The coexistence of quasi-periodic and blow-up solutions in a superlinear Duffing equation

TL;DR

This paper constructs a specific superlinear Duffing equation with a positive periodic coefficient that admits both finite-time blow-up solutions and infinitely many subharmonic and quasi-periodic solutions, revealing complex coexistence phenomena.

Contribution

It demonstrates the coexistence of blow-up and quasi-periodic solutions in a superlinear Duffing equation with a specially designed periodic coefficient.

Findings

Existence of solutions that blow up in finite time.

Presence of infinitely many subharmonic and quasi-periodic solutions.

Construction of a periodic coefficient function enabling this coexistence.

Abstract

In this paper we will construct a continuous positive periodic function $p(t)$ such that the corresponding superlinear Duffing equation $$ x"+a(x)\,x^{2n+1}+p(t)\,x^{2m+1}=0,\ \ \ \ n+2\leq 2m+1<2n+1 $$ possesses a solution which escapes to infinity in some finite time, and also has infinitely many subharmonic and quasi-periodic solutions, where the coefficient $a(x)$ is an arbitrary positive smooth periodic function defined in the whole real axis.

Equations

$$x^{\prime \prime }+a(x)x^{2n+1}+p(t)x^{2m+1}=0,n+2\le 2m+1<2n+1$$

$$x^{\prime \prime }+V_x(x,t)=0$$

$$x^{\prime \prime }+x^{2n+1}+\sum _{k=0}^{2n}{p}_k(t)x^k=0,$$

$$x^{\prime \prime }+x^{2n+1}=0$$

$$x^{\prime \prime }+2x^3=p(t)=p(t+1)$$

$$x^{\prime \prime }+x^{2n+1}+p(t)x^{2m+1}=0$$

$$x^{\prime \prime }+x^{2n+1}+p(t)x^i=0$$

$$x^{\prime \prime }+a(x)x^{2n+1}+p(t)x^{2m+1}=0,n+2\le 2m+1<2n+1,$$

$$x^{\prime \prime }+p(t)x^{2m+1}+e(t,x)=0,m\ge 1,$$

$$x^{\prime }=\frac{\partial H}{\partial y},y^{\prime }=-\frac{\partial H}{\partial x},$$

$$H(x,y,t)=\frac{1}{2}{y}^2+G(x)+\frac{p(t)}{2m+2}{x}^{2m+2}$$

$$G(x)={\int }_0^{x}a(s)s^{2n+1}ds.$$

$$x^{\prime }=y,y^{\prime }=-G^{\prime }(x).$$

$${\Gamma }_h:\frac{1}{2}{y}^2+G(x)=h.$$

$$I=I(h)={\int }_{{\Gamma }_h}\sqrt{2(h-G(x))}dx.$$

$$S(x,I)=\left\{\begin{array}[]{l}\displaystyle\int_{x_{-}}^{x}\sqrt{2(h(I)-G(s))}ds,\ \ \ \ \ \ \ \,y\geq 0,\\[17.07182pt] I-\displaystyle\int_{x_{-}}^{x}\sqrt{2(h(I)-G(s))}ds,\ \ y<0,\end{array}\right.$$

$$y=\frac{\partial S}{\partial x},\theta =\frac{\partial S}{\partial I}.$$

$$\theta=\left\{\begin{array}[]{l}h^{\prime}(I)\displaystyle\int_{x_{-}}^{x}\frac{ds}{\sqrt{2(h(I)-G(s))}},\ \ \ \ \ \ \ \ \ \ y\geq 0,\\[17.07182pt] 1-h^{\prime}(I)\displaystyle\int_{x_{-}}^{x}\frac{ds}{\sqrt{2(h(I)-G(s))}},\ \ \ \ \ \,y<0.\end{array}\right.$$

$$\Psi :(\theta ,I)\to (x,y),$$

$$H_1=H\circ \Psi =h(I)+\frac{p(t)}{2m+2}x(I,\theta {)}^{2m+2},$$

$$\left\{\begin{array}[]{l}\frac{\displaystyle{d\theta}}{\displaystyle{dt}}=h^{\prime}(I)+p(t)x(I,\theta)^{2m+1}\partial_{I}x(I,\theta),\\[11.38092pt] \frac{\displaystyle{dI}}{\displaystyle{dt}}=-p(t)x(I,\theta)^{2m+1}\partial_{\theta}x(I,\theta).\end{array}\right.$$

