# 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 [110]). For example, in 2009, Xiao and Liu [7] studied the Rayleigh type p-Laplacian equation with a deviating argument of the form

$( ϕ p ( x ′ ( t ) ) ) ′ +f ( t , x ′ ( t ) ) +g ( t , x ( t − τ ( t ) ) ) =e(t).$

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

$( ϕ p ( x ′ ( t ) ) ) ′ +f ( t , x ′ ( t ) ) +g ( t , x ( t ) ) =e(t).$

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

$( ϕ ( x ′ ( t ) ) ) ′ +f ( t , x ′ ( t ) ) +g ( t , x ( t ) ) =e(t),$
(1.1)

where the function $ϕ:R→R$ is continuous and $ϕ(0)=0$. $f,g∈Car(R×R,R)$ is an $L p$-Carathéodory function and $p= m m − 1$, $m≥2$, which means it is measurable in the first variable and continuous in the second variable. For every $0, there exists $h r , s ∈ L p [0,T]$ such that $|g(t,x(t))|≤ h r , s$ for all $x∈[r,s]$ and a.e. $t∈[0,T]$; and f, g is a T-periodic function about t and $f(t,0)=0$. $e∈ L p ([0,T],R)$ and is T-periodic.

Here $ϕ:R→R$ is a continuous function and $ϕ(0)=0$, which satisfies

(A1) $(ϕ( x 1 )−ϕ( x 2 ))( x 1 − x 2 )>0$ for $∀ x 1 ≠ x 2$, $x 1 , x 2 ∈R$;

(A2) there exists a function $α:[0,+∞]→[0,+∞]$, $α(s)→+∞$ as $s→+∞$, such that $ϕ(x)⋅x≥α(|x|)|x|$ for $∀x∈R$.

It is easy to see that ϕ represents a large class of nonlinear operators, including $ϕ p :R→R$ is a p-Laplacian, i.e., $ϕ p (x)= | x | p − 2 x$ for $x∈R$.

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 $ϕ p$ to get a priori solutions, $∫ 0 T ( ϕ p ′ ( x ( t ) ) ) ′ x(t)dt=− ∫ 0 T | x ′ ( t ) | p dt$, 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

which is a Banach space endowed with the norm $∥⋅∥$; define $∥x∥=max{ | x | 0 , | x ′ | 0 }$ for all x, and

$| x | 0 = max t ∈ [ 0 , T ] |x(t)|,| x ′ | 0 = max t ∈ [ 0 , T ] | x ′ (t)|.$

For the T-periodic boundary value problem

$( ϕ ( x ′ ( t ) ) ) ′ = f ˜ ( t , x , x ′ ) ,$
(2.1)

here $f ˜ :[0,T]×R×R→R$ is assumed to be Carathéodory.

Lemma 2.1 (Manásevich-Mawhin [11])

Let Ω be an open bounded set in $C T 1$. If

1. (i)

for each $λ∈(0,1)$, the problem

$( ϕ ( x ′ ) ) ′ =λ f ˜ ( t , x , x ′ ) ,x(0)=x(T), x ′ (0)= x ′ (T)$

has no solution on Ω;

1. (ii)

the equation

$F(a):= 1 T ∫ 0 T f ˜ ( t , x , x ′ ) dt=0$

has no solution on $∂Ω∩R$;

1. (iii)

the Brouwer degree of F

$deg{F,Ω∩R,0}≠0.$

Then the periodic boundary value problem (2.1) has at least one periodic solution on $Ω ¯$.

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

Proof Since $ϕ(x)$ is bounded, then there exists a positive constant N such that $|ϕ(x)|≤N$. From (A2), we have $α(|x|)|x|≤ϕ(x)⋅x≤|ϕ(x)|⋅|x|≤N|x|$. Hence, we can get $α(|x|)≤N$ for all $x∈R$. If x is not bounded, then from the definition of α, we get $α(|x|)>N$ for some $x∈R$, 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, $x 1 , x 2 ∈R$, $x 1 ≠ x 2$.

