## Boundary Value Problems

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# Positive periodic solutions for p-Laplacian neutral differential equations with a singularity

Boundary Value Problems20172017:54

DOI: 10.1186/s13661-017-0790-0

Accepted: 6 April 2017

Published: 18 April 2017

## Abstract

In this paper, we study the positive periodic solutions of a kind of p-Laplacian neutral differential equation with a singularity. By applying the continuation theorem and some analytic techniques, we shall establish several new criteria for the existence of positive periodic solutions for the considered problem. Some recent results in the literature are generalized and improved. Three examples are given to illustrate the effectiveness of our results.

### Keywords

neutral differential equations positive periodic solutions continuation theorem p-Laplacian singularity

## 1 Introduction

In the past years, the study of periodic solutions for some types of p-Laplacian neutral differential equations has attracted much attention from researchers; see [112] and the references cited therein. For example, in [11] and [12], Zhu and Lu studied the periodic solutions for p-Laplacian neutral functional differential equations as follows:
$$\bigl(\varphi_{p} \bigl(x(t)-cx(t-\tau) \bigr)' \bigr)'=g \bigl(t,x \bigl(t-\tau(t) \bigr) \bigr)+e(t)$$
(1.1)
and
$$\bigl(\varphi_{p} \bigl(x(t)-cx(t-\delta) \bigr)' \bigr)'=f \bigl(t,x(t) \bigr)x'(t)+ \sum ^{n}_{j=1}\beta_{j}g \bigl(t,x \bigl(t- \tau_{j}(t) \bigr) \bigr)+p(t),$$
(1.2)
respectively.
On the basis of work of [11] and [12], Wang and Zhu [7] further discussed existence of periodic solutions for a fourth-order p-Laplacian neutral functional differential equation of the form:
$$\bigl(\varphi_{p} \bigl(x(t)-cx(t-\delta) \bigr)'' \bigr)''=f \bigl(x(t) \bigr)x'(t)+g(t,x \bigl(t-\tau\bigl(t, \vert x \vert _{\infty}\bigr) \bigr)+e(t),$$
(1.3)
where $$p>1$$, $$\varphi_{p} : \mathbb {R}\rightarrow \mathbb {R}$$, $$\varphi_{p}(u)= \vert u \vert ^{p-2}u$$ for $$u\neq0$$ and $$\varphi_{p}(0)=0$$; $$f\in C(\mathbb {R},\mathbb {R})$$; $$g, \tau\in C(\mathbb {R}^{2}, \mathbb {R})$$ with $$g(t+T,\cdot)=g(t,\cdot)$$; $$\tau (t+T)\equiv\tau(t)$$; $$e\in C(\mathbb {R},\mathbb {R})$$ with $$e(t+T)\equiv e(t)$$; $$T>0$$, $$\vert c \vert \neq1$$ and δ are given constants.
In recent years, by applying the method of coincidence degree, many good results of the existence of positive periodic solutions for some types of differential equations with a singularity have been obtained; see [1322] and the references cited therein. For example, Wang in [21] studied the periodic solutions for the Liénard equation with a singularity and a deviating argument, which extends the results of Zhang in [22]
$$x''(t)+f \bigl(x(t) \bigr)x'(t)+g \bigl(t,x(t-\sigma) \bigr)=0,$$
(1.4)
where $$g: \mathbb{R}\times(0,+\infty)\rightarrow\mathbb{R}$$ is an $$L^{2}$$-Carathéodory funtion, $$g(t,x)$$ is a T-periodic function in the first argument and can be singular at $$x=0$$.
As usual, we say that g has a singularity of repulsive type, if
$$\lim_{x\rightarrow0^{+}} g(t,x)=-\infty, \quad \text{uniformly for a.e. } t \in(- \infty, +\infty),$$
and g has a singularity of attractive type, if
$$\lim_{x\rightarrow0^{+}} g(t,x)=+\infty, \quad \text{uniformly for a.e. } t \in(- \infty, +\infty).$$
Compared with the classical p-Laplacian neutral differential problems or the singular problems, p-Laplacian neutral differential problems with singular effects have been scarcely studied, not to mention the high-order p-Laplacian neutral differential equations with a singularity. This motivated us to carry out a study. In this paper, we consider the following high-order p-Laplacian neutral differential equation with a singularity and a deviating argument:
$$\bigl[\varphi_{p} \bigl( \bigl(x(t)-cx(t-\gamma) \bigr)^{(m)} \bigr) \bigr]^{(m)}+f \bigl(x(t) \bigr) x'(t)+g \bigl(t,x(t- \tau) \bigr)=e(t),$$
(1.5)
where $$\varphi_{p}:\mathbb{R}\rightarrow\mathbb{R}$$, $$\varphi_{p}(u)= \vert u \vert ^{p-2}u$$, $$p\geq2$$; m is a positive integer; c is a constant with $$\vert c \vert <1$$; $$0\leq\gamma, \tau< T$$; $$f:\mathbb{R}\rightarrow\mathbb{R}$$ is continuous, $$g: [0,T]\times(0,+\infty)\rightarrow\mathbb{R}$$ is a continuous function and can be singular at $$u=0$$, $$e(t)$$ is T-periodic with $$\int_{0}^{T} e(t)\,dt=0$$.

### Remark 1.1

Obviously, if there is no singularity, (1.1) is the specially case of (1.5). The equations in [1, 36, 10] are also the specially cases of (1.5) if $$m\equiv1$$ and there are no singularities. So we extend the result in [1, 37, 911]. Therefore, we extend the corresponding results in literature to the high-order case and the singular case.

The paper is organized as follows. Section 2 is devoted to introducing some definitions and recalling some preliminary results that will be extensively used. The existence results will be obtained in Section 3. Finally, two examples are given to illustrate the effectiveness of our result in Section 4.

## 2 Preliminaries

Let
$$C_{T}= \bigl\{ x\mid x \in C(\mathbb{R},\mathbb{R}), x(t+T)\equiv x(t) \bigr\}$$
with the norm $$\vert x \vert _{0}=\max_{t\in[0,T]} \vert x(t) \vert$$, and
$$C^{1}_{T}= \bigl\{ x\mid x \in C^{1}(\mathbb{R}, \mathbb{R}), x(t+T)\equiv x(t) \bigr\}$$
with the norm $$\Vert x \Vert =\max\{ \vert x \vert _{0}, \vert x' \vert _{0} \}$$.
Define a linear operator
$$A:\quad C_{T}\rightarrow C_{T}, \quad\quad (Ax) (t)=x(t)-c x(t-\gamma).$$

### Lemma 2.1

[23]

If $$\vert c \vert < 1$$, then A has continuous bounded inverse on X, and
1. (1)

$$\Vert A^{-1} x \Vert \leq\frac{ \vert x \vert _{0}}{ \vert 1- \vert c \vert \vert }$$, $$\forall x\in X$$;

2. (2)

$$\int_{0}^{T} \vert (A^{-1}f)(t) \vert \,dt \leq\frac {1}{1- \vert c \vert }\int_{0}^{T} \vert f(t) \vert \,dt$$, $$\forall f\in X$$.

