Multiplicity and nonexistence of positive solutions to impulsive Sturm–Liouville boundary value problems

Abstract

In this paper, we study the existence, nonexistence, and multiplicity of positive solutions to a nonlinear impulsive Sturm–Liouville boundary value problem with a parameter. By using a variational method, we prove that the problem has at least two positive solutions for the parameter $\lambda \in (0,\Lambda )$, one positive solution for $\lambda =\Lambda$, and no positive solution for $\lambda >\Lambda$, where $\Lambda >0$ is a constant.

1 Introduction

In this paper, we investigate the following nonlinear impulsive Sturm–Liouville boundary value problem:

$$ \textstyle\begin{cases} -Lu(t)=f(t,u(t)),\quad t\in J/\{t_{1},t_{2},\ldots,t_{p}\}, \\ \Delta (h(t_{k})u'(t_{k}))=-\lambda I_{k}(u(t_{k})),\quad k=1,2,\ldots,p, \\ R_{1}(u)=R_{2}(u)=0, \end{cases} $$

(1.1)

where $J=[0,1]$, p is a positive integer, $0=t_{0}< t_{1}< t_{2}<\cdots<t_{p}<t_{p+1}=1$, $Lu=(h(t)u'(t))'-q(t)u(t)$, $R_{1}(u)=\alpha u'(0)-\beta u(0), R_{2}(u)=\gamma u'(1)+\sigma u(1)$, $\alpha,\beta ,\gamma ,\sigma \in \mathbb{R}$, $\Delta (h(t_{k})u'(t_{k}))=h(t_{k}^{+})u'(t_{k}^{+})-h(t_{k}^{-})u'(t_{k}^{-})$, $u'(t_{k}^{+})$ and $u'(t_{k}^{-})$ represent the right limit and the left limit of $u'(t)$ at $t_{k}$, respectively, and λ is a positive parameter.

The following conditions are assumed:

$(H_{1})$:

$h\in C^{1}(J), q\in C(J), h>0,q>0$ for all $t\in J$, $\alpha ,\beta ,\gamma ,\sigma \geq 0$, $\alpha ^{2}+\beta ^{2}>0,\gamma ^{2}+\sigma ^{2}>0$, and the linear equation (2.1) has only a trivial solution.

$(H_{2})$:

$f\in C(J\times \mathbb{R}^{+},\mathbb{R}^{+})$, where $\mathbb{R}^{+}=(0,+\infty )$. $f(t,x)=o(x)$ as $x\rightarrow 0^{+}$ uniformly to $t\in J$, and there exist constants $\mu >2$ and $r>0$ such that for $x\geq r, t\in J$,

$$ \mu F(t,x)\leq xf(t,x), $$

where $F(t,x)=\int _{0}^{x}f(t,s)\,ds$.

$(H_{3})$:

$I_{k}\in C([0,+\infty ),[0,+\infty ))$. For any $k\in \{1,2,\ldots, p\}$, $I_{k}(x)\rightarrow +\infty$ as $x\rightarrow +\infty$ or $\sup_{x>0}I_{k}(x)<\infty$, and there exist $1\leq K\leq p$, $0\leq \tau _{0}<1, \kappa >0$ such that $I_{K}(x)\geq \kappa x^{\tau _{0}}$ for $x>0$.

It should be noted that impulsive differential equations are important models described by phenomena with abrupt changes in their states. Such models have considerable popularity in physics, population dynamics, ecology, industrial robotics, economics, biotechnology, and so on, see [1–3].

In recent years, boundary value problems of impulsive differential equations have been researched extensively, see for example [4–18] and the references therein. In [19], Tian and Ge studied the special cases of (1.1) without parameters

$$ \textstyle\begin{cases} -u''(t)-\mu u(t)=g(t,u(t)),\quad t\in J/\{t_{1},t_{2},\ldots,t_{p}\}, \\ \Delta (h(t_{k})u'(t_{k}))=s_{k}(u(t_{k})),\quad k=1,2,\ldots,p, \\ R_{1}(u)=R_{2}(u)=0, \end{cases} $$

(1.2)

where $\mu \in \mathbb{R}$, g and $s_{k}$ are of superlinear growth or sublinear growth, the authors showed the existence of one or two positive solutions for (1.2). In [20], authors obtained the existence of a sign-changing solution and multiple solutions of

$$ \textstyle\begin{cases} -Lu(t)=g(t,u(t)), \quad t\in J/\{t_{1},t_{2},\ldots,t_{p}\}, \\ \Delta (h(t_{k})u'(t_{k}))=s_{k}(u(t_{k})), \quad k=1,2,\ldots,p, \\ R_{1}(u)=R_{2}(u)=0. \end{cases} $$

(1.3)

The technical approach makes use of the minimax method.

In [21], by using a three critical point theorem, authors considered the following fourth order impulsive differential equations with two control parameters:

$$ \textstyle\begin{cases} u^{(4)}(t)+Au''(t)+Bu(t)=\lambda f(t,u(t))+\mu g(t,u(t)),\quad t\neq t_{k}, t\in J, \\ \Delta (u''(t_{k}))=I_{1k}(u'(t_{k})),\quad k=1,2,\ldots,p, \\ \Delta (u'''(t_{k}))=I_{2k}(u(t_{k})),\quad k=1,2,\ldots,p, \\ u(0)=u(T)=u''(0)=u''(T)=0. \end{cases} $$

(1.4)

They established the sharp bounds of the parameters λ and μ for which problem (1.4) admits at least three solutions. For the existence and multiplicity results of solutions of impulsive boundary value problems obtained by using variational methods, we refer to [22–31].

To our knowledge, there are few studies on the connection between the number of solutions and the given parameter for impulsive differential equations. The aim of this paper is to show the multiplicity, existence, and nonexistence of positive solutions for various values of the given parameter. Our result shows that the number of positive solutions of (1.1) changes if the parameter crosses a certain threshold.

This paper is organized as follows. In Sect. 2, we recall some preliminary results. In Sect. 3, by using the mountain pass principle, we prove that (1.1) has at least two positive solutions if λ is sufficiently small, and no solution if λ is sufficiently large, see Theorem 3.1. Throughout the paper, the symbols $C_{1}, C_{2}, \ldots$ denote various positive constants whose exact values are not essential to the analysis of the problem. In addition, without loss of generality, we may assume that $f(t,x)=0,I_{k}(x)=0(1\leq k\leq p)$ for $x<0$ as we only consider positive solutions.

2 Basic lemma

Let $G(t,s)$ be Green’s function of the problem

$$ \textstyle\begin{cases} -Lu=0, \\ R_{1}(u)=R_{2}(u)=0. \end{cases} $$

(2.1)

From [32], $G(t,s)$ can be written as

$$ G(t,s)=\frac{1}{\omega}\textstyle\begin{cases} m(t)n(s), & 0\leq t\leq s \leq 1, \\ m(s)n(t), & 0\leq s\leq t \leq 1. \end{cases} $$

(2.2)

Lemma 2.1

The function $G(t,s)$ defined by (2.2) has the following properties:

$(1)$ $m\in C^{2}(J,\mathbb{R})$ is increasing and $m(t)>0, t\in (0,1]$.

$(2)$ $n\in C^{2}(J,\mathbb{R})$ is decreasing and $n(t)>0, t\in [0,1)$.

$(3)$ $Lm\equiv 0, m(0)= \alpha , m'(0)= \beta$.

$(4)$ $Ln\equiv 0, n(1)= \gamma , n'(1)= -\sigma$.

$(5)$ ω is a positive constant and $p(t)(m'(t)n(t)-m(t)n'(t))\equiv \omega$.

$(6)$ $G(t,s)$ is continuous and symmetric in $\{(t,s): 0\leq t \leq s\leq 1\} \times \{(t,s): 0\leq s \leq t\leq 1 \}$.

$(7)$ $G(t,s)$ has continuous partial derivatives in $\{(t,s): 0\leq t \leq s\leq 1\}, \{(t,s): 0\leq s \leq t\leq 1 \}$.

$(8)$ For each fixed $s\in [0,1], G(t,s)$ satisfies $LG(t,s)=0$ for $t\neq s, t\in J$ and $R_{1}(G)=R_{2}(G)=0$.

