limit cycle bifurcations of two kinds of polynomial differential systems

December 9, 2011 14:10 WSPC/S0218-1274 03057

International Journal of Bifurcation and Chaos, Vol. 21, No. 11 (2011) 3341–3357c© World Scientific Publishing CompanyDOI: 10.1142/S021812741103057X

LIMIT CYCLE BIFURCATIONS OF TWO KINDSOF POLYNOMIAL DIFFERENTIAL SYSTEMS*

PEIPEI ZUO and MAOAN HAN†The Institute of Mathematics, Shanghai Normal University,

Shanghai 200234, P. R. China†[email protected]

Received October 2, 2010; Revised April 6, 2011

In this paper, by using qualitative analysis and the first-order Melnikov function method, weconsider two kinds of polynomial systems, and study the Hopf bifurcation problem, obtainingthe maximum number of limit cycles.

Keywords : Limit cycle; bifurcation; Melnikov function.

1. Introduction and Main Results

As we know, an important problem in studying thereal polynomial differential systems is to find thenumber of limit cycles and their distribution. Atypical way to produce limit cycles is to perturba family of periodic orbits by using the first-orderMelnikov function. There have been many works onthis aspect. Below, we list some of them. Let Pn andQn be two polynomials of degree n. It is well knownthat the system

x = −y + εPn(x, y), y = x + εQn(x, y)

has at most [(n− 1)/2] limit cycles up to first-orderin ε using the first-order Melnikov method, where[ · ] denotes the integer part function, see [Giaco-mini et al., 1996] for instance. Llibre et al. [2001]considered the system

x = −y(1 + x) + εPn(x, y),

y = x(1 + x) + εQn(x, y)

and proved that it has at most n limit cycles upto first-order in ε. Xiang and Han [2004] consideredthe system

x = −y(1 + x)(2 + x) + εPn(x, y),

y = x(1 + x)(2 + x) + εQn(x, y)

and proved that it has at most 2n+2− (−1)n limitcycles up to first-order in ε. Later Buica and Llibre[2007] studied the system

x = −y(x + a)(y + b) + εPn(x, y),

y = x(x + a)(y + b) + εQn(x, y)

and obtained that up to first-order in ε, the upperbound for the number of limit cycles is 3[(n − 1)/2] + 4 if a �= b, and 2[(n− 1)/2] + 2 if a = b. For thesystem

x = y(x2 − a2)(y2 − b2) + εPn(x, y),

y = −x(x2 − a2)(y2 − b2) + εQn(x, y),

Atabaigi et al. [2009a, 2009b] proved that up tofirst-order in ε, the upper bound for the numberof limit cycles is (3/2)(n + sin2(nπ/2)) + 1 if a �= b,and n − 1 if a = b.

Recently, Atabaigi et al. [2009a, 2009b] studiedthe perturbed system

x = y(1 + x4),

y = −x(1 + x4) + ε

l∑j=0

ajxjy2m−1

(1)

∗The project was supported by National Natural Science Foundation of China (10971139).†Author for correspondence

3341

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http://dx.doi.org/10.1142/S021812741103057X

December 9, 2011 14:10 WSPC/S0218-1274 03057

3342 P. Zuo & M. Han

for 0 < ε � 1, where l = 2n + 2 or 2n + 3, m, nare arbitrary positive integers and a0, a1, . . . , al arereal, obtaining the following Theorem A.

Theorem A [Atabaigi et al., 2009a, 2009b]. Anupper bound for the number of limit cycles of system(1) that bifurcate from the period annulus of system(1) |ε=0 up to first-order in ε is min{N1, N2, N3},where

N1 = 4m + 2n − 2 + sin2 nπ

2

+

m + n − 3 + sin2 nπ

24

,

N2 = 3m + 2n − 2 − sin2 nπ

2+[m + n − 1

2

],

N3 = 5m + 2n − 2 + sin2 nπ

2.

Moreover, there is a system (1) with at least 3m +n − 2 limit cycles.

From the proof of the above theorem, one canfind that the first part of it (i.e. the upper boundresult) is also true for the following perturbedsystem

x = y(1 + x4),

y = −x(1 + x4) + εl∑

j=0

m∑i=1

aijxjy2i−1

(2)

for 0 < ε � 1, where l = 2n + 2 or 2n + 3, m ≥ 1,n ≥ 0, and ai,j (1 ≤ i ≤ m, 0 ≤ j ≤ l) are real.

Also, it is not necessary to require n ≥ 1. Thatis, the proof of the first part of Theorem A givenin [Atabaigi et al., 2009a, 2009b] remains true forthe case n = 0. However, we will give an exam-ple to show that the second part of Theorem A isfalse for the case m = 2 and n = 0. Then a prob-lem arises naturally: How many limit cycles can wehave for system (1)? A similar problem occurs forsystem (2).

Our main results in this paper are the following.

Theorem B. For any m ≥ 1, n ≥ 0, the system (1)has Hopf cyclicity n + 1 at the origin. Therefore,there is a system of the form (1) which has at leastn + 1 limit cycles.

Theorem C. For any m ≥ 1, n ≥ 0, the system (2)has Hopf cyclicity at least 3m + n− 2 at the origin.Therefore, there is a system of the form (2) whichhas at least 3m + n − 2 limit cycles.

The paper is organized as follows. In Sec. 2, weestablish some preliminary lemmas and theoremswhich are necessary to prove the main results. InSec. 3, we prove our main results, studying the Hopfbifurcation of limit cycles for systems (1) and (2),and obtaining their Hopf cyclicity. Then in Sec. 4,we will give an example to show that the secondpart of Theorem A is false.

2. Preliminary Lemmas andTheorems

In this section, we consider a general near-Hamiltonsystem of the form{

x = Hy + εP (x, y, δ),

y = −Hx + εQ(x, y, δ),(3)

where P, Q and H are Cω functions, ε is a smallparameter, δ ∈ D ⊆ R

N with D compact. Supposefor (x, y) near the origin

H(x, y) =12(x2 + y2) + O(|x, y|3),

P (0, 0, δ) = Q(0, 0, δ) = 0.(4)

For ε = 0, (3) becomes a Hamiltonian system withthe Hamiltonian function H(x, y), having a centerat the origin. In fact, for 0 < h � 1, the equa-tion H(x, y) = h defines a periodic orbit Lh of (3)(ε = 0) surrounding the origin and shrinking to theorigin as h → 0. Let Lh intersect the positive x-axisat A(h) = (a(h), 0). Let B(h, ε, δ) denote the firstintersection point of the positive orbit of (3) start-ing at A(h) with the positive x-axis. Then by (3)and (4) we have

H(B) − H(A) =∫

ABdH = ε[M(h, δ) + O(ε)],

(5)

where M(h, δ) =∮Lh

(QHy +PHx)|ε=0dt. We call M

the first-order Melnikov function or Abelian integralof (3), which plays an important role in the studyof limit cycle bifurcation.