### Lemma 2.2

[24]

Let $$T>0$$ be a constant, $$x\in C^{m}(\mathbb{R},\mathbb{R})$$, $$m\geq2$$, and $$x(t+T)\equiv x(t)$$, $$\vert x^{(i)} \vert _{0}=\max_{t\in [0,T]} \vert x^{(i)}(t) \vert$$, then there are $$M_{i}(m)>0$$ independent of x such that
$$\bigl\vert x^{(i)} \bigr\vert _{0}\leq M_{i}(m) \int_{0}^{T} \bigl\vert x^{(m)}(t) \bigr\vert \,dt,\quad i=1,2,\ldots,m-1,$$
where if m is an even integer,
\begin{aligned} M_{i}(m)= \textstyle\begin{cases} M_{2s-1}(m)=T^{m-2s}\sqrt{-B_{2m-4s}/12(2m-4s)!}, &s=1,2,\ldots,\frac {m}{2}-1,\\ M_{2s}(m)=\frac{(-1)^{\frac{m-2s}{2}+1} T^{m-2s-1}B_{m-2s}}{(m-2s)!}, &s=1,\ldots,\frac{m}{2}-1,\\ M_{m-1}(m)=\frac{1}{2}, \end{cases}\displaystyle \end{aligned}
if m is an odd integer,
\begin{aligned} M_{i}(m)= \textstyle\begin{cases} M_{2s+1}(m)=\frac{(-1)^{\frac{m-2s-1}{2}+1} T^{m-2s-2}B_{m-2s-1}}{(m-2s-1)!}, &s=0,1,\ldots,\frac{m+1}{2}-2,\\ M_{2s}(m)=T^{m-2s-1}\sqrt{-B_{2m-4s-2}/12(2m-4s-2)!}, &s=1,2,\ldots ,\frac{m+1}{2}-2,\\ M_{m-1}(m)=\frac{1}{2}, \end{cases}\displaystyle \end{aligned}
and $$B_{m-2s}$$, $$B_{2m-4s}$$, $$B_{m-2s-1}$$, $$B_{2m-4s-2}$$ are Bernoulli numbers, which can be calculated by the following recursion formula:
$$B_{0}=1, \quad\quad B_{p}=\frac{-\sum^{p-1}_{i=0}C^{i}_{p+1}B_{i}}{p+1},$$
where $$C^{i}_{p+1}$$ is the number of combinations.

### Lemma 2.3

[25]

Let X and Y be two real Banach spaces, $$L: D(L)\subset X \rightarrow Y$$ be a Fredholm operator with index zero, $$\Omega\subset X$$ be an open bounded set, and $$N:\bar{\Omega}\subset X\rightarrow Y$$ be L-compact on Ω̄. Suppose that all of the following conditions hold:
1. (1)

$$Lx\neq\lambda Nx$$, $$\forall x \in\partial\Omega\cap D(L)$$, $$\forall \lambda\in(0,1)$$;

2. (2)

$$QNx\neq0$$, $$\forall x\in\partial\Omega\cap\ker L$$;

3. (3)

$$\deg\{JQN,\Omega\cap\ker L,0\}\neq0$$, where $$J: \operatorname {Im}Q\rightarrow \ker L$$ is an homeomorphism map.

Then the equation $$Lx=Nx$$ has at least one solution on $$D(L)\cap\bar {\Omega}$$.
In order to use the continuation theorem to study the positive T-periodic solutions for equation (1.5), we consider the following system:
$$\textstyle\begin{cases}u^{(m)}(t)= [A^{-1}\varphi_{q}(v)](t),\\ v^{(m)}(t)=-f(u(t))u'(t)-g(t,u(t-\tau))+e(t), \end{cases}$$
(2.1)
where $$q>1$$ is a constant with $$1/p+1/q=1$$. Clearly, if $$x(t)=(u(t),v(t))^{\top}$$ is a T-periodic solution of system (2.1), then $$u(t)$$ must be a T-periodic solution of equation (1.5). Thus, the problem of finding a positive T-periodic solution for (1.5) reduces to finding one for (2.1).
Define
$$X= \bigl\{ x=(u,v)^{\top}\in C^{1} \bigl(\mathbb{R}, \mathbb{R}^{2} \bigr):x(t+T)\equiv x(t) \bigr\}$$
with the norm $$\Vert x \Vert _{X}=\max\{ \Vert u \Vert ; \Vert v \Vert \}$$,
$$Y= \bigl\{ x=(u,v)^{\top}\in C \bigl(\mathbb{R},\mathbb{R}^{2} \bigr):x(t+T)\equiv x(t) \bigr\}$$
with the norm $$\Vert x \Vert _{Y}=\max\{ \vert u \vert _{0}; \vert v \vert _{0}\}$$. Clearly, X and Y are two Banach spaces.
Define the linear operator
$$L:\quad D(L)\subset X \rightarrow Y, \quad\quad Lx= \bigl(u^{(m)}, v^{(m)} \bigr)^{\top},$$
(2.2)
and
\begin{aligned} &N: \quad D(N)\subset X \rightarrow Y, \\ &(Nx) (t)= \begin{pmatrix} [A^{-1}\varphi_{q}(v)](t) \\ -f(u(t))u'(t)-g(t,u(t-\tau))+e(t) \end{pmatrix} , \quad \forall t\in \mathbb{R}, \end{aligned}
(2.3)
where
\begin{aligned}& D(L)= \bigl\{ x\in C^{m} \bigl(\mathbb{R},\mathbb{R}^{2} \bigr), x(t+T)\equiv x(t) \bigr\} , \\& D(N)= \bigl\{ x=(u,v)^{\top}\in X:u(t)>0,t\in[0, T] \bigr\} . \end{aligned}
Then we can see that equation (2.1) can be converted to the abstract equation $$Lx= Nx$$. Moreover, from the definition of L, we can have
$$\ker L= \mathbb{R}^{2}, \quad\text{and} \quad \operatorname {Im}L= \biggl\{ y\in Y, \int_{0}^{T}y(s)\,ds=0 \biggr\} .$$
Clearly, L is a Fredholm operator with index zero.
Let projectors $$P: X \rightarrow\ker L$$ and $$Q: Y \rightarrow Y/\operatorname{Im}L$$ be defined by
$$Px=x(0); \quad\quad Qy=\frac{1}{T} \int_{0}^{T}y(s)\,ds,$$
and let K represent the inverse of $$L_{\ker P\cap D(L)}$$. Obviously, $$\ker Q =\operatorname {Im}L = \mathbb{R}^{2}$$ and
$$(Ky) (t)=\sum^{m-1}_{i=1} \frac{1}{i!}x^{(i)}(0)t^{i}+\frac{1}{(m-1)!} \int_{0}^{t}(t-s)^{m-1}y(s) \,ds,$$
(2.4)
where $$x^{(i)}(0)$$ ($$i=1,2,\ldots,m-1$$) are defined by the equation
$$\mathbf{A}\mathbf{X}=\mathbf{D},$$
where
$$\mathbf{A} = \begin{pmatrix} 1 & 0 & 0 & \ldots& 0 & 0\\ c_{1} & 1 & 0 & \ldots&0 &0 \\ c_{2} & c_{1} & 1& \ldots&0 &0\\ \vdots& \vdots& \vdots& \ddots& \vdots&\vdots\\ c_{m-3} & c_{m-4} & c_{m-5} & \ldots&1 & 0\\ c_{m-2} & c_{m-3} & c_{m-4} & \ldots&c_{1} & 1 \end{pmatrix} ,$$
$$\mathbf{X}= (x^{(m-1)}(0), x^{(m-2)}(0),\ldots,x''(0),x'(0) )^{\top}$$, $$\mathbf{D}= (d_{1},d_{2},\ldots,d_{m-2},d_{m-1} )^{\top}$$, $$d_{i}=-\frac{1}{i!T}\int_{0}^{T}(T-s)^{i}y(s)\,ds$$, $$i=1,2,\ldots,m-1$$, and $$c_{j}=\frac{T^{j}}{(j+1)!}$$, $$j=1,2,\ldots,m-2$$.
For the sake of convenience, we list the following assumptions:
[$$H_{1}$$]:
There exist positive constants $$D_{1}$$ and $$D_{2}$$ with $$D_{1}< D_{2}$$ such that
1. (1)
for each positive continuous T-periodic function $$x(t)$$ satisfying $$\int_{0}^{T} g(t, x(t))\,dt=0$$, then there exists a positive point $$\sigma\in[0,T]$$ such that
$$D_{1}\leq x(\sigma)\leq D_{2};$$

