In this section, we give some notations and lemmas which will be used in this paper. Let
$$C_{T}=\bigl\{ x\mid x\in C({R},{R}), x(t+T)\equiv x(t), \forall t\in {R}\bigr\} $$
with the norm
$$\vert \varphi \vert _{0}=\max_{t\in [0,T]} \bigl\vert \varphi (t) \bigr\vert , \quad \forall \varphi \in C_{T} $$
and
$$C_{T}^{1}=\bigl\{ x\mid x\in C^{1}({R},{R}), x(t+T)\equiv x(t), \forall t \in {R}\bigr\} $$
with the norm
$$\vert \varphi \vert _{\infty }=\max_{t\in [0,T]}{\bigl\{ \vert \varphi \vert _{0}, \bigl\vert \varphi ' \bigr\vert _{0}}\bigr\} , \quad \forall \varphi \in C_{T}^{1}. $$
Clearly, \(C_{T}\) and \(C_{T}^{1}\) are Banach spaces. For each \(\phi \in C_{T}\) with \(y\in L([0,T],{R})\), let
$$\phi _{+}(t)=\max \bigl\{ \phi (t),0\bigr\} , \quad\quad \phi _{-}(t)=\max \bigl\{ -\phi (t),0\bigr\} , \quad\quad \overline{\phi }= \frac{1}{T} \int _{0}^{T}\phi (s)\,ds. $$
Clearly, for \(t\in {R}\), \(\phi (t)=\phi _{+}(t)-\phi _{-}(t)\), \(\overline{ \phi }=\overline{\phi _{+}}-\overline{\phi _{-}}\).
Since p-Laplacian \((\phi _{p}(s'))\) (\(p\neq 2\)) in (1.1) is a nonlinear operator, the famous Mawhin’s continuation theorem [27] cannot be directly applied to (1.1). Fortunately, Manásevich and Mawhin [1] obtained the following continuation theorem for nonlinear systems with p-Laplacian-like operators.
Theorem 2.1
Assume that
Ω
is an open bounded set in
\(C_{T}\)
such that the following conditions hold.
-
(1)
For each
\(\lambda \in (0,1)\), the problem
$$\bigl(\phi _{p}\bigl(u'\bigr)\bigr)'= \lambda f\bigl(t,u,u'\bigr), \quad\quad u(0)=u(T), \quad\quad u'(0)=u'(T), $$
has no solution on
∂Ω.
-
(2)
The equation
$$\mathcal{F}(a)=\frac{1}{T} \int _{0}^{T}f(t,a,0)\,dt=0 $$
has no solution on
\(\partial \varOmega \cap {R}^{N}\).
-
(3)
The Brouwer degree
$$d_{B}\bigl(\mathcal{F},\varOmega \cap {R}^{N},0\bigr)\neq 0. $$
Then problem
$$\bigl(\phi _{p}\bigl(u'\bigr)\bigr)=f \bigl(t,u,u'\bigr), \quad\quad u(0)=u(T), \quad\quad u'(0)=u'(T), $$
has a solution in
Ω̅.