By taking a polar coordinate transformationx = r cos θ, y = r sin θ for (3), we obtain

dr

dθ= R(θ, r, ε, δ), (6)

where R is C∞ and 2π-periodic in θ withR(θ, 0, ε, δ) = 0. We can suppose

R(θ, r, ε, δ) = R1(θ, ε, δ)r + R2(θ, ε, δ)r2 + · · · .

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Limit Cycle Bifurcations of Two Kinds of Polynomial Differential Systems 3343

Let r(θ, a, ε, δ) denote the solution of (6) with initialvalue r(0, a, ε, δ) = a. Then

r(θ, a, ε, δ)

= a + ε[A1(θ, ε, δ)a + A2(θ, ε, δ)a2 + · · ·]. (7)

Let P (a, ε, δ) = r(−2π, a, ε, δ) denote the Poincaremap of (3). Then the point B in (5) is given by(P (a(h), ε, δ), 0), where a(h) =

√2h(1 + O(

√h))

satisfying H(a(h), 0) = h.By (7), it is easy to see that P (a, ε, δ) − a =

O(|εa|). By the mean value theorem, we have

H(B) − H(A) = Hx(a + θ(P − a), 0)(P − a)

= Hx(a + O(εa), 0)(P − a)

= Hx(a(1 + O(ε)), 0)(P − a)

= a[1 + O(|ε| + |a|)](P − a), (8)

where 0 < θ < 1. If we take r = a(h) and d(r,ε, δ) = P (r, ε, δ) − r, then by (5) and (8)

rd(r, ε, δ)(1 + O(|ε| + |r|)) = ε[M(h, δ) + O(ε)].(9)

The following lemma is well known, called theprinciple of symmetry.

Lemma 2.1. Let f(x, y) = Hy + εP (x, y, δ), g(x,y) = −Hx+εQ(x, y, δ). If f(−x, y) = f(x, y), g(−x,y) = −g(x, y), then the system (3) has a center atthe origin.

Lemma 2.2 [Han, 2000, 2006]. Consider system (3)satisfying (4). Then for any k ≥ 1, we have thefollowing expansion for M :

M(h, δ) = h[b0(δ) + b1(δ)h + · · · + bk(δ)hk

+ O(hk+1)], 0 < h � 1.

Further if there exists δ0 ∈ RN such that

bj(δ0) = 0 for j = 0, . . . , k − 1, bk(δ0) �= 0 and

rank(

∂b0

∂δ, . . . ,

∂bk−1

∂δ

)(δ0) = k,

then system (3) has k limit cycles near the originfor some (ε, δ) near (0, δ0).

Lemma 2.3. If H(x, y) = (1/2)(x2 + y2) and P,Q = O(|x, y|2m−1), then d(r, ε, δ) = εr2m−2d∗(r,ε, δ), where d∗ ∈ C∞ satisfying d∗(0, ε, δ) = 0.

Proof. We take a polar coordinate transformationx = r cos θ, y = r sin θ for system (3), to obtain

r = ε[cos θ · P (r cos θ, r sin θ)

+ sin θ · Q(r cos θ, r sin θ)],

θ = −1 +ε

r[cos θ · Q(r cos θ, r sin θ)

− sin θ · P (r cos θ, r sin θ)].

(10)

Then from (10), we can get a 2π-periodic equationof the form

dr

dθ= εr2m−1R(θ, r, ε, δ), R ∈ C∞. (11)

Let r(θ, r0, ε, δ) denote the solution of (11) withinitial value r(0, r0, ε, δ) = r0. We have

r(θ, r0, ε, δ) =∑j≥1

ϕj(θ, ε, δ)rj0. (12)

Since r(0, r0, ε, δ) = r0, it follows

ϕ1(0, ε, δ) = 1, ϕi(0, ε, δ) = 0 (i ≥ 2, i ∈ N).(13)

Now we substitute (12) into (11) and obtain

dr

dθ= ϕ′

1(θ, ε, δ)r0 + ϕ′2(θ, ε, δ)r2

0 + · · ·

+ ϕ′2m−1(θ, ε, δ)r2m−1

0 + · · ·= ε[ϕ2m−1

1 (θ, ε, δ)r2m−10 + O(r2m

0 )]

× [R(θ, ϕ1r0, ε, δ) + O(r20)]

= εr2m−10 [ϕ2m−1

1 (θ, ε, δ)R(θ, 0, ε, δ) + O(r0)],

which yields

ϕ′1(θ, ε, δ) = ϕ′

2(θ, ε, δ)

= · · · = ϕ′2m−2(θ, ε, δ) = 0,

ϕ′2m−1(θ, ε, δ) = εϕ2m−1

1 (θ, ε, δ)R(θ, 0, ε, δ),

ϕ′j(θ, ε, δ) = O(ε) (j ≥ 2m).

Using (13), we get

ϕ1(θ, ε, δ) = 1,

ϕ2(θ, ε, δ) = · · · = ϕ2m−2(θ, ε, δ) = 0,

ϕ2m−1(θ, ε, δ) = ε

∫ θ

0R(θ, 0, ε, δ)dθ,

ϕj(θ, ε, δ) = O(ε) (j ≥ 2m).

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Hence, (12) becomes

r(θ, r0, ε, δ) = r0 + ϕ2m−1(θ, ε, δ)r2m−10

+ εO(r2m0 ).

Since the Poincare map is P (r0, ε, δ) = r(−2π, r0,ε, δ), we have

d(r0, ε, δ)

= P (r0, ε, δ) − r0

= ε

[r2m−10

∫ −2π

0R(θ, 0, ε, δ)dθ + O(r2m

0 )],

or

d(r, ε, δ) = εr2m−2d∗(r, ε, δ),

d∗(r, ε, δ) = r

[∫ −2π

0R(θ, 0, ε, δ)dθ + O(r)

].

This ends the proof. �

The following theorem generalizes a main resulton limit cycle bifurcation in [Han, 2000].

Theorem 2.1. Suppose that there exists m ≥1, such that d(r, ε, δ) = εr2m−2d∗(r, ε, δ), whered∗(0, ε, δ) = 0. Then we have M(h, δ) =hm−1M∗(h, δ), where M∗(h, δ) = h

∑j≥0 bj(δ)hj ,

0 < h � 1. Further if the following conditions aresatisfied :

(i) there exist an integer k ≥ 1 and δ0 ∈ Rm, such

that bj(δ0) = 0, j = 0, 1, . . . k.(ii) rank (∂(b0, b1, . . . , bk)/∂(δ1, δ2, . . . , δm))|δ=δ0 =

k + 1, m ≥ k + 1.(iii) Equation (3) has a center at the origin as

bj(δ) = 0, j = 0, 1, . . . , k.

