The existence of unbounded solutions of asymmetric oscillations in the degenerate resonant case

Abstract

We prove the existence of unbounded solutions of the asymmetric oscillation in the case when each zero of the discriminative function is degenerate. This is the only case that has not been studied in the literature.

1. Introduction

We are concerned with the asymmetric oscillation

(1.1)\begin{equation} x''+ax^+{-}bx^{-}=f(t), \end{equation}

where $x^+=\max \{x,\, 0\},$ $x^{-}=\max \{-x,\, 0\},$ $a,\, b$ are two different positive constants, and $f(t)$ is a $2\pi$-periodic function.

This equation models the suspension bridge [8] and has been widely studied. Fuçik [3] and Dancer [2] studied it in their investigations of boundary value problems associated to equations with ‘jumping nonlinearities’. For recent developments, one can refer to [4, 5, 7, 16] and the references therein.

For Littlewood's boundedness problem of the asymmetric oscillation (1.1), the earliest contribution was due to Ortega [12]. In 1996, he considered the equation

\[ x''+ax^+{-}bx^{-}=1+\varepsilon h(t), \]

where the smooth function $h(t)$ is $2\pi$-periodic. He proved that if $|\varepsilon |$ is sufficiently small, then all solutions are bounded. That is, if $x(t)$ is a solution, then it is defined for all $t\in \mathbb {R}$ and

\[ \sup_{t\in\mathbb{R}}(|x(t)|+|x'(t)|)<{+}\infty. \]

This result is in contrast with the well-known phenomenon of linear resonance that occurs in the case $a=b=n^2.$ For example, for any $\varepsilon \neq 0,$ all solutions of

\[ x''+n^2x=1+\varepsilon\cos(nt) \]

are unbounded.

For the asymmetric oscillation (1.1). Let $\omega _0:=\frac {1}{2}(\frac {1}{\sqrt {a}}+\frac {1}{\sqrt {b}})$.

If $\omega _0\in \mathbb {R}\backslash \mathbb {Q},$ Ortega [14] in 2001 proved the boundedness of all solutions of equation (1.1) under the condition $\int _0^{2\pi }f(t){\rm d}t\neq 0.$

Recently, Hu et al. [6] established an invariant curve theorem and applied it to equation (1.1), then they obtained the boundedness of all solutions with $\omega _0$ satisfying the Diophantine condition, but without the assumption $\int _0^{2\pi }f(t){\rm d}t\neq 0$.

Subsequently, we [10] also proved the boundedness of all solutions of equation (1.1) without the assumption $\int _0^{2\pi }f(t){\rm d}t\neq 0$, but $\omega _0$ is assumed to satisfy an approximation function condition.

If $\omega _0\in \mathbb {Q}$, then there exist two positive integers $m$ and $n$ such that

(1.2)\begin{equation} \omega_0=\frac{m}{n}. \end{equation}

Moreover, $m$ and $n$ are relatively prime.

Denote by $C(t)$ the solution of the ‘homogeneous’ equation

\[ x''+ax^+{-}bx^{-}=0 \]

with the initial conditions $C(0)=1,\, C'(0)=0$. Then it is well known that $C(t)\in C^2(\mathbb {R})$ and can be given explicitly by the formula

\[ C(t)= \begin{array}{@{}ll} \left\{\begin{array}{@{}ll} \cos\sqrt{a}|t|, & 0\leq|t|\leq\dfrac{\pi}{2\sqrt{a}},\\ -\sqrt{\dfrac{a}{b}}\sin\sqrt{b}\left(|t|-\dfrac{\pi}{2\sqrt{a}}\right), & \dfrac{\pi}{2\sqrt{a}}<|t|\leq\dfrac{m}{n}\pi. \end{array}\right. \end{array} \]

Denote the derivative of $C$ by $S=C'$, then $S(t)\in C(\mathbb {R})$ and

1. (1) $C(-t)=C(t),\, ~S(-t)=-S(t)$;

2. (2) $C(t)$ and $S(t)$ are $2\pi \frac {m}{n}$-periodic functions;

3. (3) $S(t)^2+a\,C^+(t)^2+b\,C^{-}(t)^2\equiv a$.

For a given $2\pi$-periodic function $f(t).$ Let

\[ \Phi_f(\theta):=\int_0^{2\pi}C\left(\frac{m}{n}\theta+mt\right)f(mt){\rm d}t, \quad\theta\in\mathbb{R}, \]

and

\[ \mathcal{A}(f):=\{\theta\in\mathbb{R}:~\Phi_f(\theta)=0\}. \]

Then $\Phi _f(\theta )$ is a $2\pi$-periodic function and its derivative is

\[ \Phi_f'(\theta)=\frac{m}{n}\int_0^{2\pi}S\left(\frac{m}{n}\theta+mt\right)f(mt){\rm d}t,\quad\theta\in\mathbb{R}. \]

On the one hand, if $\mathcal {A}=\emptyset,$ Liu [11] in 1999 proved that all solutions of equation (1.1) are bounded. On the other hand, if $\mathcal {A}\neq \emptyset$ and all zeros of $\Phi _f(\theta )$ are non-degenerate, that is, $\Phi _f'(\theta )\neq 0$ for all $\theta \in \mathcal {A},$ Alonso and Ortega [1] in 1998 proved that there exists $R>0$ such that every solution of (1.1) with

\[ x(t_0)^2+x'(t_0)^2>R \]

for some $t_0\in \mathbb {R}$ is unbounded. Particularly, the proof of this result implies that if there is a non-degenerate zero of $\Phi _f(\theta )$, then there exist unbounded solutions of equation (1.1). We remark that the references [11] and [1] assume that $\Phi _f(\theta )\not \equiv 0$.

In 1998, Ortega [13] proposed an example

(1.3)\begin{equation} x''+4x^+{-}x^{-}=\lambda+\cos 4t,\quad\lambda\in\mathbb{R}. \end{equation}

In this example, $\omega _0=3/4$. Hence, the results of [1] and [11] can be applied. If $|\lambda |<1/45,$ then all solutions with large initial conditions are unbounded. If $|\lambda |>1/45,$ then all solutions are bounded.

However, when $|\lambda |=1/45,$ all zeros of $\Phi _f(\theta )$ are degenerate. Therefore, the references [1] and [11] can not be applied to this equation. In 2021, we [9] proved the existence of unbounded solutions of equation (1.3) with $\lambda =\pm 1/45$.

The main idea of [9] is as follows. First, the corresponding Poincaré map in action and angle variables can be expressed by

\[ \left\{ \begin{array}{@{}ll} \theta_1=\theta_0+2m\pi+\dfrac{\displaystyle {1}}{\displaystyle {r_0}}\mu_1(\theta_0)+\dfrac{\displaystyle {1}}{\displaystyle {r_0^2}}k_1(\theta_0)+\dfrac{\displaystyle {1}}{\displaystyle {r_0^3}}h_1(\theta_0)+g_1(\theta_0,r_0), \\ r_1=r_0+\mu_2(\theta_0)+\dfrac{\displaystyle {1}}{\displaystyle {r_0}}k_2(\theta_0)+\dfrac{\displaystyle {1}}{\displaystyle {r_0^2}}h_2(\theta_0)+g_2(\theta_0, r_0). \end{array} \right. \]

Then in equation (1.3) with $\lambda =\pm \frac {1}{45},$ for all $\theta ^*\in \mathcal {A},$ we have $\Phi _f(\theta ^*)=\Phi _f'(\theta ^*)=0.$ Thus, $\mu _1(\theta )$ has only degenerate zeros. However, in these two examples, the function $p(\theta,\, r):=\mu _1(\theta )+\frac {1}{r}k_1(\theta )+\frac {1}{r^2}h_1(\theta )$ has some non-degenerate zeros. Then an invariant set near the zero of $\mu _1(\theta )$ can be found, and each solution starting from this invariant set is unbounded.

Unfortunately, this method to find the invariant set depends on the property that the function $p(\theta,\, r)$ has non-degenerate zero, and cannot deal with the other cases, including that all zeros of $p(\theta,\, r)$ are degenerate, or $p(\theta,\, r)$ has no zero. In this paper, we will obtain the existence of unbounded solutions of equation (1.1) without considering the zero of $p(\theta,\,r).$ More precisely, we will prove

Theorem 1.1 Assume that the resonance condition (1.2) holds, $f(t)$ is a real analytic $2\pi$-periodic function such that $\Phi _f(\theta )\not \equiv 0$ and $\mathcal {A}(f)\neq \emptyset$. Then equation (1.1) has unbounded solutions.

