Existence and multiplicity of periodic solutions for a nonlinear resonance equation with singularities

Abstract

We investigate a second-order periodic system with singular potential and resonance. Utilizing the main integral method and fixed point theorems, we establish the existence and multiplicity of periodic solutions with respect to time under certain assumptions on the unbounded or oscillatory term.

1 Introduction and main result

Many scholars have investigated the singular second-order equation

$$\begin{aligned} \ddot{x}+V'(x)+g(x)=p(t) \end{aligned}$$

(1.1)

with functions $V(x)$, $g(x)$, and $p(t)$ satisfying certain restrictions. The multiplicity and existence of periodic solutions for Eq. (1.1) are discussed by utilizing the topology degree theory [1]. Qian and Torres [2] use the Poincaré–Birkhoff twist theorem to find the dynamical features of Eq. (1.1). Jiang [3] employs invariant curves and Moser’s small twist theorem to discuss the boundedness of solutions for Eq. (1.1). The unbounded and periodic solutions of Eq. (1.1) may coexist [4]. Assuming that $g(x)=0$ and $p(t+\pi )=p(t)$, Capietto et al. [5] consider Eq. (1.1) with

$$\begin{aligned} V(x)=\frac{1}{2}x_{+}^{2}+ \frac{1}{(1-x_{-}^{2})^{\nu}}-1, \end{aligned}$$

(1.2)

where $x_{+}=\max \{x,0\}$, $x_{-}=\max \{-x,0\}$, and $\nu >0$ is an integer. Using the Moser twist theorem and the Lazer–Leach assumption

$$\begin{aligned} 1+\frac{1}{2} \int _{0}^{\pi}p(t_{0}+\theta )\sin \theta \,d\theta >0 \quad\forall t_{0}\in \mathbb{R}, \end{aligned}$$

Capietto et al. [5] investigated the boundedness of solutions and quasi-periodic solutions for Eq. (1.1). Following the ideas in [5], Jiang [3] and Liu [6] and Wang and Jiang [7] discussed the boundedness of solutions for Eq. (1.1) under different conditions.

To clearly understand the objective of our work, we recall the main result about Eq. (1.1) in Lazer and Leach [8]. Suppose that $g(x)$ is smooth and bounded and that $V(x)$ satisfies the conditions

$$\begin{aligned} \lim_{x\rightarrow a_{+}}V(x)=+\infty,\qquad \lim_{x \rightarrow +\infty} \frac{2V(x)}{x^{2}}=\frac{m^{2}}{4}, \end{aligned}$$

where the domain of $V(x)$ is $(a,+\infty )$, $m>0$ is an integer, and the constant a belongs to $\in (-\infty,0)$. Let

$$\begin{aligned} g_{*}(\rho )= \int _{0}^{2\pi}g \bigl(\rho \bigl\vert \sin (mt/2) \bigr\vert \bigr) \bigl\vert \sin (mt/2) \bigr\vert \,dt,\qquad p_{*}( \theta )= \int _{0}^{2\pi}p(t+\theta ) \bigl\vert \sin (mt/2) \bigr\vert \,dt. \end{aligned}$$

Applying the analysis of phase plane and topological degree methods, it is proved in [8] that Eq.‘(1.1) possesses at least one 2π-periodic solution, provided that there is $g_{0}\in [g_{*}^{-},g_{*}^{+}]$ $(g_{*}^{-}=\liminf_{\rho \rightarrow +\infty}g_{*}(\rho ), g_{*}^{+}=\limsup_{\rho \rightarrow +\infty}g_{*}(\rho ) )$ (i.e.. $p_{*}$ has a regular value of $g_{0}$) and the number of zeros of $p_{*}-g_{0}$ in $[0,2\pi /m]$ is not 2.

In 2019, Ma [9] considered the periodic solution of Eq. (1.1). Suppose that $V'(x)=n^{2}x, p(t)\in C^{2}(\mathbb{R}/2\pi \mathbb{Z})$, and $g(x)\in C^{1}(\mathbb{R})$ with the restrictions

$$\begin{aligned} \lim_{ \vert x \vert \rightarrow +\infty}x^{-\frac{1}{2}} \bigl\vert g(x) \bigr\vert =0,\qquad \lim_{x\rightarrow +\infty}x^{\frac{1}{2}} \bigl\vert g'(x) \bigr\vert < +\infty. \end{aligned}$$

Ma [9] obtained that Eq. (1.1) possesses at least one 2π solution, provided that

$$\begin{aligned} \biggl\vert \int _{0}^{2\pi}p(t)e^{-int}\,dt \biggr\vert < \limsup_{\rho \rightarrow +\infty} \biggl\vert \int _{0}^{2\pi}g(\rho \cos \theta )\cos \theta \,d \theta \biggr\vert , \end{aligned}$$

and Eq. (1.1) possesses an unbounded sequence of 2π-periodic solutions if

$$\begin{aligned} \biggl\vert \int _{0}^{2\pi}p(t)e^{-int}\,dt \biggr\vert < {}& \min \biggl\{ \limsup_{\rho \rightarrow +\infty}| \int _{0}^{2\pi}g(\rho \cos \theta )\cos \theta \,d \theta, \\ &{} -\liminf_{\rho \rightarrow +\infty} \biggl\vert \int _{0}^{2 \pi}g(\rho \cos \theta )\cos \theta \,d \theta \biggr\vert \biggr\} . \end{aligned}$$

We observe that the function $V(x)$ is globally defined in $\mathbb{R}$ and the function $g(x)$ is unbounded or oscillatory in [9]. Note that in [8], $V(x)$ possesses a repulsive singularity at a, and $g(x)$ is bounded. Also, in [5], $V(x)$ possesses a repulsive singularity at −1, and $g(x)=0$.

When the function $V(x)$ is of the form (1.2), a natural question is to find restrictions imposed on unbounded function $g(x)$ to make Eq. (1.1) have at least one π-periodic solution and possess an unbounded sequence of π-periodic solutions. The objective of this work is to handle this problem. Precisely speaking, we investigate the existence and multiplicity of periodic solutions of the problem

$$\begin{aligned} \textstyle\begin{cases} \ddot{x}+V'(x)+g(x)=p(t), \\ V(x)=\frac{1}{2}x_{+}^{2}+\frac{1}{(1-x_{-}^{2})^{\nu}}-1, \qquad p(t) \in C^{2}(\mathbb{S}^{1}), \end{cases}\displaystyle \end{aligned}$$

(1.3)

where $\mathbb{S}^{1}=\mathbb{R}/\pi \mathbb{Z}$ and $\lim_{x\rightarrow +\infty}x^{k-\frac{1}{2}}g^{(k)}(x)=0$, $k=0,1$. Here we state that the function $g(x)$ considered in our work is different from those in [5, 8] and is the same as that in [9]. The novelty of our work is that the function $V(x)$ is of the the form (1.2), which is different from those in [8, 9].

The auxiliary equation of Eq. (1.1) takes the form

$$\begin{aligned} \ddot{x}+V'(x)+g(x)=0. \end{aligned}$$

(1.4)

The Hamiltonian function of (1.4) has the expression

$$\begin{aligned} H_{0}(x,y)=\frac{1}{2}y^{2}+V(x)+G(x), \end{aligned}$$

where $G(x)=\int _{0}^{{x}}g(s)\,ds$. For $H>0$, we denote by $\tau (H)$ the least positive period of the orbit $\Gamma _{H}:H_{0}(x,y)=H, \bar{p}=\int _{0}^{\pi}p(s)e^{is}\,ds$. Set

$$\begin{aligned} \Gamma (H)=\sqrt{H} \bigl(\tau (H)-\pi \bigr). \end{aligned}$$

(1.5)

Equation (1.4) possesses the following autonomous Hamiltonian system:

$$\begin{aligned} x'=y,\qquad y'=-V'(x)-g(x). \end{aligned}$$

Now we state the main result of our work.

Theorem 1.1

Suppose that $g(x)\in C^{1}(\mathbb{R})$, $p(t)\in C^{2}(\mathbb{R}/\pi \mathbb{Z})$, and

$$\begin{aligned} \lim_{x\rightarrow +\infty}x^{k-\frac{1}{2}}g^{(k)}(x)=0,\quad k=0,1. \end{aligned}$$

(1.6)

Then

(i) Problem (1.3) possesses at least one π-periodic solution if

$$\begin{aligned} { \vert \bar{p} \vert < \sqrt{\frac{2}{\pi}}\limsup _{H\rightarrow +\infty} \Gamma (H)}. \end{aligned}$$

(1.7)

(ii) Problem (1.3) possesses an unbounded sequence of π-periodic solutions if

$$\begin{aligned} \vert \bar{p} \vert < \sqrt{\frac{2}{\pi}}\min \Bigl\{ \limsup_{H\rightarrow + \infty}\Gamma (H), -\liminf_{H\rightarrow +\infty}\Gamma (H) \Bigr\} . \end{aligned}$$

(1.8)

In Sect. 2, we present several lemmas, and in Sect. 3, we provide the proof of Theorem 1.1.

2 Preliminaries

2.1 Action-angle coordinates

To use action-angle variables, we write the auxiliary equations

$$\begin{aligned} x'=\frac{\partial H_{1}}{\partial y},\qquad y'=- \frac{\partial H_{1}}{\partial x} \end{aligned}$$

(2.1)

with the Hamiltonian function

$$\begin{aligned} H_{1}(x,y)=\frac{1}{2}y^{2}+V(x). \end{aligned}$$

Let $T_{0}(h)$ denote the time period of the integral curve $\Gamma _{h}$ of (2.1) with $H_{1}(x,y)=h$. Denote by $I=I_{0}(h)$ the area enclosed by the closed curve $\Gamma _{h}$ for $h>0$. Assume that $-1<-\alpha _{h}<0<\beta _{h}$ satisfy $V(-\alpha _{h})=V(\beta _{h})=h$ and

$$\begin{aligned} I_{-}(h)=2 \int _{0}^{\alpha _{h}}\sqrt{2 \bigl(h-V(s) \bigr)}\,ds,\qquad T_{-}(h)=2 \int _{0}^{\alpha _{h}}\frac{1}{\sqrt{2(h-V(-s))}}\,ds. \end{aligned}$$

Thus we have

$$\begin{aligned} &I_{0}(h)=\pi h+2 \int _{0}^{\alpha _{h}}\sqrt{2 \bigl(h-V(-s) \bigr)}\,ds=\pi h+I_{-}(h), \end{aligned}$$

(2.2)

$$\begin{aligned} &T_{0}(h)=I_{0}'(h)=\pi +2 \int _{0}^{\alpha _{h}} \frac{1}{\sqrt{2(h-V(-s))}}\,ds=\pi +T_{-}(h). \end{aligned}$$

(2.3)

For conciseness in the following discussions, we always use c or C to represent positive constants.