Then Eq. (3) has Hopf cyclicity k at the originfor |ε| > 0 and |δ − δ0| being sufficiently small.Moveover, if the functions P and Q in (3) are linearin δ (that is to say, ∂P/∂δ, ∂Q/∂δ are independentof δ), then for any given N > |δ0|, there exist ε0 > 0,such that Eq. (3) has Hopf cyclicity k at the originfor 0 < |ε| < ε0, |δ| ≤ N.

Proof. By (9) we have

M(h, δ) = r2m−1d∗(r, 0, δ)(1 + O(r)), (14)

where r =√

2h(1 + O(√

h)). Thus

M(h, δ) = O(r2m) = O(hm).

Since M is analytic, we can write M(h, δ) =hm−1M∗(h, δ), where M∗(h, δ) = h

∑j≥0 bj(δ)hj .

Further, let

d∗(r, ε, δ) =∑j≥1

dj(ε, δ)rj .

Then by (14), we can obtain

2md10 = b0,

2md20 + O(d10) = 0,

2md30 + O(|d10, d20|) = b1, . . . ,

2md2k,0 + O(|d10, d20, . . . , d2k−1,0|) = 0,

2md2k+1,0 + O(|d10, d20, . . . , d2k,0|) = bk,

where dj0 = dj |ε=0. From the above, we can solve

d10 = 2−mb0,

d20 = O(b0),...

d2k,0 = O(|b0, b1, . . . , bk−1|),d2k+1,0 = 2−mbk + O(|b0, b1, . . . , bk−1|).

(15)

By conditions (i) and (ii), we can take b0, . . . , bk assmall free parameters. Then by condition (iii), wehave dj = 0 as b0 = · · · = bk = 0. Thus, we canwrite

dj = b0d∗j0 + b1d

∗j1 + · · · + bkd

∗jk, j ≥ 1. (16)

Hence, by (15), (16) and dj = dj0 +O(ε), we obtain

d1 = 2−mb0 + O(ε|b0, . . . , bk|),d2 = O(b0 + ε|b0, . . . , bk|),d3 = 2−mb1 + O(b0 + ε|b0, . . . , bk|),

...

d2k = O(|b0, . . . , bk−1| + ε|b0, . . . , bk|),d2k+1 = 2−mbk + O(|b0, . . . , bk−1| + ε|b0, . . . , bk|).

Obviously, from the equations

d2j+1 = 2−mbj + O(|b0, . . . , bj−1|+ ε|b0, b1, . . . , bk|), j = 0, . . . , k,

we can solve

bj = 2md2j+1 + O(|d1, d3, . . . , d2j−1|+ ε|d1, d3, . . . , d2k+1|), j = 0, . . . , k.

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This means that d1, d3, . . . , d2k+1 can be taken asfree parameters with the property that dj = 0 forall j ≥ 2k + 2 if d1 = d3 = · · · = d2k+1 = 0. Alsoby the property of the function d(r0, ε, δ) (see [Han,2006] for example), we have

d2 = O(d1), d4 = O(|d1, d3|), . . . ,d2k = O(|d1, d3, . . . , d2k−1|).

Therefore we can write

d∗(r, ε, δ) =k∑

j=0

r2j+1d2j+1(1 + Pj),

where Pj = O(r)∈Cw. Then the conclusion can beverified by following the idea of proving Theo-rem 1.3 in [Han, 2000]. The proof is completed. �

3. Proof of the Main Results

Now we are in position to prove our main results.

Proof of Theorem B. It is easy to see that system (1)is equivalent to

x = y,

y = −x + ε

l∑j=0

ajxjy2m−1

1 + x4 ,

(17)

for 0 < ε � 1, where l = 2n + 2 or 2n + 3, m,n are arbitrary positive integers and a0, a1, . . . , al

are real. The above system satisfies the condi-tions of Lemma 2.3. Hence, we have d(r, ε, δ) =εr2m−2d∗(r, ε, δ) (m ≥ 1), and d∗(0, ε, δ) = 0.

For ε = 0, the unperturbed system of (17) hasa family of periodic orbits given by

Lh : x2 + y2 = h or x =√

h sin t, y =√

h cos t.

Then by (7) in [Atabaigi et al., 2009a, 2009b], adirect calculation of the Melnikov function of (17) is

M(h, δ) =n+1∑k=0

a2k

m∑r=0

Crmhm−r(−1)r

∮Lh

x2(r+k)

1 + x4dt.

(18)

Further by (8) in [Atabaigi et al., 2009a, 2009b], wecan get that

Ij =∮

Lh

x2j

1 + x4dt

=∫ 2π

0

(√

h sin t)2j

1 + h2 sin4 tdt

=∫ 2π

0hj sin2j t

∞∑s=0

(−1)sh2s sin4s tdt

=∞∑

s=0

(−1)shj+2s

∫ 2π

0sin2(j+2s) tdt

=∞∑

s=0

(−1)shj+2sΓ2(j+2s), (19)

where Γ0 = 2π, and

Γ2i =∫ 2π

0sin2i tdt = 2π

(2i − 1)!!(2i)!!

for i ∈ N. Then using (19), (18) becomes

M(h, δ) =n+1∑k=0

a2k

m∑r=0

Crmhm−r(−1)rIr+k

=n+1∑k=0

a2k

m∑r=0

Crmhm−r(−1)r

×∞∑

s=0

(−1)shr+k+2sΓ2(r+k+2s)

=n+1∑k=0

a2k

∞∑s=0

hm+k+2s

×m∑

r=0

Crm(−1)r+sΓ2(r+k+2s)

= hmn+1∑k=0

a2k

∞∑s=0

hk+2s(−1)s

×m∑

r=0

Crm(−1)rΓ2(r+k+2s). (20)

For the sake of simplicity, let∑m

r=0 Crm(−1)r ×

Γ2(r+p) = Cp.Obviously, for all p ∈ N

Cp =m∑

r=0

Crm(−1)rΓ2(r+p)

=∫ 2π

0

(m∑

r=0

Crm(−1)r sin2(r+p) t

)dt

=∫ 2π

0sin2p t

(m∑

r=0

Crm(−1)r sin2r t

)dt

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=∫ 2π

0sin2p t cos2m tdt

> 0.

