In this section, we will prove the existence of Aubry-Mather sets and quasi-periodic solutions of (1.1) via the Aubry-Mather theory for reversible systems developed by Chow and Pei [8].
In order to fit into the framework of Aubry-Mather theory for reversible systems, we only need to show that the Poincaré map P has the monotone twist property around infinity, i.e.
\(\frac{\partial \theta(2\pi;\theta_{0},I_{0}) }{\partial I_{0}}<0\) if \(I_{0}\gg1\).
In the following, we will investigate the behavior of \(\frac{\partial \theta(2\pi;\theta_{0},I_{0}) }{\partial I_{0}} \) when \(I_{0}\gg1\) by some lemmas.
Similarly, for the sake of convenience in later discussions we write x, y, θ, I instead of \(x(\theta(t;\theta_{0},I_{0}),I(t;\theta_{0},I_{0}))\), \(y(\theta(t;\theta_{0},I_{0}),I(t;\theta_{0},I_{0}))\), \(\theta(t;\theta_{0},I_{0})\), \(I(t;\theta_{0},I_{0})\), respectively.
Lemma 3.1
The following limits exist uniformly on
\(t\in[0,2\pi]\):
-
(i)
\(\frac{x\psi(x)}{I^{\frac{4}{3}}}\rightarrow0\); \(\frac{x^{3}\psi(x)}{I^{{\frac{4}{3}}}}\rightarrow0\); \(\frac{x^{2}\psi'(x)}{I^{{\frac{4}{3}}}}\rightarrow0\), as
\(I_{0}\rightarrow+\infty\);
-
(ii)
\(\frac{y\psi(x)}{I^{{\frac{4}{3}}}}\rightarrow0\); \(\frac{yx\psi'(x)}{I^{{\frac{4}{3}}}}\rightarrow 0\); \(\frac{y^{2}\psi'(x)}{I^{{\frac{4}{3}}}}\rightarrow0\), as
\(I_{0}\rightarrow+\infty\).
Proof
From Remark 1.1, we note that \(\lim_{|x|\to+\infty}\psi'(x)=0\) and \(\lim_{|x|\to+\infty}\frac{\psi(x)}{x}=0\). So, given any \(\varepsilon>0\), there is a positive number \(M=M(\varepsilon)>0\), such that \(|x|\geq M\) imply
$$\bigl|\psi'(x)\bigr|\leq\varepsilon $$
and
$$\bigl|\psi(x)\bigr|\leq\varepsilon|x| $$
for \(\forall t\in [0,2\pi]\).
Let \(K_{1}(\varepsilon)=\max_{|x|\leq M}|\psi(x)|\), \(K_{2}(\varepsilon)=\max_{|x|\leq M}|\psi'(x)|\).
(i) According to (2.3) and Lemma 2.1(iv), one has