2. (2)

$$\overline{g}(x)<0$$ for all $$x\in(0,D_{1})$$ and $$\overline {g}(x)>0$$ for all $$x>D_{2}$$, where $$\overline{g}(x)=\frac{1}{T}\int_{0}^{T} g(t,x)\,dt$$, $$x>0$$.

[$$H_{2}$$]:
$$g(t,x)=g_{1}(t,x)+g_{0}(x)$$ for all $$u \in(0, +\infty )$$, where $$g_{0}: (0,+\infty)\rightarrow\mathbb{R}$$ is a continuous function, $$g_{1}: [0,T]\times[0,+\infty)\rightarrow\mathbb{R}$$ is a continuous function and
1. (1)
there exist positive constants $$m_{0}$$ and $$m_{1}$$ such that
$$g(t,x)\leq m_{0}x^{p-1}+m_{1}, \quad \text{for all } (t,x)\in[0,T]\times(0,+\infty);$$

2. (2)

$$\int_{0}^{1} g_{0}(x)\,dx=-\infty$$.

[$$H_{3}$$]:
There exist positive constants α, β such that
$$f(x)\leq\alpha x^{p-2} +\beta,\quad \text{for all } x\in(0,+\infty).$$

## 3 Main results

### Theorem 3.1

Suppose that conditions [$$H_{1}$$]-[$$H_{3}$$] hold and
$$\frac{M^{p}_{1}(m)T^{2p-1}(1+ \vert c \vert )(2m_{0}T+\alpha)}{(1- \vert c \vert )^{p}} < 1,$$
then there exist positive constants $$A_{1}$$, $$A_{2}$$, $$A_{3}$$ and ρ, which are independent of λ such that
$$A_{1}\leq u(t)\leq A_{2}, \quad\quad \vert v \vert _{0}\leq A_{3},$$
where $$u(t)$$ is any solution to the equation $$Lx = \lambda Nx$$, $$\lambda\in(0,1)$$.

