First, we consider the existence of solutions of (1.3) on \((0,\beta_{1})\).
Lemma 3.1
Assume that (H), (A), and (D) hold. Then (1.3) has at least one positive periodic solution.
Proof
Since \(f(\cdot)\) is singular at \(x=\beta_{1}^{}\), there exists sufficiently small \(\delta<(1\theta)\beta_{1}\) such that \(f(x)\geq\rho x\), for \(0<x\beta_{1}< \delta_{1}\), where ρ satisfies \(\rho m \Omega_{2}\theta\geq1\), \(\Omega_{2}\) is given as follows. Choose \(R\in(\beta_{1}\delta,\beta_{1})\). Then, for \(x\in \partial K_{R}\), one has \(x(t)\geq\theta\Vert x \Vert\geq\theta R\). If \(\theta R\geq\beta _{1}\delta\), then \(\frac{\beta_{1}\delta}{\theta}\leq R\leq\beta_{1}\). Further, we have \(\delta>(1\theta)\beta_{1}\), which yields a contradiction. Now we only consider the case \(\theta R<\beta_{1}\delta\). Let \([0,T]=\Omega_{1}\cup\Omega_{2}\), where
$$\begin{gathered} \Omega_{1}=\bigl\{ t\in[0,T]: \theta R\leq x(t)< \beta_{1} \delta\bigr\} , \\ \Omega_{2}=\bigl\{ t\in[0,T]: \beta_{1}\delta\leq x(t)\leq R\bigr\} . \end{gathered} $$
Since \(\Vert x \Vert=R\) and x is continuous, \(\Omega_{2}\) is nonempty and \(\Omega_{2}>0\). Therefore, we have
$$\begin{aligned} \Vert Ax \Vert &=\max_{t\in[0,T]} \int_{0}^{T} G(t,s)f\bigl(x(s)\bigr)\,ds \\ &\geq m\biggl[ \int_{\Omega_{1}} f\bigl(x(s)\bigr)\,ds+ \int_{\Omega_{2}} f\bigl(x(s)\bigr)\,ds\biggr] \\ &\geq m \int_{\Omega_{2}} f\bigl(x(s)\bigr)\,ds \geq\rho m \vert \Omega_{2} \vert \theta \Vert x \Vert \geq \Vert x \Vert . \end{aligned} $$
For any \(x\in\partial K_{\bar{r}}\), we have
$$\begin{aligned} \Vert Ax \Vert =&\max_{t\in[0,T]} \int_{0}^{T} G(t,s)f\bigl(x(s)\bigr)\,ds \leq \int_{0}^{T} G(t,s)\frac{\bar{r}}{MT}\,ds < \Vert x \Vert . \end{aligned}$$
From [22], it is clear that \(A:\overline{K_{R}}\backslash K_{\bar{r}}\rightarrow K\), and A is completely continuous on \(\overline{K_{R}}\backslash K_{\bar{r}}\). Therefore, by Lemma 2.1, (1.3) has at least a positive solution \(x(t)\in\overline{K_{R}}\backslash K_{\bar{r}}\). □
Lemma 3.2
Assume that (H) and (A) hold. If
\(f^{0}=0\), then (1.3) has at least one positive periodic solution.
Proof
Since \(f^{0}=0\), there exists sufficiently small \(\bar{r}>0\) such that \(f(x)\leq\epsilon x\) for \(x\in[0,\bar{r}]\), where \(\epsilon M T<1\). Then, for any \(x\in\partial K_{\bar{r}}\), we have
$$\begin{aligned} \Vert Ax \Vert =&\max_{t\in[0,T]} \int_{0}^{T} G(t,s)f\bigl(x(s)\bigr)\,ds \leq\epsilon M T \Vert x \Vert < \Vert x \Vert . \end{aligned}$$
The remainder is similar to the proof of Lemma 3.1, so we omit it. □
Lemma 3.3
Assume that (H), (A), and (D) hold. If
\(f^{0}=+\infty\), then (1.3) has at least two positive periodic solutions.
Remark
\(f^{0}=+\infty\) implies that f may be singular at \(x=0\).
Proof
Since \(f(\cdot)\) is singular at \(x=\beta_{1}^{}\), from the proof of Lemma 3.1, there exists \(R\in (\beta_{1}\delta,\beta_{1})>\bar{r}\) such that \(\Vert Ax \Vert> \Vert x \Vert\) for \(x\in \partial K_{R}\)
Since \(f^{0}=+\infty\), there exists sufficiently small \(r<\bar{r}\) such that \(f(x)\geq\varrho x\) for \(x\in[0,r]\), where \(\varrho m\theta T>1\). Then, for any \(x\in\partial K_{r}\), we have
$$\begin{aligned} \Vert Ax \Vert =&\max_{t\in[0,T]} \int_{0}^{T} G(t,s)f\bigl(x(s)\bigr)\,ds \geq\varrho m \theta T \Vert x \Vert < \Vert x \Vert . \end{aligned}$$
For any \(x\in\partial K_{\bar{r}}\), we have
$$\begin{aligned} \Vert Ax \Vert =&\max_{t\in[0,T]} \int_{0}^{T} G(t,s)f\bigl(x(s)\bigr)\,ds \leq\epsilon M T \Vert x \Vert < \Vert x \Vert . \end{aligned}$$
Further, from (A) and the Arzela–Ascoli theorem it follows that now we verify that \(A:\overline{K_{R}}\backslash K_{r}\rightarrow K\), and A is completely continuous.
Therefore, by Lemma 2.1, (1.3) has at least two positive solutions \(x_{1}(t)\in\overline{K_{R}}\backslash K_{\bar{r}}\) and \(x_{2}(t)\in \overline{K_{\bar{r}}}\backslash K_{r}\). □
Second, we give some existence results of solutions of (1.3) on \((\beta_{N},+\infty)\).
Lemma 3.4
Assume that (H), (A), and (C) hold. Then (1.3) has at least one positive periodic solution.
Remark
Since \(f(\cdot)\) is singular at \(x=\beta_{N}^{+}\), there exists sufficiently small \(\delta<\frac{1\theta}{\theta}\) such that \(f(x)\geq\rho x\) for \(0< x\beta_{N}< \delta\). Choose \(r\in(\beta_{N}, \beta_{N}+\delta)\). Then, for \(x\in \partial K_{r}\), one has \(x(t)\geq\theta\Vert x \Vert\geq\theta r\). If \(\theta r\geq\beta _{n}\), then \(\frac{\beta_{N}}{\theta}\leq r\leq\beta_{N}+\delta\), which yields a contradiction. Therefore \(\theta r<\beta_{N}\), which implies that A is not continuous on \(\overline{K_{R}}\backslash K_{r}\), where \(R>\frac{\beta_{N}}{\theta}\).
Proof
Let \(\omega=u\beta_{N}\). Then (1.3) is equivalent to the problem
$$\begin{aligned} \textstyle\begin{cases} \ddot{\omega}+a(t) \omega=f(\omega+\beta_{N})a(t)\beta_{N}=F(\omega ),\\ \omega(0)=\omega(T),\qquad \dot{\omega}(0)=\dot{\omega}(T). \end{cases}\displaystyle \end{aligned}$$
(3.1)
From (A) and (C), it follows that

