Bifurcation of Limit Cycles by Perturbing a Piecewise Linear Hamiltonian System

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 of 𝜀.

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: ̇,̇(𝑥),𝑥=𝑦+𝜀𝑝𝑥,𝑦,𝛿𝑦=−𝑔+𝜀𝑞𝑥,𝑦,𝛿 (1) where 𝜀>0 is a sufficiently small real parameter, 𝑔(𝑥)=𝑎₁𝑥+𝑎₀𝑏,𝑥≥0,₁𝑥+𝑏₀,𝑥<0 (2) with 𝑎₁, 𝑎₀, 𝑏₁, and 𝑏₀ real numbers satisfying 𝑎²₁+𝑎²₀≠0,𝑏²₁+𝑏²₀≠0, 𝑝=𝑝𝑥,𝑦,𝛿⁺𝑥,𝑦,𝛿=∑^𝑛_𝑖+𝑗=0𝑎⁺_𝑖𝑗𝑥^𝑖𝑦^𝑗𝑝,𝑥≥0,₋𝑥,𝑦,𝛿=∑^𝑛_𝑖+𝑗=0𝑎⁻_𝑖𝑗𝑥^𝑖𝑦^𝑗𝑞=𝑞,𝑥<0,𝑥,𝑦,𝛿⁺𝑥,𝑦,𝛿=∑^𝑛_𝑖+𝑗=0𝑏⁺_𝑖𝑗𝑥^𝑖𝑦^𝑗𝑞,𝑥≥0,₋𝑥,𝑦,𝛿=∑^𝑛_𝑖+𝑗=0𝑏⁻_𝑖𝑗𝑥^𝑖𝑦^𝑗,𝑥<0, (3)(4) and 𝛿=(𝑎⁺_𝑖𝑗,𝑎⁻_𝑖𝑗,𝑏⁺_𝑖𝑗,𝑏⁻_𝑖𝑗)∈𝐷⊂ℝ^{2(𝑛+1)(𝑛+2)} with 𝐷 compact. Then system (1) has two subsystems𝑥̇=𝑦+𝜀𝑝⁺,̇𝑥,𝑦,𝛿𝑦=−𝑎₁𝑥−𝑎₀+𝜀𝑞⁺,̇𝑥,𝑦,𝛿𝑥=𝑦+𝜀𝑝⁻,̇𝑥,𝑦,𝛿𝑦=−𝑏₁𝑥−𝑏₀+𝜀𝑞⁻𝑥,𝑦,𝛿, (5a)(5b) 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,𝐻⁺=1𝑥,𝑦2𝑦²+12𝑎₁𝑥²+𝑎₀𝐻𝑥,₋=1𝑥,𝑦2𝑦²+12𝑏₁𝑥²+𝑏₀𝑥. (6)

Note that the phase portrait of the linear system ̇̇𝑥=𝑦,𝑦=−𝑎𝑥−𝑏, (7) with 𝑎²+𝑏²≠0 has possibly the following four different phase portraits on the plane (see Figure 1).

(a) 𝑎=0,𝑏>0

(b) 𝑎=0,𝑏<0

(c) 𝑎>0

(d) 𝑎<0

Figure 1 

The possible phase portraits of system (7).

Then, 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 Figure 2, 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 Table 1 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 𝑥,𝑦→−𝑥,𝑦, (8) together with time rescaling 𝑑𝑡=−𝑑𝜏.

Table 1 

Coefficient conditions for phase portraits (1)–(13).

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 𝑎₁≥0,𝑎₀≥0,𝑎₀+𝑎₁𝑏>0,₁≥0,𝑏₀≤0,𝑏₀<𝑏₁. (9) Clearly, system (1) ∣_𝜀=0 satisfying (9) has a family of periodic orbits 𝐿_ℎ=𝐿⁺_ℎ∪𝐿⁻_ℎ=𝑥,𝑦∣𝐻⁺∪𝑥,𝑦=ℎ𝑥,𝑦∣𝐻⁻𝑥,𝑦=ℎ,ℎ>0, (10) such that the limit of 𝐿_ℎ as ℎ→0⁺ is the origin. The intersection points of the closed curve 𝐿_ℎ with the positive 𝑦-axis and the negative 𝑦-axis are denoted by 𝐴(ℎ)=(0,√2ℎ) and 𝐴₁(ℎ)=(0,−√2ℎ), respectively. Let 𝑀⁺(ℎ,𝛿)=𝐴𝐴₁𝑞⁺𝑑𝑥−𝑝⁺𝑀𝑑𝑦,₋()𝐴ℎ,𝛿=₁𝐴𝑞⁻𝑑𝑥−𝑝⁻𝑑𝑦,ℎ>0. (11) Then, from Liu and Han [7], the first-order Melnikove function corresponding to system (1) is 𝑀(ℎ,𝛿)=𝑀⁺(ℎ,𝛿)+𝑀⁻(ℎ,𝛿),ℎ∈(0,+∞). (12) 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 

This paper is organized as follows. In Section 2, 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 of 𝑀⁺(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 𝑀⁺(ℎ,𝛿)=𝑀⁺₁(ℎ,𝛿)+^𝑛_2𝑘=02^𝑘+1+1/2𝑎2𝑘+1^+0,2𝑘ℎ^𝑘+1/2, (13)  where 𝑀⁺₁(ℎ,𝛿)=^𝑛−1_𝑖+𝑗=0𝑝⁺_𝑖𝑗𝐴₁𝐴𝑥^𝑖+1𝑦^𝑗𝑑𝑦=^𝑛−1_𝑖+𝑗=0𝑞⁺_𝑖𝑗𝐴₁𝐴𝑥^𝑖𝑦^𝑗+1𝑑𝑥, (14) with 𝑝⁺_𝑖𝑗=𝑎⁺_𝑖+1,𝑗+𝑗+1𝑏𝑖+1^+𝑖,𝑗+1,𝑞⁺_𝑖𝑗=𝑏⁺_𝑖,𝑗+1+𝑖+1𝑎𝑗+1^+𝑖+1,𝑗. (15) (ii)  (ℎ,𝛿) in (11) can be expressed as 𝑀⁻(ℎ,𝛿)=𝑀⁻₁(ℎ,𝛿)−^𝑛_2𝑘=02^𝑘+1+1/2𝑎2𝑘+1^−0,2𝑘ℎ^𝑘+1/2, (16) where 𝑀⁻₁(ℎ,𝛿)=^𝑛−1_𝑖+𝑗=0𝑝⁻_𝑖𝑗𝐴₁𝐴𝑥^𝑖+1𝑦^𝑗𝑑𝑦=^𝑛−1_𝑖+𝑗=0𝑞⁻_𝑖𝑗𝐴₁𝐴𝑥^𝑖𝑦^𝑗+1𝑑𝑥, (17) with 𝑝⁻_𝑖𝑗=𝑎⁻_𝑖+1,𝑗+𝑗+1𝑏𝑖+1^{−𝑖,𝑗+1},𝑞⁻_𝑖𝑗=𝑏⁻_𝑖,𝑗+1+𝑖+1𝑎𝑗+1^{−𝑖+1,𝑗}. (18)

Proof. We only prove (i) since (ii) can be verified in a similar way. By (11), we obtain 𝑀⁺(ℎ,𝛿)=𝐴𝐴₁𝑞⁺𝑥,𝑦,𝛿𝑑𝑥−𝑝⁺𝑥,𝑦,𝛿𝑑𝑦+_→𝐴1𝐴𝑞⁺𝑥,𝑦,𝛿𝑑𝑥−𝑝⁺−𝑥,𝑦,𝛿𝑑𝑦_→𝐴1𝐴𝑞⁺𝑥,𝑦,𝛿𝑑𝑥−𝑝⁺=𝑥,𝑦,𝛿𝑑𝑦𝐴𝐴_1∪^→_𝐴1𝐴𝑞⁺𝑥,𝑦,𝛿𝑑𝑥−𝑝⁺+𝑥,𝑦,𝛿𝑑𝑦_→𝐴1𝐴𝑝⁺0,𝑦,𝛿𝑑𝑦, (19) which follows that by Green formula and (3) 𝑀⁺(ℎ,𝛿)=𝑀⁺₁(ℎ,𝛿)+^𝑛_2𝑘=02^𝑘+1+1/2𝑎2𝑘+1^+0,2𝑘ℎ^𝑘+1/2, (20) where 𝑀⁺₁(ℎ,𝛿)=_int𝐴𝐴₁^∪→𝐴₁^𝐴𝑝⁺_𝑥+𝑞⁺_𝑦𝑑𝑥𝑑𝑦. (21) Then, by Green formula again 𝑀⁺₁(ℎ,𝛿)=−𝐴𝐴_1∪^→_𝐴1𝐴𝑝⁺𝑥,𝑦,𝛿𝑑𝑦=𝐴𝐴_1∪^→_𝐴1𝐴𝑞⁺𝑥,𝑦,𝛿𝑑𝑥, (22) where 𝑝⁺𝑥,𝑦,𝛿=𝑝⁺𝑥,𝑦,𝛿−𝑝⁺0,𝑦,𝛿+^𝑥₀𝑞⁺_𝑦𝑢,𝑦,𝛿𝑑𝑢,𝑞⁺𝑥,𝑦,𝛿=𝑞⁺𝑥,𝑦,𝛿−𝑞⁺(𝑥,0,𝛿)+^𝑦₀𝑝⁺_𝑥(𝑥,𝑣,𝛿)𝑑𝑣. (23) By (3), (4), and the above formulas, we have 𝑝⁺𝑥,𝑦,𝛿=^𝑛_𝑖+𝑗=0𝑎⁺_𝑖𝑗𝑥^𝑖𝑦^𝑗−^𝑛_𝑗=0𝑎⁺_0𝑗𝑦^𝑗+^𝑛_𝑖+𝑗=1𝑗𝑏𝑖+1^+𝑖𝑗𝑥^𝑖+1𝑦^𝑗−1=𝑥^𝑛−1_𝑖+𝑗=0𝑎⁺_𝑖+1,𝑗+𝑗+1𝑏𝑖+1^+𝑖,𝑗+1𝑥^𝑖𝑦^𝑗=𝑥^𝑛−1_𝑖+𝑗=0𝑝⁺_𝑖𝑗𝑥^𝑖𝑦^𝑗,𝑞⁺𝑥,𝑦,𝛿=^𝑛_𝑖+𝑗=0𝑏⁺_𝑖𝑗𝑥^𝑖𝑦^𝑗−^𝑛_𝑗=0𝑏⁺_𝑖0𝑥^𝑖+^𝑛_𝑖+𝑗=1𝑖𝑎𝑗+1^+𝑖𝑗𝑥^𝑖−1𝑦^𝑗+1=𝑦^𝑛−1_𝑖+𝑗=0𝑏⁺_𝑖,𝑗+1+𝑖+1𝑎𝑗+1^+𝑖+1,𝑗𝑥^𝑖𝑦^𝑗=𝑦^𝑛−1_𝑖+𝑗=0𝑞⁺_𝑖𝑗𝑥^𝑖𝑦^𝑗. (24)(25) Combining (20)–(25) gives (13) and (14). Thus, the proof is ended.

