Existence and uniqueness of a positive periodic solution for Rayleigh type ϕ-Laplacian equation

Abstract

By using the Manásevich-Mawhin continuation theorem and some analysis skills, we establish some sufficient condition for the existence and uniqueness of positive T-periodic solutions for a generalized Rayleigh type ϕ-Laplacian operator equation. The results of this paper are new and they complement previous known results.

MSC:34K13, 34C25.

1 Introduction

During the past few years, many researchers have discussed the periodic solutions of a Rayleigh type differential equation (see [1–10]). For example, in 2009, Xiao and Liu [7] studied the Rayleigh type p-Laplacian equation with a deviating argument of the form

$${\left({\varphi }_p(x^{\prime }(t))\right)}^{\prime }+f(t,x^{\prime }(t))+g(t,x(t-\tau (t)))=e(t).$$

By using the coincidence degree theory, we establish new results on the existence of periodic solutions for the above equation. Afterward, Xiong and Shao [9] used the coincidence degree theory to establish new results on the existence and uniqueness of positive T-periodic solutions for the Rayleigh type p-Laplacian equation of the form

$${\left({\varphi }_p(x^{\prime }(t))\right)}^{\prime }+f(t,x^{\prime }(t))+g(t,x(t))=e(t).$$

In this paper, we consider the following Rayleigh type ϕ-Laplacian operator equation:

$${\left(\varphi (x^{\prime }(t))\right)}^{\prime }+f(t,x^{\prime }(t))+g(t,x(t))=e(t),$$

(1.1)

where the function $\varphi :\mathbb{R}\to \mathbb{R}$ is continuous and $\varphi (0)=0$. $f,g\in Car(\mathbb{R}\times \mathbb{R},\mathbb{R})$ is an $L^p$-Carathéodory function and $p=\frac{m}{m-1}$, $m\ge 2$, which means it is measurable in the first variable and continuous in the second variable. For every $0<r<s$, there exists $h_{r,s}\in L^p[0,T]$ such that $|g(t,x(t))|\le h_{r,s}$ for all $x\in [r,s]$ and a.e. $t\in [0,T]$; and f, g is a T-periodic function about t and $f(t,0)=0$. $e\in L^p([0,T],\mathbb{R})$ and is T-periodic.

Here $\varphi :\mathbb{R}\to \mathbb{R}$ is a continuous function and $\varphi (0)=0$, which satisfies

(A₁) $(\varphi (x_1)-\varphi (x_2))(x_1-x_2)>0$ for $\mathrm{\forall }{x}_1\ne x_2$, $x_1,x_2\in \mathbb{R}$;

(A₂) there exists a function $\alpha :[0,+\mathrm{\infty }]\to [0,+\mathrm{\infty }]$, $\alpha (s)\to +\mathrm{\infty }$ as $s\to +\mathrm{\infty }$, such that $\varphi (x)\cdot x\ge \alpha (|x|)|x|$ for $\mathrm{\forall }x\in \mathbb{R}$.

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

We know that the study on ϕ-Laplacian is relatively infrequent, the main difficulty lies in the fact that the ϕ-Laplacian operator typically possesses more uncertainty than the p-Laplacian operator. For example, the key step for ${\varphi }_p$ to get a priori solutions, ${\int }_0^T{({\varphi }_p^{\prime }(x(t)))}^{\prime }x(t)dt=-{\int }_0^T{|x^{\prime }(t)|}^{p}dt$, is no longer available for general ϕ-Laplacian. So, we need to find a new method to get over it.

By using the Manásevich-Mawhin continuation theorem and some analysis skills, we establish some sufficient condition for the existence of positive T-periodic solutions of (1.1). The results of this paper are new and they complement previous known results.

