Existence and Multiplicity Results for Generalized Laplacian Problems with a Parameter

Abstract

We study the homogeneous Dirichlet boundary value problem of generalized Laplacian equations with a singular weight which may not be integrable. Some existence, multiplicity and nonexistence results of positive solutions under two different asymptotic behaviors of the nonlinearity at 0 and $\infty$ are established in terms of different ranges of a parameter. Our approach is based on the fixed point theorem of expansion/compression of a cone and Schauder’s fixed point theorem.

1 Introduction

In this paper, we study the following scalar differential problem

$$\begin{aligned} \left\{ \begin{array}{ll} -\varphi (u')'= \lambda h(t)f(u),\quad t\in (0,1),\\ u(0)= 0 = u(1), \end{array} \right. \end{aligned}$$

(1)

where $\varphi : {\mathbb {R}} \rightarrow {\mathbb {R}}$ is an odd increasing homeomorphism, $\lambda$ is a positive real parameter, $h \not \equiv 0$ on any subinterval in (0, 1). We also assume that

(A):

There exist an increasing homeomorphism $\psi$ from $(0,\infty )$ onto $(0,\infty )$ and a function $\gamma$ from $(0,\infty )$ into $(0,\infty )$ such that

$$\begin{aligned} \psi (\sigma )\le \frac{\varphi (\sigma x)}{\varphi (x)}\le \gamma (\sigma ),\quad \ \ \mathrm {for \ \ all}\quad \ \ \sigma >0, x \in {\mathbb {R}}/\{0\}. \end{aligned}$$

(H):

$h:(0,1)\rightarrow [0,\infty )$ is locally integrable satisfying

$$\begin{aligned} \int _{0}^{\frac{1}{2}}\psi ^{-1}\left( \int _{s}^{\frac{1}{2}}h(\tau )\mathrm{d}\tau \right) \mathrm{d}s+\int _{\frac{1}{2}}^{1}\psi ^{-1}\left( \int _{\frac{1}{2}}^{s}h(\tau )\mathrm{d}\tau \right) \mathrm{d}s<\infty . \end{aligned}$$

$(F_1)$:

$f:[0,\infty )\rightarrow [0,\infty )$ is continuous.

$(F_2)$:

$f(u)>0$, for $u>0$.

We note that condition (A) on $\varphi$ is more general than the condition given by Wang in [15, 16], which was first introduced by Xu and Lee [17]. Condition (H) indicates that the weight function h(t) may have a singularity at $t=0$ and/or $t=1$. For convenience, we introduce a new class of weight functions. For a bijection $\iota : {\mathbb {R}}\rightarrow {\mathbb {R}}$, define ${\mathcal {H}}_\iota$ to be the subset of $L_{loc}^{1}((0,1),[0,\infty ))$ given by

$$\begin{aligned} {\mathcal {H}}_\iota \,{=} \left\{ g \in L_{loc}^{1}((0,1),[0,\infty )) \;\Big |\; \int _{0}^{\frac{1}{2}} \iota ^{-1} \left( \int _{s}^{\frac{1}{2}} g(\tau )\mathrm{d} \tau \right) \mathrm{d}s + \int _{\frac{1}{2}}^{1} \iota ^{-1} \left( \int _{\frac{1}{2}}^{s} g(\tau )\mathrm{d} \tau \right) \mathrm{d}s < \infty \right\} . \end{aligned}$$

By the notation, condition (H) means $h \in {\mathcal {H}}_\psi .$

As an example of conditions (A) and (H), define $\varphi : {\mathbb {R}}\rightarrow {\mathbb {R}}$ to be an odd function with

$$\begin{aligned} \varphi (x)=x^{2}+x,\quad \,\,x\ge 0. \end{aligned}$$

Then, $\varphi$ satisfies (A) with functions $\psi$ and $\gamma$ given as

$$\begin{aligned} \psi (\sigma ) = \left\{ \begin{array}{ll} \sigma ^{2},\quad &{} \quad \mathrm{if} \ 0 <\sigma \le 1,\\ \sigma ,\quad &{} \quad \mathrm{if}\ \sigma >1, \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} \gamma (\sigma ) = \left\{ \begin{array}{ll} 1, &{}\quad \mathrm{if}\ 0 <\sigma \le 1,\\ \sigma ^{2}, &{}\quad \mathrm{if}\ \sigma >1. \end{array} \right. \end{aligned}$$

Moreover, for $h(t)=t^{-\frac{3}{2}}$, we can easily check h is singular at $t=0$ and $h\in {\mathcal {H}}_\psi$.

Due to a wide range of applications in mathematics and physics, p-Laplacian or more generalized Laplacian problems like (1) have been extensively investigated by many researchers in recent years. There is a vast literature dealing with existence, multiplicity and nonexistence of positive solutions for these problems under various assumptions on weight functions and nonlinearities (see [1,2,3,4,5, 7, 10,11,12,13,14,15,16,17,18] and the references therein). Among these works mentioned above, different asymptotic behaviors of the nonlinearities at 0 and/or $\infty$ affect the number of positive solutions. Let us introduce notations

$$\begin{aligned} f_0:=\lim _{u\rightarrow 0}\frac{f(u)}{\varphi (u)},\ \ \ \ f_\infty :=\lim _{u\rightarrow \infty }\frac{f(u)}{\varphi (u)}. \end{aligned}$$

When $\varphi (x)=\varphi _p(x):=|x|^{p-2}x$, $p>1$, Agarwal et al.[1] and Sánchez [14] proved the multiplicity of positive solutions of problem (1) for $\lambda$ belonging to some open interval if either $f_0=f_\infty =0$ or $f_0=f_\infty =\infty$. Later, Wang [16] extended the multiplicity results in [1, 14] to a scalar $\varphi$-Laplacian problem with $h\in C[0,1]$. Recently, Xu and Lee [18] derived some explicit intervals for $\lambda$ such that the singularly weighted $\varphi$-Laplacian problem (1) has at least one positive solution if $0<f_0,f_\infty <\infty$. For more on existence of positive solutions of p-Laplcain or more generalized Laplacian systems, we refer to [5, 7, 11,12,13, 15, 17, 18] and the references therein for details.

Our goal is to extend the previous results of [1, 14, 16] to a singularly weighted $\varphi$-Laplacian problem (1) for the cases $f_0=f_\infty =0$ and $f_0=f_\infty =\infty$. We not only obtain the multiplicity results of positive solutions but also obtain the global results of positive solutions for problem (1) with respect to the parameter $\lambda$. More precisely, our main results can be listed as follows.

Theorem 1

Assume that (A), (H), ($F_1$) and ($F_2$) hold. If $f_0=f_\infty =0$, then there exist $\lambda ^*\ge \lambda _*>0$ such that problem (1) has at least two positive solutions for $\lambda >\lambda ^{*}$, one positive solution for $\lambda \in [\lambda _*,\lambda ^{*}]$, and no positive solution for $\lambda \in (0,\lambda _*).$

Theorem 2

Assume that (A), (H), ($F_1$) and ($F_2$) hold. If $f_0=f_\infty =\infty$, then there exist $\lambda ^*\ge \lambda _*>0$ such that problem (1) has at least two positive solutions for $\lambda \in (0,\lambda _*)$, one positive solution for $\lambda \in [\lambda _*,\lambda ^{*}]$, and no positive solution for $\lambda >\lambda ^{*}.$

Remark 1

The condition $f_0=\infty$ implies that $f(0)\ge 0$. If $f(0)>0$, then we can get $\lambda _*=\lambda ^*$ in Theorem 2. The proof can be easily completed by arguments under to there in [10, 12].

For the proofs, we employ a newly developed solution operator for (1) introduced by Xu and Lee [17] and then we make use of the fixed point theorem of expansion/compression of a cone and Schauder’s fixed point theorem for the main results.

Our paper is organized as follows. In Sect. 2, we establish a solution operator for problem (1) and show some preliminary results. In Sects. 3 and 4, we prove the main results and give some examples, respectively.

2 Preliminaries

The main condition on the weight function h in problem (1) is that h belongs to ${\mathcal {H}}_\psi$-class which includes singular functions specially on the boundary, i.e., h may not be integrable near the boundary, $t=0$ and/or $t=1$. In this case, the solution u may not be of $C^{1}[0,1]$-class. So by a solution to problem (1), we understand a function $u \in C_0[0,1]\cap C^{1}(0,1)$ with $\varphi (u')$ absolutely continuous and which satisfies problem (1).

Basic tools for proving our main results are the following two well-known fixed point theorems.

Lemma 1

([6, 8], Fixed point theorem of expansion/compression of a cone) Let E be a Banach space and let K be a cone in E. Assume that $\varOmega _1$ and $\varOmega _2$ are open subsets of E with $0\in \varOmega _1$, $\overline{\varOmega _1}\subset \varOmega _2$. Assume that $T: K\cap (\overline{\varOmega _2}{\setminus } \varOmega _1 ) \rightarrow K$ is completely continuous such that either

$\Vert Tu\Vert \le \Vert u\Vert$, for $u\in K\cap \partial \varOmega _1$ and $\Vert Tu\Vert \ge \Vert u\Vert$, for $u\in K\cap \partial \varOmega _2$ or

$\Vert Tu\Vert \ge \Vert u\Vert$, for $u\in K\cap \partial \varOmega _1$ and $\Vert Tu\Vert \le \Vert u\Vert$, for $u\in K\cap \partial \varOmega _2$,

Then, T has a fixed point in $K\cap ( \overline{\varOmega _2}{\setminus } \varOmega _1)$.

