Positive periodic solution for high-order p-Laplacian neutral differential equation with singularity

Abstract

In this paper, we consider the following high-order p-Laplacian neutral differential equation with singularity:

$$ \bigl(\varphi_{p}\bigl(x(t)-cx(t-\tau)\bigr)^{(n)} \bigr)^{(m)}+f\bigl(x(t)\bigr)x'(t)+g\bigl(t,x(t-\sigma) \bigr)=e(t). $$

By applications of coincidence degree theory and some analysis techniques, sufficient conditions for the existence of positive periodic solutions are established.

1 Introduction

In this paper, we consider the following high-order p-Laplacian neutral differential equation with singularity:

$$ \bigl(\varphi_{p}\bigl(x(t)-cx(t-\tau) \bigr)^{(n)}\bigr)^{(m)}+f\bigl(x(t)\bigr)x'(t)+g \bigl(t,x(t-\sigma)\bigr)=e(t), $$

(1.1)

where $p\geq2$, $\varphi_{p}(x)=|x|^{p-2}x$ for $x\neq0$ and $\varphi_{p}(0)=0$; $g:[0,T]\times(0,\infty)\rightarrow\mathbb{R}$ is an $L^{2}$-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 $h_{r,s}\in L^{2}[0,T]$ such that $|f(t,x(t))|\leq h_{r,s}$ for all $x\in[r,s]$ and a.e. $t\in[0,T]$. $g(t,x)$ being singular at 0 means that $g(t,x)$ becomes unbounded when $x\rightarrow0^{+}$. τ and σ are constants and $0\leq \tau, \sigma< T$; $e:\mathbb{R}\rightarrow\mathbb{R}$ is a continuous periodic function with $e(t+T)\equiv e(t)$ and $\int^{T}_{0}e(t)\,dt=0$. T is a positive constant, c is a constant and $|c|\neq1$; n, m are positive integers.

Generally speaking, differential equations with singularities have been considered from the very beginning of the discipline. The main reason is that singular forces are ubiquitous in applications, gravitational and electromagnetic forces being the most obvious examples. In 1986, Lazer and Solimini [1] discussed the second-order singular equation

$$ u''+\frac{1}{u^{\alpha}}=h(t), $$

(1.2)

and they showed that, if $h(t)$ was continuous and T-periodic, then for all $\alpha>0$ a positive T-periodic solution existed if and only if $h(t)$ had a positive mean value. Afterwards, they studied the singular equation

$$ u''-\frac{1}{u^{\alpha}}=h(t), $$

(1.3)

and they found that if $\alpha\geq1$, a positive T-periodic solution existed if and only if $h(t)$ had a negative mean value. This last result was best possible in that for any α, $0 <\alpha< 1$, h can be chosen so that h had a negative mean value and the equation had no T-periodic solution.

Lazer and Solimini’s work has attracted the attention of many specialists in differential equations. More recently, the method of lower and upper solutions [2–4], the Poincaré-Birkhoff twist theorem [5–7], topological degree theory [8, 9], the Schauder fixed point theorem [10–12], the Leray-Schauder alternative principle [13–15], the Krasnoselskii fixed point theorem in a cone [16, 17], and the fixed point index theory [18] have been employed to investigate the existence of positive periodic solutions of singular second-order, third-order, and fourth-order differential equations.

However, the singular differential equation (1.1), in which there are p-Laplacian and high-order cases, has not attracted much attention in the literature. There are not so many results concerning the existence of a positive periodic solution for (1.1) even when we have a neutral operator. In this paper, we try to fill gap and establish the existence of a positive periodic solution of (1.1) using coincidence degree theory. Our new results generalize in several aspects some recent results contained in [8].

In what follows, we need the notations:

$$|u|_{\infty}=\max_{t\in[0,T]}\bigl|u(t)\bigr|, \quad |u|_{0}=\min _{t\in[0,T]}\bigl|u(t)\bigr|,\quad |u|_{p}= \biggl( \int^{T}_{0}|u|^{p}\,dt \biggr)^{\frac{1}{p}},\quad \bar{h}=\frac{1}{T} \int^{T}_{0}h(t)\,dt. $$

2 Preparation

Let $C_{T}=\{\phi\in C(\mathbb{R},\mathbb{R}):\phi(t+T)\equiv \phi(t)\}$ with the norm $|\phi|_{\infty}=\max_{t\in[0,T]}|\phi(t)|$. Define operators A as follows:

$$A:C_{T}\rightarrow C_{T},\quad (Ax) (t)=x(t)-cx(t-\tau). $$

Lemma 2.1

(see [19])

If $|c|\neq1$, then the operator A has a continuous inverse $A^{-1}$ on $C_{T}$, satisfying:

• (1)
  $$\bigl[A^{-1}f \bigr](t)= \textstyle\begin{cases}f(t)+ \sum^{\infty}_{j=1}c^{j}f(t-j\tau),& \textit{for } |c|< 1, \forall f\in C_{T},\\ -\frac{f(t+\tau)}{c}-\sum^{\infty}_{j=1}\frac {1}{c^{j+1}}f(t+(j+1)\tau), & \textit{for } |c|>1, \forall f\in C_{T}. \end{cases} $$

• (2) $\vert [A^{-1}f ](t)\vert \leq\frac{|f|_{\infty}}{|1-|c||}$, $\forall f\in C_{T}$.

• (3) $\int^{T}_{0}\vert [A^{-1}f ](t)\vert \,dt\leq\frac{1}{|1-|c||}\int ^{T}_{0}|f(t)|\,dt$, $\forall f\in C_{T}$.

Let X and Y be real Banach spaces and $L:D(L)\subset X\rightarrow Y$ be a Fredholm operator with index zero, here $D(L)$ denotes the domain of L. This means that ImL is closed in Y and $\dim \operatorname{Ker} L=\dim(Y/\operatorname{Im} L)<+\infty$. Consider supplementary subspaces $X_{1}$, $Y_{1}$ of X, Y, respectively, such that $X=\operatorname{Ker} L \oplus X_{1}$, $Y=\operatorname{Im} L\oplus Y_{1}$. Let $P:X\rightarrow \operatorname{Ker} L$ and $Q:Y\rightarrow Y_{1}$ denote the natural projections. Clearly, $\operatorname{Ker} L\cap(D(L)\cap X_{1})=\{0\}$ and so the restriction $L_{P}:=L|_{D(L)\cap X_{1}}$ is invertible. Let K denote the inverse of $L_{P}$.