2 Main results

For convenience, define

$$C_T^1=\{x\in C^1(\mathbb{R},\mathbb{R}):x\text{ is }T\text{-periodic}\},$$

which is a Banach space endowed with the norm $\parallel \cdot \parallel$; define $\parallel x\parallel =max\{{|x|}_0,{|x^{\prime }|}_0\}$ for all x, and

$${|x|}_0=\underset{t\in [0,T]}{max}|x(t)|,|x^{\prime }{|}_0=\underset{t\in [0,T]}{max}|x^{\prime }(t)|.$$

For the T-periodic boundary value problem

$${\left(\varphi (x^{\prime }(t))\right)}^{\prime }=\tilde{f}(t,x,x^{\prime }),$$

(2.1)

here $\tilde{f}:[0,T]\times \mathbb{R}\times \mathbb{R}\to \mathbb{R}$ is assumed to be Carathéodory.

Lemma 2.1 (Manásevich-Mawhin [11])

Let Ω be an open bounded set in $C_T^1$. If

• (i) for each $\lambda \in (0,1)$, the problem

  $${\left(\varphi \left(x^{\prime }\right)\right)}^{\prime }=\lambda \tilde{f}(t,x,x^{\prime }),x(0)=x(T),x^{\prime }(0)=x^{\prime }(T)$$

has no solution on ∂ Ω;

• (ii) the equation

  $$F(a):=\frac{1}{T}{\int }_0^T\tilde{f}(t,x,x^{\prime })dt=0$$

has no solution on $\partial \mathrm{\Omega }\cap \mathbb{R}$;

• (iii) the Brouwer degree of F

  $$deg\{F,\mathrm{\Omega }\cap \mathbb{R},0\}\ne 0.$$

Then the periodic boundary value problem (2.1) has at least one periodic solution on $\overline{\mathrm{\Omega }}$.

Lemma 2.2 If $\varphi (x)$ is bounded, then x is also bounded.

Proof Since $\varphi (x)$ is bounded, then there exists a positive constant N such that $|\varphi (x)|\le N$. From (A₂), we have $\alpha (|x|)|x|\le \varphi (x)\cdot x\le |\varphi (x)|\cdot |x|\le N|x|$. Hence, we can get $\alpha (|x|)\le N$ for all $x\in \mathbb{R}$. If x is not bounded, then from the definition of α, we get $\alpha (|x|)>N$ for some $x\in \mathbb{R}$, which is a contradiction. So x is also bounded. □

Lemma 2.3 Suppose that the following condition holds:

(A₃) $(x_1-x_2)(g(t,x_1)-g(t,x_2))<0$ for all t, $x_1,x_2\in \mathbb{R}$, $x_1\ne x_2$.

Then (1.1) has at most one T-periodic solution in $C_T^1$.

Proof Assume that $x_1(t)$ and $x_2(t)$ are two T-periodic solutions of (1.1). Then we obtain

$${(\varphi (x_1^{\prime }(t))-\varphi (x_2^{\prime }(t)))}^{\prime }+f(t,x_1^{\prime }(t))-f(t,x_2^{\prime }(t))+g(t,x_1(t))-g(t,x_2(t))=0.$$

(2.2)

Set $u(t)=x_1(t)-x_2(t)$. Now, we claim that

$$u(t)\le 0\text{for all }t\in \mathbb{R}.$$

In contrast, in view of $x_1,x_2\in C^1[0,T]$, for $t\in \mathbb{R}$, we obtain

$$\underset{t\in \mathbb{R}}{max}u(t)>0.$$

Then there must exist $t^{\ast }\in \mathbb{R}$ (for convenience, we can choose $t^{\ast }\in (0,T)$) such that

$$u\left(t^{\ast }\right)=\underset{t\in [0,T]}{max}u(t)=\underset{t\in \mathbb{R}}{max}u(t)>0,$$

which implies that

$$u^{\prime }\left(t^{\ast }\right)=x_1^{\prime }\left(t^{\ast }\right)-x_2^{\prime }\left(t^{\ast }\right)=0$$

and

$$x_1\left(t^{\ast }\right)-x_2\left(t^{\ast }\right)>0.$$

By hypothesis (A₃) and (2.2), we have

$$\begin{array}{rcl}{(\varphi \left(x_1^{\prime }\left(t^{\ast }\right)\right)-\varphi \left(x_2^{\prime }\left(t^{\ast }\right)\right))}^{\prime }& =& -[f(t^{\ast },x_1^{\prime }\left(t^{\ast }\right))-f(t^{\ast },x_2^{\prime }\left(t^{\ast }\right))]\\ -[g(t^{\ast },x_1\left(t^{\ast }\right))-g(t^{\ast },x_2\left(t^{\ast }\right))]\\ =& -[g(t^{\ast },x_1\left(t^{\ast }\right))-g(t^{\ast },x_2\left(t^{\ast }\right))]>0,\end{array}$$

