Firstly, we embed (1.1) into the following equation family with a parameter \(\lambda \in (0,1]\):
$$ { } \bigl(\phi_{p}\bigl(x'(t)\bigr) \bigr)'+\lambda f\bigl(t,x(t)\bigr)x'(t)+\lambda g \bigl(t,x(t)\bigr)=\lambda e(t). $$
(2.1)
By applications of Theorem 3.1 in [18], we obtain the following result.
Lemma 2.1
Assume that there exist positive constants
\(E_{1}\), \(E_{2}\), \(E_{3}\)
and
\(E_{1}< E_{2}\)
such that the following conditions hold:
-
(1)
We have for each possible periodic solution
x
to Eq. (2.1) that
\(E_{1}< x(t)< E_{2}\), for all
\(t\in [0,T]\)
and
\(\Vert x' \Vert < E_{3}\), here
\(\Vert x' \Vert :=\max_{t\in [0,T]}\vert x'(t) \vert \).
-
(2)
Each possible solution
C
to the equation
$$ g(t,C)-\frac{1}{T} \int^{T}_{0} e(t)\,dt=0 $$
satisfies
\(E_{1}< C< E_{2}\).
-
(3)
We have
$$ \biggl( g(t,E_{1})-\frac{1}{T} \int^{T}_{0} e(t)\,dt \biggr) \biggl( g(t,E_{2})- \frac{1}{T} \int^{T}_{0} e(t)\,dt \biggr) < 0. $$
Then Eq. (1.1) has at least one
T-periodic solution.
Proof of Theorem 1.1
Proof of Theorem 1.1
Firstly, we claim that there exists a point \(\xi \in [0,T]\) such that
$$ { } d_{1}\leq x(\xi )\leq d_{2}. $$
(2.2)
In view of \(\int^{T}_{0}x'(t)\,dt=0\), we know that there exist two points \(t_{1},~t_{2}\in [0,T]\) such that
$$ x'(t_{1})\geq 0\quad \mbox{and}\quad x'(t_{2}) \leq 0. $$
Hence, we have
$$ \phi_{p}\bigl(x'(t_{1})\bigr)\geq 0\quad \mbox{and}\quad \phi_{p}\bigl(x'(t_{2})\bigr)\leq 0. $$
Let \(t_{3}\), \(t_{4}\in [0,T]\) be, respectively, a global maximum and minimum point of \(\phi_{p}(x'(t))\); clearly, we deduce
$$\begin{aligned}& \phi_{p}\bigl(x'(t_{3}) \bigr)\geq 0, \qquad \bigl(\phi_{p}\bigl(x'(t_{3}) \bigr)\bigr)'=0. \end{aligned}$$
(2.3)
$$\begin{aligned}& \phi_{p}\bigl(x'(t_{4}) \bigr)\leq 0, \qquad \bigl(\phi_{p}\bigl(x'(t_{4}) \bigr)\bigr)'=0. \end{aligned}$$
(2.4)
From condition \((H_{5})\), we can see that the friction term f will not change sign for \((t,x)\in [0,T]\times (0,+\infty )\). Without loss of generality, suppose \(f(t,x)>0\) for \((t,x)\in [0,T]\times (0,+\infty )\) and upon substitution of Eq. (2.3) into Eq. (2.1), we obtain
$$ -\lambda g\bigl(t_{3},x(t_{3})\bigr)+\lambda e(t_{3})=\lambda f\bigl(t_{3},x(t_{3}) \bigr)x'(t _{3}). $$
Since \(\phi_{p}(x'(t_{3}))=\vert x'(t_{3}) \vert ^{p-2}x'(t_{3})\geq 0\), we know \(x'(t_{3})\geq 0\). So we get
$$ g\bigl(t_{3},x(t_{3})\bigr)-e(t_{3})\leq 0. $$
From condition \((H_{1})\), we know that
$$ { } x(t_{3})\leq d_{2}. $$
(2.5)
Similarly, from Eq. (2.4), we see that
$$ g\bigl(t_{4},x(t_{4})\bigr)-e(t_{4})\geq 0, $$
and again by condition \((H_{1})\),
$$ { } x(t_{4})\geq d_{1}. $$
(2.6)
\(x(t)\) is a continuous function in \((0,+\infty )\), from Eqs. (2.5) and (2.6), we get Eq. (2.2). Then we have