Remark 1.2 When $\Phi _f(\theta )\not \equiv 0$, if $\mathcal {A}(f)=\emptyset$, then all solutions of equation (1.1) are bounded by [11]. If $\mathcal {A}(f)\not =\emptyset$, then equation (1.1) has unbounded solutions by theorem 1.1. Therefore, theorem 1.1 together with the result of [11] completely solves Littlewood's boundedness problem for the asymmetric oscillation (1.1) in the resonance case under the assumption $\Phi _f(\theta )\not \equiv 0$.

In fact, according to the result of Alonso and Ortega in [1], here we only need to consider the situation that for all $\theta ^*\in \mathcal {A}(f),$ $\Phi _f'(\theta ^*)=0$. The main idea for proving theorem 1.1 is similar to [15] and as follows. By means of a series of transformations, the original system is transformed into a normal form, for which the twist condition is violated. Then an invariant set will be found, and each solution starting from the invariant set is unbounded.

The rest of this paper is organized in 4 sections as follows. In § 2, we will give some examples to illustrate the main theorem. Section 3 is devoted to finding the transformations and the normal form (for which the twist condition is violated). Then in § 4, we will give some properties of the discriminative function $\Phi _f(\theta ),$ which is crucial in this paper. Finally, the proof of the existence of unbounded solutions will be given in § 5.

2. Some remarks

We give several examples to illustrate theorem 1.1. The first two examples show that theorem 1.1 is applicable.

Example 2.1 For equation (1.3) with $\lambda =\pm \frac {1}{45}$, the discriminative function takes the form

\[ \Phi_f(\theta)=\left\{ \begin{array}{@{}ll} -\dfrac{4}{45}+\dfrac{4}{45}\cos3\theta, & \lambda=\dfrac{1}{45}, \\ \dfrac{4}{45}+\dfrac{4}{45}\cos3\theta, & \lambda={-}\dfrac{1}{45}. \end{array} \right. \]

In view of theorem 1.1, this equation has unbounded solutions, which is consistent with the result of [9].

Example 2.2 Consider the equation

(2.1)\begin{equation} x''+4x^+{-}x^-{=}\lambda_1+\lambda_2\cos4t+\lambda_3\sin4t. \end{equation}

The discriminative function of this equation takes the form

\[ \Phi_f(\theta)={-}4\lambda_1+\frac{4}{45}\lambda_2\cos3\theta-\frac{4}{45}\lambda_3\sin3\theta, \]

and the results of [1, 11] and theorem 1.1 can be applied.

• • When $\lambda _2=\lambda _3=0,$ the discriminative function is of the form

  \[ \Phi_f(\theta)={-}4\lambda_1. \]
  If $\lambda _1\neq 0,$ then all solutions are bounded. If $\lambda _1=0,$ then equation (2.1) becomes
  \[ x''+4x^+{-}x^-{=}0, \]
  and all solutions are bounded. Thus, when $\lambda _2=\lambda _3=0,$ all solutions of (2.1) are bounded.

• • When $\lambda _2=0,\, ~\lambda _3\neq 0,$ the discriminative function is of the form

  \[ \Phi_f(\theta)={-}4\lambda_1-\frac{4}{45}\lambda_3\sin3\theta. \]
  If $\big |\frac {\lambda _1}{\lambda _3}\big |>\frac {1}{45},$ then all solutions are bounded. If $\big |\frac {\lambda _1}{\lambda _3}\big |<\frac {1}{45},$ then all solutions with large initial conditions are unbounded. If $\big |\frac {\lambda _1}{\lambda _3}\big |=\frac {1}{45},$ then equation (2.1) has unbounded solutions.

• • When $\lambda _2\neq 0,$ the discriminative function is of the form

  \[ \Phi_f(\theta)={-}4\lambda_1+\frac{4}{45}\sqrt{\lambda_2^2+\lambda_3^2}\cos(3\theta+\alpha),~~\alpha=\arctan\frac{\lambda_3}{\lambda_2}. \]
  Thus, if $\Big |\frac {\lambda _1}{\sqrt {\lambda _2^2+\lambda _3^2}}\Big |>\frac {1}{45},$ then all solutions are bounded. If $\Big |\frac {\lambda _1}{\sqrt {\lambda _2^2+\lambda _3^2}}\Big |<\frac {1}{45},$ then all solutions with large initial conditions are unbounded. If $\Big |\frac {\lambda _1}{\sqrt {\lambda _2^2+\lambda _3^2}}\Big |=\frac {1}{45},$ then equation (2.1) has unbounded solutions.

Finally, for equation (1.1), when $\Phi _f(\theta )\equiv 0,$ there are no results which can be applied to determine the boundedness of its solutions. The following examples show that this situation can indeed happen.

Example 2.3 Consider the equation

\[ x''+ax^+{-}bx^-{=}\cos(rnt), \]

where the resonance condition (1.2) holds, $a\neq b$ and $r$ is a positive integer. Then the discriminative function is

\[ \Phi_f(\theta)=\frac{2\sqrt{a}(b-a)n}{m(r^2n^2-a)(r^2n^2-b)}\cos\left(\frac{rn\pi}{2\sqrt{a}}\right)\cos(rm\theta),\quad r^2n^2\neq a,b, \]

and it is easy to see that when $\frac {rn}{\sqrt {a}}$ is odd, we have $\Phi _f(\theta )\equiv 0.$

Example 2.4 Consider the equation

\[ x''+4x^+{-}x^-{=}\cos kt, \]

where $k$ is a positive integer. Then the discriminative function is

\[ \Phi_f(\theta)=\left\{ \begin{array}{@{}ll} 0, & k=1,2, \\ \dfrac{-4}{(k^2-4)(k^2-1)}(1+({-}1)^k) \left(\cos\left(\dfrac{k\pi}{4}+\dfrac{3k}{4}\theta\right) +\cos\left(\dfrac{3k\pi}{4}+\dfrac{3k}{4}\theta\right)\right), & k\geq 3. \end{array} \right. \]

Thus, if $k=4r,\, r=1,\,2,\,3,\,\ldots,$ then

\[ \Phi_f(\theta)=\frac{({-}1)^{r+1}4}{(4r^2-1)(16r^2-1)}\cos(3r\theta), \]

and if $k\neq 4r,\, r=1,\,2,\,3,\,\ldots,$ then $\Phi _f(\theta )\equiv 0.$

3. Transformations

We make a series of canonical transformations to obtain a normal form for which the twist condition is violated.

3.1 Action and angle variables

Let $y=x'$, then equation (1.1) is equivalent to the following planar system

(3.1)\begin{equation} \left\{ \begin{array}{@{}ll} x'=y, \\ y'={-}ax^+{+}bx^-{+}f(t). \end{array} \right. \end{equation}

The following result is standard.

Lemma 3.1 For any $(x_0,\, y_0)\in \mathbb {R}^2$ and $t_0\in \mathbb {R},$ the unique solution $z(t)=(x(t; t_0,\, x_0,\, y_0),\, y(t; t_0,\, x_0,\, y_0))$ of (3.1) satisfying $z(t_0)=(x_0,\, y_0)$ exists on the whole $t$-axis.

For $r>0,\, \theta \,(\text {mod}~2\pi ),$ define the following generalized polar coordinates $\Gamma :(r,\, \theta )\rightarrow (x,\, y)$ by

\[ \left\{ \begin{array}{@{}ll} x=\rho r^{\frac{1}{2}}C(\dfrac{m}{n} \theta),\\ y=\rho r^{\frac{1}{2}}S(\dfrac{m}{n}\theta), \end{array} \right. \]

where $\rho :=\sqrt {\frac {2n}{am}}$. It is easy to check that $\Gamma$ is a symplectic transformation.

The Hamiltonian associated to the system (3.1) is expressed in Cartesian coordinates by

\[ H(x, y, t)=\frac{\displaystyle {1}}{\displaystyle {2}}y^2+\frac{\displaystyle {a}}{\displaystyle {2}}(x^+)^2+\frac{\displaystyle {b}}{\displaystyle {2}}(x^-)^2-f(t)x. \]

In the new coordinates $(r,\, \theta ),$ it becomes

(3.2)\begin{equation} H(r, \theta, t)=\frac{n}{m}r-\rho r^{\frac{1}{2}}C\left(\frac{\displaystyle {m}}{\displaystyle {n}}\theta\right)f(t). \end{equation}

Thus, the system (3.1) is transformed into

(3.3)\begin{equation} \left\{ \begin{array}{@{}ll} \theta'=H_{r}=\dfrac{\displaystyle {n}}{\displaystyle {m}}-\dfrac{\displaystyle {1}}{\displaystyle {2}}\rho r^{-\dfrac{1}{2}}C\left(\dfrac{\displaystyle {m}}{\displaystyle {n}}\theta\right)f(t), \\ r'={-}H_{\theta}=\dfrac{\displaystyle {m}}{\displaystyle {n}}\rho r^{\dfrac{1}{2}}S\left(\dfrac{\displaystyle {m}}{\displaystyle {n}}\theta\right)f(t), \end{array} \right. \end{equation}

which is a semilinear system.

