Positive periodic solution for ϕ-Laplacian Rayleigh equation with strong singularity

Abstract

In this paper, we discuss a kind of ϕ-Laplacian Rayleigh equation with strong singularity

$$\bigl(\phi\bigl(u'(t)\bigr)\bigr)'+f \bigl(t,u'(t)\bigr)+g\bigl(u(t-\tau)\bigr)=e(t). $$

By application of the Manásevich-Mawhin continuation theorem, we obtain the existence of a positive periodic solution for this equation.

1 Introduction

In this paper, we investigate the following ϕ-Laplacian Rayleigh equation:

$$ \bigl(\phi\bigl(u'(t)\bigr)\bigr)'+f \bigl(t,u'(t)\bigr)+g\bigl(u(t-\tau)\bigr)=e(t), $$

(1.1)

where $f:\mathbb{R}\to\mathbb{R}$ is a $L^{2}$-Carathéodory function, i.e., it is measurable in the first variable and continuous in the second variable, and for every $0< r< s$ there exists $h_{r,s}\in L^{2}[0,\omega]$ such that $|g(t,x(t))|\leq h_{r,s}$ for all $x\in[r,s]$ and a.e. $t\in[0,\omega]$; f is a T-periodic function about t and $f(t,0)=0$; $g:(0,+\infty)\to \mathbb{R}$ is a continuous function that has a strong singularity at the origin;

$$ \lim_{u\to0^{+}} \int^{1}_{u} g(s)\,ds=+\infty. $$

(1.2)

$e\in L^{p}(\mathbb{R})$ is a T-periodic function and $1\leq p\leq\infty$, τ is a constant, and $0\leq\tau< T$.

Moreover, let $\phi:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function, with $\phi(0)=0$, which satisfies

(A₁):

$(\phi(u_{1})-\phi(u_{2}))(u_{1}-u_{2})>0$ for $\forall u_{1}\neq u_{2}$, $u_{1}, u_{2}\in\mathbb{R}$;

(A₂):

there exists a function $d:[0,+\infty]\rightarrow[0,+\infty]$, and $d(u)\rightarrow+\infty$ as $u\rightarrow+\infty$, such that $\phi(u)\cdot u\geq d(|u|)|u|$ for $\forall u\in\mathbb{R}$.

It is easy to see that ϕ represents a large class of nonlinear operators, including $\phi_{p}: \mathbb{R}\rightarrow\mathbb{R}$ is a p-Laplacian, i.e., $\phi_{p}(u)=|u|^{p-2}u$ for $u\in\mathbb{R}$.

As is well known, the Rayleigh equation can be derived from many fields, such as the physics, mechanics, and engineering technique fields, and an important question is whether this equation can support periodic solutions. In 1977, Gaines and Mawhin [1] introduced some continuation theorems and applied this theorem to a discussion of the existence of solutions for the Rayleigh equation [1], p.99,

$$ u''(t)+f\bigl(u'(t)\bigr)+g \bigl(t,u(t)\bigr)=0. $$

(1.3)

Gaines and Mawhin’s work has attracted the attention of many scholars in differential equations. More recently, the existence of periodic solutions for the Rayleigh equation was extensively studied (see [2–11] and the references therein). In 2001, by using the method of upper and lower solutions, Habets and Torres [3] investigated the existence 2π-periodic solutions of (1.3) by assuming that $g=g(t,u,u')$ is bounded (or bounded from below). Afterwards, by application of the time map continuation theorem, Wang [11] discussed the existence of periodic solutions of a kind of Rayleigh equation

$$u''(t)+f\bigl(u'(t)\bigr)+g\bigl(u(t) \bigr)=p(t). $$

In this direction, the researchers in [12–20] discussed the p-Laplacian Rayleigh equation. In 2006, by employing Mawhin’s continuous theorem, Cheung and Ren [12] studied how the existence of the p-Laplacian Rayleigh equation

$$\bigl(\varphi_{p}\bigl(u'(t)\bigr)\bigr)'+f \bigl(u'(t)\bigr)+\beta g\bigl(u\bigl(t-\tau(t)\bigr)\bigr)=e(t) $$

under various assumptions is obtained. Recently, Xin and Cheng [21] discussed a kind of ϕ-Laplacian Rayleigh equation,

$$ \bigl(\phi\bigl(u'(t)\bigr)\bigr)'+f \bigl(t,u'(t)\bigr)+g\bigl(t,u(t)\bigr)=e(t). $$