We acquire $c\sqrt{h}< I_{-}(h)< C\sqrt{h}$. Let $h=h_{0}(I)$ be the inverse function of $I=I_{0}(h)$. Using the expression of $V(x)$ and the definition of $I_{0}(h)$, we derive that

$$\begin{aligned} cI< h_{0}(I)< CI,\qquad \bigl\vert I^{k}h_{0}^{(k)}(I) \bigr\vert < Ch(I) \quad\text{for } k=1,2. \end{aligned}$$

Lemma 2.1

[3] For $n=0,1,2$, we have

$$\begin{aligned} \frac{d^{n}T_{-}(h)}{dh^{n}}=(-1)^{n}\frac{(2n-1)!!}{2^{n}} \frac{\sqrt{2}}{h^{(2n+1)/2}}+o \biggl(\frac{1}{h^{(2n+1)/2}} \biggr), \quad h\rightarrow +\infty, \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} (2n-1)!! =\textstyle\begin{cases} 1\cdot 3\cdot \ldots \cdot (2n-1), & n\geq 1, \\ 1, & n=0. \end{cases}\displaystyle \end{aligned} \end{aligned}$$

For $(x,y)\in (-1,+\infty )\times \mathbb{R}$, we define the transformation $\Psi _{1}:(x,y)\rightarrow (\theta,I)$ by

$$\begin{aligned} \begin{aligned} \theta (x,y)=\textstyle\begin{cases} \frac{\pi}{T_{0}(h(x,y))} (\frac{T_{-}(h)}{2}+\arcsin \frac{x}{\sqrt{2h(x,y)}} )& \text{if } x>0,y>0, \\ \frac{\pi}{T_{0}(h(x,y))} (\frac{T_{-}(h)}{2}+\pi -\arcsin \frac{x}{\sqrt{2h(x,y)}} ) & \text{if } x>0,y< 0, \\ \frac{\pi}{T_{0}(h(x,y))} (\int _{-\alpha _{h}}^{x} \frac{1}{\sqrt{2(h(x,y))+1-(1-s^{2})^{-\gamma}}}\,ds )& \text{if } x< 0,y>0, \\ \frac{\pi}{T_{0}(h(x,y))} (T_{0}(h(x,y))-\int _{-\alpha _{h}}^{x} \frac{1}{\sqrt{2(h(x,y))+1-(1-s^{2})^{-\gamma}}}\,ds ),& x< 0,y< 0, \end{cases}\displaystyle \end{aligned} \end{aligned}$$

(2.4)

and

$$\begin{aligned} I(x,y)=I_{0} \bigl(h(x,y) \bigr)=2 \int _{-\alpha _{h}}^{\beta _{h}}\sqrt{2 \bigl(h(x,y)-V(s) \bigr)}\,ds. \end{aligned}$$

Note that the first equation in (1.3) possesses the Hamiltonian system

$$\begin{aligned} x'=\frac{\partial H}{\partial y},\qquad y'=- \frac{\partial H}{\partial x} \end{aligned}$$

associated with

$$\begin{aligned} H(x,y,t)=\frac{1}{2}y^{2}+V(x)+G(x)-xp(t). \end{aligned}$$

In the new variables $(\theta,I)$, we write the Hamiltonian systems of (1.3) and (1.4) in the forms

$$\begin{aligned} \theta '=\frac{\partial H}{\partial I}, \qquad I'=- \frac{\partial H}{\partial \theta} \end{aligned}$$

(2.5)

and

$$\begin{aligned} \theta '=\frac{\partial H_{0}}{\partial I},\qquad I'=- \frac{\partial H_{0}}{\partial \theta}, \end{aligned}$$

(2.6)

respectively, where

$$\begin{aligned} H(\theta,I,t)=\pi h_{0}(I)+\pi G \bigl(x(I,\theta ) \bigr)-\pi x(I,\theta )p(t) \end{aligned}$$

(2.7)

and

$$\begin{aligned} H_{0}(\theta,I,t)=\pi h_{0}(I)+\pi G \bigl(x(I,\theta ) \bigr). \end{aligned}$$

(2.8)

For $x>0$, using (2.4) gives rise to

$$\begin{aligned} x=\sqrt{2h}\sin \biggl(\frac{T_{0}(h)}{\pi}\theta - \frac{T_{-}(h)}{2} \biggr). \end{aligned}$$

(2.9)

Using formulas (2.5)–(2.9), we want to obtain estimates of the function $G(x(I,\theta ))$. We need the following lemma.

Lemma 2.2

[3] For sufficiently large I, if $x>0$, then

$$\begin{aligned} \biggl\vert I^{k}\frac{\partial ^{k}x(I,\theta )}{\partial I^{k}} \biggr\vert \leq C \sqrt{I},\quad k\leq 2. \end{aligned}$$

For sufficiently large I, if $x<0$, then

$$\begin{aligned} \biggl\vert I^{k}\frac{\partial ^{k}x(I,\theta )}{\partial I^{k}} \biggr\vert \leq C(1+x) \leq C,\quad k\leq 2. \end{aligned}$$

To transform the first equation in problem (1.3) into a nearly integrable equation, we introduce the transformation $\Psi _{2}:(I,\theta,t)\rightarrow (H,\varphi,\eta )$,

$$\begin{aligned} H(\theta,I,t)=\pi h_{0}(I)+\pi G \bigl(x(I,\theta ) \bigr)-\pi x(I, \theta )p(t), \qquad \varphi =t,\qquad \eta =\theta, \end{aligned}$$

that is, the time and energy are the new angular and action variables, respectively.

As $I\rightarrow +\infty$, from (1.6), (2.2), and Lemma 2.2 we conclude that $H/I\rightarrow 1$ and $\frac{\partial H}{\partial I}>0$. Applying the implicit function theorem, we derive that there exists a function $R_{2}(H,t,\theta )$ belonging to $C^{2}$ with $|R_{2}|< H/2\pi$ such that the function

$$\begin{aligned} I=I_{0} \biggl(\frac{H}{\pi}+R_{2}(H,t, \theta ) \biggr) \end{aligned}$$

(2.10)

satisfies (2.7). Thus, under the transform $\Psi _{2}$, the Hamiltonian function (2.7) is transformed into the Hamiltonian function (2.10).

Note that the inverse function of $I_{0}$ is $h_{0}$. Using (2.7), we acquire

$$\begin{aligned} R_{2}(H,t,\theta )=x \biggl(I_{0} \biggl( \frac{H}{\pi}+R_{2} \biggr),\theta \biggr)p(t)-G \biggl(x \biggl(I_{0} \biggl( \frac{H}{\pi}+R_{2} \biggr),\theta \biggr) \biggr). \end{aligned}$$

(2.11)

Utilizing (1.6), Lemma 2.2, and $|R_{2}|< H/2\pi$ gives rise to

$$\begin{aligned} \bigl\vert R_{2}(H,t,\theta ) \bigr\vert \leq \epsilon (H) H^{\frac{3}{4}}, \end{aligned}$$

where (here and further) $\epsilon (H)$ stands for a nonnegative function satisfying $\lim_{H\rightarrow +\infty}\epsilon (H)=0$.

Similarly, we can prove that there is a function $R_{1}(H,\theta )$ in space $C^{2}$ with $|R_{1}|< H/2\pi$ such that

$$\begin{aligned} I=I_{0} \biggl(\frac{H}{\pi}+R_{1}(H,t) \biggr) \end{aligned}$$

(2.12)

satisfies

$$\begin{aligned} H=\pi h_{0}(I)+\pi G \bigl(x(I,\theta ) \bigr). \end{aligned}$$

(2.13)

Thus we have

$$\begin{aligned} R_{1}(H,\theta )=-G \biggl(x \biggl(I_{0} \biggl(\frac{H}{\pi}+R_{1} \biggr),\theta \biggr) \biggr) \end{aligned}$$

(2.14)

and

$$\begin{aligned} \bigl\vert R_{1}(H,\theta ) \bigr\vert \leq \epsilon (H) H^{\frac{3}{4}}. \end{aligned}$$

Let $\Xi =\{\theta \in \mathbb{S}^{1}: \sin (\frac{T_{0}(h)}{\pi} \theta -\frac{T_{-}(h)}{2} )=0\}$. Obviously, by Lemma 2.1 the measure of Ξ is zero.

Lemma 2.3

If the function $R(H,t,\theta )$ belongs to $C^{2}$ and $|R(H,t,\theta )|\leq \epsilon (H) H$, provided that $h=\frac{H}{\pi}+R(H,t,\theta )$, then

$$\begin{aligned} \lim_{H\rightarrow +\infty}(\sqrt{2h})^{k-\frac{1}{2}}g^{(k)} \biggl( \sqrt{2h}\sin \biggl(\frac{T_{0}(h)}{\pi}\theta -\frac{T_{-}(h)}{2} \biggr) \biggr) \sin ^{k} \biggl(\frac{T_{0}(h)}{\pi}\theta - \frac{T_{-}(h)}{2} \biggr)=0 \end{aligned}$$

for $k=0$ and $\theta \in \mathbb{S}^{1}$ and for $k=1$ and $\theta \in \mathbb{S}^{1}\setminus \Xi$.