$$x^{2n+2}\cdot \le G(x)\le \cdot x^{2n+2},|G^{\prime }(x)|\le \cdot |x{|}^{2n+1},$$

$$|G^{\prime \prime }(x)|\le \cdot \left(|x{|}^{2n+1}+x^{2n}\right),\left|\frac{G(x)}{G^{\prime }(x)}\right|\le \cdot |x|,\left|\frac{G(x)G^{\prime \prime }(x)}{G^{\prime }(x{)}^2}\right|\le \cdot (|x|+1).$$

$$\frac{m}{2n+2}{x}^{2n+2}\le G(x)={\int }_0^{x}a(s)s^{2n+1}ds\le \frac{M}{2n+2}{x}^{2n+2};$$

$$\frac{m}{2n+2}{x}^{2n+2}\le G(x)={\int }_0^{-x}a(-s)s^{2n+1}ds\le \frac{M}{2n+2}{x}^{2n+2},$$

$$\left|\frac{G(x)G^{\prime \prime }(x)}{G^{\prime }(x{)}^2}\right|\le \cdot (|x|+1).$$

$$h^{\frac{1}{2}+\frac{1}{2n+2}}\cdot \le I(h)\le \cdot h^{\frac{1}{2}+\frac{1}{2n+2}},$$

$$h^{-\frac{1}{2}+\frac{1}{2n+2}}\cdot \le I^{\prime }(h)\le \cdot h^{-\frac{1}{2}+\frac{1}{2n+2}},$$

$$|I^{\prime \prime }(h)|\le \cdot h^{-\frac{3}{2}+\frac{1}{n+1}}.$$

$$I(h)=2\sqrt{2}{\int }_{x_{-}}^{x_{+}}\sqrt{h-G(x)}dx.$$

Full text

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possesses a solution which escapes to infinity in some finite time, and also has infinitely many subharmonic and quasi-periodic solutions, where the coefficient $a(x)$ is an arbitrary positive smooth periodic function defined in the whole real axis.

au1]sunyanmei2009@126.com

au1]xli@bnu.edu.cn

1 Introduction

In the early 1960’s, Littlewood [8] asked whether all solutions of the second order differential equation

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are bounded for all time, that is, $\displaystyle\sup_{t\in\mathbb{R}}(|x(t)|+|x^{\prime}(t)|)<+\infty$ holds for all solutions $x(t)$ of Eq.(1.1).

For the Littlewood boundedness problem, during the past years, people have paid more attention to the following equation with the polynomial potentials

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where $p_{k}(t+1)=p_{k}(t)\ (k=0,1,\cdots,2n)$, since

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is a very nice time-independent integrable system, of which all solutions are periodic. Thus if $|x|$ is large enough, Eq.(1.2) can be treated as a perturbation of an integrable system, then Moser’s twist theorem could be applied to prove the boundedness of all solutions.

The first result was due to Morris [14], who proved that all solutions of the equation with the biquadratic potential

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are bounded.

Using the famous Moser’s twist theorem [15], Diekerhoff and Zehnder [1] generalized Morris’s results to Eq.(1.2). In [1], the coefficients $p_{k}(t)$ are required to be sufficiently smooth to construct a series of variable changes to transform Eq.(1.2) into a nearly integrable systems for large energies. In fact, in [1], the smoothness on $p_{k}(t)$ depends on the index $k$.

Later, Yuan [20] proved that all solutions of Eq.(1.2) are bounded if $p_{i}(t)\in C^{2},\,n+1\leq i\leq 2n;\,p_{i}(t)\in C^{1},\,0\leq i\leq n$. Recently, we [5] obtained the same conclusion if $p_{i}(t)\in C^{1},\,n+1\leq i\leq 2n;\,p_{i}(t)\in C^{0},\,0\leq i\leq n$.

There are other results about the boundedness problem for superlinear Duffing equations during the past years, see [2, 9, 10, 17, 18] and the references therein. As for constructing unbounded solutions for superlinear Duffing equations, there also are some results ([6, 7, 12, 3, 4, 16]). Let us recall the results in [4] and [16]. Levi and You [4] proved that the equation

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with a special discontinuous coefficient $p(t)=K^{[t]mod2},\,0<K<1,\,n+2\leq 2m+1<2n+1$, possesses an oscillatory unbounded solution. In 2000, Wang [16] constructed a continuous periodic function $p(t)$ such that the corresponding equation

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possesses a solution which escapes to infinity in some finite time, where $n\geq 2$ and $n+2\leq i<2n+1$.

