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# Infinitely many singularities and denumerably many positive solutions for a second-order impulsive Neumann boundary value problem

Boundary Value Problems20172017:50

https://doi.org/10.1186/s13661-017-0784-y

• Received: 23 November 2016
• Accepted: 4 April 2017
• Published:

## Abstract

Using a fixed point theorem of cone expansion and compression of norm type and a new method to deal with the impulsive term, we prove that the second-order singular impulsive Neumann boundary value problem has denumerably many positive solutions. Noticing that $$M>0$$, our main results improve many previous results.

## Keywords

• denumerably many positive solutions
• infinitely many singularities
• Neumann impulsive boundary conditions
• cone expansion and compression

## 1 Introduction

We are concerned with the existence of denumerably many positive solutions of the second-order singular impulsive Neumann boundary value problem
$$\textstyle\begin{cases} -x''(t)+ Mx(t)=\omega(t)f(t,x(t)),\quad t\in J,\\ -\Delta x'|_{t=t_{k}}=I_{k}(x(t_{k})), \quad k=1,2,\ldots,m,\\ x'(0)=x'(1)=0, \end{cases}$$
(1.1)
where M is a positive constant, $$J=[0,1]$$, $$t_{k} \in\mathrm{R}$$, $$k =1,\ldots,m$$, $$m \in\mathrm{N,}$$ satisfy $$0=t_{0}< t_{1}<t_{2}<\cdots <t_{m}<t_{m+1}=1$$, $$-\Delta x'|_{t=t_{k}}$$ denotes the jump of $$x'(t)$$ at $$t=t_{k}$$, that is, $$-\Delta x'|_{t=t_{k}}=x'((t_{k})^{+})-x'((t_{k})^{-})$$, here $$x'((t_{k})^{+})$$ and $$x'((t_{k})^{-})$$, respectively, represent the right-hand limit and left-hand limit of $$x'(t)$$ at $$t=t_{k}$$.
In addition, ω, f and $$I_{k}$$ satisfy the following conditions:
$$(H_{1})$$

$$\omega(t)\in L^{p}[0,1]$$ for some $$p\in[1,+\infty)$$, and there exists $$N>0$$ such that $$\omega(t)\geq N$$ a.e. on J;

$$(H_{2})$$

$$f\in C(J\times\mathrm{R^{+}}, \mathrm{R^{+}})$$, $$I_{k}\in C( \mathrm{R^{+}}, \mathrm{R^{+}})$$, where $$\mathrm {R^{+}}=[0,+\infty)$$;

$$(H_{3})$$

there exists a sequence $$\{t_{i}'\}_{i=1}^{\infty}$$ such that $$t_{1}'<\delta$$, where $$\delta=\min\{t_{1},\frac{1}{2}\}$$, $$t_{i}' \downarrow t^{*}\geq0$$ and $$\lim_{t\rightarrow t_{i}'} \omega(t) =+\infty$$ for all $$i=1, 2,\dots$$.

For the case $$M=0$$ and $$I_{k}=0$$ ($$k=1,2,\ldots,m$$), problem (1.1) reduces to the problem studied by Kaufmann and Kosmatov in . By using Krasnosel’skiĭ’s fixed point theorem and Hölder’s inequality, the authors showed the existence of countably many positive solutions. The other related results can be found in . However, there are almost no papers considering second-order impulsive Neumann boundary value problem with infinitely many singularities. To identify a few, we refer the reader to  and the references therein.

The main reason is that $$M\neq0$$ in problem (1.1), which shows that the solution of problem (1.1) has no concave properties. On the other hand, under the case $$M\neq0$$ and $$\omega(t)$$ with infinitely many singularities, the properties of the corresponding Green’s function for problem (1.1) are more complicated.

Our plan of the paper is as follows: in Section 2, we collect some well-known results to be used in the subsequent sections. In particular, we also present some new properties of Green’s function under the case $$M\neq0$$ and $$\omega(t)$$ with infinitely many singularities. In Section 3, we obtain some new sufficient conditions for the existence of denumerably many positive solutions for problem (1.1). In Section 4, we give an example of a family functions $$\omega(t)$$ such that $$(H_{3})$$ holds.

## 2 Preliminaries

In this installment, we list some definitions and lemmas which are needed throughout this paper.

Let $$J'=J\setminus \{t_{1},t_{2},\ldots,t_{m} \}$$ and $$E=C[0,1]$$. We define $$\mathit {PC}^{1}[0,1]$$ in E by
$$\mathit {PC}^{1}[0,1]= \bigl\{ x\in E:x'(t)\in C(t_{k}, t_{k+1}), \exists x' \bigl(t_{k}^{-} \bigr), x' \bigl(t_{k}^{+} \bigr), k=1,2, \ldots,m \bigr\} .$$
(2.1)
Then $$\mathit {PC}^{1}[0,1]$$ is a real Banach space with norm
$$\Vert x \Vert _{\mathit {PC}^{1}}=\max \bigl\{ \Vert x \Vert _{\infty}, \bigl\Vert x' \bigr\Vert _{\infty} \bigr\} ,$$
(2.2)
where $$\Vert x \Vert _{\infty}=\sup_{t\in J} \vert x(t) \vert$$, $$\Vert x' \Vert _{\infty}=\sup_{t\in J} \vert x'(t) \vert$$.
Suppose that $$G(t,s)$$ is the Green’s function of the boundary value problem
$$-x''(t)+Mu(t)=0,\quad\quad x'(0)=x'(1)=0,$$
then
$$G(t,s)= \frac{1}{\gamma\sinh\gamma} \textstyle\begin{cases} \cosh\gamma(1-t)\cosh\gamma s,& 0\leq s\leq t\leq1,\\ \cosh\gamma(1-s)\cosh\gamma t,& 0\leq t\leq s\leq1. \end{cases}$$
(2.3)

