Existence of positive periodic solutions for Liénard equation with a singularity of repulsive type

Abstract

In this paper, the existence of positive periodic solutions is studied for Liénard equation with a singularity of repulsive type,

$$ x''(t)+f(x(t))x'(t)+\varphi (t)x^{\mu}(t)-\frac{1}{x^{\gamma}(t)}=e(t), $$

where $f:(0,+\infty )\rightarrow R$ is continuous, which may have a singularity at the origin, the sign of $\varphi (t)$, $e(t)$ is allowed to change, and μ, γ are positive constants. By using a continuation theorem, as well as the techniques of a priori estimates, we show that this equation has a positive T-periodic solution when $\mu \in [0,+\infty )$.

1 Introduction

Since singular equations have a wide range of application in physics, engineering, mechanics, and other subjects (see [1–7]), the periodic problem for a certain second order differential equation has attracted much attention from many researchers. In the past years, lots of papers (see [8–14]) were concerned with the problem of periodic solutions to the second order singular equation without the first derivative term,

$$ x''+\varphi (t)x-\frac{b(t)}{x^{\mu}}=h(t),$$

(1.1)

where $f:[0,\infty )\rightarrow \mathbb{R}$ is continuous, $\varphi ,b, h\in L^{1}[0,T]$, and $\mu >0$ is a constant. Among these papers, we notice that the coefficient function $\varphi (t)$ is required to be

$$ \varphi (t)\ge 0\quad \text{for a.e.}~ t\in [0,T].$$

(1.2)

This is because (1.2), together with other conditions, can ensure that the function $G(t,s)\ge 0$ for $(t,s)\in [0,T]\times [0,T]$, where the $G(t,s)$ is the Green function associated with the boundary value problem for Hill’s equation

$$ x''(t)+\varphi (t)x(t)=h(t),\quad x(0)=x(T),~x'(0)=x'(T).$$

The condition $G(t,s)\ge 0$ for $(t,s)\in [0,T]\times [0,T]$ is crucial for obtaining the positive periodic solutions to (1.1) by means of some fixed point theorems on cones. Beginning with the paper of Habets–Sanchez [15], many works (see [16–21]) discussed the existence of a periodic solution for Liénard equations with singularities,

$$ x''(t)+f(x(t))x'(t)+\varphi (t)x(t)-\frac{1}{x^{\gamma}(t)}=e(t), $$

(1.3)

where $\varphi (t)$ and $e(t)$ are T-periodic with $\varphi , e\in L^{1}[0,T]$, while γ is a constant with $\gamma > 0$. However, in those papers, the conditions of $\varphi (t)\ge 0$ for a.e. $t\in [0,T]$, the strong singularity $\gamma \in [1,+\infty )$, and $f(x)$ being continuous on $[0,+\infty )$ are needed. To the best of our knowledge, there are fewer papers dealing with the equation where the function $f(x)$ possesses a singularity at $x=0$. We find that Hakl, Torres, and Zamora in [22] considered the periodic problem for the singular equation of repulsive type,

$$ x''(t)+f(x(t))x'(t)+\varphi (t)x^{\mu}(t)+g(x(t))= 0, $$

(1.4)

where $\mu \in (0,1]$ is a constant, φ is a T-periodic function with $\varphi \in L^{1}([0,T], R)$, and the sign of $\varphi (t)$ can change, while $f\in C((0,+\infty ),R)$ may be singular at $x=0$ and $g\in C((0,+\infty ),R)$ has a repulsive singularity at $x=0$, i.e., $\lim _{x\rightarrow 0^{+}}g(x)=-\infty$. By using Schauder’s fixed point theorem, some results on the existence of positive T-periodic solutions were obtained. However, the strong singularity condition $\int ^{1}_{0}g(s)ds = -\infty$ is also required. In a recent paper [23], the authors consider the periodic problem to (1.4) for the special case $g(x)=\frac{1}{x^{\gamma}}$, where $\gamma \in (0,+\infty )$. But, in [23], the function $\varphi (t)$ is required to satisfy $\varphi (t)\ge 0$ a.e. $t\in [0,T]$ for the case $\mu >1$ (see Theorem 3.1, [23]). Motivated by this, in the present paper, we continue to study the periodic problem for the singular equation,