$(9)$ $G_{t}'$ has a discontinuous point of the first kind at $t=s$ and $G_{t}'(s+0,s)-G_{t}'(s-0,s)=-\frac{1}{h(s)}, s\in (0,1)$.

Let $P=\{u\in C(J), u\geq 0\}$ and $(Tu)(t)=\int _{0}^{1}G(t,s)u(s)\,ds$, then

$$ T \bigl(P/\{0\} \bigr)\subset \operatorname{int} P. $$

By the Krein–Rutman theorem, the spectral radius $r(T)>0$ is determined by a simple eigenvalue of T having an eigenfunction $\varphi _{0}(t)\in P$ with $\varphi _{0}>0, t\in (0,1)$. It is easy to check that $\lambda ^{*}=r^{-1}(T)$ is the smallest eigenvalue of the eigenvalue problem:

$$ \textstyle\begin{cases} -Lu=\lambda u, \\ R_{1}(u)=R_{2}(u)=0, \end{cases} $$

(2.3)

and $\varphi _{0}$ is eigenfunction corresponding to $\lambda _{1}$.

Definition 2.1

A function $u\in \Theta =\{x\in C(J):h x'\in C(J/\{t_{1},\ldots,t_{p}\}),x'(t_{k}^{+}),x'(t_{k}^{-})$ exists and $x'(t_{k})=x'(t_{k}^{-})\}$ is said to be a solution of (1.1) if u satisfies the equation in (1.1) for $t\in J/\{t_{1}, t_{2},\ldots,t_{p}\}$ and impulsive conditions, boundary conditions of (1.1). The function u is a positive solution of (1.1) if u is a solution of (1.1) and $u(t)>0$ for $t\in (0,1)$.

Let

$$\begin{aligned} &H^{1}(0,1)= \bigl\{ u\in L^{2}(0,1):u'\in L^{2}(0,1) \bigr\} ,\\ &H^{1}_{0}(0,1)= \bigl\{ u\in H^{1}(0,1):u(0)=u(1)=0 \bigr\} ,\\ &\Sigma _{1}= \bigl\{ u\in H^{1}(0,1):u(0)=0 \bigr\} ,\\ &\Sigma _{2}= \bigl\{ u\in H^{1}(0,1):u(1)=0 \bigr\} ,\\ &A=\textstyle\begin{cases} \frac{\sigma}{\gamma}, & \gamma \neq 0, \\ 0, & \gamma =0,\end{cases}\displaystyle \qquad B=\textstyle\begin{cases} \frac{\beta}{\alpha}, & \alpha \neq 0, \\ 0, & \alpha =0,\end{cases}\displaystyle \\ &H=\textstyle\begin{cases} H^{1}(0,1), & \alpha \neq 0,\gamma \neq 0, \\ H_{0}^{1}(0,1), & \alpha =\gamma =0, \\ \Sigma _{1}, & \alpha =0, \gamma \neq 0, \\ \Sigma _{2}, & \alpha \neq 0, \gamma =0. \end{cases}\displaystyle \end{aligned}$$

Define an inner product in H as follows:

$$ (u,v)= \int _{0}^{1} \bigl(h(t)u'v'+q(t)uv \bigr)\,dt+h(1)Au(1)v(1)+h(0)Bu(0)v(0). $$

This inner product induces the norm

$$ \Vert u \Vert = \biggl( \int _{0}^{1} \bigl(h \bigl(u' \bigr)^{2}+qu^{2} \bigr)\,dt+h(1)Au^{2}(1)+h(0)Bu^{2}(0) \biggr)^{\frac{1}{2}}. $$

It is easy to check that H with the inner product $(\cdot , \cdot )$ is a Hilbert space, and $u^{+}=\max \{u,0\}\in H, u^{-}=\max \{-u,0\}\in H$ for $u\in H$.

Define the functional $\Phi _{\lambda}$ in H by

$$ \Phi _{\lambda}(u)=\frac{1}{2} \Vert u \Vert ^{2}- \int _{0}^{1}F \bigl(t,u(t) \bigr)\,dt- \lambda \sum ^{p}_{k=1} \int _{0}^{u(t_{k})}I_{k}(s)\,ds. $$

Then $\Phi _{\lambda}\in C^{1}(H,R)$, and

$$ \bigl\langle \Phi '_{\lambda}(u),v \bigr\rangle =(u,v)- \int _{0}^{1}f \bigl(t,u(t) \bigr)v(t)\,dt- \lambda \sum^{p}_{k=1}I_{k} \bigl(u(t_{k}) \bigr)v(t_{k}). $$

Lemma 2.2

If $u\in H$, then there is $C>0$ such that

$$ \vert u \vert _{0}\leq C \Vert u \Vert ,\quad \forall u\in H, $$

where $|u|_{0}=\max_{t\in J}|u(t)|$.

Proof

Let $\|\cdot \|_{H}$ be the usual norm of $H^{1}(0,1)$. From the imbedding theorem, we know that there is $C_{0}>0$ such that

$$ \vert u \vert _{0}\leq C_{0} \Vert u \Vert _{H}. $$

Note that there exists $C>0$ such that

$$ \vert u \vert _{0}\leq C_{0} \Vert u \Vert _{H}\leq C \Vert u \Vert . $$

The claim follows. □

Lemma 2.3

The problem of finding a solution u of (1.1) is equivalent to that of finding a critical point of $\Phi _{\lambda}$, that is, $\langle \Phi '_{\lambda}(u),v \rangle =0$ for all $v\in H$.

Proof

Assume that $u\in \Theta$ is a solution of (1.1). It is easy to check that $u\in H$. For any $v\in H$,

$$ \int _{0}^{1} \bigl[- \bigl(hu' \bigr)'+qu \bigr]v\,dt= \int _{0}^{1}f(t,u)v \,dt, $$

that is,

$$\begin{aligned} &{-} \int _{0}^{t_{1}}v \bigl(hu' \bigr)'\,dt- \int _{t_{1}}^{t_{2}}v \bigl(hu' \bigr)'\,dt-\cdots- \int _{t_{p}}^{1}v \bigl(hu' \bigr)'\,dt+ \int _{0}^{1}quv \,dt= \int _{0}^{1}f(t,u)v \,dt,\\ &\int _{0}^{1}hu'v'\,dt+ \int _{0}^{1}huv \,dt+h(0)u'(0)v(0)-h(1)u'(1)v(1)\\ &\quad = \int _{0}^{1}f(t,u)v \,dt+\lambda \sum ^{p}_{k=1}I_{k} \bigl(u(t_{k}) \bigr)v(t_{k}),\\ &(u,v)= \int _{0}^{1}f(t,u)v \,dt+\lambda \sum ^{p}_{k=1}I_{k} \bigl(u(t_{k}) \bigr)v(t_{k}). \end{aligned}$$

Hence, u is a critical point of $\Phi _{\lambda}$ in H.

Let $u\in H$ be a critical point of $\Phi _{\lambda}$, then $\langle \Phi '_{\lambda}(u),v\rangle =0$ for any $v\in H$. Without generality, we have

$$ \begin{aligned} 0&= \int _{t_{k}}^{t_{k+1}} \bigl(hu'v'+quv-f(t,u)v \bigr)\,dt \\ &= \int _{t_{k}}^{t_{k+1}}hu'v' \,dt+ \int _{t_{k}}^{t_{k+1}}v \,d \biggl( \int _{t}^{t_{k}} \bigl(q(s)u(s)-f \bigl(s,u(s) \bigr) \bigr)\,ds \biggr) \\ &= \int _{t_{k}}^{t_{k+1}} \biggl[h(t)u'- \int _{t}^{t_{k}} \bigl(q(s)u(s)-f \bigl(s,u(s) \bigr) \bigr)\,ds \biggr]v'(t)\,dt. \end{aligned} $$

Hence,

$$ h(t)u'(t)- \int _{t}^{t_{k}} \bigl(q(s)u(s)-f \bigl(s,u(s) \bigr) \bigr)\,ds\equiv C, \quad\text{a.e } t\in (t_{k},t_{k+1}). $$

So, $hu'$ has a weak derivative satisfying

$$ \bigl(hu' \bigr)'-q(t)u(t)+f \bigl(t,u(t) \bigr)=0, \quad\text{a.e. } t\in (t_{k},t_{k+1}). $$

From the continuity of $h,q, f,u$, we see that $(hu')'$ exists for $t\in (t_{k},t_{k+1})$. Hence u satisfies the equality in (1.1).