$$\begin{aligned} \Vert x \Vert &=\max_{t\in [0,T]}\bigl\vert x(t) \bigr\vert =\max_{t\in [\xi ,\xi +T]} \bigl\vert x(t) \bigr\vert \\ &=\max_{t\in [\xi ,\xi +T]}\biggl\vert \frac{1}{2} \bigl( x(t)+x(t-T) \bigr) \biggr\vert \\ &=\max_{t\in [\xi ,\xi +T]}\biggl\vert \frac{1}{2} \biggl( \biggl( x(\xi )+ \int ^{t}_{\xi }x'(s)\,ds \biggr) + \biggl( x(\xi )- \int^{\xi }_{t-T}x'(s)\,ds \biggr) \biggr) \biggr\vert \\ & \leq \max_{t\in [\xi ,\xi +T]} \biggl\{ d_{2}+\frac{1}{2} \biggl( \int ^{t}_{\xi }\bigl\vert x'(s) \bigr\vert \,ds+ \int^{\xi }_{t-T}\bigl\vert x'(s) \bigr\vert \,ds \biggr) \biggr\} \\ &\leq d_{2}+\frac{1}{2} \int^{T}_{0}\bigl\vert x'(s) \bigr\vert \,ds. \end{aligned}$$
(2.7)
Multiplying both sides of Eq. (2.1) by \(x(t)\) and integrating over the interval \([0,T]\), it is clear that
$$\begin{aligned} &\int^{T}_{0}\bigl(\phi_{p} \bigl(x'(t)\bigr)\bigr)'x(t)\,dt+\lambda \int^{T}_{0}f\bigl(t,x(t)\bigr)x'(t)x(t)\,dt+ \lambda \int^{T}_{0}g\bigl(t,x(t)\bigr)x(t)\,dt \\ &\quad =\lambda \int^{T}_{0}e(t)x(t)\,dt. \end{aligned}$$
(2.8)
Substituting \(\int^{T}_{0}(\phi_{p}(x'(t)))'x(t)\,dt=-\int^{T}_{0}\vert x'(t) \vert ^{p}\,dt\) into Eq. (2.8), we arrive at
$$\begin{aligned}& \int^{T}_{0}\bigl\vert x'(t) \bigr\vert ^{p}\,dt \\& \quad = \lambda \int^{T}_{0}f\bigl(t,x(t)\bigr)x'(t)x(t)\,dt+ \lambda \int^{T}_{0}g\bigl(t,x(t)\bigr)x(t)\,dt+\lambda \int^{T}_{0}e(t)x(t)\,dt \\& \quad \leq \int^{T}_{0}\bigl\vert f\bigl(t,x(t)\bigr) \bigr\vert \bigl\vert x'(t) \bigr\vert \bigl\vert x(t) \bigr\vert \,dt+ \int^{T}_{0}\bigl\vert g\bigl(t,x(t)\bigr) \bigr\vert \bigl\vert x(t) \bigr\vert \,dt+ \int^{T}_{0}\bigl\vert e(t) \bigr\vert \bigl\vert x(t) \bigr\vert \,dt \\& \quad \leq \Vert x \Vert \int^{T}_{0}\bigl\vert f\bigl(t,x(t)\bigr) \bigr\vert \bigl\vert x'(t) \bigr\vert \,dt+\Vert x \Vert \int^{T}_{0}\bigl\vert g\bigl(t,x(t)\bigr) \bigr\vert \,dt+ \Vert x \Vert \int^{T}_{0}\bigl\vert e(t) \bigr\vert \,dt. \end{aligned}$$
(2.9)
From condition \((H_{2})\), Eq. (2.9) and the Hölder inequality, we can observe that
$$\begin{aligned}& \int^{T}_{0}\bigl\vert x'(t) \bigr\vert ^{p}\,dt \\& \quad \leq m\Vert x \Vert \int^{T}_{0}\bigl\vert x(t) \bigr\vert ^{p-2}\bigl\vert x'(t) \bigr\vert \,dt+n \Vert x \Vert \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt+\Vert x \Vert \int^{T}_{0}\bigl\vert g\bigl(t,x(t)\bigr) \bigr\vert \,dt \\& \quad \quad {}+\Vert x \Vert T^{ \frac{1}{2}} \biggl( \int^{T}_{0}\bigl\vert e(t) \bigr\vert ^{2}\,dt \biggr) ^{\frac{1}{2}} \\& \quad \leq m\Vert x \Vert ^{p-1} \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt+n\Vert x \Vert \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt \\& \quad \quad {}+ \Vert x \Vert \int^{T}_{0}\bigl\vert g\bigl(t,x(t)\bigr) \bigr\vert \,dt+T^{\frac{1}{2}}\Vert x \Vert \Vert e \Vert _{2}, \end{aligned}$$
(2.10)
where \(\Vert e \Vert := ( \int^{T}_{0}\vert e(t) \vert ^{2}\,dt ) ^{\frac{1}{2}}\).