$$ x''(t)+f(x(t))x'(t)+\varphi (t)x^{\mu}(t)-\frac{1}{x^{\gamma}}=e(t), $$

(1.5)

where f, φ are as same as those in (1.4); $\mu >0$ and $\gamma >0$ are constants, e is a T-periodic function with $e\in L^{1}([0,T],R)$, and $\int _{0}^{T}e(s)ds=0$. By means of a continuation theorem of coincidence degree principle developed by Manásevich and Mawhin, as well as the techniques of a priori estimates, some new results on the existence of positive periodic solutions are obtained. The interesting point in this paper is that the function $f(x)$ has a singularity at $x=0$, the sign of $\varphi (t)$ is allowed to change, and $\mu ,\gamma \in (0,+\infty )$. Compared with [22], we allow the singular term $\frac{1}{x^{\gamma}}$ to have a weak singularity, i.e., $\gamma \in (0,1)$. Also, for the case of $\mu >1$, the sign of $\varphi (t)$ is allowed to change, which is essentially different from the condition $\varphi (t)\ge 0$ for a.e. $t\in [0,T]$ in [23].

2 Essential definitions and lemmas

Throughout this paper, let $C_{T}=\{x\in C(R,R) :x(t+T)=x(t),\forall t\in R\}$ with the norm $|x|_{\infty} = \max _{t\in [0,T]}|x(t)|$. Clearly, $C_{T}$ is a Banach space. For any T-periodic function $x(t)$, we denote $\bar{x}=\frac{1}{T}\int _{0}^{T}x(s)ds$, $x_{+}(t)= \max \{ x(t),0\}$, and $x_{-}= -\min \{x(t),0\}$. Thus, $x(t)=x_{+}(t)- x_{-}(t)$ for all $t\in R$, and $\overline{x}= \overline{x_{+}}-\overline{x_{-}}$. Furthermore, for each $u\in C_{T}$, let $\|u\|_{p}=(\int ^{T}_{0}|u(s)|^{p}ds)^{\frac{1}{p}}$, $p\in [1,+\infty )$.

Lemma 2.1

([24])

Assume that there exit positive constants $M_{0}$ and $M_{1}$, with $0 < M_{0} < M_{1}$, such that the following conditions hold:

(1) for each $\lambda \in (0,1]$, each possible positive T-periodic solution u to the equation

$$ x''(t)+ \lambda f(x(t))x'(t)+ \lambda \varphi (t)x^{\mu}(t)- \frac{\lambda}{x^{\gamma}(t)}= 0 $$

satisfies the inequality $M_{0}< u(t)< M_{1}$ for all $t\in [0,T]$;

(2) each possible solution $c\in (0,+\infty )$ to the equation

$$ \frac{1}{c^{\gamma}}- \overline{\varphi}c^{\mu} = 0 $$

satisfies the inequality $M_{0}< c< M_{1}$;

(3) the inequality

$$ \Big(\frac{1}{M^{\gamma}_{0}}-\overline{\varphi}M_{0}^{\mu}\Big)\Big( \frac{\overline{\alpha}}{M^{\gamma}_{1}}-\overline{\varphi}M_{1}^{\mu} \Big) < 0 $$

holds.

Then equation has at least one positive T-periodic solution $u(t)$ such that $M_{0}< u(t)< M_{1}$ for all $t\in [0,T]$.

Lemma 2.2

([22])

Let $u(t):[0,\omega ]\rightarrow R$ be an arbitrary absolutely continuous function with $u(0)=u(\omega )$. Then the inequality

$$ \Big(\max _{t\in [0,\omega ]} u(t)-\min _{t\in [0,\omega ]} u(t)\Big)^{2} \leq \frac{\omega}{4}\int ^{\omega}_{4}|u'(s)|^{2}ds $$

holds.