Noting that

$$ \begin{aligned} 0={}&(u,v)- \int _{0}^{1}f \bigl(t,u(t) \bigr)v(t)\,dt-\lambda \sum^{p}_{k=1}I_{k} \bigl(u(t_{k}) \bigr)v(t_{k}) \\ ={}&\sum_{k=0}^{p}h(t)u'(t)v(t)|_{t_{k}^{+}}^{t_{k+1}}+ \int _{0}^{1}(-Lu)v\,dt +h(1)Au(1)v(1)+h(0)Bu(0)v(0) \\ &{}- \int _{0}^{1}f \bigl(t,u(t) \bigr)v(t)\,dt-\lambda \sum^{p}_{k=1}I_{k} \bigl(u(t_{k}) \bigr)v(t_{k}) \\ ={}& \int _{0}^{1} \bigl(-Lu-f(t,u) \bigr)v\,dt+ h(1) \bigl[u'(1)+Au(1) \bigr]v(1)+h(0) \bigl[-u'(0)+Bu(0) \bigr]v(0) \\ &{}-\sum_{k=1}^{p} \bigl[\Delta \bigl(h(t_{k})u'(t_{k}) \bigr)+\lambda I_{k} \bigl(u(t_{k}) \bigr) \bigr]v(t_{k}), \end{aligned} $$

we get

$$\begin{aligned} & h(1) \bigl[u'(1)+Au(1) \bigr]v(1)+h(0) \bigl[-u'(0)+Bu(0) \bigr]v(0) \\ &\quad{}-\sum_{k=1}^{p} \bigl[\Delta \bigl(h(t_{k})u'(t_{k}) \bigr)+\lambda I_{k} \bigl(u(t_{k}) \bigr) \bigr]v(t_{k})=0. \end{aligned}$$

(2.4)

Next, we show that u satisfies the impulsive conditions in (1.1). If this is not the case, without loss of generality, we may assume that there exists $i\in \{1,2,\ldots,p\}$ such that

$$ \Delta \bigl(h(t_{i})u'(t_{i}) \bigr)+\lambda I_{i} \bigl(u(t_{i}) \bigr)\neq 0. $$

Let $v=\Pi _{j=0,j\neq i}^{p+1}(t-t_{j})$, then by (2.4),

$$ \begin{aligned} &h(1) \bigl[u'(1)+Au(1) \bigr]v(1)+h(0) \bigl[-u'(0)+Bu(0) \bigr]v(0)\\ &\qquad{}-\sum^{p}_{k=1} \bigl[ \Delta \bigl(h(t_{k})u'(t_{k}) \bigr)+ \lambda I_{k} \bigl(u(t_{k}) \bigr) \bigr]v(t) \\ &\quad=- \bigl[\Delta \bigl(h(t_{i})u'(t_{i}) \bigr)+\lambda I_{i} \bigl(u(t_{i}) \bigr) \bigr]v(t_{i}) \neq 0, \end{aligned} $$

which contradicts (2.4). So u satisfies the impulsive conditions in (1.1). Similarly, u satisfies the boundary conditions. □

Lemma 2.4

The function $u\in \Theta$ is a solution of (1.1), then u satisfies

$$ u(t)= \int _{0}^{1}G(t,s)f \bigl(s,u(s) \bigr)\,ds+\lambda \sum_{k=1}^{p}G(t,t_{k})I_{k} \bigl(u(t_{k}) \bigr). $$

Proof

Let $g\in C(J), d_{k}\in \mathbb{R}$ $(1\leq k\leq p)$ and consider the equation

$$ \textstyle\begin{cases} -Lu=g(t), \quad t\neq t_{k}, \\ \Delta (h(t_{k})u'(t_{k}))=-d_{k}, \quad k=1,2,\ldots,p, \\ R_{1}(u)=R_{2}(u)=0. \end{cases} $$

(2.5)

We only need to show that the solution u of (2.5) satisfies

$$ u(t)= \int _{0}^{1}G(t,s)g(s)\,ds+\sum ^{p}_{k=1}G(t,t_{k})d_{k}. $$

It is easy to check that (2.5) has a unique solution. Let

$$ S_{1}(t)= \int _{0}^{1}G(t,s)g(s)\,ds,\qquad S_{2}(t)=\sum ^{p}_{k=1}G(t,t_{k})d_{k},\qquad S(t)=S_{1}(t)+S_{2}(t). $$

It follows from Theorem 3.1.1 in [32] that $S_{1}\in C^{2}(J)$ and

$$ -LS_{1}=g(t),\qquad R_{1}(S_{1})=R_{2}(S_{1})=0. $$

Hence,$-\Delta (h(t_{k})S_{1}'(t_{k}))=0$ for $1\leq k\leq p$. From $(8)$ and $(9)$ of Lemma 2.1, one easily shows that $S_{2}\in C(J)$ and

$$ \textstyle\begin{cases} -LS_{2}=0, \quad t\neq t_{k}, \\ -\Delta (h(t_{k})S_{2}'(t_{k}))=d_{k}, \quad k=1,2,\ldots,p, \\ R_{1}(u)=R_{2}(u)=0, \end{cases} $$

(2.6)

which implies that $(hS_{2})'=qS_{2}\in C(J/\{t_{1},\ldots,t_{p}\})$. Hence, S is the solution of (2.5). □

Corollary 2.1

Let $g\geq 0$ for $t\in J$ and $d_{k}\geq 0$ for $1\leq k\leq p$, then the solution of (2.5) is positive if $g\not \equiv 0$ or $d_{k}\not \equiv 0$.

3 Main result

In this section, we give our main result. Firstly, we need to prove some lemmas.

Lemma 3.1

Let $(H_{1})-(H_{3})$ hold. Then the problem

$$ \textstyle\begin{cases} -Lu=f(t,u(t)), \quad t\neq t_{k}, \\ -\Delta (h(t_{k})u(t_{k}))=\tau \in \mathbb{R^{+}}, \quad k=1,2,\ldots,p, \\ R_{1}(u)=R_{2}(u)=0 \end{cases} $$

(3.1)

has a positive solution for sufficiently small $\tau >0$.

Proof

Let

$$\begin{aligned} &\widetilde{C}=\inf \biggl\{ C: \int _{0}^{1}u^{2}\,dt\leq C \Vert u \Vert , u\in H \biggr\} ,\\ &\widehat{C}=\inf \bigl\{ C: \vert u \vert _{0}\leq C \Vert u \Vert , \forall u\in H \bigr\} . \end{aligned}$$

By the imbedding theorem, $\widetilde{C}>0, \widehat{C}>0$. From $(H_{2})$, there exist $a>0,b>0,\sigma >0$ such that

$$\begin{aligned} &F(t,\varepsilon )\leq \frac{1}{4\widetilde{C}}\varepsilon ^{2}, \quad\forall t\in J, \forall \varepsilon < \sigma , \end{aligned}$$

(3.2)

$$\begin{aligned} & F(t,s)\geq as^{\mu}-b,\quad \forall t\in J, \forall s\geq 0. \end{aligned}$$

(3.3)

Consider the functional

$$ J(u)=\frac{1}{2} \Vert u \Vert ^{2}- \int _{0}^{1}F \bigl(t,u(t) \bigr)\,dt-\tau \sum _{k=1}^{p}u(t_{k}), $$

whose critical point is a solution of (3.1). If $u\neq 0$ is a solution of (3.1), from Corollary 2.1, u is a positive solution of (3.1).