Proof

From (1.6), for any $\epsilon >0$, there exist constants $M_{k}>0$ such that

$$\begin{aligned} x^{k-\frac{1}{2}}g^{(k)}(x)< \epsilon, \quad x>M_{k}. \end{aligned}$$

Since $\lim_{H\rightarrow +\infty}h(H,t,\theta )\rightarrow + \infty$ uniformly for t and $\theta \setminus \Xi$, there exist positive numbers $M_{k}>0$ such that for $H>M_{k}$,

$$\begin{aligned} \frac{1}{\sqrt[4]{2h}}\max_{ \vert x \vert \leq M_{k}} \bigl\vert x^{k}g^{(k)}(x) \bigr\vert < \epsilon. \end{aligned}$$

Thus, for $|\sqrt{2h}\sin (\frac{T_{0}(h)}{\pi}\theta - \frac{T_{-}(h)}{2} ) |>M_{k}$, we have

$$\begin{aligned} \biggl\vert (\sqrt{2h})^{k-\frac{1}{2}}g^{(k)} \biggl(\sqrt{2h}\sin \biggl( \frac{T_{0}(h)}{\pi}\theta -\frac{T_{-}(h)}{2} \biggr) \biggr)\sin ^{k} \biggl(\frac{T_{0}(h)}{\pi}\theta -\frac{T_{-}(h)}{2} \biggr) \biggr\vert \leq \epsilon, \end{aligned}$$

and for $|\sqrt{2h}\sin (\frac{T_{0}(h)}{\pi}\theta - \frac{T_{-}(h)}{2} ) |\leq M_{k}$, we have

$$\begin{aligned} & \biggl\vert (\sqrt{2h})^{k-\frac{1}{2}}g^{(k)} \biggl(\sqrt{2h}\sin \biggl( \frac{T_{0}(h)}{\pi}\theta -\frac{T_{-}(h)}{2} \biggr) \biggr)\sin ^{k} \biggl(\frac{T_{0}(h)}{\pi}\theta -\frac{T_{-}(h)}{2} \biggr) \biggr\vert \\ &\quad \leq \frac{1}{\sqrt[4]{2h}}\max_{ \vert x \vert \leq M_{k}} \bigl\vert x^{k}g^{(k)}(x) \bigr\vert < \epsilon, \end{aligned}$$

which ends the proof. □

For $r\in \mathbb{R}$, when $H\gg 1$, we write $u=u(H,t,\theta )\in C^{2}(r,\epsilon )$ if $|\partial _{H}^{k}u|\leq \epsilon (H)H^{r-k}$ for $k=0,1,2$ and $|\partial _{H}^{k}\partial _{t}u|\leq CH^{\frac{1}{2}-k}$ for $k=0,1$.

Lemma 2.4

Let $h=\frac{H}{\pi}+u(H,t,\theta )$ with $u\in C(\frac{3}{4},\epsilon )$. Then

$$\begin{aligned} \bigl\vert g^{(k)} \bigl(x \bigl(I_{0}(h), \theta \bigr) \bigr) \bigl(\partial _{H}x \bigl(I_{0}(h), \theta \bigr) \bigr)^{k+1} \bigr\vert \leq \epsilon (H) H^{\frac{3}{4}-(k+1)} \end{aligned}$$

(2.15)

for $k=0,1$ and $\theta \in \mathbb{S}^{1}\setminus \Xi$.

Proof

When $x>0$, by a direct computation we have

$$\begin{aligned} & g(x)\partial _{H}x \\ &\quad=g \biggl(\sqrt{2h}\sin \biggl(\frac{T_{0}(h)}{\pi}\theta - \frac{T_{-}(h)}{2} \biggr) \biggr)\frac{1}{\sqrt{2h}} \biggl(\frac{1}{\pi}+ \partial _{H} u \biggr)\sin \biggl(\frac{T_{0}(h)}{\pi}\theta - \frac{T_{-}(h)}{2} \biggr) \\ &\qquad{} + g \biggl(\sqrt{2h}\sin \biggl(\frac{T_{0}(h)}{\pi}\theta - \frac{T_{-}(h)}{2} \biggr) \biggr)\sqrt{2h}\cos \biggl(\frac{T_{0}(h)}{\pi} \theta -\frac{T_{-}(h)}{2} \biggr) \\ &\qquad{} \times \biggl[\frac{1}{\pi}T_{0}^{\prime }(h) \theta + \frac{T_{-}'(h)}{2} \biggr]\frac{1}{\pi}+\partial _{H}u) \end{aligned}$$

and

$$\begin{aligned} &g'(x) (\partial _{H}x)^{2} \\ &\quad=g' \biggl(\sqrt{2h}\sin \biggl(\frac{T_{0}(h)}{\pi}\theta - \frac{T_{-}(h)}{2} \biggr) \biggr)\frac{1}{2h} \biggl(\frac{1}{\pi}+ \partial _{H} u \biggr)^{2} \sin ^{2} \biggl( \frac{T_{0}(h)}{\pi}\theta -\frac{T_{-}(h)}{2} \biggr) \\ &\qquad{} + g' \biggl(\sqrt{2h}\sin \biggl(\frac{T_{0}(h)}{\pi} \theta - \frac{T_{-}(h)}{2} \biggr) \biggr)2h\cos ^{2} \biggl( \frac{T_{0}(h)}{\pi} \theta -\frac{T_{-}(h)}{2} \biggr) \\ &\qquad{} \times \biggl[\frac{1}{\pi}T_{0}^{\prime }(h) \theta + \frac{T_{-}'(h)}{2} \biggr]^{2} \biggl(\frac{1}{\pi}+\partial _{H}u \biggr)^{2} \\ &\qquad{} +2g' \biggl(\sqrt{2h}\sin \biggl(\frac{T_{0}(h)}{\pi} \theta - \frac{T_{-}(h)}{2} \biggr) \biggr) \biggl(\frac{1}{\pi}+\partial _{H} u \biggr)^{2} \sin \biggl(\frac{T_{0}(h)}{\pi}\theta - \frac{T_{-}(h)}{2} \biggr) \\ &\qquad{}\times \cos \biggl(\frac{T_{0}(h)}{\pi}\theta - \frac{T_{-}(h)}{2} \biggr) \cdot \biggl[\frac{1}{\pi}T_{0}^{\prime }(h)\theta + \frac{T_{-}'(h)}{2} \biggr]. \end{aligned}$$

From (2.3), Lemma 2.1, and Lemma 2.3 we obtain (2.15). When $x<0$, using Lemma 2.2, we acquire that (2.15) holds. The proof is finished. □

Lemma 2.5

For all $\epsilon >0$, t, and $\theta \in \mathbb{S}^{1}\setminus \Xi$, as $H\rightarrow +\infty$, we have

$$\begin{aligned} & \bigl\vert \partial _{H}^{k}R_{2}(H,t,\theta ) \bigr\vert \leq \epsilon (H) H^{ \frac{3}{4}-k} \quad \textit{for } k\leq 2, \\ & \bigl\vert \partial _{H}^{k}\partial _{t}^{l}R_{2}(H,t,\theta ) \bigr\vert \leq C H^{ \frac{1}{2}-k}\quad \textit{for } k+l\leq 2, l\geq 1. \end{aligned}$$

Proof

(i) When $k+l=0$, the conclusion follows from (1.6), (2.2), (2.11), and Lemma 2.2.

(ii) When $k+l=1$, define

$$\begin{aligned} \triangle ={}&1-\partial _{I}x \biggl(I_{0} \biggl( \frac{H}{\pi}+R_{2} \biggr),\theta \biggr)T_{0} \biggl( \frac{H}{\pi}+R_{2} \biggr)p(t) \\ &{}+g(x)\partial _{I}x \biggl(I_{0} \biggl(\frac{H}{\pi}+R_{2} \biggr), \theta \biggr)T_{0} \biggl(\frac{H}{\pi}+R_{2} \biggr). \end{aligned}$$

For $|\triangle |\geq \frac{1}{2}$ and $H\gg 1$, we get

$$\begin{aligned} &\Delta \cdot \partial _{H}R_{2}(H,t,\theta )=- \frac{1}{\pi}( \triangle -1), \end{aligned}$$

(2.16)

$$\begin{aligned} & \Delta \cdot \partial _{t}R_{2}(H,t,\theta )=x \biggl(I_{0} \biggl(\frac{H}{\pi}+R_{2} \biggr), \theta \biggr)p'(t). \end{aligned}$$

(2.17)

Using Lemma 2.2 and (1.6) yields

$$\begin{aligned} \biggl\vert -\frac{1}{\pi}(\triangle -1) \biggr\vert \leq \epsilon (H) H^{\frac{3}{4}-1}. \end{aligned}$$

Applying Lemma 2.2 gives rise to

$$\begin{aligned} \biggl\vert x \biggl(I_{0} \biggl(\frac{H}{\pi}+R_{2} \biggr),\theta \biggr)p'(t) \biggr\vert \leq C\sqrt{H}. \end{aligned}$$

Thus $|\partial _{H}R_{2}(H,t,\theta )|\leq \epsilon (H) H^{\frac{3}{4}-1}$ and $|\partial _{t}R_{2}(H,t,\theta )|\leq C H^{\frac{1}{2}}$.