In this paper we consider the following second order differential equation

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where the coefficient $a(x)$ is an arbitrary positive smooth periodic function defined in the whole real axis, will construct a continuous positive periodic function $p(t)$ and obtain the coexistence of quasi-periodic solutions and blow-up phenomena for the corresponding equation (1.3). More precisely, we will prove

Theorem 1.1

There exists a continuous positive periodic function $p(t)$ such that the corresponding equation (1.3) possesses a solution which escapes to infinity in some finite time, and also has infinitely many subharmonic and quasi-periodic solutions.

Firstly, we will employ the idea in [16] to construct a continuous positive periodic function $p(t)$ such that the corresponding equation (1.3) possesses a solution which escapes to infinity in some finite time. Here, we will construct the positive periodic function $p(t)$ and the blow up solution $x(t)$ simultaneously.

First of all, we observe that during the time when the curve spirals once around the origin, the action variable $I$ increases at some times and decreases at other times after the action-angle variables $(I,\theta)$ are introduced. Therefore we do not know whether the increment of $I$ is positive or negative. However we can construct a time $t_{1}\ll 1$ and modify $p(t)\equiv 1$ on $[0,1]$ so that the increment of $I$ on this time interval $[0,t_{1}]$ is positive and equals to $O\left(\frac{1}{\tau}I_{0}^{\frac{2m-n+2}{n+2}}\right)$ if the initial point \big{(}I(0),\theta(0)\big{)}=(I_{0},0) is far enough from the origin, where the ”jump” $\frac{1}{\tau}\,(0<\frac{1}{\tau}<1)$ is critical to modify $p(t)$ and to our estimations.

Inductively, we can construct a series of times $t_{1},t_{2},\cdots,t_{i},t_{i+1},\cdots$ and modify $p(t)$ on $[t_{i},t_{i+1}],i=1,2,\cdot\cdot\cdot,\,$ so that on every such interval $[t_{i},t_{i+1}],$ the increment is positive and at least $O\left(\frac{1}{\tau}I_{0}^{\frac{2m-n+2}{n+2}}\right).$

Hence, we can construct a time $T_{1}\leq\frac{1}{\tau^{\prime}}<1,$ so that the curve spirals at least $\left[\frac{1}{\tau^{\prime}}I_{0}^{\frac{n}{n+2}}\right]$ times around the origin on the interval $[0,T_{1}]$ and $I_{1}:=I(T_{1})>I_{0}+\frac{c}{\tau\tau^{\prime}}I_{0}^{\frac{2m+2}{n+2}}$ with $c>0$ independent of induction steps, where $\frac{2m+2}{n+2}>1$ and sufficiently large $\tau^{\prime}$ is used to ensure the blow up time not more than $1$. This complete an induction step: during the interval of time $[0,T_{1}],\,I$ increases from $I_{0}$ to $I_{1}$.

Inductively, a series of times $T_{1},T_{2},\cdots,T_{i},T_{i+1},\cdots$ are constructed such that during the interval of time $[T_{k},T_{k+1}],\,I$ increases from $I_{k}$ to $I_{k+1},$ where $I_{k+1}>I_{k}+\frac{c}{(\tau\tau^{\prime})^{k}}I_{k}^{\frac{2m+2}{n+2}}$ with the jump $\frac{1}{\tau^{k}},$ where $T_{k+1}-T_{k}\leq\frac{1}{\tau^{\prime k}}$. The reason that the jump is less and less is that we have to assure $p(t)$ is continuous. Because the exponent $\frac{2m+2}{n+2}>1,$ the less and less jump cannot stop the rapid increase of $I$. If $\frac{1}{\tau^{\prime}}$ is chosen small enough, we will find that $T_{k}\rightarrow T_{\infty}<1$ as $k\rightarrow\infty$ and $I(t)\rightarrow+\infty$ as $t\rightarrow T_{\infty}.$