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

$( ϕ ( x 1 ′ ( t ) ) − ϕ ( x 2 ′ ( t ) ) ) ′ +f ( t , x 1 ′ ( t ) ) −f ( t , x 2 ′ ( t ) ) +g ( t , x 1 ( t ) ) −g ( t , x 2 ( t ) ) =0.$
(2.2)

Set $u(t)= x 1 (t)− x 2 (t)$. Now, we claim that

In contrast, in view of $x 1 , x 2 ∈ C 1 [0,T]$, for $t∈R$, we obtain

$max t ∈ R u(t)>0.$

Then there must exist $t ∗ ∈R$ (for convenience, we can choose $t ∗ ∈(0,T)$) such that

$u ( t ∗ ) = max t ∈ [ 0 , T ] u(t)= max t ∈ R u(t)>0,$

which implies that

$u ′ ( t ∗ ) = x 1 ′ ( t ∗ ) − x 2 ′ ( t ∗ ) =0$

and

$x 1 ( t ∗ ) − x 2 ( t ∗ ) >0.$

By hypothesis (A3) and (2.2), we have

$( ϕ ( x 1 ′ ( t ∗ ) ) − ϕ ( x 2 ′ ( t ∗ ) ) ) ′ = − [ f ( t ∗ , x 1 ′ ( t ∗ ) ) − f ( t ∗ , x 2 ′ ( t ∗ ) ) ] − [ g ( t ∗ , x 1 ( t ∗ ) ) − g ( t ∗ , x 2 ( t ∗ ) ) ] = − [ g ( t ∗ , x 1 ( t ∗ ) ) − g ( t ∗ , x 2 ( t ∗ ) ) ] > 0 ,$

and there exists $ε>0$ such that $( ϕ ( x 1 ′ ( t ) ) − ϕ ( x 2 ′ ( t ) ) ) ′ >0$ for all $t∈( t ∗ −ε, t ∗ ]$. Therefore, $ϕ( x 1 ′ (t))−ϕ( x 2 ′ (t))$ is strictly increasing for $t∈( t ∗ −ε, t ∗ ]$, which implies that

From (A1) we get

This contradicts the definition of $t ∗$. Thus,

By using a similar argument, we can also show that

$x 2 (t)− x 1 (t)≤0.$

Therefore, we obtain

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 $t∈R$, $g(t,x)−e(t)>0$ for $x≤0$ and $t∈R$;

(H2) there exist constants $σ>0$ and $m≥2$ such that $f(t,u)u≥σ | u | m$ for $(t,u)∈[0,T]×R$;

(H3) there exist positive constants ρ and γ such that $|f(t,u)|≤ρ | u | m − 1 +γ$ for $(t,u)∈[0,T]×R$;

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

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 $σ− α T m − 1 2 m − 1 >0$.

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

$( ϕ ( x ′ ( t ) ) ) ′ +λf ( t , x ′ ( t ) ) +λg ( t , x ( t ) ) =λe(t).$
(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 $x(t)∈ C T 1$ be an arbitrary solution of (2.3) with period T. As $x(0)=x(T)$, there exists $t 0 ∈[0,T]$ such that $x ′ ( t 0 )=0$, while $ϕ(0)=0$, we see

$| ϕ ( x ′ ( t ) ) | = | ∫ t 0 t ( ϕ ( x ′ ( s ) ) ) ′ d s | ≤ λ ∫ 0 T | f ( t , x ′ ( t ) ) | d t + λ ∫ 0 T | g ( t , x ( t ) ) | d t + λ ∫ 0 T | e ( t ) | d t ,$
(2.4)

where $t∈[ t 0 , t 0 +T]$.

We claim that there is a constant $ξ∈R$ such that

$|x(ξ)|≤D.$
(2.5)

Let $t ¯$, $t ̲$ be, respectively, the global maximum point and the global minimum point of $x(t)$ on $[0,T]$; then $x ′ ( t ¯ )=0$, and we claim that

$( ϕ ( x ′ ( t ¯ ) ) ) ′ ≤0.$
(2.6)

Assume, by way of contradiction, that (2.6) does not hold. Then $( ϕ ( x ′ ( t ¯ ) ) ) ′ >0$ and there exists $ε>0$ such that $( ϕ ( x ′ ( t ) ) ) ′ >0$ for $t∈( t ¯ −ε, t ¯ +ε)$. Therefore $ϕ( x ′ (t))$ is strictly increasing for $t∈( t ¯ −ε, t ¯ +ε)$. From (A1) we know that $x ′ (t)$ is strictly increasing for $t∈( t ¯ −ε, t ¯ +ε)$. This contradicts the definition of $t ¯$. Thus, (2.6) is true. From $f(t,0)=0$, (2.3) and (2.6), we have

$g ( t ¯ , x ( t ¯ ) ) −e( t ¯ )≥0.$
(2.7)

Similarly, we get

$g ( t ̲ , x ( t ̲ ) ) −e( t ̲ )≤0.$
(2.8)

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

$x( t ¯ )≤D,x( t ̲ )>0.$

Case (1): If $x( t ̲ )∈(0,D)$, define $ξ= t ¯$, obviously, $|x(ξ)|≤D$.