Then by Lemma 2.2, (20) becomes

M(h, δ) = hmn+1∑k=0

a2k

∞∑s=0

hk+2s(−1)sCk+2s

= hm[b0(δ) + b1(δ)h + · · · + bn+1(δ)hn+1

+ bn+2(δ)hn+2 + bn+3(δ)hn+3 + · · ·],where

b0(δ) = a0C0,

b1(δ) = a2C1,

b2(δ) = (−a0 + a4)C2,

b3(δ) = (−a2 + a6)C3,

b4(δ) = (a0 − a4 + a8)C4,

b5(δ) = (a2 − a6 + a10)C5,...

bn+1(δ) =

n+12∑

t=0

(−1)n+1

2−ta4tCn+1, for n odd,

n2∑

t=0

(−1)n2−ta4t+2Cn+1, for n even.

(21)

Now without loss of generality, we proceed our dis-cussion for n odd (the case of n even can be donesimilarly). In this case, we have

bn(δ) =

n−12∑

t=0

(−1)n−1

2−ta4t+2Cn,

bn+2(δ) =

n+12∑

t=0

(−1)n+1

2−ta4t+2Cn+2

=

n−12∑

t=0

(−1)n+1

2−ta4t+2Cn+2,

bn+3(δ) =

n+32∑

t=0

(−1)n+3

2−ta4tCn+3

=

n+12∑

t=0

(−1)n+3

2−ta4tCn+3.

(22)

For 1 ≤ k ≤ n + 2, set

δk =

k−12∑

l=0

(−1)k−1

2−la4l, (if k is odd).

k2−1∑

l=0

(−1)k2−1−la4l+2, (if k is even).

(23)

Then

∂(δ1, δ2, . . . , δn+2)∂(a0, a2, a4, . . . , a2n+2)

=

1 0 0 0 · · · 0

0 1 0 0 · · · 0

−1 0 1 0 · · · 0

0 −1 0 1 · · · 0...

......

......

1 0 −1 0 · · · 1

,

which yields

det∂(δ1, δ2, . . . , δn+2)∂(a0, a2, . . . , a2n+2)

= 1.

Thus, using (23), it follows from (21) and (22)

bj(δ) = δj+1Cj , j = 0, 1, . . . , n + 1,

bn+2(δ) = −δn+1Cn+2,

bn+3(δ) = −δn+2Cn+3.

(24)

Noting Cp > 0 (p ∈ N), we have

rank∂(b0(δ), b1(δ), . . . , bn+1(δ))

∂(δ1, δ2, . . . , δn+2)

= rank

C0 0 0 · · · 0

0 C1 0 · · · 0

0 0 C2 · · · 0...

......

...

0 0 0 · · · Cn+1

= n + 2.

Therefore

det∂(b0(δ), b1(δ), . . . , bn+1(δ))

∂(a0, a2, . . . , a2n+2)> 0.

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From (23) and (24), we have b0(δ) = b1(δ) = · · · =bn+1(δ) = 0 if and only if a0 = a2 = · · · = a2n+2 = 0.If a0 = a2 = · · · = a2n+2 = 0, (17) becomes

x = y,

y = −x + ε

n+1∑k=0

a2k+1x2k+1y2m−1

1 + x4,

which has a center at the origin by Lemma 2.1.By Theorem 2.1, the proof is completed. �

Proof of Theorem C. Just like the proof of The-orem B, we still use the first-order Melnikov

function method. It is easy to see that thesystem (2) is equivalent to

x = y,

y = −x + ε

l∑j=0

m∑i=1

ai,jxjy2i−1

1 + x4,

(25)

for 0 < ε � 1, where l = 2n + 2 or 2n + 3,m ≥ 1, n ≥ 0 and ai,j (1 ≤ i ≤ m, 0 ≤ j ≤ l)are real.

By (3), (19) and Lemma 2.1 in [Atabaigi et al.,2009a, 2009b], for the Melnikov function of (25)we have

M(h, δ) =∮

Lh

l∑j=0

m∑i=1

aijxjy2i

1 + x4dt =

l∑j=0

m∑i=1

aij

∮Lh

xjy2i

1 + x4dt

=l∑

j=0

m∑i=1

aij

∮Lh

xj(h − x2)i

1 + x4dt =

l∑j=0

m∑i=1

aij

∮Lh

xji∑

r=0

Cri h

i−r(−1)rx2r

1 + x4dt

=l∑

j=0

m∑i=1

aij

i∑r=0

Cri h

i−r(−1)r∮

Lh

x2r+j

1 + x4dt (let j = 2k)

=n+1∑k=0

m∑i=1

ai,2k

i∑r=0

Crih

i−r(−1)r∮

Lh

x2(r+k)

1 + x4dt =

m∑i=1

n+1∑k=0

ai,2k

i∑r=0

Crih

i−r(−1)rIr+k

=m∑

i=1

n+1∑k=0

ai,2k

i∑r=0

Cri h

i−r(−1)r∞∑

s=0

(−1)shr+k+2sΓ2(r+k+2s)

=m∑

i=1

n+1∑k=0

ai,2k

∞∑s=0

hi+k+2si∑

r=0

Cri (−1)r+sΓ2(r+k+2s).

We can write M(h, δ) =∑m

i=1 Mi(h, δ), where

Mi(h, δ) =n+1∑k=0

ai,2k

∞∑s=0

hi+k+2si∑

r=0

Cri (−1)r+sΓ2(r+k+2s)

= hi

[n+1∑k=0

ai,2k

∞∑s=0

hk+2s(−1)si∑

r=0

Cri (−1)rΓ2(r+k+2s)

]. (26)

As before, we have∑i

r=0 Cri (−1)rΓ2(r+k+2s) ≡ λi,i+k+2s > 0.

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For i, j ∈ N and i < j, we have

λi,j =i∑

r=0

Cri (−1)rΓ2(r+j−i)

=∫ 2π

0

(i∑

r=0

Cri (−1)r sin2(r+j−i) t

)dt

=∫ 2π

0sin2(j−i) t

(i∑

r=0

Cri (−1)r sin2r t

)dt

=∫ 2π

0sin2(j−i) t cos2i tdt

= 2π(2i − 1)!!(2j − 2i − 1)!!

(2j)!!, (27)

which follows

λi,j + λi+1,j = λi,j−1. (28)

Particularly, we have

λ1,j = 2π(2j − 3)!!