### Proof

Consider the following operator equation:
$$Lx=\lambda Nx, \quad \lambda\in(0,1),$$
where L and N is defined by (2.2) and (2.3), respectively.
Define
$$\Omega_{1}= \Bigl\{ (u,v)^{\top}\in X: \min _{t\in[0,T]}u(t)>0, Lx=\lambda Nx ,\lambda\in(0,1) \Bigr\} .$$
If $$x=(u,v)^{\top}\in\Omega_{1}$$, then $$(u,v)$$ satisfies
$$\textstyle\begin{cases}u^{(m)}(t)=\lambda[A^{-1}\varphi_{q}(v)](t),\\ v^{(m)}(t)=-\lambda f(u(t))u'(t)-\lambda g(t,u(t-\tau))+ \lambda e(t). \end{cases}$$
(3.1)
From the first equation of (3.1), we get $$v(t)=\varphi _{p}(\lambda^{-1}(Au)^{(m)}(t))$$, and combining with the second equation of (3.1) yields
$$\bigl(\varphi_{p} \bigl((Au)^{(m)}(t) \bigr) \bigr)^{(m)}+\lambda^{p }f \bigl(u(t) \bigr)u'(t)+ \lambda^{p}g \bigl(t,u(t-\tau) \bigr)=\lambda^{p}e(t).$$
(3.2)
Integrating equation (3.2) on the interval $$[0,T]$$, we have
$$\int_{0}^{T}g \bigl(t, u(t-\tau) \bigr)\,dt=0.$$
(3.3)
It follows from condition (1) in assumption $$[H_{1}]$$ that there exist positive constants $$D_{1}$$, $$D_{2}$$ and $$\sigma\in[0,T]$$ such that
$$D_{1}\leq u(\sigma)\leq D_{2},$$
(3.4)
then we get
\begin{aligned}[b] \vert u \vert _{0} &=\max _{t\in[0,T]} \bigl\vert u(t) \bigr\vert \leq\max _{t\in[0,T]} \biggl\vert u( \sigma)+ \int_{\sigma}^{t}u'(s)\,ds \biggr\vert \\ &\leq D_{2}+ \int_{0}^{T} \bigl\vert u'(s) \bigr\vert \,ds. \end{aligned}
(3.5)
Multiplying both sides of (3.2) by $$(Au)(t)$$ and integrating on the interval $$[0,T]$$, we get
\begin{aligned} \int_{0}^{T} \bigl\vert (Au)^{(m)}(t) \bigr\vert ^{p}\,dt &\leq \bigl(1+ \vert c \vert \bigr) \vert u \vert _{0} \int_{0}^{T} \bigl\vert f \bigl(u(t) \bigr) \bigr\vert \bigl\vert u' \bigr\vert \,dt \\ &\quad{} + \bigl(1+ \vert c \vert \bigr) \vert u \vert _{0} \int_{0}^{T} \bigl\vert g \bigl(t,u(t-\tau) \bigr) \bigr\vert \,dt \\ &\quad{} + \bigl(1+ \vert c \vert \bigr) \vert u \vert _{0} \int_{0}^{T} \bigl\vert e(t) \bigr\vert \,dt, \end{aligned}
which combining with [$$H_{3}$$] yields
\begin{aligned}[b] \int_{0}^{T} \bigl\vert (Au)^{(m)}(t) \bigr\vert ^{p}\,dt &\leq \bigl(1+ \vert c \vert \bigr) \vert u \vert _{0} \int_{0}^{T} \bigl(\alpha \bigl\vert u(t) \bigr\vert ^{p-2}+\beta \bigr) \bigl\vert u'(t) \bigr\vert \,dt \\ &\quad{} + \bigl(1+ \vert c \vert \bigr) \vert u \vert _{0} \int_{0}^{T} \bigl\vert g \bigl(t,u(t-\tau) \bigr) \bigr\vert \,dt+ \bigl(1+ \vert c \vert \bigr) \vert u \vert _{0} \vert e \vert _{0}T \\ &\leq \bigl(1+ \vert c \vert \bigr)\alpha \vert u \vert ^{p-1}_{0} \int_{0}^{T} \bigl\vert u'(t) \bigr\vert \,dt+ \bigl(1+ \vert c \vert \bigr)\beta \vert u \vert _{0} \int_{0}^{T} \bigl\vert u'(t) \bigr\vert \,dt \\ &\quad{} + \bigl(1+ \vert c \vert \bigr) \vert u \vert _{0} \int_{0}^{T} \bigl\vert g \bigl(t,u(t-\tau) \bigr) \bigr\vert \,dt+ \bigl(1+ \vert c \vert \bigr) \vert u \vert _{0} \vert e \vert _{0}T. \end{aligned}
(3.6)
Write
\begin{aligned}& I_{+}= \bigl\{ t\in[0,T]:g \bigl(t, u(t-\tau) \bigr)\geq0 \bigr\} ; \\& I_{-}= \bigl\{ t\in[0,T]:g \bigl(t, u(t-\tau) \bigr)\leq0 \bigr\} . \end{aligned}
Then it follows from (3.3) and [$$H_{2}$$](1) that
\begin{aligned}[b] \int_{0}^{T} \bigl\vert g \bigl(t,u(t-\tau) \bigr) \bigr\vert \,dt &= \int_{I_{+}}g \bigl(t,u(t-\tau) \bigr)\,dt- \int_{I_{-}}g \bigl(t,u(t-\tau) \bigr)\,dt \\ &=2 \int_{I_{+}}g \bigl(t,u(t-\tau) \bigr)\,dt \\ &\leq 2m_{0} \int_{0}^{T} u^{p-1}(t-\tau)\,dt+2 \int_{0}^{T}m_{1}\,dt \\ &\leq 2m_{0}T \vert u \vert ^{p-1}_{0}+2Tm_{1}. \end{aligned}
(3.7)
Substituting (3.7) into (3.6), combining with (3.5) and Lemma 2.2, we can have
\begin{aligned}& \int_{0}^{T} \bigl\vert (Au)^{(m)}(t) \bigr\vert ^{p}\,dt \\& \quad \leq 2 \bigl(1+ \vert c \vert \bigr)m_{0}T \vert u \vert ^{p}_{0}+ \bigl(1+ \vert c \vert \bigr)\alpha \vert u \vert ^{p-1}_{0} \int_{0}^{T} \bigl\vert u'(t) \bigr\vert \,dt \\& \quad\quad{} + \bigl(1+ \vert c \vert \bigr)\beta \vert u \vert _{0} \int_{0}^{T} \bigl\vert u'(t) \bigr\vert \,dt +2 \bigl(1+ \vert c \vert \bigr)m_{1}T \vert u \vert _{0}+ \bigl(1+ \vert c \vert \bigr) \vert e \vert _{0}T \vert u \vert _{0} \\& \quad \leq 2 \bigl(1+ \vert c \vert \bigr)m_{0}T^{p+1}M^{p}_{1}(m) \biggl(\frac{D_{2}}{TM_{1}(m)}+ \int_{0}^{T} \bigl\vert u^{(m)}(t) \bigr\vert \,dt \biggr)^{p} \\& \quad\quad{} + \bigl(1+ \vert c \vert \bigr)\alpha T^{p}M^{p}_{1}(m) \biggl(\frac{D_{2}}{TM_{1}(m)}+ \int_{0}^{T} \bigl\vert u^{(m)}(t) \bigr\vert \,dt \biggr)^{p-1} \cdot \int_{0}^{T} \bigl\vert u^{(m)}(t) \bigr\vert \,dt \\& \quad\quad{} + \bigl(1+ \vert c \vert \bigr)\beta T^{2}M^{2}_{1}(m) \biggl( \frac{D_{2}}{TM_{1}(m)}+ \int_{0}^{T} \bigl\vert u^{(m)}(t) \bigr\vert \,dt \biggr)\cdot \int_{0}^{T} \bigl\vert u^{(m)}(t) \bigr\vert \,dt \\& \quad\quad{} +2 \bigl(1+ \vert c \vert \bigr)m_{1}T^{2}M_{1}(m) \biggl( \frac{D_{2}}{TM_{1}(m)}+ \int_{0}^{T} \bigl\vert u^{(m)}(t) \bigr\vert \,dt \biggr) \\& \quad\quad{} + \bigl(1+ \vert c \vert \bigr) \vert e \vert _{0}T^{2}M_{1}(m) \biggl(\frac{D_{2}}{TM_{1}(m)}+ \int_{0}^{T} \bigl\vert u^{(m)}(t) \bigr\vert \,dt \biggr). \end{aligned}
(3.8)
It follows from conclusion (2) of Lemma 2.1 and by applying the Hölder inequality that
\begin{aligned}[b] \int_{0}^{T} \bigl\vert u^{(m)}(t) \bigr\vert \,dt & = \int_{0}^{T} \bigl\vert \bigl(A^{-1}(Au)^{(m)} \bigr) (t) \bigr\vert \,dt \\ &\leq \frac{\int_{0}^{T} \vert (Au )^{(m)}(t) \vert \,dt}{ 1- \vert c \vert } . \end{aligned}
(3.9)
From the above inequality, we consider the following two cases:

### Case 1

If $$\int_{0}^{T} \vert (Au )^{(m)}(t) \vert \,dt=0$$, then $$\int_{0}^{T} \vert u^{(m)} (t) \vert \,dt=0$$, it follows from (3.5) and Lemma 2.2 that
\begin{aligned}[b] \vert u \vert _{0} &\leq D_{2}+ \int_{0}^{T} \bigl\vert u'(s) \bigr\vert \,ds\leq D_{2}+T \bigl\vert u' \bigr\vert _{0} \\ &\leq D_{2}+TM_{1}(m) \int_{0}^{T} \bigl\vert u^{(m)}(t) \bigr\vert \,dt \\ &=D_{2}. \end{aligned}
(3.10)