Once we have found the continuous positive periodic function $p(t)$ such that the corresponding equation (1.3) possesses a solution which escapes to infinity in some finite time, the remain thing is to apply the result in [11]. To this end, we first introduce this result. Consider the conservative system

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where $p(t)$ is a continuous and $1$-periodic function in the time $t$, $e(t,x)$ is also $1$-periodic in the time $t$ and dominated by the power $x^{2m+2}$ in a neighborhood of $x=0$. Liu in [11] proved that if $\int_{0}^{1}p(t)dt\neq 0$, then the trivial solution $x=0$ of Eq.(1.4) is stable in the Liapunov sense if and only if $\int_{0}^{1}p(t)dt>0$ by Moser’s twist theorem. Moreover, by the same argument in [1], if $\int_{0}^{1}p(t)dt>0$, then Eq.(1.4) also has infinitely many subharmonic and quasi-periodic solutions with small amplitudes. Compared Eq.(1.3) with Eq.(1.4), since $p(t)$ is positive, then Eq.(1.3) also has infinitely many subharmonic and quasi-periodic solutions with small amplitudes.

Therefore, if we find a continuous positive periodic function $p(t)$ such that the corresponding equation (1.3) possesses a blow-up solution, then such equation also has infinitely many subharmonic and quasi-periodic solutions with small amplitudes simultaneously.

Similar to the above, if we modify $p(t)\equiv 0$ in $[0,1]$, we can construct a continuous non-positive periodic function $p(t)$ with $\int_{0}^{1}p(t)dt<0$ such that the corresponding equation (1.3) possesses a solution which escapes to infinity in some finite time, and at the same time, by Liu’s result in [11], the trivial solution $x=0$ of such equation is not stable. Therefore we can obtain

Theorem 1.2

There exists a continuous non-positive periodic function $p(t)$ such that the corresponding equation (1.3) possesses a solution which escapes to infinity in some finite time, and the trivial solution $x=0$ is not stable.

Finally we remark that the authors in [19] also obtained the coexistence of quasi-periodic solutions and blow-up phenomena in a class of higher dimensional Duffing-type equations, and the author [13] obtained the coexistence of bounded and unbounded motions in a bouncing ball model.

The paper is organized as follows. The action-angle variables are introduced in Section 2. In Section 3 we first prove some Lemmas which will be useful later. After that, we will construct a continuous positive periodic function $p(t)$ and a series of times $T_{k}$, then obtain an unbounded solution of equation (1.3) and finish the proof of Theorem 1.1.

2 Action-angle variables

In this section we first introduce action and angle variables. Let $y=x^{\prime}$, then Eq.(1.3) is equivalent to the following Hamiltonian system

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where the Hamiltonian is

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with

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In order to introduce action and angle variables, we consider the auxiliary autonomous system

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Since $a(x)>0$ for all $x\in\mathbb{R}$, then $G(x)>0$ for all $x\neq 0$, and all solutions of this system are periodic. For every $h>0$, denote by $I(h)$ the area enclosed by the closed curve

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That is,

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Let $h=h(I)$ be the inverse function of $I=I(h).$ Define

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where $x_{-}=x_{-}(I)<0$ is determined uniquely by $G(x_{-})=h(I).$

Now we introduce the well-known action-angle transformation

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Then

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Denote

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then under $\Psi$, the Hamiltonian $H$ of (2.1) is transformed into

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and the corresponding Hamiltonian system is

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In the following, we do not attempt to obtain estimates with particularly sharp constants. Indeed, we suppress all constants, and use the notations $u\leq\cdot\ v$ and $u\ \cdot\leq v$ to indicate that $u\leq c\,v$ and $c\,u\leq v$, respectively, with some constant $c>0$.

Now we give some estimates on $h(I)$ and $x(I,\theta)$. For this purpose, we first give some properties of the potential function $G$.

Lemma 2.1

For all $x\neq 0$, we have

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Proof. Since the periodic function $a(x)$ is positive, then there exist two positive constants $m,M$ such that $0<m\leq a(x)\leq M$ for all $x\in\mathbb{R}$ and from the expression (2.2) of $G(x)$, we know that for $x>0$,

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and if $x<0$,

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which yields the first inequality of this lemma.