Then, using Lemma 2 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 𝑀⁺(ℎ,𝛿)=ℎ^𝑛1/2_𝑖+2𝑘=0𝐵⁺_𝑖,2𝑘ℎ^𝑖+𝑘, (26) where 𝐵⁺_0,2𝑘=2^𝑘+1+1/2𝑎⁺_0,2𝑘,𝐵2𝑘+1_+𝑖,2𝑘=2^𝑘+1+1/2𝑎^𝑖₀𝑎⁺_𝑖,2𝑘+2𝑘+1𝑖𝑏⁺_{𝑖−1,2𝑘+1}×₀^𝜋/2sin^2𝑘𝜃cos^2𝑖+1𝜃d𝜃,1≤𝑖≤𝑛. (27)(ii)  If b₁=0,b₀<0, then we have 𝑀⁻(ℎ,𝛿)=ℎ^𝑛1/2_𝑖+2𝑘=0B⁻_𝑖,2𝑘ℎ^𝑖+𝑘, (28) where 𝐵⁻_0,2𝑘2=−^𝑘+1+1/2𝑎⁻_0,2𝑘,𝐵2𝑘+1_{−𝑖,2𝑘}=−2^𝑘+1+1/2𝑏^𝑖₀𝑎⁻_𝑖,2𝑘+2𝑘+1𝑖𝑏⁻_{𝑖−1,2𝑘+1}×₀^𝜋/2sin^2𝑘𝜃cos^2𝑖+1𝜃d𝜃,1≤𝑖≤𝑛. (29)

Proof. Note that along AA₁,𝑥=ℎ/𝑎₀−(1/2𝑎₀)𝑦². Then, inserting it into (14) follows that 𝑀⁺₁(ℎ,𝛿)=^𝑛−1_𝑖+𝑗=0𝑝⁺_𝑖𝑗^√_−√^2ℎ_2ℎℎ𝑎₀−12𝑎₀𝑦^2𝑖+1𝑦^𝑗=𝑑𝑦_{𝑛−1𝑖+2𝑘=0}𝑝⁺_𝑖,2𝑘2^𝑖𝑎₀^𝑖+1√₀^2ℎ2ℎ−𝑦^2𝑖+1𝑦^2𝑘𝑑𝑦. (30) Let 𝑦=√2ℎsin𝜃. Then we have 𝑑𝑦=√2ℎcos𝜃𝑑𝜃 and the above integral can be carried into ^√₀^2ℎ2ℎ−𝑦^2𝑖+1𝑦^2𝑘𝑑𝑦=(2ℎ)^{𝑖+1+𝑘+(1/2)}×₀^𝜋/2sin^2𝑘𝜃cos^2(𝑖+1)+1𝜃d𝜃. (31) Thus, using (30) and the above equation we can write (13) as 𝑀⁺(ℎ,𝛿)=^𝑛−1_𝑖+2𝑘=02^𝑘+1+1/2𝑝⁺_𝑖,2𝑘𝑎₀^𝑖+1×₀^𝜋/2sin^2𝑘𝜃cos^2(𝑖+1)+1𝜃𝑑𝜃×ℎ^{𝑖+1+𝑘+1/2}+^𝑛_2𝑘=02^𝑘+1+1/2𝑎⁺_0,2𝑘ℎ2𝑘+1^𝑘+1/2=ℎ^𝑛1/2_𝑖+2𝑘=0𝐵⁺_𝑖,2𝑘ℎ^𝑖+𝑘, (32) where 𝐵⁺_0,2𝑘=2^𝑘+1+1/2𝑎⁺_0,2𝑘,𝐵2𝑘+1_+𝑖,2𝑘=2^𝑘+1+1/2𝑝⁺_{𝑖−1,2𝑘}𝑎^𝑖₀₀^𝜋/2sin^2𝑘𝜃cos^2𝑖+1𝜃𝑑𝜃,1≤𝑖≤𝑛, (33) 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 𝑀⁺(ℎ,𝛿)=√ℎ^𝑛_𝑖+2𝑘=0𝐴⁺_𝑖,2𝑘√ℎ^𝑖+2𝑘, (34) where 𝐴⁺_0,2𝑘=2√2^2𝑘+1𝑎⁺_0,2𝑘,𝐴2𝑘+1_+𝑖,2𝑘=2√2^2𝑘+1+𝑖√𝑎₁^𝑖𝑏⁺_{𝑖−1,2𝑘+1}+𝑖𝑎2𝑘+1^+𝑖,2𝑘×₀^𝜋/2sin^𝑖−1𝜃cos^2𝑘+2𝜃𝑑𝜃,𝑖≥1. (35) (ii) If a₁>0,a₀≠0,M⁺(ℎ,𝛿) can be written as 𝑀⁺(ℎ,𝛿)=√ℎ^𝑛−1_𝑖+2𝑘=0𝑏⁺_𝑖,2𝑘+1+𝑖+1𝑎2𝑘+1^{+𝑖+1,2𝑘}𝜙⁺_𝑖𝑘(ℎ)+^𝑛_2𝑘=02^𝑘+1+1/2𝑎⁺_0,2𝑘ℎ2𝑘+1^𝑘+^𝑛−1_𝑖+2𝑘=0𝑏⁺_𝑖,2𝑘+1+𝑖+1𝑎2𝑘+1^{+𝑖+1,2𝑘}×^𝑖_{𝑟=0,𝑟𝑒𝑣𝑒𝑛}𝛼⁺_𝑖𝑟𝑘𝑎2ℎ+²⁰𝑎₁^𝑘+𝑟/2𝐼⁺₀₀(ℎ,𝛿), (36) or 𝑀⁺(ℎ,𝛿)=√ℎ𝜓⁺_[]𝑛/2()+𝑎ℎ,𝛿2ℎ+²⁰𝑎₁𝜑⁺_{)][(𝑛−1/2}𝑎2ℎ+²⁰𝑎₁×𝜋,𝛿2𝑎−arcsin⁰2𝑎₁ℎ+𝑎²₀, (37) where 𝐼⁺₀₀(ℎ,𝛿)=^{2ℎ+𝑎20/𝑎1}_{𝑎0/√𝑎1}𝑎2ℎ+²⁰𝑎₁−𝑣²𝑑𝑣, (38) each 𝛼⁺_𝑖𝑟𝑘 is a nonzero constant and 𝜙⁺_𝑖𝑘,𝜓⁺_[𝑛/2],𝜑⁺_{[(𝑛−1)/2]} are polynomials of degree 𝑘+[(𝑖+1)/2],[𝑛/2],[(𝑛−1)/2], respectively.