(1.4)

By using the Manásevich-Mawhin continuation theorem and some analysis techniques, the authors established a sufficient condition for the existence and uniqueness of positive periodic solutions for (1.4).

In the above papers, the authors investigated several kinds of Rayleigh and p-Laplacian Rayleigh equations. However, as far as we know, the study of periodic solutions for the ϕ-Laplacian differential equation with strong singularity is relatively rare. In this paper, we try to fill this gap and establish the existence of positive periodic solutions of (1.1) by employing the Manásevich-Mawhin continuation theorem. Finally, a numerical example demonstrates the validity of the method.

2 Positive periodic solution for (1.1)

In this section, we will consider the existence of positive periodic solution for (1.1) with strong singularity. First of all, we embed equation (1.1) into the following equation family with a parameter $\lambda\in(0,1]$:

$$ \bigl(\phi\bigl(u'(t)\bigr)\bigr)'+ \lambda f\bigl(t,u'(t)\bigr)+\lambda g\bigl(u(t-\tau)\bigr)=\lambda e(t). $$

(2.1)

The following lemma is a consequence of Theorem 3.1 of [22].

Lemma 2.1

Assume that there exist positive constants $E_{1}$, $E_{2}$, $E_{3}$, and $E_{1}< E_{2}$ such that the following conditions hold:

• (1) Each possible periodic solution u to equation (2.1) such that $E_{1}< u(t)< E_{2}$, for all $t\in[0,T]$ and $\|u'\|< E_{3}$, here $\|u'\|:=\max_{t\in[0,T]}|u'(t)|$.

• (2) Each possible solution C to the equation

  $$g(C)-\frac{1}{T} \int^{T}_{0} e(t)\,dt=0 $$

  satisfies $E_{1}< C< E_{2}$.

• (3) We have

  $$\biggl(g(E_{1})-\frac{1}{T} \int^{T}_{0} e(t)\,dt \biggr) \biggl(g(E_{2})- \frac {1}{T} \int^{T}_{0} e(t)\,dt \biggr)< 0. $$

Then (1.1) has at least one T-periodic solution.

For the sake of convenience, we list the following assumptions which will be used repeatedly in the sequel:

(H₁):

There exist constants $0< d_{1}< d_{2}$ such that $g(u)-e(t)>0$ for $u\in(0,d_{1})$ and $g(u)-e(t)<0$ for $u\in(d_{2},+\infty)$.

(H₂):

There exist positive constants a, b such that

$$ g(u)\leq a u+b, \quad \mbox{for all } u>0. $$

(2.2)

(H₃):

There exist constants α and $m>1$ such that

$$f(t,u)u\geq\alpha|u|^{m}, \quad \mbox{for } (t,u)\in[0,T]\times \mathbb{R}. $$

(H₄):

There exist positive constants β and γ such that

$$\bigl\vert f(t,u)\bigr\vert \leq\beta |u|^{m-1}+\gamma,\quad \mbox{for } (t,u)\in[0,T]\times\mathbb{R}. $$

Lemma 2.2

Assume that (A₁) and (H₁) hold. Then there exists a point $t_{1}\in[0,T]$ such that

$$ d_{1}< u(t_{1})\leq d_{2}. $$

(2.3)

Proof

Let $\underline{t}$, t̅, respectively, be the global minimum point and the global maximum point $u(t)$ on $[0,T]$; then $u'(\underline{t})=0$ and $u'(\overline{t})=0$, and we claim that

$$ \bigl(\phi\bigl(u'(\underline{t})\bigr) \bigr)'\geq0. $$

(2.4)