If $x>0$, then $mx^{2n+1}\leq G^{\prime}(x)=a(x)x^{2n+1}\leq Mx^{2n+1}$; if $x<0$, then $Mx^{2n+1}\leq G^{\prime}(x)=a(x)x^{2n+1}\leq mx^{2n+1}$. Combining this two inequalities, one can obtain the second inequality.

Since $G^{\prime\prime}(x)=a^{\prime}(x)x^{2n+1}+(2n+1)a(x)x^{2n}$, if we let $m_{1}\leq a^{\prime}(x)\leq M_{1}$ for all $x\in\mathbb{R}$, where $m_{1}<0,M_{1}>0$ are two constants, then $m_{1}x^{2n+1}+(2n+1)mx^{2n}\leq G^{\prime\prime}(x)\leq M_{1}x^{2n+1}+(2n+1)Mx^{2n}$ for $x>0$, and $M_{1}x^{2n+1}+(2n+1)mx^{2n}\leq G^{\prime\prime}(x)\leq m_{1}x^{2n+1}+(2n+1)Mx^{2n}$ for $x<0$, therefore $|G^{\prime\prime}(x)|\leq\max\{|m_{1}|,M_{1}\}|x|^{2n+1}+(2n+1)Mx^{2n}\leq\cdot\ \left(|x|^{2n+1}+x^{2n}\right)$, which is the third inequality.

For $x>0$, $\frac{m}{M}x\leq\frac{G(x)}{G^{\prime}(x)}\leq\frac{M}{m}x$, and if $x<0$, $\frac{M}{m}x\leq\frac{G(x)}{G^{\prime}(x)}\leq\frac{m}{M}x$, then the fourth inequality holds.

If $x>0$, $\frac{mm_{1}}{(2n+2)M^{2}}x+\frac{(2n+1)m^{2}}{(2n+2)M^{2}}\leq\frac{G(x)G^{\prime\prime}(x)}{G^{\prime}(x)^{2}}\leq\frac{MM_{1}}{(2n+2)m^{2}}x+\frac{(2n+1)M^{2}}{(2n+2)m^{2}}$, and if $x<0$, $\frac{mM_{1}}{(2n+2)M^{2}}x+\frac{(2n+1)m^{2}}{(2n+2)M^{2}}\leq\frac{G(x)G^{\prime\prime}(x)}{G^{\prime}(x)^{2}}\leq\frac{Mm_{1}}{(2n+2)m^{2}}x+\frac{(2n+1)M^{2}}{(2n+2)m^{2}}$, hence

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Up to now, we have finished the proof of the lemma. $\square$

Lemma 2.2

For sufficiently large $h>0$, we have

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[TABLE]

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Proof. Let $x_{-}<0<x_{+}$ defined by $G(x_{-})=G(x_{+})=h$ for any $h>0$, then

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By Lemma 2.1, we know that

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which implies that

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On the other hand, if we let $\bar{x}>0$ determined by $G(\bar{x})=\frac{h}{2}$, then

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and for $0\leq s\leq\bar{x}$,

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thus

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Now we prove the second inequality. From the expression of $I(h)$, we know that

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The second term can be rewritten as follows

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For $0\leq x\leq\bar{x}$, we have

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which together with (2.4) implies that

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If $\bar{x}\leq x\leq x_{+}$, then

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and

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which implies that

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Combining the last equation with the fact that

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we obtain

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and thus

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Similarly, one can prove that

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which completes the proof of the second inequality.

Finally we prove the estimate on $I^{\prime\prime}(h)$. From [2], we know that

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By Lemma 2.1, for sufficiently large $h>0$, we have

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and thus

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which finishes the proof of this lemma. $\square$

Since $h=h(I)$ is the inverse function of $I=I(h)$, we immediately obtain

Lemma 2.3

For sufficiently large $I>0$, we have

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Now we give some estimates on $x(I,\theta)$.

Lemma 2.4

For sufficiently large $I>0$, we have

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[TABLE]

Proof. The inequality $|x|\leq\cdot\ I^{\frac{1}{n+2}}$ is obvious. According to the definition of $\theta$, when $y>0$ (or $0<\theta<\frac{1}{2}$), we have

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Differentiating the above equality with respect to $\theta$ yields that

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which implies that

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Therefore, $\partial_{\theta}x>0$ for $0<\theta<\frac{1}{2}$, and by Lemma 2.3, $\partial_{\theta}x\leq\cdot\ I^{\frac{1}{n+2}}$ holds.