$$\begin{aligned}& \begin{aligned}[b] \biggl|\frac{x\psi(x)}{I^{{\frac{4}{3}}}}\biggr|&\leq\frac{MK_{1}(\varepsilon )}{I^{{\frac{4}{3}}}}+ \frac{ \varepsilon x^{2}}{I^{{\frac{4}{3}}}} \leq\frac{MK_{1}(\varepsilon )}{I^{{\frac{4}{3}}}}+\frac{ \varepsilon(\lambda I)^{{\frac{2}{3}}}C^{2}_{\infty}}{I^{{\frac{4}{3}}}} \\ &\leq\frac{MK_{1}(\varepsilon)}{I^{{\frac{4}{3}}}}+\frac{ \varepsilon\lambda^{{\frac{2}{3}}} }{I^{{\frac{2}{3}}}}; \end{aligned} \\& \begin{aligned}[b] \biggl|\frac{x^{3}\psi(x)}{I^{{\frac{4}{3}}}}\biggr|&\leq\frac {M^{3}K_{1}(\varepsilon)}{I^{{\frac{4}{3}}}}+ \frac{ \varepsilon x^{4}}{I^{{\frac{4}{3}}}} \leq\frac{M^{3}K_{1}(\varepsilon )}{I^{{\frac{4}{3}}}}+\frac{ \varepsilon(\lambda I)^{{\frac{4}{3}}}C^{4}_{\infty}}{I^{{\frac{4}{3}}}} \\ &\leq\frac{M^{3}K_{1}(\varepsilon)}{I^{{\frac{4}{3}}}}+\varepsilon C^{4}_{\infty}\lambda ^{{\frac{4}{3}}}; \end{aligned} \\& \begin{aligned}[b] \biggl|\frac{x^{2}\psi'(x)}{I^{\frac{4}{3}}}\biggr|&\leq\frac {M^{2}K_{2}(\varepsilon)}{I^{\frac{4}{3}}}+ \frac{ \varepsilon x^{2}}{I^{\frac{4}{3}}} \leq\frac{M^{2}K_{2}(\varepsilon )}{I^{\frac{4}{3}}}+\frac{ \varepsilon(\lambda I)^{\frac{2}{3}}C^{2}_{\infty}}{I^{\frac{4}{3}}} \\ &\leq\frac{M^{2}K_{2}(\varepsilon)}{I^{\frac{4}{3}}}+\frac{ \varepsilon C^{2}_{\infty}\lambda^{\frac{2}{3}} }{I^{\frac{2}{3}}}. \end{aligned} \end{aligned}$$
Then, by using Lemma 2.3(i), given \(\bar{I}>0\), choose \(I_{0}\) so that \(I_{0}\geq\bar{I}\), provided
$$I(t)^{\frac{4}{3}}> \max \biggl\{ \frac{MK_{1}(\varepsilon)}{\varepsilon};\frac {M^{3}K_{1}(\varepsilon)}{\varepsilon}; \frac{M^{2}K_{2}(\varepsilon )}{\varepsilon};1 \biggr\} , $$
we have
$$\biggl|\frac{x\psi(x)}{I^{\frac{4}{3}}}\biggr|< \varepsilon+\lambda^{\frac {2}{3}} \varepsilon;\qquad \biggl| \frac{x^{3}\psi(x)}{I^{\frac{4}{3}}}\biggr|< \varepsilon+\lambda ^{\frac{4}{3}}C^{4}_{\infty}\varepsilon;\qquad \biggl|\frac{x^{2}\psi'(x)}{I^{\frac{4}{3}}}\biggr|< \varepsilon+\lambda^{\frac {2}{3}}C^{2}_{\infty}\varepsilon. $$
Since \(\varepsilon>0\) is arbitrary the proof of (i) is complete.
(ii) By (2.3), (H3), Lemma 2.1(iv), and \(\psi'(0)=0\), it follows that
$$\begin{aligned}& \biggl|\frac{y\psi(x)}{I^{\frac{4}{3}}}\biggr| \leq \frac{|y|K_{1}(\varepsilon )}{I^{\frac{4}{3}}}+\frac{\varepsilon|y||x|}{I^{\frac{4}{3}}} \leq \frac{(\lambda I)^{\frac{2}{3}}S_{\infty}K_{1}(\varepsilon)}{I^{\frac{4}{3}}}+\frac{\varepsilon\lambda IC_{\infty}S_{\infty}}{I^{\frac{4}{3}}} \\& \hphantom{ \biggl|\frac{y\psi(x)}{I^{\frac{4}{3}}} \biggr|} \leq\frac{\lambda^{\frac{2}{3}}S_{\infty}K_{1}(\varepsilon)}{I^{ \frac{2}{3}}}+\frac{\lambda C_{\infty}S_{\infty}\varepsilon}{I^{ \frac {1}{3}}}; \\& \biggl|\frac{yx\psi'(x)}{I^{\frac{4}{3}}}\biggr|\leq \frac{\mu|y|}{I^{\frac{4}{3}}} \leq \frac{\mu(\lambda I)^{\frac{2}{3}}S_{\infty}}{I^{\frac{4}{3}}}\leq \frac{\mu\lambda^{\frac{2}{3}} S_{\infty}}{I^{ \frac{2}{3}}}; \\& \biggl|\frac{y^{2}\psi'(x)}{I^{\frac{4}{3}}}\biggr|\leq\biggl|\frac{y^{2}\psi '(0)}{I^{\frac{4}{3}}}\biggr|+\frac{y^{2}|x\psi'(x)|}{|x|I^{\frac{4}{3}}} \leq \frac{y^{2}|x\psi'(x)|}{|x|I^{\frac{4}{3}}}, \end{aligned}$$
so we have
$$\begin{aligned}& \biggl|\frac{y^{2}\psi'(x)}{I^{\frac{4}{3}}}\biggr|\leq\frac{y^{2}|x\psi '(x)|}{I^{\frac{4}{3}}(\lambda I)^{\frac{1}{3}} C_{\infty}} \leq\frac{(\lambda I)^{\frac{4}{3}}S^{2}_{\infty}\mu}{I^{\frac{4}{3}}(\lambda I)^{\frac {1}{3}} C_{\infty}} = \frac{\lambda S^{2}_{\infty}\mu}{ I^{\frac{1}{3}} C_{\infty}}\leq\frac{\lambda S^{2}_{\infty}\mu}{ I^{\frac{1}{3}} C_{\infty}}. \end{aligned}$$
Consequently, by using Lemma 2.3(i), given \(\bar{I}>0\), choose \(I_{0}\) so that \(I_{0}\geq \bar{I}\), provided
$$I(t)^{\frac{2}{3}}> \max \biggl\{ \frac{ \lambda^{\frac{2}{3}}S_{\infty}K_{1}(\varepsilon)}{\varepsilon} ; \frac{\mu\lambda^{\frac{2}{3}} S_{\infty}}{\varepsilon} ; \biggl(\frac{\lambda S^{2}_{\infty}\mu}{ C_{\infty}\varepsilon} \biggr)^{2};1 \biggr\} , $$
we have
$$\biggl|\frac{y\psi(x)}{I^{\frac{4}{3}}}\biggr|< (1+\lambda C_{\infty}S_{\infty}) \varepsilon;\qquad \biggl|\frac{yx\psi'(x)}{I^{\frac{4}{3}}}\biggr|< \varepsilon;\qquad \biggl|\frac{y^{2}\psi'(x)}{I^{\frac{4}{3}}}\biggr|< \varepsilon. $$
Since \(\varepsilon>0\) is arbitrary, (ii) is proved. □
Lemma 3.2
The following limits hold uniformly over
\(t\in[0,2\pi]\):
-
(i)
\(\frac{yq'(y)xf(x)}{I^{\frac{4}{3}}}\rightarrow0\); \(\frac{yq(y)f(x)}{I^{\frac{4}{3}}}\rightarrow0\); \(\frac{q(y)f(x)x^{3}}{I^{\frac{4}{3}}}\rightarrow0\), as
\(I_{0}\rightarrow+\infty\);
-
(ii)
\(\frac{yq(y)xf'(x)}{I^{\frac{4}{3}}}\rightarrow0\); \(\frac{q(y)x^{2}f'(x)}{I^{\frac{4}{3}}}\rightarrow0\); \(\frac{y^{2}q(y)f'(x)}{I^{\frac{4}{3}}}\rightarrow0\); \(\frac{ q'(y)f(x)x^{4}}{I^{\frac{4}{3}}}\rightarrow0\), as
\(I_{0}\rightarrow+\infty\).
