A Lazer-Leach-type condition for singular differential equations with a deviating argument at resonance

Abstract

In this paper, we establish a Lazer-Leach-type condition depending on the delay for the existence of positive periodic solutions for singular differential equations with a deviating argument at resonance. The proof of the main result is based on the phase-plane analysis and topological degree methods.

1 Introduction

In this paper, we deal with the following delay differential equations at resonance:

$$ x''+\frac{1}{4}n^{2}x+g \bigl(x(t),x(t-\tau) \bigr)=p(t), $$

(1.1)

where $g:(0,\infty)\times\mathbb{R}\rightarrow\mathbb{R}$ is continuous, $\tau\geqslant0$ is a constant, $p(t)$ is continuous and 2π-periodic, and the function g has a singularity of repulsive type at the origin for its first variable, that is, $\lim_{s_{1}\to \infty}g(s_{1},s_{2})=-\infty$.

The periodic problems of singular differential equations had attracted the attentions of many researchers during more than the last two decades because of their background in applied science [1–17]. A landmark work on mathematical treatment of the differential equations with singularities, as we all know, is done by Lazer and Solimini [11]. From then on, some classical mathematical tools were used successfully to study these singular equations, such as Mawhin’s continuation theorem in the coincidence degree theory [17], the method of upper and lower solutions [8], some fixed point theorems in cones [2], the Poincaré-Birkhoff theorem [14], the phase-plane analysis and topological degree methods [12], and so on.

Wang and Ma [14] first studied the resonant singular equation

$$ x''+\frac{1}{4}n^{2}x+g(x)=p(t), $$

(1.2)

where g has a singularity and satisfies

$$ \lim_{x\rightarrow\infty}g(x)=g(+\infty). $$

They obtained the existence of 2π-periodic solutions of (1.2) under the following so-called Lazer-Leach-type condition:

$$ 4g(+\infty)- \int_{0}^{2\pi}p(t) \biggl\vert \sin\biggl(\theta+ \frac {nt}{2}\biggr) \biggr\vert \,dt\neq0 \quad \text{for all } \theta\in \mathbb{R}. $$

(1.3)

Note that they perfectly answered the open problem raised by del Pino and Manasevich [4].

After that, Wang [18] discussed the periodic problem of the resonant Liénard equation with constant delay but without singularity and established some Lazer-Leach-type conditions depending on the delay. A natural and delicate idea is that the delay may affect the existence of periodic solutions not only for the equations in [18], but also for many other kinds of equations. Based on this, in the present paper, we consider equation (1.1) and look for Lazer-Leach-type conditions. The main difficulty to overcome is the coexistence of the singularity and delay. A possible way for us is to use the phase-plane analysis and topological degree methods, also used in [12], and give the following fundamental hypotheses.

(H₁):

The variables of g are separable, that is, there exist two functions $g_{0}$ and $g_{1}$ such that $g (x(t),x(t-\tau ) )=g_{0}(x(t))+g_{1}(x(t-\tau))$. Moreover, $g_{0}$ is bounded on $[1,+\infty)$, $\lim_{x\rightarrow{+\infty}}g_{0}(x)=0$, and $\lim_{x\rightarrow{+\infty}}g_{1}(x)=g(+\infty)$ is finite.

(H₂):

For all $\theta\in\mathbb{R}$,

$$ 4\cos\frac{n\tau}{2}g(+\infty) \neq \int_{0}^{2\pi}p(t) \biggl\vert \sin\biggl( \frac{n}{2}t+\theta\biggr) \biggr\vert \,dt. $$

(1.4)

(H₃):

Put $G_{0}(x)=\int_{1}^{x}g_{0}(s)\,ds$. It satisfies

$$ \lim_{x\rightarrow0^{+}}g_{0}(x)=-\infty, \qquad \lim_{x\rightarrow 0^{+}}G_{0}(x)=+\infty. $$

(1.5)

Furthermore, there exist $0<\varepsilon_{0}\leqslant1$, $0< l\leqslant 1$, and $0< L<\infty$ such that

$$ \frac{-g_{0}(x)}{G_{0}^{1+l}(x)}>L \quad \text{for all } x\in (0, \varepsilon_{0}). $$

(1.6)

We now state our main theorem.

Theorem 1.1

Let (H₁), (H₂), and (H₃) hold. Then (1.1) has at least one periodic solution.

Remark 1.2

(H₃) is a universal strong singularity condition. We give some examples to show that (H₃) can be satisfied.

• $g_{0}(x)=-\frac{1}{x}$. Then $G_{0}(x)=-\ln x$, and (1.5) holds. Let $\varepsilon_{0}=e^{-2}$ and $l=L=1$. By using a short calculation we can obtain that (1.6) holds.

• $g_{0}(x)=-\frac{1}{x^{3}}$. Then $G_{0}(x)=\frac {1}{2x^{2}}-\frac{1}{2}$, and (1.5) holds. Let $\varepsilon _{0}=L=1$ and $l=\frac{1}{2}$. We can also check that (1.6) holds.

• $g_{0,1}(x)+g_{0,2}(x)$, where the functions $g_{0,i}(x)$ ($i=1,2$) satisfy (1.5) and (1.6). In fact, it is easy to see that (1.5) holds for $g_{0,1}(x)+g_{0,2}(x)$ and $G_{0,1}(x)+G_{0,2}(x)$, where $G_{0,i}(x)=\int_{1}^{x}g_{0,i}(s)\,ds$ ($i=1,2$). On the other hand, we assume that there exist $0<\varepsilon _{0,i}\leqslant1$, $0< l_{i}\leqslant1$, and $0< L_{i}\leqslant\infty$ such that

  $$ \frac{-g_{0,i}(x)}{G_{0,i}^{1+l}(x)}>L_{i} \quad \text{for all } x\in(0, \varepsilon_{0,i}). $$

  Without loss of generality, we assume also that $G_{0,i}(x)>1$ and $g_{0,i}(x)<-1$ for all $x\in(0,\varepsilon_{0,i})$. Set $\varepsilon _{0}=\min\{\varepsilon_{0,1},\varepsilon_{0,2}\}$. For all $x\in (0,\varepsilon_{0})$, if $G_{0,1}(x)>G_{0,2}(x)$, then

  $$ \frac{-g_{0,1}(x)-g_{0,2}(x)}{[G_{0,1}(x)+G_{0,2}(x)]^{1+l}}>\frac {-g_{0,1}(x)}{[2G_{0,1}(x)]^{1+l}}>\frac{L_{1}}{2^{1+l}}; $$

  otherwise,

  $$ \frac{-g_{0,1}(x)-g_{0,2}(x)}{[G_{0,1}(x)+G_{0,2}(x)]^{1+l}}>\frac {-g_{0,2}(x)}{[2G_{0,2}(x)]^{1+l}}>\frac{L_{2}}{2^{1+l}}. $$

  Hence (1.6) holds.

