Open Access

Existence of positive solutions for fractional differential equation with integral boundary conditions on the half-line

Boundary Value Problems20162016:104

https://doi.org/10.1186/s13661-016-0614-7

Received: 4 December 2015

Accepted: 13 May 2016

Published: 23 May 2016

Abstract

This paper considers the existence of positive solutions for fractional-order nonlinear differential equation with integral boundary conditions on the half-infinite interval. By using the fixed point theorem in a cone, sufficient conditions for the existence of at least one or at least two positive solutions of a boundary value problem are established. These theorems also reveal the properties of solutions on the half-line.

Keywords

Caputo derivative integral boundary conditions half-line positive solutions fixed point theorems

MSC

34B18 34B10 26A33

1 Introduction

Boundary value problems are often studies in the areas of applied mathematics and physics. With the development of technology, applications of boundary value problems on the infinite interval attract increasing attention; see [14] and the references therein. Recently, fractional differential equations have also aroused great interest; see [58]. At the same time, the existence of positive solutions for nonlinear fractional differential equation boundary value problems have been widely studied by many authors; see [917] and the references therein.

In [3], the authors, using fixed point theorems in a cone, established the existence of one positive solution and three positive solutions for the following second-order nonlinear boundary value problems with integral boundary conditions on an infinite interval:
$$ \textstyle\begin{cases} \frac{1}{p(t)}(p(t)u'(t))'+f(t,u(t))=0,\quad t\in(0,{+\infty}), \\ a_{1}u(0)-b_{1}\lim_{t\to0^{+}}p(t)u'(t)=\int_{0}^{+\infty} g_{1}(u(s))\psi (s)\,\mathrm{d}s, \\ a_{2}\lim_{t\to{+\infty}}u(t)+b_{2}\lim_{t\to{+\infty}}p(t)u'(t)=\int _{0}^{+\infty} g_{2}(u(s))\psi(s)\,\mathrm{d}s, \end{cases} $$
where \(f \in C((0,+\infty)\times[0,+\infty)\times \mathbb{R},[0,+\infty))\), f may be singular at \(t=0\), \(g_{1},g_{2}:[0,+\infty)\rightarrow[0,+\infty)\) and \(\psi:[0,+\infty)\rightarrow(0,+\infty) \) are continuous, \(\int _{0}^{+\infty} \psi(s)\,\mathrm{d}s<+\infty\), \(p\in C[0,+\infty)\cap C^{1}(0,+\infty)\) with \(p(t)>0\) on \((0,+\infty)\) and \(\int_{0}^{+\infty}\frac {\,\mathrm{d}s}{p(s)}<+\infty\), \(a_{1}+a_{2}>0\), and \(b_{i}>0\) for \(i=1,2\). Iterative schemes for approximating the solutions of a nonlinear fractional boundary value problem on the half-line were presented in [15]. The authors, based on the monotone iterative technique, obtained the existence of positive solutions of the following fractional boundary value problem:
$$ \textstyle\begin{cases} D_{0+}^{\alpha}u(t)+q(t)f(t,u(t))=0,\quad t\in(0,{+\infty}), \\ u(0)=0,\qquad \lim_{t\to{+\infty}} D_{0+}^{\alpha-1}u(t)=\lambda>0, \end{cases} $$
where \(1<{\alpha}<2\), and \(D_{0+}^{\alpha} \) is the standard Riemann-Liouville fractional derivative. For an overview of the literature on differential equations boundary value problems, see [68] and the references therein.
Motivated by all the works mentioned, we study the following fractional boundary value problem on the half-line:
$$ \textstyle\begin{cases} \frac{1}{p(t)}(p(t) \,{}^{\mathrm{C}} D_{0+}^{\alpha}u(t))'+f(t,u(t))=0,\quad t\in (0,{+\infty}), \\ a_{1}u(0)-b_{1}\lim_{t\to0^{+}}p(t) \,{}^{\mathrm{C}} D_{0+}^{\alpha}u(t)=\int _{0}^{+\infty} g_{1}(u(s))\psi_{1}(s)\,\mathrm{d}s, \\ a_{2}\lim_{t\to{+\infty}}u(t)+b_{2}\lim_{t\to{+\infty}}p(t) \,{}^{\mathrm{C}} D_{0+}^{\alpha}u(t)=\int_{0}^{+\infty} g_{2}(u(s))\psi_{2}(s)\,\mathrm{d}s, \end{cases} $$
(1.1)
where \({}^{\mathrm{C}} D^{\alpha}\) is the Caputo fractional derivative of order \(\alpha\in(0,1)\), \(p\in C^{1}([0,{+\infty}),(0,{+\infty}) )\), \(f:(0,{+\infty})\times[0,{+\infty})\rightarrow[0,{+\infty}) \) is a continuous function and may be singular at \(t=0\); \(a_{i}>0\), \(b_{i}>0 \), \(g_{i}\in C([0,{+\infty}),[0,{+\infty}))\), and \(\psi_{i}\in L^{1}([0,{+\infty}))\) is nonnegative for \(i=1,2\).
We assume that the following conditions are satisfied:
  1. (H0)
    \(\lim_{t\to{+\infty}}\int_{0}^{t}\frac{(t-s)^{\alpha -1}}{p(s)}\,\mathrm{d}s<{+\infty}\), \(\frac{b_{2}}{a_{2}}>M\), where
    $$ M= \frac{1}{\Gamma(\alpha)} \sup_{t\in[0,+\infty)} \int_{0}^{t}\frac {(t-s)^{\alpha-1}}{p(s)}\,\mathrm{d}s. $$
    (1.2)
     