Remark 2.3

If $\overline{\varphi}>0$, then there are constants $C_{1}$ and $C_{2}$ with $0< C_{1}< C_{2}$ such that

$$ \frac{1}{x^{\gamma}}-\overline{\varphi}x^{\mu}>0\quad \forall x\in (0,C_{1}) $$

(2.1)

and

$$ \frac{1}{x^{\gamma}}-\overline{\varphi}x^{\mu}< 0\quad \forall x\in (C_{2},+ \infty ). $$

(2.2)

Now, we embed equation (1.5) into the following equation family with a parameter $\lambda \in (0,1]$:

$$ x''(t)+ \lambda f(x(t))x'(t)+ \lambda \varphi (t)x^{\mu}(t)- \lambda \frac{1}{x^{\gamma}(t)}= \lambda e(t). $$

(2.3)

Let

$$ D = \Big\{ x\in C^{1}_{T}: x''(t)+ \lambda f(x(t))x'(t)+ \lambda \varphi (t)x^{\mu}(t)- \lambda \frac{1}{x^{\gamma}(t)}= \lambda e(t), \lambda \in (0,1]\Big\} , $$

(2.4)

and

$$ F(x)=\int ^{x}_{1}f(s)ds, G(x)=\int ^{x}_{1}s^{\gamma}f(s)ds, \quad x \in (0,+\infty ). $$

(2.5)

Lemma 2.4

Assume $\overline{\varphi}>0$ and $\overline{e}=0$, then there are two constants $\tau _{1}, \tau _{2} \in [0,T]$ for each $u\in D$, such that

$$ u(\tau _{1})\leq \max \Big\{ 1, \Big(\frac{2}{\overline{\varphi}}\Big)^{ \frac{1}{\mu}} \Big\} :=A_{0} $$

(2.6)

and

$$ u(\tau _{2})\geq \min \Big\{ 1, \Big(\frac{1}{\overline{\varphi _{+}}} \Big)^{\frac{1}{\gamma}} \Big\} :=A_{1}. $$

(2.7)

Proof

Let $u \in D$, then

$$ u''(t)+ \lambda f(u(t))u'(t)+ \lambda \varphi (t)u^{\mu}(t)- \frac{\lambda}{u^{\gamma}(t)}= \lambda e(t). $$

(2.8)

Dividing both sides of (2.8) by $u^{\mu}(t)$ and integrating over the interval $[0,T]$, we obtain

$$ \int ^{T}_{0}\frac{u''(t)}{u^{\mu}(t)}dt +\lambda T\overline{\varphi} = \lambda \int ^{T}_{0}\frac{1}{u^{\mu +\gamma}(t)}dt + \lambda \int ^{T}_{0} \frac{e(t)}{u^{\mu}(t)}dt. $$

Since the inequality $\int ^{T}_{0}\frac{u''(t)}{u^{\mu}(t)}dt \geq 0$ holds, it is easy to see that

$$ \lambda T \overline{\varphi} \leq \lambda T \frac{1}{u^{\mu +\gamma}(\xi )} +\lambda T \frac{\overline{e_{+}}}{u^{\mu}(\tau _{1})}, $$

i.e.,

$$ 0 < \overline{\varphi} \leq \frac{1}{u^{\mu +\gamma}(\xi )} + \frac{1}{u^{\mu}(\tau _{1})}. $$

(2.9)

From this, we can verify (2.6). In fact, if (2.6) does not hold, then

$$ u(t)> \max \Big\{ 1,\Big(\frac{2}{\overline{\varphi}}\Big)^{ \frac{1}{\mu + \gamma}}\Big\} , \quad \forall t\in [0,T], $$

(2.10)

which together with (2.9) gives

$$ 0 < \overline{\varphi} \leq \frac{1}{u^{\mu}(\tau _{1})} + \frac{\overline{\varphi}}{2}, $$

i.e.,

$$ u(\tau _{1}) < \Big(\frac{2}{\overline{\varphi}}\Big)^{\frac{1}{\mu}}. $$

(2.11)

On the other hand, (2.10) implies that $u(\tau _{1}) > 1$. It follows from (2.11) that $(\frac{2}{\overline{\varphi}})^{\frac{1}{\mu}} > 1$, i.e., $\frac{2}{\overline{\varphi}} >1$. By using (2.11) again, we get

$$ u(\tau _{1}) < \Big(\frac{2}{\overline{\varphi}}\Big)^{ \frac{1}{\mu + \gamma}}, $$

(2.12)

which contradicts with (2.10), verifying (2.6).