and there exists $\epsilon >0$ such that ${(\varphi (x_1^{\prime }(t))-\varphi (x_2^{\prime }(t)))}^{\prime }>0$ for all $t\in (t^{\ast }-\epsilon ,t^{\ast }]$. Therefore, $\varphi (x_1^{\prime }(t))-\varphi (x_2^{\prime }(t))$ is strictly increasing for $t\in (t^{\ast }-\epsilon ,t^{\ast }]$, which implies that

$$\varphi (x_1^{\prime }(t))-\varphi (x_2^{\prime }(t))<\varphi \left(x_1^{\prime }\left(t^{\ast }\right)\right)-\varphi \left(x_2^{\prime }\left(t^{\ast }\right)\right)=0\text{for all }t\in (t^{\ast }-\epsilon ,t^{\ast }).$$

From (A₁) we get

$$u^{\prime }(t)=x_1^{\prime }(t)-x_2^{\prime }(t)<0\text{for all }t\in (t^{\ast }-\epsilon ,t^{\ast }).$$

This contradicts the definition of $t^{\ast }$. Thus,

$$u(t)=x_1(t)-x_2(t)\le 0\text{for all }t\in \mathbb{R}.$$

By using a similar argument, we can also show that

$$x_2(t)-x_1(t)\le 0.$$

Therefore, we obtain

$$x_1(t)\equiv x_2(t)\text{for all }t\in \mathbb{R}.$$

Hence, (1.1) has at most one T-periodic solution in $C_T^1$. The proof of Lemma 2.3 is now complete. □

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

(H₁) there exists a positive constant D such that $g(t,x)-e(t)<0$ for $x>D$ and $t\in \mathbb{R}$, $g(t,x)-e(t)>0$ for $x\le 0$ and $t\in \mathbb{R}$;

(H₂) there exist constants $\sigma >0$ and $m\ge 2$ such that $f(t,u)u\ge \sigma {|u|}^m$ for $(t,u)\in [0,T]\times \mathbb{R}$;

(H₃) there exist positive constants ρ and γ such that $|f(t,u)|\le \rho {|u|}^{m-1}+\gamma$ for $(t,u)\in [0,T]\times \mathbb{R}$;

(H₄) there exist positive constants α, β, B such that

$$|g(t,x)|\le \alpha {|x|}^{m-1}+\beta \text{for }|x|\ge B\text{ and }t\in \mathbb{R}.$$

By using Lemmas 2.1-2.3, we obtain our main results.

Theorem 2.1 Assume that conditions (H₁)-(H₄) and (A₃) hold. Then (1.1) has a unique positive T-periodic solution if $\sigma -\frac{\alpha T^{m-1}}{2^{m-1}}>0$.

Proof Consider the homotopic equation of (1.1) as follows:

$${\left(\varphi (x^{\prime }(t))\right)}^{\prime }+\lambda f(t,x^{\prime }(t))+\lambda g(t,x(t))=\lambda e(t).$$

(2.3)

By Lemma 2.3, it is easy to see that (1.1) has at most one T-periodic solution in $C_T^1$. Thus, to prove Theorem 2.1, it suffices to show that (1.1) has at least one T-periodic solution in $C_T^1$. To do this, we are going to apply Lemmas 2.1 and 2.2. Firstly, we will claim that the set of all possible T-periodic solutions of (2.3) is bounded. Let $x(t)\in C_T^1$ be an arbitrary solution of (2.3) with period T. As $x(0)=x(T)$, there exists $t_0\in [0,T]$ such that $x^{\prime }(t_0)=0$, while $\varphi (0)=0$, we see

$$\begin{array}{rcl}\left|\varphi (x^{\prime }(t))\right|& =& \left|{\int }_{t_0}^t{\left(\varphi (x^{\prime }(s))\right)}^{\prime }ds\right|\\ \le & \lambda {\int }_0^T\left|f(t,x^{\prime }(t))\right|dt+\lambda {\int }_0^T\left|g(t,x(t))\right|dt+\lambda {\int }_0^T|e(t)|dt,\end{array}$$

(2.4)

where $t\in [t_0,t_0+T]$.