### Lemma 2.1

By the definition of $$G(t,s)$$ and the properties of $$sinhx$$ and $$coshx$$, we have the following results.
1. (a)
For any $$t, s\in J$$, there is
$$A=\frac{1}{\gamma\sinh\gamma}\leq G(t,s)\leq\frac{\cosh\gamma}{\gamma \sinh\gamma}=B.$$
(2.4)
Then it follows from (2.4) that
$$A\leq G(t,s)\leq G(s,s)\leq B.$$

2. (b)
For any $$\tau\in(0,\delta)$$,
$$\frac{D'_{k}}{\gamma\sinh\gamma}\leq G(t,s)\leq\frac{\cosh\gamma(1-\tau )\cosh\gamma\tau'_{k}}{\gamma\sinh\gamma}, \quad \forall t\in \bigl[ \tau,\tau '_{k} \bigr], s\in J,$$
(2.5)
where
$$\tau'_{k}=\max\{1-\tau, 1-t_{k}\},\quad\quad D'_{k}=\max \bigl\{ \cosh\gamma\tau, \cosh\gamma \bigl(1- \tau'_{k} \bigr) \bigr\} , \quad k=1,2,3,\ldots, m.$$

3. (c)
$$G_{t}'(t,s)= \frac{1}{\sinh\gamma} \textstyle\begin{cases} -\sinh\gamma(1-t)\cosh\gamma s,& 0\leq s\leq t\leq1,\\ \sinh\gamma(1-s)\cosh\gamma t,& 0\leq t\leq s\leq1, \end{cases}$$
(2.6)
and
$$\max_{t,s\in J,t\neq s} \bigl\vert G_{t}'(t,s) \bigr\vert \leq \sinh\gamma.$$
(2.7)

### Proof

We can get equations (2.4)-(2.7) by the definition of $$G(t,s)$$, so we omit it here. □

To establish the existence of positive solutions to problem (1.1), for a fixed $$\tau\in(0,\delta)$$, we construct the cone $$K_{\tau}$$ in $$\mathit {PC}^{1}[0,1]$$ by
$$K_{\tau}= \Bigl\{ x \in \mathit {PC}^{1}[0,1]: x(t) \geq0,t\in J, \min _{t\in[\tau ,\tau'_{k}]}x(t) \geq\sigma_{k} \Vert x \Vert _{\mathit {PC}^{1}} \Bigr\} ,$$
(2.8)
where
\begin{aligned}& \sigma_{k}=\frac{D'_{k}}{ \rho\gamma\sinh\gamma}, \quad k=1,2,\ldots,m, \end{aligned}
(2.9)
\begin{aligned}& \rho=\max\{B, \sinh\gamma\}. \end{aligned}
(2.10)
It is easy to see $$K_{\tau}$$ is a closed convex cone of $$\mathit {PC}^{1}[0,1]$$.
Let $$\{\tau_{i}\}_{i=1}^{\infty}$$ be such that $$t_{i+1}'<\tau _{i}<t_{i}'$$, $$i=1,2,\dots$$. Then for any $$i\in\mathrm{N}$$, we define the cone $$K_{\tau_{i}}$$ by
$$K_{\tau_{i}}= \Bigl\{ x(t)\in \mathit {PC}^{1}[0,1]: x(t)\geq0,t\in J,\min _{t\in[\tau_{i},\tau'_{ik}]}x(t)\geq\sigma_{ik} \Vert x \Vert _{\mathit {PC}^{1}} \Bigr\} ,$$
(2.11)
where
\begin{aligned}& \tau'_{ik}=\max\{1-\tau_{i},1-t_{k} \}, \quad\quad \sigma_{ik}=\frac{D'_{ik}}{ \rho\gamma\sinh\gamma}, \end{aligned}
(2.12)
\begin{aligned}& D'_{ik}=\max \bigl\{ \cosh\gamma\tau_{i}, \cosh \gamma \bigl(1-\tau'_{k} \bigr) \bigr\} , \quad i=1,2, \ldots, k=1,2,\ldots,m. \end{aligned}
(2.13)
It is easy to see $$K_{\tau_{i}}$$ is a closed convex cone of $$\mathit {PC}^{1}[0,1]$$.

### Remark 2.1

For any $$i=1,2,\ldots$$ , $$k=1,2,\ldots,m$$, it follows from the definition of $$\sigma_{k}$$ and $$\sigma_{ik}$$ that $$0<\sigma _{k},\sigma_{ik} <1$$.

### Lemma 2.2

If $$(H_{1})$$-$$(H_{3})$$ hold, then problem (1.1) has a unique solution x given by
$$x(t) = \int_{0}^{1}G(t,s)\omega(s)f \bigl(s,x(s) \bigr) \,ds+ \sum_{k=1}^{m}G(t,t_{k})I_{k} \bigl(t_{k},x(t_{k}) \bigr).$$

### Proof

The proof is similar to that of Lemma 2.4 in . □

### Definition 2.1

A function $$x(t)$$ is said to be a solution of problem (1.1) on J if:
1. (i)

$$x(t)$$ is absolutely continuous on each interval $$(0,t_{1}]$$ and $$(t_{k},t_{k+1}]$$, $$k =1,2,\ldots,n$$;

2. (ii)

for any $$k =1,2,\ldots,m$$, $$x(t_{k}^{ +})$$, $$x(t_{k}^{-})$$ exist;

3. (iii)

$$x(t)$$ satisfies (1).

