Multiplicity of Positive Solutions to Nonlocal Boundary Value Problems with Strong Singularity

Abstract

In this paper, we consider generalized Laplacian problems with nonlocal boundary conditions and a singular weight, which may not be integrable. The existence of two positive solutions to the given problem for parameter $\lambda$ belonging to some open interval is shown. Our approach is based on the fixed point index theory.

1. Introduction

Consider the following singular $\phi$-Laplacian problem:

$${\left(q\left(t\right)\phi \left(u^{\prime }\left(t\right)\right)\right)}^{\prime }+\lambda h\left(t\right)f\left(u\left(t\right)\right)=0,t\in (0,1),$$

(1)

$$u\left(0\right)={\int }_0^{1}u\left(r\right)d{\alpha }_1\left(r\right),u\left(1\right)={\int }_0^{1}u\left(r\right)d{\alpha }_2\left(r\right),$$

(2)

where $\phi :\mathbb{R}\to \mathbb{R}$ is an odd increasing homeomorphism, $q\in C\left(\right[0,1],(0,\infty \left)\right)$, $\lambda \in {\mathbb{R}}_{+}:=[0,\infty )$ is a parameter, $f\in C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$, $h\in C((0,1),{\mathbb{R}}_{+})$, and the integrator functions ${\alpha }_i$ ($i=1,2$) are nondecreasing on $[0,1]$.

All integrals in (2) are meant in the sense of Riemann–Stieltjes. Throughout this paper, we assume the following hypotheses:

$\left(F_1\right)$

There exist increasing homeomorphisms ${\psi }_1,{\psi }_2:[0,\infty )\to [0,\infty )$ such that:

$$\phi \left(x\right){\psi }_1\left(y\right)\le \phi \left(yx\right)\le \phi \left(x\right){\psi }_2\left(y\right)\mathrm{for}\mathrm{all}x,y\in [0,\infty ).$$

(3)

$\left(F_2\right)$

For $i=1,2,$${\widehat{\alpha }}_i:={\alpha }_i\left(1\right)-{\alpha }_i\left(0\right)\in [0,1)$.

Let $\xi :[0,\infty )\to [0,\infty )$ be an increasing homeomorphism. Then, we denote by ${\mathcal{H}}_{\xi }$ the set:

$$\left\{g\in C((0,1),{\mathbb{R}}_{+}):{\int }_0^1{\xi }^{-1}\left(\left|{\int }_s^{\frac{1}{2}}g\left(\tau \right)d\tau \right|\right)ds<\infty \right\}.$$

It is well known that if $\left(F_1\right)$ is assumed, then:

$${\phi }^{-1}\left(x\right){\psi }_2^{-1}\left(y\right)\le {\phi }^{-1}\left(xy\right)\le {\phi }^{-1}\left(x\right){\psi }_1^{-1}\left(y\right)\mathrm{for}\mathrm{all}x,y\in {\mathbb{R}}_{+}$$

(4)

and

$$L^1(0,1)\cap C(0,1)⊊{\mathcal{H}}_{{\psi }_1}\subseteq {\mathcal{H}}_{\phi }\subseteq {\mathcal{H}}_{{\psi }_2}$$

(see, e.g., ([1], Remark 1)).

It is not hard to see that any function of the form

$$\phi \left(s\right)=\sum _{k=1}^n{\left|s\right|}^{p_k-2}s$$

satisfies the assumption $\left(F_1\right)$ with ${\psi }_1\left(s\right)=min\{s^{p_n-1},s^{p_1-1}\}$ and ${\psi }_2\left(s\right)=max\{s^{p_n-1},s^{p_1-1}\}$ for $s\in {\mathbb{R}}_{+}$ (see, e.g., [1,2]). Here, $n\in \mathbb{N}$, $p_k\in (1,\infty )$ for $1\le k\le n$ and $p_i\le p_j$ for $1\le i\le j\le n$. If $n=1$, it follows that $\phi \left(s\right)={\left|s\right|}^{p-2}s$ for some $p\in (1,\infty )$, that is, Equation (1) becomes the classical p-Laplacian one.

The study of problems with nonlocal boundary conditions is motivated by a variety of applications such as beam deflection [3], chemical reactor theory [4], and thermostatics [5]. For this reason, the existence of positive solutions for nonlocal boundary value problems has been extensively studied. For example, Liu [6] studied the multi-point boundary value problem, which is a special case of problem (1)–(2) with $\lambda =1$. Under various assumptions of the nonlinearity f, the existence of positive solutions was shown. Bachouche, Djebali and Moussaoui [7] proved, under suitable assumptions of the nonlinearity $f=f(t,u,u^{\prime })$ satisfying the $L^1$-Carathéodory condition, several existence results for positive solutions to $\phi$-Laplacian boundary value problems involving linear bounded operators in the boundary conditions. Yang [8], by using the Avery– Peterson fixed point theorem, obtained the existence of at least three positive solutions to the p-Laplacian equation with integral boundary conditions. Goodrich [9] studied perturbed Volterra integral operator equations and, as an application, established the existence of at least one positive solution to the p-Laplacian differential equation with nonlocal boundary conditions. Jeong and Kim [10] obtained sufficient conditions on the nonlinearity f for the existence of multiple positive solutions to problem (1)–(2) with $\lambda =1$. For the nonlinearity $f=f(t,s)$ satisfying $f(t,0)\not\equiv 0$, Kim [11] showed the existence, nonexistence and multiplicity of positive solutions to problem (1)–(2) by investigating the shape of the unbounded solution continuum. For the historical development of the theory of the problems with nonlocal boundary conditions, we refer the reader to the survey papers [12,13,14,15].