Integrating both sides of (2.8) over the interval $[0,T]$, we obtain

$$ \int ^{T}_{0}\varphi (t)u^{\mu}(t)dt -\int ^{T}_{0} \frac{1}{u^{\gamma}(t)}dt =\int ^{T}_{0}e(t)dt. $$

Since $\int ^{T}_{0}e(t)dt=T\bar{e}=0$, it follows that $\int ^{T}_{0}\varphi (t)u^{\mu}(t)dt = \int ^{T}_{0} \frac{1}{u^{\gamma}(t)}dt$. If

$$ u(t)< 1 \quad \forall t\in [0,T], $$

(2.13)

then

$$ \int ^{T}_{0}\frac{1}{u^{\gamma}(t)}dt\leq \int ^{T}_{0}\varphi _{+}(t)u^{ \mu}(t)dt \leq T\overline{\varphi _{+}}. $$

By using the mean value theorem for integrals, we get that there is a point $\xi \in [0,T]$ such that

$$ \frac{T}{u^{\gamma}(\xi )} \leq T\overline{\varphi _{+}}, $$

i.e.,

$$ u(\xi ) \geq \Big(\frac{1}{\overline{\varphi _{+}}}\Big)^{ \frac{1}{\gamma}}. $$

(2.14)

Thus (2.7) immediately follows from (2.13) and (2.14). □

Lemma 2.5

Assume $\overline{\varphi}>0$ and $\overline{e}=0$ for a.e. $t\in [0,T]$ and suppose that the following assumptions:

$$ B _{0}=\inf _{[A_{1},+\infty )}H(x) >-\infty $$

(2.15)

and

$$ \lim _{s\rightarrow 0^{+}}\Big(F(s)+\frac{T}{s^{\gamma}}\Big)< B_{0}-T \overline{e_{+}} $$

(2.16)

hold, where $H(x)=F(x)-T\overline{\varphi _{-}}x^{\mu}$. Then there is a constant $\gamma _{0}>0$ such that

$$ \min _{t\in [0,T]}u(t) \geq \gamma _{0}, \quad \textit{uniformly for} \ u \in D. $$

(2.17)

Proof

Let $u\in D$, then u satisfies

$$ u''(t)+ \lambda f(u(t))u'(t)+ \lambda \varphi (t)u^{\mu}(t)- \lambda \frac{1}{u^{\gamma}(t)}= \lambda e(t). $$

(2.18)

Since $u\in D$, it is easy to see that there are two points $t_{1}, t_{2}\in R$ such that $u(t_{1})=\max _{t\in [0,T]}u(t)$, $u(t_{2})=\min _{t\in [0,T]}u(t)$, and $0< t_{1}-t_{2}\le T$. By integrating (2.18) over the interval $[t_{2},t_{1}]$, we get

$$ \begin{aligned} F(u(t_{1}))&= F(u(t_{2}) +\int ^{t_{1}}_{t_{2}} \frac{1}{u^{\gamma}(t)}dt-\int ^{t_{1}}_{t_{2}}\varphi (t)u^{\mu}(t)dt + \int ^{t_{1}}_{t_{2}}e(t)dt \\ &\leq F(u(t_{2}))+ \frac{T}{u^{\gamma}(t_{2})} + T \overline{\varphi _{-}}u^{\mu}(t_{1}) + T\overline{e_{+}}, \end{aligned} $$

and then

$$ \begin{aligned} F(u(t_{2}))+ \frac{T}{u^{\gamma}(t_{2})}&\geq F(u(t_{1}))-T \overline{\varphi _{-}}u^{\mu}(t_{1}) - T\overline{e_{+}} \\ &\geq \inf _{[A_{1},+\infty )}H(x)-T\overline{e_{+}} \\ &= B_{0}- T\overline{e_{+}}. \end{aligned} $$

(2.19)

Assumption (2.16) ensures that there is a constant $\gamma _{0}> 0$ such that

$$ F(s)+\frac{T}{s^{\gamma}} < B_{0}-T\overline{e_{+}}, \quad \text{for} \ s\in (0,\gamma _{0}). $$

(2.20)

Combining (2.19) with (2.20), we get that

$$ \min _{t\in [0,T]}u(t)= u(t_{2})\geq \gamma _{0}. $$

(2.21)

 □

Lemma 2.6

Assume $\overline{\varphi}>0$ and $\overline{e}=0$ for a.e. $t\in [0,T]$ and suppose that the following assumptions:

$$\begin{aligned} &B _{0}=\inf _{[A_{1},+\infty )}H(x) >-\infty , \end{aligned}$$