In fact, if (2.4) does not hold, then $(\phi(u'(\underline{t})))'<0$ and there exists $\varepsilon>0$ such that $(\phi(u'(t)))'<0$ for $t\in(\underline{t}-\varepsilon,\underline{t}+\varepsilon)$. Therefore, $\phi(u'(t))$ is strictly decreasing for $t\in(\underline{t}-\varepsilon,\underline{t}+\varepsilon)$. From (A₁), we know that $u'(t)$ is strictly decreasing for $t\in(\underline{t}-\varepsilon,\underline{t}+\varepsilon)$. This contradicts the definition of $\underline{t}$. Thus, (2.4) is true. From $f(t,0)=0$, (2.1) and (2.4), we have

$$ g\bigl(u(\underline{t}-\tau)\bigr)-e(\underline{t})\leq0. $$

(2.5)

Similarly, we can get

$$ g\bigl(u(\overline{t}-\tau)\bigr)-e(\overline{t})\geq0. $$

(2.6)

From (H₁), (2.5), and (2.6), we have

$$u(\underline{t}-\tau)\geq d_{1}\quad \mbox{and}\quad u(\overline{t}- \tau)\leq d_{2}. $$

In view of u being a continuous function, there exists a point $t_{1}\in[0,T]$, such that

$$d_{1}\leq u(t_{1})\leq d_{2}. $$

 □

Lemma 2.3

Assume that (A₁) and (H₁), (H₃) hold. Then there exists a positive constant $M_{1}$ such that

$$ u(t)< M_{1}. $$

(2.7)

Proof

Multiplying both sides of (2.1) by $u'(t)$ and integrating over the interval $[0,T]$, we have

$$\begin{aligned}& \int^{T}_{0}\bigl(\phi\bigl(u'(t)\bigr) \bigr)'u'(t)\,dt+\lambda \int^{T}_{0}f\bigl(t,u'(t) \bigr)u'(t)\,dt+\lambda \int^{T}_{0}g\bigl(u(t-\tau)\bigr)u'(t) \,dt \\& \quad =\lambda \int^{T}_{0}e(t)u'(t)\,dt. \end{aligned}$$

(2.8)

Moreover, we have

$$\begin{aligned} \int^{T}_{0}\bigl(\phi\bigl(u'(t)\bigr) \bigr)'u'(t)\,dt =& \int^{T}_{0}u'(t)d\bigl(\phi \bigl(u'(t)\bigr)\bigr) \\ =& \bigl[\phi \bigl(u'(t) \bigr)u'(t) \bigr]^{T}_{0}- \int^{T}_{0}\phi\bigl(u'(t)\bigr) \,du'(t)=0 \end{aligned}$$

(2.9)

and

$$\begin{aligned} \int^{T}_{0}g\bigl(u(t-\tau)\bigr)u'(t) \,dt =& \int^{T}_{0}g\bigl(u(t-\tau)\bigr)\,du(t) \\ =& \int^{T}_{0}g\bigl(u(t-\tau )\bigr)\,du(t-\tau)=0, \end{aligned}$$

(2.10)

since $du(t)=\frac{du(t-\tau)}{d(t-\tau)}\,dt=du(t-\tau)$.

Substituting (2.9) and (2.10) into (2.8), we have

$$ \int^{T}_{0}f\bigl(t,u'(t) \bigr)u'(t)\,dt= \int^{T}_{0}e(t)u'(t)\,dt. $$

(2.11)

Thus, we have

$$\biggl\vert \int^{T}_{0} f\bigl(t,u'(t) \bigr)u'(t)\,dt\biggr\vert =\biggl\vert \int^{T}_{0}e(t)u'(t)\,dt\biggr\vert . $$