Taking $\tau <\frac{\sigma}{8\widehat{C}p}$, for $\|u\|=8p\tau$, we have

$$ \begin{aligned} J(u)&=\frac{1}{2} \Vert u \Vert ^{2}- \int _{0}^{1}F(t,u)\,dt-\tau \sum _{k=1}^{p}u(t_{k}) \\ &\geq \frac{1}{2} \Vert u \Vert ^{2}-\frac{1}{4\widetilde{C}} \int _{0}^{1}u^{2}\,dt- \tau p \vert u \vert _{0} \\ &\geq \Vert u \Vert \biggl(\frac{1}{4} \Vert u \Vert -\tau p \widehat{C} \biggr)=8(p \widehat{C}\tau )^{2}>0. \end{aligned} $$

For any $u^{+}\neq 0$,

$$ \begin{aligned} J \bigl(tu^{+} \bigr)&=\frac{t^{2}}{2} \bigl\Vert u^{+} \bigr\Vert ^{2}- \int _{0}^{1}F \bigl(s,tu^{+} \bigr)\,ds- \tau t\sum_{k=1}^{p}u^{+}(t_{k}) \\ &=\frac{t^{2}}{2} \bigl\Vert u^{+} \bigr\Vert ^{2}-at^{\mu} \int _{0}^{1} \bigl(u^{+} \bigr)^{\mu }\,ds+b- \tau t\sum_{k=1}^{p}u^{+}(t_{k}) \\ &\rightarrow -\infty \end{aligned} $$

as $t\rightarrow +\infty$ since $\mu >2$. Hence, there exists $e\in H$ with $\|e\|>8p\tau$ such that $J(e)<0$.

It remains to check that J satisfies the PS condition. Let $\{u_{n}\}\subset H$ such that $|J(u_{n})|\leq M$ and $|J'(u_{n})|\rightarrow$ as $n\rightarrow \infty$. Then

$$ \begin{aligned} & M+\frac{1}{\mu} \Vert u_{n} \Vert \\ &\quad\geq J(u_{n})-\frac{1}{\mu}J'(u_{n})u_{n} \\ &\quad= \biggl(\frac{1}{2}-\frac{1}{\mu} \biggr) \Vert u_{n} \Vert ^{2}- \int _{0}^{1} \biggl[F(t,u_{n})- \frac{1}{\mu}f(t,u_{n})u_{n} \biggr]\,dt-\tau \biggl(1- \frac{1}{\mu} \biggr)\sum^{p}_{k=1}u_{n}(t_{k}) \\ &\quad\geq \biggl(\frac{1}{2}-\frac{1}{\mu} \biggr) \Vert u_{n} \Vert ^{2}-C_{2} \Vert u_{n} \Vert -C_{1} \end{aligned} $$

for some constants $C_{1}>0$ and $C_{2}>0$. So, $\{u_{n}\}$ is bounded in H. Considering a subsequence if necessary, we may assume that $u_{n}\rightharpoonup u$ in H. Thus,

$$ \begin{aligned} 0&\leftarrow (J'(u_{n})-J'(u), u_{n}-u) \\ &= \Vert u_{n}-u \Vert ^{2}- \int _{0}^{1}[f(t,u_{n})-f(t,u)](u_{n}-u)\,dt-\tau \sum^{p}_{k=1}(u_{n}(t_{k})-u(t_{k})). \end{aligned} $$

Noting that $u_{n}\rightharpoonup u$ in H implies $u_{n}\rightarrow u$ in $C(J)$, we have

$$ \int _{0}^{1} \bigl[f(t,u_{n})-f(t,u) \bigr](u_{n}-u)\,dt\rightarrow 0, \sum^{p}_{k=1} \bigl(u_{n}(t_{k})-u(t_{k}) \bigr) \rightarrow 0 $$

as $n\rightarrow \infty$. Hence, $\|u_{n}-u\|\rightarrow 0$, that is, $u_{n}\rightarrow u$ in H. Then the mountain pass theorem implies that J has the critical point u with $J(u)>0$. Clearly, $u\not \equiv 0$. □

Let $\Lambda _{1}=\{\lambda >0, \text{(1.1)}$ has positive solution} and $\Lambda =\sup{\Lambda _{1}}$.

Lemma 3.2

There exists $\bar{\lambda}>0$ such that (1.1) has at least a positive solution $u_{\lambda}$ for $\lambda \in (0,\bar{\lambda})$ and $u_{\lambda}$ is the local minimum of $\Phi _{\lambda}$ with $\Phi _{\lambda}(u_{\lambda})<0$. Moreover, $u_{\lambda _{1}}\leq u_{\lambda _{2}}$ for $0<\lambda _{1}\leq \lambda _{2}<\bar{\lambda}$.

Proof

Let $\tau >0$ be sufficiently small and $x_{\tau}$ be the positive solution of (3.1) in Lemma 3.1. Consider the equation

$$ \textstyle\begin{cases} -Lu=f(t,T_{x_{\tau}}(u)), \quad t\neq t_{k}, \\ -\Delta h(t_{k})u(t_{k})=\lambda _{0}I_{k}(T_{x_{\tau}(t_{k})}(u(t_{k}))),\quad k=1,2,\ldots,p, \\ R_{1}(u)=R_{2}(u)=0, \end{cases} $$

(3.4)

where

$$\begin{aligned} &\lambda _{0}= \frac{\tau}{\max_{1\leq k\leq p}(I_{k}(x_{\tau}(t_{k})))},\\ & T_{\alpha}(u)=\textstyle\begin{cases} \alpha , & u> \alpha , \\ u, &0\leq u\leq \alpha , \quad \text{if } \alpha >0 \\ 0, & u< 0. \end{cases}\displaystyle \end{aligned}$$

Obviously, the solution of (3.4) is equivalent to the critical point of $\phi _{\lambda _{0},T_{x_{\tau}}}$, where

$$\begin{aligned} &\phi _{\lambda ,T_{x_{\tau}}}(u)=\frac{1}{2} \Vert u \Vert ^{2}- \int _{0}^{1}F_{T_{x_{ \tau}}} \bigl(t,u(t) \bigr) \,dt- \lambda \sum^{p}_{k=1}I_{k}^{T_{x_{\tau}(t_{k})}} \bigl(u(t_{k}) \bigr),\\ &F_{T_{x_{\tau}}}(t,u)= \int _{0}^{u}f \bigl(t,T_{x_{\tau}}(s) \bigr) \,ds, \qquad I_{k}^{T_{x_{ \tau}(t_{k})}}(u)= \int _{0}^{u}I_{k} \bigl(T_{x_{\tau}(t_{k})}(s) \bigr)\,ds. \end{aligned}$$

Since $T_{x_{\tau}}(s),T_{x_{\tau}(t_{k})}(s)$ are bounded, we obtain that for any $u\in H$,

$$\begin{aligned} &\bigl\vert F_{T_{x_{\tau}}}(t,u) \bigr\vert \leq C_{3} \vert u \vert _{0}\leq C_{4} \Vert u \Vert ,\\ &\bigl\vert I_{K}^{T_{x_{\tau}(t_{k})}} \bigl(u(t_{k}) \bigr) \bigr\vert \leq C_{5} \vert u \vert _{0}\leq C_{6} \Vert u \Vert , \end{aligned}$$

which imply that $\phi _{\lambda _{0},T_{x_{\tau}}(u)}\rightarrow +\infty$ as $\|u\|\rightarrow +\infty$. In addition, $\phi _{\lambda _{0},T_{x_{\tau}}(u)}$ is sequentially weakly lower semicontinuous. It follows that there exists $u_{\lambda _{0}}\in H$ such that

$$ \phi _{\lambda _{0},T_{x_{\tau}}}(u_{\lambda _{0}})=\inf \bigl\{ \phi _{ \lambda _{0},T_{x_{\tau}}}(u):u\in H \bigr\} =m_{\lambda _{0}}. $$