Remark 1.3

When $\tau=0$, condition (1.4) degenerates to condition (1.3). Therefore Theorem 1.1 generalizes the result in [14]. Moreover, the delay τ may affect the existence of periodic solutions.

This paper is structured into three sections. Section 2 is devoted to the proof of a useful lemma. In Section 3, we state some lemmas to prove the main theorem.

2 Preliminary lemma

To use the phase-plane analysis and topological degree methods, we embed (1.1) into a family of equations with one parameter $\lambda\in[0,1]$,

$$ x''+\frac{1}{4}n^{2}x+(1- \lambda) \biggl(-1-\frac{1}{x^{3}}\biggr)+\lambda g \bigl(x(t),x(t-\tau) \bigr)= \lambda p(t). $$

(2.1)

Now, we give the following fundamental lemma.

Lemma 2.1

Suppose that there exist three positive constants $M_{0}$, $M_{1}$, and $M_{2}$ such that, for any 2π-periodic solution $x(t)$ of (2.1),

$$ M_{0}< x(t)< M_{1} \quad \textit{for all } t\in \mathbb{R} $$

and

$$ \bigl\Vert x' \bigr\Vert _{\infty}\triangleq \max_{t\in[0,2\pi]} \bigl\vert x'(t) \bigr\vert < M_{2}. $$

Then Eq. (1.1) has at least one 2π-periodic solution.

Since the proof is similar as that in [12], we omit it.

Remark 2.2

In fact, Lemma 2.1 is also valid if we embed (1.1) into the following family of equations with one parameter $\lambda\in[0,1]$:

$$ x''+\frac{1}{4}n^{2}x+(1- \lambda) \biggl(1-\frac{1}{x^{3}}\biggr)+\lambda g \bigl(x(t),x(t-\tau) \bigr)= \lambda p(t). $$

(2.2)

3 The proof of the main theorem

Condition (H₂) reduces to

($\mathrm{H}'_{2}$):

for all $\theta\in\mathbb{R}$,

$$ 4\cos\frac{n\tau}{2}g(+\infty) < \int_{0}^{2\pi}p(t) \biggl\vert \sin \biggl( \frac{n}{2}t+\theta\biggr) \biggr\vert \,dt $$

or

($\mathrm{H}''_{2}$):

for all $\theta\in\mathbb{R}$,

$$ 4\cos\frac{n\tau}{2}g(+\infty) > \int_{0}^{2\pi}p(t) \biggl\vert \sin \biggl( \frac{n}{2}t+\theta\biggr) \biggr\vert \,dt. $$

In this section, we always assume that ($\mathrm{H}'_{2}$) holds. The argument for ($\mathrm{H}''_{2}$) is similar.

We first suppose that the sequence $\{(x_{k},y_{k})\}_{k=1}^{\infty}$ satisfies

$$ x'_{k}=y_{k},\qquad y'_{k}=-\frac{1}{4}n^{2}x_{k}-g \bigl(x_{k},x_{k}(t-\tau),\lambda_{k}\bigr) $$

(3.1)

with $\|x_{k}\|_{\infty}+\|y_{k}\|_{\infty}\rightarrow\infty$ as $k\rightarrow\infty$, where

$$ g\bigl(x_{k},x_{k}(t-\tau),\lambda_{k} \bigr)=(1-\lambda_{k}) \biggl(-1-\frac {1}{x_{k}^{3}}\biggr)+ \lambda_{k} g_{0}(x_{k})+\lambda_{k} g_{1}\bigl(x_{k}(t-\tau)\bigr)-\lambda_{k} p(t). $$

In this position, we only consider $\lambda_{k}\to\lambda_{0}\in[0,1]$. Even if $\lambda_{k}$ has no limit, we can consider its convergent subsequence $\{\lambda_{k_{i}}\}$ and the corresponding solution sequence $\{(x_{k_{i}},y_{k_{i}})\}$ for (3.1) because the sequence $\{\lambda_{k}\}$ is bounded. For simplicity, we omit these discussions. It is easy to see that $\|x_{k}\|_{\infty}+\|y_{k}\|_{\infty}\rightarrow \infty$ is equivalent to $\|x_{k}\|_{\infty}\rightarrow\infty$ and $\| y_{k}\|_{\infty}\rightarrow\infty$ as $k\rightarrow\infty$. Define

$$ g_{0}(x_{k},\lambda_{k})=-(1- \lambda_{k})\frac{1}{x_{k}^{3}}+\lambda_{k} g_{0}(x_{k}) $$

and

$$ g_{1}\bigl(x_{k}(t-\tau),\lambda_{k}\bigr)=-(1- \lambda_{k})+\lambda_{k} g_{1} \bigl(x_{k}(t-\tau)\bigr). $$

Obviously, $g_{0}(x_{k},\lambda_{k})$ satisfies condition (H₃).

Take the transformation

$$ x_{k}=1+r_{k}\cos\theta_{k}, \qquad y_{k}=\frac{n}{2}r_{k}\sin\theta_{k}. $$

Then system (3.1) is equivalent to the following system:

$$ \textstyle\begin{cases} \frac{d\theta_{k}}{dt}=-\frac{n}{2}-\frac{2}{nr_{k}}g(1+r_{k}\cos\theta _{k},1+r_{k}(t-\tau)\cos\theta_{k}(t-\tau),\lambda_{k})\cos\theta _{k}+\frac{n}{2r_{k}}\cos\theta_{k}, \\ \frac{dr_{k}}{dt}=-\frac{n}{2}\sin\theta_{k}-\frac{2}{n}g(1+r_{k}\cos \theta_{k},1+r_{k}(t-\tau)\cos\theta_{k}(t-\tau),\lambda_{k})\sin\theta_{k}. \end{cases} $$

(3.2)

Lemma 3.1

Assume that (H₁) and (H₃) hold. For k large enough, we have

$$ \theta'_{k}(t)< 0 \quad \forall t\in\mathbb{R}. $$

Proof

By (H₃) and the definition of $g(x_{k},x_{k}(t-\tau),\lambda_{k})$ we have that there exists $0<\delta <1$ such that

$$ g\bigl(x_{k},x_{k}(t-\tau),\lambda_{k}\bigr)- \frac{n^{2}}{4}< 0. $$

Noticing that $\cos\theta_{k}<0$ for $0< x_{k}\leqslant\delta<1$, we get, for $0< x_{k}\leqslant\delta$,

$$ \frac{d\theta_{k}}{dt}\leqslant-\frac{n}{2}. $$

(3.3)