In this paper, we show the existence of two positive solutions to nonlocal boundary value problems (1)–(2) for $\lambda$ belonging to some open interval in the case when either $f_0=f_{\infty }=\infty$ or $f_0=f_{\infty }=0$. Here,

$$f_0:=\underset{s\to 0}{lim}\frac{f\left(s\right)}{\phi \left(s\right)}\mathrm{and}{f}_{\infty }:=\underset{s\to \infty }{lim}\frac{f\left(s\right)}{\phi \left(s\right)}.$$

For problems with zero Dirichlet boundary conditions, that is, ${\widehat{\alpha }}_1={\widehat{\alpha }}_2=0$, there have been several works for problems with such assumptions on the nonlinearity f. For example, when $\phi \left(s\right)={\left|s\right|}^{p-2}s$ for some $p\in (1,\infty )$, $q\equiv 1$ and $h\in {\mathcal{H}}_{\phi }$, Agarwal, Lü and O’Regan [16] investigated the existence of two positive solutions to problem (1)–(2). After that, Wang [17] obtained the same multiplicity results in [16] for generalized $\phi$-Laplacian problems with the assumptions that $\phi$ satisfies $\left(F_1\right)$ and $h\in C[0,1]$. Recently, Lee and Xu [18] extended the result of [17] to the singularly weighed $\phi$-Laplacian problem under the assumptions that $q\equiv 1$ and $h\in {\mathcal{H}}_{{\psi }_1}$, that is, h may be singular at $t=0$ and/or $t=1$.

The aim of this paper is to generalize the results for the previous papers [16,17,18]. The main result is stated as follows:

Theorem 1.

Assume that $\left(F_1\right)$, $\left(F_2\right)$ and $h\in {\mathcal{H}}_{{\psi }_1}\backslash \left\{0\right\}$ hold.

(1)

If $f_0=f_{\infty }=\infty ,$ then there exist ${\lambda }^{\ast }\in (0,\infty )$ and $m^{\ast }\in (0,\infty )$ such that problem (1) has two positive solutions $u_1\left(\lambda \right)$ and $u_2\left(\lambda \right)$ for any $\lambda \in (0,{\lambda }^{\ast })$. Moreover, $u_1\left(\lambda \right)$ and $u_2\left(\lambda \right)$ can be chosen with the property that:

$0<\parallel u_1{\left(\lambda \right)\parallel }_{\infty }<m^{\ast }<{\parallel u_2\left(\lambda \right)\parallel }_{\infty }$, ${\displaystyle \underset{\lambda \to 0}{lim}{\parallel u_1\left(\lambda \right)\parallel }_{\infty }=0}$ and ${\displaystyle \underset{\lambda \to 0}{lim}{\parallel u_2\left(\lambda \right)\parallel }_{\infty }=\infty }$.

(2)

If $f_0=f_{\infty }=0,$ then there exist ${\lambda }_{\ast }\in (0,\infty )$ and $m_{\ast }\in (0,\infty )$ such that (1) has two positive solutions $u_1\left(\lambda \right)$ and $u_2\left(\lambda \right)$ for any $\lambda \in ({\lambda }_{\ast },\infty )$. Moreover, $u_1\left(\lambda \right)$ and $u_2\left(\lambda \right)$ can be chosen with the property that:

$0<\parallel u_1{\left(\lambda \right)\parallel }_{\infty }<m_{\ast }<{\parallel u_2\left(\lambda \right)\parallel }_{\infty }$, ${\displaystyle \underset{\lambda \to \infty }{lim}{\parallel u_1\left(\lambda \right)\parallel }_{\infty }=0}$ and ${\displaystyle \underset{\lambda \to \infty }{lim}{\parallel u_2\left(\lambda \right)\parallel }_{\infty }=\infty }$.

The rest of this paper is organized as follows. In Section 2, preliminary results which are essential for proving Theorem 1 are provided. In Section 3, the proof of Theorem 1 is given. Finally, the summary of this paper is provided in Section 4.

2. Preliminaries

Throughout this section, we assume that $\left(F_1\right),\left(F_2\right)$ and $h\in {\mathcal{H}}_{\phi }\backslash \left\{0\right\}$ hold. For convenience, we use some notations which were used by Jeong and Kim ([10]).