Define an operator $$T: K_{\tau} \to \mathit {PC}^{1}[0,1]$$ by
$$(Tx) (t) = \int_{0}^{1}G(t,s)\omega(s)f \bigl(s,x(s) \bigr) \,ds+ \sum_{k=1}^{m}G(t,t_{k})I_{k} \bigl(t_{k},x(t_{k}) \bigr).$$
(2.14)
From (2.14), we know that $$x(t)\in \mathit {PC}^{1}[0,1]$$ is a solution of problem (1.1) if and only if x is a fixed point of the operator T. Also, for a positive number r, define $$\Omega_{r}$$ by
$$\Omega_{r}= \bigl\{ x\in \mathit {PC}^{1}[0,1]: \Vert x \Vert _{\mathit {PC}^{1}}< r \bigr\} .$$
Note that $$\partial\Omega_{r}= \{x\in \mathit {PC}^{1}[0,1]: \Vert x \Vert _{\mathit {PC}^{1}}=r \}$$ and $$\bar{\Omega}_{r}= \{x\in \mathit {PC}^{1}[0,1]: \Vert x \Vert _{\mathit {PC}^{1}}\leq r \}$$.

### Definition 2.2

An operator is called completely continuous if it is continuous and maps bounded sets into pre-compact sets.

### Lemma 2.3

Assume that $$(H_{1})$$-$$(H_{3})$$ hold. Then $$T(K_{\tau })\subset K_{\tau}$$ and $$T: K_{\tau} \to K_{\tau}$$ is a completely continuous.

### Proof

For $$t\in J$$, $$x\in K_{\tau}$$, it follows from ((2.5)) and (2.14) that
\begin{aligned}[b] (Tx) (t) &= \int_{0}^{1}G(t,s)\omega(s)f \bigl(s,x(s) \bigr) \,ds+ \sum_{k=1}^{m}G(t,t_{k})I_{k} \bigl(t_{k},x(t_{k}) \bigr) \\ & \leq B \Biggl[ \int_{0}^{1}\omega(s)f \bigl(s,x(s) \bigr)\,ds + \sum_{k=1}^{m}I_{k} \bigl(t_{k},x(t_{k}) \bigr) \Biggr]. \end{aligned}
(2.15)
On the other hand, it follows from (2.6), (2.7) and (2.14) that
\begin{aligned}[b] \bigl\vert (Tx)'(t) \bigr\vert &= \Biggl\vert \int _{0}^{1}G_{t}'(t,s) \omega(s)f \bigl(s,x(s) \bigr)\,ds+ \sum_{k=1}^{m}G_{t}'(t,t_{k})I_{k} \bigl(t_{k},x(t_{k}) \bigr) \Biggr\vert \\ &\leq \int_{0}^{1} \bigl\vert G_{t}'(t,s) \bigr\vert \omega(s)f \bigl(s,x(s) \bigr)\,ds+ \sum _{k=1}^{m} \bigl\vert G_{t}'(t,t_{k}) \bigr\vert I_{k} \bigl(t_{k},x(t_{k}) \bigr) \\ &\leq \sinh\gamma \Biggl[ \int_{0}^{1}\omega(s)f \bigl(s,x(s) \bigr)\,ds+ \sum _{k=1}^{m}I_{k} \bigl(t_{k},x(t_{k}) \bigr) \Biggr]. \end{aligned}
(2.16)
For any $$t\in J$$, combined with (2.15) and (2.16), we have
$$\Vert Tx \Vert _{\mathit {PC}^{1}} \leq\rho \Biggl[ \int_{0}^{1}\omega(s)f \bigl(s,x(s) \bigr)\,ds+ \sum _{k=1}^{m}I_{k} \bigl(t_{k},x(t_{k})\bigr) \Biggr].$$
(2.17)
Then, by (2.5), (2.8) and (2.17), we have
\begin{aligned}[b] \min_{t\in[\tau,\tau_{k}]}(Tx) (t) &=\min _{t\in[\tau,\tau_{k}]} \Biggl[ \int_{0}^{1}G(t,s)\omega(s)f \bigl(s,x(s) \bigr) \,ds+ \sum_{k=1}^{m}G(t,t_{k})I_{k} \bigl(t_{k},x(t_{k}) \bigr) \Biggr] \\ &\geq\frac{D'_{k}}{\gamma\sinh\gamma} \Biggl[ \int_{0}^{1}\omega(s)f\bigl(s,x(s)\bigr)\,ds+\sum _{k=1}^{m}I_{k} \bigl(t_{k},x(t_{k})\bigr) \Biggr] \\ &\geq\frac{D'_{k}}{\rho\gamma\sinh\gamma} \rho \Biggl[ \int_{0}^{1}\omega(s)f\bigl(s,x(s)\bigr)\,ds+\sum _{k=1}^{m}I_{k} \bigl(t_{k},x(t_{k})\bigr) \Biggr] \\ &\geq\sigma_{k} \Vert Tx \Vert _{\mathit {PC}^{1}}. \end{aligned}
(2.18)
Evidently, $$T(K_{\tau})\subset K_{\tau}$$.