Since $\|x_{k}\|_{\infty}\rightarrow\infty$ and $\|y_{k}\|_{\infty }\rightarrow\infty$ as $k\rightarrow\infty$, we have that if $x_{k}>\delta$, then

$$ r_{k}\rightarrow\infty \quad \text{as } k\rightarrow\infty. $$

Meanwhile, if $x_{k}>\delta$, by (H₁), then we get that there exists $M>0$ such that

$$ \bigl\vert g\bigl(x_{k},x_{k}(t-\tau), \lambda_{k}\bigr) \bigr\vert < M. $$

Hence, for k large enough, if $x_{k}>\delta$, then

$$ \biggl\vert -\frac{2}{nr_{k}}g\bigl(1+r_{k}\cos \theta_{k},1+r_{k}(t-\tau)\cos\theta _{k}(t- \tau),\lambda_{k}\bigr)\cos\theta_{k}+\frac{n}{2r_{k}}\cos \theta _{k} \biggr\vert < \frac{n}{4}, $$

and then

$$ \frac{d\theta_{k}}{dt}\leqslant-\frac{n}{2}+\frac{n}{4}=- \frac{n}{4}. $$

(3.4)

From (3.3) and (3.4) we obtain the conclusion of Lemma 3.1. □

From Lemma 3.1 we conclude that, for k large enough, the solution $(x_{k}(t),x_{k}(t))$ of (3.1) makes clockwise rotations around the point $(1,0)$. Without loss of generality, we take the initial point $(x_{k}(t_{0}^{m,k}),y_{k}(t_{0}^{m,k}))$ of the mth rotation that satisfies

$$ x_{k}\bigl(t_{0}^{m,k}\bigr)=1, \qquad y_{k}\bigl(t_{0}^{m,k}\bigr)=x'_{k} \bigl(t_{0}^{m,k}\bigr)>0, $$

and

$$ \theta_{k}\bigl(t_{0}^{m,k}\bigr)= \frac{\pi}{2}-2(m-1)\pi, $$

where $m=1,2,\ldots$ .

Let $(x_{k}(t),y_{k}(t))$ take exactly one rotation from $t_{0}^{m,k}$. Then there are two points where the curve $(x_{k}(t),y_{k}(t))$ meets the line $x_{k}=1$ and two points where the curve $(x_{k}(t),y_{k}(t))$ meets the $x_{k}$-axis. We denote them by $(x_{k}(t_{1}^{m,k}),0)$, $(1,y_{k}(t_{2}^{m,k}))$, $(x_{k}(t_{3}^{m,k}),0)$, and $(1,y_{k}(t_{4}^{m,k}))$, where $x_{k}(t_{1}^{m,k})>1$, $y_{k}(t_{2}^{m,k})<0$, $0< x_{k}(t_{3}^{m,k})<1$, and $y_{k}(t_{4}^{m,k})=y_{k}(t_{0}^{m+1,k})>0$.

Define

$$ \mathcal{R}_{k}^{2}(t)=\bigl(x_{k}(t)-1 \bigr)^{2}+\frac{4}{n^{2}}y_{k}^{2}(t)+ \int _{t_{0}^{1,k}}^{t}\frac{8}{n^{2}}y_{k}(s) \biggl(g\bigl(x_{k}(s),x_{k}(s-\tau ),\lambda_{k} \bigr)+\frac{1}{4}n^{2} \biggr)\,ds. $$

(3.5)

Since $\frac{d\mathcal{R}_{k}^{2}}{dt}\equiv0$, we get $\mathcal {R}_{k}^{2}(t)=\mathcal{R}_{k}^{2}$, where $\mathcal{R}_{k}>0$. Obviously, we also get that

$$ \mathcal{R}_{k}^{2}(t)=x_{k}^{2}(t)+ \frac{4}{n^{2}}y_{k}^{2}(t)+ \int _{t_{0}^{1,k}}^{t}\frac{8}{n^{2}}y_{k}(s)g \bigl(x_{k}(s),x_{k}(s-\tau),\lambda_{k}\bigr) \,ds-1 $$

and define

$$ G_{k}(t)= \int_{t_{0}^{1,k}}^{t}\frac{8}{n^{2}}y_{k}(s) \biggl(g\bigl(x_{k}(s),x_{k}(s-\tau),\lambda_{k} \bigr)+\frac{1}{4}n^{2} \biggr)\,ds. $$

(3.6)

Therefore we have $\mathcal{R}_{k}^{2}=r_{k}^{2}(t)+G_{k}(t)$.

Lemma 3.2

Assume that (H₁) and (H₃) hold. Then

$$ \lim_{k\rightarrow\infty}x_{k}\bigl(t_{1}^{m,k} \bigr)=+\infty \quad \textit {and} \quad \lim_{k\rightarrow\infty}x_{k} \bigl(t_{3}^{m,k}\bigr)=0 $$

for $m=1,2,\ldots$ .

Proof

Without loss of generality, we assume that there exist a positive integer $m^{*}$ such that

$$ x_{k}\bigl(t_{1}^{m^{*},k}\bigr)=\max _{t\in\mathbb{R}}x_{k}(t). $$

Noticing that $\lim_{k\rightarrow\infty}\max_{t\in\mathbb {R}}x_{k}(t)\rightarrow\infty$, we get $\lim_{k\rightarrow\infty }x_{k}(t_{1}^{m^{*},k})\rightarrow\infty$. We will prove that $\lim_{k\rightarrow\infty}x_{k}(t_{3}^{m^{*},k})=0$ and $\lim_{k\rightarrow \infty}x_{k}(t_{1}^{m^{*}+1,k})=+\infty$, and the others are similar.

We first prove that $\lim_{k\rightarrow\infty}x_{k}(t_{3}^{m^{*},k})=0$. Assume by contradiction that there exists a constant $0< c<1$ such that $x_{k}(t_{3}^{m^{*},k})>c$ for $k\in\mathbb{N}^{+}$. Thus, if $t\in [t_{1}^{m^{*},k},t_{3}^{m^{*},k}]$, then $g(x_{k},x_{k}(t-\tau),\lambda_{k})$ is bounded. Hence, from the second equality of (3.2) we obtain $r_{k}(t_{3}^{m^{*},k})=r_{k}(t_{1}^{m^{*},k})+O(1)$. Therefore, $r_{k}(t_{3}^{m^{*},k})\rightarrow\infty$, which contradicts the fact $0< x_{k}(t_{3}^{m^{*},k})<1$ and $x'_{k}(t_{3}^{m^{*},k})=0$. Consequently, $\lim_{k\rightarrow\infty}x_{k}(t_{3}^{m^{*},k})=0$.