Let Ω be an open bounded subset of X with $D(L)\cap\Omega\neq\emptyset$. A map $N:\overline{\Omega}\rightarrow Y$ is said to be L-compact in Ω̅ if $QN(\overline{\Omega})$ is bounded and the operator $K(I-Q)N:\overline{\Omega}\rightarrow X$ is compact.

Lemma 2.2

(Gaines and Mawhin [20])

Suppose that X and Y are two Banach spaces, and $L:D(L)\subset X\rightarrow Y$ is a Fredholm operator with index zero. Let $\Omega\subset X$ be an open bounded set and $N:\overline{\Omega}\rightarrow Y$ be L-compact on Ω̅. Assume that the following conditions hold:

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

• (2) $Nx\notin \operatorname{Im} L$, $\forall x\in\partial\Omega\cap \operatorname{Ker} L$;

• (3) $\deg\{JQN,\Omega\cap \operatorname{Ker} L,0\}\neq0$, where $J:\operatorname{Im} Q\rightarrow \operatorname{Ker} L$ is an isomorphism.

Then the equation $Lx=Nx$ has a solution in $\overline{\Omega}\cap D(L)$.

In order to apply coincidence degree theorem, we rewrite (1.1) in the form

$$ \textstyle\begin{cases} (Ax_{1})^{(n)}(t)=\varphi_{q}(x_{2}(t)),\\ x_{2}^{(m)}(t)=-f(x_{1}(t))x_{1}'(t)-g(t,x_{1}(t-\sigma))+e(t), \end{cases} $$

(2.1)

where $\frac{1}{p}+\frac{1}{q}=1$. Clearly, if $x(t)=(x_{1}(t),x_{2}(t))^{\top}$ is a T-periodic solution to (2.1), then $x_{1}(t)$ must be a T-periodic solution to (1.1). Thus, the problem of finding a T-periodic solution for (1.1) reduces to finding one for (2.1).

Now, set $X=\{x=(x_{1}(t),x_{2}(t))\in C(\mathbb{R},\mathbb{R}^{2}): x(t+T)\equiv x(t)\}$ with the norm $|x|_{\infty}=\max\{|x_{1}|_{\infty},|x_{2}|_{\infty}\}$; $Y=\{x=(x_{1}(t),x_{2}(t))\in C^{1}(\mathbb{R},\mathbb{R}^{2}): x(t+T)\equiv x(t)\}$ with the norm $\|x\|=\max\{|x|_{\infty},|x'|_{\infty}\}$. Clearly, X and Y are both Banach spaces. Meanwhile, define

$$L:D(L)=\bigl\{ x\in C^{n+m}\bigl(\mathbb{R},\mathbb{R}^{2} \bigr): x(t+T) = x(t), t \in \mathbb{R}\bigr\} \subset X\rightarrow Y $$

by

$$ (Lx) (t)= \begin{pmatrix} (A x_{1})^{(n)}(t)\\ x_{2}^{(m)}(t) \end{pmatrix} $$

and $N: X\rightarrow Y$ by

$$ { } (Nx) (t)= \begin{pmatrix} \varphi_{q}(x_{2}(t))\\ -f(x_{1}(t))x_{1}'(t)-g(t,x_{1}(t-\sigma))+e(t) \end{pmatrix}. $$

(2.2)

Then (2.1) can be converted into the abstract equation $Lx=Nx$. From the definition of L, one can easily see that

$$\operatorname{Ker} L \cong\mathbb {R}^{2},\qquad \operatorname{Im} L= \left \{y\in Y: \large \int_{0}^{T} \begin{pmatrix} y_{1}(s)\\ y_{2}(s) \end{pmatrix}\,ds= \begin{pmatrix} 0\\ 0 \end{pmatrix} \right\}. $$

So L is a Fredholm operator with index zero. Let $P:X\rightarrow \operatorname{Ker} L$ and $Q:Y\rightarrow \operatorname{Im} Q\subset\mathbb {R}^{2}$ be defined by

$$Px= \begin{pmatrix} (Ax_{1})(0)\\ x_{2}(0) \end{pmatrix} ;\qquad Qy=\frac{1}{T} \int_{0}^{T} \begin{pmatrix} y_{1}(s)\\ y_{2}(s) \end{pmatrix}\,ds, $$

then $\operatorname{Im} P=\operatorname{Ker} L$, $\operatorname{Ker}Q= \operatorname{Im} L$. Setting $L_{P}=L|_{D(L)\cap \operatorname{Ker} P}$ and writing $L_{P}^{-1}: \operatorname{Im} L\rightarrow D(L)$ to denote the inverse of $L_{P}$, then

$$\begin{aligned}& \bigl[L_{P}^{-1}y\bigr](t)= \begin{pmatrix} (A^{-1}Gy_{1})(t)\\ (Gy_{2})(t) \end{pmatrix} , \\& \begin{aligned} &[Gy_{1}](t)=\sum _{i=1}^{n-1}\frac{1}{i!}(Ax_{1})^{(i)}(0)t^{i}+ \frac {1}{(n-1)!} \int^{t}_{0}(t-s)^{n-1}y_{1}(s)\,ds,\\ &[Gy_{2}](t)=\sum_{i=1}^{m-1} \frac{1}{i!}x_{2}^{(i)}(0)t^{i}+ \frac {1}{(m-1)!} \int_{0}^{t}(t-s)^{m-1}y_{2}(s)\,ds, \end{aligned} \end{aligned}$$

(2.3)

where $(Ax_{1})^{(i)}(0)$, $i=1,2,\ldots,n-1$ are defined by the following:

$$E_{1}Z=B, \quad\mbox{where } E_{1}= \left.\begin{pmatrix} 1 & 0 & 0 & \cdots& 0 & 0 \\ c_{1} & 1 & 0 & \cdots& 0 & 0 \\ c_{2} & c_{1} & 1 & \cdots& 0 & 0 \\ \cdots\\ c_{n-3} & c_{n-4} & c_{n-5} & \cdots& 1 & 0 \\ c_{n-2} & c_{n-3} & c_{n-4} & \cdots& c_{1} &0 \end{pmatrix} \right._{(n-1)\times(n-1)}. $$