3.2 A sublinear system

First, introduce a new time variable $\vartheta$ by $t=m\vartheta$ to eliminate the denominator of the linear part of (3.2). Then, the system (3.3) is transformed into

(3.4)\begin{equation} \left\{ \begin{array}{@{}ll} \dfrac{\displaystyle {{\rm d}\theta}}{\displaystyle {{\rm d}\vartheta}}=\dfrac{\displaystyle {\partial}}{\displaystyle {\partial r}}\mathcal{H}(r,\theta, \vartheta), \\ \dfrac{\displaystyle {{\rm d}r}}{\displaystyle {{\rm d}\vartheta}}={-}\dfrac{\displaystyle {\partial}}{\displaystyle {\partial\theta}}\mathcal{H}(r,\theta, \vartheta), \end{array} \right. \end{equation}

where

(3.5)\begin{equation} \mathcal{H}(r,\theta, \vartheta)=nr-m\rho r^{\frac{1}{2}}C\left(\frac{m}{n}\theta\right)f(m\vartheta). \end{equation}

For convenience, we can rewrite (3.5) and (3.4) respectively as

(3.6)\begin{equation} H(r, \theta, t)=nr-m\rho r^{\frac{1}{2}}C\left(\frac{m}{n}\theta\right)f(mt), \end{equation}

and

\[ \left\{ \begin{array}{@{}ll} \dfrac{\displaystyle {{\rm d}\theta}}{\displaystyle {{\rm d}t}}=\dfrac{\displaystyle {\partial}}{\displaystyle {\partial r}}H(r,\theta, t), \\ \dfrac{\displaystyle {{\rm d}r}}{\displaystyle {{\rm d}t}}={-}\dfrac{\displaystyle {\partial}}{\displaystyle {\partial\theta}}H(r,\theta, t). \end{array} \right. \]

Since $m$ is a positive integer, then the new Hamiltonian $H(r,\, \theta,\, t)$ in (3.6) is $2\pi$-periodic in $\theta$ and $t.$

Next we introduce a rotation transformation to eliminate the linear part of the Hamiltonian (3.6), which helps us to obtain a sublinear system.

Define the rotation transformation $\Phi _1: (r_1,\, \theta _1,\, t)\rightarrow (r,\, \theta,\, t)$ by

\[ \left\{ \begin{array}{@{}ll} \theta=\theta_1+nt, \\ r=r_1. \end{array} \right. \]

Then under $\Phi _1,$ the original semilinear system determined by the Hamiltonian (3.6) is transformed into a sublinear system given by the following Hamiltonian

(3.7)\begin{equation} H_1(r_1, \theta_1, t)={-}m\rho r_1^{\frac{1}{2}}C\left(\frac{m}{n}\theta_1+mt\right)f(mt). \end{equation}

It is worth to point out that the above transformations preserve the periodicity and boundedness of solutions. In fact, if $(r_1(t+2\pi ),\, \theta _1(t+2\pi ))=(r_1(t),\, \theta _1(t))$, then for the Hamiltonian (3.5), $r(\vartheta +2\pi )=r_1(\vartheta +2\pi )=r_1(\vartheta )=r(\vartheta )$, and $\theta (\vartheta +2\pi )=\theta _1(\vartheta +2\pi )+n(\vartheta +2\pi )=\theta _1(\vartheta )+n\vartheta +2n\pi =\theta (\vartheta )+2n\pi$. Since $t=m\vartheta,$ for the Hamiltonian (3.2), we have $r(\frac {1}{m}t+2\pi )=r(\frac {1}{m}t),$ and $\theta (\frac {1}{m}t+2\pi )=\theta (\frac {1}{m}t)+2n\pi.$ Thus for the original system (3.1), $x(\frac {1}{m}t+2\pi )=\rho r(\frac {1}{m}t+2\pi )^{\frac {1}{2}}C(\frac {m}{n}\theta (\frac {1}{m}t+2\pi ))=\rho r(\frac {1}{m}t)^{\frac {1}{2}}C(\frac {m}{n}\theta (\frac {1}{m}t)+2m\pi )=x(\frac {1}{m}t)$, which leads to ${x(t+2\pi )=x(t)}$. Thus, the periodicity is preserved. Similarly, it is easy to verify that the boundedness is also preserved.

3.3 The normal form without the twist condition

To reduce the power of term containing $t$ in the Hamiltonian (3.7), we make the transformation $\Phi _2: (r_2,\, \theta _2,\, t)\rightarrow (r_1,\, \theta _1,\, t)$ given by

\[ \left\{ \begin{array}{@{}ll} \theta_2=\theta_1+\dfrac{\partial S_2}{\partial r_2}(r_2, \theta_1, t), \\ r_1=r_2+\dfrac{\partial S_2}{\partial \theta_1}(r_2, \theta_1, t) \end{array} \right. \]

with the generating function $S_2(r_2,\, \theta _1,\, t)$ determined by

\[ S_2(r_2, \theta_1, t)={-}m\rho r_2^{\frac{1}{2}}\int_0^t[Cf]\left(\frac{m}{n}\theta_1\right)-C\left(\frac{m}{n}\theta_1+ms\right)f(ms){\rm d}s, \]

where

\[ [Cf]\left(\frac{m}{n}\theta_1\right)=\frac{1}{2\pi}\int_0^{2\pi}C\left(\frac{m}{n}\theta_1+mt\right)f(mt){\rm d}t. \]

Under $\Phi _2,$ the Hamiltonian $H_1$ in (3.7) is transformed into

\begin{align*} H_2(r_2, \theta_2, t)& ={-}m\rho \left(r_2+\frac{\partial S_2}{\partial \theta_1}\right)^{\frac{1}{2}}C\left(\frac{m}{n}\theta_1+mt\right)f(mt)+\frac{\partial S_2}{\partial t}\\ & ={-}m\rho r_2^{\frac{1}{2}}C\left(\frac{m}{n}\theta_1+mt\right)f(mt)\\ & \quad-m\rho C\left(\frac{m}{n}\theta_1+mt\right)f(mt)\int_0^1\frac{1}{2}\left(r_2+\mu\frac{\partial S_2}{\partial \theta_1}\right)^{-\frac{1}{2}}\frac{\partial S_2}{\partial \theta_1}{\rm d}\mu\\ & \quad+\frac{\partial S_2}{\partial t}. \end{align*}

It is obvious that

\[{-}m\rho r_2^{\frac{1}{2}}C\left(\frac{m}{n}\theta_1+mt\right)f(mt)+\frac{\partial S_2}{\partial t}={-}m\rho r_2^{\frac{1}{2}}[Cf]\left(\frac{m}{n}\theta_1\right), \]

and thus,

\begin{align*} H_2(r_2, \theta_2, t)=& -m\rho r_2^{\frac{1}{2}}[Cf]\left(\frac{m}{n}\theta_1\right)\\ & \quad-m\rho C\left(\frac{m}{n}\theta_1+mt\right)f(mt)\int_0^1\frac{1}{2}\left(r_2+\mu\frac{\partial S_2}{\partial \theta_1}\right)^{-\frac{1}{2}}\frac{\partial S_2}{\partial \theta_1}{\rm d}\mu. \end{align*}

Then by $\theta _2=\theta _1+\frac {\partial S_2}{\partial r_2}(r_2,\, \theta _1,\, t),$ we get