Proof. Since 𝑦=±2ℎ−𝑎₁𝑥²−2𝑎₀𝑥 along the curve 𝐴𝐴₁,𝑀⁺₁(ℎ,𝛿) in (14) becomes 𝑀⁺₁()=ℎ,𝛿_{𝑛−1𝑖+𝑗=0}𝑞⁺_𝑖𝑗𝐴𝐴₁𝑥^𝑖𝑦^𝑗+1=𝑑𝑥_{𝑛−1𝑖+𝑗=0}𝑞⁺_𝑖𝑗^{(−𝑎0+2𝑎1ℎ+𝑎20)/𝑎1}₀𝑥^𝑖2ℎ−𝑎₁𝑥²−2𝑎₀𝑥^(𝑗+1)/2+𝑑𝑥_{0(−𝑎0+2𝑎1ℎ+𝑎20)/𝑎1}𝑥^𝑖(−1)^𝑗+12ℎ−𝑎₁𝑥²−2𝑎₀𝑥^(𝑗+1)/2=𝑑𝑥_{𝑛−1𝑖+2𝑘=0}2𝑞⁺_𝑖,2𝑘𝐼⁺_𝑖,2𝑘(ℎ,𝛿), (39) where 𝐼⁺_𝑖,2𝑘()=ℎ,𝛿_{(−𝑎0+2𝑎1ℎ+𝑎20)/𝑎10}𝑥^𝑖𝑎2ℎ+²⁰𝑎₁−𝑎₁𝑎𝑥+⁰𝑎₁^2𝑘+1/2𝑑𝑥. (40) Let 𝑣=√𝑎₁(𝑥+𝑎₀/𝑎₁). Then, we have 𝑑𝑣=√𝑎₁𝑑𝑥 and the above equation becomes 𝐼⁺_𝑖,2𝑘()=1ℎ,𝛿√𝑎₁^{𝑖+12ℎ+𝑎20/𝑎1𝑎}₀^/√𝑎₁𝑎𝑣−⁰√𝑎₁^𝑖×𝑎2ℎ+²⁰𝑎₁−𝑣^2𝑘+(1/2)𝑑𝑣. (41)
For 𝑎₀=0, make the transformation 𝑣=√2ℎsin𝜃. Then, we have by (41) 𝐼⁺_𝑖,2𝑘()=√ℎ,𝛿2ℎ^{2𝑘+1+𝑖+1}√𝑎₁^{𝑖+10𝜋/2}sin^𝑖𝜃cos^2𝑘+2𝜃𝑑𝜃. (42) Substituting the above formula into (37), together with (13), gives that 𝑀⁺(ℎ,𝛿)=^𝑛−1_𝑖+2𝑘=02𝑞⁺_𝑖,2𝑘√𝑎₁^𝑖+1×₀^𝜋/2sin^𝑖𝜃cos^2𝑘+2𝜃𝑑𝜃×√2ℎ^{2𝑘+1+𝑖+1}+^𝑛_2𝑘=02𝑎⁺_0,2𝑘√2𝑘+12ℎ^2𝑘+1=^𝑛_𝑖+2𝑘=0𝐴⁺_𝑖,2𝑘√ℎ^2𝑘+1+𝑖, (43) where 𝐴⁺_0,2𝑘=2√2^2𝑘+1𝑎⁺_0,2𝑘,𝐴2𝑘+1_+𝑖,2𝑘=2√2^2𝑘+1+𝑖𝑞⁺_{𝑖−1,2𝑘}√𝑎₁^𝑖×₀^𝜋/2sin^𝑖−1𝜃cos^2𝑘+2𝜃𝑑𝜃,1≤𝑖≤𝑛. (44) Thus, by (15) and the above discussion we know that (i) holds.
For 𝑎₀≠0, (41) can be represented as 𝐼⁺_𝑖,2𝑘()=1ℎ,𝛿√𝑎₁^𝑖𝑖+1_𝑟=0𝐶^𝑟_𝑖−𝑎₀√𝑎₁^𝑖−𝑟𝐼⁺_𝑟𝑘(ℎ,𝛿), (45) where 𝐼⁺_𝑟𝑘(ℎ,𝛿)=^{2ℎ+𝑎20/𝑎1}_{𝑎0/√𝑎1}𝑣^𝑟𝑎2ℎ+²⁰𝑎₁−𝑣^2𝑘+1/2𝑑𝑣. (46) Recall that 𝑣^𝑟𝑎2ℎ+²⁰𝑎₁−𝑣^2𝑘+1/2=𝑣𝑑𝑣_𝑟+12ℎ+𝑎²₀/𝑎₁−𝑣^2𝑘+1/2+()2𝑘+2+𝑟2𝑘+12ℎ+𝑎²₀/𝑎₁×2𝑘+2+𝑟𝑣^𝑟𝑎2ℎ+²⁰𝑎₁−𝑣^2𝑘−1/2𝑑𝑣. (47) Then, by (46) and the above equation we obtain that 𝐼⁺_𝑟𝑘()𝑎ℎ,𝛿=−^0𝑟+1(2ℎ)^𝑘+1/2𝑎₁^(𝑟+1)/2()+()2𝑘+2+𝑟2𝑘+12ℎ+𝑎²₀/𝑎₁2𝑘+2+𝑟𝐼⁺_{𝑟,𝑘−1}(ℎ,𝛿),𝑘≥1,𝑟≥0. (48) It follows that 𝐼⁺_𝑟𝑘()𝑎ℎ,𝛿=−^0𝑟+1𝑎₁^(𝑟+1)/2√2ℎ𝜑⁺_𝑟𝑘(ℎ)+𝛼⁺_𝑟𝑘𝑎2ℎ+²⁰𝑎₁^𝑘𝐼⁺_𝑟0(ℎ,𝛿),𝑘≥1,𝑟≥0, (49) where 𝛼⁺_𝑟𝑘=(2𝑘+1)(2𝑘−1)(2𝑘−3)×⋯×3()(),𝜑2𝑘+2+𝑟)(2𝑘+𝑟)(2𝑘−2+𝑟×⋯×4+𝑟_+𝑟𝑘(ℎ)=(2ℎ)^𝑘+2𝑘+2+𝑟2𝑘+1()×𝑎2𝑘+2+𝑟)(2𝑘+𝑟2ℎ+²⁰𝑎₁(2ℎ)^𝑘−1+(2𝑘+1)(2𝑘−1)()×𝑎2𝑘+2+𝑟)(2𝑘+𝑟)(2𝑘−2+𝑟2ℎ+²⁰𝑎₁²(2ℎ)^𝑘−2+()+⋯2𝑘+1)(2𝑘−1×⋯×5()()×𝑎2𝑘+2+𝑟)(2𝑘+𝑟)(2𝑘−2+𝑟×⋯×4+𝑟2ℎ+²⁰𝑎₁^𝑘−12ℎ, (50) which is a polynomial of degree 𝑘 in ℎ. For convenience, introduce 𝜑⁺_𝑟𝑘(ℎ)=𝜑⁺_𝑟𝑘(ℎ),𝑘≥1,0,𝑘=0,𝛼⁺_𝑟𝑘=𝛼⁺_𝑟𝑘,𝑘≥1,1,𝑘=0. (51) Then, combining (49) and (51) gives that 𝐼⁺_𝑟𝑘()𝑎ℎ,𝛿=−^0𝑟+1𝑎₁^(𝑟+1)/2√2ℎ𝜑⁺_𝑟𝑘(ℎ)+𝛼⁺_𝑟𝑘𝑎2ℎ+²⁰𝑎₁^𝑘𝐼⁺_𝑟0(ℎ,𝛿),𝑘≥0,𝑟≥0. (52) Further, by using the formula 𝑣^𝑟𝑎2ℎ+²⁰𝑎₁−𝑣^21/2=𝑑𝑣−𝑣^𝑟−12ℎ+𝑎²₀/𝑎₁−𝑣^23/2+(𝑟+2𝑟−1)2ℎ+𝑎²₀/𝑎₁×𝑣𝑟+2_𝑟−2𝑎2ℎ+²⁰𝑎₁−𝑣^21/2𝑑𝑣, (53) we have that 𝐼⁺_𝑟0()=𝑎ℎ,𝛿^0𝑟−1(2ℎ)^3/2𝑎₁^(𝑟−1)/2()+𝑟+2𝑟−1𝑎𝑟+22ℎ+²⁰𝑎₁𝐼⁺_𝑟−2,0(ℎ,𝛿),𝑟≥1. (54) It follows that 𝐼⁺_𝑟0(ℎ,𝛿)=(2ℎ)^3/2𝜑⁺_𝑟(ℎ)+𝛼⁺_𝑟𝑎2ℎ+²⁰𝑎₁^𝑟/2𝐼₀₀(ℎ,𝛿),𝑟≥1, (55) where 𝛼⁺_𝑟=()0,𝑟odd,𝑟−1)(𝑟−3×⋯×3×1()𝑟()𝜑𝑟+2𝑟−2×⋯×6×4,𝑟even,_+𝑟(ℎ)=𝑎₀^𝑟−1(𝑟+2)𝑎₁^(𝑟−1)/2+(𝑟−1)𝑎₀^𝑟−3(𝑟+2)𝑟𝑎₁^(𝑟−3)/2𝑎2ℎ+²⁰𝑎₁+(𝑟−1)(𝑟−3)𝑎₀^𝑟−5(𝑟+2)𝑟(𝑟−2)𝑎₁^(𝑟−5)/2𝑎2ℎ+²⁰𝑎₁²+()()+⋯𝑟−1)(𝑟−3×⋯×𝑟+1−2𝑟−12×𝑎₀^{𝑟−1−2[(𝑟−1)/2]}×()𝑟()()𝑟+2𝑟−2×⋯×𝑟+2−2𝑟−22×𝑎₁^{(𝑟−1)/2−[(𝑟−1)/2]−1}×𝑎2ℎ+²⁰𝑎₁^{[(𝑟−1)/2]}, (56) which is a polynomial of degree [(𝑟−1)/2] in ℎ. Let 𝜑⁺_𝑟(ℎ)=𝜑⁺_𝑟(ℎ),𝑟≥1,0,𝑟=0,𝛼⁺_𝑟=𝛼⁺_𝑟,𝑟≥1,1,𝑟=0. (57) Then, we have that by (55) and the above 𝐼⁺_𝑟0(ℎ,𝛿)=(2ℎ)^3/2𝜑⁺_𝑟(ℎ)+𝛼⁺_𝑟𝑎2ℎ+²⁰𝑎₁^𝑟/2𝐼⁺₀₀(ℎ,𝛿),𝑟≥0. (58) Substituting the above equation into (52), one can find that 𝐼⁺_𝑟𝑘(ℎ,𝛿)=√ℎ𝜓⁺_𝑟𝑘(ℎ)+𝛼_𝑟𝑘𝛼_𝑟𝑎2ℎ+²⁰𝑎₁^𝑘+𝑟/2×𝐼⁺₀₀(ℎ,𝛿),𝑟≥0,𝑘≥0, (59) where 𝛼_𝑟=0 for 𝑟 odd, 𝛼_𝑟>0 for 𝑟 even, and 𝜓⁺_𝑟𝑘()√ℎ,𝛿=−2𝑎₀^𝑟+1𝑎₁^(𝑟+1)/2𝜑⁺_𝑟𝑘(ℎ)+2√2ℎ𝛼_𝑟𝑘𝑎2ℎ+²⁰𝑎₁^𝑘𝜑⁺_𝑟(ℎ), (60) which is a polynomial of degree 𝑘+[(𝑟+1)/2] in ℎ. Combining (37), (45), and (59) gives that 𝑀⁺₁(ℎ,𝛿)=√ℎ^𝑛−1_𝑖+2𝑘=02𝑞⁺_𝑖,2𝑘√𝑎₁^𝑖𝑖+1_𝑟=0𝐶^𝑟_𝑖−𝑎₀√𝑎₁^𝑖−𝑟𝜓⁺_𝑟𝑘(ℎ)+^𝑛−1_𝑖+2𝑘=02𝑞⁺_𝑖,2𝑘√𝑎₁^𝑖+1×^𝑖_{𝑟=0,𝑟even}𝐶^𝑟_𝑖𝛼_𝑟𝑘𝛼_𝑟−𝑎₀√𝑎₁^𝑖−𝑟×𝑎2ℎ+²⁰𝑎₁^𝑘+𝑟/2𝐼⁺₀₀(ℎ,𝛿), (61) which implies (35), together with (13) and (15).
Note that √𝑎²−𝑥²𝑑𝑥=12𝑥√𝑎²−𝑥²+𝑎²𝑥arcsin|𝑎|. (62) Then, we have 𝐼⁺₀₀()𝑎ℎ,𝛿=−⁰2√𝑎₁√2ℎ+12𝑎2ℎ+²⁰𝑎₁×𝜋2𝑎−arcsin⁰2𝑎₁ℎ+𝑎²₀. (63) Inserting the above formula into (35), we can obtain (36). Hence, the proof is finished.