Lemma 2

([9], Schauder’s fixed point theorem) Let X be a Banach space and let M be a closed, convex and bounded set in X. Assume that $T: M \rightarrow M$ is completely continuous. Then, T has a fixed point in M.

To set up the solution operator for (1), let $C_0[0,1]$ be the Banach space with norm $\Vert u\Vert _\infty =\max _{t\in [0,1]}|u(t)|$ and define a cone K by taking

$$\begin{aligned} K=\left\{ u\in C_0[0,1] \;| \ u\; \mathrm{is\; concave \; on} \; [0,1]\right\} . \end{aligned}$$

Let us consider the following Poisson-type problem

$$\begin{aligned} -\varphi (w')'= & {} g(t),\quad \ \ t\in (0,1) \ \ \ \ \ \ \ \ \ \ \end{aligned}$$

(2)

$$\begin{aligned} w(0)= & {} w(1)=0, \end{aligned}$$

(3)

where $\varphi$ satisfies (A) and $g\in {\mathcal {H}}_\varphi$. Note from condition (A) that ${\mathcal {H}}_\psi \subset {\mathcal {H}}_\varphi$. The first step to derive a corresponding operator equation is usually to integrate the equation on the interval [0, s] in (2) for $s>0.$ But as we remark, g may not be integrable on the interval [0, s]. It seems a small barrier at the start and we try the following process. Let w be a solution of (2), (3). Then, integrating both sides of (2) on the interval $[s,\frac{1}{2}]$ for $s\in (0,\frac{1}{2}]$ and $[\frac{1}{2},s]$ for $s\in [\frac{1}{2},1)$, respectively, we find that (2), (3) are equivalent to

$$\begin{aligned} \left\{ \begin{array}{ll} w'(s)=\varphi ^{-1}\left( a+\int _{s}^{\frac{1}{2}}g(\tau )\mathrm{d}\tau \right) ,\quad \; w(0)=0,\quad \; s\in (0,\frac{1}{2}],\\ w'(s)=\varphi ^{-1}\left( a-\int _{\frac{1}{2}}^{s}g(\tau )\mathrm{d}\tau \right) ,\quad \;w(1)=0,\quad \; s\in [\frac{1}{2},1), \end{array} \right. \end{aligned}$$

(4)

where $a=\varphi (w'(\frac{1}{2}))$. The important step is to show the fact

$$\begin{aligned} \varphi ^{-1}\left( a+\int _{s}^{\frac{1}{2}}g(\tau )\mathrm{d}\tau \right) \in L^{1}(0,\frac{1}{2}] \end{aligned}$$

and it is not obvious. One may refer to Xu and Lee [17] for the proof. Now we may integrate both sides of (4) on the interval [0, t] for $t\in [0,\frac{1}{2}]$ and on the interval [t, 1] for $t\in [\frac{1}{2},1]$, respectively. And we get

$$\begin{aligned} w(t)= \left\{ \begin{array}{ll} \int _0^{t}\varphi ^{-1}\left( a+\int _{s}^{\frac{1}{2}}g(\tau )\mathrm{d}\tau \right) \mathrm{d}s,\quad \ \ \ \ t\in [0,\frac{1}{2}],\\ \int _t^{1}\varphi ^{-1}\left( -a+\int _{\frac{1}{2}}^{s}g(\tau )\mathrm{d}\tau \right) \mathrm{d}s,\quad \;\ t\in [\frac{1}{2},1]. \end{array} \right. \end{aligned}$$

To check $w(\frac{1}{2}^{-})=w(\frac{1}{2}^{+})$, define for $a\in {\mathbb {R}}$,

$$\begin{aligned} G(a)=\int _0^{\frac{1}{2}}\varphi ^{-1}\left( a+\int _{s}^{\frac{1}{2}}g(\tau )\mathrm{d}\tau \right) \mathrm{d}s- \int _{\frac{1}{2}}^{1}\varphi ^{-1}\left( -a+\int _{\frac{1}{2}}^{s}g(\tau )\mathrm{d}\tau \right) \mathrm{d}s. \end{aligned}$$

(5)

Then, the function $G: {\mathbb {R}}\rightarrow {\mathbb {R}}$ is well defined and has a unique zero $a=a(g)$ in ${\mathbb {R}}$ (See Xu and Lee [17] for the proof). This implies $w(\frac{1}{2}^{-})=w(\frac{1}{2}^{+})$. Consequently, if $\varphi$ satisfies (A) and $g\in {\mathcal {H}}_\varphi$, then the solution w of (2), (3) can be represented by

$$\begin{aligned} w(t)= \left\{ \begin{array}{ll} \int _0^{t}\varphi ^{-1}\left( a(g)+\int _{s}^{\frac{1}{2}}g(\tau )\mathrm{d}\tau \right) \mathrm{d}s,\quad \ \ \ \ t\in [0,\frac{1}{2}],\\ \int _t^{1}\varphi ^{-1}\left( -a(g)+\int _{\frac{1}{2}}^{s}g(\tau )\mathrm{d}\tau \right) \mathrm{d}s,\quad \;\;t\in [\frac{1}{2},1], \end{array} \right. \end{aligned}$$

(6)

where $a(g)\in {\mathbb {R}}$ uniquely satisfies

$$\begin{aligned} \int _0^{\frac{1}{2}}\varphi ^{-1}\left( a(g)+\int _{s}^{\frac{1}{2}}g(\tau )\mathrm{d}\tau \right) \mathrm{d}s= \int _{\frac{1}{2}}^{1}\varphi ^{-1}\left( -a(g)+\int _{\frac{1}{2}}^{s}g(\tau )\mathrm{d}\tau \right) \mathrm{d}s. \end{aligned}$$

Replacing g(t) with $\lambda h(t) f(u(t))$ in (2), (3), for $\lambda >0$, $u\in K$, we define

$$\begin{aligned} T_\lambda (u)(t) = \left\{ \begin{array}{ll} \int _0^t \varphi ^{-1}\left( a(\lambda h f(u))+ \int _s^{\frac{1}{2}}\lambda h(\tau ) f(u(\tau )) \mathrm{d}\tau \right) \mathrm{d}s,&{}\quad t\in [0,\frac{1}{2}],\\ \int _t^1 \varphi ^{-1}\left( -a(\lambda hf(u)) + \int _{\frac{1}{2}}^s \lambda h(\tau )f(u(\tau )) \mathrm{d}\tau \right) \mathrm{d}s, &{}\quad t\in [\frac{1}{2}, 1], \end{array} \right. \end{aligned}$$

where $a(\lambda h f(u))\in {\mathbb {R}}$ uniquely satisfies

$$\begin{aligned}&\int _0^{\frac{1}{2}}\varphi ^{-1}\left( a(\lambda h f(u))+\int _{s}^{\frac{1}{2}}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau \right) \mathrm{d}s\\&\quad =\int _{\frac{1}{2}}^{1}\varphi ^{-1}\left( -a(\lambda h f(u))+\int _{\frac{1}{2}}^{s}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau \right) \mathrm{d}s. \end{aligned}$$

One may show that $T_\lambda : K \rightarrow K$ is completely continuous (See Lemma 11 in Xu and Lee [17] for details). Thus, we see that u is a positive solution of (1) if and only if

$$\begin{aligned} u=T_\lambda (u) \ \mathrm{on} \ K. \end{aligned}$$

We finally give some remarks and lemma for later use.

Remark 2

From condition (A), we get

$$\begin{aligned} \sigma x\le \varphi ^{-1} [ \gamma (\sigma )\varphi (x) ], \end{aligned}$$

and

$$\begin{aligned} \varphi ^{-1}[\sigma \varphi (x)]\le \psi ^{-1}(\sigma ) x,\quad \ \mathrm{for} \quad \ \sigma \ \mathrm{and} \ x>0. \end{aligned}$$

Remark 3

Let $h\in L_{loc}^{1}((0,1),\mathbb {R_+})$. Then, for any fixed $s\in (0,\frac{1}{2})$, we know $\int _{s}^{\frac{1}{2}}h(\tau )\mathrm{d}\tau <\infty$. Applying $\sigma =\int _{s}^{\frac{1}{2}}h(\tau )\mathrm{d}\tau$ and $x=\varphi ^{-1}(1)$ in Remark 2, we get

$$\begin{aligned} \varphi ^{-1}\left( \int _{s}^{\frac{1}{2}}h(\tau )\mathrm{d}\tau \right) \le \varphi ^{-1}(1)\psi ^{-1}\left( \int _{s}^{\frac{1}{2}}h(\tau )\mathrm{d}\tau \right) . \end{aligned}$$

This implies ${\mathcal {H}}_\psi \subset {\mathcal {H}}_\varphi$.

