Positive solutions to one-dimensional quasilinear impulsive indefinite boundary value problems

Abstract

Consider the one-dimensional quasilinear impulsive boundary value problem involving the p-Laplace operator

$$ \textstyle\begin{cases} -(\phi_{p}(u'))'=\lambda \omega (t)f(u), \quad 0< t< 1, \\ -\Delta u|_{t=t_{k}}=\mu I_{k}(u(t_{k})), \quad k=1,2,\ldots,n, \\ \Delta u'|_{t=t_{k}}=0, \quad k=1,2,\ldots,n, \\ u'(0)=0, \qquad u(1)=\int_{0}^{1}g(t)u(t)\,dt, \end{cases} $$

where $\lambda, \mu >0$ are two positive parameters, $\phi_{p}(s)$ is the p-Laplace operator, i.e., $\phi_{p}(s)=|s|^{p-2}s$, $p>1$, $\omega (t)$ changes sign on $[0,1]$. Several new results are obtained for the above quasilinear indefinite problem.

1 Introduction

Impulsive differential equation is regarded as a critical mathematical tool to provide a natural description of observed evolution processes (see [1,2,3,4]). So the consideration of impulsive differential equations has gained prominence and many authors have begun to take a great interest in the subject of impulsive differential equations, for example, see [5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22] and the references cited therein.

Meanwhile, the p-Laplace operator equation is a typical quasilinear operator equation, which comes naturally from glaciology, nonlinear flow laws, and non-Newtonian mechanics (see [23, 24]). Recently, various existence, multiplicity, and uniqueness results of positive solutions for differential equations with one-dimensional p-Laplace operator have been considered [25,26,27,28,29,30,31,32,33]. Specially, Zhang and Ge [34] investigated the following second order one-dimensional p-Laplace operator equation

$$ \textstyle\begin{cases} -(\phi_{p}(u^{\prime}(t)))^{\prime}= f(t,u(t)), \quad t\neq t_{k}, t\in (0,1), \\ \triangle u|_{t= t_{k}}=I_{k}(u(t_{k})), \quad k=1,2,\ldots,n, \\ u(0)=\sum_{i=1}^{m-2}a_{i}u(\xi_{i}), \quad u^{\prime}(1)=0, \end{cases} $$

(1.1)

where $\phi_{p}(s)$ is p-Laplace operator, i.e., $\phi_{p}(s)=|s|^{p-2}s$, $p>1$, $(\phi_{p})^{-1}=\phi_{q}$, $\frac{1}{p}+ \frac{1}{q}=1$, $t_{k}$ ($k=1,2,\dots,n$, where n is a fixed positive integer) are fixed points with $0< t_{1}< t_{2}<\cdots <t_{k}<\cdots <t _{n}<1$, $\xi_{i}$ $(i=1,2,\ldots, m-2)\in (0,1)$ is given $0<\xi_{1}< \xi_{2}<\cdots <\xi_{m-2}<1$ and $\xi_{i}\neq t_{k}$, $i=1,2,\dots,m-2$, $k=1,2,\dots,n$, $\Delta u|_{t=t_{k}}$ denotes the jump of $u(t)$ at $t=t_{k}$, i.e.,

$$ \Delta u|_{t=t_{k}}=u\bigl(t_{k}^{+}\bigr)-u \bigl(t_{k}^{-}\bigr), $$

where $u(t_{k}^{+})$ and $u(t_{k}^{-})$ represent the right-hand limit and left-hand limit of $u(t)$ at ${t=t_{k}}$, respectively. Applying the classical fixed-point index theorem for compact maps, the authors got several new multiplicity results of positive solutions.

On the other hand, we observe that many authors (see [35,36,37,38,39,40,41,42,43,44,45,46,47,48,49]) have paid more attention to a class of boundary value problems involving integral boundary conditions, which contains two-point, three-point, and general multi-point boundary value problems as exceptional cases, see [50,51,52,53,54,55,56,57,58] and the references cited therein.

However, in literature there are almost no papers on multiple positive solutions for second order impulsive nonlocal indefinite boundary value problems with one-dimensional p-Laplace operator and multiple parameters. More precisely, the study of $\lambda >0$, $\mu >0$, $p \not \equiv 2$, $I_{k}\neq 0$ ($k=1,2,\ldots,n$) and ω changes sign is still open for the second order nonlocal boundary value problem

$$ \textstyle\begin{cases} -(\phi_{p}(u'))'=\lambda \omega (t)f(u), \quad 0< t< 1, \\ -\Delta u|_{t=t_{k}}=\mu I_{k}(u(t_{k})), \quad k=1,2,\ldots,n, \\ \Delta u'|_{t=t_{k}}=0, \quad k=1,2,\ldots,n, \\ u'(0)=0, \quad u(1)=\int_{0}^{1}g(t)u(t)\,dt, \end{cases} $$

(1.2)

where $\lambda >0$ and $\mu >0$ are two parameters, $\omega (t)$ may change sign, $\phi_{p}(s)$ is a p-Laplace operator, i.e., $\phi _{p}(s)=|s|^{p-2}s$, $p>1$, $(\phi_{p})^{-1}=\phi_{q}$, $\frac{1}{p}+ \frac{1}{q}=1$. $t_{k}$ ($k=1,2,\ldots,n$) (where n is a fixed positive integer) are fixed points with $0=t_{0}< t_{1}< t_{2}<\cdots <t_{k}< \cdots <t_{n}<t_{n+1}=1$, $\Delta u|_{t=t_{k}}$ denotes the jump of $u(t)$ at $t=t_{k}$, i.e., $\Delta u|_{t=t_{k}}=u(t_{k}^{+})-u(t_{k} ^{-})$, where $u(t_{k}^{+})$ and $u(t_{k}^{-})$ represent the right-hand limit and left-hand limit of $u(t)$ at $t=t_{k}$, respectively.

In addition, set $J=[0,1]$, $R_{+}=[0,+\infty)$, $R=(-\infty,+\infty)$, and let ω, f, $I_{k}$, and g satisfy the following conditions:

($H_{1}$):

$\omega:J\rightarrow R$ is continuous, and there exists a constant $\xi \in (0,1)$ such that

$$ \omega (t)\geq 0,\quad t\in [0,\xi ],\qquad \omega (t)\leq 0,\quad t\in [ \xi,1]. $$

Moreover, $\omega (t)$ does not vanish identically on any subinterval of J.

($H_{2}$):

$f:R_{+}\rightarrow R_{+}$ is continuous, and $f(u)>0$ for all $u>0$, there exists $0< c\leq 1$ such that

$$ f(x)\geq c\psi (x),\quad x\in R_{+}, $$

where $\psi (x)=\max \{f(y):0\leq y\leq x\}$;

($H_{3}$):

$I_{k}\in C(R_{+},R_{+})$, and $I_{k}(u)>0$ for all $u>0$.

