existence conditions of thirteen limit cycles in a cubic system
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International Journal of Bifurcation and Chaos, Vol. 20, No. 8 (2010) 2569–2577 c World Scientific Publishing Company DOI: 10.1142/S0218127410027209
EXISTENCE CONDITIONS OF THIRTEEN LIMIT CYCLES IN A CUBIC SYSTEM∗
Int. J. Bifurcation Chaos 2010.20:2569-2577. Downloaded from www.worldscientific.com by UNIVERSITY OF WESTERN ONTARIO WESTERN LIBRARIES on 07/25/12. For personal use only.
JUNMIN YANG and MAOAN HAN† Department of Mathematics, Shanghai Normal University, Shanghai 200234, P. R. China †[email protected] JIBIN LI Department of Mathematics, Zhejiang Normal University, Jinhua, Zhejiang 321004, P. R. China PEI YU Department of Applied Mathematics, The University of Western Ontario London, Ontario, Canada N6A 5B7 Received November 5, 2009; Revised December 21, 2009 In this paper, we study a cubic system and obtain a concrete condition under which the cubic system has 13 limit cycles. Keywords: Cubic; limit cycle; bifurcation; polynomial system.
1. Introduction and Main Result As we know, the second part of the Hilbert problem is to find the maximal number and relative locations of limit cycles of polynomial systems of degree n. Let H(n) denote this number, which is called the Hilbert number. Then the problem of finding H(n) is divided into two parts: find an upper and lower bounds of it. For the upper bound there are ´ important works of Ecalle [1990] and IIyashenko and Yakovenko [1991]. However, if H(n) < ∞ holds or not is still open, even for the case n = 2. On the other hand, many works have been done on the lower bound, especially for quadratic and cubic systems. See [Li, 2003] for a detailed introduction to recent advancement of the problem. For example, Bautin [1952] proved H(2) ≥ 3 by studying Hopf
bifurcation. Chen and Wang [1979] and Shi [1980] separately proved H(2) ≥ 4. Li and Huang [1987] first found a cubic system having 11 limit cycles, giving H(3) ≥ 11. Li and Liu [1991], and Liu et al. [2003] respectively found more cubic systems having 11 limit cycles with the same distribution. Later, Han et al. [2004] and Han et al. [2004, 2005] used the method of stability-changing of a homoclinic loop to give more cubic systems having 11 limit cycles with two different distributions. Then Zhang et al. [2005] studied an asymmetric cubic system and found three different distributions of 11 limit cycles. Yu and Han [2004, 2005a, 2005b] proved further H(3) ≥ 12 by studying Hopf bifurcation in some centrally symmetric cubic systems. Recently, Liu and Li [2008] obtained a sufficient condition for
∗
The project was supported by National Natural Science Foundation of China (10971139), the Shanghai Leading Academic Discipline Project (S30405) and the Leading Academic Discipline Project of SHNU (DZL707). † Author for correspondence 2569
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the existence of 13 limit cycles in this kind of cubic systems. The 13 limit cycles have the distribution: one large limit cycle bifurcated from the equator surrounds 12 small limit cycles which are bifurcated from two symmetric foci. Then in the same year, Li et al. [2009] considered a cubic system of the form x˙ = y(y 2 − k2 ), y˙ = x(x + 1)(x − λ)
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+ εy(α1 + α2 x + α3
(1) x2
+ α4
y 2 ),
where 0 < λ < 1, k > 10, and α1 , α2 , α3 and α4 are parameters, and proved that the system can also have 13 limit cycles if k is sufficiently large. The limit cycles are obtained by proving the existence of zeros of Melnikov functions based on some known results, and present a new distribution. Both works of Liu and Li [2008], Li et al. [2009] are the best results so far for cubic systems and are very important, yielding H(3) ≥ 13. In this paper, motivated by the work of Li et al. [2009], we consider the following cubic system
Fig. 1.
The distribution of the 13 limit cycles of system (2).
with K0 ≈
+
x˙ = 2y(y 2 − k2 ), y˙ = −(x3 + bx2 − x)
(2)
− εy(δ1 + δ2 x + δ3 x2 + δ4 y 2 ), where k > 0, b > 0, and δ1 , δ2 , δ3 and δ4 are parameters. It is easy to see that (1) and (2) are equivalent. We use the method developed in Han and Chen [2000], Han et al. [2008], Yang and Han [2007] to prove that system (2) can have 13 limit cycles. The main purpose is to give a concrete condition for the existence of 13 limit cycles, which can be taken as an improvement of the work by Li et al. [2009] in two aspects: a simple and concrete condition for the existence of 13 limit cycles is given; on the other hand, the proof method (i.e. the way to find zeros of Melnikov functions) is simpler and more direct. The condition here is also much simpler than that of Liu and Li [2008]. More precisely, we have the following main result. Theorem 1. System (2) has 13 limit cycles if
k = 100,
7 b= , 2
0 < k1 δ4 < δ3 < k2 δ4
(3)
where k1 ≈ 0.75249999999975,
77514906777564200000 δ4 8051754528899929
K1 ≈
2127709682470669621 δ3 , 603881589667494675
38757455034549104 , 120776317933498935
δ1∗ = −30000δ4 .
In particular, if δ3 = 20000,
δ4 =
8 × 106 301
(5)
and (4) holds with K0 ≈
265838587879011391034962400 , 1038676334228090841
K1 ≈
38757455034549104 120776317933498935
δ1∗ = −
(6)
24 × 1010 301
then system (2) has 13 limit cycles. Remark. The distribution of the 13 limit cycles
under the condition of the above theorem is just the same as obtained in Li et al. [2009] as shown in Fig. 1.
2. Proof of the Main Result k2 ≈ 0.75250000000036,
and 0 < ε K0 + K1 δ1 − δ2 δ1∗ − δ1 1
(4)
Consider system (2). For ε = 0, (2) is Hamiltonian with H(x, y) =
1 1 b 1 4 y − k2 y 2 + x4 − x2 + x3 , 2 4 2 3
(7)
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and has five centers A1 , A2 , A3 , A4 , O (the origin) and four saddles S0 , S1 , S2 , S3 , where A1 = (x1 (b), k) S0 = (0, k),
A4 = (x2 (b), −k),
h1 = −
3673 455 √ − 65, 384 384
h2 = −
3673 455 √ + 65 384 384
(8)
S1 = (x1 (b), 0),
S2 = (x2 (b), 0),
(11)
S3 = (0, −k)
√ and xi (b) = (−b + (−1)i b2 + 4)/2, i = 1, 2. Let Int. J. Bifurcation Chaos 2010.20:2569-2577. Downloaded from www.worldscientific.com by UNIVERSITY OF WESTERN ONTARIO WESTERN LIBRARIES on 07/25/12. For personal use only.
h0 = −50000000,
A2 = (x2 (b), k),
A3 = (x1 (b), −k),
α1 = H(A1 ) = H(A3 ), α2 = H(A2 ) = H(A4 ), h0 = H(S0 ) = H(S3 ),
(9)
h1 = H(S1 ), h2 = H(S2 ). For example, for k = 100, b = 7/2, then (8) and (9) become √ 7 65 , 100 , A1 = − − 4 4 √ 7 65 , 100 , A2 = − + 4 4 √ 7 65 , −100 , A3 = − − 4 4 √ 7 65 , −100 , A4 = − + (10) 4 4
respectively. It is easy to see that α1 < α2 < h0 < h1 < h2 < 0. Then for each of h1 and h2 , the equation H(x, y) = hj defines a double homoclinic loop passing through Sj , denoted by Lj , and for h0 , the equation H(x, y) = h0 with y > 0 defines a double homoclinic loop passing through S0 , denoted by L0 . Let L01 = L0 |x≤0 , L02 = L0 |x≥0 , L11 = L1 |y≥0 , L12 = L1 |y≤0 , and L21 be the homoclinic loop passing through S2 and surrounding a unique singular point at the origin, and L22 be the homoclinic loop passing through S2 and surrounding all other singular points. Then Lj = Lj1 ∪ Lj2 , j = 0, 1, 2. See Fig. 2. Introduce L± h : H(x, y) = h,
h ∈ (h0 , h1 ), ±y > 0,
L± jh : H(x, y) = h,
h ∈ (αj , h0 ), j = 1, 2
˜ 1h : H(x, y) = h, L
±y > 0 ˜ 1h ⊂ Int L21 , h ∈ (h2 , 0), L
˜ 2h : H(x, y) = h, L
˜ 2h ⊃ L22 , h > h2 , Int L
˜ 2h surrounding all the singular points. In with L the following, we fix (k, b, δ3 , δ4 ) satisfying (3). Then
S0 = (0, 100), √ 7 65 ,0 , S1 = − − 4 4 √ 7 65 ,0 , S2 = − + 4 4 S3 = (0, −100) and α1 = −
19200003673 455 √ − 65, 384 384
α2 = −
19200003673 455 √ + 65, 384 384
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Fig. 2.