Lemma 2.1
([4])
Let
\(u\in C([0,\omega ],{R})\)
be an arbitrary absolutely continuous function with
\(u(0)=u(\omega )\). Then the inequality
$$\bigl(\max u(t)-\min u(t)\bigr)^{2}\leq \frac{\omega }{4} \int _{0}^{\omega } \bigl\vert u'(s) \bigr\vert ^{2}\,ds $$
holds. Throughout this paper, assume that
$$\overline{\alpha _{1}}, \overline{\alpha _{2}}>0. $$
Now, consider the equation
$$ \bigl(\phi _{p}\bigl(x'(t)\bigr) \bigr)'+ \lambda \bigl[f\bigl(x(t)\bigr)x'(t)+\alpha _{1}(t)g \bigl(x(t)\bigr)\bigr]= \lambda \frac{\alpha _{2}(t)}{x^{\mu }}, \quad \lambda \in (0,1]. $$
(2.1)
Let
$$\begin{aligned} \varOmega ={}&\biggl\{ x\in C_{T}^{1}: \bigl(\phi _{p}\bigl(x'(t)\bigr) \bigr)'+\lambda \bigl[f\bigl(x(t)\bigr)x'(t)+ \alpha _{1}(t)g\bigl(x(t) \bigr)\bigr]= \lambda \frac{\alpha _{2}(t)}{x^{\mu }}, \\&\lambda \in (0,1], x(t)>0, t\in [0,T] \biggr\} . \end{aligned}$$
Lemma 2.2
Assume that there exist positive constants
\(g_{L}\)
and
\(g_{M}\)
such that
$$g_{L}\leq g(u)\leq g_{M}, \quad \forall u\in {R}. $$
Furthermore, assume
\(\overline{(\alpha _{2})_{+}}>0\), \(g_{L}\overline{( \alpha _{1})_{+}}-g_{M}\overline{(\alpha _{1})_{-}}>0\). Then, for each
\(u\in \varOmega \), there are constants
\(\eta _{1},\eta _{2}\in [0,T]\)
such that
$$u(\eta _{1}) \leq \biggl(\frac{\overline{(\alpha _{2})_{+}}}{g_{L}\overline{( \alpha _{1})_{+}}-g_{M}\overline{(\alpha _{1})_{-}}} \biggr)^{\frac{1}{ \mu }}:=A_{1} $$
and
$$u(\eta _{2})\geq \biggl(\frac{\overline{\alpha _{2}}}{g_{M}\overline{( \alpha _{1})_{+}}} \biggr)^{\frac{1}{\mu }}:=A_{2}. $$
Proof
Let \(u\in \varOmega \), we have
$$ \bigl(\phi _{p}\bigl(u'(t)\bigr) \bigr)'+ \lambda \bigl[f(u)u'+\alpha _{1}(t)g\bigl(u(t)\bigr)\bigr]= \lambda \frac{\alpha _{2}(t)}{u^{\mu }}. $$
(2.2)
Integrating (2.2) over \([0,T]\), we have
$$\int _{0}^{T}\alpha _{1}(t)g\bigl(u(t) \bigr)\,dt= \int _{0}^{T}\frac{\alpha _{2}(t)}{u ^{\mu }}\,dt $$
and
$$\int _{0}^{T}(\alpha _{1})_{+}(t)g \bigl(u(t)\bigr)\,dt- \int _{0}^{T}(\alpha _{1})_{-}(t)g \bigl(u(t)\bigr)\,dt \leq \int _{0}^{T}\frac{(\alpha _{2})_{+}(t)}{u^{\mu }}\,dt. $$
In view of mean value theorem of integrals, there exists \(\eta _{1} \in [0,T]\) such that
$$g_{L}T\overline{(\alpha _{1})_{+}}-g_{M}T \overline{(\alpha _{1})_{-}} \leq T\frac{\overline{(\alpha _{2})_{+}}}{u^{\mu }(\eta _{1})}, $$
i.e.,
$$u(\eta _{1}) \leq \biggl(\frac{\overline{(\alpha _{2})_{+}}}{g_{L}\overline{( \alpha _{1})_{+}}-g_{M}\overline{(\alpha _{1})_{-}}} \biggr)^{\frac{1}{ \mu }}:=A_{1}. $$
Multiplying both sides of (2.2) by \(u^{\mu }\) and integrating it over \([0,T]\), we have