The usual maximum norm in a Banach space $C[0,1]$ is denoted by:

$${\displaystyle {\parallel u\parallel }_{\infty }:=\underset{t\in [0,1]}{max}\left|u\left(t\right)\right|\mathrm{for}u\in C[0,1],}$$

and let

$${\alpha }_h:=inf\{x\in (0,1):h\left(x\right)>0\},{\beta }_h:=sup\{x\in (0,1):h\left(x\right)>0\},$$

$${\displaystyle {\overline{\alpha }}_h:=sup\{x\in (0,1):h\left(y\right)>0}\mathrm{for}\mathrm{all}y\in ({\alpha }_h,x)\},$$

$${\displaystyle {\overline{\beta }}_h:=inf\{x\in (0,1):h\left(y\right)>0}\mathrm{for}\mathrm{all}y\in (x,{\beta }_h)\},$$

$${\displaystyle {\gamma }_h^1:=\frac{1}{4}(3{\alpha }_h+{\overline{\alpha }}_h)}\mathrm{and}{\displaystyle {\gamma }_h^2:=\frac{1}{4}({\overline{\beta }}_h+3{\beta }_h).}$$

Then, since $h\in C((0,1),{\mathbb{R}}_{+})\backslash \left\{0\right\}$, we have two cases, either:

$$(\mathrm{i}){\displaystyle 0\le {\alpha }_h<{\overline{\alpha }}_h\le {\overline{\beta }}_h<{\beta }_h\le 1}$$

or

$$(\mathrm{ii}){\displaystyle 0\le {\alpha }_h={\overline{\beta }}_h<{\beta }_h\le 1}\mathrm{and}{\displaystyle 0\le {\alpha }_h<{\overline{\alpha }}_h={\beta }_h\le 1}$$

Consequently,

$${\displaystyle h\left(t\right)>0\mathrm{for}t\in ({\alpha }_h,{\overline{\alpha }}_h)\cup ({\overline{\beta }}_h,{\beta }_h),\mathrm{and}0\le {\alpha }_h<{\gamma }_h^1<{\gamma }_h^2<{\beta }_h\le 1.}$$

(5)

Let ${\displaystyle {\rho }_h:={\rho }_{1}min\{{\gamma }_h^1,1-{\gamma }_h^2\}\in (0,1)}$, where

$${\displaystyle q_0:=\underset{t\in [0,1]}{min}q\left(t\right)>0}\mathrm{and}{\displaystyle {\rho }_1:={\psi }_2^{-1}\left(\frac{1}{{\parallel q\parallel }_{\infty }}\right){\left[{\psi }_1^{-1}\left(\frac{1}{q_0}\right)\right]}^{-1}\in (0,1].}$$

Then

$$\mathcal{K}:=\{u\in C([0,1],{\mathbb{R}}_{+}):u\left(t\right)\ge {\rho }_h\parallel u{\parallel }_{\infty }\mathrm{for}t\in [{\gamma }_h^1,{\gamma }_h^2]\}$$

is a cone in $C[0,1].$ For $r>0,$ let:

$${\mathcal{K}}_r:={\{u\in \mathcal{K}:\parallel u\parallel }_{\infty }<r\},\partial {\mathcal{K}}_r:={\{u\in \mathcal{K}:\parallel u\parallel }_{\infty }=r\}$$

and ${\overline{\mathcal{K}}}_r:={\mathcal{K}}_r\cup \partial {\mathcal{K}}_r$. Let

$${\displaystyle C_1:={\psi }_2^{-1}\left(\frac{1}{{\parallel q\parallel }_{\infty }}\right)min\left\{{\int }_{{\gamma }_h^1}^{{\gamma }_h}{\psi }_2^{-1}\left({\int }_s^{{\gamma }_h}h\left(\tau \right)d\tau \right)ds,{\int }_{{\gamma }_h}^{{\gamma }_h^2}{\psi }_2^{-1}\left({\int }_{{\gamma }_h}^{s}h\left(\tau \right)d\tau \right)ds\right\};}$$

$${\displaystyle C_2:={\psi }_1^{-1}\left(\frac{1}{q_0}\right)max\left\{A_1{\int }_0^{{\gamma }_h}{\psi }_1^{-1}\left({\int }_s^{{\gamma }_h}h\left(\tau \right)d\tau \right)ds,A_2{\int }_{{\gamma }_h}^1{\psi }_1^{-1}\left({\int }_{{\gamma }_h}^{s}h\left(\tau \right)d\tau \right)ds\right\}.}$$

Here, ${\displaystyle {\gamma }_h:=\frac{{\gamma }_h^1+{\gamma }_h^2}{2}}$ and $A_i:={(1-{\widehat{\alpha }}_i)}^{-1}\ge 1$ for $i=1,2.$ Clearly, by (5),

$$C_1>0\mathrm{and}{C}_2>0.$$

Define continuous functions $f_{\ast },f^{\ast }:{\mathbb{R}}_{+}\to {\mathbb{R}}_{+}$ by, for $m\in {\mathbb{R}}_{+}$,

$$f_{\ast }\left(m\right):=min\{f\left(y\right):{\rho }_{h}m\le y\le m\}\mathrm{and}{f}^{\ast }\left(m\right):=max\{f\left(y\right):0\le y\le m\}.$$