### Case 2

If $$\int_{0}^{T} \vert (Au )^{(m)}(t) \vert \,dt>0$$, then substituting (3.9) into (3.8), we can have
\begin{aligned}& \int_{0}^{T} \bigl\vert (Au)^{(m)}(t) \bigr\vert ^{p}\,dt \\& \quad \leq 2 \bigl(1+ \vert c \vert \bigr)m_{0}T^{p+1}M^{p}_{1}(m) \biggl(\frac{D_{2}}{TM_{1}(m)}+\frac{\int_{0}^{T} \vert (Au )^{(m)}(t) \vert \,dt}{1- \vert c \vert } \biggr)^{p} \\& \quad\quad{} + \bigl(1+ \vert c \vert \bigr)\alpha T^{p}M^{p}_{1}(m) \biggl(\frac{D_{2}}{TM_{1}(m)}+ \frac{\int_{0}^{T} \vert (Au )^{(m)}(t) \vert \,dt}{1- \vert c \vert } \biggr)^{p-1}\cdot \frac{\int_{0}^{T} \vert (Au )^{(m)}(t) \vert \,dt}{1- \vert c \vert } \\& \quad\quad{} + \bigl(1+ \vert c \vert \bigr)\beta T^{2}M^{2}_{1}(m) \biggl( \frac{D_{2}}{TM_{1}(m)}+\frac{\int_{0}^{T} \vert (Au )^{(m)}(t) \vert \,dt}{1- \vert c \vert } \biggr)\cdot\frac {\int_{0}^{T} \vert (Au )^{(m)}(t) \vert \,dt}{1- \vert c \vert } \\& \quad\quad{} +2 \bigl(1+ \vert c \vert \bigr)m_{1}T^{2}M_{1}(m) \biggl( \frac{D_{2}}{TM_{1}(m)}+\frac{\int_{0}^{T} \vert (Au )^{(m)}(t) \vert \,dt}{1- \vert c \vert } \biggr) \\& \quad\quad{} + \bigl(1+ \vert c \vert \bigr) \vert e \vert _{0}T^{2}M_{1}(m) \biggl(\frac{D_{2}}{TM_{1}(m)}+ \frac{\int_{0}^{T} \vert (Au )^{(m)}(t) \vert \,dt}{1- \vert c \vert } \biggr). \end{aligned}
(3.11)
Furthermore,
\begin{aligned}[b] & \biggl(\frac{D_{2}}{TM_{1}(m)}+\frac{\int _{0}^{T} \vert (Au )^{(m)}(t) \vert \,dt}{1- \vert c \vert } \biggr)^{p} \\ &\quad =\frac{1}{(1- \vert c \vert )^{p}} \biggl( \int_{0}^{T} \bigl\vert (Au)^{(m)}(t) \bigr\vert \,dt \biggr)^{p} \biggl(1+\frac{\frac{D_{2}}{TM_{1}(m)}(1- \vert c \vert )}{\int_{0}^{T} \vert (Au)^{(m)}(t) \vert \,dt} \biggr)^{p}. \end{aligned}
(3.12)
By classical elementary inequalities, we see that there exists a $$l(p)>0$$ which is dependent on p only, such that
$$(1+x)^{p}< 1+(1+p)x, \quad x\in\bigl(0,l(p) \bigr].$$
(3.13)
If $$\frac{\frac{D_{2}}{TM_{1}(m)}(1- \vert c \vert )}{\int _{0}^{T} \vert (Au)^{(m)}(t) \vert \,dt}>l(p)$$, then $$\int _{0}^{T} \vert (Au)^{(m)}(t) \vert \,dt<\frac{\frac {D_{2}}{TM_{1}(m)}(1- \vert c \vert )}{l(p)}$$. It follows from (3.5), (3.9) and Lemma 2.2 that
$$\vert u \vert _{0} \leq D_{2}+TM_{1}(m) \int_{0}^{T} \bigl\vert u^{(m)}(t) \bigr\vert \leq D_{2}+\frac{ D_{2}}{l(p)}:=M_{1}.$$
(3.14)
If $$\frac{\frac{D_{2}}{TM_{1}(m)}(1- \vert c \vert )}{\int _{0}^{T} \vert (Au)^{(m)}(t) \vert \,dt}\leq l(p)$$, then it follows from (3.12) and (3.13) that
\begin{aligned}[b] & \biggl(\frac{D_{2}}{TM_{1}(m)}+\frac{\int _{0}^{T} \vert (Au)^{(m)}(t) \vert \,dt}{1- \vert c \vert } \biggr)^{p} \\ &\quad \leq\frac{1}{(1- \vert c \vert )^{p}} \biggl( \int_{0}^{T} \bigl\vert (Au)^{(m)} (t) \bigr\vert \,dt \biggr)^{p} \biggl(1+\frac{(p+1)\frac{D_{2}}{TM_{1}(m)}(1- \vert c \vert )}{\int_{0}^{T} \vert (Au)^{(m)}(t) \vert \,dt} \biggr) \\ &\quad = \frac{ ( \int_{0}^{T} \vert (Au)^{(m)}(t) \vert \,dt )^{p}}{(1- \vert c \vert )^{p}}+\frac{(p+1)\frac {D_{2}}{TM_{1}(m)}}{(1- \vert c \vert )^{p-1}}\cdot \biggl( \int_{0}^{T} \bigl\vert (Au)^{(m)}(t) \bigr\vert \,dt \biggr)^{p-1}. \end{aligned}
(3.15)
Substituting (3.15) into (3.11) and by applying the Hölder inequality, we can see that
\begin{aligned}& \int_{0}^{T} \bigl\vert (Au)^{(m)}(t) \bigr\vert ^{p}\,dt \\& \quad \leq \frac{M^{p}_{1}(m)T^{2p-1}(1+ \vert c \vert )(2m_{0}T+\alpha)}{(1- \vert c \vert )^{p}} \int_{0}^{T} \bigl\vert (Au)^{(m)}(t) \bigr\vert ^{p}\,dt \\& \quad\quad{} +\frac{ T^{\frac{2p^{2}-3p+1}{p}} M^{p-1}_{1}(m)D_{2}(1+ \vert c \vert ) [2m_{0}T(p+1)+\alpha p] }{(1- \vert c \vert )^{p-1}} \cdot \biggl( \int_{0}^{T} \bigl\vert (Au)^{(m)}(t) \bigr\vert ^{p} \,dt \biggr)^{\frac{p-1}{p}} \\& \quad\quad{} + \frac{(1+ \vert c \vert )\beta T^{4p-2} M^{2}_{1}(m)}{(1- \vert c \vert )^{2}} \biggl( \int_{0}^{T} \bigl\vert (Au)^{(m)}(t) \bigr\vert ^{p} \,dt \biggr)^{\frac{2}{p}} \\& \quad\quad{} + \frac{(1+ \vert c \vert )T^{\frac{2p-1}{p}}M_{1}(m) [\beta D_{2}+2m_{1}T+ \vert e \vert _{0} T] }{1- \vert c \vert }\cdot \biggl( \int_{0}^{T} \bigl\vert (Au)^{(m)}(t) \bigr\vert ^{p} \,dt \biggr)^{\frac{1}{p}} \\& \quad\quad{} + \bigl(1+ \vert c \vert \bigr)TD_{2} \bigl(2m_{1}+ \vert e \vert _{0} \bigr). \end{aligned}
(3.16)
It follows from $$\frac{M^{p}_{1}(m)T^{2p-1}(1+ \vert c \vert )(2m_{0}T+\alpha)}{(1- \vert c \vert )^{p}} <1$$ and $$p\geq2$$ that there exists a positive constant $$M_{2}>0$$ such that
$$\int_{0}^{T} \bigl\vert (Au)^{(m)}(t) \bigr\vert ^{p}\,dt\leq M_{2}.$$
Then by (3.9), we can see that
\begin{aligned}[b] \int_{0}^{T} \bigl\vert u^{(m)}(t) \bigr\vert \,dt &\leq \frac{\int_{0}^{T} \vert (Au )^{(m)}(t) \vert \, dt}{1- \vert c \vert } \\ &\leq \frac{T^{\frac{p-1}{p}} (\int_{0}^{T} \vert (Au )^{(m)}(t) \vert ^{p}\,dt )^{\frac{1}{p}}}{1- \vert c \vert } \\ & \leq\frac{T^{\frac{p-1}{p}} M_{2} ^{\frac{1}{p}}}{1- \vert c \vert } :=M_{3}, \end{aligned}
(3.17)
which together with (3.5) and Lemma 2.2 yields
\begin{aligned} \vert u \vert _{0} &\leq D_{2}+ \int_{0}^{T} \bigl\vert u'(s) \bigr\vert \,ds\leq D_{2}+T \bigl\vert u' \bigr\vert _{0} \\ &\leq D_{2}+TM_{1}(m) \int_{0}^{T} \bigl\vert u^{(m)}(t) \bigr\vert \,dt \\ &\leq D_{2}+ TM_{1}(m)M_{3}:=M_{4}. \end{aligned}
Therefore, in both Case 1 and Case 2, we obtain
$$\vert u \vert _{0}\leq M_{4}.$$
(3.18)
From the second equation of (3.1), we can get
\begin{aligned} \int_{0}^{T} \bigl\vert v^{(m)}(t) \bigr\vert \,dt &\leq \lambda \biggl[ \int_{0}^{T} \bigl\vert f \bigl(u(t) \bigr) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt+ \int_{0}^{T} \bigl\vert g \bigl(t,u(t-\tau) \bigr) \bigr\vert \,dt \\ &\quad{} + \int_{0}^{T} \bigl\vert e(t) \bigr\vert \,dt \biggr], \end{aligned}
from which by applying [$$H_{3}$$] and (3.7), we have