Integrating over the interval \([0,T]\) for Eq. (2.1), we conclude that
$$ { } \int^{T}_{0}f\bigl(t,x(t)\bigr)x'(t)\,dt+ \int^{T}_{0}g\bigl(t,x(t)\bigr)\,dt= \int^{T}_{0}e(t)\,dt, $$
(2.11)
From Eq. (2.11), conditions \((H_{2})\) and \((H_{3})\), we have
$$\begin{aligned} \int^{T}_{0}\bigl\vert g\bigl(t,x(t)\bigr) \bigr\vert \,dt& =\int_{g(t,x(t))\geq 0}g\bigl(t,x(t)\bigr)\,dt- \int_{g(t,x(t))\leq 0}g\bigl(t,x(t)\bigr)\,dt \\ & = 2\int_{g(t,x(t))\geq 0}g^{+}\bigl(t,x(t)\bigr)\,dt+ \int^{T}_{0}f\bigl(t,x(t)\bigr)x'(t)\,dt- \int^{T}_{0}e(t)\,dt \\ & \leq 2a\int^{T}_{0}x^{p-1}(t)\,dt+2bT+ \int^{T}_{0}\bigl\vert f\bigl(t,x(t)\bigr) \bigr\vert \bigl\vert x'(t) \bigr\vert \,dt+ \int^{T}_{0}\bigl\vert e(t) \bigr\vert \,dt \\ & \leq 2a\Vert x \Vert ^{p-1}T+2bT+m\Vert x \Vert ^{p-2} \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt+n \int^{T} _{0}\bigl\vert x'(t) \bigr\vert \,dt \\ &\quad {}+T^{\frac{1}{2}} \biggl(\int^{T}_{0}\bigl\vert e(t) \bigr\vert ^{2}\,dt \biggr) ^{\frac{1}{2}}, \end{aligned}$$
(2.12)
where \(g^{+}(t,x):=\max \{0,g(t,x)\}\). Substituting Eqs. (2.12) and (2.7) into (2.10), we deduce
$$\begin{aligned} &\int^{T}_{0}\bigl\vert x'(t) \bigr\vert ^{p}\,dt \\ &\quad \leq 2m\Vert x \Vert ^{p-1} \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt +2n \Vert x \Vert \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt+2a\Vert x \Vert ^{p}T+2bT\Vert x \Vert +2T^{\frac{1}{2}}\Vert x \Vert \Vert e \Vert _{2} \\ &\quad \leq 2m \biggl( d_{2}+\frac{1}{2} \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt \biggr) ^{p-1} \int ^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt \\ &\quad \quad {}+2n \biggl( d_{2}+\frac{1}{2} \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt \biggr) \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt \\ &\quad \quad {}+2aT \biggl( d_{2}+\frac{1}{2} \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt \biggr) ^{p}+\bigl(2bT+2T ^{\frac{1}{2}}\bigr) \biggl( d_{2}+\frac{1}{2} \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt \biggr) \\ & \quad \leq \frac{1}{2^{p-2}}m \biggl( 1+\frac{2d_{2}}{\int^{T}_{0}\vert x'(t) \vert \,dt} \biggr) ^{p-1} \biggl( \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt \biggr) ^{p} \\ &\quad \quad {} +\frac{1}{2^{p-1}}aT \biggl( 1+ \frac{2d_{2}}{\int^{T}_{0}\vert x'(t) \vert \,dt} \biggr) ^{p} \biggl( \int^{T} _{0}\bigl\vert x'(t) \bigr\vert \,dt \biggr) ^{p} \\ &\quad \quad {}+n \biggl( \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt \biggr) ^{2}+\bigl(bT+T^{\frac{1}{2}}\bigr) \int ^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt+2d_{2}\bigl(n+bT+T^{\frac{1}{2}}\bigr). \end{aligned}$$
(2.13)
Next, we introduce classical elementary inequality (see (3.10) in [7]), there exists a \(\delta (p)>0\) which is dependent on p only,