$Z=((Ax_{1})^{(n-1)}(0),\ldots,(Ax_{1})''(0),(Ax_{1})'(0))^{\top }$, $B=(b_{1},b_{2},\ldots,b_{n-1})^{\top}$, $b_{i}=-\frac{1}{i!T}\int ^{T}_{0}(T-s)^{i} y_{1}(s)\,ds$, and $c_{j}=\frac{T^{j}}{(j+1)!}$, $j=1,2,\ldots,n-2$. $x_{2}^{(i)}(0)$, $i=1,2,\ldots,m-1$, are determined by the equation

$$E_{2}W=F, \quad\mbox{where } E_{2}= \left.\begin{pmatrix} 1 & 0 & 0 & \cdots& 0 & 0 \\ c_{1} & 1 & 0 & \cdots& 0 & 0 \\ c_{2} & c_{1} & 1 & \cdots& 0 & 0 \\ \cdots\\ c_{m-3} & c_{m-4} & c_{m-5} & \cdots& 1 & 0 \\ c_{m-2} & c_{m-3} & c_{m-4} & \cdots& c_{1} &0 \end{pmatrix} \right._{(m-1)\times(m-1)}. $$

$W=((x_{2})^{(m-1)}(0),\ldots,(x_{2})''(0),(x_{2})'(0))^{\top }$, $F=(d_{1},d_{2},\ldots,d_{n-1})^{\top}$, $d_{i}=-\frac{1}{i!T}\int ^{T}_{0}(T-s)^{i}y_{2}(s)\,ds$, and $c_{j}=\frac{T^{j}}{(j+1)!}$, $j=1,2,\ldots,m-2$.

From (2.2) and (2.3), it is clear that QN and $K(I-Q)N$ are continuous, $QN(\overline{\Omega})$ is bounded and then $K(I-Q)N(\overline{\Omega})$ is compact for any open bounded $\Omega\subset X$, which means N is L-compact on Ω̄.

3 Existence of positive periodic solutions for (1.1)

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 if x is a positive continuous T-periodic function satisfying

$$\int^{T}_{0}g\bigl(t,x(t)\bigr)\,dt=0, $$

then

$$D_{1}\leq x(\tau)\leq D_{2} $$

for some $\tau\in[0,T]$.

(H₂):

$\bar{g}(x)<0$ for all $x \in(0,D_{1})$, and $\bar{g}(x)>0$ for all $x>D_{2}$.

(H₃):

Assume that

$$ \psi(t)=\lim_{x\rightarrow+\infty}\sup\frac{g(t,x)}{x^{p-1}}, $$

exist uniformly a.e. $t\in[0,T]$, i.e., for any $\varepsilon>0$ there is $g_{\varepsilon}\in L^{2}(0,T)$ such that

$$g(t,x)\leq\bigl(\psi(t)+\varepsilon\bigr)x^{p-1}+g_{\varepsilon}(t) $$

for all $x>0$ and a.e. $t\in[0,T]$. Moreover, $\psi\in C(\mathbb{R},\mathbb{R})$ and $\psi(t+T)=\psi(t)$.

(H₄):

$g(t,x)=g_{0}(x)+g_{1}(t,x)$, where $g_{0}\in C((0,\infty);\mathbb{R})$ and $g_{1}:[0,T]\times[0,\infty)\rightarrow\mathbb{R}$ is an $L^{2}$-Carathéodory function.

(H₅):

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

(H₆):

There exist two positive constants a, b such that

$$\bigl|f\bigl(x(t)\bigr)\bigr|\leq a|x_{1}|^{p-2}+b,\quad \forall x\in \mathbb{R}. $$

Theorem 3.1

Assume that conditions (H₁)-(H₆) hold. Suppose one of the following conditions is satisfied:

• (i) $p>2$ and $\frac{(a+|\psi|_{\infty}T)T^{p}}{2^{p}|1-|c||^{p-1}} (\frac{T}{2\pi } )^{(n-1)(p-1)+(m-2)}<1$;

• (ii) $p=2$ and $\frac{(a+|\psi|_{\infty}T)T^{p}}{2^{p}|1-|c||^{p-1}} (\frac{T}{2\pi} )^{(n-1)(p-1)+(m-2)}+ \frac{bT^{2} (\frac{T}{2\pi} )^{n+m-3}}{4|1-|c||}<1$.

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

Proof

Consider the equation

$$Lx=\lambda Nx,\quad \lambda\in(0,1). $$

Set $\Omega_{1}=\{x:Lx=\lambda Nx,\lambda\in (0,1)\}$. If $x(t)=(x_{1}(t),x_{2}(t))^{\top}\in\Omega_{1}$, then

$$ \textstyle\begin{cases} (Ax_{1})^{(n)}(t)=\lambda\varphi_{q}(x_{2}(t)),\\ x_{2}^{(m)}(t)=-\lambda f(x_{1}(t))x_{1}'(t)-\lambda g(t,x_{1}(t-\sigma))+\lambda e(t). \end{cases} $$

(3.1)

Substituting $x_{2}(t)=\lambda^{1-p}\varphi_{p}[(Ax_{1})^{(n)}(t)]$ into the second equation of (3.1)

$$ \bigl(\varphi_{p}(Ax_{1})^{(n)}(t) \bigr)^{(m)}+\lambda^{p}f\bigl(x_{1}(t) \bigr)x_{1}'(t)+\lambda ^{p}g \bigl(t,x_{1}(t-\sigma)\bigr)=\lambda^{p}e(t). $$

(3.2)

Integrating of both sides of (3.2) from 0 to T, we have

$$ \int^{T}_{0}g\bigl(t,x_{1}(t-\sigma) \bigr)\,dt=0. $$

(3.3)

In view of (H₁), there exist positive constants $D_{1}$, $D_{2}$, and $\xi\in[0,T]$ such that

$$D_{1} \leq x_{1}(\xi)\leq D_{2}. $$

Then we have

$$\bigl|x_{1}(t)\bigr|=\biggl\vert x_{1}(\xi)+ \int^{t}_{\xi}x_{1}'(s)\,ds \biggr\vert \leq D_{2}+ \int^{t}_{\xi}\bigl|x_{1}'(s)\bigr|\,ds,\quad t \in[\xi,\xi+T], $$

and

$$\bigl|x_{1}(t)\bigr|=\bigl|x_{1}(t-T)\bigr|=\biggl\vert x_{1}(\xi)- \int_{t-T}^{\xi}x_{1}'(s)\,ds \biggr\vert \leq D_{2} + \int_{t-T}^{\xi}\bigl|x_{1}'(s)\bigr|\,ds,\quad t\in[\xi,\xi+T]. $$

Combing the above two inequalities, we obtain

$$\begin{aligned} |x_{1}|_{\infty}&=\max _{t\in[0,T]}\bigl|x_{1}(t)\bigr|=\max_{t\in[\xi,\xi+T]}\bigl|x_{1}(t)\bigr| \\ &\leq\max_{t\in[\xi,\xi+T]} \biggl\{ D_{2}+\frac{1}{2} \biggl( \int^{t}_{\xi}\bigl|x_{1}'(s)\bigr|\,ds+ \int^{\xi}_{t-T}\bigl|x_{1}'(s)\bigr|\,ds \biggr) \biggr\} \\ &\leq D_{2}+\frac{1}{2} \int^{T}_{0}\bigl|x_{1}'(s)\bigr|\,ds. \end{aligned}$$

