Positive solutions for nonlocal differential equations with concave and convex coefficients

Abstract

In this paper, we study the positive solutions for nonlocal differential equations with concave and convex coefficients:

$$\begin{aligned} -A\left( \int ^1_0 (u^p(s)+u^q(s))ds\right) u''(t)=f(t,u(t)),\quad t\in (0,1), \end{aligned}$$

where $0<p<1\le q.$ Using the fixed point index theory and fixed point theorems on cones, existence and multiplicity results are obtained, when the nonlinear term f(t, x) is continuous, has a singularity at $x=0$, changes sign, respectively.

1 Introduction

In this paper, we are concerned with the existence and multiplicity of positive solutions for the following nonlocal differential equation

$$\begin{aligned} -A\left( \int ^1_0 (u^p(s)+u^q(s))ds\right) u''(t)= f(t,u(t)), \quad t\in (0,1), \end{aligned}$$

(1.1)

where $0<p<1\le q$, the function $A:\mathbb {R}\rightarrow \mathbb {R}$ is continuous. Note the differential Eq. (1.1) is nonlocal due to the coefficient function $A\left( \int ^1_0 (u^p(s)+u^q(s))ds\right)$ involving an integral, and $u^p$ is a concave function and $u^q$ is a convex function.

In order to better study the positive solutions for nonlocal differential Eq. (1.1), we convert Eq. (1.1) coupled with different nonlocal boundary conditions into the following integral equation

$$\begin{aligned} u(t)= & \gamma (t)H(\varphi (u))+\int ^1_0\left( A\left( \int ^1_0 (u^p(r)+u^q(r))dr\right) \right) ^{-1}\nonumber \\ & G(t,s)f(s,u(s))ds,\quad t\in [0,1], \end{aligned}$$

(1.2)

where functions $\gamma , H$ and $\varphi$ are involved in boundary data, and then investigate the existence of positive solutions to the Eq. (1.2).

Nonlocal differential equations as (1.1) arise in various areas of physics and applied mathematics (see [1,2,3, 7, 9] and the references therein). The nonlocal differential equations and nonlocal boundary conditions are intensively studied by many scholars; for example, see [4,5,6, 8, 10, 12,13,14,15,16,17,18,19,20,21,22, 24,25,33].

Recently, Goodrich [14] studied the following nonlocal differential equation Dirichlet boundary value problems:

$$\begin{aligned} \left\{ \begin{aligned}&-A\left( \int ^1_0 |u(s)|^qds\right) u''(t)=\lambda f(t,u(t)),\quad t\in (0,1),\\&u(0)=u(1)=0, \end{aligned}\right. \end{aligned}$$

(1.3)

where $q\ge 1,\ \lambda >0,\ A:[0,+\infty )\rightarrow \mathbb {R}$ is continuous, and $f:[0,1]\times [0,+\infty )\rightarrow [0,+\infty )$ is continuous. Under weak condition on the coefficient function A, the author established the existence of at least one positive solution by using a nonstandard order cone and fixed point index theory. In fact, (1.3) is a special case of the nonlocal elliptic PDE

$$\begin{aligned} -A\left( \int _\Omega |u|^q d\textbf{s}\right) \Delta u(\textbf{x})=\lambda g(u(\textbf{x})),\quad \textbf{x}\in \Omega , \end{aligned}$$

subject to $u(\textbf{x})\equiv 0$, for $\textbf{x}\in \partial \Omega$, where $\Omega$ is an annular region. There are also many variants on (1.3). For example, the case u in coefficient function A of (1.2) is replaced by $u'$ is a one-dimensional Kirchhoff-type problem (see [4, 13, 29]).

In [18], Goodrich discussed the following nonlocal differential equation:

$$\begin{aligned} -A\left( \int ^1_0 (g\circ u)(s)ds\right) u''(t)=\lambda f(t,u(t)),\quad t\in (0,1), \end{aligned}$$

(1.4)

where $A: [0,+\infty ) \rightarrow \mathbb {R}$ is continuous, $g:[0,+\infty )\rightarrow [0,+\infty )$ is continuous concave strictly increasing, and $f:[0,1]\times [0,+\infty )\rightarrow [0,+\infty )$ is continuous. In contrast to [14], g is concave in the coefficient function. A model case of (1.4) is the nonlocal differential equation

$$\begin{aligned} -A\left( \int ^1_0 |u(s)|^pds\right) u''(t)=\lambda f(t,u(t)),\quad t\in (0,1),\quad 0<p<1. \end{aligned}$$

In [20], Goodrich considered the existence result of positive solution for the following perturbed Hammerstein integral equation:

$$\begin{aligned} u(t)= & \gamma (t)H(\varphi (u))+\lambda \int ^1_0\left( A\left( \int ^1_0 |u(\xi )|^q d\xi \right) \right) ^{-1}\nonumber \\ & G(t,s)f(s,u(s))ds, \quad t\in (0,1), \end{aligned}$$

(1.5)

where $q\ge 1$, $A:[0,+\infty )\rightarrow \mathbb {R},\ G:[0,1]\times [0,1]\rightarrow [0,+\infty ),\ f:[0,1]\times [0,+\infty )\rightarrow [0,+\infty ),\ H:[0,+\infty )\rightarrow [0,+\infty )$, and $\gamma :[0,1]\rightarrow [0,1]$ are continuous, and $\varphi (u)=\int ^1_0 u(s)d\alpha (s)$, where $\alpha$ is of bounded variation and monotone increasing on [0, 1]. Solutions of (1.5) can be associated to solutions of a boundary value problem, which possess two nonlocal elements. The nonlocality is embodied in differential equation itself and boundary condition.

More recently, Goodrich [17] studied the following nonlocal convolution-type differential equations

$$\begin{aligned} -A((a*(g\circ u))(1))u''(t)=\lambda f(t,u(t)),\quad t\in (0,1), \end{aligned}$$

(1.6)

where g is a continuous function, and there exist constants $c_1, c_2\in (0,\infty )$ and $c_3\in [0, \infty )$ such that

$$\begin{aligned} c_1u^p\le g(u)\le c_2(c_3+u^q),\quad u\ge 0,\ 1\le p\le q<+\infty . \end{aligned}$$

The author established the existence of positive solution by using fixed point index theory. If we choose $a(x)\equiv 1,\ g(u)=u^p+u^q$ and $\lambda =1$, then convolution-type Eq. (1.6) reduces to the nonlocal differential Eq. (1.1). We note that Goodrich only consider the case $1\le p< q$.

Greatly inspired by above works, in this paper, we study the positive solutions for nonlocal differential Eq. (1.1) with concave and convex coefficients. Firstly, we establish the existence and multiplicity results of Eq. (1.2) when the nonlinear term f is continuous. Secondly, we obtain the existence of positive solutions of Eq. (1.2) when f(t, x) is singular at $x=0$. Finally, we discuss the existence result when f changes sign.

2 Multiple positive solutions for Eq. (1.2)

In this section, by using the fixed point index theory in cones, we give the existence and multiplicity results of positive solutions for integral Eq. (1.2), where $0< p< 1\le q$, the functions $f:[0,1]\times [0,+\infty )\rightarrow [0,+\infty )$ and $H:[0,+\infty )\rightarrow [0,+\infty )$ are continuous. $\varphi (u)=\int ^1_0 u(s)d\alpha (s)$, where $\alpha :[0,1]\rightarrow \mathbb {R}$ is of bounded variation and monotonically increasing on [0, 1].

For example, when $\gamma (t)=1-t,$ and

$$\begin{aligned} G(t,s)=\left\{ \begin{aligned}&t(1-s),\quad 0\le t\le s\le 1, \\&s(1-t),\quad 0\le s\le t\le 1, \end{aligned}\right. \end{aligned}$$

a solution of the integral Eq. (1.2) is equivalent to a solution of the boundary value problem of differential equation

$$\begin{aligned} \left\{ \begin{aligned}&-A\left( \int ^1_0 (u^p(s)+u^q(s))ds\right) u''(t)=f(t,u(t)), \quad t\in (0,1), \\&u(0)=H(\varphi (u)),\quad u(1)=0. \end{aligned}\right. \end{aligned}$$

Set $E=C([0,1])$. Then E is a Banach space with the norm $\Vert u\Vert =\underset{t\in [0,1]}{\sup }|u(t)|$. We define the cone

$$\begin{aligned} P=\left\{ u\in E: u(t)\ge 0,\ t\in [0,1],\ \min _{t\in [c,d]} u(t)\geqslant \eta _0\Vert u\Vert \right\} , \end{aligned}$$

where $0\le c<d\le 1$, and the constant $\eta _0$ will be given in $(H_1).$ In addition, for $\rho >0$, the sets $\widehat{V}_\rho$ and $\Omega _\rho$ are given by

$$\begin{aligned} \widehat{V}_\rho =\left\{ u\in P: \int ^1_0 (u^p(s)+u^q(s))ds<\rho \right\} \end{aligned}$$

and

$$\begin{aligned} \Omega _\rho =\{u\in P:\Vert u\Vert <\rho \}. \end{aligned}$$

For $H:[0,+\infty )\rightarrow [0,+\infty )$ a continuous function and given numbers $0\le a<b<+\infty$, we denote

$$\begin{aligned} H_{[a,b]}^{m}=\underset{y\in [a,b]}{\min }\ H(y), \quad H_{[a,b]}^{M}=\underset{y\in [a,b]}{\max }\ H(y). \end{aligned}$$

For $f:[0,1]\times [0,+\infty )\rightarrow [0,+\infty )$ a continuous function and given numbers $0\le a_1<b_1\le 1$ and $0\le a_2<b_2<+\infty$, then by $f_{[a_1,b_1]\times [a_2,b_2]}^{m}$ and $f_{[a_1,b_1]\times [a_2,b_2]}^{M}$ we will denote, respectively, the numbers

$$\begin{aligned} & f_{[a_1,b_1]\times [a_2,b_2]}^{m}=\underset{(t,x)\in [a_1,b_1]\times [a_2,b_2]}{\min }\ f(t,x),\\ & f_{[a_1,b_1]\times [a_2,b_2]}^{M}=\underset{(t,x)\in [a_1,b_1]\times [a_2,b_2]}{\max }\ f(t,x). \end{aligned}$$

Denote by $\textbf{1}$ the function $\textbf{1}: \mathbb {R} \rightarrow \{1\}$. Similarly, the notation $\textbf{0}$ will denote the function that is identically zero. Define the operator $T: E\rightarrow E$ by

$$\begin{aligned} \begin{aligned} (Tu)(t)&=\gamma (t)H(\varphi (u)) +\int ^1_0\left( A\left( \int ^1_0 (u^p(r)+u^q(r))dr\right) \right) ^{-1}\\&\quad G(t,s)f(s,u(s))ds,\ t\in [0,1]. \end{aligned} \end{aligned}$$

A fixed point of T is a solution of nonlocal differential Eq. (1.1) equipped with some boundary data.