From (H₃), we can get

$$\biggl\vert \int^{T}_{0} f\bigl(t,u'(t) \bigr)u'(t)\,dt\biggr\vert \geq\alpha \int^{T}_{0}\bigl\vert u'(t)\bigr\vert ^{m}\,dt. $$

Therefore, we can get

$$\begin{aligned} \alpha \int^{T}_{0}\bigl\vert u'(t)\bigr\vert ^{m}\,dt&\leq \int^{T}_{0}\bigl\vert e(t)\bigr\vert \bigl\vert u'(t)\bigr\vert \,dt \\ &\leq \biggl( \int^{T}_{0}\bigl\vert e(t)\bigr\vert ^{\frac{m}{m-1}}\,dt \biggr)^{\frac{m-1}{m}} \biggl( \int^{T}_{0}\bigl\vert u'(t)\bigr\vert ^{m}\,dt \biggr)^{\frac{1}{m}} \\ &=\|e\|_{\frac{m}{m-1}} \biggl( \int^{T}_{0}\bigl\vert u'(t)\bigr\vert ^{m}\,dt \biggr)^{\frac{1}{m}}, \end{aligned}$$

where $\|e\|_{\frac{m}{m-1}}= (\int^{T}_{0}|e(t)|^{\frac{m}{m-1}}\,dt )^{\frac{m-1}{m}}$. It is easy to see that there exists a positive constant $M_{1}'$ (independent of λ) such that

$$ \int^{T}_{0}\bigl\vert u'(t)\bigr\vert ^{m}\,dt\leq M_{1}'. $$

(2.12)

From Lemma 2.2 and the Hölder inequality, we have

$$\begin{aligned} u(t)&=\biggl\vert \int^{t}_{t_{1}}u'(t)\,dt\biggr\vert \leq u(t_{1})+ \int^{T}_{0}\bigl\vert u'(t)\bigr\vert \,dt \\ &\leq d_{2}+ \int^{T}_{0}\bigl\vert u'(t)\bigr\vert \,dt \\ &\leq d_{2}+T^{\frac{m-1}{m}} \biggl( \int^{T}_{0}\bigl\vert u'(t)\bigr\vert ^{m}\,dt \biggr)^{\frac{1}{m}} \\ &\leq d_{2}+T^{\frac{m-1}{m}} \bigl(M_{1}' \bigr)^{\frac{1}{m}}:=M_{1}. \end{aligned}$$

 □

Lemma 2.4

Assume that (A₁), (A₂), and (H₁)-(H₄) hold. Then there exists a positive constant $M_{2}$ such that

$$ \bigl\Vert u'\bigr\Vert < M_{2}. $$

(2.13)

Proof

Integrating both sides of (2.1) over $[0,T]$, we have

$$ \int^{T}_{0}\bigl[f\bigl(t,u'(t) \bigr)+g\bigl(u(t-\tau)\bigr)-e(t)\bigr]\,dt=0. $$

(2.14)

Therefore, we get from (2.7), (2.14), (H₂), and (H₄)

$$\begin{aligned} \int^{T}_{0}\bigl\vert g\bigl(u(t-\tau)\bigr)\bigr\vert \,dt =& \int_{g(u(t-\tau))\geq0}g\bigl(u(t-\tau)\bigr)\,dt- \int _{g(u(t-\tau))\leq0}g\bigl(u(t-\tau)\bigr)\,dt \\ =&2 \int_{g(u(t-\tau))\geq0}g\bigl(u(t-\tau)\bigr)\,dt+ \int^{T}_{0}f\bigl(t,u'(t)\bigr)\,dt- \int ^{T}_{0}e(t)\,dt \\ \leq&2 \int_{g(u(t-\tau))\geq0}\bigl(ax(t-\tau)+b\bigr)\,dt+ \int ^{T}_{0}\bigl\vert f\bigl(t,u'(t) \bigr)\bigr\vert \,dt+ \int^{T}_{0}\bigl\vert e(t)\bigr\vert \,dt \\ \leq&2a \int^{T}_{0}\bigl\vert x(t-\tau)\bigr\vert \,dt+2bT+\beta \int^{T}_{0}\bigl\vert u'(t)\bigr\vert ^{m-1}\,dt+\gamma T +\|e\|_{2}T^{\frac{1}{2}} \\ \leq&2aM_{1}T+2bT+\beta T^{\frac{1}{m}} \biggl( \int^{T}_{0}\bigl\vert u'(t)\bigr\vert ^{m}\,dt \biggr)^{\frac{m-1}{m}}+\gamma T +\|e \|_{2}T^{\frac{1}{2}} \\ \leq&2aM_{1}T+2bT+\beta T^{\frac{1}{m}}M_{1}^{\prime\frac{m-1}{m}}+ \gamma T +\|e\|_{2}T^{\frac{1}{2}}. \end{aligned}$$

