## 1 Introduction

In this paper, we consider the following p-Laplacian Liénard type differential equation with singularity and deviating argument:

$$\bigl(\varphi_{p}\bigl(x'(t)\bigr) \bigr)'+f\bigl(x(t)\bigr)x'(t)+g\bigl(t,x(t-\sigma) \bigr)=e(t),$$
(1.1)

where $$\varphi_{p}:\mathbb{R}\rightarrow\mathbb{R}$$ is given by $$\varphi _{p}(s)=\vert s\vert ^{p-2}s$$, here $$p>1$$ is a constant, f is continuous function; g is a continuous function defined on $$\mathbb{R}^{2}$$ and periodic in t with $$g(t,\cdot)=g(t+T,\cdot)$$, g has a singularity at $$x=0$$; σ is a constant and $$0\leq \sigma< T$$; $$e:\mathbb{R}\rightarrow\mathbb{R}$$ are continuous periodic functions with $$e(t+T)\equiv e(t)$$ and $$\int^{T}_{0}e(t)\,dt=0$$.

As is well known, the existence of periodic solutions for Liénard type differential equations was extensively studied (see [110] and the references therein). In recent years, there also appeared some results on a Liénard type differential equation with singularity; see [11, 12]. In 1996, using coincidence degree theory, Zhang considered the existence of T-periodic solutions for the scalar Liénard equation

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

when g becomes unbounded as $$x\rightarrow0^{+}$$. The main emphasis was on the repulsive case, i.e. when $$g(t,x)\rightarrow+\infty$$, as $$x\rightarrow0^{+}$$. Afterwards, Wang [12] studied the existence of periodic solutions of the Liénard equation with a singularity and a deviating argument,

$$x''(t)+ f\bigl(x(t)\bigr)x'(t)+ g \bigl(t,x(t-\sigma)\bigr)=0,$$

where σ is a constant. When g has a strong singularity at $$x = 0$$ and satisfies a new small force condition at $$x =\infty$$, the author proved that the given equation has at least one positive T-periodic solution.

However, the Liénard type differential equation (1.1), in which there is a p-Laplacian Liénard type differential equation, has not attracted much attention in the literature. There are not so many existence results for (1.1) even as regards the p-Laplacian Liénard type differential equation with singularity and deviating argument. In this paper, we try to fill this 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 [11, 12].

## 2 Preparation

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 2.1

(Gaines and Mawhin [13])

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. (1)

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

2. (2)

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

3. (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 $$\overline{\Omega}\cap D(L)$$.

For the sake of convenience, throughout this paper we will adopt the following notation:

\begin{aligned}& \vert u\vert _{\infty}=\max_{t\in[0,T]}\bigl\vert u(t) \bigr\vert ,\qquad \vert u\vert _{0}=\min_{t\in[0,T]}\bigl\vert u(t)\bigr\vert , \\& \vert u\vert _{p}= \biggl(\int^{T}_{0}\vert u\vert ^{p}\,dt \biggr)^{\frac{1}{p}}, \qquad \bar{h}=\frac{1}{T} \int^{T}_{0}h(t)\,dt. \end{aligned}

### Lemma 2.2

([14])

If $$\omega\in C^{1}(\mathbb{R},\mathbb{R})$$ and $$\omega(0)=\omega(T)=0$$, then

$$\int^{T}_{0}\bigl\vert \omega(t)\bigr\vert ^{p}\,dt\leq \biggl(\frac{T}{\pi_{p}} \biggr)^{p} \int^{T}_{0}\bigl\vert \omega'(t) \bigr\vert ^{p}\,dt,$$

where $$1\leq p<\infty$$, $$\pi_{p}=2\int^{(p-1)/p}_{0}\frac{ds}{(1-\frac{s^{p}}{p-1})^{1/p}}=\frac {2\pi(p-1)^{1/p}}{p\sin(\pi/p)}$$.