Similar to Lemma 4, 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 𝑀⁻(ℎ,𝛿)=√ℎ^𝑛_𝑖+2𝑘=0𝐴⁻_𝑖,2𝑘√ℎ^𝑖+2𝑘, (64) where 𝐴⁻_0,2𝑘2√=−2^2𝑘+1𝑎⁻_0,2𝑘,𝐴2𝑘+1_{−𝑖,2𝑘}=2√2^2𝑘+1+𝑖𝑏₁^𝑖𝑏⁻_{𝑖−1,2𝑘+1}+𝑖𝑎2𝑘+1^{−𝑖,2𝑘}×⁰_−𝜋/2sin^𝑖−1𝜃cos^2𝑘+2𝜃d𝜃,1≤𝑖≤𝑛. (65)(ii) If 𝑏₁>0,𝑏₀≠0,𝑀⁻(ℎ,𝛿) in (11) has the form 𝑀⁻(ℎ,𝛿)=√ℎ^𝑛−1_𝑖+2𝑘=0𝑏⁻_𝑖,2𝑘+1+𝑖+1𝑎2𝑘+1^{−𝑖+1,2𝑘}𝜙⁻_𝑖𝑘(ℎ)−^𝑛_2𝑘=02^𝑘+1+1/2𝑎⁻_0,2𝑘ℎ2𝑘+1^𝑘+^𝑛−1_𝑖+2𝑘=0𝑏⁻_𝑖,2𝑘+1+𝑖+1𝑎2𝑘+1^{−𝑖+1,2𝑘}×^𝑖_{𝑟=0,𝑟𝑒𝑣𝑒𝑛}𝛼⁻_𝑖𝑟𝑘𝑏2ℎ+²⁰𝑏₁^𝑘+𝑟/2𝐼⁻₀₀(ℎ,𝛿), (66) or 𝑀⁻(ℎ,𝛿)=√ℎ𝜓⁻_[]𝑛/2()+𝑏ℎ,𝛿2ℎ+²⁰𝑏₁𝜑⁻_{[](𝑛−1)/2}×𝑏2ℎ+²⁰𝑏₁𝜋,𝛿2𝑏+arcsin⁰2𝑏₁ℎ+𝑏²₀, (67) where 𝐼⁻₀₀(ℎ,𝛿)=^𝑏0/𝑏1_{−2ℎ+𝑏20/𝑏1}𝑏2ℎ+²⁰𝑏₁−𝑣²𝑑𝑣, (68) each 𝛼⁻_𝑖𝑟𝑘 is nonzero constant and 𝜙⁻_𝑖𝑘,𝜓⁻_[𝑛/2],𝜑⁻_{[(𝑛−1)/2]} are polynomials of degree 𝑘+[(𝑖+1)/2],[𝑛/2],[(𝑛−1)/2], respectively.

3. Proof of Theorem 1

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 Subcases 1, 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. 𝑎₁=𝑏₁=0,𝑎₀>0,𝑏₀<0. From (12) and Lemma 3, one can obtain that 𝑀(ℎ,𝛿)=𝑀⁺(ℎ,𝛿)+𝑀⁻(ℎ,𝛿)=ℎ^𝑛1/2_𝑖+2𝑘=0𝐵⁺_𝑖,2𝑘ℎ^𝑖+𝑘+ℎ^𝑛1/2_𝑖+2𝑘=0𝐵⁻_𝑖,2𝑘ℎ^𝑖+𝑘=ℎ^𝑛1/2_𝑖+2𝑘=0𝐵⁺_𝑖,2𝑘+𝐵⁻_𝑖,2𝑘ℎ^𝑖+𝑘, (69) 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 𝑎⁻_𝑖𝑗=𝑏^±_𝑖𝑗=0,𝑎⁺_𝑖𝑗=0,𝑗≥1. Then, by (27) and (29), (69) has the form 𝑀(ℎ,𝛿)=ℎ^𝑛1/2_𝑖=0𝐵⁺_𝑖0ℎ^𝑖, (70) where 𝐵⁺₀₀=2^1+1/2𝑎⁺₀₀,𝐵⁺_𝑖0=2^1+1/2𝑎^𝑖₀(2𝑖)!!()𝑎2𝑖+1!!^+𝑖0,𝑖≥1. (71) Hence, using (70) we can take 𝑎⁺_𝑖0,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. 𝑎₁>0,𝑏₁>0,𝑎₀=𝑏₀=0 Similar to the above and using (32) and (64), 𝑀(ℎ,𝛿) in (12) has the expression of the form 𝑀(ℎ,𝛿)=^𝑛_𝑖+2𝑘=0𝐴⁺_𝑖,2𝑘√ℎ^2𝑘+1+𝑖+^𝑛_𝑖+2𝑘=0𝐴⁻_𝑖,2𝑘√ℎ^2𝑘+1+𝑖=√ℎ^𝑛_𝑙=0𝐴_𝑙√ℎ^𝑙, (72) where 𝐴_𝑙=∑_{𝑖+2𝑘=𝑙}(𝐴⁺_𝑖,2𝑘+𝐴⁻_𝑖,2𝑘). Further, taking 𝑎⁻_𝑖𝑗=𝑏^±_𝑖𝑗=0,𝑎⁺_𝑖𝑗=0,𝑗≥1, then, by (34) and (65), 𝑀(ℎ,𝛿) in (72) becomes 𝑀(ℎ,𝛿)=ℎ^𝑛1/2_𝑖=0𝐴⁺_𝑖0√ℎ^𝑖, (73) where 𝐴⁺₀₀=2𝑎⁺₀₀,𝐴2𝑘+1_+𝑖0=2𝑖𝑎⁺_𝑖0√𝑎₁^𝑖0𝜋/2sin^𝑖−1𝜃cos²𝜃d𝜃,𝑖≥1. (74) Thus, from (72) and (73), we can discuss similar to Subcase 1. This finishes the proof.