(2.15)

As $u(0)=u(T)$, there exists $t_{2}\in[0,T]$ such that $u'(t_{2})=0$, while $\phi(0)=0$, and we have

$$\begin{aligned} \bigl\vert \phi\bigl(u'(t)\bigr)\bigr\vert =& \biggl\vert \int^{t}_{t_{2}}\bigl(\phi\bigl(u'(s)\bigr) \bigr)'\,ds\biggr\vert \\ \leq& \lambda \int^{T}_{0}\bigl\vert f\bigl(t,u'(t) \bigr)\bigr\vert \,dt+\lambda \int^{T}_{0}\bigl\vert g\bigl(u(t-\tau)\bigr)\bigr\vert \,dt+\lambda \int^{T}_{0}\bigl\vert e(t)\bigr\vert \,dt, \end{aligned}$$

(2.16)

where $t\in[t_{2},t_{2} +T]$. In view of (2.7), (2.12), (2.15), (2.16), and (H₄), we have

$$\begin{aligned} \bigl\Vert \phi\bigl(u'\bigr)\bigr\Vert =&\max _{t\in[0,T]}\bigl\{ \bigl\vert \phi\bigl(u'(t)\bigr) \bigr\vert \bigr\} \\ =&\max_{t\in[t_{2},t_{2}+T]} \biggl\{ \biggl\vert \int^{t}_{t_{0}}\bigl(\phi\bigl(u'(s)\bigr) \bigr)'\,ds \biggr\vert \biggr\} \\ \leq& \int^{T}_{0}\bigl\vert f\bigl(t,u'(t) \bigr)\bigr\vert \,dt+ \int^{T}_{0}\bigl\vert g\bigl(u(t-\tau)\bigr)\bigr\vert \,dt+ \int^{T}_{0}\bigl\vert e(t)\bigr\vert \,dt \\ \leq&\beta \int^{T}_{0}\bigl\vert u'(t)\bigr\vert ^{m-1}\,dt+\gamma T+ \int^{T}_{0}\bigl\vert g\bigl(u(t-\tau)\bigr)\bigr\vert \,dt+ \int ^{T}_{0}\bigl\vert e(t)\bigr\vert \,dt \\ \leq&2 \bigl(aM_{1}T+bT+\beta T^{\frac{1}{m}} M_{1}^{\prime \frac{m-1}{m}}+ \gamma T+\|e\|_{2}T^{\frac{1}{2}} \bigr):=M_{2}'. \end{aligned}$$

(2.17)

We claim that there exists a positive constant $M_{2}>M_{2}'+1$ such that, for all $t\in\mathbb{R}$,

$$ \bigl\Vert u'\bigr\Vert \leq M_{2}. $$

(2.18)

In fact, if $u'$ is not bounded, then from the definition of d, there exists a positive constant $M_{2}''$ such that $d(|u'|)>M_{2}''$ for some $u'\in\mathbb{R}$. However, from (A₂), we have

$$d\bigl(\bigl\vert u'\bigr\vert \bigr)\bigl\vert u'\bigr\vert \leq\phi\bigl(u'\bigr)u' \leq\bigl\vert \phi\bigl(u'\bigr)\bigr\vert \bigl\vert u'\bigr\vert \leq M_{2}' \bigl\vert u'\bigr\vert . $$