Lemma 3

[15] Let $w\in C_{0}[0,1]\cap C^{1}(0,1)$ satisfy $\varphi (w')'\le 0$ on (0, 1). Then, w is concave on [0, 1] and $\min _{t\in [\frac{1}{4},\frac{3}{4}]}w(t)\ge \frac{1}{4}\Vert w\Vert _{\infty }$, where $\Vert w\Vert _{\infty }$ is the supremum norm of w.

3 Proof of Main Results

In this section, we prove Theorems 1 and 2. For this, we need to give some lemmas which will play a crucial role.

Lemma 4

Assume that (A), (H), ($F_1$) and ($F_2$) hold. If $f_0=f_\infty =0$, then there exists $\lambda _0>0$ such that (1) has at least two positive solutions for $\lambda >\lambda _0$.

Proof

For any $r>0$, define

$$\begin{aligned} {\hat{m}}_r:=\min \{f(x)\ | \ \frac{r}{4}\le x \le r\}. \end{aligned}$$

We see that ${\hat{m}}_r>0$, by ($F_2$). Let $u\in K\cap \partial B_r$; then, by Lemma 3, for $t\in [\frac{1}{4},\frac{3}{4}]$,

$$\begin{aligned} r=\Vert u\Vert _\infty \ge u(t)\ge \frac{1}{4}\Vert u\Vert _\infty =\frac{r}{4}, \end{aligned}$$

and

$$\begin{aligned} f(u(r))\ge {\hat{m}}_r. \end{aligned}$$

(7)

For simplicity, denote $a_{\lambda ,u}\triangleq a(\lambda hf(u))$. Then, for $u\in K\cap \partial B_r$, we get

$$\begin{aligned} 2T_\lambda (u)\left( \frac{1}{2}\right)= & {} \int _{0}^{\frac{1}{2}}\varphi ^{-1}\left( a_{\lambda ,u}+\int _{s}^{\frac{1}{2}}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau \right) \mathrm{d}s\\&\quad + \int _{\frac{1}{2}}^{1}\varphi ^{-1}\left( -a_{\lambda ,u}+\int _{\frac{1}{2}}^{s}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau \right) \mathrm{d}s. \end{aligned}$$

If $a_{\lambda ,u}\ge 0$, then

$$\begin{aligned}&\int _{0}^{\frac{1}{2}}\varphi ^{-1}\left( a_{\lambda ,u}+\int _{s}^{\frac{1}{2}}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau \right) \mathrm{d}s\\&\quad \ge \int _{0}^{\frac{1}{2}}\varphi ^{-1}\left( \int _{s}^{\frac{1}{2}}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau \right) \mathrm{d}s, \end{aligned}$$

and by the definition of $a_{\lambda ,u}$,

$$\begin{aligned}&\int _{\frac{1}{2}}^{1}\varphi ^{-1}\left( -a_{\lambda ,u}+\int _{\frac{1}{2}}^{s}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau \right) \mathrm{d}s\\&\quad = \int _{0}^{\frac{1}{2}}\varphi ^{-1}\left( a_{\lambda ,u}+\int _{s}^{\frac{1}{2}}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau \right) \mathrm{d}s \ge 0. \end{aligned}$$

Thus,

$$\begin{aligned} 2T_\lambda (u)\left( \frac{1}{2}\right) \ge \int _{0}^{\frac{1}{2}}\varphi ^{-1}\left( a_{\lambda ,u}+\int _{s}^{\frac{1}{2}}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau \right) \mathrm{d}s. \end{aligned}$$

If $a_{\lambda ,u}< 0$, then $-a_{\lambda ,u}> 0$ and

$$\begin{aligned}&\int _{\frac{1}{2}}^{1}\varphi ^{-1}\left( -a_{\lambda ,u}+\int _{\frac{1}{2}}^{s}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau \right) \mathrm{d}s\\&\quad \ge \int _{\frac{1}{2}}^{1}\varphi ^{-1}\left( \int _{\frac{1}{2}}^{s}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau \right) \mathrm{d}s, \end{aligned}$$

and by the same argument, we get

$$\begin{aligned} 2T_\lambda (u)\left( \frac{1}{2}\right) \ge \int _{\frac{1}{2}}^{1}\varphi ^{-1}\left( \int _{\frac{1}{2}}^{s}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau \right) \mathrm{d}s. \end{aligned}$$

Thus, we obtain

$$\begin{aligned}&2T_\lambda (u)(\frac{1}{2})\\&\quad \ge \min \left\{ \int _{0}^{\frac{1}{2}}\varphi ^{-1}\left( \int _{s}^{\frac{1}{2}}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau \right) \mathrm{d}s, \int _{\frac{1}{2}}^{1}\varphi ^{-1}\left( \int _{\frac{1}{2}}^{s}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau \right) \mathrm{d}s \right\} . \end{aligned}$$

By using (7), we get

$$\begin{aligned}&2\Vert T_\lambda (u)\Vert _\infty \ge 2T_\lambda (u)(\frac{1}{2})\\&\quad \ge \min \left\{ \int _{0}^{\frac{1}{2}}\varphi ^{-1}\left( \int _{s}^{\frac{1}{2}}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau \right) \mathrm{d}s, \int _{\frac{1}{2}}^{1}\varphi ^{-1}\left( \int _{\frac{1}{2}}^{s}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau \right) \mathrm{d}s \right\} \\&\quad \ge \min \left\{ \int _{0}^{\frac{1}{4}}\varphi ^{-1}\left( \int _{s}^{\frac{1}{2}}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau \right) \mathrm{d}s, \int _{\frac{3}{4}}^{1}\varphi ^{-1}\left( \int _{\frac{1}{2}}^{s}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau \right) \mathrm{d}s \right\} \\&\quad \ge \min \left\{ \int _{0}^{\frac{1}{4}}\varphi ^{-1}\left( \int _{\frac{1}{4}}^{\frac{1}{2}}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau \right) \mathrm{d}s, \int _{\frac{3}{4}}^{1}\varphi ^{-1}\left( \int _{\frac{1}{2}}^{\frac{3}{4}}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau \right) \mathrm{d}s \right\} \\&\quad \ge \min \left\{ \int _{0}^{\frac{1}{4}}\varphi ^{-1}\left( \lambda {\hat{m}}_r\int _{\frac{1}{4}}^{\frac{1}{2}} h(\tau )\mathrm{d}\tau \right) \mathrm{d}s, \int _{\frac{3}{4}}^{1}\varphi ^{-1}\left( \lambda {\hat{m}}_r \int _{\frac{1}{2}}^{\frac{3}{4}} h(\tau )\mathrm{d}\tau \right) \mathrm{d}s \right\} \\&\quad = \frac{1}{4}\varphi ^{-1}\left( \lambda {\hat{m}}_r \min \{\int _{\frac{1}{4}}^{\frac{1}{2}} h(\tau )\mathrm{d}\tau ,\int _{\frac{1}{2}}^{\frac{3}{4}} h(\tau )\mathrm{d}\tau \}\right) \triangleq \frac{1}{4}\varphi ^{-1}(\lambda {\hat{m}}_r \varGamma ). \end{aligned}$$

Define

$$\begin{aligned} p(r):= \frac{\varphi (8r)}{{\hat{m}}_r \varGamma }; \end{aligned}$$

then $p:(0,\infty )\rightarrow (0,\infty )$ is continuous. Since $f_0=f_\infty =0$, we get

$$\begin{aligned} \lim _{r\rightarrow 0}p(r)=\lim _{r\rightarrow \infty }p(r)=0. \end{aligned}$$

Thus, there exists $r^{*}\in (0,\infty )$ such that $p(r^{*})=\inf \{p(r) \ | \ r>0\}\triangleq \lambda _0$. Now for any $\lambda >\lambda _0$, there exist $r_1$, $r_2>0$ such that $0<r_1<r^{*}<r_2<\infty$ with $p(r_1)=p(r_2)=\lambda$. Therefore, if $u\in K\cap \partial B_{r_1}$, then for any $\lambda >\lambda _0$,

$$\begin{aligned} 2\Vert T_\lambda (u)\Vert _\infty \ge 2T_\lambda (u)(\frac{1}{2})\ge \frac{1}{4}\varphi ^{-1}\left( \frac{\varphi (8r_1)}{{\hat{m}}_{r_1}\varGamma } {\hat{m}}_{r_1}\varGamma \right) =2r_1=2\Vert u\Vert _\infty , \end{aligned}$$

i.e.,

$$\begin{aligned} \Vert T_\lambda (u)\Vert _\infty \ge \Vert u\Vert _\infty ,\quad \ \ \mathrm{for}\quad \ u\in K\cap \partial B_{r_1},\quad \ \lambda >\lambda _0. \end{aligned}$$

(8)

Similarly,

$$\begin{aligned} \Vert T_\lambda (u)\Vert _\infty \ge \Vert u\Vert _\infty ,\quad \ \ \mathrm{for}\quad \ u\in K\cap \partial B_{r_2},\quad \ \lambda >\lambda _0. \end{aligned}$$

(9)