Next, we prove that the operator $$T: K_{\tau}\to K_{\tau}$$ is a completely continuous.

It is obvious that T is continuous.

Let $$B_{d}=\{x\in \mathit {PC}^{1}[0,1] \mid \Vert x \Vert _{\mathit {PC}^{1}}\le d\}$$ be bounded set. Then, for all $$x\in B_{d}$$, by the definition of $$\Vert Tx \Vert _{\infty}$$, $$\Vert Tx' \Vert _{\infty}$$, $$\Vert Tx \Vert _{\mathit {PC}^{1}}$$, we have
\begin{aligned}& \begin{aligned} \Vert Tx \Vert _{\infty}&=\sup _{t\in J} \bigl\vert Tx(t) \bigr\vert \\ &\le B \Biggl[ \int_{0}^{1}\omega(s)f \bigl(s,x(s) \bigr)\,ds+ \sum _{k=1}^{m}I_{k} \bigl(t_{k},x(t_{k}) \bigr) \Biggr] \\ &\le B \bigl( \Vert \omega \Vert _{1}L+mL^{*} \bigr) \\ &=\Gamma_{0}, \end{aligned} \\& \begin{aligned} \bigl\Vert Tx' \bigr\Vert _{\infty}&=\sup_{t\in J} \bigl\vert Tx'(t) \bigr\vert \\ &\leq \sinh\gamma \Biggl[ \int_{0}^{1}\omega(s)f \bigl(s,x(s) \bigr)\,ds+ \sum _{k=1}^{m}I_{k} \bigl(t_{k},x(t_{k}) \bigr) \Biggr] \\ &\le\sinh\gamma \bigl( \Vert \omega \Vert _{1}L+mL^{*} \bigr) \\ &=\Gamma_{1}, \end{aligned} \end{aligned}
and
$$\Vert Tx \Vert _{\mathit {PC}^{1}}=\max \bigl\{ \Vert Tx \Vert _{\infty}, \bigl\Vert Tx' \bigr\Vert _{\infty} \bigr\} \le\max \{\Gamma_{0}, \Gamma_{1} \},$$
where
\begin{aligned}& L=\max_{t\in J, x\in K_{\tau}, \Vert x \Vert _{\mathit {PC}^{1}}\le d}f(t,x), \quad\quad L^{*}=\max \{L_{k}, k=1,2, \ldots,m\}, \\& L_{k}=\max_{t\in J, x\in K_{\tau}, \Vert x \Vert _{\mathit {PC}^{1}}\le d}I_{k} \bigl(t_{k},x(t_{k}) \bigr). \end{aligned}
Therefore $$T(B_{d})$$ is uniformly bounded.
On the other hand, for all $$t_{1}, t_{2}\in J_{k}$$ with $$t_{1}< t_{2}$$, we have
$$\bigl\vert (Tx) (t_{1})-(Tx) (t_{2}) \bigr\vert = \biggl\vert \int _{t_{1}}^{t_{2}}(Tx)'(t)\,dt \biggr\vert \le\Gamma_{1} \vert t_{1}-t_{2} \vert \rightarrow0 \quad (t_{1}\rightarrow t_{2}).$$
Noting (2.7), we know that $$G'(t,s)$$ is a constant and
\begin{aligned} \bigl\vert (Tx)'(t_{1})-(Tx)'(t_{2}) \bigr\vert &= \Biggl\vert \int _{0}^{1} \bigl[G_{t}'(t_{1},s)-G_{t}'(t_{2},s) \bigr]\omega(s)f \bigl(s,x(s) \bigr)\,ds \\ &\quad{} +\sum_{k=1}^{n} \bigl[G_{t}'(t_{1},t_{k})-G_{t}'(t_{2},t_{k}) \bigr]I_{k} \bigl(t_{k},x(t_{k}) \bigr) \Biggr\vert \\ & \leq \int_{0}^{1} \bigl\vert G_{t}'(t_{1},s)-G_{t}'(t_{2},s) \bigr\vert \omega(s)f \bigl(s,x(s) \bigr)\,ds \\ &\quad{} +\sum_{k=1}^{n} \bigl\vert G_{t}'(t_{1},t_{k})-G_{t}'(t_{2},t_{k}) \bigr\vert I_{k} \bigl(t_{k},x(t_{k}) \bigr) \rightarrow0 \quad (t_{1}\rightarrow t_{2}), \end{aligned}
which shows that $$T(B_{d})$$ is equicontinuous. The Arzelà-Ascoli theorem implies that T is completely continuous, and the lemma is proved. □

### Lemma 2.4

Hölder

Let $$e\in L^{p}[a,b]$$ with $$p>1$$, $$h\in L^{q}[a,b]$$ with $$q>1$$ and $$\frac{1}{p}+\frac{1}{q}=1$$. Then $$eh\in L^{1}[a,b]$$ and
$$\Vert eh \Vert _{1}\le \Vert e \Vert _{p} \Vert h \Vert _{q}.$$
Let $$e\in L^{1}[a,b]$$, $$h\in L^{\infty}[a,b]$$. Then $$eh\in L^{1}[a,b]$$ and
$$\Vert eh \Vert _{1}\le \Vert e \Vert _{1} \Vert h \Vert _{\infty}.$$