Then we can get

$$d\bigl(\bigl\vert u'\bigr\vert \bigr)\leq M_{2}', \quad \mbox{for all } u\in\mathbb{R}, $$

which is a contradiction. So, (2.18) holds. □

Lemma 2.5

Assume that (A₁), (A₂), and (H₁)-(H₄) hold. Then there exists a positive constant $M_{3}$ such that

$$ u(t)\geq M_{3}. $$

(2.19)

Proof

From (2.1), we have

$$ \bigl(\phi\bigl(u'(t+\tau)\bigr)\bigr)'+ \lambda f\bigl(t+\tau,u(t+\tau)\bigr)+\lambda g\bigl(u(t)\bigr)=\lambda e(t+\tau). $$

(2.20)

Multiplying both sides of (2.20) by $u'(t)$ and integrating on $[\xi,t]$, here $\xi\in[0,T]$, we get

$$\begin{aligned} \lambda \int^{u(t)}_{u(\xi)}g(u)\,du =&\lambda \int^{t}_{\xi}g\bigl(u(s)\bigr)u'(s)\,ds \\ =&- \int^{t}_{\xi}\bigl(\phi\bigl(u'(s+\tau) \bigr)\bigr)'u'(s)\,ds-\lambda \int^{t}_{\xi}f\bigl(s+\tau ,u(s+\tau) \bigr)u'(s)\,ds \\ &{}+\lambda \int^{t}_{\xi}e(s+\tau)u'(s)\,ds. \end{aligned}$$

(2.21)

By (2.13) and (2.17), we can get

$$\begin{aligned}& \biggl\vert \int^{t}_{\xi}\bigl(\phi\bigl(u'(t+\tau) \bigr)\bigr)'u'(s)\,ds\biggr\vert \\& \quad \leq \int^{t}_{\xi }\bigl\vert \bigl(\phi \bigl(u'(s+\tau)\bigr)\bigr)'\bigr\vert \bigl\vert u'(s)\bigr\vert \,ds \\& \quad \leq \bigl\Vert u'\bigr\Vert \int^{T}_{0}\bigl\vert \bigl(\phi \bigl(u'(t+\tau)\bigr)\bigr)'\bigr\vert \,dt \\& \quad \leq \lambda\bigl\Vert u'\bigr\Vert \biggl( \int^{T}_{0}\bigl\vert f\bigl(t,u'(t) \bigr)\bigr\vert \,dt+ \int^{T}_{0}\bigl\vert g\bigl(u(t-\tau)\bigr)\bigr\vert \,dt+ \int ^{T}_{0}\bigl\vert e(t)\bigr\vert \,dt \biggr) \\& \quad \leq 2\lambda M_{2}\bigl(aM_{1}T+bT+ \beta T^{\frac{1}{m}}M_{1}^{\prime \frac{m-1}{m}}+\gamma T+T^{\frac{1}{2}}\|e \|_{2}\bigr). \end{aligned}$$

Moreover, from (H₃) and (2.18), we have

$$\begin{aligned}& \biggl\vert \int^{t}_{\xi}f\bigl(s+\tau,u(s+\tau) \bigr)u'(s)\,ds\biggr\vert \leq \int^{T}_{0}\bigl\vert f\bigl(s+\tau,u(s+\tau) \bigr)\bigr\vert \bigl\vert u'(s)\bigr\vert \,ds \leq M_{2}\bigl(\beta M_{2}^{m-1}T+\gamma T\bigr), \\& \biggl\vert \int^{t}_{\xi}e(t+\tau)u'(t)\,dt\biggr\vert \leq M_{2}\sqrt{T}\|e\|_{2}. \end{aligned}$$