Proof
(i) By Remark 1.2, Remark 1.3, (2.3), and Lemma 2.1(iv), we know that
$$\begin{aligned}& \begin{aligned}[b] \biggl|\frac{yq'(y)xf(x)}{I^{\frac{4}{3}}}\biggr|&\leq \frac{ab(1+|y|^{\sigma})(1+|x|^{\gamma})}{I^{\frac{4}{3}}} \\ &\leq\frac{ab}{I^{\frac{4}{3}}}+\frac{ab\lambda^{\frac{\gamma }{3}}C_{\infty}^{\gamma} }{I^{\frac{4-\gamma}{3}}}+\frac{ab\lambda ^{\frac{2\sigma}{3}}S_{\infty}^{\sigma} }{I^{\frac{4-2\sigma }{3}}}+ \frac{ab\lambda^{\frac{2\sigma+\gamma}{3}}C_{\infty}^{\gamma} S_{\infty}^{\sigma}}{I^{\frac{4-2\sigma-\gamma}{3}}}; \end{aligned} \\& \begin{aligned}[b] \biggl|\frac{yq(y)f(x)}{I^{\frac{4}{3}}}\biggr| &\leq \frac{ab|y|(1+|y|^{\sigma })(1+|x|^{\gamma})}{I^{\frac{4}{3}}} \\ &\leq\frac{ab\lambda^{\frac{2}{3}}S_{\infty}}{I^{\frac{2}{3}}}+\frac {ab\lambda^{\frac{2+\gamma}{3}}S_{\infty}C_{\infty}^{\gamma} }{I^{\frac {2-\gamma}{3}}}+\frac{ab\lambda^{\frac{2+2\sigma}{3}}S_{\infty}^{\sigma+ 1 } }{I^{\frac{2-2\sigma}{3}}}+ \frac{ab\lambda^{\frac {2+2\sigma+\gamma}{3}}C_{\infty}^{ \gamma} S_{\infty}^{\sigma +1}}{I^{\frac{2-2\sigma-\gamma}{3}}} ; \end{aligned} \\& \begin{aligned}[b] \biggl|\frac{q(y)f(x)x^{3}}{I^{\frac{4}{3}}}\biggr| &\leq\frac{ab(1+|y|^{\sigma })(1+|x|^{\gamma})|x|^{3}}{I^{\frac{4}{3}}} \\ &\leq\frac{ab\lambda C_{\infty}^{3} }{I^{\frac{1}{3}}}+\frac {ab\lambda^{\frac{\sigma+3}{3}}S_{\infty}^{\sigma}C_{\infty}^{3} }{I^{\frac{1-\gamma}{3}}}+\frac{ab\lambda^{\frac{3+\gamma }{3}}C_{\infty}^{3+\gamma} }{I^{\frac{1-2\sigma}{3}}}+ \frac{ab\lambda ^{\frac{2\sigma+\gamma+3}{3}}C_{\infty}^{ \gamma+3} S_{\infty}^{\sigma }}{I^{\frac{1-2\sigma-\gamma}{3}}}. \end{aligned} \end{aligned}$$
Noting that \(0<\gamma<1\), \(0<\sigma<\frac{1}{2} \), and \(2\sigma +\gamma<1\), it follows that the conclusions of (i) are established by virtue of Lemma 2.2 and Lemma 2.3(i) as \(I_{0}\gg1\).
(ii) From \(f'(0)=q'(0)=0\), Remark 1.2, Remark 1.3, (2.3), and Lemma 2.1(iv), we get
$$\begin{aligned}& \begin{aligned}[b] \biggl|\frac{yq(y)xf'(x)}{I^{\frac{4}{3}}}\biggr| &\leq\frac {ab|y|(1+|y|^{\sigma})(1+|x|^{\gamma})}{I^{\frac{4}{3}}} \\ &\leq\frac{ab\lambda^{\frac{2}{3}}S_{\infty}}{I^{\frac{2}{3}}}+\frac {ab\lambda^{\frac{2+\gamma}{3}}S_{\infty}C_{\infty}^{\gamma} }{I^{\frac {2-\gamma}{3}}}+\frac{ab\lambda^{\frac{2+2\sigma}{3}}S_{\infty}^{\sigma+ 1 } }{I^{\frac{2-2\sigma}{3}}}+ \frac{ab\lambda^{\frac {2+2\sigma+\gamma}{3}}C_{\infty}^{ \gamma} S_{\infty}^{\sigma +1}}{I^{\frac{2-2\sigma-\gamma}{3}}} ; \end{aligned} \\& \begin{aligned}[b] \biggl|\frac{q(y)x^{2}f'(x)}{I^{\frac{4}{3}}}\biggr| &\leq\frac{ab(1+|y|^{\sigma })(1+|x|^{\gamma})|x|}{I^{\frac{4}{3}}} \\ &\leq\frac{ab\lambda^{\frac{1}{3}}C_{\infty}}{I}+\frac{ab\lambda ^{\frac{3+\gamma}{3}}C_{\infty}^{1+\gamma} }{I^{\frac{3-\gamma }{3}}}+\frac{ab\lambda^{\frac{1+2\sigma}{3}}C_{\infty}S_{\infty}^{\sigma} }{I^{\frac{3-2\sigma}{3}}}+ \frac{ab\lambda^{\frac {2\sigma+3+\gamma}{3}}C_{\infty}^{1+ \gamma} S_{\infty}^{\sigma }}{I^{\frac{3-2\sigma-\gamma}{3}}}; \end{aligned} \\& \begin{aligned}[b] \biggl|\frac{y^{2}q(y)f'(x)}{I^{\frac{4}{3}}}\biggr| &\leq\frac{|y^{2}q(y) \cdot f' (0)|}{I^{\frac{4}{3}}}+ \frac{|y^{2}q(y)xf'(x)|}{|x|I^{\frac{4}{3}}} \\ &\leq \frac{ab(1+|y|^{\sigma})|y|^{2}(1+|x|^{\gamma})}{|x|I^{\frac{4}{3}}}; \end{aligned} \\& \begin{aligned}[b] \biggl|\frac{ q'(y)f(x)x^{4}}{I^{\frac{4}{3}}}\biggr|&\leq\biggl|\frac{ q'(0)f(x)x^{4}}{I^{\frac{4}{3}}}\biggr|+ \frac{|y q'(y)f(x)x^{4}|}{|y|I^{\frac {4}{3}}} \\ &\leq \frac{ab(1+|y|^{\sigma})(1+|x|^{\gamma})|x|^{4}}{|y|I^{\frac{4}{3}}}, \end{aligned} \end{aligned}$$
then we further obtain
$$\begin{aligned}& \begin{aligned}[b] \biggl|\frac{y^{2}q(y)f'(x)}{I^{\frac{4}{3}}}\biggr| \leq{}& \frac{ab(1+|y|^{\sigma})|y|^{2}(1+|x|^{\gamma})}{\lambda^{\frac {1}{3}} I^{\frac{5}{3}}C_{\infty}} \\ \leq{}&\frac{ab\lambda C_{\infty}^{-1 } S_{\infty}^{2}}{I^{\frac {1}{3}}}+\frac{ab\lambda^{\frac{3+\gamma}{3}}C_{\infty}^{\gamma-1 } S_{\infty}^{2}}{I^{\frac{1-\gamma}{3}}}+\frac{ab\lambda^{\frac {3+2\sigma}{3}}C_{\infty}^{-1 } S_{\infty}^{2+\sigma}}{I^{\frac {1-2\sigma}{3}}}\\ &{}+ \frac{ab\lambda^{\frac{3+2\sigma+\gamma }{3}}C_{\infty}^{ \gamma-1 } S_{\infty}^{2+\sigma}}{I^{\frac{1-2\sigma -\gamma}{3}}}; \end{aligned} \\& \begin{aligned}[b] \biggl|\frac{ q'(y)f(x)x^{4}}{I^{\frac{4}{3}}}\biggr|\leq{}&\frac{ab(1+|y|^{\sigma })(1+|x|^{\gamma})|x|^{4}}{I^{\frac{4}{3}}\lambda^{\frac{2}{3}} I^{\frac{2}{3}}S_{\infty}} \\ \leq{}&\frac{ab\lambda^{\frac{2}{3}}C_{\infty}^{ 4 } S_{\infty}^{-1 }}{I^{\frac{2}{3}}}+\frac{ab\lambda^{\frac{2+\gamma}{3}}C_{\infty}^{ 4+\gamma} S_{\infty}^{-1 }}{I^{\frac{2-\gamma}{3}}}+\frac{ab\lambda ^{\frac{2+2\sigma}{3}}C_{\infty}^{ 4 } S_{\infty}^{\sigma-1 }}{I^{\frac {2-2\sigma}{3}}}\\ &{}+ \frac{ab\lambda^{\frac{2\sigma+2+\gamma }{3}}C_{\infty}^{ 4+\gamma} S_{\infty}^{\sigma-1 }}{I^{\frac{2-2\sigma -\gamma}{3}}}. \end{aligned} \end{aligned}$$