(3.8)\begin{equation} H_2(r_2, \theta_2, t)={-}m\rho r_2^{\frac{1}{2}}[Cf]\left(\frac{m}{n}\theta_2\right)+P_1(\theta_2, t)+P_2(r_2, \theta_2, t), \end{equation}

where

\begin{align*} P_1(\theta_2,t)& ={-}\frac{1}{4\pi}m^3n^{{-}1}\rho^2\int_0^{2\pi}S\left(\frac{m}{n}\theta_2+mt\right)f(mt)\int_0^t[Cf]\left(\frac{m}{n}\theta_2\right)\\& \quad -C\left(\frac{m}{n}\theta_2+ms\right)f(ms){\rm d}s{\rm d}t\\ & \quad+\frac{1}{2}m^3n^{{-}1}\rho^2 C\left(\frac{m}{n}\theta_2+mt\right)f(mt)\int_0^t[Sf]\left(\frac{m}{n}\theta_2\right)\\ & \quad -S\left(\frac{m}{n}\theta_2+ms\right)f(ms){\rm d}s, \end{align*}

and

\begin{align*} & P_2(r_2, \theta_2, t)=\frac{1}{4\pi}m^4n^{{-}2}\rho^2\int_0^{2\pi}S\left(\frac{m}{n}\theta_2+mt\right)f(mt)\int_0^t\\ & \quad\left(\frac{1}{2\pi}\int_0^{2\pi} \int_0^1S\left(\frac{m}{n}\theta_2+ms-\mu\frac{m}{n}\frac{\partial S_2}{\partial r_2}\right)\frac{\partial S_2}{\partial r_2}f(ms){\rm d}\mu {\rm d}s\right.\\ & \qquad\left.-\int_0^1S\left(\frac{m}{n}\theta_2+ms-\mu\frac{m}{n}\frac{\partial S_2}{\partial r_2}\right)\frac{\partial S_2}{\partial r_2}f(ms){\rm d}\mu\right){\rm d}s{\rm d}t\\ & \qquad-\frac{1}{2\pi}m^3n^{{-}2}\rho r_2^{\frac{1}{2}}\int_0^{2\pi}\int_0^1\int_0^1S'\left(\frac{m}{n}\theta_2+mt-s\mu\frac{m}{n}\frac{\partial S_2}{\partial r_2}\right)\mu\left(\frac{\partial S_2}{\partial r_2}\right)^2f(mt){\rm d}s{\rm d}\mu {\rm d}t\\ & \qquad-\frac{1}{2}m^4n^{{-}2}\rho^2C\left(\frac{m}{n}\theta_2+mt\right)f(mt)\\ & \qquad\quad \int_0^t\left(\frac{1}{2\pi}\int_0^{2\pi}\int_0^1S'\left(\frac{m}{n}\theta_2+ms-\mu\frac{m}{n}\frac{\partial S_2}{\partial r_2}\right)\frac{\partial S_2}{\partial r_2}f(ms){\rm d}\mu {\rm d}s\right.\\ & \qquad\left.-\int_0^1S'\left(\frac{m}{n}\theta_2+ms-\mu\frac{m}{n}\frac{\partial S_2}{\partial r_2}\right)\frac{\partial S_2}{\partial r_2}f(ms){\rm d}\mu\right){\rm d}s\\ & \qquad+\frac{1}{2}m^2n^{{-}1}\rho r_2^{-\frac{1}{2}}\int_0^1S\left(\frac{m}{n}\theta_2+mt-\mu\frac{m}{n}\frac{\partial S_2}{\partial r_2}\right)f(mt)\frac{\partial S_2}{\partial r_2}\frac{\partial S_2}{\partial \theta_1}{\rm d}\mu\\ & \qquad+\frac{1}{4}m\rho C\left(\frac{m}{n}\theta_2-\frac{m}{n}\frac{\partial S_2}{\partial r_2}+mt\right)f(mt)\int_0^1\int_0^1\left(r_2+s\mu \frac{\partial S_2}{\partial \theta_1}\right)^{-\frac{3}{2}}\mu\left(\frac{\partial S_2}{\partial \theta_1}\right)^2{\rm d}s{\rm d}\mu \end{align*}

with

\[ [Sf]\left(\frac{m}{n}\theta\right)=\frac{1}{2\pi}\int_0^{2\pi}S\left(\frac{m}{n}\theta+mt\right)f(mt){\rm d}t. \]

Then we have the following estimates, and the proof is elementary.

Lemma 3.2 For $r_2$ large enough, $\theta _2,\, t\in \mathbb {S}^1=\mathbb {R}/(2\pi \mathbb {Z}),$ we have

\[ |\partial_{r_2}^k\partial_{\theta_1}^jS_2(r_2, \theta_1, t)|\leq Cr_2^{\frac{1}{2}-k}, ~~k+j\leq 3, \]

and

\begin{align*} & |\partial_{\theta_2}^jP_1(\theta_2, t)|\leq C,\quad j\leq 1,\\ & |\partial_{r_2}^k\partial_{\theta_2}^jP_2(r_2, \theta_2, t)|\leq Cr_2^{-\frac{1}{2}-k},\quad k+j\leq 1, \end{align*}

where $C$ is a constant larger than 1.

Finally, with the definitions of $\Phi _f(\theta )$ and $[Cf](\frac {m}{n}\theta ),$ $[Sf](\frac {m}{n}\theta )$, the Hamiltonian $H_2(r_2,\, \theta _2,\, t)$ in (3.8) can be rewritten as

(3.9)\begin{equation} H(r, \theta, t)={-}\frac{1}{2\pi}m\rho r^{\frac{1}{2}}\Phi_f(\theta)+P_1(\theta, t)+P_2(r, \theta, t), \end{equation}

and for $r$ large enough, $\theta,\, t\in \mathbb {S}^1,$ one has

\begin{align*} & |\partial_{\theta}^jP_1(\theta, t)|\leq C,\quad j\leq 1,\\ & |\partial_{r}^k\partial_{\theta}^jP_2(r, \theta, t)|\leq Cr^{-\frac{1}{2}-k},\quad k+j\leq 1, \end{align*}

where $C$ is a constant larger than 1.

4. Some properties of $\Phi _f(\theta )$

We present several lemmas for the discriminative function $\Phi _f(\theta )$, which will be used in the proof of the existence of unbounded solutions.

First, we prove that under the assumptions of theorem 1.1, $\Phi _f(\theta )$ is an analytic function, and thus for any $\theta ^*\in \mathcal {A}$, there exists an integer $k\geq 2$ such that ${\Phi _f^{(k)}(\theta ^*)\neq 0}$.

Lemma 4.1 Under the assumptions of theorem 1.1, for any $\theta ^*\in \mathcal {A},$ there exists an integer $k\geq 2$ such that $\Phi _f(\theta ^*)=\Phi _f'(\theta ^*)=\cdots =\Phi _f^{(k-1)}(\theta ^*)=0,$ $\Phi _f^{(k)}(\theta ^*)\neq 0.$

Proof. Since $f$ is real analytic and $2\pi$-periodic in $t$, then it can be written as a uniformly convergent Fourier series

\[ f(t)=\sum_{k\in\mathbb{Z}}f_ke^{ikt}, \quad t\in\mathbb{R}, \]

where the Fourier coefficients

\[ f_k=\frac{1}{2\pi}\int_0^{2\pi}f(t)e^{{-}ikt}{\rm d}t,~~k\in\mathbb{Z}. \]

Moreover, $f$ can be analytically extended into a complex domain ${\{t\in \mathbb {C}:|\mathrm {Im} t|\leq r\}}$, with $r>0$ a small constant, and we have $|f_k|\leq \|f\|_re^{-|k|r},$ where $\|f\|_r=\sup _{|\mathrm {Im} t|\leq r}|f(t)|.$

Then it is obvious that the series

\begin{align*} C\left(\frac{m}{n}\theta+mt\right)\left(\sum_{k\in\mathbb{Z}}f_ke^{ikmt}\right)& =\sum_{k\in\mathbb{Z}}f_kC\left(\frac{m}{n}\theta+mt\right)e^{ikmt}\\ & =\sum_{k\in\mathbb{Z}}f_kC\left(\frac{m}{n}\theta+mt\right)(\cos(kmt)+i\sin(kmt)) \end{align*}

is also uniformly convergent to $C(\frac {m}{n}\theta +mt)f(mt)$. Thus,

\begin{align*} \Phi_f(\theta)& =\int_0^{2\pi}C\left(\frac{m}{n}\theta+mt\right)f(mt){\rm d}t\\ & =\sum_{k\in\mathbb{Z}}f_k\int_0^{2\pi}C\left(\frac{m}{n}\theta+mt\right)\cos(kmt){\rm d}t+i\sum_{k\in\mathbb{Z}}f_k\int_0^{2\pi}C\left(\frac{m}{n}\theta+mt\right)\sin(kmt){\rm d}t\\ & :=\sum_{k\in\mathbb{Z}}f_k\Phi_{ak}(\theta)+i\sum_{k\in\mathbb{Z}}f_k\Phi_{bk}(\theta), \end{align*}

where

\begin{align*} & \Phi_{ak}(\theta)=\int_0^{2\pi}C\left(\frac{m}{n}\theta+mt\right)\cos(kmt){\rm d}t,\\ & \Phi_{bk}(\theta)=\int_0^{2\pi}C\left(\frac{m}{n}\theta+mt\right)\sin(kmt){\rm d}t. \end{align*}