Let $\xi >0$ be sufficiently small such that $\xi \varphi _{0}(t_{K})\leq x_{\tau}(t_{K})$. We obtain

$$ \begin{aligned} \phi _{\lambda _{0},T_{x_{\tau}}}(\xi \varphi _{0}) &= \frac{\xi ^{2}}{2} \Vert u \Vert ^{2}- \int _{0}^{1}F_{T_{x_{\tau}}}(t,\xi \varphi _{0})\,dt-\lambda _{0}\sum_{k=0}^{p}I_{k}^{T_{x_{\tau}}} \bigl(\xi \varphi _{0}(t_{k}) \bigr), \\ &\leq \frac{\xi ^{2}}{2} \Vert u \Vert ^{2}-\lambda _{0} \int ^{\xi \varphi _{0}(t_{K})}_{0}I_{K} \bigl(T_{x_{ \tau}(t_{K})}(s) \bigr)\,ds, \\ &\leq \frac{\xi ^{2}}{2} \Vert u \Vert ^{2}- \frac{\kappa \lambda _{0}}{1+\tau _{0}} \bigl(\xi \varphi _{0}(t_{K}) \bigr)^{\tau _{0}+1}:=g(\xi ), \end{aligned} $$

where we use the fact that

$$ F_{T_{x_{\tau}}}(t,\xi \varphi _{0})\geq 0,\qquad I_{k}^{T_{x_{\tau}(t_{k})}} \bigl( \xi \varphi _{0}(t_{k}) \bigr)\geq 0\quad (k\neq K). $$

Clearly,

$$ m_{\lambda _{0}}\leq \min_{\xi >0}g(\xi )=\frac{\tau _{0}-1}{2} \bigl(\kappa \lambda _{0}\varphi _{0}^{1+\tau _{0}}(t_{K}) \bigr)^{ \frac{1}{1-\tau _{0}}}< 0=\phi _{\lambda _{0},T_{x_{\tau}}}(0), $$

which implies that $u_{\lambda _{0}}\not \equiv 0$. It is easy to show that $u_{\lambda _{0}}\geq 0$ and

$$ (u_{\lambda _{0}},v)= \int _{0}^{1}f \bigl(t,T_{x_{\tau}} \bigl(u_{\lambda _{0}}(t) \bigr) \bigr)v(t)+ \lambda _{0}\sum ^{p}_{k=1}I_{k} \bigl(T_{x_{\tau}(t_{k})} \bigl(u_{ \lambda _{0}}(t_{k}) \bigr) \bigr)v(t_{k}). $$

(3.5)

Choosing $v=(u_{\lambda _{0}}-x_{\tau})^{+}\in H$ as a test function, we have

$$\begin{aligned} & \bigl(u_{\lambda _{0}},(u_{\lambda _{0}}-x_{\tau})^{+} \bigr)\\ &\quad= \int _{0}^{1}f \bigl(t,T_{x_{ \tau}}(u_{\lambda _{0}}) \bigr) (u_{\lambda _{0}}-x_{\tau})^{+}\,dt+ \lambda _{0} \sum_{k=1}^{p}I_{k} \bigl(T_{x_{\tau}(t_{k})} \bigl(u_{\lambda _{0}}(t_{k}) \bigr) \bigr) (u_{ \lambda _{0}}-x_{\tau})^{+}(t_{k}) \\ &\quad\leq \int _{0}^{1}f(t,x_{\tau}) (u_{\lambda _{0}}-x_{\tau})^{+}\,dt+ \lambda _{0} \sum_{k=1}^{p}I_{k} \bigl(x_{\tau}(t_{k}) \bigr) (u_{\lambda _{0}}-x_{ \tau})^{+}(t_{k}) \\ &\quad\leq \int _{0}^{1}f(t,x_{\tau}) (u_{\lambda _{0}}-x_{\tau})^{+}\,dt+ \tau \sum _{k=1}^{p}(u_{\lambda _{0}}-x_{\tau})^{+}(t_{k})= \bigl(x_{\tau},(u_{ \lambda _{0}}-x_{\tau})^{+} \bigr) \\ &\quad\Rightarrow \quad\bigl(u_{\lambda _{0}}-x_{\tau},(u_{\lambda _{0}}-x_{\tau})^{+} \bigr) \leq 0, \\ &\quad\Rightarrow \quad\bigl\Vert (u_{\lambda _{0}}-x_{\tau})^{+} \bigr\Vert \leq 0, \\ &\quad\Rightarrow\quad u_{\lambda _{0}}\leq x_{\tau}, \\ &\quad\Rightarrow\quad \textstyle\begin{cases} -Lu_{\lambda _{0}}=f(t,u_{\lambda _{0}}), \\ -\Delta (h(t_{k})u_{\lambda _{0}}'(t_{k}))=\lambda _{0} I_{k}(u_{ \lambda _{0}}(t_{k})), \\ R_{1}(u_{\lambda _{0}})=R_{2}(u_{\lambda _{0}})=0. \end{cases}\displaystyle \end{aligned}$$

Next we show that $u_{\lambda _{0}}$ is a positive solution of (1.1) with $\lambda =\lambda _{0}$. Since $u_{\lambda _{0}}\not \equiv 0$ and $u_{\lambda _{0}}\geq 0$, we may assume that there exists $t^{*}\in (0,1)$ such that $u_{\lambda _{0}}(t^{*})>0$. Because $u_{\lambda _{0}}$ is continuous, there exists an open interval $D\subset J$ with $t^{*}\in D$ such that $u_{\lambda _{0}}(t)>0$ for all $t\in D$. Hence $f(t,u_{\lambda _{0}}(t))>0$ for $t\in D$. From Lemma 2.3, we obtain

$$ \begin{aligned} u_{\lambda _{0}}(t)&= \int _{0}^{1}G(t,s)f \bigl(s,u_{\lambda _{0}}(s) \bigr)\,ds+ \lambda _{0}\sum^{p}_{k=1}I_{k} \bigl(u_{\lambda}(t_{k}) \bigr) \\ &\geq \int _{D}G(t,s)f \bigl(s,u_{\lambda}(s) \bigr)\,ds>0 \end{aligned} $$

for $t\in (0,1)$.

Assume that $\mu \in (0,\lambda )$ and $u_{\lambda}$ is a positive solution of (1.1) with the parameter λ. We consider the functional $\phi _{\mu ,T_{u_{\lambda}}}$. By using a similar reasoning as above, one may obtain that $\phi _{\mu ,T_{u_{\lambda}}}$ has the critical point $u_{\mu}\leq u_{\lambda}$, which is a positive solution of (1.1) with μ and the local minimum of $\Phi _{\mu}$ with $\Phi _{\mu}(u_{\mu})<0$. □

Lemma 3.3

$0<\Lambda <+\infty$.

Proof

Clearly, $\Lambda _{1}\neq \emptyset$. Let $u_{\lambda}$ be a positive solution of (1.1), then

$$ (u_{\lambda},v)= \int _{0}^{1}f(t,u_{\lambda})v \,dt+\lambda \sum_{k=1}^{p}I_{k} \bigl(u_{ \lambda}(t_{k}) \bigr)v(t_{k}), \quad\forall v\in H. $$

(3.6)

Note that $\varphi _{0}$ is the solution of (2.3) with $\lambda =\lambda ^{*}$, which satisfies

$$ (\varphi _{0},v)=\lambda ^{*} \int _{0}^{1}\varphi _{0}(t)v(t) \,dt,\quad \forall v\in H. $$

(3.7)

From (3.6) and (3.7), we have

$$ \lambda ^{*} \int _{0}^{1}\varphi _{0}u_{\lambda } \,dt= \int _{0}^{1}f(t,u_{ \lambda})\varphi _{0}\,dt+\lambda \sum_{k=1}^{p}I_{k} \bigl(u_{\lambda}(t_{k}) \bigr) \varphi _{0}(t_{k}). $$

(3.8)

By $(H_{2})$ and (3.3), there exists $C_{7}>0$ such that

$$ f(t,s)\geq \bigl(1+\lambda ^{*} \bigr)s-C_{7}, \quad\forall t\in J, \forall s \geq 0. $$

Hence,

$$\begin{aligned} &\int _{0}^{1}\varphi _{0} u_{\lambda }\,dt\leq C_{8}, \end{aligned}$$

(3.9)

$$\begin{aligned} &\lambda \kappa u_{\lambda}^{\tau _{0}}(t_{K}) \varphi _{0}(t_{K})\leq \lambda ^{*} \int _{0}^{1}\varphi _{0} u_{\lambda }\,dt\leq \lambda ^{*}C_{8}. \end{aligned}$$

(3.10)

