Now, we embed Eq. (1.3) into the following equations family with a parameter \(\lambda\in(0,1]\):
$$ \bigl( x(t)-cx(t-\sigma)\bigr)''+ \lambda f\bigl(x(t)\bigr)x'(t)- \lambda\varphi(t)x^{ \mu}(t)+ \lambda\frac{\alpha(t)}{x^{\gamma}(t)}= \lambda e(t). $$
(3.1)
Let
$$ D = \bigl\{ x\in C^{1}_{T}: Lx=\lambda Nx,\lambda\in(0,1]\bigr\} . $$
(3.2)
Theorem 3.1
Assume \(\overline{\varphi}>0\), \(\alpha(t)>0\), and \(\overline{e}=0\), then there are two constants \(\tau_{0}, \tau_{1} \in[0,T]\) for each \(u\in D\) such that
$$ u(\tau_{0})\leq\max\biggl\{ 1, \biggl( \frac{\overline{\alpha}}{\overline{\varphi}}\biggr)^{\frac{1}{\mu}} \biggr\} :=A_{0} $$
(3.3)
and
$$ u(\tau_{1})\geq\min\biggl\{ 1, \biggl( \frac{\overline{\alpha}}{\overline{\varphi}}\biggr)^{\frac{1}{\mu}} \biggr\} :=A_{1}. $$
(3.4)
Proof
Let \(u \in D\), then
$$ (Au)''(t)+ \lambda f\bigl(u(t) \bigr)u'(t)- \lambda\varphi(t)u^{\mu}(t)+ \frac{\lambda\alpha(t)}{u^{\gamma}(t)}= \lambda e(t). $$
(3.5)
Integrating both sides of Eq. (3.5) over the interval \([0,T]\), we obtain that
$$ \int^{T}_{0}\varphi(t) u^{\mu}(t)\,dt= \int^{T}_{0} \frac{\alpha(t)}{u^{\gamma}(t)}\,dt. $$
(3.6)
If \(u(t)> 1\), combine with the mean value theorem of integrals and \(\varphi(t)>0\), then
$$ u^{\mu}(\tau)T\overline{\varphi}= \int^{T}_{0}\varphi(t) u^{\mu}(t)\,dt= \frac{T \overline{\alpha}}{u^{\gamma}(\xi)} < T, $$
which yields
$$ u(\tau)< \biggl(\frac{\overline{\alpha}}{\overline{\varphi}}\biggr)^{ \frac{1}{\mu}}. $$
So there exists a point \(\tau_{0}\in[0,T]\) such that
$$ u(\tau_{0})\leq\max\biggl\{ 1, \biggl( \frac{\overline{\alpha}}{\overline{\varphi}}\biggr)^{\frac{1}{\mu}} \biggr\} := A_{0}. $$
(3.7)
On the other hand, when \(\varphi(t)>0\) for every \(\mu\in D\), there always exists a point \(\tau_{1}\in[0,T]\) such that
$$ u(\tau_{1})\geq\min\biggl\{ 1, \biggl( \frac{\overline{\alpha}}{\overline{\varphi}}\biggr)^{\frac{1}{\mu}} \biggr\} := A_{1}. $$
(3.8)
□
Theorem 3.2
Suppose that assumptions of [\(\mathrm{H}_{1}\)] and [\(\mathrm{H}_{2}\)] hold, and \(\overline{\varphi}>0\), \(\overline{e}=0\), then Eq. (2.1) has at least one positive T-periodic solution.