  2. (H1)
    There exist functions \(h\in C([0,{+\infty}),[0,{+\infty}))\) and \(v\in C((0,{+\infty}),(0,{+\infty}))\) such that
    $$f(t,u)\leq v(t)h(u),\quad t\in(0,{+\infty}); \qquad \int_{0}^{+\infty}p(s)v(s)\,\mathrm{d}s< {+\infty}. $$
     

2 Preliminaries

In this section, we present some useful definitions and the related theorems.

Definition 2.1

(See [5, 7])

Let \(\alpha>0\). For a function \(u:(0,+\infty)\rightarrow\mathbb{R}\), the Riemann-Liouville fractional integral operator of order α of u is defined by
$$I^{\alpha}_{0+} u(t)=\frac{1}{\Gamma(\alpha)} \int^{t}_{0}(t-s)^{\alpha -1}u(s)\,\mathrm{d}s, $$
provided that the integral exists.

Definition 2.2

(See [5, 7])

The Caputo derivative of order α for a function \(u:(0,+\infty )\rightarrow\mathbb{R}\) is given by
$${}^{\mathrm{C}} D^{\alpha}_{0+} u(t)=\frac{1}{\Gamma(n-\alpha)} \int^{t}_{0}\frac{u^{(n)}(s)}{(t-s)^{\delta+1-n}}\,\mathrm{d}s, $$
provided that the right side is pointwise defined on \((0, +\infty)\), where \(n =[\alpha]+1\) and \(n-1<\alpha<n\).

If \(\alpha=n\), then \({}^{\mathrm{C}} D^{\alpha}_{0+} u(t)=u^{(n)}(t)\).

Lemma 2.1

(See [7])

Let \(\alpha>0\). Then the differential equation
$${}^{\mathrm{C}} D_{0+}^{\alpha}h(t)=0 $$
has solutions
$$h(t)=c_{0}+c_{1}t+c_{2}t^{2}+ \cdots+c_{n-1}t^{n-1},\quad c_{i}\in\mathbb{R}, i=0,1,2,\ldots,n-1, $$
where n is the smallest integer greater than or equal to α.