Define $R_1,R_2:(0,\infty )\to (0,\infty )$ by:

$${\displaystyle R_1\left(m\right):=\frac{1}{f_{\ast }\left(m\right)}\phi \left(\frac{m}{C_1}\right)}\mathrm{and}{\displaystyle R_2\left(m\right):=\frac{1}{f^{\ast }\left(m\right)}\phi \left(\frac{m}{C_2}\right)}\mathrm{for}m\in (0,\infty ).$$

By (4) and $\left(F_2\right)$, ${\psi }_2^{-1}\left(y\right)\le {\psi }_1^{-1}\left(y\right)$ for all $y\in {\mathbb{R}}_{+}$ and $A_i={(1-{\widehat{\alpha }}_i)}^{-1}\ge 1$ for $i=1,2.$ Consequently, $0<C_1<C_2$ and

$$0<R_2\left(m\right)<R_1\left(m\right)\mathrm{for}\mathrm{all}m\in (0,\infty ).$$

(6)

Remark 2.

(1) For any $L\in C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$, let ${\displaystyle L_c:=\underset{m\to c}{lim}\frac{L\left(m\right)}{\phi \left(m\right)}}$ for $c\in \{0,\infty \}$. Then it is easy to prove that:

$${\left(f_{\ast }\right)}_c={\left(f^{\ast }\right)}_c=0\mathrm{if}{f}_c=0,\mathrm{and}{\left(f_{\ast }\right)}_c={\left(f^{\ast }\right)}_c=\infty \mathrm{if}{f}_c=\infty .$$

(7)

For the reader’s convenience, we give the proof for the case ${\left(f_{\ast }\right)}_{\infty }={\left(f^{\ast }\right)}_{\infty }=0$ if $f_{\infty }=0$. The proofs for other cases are similar. Indeed, let $ϵ>0$ be given and let $f_{\infty }=0$ be assumed. Then, there exists $M>0$ such that:

$$\frac{f\left(s\right)}{\phi \left(s\right)}<ϵ\mathrm{for}\mathrm{all}s\ge M,$$

(8)

and

$$f^{\ast }\left(s\right)\le f^{\ast }\left(M\right)+f\left(x_{M,s}\right)fors\ge M.$$

Here $x_{M,s}$ is the point in $[M,s]$ satisfying

$$f\left(x_{M,s}\right)=max\{f\left(x\right):M\le x\le s\}.$$

By (8), for $s\ge M,$

$$0\le \frac{f_{\ast }\left(s\right)}{\phi \left(s\right)}\le \frac{f^{\ast }\left(s\right)}{\phi \left(s\right)}\le \frac{f^{\ast }\left(M\right)}{\phi \left(s\right)}+\frac{f\left(x_{M,s}\right)}{\phi \left(x_{M,s}\right)}\le \frac{f^{\ast }\left(M\right)}{\phi \left(s\right)}+ϵ,$$

which implies

$$0\le \underset{s\to \infty }{lim\; sup}\frac{f_{\ast }\left(s\right)}{\phi \left(s\right)}\le \underset{s\to \infty }{lim\; sup}\frac{f^{\ast }\left(s\right)}{\phi \left(s\right)}\le ϵ.$$

(9)

Consequently, ${\left(f_{\ast }\right)}_{\infty }={\left(f^{\ast }\right)}_{\infty }=0$, since (9) is true for all $ϵ>0$.

(2) By (3) and (7), for $i\in \{1,2\}$,

$$\underset{m\to 0^{+}}{lim}{R}_i\left(m\right)=0if{f}_0=\infty ,and\underset{m\to \infty }{lim}{R}_i\left(m\right)=0if{f}_{\infty }=\infty ;$$

(10)

$$\underset{m\to 0^{+}}{lim}{R}_i\left(m\right)=\infty if{f}_0=0,and\underset{m\to \infty }{lim}{R}_i\left(m\right)=\infty if{f}_{\infty }=0.$$

(11)

For $g\in {\mathcal{H}}_{\phi }$, consider the following problem:

$$\left\{\begin{array}{c}{\left(q\left(t\right)\phi \left(u^{\prime }\left(t\right)\right)\right)}^{\prime }+g\left(t\right)=0,t\in (0,1),\hfill \\ u\left(0\right)={\int }_0^{1}u\left(r\right)d{\alpha }_1\left(r\right),u\left(1\right)={\int }_0^{1}u\left(r\right)d{\alpha }_2\left(r\right).\hfill \end{array}\right.$$

(12)

Define a function $T:{\mathcal{H}}_{\phi }\to C[0,1]$ by $T\left(0\right)=0$ and, for $g\in {\mathcal{H}}_{\phi }\backslash \left\{0\right\}$,

$$T\left(g\right)\left(t\right)=\left\{\begin{array}{cc}{A}_1{\int }_0^1{\int }_0^r{I}_g(s,\sigma )dsd{\alpha }_1\left(r\right)+{\int }_0^t{I}_g(s,\sigma )ds,\hfill & \mathrm{if}0\le t\le \sigma ,\hfill \\ -A_2{\int }_0^1{\int }_r^1{I}_g(s,\sigma )dsd{\alpha }_2\left(r\right)-{\int }_t^1{I}_g(s,\sigma )ds,\hfill & \mathrm{if}\sigma \le t\le 1,\hfill \end{array}\right.$$