$$ (1+x)^{p}\leq 1+(1+p)x, \quad \mbox{for } x\in \bigl[0,\delta (p)\bigr]. $$
(2.14)
Then we consider the following two cases:
Case 1. If \(\frac{2d_{2}}{\int^{T}_{0}\vert x_{1}'(t) \vert \,dt}>\delta (p)\), then it is obvious that
$$ \int^{T}_{0}\bigl\vert x_{1}'(t) \bigr\vert \,dt< \frac{2d_{2} }{\delta (p)}. $$
From Eq. (2.7), we deduce
$$\begin{aligned} \Vert x \Vert & \leq d_{2}+\frac{1}{2} \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt \end{aligned}$$
(2.15)
$$\begin{aligned} & \leq d_{2}+\frac{2d_{2}}{\delta (p)}:=M_{1}''. \end{aligned}$$
(2.16)
Case 2. If \(\frac{2d_{2}}{\int^{T}_{0}\vert x_{1}'(t) \vert \,dt}\leq \delta (p)\), from Eq. (2.14), we obtain
$$ { } \biggl( 1+\frac{2d_{2}}{\int^{T}_{0}\vert x'(t) \vert \,dt} \biggr) ^{p-1}\leq 1+ \frac{2d _{2} p}{\int^{T}_{0}\vert x'(t) \vert \,dt} $$
(2.17)
and
$$ { } \biggl( 1+\frac{2d_{2}}{\int^{T}_{0}\vert x'(t) \vert \,dt} \biggr) ^{p}\leq 1+ \frac{2d _{2} (p+1)}{\int^{T}_{0}\vert x'(t) \vert \,dt}. $$
(2.18)
Substituting Eqs. (2.17) and (2.18) into (2.13), we have
$$\begin{aligned} \int^{T}_{0}\bigl\vert x'(t) \bigr\vert ^{p}\,dt & \leq \frac{1}{2^{p-2}}m \biggl( 1+\frac{2d_{2} p}{\int^{T}_{0}\vert x'(t) \vert \,dt} \biggr) \biggl(\int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt \biggr) ^{p} \\ &\quad {}+\frac{1}{2^{p-1}}aT \biggl( 1+\frac{2d_{2}(p+1)}{\int^{T}_{0}\vert x'(t) \vert \,dt} \biggr) \biggl( \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt \biggr) ^{p} \\ &\quad {}+n \biggl( \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt \biggr) ^{2}+\bigl(bT+T^{\frac{1}{2}}\bigr) \int ^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt +N_{1} \\ & = \frac{m}{2^{p-2}} \biggl( \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt \biggr) ^{p}+\frac{md _{2} p}{2^{p-3}} \biggl( \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt \biggr) ^{p-1} \\ &\quad {}+ \frac{aT}{2^{p-1}} \biggl( \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt \biggr) ^{p} \\ &\quad {}+\frac{aTd_{2}(p+1)}{2^{p-2}} \biggl( \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt \biggr) ^{p-1}+n \biggl( \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt \biggr) ^{2} \\ &\quad {}+\bigl(bT+T^{\frac{1}{2}}\bigr) \int^{T} _{0}\bigl\vert x'(t) \bigr\vert \,dt+N_{1} 1\\ & = \frac{1}{2^{p-2}}(2aT+m) \biggl( \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt \biggr) ^{p} \\ &\quad {}+ \frac{1}{2^{p-3}}\bigl(2aTd_{2}(p+1)+md_{2} p\bigr) \biggl( \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt \biggr) ^{p-1} \\ &\quad {}+n \biggl( \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt \biggr) ^{2}+\bigl(bT+T^{\frac{1}{2}}\bigr) \int ^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt +N_{1}, \end{aligned}$$
(2.19)