For any $\lambda >\lambda _0$, we may choose $\epsilon \ (=\epsilon (\lambda ))>0$ sufficiently small so that

$$\begin{aligned} \psi ^{-1}(\lambda \epsilon )\varUpsilon \le 1, \end{aligned}$$

where

$$\begin{aligned} \varUpsilon =\max \left\{ \int _{0}^{\frac{1}{2}}\psi ^{-1}\left( \int _{s}^{\frac{1}{2}} h(\tau )\mathrm{d}\tau \right) \mathrm{d}s, \int _{\frac{1}{2}}^{1}\psi ^{-1}\left( \int _{\frac{1}{2}}^{s} h(\tau )\mathrm{d}\tau \right) \mathrm{d}s \right\} . \end{aligned}$$

Since $f_0=0$, there exists $r_{3}^{*} \ (=r_{3}^{*}(\epsilon ))>0$ such that for $0<x\le r_{3}^{*}$,

$$\begin{aligned} f(x)\le \epsilon \varphi (x). \end{aligned}$$

Take $0<r_3<\min \{r_1,r_{3}^{*}\}$. Then, for $u\in K\cap \partial B_{r_3}$, we get

$$\begin{aligned} f(u(t))\le \epsilon \varphi (\Vert u(t)\Vert )\le \epsilon \varphi (r_3). \end{aligned}$$

(10)

Since $f_\infty =0$, we define a function ${\hat{f}}:[0,\infty )\rightarrow [0,\infty )$ by

$$\begin{aligned} {\hat{f}}(t):=\max \{f(x)\ | \ 0\le x\le t\}. \end{aligned}$$

By Lemma 2.8 in Wang [15], we have

$$\begin{aligned} {\hat{f}}_\infty :=\lim _{t\rightarrow \infty }\frac{{\hat{f}}(t)}{\varphi (t)}=f_\infty =0. \end{aligned}$$

Since ${\hat{f}}_\infty =0$, then for $\epsilon$ given above, there exists $r_{4}^{*} \ (=r_{4}^{*}(\epsilon ))>0$ such that for $t\ge r_{4}^{*}$,

$$\begin{aligned} {\hat{f}}(t)\le \epsilon \varphi (t). \end{aligned}$$

Take $r_4>\max \{r_2,r_{4}^{*}\}$. Then, for $u\in K\cap \partial B_{r_4}$, we get

$$\begin{aligned} f(u(t))\le {\hat{f}}(r_4)\le \epsilon \varphi (r_4). \end{aligned}$$

(11)

Since $T_\lambda (u)\in C_0[0,1]\cap C^1(0,1)$ for $u\in K \cap \partial B_{r_j}$ ($j=3,4$), there exists a unique $\sigma _j\in (0,1)$ such that

$$\begin{aligned} T_\lambda (u)(\sigma _j)=\max _{t\in [0,1]}T_\lambda (u)(t),\quad \ \ \ \ T_{\lambda }(u)^{'}(\sigma _j)=0. \end{aligned}$$

We first consider the case $\sigma _j\in (0,\frac{1}{2}]$.

$$\begin{aligned} 0=T_{\lambda }(u)^{'}(\sigma _j)=\varphi ^{-1}\left( a_{\lambda ,u}+\int _{\sigma _j}^{\frac{1}{2}}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau \right) . \end{aligned}$$

Since $\varphi$ is an odd homeomorphism, $a_{\lambda ,u}=-\int _{\sigma _j}^{\frac{1}{2}}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau$. Applying (10), (11) and Remark 2 with $\sigma =\lambda \epsilon$, $x=\varphi ^{-1}\left( \varphi (r_j)\int _s^{\frac{1}{2}}h(\tau )\mathrm{d}\tau \right)$ and then $\sigma =\int _s^{\frac{1}{2}}h(\tau )\mathrm{d}\tau$, $x=r_j$ consecutively, we obtain

$$\begin{aligned} \Vert T_\lambda (u)\Vert _\infty= & {} T_\lambda (u)(\sigma _j)=\int _0^{\sigma _j}\varphi ^{-1}\left( a_{\lambda ,u}+\int _s^{\frac{1}{2}}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau \right) \mathrm{d}s\\&\quad =\int _0^{\sigma _j}\varphi ^{-1}\left( -\int _{\sigma _j}^{\frac{1}{2}}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau +\int _s^{\frac{1}{2}}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau \right) \mathrm{d}s\\&\quad = \int _0^{\sigma _j}\varphi ^{-1}\left( \int _s^{\sigma _j}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau \right) \mathrm{d}s \\&\quad \le \int _0^{\frac{1}{2}}\varphi ^{-1}\left( \int _s^{\frac{1}{2}}\lambda h(\tau )f(u(\tau ))\mathrm{d}\tau \right) \mathrm{d}s \\&\quad \le \int _0^{\frac{1}{2}}\varphi ^{-1}\left( \lambda \epsilon \varphi (r_j)\int _s^{\frac{1}{2}} h(\tau )\mathrm{d}\tau \right) \mathrm{d}s \\&\quad \le \psi ^{-1}(\lambda \epsilon ) \int _0^{\frac{1}{2}}\varphi ^{-1}\left( \varphi (r_j)\int _s^{\frac{1}{2}} h(\tau )\mathrm{d}\tau \right) \mathrm{d}s \\&\quad \le \psi ^{-1}(\lambda \epsilon ) \left[ \int _0^{\frac{1}{2}}\psi ^{-1}\left( \int _s^{\frac{1}{2}} h(\tau )\mathrm{d}\tau \right) \mathrm{d}s\right] r_j. \end{aligned}$$

Similarly, for the case $\sigma _j\in [\frac{1}{2},1)$, we get

$$\begin{aligned} \Vert T_\lambda (u)\Vert _\infty \le \psi ^{-1}(\lambda \epsilon ) [\int _0^{\frac{1}{2}}\psi ^{-1}\left( \int _s^{\frac{1}{2}} h(\tau )\mathrm{d}\tau \right) \mathrm{d}s] r_j. \end{aligned}$$

Combining the above two inequalities and using the choice of $\epsilon$, we get

$$\begin{aligned} \Vert T_\lambda (u)\Vert _\infty \le \psi ^{-1}(\lambda \epsilon )\varUpsilon r_j \le r_j=\Vert u\Vert _\infty ,\quad \ \ \mathrm{for}\quad \ u\in K\cap \partial B_{r_j} (j=3,4). \end{aligned}$$

(12)

Combining (8), (9) and (12), we conclude that problem (1) has at least two positive solutions $u_1,u_2$ with $r_3\le \Vert u_1\Vert _\infty \le r_1<r_2\le \Vert u_2\Vert _\infty \le r_4$, for $\lambda >\lambda _0$. $\square$

Lemma 5

Assume that (A), (H) and ($F_1$) hold. If $f_0=f_\infty =0$, then there exists ${\bar{\lambda }}>0$ such that if (1) has a positive solution at $\lambda$, then $\lambda \ge {\bar{\lambda }}$.

Proof

Let $f_0=f_\infty =0<\infty$, then there exist positive numbers $\beta _1,\beta _2,R_1,R_2$ such that $R_1<R_2$, $\beta _1>f_0$, $\beta _2>f_\infty$,

$$\begin{aligned} f(x)\le \beta _1 \varphi (x),\quad \ \ \mathrm{for}\quad \ \ 0\le x\le R_1, \end{aligned}$$

and

$$\begin{aligned} f(x)\le \beta _2 \varphi (x),\quad \ \ \mathrm{for}\quad \ \ x\ge R_2. \end{aligned}$$

Let

$$\begin{aligned} \beta =\max \left\{ \beta _1,\beta _2,\max \{\frac{f(x)}{\varphi (x)}\ | \ R_1\le x\le R_2\}, \min \{\beta \ | \ \beta >0, \frac{\psi (\frac{1}{\varUpsilon })}{\beta }<\lambda _0\}\right\} . \end{aligned}$$

Thus, we have

$$\begin{aligned} f(x)\le \beta \varphi (x),\quad \ \ \mathrm{for}\quad \ \ x\in [0,\infty ). \end{aligned}$$

(13)

Indeed, on the contrary, suppose that (1) has a positive solution v for $0<\lambda <{\bar{\lambda }}$, where ${\bar{\lambda }}=\frac{\psi (\frac{1}{\varUpsilon })}{\beta }$. Since $v(t)=T_\lambda (v)(t)$ for $t\in [0,1]$, applying the same argument in the proof of Lemma 4 with aid of (13) and Remark 2 with $\sigma =\lambda \beta$, $x=\varphi ^{-1}\left( \varphi (\Vert u\Vert _\infty )\int _s^{\frac{1}{2}}h(\tau )\mathrm{d}\tau \right)$ and $\sigma =\int _s^{\frac{1}{2}}h(\tau )\mathrm{d}\tau$, $x=\Vert v\Vert _\infty$ consecutively, we get for $0<\lambda <{\bar{\lambda }}$,

$$\begin{aligned} \Vert v\Vert _\infty =\Vert T_\lambda (v)\Vert _\infty \le \psi ^{-1}(\lambda \beta )\varUpsilon \Vert v\Vert _\infty <\Vert v\Vert _\infty , \end{aligned}$$

which is a contradiction. $\square$

Lemma 6

Assume that (A), (H), ($F_1$) and $f_\infty =0$ hold. If (1) has a positive solution at $\lambda ={\hat{\lambda }}$, then (1) has at least one positive solution for $\lambda >{\hat{\lambda }}$.