\((a_{1})\)
:

\(\lim_{\omega\rightarrow0^{+}}F(\omega) =+\infty\);

\((a_{2})\)
:

let \(\widetilde{r}=\bar{R}\beta_{N}\), \(F(\omega)\geq0\), for \(\omega\in(0,\widetilde{r})\);

\((a_{3})\)
:

there exists sufficiently small \(\overline {r}<\widetilde{r}\) such that
$$F(\omega)>L \Vert \omega \Vert ,\quad \mbox{for }\omega\in (0,\overline{r}), $$
where \(LmT>1\).
Define the operator \(A:K\rightarrow E\) by
$$Tx(t)= \int_{0}^{T} G(t,s)\bigl[f\bigl(\omega(s)+ \beta_{N}\bigr)a(t)\beta_{N}\bigr]\,ds, \quad 0\leq t\leq T. $$
Then, for any \(\omega\in\partial K_{\widetilde{r}}\), from (ii) of (C), we have
$$\begin{aligned} \max_{\omega\in[\theta\widetilde{r},\widetilde{r}]}\bigl[f\bigl(\omega (s)+\beta_{N} \bigr)a(t)\beta_{N}\bigr] \leq& \max_{\omega\in[\theta\widetilde{r},\widetilde{r}]}\bigl[f \bigl(\omega (s)+\beta_{N}\bigr)a_{*}\beta_{N}\bigr] \\ =& \max_{x\in[\theta\bar{R}+(1\theta)\beta_{N}, \bar{R}]}f(x)a_{*}\beta_{N}< \frac{\widetilde{r}}{MT}. \end{aligned}$$
Further, we have
$$\begin{aligned} \Vert T\omega \Vert =&\max_{t\in[0,T]}T\omega(t)=\max _{t\in[0,T]} \int_{0}^{T} G(t,s)\bigl[f\bigl(\omega(s)+ \beta_{N}\bigr)a(t)\beta_{N}\bigr]\,ds \\ \leq& M T \max_{\omega\in[\theta\widetilde{r},\widetilde{r}]}\bigl[f\bigl(\omega (s)+ \beta_{N}\bigr)a(t)\beta_{N}\bigr] < \widetilde{r}= \Vert \omega \Vert . \end{aligned}$$
For \(\omega\in \partial K_{\bar{r}}\), we have
$$\begin{aligned} \Vert T\omega \Vert &=\max_{t\in[0,T]}T\omega(t)=\max _{t\in[0,T]} \int_{0}^{T} G(t,s)\bigl[f\bigl(\omega(s)+ \beta_{N}\bigr)a(t)\beta_{N}\bigr]\,ds \\ &\geq m T L \Vert \omega \Vert > \Vert \omega \Vert . \end{aligned} $$
From \((f_{5})\) and the Arzela–Ascoli theorem it follows that T is completely continuous on \(\overline{K_{\widetilde{r}}}\backslash K_{\bar{r}}\). Therefore, by Lemma 2.1, (3.1) has at least a positive solution \(\omega(t)\in\overline{K_{\widetilde{r}}}\backslash K_{\bar{r}}\). Namely, (1.3) has at least a positive solution \(x(t)=\omega(t)+\beta_{N}\) satisfying \(\Vert x(t) \Vert<\beta_{N}+\widetilde{r}=\bar{R}\). □
Corollary 3.5
Assume that (H) and (A) hold. In addition,

\((f_{1})\)
:

\(f(\cdot)\)
is continuous on
\((\beta_{N},+\infty)\)
and
\(f(x)\geq a^{*}\beta_{N}\)
on
\((\beta_{N},+\infty)\).
If
\(f^{\infty}=0\), then (1.3) has at least one positive periodic solution.
Proof
