# Bifurcation of Limit Cycles by Perturbing a Piecewise Linear Hamiltonian System

Authors: Yanqin Xiong; Maoan Han
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

This paper concerns limit cycle bifurcations by perturbing a piecewise linear Hamiltonian system. We first obtain all phase portraits of the unperturbed system having at least one family of periodic orbits. By using the first-order Melnikov function of the piecewise near-Hamiltonian system, we investigate the maximal number of limit cycles that bifurcate from a global center up to first order ofε.


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## Body

## 1. Introduction and Main Results

Recently, piecewise smooth dynamical systems have been well concerned, especially in the scientific problems and engineering applications. For example, see the works of Filippov [1], Kunze [2], di Bernardo et al. [3], and the references therein. Because of the variety of the nonsmoothness, there can appear many complicated phenomena in piecewise smooth dynamical systems such as stability (see [4, 5]), chaos (see [6]), and limit cycle bifurcation (see [7–10]). Here, we are more concerned with bifurcation of limit cycles in a perturbed piecewise linear Hamiltonian system:
(1)x˙=y+εp(x,y,δ),y˙=-g(x)+εq(x,y,δ),
where ε>0 is a sufficiently small real parameter,
(2)g(x)={a1x+a0,x≥0,b1x+b0,x<0
with a1,  a0,  b1,  and b0 real numbers satisfying a12+a02≠0,b12+b02≠0,
(3)p(x,y,δ)={p+(x,y,δ)=∑i+j=0naij+xiyj,x≥0,p-(x,y,δ)=∑i+j=0naij-xiyj,x<0,(4)q(x,y,δ)={q+(x,y,δ)=∑i+j=0nbij+xiyj,x≥0,q-(x,y,δ)=∑i+j=0nbij-xiyj,x<0,
and δ=(aij+,aij-,bij+,bij-)∈D⊂ℝ2(n+1)(n+2) with D compact. Then system (1) has two subsystems(5a)x˙=y+εp+(x,y,δ),y˙=-a1x-a0+εq+(x,y,δ),(5b)x˙=y+εp-(x,y,δ),y˙=-b1x-b0+εq-(x,y,δ),
which are called the right subsystem and the left subsystem, respectively. For ε=0, systems (5a) and (5b) are Hamiltonian with the Hamiltonian functions, respectively,(6)H+(x,y)=12y2+12a1x2+a0x,H-(x,y)=12y2+12b1x2+b0x.Note that the phase portrait of the linear system(7)x˙=y,y˙=-ax-b,
with a2+b2≠0 has possibly the following four different phase portraits on the plane (see Figure 1).The possible phase portraits of system (7).
(a)
a
=
0
,
b
>
0
(b)
a
=
0
,
b
<
0
(c)
a
>
0
(d)
a
<
0Then, one can find that system (1) ∣ε=0 can have 13 different phase portraits (see Figure 2) when at least one family of periodic orbits appears.Figure 2
The possible phase portraits of system (1) ∣ε=0. (1) GC. (2) HoC+. (3) HoC-. (4) HoC-C+. (5) HoS+. (6) HoS-. (7) HoC-L+. (8) HoC+L-. (9) HoC-HoS+. (10) HoC+HoS-. (11) HeS-S+. (12) HeS+. (13) HeS-.We remark that in Figure2,GC: global center,Ho: homoclinic,He: heteroclinic,C+: center in the region {(x,y)∣x>0},C-: center in the region {(x,y)∣x<0},S+: saddle in the region {(x,y)∣x>0},S-: saddle in the region {(x,y)∣x<0},L+: curvilinear or straightline in the region {(x,y)∣x>0},L-: curvilinear or straightline in the region  {(x,y)∣x<0}.It is easy to obtain the following Table1 which shows conditions for each possible phase portrait appearing above. Also, cases (3), (5), (7), (9), and (13) in Figure 2 are equivalent to cases (2), (6), (8), (10), and (12), respectively, by making the transformation
(8)(x,y)→(-x,y),
together with time rescaling dt=-dτ.Table 1
Coefficient conditions for phase portraits (1)–(13).
Coefficientconditions
a
1
≥
0
,
a
0
≥
0
a
1
>
0
a
1
≤
0
,
a
0
≤
0
a
1
<
0
a
1
+
a
0
>
0
a
0
<
0
a
1
+
a
0
<
0
a
0
>
0
b
1
≥
0
,
b
0
≤
0
,
b
1
>
b
0
(1)
(2)
(5)
b
1
>
0
,
b
0
>
0
(3)
(4)
(7)
(9)
b
1
≤
0
,
b
0
≥
0
,
b
0
>
b
1
(8)
(11) (a02b1=a1b02),
b
1
<
0
,
b
0
<
0
(6)
(10)
(12) (a02b1>a1b02),
(13) (a02b1<a1b02)The authors Liu and Han [7] studied system (1) in a subcase of the case (1) of Figure 2 by taking a1=b1=1,a0=b0=0. By using the first order Melnikov function, they proved that the maximal number of limit cycles on Poincaré bifurcations is n up to first-order in ε. The authors Liang et al. [8] considered system (1) in the case (5) of Figure 2 by taking a1=-1,  a0=1,b1=1, and b0=0. By using the same method, they gave lower bounds of the maximal number of limit cycles in Hopf, and Homoclinic bifurcations, and derived an upper bound of the maximal number of limit cycles bifurcating from the periodic annulus between the center and the Homoclinic loop up to the first-order in ε. Clearly, the maximal number of limit cycles in the case (7) or (8) of Figure 2 is [(n-1)/2] on Poincaré, Hopf and Homoclinic bifurcations up to first-order in ε, by using the first order Melnikov function.This paper focuses on studying the limit cycle bifurcations of system (1) in the case (1) of Figure 2 by using the first order Melnikov function. That is, system (1) satisfies
(9)a1≥0,a0≥0,a0+a1>0,b1≥0,b0≤0,b0<b1.
Clearly, system (1) ∣ε=0 satisfying (9) has a family of periodic orbits
(10)Lh=Lh+∪Lh-={(x,y)∣H+(x,y)=h}∪{(x,y)∣H-(x,y)=h},h>0,
such that the limit of Lh as h→0+ is the origin. The intersection points of the closed curve Lh with the positive y-axis and the negative y-axis are denoted by A(h)=(0,2h) and A1(h)=(0,-2h), respectively. Let
(11)M+(h,δ)=∫AA1^q+dx-p+dy,M-(h,δ)=∫A1A^q-dx-p-dy,h>0.
Then, from Liu and Han [7], the first-order Melnikove function corresponding to system (1) is
(12)M(h,δ)=M+(h,δ)+M-(h,δ),h∈(0,+∞).
Let Z(n) denote the maximal number of zeros of  M(h,δ) for h>0 and N(n) the cyclicity of system (1) at the origin. Then, we can obtain the following.Theorem 1.