Subcase 3. 𝑎₁>0,𝑏₀<0,𝑎₀=𝑏₁=0 By Lemmas 3, and 4 and (12), we can have that 𝑀(ℎ,𝛿)=^𝑛_𝑖+2𝑘=0𝐴⁺_𝑖,2𝑘√ℎ^2𝑘+1+𝑖+ℎ^𝑛1/2_𝑖+2𝑘=0𝐵⁻_𝑖,2𝑘ℎ^𝑖+𝑘=ℎ^1/2𝑛_𝑖+2𝑘=0𝐴⁺_𝑖,2𝑘√ℎ^𝑖+2𝑘+^𝑛_𝑖+2𝑘=0𝐵⁻_𝑖,2𝑘ℎ^𝑖+𝑘=ℎ^1/2𝑀^∗(ℎ,𝛿). (75) 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 𝑀^∗(ℎ,𝛿)=^𝑛_𝑖+2𝑘=0𝐴⁺_𝑖,2𝑘𝜆^𝑖+2𝑘+^𝑛_𝑖+2𝑘=0𝐵⁻_𝑖,2𝑘𝜆^2𝑖+2𝑘=^2𝑙_𝑗=0𝐶_𝑗𝜆^𝑗+^𝑙_𝑗=1𝐶_2𝑙+𝑗𝜆^2𝑗+2𝑙≜𝑀(𝜆,𝛿), (76) where 𝐶_𝑗=_{𝑖+2𝑘=𝑗}𝐴⁺_𝑖,2𝑘+_{2𝑖+2𝑘=𝑗}𝐵⁻_𝑖,2𝑘𝐶,𝑗=0,1,2,…,2𝑙,_2𝑙+𝑗=_{𝑖+𝑘=𝑗+𝑙}𝐵⁻_𝑖,2𝑘,𝑗=1,2,…,𝑙. (77) 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 𝑑^2𝑙𝑀(𝜆,𝛿)/𝑑𝜆^2𝑙 has at most 𝑙 zeros for 𝜆∈(0,+∞). From (76), we can have that 𝑑^2𝑙𝑀(𝜆,𝛿)𝑑𝜆^2𝑙=𝐶_2𝑙(2𝑙)!+^𝑙_𝑗=1𝐶_2𝑙+𝑗𝐴^2𝑙_2𝑙+2𝑗𝜆^2𝑗, (78) which shows that 𝑑^2𝑙𝑀(𝜆,𝛿)/𝑑𝜆^2𝑙 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 𝑏^±_𝑖𝑗=0,𝑎^±_𝑖𝑗=0,𝑗≥1,𝑎⁻_𝑖0=0,0≤𝑖≤𝑙, and 𝑎⁻_2𝑙≠0. Then 𝑀^∗(ℎ,𝛿) in (75) can be expressed as by (29) and (34) 𝑀^∗(ℎ,𝛿)=^2𝑙_𝑖=0𝐴⁺_𝑖0√ℎ^𝑖+^2𝑙_𝑖=𝑙+1𝐵⁻_𝑖0√ℎ^2𝑖=𝐴⁺₀₀+𝐴⁺₁₀√ℎ+𝐴⁺₂₀√ℎ²+⋯+𝐴⁺_2𝑙−1,0√ℎ^2𝑙−1+𝐴⁺_2𝑙,0√ℎ^2𝑙+𝐵⁻_𝑙+1,0√ℎ^2𝑙+2+𝐵⁻_𝑙+2,0√ℎ^2𝑙+4+⋯+𝐵⁻_2𝑙,0√ℎ^4𝑙, (79) where 𝐴⁺₀₀=2√2𝑎⁺₀₀,𝐴⁺_𝑖0=2√2^𝑖+1√𝑎₁^𝑖𝑖𝑎⁺_𝑖0𝐵,𝑖=1,2,…,2𝑙,_−𝑖0=−2√2𝑏^𝑖₀𝑎⁻_𝑖00^𝜋/2cos^2𝑖+1=√𝜃𝑑𝜃−22𝑏^𝑖₀(2𝑖)!!()𝑎2𝑖+1!!^−𝑖0,𝑖=𝑙+1,…,2𝑙. (80) Thus, by changing the sign of 𝑎⁻_2𝑙,0,𝑎⁻_2𝑙−1,0,…,𝑎⁻_𝑙+1,0, 𝑎⁺_2𝑙,0,𝑎⁺_2𝑙−1,0,…,𝑎⁺₀₀ in turn such that 𝑎⁻_𝑖−1,0𝑎⁻_𝑖0𝑎<0,𝑖=2𝑙,2𝑙−1,…,𝑙+2,_−𝑙+1𝑎⁺_2𝑙,0>0,𝑎⁺_𝑖−1,0𝑎⁺_𝑖0𝑎<0,𝑖=2𝑙,2𝑙−1,…,1,0<⁺₀₀≪𝑎⁺₁₀≪⋯≪𝑎⁺_2𝑙,0≪𝑎⁻_𝑙+1≪𝑎⁻_𝑙+2≪𝑎⁻_2𝑙−1≪1, (81) we can find 3𝑙 simply positive zeros ℎ₁,ℎ₂,…,ℎ_3𝑙 with 0<ℎ_3𝑙<ℎ_3𝑙−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. 𝑎₁>0,𝑎₀>0,𝑏₁=0,𝑏₀<0 From (12) and Lemmas 3 and 4, we get that 𝑀(ℎ,𝛿)=√ℎ𝜓⁺_[]𝑛/2()+𝑎ℎ,𝛿2ℎ+²⁰𝑎₁𝜑⁺_{[](𝑛−1)/2}×𝑎2ℎ+²⁰𝑎₁𝑎,𝛿𝜋−2arcsin⁰2𝑎₁ℎ+𝑎²₀+√ℎ^𝑛_𝑖+2𝑘=0𝐵⁻_𝑖,2𝑘ℎ^𝑖+𝑘=√ℎ𝑓_𝑛()+𝑎ℎ,𝛿2ℎ+²⁰𝑎₁𝜑⁺_{[](𝑛−1)/2}×𝑎2ℎ+²⁰𝑎₁𝑎,𝛿𝜋−2arcsin⁰2𝑎₁ℎ+𝑎²₀, (82) where 𝑓_𝑛(ℎ,𝛿)=𝜓⁺_[]𝑛/2(ℎ,𝛿)+^𝑛_𝑖+2𝑘=0𝐵⁻_𝑖,2𝑘ℎ^𝑖+𝑘, (83) which is a polynomial of degree 𝑛 in ℎ. Let 𝜆=√ℎ. Then ℎ=𝜆²,𝜆∈(0,+∞), and (82) becomes 𝑀(ℎ,𝛿)=𝜆𝑓_𝑛𝜆²()𝜋+𝑣𝜆,𝛿2𝑎−arcsin⁰2𝑎₁𝜆²+𝑎²₀≜𝑀(𝜆,𝛿), (84) where 𝑣(𝜆,𝛿)=2𝜆²+𝑎²₀𝑎₁𝜑⁺_{[](𝑛−1)/2}2𝜆²+𝑎²₀𝑎₁,𝛿. (85) One can see that (𝑑/𝑑𝜆)(𝑀(𝜆,𝛿)/𝑣(𝜆,𝛿))=𝑢(𝜆,𝛿)/𝑣²(𝜆,𝛿), where 𝑢(𝜆,𝛿)=2𝜆²+𝑎²₀𝑎₁𝜑⁺_{)][(𝑛−1/2}2𝜆²+𝑎²₀𝑎₁×𝑑,𝛿𝑑𝜆𝜆𝑓_𝑛𝜆²,𝛿−𝜆𝑓_𝑛𝜆²×𝑑,𝛿𝑑𝜆2𝜆²+𝑎²₀𝑎₁𝜑⁺_{)][(𝑛−1/2}2𝜆²+𝑎²₀𝑎₁+√,𝛿2𝑎₀√𝑎₁2𝜆²+𝑎²₀𝑎₁×𝜑⁺_{[](𝑛−1)/2}2𝜆²+𝑎²₀𝑎₁,𝛿². (86) 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 deg𝑣=2𝑛+12,deg𝑢=2𝑛+𝑛+12, (87) and they are even functions in 𝜆. Therefore, #{𝜆∈(0,+∞)∣𝑣(𝜆,𝛿)}≤𝑛−12,#{𝜆∈(0,+∞)∣𝑢(𝜆,𝛿)}≤𝑛+𝑛+12. (88) Then, from [8], we can obtain that #()∣𝑀(){()({()(≤𝜆∈0,+∞𝜆,𝛿=0≤#𝜆∈0,+∞∣𝑣𝜆,𝛿)}+#𝜆∈0,+∞∣𝑢𝜆,𝛿)}+1𝑛−12+𝑛+𝑛+12+1=𝑛+2𝑛+12, (89) which implies that 𝑍(𝑛)≤𝑛+2[(𝑛+1)/2]. Now, we verify 𝑍(𝑛)≥𝑛+[(𝑛+1)/2].
Make the transformation 𝑢=𝑣²/2ℎ−𝑎²₀/2𝑎₁ℎ. Then 𝐼⁺₀₀(ℎ,𝛿) in (35) becomes 𝐼⁺₀₀()=()ℎ,𝛿2ℎ^3/2𝑎²₀/𝑎₁¹⁰𝑢1−𝑢^21/21+2ℎ𝑎₁𝑢²/𝑎²₀𝑑𝑢, (90) which follows that as ℎ>0 small 𝐼⁺₀₀()=()ℎ,𝛿2ℎ^3/2𝑎²₀/𝑎₁¹⁰𝑢1−𝑢^21/2×1+^+∞_𝑚=1(−1)^𝑚(2𝑚−1)!!(2𝑚)!!2𝑎₁𝑎²₀^𝑚ℎ^𝑚𝑢^2𝑚𝑑𝑢. (91) Note that ¹₀𝑢^2𝑚+11−𝑢^21/2=𝑑𝑢_0𝜋/2sin^2𝑚+1𝜃cos²()=()𝜃𝑑𝜃Let𝑢=sin𝜃2𝑚!!(2𝑚+3)!!,𝑚=0,1,2,…. (92) Inserting the above formula into (91) gives that 𝐼₀₀(ℎ,𝛿)=2√ℎ2𝑎₁𝑎²₀^+∞_𝑚=0𝐶_𝑚2𝑎₁𝑎²₀^𝑚ℎ^𝑚+1, (93) where 𝐶_𝑚=(−1)^𝑚(2𝑚+1)(2𝑚+3),𝑚≥0. (94) Take 𝑏⁻_𝑖𝑗=𝑎⁺_𝑖𝑗=0,𝑎⁻_𝑖𝑗=0,𝑗≥1,𝑏⁺_𝑖𝑗=0,𝑖≥1. Then, by (28), (35), and (93), we can obtain that for ℎ>0 small 𝑀(ℎ,𝛿)=√ℎ^𝑛_𝑖=0−2√2(2𝑖)!!