### Lemma 2.3

If $$x\in C^{1}(\mathbb{R},\mathbb{R})$$ with $$x(t+T)=x(t)$$, and $$t_{0}\in[0,T]$$ such that $$\vert x(t_{0})\vert < d$$, then

$$\biggl( \int^{T}_{0}\bigl\vert x(t)\bigr\vert ^{p}\,dt \biggr)^{\frac{1}{p}}\leq \biggl(\frac{T}{\pi _{p}} \biggr) \biggl( \int^{T}_{0}\bigl\vert x'(t)\bigr\vert ^{p}\,dt \biggr)^{\frac{1}{p}}+dT^{\frac{1}{p}}.$$

### Proof

Let $$\omega(t)=x(t+t_{0})-x(t_{0})$$, and then $$\omega(0)=\omega(T)=0$$. By Lemma 2.2 and Minkowski’s inequality, we have

\begin{aligned} \biggl( \int^{T}_{0}\bigl\vert x(t)\bigr\vert ^{p}\,dt \biggr)^{\frac{1}{p}}&= \biggl( \int^{T}_{0}\bigl\vert \omega (t)+x(t_{0}) \bigr\vert ^{p}\,dt \biggr)^{\frac{1}{p}} \\ &\leq \biggl( \int^{T}_{0}\bigl\vert \omega(t)\bigr\vert ^{p}\,dt \biggr)^{\frac{1}{p}}+ \biggl( \int ^{T}_{0}\bigl\vert x(t_{0})\bigr\vert ^{p}\,dt \biggr)^{\frac{1}{p}} \\ &\leq \biggl(\frac{T}{\pi_{p}} \biggr) \biggl( \int^{T}_{0}\bigl\vert \omega'(t) \bigr\vert ^{p}\,dt \biggr)^{\frac{1}{p}}+dT^{\frac{1}{p}} \\ &= \biggl(\frac{T}{\pi_{p}} \biggr) \biggl( \int^{T}_{0}\bigl\vert x'(t)\bigr\vert ^{p}\,dt \biggr)^{\frac {1}{p}}+dT^{\frac{1}{p}}. \end{aligned}

This completes the proof of Lemma 2.3. □

In order to apply the topological degree theorem to study the existence of a positive periodic solution for (1.1), we rewrite (1.1) in the form

$$\textstyle\begin{cases} x_{1}'(t)=\varphi_{q}(x_{2}(t)),\\ x_{2}'(t)=-f(x_{1}(t))x_{1}'(t)-g(t,x_{1}(t-\sigma))+e(t), \end{cases}$$
(2.1)

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

Now, set $$X=Y=\{x=(x_{1}(t),x_{2}(t))\in C^{1}(\mathbb{R},\mathbb{R}^{2}): x(t+T)\equiv x(t)\}$$ with the norm $$\Vert x\Vert =\max\{\vert x_{1}\vert _{\infty}, \vert x_{2}\vert _{\infty}\}$$. Clearly, X and Y are both Banach spaces. Meanwhile, define

$$L:D(L)=\bigl\{ x\in C^{1}\bigl(\mathbb{R},\mathbb{R}^{2} \bigr): x(t+T) = x(t), t \in\mathbb {R}\bigr\} \subset X\rightarrow Y$$

by

$$(Lx) (t)=\begin{pmatrix} x_{1}'(t)\\x_{2}'(t) \end{pmatrix}$$

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 to the abstract equation $$Lx=Nx$$. From the definition of L, one can easily see that

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

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 $$\operatorname {Im}P=\operatorname {Ker}L$$, $$\operatorname {Ker}Q=\operatorname {Im}L$$. Let K denote the inverse of $$L|_{\operatorname {Ker}p\cap D(L)}$$. It is easy to see that $$\operatorname {Ker}L=\operatorname {Im}Q=\mathbb{R}^{2}$$ and

$$[Ky](t)= \int^{T}_{0}G(t,s)y(s)\,ds,$$

where

$$G(t,s)= \textstyle\begin{cases} \frac{s}{T}, &0\leq s < t\leq T;\\ \frac{s-t}{T}, &0\leq t\leq s \leq T. \end{cases}$$
(2.3)

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 Ω̅.

## 3 Main results

Assume that

$$\psi(t)=\lim_{x\rightarrow+\infty}\sup\frac{g(t,x)}{x^{p-1}}$$
(3.1)

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

$$g(t,x)\leq\bigl(\psi(t)+\varepsilon\bigr)x+g_{\varepsilon}(t),$$
(3.2)

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

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

(H1) (Balance condition) 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) (Degree condition) $$\bar{g}(x)<0$$ for all $$x \in(0,D_{1})$$, and $$\bar{g}(x)>0$$ for all $$x>D_{2}$$.