Proof

Let ${\hat{u}}$ be a positive solution of (1) at $\lambda ={\hat{\lambda }}$ and let $\lambda >{\hat{\lambda }}$ be fixed. Let us consider the modified problem

$$\begin{aligned} \left\{ \begin{array}{ll} -\varphi (u')'= \lambda h(t)f^*(u),\quad t\in (0,1),\\ u(0)= 0 = u(1), \end{array} \right. \end{aligned}$$

(14)

where $f^*:[0,\infty )\rightarrow [0,\infty )$ is defined by $f^*(u)=f(\gamma (u))$ with

$$\begin{aligned} \gamma (u)= \left\{ \begin{array}{ll} u,\quad \ \ \mathrm{if }\quad \ u\ge {\hat{u}},\\ {\hat{u}},\quad \ \ \mathrm{if }\quad \ 0\le u< {\hat{u}}. \end{array} \right. \end{aligned}$$

First, let us show that (14) has at least one positive solution. Since $f_\infty =0$, we can easily get

$$\begin{aligned} f_\infty ^*:=\lim _{u\rightarrow \infty }\frac{f^*(u)}{\varphi (u)}=\lim _{u\rightarrow \infty }\frac{f(u)}{\varphi (u)}=f_\infty =0. \end{aligned}$$

Define $T_\lambda ^*$ to be the same as $T_\lambda$ replacing f by $f^*$. Then, $T_{\lambda }^*: K\rightarrow K$ is also completely continuous. For fixed $\lambda >{\hat{\lambda }}$, we can choose $\epsilon \ (=\epsilon (\lambda ))>0$ so that $\psi ^{-1}(\lambda \epsilon )\varUpsilon \le 1$. Since $f_\infty ^*=0$, there exists $r \ (=r(\epsilon ))>0$ such that for $x\ge r$, we have

$$\begin{aligned} f^*(x)\le \epsilon \varphi (x). \end{aligned}$$

Here we consider two cases.

Case 1$f^*$ is bounded on $[0,\infty )$.

Then, there exists $M>0$ such that for $x\ge 0$, we have

$$\begin{aligned} f^*(x)\le M. \end{aligned}$$

Choose $R> \max \{\Vert {\hat{u}}\Vert _\infty ,\varphi ^{-1}(\lambda M)\varUpsilon \}$. Then, for $u\in K\cap B_R$, we have

$$\begin{aligned} 0\le u(t)\le \Vert u\Vert _\infty \le R, \end{aligned}$$

and

$$\begin{aligned} f^*(u(t))\le M. \end{aligned}$$

Applying the similar arguments in Lemma 4, we can get

$$\begin{aligned} \Vert T_\lambda ^*(u)\Vert _\infty \le \varphi ^{-1}(\lambda M)\varUpsilon \le R,\quad \ \ \mathrm{for}\quad \ u\in K\cap B_R. \end{aligned}$$

Case 2$f^*$ is unbounded on $[0,\infty )$.

Then, there exists $R> \max \{\Vert {\hat{u}}\Vert _\infty ,r\}$ such that for $x\in [0,R]$, we have

$$\begin{aligned} f^*(x)\le f^*(R). \end{aligned}$$

Thus, if $u\in K\cap B_R$, we have

$$\begin{aligned} 0\le u(t)\le \Vert u\Vert _\infty \le R, \end{aligned}$$

and

$$\begin{aligned} f^*(u(t))\le f^*(R)\le \epsilon \varphi (R), \end{aligned}$$

for $t\in [0,1].$ Applying arguments under to there in the proof of Lemma 4, we can get

$$\begin{aligned} \Vert T_\lambda ^*(u)\Vert _\infty \le \psi ^{-1}(\lambda \epsilon ) \varUpsilon R\le R,\quad \ \ \mathrm{for}\quad \ u\in K\cap B_R. \end{aligned}$$

By Lemma 2, we conclude that (14) has a solution $u\in K\cap B_R$.

Next, we show that if u is a solution of (14), then $u(t)\ge {\hat{u}}(t)$, for $t\in [0,1]$.

If it is true, then (14) and (1) are equivalent and the proof can be done. By contradiction, assume that $u(t) \not \ge {\hat{u}}(t)$, for $t\in [0,1]$. Then, by the boundary values of u and ${\hat{u}}$, there exist $T_1$, $T_2\in (0,1)$ such that

$$\begin{aligned} u(t)-{\hat{u}}(t)<0\ \quad \mathrm{for}\quad \ t\in (T_1,T_2),\quad \ \mathrm{and} \quad \ u(T_1)-{\hat{u}}(T_1)=u(T_2)-{\hat{u}}(T_2)=0. \end{aligned}$$

For $t\in (T_1,T_2)$, we have

$$\begin{aligned} -\varphi (u^{'}(t))^{'}=\lambda h(t)f^*(u(t))=\lambda h(t)f({\hat{u}}(t))\ge {\hat{\lambda }}h(t)f({\hat{u}}(t))=-\varphi ({\hat{u}}^{'}(t))^{'}, \end{aligned}$$

i.e.,

$$\begin{aligned} \varphi (u^{'}(t))^{'} \le \varphi ({\hat{u}}^{'}(t))^{'}. \end{aligned}$$

(15)

Since $u-{\hat{u}}\in C_0[T_1,T_2]$, there exists $t_0 \in (T_1,T_2)$ and $0<\delta <T_2-t_0$ such that

$$\begin{aligned} u(t_0)-{\hat{u}}(t_0)=\min _{t\in [T_1,T_2]}\{u(t)-{\hat{u}}(t)\}, \end{aligned}$$

and

$$\begin{aligned} \ u'(t_0)-{\hat{u}}'(t_0)=0,\quad \ \ u'(t)-{\hat{u}}'(t)>0,\quad \ t\in (t_0. t_0+\delta ). \end{aligned}$$

Integrating both sides of (15) from $t_0$ to $t\in (t_0,t_0+\delta )$, we get

$$\begin{aligned} \varphi (u'(t))-\varphi (u'(t_0))\le \varphi ({\hat{u}}'(t))-\varphi ({\hat{u}}'(t_0)). \end{aligned}$$

Since $\varphi$ is increasing, we have $u'(t)\le {\hat{u}}'(t)$, $t\in (t_0,t_0+\delta )$, which is a contradiction. $\square$

Lemma 7

Assume that (A), (H), ($F_1$) and $f_\infty =0$ hold. Let I be a compact interval of $(0,\infty )$. Then, there exists a constant $b_I>0$ such that all possible solutions u of (1) with $\lambda \in I$ satisfy $\Vert u\Vert _{\infty }<b_I$.

Proof

Suppose to the contrary that there exists a sequence $\{u_n\}$ of positive solutions of (1) at $\lambda =\lambda _n$ with $\{\lambda _n\}\subset I=[\alpha ,\beta ]\subset (0,\infty )$ and $\Vert u_n\Vert _\infty \rightarrow \infty$ as $n \rightarrow \infty$. Choose $\epsilon >0$ sufficiently small so that $\psi ^{-1}(\beta \epsilon )\varUpsilon <1$. Using the fact that $f_\infty =0$, we may define a function ${\hat{f}}:[0,\infty )\rightarrow [0,\infty )$ by

$$\begin{aligned} {\hat{f}}(t):=\max \{ f(x) \ | \ 0\le x\le t\}. \end{aligned}$$

Then, by Lemma 2.8 in Wang [15], we have

$$\begin{aligned} {\hat{f}}_\infty := \lim _{t\rightarrow \infty }\frac{{\hat{f}}(t)}{\varphi (t)}=f_\infty =0. \end{aligned}$$

Thus, for $\epsilon$ given above, there exists $r \ (=r(\epsilon ))>0$ such that for $t\ge r$,

$$\begin{aligned} {\hat{f}}(t)\le \epsilon \varphi (t). \end{aligned}$$

From the assumption, we can get $\Vert u_n\Vert _\infty >r$, for sufficiently large n. Thus, for sufficiently large n, we have

$$\begin{aligned} f(u_n(t))\le {\hat{f}}(\Vert u_n\Vert _\infty )\le \epsilon \varphi (\Vert u_n\Vert _\infty ). \end{aligned}$$

(16)

Since $v_n(t)=T_{\lambda _n}(v_n(t))$, for $t\in [0,1]$, applying the same argument in the proof of Lemma 4 with aid of (16) and Remark 2 with $\sigma =\lambda _n \epsilon$,