The periodicity of $C$ and $f$ yields that

\begin{align*} \Phi_{ak}(\theta)& =\int_0^{2\pi}C\left(\frac{m}{n}\theta+mt\right)\cos(kmt){\rm d}t=\int_0^{2\pi}C(mt)\cos\left(kmt-k\frac{m}{n}\theta\right){\rm d}t\\ & =\sum_{l=0}^{n-1}\int_{\frac{l}{m}(\frac{\pi}{\sqrt{a}}+\frac{\pi}{\sqrt{b}})}^{\frac{l+1}{m}(\frac{\pi}{\sqrt{a}}+\frac{\pi}{\sqrt{b}})}C(mt)\cos\left(kmt-k\frac{m}{n}\theta\right){\rm d}t, \end{align*}

where

\begin{align*} & \int_{\frac{l}{m}(\frac{\pi}{\sqrt{a}}+\frac{\pi}{\sqrt{b}})}^{\frac{l+1}{m}(\frac{\pi}{\sqrt{a}}+\frac{\pi}{\sqrt{b}})}C(mt)\cos\left(kmt-k\frac{m}{n}\theta\right){\rm d}t\\ & \quad=\int_{\frac{l}{m}(\frac{\pi}{\sqrt{a}}+\frac{\pi}{\sqrt{b}})}^{\frac{l}{m}(\frac{\pi}{\sqrt{a}}+\frac{\pi}{\sqrt{b}})+\frac{\pi}{2m\sqrt{a}}}({-}1)^l\cos\left(\sqrt{a}mt-l\sqrt{\frac{a}{b}}\pi\right)\cos\left(kmt-k\frac{m}{n}\theta\right){\rm d}t\\ & \qquad+\int_{\frac{l}{m}(\frac{\pi}{\sqrt{a}}+\frac{\pi}{\sqrt{b}})+\frac{\pi}{2m\sqrt{a}}}^{\frac{l}{m}(\frac{\pi}{\sqrt{a}}+\frac{\pi}{\sqrt{b}})+\frac{\pi}{2m\sqrt{a}}+\frac{\pi}{m\sqrt{b}}}({-}1)^{l+1}\\ & \qquad\quad \sqrt{\frac{a}{b}}\sin\left(\sqrt{b}mt-(2l+1)\frac{\sqrt{b}}{2\sqrt{a}}\pi\right)\cos\left(kmt-k\frac{m}{n}\theta\right){\rm d}t\\ & \qquad+\int_{\frac{l}{m}(\frac{\pi}{\sqrt{a}}+\frac{\pi}{\sqrt{b}})+\frac{\pi}{2m\sqrt{a}}+\frac{\pi}{m\sqrt{b}}}^{\frac{l+1}{m}(\frac{\pi}{\sqrt{a}}+\frac{\pi}{\sqrt{b}})}({-}1)^{l+1}\cos\left(\sqrt{a}mt-(l+1)\sqrt{\frac{a}{b}}\pi\right)\cos\left(kmt-k\frac{m}{n}\theta\right){\rm d}t. \end{align*}

By some calculations, when $k=\pm \sqrt {a},$ we get

\begin{align*} \Phi_{ak}(\theta)& =\frac{1}{2m\sqrt{a}}\sin\left(\sqrt{a}\frac{m}{n}\theta\right)+\frac{\pi}{4m\sqrt{a}}\cos\left(\sqrt{a}\frac{m}{n}\theta\right)-\frac{\sqrt{a}}{m(b-a)}\sin\left(\sqrt{a}\frac{m}{n}\theta\right)\\ & \quad+\sum_{l=0}^{n-2}({-}1)^{l+1}\frac{\pi}{2m\sqrt{a}}\cos\left((l+1)\sqrt{\frac{a}{b}}\pi-\sqrt{a}\frac{m}{n}\theta\right)\\ & \quad+({-}1)^n\frac{1}{2m\sqrt{a}}\sin\left(n\sqrt{\frac{a}{b}}\pi-\sqrt{a}\frac{m}{n}\theta\right)\\ & \quad +({-}1)^n\frac{\pi}{4m\sqrt{a}}\cos\left(n\sqrt{\frac{a}{b}}\pi-\sqrt{a}\frac{m}{n}\theta\right)\\ & \quad+({-}1)^{n+1}\frac{\sqrt{a}}{m(b-a)}\sin\left(n\sqrt{\frac{a}{b}}\pi-\sqrt{a}\frac{m}{n}\theta\right). \end{align*}

When $k=\pm \sqrt {b},$ similarly we have

\begin{align*} \Phi_{ak}(\theta)& ={-}\frac{\sqrt{b}}{m(a-b)}\sin\left(\sqrt{b}\frac{m}{n}\theta\right)+\frac{\sqrt{b}}{m(a-b)}({-}1)^{n-1}\sin\left(n\sqrt{\frac{b}{a}}\pi-\sqrt{b}\frac{m}{n}\theta\right)\\ & \quad+\sum_{l=0}^{n-1}({-}1)^l\frac{\sqrt{a}\pi}{2mb}\sin\left(l\sqrt{\frac{b}{a}}\pi+\frac{\sqrt{b}\pi}{2\sqrt{a}}-\sqrt{b}\frac{m}{n}\theta\right). \end{align*}

When $k\neq \pm \sqrt {a},\, \pm \sqrt {b},$ we also obtain

\begin{align*} \Phi_{ak}(\theta)& =\sum_{l=0}^{n-1}\frac{\sqrt{a}(b-a)}{m(k^2-a)(k^2-b)}\left(\cos\left(lk\left(\frac{\pi}{\sqrt{a}}+\frac{\pi}{\sqrt{b}}\right)+\frac{k\pi}{2\sqrt{a}}-k\frac{m}{n}\theta\right)\right.\\ & \quad\left.+\cos\left(lk\left(\frac{\pi}{\sqrt{a}}+\frac{\pi}{\sqrt{b}}\right)+\frac{k\pi}{2\sqrt{a}}+\frac{k\pi}{\sqrt{b}}-k\frac{m}{n}\theta\right)\right). \end{align*}

Thus, for all $k\in \mathbb {Z},$ $\Phi _{ak}(\theta )$ can be analytically extended to $\tilde {\Phi }_{ak}(\theta )$ in $\{\theta \in \mathbb {C}: |\operatorname {Im}\theta |< r'\},$ with $r'\leq r\frac {n}{m}$, and it is easy to see that

\[ \big|\tilde{\Phi}_{ak}(\theta)\big|\leq\left\{ \begin{array}{@{}ll} \left(\dfrac{\displaystyle {2+n\pi}}{\displaystyle {2m\sqrt{a}}}+\dfrac{\displaystyle {2\sqrt{a}}}{\displaystyle {m|b-a|}}\right)e^{\sqrt{a}\dfrac{m}{n}r'}, & k={\pm}\sqrt{a}; \\ \left(\dfrac{\displaystyle {2\sqrt{b}}}{\displaystyle {m|b-a|}}+\dfrac{\displaystyle {\sqrt{a}n\pi}}{\displaystyle {2mb}}\right)e^{\sqrt{b}\dfrac{m}{n}r'}, & k={\pm}\sqrt{b}; \\ \dfrac{\displaystyle {2\sqrt{a}|b-a|n}}{\displaystyle {m|k^2-a||k^2-b|}}e^{|k|\dfrac{m}{n}r'}, & k\neq{\pm}\sqrt{a}, \pm\sqrt{b}. \end{array} \right. \]