We claim that there is a constant $\xi \in \mathbb{R}$ such that

$$|x(\xi )|\le D.$$

(2.5)

Let $\overline{t}$, $\underline{t}$ be, respectively, the global maximum point and the global minimum point of $x(t)$ on $[0,T]$; then $x^{\prime }(\overline{t})=0$, and we claim that

$${\left(\varphi (x^{\prime }(\overline{t}))\right)}^{\prime }\le 0.$$

(2.6)

Assume, by way of contradiction, that (2.6) does not hold. Then ${(\varphi (x^{\prime }(\overline{t})))}^{\prime }>0$ and there exists $\epsilon >0$ such that ${(\varphi (x^{\prime }(t)))}^{\prime }>0$ for $t\in (\overline{t}-\epsilon ,\overline{t}+\epsilon )$. Therefore $\varphi (x^{\prime }(t))$ is strictly increasing for $t\in (\overline{t}-\epsilon ,\overline{t}+\epsilon )$. From (A₁) we know that $x^{\prime }(t)$ is strictly increasing for $t\in (\overline{t}-\epsilon ,\overline{t}+\epsilon )$. This contradicts the definition of $\overline{t}$. Thus, (2.6) is true. From $f(t,0)=0$, (2.3) and (2.6), we have

$$g(\overline{t},x(\overline{t}))-e(\overline{t})\ge 0.$$

(2.7)

Similarly, we get

$$g(\underline{t},x(\underline{t}))-e(\underline{t})\le 0.$$

(2.8)

In view of (H₁), (2.7) and (2.8) imply that

$$x(\overline{t})\le D,x(\underline{t})>0.$$

Case (1): If $x(\underline{t})\in (0,D)$, define $\xi =\overline{t}$, obviously, $|x(\xi )|\le D$.

Case (2): If $x(\underline{t})\ge D$, from $x(\overline{t})\le D$, we know $x(\overline{t})=x(\underline{t})$. Define $\xi =\overline{t}$, we have $|x(\xi )|=D$. This proves (2.5).

Then we have

$$|x(t)|=|x(\xi )+{\int }_{\xi }^t{x}^{\prime }(s)ds|\le D+{\int }_{\xi }^t|x^{\prime }(s)|ds,t\in [\xi ,\xi +T]$$

and

$$|x(t)|=|x(t-T)|=|x(\xi )-{\int }_{t-T}^{\xi }{x}^{\prime }(s)ds|\le D+{\int }_{t-T}^{\xi }|x^{\prime }(s)|ds,t\in [\xi ,\xi +T].$$

Combining the above two inequalities, we obtain

$$\begin{array}{rl}{|x|}_0& =\underset{t\in [0,T]}{max}|x(t)|=\underset{t\in [\xi ,\xi +T]}{max}|x(t)|\\ \le \underset{t\in [\xi ,\xi +T]}{max}\{D+\frac{1}{2}\left({\int }_{\xi }^t\right|x^{\prime }(s)|ds+{\int }_{t-T}^{\xi }|x^{\prime }(s)\left|ds\right)\}\\ \le D+\frac{1}{2}{\int }_0^T|x^{\prime }(s)|ds.\end{array}$$

(2.9)

Since $x^{\prime }(t)$ is T-periodic, multiplying $x^{\prime }(t)$ and (2.3) and then integrating it from 0 to T, we have

$$\begin{array}{rl}0& ={\int }_0^T{\left(\varphi (x^{\prime }(t))\right)}^{\prime }{x}^{\prime }(t)dt\\ =-\lambda {\int }_0^{T}f(t,x^{\prime }(t))x^{\prime }(t)dt-\lambda {\int }_0^{T}g(t,x(t))x^{\prime }(t)dt+\lambda {\int }_0^{T}e(t)x^{\prime }(t)dt.\end{array}$$