In addition, by Lemma 2.3 and $(H_{3})$, we have

$$\begin{aligned} &u_{\lambda}(t_{K})\geq \lambda \sum ^{p}_{k=1}G(t_{k},t_{k})I_{k} \bigl(u_{ \lambda}(t_{k}) \bigr) \geq \lambda \kappa G(t_{K},t_{K}) u_{\lambda}^{ \tau _{0}}(t_{K}),\\ &u_{\lambda}(t_{K})\geq \bigl(\lambda \kappa G(t_{K},t_{K}) \bigr)^{ \frac{1}{1-\tau _{0}}}, \end{aligned}$$

which implies that

$$ \lambda \leq \kappa ^{-1} \biggl( \frac{\lambda ^{*} C_{8}}{\varphi _{0}(t_{K})} \biggr)^{1-\tau _{0}}G^{-1}(t_{K},t_{K})< + \infty. $$

Hence, $\Lambda <+\infty$. □

Remark 3.1

Since $\lambda ^{*}\int _{0}^{1}\varphi _{0}u_{\lambda }\,dt\geq \lambda \sum_{k=1}^{p}I_{k}(u_{\lambda}(t_{k}))\varphi _{0}(t_{k})$, there exists $M>0$ independent of λ such that $I_{k}(u_{\lambda}(t_{k}))< M$ for $\forall 1\leq k\leq p$. By $(H_{3})$, there exists $\overline{C}>0$ independent of λ such that for any $1\leq k\leq p$,

$$ \int _{0}^{u_{\lambda}(t_{k})}I_{k}(t)\,dt\leq \overline{C} \Vert u_{\lambda} \Vert . $$

Lemma 3.4

$\Lambda \in \Lambda _{1}$.

Proof

Let $\{\lambda _{n}\}\in \Lambda _{1}$ be an increasing sequence such that $\lambda _{n}\rightarrow \Lambda$ as $n\rightarrow \infty$. For every $n\geq 1$, one can find $u_{n}\in H$ such that $u_{n}$ is a positive solution of (1.1) with $\lambda =\lambda _{n}$. Since $f\geq 0,I_{k}\geq 0$ and λ is increasing, if $m>n$,

$$ \textstyle\begin{cases} -Lu_{m}=f(t,u_{m}), \quad t\neq t_{k}, \\ -\Delta (h(t_{k})u_{m}'(t_{k}))=\lambda _{m} I_{k}(u_{m}(t_{k}))\geq \lambda _{n} I_{k}(u_{m}(t_{k})), \quad k=1,2,\ldots,p. \end{cases} $$

Consider the functional $\phi _{\lambda _{n},T_{u_{m}}}$. Similar to Lemma 3.2, we obtain that $\phi _{\lambda _{n},T_{u_{m}}}$ has a critical point $u_{\lambda _{n}}\leq u_{m}$, which is a local minimum of $\phi _{\lambda _{n}}$ with $\Phi _{\lambda _{n}}(u_{\lambda _{n}})<0$. Hence, without loss of generality, we may assume that for all $n\geq 1$,

$$ \Phi _{\lambda _{n}}(u_{n})\leq \frac{\tau _{0}-1}{2} \bigl(\kappa \lambda _{n}\varphi _{0}^{1+\tau _{0}}(t_{K}) \bigr)^{ \frac{1}{1-\tau _{0}}}< 0. $$

Hence,

$$\begin{aligned} &(u_{n},u_{n})= \int _{0}^{1}f(t,u_{n})u_{n} \,dt+\lambda _{n}\sum_{k=1}^{p}I_{k} \bigl(u_{n}(t_{k}) \bigr)u_{n}(t_{k}),\\ &(u_{n},u_{n})\leq 2 \int _{0}^{1}F(t,u_{n})\,dt+2\lambda _{n}\sum_{k=1}^{p} \int _{0}^{u_{n}(t_{k})}I_{k}(s)\,ds,\\ &\int _{0}^{1} \bigl[f(t,u_{n})u_{n}-2F(t,u_{n}) \bigr]\,dt\leq \lambda _{n} \Biggl[ \sum_{k=1}^{p}2 \int _{0}^{u_{n}(t_{k})}I_{k}(s) \,ds-I_{k} \bigl(u_{n}(t_{k}) \bigr)u_{n}(t_{k}) \Biggr]. \end{aligned}$$

From Remark 3.1 and $(H_{2})$, we have

$$\begin{aligned} &\begin{aligned} &\biggl(1-\frac{2}{\mu} \biggr) \int _{0}^{1}f(t,u_{n})u_{n} \,dt+2 \int _{u_{n}\geq r} \biggl[\frac{f(t,u_{n})u_{n}}{\mu}-F(t,u_{n}) \biggr]\,dt \\ &\quad{}+2 \int _{u_{n}\leq r} \biggl[\frac{f(t,u_{n})u_{n}}{\mu}-F(t,u_{n}) \biggr]\,dt\leq C_{9} \Vert u_{n} \Vert , \end{aligned} \\ &\int _{0}^{1}f(t,u_{n})u_{n} \leq C_{10} \Vert u_{n} \Vert +C_{11}. \end{aligned}$$

Hence

$$ \Vert u \Vert ^{2}=(u_{n},u_{n})\leq C_{12} \Vert u_{n} \Vert +C_{13}, $$

which implies that $\{u_{n}\}$ is bounded in H. Up to a subsequence, we have

$$ u_{n}\rightharpoonup \hat{u}\in H \quad\text{in } H,\qquad u_{n} \rightarrow \hat{u}\in H\quad \text{in } C(J). $$

It follows that for any $v\in H$,

$$\begin{aligned} &(u_{n},v)\rightarrow (\hat{u},v),\\ &\int _{0}^{1}f(t,u_{n})v\,dt\rightarrow \int _{0}^{1}f(t,\hat{u})v\,dt,\qquad I_{k} \bigl(u_{n}(t_{k}) \bigr)\rightarrow I_{k} \bigl( \hat{u}(t_{k}) \bigr). \end{aligned}$$

Combining with $\langle \Phi '_{\lambda _{n}}(u_{n}),v\rangle =0$ and $\lambda _{n}\rightarrow \Lambda$, we have

$$ \bigl\langle \phi '_{\lambda}(\hat{u}),v \bigr\rangle =( \hat{u},v)- \int _{0}^{1}f(t, \hat{u})v \,dt-\Lambda \sum _{k=1}^{p} I_{k} \bigl( \hat{u}(t_{k}) \bigr)v(t_{k})=0. $$

Hence, û is a solution of (1.1) with $\lambda =\Lambda$. Finally, we show that $\hat{u}>0$ for $t\in (0,1)$. Clearly, $\hat{u}\geq 0$ since $u_{n}\geq 0$. In addition,

$$\begin{aligned} & (u_{n},u_{n}-\hat{u})= \int _{0}^{1}f(t,u_{n}) (u_{n}-\hat{u})\,dt+ \lambda _{n}\sum _{k=1}^{p}I_{k} \bigl(u_{n}(t_{k}) \bigr) \bigl(u_{n}(t_{k})-\hat{u}(t_{k}) \bigr) \\ &\phantom{(u_{n},u_{n}-\hat{u})}\rightarrow 0, \\ &(u_{n},\hat{u})\rightarrow (\hat{u},\hat{u}), \end{aligned}$$

and therefore, $\|u_{n}\|\rightarrow \|\hat{u}\|$. Hence,

$$ \Phi _{\Lambda}(\hat{u})\leftarrow \Phi _{\lambda _{n}}(u_{n})\leq ( \kappa \lambda _{n}\varphi _{0}^{1+\tau _{0}}(t_{k}))^{ \frac{1}{1-\tau _{0}}}\frac{\tau _{0}-1}{2}< 0, $$

which implies that $\hat{u}\not \equiv 0$, which is the positive solution of (1.1) with $\lambda =\Lambda$. □

Define

$$\begin{aligned} &f_{0}(t,x)=\textstyle\begin{cases} f(t,u_{\lambda}), & x< u_{\lambda}, \\ f(t,x), & x\geq u_{\lambda},\end{cases}\displaystyle \qquad i_{k}(x)=\textstyle\begin{cases} I_{k}(u_{\lambda}), & x< u_{\lambda}, \\ I_{k}(x), & x\geq u_{\lambda}.\end{cases}\displaystyle \\ &\Phi _{0}(u)=\frac{1}{2} \Vert u \Vert ^{2}- \int _{0}^{1}F_{0}(t,u)\,dt-\lambda \sum _{k=1}^{p}\widetilde{I}_{k} \bigl(u(t_{k}) \bigr),\\ &F_{0}(t,x)= \int _{0}^{x}f_{0}(t,s)\,ds,\qquad \widetilde{I}_{k}(x)= \int _{0}^{x} i_{k}(s)\,ds, \end{aligned}$$

where $u_{\lambda}$ is the local minimum of $\Phi _{\lambda}$ with $\Phi _{\lambda}(u_{\lambda})<0$ obtained in Lemma 3.2.