Similarly, for all $k\in \mathbb {Z},$ $\Phi _{bk}(\theta )$ can also be analytically extended to $\tilde {\Phi }_{bk}(\theta )$ in $\{\theta \in \mathbb {C}: |\operatorname {Im}\theta |< r'\},$ and

\[ \big|\tilde{\Phi}_{bk}(\theta)\big|\leq\left\{ \begin{array}{@{}ll} \left(\dfrac{\displaystyle {2+n\pi}}{\displaystyle {2m\sqrt{a}}}+\dfrac{\displaystyle {2\sqrt{a}}}{\displaystyle {m|b-a|}}\right)e^{\sqrt{a}\dfrac{m}{n}r'}, & k={\pm}\sqrt{a}; \\ \left(\dfrac{\displaystyle {2\sqrt{b}}}{\displaystyle {m|b-a|}}+\dfrac{\displaystyle {\sqrt{a}n\pi}}{\displaystyle {2mb}}\right)e^{\sqrt{b}\dfrac{m}{n}r'}, & k={\pm}\sqrt{b}; \\ \dfrac{\displaystyle {2\sqrt{a}|b-a|n}}{\displaystyle {m|k^2-a||k^2-b|}}e^{|k|\dfrac{m}{n}r'}, & k\neq{\pm}\sqrt{a}, \pm\sqrt{b}. \end{array} \right. \]

Then with $|f_k|\leq \|f\|_re^{-|k|r},$ where $\|f\|_r=\sup _{|\mathrm {Im} t|\leq r}|f(t)|,$ and since $r'\leq r\frac {n}{m}$, we have

\[ |f_k\tilde{\Phi}_{ak}(\theta)|,~|if_k\tilde{\Phi}_{bk}(\theta)|\leq\left\{ \begin{array}{@{}ll} \|f\|_r\left(\dfrac{\displaystyle {2+n\pi}}{\displaystyle {2m\sqrt{a}}}+\dfrac{\displaystyle {2\sqrt{a}}}{\displaystyle {m|b-a|}}\right), & k={\pm}\sqrt{a}; \\ \|f\|_r\left(\dfrac{\displaystyle {2\sqrt{b}}}{\displaystyle {m|b-a|}}+\dfrac{\displaystyle {\sqrt{a}n\pi}}{\displaystyle {2mb}}\right), & k={\pm}\sqrt{b}; \\ \|f\|_r\dfrac{\displaystyle {2\sqrt{a}|b-a|n}}{\displaystyle {m|k^2-a||k^2-b|}}, & k\neq{\pm}\sqrt{a}, \pm\sqrt{b}. \end{array} \right. \]

By Weierstrass M-test, since the series $\sum \limits _{k\in \mathbb {Z}\atop k\neq \pm \sqrt {a},\, \pm \sqrt {b}}\frac {1}{|k^2-a||k^2-b|}$ is convergent, then in the domain $\{\theta \in \mathbb {C}: |\operatorname {Im}\theta |< r'\},$ the series

\[ \sum_{k\in\mathbb{Z}}f_k\tilde{\Phi}_{ak}(\theta)+i\sum_{k\in\mathbb{Z}}f_k\tilde{\Phi}_{bk}(\theta) \]

uniformly converge to $\tilde {\Phi }_f(\theta )$, which is a complex extension of $\Phi _f(\theta ).$ Since all $\tilde {\Phi }_{ak}(\theta ),\, \tilde {\Phi }_{bk}(\theta )$ are analytic, then by Weierstrass's Theorem, $\tilde {\Phi }_f(\theta )$ is also analytic in the domain $\{\theta \in \mathbb {C}: |\operatorname {Im}\theta |< r'\}$.

Finally, under the assumptions of theorem 1.1, for any $\theta ^*\in \mathcal {A},$ we have $\tilde {\Phi }_f(\theta ^*)=\Phi _f(\theta ^*)=0,$ $\tilde {\Phi }'_f(\theta ^*)=\Phi '_f(\theta ^*)=0.$ Then with the isolation of zeros for analytic functions, for any $\theta ^*\in \mathcal {A},$ there exists an integer $k\geq 2$ such that $\tilde {\Phi }_f(\theta ^*)=\tilde {\Phi }_f'(\theta ^*)=\cdots =\tilde {\Phi }_f^{(k-1)}(\theta ^*)=0,$ $\tilde {\Phi }_f^{(k)}(\theta ^*)\neq 0.$ Thus, $\Phi _f(\theta ^*)=\Phi _f'(\theta ^*)=\cdots =\Phi _f^{(k-1)}(\theta ^*)=0,$ $\Phi _f^{(k)}(\theta ^*)\neq 0.$

By lemma 4.1, choose some $\theta ^*\in \mathcal {A}.$ Without loss of generality, we can assume that $\Phi _f^{(k)}(\theta ^*)>0,$ otherwise, make a time change $t\mapsto -t.$

In the following, for a fixed $\theta ^*\in \mathcal {A}$ and the corresponding integer $k\geq 2$, some estimates of $\Phi _f(\theta )$ near $\theta ^*$ are given.

Lemma 4.2 Assume that there exist $\theta ^*\in \mathbb {R}$ and $2\leq k\in \mathbb {N}$ such that $\Phi _f(\theta ^*)=\Phi _f'(\theta ^*)=\cdots =\Phi _f^{(k-1)}(\theta ^*)=0,$ $\Phi _f^{(k)}(\theta ^*)>0.$ Then there exists $\delta _1>0$ such that for all $\theta : 0<\theta -\theta ^*\leq \delta _1$, one has $\Phi _f(\theta )>0,\, \Phi _f'(\theta )>0.$

Proof. This lemma can be easily proved by properties of the derivative, so we omit the details here.

Let $\tau =\theta -\theta ^*.$ Then $\Phi _f(\theta )=\Phi _f(\tau +\theta ^*),$ and we get the following lemma.

Lemma 4.3 Assume that there exists $2\leq k\in \mathbb {N}$ such that $\Phi _f(\theta ^*)=\Phi _f'(\theta ^*)=\cdots =\Phi _f^{(k-1)}(\theta ^*)=0,$ $\Phi _f^{(k)}(\theta ^*)>0.$ Then there exists $\delta _1>0$ such that for all ${\tau : 0<\tau \leq \delta _1}$, we have $\Phi _f(\tau +\theta ^*)>0,\, \Phi _f'(\tau +\theta ^*)>0.$

Lemma 4.4 Assume that the function $g(x)$ is analytic at $x=0,$ and ${g^{(j)}(0)=0,\,~j=0,\,1,\,\ldots,\,k-1,\, ~g^{(k)}(0)>0}$. Then there exists $\delta _2>0$ such that for all $x: 0\leq x\leq \delta _2,$ one has

\[ g(x)\in\Big[\frac{c_1}{k!}x^kg^{(k)}(0),~ \frac{c_2}{k!}x^kg^{(k)}(0)\Big], \]

where

\[ c_1=\frac{6k+10}{6k+11}<1,~c_2=\frac{6k+12}{6k+11}>1. \]

Proof. On the one hand, let

\[ h_1(x)=g(x)-\frac{c_1}{k!}x^kg^{(k)}(0). \]

Then $h_1(0)=h_1'(0)=\cdots =h_1^{(k-1)}(0)=0$, and $h_1^{(k)}(0)=(1-c_1)g^{(k)}(0)=\frac {1}{6k+11} g^{(k)}(0)>0,$ so there exists $\delta _3>0$ such that for all $x: 0\leq x\leq \delta _3,$ we have $h_1(x)\geq 0,$ which leads to

\[ g(x)\geq \frac{c_1}{k!}x^kg^{(k)}(0). \]

On the other hand, let

\[ h_2(x)=g(x)-\frac{c_2}{k!}x^kg^{(k)}(0). \]

Then $h_2(0)=h_2'(0)=\cdots =h_2^{(k-1)}(0)=0$, and $h_2^{(k)}(0)=(1-c_2)g^{(k)}(0)=\frac {-1}{6k+11} g^{(k)}(0)<0,$ so there exists $\delta _4>0$ such that for all $x: 0\leq x\leq \delta _4,$ we have $h_2(x)\leq 0,$ which leads to

\[ g(x)\leq \frac{c_2}{k!}x^kg^{(k)}(0). \]

Let $\delta _2=\min \{\delta _3,\, \delta _4\}>0$. Then for all $x: 0\leq x\leq \delta _2,$ we have

\[ g(x)\in\Big[\frac{c_1}{k!}x^kg^{(k)}(0),~ \frac{c_2}{k!}x^kg^{(k)}(0)\Big], \]

where

\[ c_1=\frac{6k+10}{6k+11}<1,~~c_2=\frac{6k+12}{6k+11}>1. \]

Applying lemmas 4.1–4.4 to $\Phi _f(\tau +\theta ^*)$ and $\Phi _f'(\tau +\theta ^*).$ Then we can easily obtain the following result.