(3.4)

Since

$$(Ax_{1})^{(n)}(t)=\bigl(x_{1}(t)-cx_{1}(t- \sigma )\bigr)^{(n)}=x_{1}^{(n)}(t)-cx_{1}^{(n)}(t- \sigma)=Ax_{1}^{(n)}(t), $$

from Lemma 2.1 and the first equation of (3.1), we have

$$\begin{aligned} \bigl|x_{1}^{(n)}\bigr|_{\infty}&= \max_{t\in[0,T]}\bigl|A^{-1}Ax_{1}^{(n)}(t)\bigr| \\ &\leq\frac{\max_{t\in[0,T]}|(Ax_{1})^{(n)}(t)|}{|1-|c||} \\ &\leq\frac{\varphi_{q}(|x_{2}|_{\infty})}{|1-|c||}. \end{aligned}$$

(3.5)

On the other hand, from $x_{2}^{(m-2)}(0)=x_{2}^{(m-2)}(T)$, there exists a point $t_{1}\in[0,T]$ such that $x_{2}^{(m-1)}(t_{1})=0$, which together with the integration of the second equation of (3.1) on the interval $[0,T]$ yields

$$\begin{aligned} 2\bigl|x_{2}^{(m-1)}(t)\bigr| &\leq 2 \biggl(x_{2}^{(m-1)}(t_{1})+\frac{1}{2} \int^{T}_{0}\bigl|x_{2}^{(m)}(t)\bigr|\,dt \biggr) \\ &\leq\lambda \int^{T}_{0}\bigl|-f\bigl(x_{1}(t) \bigr)x_{1}'(t)-g\bigl(t,x_{1}(t-\sigma) \bigr)+e(t)\bigr|\,dt \\ &\leq \int^{T}_{0}\bigl|f\bigl(x_{1}(t) \bigr)\bigr|\bigl|x_{1}'(t)\bigr|\,dt+ \int^{T}_{0}\bigl|g\bigl(t,x_{1}(t-\sigma) \bigr)\bigr|\,dt+ \int ^{T}_{0}\bigl|e(t)\bigr|\,dt. \end{aligned}$$

(3.6)

Write

$$I_{+}=\bigl\{ t\in[0,T]:g\bigl(t,x_{1}(t-\sigma)\bigr)\geq0\bigr\} ;\qquad I_{-}= \bigl\{ t\in [0,T]:g\bigl(t,x_{1}(t-\sigma)\bigr)\leq0\bigr\} . $$

Then we get from (H₃) and (3.3) that

$$\begin{aligned} \int^{T}_{0}\bigl|g\bigl(t,x_{1}(t-\sigma) \bigr)\bigr|\,dt&= \int_{I_{+}}g\bigl(t,x_{1}(t-\sigma)\bigr)\,dt- \int _{I_{-}}g\bigl(t,x_{1}(t-\sigma)\bigr)\,dt \\ &=2\int_{I_{+}}g\bigl(t,x_{1}(t-\sigma)\bigr)\,dt \\ &\leq2 \int_{I_{+}}\bigl(\bigl(\psi(t)+\varepsilon\bigr)x_{1}^{p-1}(t- \sigma)+g_{\varepsilon}(t)\bigr)\,dt \\ &\leq2\bigl(|\psi|_{\infty}+\varepsilon\bigr) \int^{T}_{0}\bigl|x_{1}(t)\bigr|^{p-1}\,dt+2 \int ^{T}_{0}\bigl|g_{\varepsilon}(t)\bigr|\,dt. \end{aligned}$$

(3.7)

Substituting (3.4) and (3.7) into (3.6), and from (H₅), we have

$$\begin{aligned} 2\bigl|x_{2}^{(m-1)}(t)\bigr| \leq{}&a \int^{T}_{0}\bigl|x_{1}(t)\bigr|^{p-2}\bigl|x_{1}'(t)\bigr|\,dt+b \int ^{T}_{0}\bigl|x_{1}'(t)\bigr|\,dt \\ &{}+2\bigl(|\psi|_{\infty}+\varepsilon\bigr) \int^{T}_{0}\bigl|x_{1}(t)\bigr|^{p-1}\,dt+2 \int ^{T}_{0}\bigl|g_{\varepsilon}(t)\bigr|\,dt+ \int^{T}_{0}\bigl|e(t)\bigr|\,dt \\ \leq{}&a \biggl(D_{2}+\frac{1}{2} \int^{T}_{0}\bigl|x_{1}'(t)\bigr|\,dt \biggr)^{p-2} \int ^{T}_{0}\bigl|x_{1}'(t)\bigr|\,dt+b \int^{T}_{0}\bigl|x_{1}'(t)\bigr|\,dt \\ &{}+2\bigl(|\psi|_{\infty}+\varepsilon\bigr)T \biggl(D_{2}+ \frac{1}{2} \int ^{T}_{0}\bigl|x_{1}'(t)\bigr|\,dt \biggr)^{p-1}+2T^{\frac{1}{2}}|g_{\varepsilon}|_{2}+|e|_{\infty}T. \end{aligned}$$

(3.8)