(iii) When $k+l=2$, differentiating both sides of (2.16) with respect to H and t, respectively, we acquire

$$\begin{aligned} &\partial _{H}\triangle \cdot \partial _{H}R_{2}+ \triangle \partial _{H}^{2}R_{2}=- \frac{1}{\pi}\partial _{H}\triangle, \\ &\partial _{t}\triangle \cdot \partial _{H}R_{2}+ \triangle \partial _{H} \partial _{t}R_{2}=- \frac{1}{\pi}\partial _{t}\triangle. \end{aligned}$$

Differentiating both sides of (2.17) about t gives

$$\begin{aligned} \partial _{t}\triangle \cdot \partial _{t}R_{2}+ \triangle \partial _{t}^{2}R_{2}={}& \partial _{I}x \biggl(I_{0} \biggl(\frac{H}{\pi}+R_{2} \biggr),\theta \biggr)T_{0} \biggl(\frac{H}{\pi}+R_{2} \biggr) \partial _{t}R_{2}p'(t)\\ &{}+x \biggl(I_{0} \biggl(\frac{H}{\pi}+R_{2} \biggr),\theta \biggr)p''(t). \end{aligned}$$

A direct computation yields

$$\begin{aligned} \partial _{H}\Delta ={}&\partial _{I}^{2}x \biggl(I_{0} \biggl(\frac{H}{\pi}+R_{2} \biggr), \theta \biggr)T_{0}^{2} \biggl(\frac{H}{\pi}+R_{2} \biggr) \biggl(\frac{1}{\pi}+\partial _{H}R_{2} \biggr)p(t) \\ &{}+\partial _{I}x \biggl(I_{0} \biggl(\frac{H}{\pi}+R_{2} \biggr),\theta \biggr)\partial _{H}T_{-} \biggl( \frac{H}{\pi}+R_{2} \biggr) \biggl(\frac{1}{\pi}+\partial _{H}R_{2}(H,t,\theta ) \biggr)p(t) \\ &{}+g' \biggl(x \biggl(I_{0} \biggl(\frac{H}{\pi}+R_{2} \biggr),\theta \biggr) \biggr) (\partial _{I}x \biggl(I_{0} \biggl( \frac{H}{\pi}+R_{2} \biggr)T_{0} \biggl( \frac{H}{\pi}+R_{2} \biggr) \biggr)^{2} \biggl( \frac{1}{\pi}+ \partial _{H}R_{2} \biggr) \\ &{}+g \biggl(x \biggl(I_{0} \biggl(\frac{H}{\pi}+R_{2} \biggr),\theta \biggr) \biggr) (\partial _{I}^{2}x(I_{0} \biggl( \frac{H}{\pi}+R_{2} \biggr)T_{0}^{2} \biggl(\frac{H}{\pi}+R_{2} \biggr) \biggl(\frac{1}{\pi}+ \partial _{H}R_{2} \biggr) \\ &{}+g \biggl(x \biggl(I_{0} \biggl(\frac{H}{\pi}+R_{2} \biggr),\theta \biggr) \biggr) (\partial _{I}x(I_{0} \biggl( \frac{H}{\pi}+R_{2} \biggr)\partial _{H}T_{-} \biggl(\frac{H}{\pi}+R_{2} \biggr) \biggl( \frac{1}{\pi}+ \partial _{H}R \biggr). \end{aligned}$$

From Lemmas 2.1–2.5 we have $| \partial _{H}\Delta |<\epsilon (H) H^{-\frac{5}{4}}$ and $|\partial _{H}R_{2}(H,t,\theta )|<\epsilon (H)H^{\frac{3}{4}-1}$.

Similarly to the above estimates, we have

$$\begin{aligned} & \bigl\vert \partial _{H}^{2}R_{2}(H,t,\theta ) \bigr\vert < \epsilon (H) H^{\frac{3}{4}-2},\qquad \bigl\vert \partial _{H}\partial _{t}R_{2}(H,t,\theta ) \bigr\vert < CH^{\frac{1}{2}-1}, \\ & \bigl\vert \partial _{t}^{2}R_{2}(H,t, \theta ) \bigr\vert < C H^{\frac{1}{2}-1}. \end{aligned}$$

The proof is finished. □

Using (2.14) and (2.15), similarly to the proof of Lemma 2.5, we acquire the conclusion.

Lemma 2.6

For all $\epsilon >0$, t, and $\theta \in \mathbb{S}^{1}\setminus \Xi$, as $H\rightarrow +\infty$, we have

$$\begin{aligned} \bigl\vert \partial _{H}^{k}R_{1}(H, \theta ) \bigr\vert \leq \epsilon (H) H^{ \frac{3}{4}-k} \quad\textit{for } k\leq 2. \end{aligned}$$

(2.18)

Next we rewrite (2.5) with new variables as a nearly integrable system. To handle this process, we apply

$$\begin{aligned} & R_{12}=R_{2}(H,t,\theta )-R_{1}(H,\theta ), \qquad R_{\mu}=\mu R_{2}(H,t, \theta )+(1-\mu )R_{1}(H, \theta ), \\ & h_{1}=\frac{H}{\pi}+R_{\mu}(H,t,\theta ), \qquad h_{2}= \frac{H}{\pi}+\nu R_{2}(H,t,\theta ) \end{aligned}$$

to express

$$\begin{aligned} R(H,t,\theta )=I_{0} \biggl(\frac{H}{\pi}+R_{2}(H,t, \theta ) \biggr)-I_{0} \biggl(\frac{H}{\pi}+R_{1}(H, \theta ) \biggr)-\pi x \biggl(I_{0} \biggl(\frac{H}{\pi} \biggr), \theta \biggr)p(t). \end{aligned}$$

(2.19)

Thus we obtain

$$\begin{aligned} &R(H,t,\theta ) \\ &\quad =\pi (R_{2}-R_{1})+I_{-} \biggl( \frac{H}{\pi}+R_{2} \biggr)-I_{-} \biggl( \frac{H}{\pi}+R_{1}(H,\theta ) \biggr)-\pi x \biggl(I_{0} \biggl(\frac{H }{\pi} \biggr),\theta \biggr)p(t) \\ &\quad =\pi \biggl[x \biggl(I_{0} \biggl(\frac{H}{\pi}+R_{2} \biggr),\theta \biggr)-x \biggl(I_{0} \biggl( \frac{H}{\pi} \biggr), \theta \biggr) \biggr]p(t)+\pi \biggl[G \biggl(x \biggl(I_{0} \biggl( \frac{H}{\pi}+R_{1} \biggr), \theta \biggr) \biggr) \\ &\qquad{} - G \biggl(x \biggl(I_{0} \biggl(\frac{H}{\pi}+R_{2} \biggr),\theta \biggr) \biggr) \biggr]+I_{-} \biggl( \frac{H}{\pi}+R_{2} \biggr)-I_{-} \biggl( \frac{H}{\pi}+R_{1}(H,\theta ) \biggr) \\ &\quad =R^{1}(H,t,\theta )+R^{2}(H,t,\theta )+R^{3}(H,t, \theta ), \end{aligned}$$

where

$$\begin{aligned} &R^{1}=\pi \biggl[ \int _{0}^{1}\partial _{I}x \bigl(I_{0}(h_{2}),\theta \bigr)T_{0}(h_{2})R_{2} \,d \nu \biggr]p(t), \\ &R^{2}=\pi \int _{0}^{1}g \bigl(x \bigl(I_{0}(h_{1}), \theta \bigr) \bigr)\partial _{I}x \bigl(I_{0}(h_{1}), \theta \bigr)T_{0}(h_{1})R_{12}\,d\mu, \\ &R^{3}= \int _{0}^{1}T_{-}(h_{1})R_{12} \,d\mu. \end{aligned}$$

Lemma 2.7

For all $\epsilon >0$, t, and $\theta \in \mathbb{S}^{1}\setminus \Xi$, as $H\rightarrow +\infty$, the function $R(H,t,\theta )$ possesses the property

$$\begin{aligned} \bigl\vert \partial _{H}^{k}\partial _{t}^{l}R(H,t,\theta ) \bigr\vert \leq \epsilon (H) H^{\frac{1}{4}-k} \quad\textit{for } k+l\leq 1. \end{aligned}$$

(2.20)

Proof

(i) When $k+l=0$, using Lemmas 2.1 and 2.5 and (2.18), we obtain (2.20).

(ii) When $k=1$ and $l=0$,

$$\begin{aligned} \partial _{H}R^{1}={}&\pi \biggl[ \int _{0}^{1}\partial _{I}^{2}x \bigl(I_{0}(h_{2}), \theta \bigr)T_{0}^{2}(h_{2}) \biggl(\frac{1}{\pi}+\nu \partial _{H}R_{2} \biggr)R_{2}\,d \nu \biggr] p(t) \\ &{}+\pi \biggl[ \int _{0}^{1}\partial _{I}x \bigl(I_{0}(h_{2}),\theta \bigr)T_{0}'(h_{2}) \biggl( \frac{1}{\pi}+\nu \partial _{H}R_{2} \biggr)R_{2}\,d\nu \biggr] p(t) \\ &{}+\pi \biggl[ \int _{0}^{1}\partial _{I}x \bigl(I_{0}(h_{2}),\theta \bigr)T_{0}(h_{2}) \partial _{H}R_{2})\,d\nu \biggr]p (t), \\ \partial _{H}R^{2}={}&\pi \biggl[ \int _{0}^{1}g' \bigl(x \bigl(I_{0}(h_{1}),\theta \bigr) \bigr) \bigl( \partial _{I}x \bigl(I_{0}(h_{1}),\theta \bigr)T_{0}(h_{1}) \bigr)^{2} \biggl( \frac{1}{\pi}+ \partial _{H}R_{\mu} \biggr)R_{12} \,d\mu \biggr] \\ &{}+\pi \biggl[ \int _{0}^{1}g \bigl(x \bigl(I_{0}(h_{1}), \theta \bigr) \bigr)\partial _{I}x \bigl(I_{0}(h_{1}), \theta \bigr)T_{0}'(h_{1}) \biggl( \frac{1}{\pi}+\partial _{H}R_{\mu} \biggr)R_{12} \,d\mu \biggr] \\ &{}+\pi \biggl[ \int _{0}^{1}g \bigl(x \bigl(I_{0}(h_{1}), \theta \bigr) \bigr)\partial _{I}x \bigl(I_{0}(h_{1}), \theta \bigr)T_{0}(h_{1})\partial _{H}R_{12} \,d\mu \biggr] \\ &{}+\pi \biggl[ \int _{0}^{1}g \bigl(x \bigl(I_{0}(h_{1}), \theta \bigr) \bigr)\partial _{I}^{2}x \bigl(I_{0}(h_{1}), \theta \bigr)T_{0}^{2}(h_{1}) \biggl( \frac{1}{\pi}+\partial _{H}R_{\mu} \biggr)R_{12} \,d \mu \biggr], \end{aligned}$$

and

$$\begin{aligned} \partial _{H}R^{3}= \biggl[ \int _{0}^{1}T_{-}'(h_{1}) \biggl(\frac{1}{\pi}+ \partial _{H}R_{\mu} \biggr)R_{12}\,d\mu + \int _{0}^{1}T_{-}(h_{1}) \partial _{H}R_{12}\,d \mu \biggr]. \end{aligned}$$

We derive (2.20) from Lemmas 2.2 and 2.4–2.6.