($H_{4}$):

$g\in L^{1}[0,1]$ is nonnegative and $\eta \in [0,1)$, where

$$ \eta = \int_{0}^{1}g(s)\,ds. $$

(1.3)

($H_{5}$):

There exist $0<\theta_{1}\leq +\infty$, $\theta_{1}\neq p-1$, $0<\theta_{2}\leq +\infty$, $\theta_{2}\neq 1$, and $k_{1},k_{2},k _{3},k_{4}>0$ such that

$$ k_{1}u^{\theta_{1}}\leq f(u)\leq k_{2}u^{\theta_{1}}, \qquad k_{3}u^{\theta _{2}}\leq I_{k}(u)\leq k_{4}u^{\theta_{2}}. $$

($H_{6}$):

There exists a number $0<\sigma <\xi$ such that

$$ c^{2}k_{1}\sigma^{\theta_{1}} \int_{\sigma }^{\xi }\omega^{+}(t)\,dt \geq k_{2}\xi^{\theta_{1}} \int_{\xi }^{1}\omega^{-}(t)\,dt. $$

We define $\omega^{+}(t)=\max \{\omega (t),0\}$, $\omega^{-}(t)=- \min \{\omega (t),0\}$. Then $\omega (t)=\omega^{+}(t)-\omega^{-}(t)$.

It is well accepted that the fixed point theorem in a cone is crucial in showing the existence of positive solutions of various boundary value problems for second order differential equations.

Lemma 1.1

(Theorem 2.3.4 of [59])

Let $\varOmega_{1}$ and $\varOmega_{2}$ be two bounded open sets in a real Banach space E such that $0 \in \varOmega_{1}$ and $\bar{\varOmega }_{1}\subset \varOmega_{2}$. Let the operator $T: P\cap (\bar{\varOmega }_{2}\backslash \varOmega_{1})\rightarrow P$ be completely continuous, where P is a cone in E. Suppose that one of the two conditions

• (i) $\|Tx\|\leq \|x\|$, $\forall x\in P\cap \partial \varOmega_{1}$ and $\|Tx\|\geq \|x\|$, $\forall x\in P\cap \partial \varOmega_{2}$,

or

• (ii) $\|Tx\|\geq \|x\|$, $\forall x\in P\cap \partial \varOmega_{1}$, and $\|Tx\|\leq \|x\|$, $\forall x\in P\cap \partial \varOmega_{2}$,

is satisfied. Then T has at least one fixed point in $P\cap (\bar{ \varOmega }_{2}\backslash \varOmega_{1})$.

This paper is organized in the following fashion. In Sect. 2, we present some lemmas to be used in the subsequent sections. Section 3 is devoted to proving the multiplicity of positive solutions for problem (1.2), and we give an example to illustrate the main results in the final section.

2 Preliminaries

Let $J'=J\backslash \{t_{1},t_{2},\ldots,t_{n}\}$. The basic space used in this paper $PC[0,1]=\{u|u:J\rightarrow R { }\mbox{is continuous at } t\neq t_{k}, \mbox{left continuous at }t=t_{k}, \mbox{and }u(t_{k}^{+})\mbox{ exists}, k=1,2,\ldots,n\}$. Then $PC[0,1]$ is a real Banach space with the norm $\|\cdot \|_{PC}$ defined by $\|u\|_{PC}=\sup_{t\in J}|u(t)|$. By a solution of (1.2), we mean that a function $u\in PC[0,1]\cap C ^{2}(J')$ which satisfies (1.2).

In these main results, we will make use of the following lemmas.

Lemma 2.1

Assume that ($H_{1}$)–($H_{4}$) hold. Then $u\in PC[0,1]\cap C^{2}(J')$ is a solution of problem (1.2) if and only if $u\in PC[0,1]$ is a solution of the following impulsive integral equation:

$$\begin{aligned} u(t) =&\frac{1}{1-\eta }\biggl[ \int_{0}^{1}g(t) \int_{t}^{1}\phi_{q} \biggl( \int _{0}^{s}\lambda \omega (\tau)f\bigl(u(\tau) \bigr)\,d\tau \biggr)\,ds\,dt \\ &{}+\mu \int _{0}^{1}g(t) \biggl( \sum _{t\leq t_{k}}I_{k}\bigl(u(t_{k})\bigr) \biggr)\,dt\biggr] \\ &{}+ \int_{t}^{1}\phi_{q} \biggl( \int_{0}^{s}\lambda \omega (\tau)f\bigl(u( \tau) \bigr)\,d\tau \biggr)\,ds+\mu \sum_{t\leq t_{k}}I_{k} \bigl(u(t_{k})\bigr). \end{aligned}$$

(2.1)

Proof

The proof is similar to that of Lemma 3.1 in [38]. □

To establish the existence of multiple positive solutions in $PC[0,1]\cap C^{2}(J')$ of problem (1.2), we denote

$$ PC^{+}[0,1]= \Bigl\{ u\in PC[0,1]: \min_{t\in J}u(t) \geq 0 \Bigr\} , $$

and a cone K in $PC[0,1]$ by

$$ K= \bigl\{ u\in PC^{+}[0,1]:u\mbox{ is concave on } [0,\xi ], \mbox{and } u \mbox{ is convex on }[\xi,1] \bigr\} . $$

(2.2)

Let $R>r>0$, define $K_{r}=\{u\in K:\|u\|< r\}$, $K_{R,r}=\{u\in K:r<\|u \|<R\}$. Note that $\partial K_{r}=\{u\in K:\|u\|=r\}$, $\overline{K} _{R,r}=\{u\in K:r\leq \|u\|\leq R\}$.

We define a map $T:K \rightarrow PC[0,1]$ by

$$\begin{aligned} (Tu) (t) =&\frac{1}{1-\eta }\biggl[ \int_{0}^{1}g(t) \int_{t}^{1}\phi_{q} \biggl( \int_{0}^{s}\lambda \omega (\tau)f\bigl(u(\tau) \bigr)\,d\tau \biggr)\,ds\,dt \\ &{}+ \mu \int_{0}^{1}g(t) \biggl( \sum _{t\leq t_{k}}I_{k}\bigl(u(t_{k})\bigr) \biggr)\,dt\biggr] \\ &{}+ \int_{t}^{1}\phi_{q} \biggl( \int_{0}^{s}\lambda \omega (\tau)f\bigl(u( \tau) \bigr)\,d\tau \biggr)\,ds+\mu \sum_{t\leq t_{k}}I_{k} \bigl(u(t_{k})\bigr), \end{aligned}$$

(2.3)

where η is defined in (1.3).

Lemma 2.2

From (2.1), we know that $u\in PC[0,1]$ is a solution of problem (1.2) if and only if u is a fixed point of the map T.

Lemma 2.3

Assume that ($H_{1}$)–($H_{6}$) hold. Then we have $T(K)\subset K$, and $T:K\rightarrow K$ is completely continuous.

Proof

From (2.3), we know that

$$ (Tu)'(t)=-\phi_{q} \biggl( \int_{0}^{t}\lambda \omega (s)f\bigl(u(s)\bigr)\,ds \biggr). $$

(2.4)

Define $q(t):J\rightarrow J$ as follows:

$$ q(t)=\min \biggl\{ \frac{t}{\xi },\frac{1-t}{1-\xi } \biggr\} , $$

and $\min_{\sigma \leq t\leq \xi }q(t)=\frac{\sigma }{\xi }$, $\max_{\xi \leq t\leq 1}q(t)=1$.