Proof
Suppose \(u\in D\), then Eq. (3.5) holds. By multiplying both sides of Eq. (3.5) by \(u(t)\) and integrating it over the interval \([0,T]\), we get
$$ \int^{T}_{0}(Au)''(t)u(t) \,dt= \lambda \int^{T}_{0}\varphi(t)u^{\mu}(t)u(t) \,dt- \lambda \int^{T}_{0}\frac{u(t) \alpha(t)}{u^{\gamma}(t)}\,dt+\lambda \int^{T}_{0} e(t)u(t)\,dt. $$
Since
$$ \int^{T}_{0}(Au)''(t)u(t) \,dt= - \int^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\,dt + c \int ^{T}_{0}u'(t)u'(t- \sigma)\,dt, $$
it is easy to verify that
$$\begin{aligned} \int^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\,dt ={}&c \int^{T}_{0}u'(t)u'(t- \sigma)\,dt \\ &{}+\lambda \int^{T}_{0}\biggl(\frac{\alpha(t)}{u^{\gamma}(t)}- \varphi(t)u^{ \mu}(t)\biggr)u(t)\,dt-\lambda \int^{T}_{0} e(t)u(t)\,dt. \end{aligned}$$
Let \(A_{2}=(\frac{\alpha_{m}}{|\varphi|_{\infty}})^{ \frac{1}{\mu+ \gamma}}\) and \(A_{3}=(\frac{|\alpha|_{\infty}}{\varphi_{m}})^{ \frac{1}{\mu+ \gamma}}\). Set \(E_{1}=\{ t\in[0,T]: 0< u(t)< A_{2}\}\), \(E_{2}=\{ t\in[0,T]: A_{2}\leq u(t)\leq A_{3}\}\), \(E_{3}=\{ t\in[0,T]: u(t)>A_{3}\}\), and \(E_{1}\bigcup E_{2}\bigcup E_{3} = [0,T]\). We can obtain
$$\begin{aligned} \int^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\,dt ={}&c \int^{T}_{0}u'(t)u'(t- \sigma)\,dt+ \lambda \int_{E_{1}}\biggl(\frac{\alpha(t)}{u^{\gamma}(t)}-\varphi(t)u^{ \mu}(t) \biggr)u(t)\,dt \\ &{}+\lambda \int_{E_{2}}\biggl(\frac{\alpha(t)}{u^{\gamma}(t)}-\varphi(t)u^{ \mu}(t) \biggr)u(t)\,dt+\lambda \int_{E_{3}}\biggl(\frac{\alpha(t)}{u^{\gamma}(t)} \\ &{}-\varphi(t)u^{\mu}(t)\biggr)u(t)\,dt-\lambda \int^{T}_{0} e(t)u(t)\,dt \\ \leq{}& \vert c \vert \int^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\,dt + \biggl\vert \int_{E_{1}}\biggl( \frac{\alpha(t)}{u^{\gamma}(t)}-\varphi(t)u^{\mu}(t) \biggr)u(t)\,dt \biggr\vert \\ &{}+ \int_{E_{2}} \biggl\vert \biggl(\frac{\alpha(t)}{u^{\gamma}(t)}- \varphi(t)u^{\mu }(t)\biggr)u(t) \biggr\vert \,dt + T \Vert u \Vert _{\infty}\overline{e_{-}}. \end{aligned}$$
It follows that
$$\begin{aligned} \bigl(1- \vert c \vert \bigr) \int^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\,dt\leq{}& A_{2} \int_{E_{1}}\biggl( \frac{\alpha(t)}{u^{\gamma}(t)}-\varphi(t)u^{\mu}(t) \biggr)\,dt \\ &{}+ A_{3}N_{0} + T \Vert u \Vert _{\infty}\overline{e_{-}}, \end{aligned}$$
(3.9)
where \(N_{0}=T \max_{A_{2}< x< A_{3},t\in[0,T]}\{ | \frac{\alpha(t)}{x^{\gamma}}-\varphi(t)x^{\mu}|\}\). On the other hand,
$$\biggl( \int_{E_{1}} + \int_{E_{2}}+ \int_{E_{3}}\biggr) \biggl( \frac{\alpha(t)}{u^{\gamma}(t)}- \varphi(t)u^{\mu}(t)\biggr)\,dt)=0. $$
Combining it with Eq. (3.9) we get that