(H3) (Decomposition condition) $$g(t,x)=g_{0}(x)+g_{1}(t,x)$$, 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, i.e. it is measurable in the first variable and continuous in the second variable, and for any $$b>0$$ there is $$h_{b}\in L^{2}(0,T;\mathbb{R}_{+})$$ such that

$$\bigl\vert g_{1}(t,x)\bigr\vert \leq h_{b}(t),\quad \mbox{a.e. } t \in[0,T], \forall 0\leq x\leq b.$$

(H4) (Strong force condition at $$x=0$$) $$\int^{1}_{0}g_{0}(x)\,dx=-\infty$$.

### Theorem 3.1

Assume that conditions (H1)-(H4) hold. Suppose the following condition is satisfied:

(H5) $$(\frac{T}{\pi_{p}} )^{p}\vert \psi \vert _{\infty}<1$$.

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

$$\textstyle\begin{cases} x_{1}'(t)=\lambda\varphi_{q}(x_{2}(t)),\\ x_{2}'(t)=-\lambda f(x_{1}(t))x_{1}'(t) -\lambda g(t,x_{1}(t-\sigma))+\lambda e(t). \end{cases}$$
(3.3)

Substituting $$x_{2}(t)=\frac{1}{\lambda^{p-1}}\varphi_{p}(x_{1}'(t))$$ into the second equation of (3.3)

$$\bigl(\varphi_{p}\bigl(x_{1}'(t) \bigr)\bigr)'+\lambda^{p}f\bigl(x_{1}(t) \bigr)x_{1}'(t)+\lambda ^{p}g \bigl(t,x_{1}(t-\sigma)\bigr)=\lambda^{p}e(t).$$
(3.4)

Integrating both sides of (3.4) over $$[0,T]$$, we have

$$\int^{T}_{0}g\bigl(t,x_{1}(t-\sigma)\bigr) \,dt=0.$$
(3.5)

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

$$D_{1}\leq x_{1}(\xi)\leq D_{2}.$$
(3.6)

Then we have

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

and

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

Combining the above two inequalities, we obtain

\begin{aligned}[b] \vert x_{1}\vert _{\infty}&=\max_{t\in[0,T]}\bigl\vert x_{1}(t)\bigr\vert =\max_{t\in[\xi,\xi+T]}\bigl\vert x_{1}(t)\bigr\vert \\ &\leq\max_{t\in[\xi,\xi+T]} \biggl\{ D_{2}+\frac{1}{2} \biggl( \int^{t}_{\xi}\bigl\vert x_{1}'(s) \bigr\vert \,ds+ \int^{\xi}_{t-T}\bigl\vert x_{1}'(s) \bigr\vert \,ds \biggr) \biggr\} \\ &\leq D_{2}+\frac{1}{2} \int^{T}_{0}\bigl\vert x_{1}'(s) \bigr\vert \,ds. \end{aligned}
(3.7)

Multiplying both sides of (3.4) by $$x_{1}(t)$$ and integrating over the interval $$[0,T]$$, we get

\begin{aligned} &\int^{T}_{0}\bigl(\varphi_{p} \bigl(x_{1}'(t)\bigr)\bigr)'x_{1}(t) \,dt+\lambda^{p} \int ^{T}_{0}f\bigl(x_{1}(t) \bigr)x_{1}'(t)x_{1}(t)\,dt+\lambda^{p} \int^{T}_{0}g\bigl(t,x_{1}(t-\sigma ) \bigr)x_{1}(t)\,dt \\ &\quad =\lambda^{p} \int^{T}_{0}e(t)x_{1}(t)\,dt. \end{aligned}
(3.8)

Substituting $$\int^{T}_{0}(\varphi_{p}(x_{1}'(t)))'x_{1}(t)\,dt=-\int^{T}_{0}\vert x_{1}'(t)\vert ^{p}\,dt$$, $$\int ^{T}_{0}f(x_{1}(t))x_{1}'(t)x_{1}(t)\,dt=0$$ into (3.8), we have

$$\int^{T}_{0}\bigl\vert x_{1}'(t) \bigr\vert ^{p}d=\lambda^{p} \int^{T}_{0}g\bigl(t,x_{1}(t-\sigma) \bigr)x_{1}(t)\, dt-\lambda^{p} \int^{T}_{0}e(t)x_{1}(t)\,dt.$$
(3.9)