$$\int _{0}^{T} \bigl(\phi _{p} \bigl(u'(t)\bigr) \bigr)'u^{\mu }(t)\,dt+ \int _{0}^{T}\alpha _{1}(t)g\bigl(u(t) \bigr)u^{\mu }(t)\,dt= \int _{0}^{T}\alpha _{2}(t)\,dt. $$
Since \(\int _{0}^{T} (\phi _{p}(u'(t)) )'u^{\mu }(t)\,dt =-\mu \int _{0}^{T} \vert u'(t) \vert ^{p-2}u^{\mu -1} \vert u'(t) \vert ^{2}\,dt\leq 0\), thus
$$g_{M} \int _{0}^{T}(\alpha _{1})_{+}(t)u^{\mu }(t) \,dt\geq T\overline{ \alpha _{2}}. $$
From mean value theorem of integrals, there exists \(\eta _{2}\in [0,T]\) such that
$$u(\eta _{2})\geq \biggl(\frac{\overline{\alpha _{2}}}{g_{M}\overline{( \alpha _{1})_{+}}} \biggr)^{\frac{1}{\mu }}:=A_{2}. $$
□
Lemma 2.3
Let
\(g(u)\)
satisfy the conditions of Lemma 2.2. Let
$$ F(x)= \int _{1}^{x}f(s)\,ds, \quad\quad K_{0}:= \sup_{s\in [A_{2},+\infty )}F(s)< +\infty $$
(2.3)
and
$$ \lim_{s\rightarrow 0^{+}} \biggl(F(s) - \frac{T \overline{(\alpha _{2})_{+}}}{s^{\mu }}-g_{M}T \overline{(\alpha _{1})_{-}} \biggr)>K_{0}, $$
(2.4)
where
\(A_{2}\)
is defined in Lemma 2.2. Then there exists a constant
\(\gamma _{0}>0\)
such that
$$\min_{t\in [0,T]} u(t)\geq \gamma _{0} \quad \textit{for }u \in \varOmega . $$
Proof
Let \(u\in \varOmega \), then u satisfies (2.2). There exist \(t_{1},t_{2}\in {R}\) such that \(t_{2}-t_{1}\in (0,T)\) and
$$u(t_{1})=\max_{t\in [0,T]}u(t), \quad\quad u(t_{2})=\min_{t\in [0,T]}u(t). $$
From (2.3), the definitions of \(A_{2}\) and \(u(t_{1})\), we have
$$A_{2}\leq u(t_{1})< +\infty $$
and
$$ F\bigl(u(t_{1})\bigr)\leq \sup_{s\in [A_{2},+\infty )}F(s):=K_{0}. $$
(2.5)
Integrating (2.2) over \([t_{1},t_{2}]\), we have
$$ \int _{t_{1}}^{t_{2}}f\bigl(u(t)\bigr)u'(t) \,dt+ \int _{t_{1}}^{t_{2}}\alpha _{1}(t)g\bigl(u(t) \bigr)\,dt = \int _{t_{1}}^{t_{2}}\frac{\alpha _{2}(t)}{u^{\mu }(t)}\,dt. $$
(2.6)
From the definitions of \(u(t_{1})\), (2.5), and (2.6), we obtain that
$$\begin{aligned} F\bigl(u(t_{2})\bigr) &=F\bigl(u(t_{1})\bigr)+ \int _{t_{1}}^{t_{2}} \frac{\alpha _{2}(t)}{u ^{\mu }(t)}\,dt- \int _{t_{1}}^{t_{2}} \alpha _{1}(t)g\bigl(u(t) \bigr)\,dt \\ &< K_{0}+ \int _{0}^{T}\frac{(\alpha _{2})_{+}(t)}{u^{\mu }(t)}\,dt +g_{M} \int _{0}^{T}(\alpha _{1})_{-}(t) \,dt \\ &\leq K_{0}+\frac{T\overline{(\alpha _{2})_{+}}}{u^{\mu }(t_{2})}+g _{M}T\overline{(\alpha _{1})_{-}} \end{aligned}$$
and
$$ F\bigl(u(t_{2})\bigr)-\frac{T\overline{(\alpha _{2})_{+}}}{u^{\mu }(t_{2})}-g _{M}T \overline{(\alpha _{1})_{-}}\leq K_{0}. $$
(2.7)
In view of (2.4), there exists a constant \(\gamma _{0}>0\) such that