$x=\varphi ^{-1}\left( \varphi (\Vert v_n\Vert _\infty )\int _s^{\frac{1}{2}}h(\tau )\mathrm{d}\tau \right)$ and $\sigma =\int _s^{\frac{1}{2}}h(\tau )\mathrm{d}\tau$, $x=\Vert v_n\Vert _\infty$ consecutively, and using the fact $\psi ^{-1}(\beta \epsilon )\varUpsilon <1$, we get for $\lambda _n\in I$ with sufficiently large n,

$$\begin{aligned} \Vert v_n\Vert _\infty =\Vert T_{\lambda _n}(v_n)\Vert _\infty \le \psi ^{-1}(\lambda _n \epsilon )\varUpsilon \Vert v_n\Vert _\infty \le \psi ^{-1}(\beta \epsilon )\varUpsilon \Vert v_n\Vert _\infty < \Vert v_n\Vert _\infty , \end{aligned}$$

which is a contradiction. $\square$

Proof of Theorem 1

Define

$$\begin{aligned}&\lambda ^*:=\inf \{{\tilde{\lambda }} \ | \ (1)\ \mathrm{has \ at \ least \ two \ positive \ solutions \ for} \ \lambda >{\tilde{\lambda }}\}. \end{aligned}$$

(17)

$$\begin{aligned}&\lambda _*:=\inf \{\lambda \ | \ (1) \ \mathrm{has \ at \ least \ one \ positive \ solution}\}. \end{aligned}$$

(18)

By Lemmas 4 and 5, $\lambda ^*$ and $\lambda _*$ are both well-defined and $\lambda ^*\ge \lambda _*\ge {\bar{\lambda }}>0$. By the definitions of $\lambda ^*$ and $\lambda _*$, and Lemma 6, we get that (1) has at least two positive solutions for $\lambda >\lambda ^*$, one positive solution for $\lambda \in (\lambda _*,\lambda ^*]$, and no positive solution for $\lambda \in (0,\lambda _*)$. Finally, it is enough to show that (1) has at least one positive solution at $\lambda =\lambda _*$. By the definition of $\lambda _*$ and Lemma 5, we can choose a sequence $\{\lambda _n\}$ with ${\bar{\lambda }}\le \lambda _*<\lambda _n\le 2\lambda _*$ such that $\lambda _n \rightarrow \lambda _*$ as $n\rightarrow \infty$. Then, by Lemma 7, take $I=[{\bar{\lambda }},2\lambda _*]$; there exists $b_I>0$ such that the corresponding solutions $u_n$ satisfying $\Vert u_n\Vert _\infty <b_I$, i.e., $\{u_n\}$ is bounded. Since $T_{\lambda _n}$ is completely continuous, we know that $\{T_{\lambda _n}(u_n)\}$ is equicontinuous. So is $\{u_n\}$, since $u_n = T_{\lambda _n}(u_n).$ By the Ascoli–Arzel$\grave{a}$ theorem, $\{u_n\}$ is relatively compact. Hence, there exists a convergent subsequence $\{u_n\}$ (still denoted by $\{u_n\}$) and $u_*\in K$ such that $u_n \rightarrow u_*$ as $n\rightarrow \infty$. Since $u_n=T_{\lambda _n}(u_n)$, by the Lebesgue Dominated Convergence Theorem, we get $u_*=T_{\lambda _*}(u_*)$ which implies that $u_*$ is a solution of (1) at $\lambda =\lambda _*$. Moreover, by $f_0=0$ and applying a similar argument as in the proof of Lemma 7, we see that $u_*\not \equiv 0$. Consequently $u_*$ is a positive solution of (1) at $\lambda =\lambda _*$. $\square$

Lemma 8

Assume that (A), (H), ($F_1$) and ($F_2$) hold. If $f_0=f_\infty =\infty$, then there exists $\lambda _0>0$ such that (1) has at least two positive solutions for $\lambda \in (0,\lambda _0)$.

Proof

For any $r>0$, define

$$\begin{aligned} {\hat{M}}_r = \max \{f(x) \ | \ 0\le x\le r\}. \end{aligned}$$

Then, by ($F_2$), ${\hat{M}}_r>0$. Let $u\in K\cap \partial B_r$, then for $t\in [0,1]$,

$$\begin{aligned} 0\le u(t)\le \Vert u\Vert _\infty =r, \end{aligned}$$

and

$$\begin{aligned} f(u(t))\le {\hat{M}}_r. \end{aligned}$$

(19)

Since $T_\lambda (u)\in C_0[0,1]\cap C^1 (0,1)$ for $u\in K\cap \partial B_r$, there exists a unique $\sigma \in (0,1)$ such that

$$\begin{aligned} T_\lambda (u)(\sigma )=\max _{t\in [0,1]}T_\lambda (u(t))\quad \ \ \mathrm{and}\quad \ \ T_\lambda (u)'(\sigma )=0. \end{aligned}$$

We also consider the two cases $\sigma \in (0,\frac{1}{2}]$ and $\sigma \in [\frac{1}{2},1).$ By the same argument as in the proof of Lemma 4 with aid of (19), we get

$$\begin{aligned} \Vert T_\lambda (u)\Vert _\infty \le \varphi ^{-1}(\lambda {\hat{M}}_r)\varUpsilon . \end{aligned}$$

Define

$$\begin{aligned} q(r):=\frac{\varphi (\frac{r}{\varUpsilon })}{{\hat{M}}_r}; \end{aligned}$$

then $q:(0,\infty )\rightarrow (0,\infty )$ is continuous. Since $f_0=f_\infty =\infty$, we get

$$\begin{aligned} \lim _{r\rightarrow 0}q(r)=\lim _{r\rightarrow \infty }q(r)=0. \end{aligned}$$

Therefore, we may choose $r^*\in (0,\infty )$ satisfying $q(r^*)=\sup \{ q(r) \ | \ r>0\}\triangleq \lambda _0,$ and for any $\lambda \in (0,\lambda _0)$, there exist $r_1$, $r_2>0$ such that $0<r_1<r^*<r_2<\infty$ with $q(r_1)=q(r_2)=\lambda$. Therefore, if $u\in K\cap \partial B_{r_1}$, then for $\lambda \in (0,\lambda _0)$, we obtain

$$\begin{aligned} \Vert T_\lambda (u)\Vert _\infty \le \varphi ^{-1}\left( \frac{\varphi (\frac{r_1}{\varUpsilon })}{{\hat{M}}_{r_1}}{\hat{M}}_{r_1}\right) \varUpsilon =r_1=\Vert u\Vert _\infty , \end{aligned}$$

(20)

for $u\in K\cap \partial B_{r_1}.$ Similarly,

$$\begin{aligned} \Vert T_\lambda (u)\Vert _\infty \le \Vert u\Vert _\infty , \end{aligned}$$

(21)

for $u\in K\cap \partial B_{r_2}.$ Let $\lambda \in (0,\lambda _0)$ and take $M=\frac{\gamma (32)}{\lambda \varGamma }>0$. Then, by the fact that $f_0=\infty$, there exists $r_M>0$ such that for $0 \le x\le r_M$, we have

$$\begin{aligned} f(x)\ge M\varphi (x). \end{aligned}$$

If $u\in K$ with $\Vert u\Vert _\infty \le r_M$, then by Lemma 3, we get

$$\begin{aligned} 0\le u(t)\le \Vert u\Vert _\infty \le r_M, \end{aligned}$$

and

$$\begin{aligned} f(u(t))\ge M\varphi (u(t))\ge M\varphi (\frac{1}{4}\Vert u\Vert _\infty ), \end{aligned}$$

for $t\in [\frac{1}{4},\frac{3}{4}].$ Take $0<r_3<\min \{r_1,r_M\}$. Then, for $u\in K\cap \partial B_{r_3}$, we get

$$\begin{aligned} f(u(t))\ge M\varphi (u(t))\ge M\varphi (\frac{1}{4}\Vert u\Vert _\infty ), \end{aligned}$$

(22)

for $t\in [\frac{1}{4},\frac{3}{4}].$ Since $f_\infty =\infty$, for M given above, we may choose $R_M>0$ such that $x\ge R_M$ implies $f(x)\ge M\varphi (x).$ If $u\in K$ with $\Vert u\Vert _\infty \ge 4R_M$, then by Lemma 3

$$\begin{aligned} u(t)\ge \min _{t\in [\frac{1}{4},\frac{3}{4}]}u(t)\ge \frac{1}{4}\Vert u\Vert _\infty \ge R_M, \end{aligned}$$

and

$$\begin{aligned} f(u(t))\ge M\varphi (u(t))\ge M\varphi (\frac{1}{4}\Vert u\Vert _\infty ), \end{aligned}$$

for $t\in [\frac{1}{4},\frac{3}{4}].$ Take $r_4>\max \{r_2,4R_M\}$. Then, for $u\in K\cap \partial B_{r_4}$, we get

$$\begin{aligned} f(u(t))\ge M\varphi (u(t))\ge M\varphi (\frac{1}{4}\Vert u\Vert _\infty ), \ \end{aligned}$$

(23)

for $t\in [\frac{1}{4},\frac{3}{4}].$ We also consider the two cases $a_{\lambda ,u}\ge 0$ and $a_{\lambda ,u}<0$. Applying the same argument as in the proof of Lemma 4 with aid of (22), (23) and the definition of M, we get

$$\begin{aligned} 2\Vert T_\lambda (u)\Vert _\infty \ge 2T_\lambda (u)(\frac{1}{2})\ge \frac{1}{4}\varphi ^{-1}\left( \lambda M\varphi (\frac{1}{4}\Vert u\Vert _\infty )\varGamma \right) \ge \frac{1}{4}\varphi ^{-1}\left( \gamma (32)\varphi (\frac{1}{4}\Vert u\Vert _\infty )\right) . \end{aligned}$$

Applying Remark 2 with $\sigma =32$ and $x=\frac{1}{4}\Vert u\Vert _\infty$, we get

$$\begin{aligned} 2\Vert T_\lambda (u)\Vert _\infty \ge \frac{1}{4}\times 32\times \frac{1}{4}\Vert u\Vert _\infty =2\Vert u\Vert _\infty . \end{aligned}$$

This implies that

$$\begin{aligned} \Vert T_\lambda (u)\Vert _\infty \ge \Vert u\Vert _\infty ,\quad \ \ \mathrm{for}\quad \ u\in K\cap \partial B_{r_j} (j=3,4). \end{aligned}$$

(24)

Combining (20), (21) and (24), we conclude that (1) has at least two positive solutions $u_1$ and $u_2$ with $r_3\le \Vert u_1\Vert _\infty \le r_1<r_2\le \Vert u_2\Vert _\infty \le r_4$, for $\lambda \in (0,\lambda _0)$. $\square$

Lemma 9

Assume that (A), (H) and ($F_1$) hold. If $f_0=f_\infty =\infty$, then there exists ${\bar{\lambda }}>0$ such that if (1) has a positive solution at $\lambda$, then $\lambda \le {\bar{\lambda }}$.