(2.22)

$$\begin{aligned} &\lim _{s\rightarrow 0^{+}}\Big(F(s)+\frac{T}{s^{\gamma}}\Big)< B_{0}-T \overline{e_{+}}, \end{aligned}$$

(2.23)

and

$$ \lim _{s\rightarrow +\infty}(F(s)-T\overline{\varphi _{+}}s^{\mu}) = + \infty $$

(2.24)

hold. Then, there exists a constant $\gamma _{1}> 0$ such that

$$ \max _{t\in [0,T]}u(t) \leq \gamma _{1}, \quad \textit{uniformly for} \ u \in D. $$

(2.25)

Proof

Since $u\in D$, the function u satisfies (2.18). Then there are two points $t_{1}, t_{2}\in R$ such that $u(t_{1})=\max _{t\in [0,T]}u(t)$, $u(t_{2})=\min _{t\in [0,T]}u(t)$, and $0< t_{2}-t_{1}< T$. By integrating over the interval $[t_{1},t_{2}]$, we get

$$ \begin{aligned} F(u(t_{1}))&= F(u(t_{2}) -\int ^{t_{2}}_{t_{1}} \frac{1}{u^{\gamma}(t)}dt +\int ^{t_{2}}_{t_{1}}\varphi (t)u^{\mu}(t)dt - \int ^{t_{2}}_{t_{1}}e(t)dt \\ &\leq F(u(t_{2}))+ T\overline{\varphi _{+}}u^{\mu}(t_{1}) + T \overline{e_{-}}, \end{aligned} $$

thus, by the assumptions of (2.6), (2.22), and (2.24), according to the proof of Lemma 2.4, we obtain

$$ \gamma _{0}\leq u(t_{2})=\min _{t\in [0,T]}u(t)\leq A_{0}. $$

(2.26)

So, we have

$$ \begin{aligned} F(u(t_{1}))-T\overline{\varphi _{+}}u^{\mu}(t_{1})& \leq F(u(t_{2}))+ T\overline{e_{-}} \\ &\leq \max _{x\in [\gamma _{0},A_{0}]}F(x)+T\overline{e_{-}}. \end{aligned} $$

(2.27)

Assumption (2.24) now ensures that there is a constant $\gamma _{1}> \gamma _{0}> 0$ such that

$$ F(s)-T\overline{\varphi _{+}}s^{\mu}> \max _{x\in [\gamma _{0},A_{0}]}F(x)+T \overline{e_{-}} \quad \text{for all}\ s\in (\gamma _{1}, +\infty ). $$

(2.28)

Therefore, (2.27) and (2.28) imply

$$ \max _{t\in [0,T]}u(t) =u(t_{1})\leq \gamma _{1}, \quad \text{uniformly for} \ u\in D. $$

(2.29)

 □

Lemma 2.7

Assume $\overline{\varphi}>0$ and $\overline{e}=0$ for a.e. $t\in [0,T]$ and suppose that the following assumptions:

$$ C_{0}=\sup _{[A_{1},+\infty )}H_{1}(x) < +\infty $$

(2.30)

and

$$ \lim _{s\rightarrow 0^{+}}(F(s)- \frac{T}{s^{\gamma}})> C_{0}+ T \overline{e_{+}} $$

(2.31)

hold, where $H_{1}(x)=F(x)+T\overline{\varphi _{-}}x^{\mu}$. Then there is a constant $\gamma _{2}>0$ such that

$$ \min _{t\in [0,T]}u(t) \geq \gamma _{2}, \quad \textit{uniformly for} \ u \in D. $$

(2.32)

Proof

Since $u\in D$, it is easy to see that there exist two points $t_{1}, t_{2}\in R$ such that $u(t_{1})=\max _{t\in [0,T]}u(t)$, $u(t_{2})=\min _{t\in [0,T]}u(t)$, and $0< t_{2}-t_{1}< T$. By integrating over the interval $[t_{1},t_{2}]$, we get

$$ \begin{aligned} F(u(t_{2}))&= F(u(t_{1}) +\int ^{t_{2}}_{t_{1}} \frac{1}{u^{\gamma}(t)}dt -\int ^{t_{2}}_{t_{1}}\varphi (t)u^{\mu}(t)dt +\int ^{t_{2}}_{t_{1}}e(t)dt \\ &\leq F(u(t_{1}))+\frac{T}{u^{\gamma}(t_{2})}+ T \overline{\varphi _{-}}u^{\mu}(t_{1}) + T\overline{e_{+}}, \end{aligned} $$