$$ F(s) - \frac{T\overline{(\alpha _{2})_{+}}}{s^{\mu }}-g_{M}T\overline{( \alpha _{1})_{-}}>K_{0}, \quad \forall s\in (0,\gamma _{0}). $$
(2.8)
By (2.7) and (2.8), we have
$$\min_{t\in [0,T]} u(t)\geq \gamma _{0} \quad \text{for }u \in \varOmega . $$
□
Lemma 2.4
Let
\(g(u)\)
satisfy the condition of Lemma 2.2. Let
$$ G(x)= \int _{1}^{x}s^{\mu }f(s)\,ds, \quad\quad \lim_{s\rightarrow +\infty } \bigl(G(s) -g_{M}T\overline{\alpha _{1}}s ^{\mu } \bigr)=+\infty $$
(2.9)
and
$$ \lim_{s\rightarrow 0^{+}}G(s)< \rho +T\overline{\alpha _{2}}, $$
(2.10)
where
$$\rho =\inf_{s\in [A_{2},+\infty )} \bigl(G(s) -g_{M}T\overline{\alpha _{1}}s^{\mu } \bigr), $$
\(A_{2}\)
is defined in Lemma 2.2. Then there exist constants
\(\gamma _{2}>\gamma _{1}>0\)
such that
$$\min_{t\in [0,T]} u(t)\geq \gamma _{1} \quad \textit{for } u \in \varOmega $$
and
$$\max_{t\in [0,T]}u(t)\leq \gamma _{2} \quad \textit{for }u \in \varOmega . $$
Proof
Let \(u\in \varOmega \), then u satisfies (2.2). There exist \(t_{1},t_{2}\in {R}\) such that \(t_{2}-t_{1}\in (0,T)\) and
$$u(t_{1})=\max_{t\in [0,T]}u(t), \quad\quad u(t_{2})=\min_{t\in [0,T]}u(t). $$
Multiplying (2.2) by \(u^{\mu }(t)\), and then integrating it over the interval \([t_{1}, t_{2}]\), we have
$$ \begin{aligned}[b] & \int _{t_{1}}^{t_{2}} \bigl(\phi _{p} \bigl(u'(t)\bigr) \bigr)'u^{\mu }(t)\,dt+ \int _{t _{1}}^{t_{2}}f\bigl(u(t)\bigr)u'(t)u^{\mu }(t) \,dt + \int _{t_{1}}^{t_{2}}\alpha _{1}(t)g\bigl(u(t) \bigr)u^{\mu }(t)\,dt\\&\quad = \int _{t_{1}}^{t_{2}}\alpha _{2}(t) \,dt. \end{aligned} $$
(2.11)
From \(\int _{t_{1}}^{t_{2}} (\phi _{p}(u'(t)) )'u^{\mu }(t)\,dt<0\), (2.9), and (2.11), we have
$$\begin{aligned} \begin{aligned}[b] G\bigl(u(t_{2})\bigr)&\geq G\bigl(u(t_{1}) \bigr)- \int _{t_{1}}^{t_{2}}\alpha _{1}(t)g\bigl(u(t) \bigr)u ^{\mu }(t)\,dt+ \int _{t_{1}}^{t_{2}}\alpha _{2}(t)\,dt \\ &\geq G\bigl(u(t_{1})\bigr)-g_{M}T\overline{\alpha _{1}}u^{\mu }(t_{1})+T\overline{ \alpha _{2}} \geq \rho +T\overline{\alpha _{2}}. \end{aligned} \end{aligned}$$
(2.12)
By (2.10) there exists a constant \(\gamma _{1}>0\) such that
$$ G(s)< \rho +T\overline{\alpha _{2}}, \quad \forall s\in (0,\gamma _{1}). $$
(2.13)
By (2.12) and (2.13), we have
$$ \min_{t\in [0,T]} u(t)\geq \gamma _{1} \quad \text{for }u \in \varOmega . $$
(2.14)
From Lemma 2.2 and (2.14), we have
$$ \gamma _{1}\leq u(t_{2})\leq A_{1}. $$
(2.15)
By (2.12) we have
$$ G\bigl(u(t_{1})\bigr)-g_{M}T\overline{\alpha _{1}}u^{\mu }(t_{1}) \leq G\bigl(u(t_{2}) \bigr)-T\overline{ \alpha _{2}} \leq \max_{\gamma _{1}\leq s\leq A_{1}}G(s)-T \overline{ \alpha _{2}}. $$
(2.16)