Four double homoclinic loops of (2)|ε=0 .
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correspondingly we have the following Melnikov functions ± q(x, y, δ1 , δ2 )dx, h ∈ (h0 , h1 ), M (h, δ1 , δ2 ) = Mj± (h, δ1 , δ2 )
=
q(x, y, δ1 , δ2 )dx,
h ∈ (αj , h0 ), j = 1, 2,
˜ 2 (h, δ1 , δ2 ) = M
Since H(x, y) is even in y and q is odd in y, we have (see [Li et al., 2009]) M + (h, δ1 , δ2 ) = M − (h, δ1 , δ2 ), − M+ j (h, δ1 , δ2 ) = Mj (h, δ1 , δ2 ),
L± h
L± jh
˜ 1 (h, δ1 , δ2 ) = M Int. J. Bifurcation Chaos 2010.20:2569-2577. Downloaded from www.worldscientific.com by UNIVERSITY OF WESTERN ONTARIO WESTERN LIBRARIES on 07/25/12. For personal use only.
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˜ 1h L
˜ 2h L
q(x, y, δ1 , δ2 )dx,
h2 < h < 0,
q(x, y, δ1 , δ2 )dx,
h > h2 , (12)
where q(x, y, δ1 , δ2 ) = −y(δ1 + δ2 x + δ3 x2 + δ4 y 2 ).
(13)
j = 1, 2.
(14)
Thus, we will only consider the zeros of M + , M + j ˜ j , j = 1, 2. and M For convenience, we let M (h, δ1 , δ2 ) = M + (h, δ1 , δ2 ), Mj (h, δ1 , δ2 ) = Mj+ (h, δ1 , δ2 ),
j = 1, 2.
(15)
We will prove Theorem 1 by finding zeros of the ˜ 2 . First, on their anafunctions M, M1 , M2 and M lytical property at the endpoints of their domain, by Remark 1.4 in [Han & Chen, 2000] or Theorem 2.2 in [Han et al., 2008], Theorem 1.2 in [Han, 2000] and Lemma 2.9 in [Han, 2006] we have Lemma 1. Let (3) be satisfied by δ3 and δ4 con-
stant. Then M (h, δ1 , δ2 ) = c0 + 2c1 (h − h0 )ln|h − h0 | + c2 (h − h0 ) + O(|h − h0 |2 ln|h − h0 |),
0 < h − h0 1,
M (h, δ1 , δ2 ) = c0 + O(|h − h1 |ln|h − h1 |),
0 < h1 − h 1,
Mj (h, δ1 , δ2 ) = c0j + c1 (h − h0 )ln|h − h0 | + c2j (h − h0 ) + O(|h − h0 |2 ln|h − h0 |),
M1 (h, δ1 , δ2 ) = b01 (h − α1 ) + O((h − α1 )2 ),
0 < h − α1 1,
˜ j (h, δ1 , δ2 ) = c˜0j + O(|h − h2 |ln|h − h2 |), M
j = 1, 2, 0 < h − h2 1,
˜ 1 (h, δ1 , δ2 ) = ˜b01 h + O(h2 ), M
0 < −h 1,
where c0j = Mj (h0 , δ1 , δ2 ) q(x, y, δ1 , δ2 )dx, = L0j
c˜0j =
1 δ1 + 150δ4 , 200
L0j
L11
j = 1, 2, q(x, y, δ1 , δ2 )dx,
j = 1, 2.
√
√ 2π 2π qy (O, δ1 , δ2 ) = − δ1 , 100 100 √ 2πqy (A1 , δ1 , δ2 ) = √ 50 65 + 7 65 √ √ 1 7 + 65 2 2π − δ1 + δ2 = √ 100 400 65 + 7 65 √ 57 + 7 65 δ3 − 300δ4 . − 800
˜b01 =
c2 = c2 + O(c1 ), c2 = c21 + c22 , c2j = [qy (x, y, δ1 , δ2 ) − qy (0, 100, δ1 , δ2 )]dt,
c0 =
q(x, y, δ1 , δ2 )dx,
(17) and
L2j
j = 1, 2,
c0 = c01 + c02 , c1 =
(16)
j = 1, 2, 0 < h0 − h 1,
b01
(18)
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Under the condition (3), (7) becomes
1 y1 = 6
1 1 7 1 H(x, y) = x4 + y 4 − 10000y 2 − x2 + x3 . 4 2 2 6
2573
360000 − 6x −18x2 − 84x + 36,
√ 7 67 , x ˆ2 = − + 3 3 1 360000 + 6x −18x2 − 84x + 36. y2 = 6 (20)
By (17) we have c0j = −(δ1 Ij1 + δ2 Ij2 + δ3 Ij3 + δ4 Ij4 ), j = 1, 2, c2j = − (δ2 x + δ3 x2 + 3δ4 (y 2 − 104 ))dt L0j
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=−
L0j
δ2 x + δ3 x2 + 3δ4 (y 2 − 104 ) dx 2y(y 2 − 104 )
= −(δ2 Jj2 + δ3 Jj3 + 3δ4 Jj4 ),
(19)
By Maple we can obtain I11 ≈ 0.19336179335719, I12 ≈ −0.60397799779263,
j = 1, 2,
I13 ≈ 2.1276842302270,
where Iji =
L0j
Ij4 =
L0j
xi−1 ydx =
x ˆj
0 x ˆj
I14 ≈ 5800.8535538510,
xi−1 (y1 − y2 )dx,
I21 ≈ 0.42548181555552 × 10−3 , I22 ≈ 0.96408125135325 × 10−4 ,
j = 1, 2, i = 1, 2, 3,
y 3 dx =
0
(y13 − y23 )dx,
I23 ≈ 0.25452243669621 × 10−4 ,
j = 1, 2,
I24 ≈ 12.764454466315, J12 ≈ −0.036722972448375,
xi−1 dx 2 L0j 2y(y − 10000) 0 xi−1 xi−1 − dx, = 2 2y2 (y22 − 10000) x ˆj 2y1 (y1 − 10000)
Jji =
Jj4 =
L0j
j = 1, 2, i = 2, 3, 0 1 1 1 dx = − dx, j = 1, 2 2y 2y2 x ˆj 2y1
(21)
J13 ≈ 0.10568693840688, J14 ≈ −0.96680909021854 × 10−5 , J22 ≈ 0.77058591984887 × 10−2 , J23 ≈ 0.20196618711375 × 10−2 , J24 ≈ −0.21274090779586 × 10−7 . By (17) again, we have
with
c0 = −(δ1 I 1 + δ2 I 2 + δ3 I 3 + δ4 I 4 ), c˜01 = −(δ1 I˜1 + δ2 I˜2 + δ3 I˜3 + δ4 I˜4 ),
x2 , 100) = h0 , H(ˆ x1 , 100) = H(ˆ √ 7 67 , x ˆ1 = − − 3 3 Ii =
L11
I4 =
L11
c˜02 = −(δ1 J˜1 + δ2 J˜2 + δ3 J˜3 + δ4 J˜4 ),
xi−1 ydx = 3
y dx =
x2
x1
where x2 x1
xi−1 (y 1 − y 2 )dx,
i = 1, 2, 3,
(y 31 − y32 )dx,
H(x1 , 100) = H(x1 , 100) = h1 , x1 ≈ −120.10904980338, x2 ≈ 117.77510233465, √ 1 y1 = 1440000 + 6 57599988981 − 1365 65 + 576x2 − 1344x3 − 288x4 , 12 √ 1 y2 = 1440000 − 6 57599988981 − 1365 65 + 576x2 − 1344x3 − 288x4 , 12
(22)
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and
I˜i =
i−1
L21
x
J˜i =
ydx = 2
i−1
L22
x
J˜4 =
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L22
x ˜3
y dx = 2 3
x
x ˜4
ydx = 2 x ˜5
x ˜1
i−1
x ˜5 x ˜1
y˜2 dx,
i−1
x
y˜13 dx
I˜4 =
y˜1 dx +
x ˜1
+ x ˜2
x ˜1
x ˜2