The following assumptions are used in this section.

$(H_1)$ $G:[0,1]\times [0,1]\rightarrow [0,+\infty )$ is continuous, and there exist a set $[c,d]\subseteq [0,1]$ and a constant $\eta _0:=\eta _0(c,d)\in (0,1]$ such that

$$\begin{aligned} \min _{t\in [c,d]}G(t,s)\geqslant \eta _0\mathcal {G}(s),\quad s\in [0,1], \end{aligned}$$

where $\mathcal {G}(s)=\underset{t\in [0,1]}{\max }\ G(t,s)$;

$(H_2)$ $\gamma :[0,1]\rightarrow [0,1]$ is continuous, and $\underset{t\in [c,d]}{\min }\ \gamma (t)\ge \eta _0\Vert \gamma \Vert$, where c, d and $\eta _0$ are the same as we defined in $(H_1);$

$(H_3)$ $A:[0,+\infty )\rightarrow \mathbb {R}$ is continuous, and there exist $0\le \rho _1\le \rho _2$ such that $A(t)>0$ for $t\in [\rho _1,\rho _2]$.

Denote

$$\begin{aligned} Q_1=\underset{t\in [\rho _1,\rho _2]}{\min }A(t),\quad Q_2=\underset{t\in [\rho _1,\rho _2]}{\max }A(t). \end{aligned}$$

Remark 2.1

From the definitions of $\widehat{V}_\rho$ and $\Omega _\rho$, it is clear that $\widehat{V}_\rho$ and $\Omega _\rho$ are relatively open sets in P.

Lemma 2.2

For each fixed $\rho >0$, it holds that

$$\begin{aligned} \overline{\Omega _{M_{\rho }}}\subseteq \overline{\widehat{V}_\rho }\subseteq \overline{\Omega _{N_{\rho }}}, \end{aligned}$$

where $M_{\rho }\in \left( 0,\rho ^\frac{1}{q}\right)$ is the unique positive solution of $x^p+x^q=\rho$ and $N_{\rho }\in \left( 0,\frac{1}{\eta _0}(\frac{\rho }{d-c})^\frac{1}{q}\right)$ is the unique positive solution of $(\eta _0x)^p+(\eta _0x)^q=\frac{\rho }{d-c}$.

Proof

For any $u\in \overline{\Omega _{M_{\rho }}}$, we have $\Vert u\Vert \le M_{\rho }.$ Thus,

$$\begin{aligned} \int ^1_0 (u^p(t)+u^q(t))dt\le \Vert u\Vert ^p+\Vert u\Vert ^q\le M_{\rho }^{p}+M_{\rho }^{q}=\rho . \end{aligned}$$

Then we obtain $u\in \overline{\widehat{V}_\rho }$, i.e., $\overline{\Omega _{M_{\rho }}}\subseteq \overline{\widehat{V}_\rho }.$

Next, we prove $\overline{\widehat{V}_\rho }\subseteq \overline{\Omega _{N_{\rho }}}$. In fact, for any $u\in \overline{\widehat{V}_\rho }$, we have

$$\begin{aligned} \begin{aligned} \rho&\ge \int ^1_0 (u^p(t)+u^q(t))dt\ge \int ^d_c ((\eta _0\Vert u\Vert )^p+(\eta _0\Vert u\Vert )^q)dt\\&=(d-c)((\eta _0\Vert u\Vert )^p+(\eta _0\Vert u\Vert )^q). \end{aligned} \end{aligned}$$

Then

$$\begin{aligned} \frac{\rho }{d-c}\ge (\eta _0\Vert u\Vert )^p+(\eta _0\Vert u\Vert )^q. \end{aligned}$$

Let

$$\begin{aligned} h(x)=(\eta _0x)^p+(\eta _0x)^q-\frac{\rho }{d-c}. \end{aligned}$$

It is easy to see that h(x) is a strictly monotone increasing function on $[0,+\infty ).$ Thus, we obtain $\Vert u\Vert \le N_{\rho }$. Therefore, $u\in \overline{\Omega _{N_{\rho }}}$, i.e., $\overline{\widehat{V}_\rho }\subseteq \overline{\Omega _{N_{\rho }}}$. This completes the proof. $\square$

By using standard arguments, we obtain the following result.

Lemma 2.3

Assume that $(H_1)$-$(H_3)$ hold. Then $T:\overline{\widehat{V}_{\rho _2}}{\setminus }\widehat{V}_{\rho _1} \rightarrow P$ is completely continuous.

Lemma 2.4

[23] Let U be a bounded open set, and with K a cone in a real Banach space X. Suppose both that $U_K=U\cap K\supseteq \{0\}$ and that $\overline{U_K}\ne K.$ Assume that $T:\overline{U_K} \rightarrow K$ is completely continuous such that $x\ne Tx,$ for any $x\in \partial U_K$. Then the fixed point index $i_K(T, U_K)$ has the following properties.

• (1) If there exists $e \in K\setminus \{0\}$ such that $x\ne Tx+\lambda e$ for each $x\in \partial U_K$ and $\lambda >0,$ then $i_K(T, U_K)=0$.

• (2) If $\mu x\ne Tx$ for $x\in \partial U_K$ and $\mu \ge 1$, then $i_K(T, U_K)=1$.

• (3) Let $U_K^1$ be a open set in X with $U_K^1\subseteq U_K$. If $i_K(T, U_K)=0$ and $i_K(T, U_K^1)=1,$ then T has a fixed point in $U_K\setminus \overline{U_K^1}$. The same result holds if $i_K(T, U_K)=1$ and $i_K(T, U_K^1)=0$.

Theorem 2.5

Assume that $(H_1)-(H_3)$ hold. If

$$\begin{aligned} & (H_4) \int _0^1\left[ \left( {\frac{1}{A(\rho _1)}} {f_{[c,d]\times [{{\eta _0}{M_{\rho _1}}},{N_{\rho _1}}]}^m} {\int _c^d}G(t,s)ds\right) ^p\right. \\ & \quad \left. +\left( {{\frac{1}{A(\rho _1)}} {f_{[c,d]\times [{{\eta _0}{M_{\rho _1}}},{N_{\rho _1}}]}^m} \int _c^dG(t,s)ds}\right) ^q\right] dt > \rho _1;\\ & (H_5)\int _0^1\left[ \left( H_{[0,N_{\rho _2}\varphi (\textbf{1})]}^M + {\frac{1}{A(\rho _2)} f_{[0,1]\times [0,N\rho _2]}^M} \int _0^1G(t,s)ds\right) ^p \right. \\ & \quad \left. +\left( H_{[0,N_{\rho _2}\varphi (\textbf{1})]}^M + {\frac{1}{A(\rho _2)} f_{[0,1]\times [0,N\rho _2]}^M} \int _0^1G(t,s)ds\right) ^q\right] dt < \rho _2, \end{aligned}$$

then Eq. (1.2) has at least one positive solution u with $M_{\rho _1}\le \Vert u\Vert \le N_{\rho _2}$.

Proof

Clearly, $\textbf{0}\in \widehat{V}_{\rho _1}\subseteq \widehat{V}_{\rho _2}$ and $c \textbf{1}\in \widehat{V}_{\rho _1}$ for small constant $c>0$, so $\widehat{V}_{\rho _1}{\setminus } \{\textbf{0}\}\ne \emptyset .$ From Lemma 2.3 and the extension theorem of a completely continuous operator, there exists $\widetilde{T}:\widehat{V}_{\rho _2}\rightarrow P$, which is still completely continuous. Without loss of the generality, we still write it as T.

Now, we claim that $u\ne Tu+\mu \textbf{1}$ for any $\mu >0$ and $u\in \partial \widehat{V}_{\rho _1}$. Suppose that T has no fixed points on $\partial \widehat{V}_{\rho _1}$ (otherwise, the proof is finished). Assume by contradiction that there exist $u\in \partial \widehat{V}_{\rho _1}$ and $\mu >0$ such that $u=Tu+\mu \textbf{1}.$ It follows from Lemma 2.2 that

$$\begin{aligned} M_{\rho _1}\le \Vert u\Vert \le N_{\rho _1}, \quad \forall \ u\in \partial \widehat{V}_{\rho _1}. \end{aligned}$$

Therefore, for $s\in [c,d],$ we have

$$\begin{aligned} \eta _0 M_{\rho _1}\le \eta _0\Vert u\Vert \le u(s)\le \Vert u\Vert \le N_{\rho _1}. \end{aligned}$$

Then for $u\in \partial \widehat{V}_{\rho _1}$,

$$\begin{aligned} (Tu)(t)\ge \frac{1}{A(\rho _1)}f_{[c,d]\times [\eta _0 M_{\rho _1}, N_{\rho _1}]}^{m}\int ^d_cG(t,s)ds,\quad t\in [0,1]. \end{aligned}$$