In the same way, by the facts \(0<\gamma<1\), \(0<\sigma<\frac{1}{2} \), and \(2\sigma+\gamma<1\), we can draw the conclusions of (ii) in view of Lemma 2.2 and Lemma 2.3(i) as \(I_{0}\gg1\). □
For any \(t\in[0,2\pi]\), we put
$$\begin{aligned} a_{1}(t) =&\frac{\partial\Phi_{1}}{\partial I} \\ =&\frac{1}{9}\alpha\lambda^{\frac{4}{3}}I^{-\frac{2}{3}}- \frac{ x(2(\psi(x)-p(t))- x\psi'(x))}{9I^{2}} + \frac{ x^{2} f'(x)q(y)}{9I^{2}}+\frac{2 xf(x)( yq'(y)-q(y))}{9I^{2}}; \\ a_{2}(t) =&\frac{\partial\Phi_{1}}{\partial\theta} \\ =&\frac{ y(x \psi'(x)+\psi(x)-p(t))}{3\omega\lambda^{\frac {1}{3}}I^{\frac{4}{3}}}+\frac{ yq(y)(f(x)+x f'(x))}{3\omega\lambda ^{\frac{1}{3}}I^{\frac{4}{3}}} +\frac{ q'(y) f(x)( \alpha{x^{+}}^{4}+\beta{x^{-}}^{4})}{3\omega\lambda ^{\frac{1}{3}}I^{\frac{4}{3}}}; \\ a_{3}(t) =&\frac{\partial\Phi_{2}}{\partial I} \\ =&-\frac{ y(x \psi'(x)+\psi(x)-p(t))}{3\omega\lambda^{\frac {1}{3}}I^{\frac{4}{3}}}-\frac{ yq(y)(f(x)+ x f'(x))+2 y^{2}q'(y)f(x)}{ 3\omega\lambda^{\frac{1}{3}}I^{\frac{4}{3}}}; \\ a_{4}(t) =&\frac{\partial\Phi_{2}}{\partial\theta} \\ =&\frac{(\beta{x^{-}}^{3}-\alpha{x^{+}}^{3})(\psi (x)-p(t)+f(x)q(y)+yf(x)q'(y))}{\omega^{2}( \lambda I)^{\frac{2}{3}}} +\frac{y^{2}(\psi'(x)+f'(x)q(y))}{\omega^{2}( \lambda I)^{\frac{2}{3}}}. \end{aligned}$$
As a consequence of Lemma 2.3(i), Lemma 3.1, and Lemma 3.2, we have the following.
Lemma 3.3
For
\(I_{0}\)
large enough and
\(\forall t,s \in[0,2\pi]\), the following conclusions hold:
-
(i)
\(a_{1}(t)=o(\frac{1}{I^{\frac{1}{3}}_{0}})\);
-
(ii)
\(a_{2}(t)=o(1)\), \(a_{3}(t)=o(1)\);
-
(iii)
\(a_{1}(t)\cdot a_{4}(s)=o(1)\).
For
\(\forall t\in[0,2\pi]\), consider the variational equation of (2.4) with respect to the initial value
\(I_{0}\), we have
$$ \dot{\theta}_{I_{0}}=a_{1}(t) \frac{\partial I}{\partial I_{0}}+ a_{2}(t)\frac{\partial\theta}{\partial I_{0}},\qquad \dot{I}_{I_{0}}= a_{3}(t) \frac{\partial I}{\partial I_{0}}+ a_{4}(t) \frac{\partial\theta}{\partial I_{0}}. $$
(3.1)
Combining the previous estimates, we have the following.