y˜23 dx
3
L21
y dx = 2
i−1
x
y˜2 dx +
x ˜4
+ x ˜5
y˜23 dx
x ˜3
x ˜4
i−1
x
x ˜5
y˜23 dx,
x ˜4
i = 1, 2, 3,
y˜2 dx ,
i = 1, 2, 3,
,
x2 , 0) = H(˜ x3 , 0) = H(˜ x4 , 0) = H(˜ x5 , 100) = h2 , x ˜1 < x ˜2 < x ˜3 < x ˜4 < x ˜5 , H(˜ x1 , 100) = H(˜ √ √ 65 1 7 − ˜2 = − − 49 + 21 65, x ˜1 ≈ −120.1090612, x 12 4 6 √ √ 7 7 1√ 65 1 + 49 + 21 65, x ˜4 = − + 65, x ˜5 ≈ 117.7751137, x ˜3 = − − 12 4 6 4 4 √ 1 1440000 + 6 57599988981 + 1365 65 + 576x2 − 1344x3 − 288x4 , y˜1 = 12 √ 1 1440000 − 6 57599988981 + 1365 65 + 576x2 − 1344x3 − 288x4 . y˜2 = 12 Also by Maple, we get where I 1 ≈ 26426.695431194,
K 0 = −(I13 + I23 )δ3 − (I14 + I24 )δ4
I 2 ≈ −30847.384030487,
≈−
I 3 ≈ 93503233.017653, I 4 ≈ 594536975.78668,
−
I˜1 ≈ −0.6165686494 × 10−3 , I˜2 ≈ −0.2266214100 × 10−4 , I˜3 ≈ −0.6263873180 × 10−5 ,
(23)
I˜4 ≈ −0.5273174788 × 10−9 , J˜1 ≈ 0.5285407076 × 105 , J˜2 ≈ −0.6169553114 × 105 , J˜3 ≈ 0.1870066096 × 109 ,
2127709682470669621 δ3 1018 1162723601663463 δ4 , 2 × 1011
K 1 = −(I11 + I21 ) ≈ − K 2 = −(I12 + I22 ) ≈
c0 = K 0 + K 1 δ1 + K 2 δ2 , c1 =
1 δ1 + 150δ4 , 200
24155263586699787 . 4 × 1016
Therefore we have Lemma 2. Let (3) be satisfied. Then c0 ≥ 0 if and
only if
J˜4 ≈ 0.1189074254 × 1010 . Now substituting (21) into (19) and by (17) we get
2422340939659319 , 125 × 1014
δ2 ≥ K0 + K1 δ1 , and c1 ≥ 0 if and only if δ1 ≥ −30000δ4 , where K0 = − +
77514906777564200000 K0 δ4 ≈ 8051754528899929 K2 2127709682470669621 δ3 , 603881589667494675
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K1 = −
which together with (3) follows that
K1 38757455034549104 . ≈ 120776317933498935 K2
c01 > 0,
(24) Let δ1∗ = −30000δ4 , δ2∗ = K0 + K1 δ1∗ . Then by (24) δ2∗ ≈ −
Next, we study the property of the functions ˜ 2 as (δ1 , δ2 ) = (δ∗ , δ∗ ). In this M, M1 , M2 and M 1 2 case, by (18), (19) and (21)–(24), we have
c2 < 0,
c˜02 > 0,
c˜01
b01 < 0, c0 > 0, < 0, ˜b01 > 0.
(26)
Hence, by (16) and (26) we have
3291534008000 δ4 8051754528899929
2127709682470669621 δ3 . + 603881589667494675 Int. J. Bifurcation Chaos 2010.20:2569-2577. Downloaded from www.worldscientific.com by UNIVERSITY OF WESTERN ONTARIO WESTERN LIBRARIES on 07/25/12. For personal use only.
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M (h, δ1∗ , δ2∗ ) < 0
for 0 < h − h0 1,
M (h, δ1∗ , δ2∗ ) > 0
for 0 < h1 − h 1,
M1 (h, δ1∗ , δ2∗ ) > 0
for 0 < h0 − h 1,
M1 (h, δ1∗ , δ2∗ ) < 0
for 0 < h − α1 1,
˜ 2 (h, δ∗ , δ∗ ) > 0 M 1 2
for 0 < h − h2 1.
(27)
Then we can prove c01
16007678383413707737 δ4 ≈− 402587726444996450000000000 +
c02 ≈
Lemma 3. Let (3) be satisfied. Then
7349954756236336535315248941 δ3 , 20129386322249822500000000000000
16007678383413707737 δ4 402587726444996450000000000 −
7349954756236336535315248941 δ3 , 20129386322249822500000000000000
69269175158411302787920692328991 δ4 c2 ≈ 4025877264449964500000000000000000000 16510200872197961063715188463601 δ3 , 3019407948337473375000000000000000 √ (28248237268876517 65 √ − 4633315056457821781) 2 ≈ √ δ3 483105271733995740000 65 + 7 65 √ √ (8228835020 65 + 57601845140) 2 δ4 , − √ 8051754528899929 65 + 7 65 −
b01
c0 ≈ 198263874.53882 δ4 − 93394545.687548δ3 , c˜02 ≈ 396547843.8 δ4 − 186789232.3δ3 , √ ˜b01 = 300 2 δ4 π,
(1) the function M1 (h, δ1∗ , δ2∗ ) has a zero h∗0 ∈ (α1 , h0 ); (2) M (h, δ1∗ , δ2∗ ) has three zeros h∗j ∈ (h0 , h1 ), j = 3, 4, 5, with h∗3 < h∗4 < h∗5 ; ˜ ∗ ∈ (h2 , +∞); ˜ 2 (h, δ∗ , δ∗ ) has a zero h (3) M 1 2 0 (4) all of these zeros are of an odd multiplicity. Proof. First by (27), the function M1 (h, δ1∗ , δ2∗ )
has a zero h∗0 ∈ (α1 , h0 ). Similarly, the function M (h, δ1∗ , δ2∗ ) also has a zero in (h0 , h1 ). To find more zeros of M , let us calculate their values at some points. By (12), (13) and (15) we have (δ1 + δ2 x + δ3 x2 + δ4 y 2 )ydx M (h, δ1 , δ2 ) = − Lh
= −(δ1 Iˆ1 + δ2 Iˆ2 + δ3 Iˆ3 + δ4 Iˆ4 ), ˜ 2 (h, δ1 , δ2 ) = − M
h > h2 , (25) where
Lh
Iˆ4 =
xj−1 ydx = 3
Lh
y dx =
x2 (h) x1 (h)
x2 (h) x1 (h)
xj−1 (ˆ y1 − yˆ2 )dx,
(ˆ y13 − yˆ23 )dx,
(δ1 + δ2 x + δ3 x2 + δ4 y 2 )ydx
= −(δ1 Jˆ1 + δ2 Jˆ2 + δ3 Jˆ3 + δ4 Jˆ4 ),
c˜01 ≈ −18.49704737 δ4 + 0.00008611108867δ3
Iˆj =
˜ 2h L
h0 < h < h1 ,
j = 1, 2, 3,
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Jˆj =
˜ 2h L
Jˆ4 =
xj−1 ydx = 2
˜ 2h L
1 yˆ1 = 6
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1 yˆ2 = 6
y 3 dx = 2
x ˜4 (h)
x ˜1 (h) x ˜4 (h)
x ˜1 (h)
xj−1 yˆ1 dx +
yˆ13 dx +
x ˜3 (h)
x ˜4 (h)
x ˜3 (h) x ˜4 (h)
yˆ23 dx +
x ˜1 (h) x ˜2 (h)
x ˜1 (h) x ˜2 (h)
360000 − 6 −18x4 − 84x3 + 36x2 + 3600000000 + 72h.