and then

$$ \begin{aligned} F(u(t_{2}))-\frac{T}{u^{\gamma}(t_{2})} &\leq F(u(t_{1}))+ T \overline{\varphi _{-}}u^{\mu}(t_{1}) + T\overline{e_{+}} \\ &\le \sup _{[A_{1},+\infty )}H_{1}(x)+ T\overline{e_{+}} \\ &=C_{0}+ T\overline{e_{+}}. \end{aligned} $$

(2.33)

Assumption (2.31) ensures that there is a constant $\gamma _{2}> 0$ such that

$$ F(s)- \frac{T}{s^{\gamma}} > C_{0} + T\overline{e_{+}}, \quad \text{for} \ s\in (0,\gamma _{2}). $$

So, it is easy to see from (2.33) that

$$ u(t_{2})=\min _{t\in [0,T]}u(t) \geq \gamma _{2}. $$

(2.34)

 □

Lemma 2.8

Assume $\overline{\varphi}>0$ and $\overline{e}=0$ for a.e. $t\in [0,T]$ and suppose that the following assumptions:

$$ C_{0}=\sup _{[A_{1},+\infty )}H_{1}(x) < +\infty , \quad H_{1}(x)=F(x)+T \overline{\varphi _{-}}x^{\mu}, $$

(2.35)

as well as

$$ \lim _{s\rightarrow 0^{+}}(F(s)- \frac{T}{s^{\gamma}})> C_{0}+ T \overline{e_{+}} $$

(2.36)

and

$$ \lim _{s\rightarrow +\infty}(F(s)+T\overline{\varphi _{-}}s^{\mu}+ \frac{T}{s^{\gamma}})= -\infty , $$

(2.37)

hold. Then, there exists a constant $\gamma _{3}> 0$ such that

$$ \max _{t\in [0,T]}u(t) \leq \gamma _{3}, \quad \textit{uniformly for} \ u \in D. $$

(2.38)

Proof

Let $u\in D$, then u satisfies (2.18). Let $t_{1}$ and $t_{2}$ be defined as in the proof of Lemma 2.6, that is, $u(t_{1})=\max _{t\in [0,T]}u(t)$, $u(t_{2})=\min _{t\in [0,T]}u(t)$, and $0< t_{2}-t_{1}< T$. By integrating over the interval $[t_{1},t_{2}]$, we get

$$ \begin{aligned} F(u(t_{1}))&= F(u(t_{2}) -\int ^{t_{2}}_{t_{1}} \frac{1}{u^{\gamma}(t)}dt +\int ^{t_{2}}_{t_{1}}\varphi (t)u^{\mu}(t)dt - \int ^{t_{2}}_{t_{1}}e(t)dt \\ &\geq F(u(t_{2}))- \frac{T}{u^{\gamma}(t_{1})}- T \overline{\varphi _{-}}u^{\mu}(t_{1}) -T\overline{e_{+}}. \end{aligned} $$

(2.39)

Thus, by the assumptions of (2.6), (2.35), and (2.36), and according to the proof of Lemma 2.6, we have

$$ \gamma _{2}\leq u(t_{2})=\min _{t\in [0,T]}u(t)\leq A_{0}, $$

(2.40)

which together with (2.39) yields

$$ \begin{aligned} F(u(t_{1})) + T\overline{\varphi _{-}}u^{\mu}(t_{1})+ \frac{T}{u^{\gamma}(t_{1})}&= F(u(t_{2})-T\overline{e_{+}} \\ &\geq \min _{x\in [\gamma _{2}, A_{0}]}F(x)- T\overline{e_{+}}. \end{aligned} $$

(2.41)

On the other hand, assumption (2.37) gives that there exits a constant $\gamma _{3}> 0$ such that

$$ F(s) + T\overline{\varphi _{-}}s^{\mu}+ \frac{T}{s^{\gamma}} < \min _{x \in [\gamma _{2}, A_{0}]}F(x)- T\overline{e_{+}}, \quad s\in (\gamma _{3},+ \infty ). $$

(2.42)

Combining (2.41) with (2.42), we get that

$$ u(t_{1})= \max _{t\in [0,T]}u(t) \leq \gamma _{3}. $$

(2.43)

 □

3 Main results

Theorem 3.1

Assume $\overline{\varphi}>0$ and $\overline{e}=0$ for a.e. $t\in [0,T]$ and suppose that the assumptions of (2.15) and (2.16) in Lemma 2.4, as well as the assumption (2.24) in Lemma 2.5, hold. Then for each $\mu \in [0, +\infty )$, equation (1.5) has at least one positive T-periodic solution.