It follows that

$$\begin{aligned} \rho _1&= \int ^1_0 (u^p+u^q)(t)dt =\int ^1_0 ((Tu+\mu \textbf{1})^p+(Tu+\mu \textbf{1})^q)(t)dt \\&\ge \int ^1_0 ((Tu)^p+(Tu)^q)(t)dt\ge \int ^1_0\left[ \left( \frac{1}{A(\rho _1)}f_{[c,d]\times [\eta _0 M_{\rho _1}, N_{\rho _1}]}^{m}\int ^d_cG(t,s)ds\right) ^p\right. \\&\quad \left. +\left( \frac{1}{A(\rho _1)}f_{[c,d]\times [\eta _0 M_{\rho _1}, N_{\rho _1}]}^{m}\int ^d_cG(t,s)ds\right) ^q \right] dt, \end{aligned}$$

which is a contradiction to $(H_4)$. So we get

$$\begin{aligned} i_K(T, \widehat{V}_{\rho _1})=0. \end{aligned}$$

We next show that $\mu u\ne Tu$ for any $u\in \partial \widehat{V}_{\rho _2}$ and $\mu \ge 1$. If otherwise, there exist $u\in \partial \widehat{V}_{\rho _2}$ and $\mu \ge 1$ such that $\mu u=Tu$. For $u\in \partial \widehat{V}_{\rho _2}$, it is easy to check that $M_{\rho _2}\le \Vert u\Vert \le N_{\rho _2}$ and $0\le u(t)\le \Vert u\Vert \le N_{\rho _2}$ for $t\in [0,1]$, and $0\le \varphi (u)\le \Vert u\Vert \varphi (\textbf{1})\le N_{\rho _2}\varphi (\textbf{1})$. It follows that

$$\begin{aligned} (Tu)(t)\le H_{[0,N_{\rho _2}\varphi (\mathbf {1)}]}^{M}+\frac{1}{A(\rho _2)}f_{[0,1]\times [0, N_{\rho _2}]}^{M}\int ^1_0G(t,s)ds,\quad t\in [0,1]. \end{aligned}$$

Then we deduce that

$$\begin{aligned} \rho _2&= \int ^1_0 (u^p+u^q)(t)dt \le \int ^1_0((\mu u)^p+(\mu u)^q)(t)dt \\&=\int ^1_0 ((Tu)^p+(Tu)^q)(t)dt\\&\le \int ^1_0\left[ \left( H_{[0,N_{\rho _2}\varphi (\textbf{1})]}^{M}+\frac{1}{A(\rho _2)}f_{[0,1]\times [0, N_{\rho _2}]}^{M}\int ^1_0G(t,s)ds\right) ^p\right. \\&\quad \left. +\left( H_{[0,N_{\rho _2}\varphi (\mathbf {1)}]}^{M}+\frac{1}{A(\rho _2)}f_{[0,1]\times [0, N_{\rho _2}]}^{M}\int ^1_0G(t,s)ds\right) ^q \right] dt, \end{aligned}$$

which contradicts $(H_5)$. So we obtain

$$\begin{aligned} i_K(T, \widehat{V}_{\rho _2})=1. \end{aligned}$$

It follows from Lemma 2.4 that T has a fixed point $u\in \widehat{V}_{\rho _2}{\setminus }\overline{\widehat{V}_{\rho _1}}$ with $M_{\rho _1}\le \Vert u\Vert \le N_{\rho _2}$, and u is a positive solution of the Eq. (1.2). $\square$

Corollary 2.6

If we reverse $\rho _1$ and $\rho _2$ in Theorem 2.5, then $i_K(T, \widehat{V}_{\rho _1})=1$, $i_K(T, \widehat{V}_{\rho _2})=0$. We can get the same result as Theorem 2.5.

Next, we prove the following multiplicity results.

Theorem 2.7

Assume that $(H_1){-}(H_4)$ hold. If

$$\begin{aligned} \begin{aligned} (H_6)\ &N_{\rho _1}< M_{\rho _2};\\ (H_7) \ &\int _0^1\left[ \left( {\frac{1}{A(\rho _2)}} {f_{[c,d]\times [{{\eta _0}{M_{\rho _2}}},{N_{\rho _2}}]}^m} \int _c^d {G(t,s)ds}\right) ^p\right. \\&\left. + \left( {{\frac{1}{A(\rho _2)}} {f_{[c,d]\times [{{\eta _0}{M_{\rho _2}}},{N_{\rho _2}}]}^m} \int _c^dG(t,s)ds}\right) ^q\right] dt > \rho _2; \\ (H_8)\ &H_{[0,N_{\rho _1}\varphi (\textbf{1})]}^M + \frac{1}{Q_1} f_{[0,1]\times [0,N_{\rho _1}]}^M {\int _0^1\mathcal {G}(s)ds} < N_{\rho _1}, \end{aligned} \end{aligned}$$

then Eq. (1.2) has at least two positive solutions.

Proof

Firstly, we prove that T has a fixed point which is either on $\partial \widehat{V}_{\rho _1}$ or in $\Omega _{N_{\rho _1}}\setminus \overline{\widehat{V}_{\rho _1}}$. If $x\ne Tx$ for $x\in \partial \widehat{V}_{\rho _1}$, by Theorem 2.5, we obtain

$$\begin{aligned} i_K(T, \widehat{V}_{\rho _1})=0. \end{aligned}$$

We next prove that $\Vert Tu\Vert <\Vert u\Vert$ for any $u\in \partial \Omega _{N_{\rho _1}}$. From Lemma 2.2, it follows that

$$\begin{aligned} \partial \Omega _{N_{\rho _1}}\subseteq \overline{\Omega _{N_{\rho _1}}}\subseteq \overline{\Omega _{ M_{\rho _2}}}\subseteq \overline{\widehat{V}_{\rho _2}}. \end{aligned}$$

Then

$$\begin{aligned} \int ^1_0 (u^p(r)+u^q(r))dr\le \rho _2. \end{aligned}$$

Owing to

$$\begin{aligned} \int ^1_0 (u^p(r)+u^q(r))dr\ge \int ^d_c (\eta _0\Vert u\Vert )^p+(\eta _0\Vert u\Vert )^qdr=\rho _1, \end{aligned}$$

we have

$$\begin{aligned} Q_1\le A\left( \int ^1_0 (u^p(r)+u^q(r))dr\right) \le Q_2. \end{aligned}$$

Since $0\le \varphi (u)\le \Vert u\Vert \varphi (\textbf{1})=N_{\rho _1}\varphi (\textbf{1})$, we have

$$\begin{aligned} \begin{aligned} \Vert Tu\Vert \le H_{[0,N_{\rho _1}\varphi (\mathbf {1)}]}^{M}+\frac{1}{Q_1}f_{[0,1]\times [0, N_{\rho _1}]}^{M}\int ^1_0\mathcal {G}(s)ds<N_{\rho _1}=\Vert u\Vert . \end{aligned} \end{aligned}$$

Therefore, it is obvious that $Tu\ne u$ for $u\in \partial \Omega _{N_{\rho _1}}$, and $\mu u\ne Tu$ for $u\in \partial \Omega _{N_{\rho _1}}$ and $\mu \ge 1$. By Lemma 2.4,

$$\begin{aligned} i_K(T, \Omega _{N_{\rho _1}})=1. \end{aligned}$$

Since $\widehat{V}_{\rho _1}\subseteq \Omega _{N_{\rho _1}}$, by Lemma 2.4, T has a fixed point in $\Omega _{N_{\rho _1}}{\setminus }\overline{\widehat{V}_{\rho _1}}$. So T has a fixed point which is either on $\partial \widehat{V}_{\rho _1}$ or in $\Omega _{N_{\rho _1}}{\setminus }\overline{\widehat{V}_{\rho _1}}$.

On the other hand, we prove T has a fixed point which is either on $\partial \widehat{V}_{\rho _2}$ or in ${\widehat{V}}_{\rho _2}{\setminus } \overline{\Omega _{N_{\rho _1}}}$. If $x\ne Tx,\ x\in \partial \widehat{V}_{\rho _2}$, by Theorem 2.5, we conclude that

$$\begin{aligned} i_K(T, \widehat{V}_{\rho _2})=0. \end{aligned}$$

It follows from Lemma 2.4 that T has a fixed point in ${\widehat{V}}_{\rho _2}{\setminus } \overline{\Omega _{N_{\rho _1}}}$. So T has a fixed point which is either on $\partial \widehat{V}_{\rho _2}$ or in ${\widehat{V}}_{\rho _2}{\setminus } \overline{\Omega _{N_{\rho _1}}}$.

Therefore, T has at least two fixed points $u_1$ and $u_2$ with $0< M_{\rho _1}\le \Vert u_1\Vert \le N_{\rho _1}< M_{\rho _2}<\Vert u_2\Vert \le N_{\rho _2}$, i.e., Eq. (1.2) has at least two positive solutions.$\square$

Theorem 2.8

Assume that $(H_1)$-$(H_3), (H_5)$ and $(H_6)$ hold. If

$$\begin{aligned} \begin{aligned} (H_9)\ &\int _0^1\left[ \left( H_{[0,N_{\rho _1}\varphi (\textbf{1})]}^M + {\frac{1}{A(\rho _1)} f_{[0,1]\times [0,N_{\rho _1}]}^M}{\int _0^1 {G(t,s)ds}}\right) ^p\right. \\&+ \left. \left( H_{[0,N_{\rho _1}\varphi (\textbf{1})]}^M + {\frac{1}{A(\rho _1)}f_{[0,1]\times [0,N_{\rho _1}]}^M} \int _0^1G(t,s)ds\right) ^q\right] dt < \rho _1;\\ (H_{10})\ &\frac{ \eta _0}{Q_2} f_{[c,d]\times [{{\eta _0}{N_{\rho _1}}},{N_{\rho _1}}]}^m \int _c^d\mathcal {G}(s)ds > N_{\rho _1}, \end{aligned} \end{aligned}$$

then Eq. (1.2) has at least two positive solutions.

Proof

Similar to the proof of Theorem 2.5, we obtain that $i_K(T, \widehat{V}_{\rho _1})=1$ and $i_K(T, \widehat{V}_{\rho _2})=1.$ So it suffices to show that $i_K(T, \Omega _{N_{\rho _1}})=0$. Assume that there exist $u\in \partial \Omega _{N_{\rho _1}}$ and $\mu >0$ such that $u=Tu+\mu \textbf{1}.$ From Theorem 2.7, we have $Q_1\le A\left( \int ^1_0 (u^p(r)+u^q(r))dr\right) \le Q_2$. Then we have

$$\begin{aligned} \begin{aligned} N_{\rho _1}&=\Vert u\Vert =\Vert Tu+\mu \textbf{1}\Vert \ge (Tu)(c)\ge \frac{\eta _0}{Q_2} f_{[c,d]\times [{{\eta _0}{N_{\rho _1}}},{N_{\rho _1}}]}^m \int _c^d\mathcal {G}(s)ds, \end{aligned} \end{aligned}$$

which contradicts the assumption. Then for all $u\in \partial \Omega _{N_{\rho _1}}$ and $\mu >0$, we have $u\ne Tu+\mu \textbf{1}$. It follows from Lemma 2.4 that $i_K(T, \Omega _{N_{\rho _1}})=0.$ Then Eq. (1.2) has at least two positive solutions.$\square$

By arguments similar to Theorems 2.5, 2.7 and 2.8, we have the following results.