Firstly, for any $u\in K$, we have

$$ \int_{0}^{1}\omega (s)f\bigl(u(s)\bigr)\,ds\geq \int_{0}^{\sigma }\omega^{+}(s)f\bigl(u(s) \bigr)\,ds. $$

(2.5)

In fact, by (2.2), we know that $u(t)\geq 0$. Since $u\in K$, $u(0) \geq 0$, and $u(1)\geq 0$, we have

$$\begin{aligned}& \frac{u(t)-u(0)}{t-0}\geq \frac{u(\xi)-u(0)}{\xi -0},\quad t\in [0,\xi ] \quad \Rightarrow \quad u(t)\geq q(t)u(\xi),\quad t\in [0,\xi ], \\& \frac{u(t)-u(1)}{t-1}\geq \frac{u(\xi)-u(1)}{\xi -1},\quad t\in [\xi,1] \quad \Rightarrow \quad u(t)\leq q(t)u(\xi),\quad t\in [\xi,1]. \end{aligned}$$

As we all know, ψ is nondecreasing on J, so we have

$$ \psi \bigl(u(t)\bigr)\geq \psi \bigl(q(t)u(\xi)\bigr),\quad t\in [0,\xi ], \qquad \psi \bigl(u(t)\bigr) \leq \psi \bigl(q(t)u(\xi)\bigr),\quad t\in [\xi,1]. $$

So, it follows from ($H_{5}$) and ($H_{6}$) that

$$\begin{aligned}& \int_{0}^{1}\omega (s)f\bigl(u(s)\bigr)\,ds- \int_{0}^{\sigma }\omega^{+}(s)f\bigl(u(s) \bigr)\,ds \\& \quad = \int_{\sigma }^{\xi }\omega^{+}(s)f\bigl(u(s) \bigr)\,ds- \int_{\xi }^{1}\omega^{-}(s)f\bigl(u(s) \bigr)\,ds \\& \quad \geq c \int_{\sigma }^{\xi }\omega^{+}(s)\psi \bigl(u(s) \bigr)\,ds- \int_{\xi }^{1} \omega^{-}(s)\psi \bigl(u(s) \bigr)\,ds \\& \quad \geq c \int_{\sigma }^{\xi }\omega^{+}(s)\psi \bigl(q(s)u(\xi)\bigr)\,ds- \int_{ \xi }^{1}\omega^{-}(s)\psi \bigl(q(s)u( \xi)\bigr)\,ds \\& \quad \geq c \int_{\sigma }^{\xi }\omega^{+}(s)f\bigl(q(s)u( \xi)\bigr)\,ds-\frac{1}{c} \int_{\xi }^{1}\omega^{-}(s)f\bigl(q(s)u(\xi) \bigr)\,ds \\& \quad \geq ck_{1}u^{\theta }(\xi)\frac{\sigma^{\theta }}{\xi^{\theta }} \int_{\sigma }^{\xi }\omega^{+}(s)\,ds- \frac{1}{c}k_{2}u^{\theta }( \xi) \int_{\xi }^{1}\omega^{-}(s)\,ds \\& \quad \geq u^{\theta }(\xi) \biggl( ck_{1}\frac{\sigma^{\theta }}{\xi^{\theta }} \int_{\sigma }^{\xi }\omega^{+}(s)\,ds- \frac{1}{c}k_{2} \int_{\xi } ^{1}\omega^{-}(s)\,ds \biggr) \\& \quad \geq 0. \end{aligned}$$

Secondly, if $t\in [0,\xi ]$, we have

$$ \int_{0}^{t}\omega (s)f\bigl(u(s)\bigr)\,ds= \int_{0}^{t}\omega^{+}(s)f\bigl(u(s) \bigr)\,ds \geq 0. $$

Since $p, q>1$, we get

$$\begin{aligned} (Tu)''(t) =& \biggl( -\phi_{q} \biggl( \int_{0}^{t}\lambda \omega (s)f\bigl(u(s)\bigr)\,ds \biggr) \biggr) ' \\ =& \biggl( - \biggl( \int_{0}^{t}\lambda \omega^{+}(s)f \bigl(u(s)\bigr)\,ds \biggr) ^{q-1} \biggr) ' \\ =&-(q-1) \biggl( \int_{0}^{t}\lambda \omega^{+}(s)f \bigl(u(s)\bigr)\,ds \biggr) ^{q-2} \lambda \omega^{+}(t)f \bigl(u(t)\bigr) \\ \leq& 0. \end{aligned}$$

If $t\in [\xi,1]$, then we have

$$\begin{aligned} \int_{0}^{t}\omega (s)f\bigl(u(s)\bigr)\,ds =& \int_{0}^{\xi }\omega^{+}(s)f\bigl(u(s) \bigr)\,ds- \int_{\xi }^{t}\omega^{-}(s)f\bigl(u(s) \bigr)\,ds \\ \geq& \int_{0}^{\xi }\omega^{+}(s)f\bigl(u(s) \bigr)\,ds- \int_{\xi }^{1}\omega^{-}(s)f\bigl(u(s) \bigr)\,ds \\ =& \int_{0}^{1}\omega (s)f\bigl(u(s)\bigr)\,ds \\ \geq& \int_{0}^{\sigma }\omega^{+}(s)f\bigl(u(s) \bigr)\,ds \\ \geq& 0. \end{aligned}$$

And then, for $t\in [\xi,1]$, it follows from $p,q>1$ that

$$\begin{aligned} (Tu)''(t) =& \biggl( -\phi_{q} \biggl( \int_{0}^{t}\lambda \omega (s)f\bigl(u(s)\bigr)\,ds \biggr) \biggr) ' \\ =& \biggl( -\phi_{q} \biggl( \int_{0}^{\xi }\lambda \omega^{+}(s)f \bigl(u(s)\bigr)\,ds- \int_{\xi }^{t}\lambda \omega^{-}(s)f \bigl(u(s)\bigr) \biggr) \biggr) ' \\ =& \biggl( - \biggl( \int_{0}^{\xi }\lambda \omega^{+}(s)f \bigl(u(s)\bigr)\,ds- \int_{ \xi }^{t}\lambda \omega^{-}(s)f \bigl(u(s)\bigr) \biggr) ^{q-1} \biggr) ' \\ =&-(q-1) \biggl( \int_{0}^{\xi }\lambda \omega^{+}(s)f \bigl(u(s)\bigr)\,ds- \int_{ \xi }^{t}\lambda \omega^{-}(s)f \bigl(u(s)\bigr) \biggr) ^{q-2} \bigl( -\lambda \omega^{-}(t)f \bigl(u(t)\bigr) \bigr) \\ \geq& 0. \end{aligned}$$

Moreover, by direct calculating, we get $(Tu)(t)\geq 0$ for $t\in J$, $(Tu)''(t)\leq 0$ for $t\in [0,\xi ]$, and $(Tu)''(t)\geq 0$ for $t\in [\xi,1]$. Thus, $T(K)\subset K$.

