 x3 − B2 216xy +A4 18 xy2 − −8B1 + B2 216 y3) , dy dt = x + −8B1 + B2 72√3 x 3 + A4 3√3x 2y + B2 72√3xy 2 + A4 18√3y 3. (38) System (38) has an analytic first integral H(x, y) = 2x2 + 1 2y − −8B1 − B2 864 y4 + √3 (−8B1 + B2) 864 x4 + √3 432B2xy + √3A4 9 x3y. (39) When condition (35) holds, system (18) could be written as dx dt = −y + √3A4 108 x3 − √3A4 108 xy2, dy dt = x + 7√3A4 108 x2y − √3A4 36 y3. (40) System (40) has an analytic first integral H(x, y) = 1 2x + 1 2y − √3A4 108 x3y − √3A4 36 xy3. (41) Next, wewill prove that when the critical pointO(0, 0) is a 5-order weak focus, the perturbed system of (15) can generate 5 limit cycles enclosing the origin of perturbation system (15). Using the fact λ1 = λ2 = λ3 = λ4 = 0, λ5 / = 0, (42) we obtain the following. Abstract and Applied Analysis 5and Applied Analysis 5 Theorem 10. The origin of system (18) is a 5-order weak focus if and only if one of the following conditions is satisfied: A2=0, A3= 3 − k 2 A4, B3=− 3 − k 3 B1, B2=0, A4B2 1 (3A4 − 2B1) (3A4 + 2B1) (k − 3) / = 0, (43) k = 1, A3 = A4, B3 = −2 3B1, B2 = 0, A4B1 (3A4 − 2B1) (3A4 + 2B1) / = 0. (44) We next study the perturbed system of (15) as follows: dx dt = δx − y − √2k + 1(A30x3 + A21 √2k + 1x 2y + A12 2k + 1xy + A03 (2k + 1)3/2) , dy dt = δy + x − (B30x3 + B21 √2k + 1x 2y + B12 2k + 1xy + B03 (2k + 1)3/2) . (45) Theorem 11. If the origin of system (15) is a 5-order weak focus, for 0 < δ ≪ 1, making a small perturbation to the coefficients of system (15), then, for system (45), in a small neighborhood of the origin, there exist exactly 5 small amplitude limit cycles enclosing the origin O(0, 0). Proof. It is easy to check that when condition (43) or (44) holds, ∂ (λ1, λ2, λ3, λ4) ∂ (A2, A3, B3, B2) / = 0. 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