(2.10)

In view of (2.10), we have

$$|{\int }_0^{T}f(t,x^{\prime }(t))x^{\prime }(t)dt|=|-{\int }_0^{T}g(t,x(t))x^{\prime }(t)dt+{\int }_0^{T}e(t)x^{\prime }(t)dt|.$$

From (H₂), we know

$$|{\int }_0^{T}f(t,x^{\prime }(t))x^{\prime }(t)dt|\ge \sigma {\int }_0^T|x^{\prime }(t){|}^{m}dt.$$

Set

$$E_1=\{t\in [0,T]\mid |x(t)|\le B\},E_2=\{t\in [0,T]\mid |x(t)|\ge B\}.$$

From (H₄), we have

$$\begin{array}{c}\sigma {\int }_0^T|x^{\prime }(t){|}^{m}dt\hfill \\ \le {\int }_{E_1+E_2}\left|g(t,x(t))\right||x^{\prime }(t)|dt+{\int }_0^T|e(t)||x^{\prime }(t)|dt\hfill \\ \le {\left({\int }_{E_1}\right|g(t,x(t)){|}^{\frac{m}{m-1}}dt)}^{\frac{m-1}{m}}{\left({\int }_0^T\right|x^{\prime }(t){|}^{m}dt)}^{\frac{1}{m}}+\alpha {\int }_0^T|x(t){|}^{m-1}|x^{\prime }(t)|dt\hfill \\ +\beta {\int }_0^T|x^{\prime }(t)|dt+{\int }_0^T|e(t)||x^{\prime }(t)|dt\hfill \\ \le {|g_B|}_{\frac{m}{m-1}}{\left({\int }_0^T\right|x^{\prime }(t){|}^{m}dt)}^{\frac{1}{m}}+\alpha {(D+\frac{1}{2}{\int }_0^T|x^{\prime }(t)\left|dt\right)}^{m-1}{\int }_0^T|x^{\prime }(t)|dt\hfill \\ +\beta T^{\frac{m-1}{m}}{\left({\int }_0^T\right|x^{\prime }(t){|}^{m}dt)}^{\frac{1}{m}}+{\left({\int }_0^T\right|e(t){|}^{\frac{m}{m-1}})}^{\frac{m-1}{m}}{\left({\int }_0^T\right|x^{\prime }(t){|}^{m}dt)}^{\frac{1}{m}}\hfill \\ ={|g_B|}_{\frac{m}{m-1}}{\left({\int }_0^T\right|x^{\prime }(t){|}^{m}dt)}^{\frac{1}{m}}+\frac{\alpha }{2^{m-1}}{(\frac{2D}{{\int }_0^T|x^{\prime }(t)|dt}+1)}^{m-1}{\left({\int }_0^T\right|x^{\prime }(t)\left|dt\right)}^m\hfill \\ +\beta T^{\frac{m-1}{m}}{\left({\int }_0^T\right|x^{\prime }(t){|}^{m}dt)}^{\frac{1}{m}}+{|e|}_{\frac{m}{m-1}}{\left({\int }_0^T\right|x^{\prime }(t){|}^{m}dt)}^{\frac{1}{m}},\hfill \end{array}$$

(2.11)

where $g_B={max}_{|x|\le B}|g(t,x(t))|$, ${|g_B|}_{\frac{m}{m-1}}={({\int }_0^T{|g_B|}^{\frac{m}{m-1}}dt)}^{\frac{m-1}{m}}$.