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

$$(1+x)^{k}\leq1+(1+k)x \quad\mbox{for } x\in[0,\delta]. $$

From (3.8), we have

$$\begin{aligned} 2\bigl|x_{2}^{(m-1)}(t)\bigr| \leq{}&\frac{a}{2^{p-2}} \biggl(\frac{2D_{2}}{\int ^{T}_{0}|x_{1}'(t)|\,dt}+1 \biggr)^{p-2} \biggl( \int^{T}_{0}\bigl|x_{1}'(t)\bigr|\,dt \biggr)^{p-1} \\ &{}+\frac{(|\psi|_{\infty}+\varepsilon)T}{2^{p-2}} \biggl(\frac{2D_{2}}{\int ^{T}_{0}|x_{1}'(t)|\,dt}+1 \biggr)^{p-1} \\ &{}\cdot \biggl( \int^{T}_{0}\bigl|x_{1}'(t)\bigr|\,dt \biggr)^{p-1} +b \int^{T}_{0}\bigl|x_{1}'(t)\bigr|\,dt+2T^{\frac{1}{2}}|g_{\varepsilon}|_{2}+|e|_{\infty}T \\ \leq{}& \biggl[\frac{a}{2^{p-2}} \biggl(1+\frac{2D_{2}(p-1)}{\int ^{T}_{0}|x_{1}'(t)|\,dt} \biggr)+ \frac{(|\psi|_{\infty}+\varepsilon)T}{2^{p-2}} \biggl(1+\frac{2D_{2}p}{\int^{T}_{0}|x_{1}'(t)|\,dt} \biggr) \biggr] \\ &{}\cdot \biggl( \int^{T}_{0}\bigl|x_{1}'(t)\bigr|\,dt \biggr)^{p-1}+b \int ^{T}_{0}\bigl|x_{1}'(t)\bigr|\,dt+2T^{\frac{1}{2}}|g_{\varepsilon}|_{2}+|e|_{\infty}T \\ ={}& \biggl(\frac{a}{2^{p-2}}+\frac{(|\psi|_{\infty}+\varepsilon )T}{2^{p-2}} \biggr) \biggl( \int^{T}_{0}\bigl|x_{1}'(t)\bigr|\,dt \biggr)^{p-1} \\ &{}+\frac{aD_{2}(p-1)+\bigl(|\psi|_{\infty}+\varepsilon\bigr)TD_{2}p}{2^{p-3}} \\ &{}\cdot \biggl( \int^{T}_{0}\bigl|x_{1}'(t)\bigr|\,dt \biggr)^{p-2} +b \int^{T}_{0}\bigl|x_{1}'(t)\bigr|\,dt+2T^{\frac{1}{2}}|g_{\varepsilon}|_{2}+|e|_{\infty}T. \end{aligned}$$

(3.9)

From the Wirtinger inequality (see [21], Lemma 2.4), we get

$$\begin{aligned} \int^{T}_{0}\bigl|x_{1}'(s)\bigr|\,ds &\leq T^{\frac{1}{2}} \biggl( \int^{T}_{0}\bigl|x_{1}'(s)\bigr|^{2}\,ds \biggr)^{\frac{1}{2}} \\ &\leq T^{\frac{1}{2}} \biggl(\frac{T}{2\pi} \biggr)^{n-1} \biggl( \int ^{T}_{0}\bigl|x_{1}^{(n)}(s)\bigr|^{2}\,ds \biggr)^{\frac{1}{2}} \\ &\leq T \biggl(\frac{T}{2\pi} \biggr)^{n-1}\bigl|x_{1}^{(n)}\bigr|_{\infty}. \end{aligned}$$

(3.10)

Substituting (3.10) into (3.9), we have

$$\begin{aligned} 2\bigl|x_{2}^{(m-1)}(t)\bigr| \leq{}&\frac{(a+(|\psi|_{\infty}+\varepsilon )T)T^{p-1}}{2^{p-2}} \biggl(\frac{T}{2\pi} \biggr)^{(n-1)(p-1)} \bigl|x_{1}^{(n)}\bigr|^{p-1}_{\infty} \\ &{}+\frac{(aD_{2}(p-1)+(|\psi|_{\infty}+\varepsilon )TD_{2}p)T^{p-2}}{2^{p-3}} \biggl(\frac{T}{2\pi} \biggr)^{(n-1)(p-2)} \bigl|x_{1}^{(n)}\bigr|^{p-2}_{\infty} \\ &{}+bT \biggl(\frac{T}{2\pi} \biggr)^{n-1}\bigl|x_{1}^{(n)}\bigr|_{\infty}+2T^{\frac {1}{2}}|g_{\varepsilon}|_{2}+|e|_{\infty}T. \end{aligned}$$

(3.11)

Substituting (3.5) into (3.11), we have

$$\begin{aligned} 2\bigl|x_{2}^{(m-1)}(t)\bigr| \leq{}&\frac{(a+(|\psi|_{\infty}+\varepsilon )T)T^{p-1}}{2^{p-2}} \biggl(\frac{T}{2\pi} \biggr)^{(n-1)(p-1)} \frac {(\varphi_{q}(|x_{2}|_{\infty}))^{p-1}}{|1-|c||^{p-1}} \\ &{}+\frac{(aD_{2}(p-1)+(|\psi|_{\infty}+\varepsilon )TD_{2}p)T^{p-2}}{2^{p-3}} \biggl(\frac{T}{2\pi} \biggr)^{(n-1)(p-2)} \frac{(\varphi_{q}(|x_{2}|_{\infty}))^{p-2}}{|1-|c||^{p-2}} \\ &{}+bT \biggl(\frac{T}{2\pi} \biggr)^{n-1}\frac{\varphi_{q}(|x_{2}|_{\infty})}{|1-|c||}+2T^{\frac{1}{2}}|g_{\varepsilon}|_{2}+|e|_{\infty}T \\ ={}&\frac{(a+(|\psi|_{\infty}+\varepsilon)T)T^{p-1}}{2^{p-2}} \biggl(\frac {T}{2\pi} \biggr)^{(n-1)(p-1)} \frac{|x_{2}|_{\infty}}{|1-|c||^{p-1}} \\ &{}+\frac{(aD_{2}(p-1)+(|\psi|_{\infty}+\varepsilon )TD_{2}p)T^{p-2}}{2^{p-3}} \biggl(\frac{T}{2\pi} \biggr)^{(n-1)(p-2)} \frac{|x_{2}|_{\infty}^{2-q}}{|1-|c||^{p-2}} \\ &{}+bT \biggl(\frac{T}{2\pi} \biggr)^{n-1}\frac{|x_{2}|_{\infty}^{q-1}}{|1-|c||}+2T^{\frac{1}{2}}|g_{\varepsilon}|_{2}+|e|_{\infty}T. \end{aligned}$$

(3.12)

Since $\int^{T}_{0}(\varphi_{q}(x_{2}(t)))\,dt=\int^{T}_{0}(Ax_{1}(t))^{(n)}(t)\,dt=0$, there exists a point $t_{2}\in[0,T]$ such that $x_{2}(t_{2})=0$. From the Wirtinger inequality, we can easily get

$$\begin{aligned} |x_{2}|_{\infty}&\leq \frac{1}{2} \int^{T}_{0}\bigl|x_{2}'(t)\bigr|\,dt \\ &\leq \frac{\sqrt{T}}{2} \biggl( \int^{T}_{0}\bigl|x_{2}'(t)\bigr|^{2} \biggr)^{\frac{1}{2}} \\ &\leq\frac{\sqrt{T}}{2} \biggl(\frac{T}{2\pi} \biggr)^{m-2} \biggl( \int ^{T}_{0}\bigl|x_{2}^{(m-1)}(t)\bigr|^{2}\,dt \biggr)^{\frac{1}{2}} \\ &\leq\frac{T}{2} \biggl(\frac{T}{2\pi} \biggr)^{m-2}\bigl|x_{2}^{(m-1)}\bigr|_{\infty}. \end{aligned}$$