Theorem 2.9

Assume that $(H_1)$–$(H_3), (H_6), (H_7), (H_9)$ and $(H_{10})$ hold. If

$$\begin{aligned} (H_{11})\ H_{[0,M_{\rho _2}\varphi (\textbf{1})]}^M + {\frac{1}{Q_1}} f_{[0,1]\times [0,M_{\rho _2}]}^M {\int _0^1\mathcal {G}(s)ds} < M_{\rho _2}, \end{aligned}$$

then Eq. (1.2) has at least three positive solutions.

Theorem 2.10

Assume that $(H_1)$-$(H_6)$ and $(H_8)$ hold. If

$$\begin{aligned} (H_{12})\ \frac{\eta _0}{Q_2} f_{[c,d]\times [{{\eta _0}{M_{\rho _2}}},{M_{\rho _2}}]}^m \int _c^d\mathcal {G}(s)ds> M_{\rho _2}, \end{aligned}$$

then Eq. (1.2) has at least three positive solutions.

Example 2.11

Let $p=\frac{1}{2},\ q=2,\ A(t)=\sin t,\ \varphi (u)=\frac{1}{2}u\left( \frac{1}{3}\right) +\frac{1}{50}u\left( \frac{1}{10}\right) ,\ \gamma (t)=1-t,\ H(t)=\frac{9}{100}\sqrt{t},\ f(t,x)=tx$. We consider the following nonlocal problem

$$\begin{aligned} \left\{ \begin{array}{ll} -\sin \left( \int ^1_0 (u^{\frac{1}{2}}(s)+u^2(s))ds\right) u''(t)=t u(t),& \quad t\in (0,1), \\ u(0)=\frac{9}{100}\sqrt{\frac{1}{2}u\left( \frac{1}{3}\right) +\frac{1}{50}u\left( \frac{1}{10}\right) },& \quad u(1)=0. \end{array}\right. \end{aligned}$$

(2.1)

Then

$$\begin{aligned} G(t,s)=\left\{ \begin{aligned}&t(1-s),\quad 0\le t\le s\le 1, \\&s(1-t),\quad 0\le s\le t\le 1, \end{aligned}\right. \end{aligned}$$

and $\mathcal {G}(s)=\underset{t\in [0,1]}{\max }\ G(t,s)=s(1-s)$. Choose $c=\frac{1}{4},\ d=\frac{3}{4}.$ Then $\eta _0=\min {\{c,1-d\}}=\frac{1}{4}.$ Obviously, $(H_1)$ and $(H_2)$ hold. Take $\rho _1=0.002,\ \rho _2=\frac{\pi }{2}$. Then $A(t)=\sin t>0$ on $[0.002,\frac{\pi }{2}],$ which implies $(H_3)$ holds. It follows from Lemma 2.2 that $M_{\rho _1}\thickapprox 4\times 10^{-6}, \ N_{\rho _1}\thickapprox 6.4\times 10^{-5}$, $M_{\rho _2}\thickapprox 0.817, \ N_{\rho _2}\thickapprox 5.598.$ Direct computation demonstrates that

$$\begin{aligned} \begin{aligned}&\int _0^1\left[ \left( {\frac{1}{A(\rho _1)}} {f_{[c,d]\times [{{\eta _0}{M_{\rho _1}}},{N_{\rho _1}}]}^m} {\int _c^d}G(t,s)ds\right) ^p\right. \\&\quad \left. +\left( {{\frac{1}{A(\rho _1)}} {f_{[c,d]\times [{{\eta _0}{M_{\rho _1}}},{N_{\rho _1}}]}^m} \int _c^dG(t,s)ds}\right) ^q\right] dt \\&\approx \int _0^1\left[ \left( {\frac{1}{\sin (0.002)}}\times \frac{1}{4}\times \frac{1}{4}\times 4\times 10^{-6}\times \int _{\frac{1}{4}}^{\frac{3}{4}} G(t,s)ds\right) ^{\frac{1}{2}}\right. \\&\quad \left. +\left( {{\frac{1}{\sin (0.002)}} \times \frac{1}{4}\times \frac{1}{4}\times 4\times 10^{-6} \times \int _{\frac{1}{4}}^{\frac{3}{4}} G(t,s)ds}\right) ^2\right] dt \\&\approx 0.0023>\rho _1, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&\int _0^1\left[ \left( H_{[0,N_{\rho _2}\varphi (\textbf{1})]}^M + {\frac{1}{A(\rho _2)} f_{[0,1]\times [0,N\rho _2]}^M} \int _0^1G(t,s)ds\right) ^p \right. \\&\quad \left. +\left( H_{[0,N_{\rho _2}\varphi (\textbf{1})]}^M + {\frac{1}{A(\rho _2)} f_{[0,1]\times [0,N\rho _2]}^M} \int _0^1G(t,s)ds\right) ^q\right] dt\\&\approx \int _0^1\left[ \left( {\frac{9}{100}}\times {\sqrt{{5.598\times {\frac{13}{25}}}}}+ 1\times 5.598\times \int _0^1G(t,s)ds\right) ^{\frac{1}{2}} \right. \\&\quad \left. +\left( {\frac{9}{100}}\times {\sqrt{{5.598\times {\frac{13}{25}}}}}+ 1\times 5.598\times \int _0^1G(t,s)ds\right) ^2\right] dt\\&\approx 1.202<\rho _2. \end{aligned} \end{aligned}$$

So assumptions $(H_4)$ and $(H_5)$ hold. By Theorem 2.5 we conclude that problem (2.1) has at least one positive solution u with $4\times 10^{-6}\le \Vert u\Vert \le 5.598$.

Remark 2.12

Note in Example 2.11 that the nonlocal coefficient function $z\mapsto \sin z$ is sign changing on $\mathbb {R}$. This is in considerable contrast to most of the the existing literature. Our main tool is topological fixed point theory in a nonstandard order cone due to Goodrich (for example, [14,17,18,20]).

Example 2.13

Let $p,\ q,\ A, \ \varphi ,\ \gamma ,\ c$ and d are same as those in Example 2.11. Take $\rho _1=0.01$ and $\rho _2=\frac{\pi }{2}$. Then $(H_1)-(H_3)$ hold. From Lemma 2.2, we know that $M_{\rho _1}\approx 0.0001,\ N_{\rho _1}\approx 0.0016, M_{\rho _2}\approx 0.817$ and $N_{\rho _2}\approx 5.598$. Obviously, $(H_6)$ holds. Let $f:[0,1]\times [0,+\infty )\rightarrow [0,+\infty )$ be defined by

$$\begin{aligned} f(t,x)=10^{-8}t+\left\{ \begin{array}{ll} \frac{1}{1250}x+\frac{3}{50000},& \quad 0\le x <0.1,\\ 140x-\frac{699993}{50000},& \quad x\ge 0.1. \end{array}\right. \end{aligned}$$

Choose $H(t)=\frac{9}{1000}\sqrt{t}$. By calculation, we obtain

$$\begin{aligned}&\int _0^1\left[ \left( {\frac{1}{A(\rho _1)}} {f_{[c,d]\times [{{\eta _0}{M_{\rho _1}}},{N_{\rho _1}}]}^m} {\int _c^d}G(t,s)ds\right) ^p\right. \\&\quad \left. +\left( {{\frac{1}{A(\rho _1)}} {f_{[c,d]\times [{{\eta _0}{M_{\rho _1}}},{N_{\rho _1}}]}^m} \int _c^dG(t,s)ds}\right) ^q\right] dt \\&\approx \int _0^1\left[ \left( {\frac{1}{\sin (0.01)}}\times \left( \frac{10^{-8}}{4}+\frac{0.0001}{5000}+\frac{3}{50000}\right) \times \int _{\frac{1}{4}}^{\frac{3}{4}} G(t,s)ds\right) ^{\frac{1}{2}}\right. \\&\quad \left. +\left( {{\frac{1}{\sin (0.01)}} \times \left( \frac{10^{-8}}{4}+\frac{0.0001}{5000}+\frac{3}{50000}\right) \times \int _{\frac{1}{4}}^{\frac{3}{4}} G(t,s)ds}\right) ^2\right] dt \\&\approx 0.016> \rho _1,\\&\int _0^1\left[ \left( {\frac{1}{A(\rho _2)}} {f_{[c,d]\times [{{\eta _0}{M_{\rho _2}}},{N_{\rho _2}}]}^m} \int _c^d {G(t,s)ds}\right) ^p\right. \\&\quad \left. + \left( {{\frac{1}{A(\rho _2)}} {f_{[c,d]\times [{{\eta _0}{M_{\rho _2}}},{N_{\rho _2}}]}^m} \int _c^dG(t,s)ds}\right) ^q\right] dt\\&\approx \int _0^1\left[ \left( {\frac{1}{\sin \frac{\pi }{2}}} \times \left( \frac{10^{-8}}{4}+140\times 0.204 -\frac{699993}{50000}\right) \times \int _{\frac{1}{4}}^{\frac{3}{4}} {G(t,s)ds}\right) ^{\frac{1}{2}}\right. \\&\quad \left. + \left( {{\frac{1}{\sin \frac{\pi }{2}}} \times \left( \frac{10^{-8}}{4}+140\times 0.204 -\frac{699993}{50000}\right) \times \int _{\frac{1}{4}}^{\frac{3}{4}}G(t,s)ds}\right) ^2\right] dt \\&\approx 1.642> \rho _2,\\&H_{[0,N_{\rho _1}\varphi (\textbf{1})]}^M + \frac{1}{Q_1} f_{[0,1]\times [0,N_{\rho _1}]}^M {\int _0^1\mathcal {G}(s)ds}\\&\approx \frac{9}{1000}\sqrt{0.0016\times \frac{13}{25}} + \frac{1}{\sin (0.01)} \times \left( 10^{-8}+\frac{1}{1250}\times 0.0016+\frac{3}{50000}\right) \\&\quad \times {\int _0^1s(1-s)ds} \approx 0.0013<0.0016\approx N_{\rho _1}. \end{aligned}$$

Therefore, $(H_4), (H_7)$ and $(H_8)$ hold. By Theorem 2.7 we conclude that problem (2.1) has at least two positive solutions.