$$\begin{aligned} &\bigl(1- \vert c \vert \bigr) \int^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\,dt \\ &\quad \leq A_{2}\biggl(- \int_{E_{2}}\biggl(\frac{\alpha(t)}{u^{\gamma}(t)}-\varphi(t)u^{ \mu}(t) \biggr)\,dt- \int_{E_{3}}\biggl(\frac{\alpha(t)}{u^{\gamma}(t)}-\varphi(t)u^{ \mu}(t) \biggr)\,dt\biggr) \\ &\qquad {} + A_{3}N_{0} + T \Vert u \Vert _{\infty}\overline{e_{-}} \\ &\quad \leq A_{2} \biggl\vert \int_{E_{2}}\biggl(\frac{\alpha(t)}{u^{\gamma}(t)}-\varphi(t)u^{ \mu}(t) \biggr)\,dt \biggr\vert + A_{2} \biggl\vert \int_{E_{3}}\biggl(\frac{\alpha(t)}{u^{\gamma}(t)}- \varphi(t)u^{\mu}(t) \biggr)\,dt \biggr\vert \\ &\qquad {} + A_{3}N_{0} + T \Vert u \Vert _{\infty}\overline{e_{-}} \\ &\quad \leq A_{2}N_{0} +A_{2} \int_{E_{3}}\frac{\alpha(t)}{u^{\gamma}(t)}\,dt +A_{2} \int_{E_{3}}\varphi(t)u^{\mu}(t)\,dt \\ &\qquad {} + A_{3}N_{0} + T \Vert u \Vert _{\infty}\overline{e_{-}} \\ &\quad \leq(A_{2}+A_{3})N_{0} + \frac{A_{2}T\overline{\alpha}}{A^{\gamma}_{3}} +A_{2}T \overline{\varphi} \Vert u \Vert ^{\mu}_{\infty}+ T \Vert u \Vert _{\infty} \overline{e_{-}}. \end{aligned}$$
(3.10)
From the condition, we see that
$$ \int^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\,dt \leq \frac{[(A_{2}+A_{3})N_{0} +\frac{A_{2}T\overline{\alpha}}{A_{3}}]}{1- \vert c \vert } + \frac{A_{2}T\overline{\varphi} \Vert u \Vert ^{\mu}_{\infty}}{1- \vert c \vert }+ \frac{T \Vert u \Vert _{\infty}\overline{e_{-}}}{1- \vert c \vert } . $$
Let \(N_{1}= \frac{[(A_{2}+A_{3})N_{0} +\frac{A_{2}T\overline{\alpha }}{A_{3}}]}{1-|c|}\), \(N_{2}= \frac{A_{2}T\overline{\varphi}}{1-|c|}\), and \(N_{3}=\frac{T\overline{e_{-}}}{1-|c|}\), we have
$$ \int^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\,dt \leq N_{1} +N_{2} \Vert u \Vert ^{\mu}_{\infty}+ N_{3} \Vert u \Vert _{\infty} . $$
(3.11)
By using Lemma 2.2, we obtain that
$$\begin{aligned} \Vert u \Vert _{\infty} & \leq A_{0} + \frac{\sqrt{T}}{2}\biggl( \int^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\biggr)^{ \frac{1}{2}} \\ &\leq A_{0} + \frac{\sqrt{T}}{2}\bigl(N_{1} +N_{2} \Vert u \Vert ^{\mu}_{\infty}+ N_{3} \Vert u \Vert _{\infty}\bigr)^{\frac{1}{2}} \\ &\leq A_{0}+ \frac{\sqrt{T}}{2}N^{\frac{1}{2}}_{1} + \frac{\sqrt{T}}{2}N^{\frac{1}{2}}_{2} \Vert u \Vert ^{\frac{\mu}{2}}_{\infty}+ \frac{\sqrt{T}}{2} N^{\frac{1}{2}}_{3} \Vert u \Vert _{\infty}^{\frac{1}{2}}. \end{aligned}$$
(3.12)
Now, we begin to estimate a priori upper bounds of \(u(t)\). In order to do this, we divide the estimation into two cases.
Case 1: \(0<\mu<2\). From Eq. (3.12), it is easy to see that there exists a constant \(M_{1}>0\) such that
$$ \Vert u \Vert _{\infty}< M_{1}. $$
Case 2: \(\mu=2\). For this case, Eq. (3.12) can be rewritten as
$$ \biggl(1- \frac{\sqrt{T}}{2}N^{\frac{1}{2}}_{2}\biggr) \Vert u \Vert _{\infty}\leq A_{0}+ \frac{\sqrt{T}}{2}N^{\frac{1}{2}}_{1}+ \frac{\sqrt{T}}{2} N^{ \frac{1}{2}}_{3} \Vert u \Vert _{\infty}^{\frac{1}{2}}, $$
which together with assumption \([H_{2}]\) yields that there exists a constant \(M_{2}> 0\) such that
$$ \Vert u \Vert _{\infty}< M_{2}. $$
Thus, in either case 1 or case 2, we have
$$ \Vert u \Vert _{\infty}< \max\{M_{1}, M_{2}\}:=M_{3}. $$
(3.13)