If \((f_{1})\) holds, then \(F(\omega)\geq\) and (i) of (C) hold. From \((f_{1})\) and \(f^{\infty}=0\) it follows that \(\lim_{\omega\rightarrow+\infty}\frac{F(\omega)}{\omega}=0\). Then there exists sufficiently large \(R>0\) such that \(F(\omega)\leq\epsilon\omega\) for \(\omega\geq R\), where \(\epsilon M T<1\). Choose \(\tilde{r}>\max\{{\frac{R}{\theta},\bar{r}}\}\), where r̄ is given in \((a_{3})\). Then, for any \(\omega\in \partial K_{\tilde{r}}\), one has \(\omega(t)\geq\theta\Vert\omega \Vert>R\), and further we have
$$\begin{aligned} \Vert T\omega \Vert =&\max_{t\in[0,T]} \int_{0}^{T} G(t,s)F\bigl(\omega(s)\bigr)\,ds \leq \epsilon M T \Vert \omega \Vert < \Vert \omega \Vert . \end{aligned}$$
Combining Lemma 3.4 and Lemma 2.1, we have that (1.3) has at least one positive periodic solution. □
Lemma 3.6
Assume that (H), (A), and (E) hold. If
\(f^{\infty}=+\infty\), then (1.3) has at least two positive periodic solutions.
Proof
It is easy to see that (E) implies (C). Then, from Lemma 3.3, we can obtain a solution \(x_{1}(t)\) satisfying \(\Vert x_{1}(t) \Vert<\bar{R}\).
For any \(x(t)\in\partial K_{\hat{R}} \), we get
$$\begin{aligned} \Vert Ax \Vert =&\max_{t\in[0,T]}Ax(t)=\max _{t\in [0,T]} \int_{0}^{T} G(t,s)f\bigl(x(s)\bigr)\,ds \\ \leq& M T \max_{x\in[\theta\hat{R},\hat{R}]}f\bigl(x(s)\bigr) < \hat{R}= \Vert x \Vert . \end{aligned}$$
By the definition of \(f^{\infty}=\infty\), there exists \(\widetilde{R}>\hat{R}\) such that \(f(x)\geq\mu x\) for \(x\geq \widetilde{R}\), where μ satisfies \(\mu m T \theta>1\).
Choose \(R=\frac{\widetilde{R}}{\theta}\). Then, for \(x\in \partial K_{R}\), one has \(x(t)\geq\theta\Vert x \Vert\geq\widetilde{R}\), and
$$\begin{aligned} \Vert Ax \Vert =&\max_{t\in[0,T]} \int_{0}^{T} G(t,s)f\bigl(x(s)\bigr)\,ds \\ \geq& \mu m T \theta \Vert x \Vert > \Vert x \Vert . \end{aligned}$$
From (A) and the Arzela–Ascoli theorem it follows that A is completely continuous on \(\overline{K_{R}}\backslash K_{\hat{R}}\). Therefore, by Lemma 2.1, (1.3) has another positive solution \(x_{2}(t)\in\overline{K_{R}}\backslash K_{\hat{R}}\). □
Finally, we are studying the existence of solutions of (1.3) on \((\beta_{i},\beta_{i+1})\), \(i=1,2\ldots,N1\).
Lemma 3.7
Assume that (H), (A), and (B) hold. Then (1.3) has at least two positive periodic solutions.
Proof
Let \(\omega=u\beta_{i}\). Then (1.3) is equivalent to the problem
$$\begin{aligned} \textstyle\begin{cases} \ddot{\omega}+a(t) \omega=f(\omega+\beta_{i})a(t)\beta _{i}=F_{i}(\omega),\\ \omega(0)=\omega(T),\qquad \dot{\omega}(0)=\dot{\omega}(T). \end{cases}\displaystyle \end{aligned}$$
(3.2)
From (A) and (B), it follows that