H(˜ xi (h), 0) = h, H(˜ xi (h), 100) = h,
i = 1, 2, h0 < h < h1 ,
j = 1, 2, 3,
Unfortunately, we cannot find a zero of ˜ M1 (h, δ1∗ , δ2∗ ) for h ∈ (h2 , 0). By Lemmas 2 and 3 we can prove Lemma 4. Let (3) be satisfied by δ3 and δ4 con-
i = 2, 3, h > h2 ,
stant. If
i = 1, 4, h > h2 .
0 < K0 + K1 δ1 − δ2 δ1∗ − δ1 1,
(29)
where δ1∗ = −30000δ4 , and K0 and K1 are given by (24), then
By Maple 10 we have M (h, δ1∗ , δ2∗ )|h=−31502882.64306 ≈ 18119541.753047δ4 − 24079125.253207δ3 ≡ M1∗ , M (h, δ1∗ , δ2∗ )|h=−31502882.64304 ≈ 18119541.753057δ4 − 24079125.253240δ3
(1) (2) (3) (4)
M (h, δ1 , δ2 ) has five zeros in (h0 , h1 ); M1 (h, δ1 , δ2 ) has a zero in (α1 , h0 ); ˜ 2 (h, δ1 , δ2 ) has a zero in (h2 , +∞); M all of these zeros are of an odd multiplicity.
Proof.
By Lemma 2, under (29) we have 0 < −c0 −c1 −c2
≡ M2∗ , ˜ 2 (h, δ∗ , δ∗ )|h=9×107 M 1 2 ≈ 410963172.382053δ4 − 645192646.536934δ3 ˜ ∗. ≡M (28) By (3) and (28), we have M1∗
xj−1 yˆ2 dx ,
yˆ23 dx ,
360000 + 6 −18x4 − 84x3 + 36x2 + 3600000000 + 72h,
H(xi (h), 100) = h,
˜1 (h) < x ˜2 (h) < 0 < and x1 (h) < 0 < x2 (h), x x ˜3 (h) < x ˜4 (h) satisfy
> 0,
M2∗
< 0,
˜∗
M < 0.
Thus, by (27), M (h, δ1∗ , δ2∗ ) has three zeros h∗3 , h∗4 and h∗5 satisfying h∗3
xj−1 yˆ2 dx +
∈ (h0 , −31502882.64306),
h∗4 ∈ (−31502882.64306, −31502882.64304), h∗5 ∈ (−31502882.64304, h1 ), ˜ ∗ ∈ (h2 , 9 × 107 ). ˜ 2 (h, δ∗ , δ∗ ) has a zero h while M 1 2 0 ∗ ∗ ˜ , h∗ , h∗ and h∗ are Note that all of the zeros h0 , h 0 3 4 5 of an odd multiplicity. The proof is completed.
which ensures that M (h, δ1 , δ2 ) has two new zeros h∗1 , h∗2 ∈ (h0 , h∗3 ) near h0 both having an odd multiplicity with h∗1 < h∗2 . At the same time, the zeros h∗3 , h∗4 and h∗5 of M (h, δ1 , δ2 ), h∗0 of M1 (h, δ1 , δ2 ) and ˜ ∗ of M ˜ 2 (h, δ1 , δ2 ) remain under (29). This ends the h 0 proof. Corollary 1. Let k = 100, b = 7/2 and (δ3 , δ4 ) satisfy (5). Then for K0 , K1 and δ1∗ satisfying (6) and δ1 and δ2 satisfying (29), the conclusions (1)–(4) of Lemma 4 hold.
In fact, under (5) we have k1 δ4 ≈ 19999.999999993,
k2 δ4 ≈ 20000.000000010.
Thus, (3) is satisfied. Hence, the corollary follows from Lemma 4. On the other hand, in this case, similar to (25) and (28) we have c01 ≈ 7.3016544125820 > 0, c2 ≈ −108.90321618344 < 0,
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b01 ≈ −147.39575898743 < 0, c0 ≈ 0.34015808347891 × 1013 > 0, c˜02 ≈ 0.680369293 × 1013 > 0, and M (h, δ1∗ , δ2∗ )|h=−31502882.64306 ≈ 0.22899937435633 > 0,
Int. J. Bifurcation Chaos 2010.20:2569-2577. Downloaded from www.worldscientific.com by UNIVERSITY OF WESTERN ONTARIO WESTERN LIBRARIES on 07/25/12. For personal use only.
M (h, δ1∗ , δ2∗ )|h=−31502882.64304 ≈ −0.16451532561954 < 0, ˜ 2 (h, δ∗ , δ∗ )|h=9×107 M 1 2 ≈ −1981243697993.1 < 0 which also yield Corollary 1 by the above discussion. Now, it is clear that Theorem 1 follows from Lemma 4, Corollary 1 and the well-known Poincare– Pontrjagin–Andronov theorem (Theorem 6.1 in [Li, 2003]).
Acknowledgments The authors would like to thank C. Li for providing them a preprint of the work [Li et al., 2009] before the preparation of the present work.
References Bautin, N. N. [1952] “On the number of limit cycles which appear with the variation of coefficients from an equilibrium position of focus or center type,” Mat. Sb (N.S.) 30, 181–196. Chen, L. & Wang, M. [1979] “The relative position and the number of limit cycles of a quadratic differential system,” Chinese Acta Math. Sinica 22, 751–758. ´ Ecalle, J. [1990] “Finitude des cycles limites et acc´el´erosommation de L’application de retor,” Lecture Notes in Mathematics 1455, 74–159. Han, M. [2000] “On Hopf cyclicity of planar systems,” J. Math. Anal. Appl. 245, 404–422. Han, M. & Chen, J. [2000] “The number of limit cycles bifurcating from a pair of homoclinic loops,” Sci. China Ser. A 30, 401–414. Han, M. [2002] “On the number and distributions of limit cycles in a cubic system,” Chin. Ann. Math. A 23, 143–152. Han, M., Wu, Y. & Bi, P. [2004] “Bifurcation of limit cycles near polycycles with n vertices,” Chaos Solit. Fract. 22, 383–394.