(iii) When $k=0$ and $l=1$,

$$\begin{aligned} \partial _{t}R^{1}={}&\pi \biggl[ \int _{0}^{1}\partial _{I}^{2}x \bigl(I_{0}(h_{2}), \theta \bigr)T_{0}^{2}(h_{2}) (\nu \partial _{t}R_{2})R_{2}\,d\nu \biggr] p(t) \\ &{}+\pi \biggl[ \int _{0}^{1}\partial _{I}x \bigl(I_{0}(h_{2}),\theta \bigr)T_{0}'(h_{2}) ( \nu \partial _{t}R_{2})R_{2}\,d\nu \biggr] p(t) \\ &{}+\pi \biggl[ \int _{0}^{1}\partial _{I}x \bigl(I_{0}(h_{2}),\theta \bigr)T_{0}(h_{2}) \partial _{t}R_{2}\,d\nu \biggr]p(t) \\ &{}+\pi \biggl[ \int _{0}^{1}\partial _{I}x \bigl(I_{0}(h_{2}),\theta \bigr)T_{0}(h_{2})R_{2} \,d \nu \biggr]p'(t), \\ \partial _{t}R^{2}={}&\pi \biggl[ \int _{0}^{1}g' \bigl(x \bigl(I_{0}(h_{1}),\theta \bigr) \bigr) \bigl[ \partial _{I}x \bigl(I_{0}(h_{1}),\theta \bigr)T_{0}(h_{1}) \bigr]^{2}(\partial _{t}R_{ \mu})R_{12}\,d\mu \biggr] \\ &{}+\pi \biggl[ \int _{0}^{1}g \bigl(x \bigl(I_{0}(h_{1}), \theta \bigr) \bigr)\partial _{I}x \bigl(I_{0}(h_{1}), \theta \bigr)T_{0}'(h_{1}) (\partial _{t}R_{\mu})R_{12}\,d\mu \biggr] \\ &{}+\pi \biggl[ \int _{0}^{1}g \bigl(x \bigl(I_{0}(h_{1}), \theta \bigr) \bigr)\partial _{I}x \bigl(I_{0}(h_{1}), \theta \bigr)T_{0}(h_{1})\partial _{t}R_{12} \,d\mu \biggr] \\ &{}+\pi \biggl[ \int _{0}^{1}g \bigl(x \bigl(I_{0}(h_{1}), \theta \bigr) \bigr)\partial _{I}^{2}x \bigl(I_{0}(h_{1}), \theta \bigr)T_{0}^{2}(h_{1})\partial _{t}R_{\mu}R_{12}\,d\mu \biggr], \end{aligned}$$

and

$$\begin{aligned} \partial _{t}R^{3}= \int _{0}^{1}T_{-}'(h_{1}) \partial _{t}R_{\mu}R_{12}\,d \mu + \int _{0}^{1}T_{-}(h_{1}) \partial _{t}R_{12}\,d\mu. \end{aligned}$$

We obtain inequality (2.20) from Lemmas 2.1–2.2 and 2.4–2.6. □

Now we rewrite (2.5) with the variables $H,t,\theta$. Utilizing (2.19) yields

$$\begin{aligned} I&=I_{0} \biggl(\frac{H}{\pi}+R_{1}(H, \theta ) \biggr)+\pi x \biggl(I_{0} \biggl(\frac{H}{\pi} \biggr), \theta \biggr)p(t)+R(H,t,\theta ) \\ &=I_{0} \biggl(\frac{H}{\pi}+R_{1}(H,\theta ) \biggr)+ \pi x(H,\theta )p(t)+ \tilde{R}(H,t,\theta ), \end{aligned}$$

(2.21)

where

$$\begin{aligned} \tilde{R}(H,t,\theta )=\pi \int _{0}^{1}\partial _{I}x \biggl(H+\mu I_{-} \biggl( \frac{H}{\pi} \biggr),\theta \biggr)p(t)I_{-} \biggl(\frac{H}{\pi} \biggr)\,d\mu +R(H,t,\theta ). \end{aligned}$$

For the new perturbation R̃, from Lemmas 2.1–2.2 and 2.7 we have

$$\begin{aligned} \bigl\vert \partial _{H}^{k}\partial _{t}^{l}\tilde{R}(H,t,\theta ) \bigr\vert \leq \epsilon (H) H^{\frac{1}{4}-k} \quad\text{for } k+l\leq 1. \end{aligned}$$

(2.22)

Lemma 2.8

If (1.6) holds, then

$$\begin{aligned} \lim_{H\rightarrow +\infty}H^{-\frac{1}{4}+k}\Gamma ^{(k)}(H)=0, \quad k=0,1. \end{aligned}$$

(2.23)

Proof

Using (2.12) and (2.13), we get

$$\begin{aligned} &\tau (H)-\pi \\ &\quad = \int _{0}^{\pi}\partial _{H}I_{0} \biggl(\frac{H}{\pi}+R_{1}(H, \theta ) \biggr)\,d\theta -\pi \\ &\quad = \int _{0}^{\pi} \frac{1}{\partial _{I}[\pi h_{0}(I)+\pi G(x(I,\theta ))]}\,d\theta - \pi \\ &\quad = \int _{0}^{\pi} \frac{1}{\frac{1}{1+\pi \partial _{H}R_{1}+\partial _{H}I_{-}(\frac{H}{\pi}+R_{1})}+ g(x)\partial _{I}x[1+\pi \partial _{H}R_{1}+\partial _{H}I_{-}(\frac{H}{\pi}+R_{1})]}\,d \theta -\pi \\ &\quad = \int _{0}^{\pi} \frac{1+\pi \partial _{H}R_{1}+\partial _{H}I_{-}(\frac{H}{\pi}+R_{1})}{1+ g(x)\partial _{I}x[1+\pi \partial _{H}R_{1}+\partial _{H}I_{-}(\frac{H}{\pi}+R_{1})]^{2}}\,d \theta -\pi. \end{aligned}$$

Applying (1.5), (2.2), and the definition of $h_{0}(I)$ gives rise to

$$\begin{aligned} &\Gamma (H) = \sqrt{H} \int _{0}^{\pi} \biggl[ \frac{\pi \partial _{H}R_{1}+\partial _{H}I_{-}(\frac{H}{\pi}+R_{1})- g(x)\partial _{I}x[1+\pi \partial _{H}R_{1}+\partial _{H}I_{-}(\frac{H}{\pi}+R_{1})]^{2}}{1+ g(x)\partial _{I}x[1+\pi \partial _{H}R_{1}+\partial _{H}I_{-}(\frac{H}{\pi}+R_{1})]^{2}} \biggr]\,d \theta. \end{aligned}$$

(2.24)

Thus

$$\begin{aligned} &H^{-\frac{1}{4}}\Gamma (H) \\ &\quad= H^{\frac{1}{4}} \int _{0}^{\pi} \biggl[ \frac{\pi \partial _{H}R_{1}+\partial _{H}I_{-}(\frac{H}{\pi}+R_{1})- g(x)\partial _{I}x[1+\pi \partial _{H}R_{1}+\partial _{H}I_{-}(\frac{H}{\pi}+R_{1})]^{2}}{ 1+g(x)\partial _{I}x[1+\pi \partial _{H}R_{1}+\partial _{H}I_{-}(\frac{H}{\pi}+R_{1})]^{2}} \biggr]\,d \theta. \end{aligned}$$

For $k=0$, we obtain (2.23) from Lemmas 2.1, 2.4, and 2.6. Note that

$$\begin{aligned} \tau '(H)={}& \int _{0}^{\pi} \frac{\pi \partial _{H}^{2}R_{1}+\partial _{H}^{2}I_{-}(\frac{H}{\pi}+R_{1})(\frac{1}{\pi}+\partial _{H}R_{1})}{1+g(x)\partial _{I}x[1+\pi \partial _{H}R_{1}+\partial _{H}I_{-}(\frac{H}{\pi}+R_{1})]^{2}}\,d \theta \\ &{} - \int _{0}^{\pi} \frac{g'(x)(\partial _{I}x)^{2}[1+\pi \partial _{H}R_{1}+\partial _{H}I_{-}(\frac{H}{\pi}+R_{1})]^{4}}{\{1+g(x)\partial _{I}x[1+\pi \partial _{H}R_{1}+\partial _{H}I_{-}(\frac{H}{\pi}+R_{1})]^{2}\}^{2}}\,d \theta \\ &{} - \int _{0}^{\pi} \frac{g(x)(\partial _{I}^{2}x)[1+\pi \partial _{H}R_{1}+\partial _{H}I_{-}(\frac{H}{\pi}+R_{1})]^{4}}{\{1+g(x)\partial _{I}x[1+\pi \partial _{H}R_{1}+\partial _{H}I_{-}(\frac{H}{\pi}+R_{1})]^{2}\}^{2}}\,d \theta \\ &{} - \int _{0}^{\pi} \frac{2[1+\pi \partial _{H}R_{1}+\partial _{H}I_{-}(\frac{H}{\pi}+R_{1})]^{2}g(x)\partial _{I} x[\pi \partial _{H}^{2}R_{1}+\partial _{H}^{2}I_{-}(\frac{H}{\pi}+R_{1})]}{\{1+g(x)\partial _{I}x[1+\pi \partial _{H}R_{1}+\partial _{H}I_{-}(\frac{H}{\pi}+R_{1})]^{2}\}^{2}}\,d \theta. \end{aligned}$$

Using Lemmas 2.1, 2.4, and 2.6, we get $H^{\frac{5}{4}}\tau '(H)\leq \epsilon (H)$. Thus we obtain

$$\begin{aligned} H^{\frac{3}{4}}\Gamma '(H)=\frac{1}{2}H^{-\frac{1}{4}}\Gamma (H)+H^{ \frac{5}{4}}\tau '(H)\leq \epsilon (H). \end{aligned}$$

Hence (2.23) holds for $k=1$. The proof is finished. □

2.2 Canonical transformations

Lemma 2.9

If (1.6) holds, then there exists a canonical transform

$$\begin{aligned} \Psi _{3}:H=\rho,\qquad t=\tau +T(\rho,\theta ) \end{aligned}$$

associated with $T(\rho,\theta +\pi )=T(\rho,\theta )$ such that the transformed system of (2.21) takes the form

$$\begin{aligned} \frac{d\rho}{d\theta}=-\partial _{\tau}\hat{I}(\rho,\tau, \theta ), \qquad\frac{d\tau}{d\theta}=\partial _{\rho}\hat{I}(\rho,\tau, \theta ), \end{aligned}$$