(2j)!!. (29)

By Lemma 2.2, (26) becomes

Mi(h, δ) = hin+1∑k=0

ai,2k

∞∑s=0

hk+2s(−1)sλi,i+k+2s

= h[bi,0(δ) + bi,1(δ)h + bi,2(δ)h2

+ · · · + bi,i−2(δ)hi−2 + bi,i−1(δ)hi−1

+ bi,i(δ)hi + bi,i+1(δ)hi+1

+ · · · + bi,i+n(δ)hi+n + · · ·]= hi[bi,i−1(δ) + bi,i(δ)h + bi,i+1(δ)h2

+ · · · + bi,i+n(δ)hn+1 + · · ·],

where

bi,0(δ) = bi,1(δ) = bi,2(δ) = · · · = bi,i−2(δ) = 0,

bi,i−1(δ) = ai,0λi,i,

bi,i(δ) = ai,2λi,i+1,

bi,i+1(δ) = (−ai,0 + ai,4)λi,i+2,

bi,i+2(δ) = (−ai,2 + ai,6)λi,i+3,

bi,i+3(δ) = (ai,0 − ai,4 + ai,8)λi,i+4,

bi,i+4(δ) = (ai,2 − ai,6 + ai,10)λi,i+5,

...

bi,i+n(δ) =

n+12∑

t=0

(−1)n+1

2−tai,4tλi,i+n+1,

for n odd,

n2∑

t=0

(−1)n2−tai,4t+2λi,i+n+1,

for n even.

(30)

Just like what we have done in the proof of The-orem B, we also proceed our discussion for n odd(the case of n even can be done similarly).

In this case, we have

bi,i+n−1(δ) =

n−12∑

t=0

(−1)n−1

2−tai,4t+2λi,i+n,

bi,i+n+1(δ) =

n+12∑

t=0

(−1)n+1

2−tai,4t+2λi,i+n+2

=

n−12∑

t=0

(−1)n+1

2−tai,4t+2λi,i+n+2,

bi,i+n+2(δ) =

n+32∑

t=0

(−1)n+3

2−tai,4tλi,i+n+3

=

n+12∑

t=0

(−1)n+3

2−tai,4tλi,i+n+3,

...

(31)

For 1 ≤ i ≤ m, 1 ≤ j ≤ n + 2, set

δi,j =

j−12∑

l=0

(−1)j−12

−lai,4l, (if j is odd),

j2−1∑

l=0

(−1)j2−1−lai,4l+2, (if j is even),

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δi = (δi,1, δi,2, . . . , δi,n+2),

ai = (ai,0, ai,2, . . . , ai,2n+2).

(32)

Then ∂(δ1, δ2, . . . , δm)/∂(a1, a2, . . . , am) is a lowertriangular matrix of order m(n + 2) with the ele-ments on the main diagonal equal to 1. Thus,

rank∂(δ1, δ2, . . . , δm)∂(a1, a2, . . . , am)

= m(n + 2). (33)

Hence, by (32), it follows from (30) and (31)

bi,0(δ) = bi,1(δ) = bi,2(δ) = · · · = bi,i−2(δ) = 0,

bi,i+j−1(δ) = δi,j+1λi,i+j, j = 0, 1, . . . , n + 1,

bi,i+j−1(δ) =

(−1)sδi,n+1λi,i+j,

j = n + 2s, s ≥ 1,

(−1)sδi,n+2λi,i+j,

j = n + 2s + 1, s ≥ 1.

(34)

In fact from Lemma 2.2, we have

Mi(h, δ) = h[bi,i−1(δ)hi−1 + bi,i(δ)hi

+ bi,i+1(δ)hi+1 + · · ·+ bi,i+n(δ)hi+n + · · ·],

M(h, δ) = h[b0(δ) + b1(δ)h + b2(δ)h2 + · · ·+ bm−1(δ)hm−1 + bm(δ)hm + · · ·+ bm(n+2)−1(δ)h

m(n+2)−1 + · · ·].

Therefore, bk(δ) =∑m

i=1 bi,k(δ), where 0 ≤ k ≤m(n + 2) − 1. Noting bi,k(δ) = 0 for 0 ≤ k ≤ i − 2,

using (34) we have for 0 ≤ k ≤ m(n + 2) − 1,1 ≤ i ≤ m, 1 ≤ j ≤ n + 2,

(1) if 0 ≤ k ≤ i − 2, then

∂bk

∂δi,j=

m∑i=1

∂bi,k

∂δi,j=

∂bi,k

∂δi,j= 0; (35)

(2) if i − 1 ≤ k ≤ n + i, then

∂bk

∂δi,j=

m∑i=1

∂bi,k

∂δi,j=

∂bi,k

∂δi,j

=

{λi,k+1, j = k − i + 2,

0, others;(36)

(3) if k ≥ n + i + 1, then

(i) for k = n + i + (2s − 1), s ≥ 1

∂bk

∂δi,j=

m∑i=1

∂bi,k

∂δi,j=

∂bi,k

∂δi,j

=

{(−1)sλi,k+1, j = n + 1,

0, others,(37)

(ii) for k = n + i + 2s, s ≥ 1

∂bk

∂δi,j=

m∑i=1

∂bi,k

∂δi,j=

∂bi,k

∂δi,j

=

{(−1)sλi,k+1, j = n + 2,

0, others.(38)

Here, we consider the case m ≥ 2 (the case m = 1is already discussed in Theorem B).

Now, in the case m ≥ 2, we devote to calculat-ing the rank of the following matrix

A =∂[b0(δ), b1(δ), b2(δ), . . . , bm−1(δ), bm(δ), . . . , bm+n(δ), . . . , bm(n+2)−1(δ)]

∂[δ1, δ2, . . . , δm],

where δ1, . . . , δm are given in (32). This is a square matrix of order m × (n + 2).For i = 1, 2, . . . ,m, let

Ai =∂[b0(δ), b1(δ), b2(δ), . . . , bm−1(δ), bm(δ), . . . , bm+n(δ), . . . , bm(n+2)−1(δ)]

∂[δi].

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First, we calculate A1 which consists of the first n + 2 columns of the matrix A. By (35)–(38), weobtain

A1 =∂[b0(δ), b1(δ), b2(δ), . . . , bm−1(δ), bm(δ), . . . , bm+n(δ), . . . , bm(n+2)−1(δ)]

∂[δ1]

=∂[b1,0(δ), b1,1(δ), b1,2(δ), . . . , b1,n(δ), b1,n+1(δ); b1,n+2(δ), b1,n+3(δ); . . . , b1,m+n(δ), . . . , b1,m(n+2)−1(δ)]

∂[δ1,1, δ1,2, . . . , δ1,n+2]

=

λ1,1 0 · · · 0 0

0 λ1,2 · · · 0 0...

......

...

0 0 · · · λ1,n+1 0

0 0 · · · 0 λ1,n+2

0 0 · · · −λ1,n+3 0

0 0 · · · 0 −λ1,n+4

......

......

.