Let (9) be satisfied. For any given n≥1,  one has Table 2.Table 2
a
1
>
0
,
a
0
=
0
a
1
=
0
,
a
0
>
0
a
1
>
0
,
a
0
>
0
b
1
>
0,
Z
(
n
)
=
n,
Z
(
n
)
=
n
+
[
n
+
1
2
],
Z
(
n
)
=
n
+
[
n
+
1
2
],
b
0
=
0
N
(
n
)
≥
n
N
(
n
)
≥
n
+
[
n
+
1
2
]
N
(
n
)
≥
n
+
[
n
+
1
2
]
b
1
=
0,
Z
(
n
)
=
n
+
[
n
+
1
2
],
Z
(
n
)
=
n,
n
+
[
n
+
1
2
]
≤
Z
(
n
)
≤
n
+
2
[
n
+
1
2
]
b
0
<
0
N
(
n
)
≥
n
+
[
n
+
1
2
]
N
(
n
)
≥
n
N
(
n
)
≥
n
+
[
n
+
1
2
],
b
1
>
0,
Z
(
n
)
=
n
+
[
n
+
1
2
],
n
+
[
n
+
1
2
]
≤
Z
(
n
)
≤
n
+
2
[
n
+
1
2
]
n
≤
Z
(
n
)
≤
n
+
[
n
+
1
2
]
,
a
0
2
a
1
=
b
0
2
b
1,
b
0
<
0
N
(
n
)
≥
n
+
[
n
+
1
2
]
N
(
n
)
≥
n
+
[
n
+
1
2
],
n
≤
Z
(
n
)
≤
n
+
2
[
n
+
1
2
]
,
a
0
2
a
1
≠
b
0
2
b
1
N
(
n
)
≥
n,This paper is organized as follows. In Section2, we will provide some preliminary lemmas, which will be used to prove the main results. In Section 3, we present the proof of Theorem 1.


## 2. Preliminary Lemmas

In this section, we will derive expressions ofM+(h,δ),M-(h,δ) in (11). First, we have the following.Lemma 2.
Suppose system (1) satisfies (9). Then,(i) M+(h,δ) in (11) can be written as
(13)M+(h,δ)=M1+(h,δ)+∑2k=0n‍2k+1+1/22k+1a0,2k+hk+1/2,where(14)M1+(h,δ)=∑i+j=0n-1‍p-ij+∫A1A^xi+1yjdy=∑i+j=0n-1‍q-ij+∫A1A^xiyj+1dx,with(15)p-ij+=ai+1,j++j+1i+1bi,j+1+,q-ij+=bi,j+1++i+1j+1ai+1,j+.(ii)  (h,δ) in (11) can be expressed as
(16)M-(h,δ)=M1-(h,δ)-∑2k=0n‍2k+1+1/22k+1a0,2k-hk+1/2,where(17)M1-(h,δ)=∑i+j=0n-1‍p-ij-∫A1A^xi+1yjdy=∑i+j=0n-1‍q-ij-∫A1A^xiyj+1dx,with(18)p-ij-=ai+1,j-+j+1i+1bi,j+1-,q-ij-=bi,j+1-+i+1j+1ai+1,j-.Proof.
We only prove (i) since (ii) can be verified in a similar way. By (11), we obtain
(19)M+(h,δ)=∫AA1^q+(x,y,δ)dx-p+(x,y,δ)dy+∫→A1Aq+(x,y,δ)dx-p+(x,y,δ)dy-∫→A1Aq+(x,y,δ)dx-p+(x,y,δ)dy=∮AA1^∪A1A→q+(x,y,δ)dx-p+(x,y,δ)dy+∫→A1Ap+(0,y,δ)dy,
which follows that by Green formula and (3)
(20)M+(h,δ)=M1+(h,δ)+∑2k=0n‍2k+1+1/22k+1a0,2k+hk+1/2,
where
(21)M1+(h,δ)=∬intAA1^∪A1A→(px++qy+)dxdy.
Then, by Green formula again
(22)M1+(h,δ)=-∮AA1^∪A1A→p-+(x,y,δ)dy=∮AA1^∪A1A→q-+(x,y,δ)dx,
where
(23)p-+(x,y,δ)=p+(x,y,δ)-p+(0,y,δ)+∫0x‍qy+(u,y,δ)du,q-+(x,y,δ)=q+(x,y,δ)-q+(x,0,δ)+∫0y‍px+(x,v,δ)dv.
By (3), (4), and the above formulas, we have
(24)p-+(x,y,δ)=∑i+j=0n‍aij+xiyj-∑j=0n‍a0j+yj+∑i+j=1n‍ji+1bij+xi+1yj-1=x∑i+j=0n-1‍(ai+1,j++j+1i+1bi,j+1+)xiyj=x∑i+j=0n-1‍p-ij+xiyj,(25)q-+(x,y,δ)=∑i+j=0n‍bij+xiyj-∑j=0n‍bi0+xi+∑i+j=1n‍ij+1aij+xi-1yj+1=y∑i+j=0n-1‍(bi,j+1++i+1j+1ai+1,j+)xiyj=y∑i+j=0n-1‍q-ij+xiyj.
Combining (20)–(25) gives (13) and (14). Thus, the proof is ended.Then, using Lemma2 and (6) we can obtain the following three lemmas.Lemma 3.
(i)
If a1=0,a0>0, then M+(h,δ) in (11) has form
(26)M+(h,δ)=h1/2∑i+2k=0n‍Bi,2k+hi+k,
where
(27)B0,2k+=2k+1+1/2a0,2k+2k+1,Bi,2k+=2k+1+1/2a0i(ai,2k++2k+1ibi-1,2k+1+)×∫0π/2‍sin2kθcos2i+1θdθ,1≤i≤n.(ii)  If b1=0,b0<0, then we have
(28)M-(h,δ)=h1/2∑i+2k=0n‍Bi,2k-hi+k,
where
(29)B0,2k-=-2k+1+1/2a0,2k-2k+1,Bi,2k-=-2k+1+1/2b0i(ai,2k-+2k+1ibi-1,2k+1-)×∫0π/2‍sin2kθcos2i+1θdθ,1≤i≤n.Proof.
Note that alongAA1^,x=h/a0-(1/2a0)y2. Then, inserting it into (14) follows that
(30)M1+(h,δ)=∑i+j=0n-1‍p-ij+∫-2h2h‍(ha0-12a0y2)i+1yjdy=∑i+2k=0n-1‍p-i,2k+2ia0i+1∫02h‍(2h-y2)i+1y2kdy.
Let y=2hsinθ. Then we have dy=2hcosθdθ and the above integral can be carried into
(31)∫02h‍(2h-y2)i+1y2kdy=(2h)i+1+k+(1/2)×∫0π/2‍sin2kθcos2(i+1)+1θdθ.