Lemma 3.4
For all
\(t \in(0,2\pi]\), \(I_{0}\rightarrow +\infty\), the following conclusions hold true:
-
(i)
\(\theta_{I_{0}}(t;\theta_{0}, I_{0})\rightarrow0\);
-
(ii)
\(I_{I_{0}}(t;\theta_{0}, I_{0})=1+o(1)\);
-
(iii)
\(\theta_{\theta_{0}}(t;\theta_{0}, I_{0})=1+o(1)\).
Proof
From the variational equations (3.1) and Lemma 3.3, one has
$$\begin{aligned}& \begin{aligned}[b] \theta_{I_{0}}(t)&=e^{\int_{0}^{t}a_{2}(s)\,ds}\int _{0}^{t} e^{-\int _{0}^{s}a_{2}(t)\,dt} a_{1}(s) I_{I_{0}}(s)\,ds \\ &= \bigl(1+o(1) \bigr)\int_{0}^{t}a_{1}(s) I_{I_{0}}(s)\,ds; \end{aligned} \\& \begin{aligned}[b] I_{I_{0}}(t)&=e^{\int_{0}^{t}a_{3}(s)\,ds} \biggl(1+\int _{0}^{t} e^{-\int _{0}^{s}a_{3}(t)\,dt} a_{4}(s) \theta_{I_{0}}(s)\,ds \biggr) \\ &=1+o(1)+ \bigl(1+o(1) \bigr)\int_{0}^{t}a_{4}(s) \biggl(\int_{0}^{s}a_{1}(t) I_{I_{0}}(t)\,dt \biggr)\,ds \\ &=1+o(1)+o(1)\int_{0}^{t}\int _{0}^{s}I_{I_{0}}(t)\,dt\,ds. \end{aligned} \end{aligned}$$
Hence, for all \(t\in(0,2\pi]\), as \(I_{0}\rightarrow+\infty\), we have \(I_{I_{0}}(t)=1+o(1)\) and \(\theta_{I_{0}}(t)=(1+o(1))\int_{0}^{t}a_{1}(s)\,ds\rightarrow0\). Thus, (i) and (ii) are proved.
To verify (iii), considering the variational equation of (2.4) about \(\theta_{0}\), one obtains
$$ \dot{\theta}_{\theta_{0}}=a_{1}(t) \frac{\partial I}{\partial\theta _{0}}+ a_{2}(t)\frac{\partial\theta}{\partial\theta_{0}}, \qquad \dot{I}_{\theta_{0}}= a_{3}(t) \frac{\partial I}{\partial\theta_{0}}+ a_{4}(t)\frac{\partial\theta}{\partial\theta_{0}}. $$
(3.2)
Using similar arguments in (ii), we deduce that \(\theta_{\theta_{0}}(t;\theta_{0}, I_{0})=1+o(1)\) for \(\forall t\in (0,2\pi]\), as \(I_{0}\rightarrow+\infty\). This completes the proof of Lemma 3.4. □
The following lemma gives an estimate of lower bound for \(a_{1}(t)\).