Then it finally follows from the Arzelà–Ascoli theorem that the operator T is completely continuous. □

From Lemma 2.3, since $(Tu)'(t)\leq 0$, then T is nonincreasing for $u\in K$. It is not difficult to see that

$$\begin{aligned} \Vert Tu \Vert _{PC} =&(Tu) (0) \\ =&\frac{1}{1-\eta }\biggl[ \int_{0}^{1}g(0) \int_{0} ^{1}\phi_{q} \biggl( \int_{0}^{s}\lambda \omega (\tau)f\bigl(u(\tau) \bigr)\,d\tau \biggr)\,ds\,dt \\ &{} +\mu \int_{0}^{1}g(0) \biggl( \sum _{t\leq t_{k}}I _{k}\bigl(u(t_{k})\bigr) \biggr)\,dt\biggr] \\ &{}+ \int_{0}^{1}\phi_{q} \biggl( \int_{0}^{s}\lambda \omega (\tau)f\bigl(u( \tau) \bigr)\,d\tau \biggr)\,ds+\mu \sum_{k=1}^{n}I_{k} \bigl(u(t_{k})\bigr). \end{aligned}$$

(2.6)

Lemma 2.4

If ($H_{1}$)–($H_{4}$) hold, then for $u\in K$ we get

$$\begin{aligned}& \Vert Tu \Vert _{PC}\leq \frac{1}{1-\eta }\phi_{q} \biggl( \int_{0}^{\xi }\lambda \omega^{+}(\tau)f \bigl(u(\tau)\bigr)\,d\tau \biggr) +\mu \frac{1}{1-\eta }\sum _{k=1}^{n}I_{k}\bigl(u(t_{k}) \bigr), \end{aligned}$$

(2.7)

$$\begin{aligned}& \Vert Tu \Vert _{PC}\geq \frac{(1-\int_{\xi }^{1}g(t)\,dt)(1-\xi)}{1-\eta }\phi _{q} \biggl( \int_{\frac{\sigma }{2}}^{\sigma }\lambda \omega^{+}(\tau)f \bigl(u( \tau)\bigr)\,d\tau \biggr) +\mu \sum_{k=1}^{n}I_{k} \bigl(u(t_{k})\bigr). \end{aligned}$$

(2.8)

Proof

By (2.6), for $u\in K$, we have

$$\begin{aligned} \Vert Tu \Vert _{PC} =&\frac{1}{1-\eta }\biggl[ \int_{0}^{1}g(t) \int_{t}^{1}\phi _{q} \biggl( \int_{0}^{s}\lambda \omega (\tau)f\bigl(u(\tau) \bigr)\,d\tau \biggr)\,ds\,dt \\ &{}+ \mu \int_{0}^{1}g(t) \biggl( \sum _{t\leq t_{k}}I_{k}\bigl(u(t_{k})\bigr) \biggr)\,dt\biggr] \\ &{} + \int_{0}^{1}\phi_{q} \biggl( \int_{0}^{s}\lambda \omega (\tau)f\bigl(u( \tau) \bigr)\,d\tau \biggr)\,ds+\mu \sum_{k=1}^{n}I_{k} \bigl(u(t_{k})\bigr) \\ \leq& \frac{1}{1-\eta }\Biggl[ \int_{0}^{1}g(t) \int_{0}^{1}\phi_{q} \biggl( \int _{0}^{s}\lambda \omega (\tau)f\bigl(u(\tau) \bigr)\,d\tau \biggr)\,ds\,dt \\ &{}+\mu \int _{0}^{1}g(t) \Biggl( \sum _{k=1}^{n}I_{k}\bigl(u(t_{k}) \bigr) \Biggr)\,dt\Biggr] \\ &{} + \int_{0}^{1}\phi_{q} \biggl( \int_{0}^{s}\lambda \omega (\tau)f\bigl(u( \tau) \bigr)\,d\tau \biggr)\,ds+\mu \sum_{k=1}^{n}I_{k} \bigl(u(t_{k})\bigr) \\ =&\frac{1}{1-\eta } \int_{0}^{1}\phi_{q} \biggl( \int_{0}^{s}\lambda \omega (\tau)f\bigl(u(\tau) \bigr)\,d\tau \biggr)\,ds+\mu \frac{1}{1-\eta }\sum_{k=1}^{n}I_{k} \bigl(u(t_{k})\bigr) \\ =&\frac{1}{1-\eta } \biggl[ \int_{0}^{\xi }\phi_{q} \biggl( \int_{0}^{s} \lambda \omega^{+}(\tau)f \bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds+ \int_{\xi }^{1} \phi_{q} \biggl( \int_{0}^{\xi }\lambda \omega^{+}(\tau)f \bigl(u(\tau)\bigr)\,d\tau \\ &{}- \int_{\xi }^{s}\lambda \omega^{-}(\tau)f \bigl(u(\tau)\bigr)\,d\tau \biggr) \biggr]\,ds+\mu \frac{1}{1-\eta }\sum _{k=1}^{n}I_{k}\bigl(u(t _{k}) \bigr) \\ \leq& \frac{1}{1-\eta }\biggl[ \int_{0}^{\xi }\phi_{q} \biggl( \int_{0}^{ \xi }\lambda \omega^{+}(\tau)f \bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds \\ &{}+ \int_{\xi } ^{1}\phi_{q} \biggl( \int_{0}^{\xi }\lambda \omega^{+}(\tau)f \bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds\biggr] \\ &{} +\mu \frac{1}{1-\eta }\sum_{k=1}^{n}I_{k} \bigl(u(t_{k})\bigr) \\ =&\frac{1}{1-\eta } \int_{0}^{1}\phi_{q} \biggl( \int_{0}^{\xi }\lambda \omega^{+}(\tau)f \bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds+\mu \frac{1}{1-\eta } \sum _{k=1}^{n}I_{k}\bigl(u(t_{k}) \bigr) \\ =&\frac{1}{1-\eta }\phi_{q} \biggl( \int_{0}^{\xi }\lambda \omega^{+}( \tau)f \bigl(u(\tau)\bigr)\,d\tau \biggr) +\mu \frac{1}{1-\eta }\sum _{k=1} ^{n}I_{k}\bigl(u(t_{k}) \bigr). \end{aligned}$$

Then (2.7) holds.