Definition 3.1

Let $\Xi \subseteq H$ be a closed set and $\varphi \in C^{1}(H,\mathbb{R})$. We say that a sequence $\{v_{n}\}\subset H$ is a $(PS)_{\Xi ,c}$ sequence of φ if

$$ \operatorname{dist}(v_{n},\Xi )\rightarrow 0, \qquad \varphi (v_{n}) \rightarrow c, \qquad \bigl\Vert \varphi '(v_{n}) \bigr\Vert \rightarrow 0 $$

as $n\rightarrow \infty$. φ satisfies the $(PS)_{\Xi ,c}$ condition if every $(PS)_{\Xi ,c}$ sequence of φ has a convergent subsequence.

Lemma 3.5

[33] Let $\varphi \in C^{1}(H,\mathbb{R})$. Consider the number

$$ c=\inf_{\gamma \in \Gamma}\max_{t\in [0,1]}\varphi \bigl(\gamma (t) \bigr), $$

where Γ is the set of all continuous paths joining two points u and v in H. Suppose that Ξ is a closed subset of H such that

$$ \Xi \cap \bigl\{ w\in H:\varphi (w)\geq c \bigr\} $$

separates u and v. If φ satisfies the $(PS)_{\Xi ,c}$ condition, then φ has a critical point of level c on Ξ.

Lemma 3.6

Suppose that Ξ is a close subset of H, then $\Phi _{0}$ satisfies the $(PS)_{\Xi ,c}$ condition for any $c\in \mathbb{R}$.

Proof

Clearly, $\Phi _{0}\in C^{1}(H,\mathbb{R})$, and there exist $C_{14}>0,C_{15}>0$ such that

$$\begin{aligned} &xf_{0}(t,x)x-\mu F_{0}(t,x)>-C_{14},\quad \forall t\in J, \forall x \in \mathbb{R}, \end{aligned}$$

(3.11)

$$\begin{aligned} & f_{0}(t,x)\geq \bigl(1+\lambda ^{*} \bigr)x-C_{14}, \quad\forall t\in J, \forall x\in \mathbb{R}, \end{aligned}$$

(3.12)

$$\begin{aligned} & F_{0}(t,x)\geq C_{15}x^{\mu}-C_{14},\quad \forall t\in J, \forall x>0, \end{aligned}$$

(3.13)

$$\begin{aligned} & f_{0}(t,x)\geq 0,\qquad i_{k}(x)\geq 0\quad (1\leq i \leq p), \forall t\in J, x \in \mathbb{R}, \end{aligned}$$

(3.14)

$$\begin{aligned} & i_{k}(x)\rightarrow +\infty \quad\text{as } x\rightarrow + \infty \quad\text{or}\quad \sup_{x>0} i_{k}(x)< +\infty ,\quad \forall 1\leq k \leq p. \end{aligned}$$

(3.15)

Assume that $\{u_{n}\}\subset H$ is a $(PS)_{\Xi ,c}$ sequence of $\Phi _{0}$, we have

$$ (u_{n},v)= \int _{0}^{1}f_{0}(t,u_{n})v \,dt+\lambda \sum_{k=1}^{p}i_{k} \bigl(u_{n}(t_{k}) \bigr)v(t_{k})+o_{n}(1),\quad \forall v\in H. $$

(3.16)

Similar to (3.8), using (3.7) and (3.16), we have

$$ \lambda ^{*} \int _{0}^{1}u_{n}\varphi _{0} \,dt= \int _{0}^{1}f_{0}(t,u_{n}) \varphi _{0} \,dt+\lambda \sum_{k=1}^{p}i_{k} \bigl(u_{n}(t_{k}) \bigr)\varphi _{0}(t_{k})+o_{n}(1). $$

(3.17)

Let $\Omega _{n}^{1}=\{t\in J: u_{n}(t)\geq 0\}, \Omega _{n}^{2}=\{t \in J: u_{n}(t)< 0\}$, then if n is sufficiently large,

$$ \begin{aligned} 1+\lambda ^{*} \int _{\Omega _{n}^{1}}u_{n}\varphi _{0}\,dt\geq \int _{ \Omega _{n}^{1}}f_{0}(t,u_{n})\varphi _{0} \,dt-\lambda ^{*} \int _{ \Omega _{n}^{2}}u_{n} \varphi _{0} \,dt \geq \int _{\Omega _{n}^{1}}f_{0}(t,u_{n}) \varphi _{0} \,dt. \end{aligned} $$

(3.18)

From (3.12), there exists $C_{16}>0$ such that

$$ 0\leq \int _{\Omega _{n}^{1}}u_{n}\varphi _{0}\,dt\leq C_{16} \quad\text{if } n \text{ is sufficiently large.} $$

(3.19)

It follows that there exist $C_{17}>0, C_{18}>0$ such that if n is sufficiently large,

$$ 0\leq i_{k} \bigl(u_{n}(t_{k}) \bigr)\leq C_{17},\quad 0\leq \widetilde{I}_{n} \bigl(u_{n}(t_{k}) \bigr) \leq \int _{0}^{u_{n}(t_{k})}i_{k}(s)\,ds\leq C_{18} \Vert u_{n} \Vert . $$

Hence, if n is sufficiently large, we have

$$ \begin{aligned} 1+c+\frac{1}{\mu} \Vert u_{n} \Vert \geq{}& \Phi _{0}(u_{n})-\frac{1}{\mu} \bigl\langle \Phi _{0}(u_{n}),u_{n} \bigr\rangle \\ ={}& \biggl(\frac{1}{2}-\frac{1}{\mu} \biggr) \Vert u_{n} \Vert - \int _{0}^{1} \biggl(F_{0}(t,u_{n})- \frac{1}{\mu}f_{0}(t,u_{n})u_{n} \biggr) \,dt \\ &{}-\lambda \sum^{p}_{k=1} \biggl[ \widetilde{I}_{k} \bigl(u_{n}(t_{k}) \bigr) - \frac{1}{\mu}i_{k} \bigl(u_{n}(t_{k}) \bigr)u_{n}(t_{k})) \biggr] \\ \geq {}&\biggl(\frac{1}{2}-\frac{1}{\mu} \biggr) \Vert u_{n} \Vert ^{2}-C_{19}-C_{20} \Vert u_{n} \Vert . \end{aligned} $$

This implies that $\{u_{n}\}\subset H$ is bounded. By a standard argument, one can show that $\{u_{n}\}$ has a convergent subsequence. □

Remark 3.2

For $\Phi _{0}$,

$$\begin{aligned} &\bigl\langle \Phi '_{0}(u_{\lambda}),v \bigr\rangle = \bigl\langle \Phi '_{\lambda}(u_{ \lambda}),v \bigr\rangle =0,\quad \forall v\in H, \end{aligned}$$

(3.20)

$$\begin{aligned} &\Phi _{0} \bigl(u_{\lambda}+v^{+} \bigr)= \Phi _{\lambda} \bigl(u_{\lambda}+v^{+} \bigr)\geq \Phi _{\lambda}(u_{\lambda})=\Phi _{0}(u_{\lambda}),\quad \forall v \in H. \end{aligned}$$

(3.21)

Theorem 3.1

There exists $0<\Lambda <+\infty$ such that (1.1) has at least two positive solutions for all $\lambda \in (0,\Lambda )$, one positive solution for $\lambda =\Lambda$, and no positive solutions for $\lambda >\Lambda$.