3 The case f(t, x) is singular at $x=0$

In this section, by applying the Guo–Krasnoselskii fixed point theorem on cones, we obtain the existence of positive solutions for Eq. (1.1) with nonlocal boundary conditions when the nonlinear term f(t, x) has singularity at $x=0$. It is worth mentioning that the restrictions on functions A and H are different from those in Sect. 2.

We continue to study the integral Eq. (1.2), where $0< p< 1\le q$, $A:[0,+\infty )\rightarrow (0,+\infty )$ is continuous and monotone increasing, $H:[0,+\infty )\rightarrow [0,+\infty )$ is continuous and bounded (we will assume that there exists $0< \overline{H}<+\infty$ such that $0\le H(x)\le \overline{H}$ for $x\in [0,+\infty )).\ \varphi (u)$ is the same as Sect. 2. $f:[0,1]\times (0,+\infty )\rightarrow [0,+\infty )$ is continuous and singular at $x=0$.

We define a cone

$$\begin{aligned} P'=\left\{ u\in E: u(t)\geqslant c(t)\Vert u\Vert , \ t\in [0,1]\right\} , \end{aligned}$$

and a set

$$\begin{aligned} \Omega _\rho =\{u\in P':\Vert u\Vert <\rho \}, \end{aligned}$$

where c(t) will be given in $(H_1)'.$

$(H_1)'\ G:[0,1]\times [0,1]\rightarrow [0,+\infty )$ is continuous and there exists a continuous function $c:[0,1]\rightarrow [0,1]$ with $0<c(t)<1$ for $t\in (0,1)$ such that

$$\begin{aligned} G(t,s)\geqslant c(t)\mathcal {G}(s), \quad t, s\in [0,1], \end{aligned}$$

where $\mathcal {G}(s)=\underset{t\in [0,1]}{\max }\ G(t,s),\ s\in [0,1]$;

$(H_2)'\ \gamma :[0,1]\rightarrow [0,1]$ is continuous and $\gamma (t)\ge c(t)\Vert \gamma \Vert$ for $t\in [0,1]$, where c(t) is the same as we defined in $(H_1)';$

$(H_3)'$ For any $0<r<R<+\infty$,

$$\begin{aligned} \lim _{m \rightarrow \infty } \sup \limits _{u \in \overline{\Omega _{R}} \backslash \Omega _{r}} \int _{e(m)} \mathcal {G}(s)f(s,u(s))ds=0, \end{aligned}$$

where $e(m)=[0,\frac{1}{m}]\cup [\frac{m-1}{m}, 1]$.

For a given $\theta \in (0,\frac{1}{2})$, denote $\eta =\min \{c(t):\theta \le t\le 1-\theta \}$.

Lemma 3.1

Assume that $(H_1)'$-$(H_3)'$ hold. Then $T:\overline{\Omega _{R}}\backslash \Omega _{r}\rightarrow P'$ is completely continuous.

Proof

For any $u\in \overline{\Omega _{R}}\backslash \Omega _{r}$, we have $r\le \Vert u\Vert \le R$, and $0\le \varphi (u)\le \Vert u\Vert \varphi (\textbf{1})\le R\varphi (\textbf{1})<+\infty$. For any $u\in \overline{\Omega _{R}}\backslash \Omega _{r},$ we have

$$\begin{aligned} \begin{aligned}0<M_1&\triangleq (1-2\theta )((\eta r)^p+(\eta r)^q)\\&\le \int ^{1-\theta }_\theta (\eta \Vert u\Vert )^p+(\eta \Vert u\Vert )^qdr\\&\le \int ^1_0(u^p(r)+u^q(r))dr\\&\le \Vert u\Vert ^p+\Vert u\Vert ^q\le R^p+R^q\triangleq M_2. \end{aligned} \end{aligned}$$

It follows from $(H_3)'$ that there exists a natural number $l>0$ such that

$$\begin{aligned} \underset{u\in \overline{\Omega _{R}}\backslash \Omega _{r}}{\sup }\ \int _{e(l)} \mathcal {G}(s)f(s,u(s))ds<1. \end{aligned}$$

For $s\in [\frac{1}{l},\frac{l-1}{l}]$, we have $\eta _1 r\le \eta _1\Vert u\Vert \le u(s)\le \Vert u\Vert \le R,$ where $\eta _1=\min \{c(t):\frac{1}{l}\le t\le \frac{l-1}{l}\}.$ Then for any $t\in [0,1]$, we have

$$\begin{aligned} Tu(t)\le & \overline{H}+(A(M_1))^{-1}\left[ \int _{e(l)} \mathcal {G}(s)f(s,u(s))ds+f_{[\frac{1}{l},\frac{l-1}{l}]\times [\eta _1 r,R]}^{M}\int ^{\frac{l-1}{l}}_\frac{1}{l}\mathcal {G}(s)ds\right] \\\le & \overline{H}+(A(M_1))^{-1}\left( 1+f_{[\frac{1}{l},\frac{l-1}{l}]\times [\eta _1 r,R]}^{M}\int ^1_0\mathcal {G}(s)ds\right) <+\infty . \end{aligned}$$

The proof of $(Tu)(t)\geqslant c(t)\Vert Tu\Vert$ is similar to the proof of Lemma 2.3, so we omit it. Thus, $T(\overline{\Omega _{R}}\backslash \Omega _{r})\subseteq P'$.

Suppose that $u_n, u_0\in \overline{\Omega _{R}}\backslash \Omega _{r}$ and $\Vert u_n-u_0\Vert \rightarrow 0\ (n\rightarrow \infty )$. Then

$$\begin{aligned} 0\le \varphi (u_n)\le R\varphi (\textbf{1})<+\infty ,\quad 0\le \varphi (u_0)\le R\varphi (\textbf{1})<+\infty , \end{aligned}$$

and

$$\begin{aligned} & 0<M_1\le \int ^1_0((u_n(r))^p+(u_n(r))^q)dr\le M_2,\\ & 0<M_1\le \int ^1_0((u_0(r))^p+(u_0(r))^q)dr\le M_2. \end{aligned}$$

By $(H_3)'$, $\forall \ \epsilon >0,$ there exists a natural number $m_0>0$ such that

$$\begin{aligned} \underset{u\in \overline{\Omega _{R}}\backslash \Omega _{r}}{\sup }\ \int _{e(m_0)}\mathcal {G}(s)f(s,u(s))ds<\frac{\epsilon A(M_1)}{6}. \end{aligned}$$

For $s\in [\frac{1}{m_0},\frac{m_0-1}{m_0}]$, we have

$$\begin{aligned} \eta _2 r\le u_n(s)\le R,\quad \eta _2 r\le u_0(s)\le R, \end{aligned}$$

where $\eta _2=\min \{c(s):\frac{1}{m_0}\le s\le \frac{m_0-1}{m_0}\}.$ Since f is uniformly continuous on $[\frac{1}{m_0},\frac{m_0-1}{m_0}]\times [\eta _2 r,R],$ then

$$\begin{aligned} \lim \limits _{n \rightarrow \infty }| f(s,u_n(s))-f(s,u_0(s))|=0. \end{aligned}$$

Applying Lebesgue dominated convergence theorem, we have

$$\begin{aligned} (A(M_1))^{-1}\int ^{\frac{m_0-1}{m_0}}_\frac{1}{m_0}\mathcal {G}(s)|f(s,u_n(s))-f(s,u_0(s))|ds\rightarrow 0 \ (n\rightarrow +\infty ). \end{aligned}$$

For the above $\epsilon >0$, there exists a natural number $N_1$ such that if $n>N_1,$ then

$$\begin{aligned} (A(M_1))^{-1}\int ^{\frac{m_0-1}{m_0}}_\frac{1}{m_0}\mathcal {G}(s)|f(s,u_n(s))-f(s,u_0(s))|ds<\frac{\epsilon }{3}. \end{aligned}$$

Since H and $\varphi$ are continuous, then there exists a natural number $N_2$ such that for $n>N_2$,

$$\begin{aligned} |H(\varphi (u_n))-H(\varphi (u_0))|<\frac{\epsilon }{3}. \end{aligned}$$

Therefore, for $n>N=\max \{N_1,N_2\},$ we have

$$\begin{aligned} \Vert Tu_n-Tu_0\Vert\le & |H(\varphi (u_n))-H(\varphi (u_0))|+2(A(M_1))^{-1} \int _{e(m_0)} \mathcal {G}(s)f(s,u_0(s))ds\\ & +(A(M_1))^{-1}\int ^{\frac{m_0-1}{m_0}}_\frac{1}{m_0}\mathcal {G}(s)|f(s,u_n(s))-f(s,u_0(s))|ds<\epsilon , \end{aligned}$$

which implies T is continuous.

Assume that B is a bounded subset in $\overline{\Omega _{R}}\backslash \Omega _{r}$, from Ascoli–Arzela theorem and the Lebesgue dominated convergence theorem, it is easy to prove that T(B) is uniformly bounded and equicontinuous. Thus, $T:\overline{\Omega _{R}}\backslash \Omega _{r}\rightarrow P'$ is completely continuous.$\square$

The main tool is the Guo–Krasnoselskii fixed point theorem on cones.

Lemma 3.2

[23] Let E be a Banach space and let P be a cone in E. Let $\Omega _1$ and $\Omega _2$ be two bounded open subsets in E such that $0 \in \Omega _1$ and $\overline{\Omega _1} \subset \Omega _2$. Let the operator $A: P \cap (\overline{\Omega _2} \backslash \Omega _1) \rightarrow P$ be completely continuous. If the following conditions are satisfied:

• (i) $\Vert Au\Vert \le \Vert u\Vert$ for any $u \in P \cap \partial \Omega _1$, $\Vert Au\Vert \ge \Vert u\Vert$ for any $u \in P \cap \partial \Omega _2$, or

• (ii) $\Vert Au\Vert \ge \Vert u\Vert$ for any $u \in P \cap \partial \Omega _1$, $\Vert Au\Vert \le \Vert u\Vert$ for any $u \in P \cap \partial \Omega _2$,

then A has at least one fixed point in $P \cap (\overline{\Omega _2} \backslash \Omega _1)$.