Case (2): If $x( t ̲ )≥D$, from $x( t ¯ )≤D$, we know $x( t ¯ )=x( t ̲ )$. Define $ξ= t ¯$, we have $|x(ξ)|=D$. This proves (2.5).

Then we have

$|x(t)|=|x(ξ)+ ∫ ξ t x ′ (s)ds|≤D+ ∫ ξ t | x ′ (s)|ds,t∈[ξ,ξ+T]$

and

$|x(t)|=|x(t−T)|=|x(ξ)− ∫ t − T ξ x ′ (s)ds|≤D+ ∫ t − T ξ | x ′ (s)|ds,t∈[ξ,ξ+T].$

Combining the above two inequalities, we obtain

$| x | 0 = max t ∈ [ 0 , T ] | x ( t ) | = max t ∈ [ ξ , ξ + T ] | x ( t ) | ≤ max t ∈ [ ξ , ξ + T ] { D + 1 2 ( ∫ ξ t | x ′ ( s ) | d s + ∫ t − T ξ | x ′ ( s ) | d s ) } ≤ D + 1 2 ∫ 0 T | x ′ ( s ) | d s .$
(2.9)

Since $x ′ (t)$ is T-periodic, multiplying $x ′ (t)$ and (2.3) and then integrating it from 0 to T, we have

$0 = ∫ 0 T ( ϕ ( x ′ ( t ) ) ) ′ x ′ ( t ) d t = − λ ∫ 0 T f ( t , x ′ ( t ) ) x ′ ( t ) d t − λ ∫ 0 T g ( t , x ( t ) ) x ′ ( t ) d t + λ ∫ 0 T e ( t ) x ′ ( t ) d t .$
(2.10)

In view of (2.10), we have

$| ∫ 0 T f ( t , x ′ ( t ) ) x ′ (t)dt|=|− ∫ 0 T g ( t , x ( t ) ) x ′ (t)dt+ ∫ 0 T e(t) x ′ (t)dt|.$

From (H2), we know

$| ∫ 0 T f ( t , x ′ ( t ) ) x ′ (t)dt|≥σ ∫ 0 T | x ′ (t) | m dt.$

Set

$E 1 = { t ∈ [ 0 , T ] ∣ | x ( t ) | ≤ B } , E 2 = { t ∈ [ 0 , T ] ∣ | x ( t ) | ≥ B } .$

From (H4), we have

$σ ∫ 0 T | x ′ ( t ) | m d t ≤ ∫ E 1 + E 2 | g ( t , x ( t ) ) | | x ′ ( t ) | d t + ∫ 0 T | e ( t ) | | x ′ ( t ) | d t ≤ ( ∫ E 1 | g ( t , x ( t ) ) | m m − 1 d t ) m − 1 m ( ∫ 0 T | x ′ ( t ) | m d t ) 1 m + α ∫ 0 T | x ( t ) | m − 1 | x ′ ( t ) | d t + β ∫ 0 T | x ′ ( t ) | d t + ∫ 0 T | e ( t ) | | x ′ ( t ) | d t ≤ | g B | m m − 1 ( ∫ 0 T | x ′ ( t ) | m d t ) 1 m + α ( D + 1 2 ∫ 0 T | x ′ ( t ) | d t ) m − 1 ∫ 0 T | x ′ ( t ) | d t + β T m − 1 m ( ∫ 0 T | x ′ ( t ) | m d t ) 1 m + ( ∫ 0 T | e ( t ) | m m − 1 ) m − 1 m ( ∫ 0 T | x ′ ( t ) | m d t ) 1 m = | g B | m m − 1 ( ∫ 0 T | x ′ ( t ) | m d t ) 1 m + α 2 m − 1 ( 2 D ∫ 0 T | x ′ ( t ) | d t + 1 ) m − 1 ( ∫ 0 T | x ′ ( t ) | d t ) m + β T m − 1 m ( ∫ 0 T | x ′ ( t ) | m d t ) 1 m + | e | m m − 1 ( ∫ 0 T | x ′ ( t ) | m d t ) 1 m ,$
(2.11)

where $g B = max | x | ≤ B |g(t,x(t))|$, $| g B | m m − 1 = ( ∫ 0 T | g B | m m − 1 d t ) m − 1 m$.