Thus, using (30) and the above equation we can write (13) as
(32)M+(h,δ)=∑i+2k=0n-1‍2k+1+1/2p-i,2k+a0i+1×∫0π/2‍sin2kθcos2(i+1)+1θdθ×hi+1+k+1/2+∑2k=0n‍2k+1+1/2a0,2k+2k+1hk+1/2=h1/2∑i+2k=0n‍Bi,2k+hi+k,
where
(33)B0,2k+=2k+1+1/2a0,2k+2k+1,Bi,2k+=2k+1+1/2p-i-1,2k+a0i∫0π/2‍sin2kθcos2i+1θdθ,1≤i≤n,
which gives (i) by (15). Thus, (i) holds and we can prove (ii) in the same way by (16)–(18). This ends the proof.Lemma 4.
Let system (5a) satisfy (3) and (4). Then (i) If a1>0,a0=0,M+(h,δ) has the expression
(34)M+(h,δ)=h∑i+2k=0n‍Ai,2k+(h)i+2k,where(35)A0,2k+=2(2)2k+1a0,2k+2k+1,Ai,2k+=2(2)2k+1+i(a1)i(bi-1,2k+1++i2k+1ai,2k+)×∫0π/2‍sini-1θcos2k+2θdθ,i≥1.(ii) If a1>0,a0≠0,M+(h,δ) can be written as
(36)M+(h,δ)=h[∑i+2k=0n-1(bi,2k+1++i+12k+1ai+1,2k+)ϕik+(h)+∑2k=0n2k+1+1/2a0,2k+2k+1hk]+∑i+2k=0n-1‍(bi,2k+1++i+12k+1ai+1,2k+)×∑r=0,reveni‍αirk+(2h+a02a1)k+r/2I-00+(h,δ),or(37)M+(h,δ)=hψ[n/2]+(h,δ)+(2h+a02a1)φ[(n-1)/2]+(2h+a02a1,δ)×(π2-arcsina02a1h+a02),where(38)I-00+(h,δ)=∫a0/a12h+a02/a1‍2h+a02a1-v2dv,eachαirk+ is a nonzero constant and ϕik+,ψ[n/2]+,φ[(n-1)/2]+ are polynomials of degree k+[(i+1)/2],[n/2],[(n-1)/2], respectively.Proof.
Sincey=±2h-a1x2-2a0x along the curve AA1^,M1+(h,δ) in (14) becomes
(39)M1+(h,δ)=∑i+j=0n-1‍q-ij+∫AA1^xiyj+1dx=∑i+j=0n-1‍q-ij+[∫0(-a0+2a1h+a02)/a1‍xi(2h-a1x2-2a0x)(j+1)/2dx]+∫(-a0+2a1h+a02)/a10‍xi(-1)j+1(2h-a1x2-2a0x)(j+1)/2dx=∑i+2k=0n-1‍2q-i,2k+Ii,2k+(h,δ),
where
(40)Ii,2k+(h,δ)=∫0(-a0+2a1h+a02)/a1‍xi[2h+a02a1-a1(x+a0a1)2]k+1/2dx.
Let v=a1(x+a0/a1). Then, we have dv=a1dx and the above equation becomes
(41)Ii,2k+(h,δ)=1(a1)i+1∫a0/a12h+a02/a1‍(v-a0a1)i×(2h+a02a1-v2)k+(1/2)dv.
Fora0=0, make the transformation v=2hsinθ. Then, we have by (41)
(42)Ii,2k+(h,δ)=(2h)2k+1+i+1(a1)i+1∫0π/2‍siniθcos2k+2θdθ.
Substituting the above formula into (37), together with (13), gives that
(43)M+(h,δ)=∑i+2k=0n-1‍2q-i,2k+(a1)i+1×∫0π/2‍siniθcos2k+2θdθ×(2h)2k+1+i+1+∑2k=0n‍2a0,2k+2k+1(2h)2k+1=∑i+2k=0n‍Ai,2k+(h)2k+1+i,
where
(44)A0,2k+=2(2)2k+1a0,2k+2k+1,Ai,2k+=2(2)2k+1+iq-i-1,2k+(a1)i×∫0π/2‍sini-1θcos2k+2θdθ,1≤i≤n.
Thus, by (15) and the above discussion we know that (i) holds.
Fora0≠0, (41) can be represented as
(45)Ii,2k+(h,δ)=1(a1)i+1∑r=0i‍Cir(-a0a1)i-rI-rk+(h,δ),
where
(46)I-rk+(h,δ)=∫a0/a12h+a02/a1‍vr(2h+a02a1-v2)k+1/2dv.
Recall that
(47)∫vr(2h+a02a1-v2)k+1/2dv=vr+1(2h+a02/a1-v2)k+1/22k+2+r+(2k+1)(2h+a02/a1)2k+2+r×∫‍vr(2h+a02a1-v2)k-1/2dv.
Then, by (46) and the above equation we obtain that
(48)I-rk+(h,δ)=-a0r+1(2h)k+1/2a1(r+1)/2(2k+2+r)+(2k+1)(2h+a02/a1)2k+2+rI-r,k-1+(h,δ),k≥1,r≥0.
It follows that
(49)I-rk+(h,δ)=-a0r+1a1(r+1)/22hφ~rk+(h)+α~rk+(2h+a02a1)kI-r0+(h,δ),k≥1,r≥0,
where
(50)α~rk+=(2k+1)(2k-1)(2k-3)×⋯×3(2k+2+r)(2k+r)(2k-2+r)×⋯×(4+r),φ~rk+(h)=(2h)k2k+2+r+2k+1(2k+2+r)(2k+r)×(2h+a02a1)(2h)k-1+(2k+1)(2k-1)(2k+2+r)(2k+r)(2k-2+r)×(2h+a02a1)2(2h)k-2+⋯+(2k+1)(2k-1)×⋯×5(2k+2+r)(2k+r)(2k-2+r)×⋯×(4+r)×(2h+a02a1)k-12h,
which is a polynomial of degree k in h. For convenience, introduce
(51)φ-rk+(h)={φ~rk+(h),k≥1,0,k=0,α-rk+={α~rk+,k≥1,1,k=0.
Then, combining (49) and (51) gives that
(52)I-rk+(h,δ)=-a0r+1a1(r+1)/22hφ-rk+(h)+α-rk+(2h+a02a1)kI-r0+(h,δ),k≥0,r≥0.
Further, by using the formula
(53)∫‍vr(2h+a02a1-v2)1/2dv=-vr-1(2h+a02/a1-v2)3/2r+2+(r-1)(2h+a02/a1)r+2×∫‍vr-2(2h+a02a1-v2)1/2dv,
we have that
(54)I-r0+(h,δ)=a0r-1(2h)3/2a1(r-1)/2(r+2)+r-1r+2(2h+a02a1)I-r-2,0+(h,δ),r≥1.