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Han, M., Zhang, T. & Zang, H. [2004] “On the number and distribution of limit cycles in a cubic system,” Int. J. Bifurcation and Chaos 12, 4285–4292. Han, M., Wu, Y. & Bi, P. [2005] “A new cubic system having eleven limit cycles,” Discr. Cont. Dyn. Syst. 12, 675–686. Han, M. [2006] “Bifurcation theory of limit cycles of planar systems,” Handbook of Differential Equations, Ordinary Differential Equations, Vol. 3, eds. Canada, A., Drabek, P. & Fonda, A. (Elsevier) Chap. 4. Han, M., Yang, J., Tarta, A. & Gao, Y. [2008] “Limit cycles near homoclinic and heteroclinic loops,” J. Dyn. Diff. Eq. 20, 923–944. IIyashenko, Yu. S. & Yakovenko, S. [1991] Finiteness for Limit Cycles, Transl. Math. Monographs, Vol. 94 (Amer. Math. Soc., Providence, RI). Li, C., Liu, C. & Yang, J. [2009] “A cubic system with thirteen limit cycles,” J. Diff. Eqs. 246, 3609–3619. Li, J. & Huang, Q. [1987] “Bifurcations of limit cycles forming compound eyes in the cubic system,” Chinese Ann. Math. Ser. B 8, 391–403. Li, J. & Liu, Z. [1991] “Bifurcation set and limit cycles forming compound eyes in a perturbed Hamiltonian system,” Publ. Math. 35, 487–506. Li, J. [2003] “Hilbert’s 16th problem and bifurcations of planar polynomial vector fields,” Int. J. Bifurcation and Chaos 13, 47–106. Liu, Z., Yang, Z. & Jiang, T. [2003] “The same distribution of limit cycles in five perturbed cubic Hamiltonian systems,” Int. J. Bifurcation and Chaos 13, 243–249. Liu, Y. & Li, J. [2008] “Z2 -Equivariant cubic systems which yield 13 limit cycles,” to appear in Qualitative Theory of Dynamical Systems. Shi, S. [1980] “A concrete example of the existence of four limit cycles for quadratic system,” Sci. Sinica 23, 153–158. Yang, J. & Han, M. [2007] “Limit cycles near a double homoclinic loop,” Ann. Diff. Eqs. 23, 536–545. Yu, P. & Han, M. [2004] “Twelve limit cycles in a cubic order planar system with Z2 -symmetry,” Commun. Pure Appl. Anal. 3, 515–526. Yu, P. & Han, M. [2005a] “Small limit cycles bifurcating from fine focus points in cubic order Z2 -equivariant vector fields,” Chaos Solit. Fract. 24, 329–348. Yu, P. & Han, M. [2005b] “Twelve limit cycles in a cubic case of the 16th Hilbert problem,” Int. J. Bifurcation and Chaos 15, 2191–2205. Zhang, T., Han, M. & Zang, H. [2005] “Perturbation from an asymmetric cubic Hamiltonian,” J. Math. Anal. Appl. 305, 617–630.
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International Journal of Bifurcation and Chaos, Vol. 20, No. 8 (2010) 2569–2577 c World Scientific Publishing Company DOI: 10.1142/S0218127410027209
EXISTENCE CONDITIONS OF THIRTEEN LIMIT CYCLES IN A CUBIC SYSTEM∗
Int. J. Bifurcation Chaos 2010.20:2569-2577. Downloaded from www.worldscientific.com by UNIVERSITY OF WESTERN ONTARIO WESTERN LIBRARIES on 07/25/12. For personal use only.
JUNMIN YANG and MAOAN HAN† Department of Mathematics, Shanghai Normal University, Shanghai 200234, P. R. China †[email protected] JIBIN LI Department of Mathematics, Zhejiang Normal University, Jinhua, Zhejiang 321004, P. R. China PEI YU Department of Applied Mathematics, The University of Western Ontario London, Ontario, Canada N6A 5B7 Received November 5, 2009; Revised December 21, 2009 In this paper, we study a cubic system and obtain a concrete condition under which the cubic system has 13 limit cycles. Keywords: Cubic; limit cycle; bifurcation; polynomial system.
1. Introduction and Main Result As we know, the second part of the Hilbert problem is to find the maximal number and relative locations of limit cycles of polynomial systems of degree n. Let H(n) denote this number, which is called the Hilbert number. Then the problem of finding H(n) is divided into two parts: find an upper and lower bounds of it. For the upper bound there are ´ important works of Ecalle [1990] and IIyashenko and Yakovenko [1991]. However, if H(n) < ∞ holds or not is still open, even for the case n = 2. On the other hand, many works have been done on the lower bound, especially for quadratic and cubic systems. See [Li, 2003] for a detailed introduction to recent advancement of the problem. For example, Bautin [1952] proved H(2) ≥ 3 by studying Hopf
bifurcation. Chen and Wang [1979] and Shi [1980] separately proved H(2) ≥ 4. Li and Huang [1987] first found a cubic system having 11 limit cycles, giving H(3) ≥ 11. Li and Liu [1991], and Liu et al. [2003] respectively found more cubic systems having 11 limit cycles with the same distribution. Later, Han et al. [2004] and Han et al. [2004, 2005] used the method of stability-changing of a homoclinic loop to give more cubic systems having 11 limit cycles with two different distributions. Then Zhang et al. [2005] studied an asymmetric cubic system and found three different distributions of 11 limit cycles. Yu and Han [2004, 2005a, 2005b] proved further H(3) ≥ 12 by studying Hopf bifurcation in some centrally symmetric cubic systems. Recently, Liu and Li [2008] obtained a sufficient condition for
∗
The project was supported by National Natural Science Foundation of China (10971139), the Shanghai Leading Academic Discipline Project (S30405) and the Leading Academic Discipline Project of SHNU (DZL707). † Author for correspondence 2569
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the existence of 13 limit cycles in this kind of cubic systems. The 13 limit cycles have the distribution: one large limit cycle bifurcated from the equator surrounds 12 small limit cycles which are bifurcated from two symmetric foci. Then in the same year, Li et al. [2009] considered a cubic system of the form x˙ = y(y 2 − k2 ), y˙ = x(x + 1)(x − λ)
Int. J. Bifurcation Chaos 2010.20:2569-2577. Downloaded from www.worldscientific.com by UNIVERSITY OF WESTERN ONTARIO WESTERN LIBRARIES on 07/25/12. For personal use only.
+ εy(α1 + α2 x + α3
(1) x2
+ α4
y 2 ),
where 0 < λ < 1, k > 10, and α1 , α2 , α3 and α4 are parameters, and proved that the system can also have 13 limit cycles if k is sufficiently large. The limit cycles are obtained by proving the existence of zeros of Melnikov functions based on some known results, and present a new distribution. Both works of Liu and Li [2008], Li et al. [2009] are the best results so far for cubic systems and are very important, yielding H(3) ≥ 13. In this paper, motivated by the work of Li et al. [2009], we consider the following cubic system
Fig. 1.
The distribution of the 13 limit cycles of system (2).
with K0 ≈
+
x˙ = 2y(y 2 − k2 ), y˙ = −(x3 + bx2 − x)
(2)
− εy(δ1 + δ2 x + δ3 x2 + δ4 y 2 ), where k > 0, b > 0, and δ1 , δ2 , δ3 and δ4 are parameters. It is easy to see that (1) and (2) are equivalent. We use the method developed in Han and Chen [2000], Han et al. [2008], Yang and Han [2007] to prove that system (2) can have 13 limit cycles. The main purpose is to give a concrete condition for the existence of 13 limit cycles, which can be taken as an improvement of the work by Li et al. [2009] in two aspects: a simple and concrete condition for the existence of 13 limit cycles is given; on the other hand, the proof method (i.e. the way to find zeros of Melnikov functions) is simpler and more direct. The condition here is also much simpler than that of Liu and Li [2008]. More precisely, we have the following main result. Theorem 1. System (2) has 13 limit cycles if
k = 100,
7 b= , 2
0 < k1 δ4 < δ3 < k2 δ4
(3)
where k1 ≈ 0.75249999999975,
77514906777564200000 δ4 8051754528899929
K1 ≈
2127709682470669621 δ3 , 603881589667494675
38757455034549104 , 120776317933498935
δ1∗ = −30000δ4 .
In particular, if δ3 = 20000,
δ4 =
8 × 106 301
(5)
and (4) holds with K0 ≈
265838587879011391034962400 , 1038676334228090841
K1 ≈
38757455034549104 120776317933498935
δ1∗ = −
(6)
24 × 1010 301
then system (2) has 13 limit cycles. Remark. The distribution of the 13 limit cycles
under the condition of the above theorem is just the same as obtained in Li et al. [2009] as shown in Fig. 1.