### Lemma 2.5

See ; fixed point theorem of cone expansion and compression of norm type

Let E be a Banach space, P be a cone in E. Assume that $$\Omega _{1}$$, $$\Omega_{2}$$ are bounded open subsets in E with $$\theta \in\Omega_{1}$$ and $$\bar{\Omega}_{1}\subset\Omega_{2}$$, where θ denotes zero operator. Suppose $$A : P \cap(\bar{\Omega}_{2} \setminus \Omega_{1})\rightarrow P$$ is completely continuous such that either
1. (i)

$$\Vert Ax \Vert \leq \Vert x \Vert$$, $$\forall x\in P\cap\partial\Omega_{1}$$; $$\Vert Ax \Vert \geq \Vert x \Vert$$, $$\forall x\in P\cap\partial\Omega_{2}$$;

2. (ii)

$$\Vert Ax \Vert \leq \Vert x \Vert$$, $$\forall x\in P\cap\partial\Omega_{2}$$; $$\Vert Ax \Vert \geq \Vert x \Vert$$, $$\forall x\in P\cap\partial\Omega_{1}$$.

Then A has a fixed point in $$P \cap(\bar{\Omega}_{2} \setminus \Omega_{1})$$.

## 3 Main results

In this section, using Lemmas 2.1-2.5, we give our main results in the case $$\omega\in L^{P}[0,1]$$; $$p>1$$, $$p=1$$ and $$p=\infty$$.

For convenience, we write
$$D=\max \bigl\{ \Vert G \Vert _{q} \Vert \omega \Vert _{p}, \Vert G \Vert _{1} \Vert \omega \Vert _{\infty }, B \Vert \omega \Vert _{1} \bigr\} ,\quad\quad \rho_{0}=\min \biggl\{ 1,\frac{A}{\sinh\gamma} \biggr\} .$$

Firstly, we consider the case $$p>1$$.

### Theorem 3.1

Assume that $$(H_{1})$$-$$(H_{3})$$ hold. Let $$\{r_{i}\} _{i=1}^{\infty}$$ and $$\{R_{i}\}_{i=1}^{\infty}$$ be such that
$$R_{i+1}< \sigma_{ik}r_{i}< r_{i}< L_{0}r_{i}< R_{i}, \quad i=1,2, \ldots, k=1,2,\ldots, m,$$
where
$$L_{0}=\max \biggl\{ \frac{\gamma\sinh\gamma}{A(N+m)D'_{k}},\frac{2\rho _{0}}{D+mB},2 \biggr\} .$$
For each natural number i, we assume that f and $$I_{k}$$ satisfy:
($$H_{4}$$):
For any $$t\in J$$, $$x\in[0,R_{i}]$$, $$f(t,x)\leq M_{0}R_{i}$$, and for any $$x\in[0,R_{i}]$$, $$k\in\{1,2,\dots,m\}$$, $$I_{k}(x(t_{k}))\leq M_{0}R_{i}$$, where
$$0< M_{0}\leq\frac{\rho_{0}}{D+mB}.$$
($$H_{5}$$):

For any $$t\in J$$, $$x\in[\sigma_{ik}r_{i},r_{i}]$$, $$f(t,x)\geq L_{0}r_{i}$$, and for any $$x\in[\sigma_{ik}r_{i},r_{i}]$$, $$k\in\{1,2,\dots,m\}$$, $$I_{k}(x)\geq L_{0}r_{i}$$.

Then problem (1.1) has denumerably many positive solutions $$\{x_{i}(t)\}_{i=1}^{\infty}$$ such that
$$r_{i}\leq \Vert x_{i} \Vert _{\mathit {PC}^{1}}\leq R_{i},\quad i=1,2,\dots.$$

### Proof

We consider the following open subset sequences $$\{\Omega_{1,i}\} _{i=1}^{\infty}$$ and $$\{\Omega_{2,i}\}_{i=1}^{\infty}$$ of $$\mathit {PC}^{1}[0,1]$$:
\begin{aligned}& \{\Omega_{1,i}\}_{i=1}^{\infty}= \bigl\{ x\in \mathit {PC}^{1}[0,1]: \Vert x \Vert _{\mathit {PC}^{1}}< R_{i} \bigr\} ; \\& \{\Omega_{2,i}\}_{i=1}^{\infty}= \bigl\{ x\in \mathit {PC}^{1}[0,1]: \Vert x \Vert _{\mathit {PC}^{1}}< r_{i} \bigr\} . \end{aligned}

Let $$\{\tau_{i}\}_{i=1}^{\infty}$$ be as in the hypothesis and note that $$0< t_{i+1}'<\tau_{i}<t_{i}'<\delta$$, $$i=1,2,\dots$$.