It follows that
(55)I-r0+(h,δ)=(2h)3/2φ~r+(h)+α~r+(2h+a02a1)r/2I-00(h,δ),r≥1,
where
(56)α~r+={0,rodd,(r-1)(r-3)×⋯×3×1(r+2)r(r-2)×⋯×6×4,reven,φ~r+(h)=a0r-1(r+2)a1(r-1)/2+(r-1)a0r-3(r+2)ra1(r-3)/2(2h+a02a1)+(r-1)(r-3)a0r-5(r+2)r(r-2)a1(r-5)/2(2h+a02a1)2+⋯+(a0r-1-2[(r-1)/2](r-1)(r-3)×⋯×(r+1-2[(r-1)2])×a0r-1-2[(r-1)/2][(r-2)2])×(a1(r-1)/2-[(r-1)/2](r+2)r(r-2)×⋯×(r+2-2[(r-2)2])×a1(r-1)/2-[(r-1)/2][(r-2)2])-1×(2h+a02a1)[(r-1)/2],
which is a polynomial of degree [(r-1)/2] in h. Let
(57)φ-r+(h)={φ~r+(h),r≥1,0,r=0,α-r+={α~r+,r≥1,1,r=0.
Then, we have that by (55) and the above
(58)I-r0+(h,δ)=(2h)3/2φ-r+(h)+α-r+(2h+a02a1)r/2I-00+(h,δ),r≥0.
Substituting the above equation into (52), one can find that
(59)I-rk+(h,δ)=hψ-rk+(h)+α-rkα-r(2h+a02a1)k+r/2×I-00+(h,δ),r≥0,k≥0,
where α-r=0 for r odd, α-r>0 for r even, and
(60)ψ-rk+(h,δ)=-2a0r+1a1(r+1)/2φ-rk+(h)+22hα-rk(2h+a02a1)kφ-r+(h),
which is a polynomial of degree k+[(r+1)/2] in h. Combining (37), (45), and (59) gives that
(61)M1+(h,δ)=h∑i+2k=0n-1‍2q-i,2k+(a1)i+1∑r=0i‍Cir(-a0a1)i-rψ-rk+(h)+∑i+2k=0n-1‍2q-i,2k+(a1)i+1×∑r=0,reveni‍Cirα-rkα-r(-a0a1)i-r×(2h+a02a1)k+r/2I-00+(h,δ),
which implies (35), together with (13) and (15).
Note that(62)∫‍a2-x2dx=12[xa2-x2+a2arcsinx|a|].
Then, we have
(63)I-00+(h,δ)=-a02a12h+12(2h+a02a1)×(π2-arcsina02a1h+a02).
Inserting the above formula into (35), we can obtain (36). Hence, the proof is finished.Similar to Lemma4, we can obtain the following lemma about M-(h,δ).Lemma 5.
Let system (5b) satisfy (3) and (4). Then
(i)
If b1>0,b0=0, M-(h,δ) in (11) has the expression
(64)M-(h,δ)=h∑i+2k=0n‍Ai,2k-(h)i+2k,
where
(65)A0,2k-=-2(2)2k+1a0,2k-2k+1,Ai,2k-=2(2)2k+1+i(b1)i(bi-1,2k+1-+i2k+1ai,2k-)×∫-π/20‍sini-1θcos2k+2θdθ,1≤i≤n.(ii) If b1>0,b0≠0,M-(h,δ) in (11) has the form
(66)M-(h,δ)=h[∑i+2k=0n-1‍(bi,2k+1-+i+12k+1ai+1,2k-)ϕik-(h)-∑2k=0n‍2k+1+1/2a0,2k-2k+1hk]+∑i+2k=0n-1‍(bi,2k+1-+i+12k+1ai+1,2k-)×∑r=0,reveni‍αirk-(2h+b02b1)k+r/2I-00-(h,δ),
or
(67)M-(h,δ)=hψ[n/2]-(h,δ)+(2h+b02b1)φ[(n-1)/2]-×(2h+b02b1,δ)(π2+arcsinb02b1h+b02),
where
(68)I-00-(h,δ)=∫-2h+b02/b1b0/b1‍2h+b02b1-v2dv,
each αirk- is nonzero constant and ϕik-,ψ[n/2]-,φ[(n-1)/2]- are polynomials of degree k+[(i+1)/2],[n/2],[(n-1)/2], respectively.


## 3. Proof of Theorem1

In this section, we will prove the main results. Obviously, under (9) there are the following 9 subcases:(1)
a1=b1=0,a0>0,b0<0,(2)
a1>0,b1>0,a0=b0=0,(3)
a1>0,b0<0,a0=b1=0,(4)
a1>0,a0>0,b1=0,b0<0,(5)
a1>0,a0>0,b1>0,b0=0,(6)
a1>0,a0>0,b1>0,b0<0,(7)
a0>0,b1>0,a1=b0=0,(8)
a1=0,a0>0,b1>0,b0<0,(9)
a1>0,a0=0,b1>0,b0<0.We only give the proof of Subcases1, 2, 3, 4, 5, and 6. And the Subcases 7, 8, and 9 can be verified, similar to Subcases 3, 4, and 5, respectively.Subcase 1.
a
1
=
b
1
=
0
,
a
0
>
0
,
b
0
<
0. From (12) and Lemma 3, one can obtain that
(69)M(h,δ)=M+(h,δ)+M-(h,δ)=h1/2∑i+2k=0n‍Bi,2k+hi+k+h1/2∑i+2k=0n‍Bi,2k-hi+k=h1/2∑i+2k=0n‍(Bi,2k++Bi,2k-)hi+k,
which implies that M(h,δ) has at most n isolated positive zeros for h>0. To show that this bound can be reached, take aij-=bij±=0,aij+=0,j≥1. Then, by (27) and (29), (69) has the form
(70)M(h,δ)=h1/2∑i=0n‍Bi0+hi,
where
(71)B00+=21+1/2a00+,Bi0+=21+1/2a0i(2i)!!(2i+1)!!ai0+,i≥1.
Hence, using (70) we can take ai0+,i=0,1,…,n as free parameters to produce n simple positive zeros of  M(h,δ) near h=0, which gives n limit cycles correspondingly near the origin. Thus, N(n)≥n in this case. This ends the proof.Subcase 2.
a
1
>
0
,
b
1
>
0
,
a
0
=
b
0
=
0 Similar to the above and using (32) and (64), M(h,δ) in (12) has the expression of the form
(72)M(h,δ)=∑i+2k=0n‍Ai,2k+(h)2k+1+i+∑i+2k=0n‍Ai,2k-(h)2k+1+i=h∑l=0n‍Al(h)l,
where Al=∑i+2k=l‍(Ai,2k++Ai,2k-). Further, taking aij-=bij±=0,aij+=0,j≥1, then, by (34) and (65),  M(h,δ) in (72) becomes
(73)M(h,δ)=h1/2∑i=0n‍Ai0+(h)i,
where
(74)A00+=2a00+2k+1,Ai0+=2iai0+(a1)i∫0π/2‍sini-1θcos2θdθ,i≥1.