Lemma 3.5
For all
\(t \in[0,2\pi]\)
and
\(I_{0}\)
large enough, there exists a constant
\(L_{d}>0\), such that
$$\bigl|a_{1}(t)\bigr|\geq \frac{L_{d}}{I^{\frac{2}{3}}(t)}. $$
Moreover,
Proof
According to condition (H3), we can assume
$$\bigl|\psi(x)\bigr|\leq\mu|x|+c_{1}, $$
where \(c_{1}>0\) is a constant. Then, by (H3) and (2.3), we have
$$\biggl|\frac{ x(2(\psi(x)-p(t))- x\psi'(x))}{9I^{2}}\biggr|\leq\biggl|\frac{2\mu \lambda^{\frac{2}{3}}C^{2}_{\infty}}{9I^{\frac{4}{3}}}+\frac {(2c_{1}+p_{\infty}+\mu)\lambda^{\frac{1}{3}}C_{\infty}}{9I^{\frac {5}{3}}}\biggr|:=a_{11}(I). $$
Similar to the proof of Lemma 3.2, we have
$$\begin{aligned}& \begin{aligned}[b] \biggl|\frac{ x^{2} f'(x)q(y)}{9I^{2}}\biggr| &\leq\biggl|\frac{ab\lambda^{\frac{1}{3}}C_{\infty}}{9I^{\frac{5}{3}}}+ \frac{ab\lambda^{\frac{3+\gamma }{3}}C_{\infty}^{1+\gamma} }{9I^{\frac{5-\gamma}{3}}}+\frac{ab\lambda ^{\frac{1+2\sigma}{3}}C_{\infty}S_{\infty}^{\sigma} }{9I^{\frac {5-2\sigma}{3}}}+\frac{ab\lambda^{\frac{2\sigma+3+\gamma }{3}}C_{\infty}^{1+ \gamma} S_{\infty}^{\sigma}}{9I^{\frac{5-2\sigma -\gamma}{3}}}\biggr| \\ &:=a_{12}(I); \end{aligned} \\& \begin{aligned}[b] \biggl|\frac{2 xf(x)( yq'(y)-q(y))}{9I^{2}}\biggr|&\leq \biggl|\frac{4ab}{9I^{2}}+ \frac {4ab\lambda^{\frac{\gamma}{3}}C_{\infty}^{\gamma} }{9I^{\frac {6-\gamma}{3}}}+\frac{4ab\lambda^{\frac{2\sigma}{3}}S_{\infty}^{\sigma} }{9I^{\frac{6-2\sigma}{3}}}+\frac{4ab\lambda^{\frac {2\sigma+\gamma}{3}}C_{\infty}^{\gamma} S_{\infty}^{\sigma}}{9I^{\frac {6-2\sigma-\gamma}{3}}}\biggr| \\ &:=a_{13}(I). \end{aligned} \end{aligned}$$
By \(0<\gamma<1\), \(0<\sigma<\frac{1}{2}\), \(2\sigma+\gamma<1\), and Lemma 2.3(i), we can choose \(I_{0}\) large enough so that
$$\frac{\alpha\lambda^{\frac{4}{3}}}{18I^{\frac {2}{3}}}-a_{11}(I)-a_{12}(I)-a_{13}(I) \geq0. $$
Hence
Further,
$$\begin{aligned} \bigl|a_{1}(t)\bigr| =&\biggl| \frac{1}{9}\alpha \lambda^{\frac{4}{3}}I^{-\frac {2}{3}}-\frac{ x(2(\psi(x)-p(t))- x\psi'(x))}{9I^{2}} \\ &{}+\frac{ x^{2} f'(x)q(y)}{9I^{2}}+ \frac{2 xf(x)( yq'(y)-q(y))}{9I^{2}}\biggr| \\ \geq& \frac{\alpha\lambda^{\frac{4}{3}}}{18I^{\frac{2}{3}}}+\frac {\alpha\lambda^{\frac{4}{3}}}{18I^{\frac {2}{3}}}-a_{11}(I)-a_{12}(I)-a_{13}(I) \\ \geq& \frac{\lambda^{\frac{4}{3}}\alpha}{18I^{\frac{2}{3}}(t)}. \end{aligned}$$
Thus, if we take \(L_{d}=\frac{\lambda^{\frac{4}{3}}\alpha}{18}\), we see \(|a_{1}(t)|\geq\frac{L_{d}}{I^{\frac{2}{3}}(t)}\). □
Therefore, combining the above discussions and Lemma 2.3(i), we see that
$$\begin{aligned} \theta_{I_{0}}(2\pi)= \bigl(1+o(1) \bigr)\int _{0}^{2\pi}a_{1}(s)\,ds \geq \bigl(1+o(1) \bigr)\frac{2\pi L_{d}}{(k_{2}I_{0})^{\frac{2}{3}}}>0 \end{aligned}$$
if \(I_{0}\) large enough.
Hence, the Poincaré map has the monotone twist property for all \(I_{0}\) large enough. This completes the proof of Theorem 1.1.