Denote

$$\begin{aligned} L=\frac{1}{A(2)}\int ^1_0 \mathcal {G}(s)ds, \quad l=\frac{\eta ^2}{A(2)}\int ^{1-\theta }_\theta \mathcal {G}(s)ds, \end{aligned}$$

where $\theta \in \left( 0,\frac{1}{2}\right)$ is given previously.

Theorem 3.3

Assume that $(H_1)'$-$(H_3)'$ hold. Further assume that the following conditions hold:

$(H_4)'$

$$\begin{aligned} 0<l^{-1}<f^0=\liminf \limits _{x \rightarrow 0}\min \limits _{t \in [0,1]}\frac{f(t,x)}{x}\le \infty ; \end{aligned}$$

$(H_5)'$

$$\begin{aligned} 0\le f^\infty =\limsup \limits _{x \rightarrow +\infty } \max \limits _{t \in [0,1]} \frac{f(t,x)}{x}\le L^{-1}. \end{aligned}$$

Then (1.2) has at least one positive solution.

Proof

From Lemma 3.1 and the extension theorem of a completely continuous operator, for any $R > 0$, there exists $\widetilde{T}: \overline{\Omega }_R \rightarrow P'$, which is still completely continuous. Without loss of the generality, we still write it as T.

By $(H_4)'$, there exist $\epsilon _1>0$ and $0< r \le 1$ such that

$$\begin{aligned} f(t,x)\ge (l^{-1}+\epsilon _1)x, \quad 0<x\le r,\ 0\le t\le 1. \end{aligned}$$

For $u\in \partial \Omega _r$, we have

$$\begin{aligned} \begin{aligned} \int ^1_0(u^p(r)+u^q(r))dr\le \Vert u\Vert ^p+\Vert u\Vert ^q=r^p+r^q\le 2. \end{aligned} \end{aligned}$$

Thus,

$$\begin{aligned} 0<A\left( \int ^1_0 (u^p(r)+u^q(r))dr\right) \le A(2). \end{aligned}$$

It follows that

$$\begin{aligned} (Tu)(t)&\ge \frac{1}{A(2)}\int ^1_0G(t,s)f(s,u(s))ds\\&\ge \frac{1}{A(2)}\int _0^1 c(t)\mathcal {G}(s)(l^{-1}+\epsilon _1)u(s)ds\\&\ge \frac{ c(t)}{A(2)}(l^{-1}+\epsilon _1)\int ^{1-\theta }_\theta \mathcal {G}(s)\eta \Vert u\Vert ds\\&\ge \frac{ \eta ^2}{A(2)}(l^{-1}+\epsilon _1)\Vert u\Vert \int ^{1-\theta }_\theta \mathcal {G}(s)ds> \Vert u\Vert , \quad t\in [\theta ,1-\theta ]. \end{aligned}$$

Therefore, $\Vert Tu\Vert \ge \Vert u\Vert$ for $u\in \partial \Omega _r$.

On the other hand, by $(H_5)'$, there exist $0<\epsilon _2<L^{-1}$ and $R_1>1$ such that

$$\begin{aligned} f(t,x)\le (L^{-1}-\epsilon _2)x,\quad x\ge R_1,\ 0\le t\le 1 \end{aligned}$$

Let

$$\begin{aligned} M_0=\overline{H}+\sup _{u\in \partial \Omega _{R_1}}\int ^1_0\left( A\left( \int ^1_0 (u^p(r)+u^q(r))dr\right) \right) ^{-1}\mathcal {G}(s)f(s,u(s))ds. \end{aligned}$$

From Lemma 3.1, we know $M_0<+\infty .$ We choose $R>\max \{R_1, \frac{1}{\epsilon _2L}M_0\}$. Let

$$\begin{aligned} D(u)=\{t\in [0,1]:u(t)>R_1\}. \end{aligned}$$

For any $u\in \partial \Omega _R$ and $t \in D(u)$, we have $R_1 < u(t)\le R,$ which implies $f(t,u(t))\le (L^{-1}-\epsilon _2)u(t).$ For $u\in \partial \Omega _R$, there exists $t_0\in [0,1]$ such that $\Vert u\Vert =u(t_0)$. Let $u_1(t)=\min \{u(t),R_1\}$. Then $u_1(t)\le R_1$ for $t\in [0,1]$ and $u_1(t_0)=\min \{u(t_0),R_1\}=\min \{\Vert u\Vert ,R_1\}=R_1$, which implies that $u_1\in \partial \Omega _{R_1}.$ Thus, for $t\in [0,1]$,

$$\begin{aligned} \begin{aligned} Tu(t)&\le \overline{H}+\int _{D(u)}\left( A\left( \int ^1_0 (u^p+u^q)(r)dr\right) \right) ^{-1}\mathcal {G}(s)f(s,u(s))ds\\&\quad +\int _{[0,1]\backslash D(u)}\left( A\left( \int ^1_0 ((u_1)^p+(u_1)^q)(r)dr\right) \right) ^{-1}\mathcal {G}(s)f(s,u_1(s))ds\\&\le \overline{H}+\frac{1}{A(2)}(L^{-1}-\epsilon _2)\int _{D(u)}\mathcal {G}(s)\Vert u\Vert ds\\&\quad +\int ^1_0\left( A\left( \int ^1_0 ((u_1)^p+(u_1)^q)(r)dr\right) \right) ^{-1}\mathcal {G}(s)f(s,u_1(s))ds\\&\le (L^{-1}-\epsilon _2)L\Vert u\Vert +M_0<\Vert u\Vert . \end{aligned} \end{aligned}$$

Therefore, $\Vert Tu\Vert \le \Vert u\Vert$ for $u\in \partial \Omega _R$.

By Lemma 3.2, we conclude that T has a fixed point $u\in \overline{\Omega _R} \backslash \Omega _r$, and Eq. (1.2) has at least one positive solution. $\square$

Example 3.4

Let $p=\frac{1}{2},\ q=2,\ A(x)=2+x,\ H(x)=3-e^{-x},\ \varphi (u)=\int ^1_0u(s)d(2s),\ f(t,x)=2-t+|\ln x|$. Consider the following nonlocal problem:

$$\begin{aligned} \left\{ \begin{array}{l} -\left( 2+\int ^1_0 (u^\frac{1}{2}(s)+u^2(s))ds\right) u''(t)=2-t+|\ln u(t)|,\quad t\in (0,1), \\ u(0)=3-e^{-\int ^1_0u(s)d(2s)},\quad u(1)=0. \end{array}\right. \end{aligned}$$

(3.1)

Obviously, f(t, x) is singular at $x=0$. It is easy to see that $\gamma (t)=1-t$, and

$$\begin{aligned} G(t,s)=\left\{ \begin{array}{ll} t(1-s),& 0\le t\le s\le 1, \\ s(1-t),& 0\le s\le t\le 1. \end{array}\right. \end{aligned}$$

Let $\mathcal {G}(s)=s(1-s),\ c(t)=t(1-t)$. Then $(H_1)'$ and $(H_2)'$ are satisfied.

For any $0<r<R<+\infty$ and $u \in \overline{\Omega _{R}} \backslash \Omega _{r}$, we have $0<rc(t)\le u(t)\le R$ for $t\in [0,1]$. Then $|\ln u(t)|\le |\ln R|+|\ln rc(t)|$. Due to $\int _0^1 |\ln c(s)|ds=\int _0^1(|\ln s|+|\ln (1-s)|)ds=2$, the absolute continuity of the integral yields that $\lim \limits _{m \rightarrow \infty } \int _{e(m)} |\ln c(s)|ds=0$. Thus,

$$\begin{aligned} \begin{aligned}&\lim _{m \rightarrow \infty } \sup _{u \in \overline{\Omega _{R}} \backslash \Omega _{r}} \int _{e(m)} \mathcal {G}(s)f(u(s))ds\\&\quad \le \lim _{m \rightarrow \infty } \int _{e(m)} (2-s+|\ln R|+|\ln rc(s)|)ds\\&\quad =(3+2|\ln R|+2|\ln r|)\lim _{m \rightarrow \infty }\frac{1}{m}+\lim _{m \rightarrow \infty } \int _{e(m)} |\ln c(s)|ds=0. \end{aligned} \end{aligned}$$

So assumption $(H_3)'$ is satisfied.

We choose $\theta =\frac{1}{4}$. Then $\eta =\frac{3}{16},\ l=\frac{(\frac{3}{16})^2}{4}\int ^\frac{3}{4}_\frac{1}{4} s(1-s)ds=\frac{11}{98304}, \ L=\frac{1}{4}\int ^1_0 s(1-s)ds=\frac{1}{24}$. Direct computation shows that

$$\begin{aligned} \liminf \limits _{x \rightarrow 0}\min \limits _{ t \in [0,1]}\frac{f(t,x)}{x}=\infty ,\quad \limsup \limits _{x \rightarrow +\infty } \max \limits _{t \in [0,1]} \frac{f(t,x)}{x}=0. \end{aligned}$$

Therefore the assumptions of Theorem 3.3 are satisfied, and nonlocal boundary value problem (3.1) has at least one positive solution.

4 The case f changes sign

In this section, we investigate nonlocal differential Eq. (1.1) subject to a specific nonlocal boundary condition

$$\begin{aligned} \alpha u(0)-\beta u'(0)=0,\quad \delta u'(1)=\varphi (u), \end{aligned}$$

(4.1)

where $\alpha>0,\ \beta>0,\ \delta >0,\ \varphi (u)$ is the same as Sect. 2. Using the fixed point theorem in double cones, we obtain the existence of positive solutions. It is worth mentioning that the nonlinearity is allowed to change sign and tend to negative infinity.

$(H)''$$f:[0,1]\times [0,+\infty )\rightarrow \mathbb {R}$ is continuous, and $f(t,0)\ge 0 \ (\not \equiv 0)$ for $t\in [0,1]$.

Let

$$\begin{aligned} \Delta =1-\int ^1_0\frac{\beta +\alpha t}{\alpha \delta }d\alpha (t), \quad H(t,s)=\frac{\beta }{\alpha }+\left\{ \begin{array}{ll} s,& \quad 0\le s \le t \le 1,\\ t,& \quad 0\le t\le s \le 1. \end{array}\right. \end{aligned}$$

By using standard arguments we obtain the following lemma.