Thus, from (72) and (73), we can discuss similar to Subcase 1. This finishes the proof.Subcase 3.
a
1
>
0
,
b
0
<
0
,
a
0
=
b
1
=
0 By Lemmas 3, and 4 and (12), we can have that
(75)M(h,δ)=∑i+2k=0n‍Ai,2k+(h)2k+1+i+h1/2∑i+2k=0n‍Bi,2k-hi+k=h1/2(∑i+2k=0n‍Ai,2k+(h)i+2k+∑i+2k=0n‍Bi,2k-hi+k)=h1/2M*(h,δ).
Let us prove that M*(h,δ) has at most n+[(n+1)/2] zeros on the open interval (0,+∞). For the purpose, let h=λ. Then, for n=2l,l≥1, M*(h,δ) in (75) has the expression
(76)M*(h,δ)=∑i+2k=0n‍Ai,2k+λi+2k+∑i+2k=0n‍Bi,2k-λ2i+2k=∑j=02l‍Cjλj+∑j=1l‍C2l+jλ2j+2l≜M-(λ,δ),
where
(77)Cj=∑i+2k=j‍Ai,2k++∑2i+2k=j‍Bi,2k-,j=0,1,2,…,2l,C2l+j=∑i+k=j+l‍Bi,2k-,j=1,2,…,l.
To prove M*(h,δ) has at most n+[(n+1)/2] zeros, it suffices to prove M-(λ,δ) has at most n+[(n+1)/2]=3l zeros for λ>0. By Rolles theorem we need only to prove that d2lM-(λ,δ)/dλ2l has at most l zeros for λ∈(0,+∞). From (76), we can have that
(78)d2lM-(λ,δ)dλ2l=C2l(2l)!+∑j=1l‍C2l+jA2l+2j2lλ2j,
which shows that d2lM-(λ,δ)/dλ2l has at most l zeros for λ>0. Thus, M(h,δ) has at most 3l zeros for h>0. To prove 3l zeros can appear, we only need to prove that M*(h,δ) in (75) can appear 3l zeros for h>0 small. Let bij±=0,aij±=0,j≥1,ai0-=0,0≤i≤l, and a2l-≠0. Then M*(h,δ) in (75) can be expressed as by (29) and (34)
(79)M*(h,δ)=∑i=02l‍Ai0+(h)i+∑i=l+12l‍Bi0-(h)2i=A00++A10+h+A20+(h)2+⋯+A2l-1,0+(h)2l-1+A2l,0+(h)2l+Bl+1,0-(h)2l+2+Bl+2,0-(h)2l+4+⋯+B2l,0-(h)4l,
where
(80)A00+=22a00+,Ai0+=2(2)i+1(a1)iiai0+,i=1,2,…,2l,Bi0-=-22b0iai0-∫0π/2‍cos2i+1θdθ=-22b0i(2i)!!(2i+1)!!ai0-,i=l+1,…,2l.
Thus, by changing the sign of a2l,0-,a2l-1,0-,…,al+1,0-, a2l,0+,a2l-1,0+,…,a00+ in turn such that
(81)ai-1,0-ai0-<0,i=2l,2l-1,…,l+2,al+1-a2l,0+>0,ai-1,0+ai0+<0,i=2l,2l-1,…,1,0<|a00+|≪|a10+|≪⋯≪|a2l,0+|≪|al+1-|≪|al+2-|≪|a2l-1-|≪1,
we can find 3l simply positive zeros h1,h2,…,h3l with 0<h3l<h3l-1<⋯<h1≪1. For n=2l+1,l=0,1,…, we can discuss in a similar way. Thus, this bound can be reached and N(n)≥n+[(n+1)/2]. The proof is finished.Subcase 4.
a
1
>
0
,
a
0
>
0
,
b
1
=
0
,
b
0
<
0 From (12) and Lemmas 3 and 4, we get that
(82)M(h,δ)=hψ[n/2]+(h,δ)+(2h+a02a1)φ[(n-1)/2]+×(2h+a02a1,δ)(π-2arcsina02a1h+a02)+h∑i+2k=0n‍Bi,2k-hi+k=hfn(h,δ)+(2h+a02a1)φ[(n-1)/2]+×(2h+a02a1,δ)(π-2arcsina02a1h+a02),
where
(83)fn(h,δ)=ψ[n/2]+(h,δ)+∑i+2k=0n‍Bi,2k-hi+k,
which is a polynomial of degree n in h. Let λ=h. Then h=λ2,λ∈(0,+∞), and (82) becomes
(84)M(h,δ)=λfn(λ2)+v(λ,δ)(π2-arcsina02a1λ2+a02)≜M~(λ,δ),
where
(85)v(λ,δ)=(2λ2+a02a1)φ[(n-1)/2]+(2λ2+a02a1,δ).
One can see that (d/dλ)(M~(λ,δ)/v(λ,δ))=u(λ,δ)/v^2(λ,δ), where
(86)u(λ,δ)=(2λ2+a02a1)φ[(n-1)/2]+(2λ2+a02a1,δ)×ddλ(λfn(λ2,δ))-λfn(λ2,δ)×ddλ[(2λ2+a02a1)φ[(n-1)/2]+(2λ2+a02a1,δ)]+2a0a1(2λ2+a02a1)×(φ[(n-1)/2]+(2λ2+a02a1,δ))2.
Denote by #{λ∈(0,+∞)∣f(λ)=0} the number of zeros of the function in the interval (0,+∞) taking into account their multiplicities. Note that
(87)degv=2[n+12],degu=2(n+[n+12]),
and they are even functions in λ. Therefore,
(88)#{λ∈(0,+∞)∣v(λ,δ)}≤[n-12],#{λ∈(0,+∞)∣u(λ,δ)}≤n+[n+12].
Then, from [8], we can obtain that
(89)#{λ∈(0,+∞)∣M~(λ,δ)=0}≤#{λ∈(0,+∞)∣v(λ,δ)}+#{λ∈(0,+∞)∣u(λ,δ)}+1≤[n-12]+n+[n+12]+1=n+2[n+12],
which implies that Z(n)≤n+2[(n+1)/2]. Now, we verify Z(n)≥n+[(n+1)/2].
Make the transformationu=v2/2h-a02/2a1h. Then I-00+(h,δ) in (35) becomes
(90)I-00+(h,δ)=(2h)3/2a02/a1∫01‍u(1-u2)1/21+(2ha1u2/a02)du,
which follows that as h>0 small
(91)I-00+(h,δ)=(2h)3/2a02/a1∫01‍u(1-u2)1/2×[1+∑m=1+∞‍(-1)m(2m-1)!!(2m)!!(2a1a02)mhmu2m]du.