2. Proof of the Main Result k2 ≈ 0.75250000000036,
and 0 < ε K0 + K1 δ1 − δ2 δ1∗ − δ1 1
(4)
Consider system (2). For ε = 0, (2) is Hamiltonian with H(x, y) =
1 1 b 1 4 y − k2 y 2 + x4 − x2 + x3 , 2 4 2 3
(7)
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Existence Conditions of Thirteen Limit Cycles in a Cubic System
and has five centers A1 , A2 , A3 , A4 , O (the origin) and four saddles S0 , S1 , S2 , S3 , where A1 = (x1 (b), k) S0 = (0, k),
A4 = (x2 (b), −k),
h1 = −
3673 455 √ − 65, 384 384
h2 = −
3673 455 √ + 65 384 384
(8)
S1 = (x1 (b), 0),
S2 = (x2 (b), 0),
(11)
S3 = (0, −k)
√ and xi (b) = (−b + (−1)i b2 + 4)/2, i = 1, 2. Let Int. J. Bifurcation Chaos 2010.20:2569-2577. Downloaded from www.worldscientific.com by UNIVERSITY OF WESTERN ONTARIO WESTERN LIBRARIES on 07/25/12. For personal use only.
h0 = −50000000,
A2 = (x2 (b), k),
A3 = (x1 (b), −k),
α1 = H(A1 ) = H(A3 ), α2 = H(A2 ) = H(A4 ), h0 = H(S0 ) = H(S3 ),
(9)
h1 = H(S1 ), h2 = H(S2 ). For example, for k = 100, b = 7/2, then (8) and (9) become √ 7 65 , 100 , A1 = − − 4 4 √ 7 65 , 100 , A2 = − + 4 4 √ 7 65 , −100 , A3 = − − 4 4 √ 7 65 , −100 , A4 = − + (10) 4 4
respectively. It is easy to see that α1 < α2 < h0 < h1 < h2 < 0. Then for each of h1 and h2 , the equation H(x, y) = hj defines a double homoclinic loop passing through Sj , denoted by Lj , and for h0 , the equation H(x, y) = h0 with y > 0 defines a double homoclinic loop passing through S0 , denoted by L0 . Let L01 = L0 |x≤0 , L02 = L0 |x≥0 , L11 = L1 |y≥0 , L12 = L1 |y≤0 , and L21 be the homoclinic loop passing through S2 and surrounding a unique singular point at the origin, and L22 be the homoclinic loop passing through S2 and surrounding all other singular points. Then Lj = Lj1 ∪ Lj2 , j = 0, 1, 2. See Fig. 2. Introduce L± h : H(x, y) = h,
h ∈ (h0 , h1 ), ±y > 0,
L± jh : H(x, y) = h,
h ∈ (αj , h0 ), j = 1, 2
˜ 1h : H(x, y) = h, L
±y > 0 ˜ 1h ⊂ Int L21 , h ∈ (h2 , 0), L
˜ 2h : H(x, y) = h, L
˜ 2h ⊃ L22 , h > h2 , Int L
˜ 2h surrounding all the singular points. In with L the following, we fix (k, b, δ3 , δ4 ) satisfying (3). Then
S0 = (0, 100), √ 7 65 ,0 , S1 = − − 4 4 √ 7 65 ,0 , S2 = − + 4 4 S3 = (0, −100) and α1 = −
19200003673 455 √ − 65, 384 384
α2 = −
19200003673 455 √ + 65, 384 384
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Fig. 2.
Four double homoclinic loops of (2)|ε=0 .
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J. Yang et al.
correspondingly we have the following Melnikov functions ± q(x, y, δ1 , δ2 )dx, h ∈ (h0 , h1 ), M (h, δ1 , δ2 ) = Mj± (h, δ1 , δ2 )
=
q(x, y, δ1 , δ2 )dx,
h ∈ (αj , h0 ), j = 1, 2,
˜ 2 (h, δ1 , δ2 ) = M
Since H(x, y) is even in y and q is odd in y, we have (see [Li et al., 2009]) M + (h, δ1 , δ2 ) = M − (h, δ1 , δ2 ), − M+ j (h, δ1 , δ2 ) = Mj (h, δ1 , δ2 ),
L± h
L± jh
˜ 1 (h, δ1 , δ2 ) = M Int. J. Bifurcation Chaos 2010.20:2569-2577. Downloaded from www.worldscientific.com by UNIVERSITY OF WESTERN ONTARIO WESTERN LIBRARIES on 07/25/12. For personal use only.
WSPC/S0218-1274
˜ 1h L
˜ 2h L
q(x, y, δ1 , δ2 )dx,
h2 < h < 0,
q(x, y, δ1 , δ2 )dx,
h > h2 , (12)
where q(x, y, δ1 , δ2 ) = −y(δ1 + δ2 x + δ3 x2 + δ4 y 2 ).
(13)
j = 1, 2.
(14)
Thus, we will only consider the zeros of M + , M + j ˜ j , j = 1, 2. and M For convenience, we let M (h, δ1 , δ2 ) = M + (h, δ1 , δ2 ), Mj (h, δ1 , δ2 ) = Mj+ (h, δ1 , δ2 ),
j = 1, 2.
(15)
We will prove Theorem 1 by finding zeros of the ˜ 2 . First, on their anafunctions M, M1 , M2 and M lytical property at the endpoints of their domain, by Remark 1.4 in [Han & Chen, 2000] or Theorem 2.2 in [Han et al., 2008], Theorem 1.2 in [Han, 2000] and Lemma 2.9 in [Han, 2006] we have Lemma 1. Let (3) be satisfied by δ3 and δ4 con-
stant. Then M (h, δ1 , δ2 ) = c0 + 2c1 (h − h0 )ln|h − h0 | + c2 (h − h0 ) + O(|h − h0 |2 ln|h − h0 |),
0 < h − h0 1,
M (h, δ1 , δ2 ) = c0 + O(|h − h1 |ln|h − h1 |),
0 < h1 − h 1,
Mj (h, δ1 , δ2 ) = c0j + c1 (h − h0 )ln|h − h0 | + c2j (h − h0 ) + O(|h − h0 |2 ln|h − h0 |),
M1 (h, δ1 , δ2 ) = b01 (h − α1 ) + O((h − α1 )2 ),
0 < h − α1 1,
˜ j (h, δ1 , δ2 ) = c˜0j + O(|h − h2 |ln|h − h2 |), M
j = 1, 2, 0 < h − h2 1,
˜ 1 (h, δ1 , δ2 ) = ˜b01 h + O(h2 ), M
0 < −h 1,
where c0j = Mj (h0 , δ1 , δ2 ) q(x, y, δ1 , δ2 )dx, = L0j
c˜0j =
1 δ1 + 150δ4 , 200
L0j
L11
j = 1, 2, q(x, y, δ1 , δ2 )dx,
j = 1, 2.
√
√ 2π 2π qy (O, δ1 , δ2 ) = − δ1 , 100 100 √ 2πqy (A1 , δ1 , δ2 ) = √ 50 65 + 7 65 √ √ 1 7 + 65 2 2π − δ1 + δ2 = √ 100 400 65 + 7 65 √ 57 + 7 65 δ3 − 300δ4 . − 800
˜b01 =
c2 = c2 + O(c1 ), c2 = c21 + c22 , c2j = [qy (x, y, δ1 , δ2 ) − qy (0, 100, δ1 , δ2 )]dt,
c0 =
q(x, y, δ1 , δ2 )dx,
(17) and
L2j
j = 1, 2,
c0 = c01 + c02 , c1 =
(16)
j = 1, 2, 0 < h0 − h 1,
b01
(18)
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Existence Conditions of Thirteen Limit Cycles in a Cubic System
Under the condition (3), (7) becomes
1 y1 = 6
1 1 7 1 H(x, y) = x4 + y 4 − 10000y 2 − x2 + x3 . 4 2 2 6
2573
360000 − 6x −18x2 − 84x + 36,
√ 7 67 , x ˆ2 = − + 3 3 1 360000 + 6x −18x2 − 84x + 36. y2 = 6 (20)
By (17) we have c0j = −(δ1 Ij1 + δ2 Ij2 + δ3 Ij3 + δ4 Ij4 ), j = 1, 2, c2j = − (δ2 x + δ3 x2 + 3δ4 (y 2 − 104 ))dt L0j
Int. J. Bifurcation Chaos 2010.20:2569-2577. Downloaded from www.worldscientific.com by UNIVERSITY OF WESTERN ONTARIO WESTERN LIBRARIES on 07/25/12. For personal use only.