(13)

where

$${\displaystyle I_g(s,x):={\phi }^{-1}\left(\frac{1}{q\left(s\right)}{\int }_s^{x}g\left(\tau \right)d\tau \right)}\mathrm{for}s,x\in (0,1)$$

and $\sigma =\sigma \left(g\right)$ is a constant satisfying:

$$A_1{\int }_0^1{\int }_0^r{I}_g(s,\sigma )dsd{\alpha }_1\left(r\right)+{\int }_0^{\sigma }{I}_g(s,\sigma )ds=-A_2{\int }_0^1{\int }_r^1{I}_g(s,\sigma )dsd{\alpha }_2\left(r\right)-{\int }_{\sigma }^1{I}_g(s,\sigma )ds.$$

(14)

For any $g\in {\mathcal{H}}_{\phi }$ and any $\sigma$ satisfying (14), $T\left(g\right)$ is monotone increasing on $[0,\sigma )$ and monotone decreasing on $(\sigma ,1]$. We notice that $\sigma =\sigma \left(g\right)$ is not necessarily unique, but $T\left(g\right)$ is independent of the choice of $\sigma$ satisfying (14) (see [10], [Remark 2]).

Lemma 3.

([10], [Lemma 2] Assume that $\left(F_1\right),\left(F_2\right)$ and $g\in {\mathcal{H}}_{\phi }$ hold. Then $T\left(g\right)$ is a unique solution to problem (12), satisfying the following properties:

$\left(i\right)$

$T\left(g\right)\left(t\right)\ge min\left\{T\right(g\left)\right(0),T(g\left)\right(1\left)\right\}\ge 0$ for $t\in [0,1]$;

$\left(ii\right)$

for any $g\neg \equiv 0$, ${max\{T\left(g\right)\left(0\right),T\left(g\right)\left(1\right)\}<\parallel T\left(g\right)\parallel }_{\infty }$;

$\left(iii\right)$

σ is a constant satisfying (14) if and only if $T\left(g\right)\left(\sigma \right)={\parallel T\left(g\right)\parallel }_{\infty }$;

$\left(iv\right)$

$T\left(g\right)\left(t\right)\ge {\rho }_{1}min\{t,1-t\}{\parallel T\left(g\right)\parallel }_{\infty }$ for $t\in [0,1]$ and $T\left(g\right)\in \mathcal{K}.$

Define a function $F:{\mathbb{R}}_{+}\times \mathcal{K}\to C(0,1)$ by

$$F(\lambda ,u)\left(t\right):=\lambda h\left(t\right)f\left(u\right(t\left)\right)\mathrm{for}(\lambda ,u)\in {\mathbb{R}}_{+}\times \mathcal{K}\mathrm{and}t\in (0,1).$$

Clearly, $F(\lambda ,u)\in {\mathcal{H}}_{\phi }$ for any $(\lambda ,u)\in {\mathbb{R}}_{+}\times \mathcal{K}$, since $h\in {\mathcal{H}}_{\phi }$. Let us define an operator $H:{\mathbb{R}}_{+}\times \mathcal{K}\to \mathcal{K}$ by

$$H(\lambda ,u):=T\left(F\right(\lambda ,u\left)\right)\mathrm{for}(\lambda ,u)\in {\mathbb{R}}_{+}\times \mathcal{K}$$

By Lemma 3 $\left(iv\right)$, $H({\mathbb{R}}_{+}\times \mathcal{K})\subseteq \mathcal{K}$, and consequently H is well defined. Moreover, u is a solution to BVP (1)–(2) if and only if $H(\lambda ,u)=u$ for some $(\lambda ,u)\in {\mathbb{R}}_{+}\times \mathcal{K}.$

Lemma 4.

([11],[Lemma 2]) Assume that $\left(F_1\right),\left(F_2\right)$ and $h\in {\mathcal{H}}_{\phi }\backslash \left\{0\right\}$ hold. Then, the operator $H:{\mathbb{R}}_{+}\times \mathcal{K}\to \mathcal{K}$ is completely continuous.

Finally, we recall a well-known theorem of the fixed point index theory.

Theorem 5.

([19,20]) Assume that, for some $m>0,$$\mathcal{H}:{\overline{\mathcal{K}}}_m\to \mathcal{K}$ is completely continuous. Then the following assertions are true:

(i) $i(\mathcal{H},{\mathcal{K}}_m,\mathcal{K})=1$ if ${\parallel \mathcal{H}\left(u\right)\parallel }_{\infty }<{\parallel u\parallel }_{\infty }$ for $u\in \partial {\mathcal{K}}_m$;

(ii) $i(\mathcal{H},{\mathcal{K}}_m,\mathcal{K})=0$ if ${\parallel \mathcal{H}\left(u\right)\parallel }_{\infty }>{\parallel u\parallel }_{\infty }$ for $u\in \partial {\mathcal{K}}_m$.

3. Proof of Theorem 1

In this section, we give the proof of Theorem 1.