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

So, from (2.11), we have

$σ ∫ 0 T | x ′ ( t ) | m d t ≤ | g B | m m − 1 ( ∫ 0 T | x ′ ( t ) | m d t ) 1 m + α 2 m − 1 ( 1 + 2 D m ∫ 0 T | x ′ ( t ) | d t ) ( ∫ 0 T | x ′ ( t ) | d t ) m + β T m − 1 m ( ∫ 0 T | x ′ ( t ) | m d t ) 1 m + | e | m m − 1 ( ∫ 0 T | x ′ ( t ) | m d t ) 1 m = | g B | m m − 1 ( ∫ 0 T | x ′ ( t ) | m d t ) 1 m + α 2 m − 1 ( ∫ 0 T | x ′ ( t ) | d t ) m + α D m 2 m − 2 ( ∫ 0 T | x ′ ( t ) | d t ) m − 1 + β T m − 1 m ( ∫ 0 T | x ′ ( t ) | m d t ) 1 m + | e | m m − 1 ( ∫ 0 T | x ′ ( t ) | m d t ) 1 m ≤ | g B | m m − 1 ( ∫ 0 T | x ′ ( t ) | m d t ) 1 m + α T m − 1 2 m − 1 ∫ 0 T | x ′ ( t ) | m d t + α D m T ( m − 1 ) 2 m 2 m − 2 ( ∫ 0 T | x ′ ( t ) | m d t ) m − 1 m + β T m − 1 m ( ∫ 0 T | x ′ ( t ) | m d t ) 1 m + | e | m m − 1 ( ∫ 0 T | x ′ ( t ) | m d t ) 1 m = α T m − 1 2 m − 1 ∫ 0 T | x ′ ( t ) | m d t + α D m T ( m − 1 ) 2 m 2 m − 2 ( ∫ 0 T | x ′ ( t ) | m d t ) m − 1 m + ( | g B | m m − 1 + β T m − 1 m + | e | m m − 1 ) ( ∫ 0 T | x ′ ( t ) | m d t ) 1 m .$

Since $σ− α T m − 1 2 m − 1 >0$, so it is easy to see that there is a constant $M 1 ′ >0$ (independent of λ) such that

$∫ 0 T | x ′ (t) | m dt≤ M 1 ′ .$

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

$| x | 0 ≤D+ 1 2 ∫ 0 T | x ′ (s)|ds≤D+ 1 2 T m − 1 m ( ∫ 0 T | x ′ ( t ) | m d t ) 1 m ≤D+ 1 2 T m − 1 m ( M 1 ′ ) 1 m := M 1 .$

In view of (2.4) and (H3), we have

$| ϕ ( x ′ ) | 0 = max t ∈ [ 0 , T ] { | ϕ ( x ′ ( t ) ) | } = max t ∈ [ t 0 , t 0 + T ] { | ∫ t 0 t ( ϕ ( x ′ ( s ) ) ) ′ d s | } ≤ ∫ 0 T | f ( t , x ′ ( t ) ) | d t + ∫ 0 T | g ( t , x ( t ) ) | d t + ∫ 0 T | e ( t ) | d t ≤ ρ ∫ 0 T | x ′ ( t ) | m − 1 d t + γ T + T 1 m ( ∫ 0 T | g ( t , x ( t ) ) | m m − 1 d t ) m − 1 m + T 1 m ( ∫ 0 T | e ( t ) | m m − 1 d t ) m − 1 m ≤ ρ T 1 m ( ∫ 0 T | x ′ ( t ) | m d t ) m − 1 m + γ T + T 1 m ( ∫ 0 T | g ( t , x ( t ) ) | m m − 1 d t ) m − 1 m + T 1 m ( ∫ 0 T | e ( t ) | m m − 1 d t ) m − 1 m ≤ ρ T 1 m ( M 1 ′ ) m − 1 m + γ T + T 1 m | g M 1 | m − 1 m + T 1 m | e | m − 1 m : = M 2 ′ ,$

where $| g M 1 |= max | x ( t ) | ≤ M 1 |g(t,x(t))|$.