From (2.5) and (2.6), we have

$$\begin{aligned} \Vert Tu \Vert _{PC} =&\frac{1}{1-\eta }\biggl[ \int_{0}^{1}g(t) \int_{t}^{1}\phi _{q}\biggl( \int_{0}^{s}\lambda \omega (\tau)f\bigl(u(\tau) \bigr)\,d\tau \biggr)\,ds\,dt \\ &{}+ \mu \int_{0}^{1}g(t) \biggl( \sum _{t\leq t_{k}}I_{k}\bigl(u(t_{k})\bigr)\biggr)\,dt \biggr] \\ &{}+ \int_{0}^{1}\phi_{q}\biggl( \int_{0}^{s}\lambda \omega (\tau)f\bigl(u( \tau) \bigr)\,d\tau \biggr)\,ds+\mu \sum_{k=1}^{n}I_{k} \bigl(u(t_{k})\bigr) \\ =&\frac{1}{1-\eta }\biggl\{ \int_{0}^{\xi }g(t)\biggl[ \int_{t}^{\xi }\phi _{q}\biggl( \int_{0}^{s}\lambda \omega^{+}(\tau)f \bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds \\ &{}+ \int_{\xi }^{1}\phi_{q}\biggl( \int_{0}^{\xi }\lambda \omega^{+}(\tau)f \bigl(u( \tau)\bigr)\,d\tau \\ &{} - \int_{\xi }^{s}\lambda \omega^{-}(\tau)f \bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds\biggr]\,dt+ \int_{\xi }^{1}g(t) \int_{t}^{1}\phi_{q}\biggl( \int _{0}^{\xi }\lambda \omega^{+}(\tau)f \bigl(u(\tau)\bigr)\,d\tau \\ &{} - \int_{\xi }^{s}\lambda \omega^{-}(\tau)f \bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds\,dt\biggr\} + \int_{0}^{\xi }\phi_{q}\biggl( \int_{0}^{s} \lambda \omega^{+}(\tau)f \bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds \\ &{}+ \int_{\xi }^{1} \phi_{q}\biggl( \int_{0}^{\xi }\lambda \omega^{+}(\tau)f \bigl(u(\tau)\bigr)\,d\tau \\ &{} - \int_{\xi }^{s}\lambda \omega^{-}(\tau)f \bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds+ \mu \frac{1}{1-\eta } \int_{0}^{1}g(t)\sum_{t\leq t_{k}}I_{k} \bigl(u(t _{k})\bigr)\,dt \\ &{}+\mu \sum_{k=1}^{n}I_{k} \bigl(u(t_{k})\bigr) \\ \geq& \frac{1}{1-\eta } \int_{0}^{\xi }g(t) \int_{\xi }^{1}\phi_{q}\biggl( \int _{0}^{\xi }\lambda \omega^{+}(\tau)f \bigl(u(\tau)\bigr)\,d\tau - \int_{\xi }^{s} \lambda \omega^{-}(\tau)f \bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds\,dt \\ &{}+ \int_{\xi }^{1}\phi_{q}\biggl( \int_{0}^{\xi }\lambda \omega^{+}( \tau)f \bigl(u(\tau)\bigr)\,d\tau - \int_{\xi }^{s}\lambda \omega^{-}(\tau)f \bigl(u( \tau)\bigr)\,d\tau \biggr)\,ds \\ &{}+\mu \sum_{k=1}^{n}I_{k} \bigl(u(t_{k})\bigr) \\ =&\frac{1-\int_{\xi }^{1}g(t)\,dt}{1-\eta } \int_{\xi }^{1}\phi_{q}\biggl( \int _{0}^{\xi }\lambda \omega^{+}(\tau)f \bigl(u(\tau)\bigr)\,d\tau - \int_{\xi }^{s} \lambda \omega^{-}(\tau)f \bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds \\ &{}+\mu \sum_{k=1}^{n}I_{k} \bigl(u(t_{k})\bigr) \\ \geq& \frac{1-\int_{\xi }^{1}g(t)\,dt}{1-\eta } \int_{\xi }^{1}\phi_{q} \biggl( \int_{0}^{\xi }\lambda \omega^{+}(\tau)f \bigl(u(\tau)\bigr)\,d\tau - \int_{\xi }^{1}\lambda \omega^{-}(\tau)f \bigl(u(\tau)\bigr)\,d\tau \biggr)\,ds \\ &{}+ \mu \sum_{k=1}^{n}I_{k} \bigl(u(t_{k})\bigr) \\ =&\frac{(1-\int_{\xi }^{1}g(t)\,dt)(1-\xi)}{1-\eta }\phi_{q}\biggl( \int _{0}^{1}\lambda \omega (\tau)f\bigl(u(\tau) \bigr)\,d\tau \biggr) +\mu \sum_{k=1}^{n}I_{k} \bigl(u(t_{k})\bigr) \\ \geq& \frac{(1-\int_{\xi }^{1}g(t)\,dt)(1-\xi)}{1-\eta }\phi_{q}\biggl( \int _{0}^{\sigma }\lambda \omega^{+}(\tau)f \bigl(u(\tau)\bigr)\,d\tau \biggr) + \mu \sum_{k=1}^{n}I_{k} \bigl(u(t_{k})\bigr) \\ \geq& \frac{(1-\int_{\xi }^{1}g(t)\,dt)(1-\xi)}{1-\eta }\phi_{q}\biggl( \int _{\frac{\sigma }{2}}^{\sigma }\lambda \omega^{+}(\tau)f \bigl(u(\tau)\bigr)\,d\tau \biggr) +\mu \sum_{k=1}^{n}I_{k} \bigl(u(t_{k})\bigr). \end{aligned}$$

Then (2.8) holds. □

3 Main results

Based on the lemmas mentioned above, we give the following theorems and their proofs.

Theorem 3.1

Assume that ($H_{1}$)–($H_{6}$) hold. If $\theta_{1}>p-1$ and $\theta_{2}>1$, there exist $\lambda_{0}>0$ and $\mu_{0}>0$ such that problem (1.2) admits two positive solutions for $\lambda \in [\lambda_{0},+\infty)$, $\mu \in [\mu_{0},+\infty)$.

Proof

Denote

$$\begin{aligned}& A_{1}=\frac{1}{\int_{0}^{\xi }\lambda \omega^{+}(\tau)\,d\tau }\phi _{p}\biggl(\frac{1-\eta }{2} \biggr),\qquad A_{2}=\frac{1-\eta }{2\mu n}, \\& B_{1}=\frac{1}{\int_{\frac{\sigma }{2}}^{\sigma }\lambda \omega^{+}( \tau)\,d\tau }\phi_{p}\biggl(\frac{1-\eta }{2\alpha (1-\xi)(1-\int_{\xi } ^{1}g(t)\,dt)} \biggr), \qquad B_{2}=\frac{1}{2n\mu \alpha }. \end{aligned}$$

On the one hand, since $\theta_{1}>p-1$ and $\theta_{2}>1$, by ($H_{5}$), we get

$$ \lim_{u\rightarrow 0}\frac{f(u)}{\phi_{p}(u)}\leq \lim_{u\rightarrow 0} \frac{k_{2}u^{\theta_{1}}}{u^{p-1}}=0, \qquad \lim_{u\rightarrow 0}\frac{I _{k}(u)}{u}\leq \lim_{u\rightarrow 0}\frac{k_{4}u^{\theta_{2}}}{u}=0. $$

Hence, there exists $r>0$ such that

$$ f(u)< A_{1}\phi_{p}(u), \qquad I_{k}(u)< A_{2}u, \quad u\in [0,r]. $$

Then from (2.7), for $u\in \partial K_{r}$, then $\|u\|_{PC}=r$ and $0\leq u(t)\leq \|u\|=r$ for all $t\in J$. It is clear that $f(u(t))< A_{1}\phi_{p}(u(t))$ and $I_{k}(u(t))< A_{2}u(t)$ for all $t\in J$. Then from (2.7), for $u\in \partial K_{r}$, we get

$$\begin{aligned} \Vert Tu \Vert _{PC} \leq& \frac{1}{1-\eta }\phi_{q} \biggl( \int_{0}^{\xi }\lambda \omega^{+}(\tau)f \bigl(u(\tau)\bigr)\,d\tau \biggr) +\mu \frac{1}{1-\eta } \sum _{k=1}^{n}I_{k}\bigl(u(t_{k}) \bigr) \\ < &\frac{1}{1-\eta }\phi_{q} \biggl( \int_{0}^{\xi }\lambda \omega^{+}( \tau)A_{1}\phi_{p}\bigl(u(\tau)\bigr)\,d\tau \biggr) +\mu \frac{1}{1-\eta } \sum_{k=1}^{n}A_{2}u(t_{k}) \\ \leq& \frac{1}{1-\eta }\phi_{q} \biggl( \int_{0}^{\xi }\lambda \omega^{+}( \tau)A_{1}\phi_{p}\bigl( \Vert u \Vert _{PC} \bigr)\,d\tau \biggr) +\mu \frac{1}{1-\eta } \sum_{k=1}^{n}A_{2} \Vert u \Vert _{PC} \\ =&\frac{ \Vert u \Vert _{PC}}{2}+\frac{ \Vert u \Vert _{PC}}{2} \\ =& \Vert u \Vert _{PC}. \end{aligned}$$

Consequently,

$$ \Vert Tu \Vert _{PC}< \Vert u \Vert _{PC}, \quad \forall u\in \partial K_{r}. $$

(3.1)

On the other hand, we denote $\delta (t)=\min \{\frac{t}{\xi },\frac{ \xi -t}{\xi } \}$, $t\in [0,\xi ]$. If $u\in K$, then u is a nonnegative function on $[0,\xi ]$. So we get

$$ u(t)\geq \delta (t) \Vert u \Vert _{PC}, \quad t\in [0,\xi ]. $$

It follows that $u(t)\geq \alpha \|u\|_{PC}$, $t\in [\frac{\sigma }{2}, \sigma ]$, where $\alpha =\min_{\frac{\sigma }{2}\leq t\leq \sigma }\delta (t)$.