For the constant $\delta >0$, which is only dependent on $k>0$, we have

$${(1+x)}^k\le 1+(1+k)x\text{for }x\in [0,\delta ].$$

So, from (2.11), we have

$$\begin{array}{c}\sigma {\int }_0^T|x^{\prime }(t){|}^{m}dt\hfill \\ \le |g_B{|}_{\frac{m}{m-1}}{\left({\int }_0^T\right|x^{\prime }(t){|}^{m}dt)}^{\frac{1}{m}}+\frac{\alpha }{2^{m-1}}(1+\frac{2Dm}{{\int }_0^T|x^{\prime }(t)|dt}){\left({\int }_0^T\right|x^{\prime }(t)\left|dt\right)}^m\hfill \\ +\beta T^{\frac{m-1}{m}}{\left({\int }_0^T\right|x^{\prime }(t){|}^{m}dt)}^{\frac{1}{m}}+|e{|}_{\frac{m}{m-1}}{\left({\int }_0^T\right|x^{\prime }(t){|}^{m}dt)}^{\frac{1}{m}}\hfill \\ =|g_B{|}_{\frac{m}{m-1}}{\left({\int }_0^T\right|x^{\prime }(t){|}^{m}dt)}^{\frac{1}{m}}+\frac{\alpha }{2^{m-1}}{\left({\int }_0^T\right|x^{\prime }(t)\left|dt\right)}^m+\frac{\alpha Dm}{2^{m-2}}{\left({\int }_0^T\right|x^{\prime }(t)\left|dt\right)}^{m-1}\hfill \\ +\beta T^{\frac{m-1}{m}}{\left({\int }_0^T\right|x^{\prime }(t){|}^{m}dt)}^{\frac{1}{m}}+|e{|}_{\frac{m}{m-1}}{\left({\int }_0^T\right|x^{\prime }(t){|}^{m}dt)}^{\frac{1}{m}}\hfill \\ \le |g_B{|}_{\frac{m}{m-1}}{\left({\int }_0^T\right|x^{\prime }(t){|}^{m}dt)}^{\frac{1}{m}}+\frac{\alpha T^{m-1}}{2^{m-1}}{\int }_0^T|x^{\prime }(t){|}^{m}dt\hfill \\ +\frac{\alpha Dm{T}^{\frac{{(m-1)}^2}{m}}}{2^{m-2}}{\left({\int }_0^T\right|x^{\prime }(t){|}^{m}dt)}^{\frac{m-1}{m}}\hfill \\ +\beta T^{\frac{m-1}{m}}{\left({\int }_0^T\right|x^{\prime }(t){|}^{m}dt)}^{\frac{1}{m}}+|e{|}_{\frac{m}{m-1}}{\left({\int }_0^T\right|x^{\prime }(t){|}^{m}dt)}^{\frac{1}{m}}\hfill \\ =\frac{\alpha T^{m-1}}{2^{m-1}}{\int }_0^T|x^{\prime }(t){|}^{m}dt+\frac{\alpha Dm{T}^{\frac{{(m-1)}^2}{m}}}{2^{m-2}}{\left({\int }_0^T\right|x^{\prime }(t){|}^{m}dt)}^{\frac{m-1}{m}}\hfill \\ +(|g_B{|}_{\frac{m}{m-1}}+\beta T^{\frac{m-1}{m}}+|e{|}_{\frac{m}{m-1}}){\left({\int }_0^T\right|x^{\prime }(t){|}^{m}dt)}^{\frac{1}{m}}.\hfill \end{array}$$

Since $\sigma -\frac{\alpha T^{m-1}}{2^{m-1}}>0$, so it is easy to see that there is a constant $M_1^{\prime }>0$ (independent of λ) such that

$${\int }_0^T|x^{\prime }(t){|}^{m}dt\le M_1^{\prime }.$$

By applying Hölder’s inequality and (2.9), we have

$${|x|}_0\le D+\frac{1}{2}{\int }_0^T|x^{\prime }(s)|ds\le D+\frac{1}{2}{T}^{\frac{m-1}{m}}{\left({\int }_0^T\right|x^{\prime }(t){|}^{m}dt)}^{\frac{1}{m}}\le D+\frac{1}{2}{T}^{\frac{m-1}{m}}{\left(M_1^{\prime }\right)}^{\frac{1}{m}}:=M_1.$$