where \(N_{1}:=2d_{2}(n+bT+T^{\frac{1}{2}})\). Applying the Hölder inequality, it is easy to verify that
$$\begin{aligned} \int^{T}_{0}\bigl\vert x'(t) \bigr\vert ^{p}\,dt& \leq \frac{1}{2^{p-2}}(2aT+m)T^{\frac{p}{q}} \int^{T}_{0}\bigl\vert x'(t) \bigr\vert ^{p}\,dt \\ &\quad {}+\frac{1}{2^{p-3}}\bigl(2aTd_{2}(p+1)+md_{2} p\bigr)T ^{\frac{p-1}{q}} \biggl( \int^{T}_{0}\bigl\vert x'(t) \bigr\vert ^{p}\,dt \biggr) ^{\frac{p-1}{p}} \\ &\quad {}+nT^{\frac{2}{q}} \biggl( \int^{T}_{0}\bigl\vert x'(t) \bigr\vert ^{p}\,dt \biggr) ^{ \frac{2}{p}}+\bigl(bT+T^{\frac{1}{2}} \bigr)T^{\frac{1}{q}} \biggl( \int^{T}_{0}\bigl\vert x'(t) \bigr\vert ^{p}\,dt \biggr) ^{\frac{1}{p}} +N_{1}. \end{aligned}$$
Case (I). If \(p>2\) and \(\frac{1}{2^{p-2}}(\frac{aT}{2}+m)T^{ \frac{p}{q}}<1\), it is easy to see that there exists a positive \(M_{1}'\) (independent of λ) such that
$$ { } \int^{T}_{0}\bigl\vert x'(t) \bigr\vert ^{p}\,dt\leq M_{1}'. $$
(2.20)
Substituting Eq. (2.20) into (2.7), and using the Hölder inequality, we see that
$$ { } \Vert x \Vert \leq d_{2}+\frac{1}{2}T^{\frac{1}{q}} \biggl( \int^{T}_{0}\bigl\vert x'(t) \bigr\vert ^{p}\,dt \biggr) ^{\frac{1}{p}}\leq d_{2}+ \frac{1}{2}T^{\frac{1}{q}} \bigl( M_{1}' \bigr) ^{\frac{1}{p}}:=M_{1}'''. $$
Take \(M:=\max \{M_{1}'',M_{1}'''\}\), we arrive at
$$ { } \Vert x \Vert \leq M_{1}. $$
(2.21)
In view of \(x(0)=x(T)\), there exists a point \(t_{0}\in [0,T]\) such that \(x'(t_{0})=0\), while \(\phi_{p}(0)=0\). Therefore, from Eqs. (2.12), (2.20), (2.21) and condition \((H_{2})\), we have
$$\begin{aligned} \bigl\vert \phi_{p} \bigl(x'(t)\bigr) \bigr\vert & = \biggl\vert \int^{t}_{t_{0}}\bigl(\phi_{p} \bigl(x'(s)\bigr)\bigr)'\,ds\biggr\vert \\ & \leq \lambda \biggl( \int^{T}_{0}\bigl\vert f\bigl(t,x(t)\bigr) \bigr\vert \bigl\vert x'(t) \bigr\vert \,dt+ \int^{T}_{0}\bigl\vert g\bigl(t,x(t)\bigr) \bigr\vert \,dt+ \int^{T}_{0}\bigl\vert e(t) \bigr\vert \,dt \biggr) \\ & \leq 2m\Vert x \Vert ^{p-2} \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt+2n \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt+2a \Vert x \Vert ^{p-1}T+2bT+2T^{\frac{1}{2}}\Vert e \Vert _{2} \\ & \leq 2mM_{1}^{p-2}T^{\frac{1}{q}}\bigl(M_{1}' \bigr)^{\frac{1}{p}}+2nT^{ \frac{1}{q}}\bigl(M_{1}' \bigr)^{\frac{1}{p}} +2aM_{1}^{p-1}T+2bT+2T^{ \frac{1}{2}} \Vert e \Vert _{2} \\ &:=M_{2}'. \end{aligned}$$
(2.22)
Besides, we claim that there exists a positive constant \(M_{2}>M_{2}'+1\) such that, for all \(t\in \mathbb{R}\)
$$ { } \bigl\Vert x' \bigr\Vert \leq M_{2}. $$
(2.23)
In fact, if \(x'\) is not bounded, there exists a positive constant \(M_{2}''\) such that
$$ \bigl\Vert x' \bigr\Vert >M_{2}''. $$
Then we can get
$$ \bigl\Vert \phi_{p}\bigl(x'\bigr) \bigr\Vert =\bigl\Vert x' \bigr\Vert ^{p-1}\geq \bigl(M_{2}'' \bigr)^{p-1}, $$
which is a contradiction. So Eq. (2.23) holds.