Proof

Due to assumptions of Lemma 2.4, we see that there are two constants $\gamma _{0}> 0$, $\gamma _{1}> 0$ such that $\min u(t) \geq \gamma _{0}$, $\max u(t)\leq \gamma _{1}$.

Now, we will show that there exists a positive constant $M > 0$ such that $\max _{t\in [0,T]} |u'(t)| \leq M$, uniformly for $u\in D$. If $u(t_{1})=\max _{t\in [0,T]}$, $t_{1}\in [0,T]$, then $u'(t_{1})=0$. Letting $t\in [0, T]$, we integrate (2.8) over the interval $[t_{1},t]$ and get

$$ \int ^{t}_{t_{1}} u''(t)dt + \lambda \int ^{t}_{t_{1}}f(u(t))u'(t)dt + \lambda \int ^{t}_{t_{1}}\varphi (t)u^{\mu}(t)dt - \lambda \int ^{t}_{t_{1}} \frac{1}{u^{\gamma}(t)}dt = \lambda \int ^{t}_{t_{1}}e(t)dt, $$

(3.1)

which yields

$$ u'(t)= \lambda \int ^{t}_{t_{1}}(-f(u(t))u'(t)-\varphi (t)u^{\mu}(t)+ \frac{1}{u^{\gamma}(t)} +e(t))dt, $$

(3.2)

and then we obtain

$$ \begin{aligned} |u'(t)| &\leq \lambda |F(u(t))- F(u(t_{1}))| + \lambda \int ^{t_{1}+T}_{t_{1}} |\frac{1}{u^{\gamma}(t)}|dt + \lambda \int ^{t_{1}+T}_{t_{1}}|e(t)|dt \\ &\quad + \lambda \int ^{t^{1}+T}_{t_{1}} |\varphi (t)u^{\mu}(t)|dt \\ &\leq 2 \max _{\gamma _{0}\leq u \leq \gamma _{1}}| F(u(t))|+ \frac{T}{\gamma _{0}^{\gamma}} + T\overline{e_{+}} + T \overline{|\varphi |}\gamma _{1}^{\mu} \\ &:= M, \quad \text{for all} \ t\in [0,T]. \end{aligned} $$

(3.3)

So, we have

$$ \max _{t\in [0,T]}| u'(t)| \leq M, \quad \text{uniformly for} \ u\in D. $$

(3.4)

Let $m_{1}=\min \{\gamma _{0}, D_{1}\}$ and $m_{2}= \{\gamma _{1}, D_{2}\}$ be two constants, where $D_{1}$ and $D_{2}$ are the constants determined in Remark 2.3. Then we get that every possible positive T-periodic solution $x(t)$ to equation (1.5) satisfies

$$ m_{1}< x(t)< m_{2},\quad |x'(t)|< M, \quad \text{for all} \ t\in [0,T]. $$

(3.5)

Furthermore, we have

$$ \Big(\overline{\varphi}m_{1}^{\mu}-\frac{1}{m_{1}^{\gamma}}\Big)\Big( \overline{\varphi}m_{2}^{\mu}-\frac{1}{m_{2}^{\gamma}}\Big) < 0, $$

(3.6)

by using Lemma 2.1, thus equation (1.5) has at least one positive T-periodic solution.

On the other hand, by Lemmas 2.6 and 2.7, we get the same conclusion as in Theorem 3.1, which can be proved similarly. Thus, the proofs are omitted. □

Theorem 3.2

Assume $\overline{\varphi}>0$ and $\overline{e}=0$ for a.e. $t\in [0,T]$ and suppose that the assumptions of (2.30) and (2.31) in Lemma 2.6, as well as the assumption (2.37) in Lemma 2.7, hold. Then for each $\mu \in [0, +\infty )$, equation (1.5) has at least one positive T-periodic solution.