Proof of Theorem 1. (1) Since $f_0=f_{\infty }=\infty ,$ from (10), it follows that, for $i=1,2,$

$$\underset{m\to 0}{lim}{R}_i\left(m\right)=\underset{m\to \infty }{lim}{R}_i\left(m\right)=0.$$

(15)

We can choose ${\lambda }^{\ast }>0$ and $m^{\ast }>0$ satisfying:

$${\lambda }^{\ast }=max\{R_2\left(m\right):m\in {\mathbb{R}}_{+}\}\mathrm{and}{R}_2\left(m^{\ast }\right)={\lambda }^{\ast }.$$

Let $\lambda \in (0,{\lambda }^{\ast })$ be fixed. By (6), there exist $m_1=m_1\left(\lambda \right),m_2=m_2\left(\lambda \right),M_1=M_1\left(\lambda \right),M_2=M_2\left(\lambda \right)$ such that:

$$m_1<m_2<m^{\ast }<M_2<M_1$$

and

$$max\{R_1\left(m_1\right),R_1\left(M_1\right)\}<\lambda <min\{R_2\left(m_2\right),R_2\left(M_2\right)\}.$$

Since $\lambda <R_2\left(m_2\right),$

$$0\le \lambda f\left(v\left(t\right)\right)\le \lambda f^{\ast }\left(m_2\right)=\frac{\lambda }{R_2\left(m_2\right)}\phi \left(\frac{m_2}{C_2}\right)<\phi \left(\frac{m_2}{C_2}\right)\mathrm{for}t\in [0,1].$$

(16)

Let $u\in \partial {\mathcal{K}}_{m_2}$ be given and let $\sigma$ be a number satisfying $H(\lambda ,u)\left(\sigma \right)={\parallel H(\lambda ,u)\parallel }_{\infty }.$ We have two cases: either (i) $\sigma \in (0,{\gamma }_h)$ or (ii) $\sigma \in [{\gamma }_h,1)$. We only give the proof for the case (i), since the case (ii) can be proved in a similar manner. First, we show that:

$${\parallel H(\lambda ,u)\parallel }_{\infty }\le A_1{\int }_0^{\sigma }{I}_{F(\lambda ,u)}(s,\sigma )\mathrm{for}s\in [0,\sigma ].$$

(17)

Since $I_{F(\lambda ,u)}(s,x)\ge 0$ for $x\ge s$ and $I_{F(\lambda ,u)}(s,x)\le 0$ for $x\le s$,

$$\begin{array}{ccc}& & {\int }_0^1{\int }_{\sigma }^r{I}_{F(\lambda ,u)}(s,\sigma )dsd{\alpha }_1\left(r\right)\hfill \\ & =& -{\int }_0^{\sigma }{\int }_r^{\sigma }{I}_{F(\lambda ,u)}(s,\sigma )dsd{\alpha }_1\left(r\right)+{\int }_{\sigma }^1{\int }_{\sigma }^r{I}_{F(\lambda ,u)}(s,\sigma )dsd{\alpha }_1\left(r\right)\le 0.\hfill \end{array}$$

Consequently,

$$\begin{array}{ccc}\hfill H(\lambda ,u)\left(\sigma \right)& =& A_1{\int }_0^1{\int }_0^r{I}_{F(\lambda ,u)}(s,\sigma )dsd{\alpha }_1\left(r\right)+{\int }_0^{\sigma }{I}_{F(\lambda ,u)}(s,\sigma )ds\hfill \\ & =& A_1\left[{\int }_0^1{\int }_0^r{I}_{F(\lambda ,u)}(s,\sigma )dsd{\alpha }_1\left(r\right)+\left(1-{\int }_0^{1}d{\alpha }_1\left(r\right)\right){\int }_0^{\sigma }{I}_{F(\lambda ,u)}(s,\sigma )ds\right]\hfill \\ & =& A_1\left[{\int }_0^1{\int }_{\sigma }^r{I}_{F(\lambda ,u)}(s,\sigma )dsd{\alpha }_1\left(r\right)+{\int }_0^{\sigma }{I}_{F(\lambda ,u)}(s,\sigma )ds\right]\hfill \\ & \le & A_1{\int }_0^{\sigma }{I}_{F(\lambda ,u)}(s,\sigma )ds.\hfill \end{array}$$

From (4), (16), (17) and the definition of $C_2$, it follows that:

$$\begin{array}{ccc}\hfill {\parallel H(\lambda ,u)\parallel }_{\infty }& \le & A_1{\int }_0^{\sigma }{\phi }^{-1}\left(\frac{1}{q\left(s\right)}{\int }_s^{\sigma }\lambda h\left(\tau \right)f\left(u\left(\tau \right)\right)d\tau \right)ds\hfill \\ & <& A_1{\int }_0^{{\gamma }_h}{\phi }^{-1}\left({\int }_s^{{\gamma }_h}h\left(\tau \right)d\tau \frac{1}{q_0}\phi \left(\frac{m_2}{C_2}\right)\right)ds\hfill \\ & \le & A_1{\int }_0^{{\gamma }_h}{\psi }_1^{-1}\left({\int }_s^{{\gamma }_h}h\left(\tau \right)d\tau \right)ds{\phi }^{-1}\left(\frac{1}{q_0}\phi \left(\frac{m_2}{C_2}\right)\right)\hfill \\ & \le & A_1{\int }_0^{{\gamma }_h}{\psi }_1^{-1}\left({\int }_s^{{\gamma }_h}h\left(\tau \right)d\tau \right)ds{\psi }_1^{-1}\left(\frac{1}{q_0}\right)\frac{m_2}{C_2}\le m_2={\parallel u\parallel }_{\infty }.\hfill \end{array}$$