Note that
(92)∫01‍u2m+1(1-u2)1/2du=∫0π/2‍sin2m+1θcos2θdθ(Letu=sinθ)=(2m)!!(2m+3)!!,m=0,1,2,….
Inserting the above formula into (91) gives that
(93)I-00(h,δ)=2h2a1a02∑m=0+∞‍Cm(2a1a02)mhm+1,
where
(94)Cm=(-1)m(2m+1)(2m+3),m≥0.
Take bij-=aij+=0,aij-=0,j≥1,bij+=0,i≥1. Then, by (28), (35), and (93), we can obtain that for h>0 small
(95)M(h,δ)=h[∑i=0n‍-22(2i)!!(2i+1)!!b0iai0-hi+∑2k=0n-1‍b0,2k+1+ψ0k+(h)+∑2k=0n-1‍2k+1α00k+2a1a02b0,2k+1+(h+a022a1)k×∑m=0+∞‍Cm(2a1a02)mhm+1]=h∑i≥0‍vihi,
where
(96)vi=-22(2i)!!(2i+1)!!b0iai0-+Li(b01+,b03+,…,b0,2[(n-1)/2]+1+),i=0,1,2,…,nvn+1+i=∑2k=0n-1‍α00k+2k+1(2a1a02)n+i-k+1/2×∑r=0k‍CkrCn+i-rb0,2k+1+,i≥0,
with Li,i=0,1,…,n being linear combination, Li(0,0,…,0)=0. One can find that
(97)∂(v0,v1,…,vn,vn+1,vn+2,…,vn+[(n+1)/2])∂(a00-,a10-,…,an0-,b01+,b03+,…,b0,2[(n-1)/2]+1+)=(A1A2022a1a02A3)≡A,
where A2 is a (n+1)×[(n+1)/2] matrix,
(98)A1=(-2200⋯00-42b00⋯000-16215b02⋯0⋮⋮⋮⋱⋮000⋯-22(2n)!!(2n+1)!!b0n),A3=(2[(n-1)/2]α000+β0,2α001+β1,22α002+β2,…,2[(n-1)/2]α00,[(n-1)/2]+β[(n-1)/2]),
with βi are [(n+1)/2]×1 matrix satisfying
(99)βi=((2a1a02)n-i∑r=0iCirCn-r(2a1a02)n+1-i∑r=0iCirCn+1-r⋮(2a1a02)n+[(n-1)/2]-i∑r=0iCirCn+[(n-1)/2]-r),i=0,1,2,…,[n-12].
Hence, we can obtain that from (97)
(100)|A|=(22a1a02)[(n+1)/2]|A1||A3|=(2α0+2a1a02)[(n+1)/2]∏i=0[(n-1)/2]2iα00i+∏i=0n-22(2i)!!(2i+1)!!b0i|B|,
where
(101)B=(β0,β1,…,β[(n-1)/2]),
and βi are given in (99). We claim that |A|≠0. We only need to prove |B|≠0 by the above formula. Using elementary transformations to |B| by multiplying ith column by (2a1/a02)i-1,i=2,3,…,[(n+1)/2], we can obtain that by (99) and (101)
(102)|B|=(2a1a02)n[(n+1)/2]|B1|,
where
(103)|B1|=|Cn∑r=01C1rCn-r⋯∑r=0[(n-1)/2]C[(n-1)/2]rCn-rCn+1∑r=01C1rCn+1-r⋯∑r=0[(n-1)/2]C[(n-1)/2]rCn+1-r⋮⋮⋱⋮Cn+[(n-1)/2]∑r=01C1rCn+[(n-1)/2]-r⋯∑r=0[(n-1)/2]C[(n-1)/2]rCn+[(n-1)/2]-r|.
Now we will use elementary transformations to B1 as follows.(1)
Add the first column multiplying by-1 to ith column,  i=2,3,…,[(n+1)/2].(2)
Add the second column multiplying by-Ci-11 to ith column, i=3,4,…,[(n+1)/2].(3)
Add the third column multiplying by-Ci-12 to ith column, i=4,5,…,[(n+1)/2]⋮
[
(
n
-
1
)
/
2
]. Add the [(n-1)/2]th column multiplying by -C[(n-1)/2][(n-3)/2] to [(n+1)/2]th column,
[
(
n
+
1
)
/
2
]. multiply ith column by (-1)i-1, i=2,3,…,[(n+1)/2].
Then,|B1| becomes, together with (94)
(104)|B1|=(-1)[(n-1)/2][(n+1)/2]/2det(β~0,β~1,…,β~[(n-1)/2])=2[(n+1)/2](-1)n[(n+1)/2]det(β-0,β-1,…,β-[(n-1)/2])≜2[(n+1)/2](-1)n[(n+1)/2]|B2|,
where
(105)β~i=((-1)n(2(n-i)+1)(2(n-i)+3)(-1)n+1(2(n+1-i)+1)(2(n+1-i)+3)⋮(-1)n+[(n-1)/2](2(n+[(n-1)/2]-i)+1)(2(n+[(n-1)/2]-i)+3)),β-i=(12(n-i)+1-12(n-i)+312(n+1-i)+1-12(n+1-i)+3⋮12(n+[(n-1)/2]-i)+1-12(n+[(n-1)/2]-i)+3),
with i=0,1,…,[(n-1)/2]. For B2 in (104) by adding a column on the left and a row on the above, we can obtain that, together with adding ith column to (i+1)th column with i=1,2,…,[(n+1)/2](106)|B2|=|111⋯112n+312n+112n-1⋯12(n-[(n-1)/2])+112n+512n+312n+1⋯12(n-[(n-1)/2])+3⋮⋮⋮⋱⋮12(n+[(n-1)/2])+312(n+[(n-1)/2])+112(n+[(n-1)/2])-1⋯12n+1|, which implies that |A3|≠0 by (102) and (104) if |B2|≠0. We claim that |B2|≠0 and(107)|B3|=|12n+312n+112n-1⋯12(n-[(n-1)/2])+112n+512n+312n+1⋯12(n-[(n-1)/2])+3⋮⋮⋱⋮⋮12(n+[(n-1)/2])+312(n+[(n-1)/2])+112(n+[(n-1)/2])-1⋯12n+112(n+[(n+1)/2])+312(n+[(n+1)/2])+112(n+[(n+1)/2])-1⋯12n+3|≠0. Now, we prove them by induction on n. For n=1,2 we have
(108)|111513|=215,|111715|=235,|15131715|=-4525,|17151917|=-42205,
which means that (106) and (107) hold for n=1,2. Suppose (106) and (107) hold for n=2l-1,2l,l≥1. That is, we have for n=2l(109)|B2|=|111⋯114l+314l+114l-1⋯12l+314l+514l+314l+1⋯12l+5⋮⋮⋮⋱⋮16l+116l-116l-3⋯14l+1|≜|C|≠0,|B3|=|14l+314l+114l-1⋯12l+314l+514l+314l+1⋯12l+5⋮⋮⋱⋮⋮16l+116l-116l-3⋯14l+116l+316l+116l-1⋯14l+3|≠0.