Lemma 4.1

Suppose that $\Delta \ne 0$. For any $g\in C([0,1])$, the nonlocal problem

$$\begin{aligned} \left\{ \begin{array}{l} -A\left( \int ^1_0 (u^p(s)+u^q(s))ds\right) u''(t)=g(t),\quad t\in (0,1), \\ \alpha u(0)-\beta u'(0)=0,\quad \delta u'(1)=\varphi (u). \end{array}\right. \end{aligned}$$

has a solution

$$\begin{aligned} \begin{aligned} u(t)=&\int ^1_0\left( A\left( \int ^1_0(u^p+u^q)(r)dr\right) \right) ^{-1}G(t,s)g(s)ds,\quad t\in [0,1], \end{aligned} \end{aligned}$$

where

$$\begin{aligned} G(t,s)=\frac{\beta +\alpha t}{\alpha \delta \Delta }\int ^1_0H(\tau ,s)d\alpha (\tau )+H(t,s). \end{aligned}$$

In the following we always assume that $\Delta >0$. Obviously, $G:[0,1]\times [0,1]\rightarrow [0,+\infty )$ is continuous.

Define two cones K and $K'$:

$$\begin{aligned} K=\{u\in E:u(t)\ge 0,\ t\in [0,1]\},\quad K'=\{u\in K: u\ \text {is}\ \text {concave} \ \text {on}\ [0,1]\}. \end{aligned}$$

Define $\acute{\alpha }:K'\rightarrow [0,+\infty )$ by

$$\begin{aligned} \acute{\alpha }(x)=\underset{t\in [\theta ,1-\theta ]}{\min }x(t),\quad \theta \in \left( 0,\frac{1}{2}\right) . \end{aligned}$$

For $\rho>0,\ a>0,\ b>0$, let

$$\begin{aligned} & \widehat{V}_\rho =\left\{ u\in K:\int ^1_0 (u^p(s)+u^q(s))ds<\rho \right\} ,\\ & K_r=\{u\in K:\Vert u\Vert<r\},\quad K'_r=\{u\in K':\Vert u\Vert<r\},\\ & K(b)=\{u\in K: \acute{\alpha }(u)<b\},\quad K_a(b)= \{u\in K: a<\Vert u\Vert ,\ \acute{\alpha }(u)<b\}. \end{aligned}$$

From the definition of $\acute{\alpha }$, we immediately obtain the following properties.

Lemma 4.2

$\acute{\alpha }$ is a continuous increasing function satisfying $\acute{\alpha }(x)\le \Vert x\Vert \le M\acute{\alpha }(x)$, where $M\ge 1$ is a constant.

Define operators $S,\ T$ and $\tilde{T}:E\rightarrow E$ as follows:

$$\begin{aligned} & Su(t)=\int ^1_0\left( A\left( \int ^1_0(u^p(r)+u^q(r))dr\right) \right) ^{-1}G(t,s)f(s,u(s))ds,\quad t\in [0,1],\\ & Tu(t)=\left( \int ^1_0\left( A\left( \int ^1_0(u^p(r)+u^q(r))dr\right) \right) ^{-1}G(t,s)f(s,u(s))ds\right) ^+,\quad t\in [0,1],\\ & \tilde{T}u(t)=\int ^1_0\left( A\left( \int ^1_0(u^p(r)+u^q(r))dr\right) \right) ^{-1}G(t,s)f^+(s,u(s))ds,\quad t\in [0,1], \end{aligned}$$

where $f^+(t,x)=\max \{f(t,x),0\}$. We define $\psi :E\rightarrow K$ by $(\psi u)(t)=\max \{u(t),0\}$. Then $T=\psi \circ S$ and u is a positive solution of problem (1.1) and (4.1) if and only if u is a positive fixed point of operator S.

By the arguments similar to Lemma 3.2 in [11], we have the following result.

Lemma 4.3

If $S:\overline{\widehat{V}_{\rho _2}}{\setminus }\widehat{V}_{\rho _1} \rightarrow E$ is completely continuous, then $T=\psi \circ S: \overline{\widehat{V}_{\rho _2}}{\setminus }\widehat{V}_{\rho _1} \rightarrow K$ is completely continuous.

Lemma 4.4

Assume that $(H)''$ and $(H_3)$ hold. Then $T:\overline{\widehat{V}_{\rho _2}}{\setminus }\widehat{V}_{\rho _1} \rightarrow K$ and $\tilde{T}:K'\cap \left( \overline{\widehat{V}_{\rho _2}}{\setminus }\widehat{V}_{\rho _1}\right) \rightarrow K'$ are completely continuous.

Proof

From the continuity of f, it is easy to show that $S:\overline{\widehat{V}_{\rho _2}}{\setminus }\widehat{V}_{\rho _1} \rightarrow E$ is completely continuous. So $T:\overline{\widehat{V}_{\rho _2}}{\setminus }\widehat{V}_{\rho _1} \rightarrow K$ is completely continuous by Lemma 4.3. By standard arguments, $\tilde{T}:K'\cap \left( \overline{\widehat{V}_{\rho _2}}{\setminus }\widehat{V}_{\rho _1}\right) \rightarrow K'$ is completely continuous.$\square$

Lemma 4.5

For any $\rho >0,\ \overline{K'_{M_{\rho }}}\subseteq \overline{\widehat{V}_\rho }\bigcap K'\subseteq \overline{K'_{E_{\rho }}}$, where $M_{\rho }\in (0,\rho ^\frac{1}{q})$ is the unique positive solution of $x^p+x^q=\rho$ and $E_{\rho }\in (0,\frac{1}{\theta }(\frac{\rho }{1-2\theta })^\frac{1}{q})$ is the unique positive solution of $(\theta x)^p+(\theta x)^q-\frac{\rho }{1-2\theta }=0$.

Proof

The proof is similar to Lemma 2.2, so we omit it. $\square$

Lemma 4.6

[11] Let X be a real Banach space with norm $\Vert \cdot \Vert$, and let $K,\ K'\subset X$ be two cones with $K'\subseteq K$. Suppose that $T:K\rightarrow K$ and $\tilde{T}:K'\rightarrow K'$ are two completely continuous operators and $\acute{\alpha }(x):K'\rightarrow [0,+\infty )$ is a continuous increasing functional satisfying $\acute{\alpha }(x)\le \Vert x\Vert \le M\acute{\alpha }(x)$ for all $x\in K'$ and for some constant $M\ge 1$. Suppose that there exist two constants $b>a>0$ such that

• (1) $\Vert \tilde{T}x\Vert <a$ for $x\in \partial K'_a$, and $\acute{\alpha }(\tilde{T}x)>b$ for $x\in \partial K'(b)$;

• (2) $Tx=\tilde{T}x$ for $x\in K'_a(b)\bigcap \{x:\tilde{T}x=x\}$.

Then T has a fixed point y in K satisfying $\Vert y\Vert >a,\ \acute{\alpha }(y)<b$.

Theorem 4.7

Assume that $(H)''$ and $(H_3)$ hold, and $\delta >\varphi (\textbf{1})$. Suppose that there exist constants $d>0, \ \theta \in (0,\frac{1}{2})$ such that $0<\left( 1+\frac{\alpha }{\beta }\right) d<E_{\rho _1}<\theta M_{\rho _2}<M_{\rho _2}$, and

• (1) $f(t,x)<\frac{E_{\rho _1}}{I}$ for $(t,x)\in [0,1]\times [0,E_{\rho _1}]$, where $I=\frac{1}{Q_1}\underset{t\in [0,1]}{\max }\int ^1_0G(t,s))ds;$

• (2) $f(t,x)>\frac{\theta M_{\rho _2}}{J}$ for $(t,x)\in [\theta ,1-\theta ]\times [\theta M_{\rho _2}, M_{\rho _2}]$, where $J=\frac{1}{Q_2}\underset{t\in [\theta ,1-\theta ]}{\min }\int ^{1-\theta }_\theta G(t,s)ds;$

• (3) $f(t,x)\ge 0$ for $(t,x)\in [0,1]\times [d,M_{\rho _2}]$.

Then boundary value problem (1.1) and (4.1) has a positive solution.

Proof

For $u\in \partial K'_{E_{\rho _1}}$ and $t\in [0,1]$, we have $0\le u(t)\le \Vert u\Vert =E_{\rho _1}$. Then

$$\begin{aligned} f^+(t,u(t))\le \frac{E_{\rho _1}}{I}. \end{aligned}$$

Since $\underset{t\in [\theta ,1-\theta ]}{\min }u(t)\ge \theta \Vert u\Vert$ and $\partial K'_{E_{\rho _1}}\subseteq \overline{K'_{E_{\rho _1}}}\subseteq \overline{K'_{M_{\rho _2}}}\subseteq \overline{\widehat{V}_{\rho _2}}\bigcap K'$, we have

$$\begin{aligned} \int ^1_0 (u^p(r)+u^q(r))dr\ge \int ^{1-\theta }_\theta [(\theta \Vert u\Vert )^p+(\theta \Vert u\Vert )^q]dr=\rho _1, \end{aligned}$$

and

$$\begin{aligned} \int ^1_0 (u^p(r)+u^q(r))dr\le \int ^1_0 (\Vert u\Vert ^p+\Vert u\Vert ^q)dr\le (M_{\rho _2})^p+(M_{\rho _2})^q=\rho _2. \end{aligned}$$

Then

$$\begin{aligned} \Vert \tilde{T}u\Vert\le & \frac{1}{Q_1}\underset{t\in [0,1]}{\max }\int ^1_0G(t,s)f^+(s,u(s))ds\\\le & \frac{1}{Q_1}\underset{t\in [0,1]}{\max }\int ^1_0G(t,s)\frac{E_{\rho _1}}{I}ds=E_{\rho _1}, \quad u\in \partial K'_{E_{\rho _1}}. \end{aligned}$$