Since $\theta_{1}>p-1$ and $\theta_{2}>1$, by ($H_{5}$), we have

$$ \lim_{u\rightarrow +\infty }\frac{f(u)}{\phi_{p}(u)}\geq \lim_{u\rightarrow +\infty } \frac{k_{1}u^{\theta_{1}}}{u^{p-1}}=+ \infty,\qquad \lim_{u\rightarrow +\infty }\frac{I_{k}(u)}{u} \geq \lim_{u\rightarrow +\infty }\frac{k_{3}u^{\theta_{2}}}{u}=+\infty. $$

Furthermore, there exists $0< r< R'$ such that

$$ f(u)\geq B_{1}\phi_{p}(u),\qquad I_{k}(u)\geq B_{2}u,\quad u\in [R',+\infty), $$

Choose $R\geq \frac{R'}{\alpha }$. Then, for any $u\in \partial K_{R}$, we have $\min_{\frac{\sigma }{2}\leq t\leq \sigma }u(t)\geq \min_{\frac{\sigma }{2}\leq t\leq \sigma }\delta (t)\|u\|_{PC}= \alpha R\geq R'$, and $f(u(t))\geq B_{1}u^{p-1}(t)$, $I_{k}(u(t)) \geq B_{2}u(t)$, $t\in [\frac{\sigma }{2},\sigma ]$.

Then by (2.8), for $u\in \partial K_{R}$, we have

$$\begin{aligned} \Vert Tu \Vert _{PC} \geq& \frac{(1-\int_{\xi }^{1}g(t)\,dt)(1-\xi)}{1-\eta }\phi _{q} \biggl( \int_{\frac{\sigma }{2}}^{\sigma }\lambda \omega^{+}(\tau)f \bigl(u( \tau)\bigr)\,d\tau \biggr) +\mu \sum_{k=1}^{n}I_{k} \bigl(u(t_{k})\bigr) \\ \geq& \frac{(1-\int_{\xi }^{1}g(t)\,dt)(1-\xi)}{1-\eta }\phi_{q} \biggl( \int _{\frac{\sigma }{2}}^{\sigma }\lambda \omega^{+}( \tau)B_{1}\phi_{p}\bigl(u( \tau)\bigr)\,d\tau \biggr) +\mu \sum _{k=1}^{n}B_{2}u(t_{k}) \\ \geq& \frac{(1-\int_{\xi }^{1}g(t)\,dt)(1-\xi)}{1-\eta }\phi_{q} \biggl( \int _{\frac{\sigma }{2}}^{\sigma }\lambda \omega^{+}( \tau)B_{1}\phi_{p}\bigl( \alpha \Vert u \Vert _{PC}\bigr)\,d\tau \biggr) +\mu \sum_{k=1}^{n}B_{2} \alpha \Vert u \Vert _{PC} \\ =&\frac{\alpha (1-\int_{\xi }^{1}g(t)\,dt)(1-\xi)}{1-\eta } \Vert u \Vert _{PC} \phi_{q} \biggl( \int_{\frac{\sigma }{2}}^{\sigma }\lambda \omega^{+}( \tau)B_{1}\,d\tau \biggr) +\mu nB_{2}\alpha \Vert u \Vert _{PC} \\ \geq& \frac{1}{2} \Vert u \Vert _{PC}+\frac{1}{2} \Vert u \Vert _{PC} \\ =& \Vert u \Vert _{PC}. \end{aligned}$$

Consequently,

$$ \Vert Tu \Vert _{PC}\geq \Vert u \Vert _{PC}, \quad \forall u\in \partial K_{R}. $$

(3.2)

In addition, choose a number $r'\in (0,r)$. Noticing that $f(u)>0$ for all $u>0$ and $I_{k}(u)>0$ for all $u>0$, we can define

$$\begin{aligned}& f_{r'}=\min \bigl\{ f(u):\alpha r'\leq u\leq r'\bigr\} , \qquad I_{kr'}=\min \bigl\{ I _{k}: \alpha r'\leq u \leq r'\bigr\} , \\& I_{r'}=\min \{I_{kr'}:k=1,2,\ldots,n\}. \end{aligned}$$

Let $\lambda_{0}=\frac{1}{\int_{\frac{\sigma }{2}}^{\sigma }\omega ^{+}(\tau)f_{r'}\,d\tau }\phi_{p} ( \frac{r'(1-\eta)}{2(1-\int_{ \xi }^{1}g(t)\,dt)(1-\xi)} )$, $\mu_{0}=\frac{r'}{2nI_{r'}}$. Thus we have

$$\begin{aligned}& \frac{(1-\int_{\xi }^{1}g(t)\,dt)(1-\xi)}{1-\eta }\phi_{q} \biggl( \int_{\frac{ \sigma }{2}}^{\sigma }\lambda_{0} \omega^{+}(\tau)f_{r'}\,d\tau \biggr) = \frac{1}{2}r', \\& \mu_{0}nI_{r'}=\frac{1}{2}r'. \end{aligned}$$

If $u\in \partial K_{r'}$, then $\|u\|_{PC}=r'$ and $\alpha r'= \min_{\frac{\sigma }{2}\leq t\leq \sigma }\delta (t)\|u\|_{PC} \leq u(t)\leq \|u\|_{PC}=r'$, $t\in [\frac{\sigma }{2},\sigma ]$. It is clear that $f(u(t))\geq f_{r'}$ and $I_{k}(u(t))\geq I_{r'}$, $t\in [\frac{ \sigma }{2},\sigma ]$.

Then from (2.8), for $u\in \partial K_{r'}$, we have

$$\begin{aligned} \Vert Tu \Vert _{PC} \geq& \frac{(1-\int_{\xi }^{1}g(t)\,dt)(1-\xi)}{1-\eta }\phi _{q} \biggl( \int_{\frac{\sigma }{2}}^{\sigma }\lambda \omega^{+}(\tau)f \bigl(u( \tau)\bigr)\,d\tau \biggr) +\mu \sum_{k=1}^{n}I_{k} \bigl(u(t_{k})\bigr) \\ \geq& \frac{(1-\int_{\xi }^{1}g(t)\,dt)(1-\xi)}{1-\eta }\phi_{q} \biggl( \int _{\frac{\sigma }{2}}^{\sigma }\lambda \omega^{+}( \tau)f_{r'}\,d\tau \biggr) + \mu \sum_{k=1}^{n}I_{r'} \\ \geq& \frac{(1-\int_{\xi }^{1}g(t)\,dt)(1-\xi)}{1-\eta }\phi_{q} \biggl( \int _{\frac{\sigma }{2}}^{\sigma }\lambda_{0} \omega^{+}(\tau)f_{r'}\,d\tau \biggr) +\mu_{0}nI_{r'} \\ =&\frac{1}{2}r'+\frac{1}{2}r' \\ =&r'= \Vert u \Vert _{PC}. \end{aligned}$$