4 Example

In this section, we present two examples to demonstrate the main results.

Example 4.1

Considering the following equation:

$$ x''(t)+ \Big[\frac{3}{x^{4}}+ \Big(\frac{25\pi}{6} + 5\Big)x^{ \frac{3}{2}}\Big]x'(t) + (1+2\cos{t})x^{\frac{3}{2}}(t)- \frac{1}{x^{2}(t)} = \sin{t}. $$

(4.1)

Corresponding to equation (1.5), in (4.1), $e(t)=\sin (t)$, $\varphi (t)=1+2\cos{t}$, $T=2\pi$. Obviously, $\overline{\varphi}=1 > 0$, and $\overline{e}=0$ for all $t \in [0,T]$ with $\overline{\varphi _{+}}= \frac{5}{6} +\frac{1}{\pi}$ and $\overline{\varphi _{-}}=\frac{1}{\pi}- \frac{1}{6}$. Since $F(x)= -\frac{1}{x^{3}} + (\frac{5\pi}{3} + 2)x^{\frac{5}{2}}$, we can easily verify that equation (4.1) satisfies

$$\begin{aligned} & B_{0}= \inf _{[A_{1},+\infty )}\Big(F(x) - T\overline{\varphi _{-}}x^{ \frac{3}{2}}\Big) > -\infty , \end{aligned}$$

(4.2)

$$\begin{aligned} &\lim _{x\rightarrow 0^{+}} (F(x) + \frac{2\pi}{x^{2}}) = -\infty , \end{aligned}$$

(4.3)

and

$$ \lim _{x\rightarrow +\infty} \Big(F(x) - T\overline{\varphi _{+}}x^{ \frac{3}{2}}\Big) = +\infty . $$

(4.4)

Obviously, (4.2), (4.3), and (4.4) imply that assumptions (2.15), (2.16), and (2.24) hold. Thus, by using Theorem 3.1, equation (4.1) has at least one positive 2π-periodic solution.

Example 4.2

Now consider

$$ x''(t)- \Big[\frac{3}{x^{4}}+\Big(5-\frac{5\pi}{6}\Big)x^{\frac{3}{2}} \Big]x'(t) + (1+2\cos{t})x^{\frac{3}{2}}(t)-\frac{1}{x^{2}(t)} = \sin{t}. $$

(4.5)

Corresponding to equation (1.5), here, $e(t)=\sin{t}$, $\varphi (t)=1+2\cos{t}$, $T=2\pi$. Clearly, $\overline{\varphi}=1 > 0$, and $\overline{e}=0$ for all $t \in [0,T]$ with $\overline{\varphi _{+}}= \frac{5}{6} +\frac{1}{\pi}$ and $\overline{\varphi _{-}}=\frac{1}{\pi}- \frac{1}{6}$. Since $F(x)= \frac{1}{x^{3}} - (2-\frac{\pi}{3})x^{\frac{5}{2}}$, we can easily verify that (4.1) satisfies

$$\begin{aligned} &C_{0}= \sup _{[A_{1},+\infty )}\Big(F(x) + T\overline{\varphi _{-}}x^{ \frac{3}{2}}\Big) < +\infty , \end{aligned}$$

(4.6)

$$\begin{aligned} &\lim _{x\rightarrow 0^{+}} \Big(F(x) - \frac{2\pi}{x^{2}}\Big) = + \infty , \end{aligned}$$

(4.7)

and

$$ \lim _{x\rightarrow +\infty} \Big(F(x) + 2\pi \overline{\varphi _{-}}x^{ \frac{3}{2}} + \frac{2\pi}{x^{2}}\Big) = -\infty . $$

(4.8)

Obviously, (4.6), (4.7), (4.8) imply that assumptions (2.30), (2.31), and (2.37) hold. Thus, by using Theorem 3.2, equation (4.5) has at least one positive 2π-periodic solution.

Remark 4.3

In (4.5), since $\mu =\frac{3}{2}>1$ and $\varphi (t)=1+2\cos t$ is a sign-changing function, the result of Example 4.2 can be obtained neither by using the main results of [23], nor by using the theorems of [23]. In this sense, the theorems of the present paper are new results on the existence of positive periodic solutions for singular Liénard equations.