Next, we prove that $\lim_{k\rightarrow\infty }x_{k}(t_{1}^{m^{*}+1,k})=+\infty$. Since $g(x_{k},x_{k}(t-\tau),\lambda_{k})$ is bounded for $t\in[t_{1}^{m^{*},k},t_{2}^{m^{*},k}]$, we have $\lim_{k\rightarrow\infty}r_{k}(t_{2}^{m^{*},k})\rightarrow\infty$. It follows from $x_{k}(t_{2}^{m^{*},k})=1$ that $\lim_{k\rightarrow\infty }y_{k}(t_{2}^{m^{*},k})\rightarrow\infty$. By (3.6) we obtain

$$\begin{aligned} G_{k}\bigl(t_{0}^{m^{*}+1,k}\bigr)-G_{k} \bigl(t_{2}^{m^{*},k}\bigr)&=G_{k}\bigl(t_{4}^{m^{*},k} \bigr)-G_{k}\bigl(t_{2}^{m^{*},k}\bigr) \\ &= \int_{t_{2}^{m^{*},k}}^{t_{4}^{m^{*},k}}\frac{8}{n^{2}}y_{k}(s) \biggl(g\bigl(x_{k}(s),x_{k}(s-\tau),\lambda_{k} \bigr)+\frac{1}{4}n^{2} \biggr)\,ds \\ &= \int_{t_{2}^{m^{*},k}}^{t_{4}^{m^{*},k}}\frac{8\lambda _{k}}{n^{2}}y_{k}(s)g_{1} \bigl(x_{k}(s-\tau)\bigr)\,ds \\ &=o\bigl(r_{k}^{2}\bigl(t_{2}^{m^{*},k} \bigr)\bigr). \end{aligned}$$

It follows from the relation $\mathcal{R}_{k}^{2}=r_{k}^{2}(t)+G_{k}(t)$ that

$$ r_{k}^{2}\bigl(t_{0}^{m^{*}+1,k} \bigr)=r_{k}^{2}\bigl(t_{2}^{m^{*},k}\bigr)- \bigl(G_{k}\bigl(t_{0}^{m^{*}+1}\bigr)-G_{k} \bigl(t_{2}^{m^{*},k}\bigr)\bigr)=r_{k}^{2} \bigl(t_{2}^{m^{*},k}\bigr)+o\bigl(r_{k}^{2} \bigl(t_{2}^{m^{*},k}\bigr)\bigr). $$

Hence, $r_{k}(t_{0}^{m^{*}+1,k})\rightarrow\infty$. By using the second equality of (3.2) again we obtain $r_{k}(t_{1}^{m^{*}+1,k})\rightarrow \infty$. Equivalently, $\lim_{k\rightarrow\infty }x_{k}(t_{1}^{m^{*}+1,k})=+\infty$. □

Denote by $\tau_{k}^{m}$ the required time for the solution $(x_{k}(t),y_{k}(t))$ to complete the mth rotation around the point $(0,1)$. Then,

$$ \tau_{k}^{m}=\bigl(t_{2}^{m,k}-t_{0}^{m,k} \bigr)+\bigl(t_{4}^{m,k}-t_{2}^{m,k}\bigr). $$

Lemma 3.3

Assume that (H₁) and (H₃) hold. Then, for k large enough,

$$ \tau^{m}_{k}=\frac{2\pi}{n}+o(1). $$

Proof

We first compute $t_{2}^{m,k}-t_{0}^{m,k}$. Since $g(x_{k},x_{k}(t-\tau),\lambda_{k})$ is bounded on the interval $[t_{0}^{m,k},t_{2}^{m,k}]$, we get from the first equality of (3.2) that

$$ \frac{dt}{d\theta_{k}}=-\frac{2}{n}+o(1), $$

which implies $t_{2}^{m,k}-t_{0}^{m,k}=\frac{2\pi}{n}+o(1)$.

Next, we estimate $(t_{4}^{m,k}-t_{2}^{m,k})$. By (3.5) we have, for $t\in[t_{2}^{m,k},t_{3}^{m,k}]$,

$$ \frac{dt}{dx_{k}}=\frac{1}{y_{k}}=-\frac{2}{n}\frac{1}{\sqrt{\mathcal {R}_{k}^{2}-(x_{k}(t)-1)^{2}-G_{k}(t)}}. $$

Furthermore,

$$ -\frac{2}{n}\,dx_{k}=\sqrt{ \mathcal{R}_{k}^{2}-\bigl(x_{k}(t)-1 \bigr)^{2}-G_{k}(t)}\,dt. $$

(3.7)

By (H₃) and Remark 1.2 we have that there exist $0<\varepsilon _{0}\leqslant1$, $0< l\leqslant1$, and $0< L<\infty$ such that

$$ \frac{-g_{0}(x_{k},\lambda_{k})}{G_{0}^{1+l}(x_{k},\lambda_{k})}>L\quad \text{for all } x\in(0,\varepsilon_{0}). $$

For above l, we can choose $t_{-}^{m,k}\in[t_{2}^{m,k},t_{3}^{m,k}]$ such that

$$ \begin{aligned} &G_{k}\bigl(t_{-}^{m,k}\bigr)=\mathcal{R}_{k}^{\frac{2}{1+\frac{l}{2}}}, \\ &\mathcal{R}_{k}^{\frac{2}{1+\frac{l}{2}}}\leqslant G_{k}(t)\leqslant \mathcal{R}_{k}^{2} \quad \text{for } t\in \bigl[t_{-}^{m,k},t_{3}^{m,k}\bigr], \end{aligned} $$

and

$$ G_{k}(t)\leqslant\mathcal{R}_{k}^{\frac{2}{1+\frac{l}{2}}}\quad \text{for } t\in\bigl[t_{2}^{m,k},t_{-}^{m,k}\bigr]. $$

Integrating over $[t_{2}^{m,k}, t_{-}^{m,k}]$ for (3.7), we have

$$ \int_{x_{k}(t_{2}^{m,k})}^{x_{k}(t_{-}^{m,k})}-\frac{2}{n}\,dx_{k}= \int _{t_{2}^{m,k}}^{t_{-}^{m,k}}\sqrt{\mathcal{R}_{k}^{2}- \bigl(x_{k}(t)-1\bigr)^{2}-G_{k}(t)}\,dt. $$

Noticing that $x_{k}(t_{2}^{m,k})\rightarrow1$ and $x_{k}(t_{-}^{m,k})\rightarrow0$, we get

$$ \int_{x_{k}(t_{2}^{m,k})}^{x_{k}(t_{-}^{m,k})}-\frac{2}{n}\,dx_{k}= \frac {2}{n}\bigl(1-x_{k}\bigl(t_{-}^{m,k}\bigr)\bigr)= \frac{2}{n}+o(1). $$