(2.25)

where

$$\begin{aligned} \hat{I}(\rho,\tau,\theta )=J(\rho )+\pi x(\rho,\theta )p(\tau )+ \hat{R}(\rho, \tau,\theta ) \end{aligned}$$

and

$$\begin{aligned} J(\rho )=\frac{1}{\pi} \int _{0}^{\pi}I_{0} \biggl( \frac{\rho}{\pi}+R_{1}( \rho,\theta ) \biggr)\,d\theta. \end{aligned}$$

For the new perturbation R̂ and for all $\epsilon >0$, t, and $\theta \in \mathbb{S}^{1}\setminus \Xi$, if $H\rightarrow +\infty$ and $k+l\leq 1$, then

$$\begin{aligned} \bigl\vert \partial _{\rho}^{k}\partial _{\tau}^{l}\hat{R}(\rho,\tau, \theta ) \bigr\vert \leq \epsilon (\rho ) \rho ^{\frac{1}{4}-k}. \end{aligned}$$

(2.26)

Proof

Define $\Psi _{3}$ implicitly by

$$\begin{aligned} \rho =H+\partial _{\tau}S(H,\tau,\theta ),\qquad t=\tau +\partial _{H}S(H, \tau,\theta ), \end{aligned}$$

where the function $S=S(H,\tau,\theta )$ will be determined later. Using $\Psi _{3}$, (2.21) becomes

$$\begin{aligned} \tilde{I}(\rho,\tau,\theta )=I_{0} \biggl(\frac{H}{\pi}+R_{1}(H, \theta ) \biggr)+ \pi x(H,\theta )p(t)+\tilde{R}(H,t,\theta )- \frac{\partial S}{\partial \theta}. \end{aligned}$$

Now we choose

$$\begin{aligned} S= \int _{0}^{\theta} \biggl[I_{0} \biggl( \frac{H}{\pi}+R_{1}(H,\theta ) \biggr)-J(H) \biggr]\,d\theta. \end{aligned}$$

Therefore $\rho =H$. Assuming that $T(H,\theta )=\partial _{H}S(H,\theta )$, we know that $\Psi _{3}$ takes the form

$$\begin{aligned} H=\rho,\qquad t=\tau +T(\rho,\theta ). \end{aligned}$$

and the function Ĩ reads as

$$\begin{aligned} \hat{I}(\rho,\tau,\theta )=J(\rho )+\pi x(\rho,\theta )p(\tau )+ \hat{R}(\rho, \tau,\theta ), \end{aligned}$$

where

$$\begin{aligned} \hat{R}(\rho,\tau,\theta )=\tilde{R} \bigl(\rho,\tau +T(\rho,\theta ), \theta \bigr)+\pi x(\rho,\theta ) \int _{0}^{1}p' \bigl(\tau +\mu T(\rho, \theta ) \bigr)T(\rho,\theta )\,d\mu. \end{aligned}$$

By a direct computation, (2.26) is derived from Lemmas 2.1, 2.2, and 2.6–2.8. □

3 Proof of main result

Now we introduce a small parameter $\delta >0$ satisfying

$$\begin{aligned} \Psi _{4}:\quad \rho =\delta ^{-2}v,\quad v\in [a,b], \end{aligned}$$

where a and b such that $b>a>0$ do not depend on $\delta >0$.

In the new variables $(v,\tau )$, (2.25) takes the form

$$\begin{aligned} \frac{dv}{d\theta}=-\frac{\partial}{\partial \tau}\hat{I}(v,\tau, \theta, \delta ), \qquad\frac{d\tau}{d\theta}= \frac{\partial}{\partial v}\hat{I}(v,\tau,\theta,\delta ), \end{aligned}$$

(3.1)

where

$$\begin{aligned} \hat{I}(v,\tau,\theta,\delta )=\delta ^{2} \bigl[J \bigl(\delta ^{-2}v \bigr)+\pi x \bigl( \delta ^{-2}v,\theta \bigr)p(\tau )+ \hat{R} \bigl(\delta ^{-2}v,\tau,\theta \bigr) \bigr]. \end{aligned}$$

Denote $\hat{R}(v,\tau,\theta,\delta )=\delta ^{2}\hat{R}(\delta ^{-2}v, \tau,\theta )$. From (2.22) we derive that for $k+l\leq 1$,

$$\begin{aligned} \delta ^{-1} \bigl\vert \partial _{v}^{k} \partial _{\tau}^{l}\hat{R}(v,\tau, \theta,\delta ) \bigr\vert \leq \epsilon \bigl(\delta ^{-2}v \bigr)v^{\frac{1}{4}-k} \quad\text{as } \delta \rightarrow 0^{+}. \end{aligned}$$

(3.2)

Because of

$$\begin{aligned} \tau \bigl(\delta ^{-2}v \bigr)=\pi +\delta v^{-\frac{1}{2}}\Gamma \bigl(\delta ^{-2}v \bigr),\qquad J' \bigl(\delta ^{-2}v \bigr)=\frac{1}{\pi}\tau \bigl(\delta ^{-2}v \bigr), \end{aligned}$$

we write system (3.1) in the form

$$\begin{aligned} \begin{aligned} \textstyle\begin{cases} \frac{dv}{d\theta}=-\pi \delta ^{2}x(\delta ^{-2}v,\theta )p'(\tau )- \partial _{\tau}\hat{R}(v,\tau,\theta,\delta ), \\ \frac{d\tau}{d\theta}=1+\frac{\delta}{\pi}v^{-\frac{1}{2}}\Gamma ( \delta ^{-2}v)+\delta ^{2}\pi \partial _{v}x(\delta ^{-2}v,\theta )p( \tau ) +\partial _{v}\hat{R}(v,\tau,\theta,\delta ). \end{cases}\displaystyle \end{aligned} \end{aligned}$$

(3.3)

Let $(v(\theta,v_{0},\tau _{0}),\tau (\theta,v_{0},\tau _{0}))$ denote the solution of (3.3) associated with the initial data

$$\begin{aligned} \bigl(v(0,v_{0},\tau _{0}),\tau (0,v_{0},\tau _{0}) \bigr)=(v_{0},\tau _{0}). \end{aligned}$$

Utilizing (3.2), we conclude that if $\delta \ll 1$, then a solution of (3.3) exists in $[0,2\pi ]$ for any $(v_{0},\tau _{0})\in [a,b]\times [0,\pi ]$. Moreover,

$$\begin{aligned} 0< \frac{1}{2}a\leq v(\theta,v_{0},\tau _{0})\leq 2b\quad \forall \theta \in [0,2\pi ]. \end{aligned}$$

Assume that the solution $(v(\theta,v_{0},\tau _{0}),\tau (\theta,v_{0},\tau _{0}))$ is of the form

$$\begin{aligned} v(\theta,v_{0},\tau _{0})=v_{0}+\delta F_{2}(\theta,v_{0},\tau _{0})),\qquad \tau ( \theta,v_{0},\tau _{0})=\tau _{0}+\theta +\delta F_{1}( \theta,v_{0},\tau _{0}). \end{aligned}$$

Then the Poincaré map of (3.3), represented by P, possesses the expression

$$\begin{aligned} P(v_{0},\tau _{0})= \bigl(v_{0}+\delta F_{2}(\pi,v_{0},\tau _{0}),\tau _{0}+ \pi +\delta F_{1}(\pi,v_{0},\tau _{0}) \bigr). \end{aligned}$$

Since $(v(\theta,v_{0},\tau _{0}),\tau (\theta,v_{0},\tau _{0}))$ is a solution of (3.3), we acquire

$$\begin{aligned} \begin{aligned} \textstyle\begin{cases} \frac{dF_{1}}{d\theta}=\frac{1}{\pi}(v_{0}+\delta F_{2})^{- \frac{1}{2}}\Gamma (\delta ^{-2}(v_{0}+\delta F_{2})) \\ \phantom{\frac{dF_{1}}{d\theta}=}{} +\pi \delta \partial _{v}x(\delta ^{-2}(v_{0}+ \delta F_{2}),\theta )p(\tau )+\delta ^{-1}\partial _{v}\hat{R}(v, \tau,\theta,\delta ), \\ \frac{dF_{2}}{d\theta}=-\pi \delta x(\delta ^{-2}(v_{0}+\delta F_{2}), \theta )p'(\tau )-\delta ^{-1}\partial _{\tau}\hat{R}(v,\tau,\theta, \delta ). \end{cases}\displaystyle \end{aligned} \end{aligned}$$

(3.4)

From (3.2) and (3.4) we derive that

$$\begin{aligned} \vert F_{1} \vert \leq C,\qquad \vert F_{2} \vert \leq C \end{aligned}$$

uniformly in $\theta \in \mathbb{S}^{1}\setminus \Xi$.