(2𝑖+1)!!𝑏^𝑖₀𝑎⁻_𝑖0ℎ^𝑖+^𝑛−1_2𝑘=0𝑏⁺_0,2𝑘+1𝜓⁺_0𝑘(ℎ)+^𝑛−1_2𝑘=02^𝑘+1𝛼⁺_00𝑘2𝑎₁𝑎²₀𝑏⁺_0,2𝑘+1𝑎ℎ+²⁰2𝑎₁^𝑘×^+∞_𝑚=0𝐶_𝑚2𝑎₁𝑎²₀^𝑚ℎ^𝑚+1=√ℎ_𝑖≥0𝑣_𝑖ℎ^𝑖, (95) where 𝑣_𝑖=−2√2(2𝑖)!!(2𝑖+1)!!𝑏^𝑖₀𝑎⁻_𝑖0+𝐿_𝑖𝑏⁺₀₁,𝑏⁺₀₃,…,𝑏⁺_{)]0,2[(𝑛−1/2+1}𝑣,𝑖=0,1,2,…,𝑛_𝑛+1+𝑖=^𝑛−1_2𝑘=0𝛼⁺_00𝑘2^𝑘+12𝑎₁𝑎²₀^{𝑛+𝑖−𝑘+1/2}×^𝑘_𝑟=0𝐶^𝑟_𝑘𝐶_{𝑛+𝑖−𝑟}𝑏⁺_0,2𝑘+1,𝑖≥0, (96) with 𝐿_𝑖,𝑖=0,1,…,𝑛 being linear combination, 𝐿_𝑖(0,0,…,0)=0. One can find that 𝜕𝑣₀,𝑣₁,…,𝑣_𝑛,𝑣_𝑛+1,𝑣_𝑛+2,…,𝑣_{𝑛+[(𝑛+1)/2]}𝜕𝑎⁻₀₀,𝑎⁻₁₀,…,𝑎⁻_𝑛0,𝑏⁺₀₁,𝑏⁺₀₃,…,𝑏⁺_{)]0,2[(𝑛−1/2+1}=𝐴₁𝐴₂022𝑎₁𝑎²₀𝐴₃≡𝐴, (97) where 𝐴₂ is a (𝑛+1)×[(𝑛+1)/2] matrix, 𝐴₁=−2√0√200⋯0−42𝑏₀√0⋯000−16215𝑏²₀√⋯0⋮⋮⋮⋱⋮000⋯−22(2𝑛)!!(2𝑛+1)!!𝑏^𝑛₀,𝐴₃=𝛼⁺₀₀₀𝛽₀,2𝛼⁺₀₀₁𝛽₁,2²𝛼⁺₀₀₂𝛽₂2,…,_{[(𝑛−1)/2]}𝛼⁺_{)]00,[(𝑛−1/2}𝛽_{[(𝑛−1)/2]}, (98) with 𝛽_i are [(𝑛+1)/2]×1 matrix satisfying 𝛽_𝑖=2𝑎₁𝑎²₀^{𝑖𝑛−𝑖}_𝑟∑₌₀𝐶^𝑟_𝑖𝐶_𝑛−𝑟2𝑎₁𝑎²₀^{𝑖𝑛+1−𝑖}_𝑟∑₌₀𝐶^𝑟_𝑖𝐶_{𝑛+1−𝑟}⋮2𝑎₁𝑎²₀^{𝑖𝑛+[(𝑛−1)/2]−𝑖}_𝑟∑₌₀𝐶^𝑟_𝑖𝐶_{𝑛+[(𝑛−1)/2]−𝑟},𝑖=0,1,2,…,𝑛−12. (99) Hence, we can obtain that from (97) |𝐴|=22𝑎₁𝑎²₀^[(𝑛+1)/2]𝐴₁𝐴₃=2𝛼⁺₀2𝑎₁𝑎²₀^{[(𝑛+1)/2][(𝑛−1)/2]}_𝑖=02^𝑖𝛼^+𝑛_00𝑖𝑖=0−2√2(2𝑖)!!(2𝑖+1)!!𝑏^𝑖₀|𝐵|, (100) where 𝐵=𝛽₀,𝛽₁,…,𝛽_{[(𝑛−1)/2]}, (101) and 𝛽_𝑖 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/𝑎²₀)^𝑖−1,𝑖=2,3,…,[(𝑛+1)/2], we can obtain that by (99) and (101) |𝐵|=2𝑎₁𝑎²₀^{𝑛[(𝑛+1)/2]}𝐵₁, (102) where 𝐵₁=𝐶∑_𝑛¹_𝑟=0𝐶^𝑟₁𝐶_𝑛−𝑟⋯∑^{[(𝑛−1)/2]}_𝑟=0𝐶^𝑟_{[(𝑛−1)/2]}𝐶_𝑛−𝑟𝐶¹_𝑛+1∑_𝑟=0𝐶^𝑟₁𝐶_{𝑛+1−𝑟}⋯∑^{[(𝑛−1)/2]}_𝑟=0𝐶^𝑟_{[(𝑛−1)/2]}𝐶_{𝑛+1−𝑟}𝐶⋮⋮⋱⋮_{1𝑛+[(𝑛−1)/2]𝑟}∑₌₀𝐶^𝑟₁𝐶_{𝑛+[(𝑛−1)/2]−𝑟}⋯∑^{[(𝑛−1)/2]}_𝑟=0𝐶^𝑟_{[(𝑛−1)/2]}𝐶_{𝑛+[(𝑛−1)/2]−𝑟}. (103) 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 −𝐶¹_𝑖−1 to 𝑖th column, 𝑖=3,4,…,[(𝑛+1)/2].(3) Add the third column multiplying by −𝐶²_𝑖−1 to 𝑖th column, 𝑖=4,5,…,[(𝑛+1)/2] ⋮
[(𝑛−1)/2]. Add the [(𝑛−1)/2]th column multiplying by −𝐶^{[(𝑛−3)/2]}_{[(𝑛−1)/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) 𝐵₁=(−1)^{[(𝑛−1)/2][(𝑛+1)/2]/2}det𝛽₀,𝛽₁,…,𝛽_{[(𝑛−1)/2]}=2^[(𝑛+1)/2](−1)^{𝑛[(𝑛+1)/2]}det𝛽₀,𝛽₁,…,𝛽_{[(𝑛−1)/2]}≜2^[(𝑛+1)/2](−1)^{𝑛[(𝑛+1)/2]}𝐵₂, (104) where 𝛽_𝑖=(−1)^𝑛(2()2())()𝑛−𝑖+1)(𝑛−𝑖+3−1^𝑛+1(2()2())⋮()𝑛+1−𝑖+1)(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()−1𝑛−𝑖+12()1𝑛−𝑖+32()−1𝑛+1−𝑖+12()⋮1𝑛+1−𝑖+32()])−1𝑛+[(𝑛−1/2−𝑖+12(𝑛+[(𝑛−1)/2]−𝑖)+3, (105) 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]𝐵₂=1111⋯112𝑛+312𝑛+1⋯12𝑛−12()1𝑛−[(𝑛−1/2])+112𝑛+512𝑛+3⋯12𝑛+12()1𝑛−[(𝑛−1/2])+3⋮⋮⋮⋱⋮2()1𝑛+[(𝑛−1/2])+32()1𝑛+[(𝑛−1/2])+12()⋯1𝑛+[(𝑛−1/2])−1,2𝑛+1 (106) which implies that |𝐴₃|≠0 by (102) and (104) if |𝐵₂|≠0. We claim that |𝐵₂|≠0 and𝐵₃=112𝑛+312𝑛+1⋯12𝑛−12()1𝑛−[(𝑛−1/2])+112𝑛+512𝑛+3⋯12𝑛+12()1𝑛−[(𝑛−1/2])+3⋮⋮⋱⋮⋮2()1𝑛+[(𝑛−1/2])+32()1𝑛+[(𝑛−1/2])+12()⋯1𝑛+[(𝑛−1/2])−112𝑛+12()1𝑛+[(𝑛+1/2])+32()1𝑛+[(𝑛+1/2])+12()⋯1𝑛+[(𝑛+1/2])−12𝑛+3≠0. (107) Now, we prove them by induction on 𝑛. For 𝑛=1,2 we have 111513=2,15111715=2,1355131715=−4,15257151917=−4,2205 (108) 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=2𝑙𝐵₂=1111⋯114𝑙+314𝑙+1⋯14𝑙−112𝑙+314𝑙+514𝑙+3⋯14𝑙+112𝑙+5⋮⋮⋮⋱⋮16𝑙+116𝑙−1⋯16𝑙−3≜|𝐶|𝐵4𝑙+1≠0,₃=114𝑙+314𝑙+1⋯14𝑙−112𝑙+314𝑙+514𝑙+3⋯14𝑙+112𝑙+5⋮⋮⋱⋮⋮16𝑙+116𝑙−1⋯16𝑙−314𝑙+116𝑙+316𝑙+1⋯16𝑙−14𝑙+3≠0. (109) Then for 𝑛=2𝑙+1, we