Similarly, when $y<0$ (or $\frac{1}{2}<\theta<1$), we have

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hence $\partial_{\theta}x<0$ and $|\partial_{\theta}x|\leq\cdot\ I^{\frac{1}{n+2}}$.

Now we prove the estimate on $\partial_{I}x$. From [2], when $y\geq 0$ (or $0\leq\theta\leq\frac{1}{2}$), we have

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where

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According to Lemmas 2.1, 2.3, $|L|\leq\cdot\ I^{-\frac{n+1}{n+2}}$ and $\left|\frac{h^{\prime}(I)}{h(I)}\frac{G(x)}{G^{\prime}(x)}\right|\leq\cdot\ I^{-\frac{n+1}{n+2}}$ hold. Also, since for $x_{-}\leq s\leq x$, $\frac{\sqrt{2(h(I)-G(x))}}{\sqrt{2(h(I)-G(s))}}\leq 1$, then $|\partial_{I}x|\leq\cdot\ I^{-\frac{n}{n+2}}$. One can obtain the same estimate for $y<0$. Thus, we have finished the proof of this lemma. $\square$

If we define $x_{1},x_{2},x_{3}$ by

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then they are bounded functions for sufficiently large $I$ and all $\theta\in\mathbb{R}$, that is, there are three positive constants $B_{1},B_{2},B_{3}$ such that

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Furthermore, there exist two positive constants $C_{1},C_{2}$ such that

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for sufficiently large $I$ and $\theta\in[\frac{1}{16},\frac{3}{16}]$. Moreover, we can rewrite system (2.3) into

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By Lemma 2.3, we know that $I^{\frac{n}{n+2}}\ \cdot\leq h^{\prime}(I)\leq\cdot\ I^{\frac{n}{n+2}}$. Also by our assumption, $2m+1\leq 2(n-1)+1$, thus $2m+1-n\leq n-1$ and the right of the first equation in (2.7) is dominated by $h^{\prime}(I)$ for sufficiently large $I$.

3 The proof of Theorem 1.1

Now we define $p(t)$ in $[0,1]$. We will construct a time $t_{1}<1$ and modify $p^{0}(t)\equiv 1$ on $[0,1]$ so that the action of one solution of (2.7) increases in $[0,t_{1}]$. We divide the construction into two steps: first, we construct a piecewise continuous function $p^{1}(t)$ so that the action of one solution of (2.7) obtains a positive increment in $[0,t_{1}]$ as we expect. Then we modify this function $p^{1}(t)$ into a continuous one in such a way that the modification does not influence the estimate we had obtained before.

Without loss of generality, we assume that the function $a(x)$ is even. Denote the corresponding Hamiltonian system (2.7) with the coefficient function $p(t)$ by $X_{p}$. Suppose the solution $(I(t),\theta(t))$ of $X_{p^{0}}$ with $(I(0),\theta(0))=(I_{0},0)$ at $t=0$ arrives at $(I_{\frac{1}{4}},\frac{1}{4})$ at $t=t_{\frac{1}{4}}\ll 1$, where $I_{0}$ is a sufficiently large constant which will be determined later. Define $p^{\frac{1}{4}}(t)$ be a piecewise continuous function as follows

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where $0<\sigma<1$ is the jump, which is used to control the increment of $I$.

Suppose the solution $(I(t),\theta(t))$ of $X_{p^{\frac{1}{4}}}$ with $(I(0),\theta(0))=(I_{0},0)$ at $t=0$ arrives at $(I_{\frac{1}{2}},\frac{1}{2})$ at $t=t_{\frac{1}{2}}\ll 1$. Define $p^{\frac{1}{2}}(t)$ be a piecewise continuous function as follows

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Suppose the solution $(I(t),\theta(t))$ of $X_{p^{\frac{1}{2}}}$ with $(I(0),\theta(0))=(I_{0},0)$ at $t=0$ arrives at $(I_{\frac{3}{4}},\frac{3}{4})$ at $t=t_{\frac{3}{4}}\ll 1$. Define $p^{\frac{3}{4}}(t)$ be a piecewise continuous function as follows