Lemma 2.9, (3.2), (3.3), and (3.4) yield

$$\begin{aligned} F_{1}(\pi,v_{0},\tau _{0})={}& \int _{0}^{\pi} \biggl[\frac{1}{\pi}(v_{0}+ \delta F_{2})^{-\frac{1}{2}}\Gamma \bigl(\delta ^{-2}(v_{0}+ \delta F_{2}) \bigr) \\ &{} +\delta \pi \partial _{v}x \bigl( \delta ^{-2}(v_{0}+ \delta F_{2}),\theta \bigr)p(\tau ) \biggr]\,d\theta +o(1), \\ F_{2}(\pi,v_{0},\tau _{0})={}&{-}\pi \int _{0}^{\pi}\delta x \bigl(\delta ^{-2}(v_{0}+ \delta F_{2}),\theta \bigr)p'(\tau )\,d\theta +o(1). \end{aligned}$$

Thus the Poincaré map P of (3.3) takes the form

$$\begin{aligned} P:\textstyle\begin{cases} \tau _{1}=\tau _{0}+\pi +\delta l_{1}(v_{0},\tau _{0},\delta )+ \delta o(1), \\ v_{1}=v_{0}+\delta l_{2}(v_{0},\tau _{0},\delta )+\delta o(1), \end{cases}\displaystyle \end{aligned}$$

where

$$\begin{aligned} l_{1}(v_{0},\tau _{0},\delta )={}& \int _{0}^{\pi} \biggl[\frac{1}{\pi}(v_{0}+ \delta F_{2})^{-\frac{1}{2}}\Gamma \bigl(\delta ^{-2}(v_{0}+ \delta F_{2}) \bigr) \\ &{} +\delta \pi \partial _{v}x \bigl(\delta ^{-2}(v_{0}+ \delta F_{2}),\theta \bigr)p(\tau ) \biggr]\,d\theta \\ ={}& v_{0}^{-\frac{1}{2}}\Gamma \bigl(\delta ^{-2}v_{0} \bigr)+ \int _{0}^{\pi} \delta \pi \partial _{v}x \bigl(\delta ^{-2}(v_{0}+\delta F_{2}),\theta \bigr)p( \tau )\,d\theta +o(1) \\ ={}& v_{0}^{-\frac{1}{2}}\Gamma \bigl(\delta ^{-2}v_{0} \bigr)+ \int _{0}^{\pi} \delta \pi \partial _{v}x \bigl(\delta ^{-2}v_{0},\theta \bigr)p(\tau _{0}+ \theta )\,d\theta +o(1) \end{aligned}$$

(3.5)

and

$$\begin{aligned} l_{2}(v_{0},\tau _{0},\delta )&=- \pi \int _{0}^{\pi}\delta x \bigl(\delta ^{-2}(v_{0}+ \delta F_{2}),\theta \bigr)p'(\tau )\,d\theta \\ & =-\pi \int _{0}^{\pi}\delta x \bigl(\delta ^{-2}v_{0},\theta \bigr)p'(t_{0}+ \theta )\,d\theta +o(1). \end{aligned}$$

(3.6)

Applying arguments similar to those in [5], we obtain the following estimates:

$$\begin{aligned} &\mathrm{mes} \biggl\{ \theta \in [0,\pi ],x \biggl(\frac{v_{0}}{\delta ^{2}},\theta \biggr)>0 \biggr\} =\pi +\delta O(1), \\ &\mathrm{mes} \biggl\{ \theta \in [0,\pi ],x \biggl(\frac{v_{0}}{\delta ^{2}},\theta \biggr)< 0 \biggr\} =\delta O(1). \end{aligned}$$

When $x<0$, we have that

$$\begin{aligned} \biggl\vert x \biggl(\frac{v_{0}}{\delta ^{2}} \biggr) \biggr\vert =O(1),\qquad \biggl\vert \partial _{I}x \biggl( \frac{v_{0}}{\delta ^{2}} \biggr) \biggr\vert = \delta ^{2}O(1). \end{aligned}$$

When $x>0$, from the definition of θ it follows that

$$\begin{aligned} x \biggl(\frac{v_{0}}{\delta ^{2}},\theta \biggr)=\sqrt{ \frac{2\rho _{0}}{\delta ^{2}\pi}}\sin \theta +O(1), \qquad\partial _{I}x \biggl( \frac{v_{0}}{\delta ^{2}},\theta \biggr)=\sqrt{ \frac{\delta ^{2}}{2\pi \rho _{0}}}\sin \theta +\delta ^{2}O(1). \end{aligned}$$

Thus from (3.5) we obtain

$$\begin{aligned} &l_{1}(v_{0},\tau _{0},\delta ) \\ & \quad=v_{0}^{-\frac{1}{2}}\Gamma \bigl(\delta ^{-2}v_{0} \bigr)+\delta \pi \int _{\{\theta \in [0,\pi ]:x>0\}}\partial _{v} x \bigl(\delta ^{-2}v_{0}, \theta \bigr)p(\tau _{0}+\theta )\,d \theta +o(1) \\ & \quad=v_{0}^{-\frac{1}{2}}\Gamma \bigl(\delta ^{-2}v_{0} \bigr)+\delta \pi \int _{\{\theta \in [0,\pi ]:x>0\}} \delta ^{-2}\sqrt{ \frac{\delta ^{2}}{2\pi v_{0}}} \sin \theta p(\tau _{0}+\theta )\,d \theta +o(1) \\ &\quad =v_{0}^{-\frac{1}{2}}\Gamma \bigl(\delta ^{-2}v_{0} \bigr)+\pi \int _{0}^{ \pi} \sqrt{\frac{1}{2\pi v_{0}}} \sin \theta p(\tau _{0}+\theta )\,d \theta +o(1) \\ &\quad =v_{0}^{-\frac{1}{2}} \biggl[\Gamma \bigl(\delta ^{-2}v_{0} \bigr)+\sqrt{ \frac{\pi}{2}} \int _{0}^{\pi} \sin \theta p(\tau _{0}+ \theta )\,d \theta \biggr]+o(1). \end{aligned}$$

(3.7)

Similarly, we derive from (3.6) that

$$\begin{aligned} l_{2}(v_{0},\tau _{0},\delta )&=- \pi \int _{0}^{\pi}\delta x \bigl(\delta ^{-2}v_{0}, \theta \bigr)p'(\tau _{0}+\theta )\,d\theta +o(1) \\ &=-\pi \delta \int _{\{\theta \in [0,\pi ]:x>0\}} x \bigl(\delta ^{-2}v_{0}, \theta \bigr)p'(\tau _{0}+\theta )\,d\theta +o(1) \\ &=-\pi \delta \sqrt{\frac{2\rho _{0}}{\delta ^{2}\pi}} \int _{\{ \theta \in [0,\pi ]:x>0\}} p'(\tau _{0}+\theta )\sin \theta \,d\theta +o(1) \\ &=-\sqrt{2\pi \rho _{0}} \int _{0}^{\pi}p'(\tau _{0}+ \theta )\sin \theta \,d\theta +o(1). \end{aligned}$$

(3.8)

We conclude that the Poincaré map P reads as

$$\begin{aligned} P:\textstyle\begin{cases} \tau _{1}=\tau _{0}+\pi +\delta l_{1}(v_{0},\tau _{0},\delta ) + \delta o(1), \\ v_{1}=v_{0}+\delta l_{2}(v_{0},\tau _{0},\delta )+\delta o(1). \end{cases}\displaystyle \end{aligned}$$

(3.9)

Using

$$\begin{aligned} \biggl\vert \int _{0}^{\pi}\sin \theta p(\tau _{0}+ \theta )\,d\theta \biggr\vert = \biggl\vert \int _{\tau _{0}}^{\tau _{0}+\pi}p(s) (\sin s\cos \tau _{0}- \cos \theta \sin \tau _{0})\,ds \biggr\vert \leq \vert \bar{p} \vert \end{aligned}$$

(3.10)

and noticing (1.7), we acquire

$$\begin{aligned} \limsup_{H\rightarrow +\infty}\Gamma (H)>\sqrt{ \frac{\pi}{2}} \vert \bar{p} \vert . \end{aligned}$$

(3.11)

Let $\varpi >0$ and

$$\begin{aligned} \varpi \leq \limsup_{H\rightarrow +\infty}\Gamma (H)-\sqrt{ \frac{\pi}{2}} \vert \bar{p} \vert . \end{aligned}$$

Since $\sqrt{H}\Gamma '(H)\rightarrow 0$ as $H\rightarrow +\infty$, there exists a number $\bar{H}>\frac{1}{b}>0$ satisfying

$$\begin{aligned} \bigl\vert \sqrt{H}\Gamma '(H) \bigr\vert \leq \min \biggl\{ \frac{b^{-\frac{1}{2}}\varpi}{4(a^{-1}-b^{-1})},\frac{\varpi}{16} \biggr\} \end{aligned}$$

(3.12)

for $H\geq \bar{H}$. Utilizing (3.11), we can choose a sequence $\{H_{m}^{1} \}_{m=1}^{\infty}$ with $\bar{H}\leq H_{m}^{1}\rightarrow +\infty$ such that

$$\begin{aligned} H_{m+1}^{1}>\frac{b}{a}H_{m}^{1} \end{aligned}$$

(3.13)

and

$$\begin{aligned} \Gamma \bigl(H_{m}^{1} \bigr)> \frac{3}{4}\varpi +\sqrt{\frac{\pi}{2}} \vert \bar{p} \vert . \end{aligned}$$

(3.14)

Take $\delta _{1m}=(bH_{m}^{1})^{-\frac{1}{2}}$. Then we have $\delta _{1m}\rightarrow 0$ as $m\rightarrow +\infty$. It follows from (3.13) and (3.14) that

$$\begin{aligned} \delta _{1m}^{-2} \bigl(a^{-1}-b^{-1} \bigr)=H_{m}^{1} \biggl(\frac{b}{a}-1 \biggr)< H_{m+1}^{1}, \end{aligned}$$

and hence $[\delta _{1m}^{-2}b^{-1},\delta _{1m}^{-2}b^{-1}+(a^{-1}-b^{-1}) \delta _{1m}^{-1}]\subset [\delta _{1m}^{-2}b^{-1},\delta _{1m}^{-2}a^{-1}] \subset [H_{m}^{1},H_{m+1}^{1}]$.

For any $H\in [\delta _{1m}^{-2}b^{-1},\delta _{1m}^{-2}b^{-1}+(a^{-1}-b^{-1}) \delta _{1m}^{-1}]$, we claim that

$$\begin{aligned} \Gamma (H)>\frac{1}{4}\varpi +\sqrt{\frac{\pi}{2}} \vert \bar{p} \vert . \end{aligned}$$

(3.15)

Indeed, suppose that there is $H_{m}^{1*} \in [\delta _{1m}^{-2}b^{-1},\delta _{1m}^{-2}b^{-1}+(a^{-1}-b^{-1}) \delta _{1m}^{-1}]$ such that

$$\begin{aligned} \Gamma \bigl(H_{m}^{1*} \bigr)\leq \frac{1}{4}\varpi +\sqrt{\frac{\pi}{2}} \vert \bar{p} \vert . \end{aligned}$$

Using (3.12), we have

$$\begin{aligned} \frac{1}{2}\varpi \leq {}& \bigl\vert \Gamma \bigl(H_{m}^{1*} \bigr)-\Gamma \bigl(H_{m}^{1} \bigr) \bigr\vert \\ ={}& \bigl\vert \Gamma ' \bigl(H_{m}^{1}+\mu \bigl(H_{m}^{1*}-H_{m}^{1} \bigr) \bigr) \bigr\vert \bigl(H_{m}^{1*}-H_{m}^{1} \bigr) \bigl(\mu \in [0,1] \bigr) \\ ={}& \frac{ \vert \sqrt{H_{m}^{1}+\mu (H_{m}^{1*}-H_{m}^{1})}\Gamma '(H_{m}^{1}+\mu (H_{m}^{1*}-H_{m}^{1})) \vert (H_{m}^{1*}-H_{m}^{1})}{\sqrt{H_{m}^{1}+\mu (H_{m}^{1*}-H_{m}^{1})}} \\ \leq {}& \frac{b^{-\frac{1}{2}}\varpi ((a^{-1}-b^{-1})\delta _{m}^{-1})}{4(a^{-1}-b^{-1})\delta _{m}^{-1}b^{-\frac{1}{2}}}, \\ \leq {}& \frac{\varpi}{4}, \end{aligned}$$

which is a contradiction. Thus (3.15) holds.