have 𝐵₂=1111⋯114𝑙+514𝑙+3⋯14𝑙+112𝑙+314𝑙+714𝑙+5⋯14𝑙+312𝑙+5⋮⋮⋮⋱⋮16𝑙+516𝑙+3⋯16𝑙+1,𝐵4𝑙+3₃=114𝑙+514𝑙+3⋯14𝑙+112𝑙+314𝑙+714𝑙+5⋯14𝑙+312𝑙+5⋮⋮⋱⋮⋮16𝑙+516𝑙+3⋯16𝑙+114𝑙+316𝑙+716𝑙+5⋯16𝑙+3.4𝑙+5 (110) Note that by the first equation of (109) there only exist 𝛼₁,𝛼₂,…,𝛼_l+1 such that 𝐵₂=1−𝛼6𝑙+5¹+𝛼6𝑙+3²𝛼6𝑙+1+⋯+^𝑙+1𝐶|=4𝑙+3|−2𝛼6𝑙+5¹+6𝑙+32𝛼₂()𝛼6𝑙+1+⋯+𝑙+1_𝑙+1𝐶|4𝑙+3| (111) since 𝛼₁+𝛼₂+⋯+𝛼_l+1=1, where 𝐶 is given in (109). If |𝐵₂|=0, then we can obtain that from (109) and (111) 𝛼₁+6𝑙+32𝛼₂()𝛼6𝑙+1+⋯+𝑙+1_𝑙+14𝑙+3=0. (112) Note that 𝛼₁+𝛼4𝑙+3²𝛼4𝑙+1+⋯+^𝑙+1=12𝑙+3,𝛼4𝑙+5₁+𝛼4𝑙+5²𝛼4𝑙+3+⋯+^𝑙+1=12𝑙+5,⋮𝛼4𝑙+7₁+𝛼6𝑙+1²𝛼6𝑙−1+⋯+^𝑙+1=14𝑙+1,6𝑙+3 (113) which follows that, together with (112) 𝛼₁+4𝑙+32𝛼₂()𝛼4𝑙+1+⋯+𝑙+1_𝑙+1𝛼2𝑙+3=0,¹+4𝑙+52𝛼₂()𝛼4𝑙+3+⋯+𝑙+1_𝑙+1⋮𝛼2𝑙+5=0,₁+6𝑙+12𝛼₂()𝛼6𝑙−1+⋯+𝑙+1_𝑙+1𝛼4𝑙+1=0,¹+6𝑙+32𝛼₂()𝛼6𝑙+1+⋯+𝑙+1_𝑙+14𝑙+3=0. (114) By the second equation in (109) and the above formula, we have 𝛼₁=𝛼₂=⋯=𝛼_l+1=0. (115) This is a contradiction with 𝛼₁+𝛼₂+⋯+𝛼_𝑙+1=1. Hence 𝛼₁/(6𝑙+3)+2𝛼₂/(6𝑙+1)+⋯+(𝑙+1)𝛼_𝑙+1/(4𝑙+3)≠0, which means that (106) holds for 𝑛=2𝑙+1. Since |𝐵₂|≠0 in (110), there only exist 𝛽₁,𝛽₂,…,𝛽_𝑙+1 such that 𝛽₁+𝛽₂+𝛽4𝑙+5³𝛽4𝑙+7+⋯+^𝑙+1=16𝑙+5,𝛽6𝑙+7₁+𝛽₂+𝛽4𝑙+3³𝛽4𝑙+5+⋯+^𝑙+1=16𝑙+3,⋮𝛽6𝑙+5₁+𝛽₂+𝛽2𝑙+3³𝛽2𝑙+5+⋯+^𝑙+1=14𝑙+3,4𝑙+5 (116) 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 𝑎⁻₀₀,𝑎⁻₁₀,…,𝑎⁻_𝑛0,𝑏⁺₀₁,𝑏⁺₀₃,…,𝑏⁺_{0,2[(𝑛−1)/2]+1} can be taken as free parameters. So we can choose these values such that 𝑣_𝑖𝑣_𝑖+1()<0,𝑖=0,1,…,𝑛+𝑛−12,0<𝑣₀≪𝑣₁≪⋯≪𝑣_{𝑛+[(𝑛+1)/2]}≪1, (117) 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. 𝑎₁>0,𝑎₀>0,𝑏₁>0,𝑏₀=0 By (36) and (67), one can see that 𝑀(ℎ,𝛿)=√ℎ𝜓⁺_[]𝑛/2()+𝑎ℎ,𝛿2ℎ+²⁰𝑎₁𝜑⁺_{)][(𝑛−1/2}×𝑎2ℎ+²⁰𝑎₁𝜋,𝛿2𝑎−arcsin⁰2𝑎₁ℎ+𝑎²₀+√ℎ^𝑛_𝑖+2𝑘=0𝐴⁻_𝑖,2𝑘√ℎ^𝑖+2𝑘=√ℎ𝑔_[𝑛/2](ℎ,𝛿)+𝑔_[(𝑛+1)/2]()+𝑎ℎ,𝛿2ℎ+²⁰𝑎₁𝜑⁺_{)][(𝑛−1/2}𝑎2ℎ+²⁰𝑎₁×𝜋,𝛿2𝑎−arcsin⁰2𝑎₁ℎ+𝑎²₀, (118) where 𝑔_[𝑛/2](ℎ,𝛿)=𝜓⁺_[]𝑛/2(ℎ,𝛿)+^[𝑛/2]_{𝑙=0𝑖+2𝑘=2𝑙}𝐴⁻_𝑖,2𝑘ℎ^l,𝑔_[(𝑛+1)/2](ℎ,𝛿)=^{[(𝑛−1)/2]}_{𝑙=0𝑖+2𝑘=2𝑙+1}𝐴⁻_𝑖,2𝑘ℎ^𝑙+1,𝑛≥1. (119) For convenience, we denote by 𝑔_𝑛 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]𝑑^𝑘𝑀(ℎ,𝛿)𝑑ℎ^𝑘=^𝑘_𝑗=0ℎ^{1/2−𝑘+𝑗}𝑔_{[𝑛/2]−𝑗}+𝑑^𝑘𝑔_[(𝑛+1)/2]𝑑ℎ^𝑘+^𝑘−2_𝑗=0𝑔_𝑗ℎ^{−𝑗−1/2}𝑎2ℎ+²⁰𝑎₁^−𝑗−1+𝑑^𝑘𝑑ℎ^𝑘𝑎2ℎ+²⁰𝑎₁𝜑⁺_{)][(𝑛−1/2}𝑎2ℎ+²⁰𝑎₁×𝜋,𝛿2𝑎−arcsin⁰2𝑎₁ℎ+𝑎²₀. (120) Now, we verify this claim by induction on 𝑘. For 𝑘=2, by (118) we can obtain that 𝑑𝑀(ℎ,𝛿)=1𝑑ℎ2ℎ^−1/2𝑔_[𝑛/2](ℎ,𝛿)+ℎ^1/2𝑑𝑔_[𝑛/2](ℎ,𝛿)+𝑑𝑑ℎ𝑔_[(𝑛+1)/2]+𝑑𝑑ℎ𝑎𝑑ℎ2ℎ+²⁰𝑎₁𝜑⁺_{)][(𝑛−1/2}𝑎2ℎ+²⁰𝑎₁×𝜋,𝛿2𝑎−arcsin⁰2𝑎₁ℎ+𝑎²₀+𝑎2ℎ+²⁰𝑎₁𝜑⁺_{)][(𝑛−1/2}×𝑎2ℎ+²⁰𝑎₁𝑎,𝛿⁰2𝑎₁√ℎ2ℎ+𝑎²₀/𝑎₁=ℎ^−1/2𝑔_[𝑛/2]+ℎ^1/2𝑔_[𝑛/2]−1+𝑑𝑔_[(𝑛+1)/2]+𝑑𝑑ℎ𝑎𝑑ℎ2ℎ+²⁰𝑎₁𝜑⁺_{)][(𝑛−1/2}𝑎2ℎ+²⁰𝑎₁×𝜋,𝛿2a−arcsin⁰2a₁ℎ+a²₀, (121) which follows that 𝑑²𝑀(ℎ,𝛿)𝑑ℎ²=ℎ^−3/2𝑔_[𝑛/2]+ℎ^−1/2𝑔_[𝑛/2]−1+ℎ^1/2𝑔_[𝑛/2]−2+𝑑²𝑔_[(𝑛+1)/2]𝑑ℎ²+𝑔₀√ℎ2ℎ+𝑎²₀/𝑎₁+𝑑²𝑑ℎ²𝑎2ℎ+²⁰𝑎₁𝜑⁺_{)][(𝑛−1/2}𝑎2ℎ+²⁰𝑎₁×𝜋,𝛿2𝑎−arcsin⁰2𝑎₁ℎ+𝑎²₀. (122) Hence, (120) holds for 𝑘=2. Suppose (120) holds for  𝑘,2≤𝑘≤[(𝑛−1)/2]−1. Then for 𝑘+1, we have 𝑑^𝑘+1𝑀(ℎ,𝛿)𝑑ℎ^𝑘+1=^𝑘_𝑗=012−𝑘+𝑗ℎ^{1/2−(𝑘+1)+𝑗}𝑔_{[𝑛/2]−𝑗}+^𝑘_𝑗=0ℎ^{1/2−𝑘+𝑗}𝑔_{[𝑛/2]−1−𝑗}+𝑑^𝑘+1𝑔_[(𝑛+1)/2]𝑑ℎ^𝑘+1+𝑑^𝑘+1𝑑ℎ^𝑘+1𝑎2ℎ+²⁰𝑎₁𝜑⁺_{)][(𝑛−1/2}𝑎2ℎ+²⁰𝑎₁×𝜋,𝛿2𝑎−arcsin⁰2𝑎₁ℎ+𝑎²₀+𝑑^𝑘𝑑ℎ^𝑘𝑎2ℎ+²⁰𝑎₁𝜑⁺_{)][(𝑛−1/2}𝑎2ℎ+²⁰𝑎₁×𝑎,𝛿₀2𝑎₁√ℎ2ℎ+𝑎²₀/𝑎₁+^𝑘−2_𝑗=0ℎ𝑎2ℎ+²⁰𝑎₁𝑑𝑔𝑑ℎ^𝑗−𝑔_𝑗21𝑗+1ℎ+𝑗+2𝑎2ℎ+²⁰𝑎₁×ℎ^𝑗+3/2𝑎2ℎ+²⁰𝑎₁^𝑗+2−1, (123) 