By Theorem 5 (i),

$$i(H(\lambda ,·),{\mathcal{K}}_{m_2},\mathcal{K})=1.$$

(18)

Let $v\in \partial {\mathcal{K}}_{m_1}$ be given. Since $\lambda >R_1\left(m_1\right)$ and ${\rho }_h{m}_1\le v\left(t\right)\le m_1$ for $t\in [{\gamma }_h^1,{\gamma }_h^2],$ and

$$\lambda f\left(v\left(t\right)\right)\ge \lambda f_{\ast }\left(m_1\right)=\frac{\lambda }{R_1\left(m_1\right)}\phi \left(\frac{m_1}{C_1}\right)>\phi \left(\frac{m_1}{C_1}\right)\mathrm{for}t\in [{\gamma }_h^1,{\gamma }_h^2].$$

(19)

Let $\sigma$ be a constant satisfying $H(\lambda ,v)\left(\sigma \right)={\parallel H(\lambda ,v)\parallel }_{\infty }.$ Then we have two cases: either (i) $\sigma \in [{\gamma }_h,1)$ or (ii) $\sigma \in (0,{\gamma }_h)$. We only give the proof for the case (i), since the case (ii) can be proved in a similar manner. By Lemma 3 (i), $H(\lambda ,v)\left(0\right)\ge 0$, and it follows from (4), (19) and the definition of $C_1$ that:

$$\begin{array}{ccc}\hfill {\parallel H(\lambda ,v)\parallel }_{\infty }& =& H(\lambda ,v)\left(0\right)+{\int }_0^{\sigma }{\phi }^{-1}\left(\frac{1}{q\left(s\right)}{\int }_s^{\sigma }\lambda h\left(\tau \right)f\left(v\left(\tau \right)\right)d\tau \right)ds\hfill \\ & >& {\int }_{{\gamma }_h^1}^{{\gamma }_h}{\phi }^{-1}\left({\int }_s^{{\gamma }_h}h\left(\tau \right)d\tau \frac{1}{{\parallel q\parallel }_{\infty }}\phi \left(\frac{m_1}{C_1}\right)\right)ds\hfill \\ & \ge & {\int }_{{\gamma }_h^1}^{{\gamma }_h}{\psi }_2^{-1}\left({\int }_s^{{\gamma }_h}h\left(\tau \right)d\tau \right)ds{\phi }^{-1}\left(\frac{1}{{\parallel q\parallel }_{\infty }}\phi \left(\frac{m_1}{C_1}\right)\right)\hfill \\ & \ge & {\int }_{{\gamma }_h^1}^{{\gamma }_h}{\psi }_2^{-1}\left({\int }_s^{{\gamma }_h}h\left(\tau \right)d\tau \right)ds{\psi }_2^{-1}\left(\frac{1}{{\parallel q\parallel }_{\infty }}\right)\frac{m_1}{C_1}\ge m_1={\parallel v\parallel }_{\infty }.\hfill \end{array}$$

By Theorem 5 (ii),

$$i(H(\lambda ,·),{\mathcal{K}}_{m_1},\mathcal{K})=0.$$

(20)

From (18), (20) and the additivity property,

$$i(H(\lambda ,·),{\mathcal{K}}_{m_2}\backslash {\overline{\mathcal{K}}}_{m_1},\mathcal{K})=-1.$$

Then there exists $u_{\lambda }^1\in {\mathcal{K}}_{m_2}\backslash {\overline{\mathcal{K}}}_{m_1}$ such that $H(\lambda ,u_{\lambda }^1)=u_{\lambda }^1$ by the solution property. Consequently, problem (1)–(2) has a positive solution $u_{\lambda }^1$ satisfying $\parallel u_{\lambda }^1{\parallel }_{\infty }\in (m_1,m_2).$

By the similar argument above, one can show the existence of another positive solution $u_{\lambda }^2$ to problem (1) satisfying $\parallel u_{\lambda }^2{\parallel }_{\infty }\in (M_2,M_1).$ Moreover, by (15), we may choose $m_2\left(\lambda \right),M_2\left(\lambda \right)$ satisfying $m_2\left(\lambda \right)\to 0$ and $M_2\left(\lambda \right)\to \infty$ as $\lambda \to 0^{+}$, and thus (1) has two positive solutions $u_{\lambda }^1,u_{\lambda }^2$ for any $\lambda \in (0,{\lambda }^{\ast })$ satisfying $\parallel u_{\lambda }^1{\parallel }_{\infty }\to 0$ and $\parallel u_{\lambda }^2{\parallel }_{\infty }\to \infty$ as $\lambda \to 0^{+}$.