On the other hand, we get

$$\begin{aligned}& \int_{t_{2}^{m,k}}^{t_{-}^{m,k}}\sqrt{\mathcal {R}_{k}^{2}-\bigl(x_{k}(t)-1\bigr)^{2}-G_{k}(t)} \,dt \\& \quad = \bigl(t_{-}^{m,k}-t_{2}^{m,k}\bigr) \int_{0}^{1} \sqrt{\mathcal{R}_{k}^{2}- \bigl(x_{k} \bigl(t_{2}^{m,k}+s \bigl(t_{-}^{m,k}-t_{2}^{m,k}\bigr) \bigr)-1 \bigr)^{2}-G_{k} \bigl(t_{2}^{m,k}+s \bigl(t_{-}^{m,k}-t_{2}^{m,k}\bigr) \bigr)}\,ds \\& \quad =\bigl(t_{-}^{m,k}-t_{2}^{m,k}\bigr) \mathcal{R}_{k} \int_{0}^{1} \sqrt{1-\frac { (x_{k} (t_{2}^{m,k}+s(t_{-}^{m,k}-t_{2}^{m,k}) )-1 )^{2}}{\mathcal{R}_{k}^{2}}- \frac{G_{k} (t_{2}^{m,k}+s(t_{-}^{m,k}-t_{2}^{m,k}) )}{\mathcal{R}_{k}^{2}}}\,ds. \end{aligned}$$

Thus we have

$$ \frac{\frac{2}{n}+o(1)}{(t_{-}^{m,k}-t_{2}^{m,k})\mathcal{R}_{k}}= \int _{0}^{1} \sqrt{1-\frac{ (x_{k} (t_{2}^{m,k}+s(t_{-}^{m,k}-t_{2}^{m,k}) )-1 )^{2}}{\mathcal{R}_{k}^{2}}- \frac{G_{k} (t_{2}^{m,k}+s(t_{-}^{m,k}-t_{2}^{m,k}) )}{\mathcal{R}_{k}^{2}}}\,ds. $$

Passing to limits and using the Lebesgue dominated convergence theorem, we get

$$ \lim_{k\rightarrow\infty}\mathcal{R}_{k}\bigl(t_{-}^{m,k}-t_{2}^{m,k} \bigr)=\frac{2}{n}, $$

which implies

$$ t_{-}^{m,k}-t_{2}^{m,k}= \frac{2}{n\mathcal{R}_{k}}+o\biggl(\frac{1}{\mathcal{R}_{k}}\biggr). $$

(3.8)

By (3.1) we get, for $t\in[t_{-}^{m,k},t_{3}^{m,k}]$,

$$\begin{aligned} \frac{dt}{dy_{k}}&=\frac{1}{-\frac{1}{4}n^{2}x_{k}-g(x_{k},x_{k}(t-\tau ),\lambda_{k})} \\ &=\frac{1}{-\frac{1}{4}n^{2}x_{k}-g_{0}(x_{k},\lambda_{k})-g_{1}(x_{k}(t-\tau ),\lambda_{k})+\lambda_{k} p(t)} \\ &=\frac{1}{-g_{0}(x_{k},\lambda_{k})+O(1)}. \end{aligned}$$

Since $x_{k}(t)\to0$ as $k\to\infty$ for $t\in[t_{-}^{m,k},t_{3}^{m,k}]$, we obtain, for k large enough, $[x_{k}(t_{3}^{m,k}), x_{k}(t_{-}^{m,k})]\subset(0,\varepsilon_{0})$. Then, for $t\in[t_{-}^{m,k},t_{3}^{m,k}]$,

$$ \frac{1}{-g_{0}(x_{k},\lambda_{k})}< \frac{1}{L}\frac {1}{G_{0}^{1+l}(x_{k},\lambda_{k})}, $$

which yields that, for $t\in[t_{-}^{m,k},t_{3}^{m,k}]$,

$$\begin{aligned} \frac{dt}{dy_{k}}&\leqslant\frac{1}{L}\frac{1}{G_{0}^{1+l}(x_{k},\lambda _{k})}+o\biggl( \frac{1}{G_{0}^{1+l}(x_{k},\lambda_{k})}\biggr) \\ &=\frac{1}{L}\frac{1}{G_{k}^{1+l}(t)}+o\biggl(\frac{1}{G_{k}^{1+l}(t)}\biggr) \\ &\leqslant\frac{1}{L}\frac{1}{\mathcal{R}_{k}^{\frac{2(1+l)}{1+\frac {l}{2}}}}+o \biggl(\frac{1}{\mathcal{R}_{k}^{\frac{2(1+l)}{1+\frac {l}{2}}}} \biggr). \end{aligned}$$

Integrating over $[t_{-}^{m,k}, t_{3}^{m,k}]$ and noticing that $y_{k}(t)\leqslant\frac{n}{2}\mathcal{R}_{k}$, we have

$$ t_{3}^{m,k}-t_{-}^{m,k}\leqslant\frac{n}{2L\mathcal{R}_{k}^{\frac {2(1+l)}{1+\frac{l}{2}}}-1}+o \biggl(\frac{1}{\mathcal{R}_{k}^{\frac {2(1+l)}{1+\frac{l}{2}}}-1} \biggr), $$

which implies that

$$ t_{3}^{m,k}-t_{-}^{m,k}=o\biggl( \frac{1}{\mathcal{R}_{k}}\biggr). $$

(3.9)

It follows from (3.8) and (3.9) that

$$ t_{3}^{m,k}-t_{2}^{m,k}= \frac{2}{n\mathcal{R}_{k}}+o\biggl(\frac{1}{\mathcal{R}_{k}}\biggr). $$

(3.10)

Using a similar argument, we can prove that $t_{4}^{m,k}-t_{3}^{m,k}=\frac {2}{n\mathcal{R}_{k}}+o(\frac{1}{\mathcal{R}_{k}})$. It follows that

$$ t_{4}^{m,k}-t_{2}^{m,k}= \frac{4}{n\mathcal{R}_{k}}+o\biggl(\frac{1}{\mathcal{R}_{k}}\biggr). $$

(3.11)

Then $t_{4}^{m,k}-t_{2}^{m,k}=o(1)$. This completes the proof. □

Since $(x_{k},y_{k})$ is 2π-periodic, by Lemma 3.3 we know that, for k sufficiently large, $(x_{k},y_{k})$ makes exactly n clockwise revolutions around the point $(1,0)$ as t varies from 0 to 2π.

We use the following lemma to estimate the maxima.

Lemma 3.4

Assume that (H₁), ($\mathrm{H}'_{2}$), and (H₃) hold. There exist two positive constants $M_{1}$ and $M_{2}$ such that, for any 2π-periodic solution $x(t)$ of Eq. (2.1),

$$ \Vert x \Vert _{\infty}< M_{1},\qquad \bigl\Vert x' \bigr\Vert _{\infty}< M_{2}. $$

(3.12)

Proof

Assume by contradiction that there exists a sequence $\{(x_{k},y_{k})\}_{k=1}^{\infty}$ satisfying the system (3.1) with $\|x_{k}\|_{\infty}+\|y_{k}\|_{\infty }\rightarrow\infty$ as $k\rightarrow\infty$. It follows that $\| x_{k}\|_{\infty}\rightarrow\infty$ and $\|y_{k}\|_{\infty}\rightarrow \infty$.