Consequently,

$$ \Vert Tu \Vert _{PC}\geq \Vert u \Vert _{PC}, \quad \forall u\in \partial K_{r'}. $$

(3.3)

Therefore, applying Lemma 1.1 to (3.1), (3.2), and (3.3) yields that T has two fixed points $u_{1}\in \overline{K}_{R}\setminus \overline{K}_{r}$ and $u_{2}\in K_{r}\setminus K_{r'}$. Thus, if $\theta_{1}>p-1$ and $\theta_{2}>1$, there exist $\lambda_{0}>0$ and $\mu_{0}>0$ such that problem (1.2) admits two positive solutions for $\lambda \in [\lambda_{0},+\infty)$ and $\mu \in [\mu_{0},+\infty)$. The proof of Theorem 3.1 is completed. □

Theorem 3.2

Assume that ($H_{1}$)–($H_{6}$) hold. If $0<\theta_{1}<p-1$ and $0<\theta_{2}<1$, there exist $\lambda^{0}>0$ and $\mu^{0}>0$ such that problem (1.2) admits two positive solutions for $\lambda \in (0,\lambda^{0}]$ and $\mu \in (0,\mu^{0}]$.

Proof

On the one hand, since $0<\theta_{1}<p-1$ and $0<\theta _{2}<1$, by ($H_{5}$), we get

$$ \lim_{u\rightarrow 0}\frac{f(u)}{\phi_{p}(u)}\geq \lim_{u\rightarrow 0} \frac{k_{1}u^{\theta_{1}}}{u^{p-1}}=+\infty,\qquad \lim_{u\rightarrow 0}\frac{I _{k}(u)}{u} \geq \lim_{u\rightarrow 0}\frac{k_{3}u^{\theta_{2}}}{u}=+ \infty. $$

Hence, there exists $r_{1}>0$ such that

$$ f(u)> B_{1} \phi_{p}(u),\qquad I_{k}(u)> B_{2}u,\quad u\in [0,r_{1}]. $$

Then we have $\min \{f(u):\alpha r_{1}\leq u\leq r_{1}\}> B_{1}\phi _{p}(u)$ and $\min \{I_{k}(u):\alpha r_{1}\leq u\leq r_{1}\}> B_{2}u$.

If $u\in \partial K_{r_{1}}$, then $\|u\|_{PC}=r_{1}$ and $\alpha r _{1}=\min_{\frac{\sigma }{2}\leq t\leq \sigma }\delta (t)\|u\| _{PC}\leq u(t)\leq \|u\|_{PC}=r_{1}$, $t\in [\frac{\sigma }{2},\sigma ]$. It is easy to see that $f(u(t))> B_{3}\phi_{p}(u(t))$, $I_{k}(u(t))> B_{4}u(t)$, $t\in [\frac{\sigma }{2},\sigma ]$. Then from (2.8), for $u\in \partial K_{r_{1}}$, similar to (3.2), we have

$$ \Vert Tu \Vert _{PC}> \Vert u \Vert _{PC}, \quad \forall u\in \partial K_{r_{1}}. $$

(3.4)

On the other hand, since $0<\theta_{1}<p-1$ and $0<\theta_{2}<1$, by ($H_{5}$), we have

$$ \lim_{u\rightarrow +\infty }\frac{f(u)}{\phi_{p}(u)}\leq \lim_{u\rightarrow +\infty } \frac{k_{2}u^{\theta }}{u^{p-1}}=0,\qquad \lim_{u\rightarrow +\infty }\frac{I_{k}(u)}{u}\leq \lim_{u\rightarrow +\infty }\frac{k_{4}u^{\theta }}{u}=0. $$

Furthermore, there exists $0< r_{1}< R'_{1}<+\infty$ such that

$$ f(u)\leq \frac{A_{1}}{2}\phi_{p}(u),\qquad I_{k}(u)\leq \frac{A_{2}}{2}u, \quad u\in \bigl[R'_{1},+\infty\bigr). $$

Let $M_{1}=\max \{f(u):0\leq u\leq R'_{1}\}$ and $M_{2}=\max \{I_{k}:0 \leq u\leq R'_{1},k=1,2,\ldots,n\}$. It implies that

$$ f(u)\leq \frac{A_{1}}{2}\phi_{p}(u)+M_{1},\qquad I_{k}(u)\leq \frac{A_{2}}{2}u+M_{2}, \quad u\in [0,+ \infty). $$

Choose $R_{1}\geq \{R'_{1},\frac{2\phi_{q}(2\int_{0}^{\xi }\lambda \omega^{+}(\tau)M_{1}\,d\tau)}{1-\eta },4\mu nM_{2}\}$. If $u\in \partial K_{R_{1}}$, then $\|u\|=R_{1}$ and $0\leq u(t)\leq R_{1}$, $t \in J$. It is easy to see that $f(u(t))\leq \frac{A_{1}}{2}\phi_{p}(u(t))+M _{1}$, $I_{k}(u(t))\leq \frac{A_{2}}{2}u(t)+M_{2}$, $t\in J$. Then from (2.7), for $u\in \partial K_{R_{1}}$, we have

$$\begin{aligned} \Vert Tu \Vert _{PC} \leq& \frac{1}{1-\eta }\phi_{q} \biggl( \int_{0}^{\xi }\lambda \omega^{+}(\tau)f \bigl(u(\tau)\bigr)\,d\tau \biggr) +\mu \sum_{k=1}^{n}I _{k}\bigl(u(t_{k})\bigr) \\ \leq& \frac{1}{1-\eta }\phi_{q} \biggl( \int_{0}^{\xi }\lambda \omega^{+}( \tau) \biggl(\frac{A_{1}}{2}\phi_{p}\bigl(u(\tau)\bigr)+M_{1} \biggr)\,d\tau \biggr) +\mu \sum_{k=1}^{n} \biggl(\frac{A_{2}}{2}u(t_{k})+M_{2}\biggr) \\ \leq& \frac{1}{1-\eta }\phi_{q} \biggl( \int_{0}^{\xi }\lambda \omega^{+}( \tau) \frac{A_{1}}{2}\phi_{p}\bigl( \Vert u \Vert _{PC} \bigr)\,d\tau + \int_{0}^{\xi }\lambda \omega^{+}( \tau)M_{1}\,d\tau \biggr) \\ &{} +\mu \sum_{k=1}^{n}\frac{A _{2}}{2} \Vert u \Vert _{PC}+\mu nM_{2} \\ \leq& \frac{1}{1-\eta }\phi_{q} \biggl( \frac{1}{2} \phi_{p}\biggl(\frac{ \Vert u \Vert _{PC}(1-\eta)}{2}\biggr)+\frac{1}{2} \phi_{p}\biggl(\frac{R_{1}(1-\eta)}{2}\biggr) \biggr) +\frac{ \Vert u \Vert _{PC}}{4}+ \frac{R_{1}}{4} \\ =&\frac{1}{2}R_{1}+\frac{1}{2}R_{1} \\ =&R_{1}= \Vert u \Vert _{PC}. \end{aligned}$$

Consequently,

$$ \Vert Tu \Vert _{PC}\leq \Vert u \Vert _{PC},\quad \forall u\in \partial K_{R_{1}}. $$

(3.5)