(3.13)

Combination of (3.13) and (3.12) implies

$$\begin{aligned} |x_{2}|_{\infty}\leq{}&\frac{T}{2} \biggl( \frac{T}{2\pi} \biggr)^{m-2}\bigl|x_{2}^{(m-1)}\bigr|_{\infty}\\ \leq&\frac{T}{4} \biggl(\frac{T}{2\pi} \biggr)^{m-2} \biggl[ \frac{(a+(|\psi|_{\infty}+\varepsilon)T)T^{p-1}}{2^{p-2}} \biggl(\frac{T}{2\pi} \biggr)^{(n-1)(p-1)} \frac{|x_{2}|_{\infty}}{|1-|c||^{p-1}}\\ &{}+\frac{(aD_{2}(p-1)+(|\psi|_{\infty}+\varepsilon )TD_{2}p)T^{p-2}}{2^{p-3}} \biggl(\frac{T}{2\pi} \biggr)^{(n-1)(p-2)} \frac{|x_{2}|_{\infty}^{2-q}}{|1-|c||^{p-2}}\\ &{} +bT \biggl(\frac{T}{2\pi} \biggr)^{n-1} \frac{|x_{2}|_{\infty}^{q-1}}{|1-|c||}+2T^{\frac{1}{2}}|g_{\varepsilon}|_{2}+|e|_{\infty}T \biggr]. \end{aligned}$$

So, we have

$$\begin{aligned} |x_{2}|_{\infty}\leq{}&\frac{(a+(|\psi|_{\infty}+\varepsilon )T)T^{p}}{2^{p}|1-|c||^{p-1}} \biggl( \frac{T}{2\pi} \biggr)^{(n-1)(p-1)+(m-2)}|x_{2}|_{\infty}\\ &{}+\frac{(aD_{2}(p-1)+\bigl(|\psi|_{\infty}+\varepsilon \bigr)TD_{2}p)T^{p-1}}{2^{p-1}|1-|c||^{p-2}} \biggl(\frac{T}{2\pi} \biggr)^{(n-1)(p-2)+(m-2)} |x_{2}|_{\infty}^{2-q} \\ &{}+\frac{bT^{2}}{4} \biggl(\frac{T}{2\pi} \biggr)^{n+m-3} \frac{|x_{2}|_{\infty}^{q-1}}{|1-|c||}+\frac{T}{4} \biggl(\frac{T}{2\pi} \biggr)^{m-2} \bigl(2T^{\frac{1}{2}}|g_{\varepsilon}|_{2}+|e|_{\infty}T \bigr). \end{aligned}$$

Case (i): If $p>2$, we can get $1< q<2$. Since ε sufficiently small, we know that

$$\frac{(a+|\psi|_{\infty}T)T^{p}}{2^{p}|1-|c||^{p-1}} \biggl(\frac{T}{2\pi } \biggr)^{(n-1)(p-1)+(m-2)}< 1, $$

there exists a positive constant $M_{1}$ such that

$$ |x_{2}|_{\infty}\leq M_{1}. $$

(3.14)

Case (ii): If $p=2$, we can get $q=2$. Since ε is sufficiently small, we know that

$$\frac{(a+|\psi|_{\infty}T)T^{p}}{2^{p}|1-|c||^{p-1}} \biggl(\frac{T}{2\pi } \biggr)^{(n-1)(p-1)+(m-2)} + \frac{bT^{2} (\frac{T}{2\pi} )^{n+m-3}}{4|1-|c||}< 1, $$

there exists a positive constant $M_{1}$ such that

$$ |x_{2}|_{\infty}\leq M_{1}. $$

On the other hand, from (3.5), we have

$$\bigl|x_{1}^{(n)}\bigr|_{\infty}\leq \frac{\varphi_{q}(|x_{2}|_{\infty})}{|1-|c||}\leq\frac{M_{1}^{q-1}}{|1-|c||}:=M_{n}'. $$

(3.15)

Since $x_{1}(0)=x_{1}(T)$, there exists a point $t_{3}\in[0,T]$ such that $x_{1}'(t_{3})=0$. From the Wirtinger inequality, we can easily get

$$\begin{aligned} \bigl|x_{1}'\bigr|_{\infty}&\leq\frac{1}{2} \int^{T}_{0}\bigl|x_{1}''(t)\bigr|\,dt \leq \frac{T^{\frac{1}{2}}}{2} \biggl( \int^{T}_{0}\bigl|x_{1}''(t)\bigr|^{2}\,dt \biggr)^{\frac {1}{2}} \\ &\leq\frac{T}{2} \biggl(\frac{T}{2\pi} \biggr)^{(n-2)}\bigl|x_{1}^{(n)}\bigr|_{\infty} \\ &\leq\frac{T}{2} \biggl(\frac{T}{2\pi} \biggr)^{n-2} M_{n}':=M_{2}. \end{aligned}$$

(3.16)

Hence, from (3.4), we have

$$ |x_{1}|_{\infty}\leq D+\frac{1}{2} \int^{T}_{0}\bigl|x_{1}'(t)\bigr|\,dt \leq D+\frac{TM_{2}}{2}:=M_{3}. $$

(3.17)

From (3.7), (3.16), and (3.17) we have

$$\begin{aligned} \bigl\vert x_{2}^{(m-1)}\bigr\vert _{\infty}&\leq \frac{1}{2}\max\biggl\vert \int ^{T}_{0}x_{2}^{(m)}(t)\,dt \biggr\vert \\ &\leq\frac{\lambda}{2} \int^{T}_{0}\bigl|-f\bigl(x_{1}(t) \bigr)x_{1}'(t)-g\bigl(t,x_{1}(t-\sigma ) \bigr)+e(t)\bigr|\,dt \\ &\leq\frac{\lambda}{2} \bigl(|f|_{M_{3}}TM_{2}+2\bigl(| \psi|_{\infty}+\varepsilon \bigr)TM_{3}^{p-1}+2 \sqrt{T}|g_{\varepsilon}|_{2} +T|e|_{\infty}\bigr):=\lambda M_{m-2}, \end{aligned}$$

where $|f|_{M_{3}}=\max_{0< x_{1}(t)\leq M_{3}}|f(x_{1}(t))|$. Since $x_{2}(0)=x_{2}(T)$, there exists a point $t_{4}\in[0,T]$ such that $x_{2}'(t_{4})=0$. From the Wirtinger inequality, we can easily get

$$\begin{aligned} \bigl|x_{2}'\bigr|_{\infty}&\leq \frac{1}{2} \int^{T}_{0}\bigl|x_{2}''(t)\bigr|\,dt\\ &\leq \frac{T^{\frac{1}{2}}}{2} \biggl( \int^{T}_{0}\bigl|x_{2}''(t)\bigr|^{2}\,dt \biggr)^{\frac {1}{2}} \\ &\leq\frac{T}{2} \biggl(\frac{T}{2\pi} \biggr)^{(m-3)}\bigl|x_{2}^{(m-1)}\bigr|_{\infty}\\ &\leq\frac{T}{2} \biggl(\frac{T}{2\pi} \biggr)^{(m-3)} \lambda M_{m-2}:=\lambda M_{2}. \end{aligned}$$