For any $$\varepsilon>0$$, there exists a function $$g_{\varepsilon}\in L^{2}(0,T)$$ such that (3.2) holds. Since $$x_{1}(t)>0$$, $$t\in[0,T]$$, it follows from (3.4) that

$$g\bigl(t,x_{1}(t-\sigma)\bigr)x_{1}(t)\leq \bigl(\psi(t)+\varepsilon\bigr)x_{1}^{p-1}(t-\sigma )x_{1}(t)+g_{\varepsilon}(t)x_{1}(t).$$
(3.10)

We infer from (3.9) and (3.10)

\begin{aligned} &\int^{T}_{0}\bigl\vert x_{1}'(t) \bigr\vert ^{p}\,dt \\ &\quad \leq\lambda^{p} \int^{T}_{0}\bigl(\psi(t)+\varepsilon \bigr)x_{1}^{p-1}(t-\sigma)x_{1}(t)\,dt + \lambda^{p} \int^{T}_{0}\bigl(g_{\varepsilon}(t)+e(t) \bigr)x_{1}(t)\,dt \\ &\quad \leq \int^{T}_{0}\bigl(\bigl\vert \psi(t)\bigr\vert + \varepsilon\bigr)\bigl\vert x_{1}^{p-1}(t-\sigma)\bigr\vert \bigl\vert x_{1}(t)\bigr\vert \,dt + \int^{T}_{0}\bigl(\bigl\vert g_{\varepsilon}(t) \bigr\vert +\bigl\vert e(t)\bigr\vert \bigr)\bigl\vert x_{1}(t) \bigr\vert \,dt \\ &\quad \leq\bigl(\vert \psi \vert _{\infty}+\varepsilon\bigr) \biggl( \int^{T}_{0}\bigl\vert x_{1}(t-\sigma) \bigr\vert ^{p}\, dt \biggr)^{\frac{p-1}{p}} \biggl( \int^{T}_{0}\bigl\vert x_{1}(t)\bigr\vert ^{p}\,dt \biggr)^{\frac{1}{p}} \\ &\qquad {}+\vert x_{1}\vert _{\infty}\biggl( \int^{T}_{0}\bigl\vert g_{\varepsilon}(t)\bigr\vert \,dt+ \int^{T}_{0}\bigl\vert e(t)\bigr\vert \, dt \biggr) \\ &\quad \leq\bigl(\vert \psi \vert _{\infty}+\varepsilon\bigr) \biggl( \int^{T}_{0}\bigl\vert x_{1}(t)\bigr\vert ^{p}\,dt \biggr) +\vert x_{1}\vert _{\infty}\biggl( \int^{T}_{0}\bigl\vert g_{\varepsilon}(t)\bigr\vert \,dt+ \int^{T}_{0}\bigl\vert e(t)\bigr\vert \, dt \biggr). \end{aligned}
(3.11)

From Lemma 2.3 and (3.7), we have

$$\biggl( \int^{T}_{0}\bigl\vert x_{1}(t)\bigr\vert ^{p} \biggr)^{\frac{1}{p}}\leq \biggl(\frac{T}{\pi _{p}} \biggr) \biggl( \int^{T}_{0}\bigl\vert x_{1}'(t) \bigr\vert ^{p}\,dt \biggr)^{\frac{1}{p}}+D_{2}T^{\frac{1}{p}}.$$
(3.12)