Case (II). If \(p=2\) and \(\frac{aT^{2}}{2}+(m+n)T<1\), it is easy to see that there exists a positive \(M_{1}'\) (independent of λ) such that
$$ { } \int^{T}_{0}\bigl\vert x'(t) \bigr\vert ^{2}\,dt\leq M_{1}'. $$
Similarly, we get \(\Vert x \Vert \leq M_{1}\), \(\Vert x' \Vert \leq M_{2}\).
On the other hand, it follows from Eq. (2.1) and \(g(t,x)=g _{0}(x)+g_{1}(t,x)\) that
$$ { } \bigl(\phi_{p}\bigl(x'(t)\bigr) \bigr)'+\lambda f\bigl(t,x(t)\bigr)x'(t))+\lambda \bigl(g_{0}\bigl(x(t)\bigr)+g _{1}\bigl(t,x(t)\bigr)\bigr)= \lambda e(t). $$
(2.24)
Let \(t\in [0,\xi ]\) be as in Eq. (2.7), for any \(\xi \leq t \leq T\). From Eqs. (2.7) and (2.23), we conclude that
Next, we will show that for any \(t\in [\xi ,T]\), there exists a constant \(D_{1}\in (0,d_{1})\), such that each positive periodic solution of (1.1) satisfies
In fact, multiplying both sides of Eq. (2.24) by \(x'(t)\) and integrating on \([\xi ,t]\), we get
$$\begin{aligned} \lambda \int^{x(t)}_{x(\xi )}g_{0}(u)\,du& = \lambda \int^{t}_{\xi }g_{0}\bigl(x(s) \bigr)x'(s)\,ds \\ & = - \int^{t}_{\xi }\bigl(\phi \bigl(x'(s) \bigr)\bigr)'x'(s)\,ds-\lambda \int^{t}_{\xi }f\bigl(s,x(s)\bigr) \bigl(x'(s)\bigr)^{2}\,ds \\ &\quad {}-\lambda \int^{t}_{\xi }g_{1}\bigl(s,x(s) \bigr)x'(s)\,ds+\lambda \int^{t}_{ \xi }e(s)x'(s)\,ds. \end{aligned}$$
Furthermore, we get
$$\begin{aligned} \lambda \biggl\vert \int^{x(t)}_{x(\xi )}g_{0}(u)\,du\biggr\vert & \leq \biggl\vert \int ^{t}_{\xi }\bigl(\phi \bigl(x'(s) \bigr)\bigr)'x'(s)\,ds\biggr\vert +\lambda \biggl\vert \int^{t}_{ \xi }f\bigl(s,x(s)\bigr) \bigl(x'(s)\bigr)^{2}\,ds\biggr\vert \\ &\quad {}+\lambda \biggl\vert \int^{t}_{\xi }g_{1}\bigl(s,x(s) \bigr)x'(s)\,ds\biggr\vert +\lambda \biggl\vert \int^{t}_{\xi }e(s)x'(s)\,ds\biggr\vert . \end{aligned}$$
(2.25)
By Eqs. (2.1), (2.22) and (2.23), we arrive at