\((F_{1})\)
:

\(F_{i}(\cdot)\) is nonnegative and continuous on \((0,\beta_{i+1}\beta_{i})\);

\((F_{2})\)
:

\(F_{i}(\cdot)\) is singular at \(x=0^{+}\), \((\beta_{i+1}\beta_{i})^{}\);

\((F_{3})\)
:

Let \(\bar{r_{i}}=\tilde{ R_{i}}\beta_{i}\in (0,\beta_{i+1}\beta_{i})\) such that \(F_{i}^{*}(\bar{r_{i}})<\frac{\bar{r_{i}}}{MT}\).
Then from Lemma 3.3 it follows that (1.3) has at least two positive periodic solutions. □
Now we give the proof of the main results.
Proof of Theorem 1.1
The number of the interval \((\beta_{i},\beta_{i+1})\) is \(N1\), then we can obtain the result from Lemma 3.7. □
Proof of Theorem 1.2
From Lemmas 3.4 and 3.7 the result follows. □
Proof of Theorem 1.3
From Lemmas 3.1 and 3.7 the result follows. □
Proof of Corollary 1.4
From Lemmas 3.2 and 3.7 the result follows. □
Proof of Theorem 1.5
From Lemmas 3.3 and 3.7 the result follows. □
Proof of Theorem 1.6
From Lemmas 3.1, 3.4, and 3.7 the result follows. □
Proof of Theorem 1.7
. From Lemmas 3.6 and 3.7 the result follows. □
Proof of Theorem 1.8
From Lemmas 3.3, 3.4, and 3.7 the result follows. □
Proof of Theorem 1.9
From Lemmas 3.1, 3.6, and 3.7 the result follows. □
Proof of Theorem 1.10
From Lemmas 3.3, 3.6, and 3.7 the result follows. □
Now we just give an example to illustrate Theorem 1.10.
Example
For convenience, we consider the following periodic boundary value problem:
$$\begin{aligned} \textstyle\begin{cases} \ddot{x}+\mu x=f(x),\\ x(0)=x(T),\qquad \dot{x}(0)=\dot{x}(T), \end{cases}\displaystyle \end{aligned}$$
(3.3)
where \(a(t)=\mu\) is a constant such that (H) holds, and \(f(x)\) is expressed by
$$\begin{aligned} f(x)= \textstyle\begin{cases} C_{01}x^{\nu_{01}}+\frac{C_{02}}{(x\beta_{1})^{\nu_{02}}},&x\in (0,\beta_{1});\\ \frac{C_{11}}{(x\beta_{1})^{\nu_{11}}}+\frac{C_{12}}{(x\beta _{2})^{\nu_{12}}}+\beta_{1}\mu,&x\in(\beta_{1},\beta_{2});\\ \frac{C_{21}}{(x\beta_{2})^{\nu_{21}}}+\frac{C_{22}}{(x\beta _{3})^{\nu_{22}}}+\beta_{2}\mu,&x\in(\beta_{2},\beta_{3});\\ \frac{C_{31}}{(x\beta_{3})^{\nu{31}}}+C_{32}x^{\nu_{32}}+\beta_{3}\mu ,&x\in(\beta_{3},+\infty), \end{cases}\displaystyle \end{aligned}$$
where