Substituting (3.7), (3.12) into (3.11), we get

\begin{aligned} &\int^{T}_{0}\bigl\vert x_{1}'(t) \bigr\vert ^{p}\,dt \\ &\quad \leq\bigl(\vert \psi \vert _{\infty}+ \varepsilon\bigr) \biggl( \biggl(\frac {T}{\pi_{p}} \biggr) \biggl( \int^{T}_{0}\bigl\vert x_{1}'(t) \bigr\vert ^{p}\,dt \biggr)^{\frac{1}{p}}+D_{2}T^{\frac {1}{p}} \biggr)^{p} \\ &\qquad {}+ \biggl(D_{2}+\frac{1}{2} \int^{T}_{0}\bigl\vert x_{1}'(t) \bigr\vert \,dt \biggr) \biggl( \int^{T}_{0}\bigl\vert g_{\varepsilon}(t)\bigr\vert \,dt+ \int^{T}_{0}\bigl\vert e(t)\bigr\vert \,dt \biggr) \\ &\quad \leq\bigl(\vert \psi \vert _{\infty}+\varepsilon\bigr) \biggl( \biggl( \frac{T}{\pi_{p}} \biggr)^{p} \int ^{T}_{0}\bigl\vert x_{1}'(t) \bigr\vert ^{p}\,dt \\ &\qquad {}+p \biggl(\frac{T}{\pi_{p}} \biggr)^{p-1} \biggl( \int^{T}_{0}\bigl\vert x_{1}'(t) \bigr\vert ^{p}\,dt \biggr)^{\frac{p-1}{p}}D_{2}T^{\frac {1}{p}} +\cdots+D_{2}^{p}T \biggr) \\ &\qquad {}+ \biggl(D_{2}+ \frac{1}{2}T^{\frac{1}{q}} \biggl( \int^{T}_{0}\bigl\vert x_{1}'(t) \bigr\vert ^{p}\,dt \biggr)^{\frac{1}{p}} \biggr) \bigl(T^{\frac{1}{2}}\bigl(\vert g_{\varepsilon} \vert _{2}+\vert e\vert _{2}\bigr) \bigr) \\ &\quad =\bigl(\vert \psi \vert _{\infty}+\varepsilon\bigr) \biggl( \frac{T}{\pi_{p}} \biggr)^{p} \int ^{T}_{0}\bigl\vert x_{1}'(t) \bigr\vert ^{p}\,dt \\ &\qquad {}+\bigl(\vert \psi \vert _{\infty}+ \varepsilon\bigr)p \biggl(\frac{T}{\pi _{p}} \biggr)^{p-1} \biggl( \int^{T}_{0}\bigl\vert x_{1}'(t) \bigr\vert ^{p}\,dt \biggr)^{\frac{p-1}{p}}D_{2}T^{\frac{1}{p}}+\cdots \\ &\qquad {}+\frac{1}{2}T^{\frac{1}{q}+\frac{1}{2}} \biggl( \int ^{T}_{0}\bigl\vert x_{1}'(t) \bigr\vert ^{p}\,dt \biggr)^{\frac{1}{p}}\bigl(\vert g_{\varepsilon} \vert _{2}+\vert e\vert _{2}\bigr) \\ &\qquad {}+ \bigl(\vert \psi \vert _{\infty}+\varepsilon\bigr)D_{2}^{p}T +T^{\frac{1}{2}}D_{2}\bigl(\vert g_{\varepsilon} \vert _{2}+\vert e\vert _{2}\bigr), \end{aligned}
(3.13)

where $$\vert g_{\varepsilon} \vert _{2}= (\int^{T}_{0}\vert g_{\varepsilon}(t)\vert ^{2}\,dt )^{\frac{1}{2}}$$. Since ε is sufficiently small, from (H5) we know that $$(\frac{T}{\pi_{p}} )^{p}\vert \psi \vert _{\infty}<1$$. So, it is easy to see that there exists a positive constant $$M'_{1}$$ such that

$$\int^{T}_{0}\bigl\vert x_{1}'(t) \bigr\vert ^{p}\,dt\leq M'_{1}.$$

From (3.7), we have

\begin{aligned}[b] \vert x_{1}\vert _{\infty}&\leq D_{2}+\frac{1}{2} \int^{T}_{0}\bigl\vert x_{1}'(t) \bigr\vert \,dt\\ &\leq D_{2}+\frac{T^{\frac{1}{q}}}{2} \biggl( \int^{T}_{0}\bigl\vert x_{1}'(t) \bigr\vert ^{p}\,dt \biggr)^{\frac {1}{p}} \\ &\leq D_{2}+\frac{T^{\frac{1}{q}}}{2} \bigl(M_{1}' \bigr)^{\frac{1}{p}}:=M_{1}. \end{aligned}
(3.14)