For $u\in \partial K'(\theta M_{\rho _2})$, we have $E_{\rho _1}<\theta M_{\rho _2}=\acute{\alpha }(u)\le \Vert u\Vert \le \frac{1}{\theta }\acute{\alpha }(u)=M_{\rho _2}$. By Lemma 4.5, we deduce that $\rho _1<\int ^1_0 (u^p(r)+u^q(r))dr\le \rho _2$, and $\theta M_{\rho _2}\le u(t)\le M_{\rho _2}$ for $t\in [\theta ,1-\theta ]$. Thus, we calculate

$$\begin{aligned} \begin{aligned} \acute{\alpha }(\tilde{T}u)&=\underset{t\in [\theta ,1-\theta ]}{\min }(\tilde{T}u)(t)>\underset{t\in [\theta ,1-\theta ]}{\min }\frac{1}{Q_2}\int ^{1-\theta }_\theta G(t,s)\frac{\theta M_{\rho _2}}{J}ds\\&=\theta M_{\rho _2}, \quad u\in \partial K'(\theta M_{\rho _2}). \end{aligned} \end{aligned}$$

On the other hand, for $u\in K'_{E_{\rho _1}}(\theta M_{\rho _2})\bigcap \{u:\tilde{T}u=u\}$, we have $\Vert u\Vert> E_{\rho _1}>\left( 1+\frac{\alpha }{\beta }\right) d,\ \acute{\alpha }(u)<\theta M_{\rho _2},\ \alpha u(0)-\beta u'(0)=0$ and $\delta u'(1)=\varphi (u).$ Since u is concave, u(t) minimizes at $t=0$ or $t=1$. Due to $\varphi (u)=\int ^1_0 u(s)d\alpha (s)\ge 0,$ we have $u'(1)=\frac{1}{\delta }\varphi (u)\ge 0.$ Therefore, $u(0)=\min _{0\le t\le 1}u(t)$ and $u(1)=\Vert u\Vert$.

We claim that $u(0)\ge d.$ Otherwise, we assume $u(0)<d$. Then there exists $t_0\in (0,1)$ such that $u'(t_0)=u(1)-u(0)=\Vert u\Vert -u(0)>\frac{\alpha }{\beta }d.$ Thanks to the fact that u is concave, $u'(0)\ge u'(t_0)>\frac{\alpha }{\beta }d.$ It follows that

$$\begin{aligned} 0=\alpha u(0)-\beta u'(0)<\alpha d-\beta \frac{\alpha }{\beta }d=0, \end{aligned}$$

which is a contradiction. Then for $t\in [0,1]$, we have $d\le u(0)\le u(t)\le \Vert u\Vert \le \frac{1}{\theta }\acute{\alpha }(u)<M_{\rho _2}$. By condition (3) of the theorem, we have $f^{+}(t,u(t))=f(t,u(t))$. Thus, for $u\in K'_{E_{\rho _1}}(\theta M_{\rho _2})\bigcap \{u:\tilde{T}u=u\}$, we obtain that $Tu=\tilde{T}u=u$. It follows from Lemma 4.6 that T has a fixed point $u_0$ in K satisfying $\Vert u_0\Vert>E_{\rho _1}>0$ and $\acute{\alpha }(u_0)<\theta M_{\rho _2}.$ Thus $E_{\rho _1}\le \Vert u_0\Vert \le M_{\rho _2},$ and $\rho _1\le \int ^1_0 (u^p_0(r)+u_0^q(r))dr\le \rho _2.$

We next show that $u_0$ is a positive solution of problem (1.1) and (4.1). It is sufficient to prove that $u_0$ is a fixed point of S. Arguing indirectly, we suppose that there exists $t_1\in (0,1)$ such that $u_0(t_1)\ne (Su_0)(t_1)$. From $u_0(t)=(Tu_0)(t)=\max \{(Su_0)(t),0\}$, it follows that $(Su_0)(t_1)<0$ and $u_0(t_1)=0$. Assume that $(t_2,t_3)$ contains $t_1$, and it is the maximum interval which satisfies $(Su_0)(t)<0$ for any $t\in (t_2,t_3).$ By $(H)''$, we get $[t_2,t_3]\ne [0,1]$. Thus $t_2>0$ or $t_3<1$.

For the case $t_3<1$, we have $(Su_0)(t)<0,\ u_0(t)=0$ for any $t\in (t_2,t_3)$, so $(Su_0)(t_3)=0$, and $(Su_0)'(t_3)\ge 0.$ Noting that

$$\begin{aligned} (Su_1)''(t)=-\left( A\left( \int ^1_0 (u^p_0(r)+u^q_0(r))ds\right) \right) ^{-1} f(t,0)\le 0,\quad t\in (t_2,t_3), \end{aligned}$$

so $(Su_0)'(t)\ge (Su_0)'(t_3)\ge 0$ for $t\in [t_2,t_3].$ It follows that $(Su_0)(t)\le (Su_0)(t_3)=0$ for $t\in [t_2,t_3]$, which implies $t_2=0$, which contradicts $(Su_0)'(0) =\frac{\alpha }{\beta }(Su_0)(0)<0$.

For the case $t_2 > 0,$ we have $u_0(t)=0$ for $t\in [t_2,t_3]$ and $(Su_0)(t_2)=0$. Then $(Su_0)'(t_2) \le 0$. From $(H_1)''$, we have

$$\begin{aligned} (Su_0)''(t)=-\left( A\left( \int _0^1(u_0^p(r)+u_0^q(r))dr\right) \right) ^{-1}f(t,0)\le 0,\quad t\in (t_2,t_3), \end{aligned}$$

and

$$\begin{aligned} (Su_0)'(t)\le (Su_0)'(t_2)\le 0, \quad t\in [t_2,t_3]. \end{aligned}$$

Then $(Su_0)(t)\le (Su_0)(t_2)=0$ for $t\in [t_2,t_3]$. Therefore, $t_3=1$ and $(Su_0)'(1) < 0$.

We claim that $(Su_0)(t)\ge 0$ for $t\in [0,t_2].$ If otherwise, there exists $t_5\in (0,t_2)$ such that $(Su_0)(t_5)<0$ and $u_0(t_5)=0$. Assume that $(t_6,t_7)\subseteq [0,t_2)$ contains $t_5$, and it is the maximum interval which satisfies $(Su_0)(t)<0$ for any $t\in (t_6,t_7).$ Due to $t_7<t_2<1,$ we can obtain a contradiction by a similar proof to the case of $t_3<1.$ Thus, $(Su_0)(t)\ge 0$ for $t\in [0,t_2]$. Explicit computations show that

$$\begin{aligned} 0>\delta (Su_0)'(1)=\varphi (Su_0)=\int _0^1(Su_0)(t)d\alpha (t) \ge \int _{t_2}^{1}(Su_0)(t)d\alpha (t), \end{aligned}$$

and

$$\begin{aligned} |\delta (Su_0)'(1)| \le \left| \int _{t_2}^{1}(Su_0)(t)d\alpha (t)\right| \le \int _{t_2}^{1}|(Su_0)(t)|d\alpha (t). \end{aligned}$$

In view of the fact that $(Su_0)'(t)\le 0$ and $(Su_0)(t)\le 0$ for $t\in [t_2,1]$, we have

$$\begin{aligned} |(Su_0)'(1)|\ge \frac{|(Su_0)(t)-(Su_0)(t_2)|}{t-t_2},\quad t\in (t_2,1). \end{aligned}$$

Thus, $|(Su_0)(t)|\le (t-t_2)|(Su_0)'(1)|\le |(Su_0)'(1)|$ for $t\in (t_2,1)$. It follows that

$$\begin{aligned} |\delta (Su_0)'(1)|\le \int _{t_2}^{1}|(Su_0)'(1)|d\alpha (t)\le \varphi (\textbf{1})|(Su_0)'(1)|. \end{aligned}$$

Hence $\delta \le \varphi (\textbf{1}),$ which is a contradiction to $\delta >\varphi (\textbf{1})$. In a conclusion, $u_0$ is a fixed point of S, i.e., $u_0$ is a positive solution of problem (1.1) and (4.1).$\square$

Example 4.8

Let $p=\frac{1}{2},\ q=2,\ \alpha =\beta =\delta =1,\ \varphi (u) = \int _0^1 u(s)\,d(\frac{1}{2}s)$. Then $\delta >\varphi (\textbf{1})$. Choose $A(t)=\sin t,\ \rho _1=0.1$ and $\rho _2=2$. Then $(H_3)$ holds and $E_{\rho _1}\approx 0.16,\ M_{\rho _2}=1$. Define the function $f:[0,1]\times [0,+\infty )\rightarrow \mathbb {R}$ by

$$\begin{aligned} f(t,x)=\frac{1}{1000}t+\left\{ \begin{array}{ll} \frac{1}{1000}x,& \quad x\in [0,0.2],\\ 15.98x^2-3.195x,& \quad x\in (0.2,1],\\ -x+13.785,& \quad x\in (1,+\infty ). \end{array}\right. \end{aligned}$$

It is easy to see that $(H)''$ holds. Consider the nonlocal problem

$$\begin{aligned} \begin{aligned} \left\{ \begin{aligned}&-\sin \left( \int ^1_0 (u^\frac{1}{2}(s)+u^2(s))ds\right) u''(t)=f(t,u(t)),\quad t\in (0,1),\\&u(0)- u'(0)=0,\quad u'(1)=\int _0^1 u(s)d\left( \frac{1}{2}s\right) . \end{aligned}\right. \end{aligned} \end{aligned}$$

(4.2)

Let $d=\frac{1}{20}>0, \ \theta =\frac{1}{4}.$ Then $0<\left( 1+\frac{\alpha }{\beta }\right) d<E_{\rho _1}<\theta M_{\rho _2}<M_{\rho _2}$. Simple calculation shows that $I=\frac{935}{9},\ J=\frac{895}{384}$,

$$\begin{aligned} & \underset{(t,x)\in [0,1]\times [0,0.16]}{\max }\ f(t,x)\approx 0.00116<\frac{0.16}{I}\approx 0.0015,\\ & \underset{(t,x)\in [\frac{1}{4},\frac{3}{4}]\times [0.25, 1]}{\min }\ f(t,x)\approx 0.2>\frac{0.2}{J_1}\approx 0.11 \end{aligned}$$

and

$$\begin{aligned} \underset{(t,x)\in [0,1]\times [\frac{1}{20},1]}{\min }\ f(t,x)\ge 0. \end{aligned}$$

So, all conditions of Theorem 4.7 are satisfied. Then nonlocal problem (4.2) has a positive solution.