Then, in a similar manner, we can calculate A2, . . . , Am. For example,

A2 =∂[b0(δ), b1(δ), b2(δ), . . . , bm−1(δ), bm(δ), . . . , bm+n(δ), . . . , bm(n+2)−1(δ)]

∂[δ2]

=

∂[b2,0(δ); b2,1(δ), b2,2(δ), . . . , b2,n(δ),

b2,n+1(δ), b2,n+2(δ); b2,n+3(δ), b2,n+4(δ); . . . , b2,m+n(δ), . . . , b2,m(n+2)−1(δ)]∂[δ2,1, δ2,2, . . . , δ2,n+2]

=

0 0 · · · 0 0

λ2,2 0 · · · 0 0

0 λ2,3 · · · 0 0...

......

...

0 0 · · · λ2,n+2 0

0 0 · · · 0 λ2,n+3

0 0 · · · −λ2,n+4 0

0 0 · · · 0 −λ2,n+5

......

......

,

Am =∂[b0(δ), b1(δ), b2(δ), . . . , bm−1(δ), bm(δ), . . . , bm+n(δ), . . . , bm(n+2)−1(δ)]

∂[δm]

=

∂[bm,0(δ), bm,1(δ), . . . ; bm,m−1(δ),

bm,m(δ), bm,m+1(δ), . . . , bm,m+n−1(δ), bm,m+n(δ), bm,m+n+1(δ), . . . , bm,m(n+2)−1(δ)]∂[δm,1, δm,2, . . . , δm,n+2]

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=

0 0 · · · 0 0...

......

...

0 0 · · · 0 0

λm,m 0 · · · 0 0

0 λm,m+1 · · · 0 0...

......

...

0 0 · · · λm,m+n 0

0 0 · · · 0 λm,m+n+1

0 0 · · · −λm,m+n+2 0

0 0 · · · 0 −λm,m+n+3

0 0 · · · λm,m+n+4 0

0 0 · · · 0 λm,m+n+5

......

......

.

Now combining A1, A2, . . . , Am together, we obtain

A = (A1, A2, . . . , Am).

We now make elementary column transforma-tions to A. By adding a nonzero constant multiple ofthe (j +1)th column of Am−1 to jth column of Am,for j = 1, . . . , n − 1, then the first n− 1 columns ofthe resulting matrix of Am become zero, while thelast three columns remain invariant. In a similarmanner, we can make the first n− 1 columns of theresulting Am−1 become zero by using the resultingAm−2. We repeat the procedure up to A2 by usingA1, finally obtaining

A → (B, 0),

where B is a matrix which is constituted byA1 and the last three columns of the resultingA2, A3, . . . , Am respectively, having m(n + 2) rowsand 3m + n − 1 columns. That is to say,

rankA ≤ 3m + n − 1 ≤ m(n + 2)

for m ≥ 2, n ≥ 1.

The above discussion suggests us to introduce thefollowing matrix

A =∂(b0(δ), b1(δ), . . . , b3m+n−2(δ))

∂(δ1, δ2, . . . , δm),

which is constituted by the first 3m + n− 1 rows ofA, and has 3m + n− 1 rows and m(n + 2) columns.Hence by the above elementary column transforma-tions, we have

A → (B, 0),

where B is a square matrix of order 3m + n − 1,consisting of the first 3m + n − 1 rows of B. Byfurther elementary column and row transformationof matrices we can obtain

B →

λ1,1

λ1,2

. . .

λ1,n

λ2,n+1

λ3,n+2

. . .

λm,m+n−1

C0

D0

,

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where C0 and D0 are square matrices of order m given by

C0 =

−λ1,m+n −λ2,m+n λ3,m+n · · · λm,m+n

λ1,m+n+2 λ2,m+n+2 −λ3,m+n+2 · · · −λm,m+n+2

−λ1,m+n+4 −λ2,m+n+4 λ3,m+n+4 · · · λm,m+n+4

......

......

λ1,m+n+2(m−1) λ2,m+n+2(m−1) −λ3,m+n+2(m−1) · · · −λm,m+n+2(m−1)

and

D0 =

λ1,m+n+1 −λ2,m+n+1 −λ3,m+n+1 · · · λm,m+n+1

−λ1,m+n+3 λ2,m+n+3 λ3,m+n+3 · · · −λm,m+n+3

λ1,m+n+5 −λ2,m+n+5 −λ3,m+n+5 · · · λm,m+n+5

......

......

−λ1,m+n+(2m−1) λ2,m+n+(2m−1) λ3,m+n+(2m−1) · · · −λm,m+n+(2m−1)

respectively. Obviously

± |B| = λ1,1λ1,2 · · ·λm,m+n−1|C0||D0|. (39)

In order to prove rank A = 3m + n − 1, it is sufficient to prove |B| �= 0. By (27), λi,j > 0 for all i, j ∈ N

and i < j. Then by (39), we need only to prove |C0| �= 0 and |D0| �= 0.In fact, by adding the (j − 1)th column to the jth column of C0 and D0 for j = m,m − 1, . . . , 2, we

obtain by (28)

C1 =

−λ1,m+n −λ1,m+n−1 λ2,m+n−1 · · · λm−1,m+n−1

λ1,m+n+2 λ1,m+n+1 −λ2,m+n+1 · · · −λm−1,m+n+1

−λ1,m+n+4 −λ1,m+n+3 λ2,m+n+3 · · · λm−1,m+n+3

......

......

λ1,m+n+2(m−1) λ1,m+n+2m−3 −λ2,m+n+2m−3 · · · −λm−1,m+n+2m−3

and

D1 =

λ1,m+n+1 −λ1,m+n −λ2,m+n · · · λm−1,m+n

−λ1,m+n+3 λ1,m+n+2 λ2,m+n+2 · · · −λm−1,m+n+2

λ1,m+n+5 −λ1,m+n+4 −λ2,m+n+4 · · · λm−1,m+n+4

......

......

−λ1,m+n+(2m−1) λ1,m+n+2m−2 λ2,m+n+2m−2 · · · −λm−1,m+n+2m−2

.

We further add the (j − 1)th column to the jth column of C1 and D1 for j = m,m− 1, . . . , 3, and use (28)again to obtain two matrices, denoted by C2 and D2 respectively. Then by repeating the same procedurewe finally obtain Cm−1 and Dm−1 as follows

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Cm−1

=

−λ1,m

+n

−λ1,m

+n−1

λ1,m

+n−2

···

λ1,n

+1

λ1,m

+n+

2λ

1,m

+n+

1−λ

1,m

+n

···

−λ1,n

+3

−λ1,m

+n+

4−λ

1,m

+n+

3λ

1,m

+n+

2···

λ1,n

+5

. . .. . .

. . .. . .