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 (3.2) and (3.6)

\begin{aligned} \int^{T}_{0}\bigl\vert g\bigl(t,x_{1}(t- \sigma)\bigr)\bigr\vert \,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(\vert \psi \vert _{\infty}+\varepsilon\bigr) \int^{T}_{0}\bigl\vert x_{1}(t)\bigr\vert ^{p-1}\,dt+2 \int ^{T}_{0}\bigl\vert g_{\varepsilon}(t)\bigr\vert \,dt \\ &\leq2\bigl(\vert \psi \vert _{\infty}+\varepsilon\bigr)TM_{1}^{p-1}+2 \sqrt{T} \vert g_{\varepsilon} \vert _{2}. \end{aligned}
(3.15)

By the second equations of (3.3) and (3.15), we obtain

\begin{aligned} &\int^{T}_{0}\bigl\vert x_{2}'(t) \bigr\vert \,dt \\ &\quad \leq\lambda \int^{T}_{0}\bigl\vert f\bigl(x_{1}(t) \bigr)\bigr\vert \bigl\vert x_{1}'(t)\bigr\vert \, dt+\lambda \int^{T}_{0}\bigl\vert g\bigl(t,x_{1}(t- \sigma)\bigr)\bigr\vert \,dt+\lambda \int^{T}_{0}\bigl\vert e(t)\bigr\vert \,dt \\ &\quad \leq\lambda \vert f\vert _{M_{1}}T^{\frac{1}{q}} \biggl( \int^{T}_{0}\bigl\vert x_{1}'(t) \bigr\vert ^{p}\,dt \biggr)^{\frac{1}{p}} +\lambda\bigl(2\bigl( \vert \psi \vert _{\infty}+\varepsilon\bigr)TM_{1}^{p-1}+2 \sqrt{T} \vert g_{\varepsilon} \vert _{2}\bigr)+\lambda\sqrt{T} \vert e\vert _{2} \\ &\quad \leq\lambda \vert f\vert _{M_{1}}T^{\frac{1}{q}} \bigl(M_{1}' \bigr)^{\frac{1}{p}} +\lambda\bigl(2\bigl( \vert \psi \vert _{\infty}+\varepsilon\bigr)TM_{1}^{p-1}+2 \sqrt{T} \vert g_{\varepsilon} \vert _{2}\bigr)+\lambda\sqrt{T} \vert e\vert _{2} \\ &\quad :=\lambda M_{2}', \end{aligned}
(3.16)

where $$\vert f\vert _{M_{1}}=\max_{0< x_{1}\leq M_{1}}\vert f(x_{1}(t))\vert$$. By the first equation of (3.3), we have

$$\int^{T}_{0}\bigl\vert x_{2}(s)\bigr\vert ^{q-2}x_{2}(s)\,ds=0,$$

which implies that there is a constant $$t_{2}\in[0,T]$$ such that $$x_{2}(t_{2})=0$$, so

$$\vert x_{2}\vert _{\infty}\leq \frac{1}{2} \int^{t_{2}}_{0}\bigl\vert x_{2}'(s)\bigr| \,ds\leq\frac{1}{2} \int ^{T}_{0}\bigl\vert x_{2}'(s)\bigr| \,ds\leq\frac{\lambda}{2}M_{2}':=\lambda M_{2}.$$
(3.17)

On the other hand, it follows from (3.4) that

$$\bigl(\varphi_{p}\bigl(x_{1}'(t+ \sigma)\bigr)\bigr)'+\lambda^{p}\bigl(f \bigl(x_{1}(t+\sigma)\bigr)x_{1}'(t+\sigma )+g \bigl(t+\sigma,x_{1}(t)\bigr)\bigr)=\lambda^{p} e(t+\sigma).$$
(3.18)

Namely,

\begin{aligned} & \bigl(\varphi_{p}\bigl(x_{1}'(t+ \sigma)\bigr)\bigr)'+\lambda^{p}f\bigl(x_{1}(t+ \sigma)\bigr)x_{1}'(t+\sigma ) \\ &\qquad{} +\lambda^{p}g_{0} \bigl(x_{1}(t)\bigr)+g_{1}\bigl(t+\sigma,x_{1}(t) \bigr)=\lambda^{p} e(t+\sigma). \end{aligned}
(3.19)