Lemma 4.5 Under the assumptions of theorem 1.1, there exists $\delta >0$ such that for all $\tau : 0<\tau \leq \delta,$ we have $\Phi _f(\tau +\theta ^*)>0,\, \Phi _f'(\tau +\theta ^*)>0,$ and

\begin{align*} & \Phi_f(\tau+\theta^*)\in\bigg[\frac{c_1}{k!}\tau^k\Phi_f^{(k)}(\theta^*), \frac{c_2}{k!}\tau^k\Phi_f^{(k)}(\theta^*)\bigg],\\ & \Phi_f'(\tau+\theta^*)\in\bigg[\frac{c_1}{(k-1)!}\tau^{k-1}\Phi_f^{(k)}(\theta^*), \frac{c_2}{(k-1)!}\tau^{k-1}\Phi_f^{(k)}(\theta^*)\bigg], \end{align*}

where $c_1=\frac {6k+10}{6k+11}<1,\,~c_2=\frac {6k+12}{6k+11}>1.$

5. The existence of unbounded solutions

In this section, we prove that the Hamiltonian system with the Hamiltonian (3.9) has unbounded solutions.

The system with the Hamiltonian (3.9) is given by

(5.1)\begin{equation} \left\{ \begin{array}{@{}ll} \dfrac{\displaystyle {{\rm d}\theta}}{\displaystyle {{\rm d}t}}=\dfrac{\displaystyle {\partial H}}{\displaystyle {\partial r}}={-}\dfrac{\displaystyle {1}}{\displaystyle {4\pi}}m\rho r^{-\dfrac{1}{2}}\Phi_f(\theta)+\partial_rP_2(r, \theta, t),\\ \dfrac{\displaystyle {{\rm d}r}}{\displaystyle {{\rm d}t}}={-}\dfrac{\displaystyle {\partial H}}{\displaystyle {\partial \theta}}=\dfrac{\displaystyle {1}}{\displaystyle {2\pi}}m\rho r^{\dfrac{1}{2}}\Phi_f'(\theta)-\partial_{\theta}P_1(\theta, t)-\partial_{\theta}P_2(r, \theta, t). \end{array} \right. \end{equation}

Let $\tau =\theta -\theta ^*.$ Then the system (5.1) is transformed into

(5.2)\begin{equation} \left\{ \begin{array}{@{}ll} \dfrac{\displaystyle {d\tau}}{\displaystyle {{\rm d}t}}={-}\dfrac{\displaystyle {1}}{\displaystyle {4\pi}}m\rho r^{-\dfrac{1}{2}}\Phi_f(\tau+\theta^*)+\partial_rP_2(r, \tau+\theta^*, t),\\ \dfrac{\displaystyle {{\rm d}r}}{\displaystyle {{\rm d}t}}=\dfrac{\displaystyle {1}}{\displaystyle {2\pi}}m\rho r^{\dfrac{1}{2}}\Phi_f'(\tau+\theta^*)-\partial_{\tau}P_1(\tau+\theta^*, t)-\partial_{\tau}P_2(r, \tau+\theta^*, t). \end{array} \right. \end{equation}

For fixed $2\leq k\in \mathbb {N}$ from lemma 4.1 and $\delta$ from lemma 4.5, choose $r^*\gg 1$ satisfying

1. (1) $r^*\gg \|f\|;$

2. (2) $2(r^*)^{-\frac {1}{3k}}<\delta$;

3. (3) $\frac {1}{3\pi }\frac {c_1}{(k-1)!}m\rho (\frac {r^*}{2})^{\frac {1}{2(2k-1)}}\Phi _f^{(k)}(\theta ^*)\geq 1,$ where $c_1=\frac {6k+10}{6k+11}.$

Give an initial point $(r(0),\, \tau (0))\in D:=\Big \{(r,\, \tau ):r^{-\frac {1}{2k-1}}\leq \tau \leq r^{-\frac {1}{3k}}\Big \}$ and $r(0)\geq r^*.$

First, the second equation in (5.2) implies that $\frac {{\rm d}r}{{\rm d}t}=O(r^{\frac {1}{2}}+1),$ thus $r(t)\geq \frac {1}{2}r^*$ for any $t\in [0,\, 2\pi ]$ by $r^*\gg \|f\|$. Also the first equation in (5.2) implies $\frac {{\rm d}\tau }{{\rm d}t}=O(r^{-\frac {1}{2}}),$ hence $0<\tau (t)\leq 2(r^*)^{-\frac {1}{3k}}$ for any $t\in [0,\,2\pi ].$ Thus, lemma 4.5 can be applied in the following.

We claim that if the initial point $(r(0),\, \tau (0))\in D$ and $r(0)\geq r^*,$ then $(r(t),\, \tau (t))\in D$ for any $t\in [0,\, 2\pi ].$ Otherwise, let $t_1:=\sup \{t: (r(s),\, \tau (s))\in D,\, ~0\leq s\leq t\}<2\pi.$ It is obvious that $(r(t_1),\, \tau (t_1))\in \partial D,$ which leads to $r(t_1)^l\tau (t_1)=1$ with $l=\frac {1}{2k-1}$ or $\frac {1}{3k}.$

By a direct computation, we get

\begin{align*} & (r(t)^l\tau(t))'|_{t=t_1}\\ & \quad=lr(t)^{l-1}r'(t)\tau(t)+r(t)^l\tau'(t)|_{t=t_1}\\ & \quad=lr(t)^{l-1}\tau(t)\left(\frac{1}{2\pi}m\rho r(t)^{\frac{1}{2}}\Phi_f'(\tau(t)+\theta^*)\right.\\& \left.\quad -\partial_{\tau}P_1(\tau(t)+\theta^*, t)-\partial_{\tau}P_2(r(t), \tau(t)+\theta^*, t)\right)\Big|_{t=t_1}\\ & \qquad+r(t)^l\left(-\frac{1}{4\pi}m\rho r(t)^{-\frac{1}{2}}\Phi_f(\tau(t)+\theta^*)+\partial_rP_2(r(t), \tau(t)+\theta^*, t)\right)\Big|_{t=t_1}\\ & \quad:=J_1+J_2. \end{align*}

Since $r(t_1)^l\tau (t_1)=1,$ $r(t)\geq \frac {1}{2}r^*$ and $0<\tau (t)\leq 2(r^*)^{-\frac {1}{3k}}<\delta$ for $t\in [0,\, 2\pi ]$, then we get

\begin{align*} J_1& =lr(t_1)^{l-1}\tau(t_1)\left(\frac{1}{2\pi}m\rho r(t_1)^{\frac{1}{2}}\Phi_f'(\tau(t_1)+\theta^*)-\partial_{\tau}P_1(\tau(t)\right.\\ & \left.\quad +\theta^*, t)|_{t=t_1}-\partial_{\tau}P_2(r(t), \tau(t)+\theta^*, t)|_{t=t_1}\right)\\ & =\frac{1}{2\pi}m\rho lr(t_1)^{-\frac{1}{2}}\Phi_f'(\tau(t_1)+\theta^*)+O(r(t_1)^{{-}1}), \end{align*}

and

\begin{align*} J_2& =r(t_1)^l\left(-\frac{1}{4\pi}m\rho r(t_1)^{-\frac{1}{2}}\Phi_f(\tau(t_1)+\theta^*)+\partial_rP_2(r(t), \tau(t)+\theta^*, t)|_{t=t_1}\right)\\ & ={-}\frac{1}{4\pi}m\rho r(t_1)^{-\frac{1}{2}}r(t_1)^l\Phi_f(\tau(t_1)+\theta^*)+O\left(r(t_1)^{-\frac{3}{2}+l}\right). \end{align*}