Then for n=2l+1, we have
(110)|B2|=|111⋯114l+514l+314l+1⋯12l+314l+714l+514l+3⋯12l+5⋮⋮⋮⋱⋮16l+516l+316l+1⋯14l+3|,|B3|=|14l+514l+314l+1⋯12l+314l+714l+514l+3⋯12l+5⋮⋮⋱⋮⋮16l+516l+316l+1⋯14l+316l+716l+516l+3⋯14l+5|.
Note that by the first equation of (109) there only exist α1,α2,…,αl+1 such that
(111)|B2|=[16l+5-(α16l+3+α26l+1+⋯+αl+14l+3)]|C|=-26l+5[α16l+3+2α26l+1+⋯+(l+1)αl+14l+3]|C|
since α1+α2+⋯+αl+1=1, where C is given in (109). If |B2|=0, then we can obtain that from (109) and (111)
(112)α16l+3+2α26l+1+⋯+(l+1)αl+14l+3=0.
Note that
(113)α14l+3+α24l+1+⋯+αl+12l+3=14l+5,α14l+5+α24l+3+⋯+αl+12l+5=14l+7,⋮α16l+1+α26l-1+⋯+αl+14l+1=16l+3,
which follows that, together with (112) (114)α14l+3+2α24l+1+⋯+(l+1)αl+12l+3=0,α14l+5+2α24l+3+⋯+(l+1)αl+12l+5=0,⋮α16l+1+2α26l-1+⋯+(l+1)αl+14l+1=0,α16l+3+2α26l+1+⋯+(l+1)αl+14l+3=0. By the second equation in (109) and the above formula, we have
(115)α1=α2=⋯=αl+1=0.
This is a contradiction with α1+α2+⋯+αl+1=1. Hence α1/(6l+3)+2α2/(6l+1)+⋯+(l+1)αl+1/(4l+3)≠0, which means that (106) holds for n=2l+1. Since |B2|≠0 in (110), there only exist β1,β2,…,βl+1 such that
(116)β1+β24l+5+β34l+7+⋯+βl+16l+5=16l+7,β1+β24l+3+β34l+5+⋯+βl+16l+3=16l+5,⋮β1+β22l+3+β32l+5+⋯+βl+14l+3=14l+5,
with β1≠0 since the last row in the second formula of (110) is linearly independent with all rows in the first formula of (110), which means that (107) holds for n=2l+1. In a similar way, we can prove (106) and (107) hold for n=2l+2. Hence, the claim holds. So, from (99) we can know that a00-,a10-,…,an0-,b01+,b03+,…,b0,2[(n-1)/2]+1+ can be taken as free parameters. So we can choose these values such that
(117)vivi+1<0,i=0,1,…,n+[(n-1)2],0<|v0|≪|v1|≪⋯≪|vn+[(n+1)/2]|≪1,
which yields that by (97) and (96) M(h,δ) can appear n+[(n+1)/2] positive zeros for h>0 small. We also can know that N(n)≥n+[(n+1)/2]. Hence, the conclusion is proved.Subcase 5.
a
1
>
0
,
a
0
>
0
,
b
1
>
0
,
b
0
=
0 By (36) and (67), one can see that
(118)M(h,δ)=hψ[n/2]+(h,δ)+(2h+a02a1)φ[(n-1)/2]+×(2h+a02a1,δ)(π2-arcsina02a1h+a02)+h∑i+2k=0n‍Ai,2k-(h)i+2k=hg[n/2](h,δ)+g-[(n+1)/2](h,δ)+(2h+a02a1)φ[(n-1)/2]+(2h+a02a1,δ)×(π2-arcsina02a1h+a02),
where
(119)g[n/2](h,δ)=ψ[n/2]+(h,δ)+∑l=0[n/2]‍∑i+2k=2l‍Ai,2k-hl,g-[(n+1)/2](h,δ)=∑l=0[(n-1)/2]‍∑i+2k=2l+1‍Ai,2k-hl+1,n≥1.
For convenience, we denote by gn any polynomial of degree n although its coefficients may be different when it appears in different place. Then, we claim that for any 2≤k≤[(n-1)/2](120)dkM(h,δ)dhk=∑j=0k‍h1/2-k+jg[n/2]-j+dkg-[(n+1)/2]dhk+∑j=0k-2‍gjh-j-1/2(2h+a02a1)-j-1+dkdhk[(2h+a02a1)φ[(n-1)/2]+(2h+a02a1,δ)]×(π2-arcsina02a1h+a02).
Now, we verify this claim by induction on k. For k=2, by (118) we can obtain that
(121)dM(h,δ)dh=12h-1/2g[n/2](h,δ)+h1/2dg[n/2](h,δ)dh+dg-[(n+1)/2]dh+ddh[(2h+a02a1)φ[(n-1)/2]+(2h+a02a1,δ)]×(π2-arcsina02a1h+a02)+(2h+a02a1)φ[(n-1)/2]+×(2h+a02a1,δ)a02a1h(2h+a02/a1)=h-1/2g[n/2]+h1/2g[n/2]-1+dg-[(n+1)/2]dh+ddh[(2h+a02a1)φ[(n-1)/2]+(2h+a02a1,δ)]×(π2-arcsina02a1h+a02),
which follows that
(122)d2M(h,δ)dh2=h-3/2g[n/2]+h-1/2g[n/2]-1+h1/2g[n/2]-2+d2g-[(n+1)/2]dh2+g0h(2h+a02/a1)+d2dh2[(2h+a02a1)φ[(n-1)/2]+(2h+a02a1,δ)]×(π2-arcsina02a1h+a02).
Hence, (120) holds for k=2. Suppose (120) holds for  k,2≤k≤[(n-1)/2]-1. Then for k+1, we have
(123)dk+1M(h,δ)dhk+1=∑j=0k‍(12-k+j)h1/2-(k+1)+jg[n/2]-j+∑j=0k‍h1/2-k+jg[n/2]-1-j+dk+1g-[(n+1)/2]dhk+1+dk+1dhk+1[(2h+a02a1)φ[(n-1)/2]+(2h+a02a1,δ)]×(π2-arcsina02a1h+a02)+dkdhk[(2h+a02a1)φ[(n-1)/2]+(2h+a02a1,δ)]×a02a1h(2h+a02/a1)+∑j=0k-2‍(h(2h+a02a1)ddhgj-gj[2(j+1)h+(j+12)(2h+a02a1)])×(hj+3/2(2h+a02a1)j+2)-1,
which implies that (120) holds for k+1. Thus, the claim is proved. Then, taking k=[(n-1)/2], we can obtain that by differentiating it
(124)d[(n+1)/2]M(h,δ)dh[(n+1)/2]=∑j=0[n/2]‍h1/2-[(n+1)/2]+jg[n/2]-j+∑j=0[(n+1)/2]-2‍gjh-j-1/2(2h+a02a1)-j-1+d[(n+1)/2]g-[(n+1)/2]dh[(n+1)/2]+d[(n+1)/2]dh[(n+1)/2]×[(2h+a02a1)φ[(n-1)/2]+(2h+a02a1,δ)]×(π2-arcsina02a1h+a02).