\begin{aligned}[b] \int_{0}^{T} \bigl\vert v^{(m)}(t) \bigr\vert \,dt &\leq \lambda \bigl[ \alpha T \vert u \vert ^{p-2}_{0} \bigl\vert u' \bigr\vert _{0}+ \beta T \bigl\vert u' \bigr\vert _{0}+ 2 m_{0}T \vert u \vert ^{p-1}_{0} \\ &\quad{} + 2 Tm_{1} + \vert e \vert _{0}T \bigr]. \end{aligned}
(3.19)
By Lemma 2.2, (3.17) and (3.18), we obtain
\begin{aligned} \int_{0}^{T} \bigl\vert v^{(m)}(t) \bigr\vert \,dt &\leq \lambda \bigl[ \alpha T \vert u \vert ^{p-2}_{0} \bigl\vert u' \bigr\vert _{0}+ \beta T \bigl\vert u' \bigr\vert _{0}+ 2 m_{0}T \vert u \vert ^{p-1}_{0} \\ &\quad{} + 2 Tm_{1} + \vert e \vert _{0}T \bigr] \\ \leq& \lambda \bigl[ \alpha TM^{p-2}_{4} M_{1}(m)M_{3}+ \beta T M_{1}(m)M_{3}+ 2m_{0}T M^{p-1}_{4} \\ &\quad{} + 2Tm_{1} + \vert e \vert _{0}T \bigr] . \end{aligned}
(3.20)
Moreover, integrating the first equation of (3.1) on the interval $$[0,T]$$, we have
$$\int_{0}^{T} \bigl\vert v(t) \bigr\vert ^{q-2}v(t)\,dt=0,$$
which implies that there exists $$\eta\in[0,T]$$ such that $$v(\eta )=0$$. Thus,
\begin{aligned} \bigl\vert v(t) \bigr\vert &= \biggl\vert \int_{\eta}^{t} v'(s)\,ds+v(\eta) \biggr\vert \leq \int_{0}^{T} \bigl\vert v'(s) \bigr\vert \,ds\leq T \bigl\vert v' \bigr\vert _{0}, \end{aligned}
by Lemma 2.2 and (3.20), we can obtain
\begin{aligned}[b] \vert v \vert _{0}&=\max _{t\in[0,T]} \bigl\vert v(t) \bigr\vert \\ &\leq T \bigl\vert v' \bigr\vert _{0}\leq TM_{1}(m) \int_{0}^{T} \bigl\vert v^{(m)}(t) \bigr\vert \,dt \\ &\leq \lambda TM_{1}(m) \bigl[ \alpha TM^{p-2}_{4} M_{1}(m)M_{3}+ \beta T M_{1}(m)M_{3}+ 2m_{0}T M^{p-1}_{4} \\ &\quad{} + 2Tm_{1} + \vert e \vert _{0}T \bigr] \\ &\leq TM_{1}(m) \bigl[ \alpha TM^{p-2}_{4} M_{1}(m)M_{3}+ \beta T M_{1}(m)M_{3}+ 2m_{0}T M^{p-1}_{4} \\ &\quad{} + 2Tm_{1} + \vert e \vert _{0}T \bigr] \\ &:=A_{3} . \end{aligned}
(3.21)
On the other hand, from the second equation of (3.1) and [$$H_{2}$$], we can see that
\begin{aligned}[b] v^{(m)}(t+\tau)&=- \lambda f \bigl(u(t+ \tau) \bigr)u'(t+\tau) \\ &\quad{} -\lambda \bigl[g_{1} \bigl(t+\tau, u(t) \bigr)+g_{0} \bigl(u(t) \bigr) \bigr] \\ &\quad{} +\lambda e(t+\tau). \end{aligned}
(3.22)
Furthermore, multiplying both sides of equation (3.22) by $$u'(t)$$, we have
\begin{aligned}[b] v^{(m)}(t+\tau)u'(t)&=- \lambda f \bigl(u(t+\tau) \bigr)u'(t+\tau)u'(t) \\ &\quad{} -\lambda \bigl[g_{1} \bigl(t+\tau, u(t) \bigr)+g_{0} \bigl(u(t) \bigr) \bigr] u'(t) \\ & \quad{} +\lambda e(t+\tau)u'(t). \end{aligned}
(3.23)
Let $$\sigma\in[0,T]$$ be as in (3.4). For any $$t\in[\sigma ,T]$$, integrating equation (3.23) on the interval $$[\sigma,t]$$, we get
\begin{aligned} \lambda \int_{u(\sigma)}^{u(t)}g_{0}(u)\,du&=\lambda \int_{\sigma}^{t}g_{0} \bigl(u(t) \bigr)u'(t)\,dt \\ &=- \int_{\sigma}^{t}v^{(m)}(t+ \tau)u'(t)\,dt - \lambda \int_{\sigma}^{t}f \bigl(u(t+\tau) \bigr)u'(t+ \tau)u'(t)\,dt \\ &\quad{} -\lambda \int_{\sigma}^{t}g_{1} \bigl(t+\tau,u(t) \bigr)u'(t)\,dt+\lambda \int_{\sigma}^{t} e(t +\tau)u'(t)\,dt, \end{aligned}
which together with (3.20) yields
\begin{aligned} &\lambda \biggl\vert \int_{u(\tau)}^{u(t)}g_{0}(u)\,du \biggr\vert \\ &\quad = \lambda \biggl\vert \int_{\tau}^{t}g_{0} \bigl(u(t) \bigr)u'(t)\,dt \biggr\vert \\ &\quad \leq \int_{0}^{T} \bigl\vert v^{(m)}(t+\tau) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt + \lambda \int_{0}^{T} \bigl\vert f \bigl(u(t+\tau) \bigr) \bigr\vert \bigl\vert u'(t+\tau) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt \\ &\quad \quad{}+\lambda \int_{0}^{T} \bigl\vert g_{1} \bigl(t+ \tau,u(t) \bigr) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt+ \lambda \int_{0}^{T} \bigl\vert e(t +\tau) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt \\ &\quad \leq\lambda \bigl\vert u' \bigr\vert _{0} \bigl[ \alpha TM^{p-2}_{4} M_{1}(m)M_{3}+ \beta T M_{1}(m)M_{3}+ 2m_{0}T M^{p-1}_{4}+ 2Tm_{1} + \vert e \vert _{0}T \bigr] \\ &\quad \quad{}+\lambda \bigl\vert u' \bigr\vert ^{2}_{0} \int_{0}^{T} \bigl\vert f \bigl(u(t+\tau) \bigr) \bigr\vert \,dt+ \lambda \bigl\vert u' \bigr\vert _{0} \int_{0}^{T} \bigl\vert g_{1} \bigl(t+ \tau,u(t) \bigr) \bigr\vert \,dt \\ &\quad \quad{}+\lambda \bigl\vert u' \bigr\vert _{0} \int_{0}^{T} \bigl\vert e(t +\tau) \bigr\vert \,dt. \end{aligned}
Furthermore, set
$$F_{M_{4}}=\max_{ \vert u \vert \leq M_{4}} \bigl\vert f(u) \bigr\vert \quad\text{and} \quad G_{M_{4}}=\max_{t\in[0,T], \vert u \vert \leq M_{4}} \bigl\vert g_{1}(t,u) \bigr\vert ,$$
then we have
\begin{aligned} \lambda \biggl\vert \int_{u(\tau)}^{u(t)}g_{0}(u)\,du \biggr\vert & \leq \lambda \bigl\vert u' \bigr\vert _{0} \bigl[ \alpha TM^{p-2}_{4} M_{1}(m)M_{3}+ \beta T M_{1}(m)M_{3} \bigr] \\ &\quad{} + \lambda \bigl\vert u' \bigr\vert _{0} \bigl[ 2m_{0}T M^{p-1}_{4}+ 2Tm_{1} + \vert e \vert _{0}T \bigr] \\ & \quad{} +\lambda \bigl\vert u' \bigr\vert ^{2}_{0} TF_{M_{4}}+\lambda \bigl\vert u' \bigr\vert _{0}TG_{M_{4}}+ \lambda \bigl\vert u' \bigr\vert _{0}T \vert e \vert _{0}, \end{aligned}
by (3.17) and Lemma 2.2, we obtain
\begin{aligned}[b] \biggl\vert \int_{u(\tau)}^{u(t)}g_{0}(u)\,du \biggr\vert & \leq M_{1}(m)M_{3} \bigl[ \alpha TM^{p-2}_{4} M_{1}(m)M_{3}+ \beta T M_{1}(m)M_{3} \bigr] \\ &\quad{} + M_{1}(m)M_{3} \bigl[2m_{0}T M^{p-1}_{4}+ 2Tm_{1} + \vert e \vert _{0}T \bigr] \\ & \quad{} + \bigl[M_{1}(m)M_{3} \bigr]^{2} TF_{M_{4}}+ M_{1}(m)M_{3}TG_{M_{4}} \\ &\quad{} + M_{1}(m)M_{3}T \vert e \vert _{0} \\ &< +\infty. \end{aligned}
(3.24)
According to condition (2) in [$$H_{2}$$], we see that there exists a constant $$M_{5}>0$$ such that, for $$t\in[\sigma,T]$$,
$$u(t)\geq M_{5}.$$
(3.25)
In a similar way, we can handle the case of $$t\in[0,\sigma]$$.
Let us define
$$0< A_{1}=\min\{D_{1},M_{5}\},\quad\text{and}\quad A_{2}=\max\{D_{2},M_{4}\}.$$
Then by (3.4), (3.10), (3.18) and (3.25), we can obtain
$$A_{1}\leq u(t)\leq A_{2}.$$
(3.26)
Clearly, $$A_{1}$$ and $$A_{2}$$ are independent of λ. Therefore, the proof of Theorem 3.1 is complete.  □