Substituting Eq. (3.13) into Eq. (3.11), we have that there exists a constant \(M_{4}>0\) such that \(\int^{T}_{0}|u'(t)|^{2}\,dt< M_{4}\). Since \(Au\in C_{T}^{1}\), there is \(t_{0}\in[0,T]\) such that \((Au)'(t_{0})=0\). From Eq. (3.5), we get
$$\begin{aligned} \bigl\vert (Au)'(t) \bigr\vert = &\lambda \biggl\vert \int^{t}_{t_{0}}\biggl[-f\bigl(u(t) \bigr)u'(t)+\varphi(t)u^{ \mu}(t)-\frac{\alpha(t)}{u^{\gamma}(t)}+e(t) \biggr]\,dt \biggr\vert \\ \leq& \int^{T}_{0} \bigl\vert f\bigl(u(t)\bigr) \bigr\vert \bigl\vert u'(t) \bigr\vert \,dt + \int^{T}_{0} \bigl\vert \varphi(t)u^{ \mu}(t) \bigr\vert \,dt \\ &{}+ \int^{T}_{0} \biggl\vert \frac{\alpha(t)}{u^{\gamma}(t)} \biggr\vert \,dt + \int^{T}_{0} \bigl\vert e(t) \bigr\vert \,dt \\ \leq& \vert f \vert _{M_{3}}T^{\frac{1}{2}}\biggl( \int^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}\biggr)^{ \frac{1}{2}}+2T\overline{ \vert \varphi \vert }M^{\mu}_{3}+T\overline{ \vert e \vert } \\ \leq &\vert f \vert _{M_{3}}T^{\frac{1}{2}}(M_{4})^{\frac{1}{2}}+2T \overline{ \vert \varphi \vert }M^{\mu}_{3}+T \overline{ \vert e \vert }:=M_{5}, \end{aligned}$$
where \(|f|_{M_{3}}:=\max_{0\le x\le M_{3}}|f(x)|\). By using Lemma 2.1, we get the inequality
$$ \bigl\vert u' \bigr\vert _{\infty}< \bigl\vert A^{-1}Au' \bigr\vert _{\infty} \leq\frac{ \vert Au' \vert _{\infty}}{1- \vert c \vert } \leq\frac{M_{5}}{1- \vert c \vert }:=M_{6}. $$
(3.14)
In what follows, we will show the estimation
$$ \min_{t\in[0,T]}u(t)>\gamma_{0},\quad \mbox{uniformly for all }u \in D, $$
(3.15)
where \(\gamma_{0}>0\) is a constant, D is determined in Eq. (3.2).
Let \(\tau_{1}\) be determined in Eq. (3.8). Multiplying both sides of Eq. (3.5) by \(u'(t)\) and integrating it over the interval \([t,\tau_{1}]\), we obtain that
$$\begin{aligned} & \int^{\tau_{1}}_{t}\biggl[(Au)''(s)u'(s)+ \lambda f\bigl(u(s)\bigr) \bigl\vert u'(s) \bigr\vert ^{2}- \lambda\varphi(s)u^{\mu}(s)u'(s)+ \frac{\lambda u'(s)\alpha(s)}{u^{\gamma}(s)}\biggr]\,ds \\ &\quad =\lambda \int^{\tau_{1}}_{t} e(s)u'(s)\,ds. \end{aligned}$$
Because of \(\int^{\tau_{1}}_{t}\frac{u'(s)\alpha(s)}{u^{\gamma(s)}}\,ds= \int^{ \tau_{1}}_{t} \frac{\alpha(s)du(s)}{u^{\gamma(s)}}= \int^{u(\tau _{1})}_{u(t)} \frac{\alpha(s)\,dv}{v^{\gamma}} \), we can get from Eq. (3.13) and Eq. (3.14) that
$$\begin{aligned} &\lambda \biggl\vert \int^{u(\tau_{1})}_{u(t)}\frac{\alpha(s)\,dv}{v^{\gamma}} \biggr\vert \\ &\quad \leq \int^{\tau_{1}}_{t} \bigl\vert (Au)''(s) \bigr\vert \bigl\vert u'(s) \bigr\vert \,ds+\lambda \int^{\tau _{1}}_{t} \bigl\vert f\bigl(u(s)\bigr) \bigr\vert \bigl\vert u'(s) \bigr\vert ^{2}\,ds \\ &\qquad {}+ \lambda \int^{\tau_{1}}_{t} \bigl\vert \varphi(s) \bigr\vert \bigl\vert u^{\mu}(s) \bigr\vert \bigl\vert u'(s) \bigr\vert \,ds + \lambda \int^{\tau_{1}}_{t} \bigl\vert e(s) \bigr\vert \bigl\vert u'(s) \bigr\vert \,ds \\ &\quad \leq \bigl\vert u' \bigr\vert _{\infty} \int^{T}_{0} \bigl\vert (Au)''(s) \bigr\vert \,ds + \lambda \bigl\vert u' \bigr\vert ^{2}_{ \infty} \int^{T}_{0} \bigl\vert f\bigl(u(s)\bigr) \bigr\vert \,ds \\ &\qquad {}+ \lambda \bigl\vert u' \bigr\vert _{\infty} \vert u \vert ^{\mu}_{\infty} \int^{T}_{0} \bigl\vert \varphi(s) \bigr\vert \,ds +\lambda \bigl\vert u' \bigr\vert _{\infty} \int^{T}_{0} \bigl\vert e(s) \bigr\vert \,ds \\ &\quad \leq M_{6} \int^{T}_{0} \bigl\vert (Au)''(s) \bigr\vert \,ds +\lambda M_{6}^{2} \vert f \vert _{M_{3}} + \lambda M_{6} M^{\mu}_{3}T \overline{ \vert \varphi \vert } +\lambda M_{6}T \overline{ \vert e \vert }. \end{aligned}$$
(3.16)
Furthermore, integrating Eq. (3.5) over the interval \([0,T]\),
i.e.,
$$\begin{aligned} \int^{T}_{0}(Au)''(t) \,dt ={}&{-} \lambda \int^{T}_{0} f\bigl(u(t)\bigr)u'(t) \,dt+ \lambda \int^{T}_{0}\varphi(t)u^{\mu}(t)\,dt \\ &{}- \lambda \int^{T}_{0}\frac{\alpha(t)}{u^{\gamma}(t)}\,dt+ \lambda \int^{T}_{0}e(t)\,dt, \end{aligned}$$
it is clear that
$$\begin{aligned} \int^{T}_{0} \bigl\vert (Au)''(t) \bigr\vert \,dt\leq{}&\lambda \int^{T}_{0} \bigl\vert f\bigl(u(t) \bigr)u'(t) \bigr\vert \,dt+ \lambda \int^{T}_{0} \bigl\vert \varphi(t) \bigr\vert \bigl\vert u^{\mu}(t) \bigr\vert \,dt \\ &{}+ \lambda \int^{T}_{0} \bigl\vert e(t) \bigr\vert \,dt \\ \leq{}&\lambda \bigl( \vert f \vert _{M_{3}}M_{6} +2T \overline{ \vert \varphi \vert }M^{\mu}_{3}+ 2T \overline{ \vert e \vert } \bigr). \end{aligned}$$
(3.17)
Now, substituting Eq. (3.17) into Eq. (3.16), we have
$$\begin{aligned} \biggl\vert \int^{u(\tau_{1})}_{u(t)}\frac{\alpha(s)\,dv}{v^{\gamma}} \biggr\vert \leq{}& M_{6}\bigl[ \vert f \vert _{M_{3}}M_{6} +2T\overline{ \vert \varphi \vert }M^{\mu}_{3}+ 2T \overline{ \vert e \vert }\bigr] \\ &{} +M_{6}^{2} \vert f \vert _{M_{3}}+M_{6} M^{\mu}_{3}T\overline{ \vert \varphi \vert } +M_{6}T \overline{ \vert e \vert }:=M_{7}, \end{aligned}$$
and so
$$ \biggl\vert \int^{u(\tau_{1})}_{u(t)}\frac{\alpha(s)\,dv}{v^{\gamma}} \biggr\vert \leq M_{7} \quad\forall t\in[\tau_{1}, \tau_{1}+T]. $$
(3.18)
On the other hand, when \(\gamma\geq1\), we have \(\int^{A_{1}}_{0}\frac{\alpha_{m}}{v^{\gamma}}\,dv = +\infty\), then there exists \(\gamma_{0}\in(0, A_{1})\) such that
$$ \biggl\vert \int^{A_{1}}_{\epsilon}\frac{\alpha_{m}\,dv}{v^{\gamma}} \biggr\vert >M_{7} \quad\forall\epsilon\in(0,\gamma_{0}]. $$
(3.19)
And if \(t^{*}\in[\tau_{1},\tau_{1}+T]\) such that \(u(t^{*})\leq\gamma_{0}\), from Eq. (3.19) we see that
$$ \biggl\vert \int^{u(\tau_{1})}_{u(t^{*})}\frac{\alpha(s)\,dv}{v^{\gamma}} \biggr\vert > \int^{A_{1}}_{u(t^{*})}\frac{\alpha_{m}\,dv}{v^{\gamma}}> M_{7}, $$
(3.20)
which contradicts Eq. (3.18). This contradiction verifies Eq. (3.15). From Eq. (3.13), Eq. (3.14), and Eq. (3.15), as well as the inequality in Remark 2.5, we can verify all the conditions of Lemma 2.4. Thus, by using Lemma 2.4, we see that Eq. (3.5) has at least one positive T-periodic solution. □