$$\begin{aligned} \biggl\vert \int^{t}_{\xi }\bigl(\phi_{p} \bigl(x'(t)\bigr)\bigr)'x'(s)\,ds\biggr\vert & \leq \bigl\Vert x' \bigr\Vert \int ^{T}_{0}\bigl\vert \bigl(\phi \bigl(x'(s)\bigr)\bigr)' \bigr\vert \,ds \\ & \leq \lambda \bigl\Vert x' \bigr\Vert \biggl( \int^{T}_{0}\bigl\vert f\bigl(s,x'(s) \bigr) \bigr\vert \,ds+ \int^{T}_{0}\bigl\vert g\bigl(s,x(s)\bigr) \bigr\vert \,ds+ \int^{T}_{0}\bigl\vert e(s) \bigr\vert \,ds \biggr) \\ & \leq 2\lambda M_{2}M_{2}' . \end{aligned}$$
Moreover, from Eq. (2.25), we deduce
$$\begin{aligned} &\biggl\vert \int^{t}_{\xi }f\bigl(s,x(s)\bigr) \bigl(x'(s)\bigr)^{2}\biggr\vert \leq \bigl\Vert x' \bigr\Vert ^{2}T\bigl(m\Vert x \Vert ^{p-2}+n\bigr)\leq M_{2}^{2}T\bigl(mM_{1}^{p-2}+n \bigr), \\ &\biggl\vert \int^{t}_{\xi }g_{1}\bigl(s,x(s) \bigr)x'(s)\,ds\biggr\vert \leq \Vert x \Vert \int^{T} _{0}\bigl\vert g_{1} \bigl(s,x(s)\bigr) \bigr\vert \,ds\leq M_{2}\sqrt{T} \Vert g_{M_{1}} \Vert _{2}, \\ & \biggl\vert \int^{t}_{\xi }e(s)x'(s)\,dt\biggr\vert \leq M_{2}\sqrt{T} \Vert e \Vert _{2}, \end{aligned}$$
where \(\Vert g_{M_{1}} \Vert :=\max_{0< x\leq M_{1}}\vert g_{1}(t,x) \vert \in L ^{2}(0,T)\). Form these inequalities we can derive from equation (2.25) that
$$ { } \biggl\vert \int^{x(t)}_{x(\xi )}g_{0}(u)\,du\biggr\vert \leq M_{2}\bigl(2M_{2}'+M _{2}T \bigl(mM_{1}^{p-2}+n\bigr)+\sqrt{T} \Vert g_{M_{1}} \Vert _{2}+\sqrt{T} \Vert e \Vert _{2}\bigr):=M _{3}. $$
(2.26)
In view of the repulsive condition \((H_{4})\) and \(x(\xi )\geq d_{1}\), there exists \(D_{1}\in (0,d_{1})\) such that
$$ \biggl\vert \int^{D_{1}}_{d_{1}}g_{0}(u)\,du\biggr\vert >M_{3}. $$
Thus, there exists a point \(\eta \in [\xi ,T]\) such that \(x(\eta ) \leq D_{1}\), then
$$ \biggl\vert \int^{x(\eta )}_{x(\xi )}g_{0}(u)\,du\biggr\vert \geq \biggl\vert \int^{D _{1}}_{d_{1}}g_{0}(u)\,du\biggr\vert >M_{3}, $$
which contradicts Eq. (2.26). Therefore, we can obtain that
$$ x(t)\geq D_{1},\quad \mbox{for all }t\in [\xi ,T]. $$
Similarly, we can consider \(t\in [0,\xi ]\).