=−
L0j
δ2 x + δ3 x2 + 3δ4 (y 2 − 104 ) dx 2y(y 2 − 104 )
= −(δ2 Jj2 + δ3 Jj3 + 3δ4 Jj4 ),
(19)
By Maple we can obtain I11 ≈ 0.19336179335719, I12 ≈ −0.60397799779263,
j = 1, 2,
I13 ≈ 2.1276842302270,
where Iji =
L0j
Ij4 =
L0j
xi−1 ydx =
x ˆj
0 x ˆj
I14 ≈ 5800.8535538510,
xi−1 (y1 − y2 )dx,
I21 ≈ 0.42548181555552 × 10−3 , I22 ≈ 0.96408125135325 × 10−4 ,
j = 1, 2, i = 1, 2, 3,
y 3 dx =
0
(y13 − y23 )dx,
I23 ≈ 0.25452243669621 × 10−4 ,
j = 1, 2,
I24 ≈ 12.764454466315, J12 ≈ −0.036722972448375,
xi−1 dx 2 L0j 2y(y − 10000) 0 xi−1 xi−1 − dx, = 2 2y2 (y22 − 10000) x ˆj 2y1 (y1 − 10000)
Jji =
Jj4 =
L0j
j = 1, 2, i = 2, 3, 0 1 1 1 dx = − dx, j = 1, 2 2y 2y2 x ˆj 2y1
(21)
J13 ≈ 0.10568693840688, J14 ≈ −0.96680909021854 × 10−5 , J22 ≈ 0.77058591984887 × 10−2 , J23 ≈ 0.20196618711375 × 10−2 , J24 ≈ −0.21274090779586 × 10−7 . By (17) again, we have
with
c0 = −(δ1 I 1 + δ2 I 2 + δ3 I 3 + δ4 I 4 ), c˜01 = −(δ1 I˜1 + δ2 I˜2 + δ3 I˜3 + δ4 I˜4 ),
x2 , 100) = h0 , H(ˆ x1 , 100) = H(ˆ √ 7 67 , x ˆ1 = − − 3 3 Ii =
L11
I4 =
L11
c˜02 = −(δ1 J˜1 + δ2 J˜2 + δ3 J˜3 + δ4 J˜4 ),
xi−1 ydx = 3
y dx =
x2
x1
where x2 x1
xi−1 (y 1 − y 2 )dx,
i = 1, 2, 3,
(y 31 − y32 )dx,
H(x1 , 100) = H(x1 , 100) = h1 , x1 ≈ −120.10904980338, x2 ≈ 117.77510233465, √ 1 y1 = 1440000 + 6 57599988981 − 1365 65 + 576x2 − 1344x3 − 288x4 , 12 √ 1 y2 = 1440000 − 6 57599988981 − 1365 65 + 576x2 − 1344x3 − 288x4 , 12
(22)
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and
I˜i =
i−1
L21
x
J˜i =
ydx = 2
i−1
L22
x
J˜4 =
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9:34
L22
x ˜3
y dx = 2 3
x
x ˜4
ydx = 2 x ˜5
x ˜1
i−1
x ˜5 x ˜1
y˜2 dx,
i−1
x
y˜13 dx
I˜4 =
y˜1 dx +
x ˜1
+ x ˜2
x ˜1
x ˜2
y˜23 dx
3
L21
y dx = 2
i−1
x
y˜2 dx +
x ˜4
+ x ˜5
y˜23 dx
x ˜3
x ˜4
i−1
x
x ˜5
y˜23 dx,
x ˜4
i = 1, 2, 3,
y˜2 dx ,
i = 1, 2, 3,
,
x2 , 0) = H(˜ x3 , 0) = H(˜ x4 , 0) = H(˜ x5 , 100) = h2 , x ˜1 < x ˜2 < x ˜3 < x ˜4 < x ˜5 , H(˜ x1 , 100) = H(˜ √ √ 65 1 7 − ˜2 = − − 49 + 21 65, x ˜1 ≈ −120.1090612, x 12 4 6 √ √ 7 7 1√ 65 1 + 49 + 21 65, x ˜4 = − + 65, x ˜5 ≈ 117.7751137, x ˜3 = − − 12 4 6 4 4 √ 1 1440000 + 6 57599988981 + 1365 65 + 576x2 − 1344x3 − 288x4 , y˜1 = 12 √ 1 1440000 − 6 57599988981 + 1365 65 + 576x2 − 1344x3 − 288x4 . y˜2 = 12 Also by Maple, we get where I 1 ≈ 26426.695431194,
K 0 = −(I13 + I23 )δ3 − (I14 + I24 )δ4
I 2 ≈ −30847.384030487,
≈−
I 3 ≈ 93503233.017653, I 4 ≈ 594536975.78668,
−
I˜1 ≈ −0.6165686494 × 10−3 , I˜2 ≈ −0.2266214100 × 10−4 , I˜3 ≈ −0.6263873180 × 10−5 ,
(23)
I˜4 ≈ −0.5273174788 × 10−9 , J˜1 ≈ 0.5285407076 × 105 , J˜2 ≈ −0.6169553114 × 105 , J˜3 ≈ 0.1870066096 × 109 ,
2127709682470669621 δ3 1018 1162723601663463 δ4 , 2 × 1011
K 1 = −(I11 + I21 ) ≈ − K 2 = −(I12 + I22 ) ≈
c0 = K 0 + K 1 δ1 + K 2 δ2 , c1 =
1 δ1 + 150δ4 , 200
24155263586699787 . 4 × 1016
Therefore we have Lemma 2. Let (3) be satisfied. Then c0 ≥ 0 if and
only if
J˜4 ≈ 0.1189074254 × 1010 . Now substituting (21) into (19) and by (17) we get
2422340939659319 , 125 × 1014
δ2 ≥ K0 + K1 δ1 , and c1 ≥ 0 if and only if δ1 ≥ −30000δ4 , where K0 = − +
77514906777564200000 K0 δ4 ≈ 8051754528899929 K2 2127709682470669621 δ3 , 603881589667494675
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K1 = −
which together with (3) follows that
K1 38757455034549104 . ≈ 120776317933498935 K2
c01 > 0,
(24) Let δ1∗ = −30000δ4 , δ2∗ = K0 + K1 δ1∗ . Then by (24) δ2∗ ≈ −
Next, we study the property of the functions ˜ 2 as (δ1 , δ2 ) = (δ∗ , δ∗ ). In this M, M1 , M2 and M 1 2 case, by (18), (19) and (21)–(24), we have
c2 < 0,
c˜02 > 0,
c˜01
b01 < 0, c0 > 0, < 0, ˜b01 > 0.
(26)
Hence, by (16) and (26) we have
3291534008000 δ4 8051754528899929
2127709682470669621 δ3 . + 603881589667494675 Int. J. Bifurcation Chaos 2010.20:2569-2577. Downloaded from www.worldscientific.com by UNIVERSITY OF WESTERN ONTARIO WESTERN LIBRARIES on 07/25/12. For personal use only.
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M (h, δ1∗ , δ2∗ ) < 0
for 0 < h − h0 1,
M (h, δ1∗ , δ2∗ ) > 0
for 0 < h1 − h 1,
M1 (h, δ1∗ , δ2∗ ) > 0
for 0 < h0 − h 1,
M1 (h, δ1∗ , δ2∗ ) < 0
for 0 < h − α1 1,
˜ 2 (h, δ∗ , δ∗ ) > 0 M 1 2
for 0 < h − h2 1.