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Suppose the solution $(I(t),\theta(t))$ of $X_{p^{\frac{3}{4}}}$ with $(I(0),\theta(0))=(I_{0},0)$ at $t=0$ arrives at $(I_{1},1)$ at $t=t_{1}\ll 1$. Define $p^{1}(t)$ be a piecewise continuous function as follows

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That is to say, the solution $(I(t),\theta(t))$ of $X_{p^{1}}$ with $(I(0),\theta(0))=(I_{0},0)$ at $t=0$ arrives at $(I_{\frac{1}{4}},\frac{1}{4})$ at $t=t_{\frac{1}{4}}$, arrives at $(I_{\frac{1}{2}},\frac{1}{2})$ at $t=t_{\frac{1}{2}}$, arrives at $(I_{\frac{3}{4}},\frac{3}{4})$ at $t=t_{\frac{3}{4}}$, arrives at $(I_{1},1)$ at $t=t_{1}$, which finishes one cycle of the construction of $p(t)$.

Now we estimate the differences $I_{1}-I_{0}$ and $t_{1}-t_{0}$.

Lemma 3.1

If $I_{0}$ is sufficiently large, then

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Proof. Because $x_{1}<0,x_{3}>0$ for $\theta\in(0,\frac{1}{4})$, then $\frac{dI}{dt}>0$ for $t\in(0,t_{\frac{1}{4}})$ and thus $I(t)$ is an increasing function in this interval. Integrating the first equation of (2.7) from $t=0$ to $t=t_{\frac{1}{4}}$ yields that

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for sufficiently large $I_{0}>0$, here we use the estimate on $h^{\prime}(I)$ in Lemma 2.3 and the bound $B_{i}$ of $x_{i}$ in (2.5).

From the second equation of (2.7), we have

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Since $x_{1}<0,x_{3}>0$ for $\theta\in(0,\frac{1}{4})$, and also integrating the above equation (3.3) from $t=0$ to $t=t_{\frac{1}{4}}$ , one can obtain

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and by (3.2), we get

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where the constant $c>0$ is given by (3.2), which implies that

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here the constant $\tilde{c}=c\frac{2m-n}{n+2}B_{1}^{2m+1}B_{2}>0$. Hence we obtain

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which leads to

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On the other hand, from the first equation of (2.7), we have

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Finally, it follows from (2.6) and (3.4) that

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Similarly, the following estimate

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holds. Combining the two inequalities above yields that

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and

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$\square$

Lemma 3.2

If $I_{0}$ is sufficiently large, then

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[TABLE]

Proof. Because $x_{1}>0,x_{3}>0$ for $\theta\in(\frac{1}{4},\frac{1}{2})$, then $\frac{dI}{dt}<0$ for $t\in(t_{\frac{1}{4}},t_{\frac{1}{2}})$ and thus $I(t)$ is an decreasing function in this interval. Integrating the first equation of (2.7) from $t=t_{\frac{1}{4}}$ to $t=t_{\frac{1}{2}}$ yields that

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From the second equation of (2.7), we have

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and thus

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By Lemma 3.1, we have

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with some constant $c>0$, which implies that

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holds for sufficiently large $I_{0}>0$. Meanwhile it follows from (3.6) that

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which together with $I_{\frac{1}{4}}>I_{0}$ implies that

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and

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Combining (3.5), (3.7) with (3.9), we obtain

[TABLE]

Also, according to (3.6), (3.8) and $I_{\frac{1}{4}}\geq I_{0}$, one can obtain the second inequality in this lemma. $\square$

Using the same method, one can prove the following result.

Lemma 3.3

If $I_{0}$ is sufficiently large, then

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Combining Lemmas 3.1, 3.2 and 3.3, we can obtain immediately the estimates on the time $t_{1}$ when the curve spirals once around the origin and the increment of the action variable $I_{1}-I_{0}$.

Lemma 3.4

If $I_{0}$ is sufficiently large, then

[TABLE]

[TABLE]

Now we modify the piecewise continuous function $p^{1}(t)$ of (3.1) into a continuous one. Being short of signs, we keep the notations unchanged in the process of modification. For example, $p^{1}(t)$ denotes the continuous function modified from the original piecewise continuous function $p^{1}(t)$.