(I)
\(C_{ij}\) are positive constants; (\(i=0,1,2,3\), \(j=1,2\));

(II)
\(a^{*}=a_{*}=\mu\);

(III)
\(\nu_{01}<1\), \(\nu_{11}, \nu_{21}, \nu_{31}>0\), \(\nu_{32}>1\), \(\nu_{02}\), \(\nu_{12}\), \(\nu_{22}\) are even.
It is obvious that (A) holds. First we show that assumption (B) holds.
$$\begin{aligned}& \mathrm{(i)}\quad f(x)a^{*}\beta_{i}\geq0\mbox{ obviously holds for }x \in (\beta_{i},\beta_{i+1}), \\& \begin{aligned} \mathrm{(ii)}\quad&\max_{x\in[\theta\bar{R_{i}}+(1\theta)\beta_{i}, \bar{R_{i}}]}f(x)< \frac{\bar{R_{i}}\beta_{i}}{MT}+a_{*}\beta_{i} \\ &\Updownarrow \\ &\max_{x\in[\theta\bar{R_{i}}+(1\theta)\beta_{i}, \bar{R_{i}}]}\biggl[\frac{C_{i1}}{(x\beta_{i})^{\nu_{i1}}}+\frac {C_{i2}}{(x\beta_{i+1})^{\nu_{i2}}} \biggr]< \frac{\bar{R_{i}}\beta_{i}}{MT}. \end{aligned} \end{aligned}$$
If \(C_{ij}\) (\(i=1,2\), \(j=1,2\)) is sufficiently small, there exists \(\tilde{R_{i}}\in(\beta_{i},\beta_{i+1})\) such that the above inequality holds.
Second, if \(C_{0j}\) (\(j=1,2\)) is sufficiently small, there exists \(\bar{r}\in(0,\beta_{1})\) such that \(f^{*}(\bar{r})<\frac{\bar{r}}{MT}\), namely (D) holds.
Third, it is clear that \(f(x)a^{*}\beta_{N}\geq 0\) for \(x\in(\beta_{N},\hat{R})\). In addition, if \(C_{3j}\) (\(j=1,2\)) is sufficiently small, there exists \(\hat{R}>\frac{\beta_{N}}{\theta}\) such that
$$\max_{x\in[\theta\hat{R}, \hat{R}]}f(x)< \frac{\bar{R}}{MT}+\min \biggl\{ a_{*} \frac{1}{MT},0\biggr\} \beta_{N}, $$
namely (E) holds.
Finally, we can verify that
$$\begin{gathered} f^{0}=\lim_{x\rightarrow 0} \frac{f(x)}{x}=\lim_{x\rightarrow 0}\frac{C_{01}x^{\nu_{01}}+\frac{C_{02}}{(x\beta_{1})^{\nu _{02}}}}{x}=+\infty, \\ f^{\infty}=\lim_{x\rightarrow \infty}\frac{f(x)}{x}=\lim _{x\rightarrow \infty}\frac{\frac{C_{31}}{(x\beta_{3})^{\nu{31}}}+C_{32}x^{\nu _{32}}+\beta_{3}\mu}{x}=+\infty. \end{gathered} $$
Therefore, by Theorem 1.10, (3.3) has at least \((2N+2)\) positive periodic solutions.