Next, it follows (3.2) that

$$\begin{aligned} &\bigl(\varphi_{p}(Ax_{1})^{(n)}(t+ \sigma)\bigr)^{(m)}+\lambda^{p}\bigl(f\bigl(x_{1}(t+ \sigma )\bigr)x_{1}'(t+\sigma)+\lambda^{p}g \bigl(t+\sigma,x_{1}(t)\bigr)\bigr) \\ &\quad=\lambda^{p} e(t+\sigma). \end{aligned}$$

(3.18)

Namely,

$$\begin{aligned} &\bigl(\varphi_{p}(Ax_{1})^{(n)}(t+ \sigma)\bigr)^{(m)}+\lambda^{p}f\bigl(x_{1}(t+\sigma )\bigr)x_{1}'(t+\sigma)+\lambda^{p}\bigl(g_{0} \bigl(x_{1}(t)\bigr)\bigr)+g_{1}\bigl(t+\sigma,x_{1}(t) \bigr) \\ &\quad=\lambda^{p} e(t+\sigma). \end{aligned}$$

(3.19)

Multiplying both sides of (3.19) by $x_{1}'(t)$, we get

$$\begin{aligned} &\bigl(\varphi_{p}(Ax_{1})^{(n)}(t+ \sigma)\bigr)^{(m)}x_{1}'(t)+ \lambda^{p}f\bigl(x_{1}(t+\sigma )\bigr)x_{1}'(t+ \sigma)x_{1}'(t) \\ &\qquad{}+\lambda^{p}g_{0} \bigl(x_{1}(t)\bigr)x_{1}'(t)+\lambda ^{p}g_{1}\bigl(t+\sigma,x_{1}(t) \bigr)x_{1}'(t) =\lambda^{p} e(t+ \sigma)x_{1}'(t). \end{aligned}$$

(3.20)

Let $\tau\in[0,T]$, for any $\tau\leq t\leq T$, we integrate (3.20) on $[\tau, t]$ and get

$$\begin{aligned} &\lambda^{p} \int^{x_{1}(t)}_{x_{1}(\tau)}g_{0}(u)\,du \\ &\quad= \lambda^{p} \int^{t}_{\tau }g_{0}\bigl(x_{1}(s) \bigr)x_{1}'(s)\,ds \\ &\quad=- \int^{t}_{\tau}\bigl(\varphi_{p}(Ax_{1})^{(n)}(s+ \sigma )\bigr)^{(m)}x_{1}'(s)\,ds- \lambda^{p} \int^{t}_{\tau}f\bigl(x_{1}(s+\sigma) \bigr)x_{1}'(s+\sigma )x_{1}'(s)\,ds \\ &\qquad{}-\lambda^{p} \int^{t}_{\tau}g_{1}\bigl(s+ \sigma,x_{1}(s)\bigr)x_{1}'(s)\,ds+ \lambda^{p} \int ^{t}_{\tau}e(s+\sigma)x_{1}'(s)\,ds. \end{aligned}$$

(3.21)

By (3.2), (3.7), (3.16), and (3.17), we have

$$\begin{aligned} &\biggl\vert \int^{t}_{\tau}\bigl(\varphi_{p}(Ax_{1})^{(n)}(s+ \sigma )\bigr)^{(m)}x_{1}'(s)\,ds\biggr\vert \\ &\quad\leq \int^{t}_{\tau}\bigl|\bigl(\varphi _{p}(Ax_{1})^{(n)}(s+ \sigma)\bigr)^{(m)}\bigr|\bigl|x_{1}'(s)\bigr|\,ds \\ &\quad\leq\bigl|x_{1}'\bigr|_{\infty}\int^{T}_{0}\bigl|\bigl(\varphi_{p}(Ax_{1})^{(n)}(t+ \sigma)\bigr)^{(m)}\bigr|\,dt \\ &\quad\leq\lambda^{p}\bigl|x_{1}'\bigr|_{\infty}\biggl( \int^{T}_{0}\bigl|f\bigl(x_{1}(t) \bigr)\bigr|\bigl|x_{1}'(t)\bigr|\,dt+ \int^{T}_{0}\bigl|g\bigl(t,x_{1}(t-\sigma ) \bigr)\bigr|\,dt+ \int^{T}_{0}\bigl|e(t)\bigr|\,dt \biggr) \\ &\quad\leq\lambda^{p} M_{2}\bigl(|f|_{M_{3}}M_{2} +2\bigl(|\psi|_{\infty}+\varepsilon\bigr) TM_{3}^{p-1}+2T^{\frac{1}{2}}\bigl|g_{\varepsilon}^{+}\bigr|_{2} +T|e|_{\infty}\bigr). \end{aligned}$$

We have

$$\begin{aligned} &\biggl\vert \int^{t}_{\tau}f\bigl(x_{1}(s+\sigma) \bigr)x_{1}'(s+\sigma)x_{1}'(s)\,ds \biggr\vert \leq |f|_{M_{3}}M_{2}^{2}T, \\ &\biggl\vert \int^{t}_{\tau}g\bigl(s+\sigma,x_{1}(s) \bigr)x_{1}'(s)\,ds\biggr\vert \leq\bigl|x_{1}'\bigr| \int ^{T}_{0}\bigl|g(t,x(t-\sigma)\bigr|\,dt\leq M_{2}^{p-1}\sqrt{T}|g_{M_{3}}|_{2}, \end{aligned}$$

where $g_{M_{3}}=\max_{0\leq x\leq M_{3}}|g_{1}(t,x)|\in L^{2}(0,T)$ is as in (H₃); we have

$$\biggl\vert \int^{t}_{\tau}e(t+\sigma)x_{1}'(t)\,dt \biggr\vert \leq M_{2}T|e|_{\infty}. $$

From these inequalities we can derive from (3.21) that

$$ \biggl\vert \int^{x_{1}(t)}_{x_{1}(\tau)}g_{0}(u)\,du\biggr\vert \leq M_{5}', $$

(3.22)

for some constant $M_{5}'$, which is independent on λ, x, and t. In view of the strong force condition (H₄), we know that there exists a constant $M_{5}>0$ such that

$$ x_{1}(t)\geq M_{5}, \quad\forall t\in[\tau,T]. $$

(3.23)

The case $t\in[0,\tau]$ can be treated similarly.