(27)
Then we can prove c01
16007678383413707737 δ4 ≈− 402587726444996450000000000 +
c02 ≈
Lemma 3. Let (3) be satisfied. Then
7349954756236336535315248941 δ3 , 20129386322249822500000000000000
16007678383413707737 δ4 402587726444996450000000000 −
7349954756236336535315248941 δ3 , 20129386322249822500000000000000
69269175158411302787920692328991 δ4 c2 ≈ 4025877264449964500000000000000000000 16510200872197961063715188463601 δ3 , 3019407948337473375000000000000000 √ (28248237268876517 65 √ − 4633315056457821781) 2 ≈ √ δ3 483105271733995740000 65 + 7 65 √ √ (8228835020 65 + 57601845140) 2 δ4 , − √ 8051754528899929 65 + 7 65 −
b01
c0 ≈ 198263874.53882 δ4 − 93394545.687548δ3 , c˜02 ≈ 396547843.8 δ4 − 186789232.3δ3 , √ ˜b01 = 300 2 δ4 π,
(1) the function M1 (h, δ1∗ , δ2∗ ) has a zero h∗0 ∈ (α1 , h0 ); (2) M (h, δ1∗ , δ2∗ ) has three zeros h∗j ∈ (h0 , h1 ), j = 3, 4, 5, with h∗3 < h∗4 < h∗5 ; ˜ ∗ ∈ (h2 , +∞); ˜ 2 (h, δ∗ , δ∗ ) has a zero h (3) M 1 2 0 (4) all of these zeros are of an odd multiplicity. Proof. First by (27), the function M1 (h, δ1∗ , δ2∗ )
has a zero h∗0 ∈ (α1 , h0 ). Similarly, the function M (h, δ1∗ , δ2∗ ) also has a zero in (h0 , h1 ). To find more zeros of M , let us calculate their values at some points. By (12), (13) and (15) we have (δ1 + δ2 x + δ3 x2 + δ4 y 2 )ydx M (h, δ1 , δ2 ) = − Lh
= −(δ1 Iˆ1 + δ2 Iˆ2 + δ3 Iˆ3 + δ4 Iˆ4 ), ˜ 2 (h, δ1 , δ2 ) = − M
h > h2 , (25) where
Lh
Iˆ4 =
xj−1 ydx = 3
Lh
y dx =
x2 (h) x1 (h)
x2 (h) x1 (h)
xj−1 (ˆ y1 − yˆ2 )dx,
(ˆ y13 − yˆ23 )dx,
(δ1 + δ2 x + δ3 x2 + δ4 y 2 )ydx
= −(δ1 Jˆ1 + δ2 Jˆ2 + δ3 Jˆ3 + δ4 Jˆ4 ),
c˜01 ≈ −18.49704737 δ4 + 0.00008611108867δ3
Iˆj =
˜ 2h L
h0 < h < h1 ,
j = 1, 2, 3,
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J. Yang et al.
Jˆj =
˜ 2h L
Jˆ4 =
xj−1 ydx = 2
˜ 2h L
1 yˆ1 = 6
Int. J. Bifurcation Chaos 2010.20:2569-2577. Downloaded from www.worldscientific.com by UNIVERSITY OF WESTERN ONTARIO WESTERN LIBRARIES on 07/25/12. For personal use only.
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1 yˆ2 = 6
y 3 dx = 2
x ˜4 (h)
x ˜1 (h) x ˜4 (h)
x ˜1 (h)
xj−1 yˆ1 dx +
yˆ13 dx +
x ˜3 (h)
x ˜4 (h)
x ˜3 (h) x ˜4 (h)
yˆ23 dx +
x ˜1 (h) x ˜2 (h)
x ˜1 (h) x ˜2 (h)
360000 − 6 −18x4 − 84x3 + 36x2 + 3600000000 + 72h.
H(˜ xi (h), 0) = h, H(˜ xi (h), 100) = h,
i = 1, 2, h0 < h < h1 ,
j = 1, 2, 3,
Unfortunately, we cannot find a zero of ˜ M1 (h, δ1∗ , δ2∗ ) for h ∈ (h2 , 0). By Lemmas 2 and 3 we can prove Lemma 4. Let (3) be satisfied by δ3 and δ4 con-
i = 2, 3, h > h2 ,
stant. If
i = 1, 4, h > h2 .
0 < K0 + K1 δ1 − δ2 δ1∗ − δ1 1,
(29)
where δ1∗ = −30000δ4 , and K0 and K1 are given by (24), then
By Maple 10 we have M (h, δ1∗ , δ2∗ )|h=−31502882.64306 ≈ 18119541.753047δ4 − 24079125.253207δ3 ≡ M1∗ , M (h, δ1∗ , δ2∗ )|h=−31502882.64304 ≈ 18119541.753057δ4 − 24079125.253240δ3
(1) (2) (3) (4)
M (h, δ1 , δ2 ) has five zeros in (h0 , h1 ); M1 (h, δ1 , δ2 ) has a zero in (α1 , h0 ); ˜ 2 (h, δ1 , δ2 ) has a zero in (h2 , +∞); M all of these zeros are of an odd multiplicity.
Proof.
By Lemma 2, under (29) we have 0 < −c0 −c1 −c2
≡ M2∗ , ˜ 2 (h, δ∗ , δ∗ )|h=9×107 M 1 2 ≈ 410963172.382053δ4 − 645192646.536934δ3 ˜ ∗. ≡M (28) By (3) and (28), we have M1∗
xj−1 yˆ2 dx ,
yˆ23 dx ,
360000 + 6 −18x4 − 84x3 + 36x2 + 3600000000 + 72h,
H(xi (h), 100) = h,
˜1 (h) < x ˜2 (h) < 0 < and x1 (h) < 0 < x2 (h), x x ˜3 (h) < x ˜4 (h) satisfy
> 0,
M2∗
< 0,
˜∗
M < 0.
Thus, by (27), M (h, δ1∗ , δ2∗ ) has three zeros h∗3 , h∗4 and h∗5 satisfying h∗3
xj−1 yˆ2 dx +
∈ (h0 , −31502882.64306),
h∗4 ∈ (−31502882.64306, −31502882.64304), h∗5 ∈ (−31502882.64304, h1 ), ˜ ∗ ∈ (h2 , 9 × 107 ). ˜ 2 (h, δ∗ , δ∗ ) has a zero h while M 1 2 0 ∗ ∗ ˜ , h∗ , h∗ and h∗ are Note that all of the zeros h0 , h 0 3 4 5 of an odd multiplicity. The proof is completed.
which ensures that M (h, δ1 , δ2 ) has two new zeros h∗1 , h∗2 ∈ (h0 , h∗3 ) near h0 both having an odd multiplicity with h∗1 < h∗2 . At the same time, the zeros h∗3 , h∗4 and h∗5 of M (h, δ1 , δ2 ), h∗0 of M1 (h, δ1 , δ2 ) and ˜ ∗ of M ˜ 2 (h, δ1 , δ2 ) remain under (29). This ends the h 0 proof. Corollary 1. Let k = 100, b = 7/2 and (δ3 , δ4 ) satisfy (5). Then for K0 , K1 and δ1∗ satisfying (6) and δ1 and δ2 satisfying (29), the conclusions (1)–(4) of Lemma 4 hold.
In fact, under (5) we have k1 δ4 ≈ 19999.999999993,
k2 δ4 ≈ 20000.000000010.
Thus, (3) is satisfied. Hence, the corollary follows from Lemma 4. On the other hand, in this case, similar to (25) and (28) we have c01 ≈ 7.3016544125820 > 0, c2 ≈ −108.90321618344 < 0,
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b01 ≈ −147.39575898743 < 0, c0 ≈ 0.34015808347891 × 1013 > 0, c˜02 ≈ 0.680369293 × 1013 > 0, and M (h, δ1∗ , δ2∗ )|h=−31502882.64306 ≈ 0.22899937435633 > 0,
Int. J. Bifurcation Chaos 2010.20:2569-2577. Downloaded from www.worldscientific.com by UNIVERSITY OF WESTERN ONTARIO WESTERN LIBRARIES on 07/25/12. For personal use only.
M (h, δ1∗ , δ2∗ )|h=−31502882.64304 ≈ −0.16451532561954 < 0, ˜ 2 (h, δ∗ , δ∗ )|h=9×107 M 1 2 ≈ −1981243697993.1 < 0 which also yield Corollary 1 by the above discussion. Now, it is clear that Theorem 1 follows from Lemma 4, Corollary 1 and the well-known Poincare– Pontrjagin–Andronov theorem (Theorem 6.1 in [Li, 2003]).
Acknowledgments The authors would like to thank C. Li for providing them a preprint of the work [Li et al., 2009] before the preparation of the present work.
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