### Theorem 3.2

Assume that all the conditions in Theorem  3.1 hold, then system (1.5) has at least one positive T-periodic solution.

### Proof

Set
$$\Omega= \biggl\{ x=(u,v)^{\top}\in X:\frac{A_{1}}{2}< u(t)< A_{2}+1, \vert v \vert _{0}< A_{3}+1 \biggr\} .$$

From (2.3) and (2.4), one can easily see that Ω is an open bounded subset of X and N is L-compact on Ω̅. Then the conditions (1) and (2) of Lemma 2.3 are satisfied.

In the following, we prove that condition (3) of Lemma 2.3 also holds.

Now, we let
$$\omega=Bx=B \begin{pmatrix}u \\v \end{pmatrix} = \begin{pmatrix}u-\frac{A_{1}+A_{2}}{2} \\v \end{pmatrix} .$$
Define a linear isomorphism
$$J:\quad \operatorname {Im}Q\rightarrow\ker L,\quad\quad J (u,v)= \begin{pmatrix}v \\ -u \end{pmatrix} ,$$
and define
$$H(\mu,x)=\mu B x+(1-\mu) JQN x, \quad \forall(x,\mu)\in(\Omega\cap\ker L) \times[0,1].$$
Then we can get
\begin{aligned} K(\mu,x)= \begin{pmatrix}\mu u-\frac{\mu(A_{1}+A_{2})}{2} \\ \mu v \end{pmatrix} +\frac{1-\mu}{T} \begin{pmatrix}\int_{0}^{T}[f(u)u'+g(t, u)]\,dt \\ (1- \vert c \vert )^{-1}\int_{0}^{T}\varphi_{q}(v)\,dt \end{pmatrix} . \end{aligned}
(3.27)
In order to prove the condition (3) of Lemma 2.3 is also satisfied, firstly, we prove that $$K(\mu,x)$$ is a homotopic mapping. By way of contradiction, i.e., suppose that there exist $$\mu_{0}\in [0,1]$$ and $$x_{0}= \bigl({\scriptsize\begin{matrix}{}u_{0}\cr v_{0} \end{matrix}} \bigr) \in\partial(\Omega\cap\ker L)$$ such that $$K(\mu_{0},x_{0})=0$$. Then, substituting $$\mu_{0}$$ and $$x_{0}$$ into (3.27), we have
$$K(\mu_{0},x_{0})= \begin{pmatrix}\mu_{0} u_{0}-\frac{\mu_{0}(A_{1}+A_{2})}{2} +(1-\mu_{0}) \overline{g}(u_{0}) \\ \mu_{0} v_{0}+\frac{1-\mu_{0}}{1- \vert c \vert }\varphi_{q}(v_{0}) \end{pmatrix} .$$
(3.28)
It follows from $$K(\mu_{0},x_{0})=0$$ that
$$\mu_{0} v_{0}+\frac{1-\mu_{0}}{1- \vert c \vert }\varphi_{q}(v_{0})=0,$$
which together with $$\mu_{0}\in[0,1]$$ gives $$v_{0}=0$$. Thus, we can get $$u_{0}=\frac{A_{1}}{2}$$ or $$A_{2}+1$$. Furthermore, it follows from [$$H_{1}$$](2) that $$\overline{g} (\frac{A_{1}}{2})<0$$ and $$\overline {g} (A_{2}+1)>0$$, substituting $$u_{0}=\frac{A_{1}}{2}$$ or $$A_{2}+1$$ into (3.28), we can obtain
$$\mu_{0} u_{0}- \frac{\mu_{0}(A_{1}+A_{2})}{2} +(1- \mu_{0}) \overline{g} (u_{0})< 0$$
(3.29)
and
$$\mu_{0} u_{0}- \frac{\mu_{0}(A_{1}+A_{2})}{2} +(1- \mu_{0}) \overline{g} (u_{0})>0.$$
(3.30)
From (3.29) and (3.30), we have $$K(\mu_{0},x_{0})\neq0$$, which is a contradiction. Therefore $$K(\mu,x)$$ is a homotopic mapping and $$x^{\top}K(\mu,x)\neq0$$. For all $$(x,\mu)\in(\partial\Omega \cap\ker L )\times[0,1]$$, we get
\begin{aligned} \deg(JQN ,\Omega\cap\ker L ,0)&=\deg \bigl(K(0,x),\Omega\cap\ker L ,0 \bigr) \\ &=\deg \bigl(K(1,x),\Omega\cap\ker L ,0 \bigr) \\ &=\deg(Bx,\Omega\cap\ker L ,0) \\ &=\sum_{x\in B^{-1}(0)} \operatorname {sgn}\bigl(\det B'(x) \bigr) \\ &=1\neq0. \end{aligned}
Thus, the condition (3) of Lemma 2.3 is also satisfied. Therefore, we can conclude that equation (1.5) has at least one positive T-periodic solution. □

## 4 Examples

In this section, we provide two examples to illustrate our main result.