(2) Since $f_0=f_{\infty }=0,$ from (11), it follows that, for $i=1,2,$

$$\underset{m\to 0}{lim}{R}_i\left(m\right)=\underset{m\to \infty }{lim}{R}_i\left(m\right)=\infty .$$

(21)

We can choose ${\lambda }_{\ast }>0$ and $m_{\ast }>0$ satisfying

$${\lambda }_{\ast }=min\{R_1\left(m\right):m\in {\mathbb{R}}_{+}\}\mathrm{and}{R}_1\left(m_{\ast }\right)={\lambda }_{\ast }.$$

Let $\lambda \in ({\lambda }_{\ast },\infty )$ be fixed. By (6), there exist $m_1=m_1\left(\lambda \right),m_2=m_2\left(\lambda \right),$$M_1=M_1\left(\lambda \right),M_2=M_2\left(\lambda \right)$ such that

$$m_2<m_1<m_{\ast }<M_1<M_2$$

and

$$max\{R_1\left(m_1\right),R_1\left(M_1\right)\}<\lambda <min\{R_2\left(m_2\right),R_2\left(M_2\right)\}$$

By the argument similar to those in the proof of (1),

$$i(H(\lambda ,·),{\mathcal{K}}_{m_1}\backslash {\overline{\mathcal{K}}}_{m_2},\mathcal{K})=i(H(\lambda ,·),{\mathcal{K}}_{M_2}\backslash {\overline{\mathcal{K}}}_{M_1},\mathcal{K})=-1.$$

Thus, problem (1)–(2) has two positive solutions $u_{\lambda }^1,u_{\lambda }^2$ for any $\lambda \in ({\lambda }_{\ast },\infty )$ satisfying $\parallel u_{\lambda }^1{\parallel }_{\infty }\in (m_2,m_1)$ and $\parallel u_{\lambda }^2{\parallel }_{\infty }\in (M_1,M_2)$. Moreover, by (21), we may choose $m_1\left(\lambda \right),M_1\left(\lambda \right)$ satisfying $m_1\left(\lambda \right)\to 0$ and $M_1\left(\lambda \right)\to \infty$ as $\lambda \to \infty$, and thus (1)–(2) has two positive solutions $u_{\lambda }^1,u_{\lambda }^2$ for any $\lambda \in ({\lambda }_{\ast },\infty )$ satisfying $\parallel u_{\lambda }^1{\parallel }_{\infty }\to 0$ and $\parallel u_{\lambda }^2{\parallel }_{\infty }\to \infty$ as $\lambda \to \infty$.

4. Conclusions

In this paper, we establish the existence of two positive solutions to nonlocal boundary value problems (1)–(2) for $\lambda$ belonging to some open interval in the case when either $f_0=f_{\infty }=\infty$ or $f_0=f_{\infty }=0$.

Let $\phi$ be an odd function satisfying $\phi \left(x\right)=x+x^2$ for $x\in {\mathbb{R}}_{+}$. Then, $\phi$ satisfies $\left(F_1\right)$ with ${\psi }_1\left(y\right)=min\{y,y^2\}$ and ${\psi }_2\left(y\right)=max\{y,y^2\}$. Define $h:(0,1)\to {\mathbb{R}}_{+}$ by:

$$h\left(t\right)=0\mathrm{for}t\in [0,\frac{1}{4}]\mathrm{and}h\left(t\right)=(t-\frac{1}{4}){(1-t)}^{-c}\mathrm{for}t\in (\frac{1}{4},1).$$

Then, since ${\psi }_1^{-1}\left(s\right)=s$ for all $s\ge 1,$$h\in {\mathcal{H}}_{{\psi }_1}\backslash L^1(0,1)$ for any $c\in [1,2)$. We give some examples for nonlinearity f to illustrate the main result (Theorem 1).

Let

$$f_1\left(s\right)=\left\{\begin{array}{cc}{s}^{\frac{1}{2}},\hfill & \mathrm{for}s\in [0,1];\hfill \\ s^3,\hfill & \mathrm{for}s\in (1,\infty )\hfill \end{array}\right. \mathrm{and} f_2\left(s\right)=s^{\frac{3}{2}}\mathrm{for}s\in {\mathbb{R}}_{+}.$$

Then,

$${\left(f_1\right)}_0={\left(f_1\right)}_{\infty }=\infty \mathrm{and}{\left(f_2\right)}_0={\left(f_2\right)}_{\infty }=0.$$

Consequently, by Theorem 1, problem (1)–(2) with $f=f_1$ has two positive solutions for all small $\lambda >0$, and problem (1)–(2) with $f=f_2$ has two positive solutions for all large $\lambda >0$.

As shown in the examples of nonlinearity $f=f\left(s\right)$ above, $f\left(0\right)$ may be 0. What this means is that nonnegative solutions may be trivial ones. The existence of an unbounded solution component to problem (1)–(2) can be obtained as in the paper [11], where the nonlinearity $f=f(t,s)$ satisfies $f(t,0)\not\equiv 0$, but we cannot get any information about positive solutions from the solution component. Thus, the fixed point index theory was used in order to show the existence of two positive solutions to problem (1)–(2).