Lemma 2.2

If (H0) holds and \(y\in C((0,{+\infty}),[0,{+\infty}))\) with \(\int _{0}^{+\infty}p(s)y(s)\,\mathrm{d}s<{+\infty}\), then the fractional boundary value problem
$$ \textstyle\begin{cases} \frac{1}{p(t)}(p(t) \,{}^{\mathrm{C}} D_{0+}^{\alpha}u(t))'+y(t)=0, \quad t\in(0,{+\infty }), \\ a_{1}u(0)-b_{1}\lim_{t\to0^{+}}p(t) \,{}^{\mathrm{C}} D_{0+}^{\alpha}u(t)=\int _{0}^{+\infty} g_{1}(u(s))\psi_{1}(s)\,\mathrm{d}s, \\ a_{2}\lim_{t\to{+\infty}}u(t)+b_{2}\lim_{t\to{+\infty}}p(t) \,{}^{\mathrm{C}} D_{0+}^{\alpha}u(t)=\int_{0}^{+\infty} g_{2}(u(s))\psi_{2}(s)\,\mathrm{d}s \end{cases} $$
(2.1)
has a unique solution
$$\begin{aligned} u(t) =& \int_{0}^{+\infty}G(t,s)p(s)y(s)\,\mathrm{d}s+F_{1}(t) \int_{0}^{+\infty} g_{1}\bigl(u(s)\bigr) \psi_{1}(s)\,\mathrm{d}s \\ &{}+F_{2}(t) \int_{0}^{+\infty} g_{2}\bigl(u(s)\bigr)\psi _{2}(s)\,\mathrm{d}s, \end{aligned}$$
where
$$\begin{aligned}& G(t,s)=\frac{1}{\rho} \textstyle\begin{cases} (b_{1}+a_{1}\int_{0}^{t}\frac{(t-r)^{\alpha-1}}{\Gamma(\alpha )p(r)}\,\mathrm{d}r ) (b_{2}+a_{2}{\lim_{\tau\to{+\infty}}}\int_{s}^{\tau}\frac{(\tau -r)^{\alpha-1}}{\Gamma(\alpha)p(r)}\,\mathrm{d}r ),& 0\leq t\leq s< {+\infty}, \\ (b_{1}+a_{1}\int_{0}^{s}\frac{(t-r)^{\alpha-1}}{\Gamma(\alpha )p(r)}\,\mathrm{d}r ) (b_{2}+a_{2}{\lim_{\tau\to{+\infty}}}\int_{s}^{\tau}\frac{(\tau -r)^{\alpha-1}}{\Gamma(\alpha)p(r)}\,\mathrm{d}r ) \\ \quad {} -b_{1}a_{2}\int_{s}^{t}\frac{(t-r)^{\alpha-1}}{\Gamma(\alpha)p(r)}\,\mathrm{d}r,& 0\leq s< t< {+\infty}, \end{cases}\displaystyle \\& \rho= a_{1}b_{2}+a_{2}b_{1}+a_{1}a_{2} \lim_{\tau\to{+\infty}} \int_{0}^{\tau}\frac {(\tau-r)^{\alpha-1}}{\Gamma(\alpha)p(r)}\,\mathrm{d}r, \\& F_{1}(t)=\frac{1}{\rho} \biggl(b_{2}+a_{2}\lim _{\tau\to{+\infty}} \int_{0}^{\tau }\frac{(\tau-r)^{\alpha-1}}{\Gamma(\alpha)p(r)}\, \mathrm{d}r-a_{2} \int_{0}^{t} \frac{(t-r)^{\alpha-1}}{\Gamma(\alpha)p(r)}\,\mathrm{d}r \biggr), \end{aligned}$$
and
$$ F_{2}(t)=\frac{1}{\rho} \biggl(b_{1}+a_{1} \int_{0}^{t}\frac{(t-r)^{\alpha -1}}{\Gamma(\alpha)p(r)}\,\mathrm{d}r \biggr). $$