Form (2.21), we have

$$\begin{aligned}& \biggl\vert \int^{u(t)}_{u(\xi)}g_{0}(u)\,du\biggr\vert \\& \quad \leq M_{2}\bigl(aM_{1}T+bT+ \beta T^{\frac{1}{m}}M_{1}^{\prime \frac{m-1}{m}}+ \gamma T+T^{\frac{1}{2}}\|e\|_{2} +\beta M_{2}^{m-1}T+ \gamma T+\sqrt{T}\|e\|_{2}\bigr) \\& \quad :=M_{3}'. \end{aligned}$$

(2.22)

From the strong force condition (1.2), we know that there exists a constant $M_{3}>0$ such that

$$ u(t)\geq M_{3}, \quad \forall t\in[\xi,T]. $$

(2.23)

Similarly, we can consider $t\in[0,\xi]$. □

By Lemmas 2.1-2.5, we obtain the following main result.

Theorem 2.1

Assume that conditions (A₁), (A₂), and (H₁)-(H₄) hold. Then (1.1) has a positive T-periodic solution.

Proof

Let $E_{1}<\min\{d_{1},M_{3}\}$, $E_{2}>\max\{d_{2}, M_{1}\}$, $E_{3}>M_{2}$ are constants, from Lemmas 2.2-2.5, we see that the periodic solution u to (2.1) satisfies

$$ E_{1}< u(t)< E_{2}, \qquad \bigl\Vert u'\bigr\Vert < E_{3}. $$

(2.24)

Then condition (1) of Lemma 2.1 is satisfied. For a possible solution D to the equation

$$g(D)-\frac{1}{T} \int^{T}_{0}e(t)\,dt=0, $$

$E_{1}< D<E_{2}$ is satisfied. Therefore, condition (2) of Lemma 2.1 holds. Finally, we consider condition (3) of Lemma 2.1 also to be satisfied. In fact, from (H₁), we have

$$g(E_{1})-\frac{1}{T} \int^{T}_{0}e(t)\,dt>0 $$

and

$$g(E_{2})-\frac{1}{T} \int^{T}_{0}e(t)\,dt< 0. $$

So condition (3) is also satisfied. By application of Lemma 2.1, we see that (1.1) has at least one positive periodic solution. □

We illustrate our results with one example.

Example 2.1

Consider the following second-order ϕ-Laplacian Rayleigh equation:

$$ \bigl(\phi\bigl(u'(t)\bigr)\bigr)'+ \bigl(20-16\sin^{2}t\bigr) \bigl(u'(t)\bigr)^{m-1}+ \frac{1}{u^{\kappa}(t)}-10u(t)=e^{\cos^{2}t}, $$

(2.25)

where $\phi(u)=ue^{|u|^{2}}$, $m>1$, and $\kappa\geq1$.

Comparing (2.25) to (1.1), it is easy to see that $g(t,u)=\frac{1}{u^{\kappa}(t)}-10u(t)$, $f(t,w)=(20-16\sin^{2}t)w^{m-1}$, $e(t)=e^{\cos^{2}t}$, $T=\pi$. Obviously, we get

$$\bigl( ue^{|u|^{2}}-ve^{|v|^{2}}\bigr) (u-v)\geq\bigl( |u|e^{|u|^{2}}-|v|e^{|v|^{2}}\bigr) \bigl(\vert u\vert -|v|\bigr) \geq0 $$

and

$$\phi(u)\cdot u=|u|^{2}e^{|u|^{2}}. $$

So, conditions (A₁) and (A₂) hold. Moreover, it is easily seen that there exists a constant $d_{2}=1$ such that (H₁) holds. As $a=10$ and $b=1$, condition (H₂) holds. Consider $f(t,w)w=(20-16\cos^{2}t)w^{m}\geq4w^{m}$, here $\alpha=4$, and $|f(t,w)|=|(20-16\cos^{2}t)w^{m-1}|\leq20|w|^{m-1}+1$, here $\beta=20$, $\gamma=1$; thus, we see that conditions (H₃) and (H₄) hold. Therefore, by Theorem 2.1, we know that (2.25) has one positive periodic solution.