For fixed i, we assume that $$x\in K_{\tau_{i}}\cap\partial\Omega _{2,i}$$, then for any $$t\in J$$
$$r_{i}= \Vert x \Vert _{\mathit {PC}^{1}}\geq x(t)\geq\min _{t\in [\tau_{i},\tau'_{i}k]}x(t) \geq\sigma_{ik} \Vert x \Vert _{\mathit {PC}^{1}}= \sigma_{ik}r_{i}.$$
Noticing (2.5) and (2.14), for all $$x\in K_{\tau_{i}}\cap\partial\Omega _{2,i}$$, by $$(H_{1})$$ and $$(H_{5})$$, we have
\begin{aligned} (Tx) (t)&= \int_{0}^{1}G(t,s)\omega(s)f \bigl(s,x(s) \bigr) \,ds+ \sum_{k=1}^{m}G(t,t_{k})I_{k} \bigl(x(t_{k}) \bigr) \\ &\geq\min_{t\in[\tau,\tau'_{k}]} \Biggl[ \int_{0}^{1}G(t,s)\omega (s)f \bigl(s,x(s) \bigr) \,ds+ \sum_{k=1}^{m}G(t,t_{k})I_{k} \bigl(x(t_{k}) \bigr) \Biggr] \\ &\geq \frac{D'_{k}}{\gamma \sinh \gamma} \Biggl[ \int_{0}^{1}\omega(s)f \bigl(s,x(s) \bigr)\,ds+ \sum _{k=1}^{m}I_{k} \bigl(x(t_{k}) \bigr) \Biggr] \\ &\geq \frac{D'_{k}}{\gamma \sinh \gamma} \Biggl[N \int_{\tau_{i}}^{\tau'_{ik}}f \bigl(s,x(s) \bigr)\,ds+ \min _{t_{k}\in[\tau_{i},\tau'_{ik}]} \sum_{k=1}^{m}I_{k} \bigl(x(t_{k}) \bigr) \Biggr] \\ &\geq\frac{D'_{k}}{\gamma \sinh \gamma} L_{0}(N+m)r_{i} \\ &\geq r_{i}= \Vert x \Vert _{\mathit {PC}^{1}}, \end{aligned}
which shows that
$$\Vert Tx \Vert _{\mathit {PC}^{1}}\geq \Vert x \Vert _{\mathit {PC}^{1}}, \quad \forall x\in K_{\tau_{i}}\cap\partial\Omega_{2,i}.$$
(3.1)

On the other hand, for all $$t\in J$$, $$x\in P_{i}\cap\partial\Omega _{1,i}$$, we have $$x(t)\leq \Vert x \Vert _{\mathit {PC}^{1}}=R_{i}$$.

Noticing (2.4) and (2.14), for all $$t\in J$$, $$x\in K_{\tau_{i}}\cap \partial\Omega_{1,i}$$, by $$(H_{4})$$, we have
\begin{aligned}[b] (Tx) (t)&= \int_{0}^{1}G(t,s)\omega(s)f \bigl(s,x(s) \bigr) \,ds+ \sum_{k=1}^{m}G(t,t_{k})I_{k} \bigl(x(t_{k}) \bigr) \\ &\leq M_{0}R_{i} \int_{0}^{1}G(s,s)\omega(s)\,ds+ M_{0}R_{i} \sum_{k=1}^{m}G(t,t_{k}) \\ &\leq M_{0}R_{i} \Vert G \Vert _{q} \Vert \omega \Vert _{p}+ M_{0}R_{i}mB \\ &\leq M_{0}(D+mB)R_{i} \\ &\leq R_{i}= \Vert x \Vert _{\mathit {PC}^{1}}. \end{aligned}
(3.2)
Moreover, by (2.6), (2.16) and $$(H_{4})$$, we have
\begin{aligned}[b] \bigl\vert (Tx)'(t) \bigr\vert & \leq \int_{0}^{1} \bigl\vert G_{t}'(t,s) \bigr\vert \omega(s)f \bigl(s,x(s) \bigr)\,ds+ \sum _{k=1}^{m} \bigl\vert G_{t}'(t,t_{k}) \bigr\vert I_{k} \bigl(x(t_{k}) \bigr) \\ & \leq \sinh\gamma \Biggl[ \int_{0}^{1}\omega(s)f \bigl(s,x(s) \bigr)\,ds+ \sum _{k=1}^{m}I_{k} \bigl(x(t_{k}) \bigr) \Biggr] \\ & \leq\frac{\sinh\gamma}{ A} \Biggl[ \int_{0}^{1}G(s,s)\omega(s)f \bigl(s,x(s) \bigr) \,ds+ \sum_{k=1}^{m}G(s,s)I_{k} \bigl(x(t_{k}) \bigr) \Biggr] \\ & \leq\frac{\sinh\gamma}{ A} \Biggl[ \int_{0}^{1} \Vert G \Vert _{q} \Vert \omega \Vert _{p}f \bigl(s,x(s) \bigr)\,ds+ B\sum _{k=1}^{m}I_{k} \bigl(x(t_{k}) \bigr) \Biggr] \\ & \leq\frac{\sinh\gamma}{ A} \bigl(M_{0}R_{i} \Vert G \Vert _{q} \Vert \omega \Vert _{p}+ BmM_{0}R_{i} \bigr) \\ & \leq M_{0}\frac{\sinh\gamma}{ A}(D+ mB)R_{i} \\ & \leq R_{i}= \Vert x \Vert _{\mathit {PC}^{1}}. \end{aligned}
(3.3)
From (3.2) and (3.3), we have
$$\Vert Tx \Vert _{\mathit {PC}^{1}}\leq \Vert x \Vert _{\mathit {PC}^{1}},\quad \forall x\in K_{\tau_{i}}\cap\partial\Omega_{1,i}.$$
(3.4)

Applying Lemma 2.5 to (3.1) and (3.4) shows that the operator T has a fixed point $$x_{i}\in K_{\tau_{i}}\cap(\bar{\Omega}_{2,i}/ \Omega_{1,i})$$ such that $$r_{i}\leq \Vert x_{i} \Vert \leq R_{i}$$. Since $$i\in\mathrm{N}$$ was arbitrary, the proof is complete. □

The following results deal with the case $$p=\infty$$.