From (3.14), (3.16), and (3.17) and (3.23), we let

$$\begin{aligned} \Omega={}&\bigl\{ x=(x_{1},x_{2})^{\top}: E_{1}\leq |x_{1}|_{\infty}\leq E_{2}, \bigl|x_{1}'\bigr|_{\infty}\leq E_{3}, |x_{2}|_{\infty}\leq E_{4} \mbox{ and}\\ &{}\bigl|x_{2}'\bigr|_{\infty}\leq E_{5} , \forall t \in[0,T]\bigr\} , \end{aligned}$$

where $0< E_{1}<\min\{M_{5}, D_{1}\}$, $E_{2}>\max\{M_{1}, D_{2}\}$, $E_{3}>M_{2}$, $E_{4}>M_{4}$, and $E_{5}>M_{3}$. $\Omega_{2}=\{x:x\in\partial\Omega\cap \operatorname{Ker}L\}$ then $\forall x\in \partial\Omega\cap \operatorname{Ker} L$

$$ QNx=\frac{1}{T} \int^{T}_{0} \begin{pmatrix} \varphi_{q}(x_{2}(t))\\ -f(x_{1}(t))x_{1}'(t)-g(t,x_{1}(t-\sigma))+e(t) \end{pmatrix}\,dt. $$

If $QNx=0$, then $x_{2}(t)=0$, $x_{1}=E_{2}$ or $-E_{2}$. But if $x_{1}(t)=E_{2}$, we know

$$0= \int^{T}_{0}\bigl\{ g(t,E_{2})-e(t)\bigr\} \,dt. $$

From assumption (H₂), we have $x_{1}(t)\leq D_{2}\leq E_{2}$, which yields a contradiction. Similarly if $x_{1}=-E_{2}$. We also have $QNx\neq0$, i.e., $\forall x\in\partial\Omega\cap \operatorname{Ker} L$, $x\notin \operatorname{Im} L$, so conditions (1) and (2) of Lemma 2.2 are both satisfied. Define the isomorphism $J:\operatorname{Im} Q\rightarrow \operatorname{Ker} L$ as follows:

$$J(x_{1},x_{2})^{\top}=(x_{2},-x_{1})^{\top}. $$

Let $H(\mu,x)=-\mu x+(1-\mu)JQNx$, $(\mu,x)\in[0,1]\times\Omega$, then $\forall (\mu,x)\in(0,1)\times(\partial\Omega\cap \operatorname{Ker} L)$,

$$ H(\mu,x)= \begin{pmatrix} -\mu x_{1}-\frac{1-\mu}{T}\int^{T}_{0}[g(t,x_{1})-e(t)]\,dt\\ -\mu x_{2}-(1-\mu)|x_{2}|^{q-2}x_{2} \end{pmatrix} . $$

We have $\int^{T}_{0}e(t)\,dt=0$. So, we can get

$$\begin{aligned} &H(\mu,x)= \begin{pmatrix}-\mu x_{1}-\frac{1-\mu}{T}\int^{T}_{0}g(t,x_{1})\,dt\\ -\mu x_{2}-(1-\mu)|x_{2}|^{q-2}x_{2} \end{pmatrix} , \quad\forall (\mu,x)\in(0,1)\times(\partial\Omega\cap \operatorname{Ker} L). \end{aligned}$$

From (H₂), it is obvious that $x^{\top}H(\mu,x)<0$, $\forall (\mu,x)\in(0,1)\times(\partial\Omega\cap \operatorname{Ker} L)$. Hence

$$\begin{aligned} \deg\{JQN,\Omega\cap \operatorname{Ker} L,0\}&=\deg\bigl\{ H(0,x),\Omega\cap \operatorname{Ker} L,0\bigr\} \\ &=\deg\bigl\{ H(1,x),\Omega\cap \operatorname{Ker} L,0\bigr\} \\ &=\deg\{I,\Omega\cap \operatorname{Ker} L,0\}\neq0. \end{aligned}$$

So condition (3) of Lemma 2.2 is satisfied. By applying Lemma 2.2, we conclude that equation $Lx=Nx$ has a solution $x=(x_{1},x_{2})^{\top}$ on $\bar{\Omega}\cap D(L)$, i.e., (1.1) has an T-periodic solution $x_{1}(t)$. □

Example 3.1

Consider the p-Laplacian type high-order neutral differential equation with singularity

$$\begin{aligned} &\bigl(\varphi_{p}\bigl(x(t)-11x(t-\tau) \bigr)'''\bigr)'''+3x^{2}(t)x'(t)+ \frac{1}{10}(\cos 2t+4)x^{3}(t-\sigma)-\frac{1}{x^{\kappa}(t-\sigma)} \\ &\quad=\sin 2t, \end{aligned}$$

(3.24)

where $\kappa\geq1$ and $p=4$, σ and τ are constants, and $0\leq\sigma, \tau< T$.

It is clear that $T=\pi$, $m=n=3$, $c=11$, $g(t,x)=\frac{1}{10}(\cos2t+4)x^{3}(t-\sigma)-\frac{1}{x^{\kappa}(t-\sigma)}$, $\psi(t)=\frac{1}{10}(\cos2t+4)$, $|\psi|_{\infty}=\frac{1}{2}$, $f(x(t))=3x^{2}(t)$, and $|f(x(t))|\leq3|x^{2}(t)|+1$; here $a=3$, $b=1$. It is obvious that (H₁)-(H₅) hold. Now we consider the assumption of the condition

$$\begin{aligned} &\frac{(a+|\psi|_{\infty}T)T^{p}}{2^{p}|1-|c||^{p-1}} \biggl(\frac{T}{2\pi} \biggr)^{(n-1)(p-1)+(m-2)} \\ &\quad=\frac{ (3+\frac{\pi}{2} )\pi^{4}}{2^{4}\times10^{3}}\times\frac {1}{2^{7}} \\ &\quad=\frac{ (3+\frac{\pi}{2} )\pi^{4}}{2^{11}\times10^{3}}< 1. \end{aligned}$$

So by Theorem 3.1, we know (3.24) has at least one positive π-periodic solution.