In view of (2.4) and (H₃), we have

$$\begin{array}{rcl}{\left|\varphi \left(x^{\prime }\right)\right|}_0& =& \underset{t\in [0,T]}{max}\left\{\right|\varphi (x^{\prime }(t))\left|\right\}\\ =& \underset{t\in [t_0,t_0+T]}{max}\left\{\right|{\int }_{t_0}^t{\left(\varphi (x^{\prime }(s))\right)}^{\prime }ds\left|\right\}\\ \le & {\int }_0^T\left|f(t,x^{\prime }(t))\right|dt+{\int }_0^T\left|g(t,x(t))\right|dt+{\int }_0^T|e(t)|dt\\ \le & \rho {\int }_0^T|x^{\prime }(t){|}^{m-1}dt+\gamma T+T^{\frac{1}{m}}{\left({\int }_0^T\right|g(t,x(t)){|}^{\frac{m}{m-1}}dt)}^{\frac{m-1}{m}}\\ +T^{\frac{1}{m}}{\left({\int }_0^T\right|e(t){|}^{\frac{m}{m-1}}dt)}^{\frac{m-1}{m}}\\ \le & \rho T^{\frac{1}{m}}{\left({\int }_0^T\right|x^{\prime }(t){|}^{m}dt)}^{\frac{m-1}{m}}+\gamma T+T^{\frac{1}{m}}{\left({\int }_0^T\right|g(t,x(t)){|}^{\frac{m}{m-1}}dt)}^{\frac{m-1}{m}}\\ +T^{\frac{1}{m}}{\left({\int }_0^T{|e(t)|}^{\frac{m}{m-1}}dt\right)}^{\frac{m-1}{m}}\\ \le & \rho T^{\frac{1}{m}}{\left(M_1^{\prime }\right)}^{\frac{m-1}{m}}+\gamma T+T^{\frac{1}{m}}{|g_{M_1}|}_{\frac{m-1}{m}}+T^{\frac{1}{m}}{|e|}_{\frac{m-1}{m}}:=M_2^{\prime },\end{array}$$

where $|g_{M_1}|={max}_{|x(t)|\le M_1}|g(t,x(t))|$.

Thus, from Lemma 2.2, we know that there exists some positive constant $M_2>M_2^{\prime }+1$ such that, for all $t\in \mathbb{R}$,

$$|x^{\prime }(t)|\le M_2.$$

Set $M=\sqrt{M_1^2+M_2^2}+1$, we have

$$\mathrm{\Omega }=\{x\in C_T^1(\mathbb{R},\mathbb{R})\mid {|x|}_0\le M+1,|x^{\prime }{|}_0\le M+1\},$$

we know that (2.4) has no solution on ∂ Ω as $\lambda \in (0,1)$ and when $x(t)\in \partial \mathrm{\Omega }\cap \mathbb{R}$, $x(t)=M+1$ or $x(t)=-M-1$, from (2.11) we know that $M+1>D$. So, from (H₁) we see that

$$\begin{array}{c}\frac{1}{T}{\int }_0^T\{g(t,M+1)-e(t)\}dt<0,\hfill \\ \frac{1}{T}{\int }_0^T\{g(t,-M-1)-e(t)\}dt>0.\hfill \end{array}$$

So condition (ii) is also satisfied. Set

$$H(x,\mu )=\mu x-(1-\mu )\frac{1}{T}{\int }_0^T\{g(t,x)-e(t)\}dt,$$

where $x\in \partial \mathrm{\Omega }\cap \mathbb{R}$, $\mu \in [0,1]$, we have

$$xH(x,\mu )=\mu x^2-(1-\mu )x\frac{1}{T}{\int }_0^T\{g(t,x)-e(t)\}dt>0,$$

and thus $H(x,\mu )$ is a homotopic transformation and

$$\begin{array}{rl}deg\{F,\mathrm{\Omega }\cap \mathbb{R},0\}& =deg\{-\frac{1}{T}{\int }_0^T\{g(t,x)-e(t)\}dt,\mathrm{\Omega }\cap \mathbb{R},0\}\\ =deg\{x,\mathrm{\Omega }\cap \mathbb{R},0\}\ne 0.\end{array}$$

So condition (iii) is satisfied. In view of Lemma 2.1, there exists at least one solution with period T.