Proof

It is well known that the fractional differential equation in (2.1) is equivalent to the integral equation
$$ p(t)\,{}^{\mathrm{C}} D_{0+}^{\alpha}u(t) =\lim _{t\rightarrow0^{+}}p(t)\,{}^{\mathrm{C}} D_{0+}^{\alpha}u(t)- \int _{0}^{t}p(s)y(s)\,\mathrm{d}s. $$
(2.2)
Hence,
$$\begin{aligned} u(t)&= \biggl(I_{0+}^{\alpha}\frac{1}{p(t)} \biggr)\lim_{t\to0^{+}}p(t) \,{}^{\mathrm{C}} D_{0+}^{\alpha}u(t) -I_{0+}^{\alpha} \biggl(\frac{1}{p(t)} \int_{0}^{t} p(s)y(s)\,\mathrm{d}s \biggr)+c_{0} \\ &= \int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)p(s)}\,\mathrm{d}s\lim _{t\to0^{+}}p(t) \,{}^{\mathrm{C}} D_{0+}^{\alpha}u(t) - \int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)p(s)} \int_{0}^{s}p(s)y(s)\,\mathrm{d}r\, \mathrm{d}s+c_{0}, \end{aligned}$$
(2.3)
where \(c_{0}\in\mathbb{R} \) is any constant. It follows from (2.2) and (2.3) that
$$\lim_{t\rightarrow+\infty}p(t)\,{}^{\mathrm{C}} D_{0+}^{\alpha}u(t) =\lim_{t\rightarrow0^{+}}p(t)\,{}^{\mathrm{C}} D_{0+}^{\alpha}u(t)- \int_{0}^{+\infty }p(s)y(s)\,\mathrm{d}s $$
and
$$\begin{aligned} \lim_{t\rightarrow+\infty}u(t) =&\lim_{t\rightarrow0^{+}}p(t)\,{}^{\mathrm{C}} D_{0+}^{\alpha}u(t) \lim_{t\rightarrow+\infty} \int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha )p(s)}\,\mathrm{d}s \\ &{}-\lim_{t\rightarrow+\infty} \int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma (\alpha)p(s)} \int_{0}^{s}p(s)y(s)\,\mathrm{d}r\, \mathrm{d}s+c_{0}. \end{aligned}$$
By the boundary conditions in (2.1) we have
$$\begin{aligned} c_{0} =& \frac{1}{\rho}\biggl[a_{2}b_{1}\lim _{t\to{+\infty}} \int_{0}^{t}\frac {(t-s)^{\alpha-1}}{\Gamma(\alpha)p(s)} \int_{0}^{s} p(r)y(r)\,\mathrm{d}r\, \mathrm{d}s \\ &{}+b_{1}b_{2} \int_{0}^{+\infty}p(s)y(s)\,\mathrm{d}s+b_{1} \int_{0}^{+\infty}g_{2}\bigl(u(s)\bigr) \psi_{2}(s)\,\mathrm{d}s \\ &{}+\biggl(a_{2}\lim_{t\to{+\infty}} \int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha )p(s)}\, \mathrm{d}s+b_{2}\biggr) \int_{0}^{+\infty} g_{1}\bigl(u(s)\bigr) \psi_{1}(s)\,\mathrm{d}s\biggr] \end{aligned}$$
and
$$\begin{aligned} \lim_{t\to0^{+}}p(t) \,{}^{\mathrm{C}} D_{0+}^{\alpha}u(t) = & \frac{1}{\rho}\biggl(a_{1}a_{2}\lim _{t\to{+\infty}} \int_{0}^{t}\frac {(t-s)^{\alpha-1}}{\Gamma(\alpha)p(s)} \int_{0}^{s} p(r)y(r)\,\mathrm{d}r\, \mathrm{d}s \\ &{}+a_{1}b_{2} \int_{0}^{+\infty}p(s)y(s)\,\mathrm{d}s \\ &{} +a_{1} \int_{0}^{+\infty}g_{2}\bigl(u(s)\bigr) \psi_{2}(s)\,\mathrm{d}s-a_{2} \int_{0}^{+\infty } g_{1}\bigl(u(s)\bigr) \psi_{1}(s)\,\mathrm{d}s\biggr). \end{aligned}$$
Substituting them into (2.3), we get