In addition, choosing a number $r'_{1}\in (0,r_{1})$, we can define

$$\begin{aligned}& f^{r'_{1}}=\max \bigl\{ f(u):0< u\leq r'_{1}\bigr\} , \qquad I_{k}^{r'_{1}}=\max \bigl\{ I _{k}(u):0< u\leq r'_{1}\bigr\} , \\& I^{r'_{1}}=\max \bigl\{ I_{k}^{r'_{1}}:k=1,2,\ldots,n\bigr\} . \end{aligned}$$

Let $\lambda^{0}=\frac{1}{\int_{0}^{\xi }\omega^{+}(\tau)f^{r'_{1}}\,d\tau }\phi_{p} ( \frac{r'_{1}(1-\eta)}{2} )$ and $\mu^{0}=\frac{r'_{1}}{2nI ^{r'_{1}}}$. It is clear that

$$ \frac{1}{1-\eta }\phi_{q} \biggl( \int_{0}^{\xi }\lambda^{0} \omega^{+}( \tau)f^{r'_{1}}\,d\tau \biggr) \leq \frac{1}{2}r'_{1}, \qquad \mu^{0}nI^{r'_{1}} \leq \frac{1}{2}r'_{1}. $$

If $u\in \partial K_{r'_{1}}$, then $\|u\|_{PC}=r'_{1}$ and $0\leq u(t)\leq \|u\|_{PC}=r'_{1}$, $t\in J$. It is clear that $f(u(t))\leq f^{r'_{1}}$, $I_{k}(u(t))\leq I^{r'_{1}}$, $t\in J$. Then from (2.7), for $u\in \partial K_{r'_{1}}$, we have

$$\begin{aligned} \Vert Tu \Vert _{PC} \leq& \frac{1}{1-\eta }\phi_{q} \biggl( \int_{0}^{\xi }\lambda \omega^{+}(\tau)f \bigl(u(\tau)\bigr)\,d\tau \biggr) +\mu \sum_{k=1}^{n}I _{k}\bigl(u(t_{k})\bigr) \\ \leq& \frac{1}{1-\eta }\phi_{q} \biggl( \int_{0}^{\xi }\lambda \omega^{+}( \tau)f^{r'_{1}}\,d\tau \biggr) +\mu \sum_{k=1}^{n}I^{r'_{1}} \\ \leq& \frac{1}{1-\eta }\phi_{q} \biggl( \int_{0}^{\xi }\lambda^{0}\omega ^{+}(\tau)f^{r'_{1}}\,d\tau \biggr) +\mu^{0}nI^{r'_{1}} \\ =&\frac{1}{2}r'_{1}+\frac{1}{2}r'_{1} \\ =&r'_{1}= \Vert u \Vert _{PC}. \end{aligned}$$

Consequently,

$$ \Vert Tu \Vert _{PC}\leq \Vert u \Vert _{PC}, \quad \forall u\in \partial K_{r'_{1}}. $$

(3.6)

Therefore, applying Lemma 1.1 to (3.4), (3.5), and (3.6) yields that T has two fixed points $u'_{1}\in \overline{K}_{R_{1}}\setminus \overline{K}_{r_{1}}$ and $u'_{2}\in K_{r_{1}}\setminus K_{r'_{1}}$. Thus, if $0<\theta_{1}<p-1$ and $0<\theta_{2}<1$, there exist $\lambda^{0}>0$ and $\mu^{0}>0$ such that problem (1.2) admits two positive solutions for $\lambda \in (0,\lambda^{0}]$ and $\mu \in (0, \mu^{0}]$. The proof of Theorem 3.2 is finished. □

Remark 3.1

If $I_{k}=0$ ($k=1,2,\ldots,n$), even for the case $g(t)\equiv 0$ on J, the results of the present paper are still novel.

Remark 3.2

Comparing with Li, Feng, and Qin [60], the main features of this paper are as follows:

• (i) $p> 1$ is considered, not only $p\equiv 2$.

• (ii) $I_{k}\neq 0$ ($k=1,2,\ldots,n$) is considered.

• (iii) The basic space $PC[0,1]$ is available, not $C[0,1]$.

4 An example

We give an example to illustrate our main conclusions.

Example 4.1

Let $p=\frac{3}{2}$, $n=1$, $t_{1}=\frac{1}{2}$. Consider the following problem:

$$ \textstyle\begin{cases} -(\phi_{p}(u'))'=\lambda \omega (t)(u+\sin u), \quad 0< t< 1, \\ -\Delta u|_{t=t_{1}}=\mu I_{1}(u(t_{1})), \\ \Delta u'|_{t=t_{1}}=0, \\ u'(0)=0, \quad u(1)=\int_{0}^{1}g(t)u(t)\,dt, \end{cases} $$

(4.1)

where

$$ \omega (t)= \textstyle\begin{cases} 12(\frac{2}{3}-t), & t\in [0,\frac{2}{3}], \\ \frac{2}{3}-t, & t\in [\frac{2}{3},1], \end{cases}\displaystyle \qquad I_{1}(u)=u^{2}, \qquad g(t)=t. $$

From the definition of $\omega (t)$ and $g(t)$, we know that $\xi =\frac{1}{2}$ and $\eta =\int_{0}^{1}t\,dt=\frac{1}{2}$. From $p=\frac{3}{2}$, we can get that $q=3$.

Since f is nondecreasing, then $c=1$. For fixed $k_{1}=1$, $k_{2}=2$, $\theta_{1}=1$, $k_{3}=k_{4}=1$, $\theta_{2}=2$, $\sigma =\frac{1}{4}$, we can prove that ($H_{5}$) holds.

In fact,

$$\begin{aligned} \frac{1}{2} \int_{\frac{1}{2}}^{\frac{2}{3}}12\biggl(\frac{2}{3}-\tau \biggr)\,d\tau =&6 \int_{\frac{1}{2}}^{\frac{2}{3}}\biggl(\frac{2}{3}-\tau\biggr)\,d\tau \\ =&6\biggl(\frac{2}{3}\tau -\frac{\tau^{2}}{2}\biggr)\bigg|_{\frac{1}{2}}^{\frac{2}{3}}\,d\tau \\ =&\frac{1}{12}, \end{aligned}$$

and

$$ 2\times \frac{2}{3} \int_{\frac{2}{3}}^{1}\biggl(\tau -\frac{2}{3}\biggr)\,d\tau = \frac{4}{3}\biggl(\frac{\tau^{2}}{2}-\frac{2}{3}\tau \biggr)\bigg|_{\frac{2}{3}}^{1}= \frac{2}{27}. $$

Obviously, $\frac{1}{12}>\frac{2}{27}$. Thus

$$ \frac{1}{2} \int_{\frac{1}{2}}^{\frac{2}{3}}12\biggl(\frac{2}{3}-\tau \biggr)\,d\tau \geq \frac{4}{3} \int_{\frac{2}{3}}^{1}\biggl(\tau -\frac{2}{3}\biggr)\,d\tau. $$

This shows that ($H_{6}$) holds.

Let $\lambda_{0}=\frac{12}{7}\sqrt{\frac{3}{13}}(\frac{1}{8}+\sin \frac{1}{8})^{-1}$, $\mu_{0}=16$. Then it follows from Theorem 3.1 that problem (4.1) admits two positive solutions for $\lambda \in [ \frac{12}{7}\sqrt{\frac{3}{13}}(\frac{1}{8}+\sin \frac{1}{8})^{-1},+ \infty)$, $\mu \in [16,\infty)$.