Let \(E_{1}<\min \{D_{1},M_{3}\}\), \(E_{2}>\max \{d_{2}, M_{1}\}\), \(E_{3}>M_{2}\) are constants, from Eqs. (2.7), (2.21) and (2.23), we see that periodic solution x to Eq. (2.1) satisfies
$$ E_{1}< x(t)< E_{2}, \qquad \bigl\Vert x' \bigr\Vert < E_{3}. $$
Then condition (1) of Lemma 2.1 is satisfied. For a possible solution C in the equation
$$ g(t,C)-\frac{1}{T} \int^{T}_{0}e(t)\,dt=0, $$
satisfies \(E_{1}< C<E_{2}\). Therefore, condition (2) of Lemma 2.1 holds. Finally, we consider that condition (3) of Lemma 2.1 is also satisfied. In fact, from \((H_{1})\), we have
$$ g(t,E_{1})-\frac{1}{T} \int^{T}_{0}e(t)\,dt< 0, $$
and
$$ g(t,E_{2})-\frac{1}{T} \int^{T}_{0}e(t)\,dt>0, $$
So condition (3) is also satisfied. Applying Lemma 2.1, we see that Eq. (1.1) has at least one positive periodic solution. □
Proof of Theorem 1.5
Proof of Theorem 1.5
Let \(\underline{t}\), t̅, respectively, the global minimum and maximum points \(x(t)\) on \([0,T]\); then \(x'(\underline{t})=0\) and \(x'(\overline{t})=0\), and we claim that
$$ { } \bigl(\phi_{p}\bigl(x'(\overline{t}) \bigr)\bigr)'\leq 0. $$
(2.27)
In fact, if Eq. (2.27) does not hold, then \((\phi_{p}(x'( \overline{t})))'>0\) and there exists \(\varepsilon >0\) such that \((\phi_{p}(x'(t)))'>0\) for \(t\in (\overline{t}-\varepsilon , \overline{t}+\varepsilon )\). Therefore \(\phi_{p}(x'(t))\) is strictly increasing for \(t\in (\overline{t}-\varepsilon ,\overline{t}+\varepsilon )\). Then we know that \(x'(t)\) is strictly increasing for \(t\in ( \overline{t}-\varepsilon ,\overline{t}+\varepsilon )\). This contradicts the definition of t̅. Thus, equation (2.27) is true. From Eqs. (2.1) and (2.27), we get
$$ { } g\bigl(\overline{t}+x(\overline{t})\bigr)-e(\overline{t})\leq 0. $$
(2.28)
Similarly, we deduce
$$ g\bigl(\underline{t}+x(\underline{t})\bigr)-e(\underline{t})\geq 0. $$
(2.29)
By condition \((H_{6})\), Eqs. (2.28) and (2.29), we see that
$$ x(\underline{t})\geq d_{3},\quad \mbox{and}\quad x(\overline{t})\leq d_{4}. $$
In view of x being a continuous function, we see that there exists a point \(\xi^{*}\in (0,T)\) such that
$$ d_{3}\leq x\bigl(\xi^{*}\bigr)\leq d_{4}. $$
(2.30)
From Eq. (2.7), we have
$$ x(t)\leq d_{4}+\frac{1}{2} \int^{T}_{0}\bigl\vert x'(t) \bigr\vert \,dt. $$
We follow the same strategy and notation as in the proof of Theorem 1.1. From Eqs. (2.11), (2.12), condition \((H_{2})\) and \((H_{8})\), we obtain
$$\begin{aligned} &\int^{T}_{0}\bigl\vert g\bigl(t,x(t)\bigr) \bigr\vert \,dt \\ & \quad = \int_{g(t,x(t))\geq 0}g\bigl(t,x(t)\bigr)\,dt- \int_{g(t,x(t))\leq 0}g\bigl(t,x(t)\bigr)\,dt \\ & \quad = -2 \int_{g(t,x(t))\leq 0}g^{-}\bigl(t,x(t)\bigr)\,dt- \int^{T}_{0}f\bigl(t,x(t)\bigr)x'(t)\,dt+ \int^{T}_{0}e(t)\,dt \\ & \quad \leq 2\beta \int^{T}_{0}x^{p-1}(t)\,dt+2\gamma T+ \int^{T}_{0}\bigl\vert f\bigl(t,x(t)\bigr) \bigr\vert \bigl\vert x'(t) \bigr\vert \,dt+ \int^{T}_{0}\bigl\vert e(t) \bigr\vert \,dt \\ & \quad \leq 2\beta T \Vert x \Vert ^{p-1}+2\gamma T+ \int^{T}_{0}\bigl\vert f\bigl(t,x(t)\bigr) \bigr\vert \bigl\vert x'(t) \bigr\vert \,dt+ \int^{T}_{0}\bigl\vert e(t) \bigr\vert \,dt, \end{aligned}$$
(2.31)
where \(g^{-}(t,x):=\min \{g(t,x),0\}\). The remaining part of the proof is the same as that of Theorem 1.1. □