### Theorem 3.2

Assume that $$(H_{1})$$-$$(H_{3})$$ hold. Let $$\{a_{i}\} _{i=1}^{\infty}$$ and $$\{b_{i}\}_{i=1}^{\infty}$$ be such that
$$a_{i+1}< \sigma_{ik}b_{i}< b_{i}< L_{0}b_{i}< a_{i}, \quad i=1,2, \ldots, k=1,2,\ldots,m.$$
For each natural number i, we assume that f and $$I_{k}$$ satisfy ($$H_{4}$$) and ($$H_{5}$$), then problem (1.1) has denumerably many positive solutions $$\{x_{i}(t)\}_{i=1}^{\infty}$$ such that
$$r_{i}\leq \Vert x_{i} \Vert _{\mathit {PC}^{1}}\leq R_{i}, \quad i=1,2,\ldots.$$

### Proof

Let $$\Vert G \Vert _{1} \Vert \omega \Vert _{\infty}$$ replace $$\Vert G \Vert _{q} \Vert \omega \Vert _{p}$$ and repeat the previous argument. □

Finally, we consider the case of $$p=1$$.

### Theorem 3.3

Assume that $$(H_{1})$$-$$(H_{3})$$ hold. Let $$\{a_{i}\} _{i=1}^{\infty}$$ and $$\{b_{i}\}_{i=1}^{\infty}$$ be such that
$$a_{i+1}< \sigma_{ik}b_{i}< b_{i}< L_{0}b_{i}< a_{i}, \quad i=1,2, \ldots, k=1,2,\ldots,m.$$
For each natural number i, we assume that f and $$I_{k}$$ satisfy ($$H_{4}$$) and ($$H_{5}$$), then the problem (1.1) has denumerably many positive solutions $$\{x_{i}(t)\}_{i=1}^{\infty}$$ such that
$$r_{i}\leq \Vert x_{i} \Vert _{\mathit {PC}^{1}}\leq R_{i}, \quad i=1,2,\dots.$$

### Proof

Similar to the proof of (3.2) and (3.3), for all $$t\in[\tau _{i},\delta-\tau_{i}]$$, $$x\in K_{\tau_{i}}\cap\partial\Omega_{1,i}$$, then $$x(t)\leq \Vert x \Vert _{\mathit {PC}^{1}}=R_{i}$$.

Since (2.4) and (2.14), for all $$x\in K_{\tau_{i}}\cap\partial\Omega _{1,i}$$, by $$(H_{4})$$, we have
\begin{aligned}[b] (Tx) (t)&= \int_{0}^{1}G(t,s)\omega(s)f \bigl(s,x(s) \bigr) \,ds+ \sum_{k=1}^{m}G(t,t_{k})I_{k} \bigl(t_{k},x(t_{k}) \bigr) \\ &\leq B \Vert \omega \Vert _{1} \int_{0}^{1}f \bigl(s,x(s) \bigr)\,ds+ B\sum _{k=1}^{m}I_{k} \bigl(t_{k},x(t_{k}) \bigr) \\ &\leq M_{0}R_{i}B \Vert \omega \Vert _{1}+ mM_{0}R_{i}B \\ &\leq M_{0}(D+mB)R_{i} \\ &\leq R_{i}= \Vert x \Vert _{\mathit {PC}^{1}}, \end{aligned}
(3.5)
and by (2.4), (2.7), (2.16) and $$(H_{4})$$,
\begin{aligned} \bigl\vert (Tx)'(t) \bigr\vert & \leq \int_{0}^{1} \bigl\vert G_{t}'(t,s) \bigr\vert \omega(s)f \bigl(s,x(s) \bigr)\,ds+ \sum _{k=1}^{m} \bigl\vert G_{t}'(t,t_{k}) \bigr\vert I_{k} \bigl(t_{k},x(t_{k}) \bigr) \\ & \leq \sinh\gamma \Biggl[ \int_{0}^{1}\omega(s)f \bigl(s,x(s) \bigr)\,ds+ \sum _{k=1}^{m}I_{k} \bigl(t_{k},x(t_{k}) \bigr) \Biggr] \\ & \leq\frac{\sinh\gamma}{ A} \Biggl[ \int_{0}^{1}G(s,s)\omega(s)f \bigl(s,x(s) \bigr) \,ds+ \sum_{k=1}^{m}G(s,s)I_{k} \bigl(t_{k},x(t_{k}) \bigr) \Biggr] \\ & \leq\frac{\sinh\gamma}{ A} \Biggl[ \int_{0}^{1}B \Vert \omega \Vert _{1}f \bigl(s,x(s) \bigr)\,ds+ B\sum_{k=1}^{m}I_{k} \bigl(t_{k},x(t_{k}) \bigr) \Biggr] \\ & \leq\frac{\sinh\gamma}{ A} \bigl(M_{0}R_{i}B \Vert \omega \Vert _{1}+ BmM_{0}R_{i} \bigr) \\ & \leq M_{0}\frac{\sinh\gamma}{A}B(D+ m)R_{i} \\ & \leq R_{i}= \Vert x \Vert _{\mathit {PC}^{1}}. \end{aligned}
(3.6)
From (3.5) and (3.6), we have
$$\Vert Tx \Vert _{\mathit {PC}^{1}}\leq \Vert x \Vert _{\mathit {PC}^{1}},\quad \forall x\in K_{\tau_{i}}\cap\partial\Omega_{1,i}.$$
Similarly to the proof of Theorem 3.1, we can finish the proof of Theorem 3.3. □

## 4 An example

From Section 3, it is not difficult to see that $$(H_{3})$$ plays an important role in the proof that problem (1.1) has denumerably many positive solutions. As an example, we consider a family of functions $$\omega(t)$$ as follows.