Proof

Let $f_0=f_\infty =\infty$, then there exist positive numbers $\eta _1$, $\eta _2$, $r_1^{'}$, $r_2^{'}$ such that $r_1^{'}<r_2^{'}$, $0<\eta _1<f_0$, $0<\eta _2<f_\infty$,

$$\begin{aligned} f(x)\ge \eta _1\varphi (x),\quad \ \ 0\le x\le r_1^{'}, \end{aligned}$$

and

$$\begin{aligned} f(x)\ge \eta _2\varphi (x),\quad \ \ x\ge r_2^{'}. \end{aligned}$$

Let

$$\begin{aligned} \eta _3=\min \left\{ \eta _1,\eta _2, \min \{\frac{f(x)}{\varphi (x)} \ | \ \frac{r_1^{'}}{4}\le x\le r_2^{'}\}, \max \{\eta \ | \ \eta>0, \frac{\gamma (32)}{\eta \varGamma }>\lambda _0\}\right\} >0. \end{aligned}$$

Then, we have

$$\begin{aligned} f(x)\ge \eta _3\varphi (x),\quad \ \ 0\le x\le r_1^{'}, \end{aligned}$$

(25)

and

$$\begin{aligned} f(x)\ge \eta _3\varphi (x),\quad \ \ x\ge \frac{r_1^{'}}{4}. \end{aligned}$$

(26)

Suppose to the contrary that (1) has a positive solution v for $\lambda >{\bar{\lambda }}$, where ${\bar{\lambda }}=\frac{\gamma (32)}{\eta _3 \varGamma }$. If $\Vert v\Vert _\infty \le r_1^{'}$, by (25) and Lemma 3,

$$\begin{aligned} f(v(t))\ge \eta _3 \varphi (v(t))\ge \eta _3 \varphi (\frac{1}{4}\Vert v\Vert _\infty ), \end{aligned}$$

(27)

for $t\in [\frac{1}{4},\frac{3}{4}].$ If $\Vert v\Vert _\infty > r_1^{'}$, then by Lemma 3 and (26), we get

$$\begin{aligned} v(t)\ge \min _{t\in [\frac{1}{4},\frac{3}{4}]}v(t)\ge \frac{1}{4}\Vert v\Vert _\infty > \frac{r_1^{'}}{4}, \end{aligned}$$

and

$$\begin{aligned} f(v(t))\ge \eta _3 \varphi (v(t))\ge \eta _3 \varphi (\frac{1}{4}\Vert v\Vert _\infty ), \end{aligned}$$

(28)

for $t\in [\frac{1}{4},\frac{3}{4}].$ Since $v=T_\lambda (v),$ applying the same argument as in the proof of Lemma 4 with the aid of (27), (28), and Remark 2 with $\sigma =32$, $x=\frac{1}{4}\Vert v\Vert _\infty$, we get

$$\begin{aligned} \Vert v\Vert _\infty= & {} \Vert T_\lambda (v)\Vert _\infty \ge \frac{1}{8}\varphi ^{-1}\left( \lambda \eta _3 \varphi (\frac{1}{4}\Vert v\Vert _\infty )\varGamma \right) \\&\quad > \frac{1}{8}\varphi ^{-1}\left( \gamma (32) \varphi (\frac{1}{4}\Vert v\Vert _\infty )\right) \ge \frac{1}{8}\times 32\times \frac{1}{4}\Vert v\Vert _\infty =\Vert v\Vert _\infty , \end{aligned}$$

for $\lambda >{\bar{\lambda }}.$ This is a contradiction. $\square$

Lemma 10

Assume that (A), (H), ($F_1$) and $f_0=\infty$ hold. If (1) has a positive solution at $\lambda ={\hat{\lambda }}$, then (1) has at least one positive solution for $\lambda \in (0,{\hat{\lambda }})$.

Proof

Let ${\hat{u}}$ be a positive solution of (1) at $\lambda ={\hat{\lambda }}$ and let $\lambda \in (0,{\hat{\lambda }})$ be fixed. Let us consider the modified problem

$$\begin{aligned} \left\{ \begin{array}{ll} -\varphi (u')'=\lambda h(t)f^*(u),\quad t\in (0,1),\\ u(0)=0=u(1), \end{array} \right. \end{aligned}$$

(29)

where $f^*:[0,\infty )\rightarrow [0,\infty )$ is defined by $f^*(u)=f(\gamma (u))$ with

$$\begin{aligned} \gamma (u)= \left\{ \begin{array}{ll} {\hat{u}},&{}\quad \mathrm{if}\ u>{\hat{u}},\\ u,&{}\quad \mathrm{if}\ 0\le u\le {\hat{u}}. \end{array} \right. \end{aligned}$$

First, we show that (29) has at least one positive solution. Define $T_\lambda ^{*}$ to be the same as $T_\lambda$ replacing f by $f^*$. Then, $T_\lambda ^{*}:K\rightarrow K$ is completely continuous. By the fact that $f^*$ is bounded, we may choose $R>0$ satisfying $\Vert T_\lambda ^{*}(u)\Vert _\infty \le R$, for $u\in K$. Thus, we obtain

$$\begin{aligned} \Vert T_\lambda ^{*}(u)\Vert _\infty \le \Vert u\Vert _\infty , \end{aligned}$$

(30)

for $u\in K\cap \partial B_R.$ Since $f_0=\infty$, we can easily get

$$\begin{aligned} f_0^{*}:= \lim _{u\rightarrow 0}\frac{f^*(u)}{\varphi (u)}=\lim _{u\rightarrow 0}\frac{f(u)}{\varphi (u)}=f_0=\infty . \end{aligned}$$

For fixed $\lambda \in (0,{\hat{\lambda }})$, we take $M=\frac{\gamma (32)}{\lambda \varGamma }>0$. Then, by the fact that $f_0^{*}=\infty$, there exists $r_M^{*}>0$ such that

$$\begin{aligned} f^{*}(x)\ge M\varphi (x), \end{aligned}$$

for $0\le x\le r_M^{*}.$ Take $0<r<\min \{R,r_M^{*}\}$. Then, for $u\in K\cap \partial B_r$, we get

$$\begin{aligned} 0\le u(t)\le \Vert u\Vert _\infty =r<r_M^{*}, \end{aligned}$$

and

$$\begin{aligned} f^{*}(u(t))\ge M\varphi (u(t))\ge M\varphi (\frac{1}{4}\Vert u\Vert _\infty ), \end{aligned}$$

for $t\in [\frac{1}{4},\frac{3}{4}].$ Applying the same argument as in the proof of Lemma 8, we can get

$$\begin{aligned} \Vert T_\lambda ^{*}(u)\Vert _\infty \ge \Vert u\Vert _\infty ,\quad \ \mathrm{for}\quad \ u\in K\cap \partial B_r. \end{aligned}$$

(31)

Combining (30) and (31), we conclude that (29) has at least one solution u with $r\le \Vert u\Vert _\infty \le R$. This implies that u must be a positive solution.

Next, we show that if u is a solution of (29), then $0\le u(t)\le {\hat{u}}(t)$, $t\in [0,1]$.

If it is true, then (29) and (1) are equivalent and the proof can be done. Clearly, $u(t)\ge 0$, for $t\in [0,1]$. Applying arguments under to there in the proof of Lemma 6, we can show that $u(t)\le {\hat{u}}(t)$, $t\in [0,1]$. $\square$

Lemma 11

Assume that (A), (H), ($F_1$) and $f_\infty =\infty$ hold. Let I be a compact interval of $(0,\infty )$. Then, there exists a constant $b_I>0$ such that all possible solutions u of (1) with $\lambda \in I$ satisfy $\Vert u\Vert _{\infty }<b_I$.