Now it is a position to apply lemma 4.5 to $J_1$ and $J_2.$ On the one hand, if $l=\frac {1}{3k},$ then we have

\begin{align*} J_1+J_2& =\frac{1}{2\pi}m\rho lr(t_1)^{-\frac{1}{2}}\Phi_f'(\tau(t_1)+\theta^*)+O(r(t_1)^{{-}1})\\ & \quad-\frac{1}{4\pi}m\rho r(t_1)^{-\frac{1}{2}}r(t_1)^l\Phi_f(\tau(t_1)+\theta^*)+O\left(r(t_1)^{-\frac{3}{2}+l}\right)\\ & \leq\frac{1}{2\pi}m\rho lr(t_1)^{-\frac{1}{2}}\frac{c_2}{(k-1)!}\tau(t_1)^{k-1}\Phi_f^{(k)}(\theta^*)+O(r(t_1)^{{-}1})\\ & \quad-\frac{1}{4\pi}m\rho r(t_1)^{-\frac{1}{2}}r(t_1)^l\frac{c_1}{k!}\tau(t_1)^k\Phi_f^{(k)}(\theta^*)+O\left(r(t_1)^{-\frac{3}{2}+l}\right)\\ & =\frac{1}{2\pi}m\rho lr(t_1)^{-\frac{1}{2}}\frac{c_2}{(k-1)!}r(t_1)^{{-}l(k-1)}\Phi_f^{(k)}(\theta^*)+O(r(t_1)^{{-}1})\\ & \quad-\frac{1}{4\pi}m\rho r(t_1)^{-\frac{1}{2}}r(t_1)^l\frac{c_1}{k!}r(t_1)^{{-}lk}\Phi_f^{(k)}(\theta^*)+O\left(r(t_1)^{-\frac{3}{2}+l}\right)\\ & =\frac{1}{2\pi}\frac{1}{(k-1)!}m\rho\left(c_2 l-\frac{c_1}{2k}\right) r(t_1)^{-\frac{1}{2}-l(k-1)}\Phi_f^{(k)}(\theta^*)\\ & \quad +O(r(t_1)^{{-}1})+O\left(r(t_1)^{-\frac{3}{2}+l}\right). \end{align*}

Since $c_1=\frac {6k+10}{6k+11}$, $c_2=\frac {6k+12}{6k+11}$ and $l=\frac {1}{3k},\, k\geq 2,$ then $c_2l-\frac {c_1}{2k}<0,$ $-\frac {3}{2}+l<-1<-\frac {1}{2}-l(k-1),$ which lead to

\[ J_1+J_2=\frac{1}{2\pi}\frac{1}{(k-1)!}m\rho\left(c_2 l-\frac{c_1}{2k}\right) r(t_1)^{-\frac{1}{2}-l(k-1)}\Phi_f^{(k)}(\theta^*)(1+o(1))<0. \]

That is, if $l=\frac {1}{3k},$ then $(r(t)^l\tau (t))'|_{t=t_1}<0.$ Therefore, there exists $t_2>t_1$ such that $r(t)^\frac {1}{3k}\tau (t)\leq 1$ for $t\in [t_1,\, t_2],$ which contradicts the definition of $t_1.$ Thus for all $t\in [0,\, 2\pi ],$ we have $\tau (t)\leq r(t)^{-\frac {1}{3k}}.$

On the other hand, if $l=\frac {1}{2k-1},$ then we have

\begin{align*} J_1+J_2& =\frac{1}{2\pi}m\rho lr(t_1)^{-\frac{1}{2}}\Phi_f'(\tau(t_1)+\theta^*)+O(r(t_1)^{{-}1})\\ & \quad-\frac{1}{4\pi}m\rho r(t_1)^{-\frac{1}{2}}r(t_1)^l\Phi_f(\tau(t_1)+\theta^*)+O\left(r(t_1)^{-\frac{3}{2}+l}\right)\\ & \geq\frac{1}{2\pi}m\rho lr(t_1)^{-\frac{1}{2}}\frac{c_1}{(k-1)!}\tau(t_1)^{k-1}\Phi_f^{(k)}(\theta^*)+O(r(t_1)^{{-}1})\\ & \quad-\frac{1}{4\pi}m\rho r(t_1)^{-\frac{1}{2}}r(t_1)^l\frac{c_2}{k!}\tau(t_1)^k\Phi_f^{(k)}(\theta^*)+O\left(r(t_1)^{-\frac{3}{2}+l}\right)\\ & =\frac{1}{2\pi}m\rho lr(t_1)^{-\frac{1}{2}}\frac{c_1}{(k-1)!}r(t_1)^{{-}l(k-1)}\Phi_f^{(k)}(\theta^*)+O(r(t_1)^{{-}1})\\ & \quad-\frac{1}{4\pi}m\rho r(t_1)^{-\frac{1}{2}}r(t_1)^l\frac{c_2}{k!}r(t_1)^{{-}lk}\Phi_f^{(k)}(\theta^*)+O\left(r(t_1)^{-\frac{3}{2}+l}\right)\\ & =\frac{1}{2\pi}\frac{1}{(k-1)!}m\rho\left(c_1 l-\frac{c_2}{2k}\right) r(t_1)^{-\frac{1}{2}-l(k-1)}\Phi_f^{(k)}(\theta^*)\\& \quad +O(r(t_1)^{{-}1})+O\left(r(t_1)^{-\frac{3}{2}+l}\right). \end{align*}

Since $c_1=\frac {6k+10}{6k+11}$, $c_2=\frac {6k+12}{6k+11}$ and $l=\frac {1}{2k-1},\, k\geq 2,$ then $c_1l-\frac {c_2}{2k}>0,$ $-\frac {3}{2}+l<-1<-\frac {1}{2}-l(k-1),$ which lead to

\[ J_1+J_2=\frac{1}{2\pi}\frac{1}{(k-1)!}m\rho\left(c_1 l-\frac{c_2}{2k}\right) r(t_1)^{-\frac{1}{2}-l(k-1)}\Phi_f^{(k)}(\theta^*)(1+o(1))>0. \]

That is, if $l=\frac {1}{2k-1},$ then $(r(t)^l\tau (t))'|_{t=t_1}>0.$ Therefore, there exists $t_2>t_1$ such that $r(t)^\frac {1}{2k-1}\tau (t)\geq 1$ for $t\in [t_1,\, t_2],$ which contradicts the definition of $t_1.$ Thus for all $t\in [0,\, 2\pi ],$ one has $\tau (t)\geq r(t)^{-\frac {1}{2k-1}}.$ The proof of the claim is completed.

Now we prove that every solution of system (5.2) with the initial point $(r(0),\, \tau (0))\in D$ and $r(0)\geq r^*$ is unbounded.

From the claim, if the initial point $(r(0),\, \tau (0))\in D$ and $r(0)\geq r^*,$ then for any $t\in [0,\, 2\pi ],$ one has $r(t)^{-\frac {1}{2k-1}}\leq \tau (t)\leq r(t)^{-\frac {1}{3k}}\leq 2(r^*)^{-\frac {1}{3k}}.$ Thus, from the second equation of (5.2), for any $t\in [0,\, 2\pi ],$ we obtain

\begin{align*} \frac{{\rm d}r}{{\rm d}t}& =\frac{1}{2\pi}m\rho r(t)^{\frac{1}{2}}\Phi_f'(\tau(t)+\theta^*)-\partial_{\tau}P_1(\tau(t)+\theta^*, t)-\partial_{\tau}P_2(r(t), \tau(t)+\theta^*, t)\\ & =\frac{1}{2\pi}m\rho r(t)^{\frac{1}{2}}\Phi_f'(\tau(t)+\theta^*)+O(1)\\ & \geq\frac{1}{2\pi}m\rho r(t)^{\frac{1}{2}}\frac{c_1}{(k-1)!}\tau(t)^{k-1}\Phi_f^{(k)}(\theta^*)+O(1)\\ & \geq\frac{1}{2\pi}m\rho r(t)^{\frac{1}{2}}\frac{c_1}{(k-1)!}r(t)^{-\frac{k-1}{2k-1}}\Phi_f^{(k)}(\theta^*)+O(1)\\ & =\frac{1}{2\pi}\frac{c_1}{(k-1)!}m\rho r(t)^{\frac{1}{2(2k-1)}}\Phi_f^{(k)}(\theta^*)+O(1)\\ & \geq\frac{1}{3\pi}\frac{c_1}{(k-1)!}m\rho r(t)^{\frac{1}{2(2k-1)}}\Phi_f^{(k)}(\theta^*). \end{align*}

Choose $r^*$ sufficiently large such that $\frac {1}{3\pi }\frac {c_1}{(k-1)!}m\rho (\frac {r^*}{2})^{\frac {1}{2(2k-1)}}\Phi _f^{(k)}(\theta ^*)\geq 1.$ Then $r(2\pi )\geq r(0)+2\pi >r^*.$

In a word, if $(r(0),\, \tau (0))\in D$ and $r(0)\geq r^*,$ then $(r(2\pi ),\, \tau (2\pi ))\in D$ and $r(2\pi )\geq r(0)+2\pi >r^*.$

Using the above argument repeatedly, if $(r(0),\, \tau (0))$ satisfies the above initial conditions, then $r(2\pi i)\geq r(0)+2\pi i$ for any $i\in \mathbb {N},$ which means that the solution $(r(t),\, \tau (t))$ is unbounded. Up to now theorem 1.1 is proved.