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 𝑑^[(𝑛+1)/2]𝑀(ℎ,𝛿)𝑑ℎ^[(𝑛+1)/2]=^[𝑛/2]_𝑗=0ℎ^{1/2−[(𝑛+1)/2]+𝑗}𝑔_{[𝑛/2]−𝑗}+^{[(𝑛+1)/2]−2}_𝑗=0𝑔_𝑗ℎ^{−𝑗−1/2}𝑎2ℎ+²⁰𝑎₁^−𝑗−1+𝑑^[(𝑛+1)/2]𝑔_[(𝑛+1)/2]𝑑ℎ^[(𝑛+1)/2]+𝑑^[(𝑛+1)/2]𝑑ℎ^[(𝑛+1)/2]×𝑎2ℎ+²⁰𝑎₁𝜑⁺_{)][(𝑛−1/2}𝑎2ℎ+²⁰𝑎₁×𝜋,𝛿2𝑎−arcsin⁰2𝑎₁ℎ+𝑎²₀. (124) One can find that 𝑑^{[(𝑛+1)/2]+1}𝑀(ℎ,𝛿)𝑑ℎ^{[(𝑛+1)/2]+1}=^[𝑛/2]_𝑗=0ℎ^{1/2−[(𝑛+1)/2]−1+𝑗}𝑔_{[𝑛/2]−𝑗}+^{[(𝑛+1)/2]−1}_𝑗=0𝑔_𝑗ℎ^{−𝑗−1/2}𝑎2ℎ+²⁰𝑎₁^−𝑗−1=ℎ^{−[(𝑛+1)/2]−1/2}×𝑎2ℎ+²⁰𝑎₁^{−[(𝑛+1)/2]}𝐹(ℎ,𝛿), (125) where 𝐹()=𝑎ℎ,𝛿2ℎ+²⁰𝑎₁^{[(𝑛+1)/2][𝑛/2]}_𝑗=0ℎ^𝑗𝑔_{[𝑛/2]−𝑗}+^{[(𝑛+1)/2]−1}_𝑗=0𝑔_𝑗ℎ^{[(𝑛+1)/2]−𝑗}×𝑎2ℎ+²⁰𝑎₁^{[(𝑛+1)/2]−𝑗−1}, (126) 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 𝑏⁻_𝑖𝑗=𝑎⁺_𝑖𝑗=0,𝑎⁻_𝑖𝑗=0,𝑗≥1,𝑏⁺_𝑖𝑗=0,𝑖≥1. Then using (49), (51), we can write 𝑀(ℎ,𝛿) in (12) as 𝑀(ℎ,𝛿)=√ℎ−2√2𝑎⁻₀₀+^𝑛_𝑖=12√2^𝑖+1𝑖𝑏₁^𝑖𝑎⁻_𝑖0×⁰_−𝜋/2sin^𝑖−1𝜃cos²𝜃𝑑𝜃√ℎ^𝑖+^𝑛−1_2𝑘=0𝑏⁺_0,2𝑘+1𝜓⁺_0𝑘(ℎ)+^𝑛−1_2𝑘=02^𝑘+1𝛼⁺_00𝑘2𝑎₁𝑎²₀𝑏⁺_0,2𝑘+1×𝑎ℎ+²⁰2𝑎₁^𝑘+∞_𝑚=0𝐶_𝑚2𝑎₁𝑎²₀^𝑚ℎ^𝑚+1. (127)
For 𝑛=2𝑙,𝑙≥1, (127) can be written as 𝑀(ℎ,𝛿)=√ℎ^2𝑙_𝑖=0𝑣_𝑖√ℎ^𝑖+_𝑖≥0𝑣_2𝑙+1+𝑖ℎ^𝑙+1+𝑖, (128) where 𝑣₀=−2√2𝑎⁻₀₀.𝑣_2𝑖+1=2√2^2𝑖+2(2𝑖+1)𝑏₁^2𝑖+1𝑎⁻_2𝑖+1,0×⁰_−𝜋/2sin^2𝑖𝜃cos²𝜃𝑑𝜃,𝑖=0,1,…,𝑙−1,𝑣_2𝑖=2√2^2𝑖+12𝑖𝑏₁^2𝑖𝑎⁻_2𝑖,0⁰_−𝜋/2sin^2𝑖−1𝜃cos²𝜃𝑑𝜃+𝐿_2𝑖𝑏⁺₀₁,𝑏⁺₀₃,…,𝑏⁺_{)]0,2[(𝑛−1/2+1},𝑖=1,2,…,𝑙,𝑣_2𝑙+1+𝑖=^𝑙−1_𝑘=02^𝑘+1𝛼⁺_00𝑘2𝑎₁𝑎²₀^{𝑙−𝑘+𝑖+1/2}×𝑏^+𝑘_{0,2𝑘+1𝑟=0}𝐶^𝑟_𝑘𝐶_{𝑙+𝑖−𝑟},𝑖=0,1,…,𝑛−12, (129) 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. 𝑎₁>0,𝑎₀>0,𝑏₁>0,𝑏₀<0. We have as the above𝑀(ℎ,𝛿)=√ℎ𝜓⁺_[]𝑛/2(ℎ,𝛿)+𝜓⁻_[]𝑛/2()+𝑎ℎ,𝛿2ℎ+²⁰𝑎₁𝜑⁺_{)][(𝑛−1/2}()×𝜋ℎ,𝛿2𝑎−arcsin⁰2𝑎₁ℎ+𝑎²₀+𝑏2ℎ+²⁰𝑏₁𝜑⁻_{)][(𝑛−1/2}()×𝜋ℎ,𝛿2𝑏+arcsin⁰2𝑏₁ℎ+𝑏²₀. (130) Similar to the Subcase 5, we can prove that for any 2≤𝑘≤[(𝑛−1)/2]𝑑^𝑘𝑀(ℎ,𝛿)𝑑ℎ^𝑘=^𝑘_𝑗=0ℎ^{1/2−𝑘+𝑗}𝑔_{[𝑛/2]−𝑗}+^𝑘−2_𝑗=0𝑔_𝑗ℎ^{−𝑗−1/2}𝑎2ℎ+²⁰𝑎₁^−𝑗−1+^𝑘−2_𝑗=0𝑔_𝑗ℎ^{−𝑗−1/2}𝑏2ℎ+²⁰𝑏₁^−𝑗−1+𝑑^𝑘𝑑ℎ^𝑘𝑎2ℎ+²⁰𝑎₁𝜑⁺_{)][(𝑛−1/2}𝑎2ℎ+²⁰𝑎₁×𝜋,𝛿2𝑎−arcsin⁰2𝑎₁ℎ+𝑎²₀+𝑑^𝑘𝑑ℎ^𝑘𝑏2ℎ+²⁰𝑏₁𝜑⁺(𝑛−1)2𝑏2ℎ+²⁰𝑎₁×𝜋,𝛿2𝑏−arcsin⁰2𝑏₁ℎ+𝑏²₀. (131) Taking 𝑘=[(𝑛−1)/2] and differentiating the above twice follow that 𝑑^{[(𝑛+1)/2]+1}𝑀(ℎ,𝛿)𝑑ℎ^{[(𝑛+1)/2]+1}=^[𝑛/2]_𝑗=0ℎ^{1/2−[(𝑛+1)/2]−1+𝑗}𝑔_{[𝑛/2]−𝑗}+^{[(𝑛+1)/2]−1}_𝑗=0𝑔_𝑗ℎ^{−𝑗−1/2}𝑎2ℎ+²⁰𝑎₁^−𝑗−1+^{[(𝑛+1)/2]−1}_𝑗=0𝑔_𝑗ℎ^{−𝑗−1/2}𝑏2ℎ+²⁰𝑏₁^−𝑗−1. (132) If 𝑎²₀/𝑎₁=𝑏²₀/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 𝑎²₀/𝑎₁≠𝑏²₀/𝑏₁, then (132) can be written as 𝑑^{[(𝑛+1)/2]+1}𝑀(ℎ,𝛿)𝑑ℎ^{[(𝑛+1)/2]+1}=ℎ^{−[(𝑛+1)/2]−1/2}𝑎2ℎ+²⁰𝑎₁^{−[(𝑛+1)/2]}×𝑏2ℎ+²⁰𝑏₁^{−[(𝑛+1)/2]}𝐹(ℎ,𝛿), (133) where 𝐹()=𝑎ℎ,𝛿2ℎ+²⁰𝑎₁^[(𝑛+1)/2]𝑏2ℎ+²⁰𝑏₁^[(𝑛+1)/2]×^[𝑛/2]_𝑗=0ℎ^𝑗𝑔_{[𝑛/2]−𝑗}+𝑏2ℎ+²⁰𝑏₁^{[(𝑛+1)/2][(𝑛+1)/2]−1}_𝑗=0𝑔_𝑗ℎ^{[(𝑛+1)/2]−𝑗}×𝑎2ℎ+²⁰𝑎₁^{[(𝑛+1)/2]−𝑗−1}+𝑎2ℎ+²⁰𝑎₁^{[(𝑛+1)/2][(𝑛+1)/2]−1}_𝑗=0𝑔_𝑗ℎ^{[(𝑛+1)/2]−𝑗}×𝑏2ℎ+²⁰𝑏₁^{[(𝑛+1)/2]−𝑗−1}, (134) 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 𝑎⁻_𝑖𝑗=𝑏⁻_𝑖𝑗=0,𝑎⁺_𝑖𝑗=𝑏⁺_𝑖𝑗=0,𝑖≥1. Then, by Lemmas 4, 5, one can see that for ℎ>0 small 𝑀(ℎ,𝛿)=√ℎ^𝑛_2𝑘=02^𝑘+1+1/2𝑎⁺_0,2𝑘ℎ2𝑘+1^𝑘+^𝑛−1_2𝑘=0𝑏⁺_0,2𝑘+1𝜓⁺_0𝑘(ℎ)+^𝑛−1_2𝑘=02^𝑘+1𝛼⁺_00𝑘2𝑎₁𝑎²₀𝑏⁺_0,2𝑘+1𝑎ℎ+²⁰2𝑎₁^𝑘×^+∞_𝑚=0𝐶_𝑚2𝑎₁𝑎²₀^𝑚ℎ^𝑚+1. (135) 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.

Acknowledgment

The project was supported by National Natural Science Foundation of China (11271261), a grant from Ministry of Education of China (20103127110001) and FP7-PEOPLE-2012-IRSES-316338.