One can find that
(125)d[(n+1)/2]+1M(h,δ)dh[(n+1)/2]+1=∑j=0[n/2]‍h1/2-[(n+1)/2]-1+jg[n/2]-j+∑j=0[(n+1)/2]-1‍gjh-j-1/2(2h+a02a1)-j-1=h-[(n+1)/2]-1/2×(2h+a02a1)-[(n+1)/2]F(h,δ),
where
(126)F(h,δ)=(2h+a02a1)[(n+1)/2]∑j=0[n/2]‍hjg[n/2]-j+∑j=0[(n+1)/2]-1‍gjh[(n+1)/2]-j×(2h+a02a1)[(n+1)/2]-j-1,
where F is a polynomial of degree [n/2]+[(n+1)/2]. Since M(0,δ)=0 from (118), it is easy to see that M(h,δ) has at most [n/2]+2[(n+1)/2]=n+[(n+1)/2] zeros for h>0 by Rolle theorem. As the above discussion, we only prove Z(n)≥n+[(n+1)/2] as h>0 small, which implies N(n)≥n+[(n+1)/2]. For the purpose, take bij-=aij+=0,aij-=0,j≥1,bij+=0,i≥1. Then using (49), (51), we can write M(h,δ) in (12) as
(127)M(h,δ)=h[-22a00-+∑i=1n‍2(2)i+1i(b1)iai0-×∫-π/20‍sini-1θcos2θdθ(h)i+∑2k=0n-1‍b0,2k+1+ψ0k+(h)+∑2k=0n-1‍2k+1α00k+2a1a02b0,2k+1+×(h+a022a1)k∑m=0+∞‍Cm(2a1a02)mhm+1].
Forn=2l,l≥1, (127) can be written as
(128)M(h,δ)=h[∑i=02l‍v-i(h)i+∑i≥0‍v-2l+1+ihl+1+i],
where
(129)v-0=-22a00-.v-2i+1=2(2)2i+2(2i+1)(b1)2i+1a2i+1,0-×∫-π/20‍sin2iθcos2θdθ,i=0,1,…,l-1,v-2i=2(2)2i+12i(b1)2ia2i,0-∫-π/20‍sin2i-1θcos2θdθ+L2i(b01+,b03+,…,b0,2[(n-1)/2]+1+),i=1,2,…,l,v-2l+1+i=∑k=0l-1‍2k+1α00k+(2a1a02)l-k+i+1/2×b0,2k+1+∑r=0k‍CkrCl+i-r,i=0,1,…,[n-12],
which implies that M(h,δ) can appear n+[(n+1)/2] zeros in h>0 small by using the same method with the Subcase 4. For n=2l+1,l=0,1,…, we can discuss by (127) in a similar way. Hence, the conclusion holds.Subcase 6.
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0. We have as the above(130)M(h,δ)=h(ψ[n/2]+(h,δ)+ψ[n/2]-(h,δ))+(2h+a02a1)φ[(n-1)/2]+(h,δ)×(π2-arcsina02a1h+a02)+(2h+b02b1)φ[(n-1)/2]-(h,δ)×(π2+arcsinb02b1h+b02).
Similar to the Subcase 5, we can prove that for any 2≤k≤[(n-1)/2](131)dkM(h,δ)dhk=∑j=0k‍h1/2-k+jg[n/2]-j+∑j=0k-2‍gjh-j-1/2(2h+a02a1)-j-1+∑j=0k-2‍gjh-j-1/2(2h+b02b1)-j-1+dkdhk[(2h+a02a1)φ[(n-1)/2]+(2h+a02a1,δ)]×(π2-arcsina02a1h+a02)+dkdhk[(2h+b02b1)φ[(n-1)2]+(2h+b02a1,δ)]×(π2-arcsinb02b1h+b02).
Taking k=[(n-1)/2] and differentiating the above twice follow that
(132)d[(n+1)/2]+1M(h,δ)dh[(n+1)/2]+1=∑j=0[n/2]‍h1/2-[(n+1)/2]-1+jg[n/2]-j+∑j=0[(n+1)/2]-1‍gjh-j-1/2(2h+a02a1)-j-1+∑j=0[(n+1)/2]-1‍gjh-j-1/2(2h+b02b1)-j-1.
If a02/a1=b02/b1, then it is easy to see that (132) has the same form with (125). Hence, we can know that M(h,δ) has at most [n/2]+2[(n+1)/2] zeros for h∈(0,+∞). If a02/a1≠b02/b1, then (132) can be written as
(133)d[(n+1)/2]+1M(h,δ)dh[(n+1)/2]+1=h-[(n+1)/2]-1/2(2h+a02a1)-[(n+1)/2]×(2h+b02b1)-[(n+1)/2]F-(h,δ),
where
(134)F-(h,δ)=(2h+a02a1)[(n+1)/2](2h+b02b1)[(n+1)/2]×∑j=0[n/2]‍hjg[n/2]-j+(2h+b02b1)[(n+1)/2]∑j=0[(n+1)/2]-1‍gjh[(n+1)/2]-j×(2h+a02a1)[(n+1)/2]-j-1+(2h+a02a1)[(n+1)/2]∑j=0[(n+1)/2]-1‍gjh[(n+1)/2]-j×(2h+b02b1)[(n+1)/2]-j-1,
where F- is a polynomial of degree [n/2]+2[(n+1)/2] in h. By Rolle theorem, we can obtain that M(h,δ) has at most [n/2]+3[(n+1)/2] zeros for h>0 since M(0,δ)=0. Now, we only need to prove Z(n)≥n. Take aij-=bij-=0,aij+=bij+=0,i≥1. Then, by Lemmas 4, 5, one can see that for h>0 small
(135)M(h,δ)=h[(2a1a02)m∑2k=0n‍2k+1+1/2a0,2k+2k+1hk+∑2k=0n-1‍b0,2k+1+ψ0k+(h)+∑2k=0n-1‍2k+1α00k+2a1a02b0,2k+1+(h+a022a1)k×∑m=0+∞‍Cm(2a1a02)mhm+1].
Similarly, we can discuss the above formula such that such that M(h,δ) can appear n zeros for h>0 small. Hence, the conclusion is proved.



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*Source: 575390-2013-02-28.xml*