In view of (2.9), there exists a constant \(\gamma _{2}>\gamma _{1}\) such that
$$ G(s)-g_{M}T\overline{\alpha _{1}}s^{\mu }> \max _{\gamma _{1}\leq s\leq A_{1}}G(s)-T\overline{\alpha _{2}}, \quad s\in ( \gamma _{2},+\infty ). $$
(2.17)
Thus, (2.6) and (2.17) imply
$$\max_{t\in [0,T]}u(t)=u(t_{1})\leq \gamma _{2}. $$
□
Lemma 2.5
Let
\(g(u)\)
satisfy the conditions of Lemma 2.2. Let
$$ F(x)= \int _{0}^{x}f(s)\,ds, \quad\quad B_{0}:=\inf_{s\in [A_{2},+\infty )}F(s)>-\infty $$
(2.18)
and
$$ \lim_{s\rightarrow 0^{+}} \biggl(F(s) + \frac{T \overline{(\alpha _{2})_{+}}}{s^{\mu }}+g_{M}T \overline{(\alpha _{1})_{-}} \biggr)< B_{0}, $$
(2.19)
where
\(A_{2}\)
is defined in Lemma 2.2. Then there exists a constant
\(\gamma _{3}>0\)
such that
$$\min_{t\in [0,T]} u(t)\geq \gamma _{3} \quad \textit{for }u \in \varOmega . $$
Proof
Let \(u\in \varOmega \), then u satisfies (2.2). There exist \(t_{1},t_{2}\in {R}\) such that \(t_{2}-t_{1}\in (0,T)\) and
$$u(t_{1})=\max_{t\in [0,T]}u(t), \quad\quad u(t_{2})=\min_{t\in [0,T]}u(t). $$
From (2.18), the definitions of \(A_{2}\) and \(u(t_{1})\), we have
$$A_{2}\leq u(t_{1})< +\infty $$
and
$$ F\bigl(u(t_{1})\bigr)\geq \inf_{s\in [A_{2},+\infty )}F(s):=B_{0}. $$
(2.20)
Integrating (2.2) over \([t_{1},t_{2}]\) we have
$$\int _{t_{1}}^{t_{2}}f\bigl(u(t)\bigr)u'(t) \,dt+ \int _{t_{1}}^{t_{2}}\alpha _{1}(t)g\bigl(u(t) \bigr)\,dt = \int _{t_{1}}^{t_{2}}\frac{\alpha _{2}(t)}{u^{\mu }(t)}\,dt $$
and
$$ F\bigl(u(t_{2})\bigr)=F\bigl(u(t_{1})\bigr)+ \int _{t_{1}}^{t_{2}}\frac{\alpha _{2}(t)}{u ^{\mu }(t)}\,dt - \int _{t_{1}}^{t_{2}}\alpha _{1}(t)g\bigl(u(t) \bigr)\,dt. $$
(2.21)
Form (2.20) and (2.21), we have
$$\begin{aligned} F\bigl(u(t_{2})\bigr) &\geq B_{0}- \int _{t_{1}}^{t_{2}}\frac{(\alpha _{2})_{-}(t)}{u ^{\mu }(t)}\,dt - \int _{t_{1}}^{t_{2}}(\alpha _{1})_{+}(t)g \bigl(u(t)\bigr)\,dt \\ &\geq B_{0}- \int _{0}^{T}\frac{(\alpha _{2})_{-}(t)}{u^{\mu }(t)}\,dt - \int _{0}^{T}(\alpha _{1})_{+}(t)g \bigl(u(t)\bigr)\,dt \\ &\geq B_{0}-g_{M}T\overline{(\alpha _{1})_{+}}- \frac{\overline{(\alpha _{2})_{-}}}{u^{\mu }(t_{2})}, \end{aligned}$$
i.e.,
$$ F\bigl(u(t_{2})\bigr)+g_{M}T\overline{(\alpha _{1})_{+}} +\frac{\overline{(\alpha _{2})_{-}}}{u^{\mu }(t_{2})}\geq B_{0}. $$
(2.22)
In view of (2.19), there exists a constant \(\gamma _{3}>0\) such that
$$ F(s) + \frac{T\overline{(\alpha _{2})_{+}}}{s^{\mu }}+g_{M}T\overline{( \alpha _{1})_{-}}< B_{0}, \quad \forall s\in (0,\gamma _{3}). $$
(2.23)
By (2.22) and (2.23), we have
$$\min_{t\in [0,T]} u(t)\geq \gamma _{3} \quad\text{for }u \in \varOmega . $$
□