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.