λ1,m

+n+

2(m

−1)

λ1,m

+n+

2m−3

−λ1,m

+n+

2m−4

···

−λ1,n

+2m−1

an

d

Dm−1

=

λ1,m

+n+

1−λ

1,m

+n

−λ1,m

+n−1

···

λ1,n

+2

−λ1,m

+n+

3λ

1,m

+n+

2λ

1,m

+n+

1···

−λ1,n

+4

λ1,m

+n+

5−λ

1,m

+n+

4−λ

1,m

+n+

3···

λ1,n

+6

. . .. . .

. . .. . .

−λ1,m

+n+

(2m−1

)λ

1,m

+n+

2m−2

λ1,m

+n+

2m−3

···

−λ1,n

+2m

.

The

refo

re,b

y(2

9)an

dus

ing

the

prop

erties

ofde

term

inan

ts,w

eha

ve

±|C|=

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣λ1,m

+n

λ1,m

+n−1

λ1,m

+n−2

···

λ1,n

+1

λ1,m

+n+

2λ

1,m

+n+

1λ

1,m

+n

···

λ1,n

+3

λ1,m

+n+

4λ

1,m

+n+

3λ

1,m

+n+

2···

λ1,n

+5

. . .. . .

. . .. . .

λ1,3

m+

n−2

λ1,3

m+

n−3

λ1,3

m+

n−4

···

λ1,2

m+

n−1

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

=(2

π)m

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣(2m

+2n

−3)

!!(2

m+

2n)!!

(2m

+2n

−5)

!!(2

m+

2n−

2)!!

(2m

+2n

−7)

!!(2

m+

2n−

4)!!

···

(2n

+1)

!!(2

n+

4)!!

(2n−

1)!!

(2n

+2)

!!

(2m

+2n

+1)

!!(2

m+

2n+

4)!!

(2m

+2n

−1)

!!(2

m+

2n+

2)!!

(2m

+2n

−3)

!!(2

m+

2n)!!

···

(2n

+5)

!!(2

n+

8)!!

(2n

+3)

!!(2

n+

6)!!

(2m

+2n

+5)

!!(2

m+

2n+

8)!!

(2m

+2n

+3)

!!(2

m+

2n+

6)!!

(2m

+2n

+1)

!!(2

m+

2n+

4)!!

···

(2n

+9)

!!(2

n+

12)!!

(2n

+7)

!!(2

n+

10)!!

. . .. . .

. . .. . .

. . .

(6m

+2n

−7)

!!(6

m+

2n−

4)!!

(6m

+2n

−9)

!!(6

m+

2n−

6)!!

(6m

+2n

−11

)!!

(6m

+2n

−8)

!!···

(4m

+2n

−3)

!!(4

m+

2n)!!

(4m

+2n

−5)

!!(4

m+

2n−

2)!!

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

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=K

0

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

11

···

1

2n+

12n

+4

2n+

52n

+8

···

4m+

2n−

34m

+2n

(2n

+3)

(2n

+1)

(2n

+6)

(2n

+4)

(2n

+7)

(2n

+5)

(2n

+10

)(2n

+8)

···

(4m

+2n

−1)

(4m

+2n

−3)

(4m

+2n

+2)

(4m

+2n

)

(2n

+5)

(2n

+3)

(2n

+1)

(2n

+8)

(2n

+6)

(2n

+4)

(2n

+9)

(2n

+7)

(2n

+5)

(2n

+12

)(2n

+10

)(2n

+8)

···

(4m

+2n

+1)

(4m

+2n

−1)

(4m

+2n

−3)

(4m

+2n

+4)

(4m

+2n

+2)

(4m

+2n

). . .

. . .. . .

(2m

+2n

−3)

(2m

+2n

−5)

···(2

n+

1)(2

m+

2n)(

2m+

2n−

2)···(2

n+

4)(2

m+

2n+

1)(2

m+

2n−

1)···(2

n+

5)(2

m+

2n+

4)(2

m+

2n+

2)···(2

n+

8)···

(6m

+2n

−7)

(6m

+2n

−9)

···(4

m+

2n−

3)(6

m+

2n−

4)(6

m+

2n−

6)···(4

m+

2n)

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

=K

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

11

···

1

12n

+4

12n

+8

···

12n

+4m

12n

+6

12n

+10

···

12n

+4m

+2

12n

+8

12n

+12

···

12n

+4m

+4

. . .. . .

. . .

12n

+2m

12n

+2m

+4

···

12n

+6m

−4∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣=( 1 2) m−

1

K

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

11

···

1

1n

+2

1n

+4

···

1n

+2m

1n

+3

1n

+5

···

1n

+2m

+1

1n

+4

1n

+6

···

1n

+2m

+2

. . .. . .

. . .

1n

+m

1n

+m

+2

···

1n

+3m

−2

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

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where

K0 = (−1)[m2

](2π)m(2n − 1)!!(2n + 2)!!

(2n + 3)!!(2n + 6)!!

· · · (4m + 2n − 5)!!(4m + 2n − 2)!!

,

K =

K0(−3)m−1m∏

k=3

(2k − 12k − 4

)m−k+1

, for m ≥ 3,

12π22∏

k=1

(2n + 4k − 5)!!(2n + 4k − 2)!!

, for m = 2.

It is clear that K �= 0. Hence, |C| �= 0.In a similar manner, we can prove |D| �= 0.

Therefore |B| �= 0. That is to say,

rank∂(b0(δ), b1(δ), . . . , b3m+n−2(δ))

∂(δ1, δ2, . . . , δm)= 3m + n − 1.

From (33)

rank∂(δ1, δ2, . . . , δm)∂(a1, a2, . . . , am)

= m(n + 2).

Therefore

rank∂(b0(δ), b1(δ), . . . , b3m+n−2(δ))

∂(a1, a2, . . . , am)= 3m + n − 1,

where ai = (ai,0, ai,2, . . . , ai,2n+2), i = 1, 2, . . . ,m.Then the conclusion follows from Lemma 2.2.

This completes the proof. �

4. The Number of Limit Cyclesof (1) When m = 2

In this section, we give a counter example to thesecond part of Theorem A for the case m = 2. Weprove

Theorem 4.1. Consider the system (1) for m = 2.Then

(i) If n = 0, up to first-order in ε the system (1)has at most one limit cycle.

(ii) If n = 1, up to first-order in ε the system (1)has at most two limit cycles.

Proof. By (20), for the first-order Melnikov func-tion of (1) in the case m = 2, we have

M(h, δ) =n+1∑k=0

a2k

2∑r=0

Cr2h2−r(−1)rIr+k

=n+1∑k=0

a2k(h2Ik − 2hIk+1 + Ik+2), (40)

where Ij =∮Lh

(x2j/(1 + x4))dt, j ∈ N. ByLemma 2.4 in [Atabaigi et al., 2009a, 2009b], wehave

I2 = 2π − I0, I3 = πh − I1,

I4 = I0 − 2π +34πh2.