Using the transformation

$$ x_{k}=1+r_{k}\cos\theta_{k},\qquad y_{k}=\frac{n}{2}r_{k}\sin\theta_{k}, $$

we have system (3.2).

Without loss of generality, we take the initial point $(x_{k}(t_{0}^{m,k}),y_{k}(t_{0}^{m,k}))$ of the mth rotation satisfying

$$ x_{k}\bigl(t_{0}^{m,k}\bigr)=1,\qquad y_{k}\bigl(t_{0}^{m,k}\bigr)=x'_{k} \bigl(t_{0}^{m,k}\bigr)>0, $$

and

$$ \theta_{k}\bigl(t_{0}^{m,k}\bigr)= \frac{\pi}{2}-2(m-1)\pi $$

for $m=1,2,\ldots,n$.

Let $(x_{k}(t),y_{k}(t))$ take exactly one rotation from $t_{0}^{m,k}$. Denote by $\tau_{k}$ the required time for the solution $(x_{k}(t),y_{k}(t))$ to complete n rotations around the point $(0,1)$. Then,

$$ \tau_{k}=\sum_{m=1}^{n} \tau_{k}^{m}=\sum_{m=1}^{n} \bigl(t_{4}^{m,k}-t_{0}^{m,k}\bigr)=\sum _{m=1}^{n}\bigl(t_{2}^{m,k}-t_{0}^{m,k} \bigr)+\sum_{m=1}^{n}\bigl(t_{4}^{m,k}-t_{2}^{m,k} \bigr). $$

We further estimate $\sum_{m=1}^{n}(t_{2}^{m,k}-t_{0}^{m,k})$. By the second equality of (3.2) we get, for $t\in(t_{0}^{m,k}, t_{2}^{m,k})$,

$$ r_{k}(t)=r_{k}\bigl(t_{0}^{m} \bigr)+O(1). $$

Hence, for all $t\in[t_{0}^{m,k}, t_{2}^{m,k}]$, $1\leqslant m\leqslant n$,

$$ r_{k}(t)=\mathcal{R}_{k}+o( \mathcal{R}_{k}). $$

(3.13)

Furthermore, $\frac{1}{r_{k}(t)}=\frac{1}{\mathcal{R}_{k}}+o(\frac {1}{\mathcal{R}_{k}})$ for $t\in[t_{0}^{m,k}, t_{2}^{m,k}]$.

By the first equality of (3.2), we obtain, for $t\in[t_{0}^{m,k}, t_{2}^{m,k}]$,

$$\begin{aligned} \frac{dt}{d\theta_{k}} =&-\frac{2}{n} \\ &{}\cdot \frac{1}{1+\frac {4}{n^{2}\mathcal{R}_{k}}g(1+r_{k}\cos\theta_{k},1+r_{k}(t-\tau)\cos\theta _{k}(t-\tau),\lambda_{k})\cos\theta_{k}+\frac{1}{\mathcal{R}_{k}}\cos \theta_{k}+o(\frac{1}{\mathcal{R}_{k}})} \\ =&-\frac{2}{n}+\frac{8}{n^{3}\mathcal{R}_{k}}g\bigl(1+r_{k}\cos\theta _{k},1+r_{k}(t-\tau)\cos\theta_{k}(t-\tau), \lambda_{k}\bigr)\cos\theta _{k} \\ &{}+\frac{2}{n\mathcal{R}_{k}}\cos \theta_{k}+o\biggl(\frac{1}{\mathcal{R}_{k}}\biggr). \end{aligned}$$

Integrating over $[-\frac{\pi}{2}-2(m-1)\pi, \frac{\pi }{2}-2(m-1)\pi]$, we get

$$\begin{aligned} \begin{aligned} &t_{2}^{m,k}-t_{0}^{m,k} \\ &\quad = \int_{-\frac{\pi}{2}-2(m-1)\pi}^{\frac{\pi }{2}-2(m-1)\pi} \biggl[\frac{2}{n}- \frac{8}{n^{3}\mathcal {R}_{k}}g\bigl(1+r_{k}\cos\theta_{k},1+r_{k}(t- \tau)\cos\theta_{k}(t-\tau ),\lambda_{k}\bigr)\cos \theta_{k} \\ &\qquad {} -\frac{2}{n\mathcal{R}_{k}}\cos\theta_{k}+o\biggl(\frac{1}{\mathcal {R}_{k}} \biggr) \biggr]\, d\theta_{k} \\ &\quad =\frac{2\pi}{n}-\frac{4}{n\mathcal{R}_{k}}- \int_{-\frac{\pi }{2}-2(m-1)\pi}^{\frac{\pi}{2}-2(m-1)\pi}\frac{8}{n^{3}\mathcal {R}_{k}} \\ &\qquad {} \cdot g\bigl(1+r_{k}\cos\theta_{k},1+r_{k}(t- \tau)\cos\theta _{k}(t-\tau),\lambda_{k}\bigr)\cos \theta_{k}\, d\theta_{k}+o\biggl(\frac{1}{\mathcal{R}_{k}}\biggr). \end{aligned} \end{aligned}$$

Recalling that $\lambda_{k}\to\lambda_{0}\in[0,1]$, we distinguish two cases.

Case 1: $\lambda_{0}=0$. In this case, recalling that $g_{0}(x_{k})+g_{1}(x_{k}(t-\tau))-p(t)$ is bounded for $\theta_{k}\in (-\frac{\pi}{2}-2(m-1)\pi,\frac {\pi}{2}-2(m-1)\pi )$, we have, for k large enough,

$$ \bigl\vert \lambda_{k}g_{0}(x_{k})+ \lambda_{k}g_{1}\bigl(x_{k}(t-\tau)\bigr)-\lambda _{k}p(t) \bigr\vert < \frac{1}{3}. $$

Then it follows from the definition of $g(1+r_{k}\cos\theta _{k},1+r_{k}(t-\tau)\cos\theta_{k}(t-\tau),\lambda_{k})$ that, for k large enough (such that $\lambda_{k}<\frac{1}{3}$) and for $\theta _{k}\in (-\frac{\pi}{2}-2(m-1)\pi,\frac{\pi}{2}-2(m-1)\pi )$,

$$ g\bigl(1+r_{k}\cos\theta_{k},1+r_{k}(t-\tau) \cos\theta_{k}(t-\tau),\lambda _{k}\bigr)< -(1- \lambda_{k})+\frac{1}{3}< -\frac{1}{3}. $$