Suppose that $x(t)$ is the T-periodic solution of (1.1). We can easily show that (2.8) also holds. Thus,

$$x(t)\ge \underset{t\in [0,T]}{min}x(t)=x(\underline{t})>0\text{for all }t\in \mathbb{R},$$

which implies that (1.1) has a unique positive solution with period T. This completes the proof. □

We illustrate our results with some examples.

Example 2.1 Consider the following second-order p-Laplacian-like Rayleigh equation:

$${\left({\varphi }_p(x^{\prime }(t))\right)}^{\prime }+(10+5{sin}^{2}t)x^{\prime }(t)-(5x(t)+{sin}^{2}t-8)=e^{{cos}^{2}t},$$

(2.12)

where ${\varphi }_p(u)={|u|}^{p-2}u$.

Comparing (2.12) to (1.1), we see that $g(t,x)=-5x(t)-{sin}^{2}t+8$, $f(t,u)=(10+5{sin}^{2}t)u$, $e(t)=e^{{cos}^{2}t}$, $T=\pi$. Obviously, we know that ${\varphi }_p$ is a homeomorphism from ℝ to ℝ satisfying (A₁) and (A₂). Consider $(x_1-x_2)(g(t,x_1)-g(t,x_2))=-5{(x_1-x_2)}^2<0$ for $x_1\ne x_2$, then (A₃) holds. Moreover, it is easily seen that there exists a constant $D=2$ such that (H₁) holds. Consider $f(t,u)u=(10+5{sin}^{2}t)u^2\ge 10u^2$, here $\sigma =10$, $m=2$, and $|f(t,u)|=|(10+5{sin}^{2}t)u|\le 15|u|+1$, here $\rho =15$, $\gamma =1$. So, we can get that conditions (H₂) and (H₃) hold. Choose $B>0$, we have $|g(t,x)|\le 5|x|+9$, here $\alpha =5$, $\beta =9$, then (H₄) holds and $\sigma -\frac{\alpha T}{2}=10-\frac{5\pi }{2}>0$. So, by Theorem 2.1, we can get that (2.12) has a unique positive periodic solution.

Example 2.2 Consider the following second-order p-Laplacian-like Rayleigh equation:

$${\left(\varphi (x^{\prime }(t))\right)}^{\prime }+(200+16{cos}^{2}t){(x^{\prime }(t))}^3-(20x^3(t)+10{cos}^2(t)-15)=e^{{sin}^{2}t},$$

(2.13)

where $\varphi (u)=u{e}^{{|u|}^2}$.

Comparing (2.13) to (1.1), we see that $g(t,x)=-20x^3-10{cos}^{2}t+15$, $f(t,v)=(200+16{cos}^{2}t)v^3$, $e(t)=e^{{sin}^{2}t}$, $T=\pi$. Obviously, we get

$$(x{e}^{{|x|}^2}-y{e}^{{|y|}^2})(x-y)\ge (|x|e^{{|x|}^2}-|y|e^{{|y|}^2})(|x|-|y|)\ge 0$$

and

$$\varphi (x)\cdot x={|x|}^2{e}^{{|x|}^2}.$$

So, we know that (A₁) and (A₂) hold. Consider $(x_1-x_2)(g(t,x_1)-g(t,x_2))=-20{(x_1-x_2)}^2(x_1^2+x_1{x}_2+x_2^2)<0$ for $x_1\ne x_2$, then (A₃) holds. Moreover, it is easily seen that there exists a constant $D=1$ such that (H₁) holds. Consider $f(t,v)v=(200+16{cos}^{2}t)v^4\ge 200v^4$, here $\sigma =200$, $m=4$, and $|f(t,v)|=|(200+16{cos}^{2}t)v^3|\le 216{|v|}^3+5$, here $\rho =216$, $\gamma =5$. So, we can get that conditions (H₂) and (H₃) hold. Choose $B>0$, we have $|g(t,x)|\le 20{|x|}^3+25$, here $\alpha =20$, $\beta =25$, then (H₄) holds and $\sigma -\frac{\alpha T^{m-1}}{2^{m-1}}=200-\frac{20\times {\pi }^3}{2^3}>0$. Therefore, by Theorem 2.1, we know that (2.13) has a unique positive periodic solution.