(41)

Letting λ =√

1 + h2 and using (21) in [Atabaigiet al., 2009a, 2009b], we get

I0 =√

2π

√λ + 1λ

,

I1 =√

2π√

λ − 1λ

, h =√

λ2 − 1.

(42)

For the case n = 0, (40) becomes

M(h, δ) = a0(h2I0 − 2hI1 + I2)

+ a2(h2I1 − 2hI2 + I3)

= a0L0(h) + a2L2(h), (43)

where

L0(h) = h2I0 − 2hI1 + I2,

L2(h) = h2I1 − 2hI2 + I3.(44)

Obviously, from (41) and (42), we obtain

L0(λ) = L0(h)

= π[2 +√

2(λ − 2)√

λ + 1],

L2(λ) = L2(h)

= π[−3√

λ2 − 1 +√

2(λ + 2)√

λ − 1],

(45)

which yields

L′0(λ) = π

[3√

22

λ√λ + 1

],

L′2(λ) = π

[−3

λ√λ − 1

(1√

λ + 1−

√2

2

)].

(46)

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Note that λ > 1 for h > 0. By (45) and (46), wehave L′

0(λ) > 0, L0(λ) > 0. It follows from (43)

M(h, δ) = L0(λ)Φ1(λ), (47)

where

Φ1(λ) = a0 + a2L2(λ)L0(λ)

.

Using (45) and (46), we further have

Φ′1(λ) = a2

(L2(λ)L0(λ)

)′= P (λ)f(λ), (48)

where

P (λ) =3√

2a2π2λ

2√

λ − 1L20(λ)

,

f(λ) = λ + 3 − 2√

2√

λ + 1.

Then we have P (λ) �= 0 for a2 �= 0, λ > 1 andf ′(λ) = 1 − (

√2/

√λ + 1) > 0 for λ > 1, implying

f(λ) > f(1) = 0.Hence, from (48), we know that Φ′

1(λ) �= 0 fora2 �= 0, λ > 1, which implies that Φ1(λ) has at mostone zero for λ > 1. From (47) and L0(λ) > 0, wecan see that up to first-order in ε, M(h, δ) has atmost one zero for h > 0.

For the case n = 1, (40) becomes

M(h, δ) =2∑

k=0

a2k(h2Ik − 2hIk+1 + Ik+2)

= a0(h2I0 − 2hI1 + I2)

+ a2(h2I1 − 2hI2 + I3)

+ a4(h2I2 − 2hI3 + I4)

= a0L0(h) + a2L2(h) + a4L4(h), (49)

where L0(h) and L2(h) are defined in (44), and

L4(h) = h2I2 − 2hI3 + I4. (50)

Obviously, from (41) and (42), we obtain

L4(λ) = L4(h)

= π

[34λ2 − 11

4−

√2(λ − 2)

√λ + 1], (51)

which satisfies

L′4(λ) = π

[3λ2

(1 −

√2√

λ + 1

)]. (52)

It follows from (49)

M(h, δ) = L0(λ)Ψ1(λ), (53)

where L0(λ) > 0 is defined in the proof of casen = 0, and

Ψ1(λ) = a0 + a2L2(λ)L0(λ)

+ a4L4(λ)L0(λ)

.

Using (45), (46), (51) and (52), we further have

Ψ′1(λ) = a2

[L2(λ)L0(λ)

]′+ a4

[L4(λ)L0(λ)

]′

= G(λ)Ψ2(λ), (54)

where

G(λ) =3√

2π2λf(λ)2√

λ − 1L20(λ)

,

Ψ2(λ) = a2 + a4

√2√

λ − 1 +14

√λ2 − 1(λ − 5)

λ + 3 − 2√

2√

λ + 1.

From f(λ) > 0 in the proof of case n = 0, we haveG(λ) > 0 for λ > 1. We further obtain

Ψ′2(λ) =

a4

4√

λ2 − 11

f2(λ)Q(λ) (55)

and

Q′(λ) =3(λ − 1)√

λ + 1R(λ), (56)

where

Q(λ) = λ3 + 6λ2 − 15λ

− 24 −√

2(3λ2 − 2λ − 17)√

λ + 1,

R(λ) = (λ + 5)√

λ + 1 −√

22

(5λ + 7).

Note that

R′(λ) =3λ + 7

2√

λ + 1− 5

√2

2,

R′′(λ) =3λ − 1

4(λ + 1)32

> 0 for λ > 1,

which implies that R′(λ) > R′(1) = 0 and R(λ) >R(1) = 0 for λ > 1. Then from (56), we haveQ(λ) > Q(1) = 0 for λ > 1. Thus it followsfrom (55) that Ψ′

2(λ) �= 0 for λ > 1 as a4 �= 0.Therefore, Ψ2(λ) has at most one zero for λ > 1which follows by (54) that Ψ′

1(λ) has at most onezero for λ > 1. Hence, Ψ1(λ) has at most two zeros

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for λ > 1. Thus, by (53) and L0(λ) > 0, we knowthat up to first-order in ε, M(h, δ) has at most twozeros for h > 0. The proof is completed. �

References

Atabaigi, A., Nyamoradi, N. & Zangeneh, H. R. Z.[2009a] “The number of limit cycles of a quintic poly-nomial system,” Comput. Math. Appl. 57, 677–684.

Atabaigi, A., Nyamoradi, N. & Zangeneh, H. R. Z.[2009b] “The number of limit cycles of a quintic poly-nomial system with center,” Nonlin. Anal.: Th. Meth.Appl. 71, 3008–3017.

Buica, A. & Llibre, J. [2007] “Limit cycles of a per-turbed cubic polynomial differential center,” ChaosSolit. Fract. 32, 1059–1069.

Giacomini, H., Llibre, J. & Viano, M. [1996] “Onthe nonexistence, existence and uniqueness of limitcycles,” Nonlinearity 9, 501–506.

Han, M. [2000] “On Hopf cyclicity of planar systems,”J. Math. Anal. Appl. 245, 404–422.

Han, M. [2006] “Bifurcation theory of limit cycles ofplanar systems,” Handbook of Differential Equa-tions, Ordinary Differential Equations, Vol. 3, eds.Canada, A., Drabek, P. & Fonda, A. (Elsevier B.V.),pp. 341–433.

Llibre, J., Perez del Rıo, J. S. & Rodrıguez, J. A.[2001] “Averaging analysis of a perturbated quadraticcenter,” Nonlin. Anal.: Th. Meth. Appl. 46, 45–51.

Xiang, G. & Han, M. [2004] “Global bifurcation of limitcycles in a family of polynomial systems,” Math. Anal.Appl. 295, 633–644.

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limit cycle bifurcations of two kinds of polynomial differential systems

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