Hence,

$$ t_{2}^{m,k}-t_{0}^{m,k}> \frac{2\pi}{n}-\frac{4}{n\mathcal{R}_{k}}+\frac {16}{3n^{3}\mathcal{R}_{k}}+o\biggl( \frac{1}{\mathcal{R}_{k}}\biggr), $$

and then

$$ \sum_{m=1}^{n} \bigl(t_{2}^{m,k}-t_{0}^{m,k} \bigr)>2\pi-\frac {4}{\mathcal{R}_{k}}+\frac{16}{3n^{2}\mathcal{R}_{k}}+o\biggl(\frac{1}{\mathcal{R}_{k}} \biggr). $$

Therefore, by (3.11) we have, for k large enough,

$$ \tau_{k}=\sum_{m=1}^{n} \bigl(t_{2}^{m,k}-t_{0}^{m,k} \bigr)+\sum _{m=1}^{n} \bigl(t_{4}^{m,k}-t_{2}^{m,k} \bigr)>2\pi, $$

which is a contradiction. Consequently, (3.12) holds.

Case 2: $\lambda_{0}>0$. Then there exists a constant $l_{0}>0$ such that, for k large enough, $\lambda _{k}\geqslant l_{0}$. From (3.11) and the first equality of (3.2) we have, for $\theta_{k}\in (-\frac{\pi}{2}-2(m-1)\pi ,\frac{\pi}{2}-2(m-1)\pi )$,

$$ t(\theta_{k})=t_{0}^{1,k}+\frac{2}{n} \biggl(\frac{\pi}{2}-\theta_{k}\biggr)-\frac {2(m-1)\pi}{n}+o(1). $$

Then, from the definition of $g(1+r_{k}\cos\theta_{k},1+r_{k}(t-\tau)\cos \theta_{k}(t-\tau),\lambda_{k})$ we obtain

$$\begin{aligned}& \int_{-\frac{\pi}{2}-2(m-1)\pi}^{\frac{\pi}{2}-2(m-1)\pi} g\bigl(1+r_{k}\cos \theta_{k},1+r_{k}(t-\tau)\cos\theta_{k}(t-\tau), \lambda _{k}\bigr)\cos\theta_{k}\, d\theta_{k} \\& \quad \leqslant\lambda_{k} \int_{-\frac{\pi}{2}-2(m-1)\pi}^{\frac{\pi }{2}-2(m-1)\pi}\bigl(g_{1} \bigl(1+r_{k}(t-\tau)\cos\theta_{k}(t-\tau)\bigr)-p(t)\bigr) \cos \theta_{k}\, d\theta_{k} \\& \quad =-\frac{n\lambda_{k}}{2} \int_{t_{0}^{1,k}+\frac{2m\pi }{n}+o(1)}^{t_{0}^{1,k}+\frac{2(m-1)\pi}{n}+o(1)} \bigl(g_{1} \bigl(x_{k}(t-\tau)\bigr)-p(t)\bigr)\cdot \biggl\vert \cos\biggl( \frac{\pi}{2}-\frac {n}{2}t\biggr) \biggr\vert \,dt \\& \quad =\frac{n\lambda_{k}}{2} \int_{\frac{2(m-1)\pi}{n}}^{\frac{2m\pi }{n}} \bigl(g_{1} \bigl(x_{k}(t-\tau)\bigr)-p(t)\bigr)\cdot \biggl\vert \cos\biggl( \frac{\pi}{2}-\frac {n}{2}t\biggr) \biggr\vert \,dt+o(1). \end{aligned}$$

Denote

$$ I_{1}=\biggl[\frac{2(m-1)\pi}{n},\frac{2m\pi}{n}\biggr]\cap\bigl\{ t:x_{k}(t-\tau )\geqslant1\bigr\} $$

and

$$ I_{0}=\biggl[\frac{2(m-1)\pi}{n},\frac{2m\pi}{n}\biggr]\cap\bigl\{ t:x_{k}(t-\tau)< 1\bigr\} . $$

Then $\operatorname{mes}(I_{0})=o(1)$. Denote again

$$ I'_{1}=\biggl[\frac{2(m-1)\pi}{n}-\tau, \frac{2m\pi}{n}-\tau\biggr]\cap\bigl\{ t:x_{k}(t)\geqslant1\bigr\} . $$

For $m=2i-1$, $i=1,2,\ldots$ , we have

$$\begin{aligned}& \int_{\frac{2(m-1)\pi}{n}}^{ \frac{2m\pi}{n}} g_{1}\bigl(x_{k}(t- \tau )\bigr)\cdot \biggl\vert \cos\biggl(\frac{\pi}{2}-\frac{n}{2}t \biggr) \biggr\vert \,dt \\& \quad = \int_{\frac{2(m-1)\pi}{n}}^{ \frac{2m\pi}{n}} g_{1}\bigl(x_{k}(t- \tau )\bigr)\cdot\sin\biggl(\frac{n}{2}t\biggr)\,dt \\& \quad = \int_{I_{1}}g_{1}\bigl(x_{k}(t-\tau)\bigr) \cdot\sin\biggl(\frac{n}{2}t\biggr)\,dt+ \int _{I_{0}}g_{1}\bigl(x_{k}(t-\tau)\bigr) \cdot\sin\biggl(\frac{n}{2}t\biggr)\,dt \\& \quad = \int_{I_{1}}g_{1}\bigl(1+r_{k}(t-\tau)\cos \theta_{k}(t-\tau)\bigr)\cdot\sin \biggl(\frac{n}{2}t\biggr) \,dt+o(1) \\& \quad = \int_{I_{1}}g_{1}\biggl(\mathcal{R}_{k}\sin \biggl(\frac{n}{2}t-\frac{n}{2}\tau \biggr)+O(1)\biggr)\cdot\sin \biggl(\frac{n}{2}t\biggr)\,dt+o(1) \\& \quad = \int_{I'_{1}}g_{1}\biggl(\mathcal{R}_{k}\sin \biggl(\frac{n}{2}t\biggr)+O(1)\biggr)\cdot\sin \biggl(\frac{n}{2}t+ \frac{n}{2}\tau\biggr)\,dt+o(1) \\& \quad =\cos\biggl(\frac{n}{2}\tau\biggr) \int_{I'_{1}}g_{1}\biggl(\mathcal{R}_{k}\sin \biggl(\frac {n}{2}t\biggr)+O(1)\biggr)\cdot\sin\biggl(\frac{n}{2}t \biggr)\,dt \\& \qquad {}+\sin\biggl(\frac{n}{2}\tau\biggr) \int_{I'_{1}}g_{1}\biggl(\mathcal{R}_{k}\sin \biggl(\frac{n}{2}t\biggr)+O(1)\biggr)\cdot\cos\biggl(\frac{n}{2}t \biggr)\,dt+o(1). \end{aligned}$$