### Example 4.1

Let $$k=m=1$$, $$t_{1}=\frac{1}{3}$$, and
$$t_{n}'=t_{1}-\sum _{i=1}^{n} \frac{1}{(i+1)(i+2)(i+3)(i+4)}, \quad n=1,2,\ldots.$$
It is easy to see that
\begin{aligned}& t_{1}'=\frac{39}{120}< \frac{1}{3}, \\& t_{n}'-t_{n+1}'= \frac{1}{(n+2)(n+3)(n+4)(n+5)}, \quad n=1,2,\ldots, \end{aligned}
and
$$t^{*}=\lim_{n\rightarrow\infty}t_{n}=t_{1}- \sum_{i=1}^{\infty}\frac{1}{(i+1)(i+2)(i+3)(i+4)} = \frac{1}{3}- \frac{1}{72}=\frac{23}{72}>\frac{1}{4},$$
where $$\sum_{i=1}^{\infty}\frac{1}{(i+1)(i+2)(i+3)(i+4)}=\frac{1}{72}$$.
Let
$$\omega(t)=\sum_{i=1}^{\infty} \omega_{n}(t), \quad t\in J,$$
where
$$\omega_{n}(t)= \textstyle\begin{cases} \frac{1}{2n^{4}(t_{n}'+t_{n+1}')},& t\in[0,\frac {t_{n}'+t_{n+1}'}{2}),\\ \frac{1}{\sqrt{t_{n}'-t}}, & t\in[\frac{t_{n}'+t_{n+1}'}{2},t_{n}'),\\ \frac{1}{\sqrt{t-t_{n}'}}, & t\in[t_{n}',\frac{t_{n}'+t_{n-1}'}{2}],\\ \frac{2}{2n^{4}(2-t_{n}'-t_{n-1}')},& t\in(\frac {t_{n}'+t_{n-1}'}{2},1]. \end{cases}$$
From $$\sum_{i=1}^{\infty}\frac{1}{n^{4}}=\frac{\pi^{4}}{90}$$ and $$\sum_{i=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{4}}{6}$$, we have
\begin{aligned} \sum_{i=1}^{\infty} \int_{0}^{1}\omega_{n}(t)\,dt &= \sum _{i=1}^{\infty} \biggl\{ \int_{0}^{(t_{n}'+t_{n+1}')/2}\frac {1}{2n^{4}(t_{n}'+t_{n+1}')}\,dt + \int_{(t_{n-1}'+t_{n}')/2}^{1}\frac {2}{2n^{4}(2-t_{n}'-t_{n-1}')}\,dt \\ &\quad{} + \int_{(t_{n}'+t_{n+1}')/2}^{t_{n}}\frac{1}{\sqrt{t_{n}'-t}}\,dt+ \int_{t_{n}}^{(t_{n-1}'+t_{n}')/2}\frac{1}{\sqrt {t-t_{n}'}}\,dt \biggr\} \\ &=\sum_{i=1}^{\infty}\frac{1}{n^{4}}+\sqrt{2} \sum_{i=1}^{\infty} \Bigl(\sqrt{ \bigl(t_{n}'-t_{n+1}' \bigr)}+\sqrt { \bigl(t_{n-1}'-t_{n}' \bigr)} \Bigr) \\ &=\frac{\pi^{4}}{90}+\sqrt{2}\sum_{i=1}^{\infty} \biggl[ \frac {1}{(n+2)(n+3)(n+4)(n+5)} \biggr]^{\frac{1}{2}} \\ &\quad{} +\sqrt{2}\sum_{i=1}^{\infty} \biggl[ \frac {1}{(n+1)(n+2)(n+3)(n+4)} \biggr]^{\frac{1}{2}} \\ &\leq\frac{\pi^{4}}{90}+\sqrt{2}\sum_{i=1}^{\infty} \frac {1}{n^{2}}+\sqrt{2}\sum_{i=1}^{\infty} \frac{1}{n^{2}} \\ &=\frac{\pi^{4}}{90}+\sqrt{2}\frac{\pi^{2}}{3}. \end{aligned}
Thus, it is easy to see
$$\int_{0}^{1}\omega(t)\,dt= \int_{0}^{1}\sum_{i=1}^{\infty} \omega _{n}(t)\,dt=\sum_{i=1}^{\infty} \int_{0}^{1}\omega_{n}(t)\,dt< \infty.$$

Therefore, $$\omega(t)\in L^{1}[0,1]$$, which satisfies condition $$(H_{3})$$.

## Declarations

### Acknowledgements

This work is sponsored by the National Natural Science Foundation of China (11301178) and the Beijing Natural Science Foundation (1163007), the Scientific Research Project of Construction for Scientific and Technological Innovation Service Capacity (71E1610973) and the teaching reform project of Beijing Information Science & Technology University (2015JGYB41). The authors are grateful to the anonymous referees for their constructive comments and suggestions, which have greatly improved this paper. 