Proof

Suppose to the contrary that there exists a sequence $\{u_n\}$ of positive solutions of (1) at $\lambda =\lambda _n$ with $\{\lambda _n\}\subset I=[\alpha ,\beta ]\subset (0,\infty )$ and $\Vert u_n\Vert _\infty \rightarrow \infty$ as $n\rightarrow \infty$. Take $M=\frac{2\gamma (32)}{\alpha \varGamma }$. Since $f_\infty =\infty$, there exists $R_M>0$ such that for $x\ge R_M$, we have

$$\begin{aligned} f(x)\ge M\varphi (x). \end{aligned}$$

From the assumption, we can get $\Vert u_n\Vert _\infty \ge 4R_M$, for sufficiently large n. Thus, by Lemma 3, we have

$$\begin{aligned} u_n(t)\ge \min _{t\in [\frac{1}{4},\frac{3}{4}]}u_n(t)\ge \frac{1}{4}\Vert u_n\Vert _\infty \ge R_M, \end{aligned}$$

and

$$\begin{aligned} f(u_n(t))\ge M\varphi (u_n(t))\ge M\varphi (\frac{1}{4}\Vert u_n\Vert _\infty ), \end{aligned}$$

(32)

for $t\in [\frac{1}{4},\frac{3}{4}]$ and sufficiently large n. Since $u_n=T_{\lambda _n}(u_n)$, applying the same argument in Lemma 4 with aid of (32) and by the definition of M and Remark 2 with $\sigma =32$, $x=\frac{1}{4}\Vert u_n\Vert _\infty$, we get for $\lambda _n\in I$ with sufficiently large n,

$$\begin{aligned} \Vert u_n\Vert _\infty= & {} \Vert T_{\lambda _n}(u_n)\Vert _\infty \ge \frac{1}{8}\varphi ^{-1}\left( \lambda _n M\varphi (\frac{1}{4}\Vert u_n\Vert _\infty )\varGamma \right) \\\ge & {} \frac{1}{8}\varphi ^{-1}\left( \alpha M\varphi (\frac{1}{4}\Vert u_n\Vert _\infty )\varGamma \right) = \frac{1}{8}\varphi ^{-1}\left( 2\gamma (32)\varphi (\frac{1}{4}\Vert u_n\Vert _\infty )\right) \\> & {} \frac{1}{8}\varphi ^{-1}\left( \gamma (32)\varphi (\frac{1}{4}\Vert u_n\Vert _\infty )\right) \ge \frac{1}{8} \times 32 \times \frac{1}{4}\Vert u_n\Vert _\infty =\Vert u_n\Vert _\infty . \end{aligned}$$

This is a contradiction. $\square$

Proof of Theorem 2

Define

$$\begin{aligned}&\lambda ^*:=\sup \{\lambda \ | \ (1) \ \mathrm{has \ at \ least \ one \ positive \ solution}\}. \end{aligned}$$

(33)

$$\begin{aligned}&\lambda _*:=\sup \{{\tilde{\lambda }} \ | \ (1) \ \mathrm{has \ at \ least \ two \ positive \ solutions \ for} \ \lambda \in (0,{\tilde{\lambda }})\}. \end{aligned}$$

(34)

By Lemmas 8 and 9, both $\lambda _*$ and $\lambda ^*$ are well-defined and $0<\lambda _*\le \lambda ^*\le {\bar{\lambda }}$. By the definitions of $\lambda _*$, $\lambda ^*$ and by Lemma 10, we conclude that (1) has at least two positive solutions for $\lambda \in (0,\lambda _*)$, one positive solution for $\lambda \in [\lambda _*,\lambda ^*)$, and no positive solution for $\lambda >\lambda ^*$. Finally, it is enough to show that (1) has at least one positive solution at $\lambda =\lambda ^*$. By the definition of $\lambda ^*$ and Lemma 9, we can choose a sequence $\{\lambda _n\}$ with $\frac{\lambda ^*}{2}\le \lambda _n< \lambda ^*\le {\bar{\lambda }}$ such that $\lambda _n \rightarrow \lambda ^*$ as $n\rightarrow \infty$. By Lemma 11 and applying similar argument in the proof of Theorem 1, we have (1) has a solution $u^*$ at $\lambda =\lambda ^*$. Moreover, using the fact $f_0=\infty$ and applying similar argument to there in the proof of Lemma 11, we see that $u^*$ is a positive solution of (1) at $\lambda =\lambda ^*$. $\square$

4 Applications

In this section, we give some examples illustrating our main results.

Example 4.1

Consider the following scalar $\varphi$-Laplacian problem

$$\begin{aligned} \left\{ \begin{array}{ll} \varphi (u')'+ \lambda t^{-\frac{3}{2}}f(u)=0,\quad t\in (0,1),\\ u(0)= 0 = u(1), \end{array} \right. \end{aligned}$$

(35)

where $\varphi (x)=|x|x+x$, $x\in {\mathbb {R}}$, and

$$\begin{aligned} f(u) = \left\{ \begin{array}{ll} u^{3}, &{}\quad \mathrm{if}\ 0\le u<1,\\ u,&{}\quad \mathrm{if}\ u\ge 1. \end{array} \right. \end{aligned}$$

We easily see that $\varphi$ is an odd increasing homeomorphism. Define functions $\psi$ and $\gamma$ by

$$\begin{aligned} \psi (\sigma ) =\left\{ \begin{array}{ll} \sigma ^{2}, &{}\quad \mathrm{if}\ 0<\sigma \le 1,\\ \sigma , &{}\quad \mathrm{if}\ \sigma >1, \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} \gamma (\sigma ) =\left\{ \begin{array}{ll} 1, &{}\quad \mathrm{if} \ 0<\sigma \le 1,\\ \sigma ^{2}, &{}\quad \mathrm{if}\ \sigma >1. \end{array} \right. \end{aligned}$$

Then, $\psi , \gamma :(0,\infty )\rightarrow (0,\infty )$ and $\psi$ is an increasing homeomorphism with

$$\begin{aligned} \psi ^{-1}(\sigma ) =\left\{ \begin{array}{ll} \sigma ^{\frac{1}{2}}, &{}\quad \mathrm{if}\ 0<\sigma \le 1,\\ \sigma , &{}\quad \mathrm{if}\ \sigma >1. \end{array} \right. \end{aligned}$$

We may see that (35) satisfies assumptions $(A), \ (H), \ (F_1)$ and $(F_2)$ (see Xu and Lee [17] for details) and exactly obtain

$$\begin{aligned} f_0= & {} \lim _{u\rightarrow 0}\frac{f(u)}{\varphi (u)}=\lim _{u\rightarrow 0}\frac{u^{3}}{u^{2}+u}=0,\\ f_\infty= & {} \lim _{u\rightarrow \infty }\frac{f(u)}{\varphi (u)}=\lim _{u\rightarrow \infty }\frac{u}{u^{2}+u}=0. \end{aligned}$$

Consequently, by Theorem 1, we see that there exists $\lambda ^{*}\ge \lambda _{*}>0$ such that (35) has at least two positive solutions for $\lambda >\lambda ^{*}$, one positive solution for $\lambda \in [\lambda _*,\lambda ^{*}]$, and no positive solution for $\lambda \in (0,\lambda _*)$.

Example 4.2

Consider the following scalar $\varphi$-Laplacian problem

$$\begin{aligned} \left\{ \begin{array}{ll} \varphi (u')'+ \lambda t^{-\frac{5}{4}}f(u)=0,\quad t\in (0,1),\\ u(0)= 0 = u(1), \end{array} \right. \end{aligned}$$

(36)

where $\varphi (x)=x^{\frac{1}{3}}$, $x\in {\mathbb {R}}$, and

$$\begin{aligned} f(u) = \left\{ \begin{array}{ll} u^{\frac{1}{4}}, &{}\quad \mathrm{if}\ 0\le u<1,\\ u^{2}, &{}\quad \mathrm{if}\ u\ge 1. \end{array} \right. \end{aligned}$$

Then, $\varphi$ is an odd increasing homeomorphism. By the homogeneity of $\varphi$, taking $\psi (\sigma )=\gamma (\sigma )\equiv \varphi (\sigma )$. We can easily check that (36) satisfies the assumptions $(A), \ (H), \ (F_1)$ and $(F_2)$ (see Xu and Lee [17] for details) and exactly obtain

$$\begin{aligned} f_0= & {} \lim _{u\rightarrow 0}\frac{f(u)}{\varphi (u)}=\lim _{u\rightarrow 0}\frac{u^{\frac{1}{4}}}{u^{\frac{1}{3}}}=\lim _{u\rightarrow 0}u^{-\frac{1}{12}}=\infty ,\\ f_\infty= & {} \lim _{u\rightarrow \infty }\frac{f(u)}{\varphi (u)}=\lim _{u\rightarrow \infty }\frac{u^{2}}{u^{\frac{1}{3}}}=\lim _{u\rightarrow \infty }u^{\frac{5}{3}}=\infty . \end{aligned}$$

Consequently, by Theorem 2, we see that there exists $\lambda ^{*}\ge \lambda _{*}>0$ such that (36) has at least two positive solutions for $\lambda \in (0,\lambda _*)$, one positive solution for $\lambda \in [\lambda _*,\lambda ^{*}]$, and no positive solution for $\lambda >\lambda ^{*}$.