### Example 4.1

Consider the following third-order p-Laplacian neutral functional differential equation:
\begin{aligned}[b] & \biggl(\varphi_{4} \biggl( \biggl(x(t)- \frac{1}{3}x \biggl(t-\frac{\pi}{4} \biggr) \biggr)^{(3)} \biggr) \biggr)^{(3)}+ \biggl(\frac{u^{4}(t)}{6+u^{2}(t)} +3 \biggr)u'(t) \\ &\quad{} +\frac{1}{64}(1+\sin4t)u ^{3}(t-\tau)-\frac{1}{u^{3}(t-\tau)}= \frac{1}{16}\sin4t. \end{aligned}
(4.1)

### Conclusion

Problem (4.1) has at least one positive $$\pi /2$$-periodic solution.

### Proof

Corresponding to (1.5), we have
\begin{aligned}& f \bigl(u(t) \bigr)=\frac{u^{4}}{6+u^{2}}+3,\quad u\in(0, +\infty), \quad\quad e(t)= \frac{1}{16}\sin4t, \\& g \bigl(t,u(t-\tau) \bigr)=\frac{1}{64}(1+\sin4t)u ^{3}(t- \tau)- \frac{1}{u^{3}(t-\tau)},\quad u\in(0, +\infty). \end{aligned}
Then we can have and choose
\begin{aligned}& m=3,\quad\quad p=4, \quad\quad \gamma=\frac{\pi}{4},\quad\quad c=\frac{1}{3}, \quad\quad T=\frac{\pi}{2}, \\& \alpha=\frac{1}{6}, \quad\quad \beta=3, \quad\quad m_{0}=\frac{1}{32},\quad\quad D_{1}=1,\quad\quad D_{2}=9. \end{aligned}
Moreover, by Lemma 2.2 we have $$M_{1}(3)=T/12$$. Thus, the conditions [$$H_{1}$$]-[$$H_{3}$$] are satisfied. Meanwhile, we also have
\begin{aligned} \frac{M^{p}_{1}(m)T^{2p-1}(1+ \vert c \vert )(2m_{0}T+\alpha )}{(1- \vert c \vert )^{p}} \approx0.026 < 1. \end{aligned}
Hence, by applying Theorem 3.1-3.2, we can see that equation (4.1) has at least one positive $$\pi/2$$-periodic solution. □

### Example 4.2

Consider the following fourth-order p-Laplacian neutral functional differential equation:
\begin{aligned}[b]& \biggl(\varphi_{4} \biggl( \biggl(x(t)- \frac{1}{10}x \biggl(t-\frac{\pi}{8} \biggr) \biggr)^{(4)} \biggr) \biggr)^{(4)}+ \biggl(\frac{u^{4}(t)}{8+u^{2}(t)} +9 \biggr)u'(t) \\ &\quad{} +\frac{1}{64}(1+\sin2t)u ^{3}(t-\tau)-\frac{1}{u^{3}(t-\tau)}= \frac{1}{32}\sin2t. \end{aligned}
(4.2)

### Conclusion

Problem (4.2) has at least one positive π-periodic solution.

### Proof

Corresponding to (1.5), we have
\begin{aligned}& f \bigl(u(t) \bigr)=\frac{u^{4}}{8+u^{2}}+9,\quad u\in(0, +\infty), \quad\quad e(t)= \frac{1}{32}\sin2t, \\& g \bigl(t,u(t-\tau) \bigr)=\frac{1}{64}(1+\sin2t)u ^{3}(t- \tau)- \frac{1}{u^{3}(t-\tau)},\quad u\in(0, +\infty). \end{aligned}
Then we can choose
\begin{aligned}& m=4, \quad\quad p=4, \quad\quad \gamma=\frac{\pi}{8}, \quad\quad c=\frac{1}{10}, \quad\quad T=\pi, \\& \alpha=\frac{1}{8}, \quad\quad \beta=9, \quad\quad m_{0}=\frac{1}{32},\quad\quad D_{1}=1, \quad\quad D_{2}=9. \end{aligned}
Moreover, by Lemma 2.2, we get $$M_{1}(4)= \frac{T^{2}}{24\sqrt {15}}$$. Thus, the conditions [$$H_{1}$$]-[$$H_{3}$$] are satisfied. Meanwhile, we also have
\begin{aligned} \frac{M^{p}_{1}(m)T^{2p-1}(1+ \vert c \vert )(2m_{0}T+\alpha )}{(1- \vert c \vert )^{p}} \approx0.207< 1. \end{aligned}
Hence, by applying Theorem 3.1-3.2, we can see that equation (4.2) has at least one positive π-periodic solution. □

### Example 4.3

Consider the following fourth-order p-Laplacian neutral functional differential equation:
\begin{aligned}[b] & \biggl(\varphi_{2} \biggl( \biggl(x(t)- \frac{1}{10}x \biggl(t-\frac{\pi}{8} \biggr) \biggr)^{(4)} \biggr) \biggr)^{(4)}+ 9 u'(t) \\ &\quad{} +\frac{1}{64}(1+\sin4t)u(t-\tau)-\frac{1}{u(t-\tau)}=\frac{1}{32} \sin4t. \end{aligned}
(4.3)

### Conclusion

Problem (4.3) has at least one positive $$\frac{\pi}{2}$$-periodic solution.

### Proof

Corresponding to (1.5), we have
\begin{aligned}& f \bigl(u(t) \bigr)= 9,\quad u\in(0, +\infty), \quad\quad e(t)=\frac{1}{32}\sin4t, \\& g \bigl(t,u(t-\tau) \bigr)=\frac{1}{64}(1+\sin4t)u(t-\tau)- \frac{1}{u(t-\tau)},\quad u\in(0, +\infty). \end{aligned}
Then we can choose
\begin{aligned}& m=4,\quad\quad p=2, \quad\quad \gamma=\frac{\pi}{8}, \quad\quad c=\frac{1}{10}, \quad\quad T=\frac{\pi}{2}, \\& \alpha=0, \quad\quad \beta=9, \quad\quad m_{0}=\frac{1}{32}, \quad\quad D_{1}=1,\quad\quad D_{2}=9. \end{aligned}
Moreover, by Lemma 2.2, we get $$M_{1}(4)= \frac{T^{2}}{24\sqrt {15}}$$. Thus, the conditions [$$H_{1}$$]-[$$H_{3}$$] are satisfied. Meanwhile, we also have
\begin{aligned} \frac{M^{p}_{1}(m)T^{2p-1}(1+ \vert c \vert )(2m_{0}T+\alpha )}{(1- \vert c \vert )^{p}} \approx0.042< 1. \end{aligned}
Hence, by applying Theorem 3.1-3.2, we can see that equation (4.3) has at least one positive $$\frac{\pi }{2}$$-periodic solution. □

## Declarations

### Acknowledgements

This work was partially supported by the Fundamental Research Funds for the Central Universities (2016B07514) and the National Natural Science Foundation of China (Grant No. 11471109).

## Authors’ Affiliations

(1)
Department of Mathematics and Physics, Hohai University
(2)
Department of Mathematics, Hunan Normal University

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