Proof

From Lemma 3.2 and Lemma 3.3, (1.1) has no solution for $\lambda >\Lambda$, at least one positive solution for $\lambda =\Lambda$ and a positive solution $u_{\lambda}$ with $\Phi _{\lambda}(u_{\lambda})<0$ for $0<\lambda <\Lambda$.

It is easy to show that $\Phi _{0}(u_{\lambda}+s\varphi _{0})\rightarrow -\infty$ as $s\rightarrow +\infty$. Noting that

$$\begin{aligned} &(u_{\lambda},\varphi _{0})= \int _{0}^{1}f(t,u_{\lambda})\varphi _{0}\,dt +\lambda \sum_{k=1}^{p}I_{k} \bigl(u_{\lambda}(t_{k}) \bigr)\varphi _{0}(t_{k})>0,\\ &\Vert u_{\lambda}+s\varphi _{0} \Vert ^{2}=(u_{\lambda}+s \varphi _{0}, u_{ \lambda}+s\varphi _{0}) = \Vert u_{\lambda} \Vert ^{2}+s(u_{\lambda},\varphi _{0})+s^{2} \Vert \varphi _{0} \Vert ^{2}\geq \Vert u_{\lambda} \Vert ^{2}+s^{2} \Vert \varphi _{0} \Vert ^{2}, \end{aligned}$$

we fix $s_{0}>0$ such that $R_{2}=:\|u_{\lambda}+s_{0}\varphi _{0}\|> R_{1}=:\|u_{\lambda}\|$, and

$$ \Phi _{0}(u_{\lambda}+s_{0}\varphi _{0})< \Phi _{\lambda}(u_{\lambda})-1. $$

Let $\Gamma =\{\xi \in C([0,1],H)|\xi (0)=u_{\lambda},\xi (1)=u_{\lambda}+s_{0} \varphi _{0}\}$, and

$$ \rho =\inf_{\xi \in \Gamma}\max_{t\in [0,1]}\Phi _{0} \bigl(\xi (t) \bigr). $$

It follows that $\rho \geq \Phi _{0}(u_{\lambda})=\Phi _{\lambda}(u_{\lambda})$. If $\rho =\Phi _{\lambda}(u_{\lambda})$, from (3.21), there exists $0<\delta <R_{2}-R_{1}$ such that $\inf \{\Phi _{0}(u)|\|u\|=R\}=\rho$ for all $R\in (R_{1},R_{1}+\delta )$. Let $\Xi =H$ if $\rho > \Phi _{\lambda}(u_{\lambda})$ and $\Xi =\{u:\|u\|=R_{1}+\delta /2\}$ if $\rho = \Phi _{\lambda}(u_{\lambda})$. Clearly,

$$ \Xi \cap \bigl\{ w\in H,\Phi _{0}(w)\geq \rho \bigr\} $$

separates $u_{\lambda}$ and $u_{\lambda}+s_{0}\varphi _{0}$. Hence, $\Phi _{0}$ has a critical point $v_{\lambda}$ such that $\Phi _{0}(v_{\lambda})=\rho$ and $v_{\lambda}\in \Xi$. If $\rho = \Phi _{\lambda}(u_{\lambda}), \|v_{\lambda}\| =R_{1}+ \delta /2>\|u_{\lambda}\|$, if $\rho >\phi _{\lambda}(u_{\lambda})$, $\Phi _{0}(v_{\lambda})=\rho >\Phi _{\lambda}(u_{\lambda})=\Phi _{0}(u_{ \lambda})$. Hence, $v_{\lambda}\not \equiv u_{\lambda}$ and

$$\begin{aligned} &(v_{\lambda},w)= \int _{0}^{1}f_{0}(t,v_{\lambda})w\,dt+ \lambda \sum_{k=1}^{p}i_{k} \bigl(v_{ \lambda}(t_{k}) \bigr)w(t_{k}), \quad\forall w\in H,\\ &(u_{\lambda},w)= \int _{0}^{1}f(t,u_{\lambda})w\,dt+\lambda \sum _{k=1}^{p}I_{k} \bigl(u_{ \lambda}(t_{k}) \bigr)w(t_{k}), \quad\forall w\in H. \end{aligned}$$

Choosing $w=(u_{\lambda}-v_{\lambda})^{+}$, we have

$$ \begin{aligned} \bigl(u_{\lambda}-v_{\lambda},(u_{\lambda}-v_{\lambda})^{+} \bigr) ={}& \int _{0}^{1} \bigl[f(t,u_{ \lambda})-f_{0}(t,v_{\lambda}) \bigr](u_{\lambda}-v_{\lambda})^{+}\,dt \\ &{}+\lambda \sum_{k=1}^{p} \bigl(I_{k} \bigl(u_{\lambda}(t_{k}) \bigr)-i_{k} \bigl(v_{ \lambda}(t_{k}) \bigr) \bigr) (u_{\lambda}-v_{\lambda})^{+}(t_{k})=0, \end{aligned} $$

which implies that $\|(u_{\lambda}-v_{\lambda})^{+}\|=0$ and $u_{\lambda}\leq v_{\lambda}$. Hence,

$$\begin{aligned} &f_{0}(t,v_{\lambda})=f(t,v_{\lambda}), \qquad i_{k} \bigl(v_{\lambda}(t_{k}) \bigr)=I_{k} \bigl(v(t_{k}) \bigr),\\ &\Phi _{0}(v_{\lambda})=\Phi _{\lambda}(v_{\lambda}),\qquad \bigl\langle \Phi '_{0}(v_{\lambda}),w \bigr\rangle = \bigl\langle \Phi _{\lambda}(v_{\lambda}),w \bigr\rangle =0,\quad \forall w \in H, \end{aligned}$$

and $v_{\lambda}$ is the second positive solution of (1.1). □

Remark 3.3

In fact, the function f satisfying $(H_{2})$ is of superlinear growth, and the impulsive function affecting the number of positive solutions is of sublinear growth.

Example 3.1

Consider the differential equation

$$ \textstyle\begin{cases} - x''(t) + x(t) = x^{2}(t), \quad t\neq 0.5, \\ x'(0.5^{+}) = x'(0.5)-\lambda , \\ x'(0) = x'(1)=0. \end{cases} $$

(3.22)

Clearly, the nonimpulsive differential equation corresponding (3.22) has a positive solution $x\equiv 1$. The results in [19] cannot be applied to (3.22) since the nonlinear function and impulsive functions in [19] are of superlinear growth or of sublinear growth.

Assume that x is a positive solution of (3.22), then

$$ - \int _{0}^{0.5} x''(t) \,dt - \int _{0.5}^{1} x''(t) \,dt + \int _{0}^{1} x(t) \,dt = \int _{0}^{1} x^{2}(t) \,dt $$

and

$$ \int _{0}^{1} x(t) \,dt = \int _{0}^{1} \bigl(x^{2}(t) + \lambda \bigr) \,dt. $$

Clearly, $x\not \equiv 1$. If $\lambda \ge 1/4$, then we have

$$ \int _{0}^{1} x(t) \,dt > 2 \sqrt{\lambda} \int _{0}^{1} x(t) \,dt \geq \int _{0}^{1} x(t) \,dt. $$

Hence, (3.22) has no positive solution if $\lambda \ge 1/4$. From Theorem 3.1, (3.22) has two positive solutions for sufficiently small $\lambda >0$. When $\lambda =0.0001$, two positive solutions of (3.22) can be found in Fig. 1.

Figure 1

Positive solutions to the equation of Example 3.22

4 Conclusion

In this paper, we discussed the existence, nonexistence, and multiplicity of positive solutions for a class of impulsive Sturm–Liouville boundary value problems with a parameter. Using the mountain pass principle, we show that the number of positive solutions depends on the change of parameters, in which sublinear impulsive perturbation plays an important role. In fact, one can prove that the nonimpulsive case of (1.1), that is, $I_{k}\equiv 0(1\leq k\leq p)$, has at least a positive solution. How does the combination of impulsive perturbation and parameter affect the behavior of the equation? We will discuss the issue in follow-up research.