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

\begin{aligned}[b] &\bigl(\varphi_{p}\bigl(x_{1}'(t+ \sigma)\bigr)\bigr)'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={}& \lambda^{p} \int^{t}_{\tau }g_{0}\bigl(x_{1}(s) \bigr)x_{1}'(s)\,ds \\ ={}&{-} \int^{t}_{\tau}\bigl(\varphi_{p} \bigl(x_{1}'(s+\sigma)\bigr)\bigr)'x_{1}'(s) \,ds-\lambda^{p} \int ^{t}_{\tau}f\bigl(x_{1}(s+\sigma) \bigr)x_{1}'(s+\sigma)x_{1}'(s)\,ds \\ &{}-\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 (3.14), (3.15), (3.16), (3.17), and (3.18), we have

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

We have

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

where $$g_{M_{1}}=\max_{0\leq x\leq M_{1}}\vert g_{1}(t,x)\vert \in L^{2}(0,T)$$ is as in (H3). We have

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

From these inequalities we can derive from (3.21) that

$$\biggl\vert \int^{x_{1}(t)}_{x_{1}(\tau)}g_{0}(u)\,du\biggr\vert \leq M_{3}',$$
(3.22)

for some constant $$M_{3}'$$ which is independent on λ, x, and t. In view of the strong force condition (H4), we know that there exists a constant $$M_{3}>0$$ such that

$$x_{1}(t)\geq M_{3},\quad \forall t\in[\tau,T].$$
(3.23)

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

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

$$\Omega=\bigl\{ x=(x_{1},x_{2})^{\top}: E_{1}\leq \vert x_{1}\vert _{\infty}\leq E_{2}, \vert x_{2}\vert _{\infty}\leq E_{3}, \forall t\in[0,T]\bigr\} ,$$

where $$0< E_{1}<\min(M_{3}, D_{1})$$, $$E_{2}>\max(M_{1}, D_{2})$$, $$E_{3}>M_{2}$$. $$\Omega_{2}=\{x:x\in\partial\Omega\cap \operatorname {Ker}L\}$$ 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 2.1 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 $$\forall (\mu,x)\in(0,1)\times(\partial\Omega\cap \operatorname {Ker}L)$$,

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

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)\vert x_{2}\vert ^{p-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 2.1 is satisfied. By applying Lemma 2.1, we conclude that the equation $$Lx=Nx$$ has a solution $$x=(x_{1},x_{2})^{\top}$$ on $$\bar{\Omega}\cap D(L)$$, i.e., (2.1) has an T-periodic solution $$x_{1}(t)$$. □

Finally, we present an example to illustrate our result.

### Example 3.1

Consider the p-Laplacian Liénard type differential equation with singularity and deviating argument:

$$\bigl(\varphi_{p}\bigl(x'(t)\bigr) \bigr)'+f\bigl(x(t)\bigr)x'(t)+\frac{1}{5}( \cos2t+2)x^{3}(t-\sigma )-\frac{1}{x^{\kappa}(t-\sigma)}=\sin 2t,$$
(3.24)

where $$\kappa\geq1$$ and $$p=4$$, f is a continuous function, σ is a constant, and $$0\leq\sigma< T$$.

It is clear that $$T=\pi$$, $$g(t,x)=\frac{1}{5}(\cos2t+2)x^{3}(t-\sigma)-\frac{1}{x^{\kappa}(t-\sigma)}$$, $$\psi(t)=\frac{1}{5}(\cos2t+2)$$. It is obvious that (H1)-(H4) hold. Now we consider the assumption (H5). Since $$\vert \psi \vert _{\infty}\leq\frac{3}{5}$$, we have

\begin{aligned} \biggl(\frac{T}{\pi_{p}} \biggr)^{p}\vert \psi \vert _{\infty}= \biggl(\frac{T}{\frac{2\pi (p-1)^{1/p}}{p\sin(\pi/p)}} \biggr)^{p}\vert \psi \vert _{\infty}\leq \biggl(\frac{\pi}{\frac{2\pi(4-1)^{1/4}}{4\sin{\pi/4}}} \biggr)^{4}\times \frac{3}{5}=\frac{4}{5}< 1. \end{aligned}

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