First we modify $p^{1}(t)$ on the interval $[t_{\frac{1}{4}},t_{\frac{1}{4}}+I_{0}^{-\eta}]$ $(\eta>\frac{n}{n+2})$ to be $\sigma(t_{\frac{1}{4}}-t)I_{0}^{\eta}+1$. It is easy to see that $\{(t,p^{1}(t)):t\in[t_{\frac{1}{4}},t_{\frac{1}{4}}+I_{0}^{-\eta}]\}$ is the line segment connecting $(t_{\frac{1}{4}},1)$ and $(t_{\frac{1}{4}}+I_{0}^{-\eta},1-\sigma)$.

In view of the mean value theorem, there must exist a unique new time $t_{\frac{1}{2}}$ such that $\theta(t_{\frac{1}{2}})=\frac{1}{2}$ if we let

[TABLE]

Similarly, there exist the unique $t_{\frac{3}{4}}$ and $t_{1}$ such that $\theta(t_{\frac{3}{4}})=\frac{3}{4}$ and $\theta(t_{1})=1$ for $X_{p^{1}}$ with

[TABLE]

Now the newest coefficient is already a continuous function. It is easy to check that Lemmas 3.1–3.4 still hold with different constants after this modification in view of $I_{0}^{-\eta}\ll I_{0}^{-\frac{n}{n+2}}.$

We will modify $p^{0}$ inductively and denote the function obtained and the corresponding solution with $(I_{0},0)$ as the initial point by $p^{i}$ and $(I^{i}(t),\theta^{i}(t))$ with $(I^{i}(t_{i}),\theta^{i}(t_{i}))=(I_{i},i)$, respectively.

Suppose that we have obtained $p^{0},p^{1},\cdots,p^{i}$. The function $p^{i+1}$ defined on $[0,1]$ is constructed by modifying $p^{i}$ on the interval $[t_{i},t_{i+1}]$, where $t_{i+1}$ satisfies $\theta^{i+1}(t_{i+1})=i+1$ in the same way as above if we regard $I_{i},t_{i}$ as $I_{0},t_{0}$. All the lemmas are true after the modification.

In the process of constructing $p^{i}$, we keep the jump $\sigma=\frac{1}{\tau}\,(\tau\geq 2)$ unchanged until $i=j_{1}$. Then we let $\sigma=\frac{1}{\tau^{2}}$ and keep it unchanged until $i=j_{2}$. Inductively, we choose $\sigma=\frac{1}{\tau^{k}}$ when $\theta\in[j_{k-1},j_{k}]$, where $j_{0}=0,\,j_{1},j_{2},\cdots$ are defined as below.

Let $j_{1}=\left[\frac{1}{\tau^{\prime}}I_{0}^{\frac{n}{n+2}}\right]$, where $[x]$ denotes the integer part of $x$ and $\tau^{\prime}>0$ is used to control time and will be determined later. It follows that

[TABLE]

On the interval $[0,T_{1}]$, since $I_{k+1}-I_{k}\geq\cdot\ \frac{1}{\tau}I_{0}^{\frac{2m+2-n}{n+2}}$ for $k=0,1,\cdots,j_{1}-1$, we have

[TABLE]

Suppose we have defined $T_{0}=0,T_{1},\cdots,T_{i}$ according to the above method and the following are tenable for $k=1,2,\cdots,i$:

[TABLE]

[TABLE]

Set

[TABLE]

similar to the above discussion, we have

[TABLE]

Consequently,

[TABLE]

if $\tau^{\prime}>0$ is sufficiently large.

Let

[TABLE]

since

[TABLE]

then $p(t)$ can be extended to a continuous positive $1$-periodic function.

Lemma 3.5

If $I_{0}$ is sufficiently large, then

[TABLE]

where the constant $l>1$.

Proof. First, by the assumption $2m+1\geq n+2$, if we let $l=\frac{2m+1}{n+2}+\frac{1}{2(n+2)}$, then $l>1$, and for sufficiently large $I_{0}$ we have

[TABLE]

If

[TABLE]

then

[TABLE]

$\square$

Proof of Theorem 1.1  By Lemma 3.5, one has

[TABLE]

Since $l>1$, then $I(t)\rightarrow+\infty$ as $t\rightarrow t_{\infty}$. Therefore, Eq.(1.3) in Theorem 1.1 possesses an unbounded solution defined in the interval $[0,T_{\infty})$, and Theorem 1.1 is proved. $\square$

Reference