Let

$$\begin{aligned} D_{1}= \bigl\{ (\rho,\tau ):\rho \in \bigl[\delta _{1m}^{-2}b^{-1}, \delta _{1m}^{-2}b^{-1}+ \bigl(a^{-1}-b^{-1} \bigr) \delta _{1m}^{-1} \bigr] \bigr\} . \end{aligned}$$

Taking $B_{1}=\{(x,y):\Psi _{4}\Psi _{3}\Psi _{2}\Psi _{1}(x,y)\subset D_{1} \}$ and using the fixed point theorem in Ding [10], we derive that (1.7) in Theorem 1.1 is a consequence of (3.7)–(3.9) and (3.10)–(3.11).

Considering (1.8), we assume that

$$\begin{aligned} \liminf_{H\rightarrow +\infty}\Gamma (H)< -\sqrt{ \frac{2}{\pi}} \vert \bar{p} \vert . \end{aligned}$$

(3.16)

Let $\hat{\omega}>0$ and

$$\begin{aligned} \hat{\omega}\leq -\sqrt{\frac{2}{\pi}} \vert \bar{p} \vert -\liminf _{H \rightarrow +\infty}\Gamma (H). \end{aligned}$$

Since $\sqrt{H}\Gamma '(H)\rightarrow 0$ as $H\rightarrow +\infty$, there exists a number $\hat{H}>\frac{1}{b}>0$ such that

$$\begin{aligned} \bigl\vert \sqrt{H}\Gamma '(H) \bigr\vert \leq \min \biggl\{ \frac{b^{-\frac{1}{2}}\hat{\omega}}{4(a^{-1}-b^{-1})}, \frac{\hat{\omega}}{16} \biggr\} \end{aligned}$$

(3.17)

for $H\geq \hat{H}$. Combining (3.16), we can find a sequence $\{H_{m}^{2} \}_{m=1}^{\infty}$ associated with $\hat{H}\leq H_{m}^{2}\rightarrow +\infty$ satisfying

$$\begin{aligned} H_{m+1}^{2}>\frac{b}{a}H_{m}^{2} \end{aligned}$$

(3.18)

and

$$\begin{aligned} \Gamma \bigl(H_{m}^{2} \bigr)\leq - \frac{3}{4}\hat{\omega}-\sqrt{\frac{2}{\pi}} \vert \bar{p} \vert . \end{aligned}$$

(3.19)

Taking $\delta _{2m}=(bH_{m}^{2})^{-\frac{1}{2}}$, we get $\delta _{2m}\rightarrow 0$ as $m\rightarrow +\infty$. It follows from (3.18) that

$$\begin{aligned} \delta _{2m}^{-2} \bigl(a^{-1}-b^{-1} \bigr)=H_{m}^{2} \biggl(\frac{b}{a}-1 \biggr)< H_{m+1}^{2}, \end{aligned}$$

and hence $H\in [\delta _{2m}^{-2}b^{-1},\delta _{2m}^{-2}b^{-1}+(a^{-1}-b^{-1}) \delta _{2m}^{-1}]\subset [H_{m}^{2},H_{m+1}^{2}]$.

For any $H\in [\delta _{2m}^{-2}b^{-1},\delta _{2m}^{-2}b^{-1}+(a^{-1}-b^{-1}) \delta _{2m}^{-1}]$, we claim that

$$\begin{aligned} \Gamma (H)< -\frac{1}{4}\hat{\omega}-\sqrt{ \frac{2}{\pi}} \vert \bar{p} \vert . \end{aligned}$$

(3.20)

Indeed, assume that there is $H_{m}^{2*}\in [\delta _{2m}^{-2}b^{-1},\delta _{2m}^{-2}b^{-1}+(a^{-1}-b^{-1}) \delta _{2m}^{-1}]$ such that

$$\begin{aligned} \Gamma \bigl(H_{m}^{2*} \bigr)\geq -\frac{1}{4} \varpi -\sqrt{\frac{2}{\pi}} \vert \bar{p} \vert . \end{aligned}$$

Then using (3.17), (3.18), and (3.19) we get

$$\begin{aligned} \frac{1}{2}\hat{\omega}\leq{}& \bigl\vert \Gamma \bigl(H_{m}^{2*} \bigr)-\Gamma \bigl(H_{m}^{2} \bigr) \bigr\vert \\ ={}& \bigl\vert \Gamma ' \bigl(H_{m}^{2}+\mu \bigl(H_{m}^{2*}-H_{m}^{2} \bigr) \bigr) \bigr\vert \bigl(H_{m}^{2*}-H_{m}^{2} \bigr) \bigl(\mu \in [0,1] \bigr) \\ ={}& \frac{ \vert \sqrt{H_{m}^{2}+\mu (H_{m}^{2*}-H_{m}^{2})}\Gamma '(H_{m}^{2}+\mu (H_{m}^{2*}-H_{m}^{2})) \vert (H_{m}^{*}-H_{m}^{2})}{\sqrt{H_{m}^{2}+\mu (H_{m}^{2*}-H_{m}^{2})}} \\ \leq{}& \frac{b^{-\frac{1}{2}}\hat{\omega}((a^{-1}-b^{-1})\delta _{m}^{-1})}{4(a^{-1}-b^{-1})\delta _{m}^{-1}b^{-\frac{1}{2}}} \\ \leq{}& \frac{\hat{\hat{\omega}}}{4}, \end{aligned}$$

which is a contradiction. Thus (3.20) is valid.

Let

$$\begin{aligned} D_{2}= \bigl\{ (\rho,\tau ):\rho \in \bigl[\delta _{2m}^{-2}b^{-1}, \delta _{2m}^{-2}b^{-1}+ \bigl(a^{-1}-b^{-1} \bigr) \delta _{2m}^{-1} \bigr] \bigr\} \end{aligned}$$

and $B_{2}=\{(x,y):\Psi _{4}\Psi _{3}\Psi _{2}\Psi _{1}(x,y)\subset D_{2} \}$.

Now we choose $\{H_{1m_{k}}\}_{1}^{+\infty}$ and $\{H_{2m_{k}}\}_{1}^{+\infty}$ such that $H_{1m_{k}}< H_{1m_{k}+1}< H_{2m_{k}}< H_{2m_{k}+1}$ and let

$$\begin{aligned} D_{3}= \bigl\{ (\rho,\tau ):\rho \in \bigl[\delta _{1m_{k}}^{-2}b^{-1}, \delta _{2m_{k}}^{-2}b^{-1}+ \bigl(a^{-1}-b^{-1} \bigr) \delta _{2m_{k}}^{-1} \bigr] \bigr\} . \end{aligned}$$

Setting $B_{3}=\{(x,y):\Psi _{4}\Psi _{3}\Psi _{2}\Psi _{1}(x,y)\subset D_{3} \}$ and using the twist theorem in Ding [11], we obtain that inequality (1.8) in Theorem 1.1 is a consequence of (3.10), (3.15), (3.7), and (3.20). The proof of Theorem 1.1 is finished.

To verify the given conditions and understand our main result, we give the following remark.

Remark 3.1

Using Lemmas 2.1, 2.4, and 2.6, Eq. (2.24) takes the form

$$\begin{aligned} \Gamma (H)=\sqrt{2}\pi ^{\frac{3}{2}}+\sqrt{H} \int _{0}^{\pi} \bigl(\pi \partial _{H}R_{1}-g(x) \partial _{I}x \bigr)\,d\theta +o(1). \end{aligned}$$

Combining (2.14) with Lemma 2.4, we have $\partial _{H}R_{1}=-g(x)\partial _{I}x+o(\frac{1}{\sqrt{H}})$. Thus we obtain

$$\begin{aligned} \Gamma (H)=\sqrt{2}\pi ^{\frac{3}{2}}-\sqrt{H} \int _{0}^{\pi}(\pi +1)g(x) \partial _{I}x) \,d\theta +o(1). \end{aligned}$$

By the results in [5], $x=\sqrt{\frac{2H}{\pi}}\sin \theta +O(1)$, $\partial _{I}x=\sqrt{\frac{1}{2\pi H}}\sin \theta +O(\frac{1}{H})$, and

$$\begin{aligned} \Gamma (H)=\sqrt{2}\pi ^{\frac{3}{2}}-\frac{1}{\sqrt{2\pi}} \int _{0}^{ \pi}(\pi +1)g \biggl(\sqrt{ \frac{2H}{\pi}}\sin \theta \biggr)\sin \theta \,d\theta +o(1). \end{aligned}$$

For $g(x)$ and $V(x)$ in problem (1.3), if $g(x)$ satisfies (1.6) and $p(t)$ satisfies (1.7), then we know that the equation

$$\begin{aligned} \ddot{x}+V'(x)+\sin \ln \bigl(1+x^{2} \bigr)=p(t) \end{aligned}$$

has at least one π-periodic solutions. Letting $p(t)$ satisfy (1.8), we conclude that the equation

$$\begin{aligned} \ddot{x}+V'(x)+\ln \bigl(1+x^{2} \bigr)\sin \ln \bigl(1+x^{2} \bigr)=p(t) \end{aligned}$$

has an unbounded sequence of π-periodic solutions.