Passing to limits, we get

$$ \lim_{k\rightarrow\infty} \int_{\frac{2(m-1)\pi}{n}}^{ \frac {2m\pi}{n}} g_{1}\bigl(x_{k}(t- \tau)\bigr)\cdot \biggl\vert \cos\biggl(\frac{\pi}{2}-\frac {n}{2}t \biggr) \biggr\vert \,dt =\frac{4}{n}\cos\biggl(\frac{n}{2}\tau \biggr)g(+\infty). $$

For $m=2i$, $i=1,2,\ldots$ , we similarly get

$$ \lim_{k\rightarrow\infty} \int_{\frac{2(m-1)\pi}{n}}^{ \frac {2m\pi}{n}} g_{1}\bigl(x_{k}(t- \tau)\bigr)\cdot \biggl\vert \cos\biggl(\frac{\pi}{2}-\frac {n}{2}t \biggr) \biggr\vert \,dt =\frac{4}{n}\cos\biggl(\frac{n}{2}\tau \biggr)g(+\infty). $$

Therefore we have

$$ \int_{\frac{2(m-1)\pi}{n}}^{ \frac{2m\pi}{n}} g_{1}\bigl(x_{k}(t- \tau )\bigr)\cdot \biggl\vert \cos\biggl(\frac{\pi}{2}-\frac{n}{2}t \biggr) \biggr\vert \,dt =\frac{4}{n}\cos\biggl(\frac{n}{2}\tau \biggr)g(+\infty)+o(1). $$

Consequently, we get

$$\begin{aligned}& \sum_{m=1}^{n} \int_{-\frac{\pi}{2}-2(m-1)\pi}^{\frac{\pi }{2}-2(m-1)\pi} g\bigl(1+r_{k}\cos \theta_{k},1+r_{k}(t-\tau)\cos\theta _{k}(t- \tau),\lambda_{k}\bigr)\cos\theta_{k}\, d \theta_{k} \\& \quad \leqslant\sum_{m=1}^{n} \frac{n\lambda_{k}}{2} \int_{\frac{2(m-1)\pi }{n}}^{ \frac{2m\pi}{n}} \bigl(g_{1} \bigl(x_{k}(t-\tau)\bigr)-p(t)\bigr)\cdot \biggl\vert \cos \biggl( \frac{\pi}{2}-\frac{n}{2}t\biggr) \biggr\vert \,dt+o(1) \\& \quad =\frac{n\lambda_{k}}{2} \biggl(4\cos\biggl(\frac{n}{2}\tau\biggr)g(+ \infty )- \int_{0}^{2\pi}p(t) \biggl\vert \sin\biggl( \frac{n}{2}t+\theta\biggr) \biggr\vert \,dt \biggr)+o(1). \end{aligned}$$

By ($\mathrm{H}'_{2}$) we obtain, for k large enough,

$$ \sum_{m=1}^{n} \int_{-\frac{\pi}{2}-2(m-1)\pi}^{\frac{\pi }{2}-2(m-1)\pi} g\bigl(1+r_{k}\cos \theta_{k},1+r_{k}(t-\tau)\cos\theta _{k}(t- \tau),\lambda_{k}\bigr)\cos\theta_{k}\, d \theta_{k}< 0. $$

Consequently, for k large enough,

$$ \sum_{m=1}^{n}\bigl(t_{2}^{m,k}-t_{0}^{m,k} \bigr)>2\pi-\frac{4}{\mathcal{R}_{k}}, $$

and then

$$ \tau_{k}>2\pi, $$

which also contradicts the 2π-periodicity of $(x_{k}(t), y_{k}(t))$. This completes the proof. □

Remark 3.5

Similarly, if (H₁), ($\mathrm{H}''_{2}$), and (H₃) hold, then the result in Lemma 3.4 is valid. Under these conditions, to estimate the maxima, we can we embed (1.1) into (2.2).

Lemma 3.6

Assume (H₁), (H₂), and (H₃) hold. Then there exists a positive constant $M_{0}$ such that, for any 2π-periodic solution $x(t)$ of Eq. (2.1),

$$ \min_{t\in\mathbb{R}} x(t)>M_{0}. $$

Proof

Let $\mathcal{R}_{k}$ be defined by (3.5). We first prove that there exists $c_{1}>0$ such that, for all $k\in\mathbb{N}^{+}$,

$$ x_{k}\bigl(t_{3}^{1,k} \bigr)>c_{1}. $$

(3.14)

Assume by contradiction that $x_{k}(t_{3}^{1,k})\rightarrow0$ (as $k\rightarrow\infty$). By Lemma 3.4 and Remark 3.5 there exist $M_{1}>0$ and $M_{2}>0$ such that

$$ \Vert x_{k} \Vert _{\infty}< M_{1}, \qquad \bigl\Vert x'_{k} \bigr\Vert _{\infty}< M_{2}. $$

Then $\mathcal{R}_{k}(t)\equiv c<+\infty$. However, by (H₃),

$$\begin{aligned} r_{k}^{2}\bigl(t_{3}^{1,k} \bigr)-r_{k}^{2}\bigl(t_{2}^{1,k}\bigr)&= \frac{8}{n^{2}} \int _{t_{2}^{1,k}}^{t_{3}^{1,k}}y_{k}g\bigl(x_{k},x_{k}(t- \tau),\lambda_{k}\bigr)\,dt \\ &= \frac{8\lambda_{k}}{n^{2}} \int_{t_{3}^{1,k}}^{t_{2}^{1,k}}y_{k}\frac {1}{x_{k}^{3}}\,dt- (1-\lambda_{k})\frac{8}{n^{2}} \int _{t_{3}^{1,k}}^{t_{2}^{1,k}}y_{k}g_{0}(x_{k}) \,dt+O(1) \\ &=\frac{8\lambda_{k}}{n^{2}}\cdot\frac{1}{x_{k}^{2}(t_{3}^{1,k})}- (1-\lambda _{k}) \frac{8}{n^{2}} \int_{x(t_{3}^{1,k})}^{1}g_{0}(x_{k}) \,dx_{k}+O(1) \\ &\rightarrow+\infty, \end{aligned}$$

which is impossible. Therefore (3.14) holds. Similarly, we can obtain

$$ x_{k}\bigl(t_{3}^{m,k}\bigr)>c_{m} $$

for $m=1,2,\ldots,n$. Consequently, we get the conclusion of Lemma 3.6. □

Proof of Theorem 1.1

The result is obtained directly by Lemma 2.1, Lemma 3.4, Remark 3.5, and Lemma 3.6. □