Thus, from Lemma 2.2, we know that there exists some positive constant $M 2 > M 2 ′ +1$ such that, for all $t∈R$,

$| x ′ (t)|≤ M 2 .$

Set $M= M 1 2 + M 2 2 +1$, we have

$Ω= { x ∈ C T 1 ( R , R ) ∣ | x | 0 ≤ M + 1 , | x ′ | 0 ≤ M + 1 } ,$

we know that (2.4) has no solution on Ω as $λ∈(0,1)$ and when $x(t)∈∂Ω∩R$, $x(t)=M+1$ or $x(t)=−M−1$, from (2.11) we know that $M+1>D$. So, from (H1) we see that

$1 T ∫ 0 T { g ( t , M + 1 ) − e ( t ) } d t < 0 , 1 T ∫ 0 T { g ( t , − M − 1 ) − e ( t ) } d t > 0 .$

So condition (ii) is also satisfied. Set

$H(x,μ)=μx−(1−μ) 1 T ∫ 0 T { g ( t , x ) − e ( t ) } dt,$

where $x∈∂Ω∩R$, $μ∈[0,1]$, we have

$xH(x,μ)=μ x 2 −(1−μ)x 1 T ∫ 0 T { g ( t , x ) − e ( t ) } dt>0,$

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

$deg { F , Ω ∩ R , 0 } = deg { − 1 T ∫ 0 T { g ( t , x ) − e ( t ) } d t , Ω ∩ R , 0 } = deg { x , Ω ∩ R , 0 } ≠ 0 .$

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,

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:

$( ϕ p ( x ′ ( t ) ) ) ′ + ( 10 + 5 sin 2 t ) x ′ (t)− ( 5 x ( t ) + sin 2 t − 8 ) = e cos 2 t ,$
(2.12)

where $ϕ p (u)= | u | p − 2 u$.

Comparing (2.12) to (1.1), we see that $g(t,x)=−5x(t)− sin 2 t+8$, $f(t,u)=(10+5 sin 2 t)u$, $e(t)= e cos 2 t$, $T=π$. Obviously, we know that $ϕ p$ is a homeomorphism from to 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 $x 1 ≠ x 2$, then (A3) holds. Moreover, it is easily seen that there exists a constant $D=2$ such that (H1) holds. Consider $f(t,u)u=(10+5 sin 2 t) u 2 ≥10 u 2$, here $σ=10$, $m=2$, and $|f(t,u)|=|(10+5 sin 2 t)u|≤15|u|+1$, here $ρ=15$, $γ=1$. So, we can get that conditions (H2) and (H3) hold. Choose $B>0$, we have $|g(t,x)|≤5|x|+9$, here $α=5$, $β=9$, then (H4) holds and $σ− α T 2 =10− 5 π 2 >0$. 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:

$( ϕ ( x ′ ( t ) ) ) ′ + ( 200 + 16 cos 2 t ) ( x ′ ( t ) ) 3 − ( 20 x 3 ( t ) + 10 cos 2 ( t ) − 15 ) = e sin 2 t ,$
(2.13)

where $ϕ(u)=u e | u | 2$.

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

$( x e | x | 2 − y e | y | 2 ) (x−y)≥ ( | x | e | x | 2 − | y | e | y | 2 ) ( | x | − | y | ) ≥0$

and

$ϕ(x)⋅x= | x | 2 e | x | 2 .$

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 $x 1 ≠ x 2$, then (A3) holds. Moreover, it is easily seen that there exists a constant $D=1$ such that (H1) holds. Consider $f(t,v)v=(200+16 cos 2 t) v 4 ≥200 v 4$, here $σ=200$, $m=4$, and $|f(t,v)|=|(200+16 cos 2 t) v 3 |≤216 | v | 3 +5$, here $ρ=216$, $γ=5$. So, we can get that conditions (H2) and (H3) hold. Choose $B>0$, we have $|g(t,x)|≤20 | x | 3 +25$, here $α=20$, $β=25$, then (H4) holds and $σ− α T m − 1 2 m − 1 =200− 20 × π 3 2 3 >0$. Therefore, by Theorem 2.1, we know that (2.13) has a unique positive periodic solution.

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## Acknowledgements

Research is supported by the National Natural Science Foundation of China (Nos. 11326124, 11271339).

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Correspondence to Yun Xin.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

YX and ZBC worked together in the derivation of the mathematical results. All authors read and approved the final manuscript.

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Xin, Y., Cheng, Z. Existence and uniqueness of a positive periodic solution for Rayleigh type ϕ-Laplacian equation. Adv Differ Equ 2014, 225 (2014). https://doi.org/10.1186/1687-1847-2014-225