$$\begin{aligned} u(t) =&\frac{1}{\rho} \int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha )p(s)}\,\mathrm{d}s \biggl(a_{1}a_{2}\lim_{t\to{+\infty}} \int_{0}^{t}\frac {(t-s)^{\alpha-1}}{\Gamma(\alpha)p(s)} \int_{0}^{s}p(r)y(r)\,\mathrm{d}r\,\mathrm{d}s \\ &{}+a_{1}b_{2} \int_{0}^{+\infty}p(s)y(s)\,\mathrm{d}s \biggr) \\ &{} +\frac{1}{\rho} \biggl(a_{2}b_{1}\lim _{t\to{+\infty}} \int_{0}^{t}\frac {(t-s)^{\alpha-1}}{\Gamma(\alpha)p(s)} \int_{0}^{s}p(r)y(r)\,\mathrm{d}r\, \mathrm{d}s+b_{1}b_{2} \int_{0}^{+\infty }p(s)y(s)\,\mathrm{d}s \biggr) \\ &{} - \int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)p(s)} \int _{0}^{s}p(r)y(r)\,\mathrm{d}r\,\mathrm{d}s \\ &{} +F_{1}(t) \int_{0}^{+\infty} g_{1}\bigl(u(s)\bigr) \psi_{1}(s)\, \mathrm{d}s+F_{2}(t) \int_{0}^{+\infty} g_{2}\bigl(u(s)\bigr) \psi_{2}(s)\,\mathrm{d}s \\ =&\frac{1}{\rho} \biggl( \int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha )p(s)}\,\mathrm{d}s \biggr) \biggl[a_{1}a_{2}\lim_{t\to{+\infty}} \int_{0}^{t} \biggl( \int_{s}^{t}\frac {(t-r)^{\alpha-1}}{\Gamma(\alpha)p(r)}\,\mathrm{d}r \biggr)p(s)y(s)\,\mathrm{d}s \\ &{}+a_{1}b_{2} \int_{0}^{+\infty}p(s)y(s)\,\mathrm{d}s \biggr] \\ &{}+\frac{1}{\rho} \biggl[a_{2}b_{1}\lim _{t\to{+\infty}} \int_{0}^{t} \biggl( \int _{s}^{t}\frac{(t-r)^{\alpha-1}}{\Gamma(\alpha)p(r)}\,\mathrm{d}r \biggr) p(s)y(s)\,\mathrm{d}s+b_{1}b_{2} \int_{0}^{+\infty}p(s)y(s)\,\mathrm{d}s \biggr] \\ &{}- \int_{0}^{t} \int_{s}^{t}\frac{(t-r)^{\alpha-1}}{\Gamma(\alpha)p(r)}\,\mathrm{d}r p(s)y(s) \,\mathrm{d}s \\ &{}+F_{1}(t) \int_{0}^{+\infty}g_{1}\bigl(u(s)\bigr) \psi_{1}(s)\,\mathrm{d}s+F_{2}(t) \int _{0}^{+\infty} g_{2}\bigl(u(s)\bigr) \psi_{2}(s)\,\mathrm{d}s. \end{aligned}$$
By (1.2), \(M<\frac{b_{2}}{a_{2}}\), and \(\int_{0}^{+\infty }p(s)y(s)\,\mathrm{d}s<{+\infty}\), we have
$$p(s)y(s) \int_{s}^{t}\frac{(t-r)^{\alpha-1}}{\Gamma(\alpha)p(r)}\,\mathrm{d}r\leq Mp(s)y(s)\in L^{1}\bigl( [0,+\infty )\bigr). $$
Hence,
$$\lim_{t\to{+\infty}} \int_{0}^{t} p(s)y(s) \int_{s}^{t}\frac{(t-r)^{\alpha -1}}{\Gamma(\alpha)p(r)}\,\mathrm{d}r\, \mathrm{d}s = \int_{0}^{+\infty} p(s)y(s)\lim_{t\to{+\infty}} \int_{s}^{t}\frac {(t-r)^{\alpha-1}}{\Gamma(\alpha)p(r)}\,\mathrm{d}r\, \mathrm{d}s. $$
Therefore, the unique solution of the fractional boundary value problem (2.1) is
$$\begin{aligned} u(t) =& \int_{0}^{+\infty}G(t,s)p(s)y(s)\,\mathrm{d}s+F_{1}(t) \int_{0}^{+\infty} g_{1}\bigl(u(s)\bigr) \psi_{1}(s)\,\mathrm{d}s \\ &{}+F_{2}(t) \int_{0}^{+\infty} g_{2}\bigl(u(s)\bigr)\psi _{2}(s)\,\mathrm{d}s. \end{aligned}$$
 □
For convenience, we denote
$$\begin{aligned}& G_{M}=\frac{(b_{1}+a_{1}M)(b_{2}+a_{2}M)}{a_{1}b_{2}+a_{2}b_{1}}, \qquad G_{m}=\frac {b_{1}(b_{2}-a_{2}M)}{a_{1}b_{2}+a_{2}b_{1}+a_{1}a_{2}M}, \\& F_{1M}=\frac{b_{2}+a_{2}M}{a_{1}b_{2}+a_{2}b_{1}}, \qquad F_{1m}=\frac {b_{2}-a_{2}M}{a_{1}b_{2}+a_{2}b_{1}+a_{1}a_{2}M}, \\& F_{2M}=\frac{b_{1}+a_{1}M}{a_{1}b_{2}+a_{2}b_{1}},\qquad F_{2m}=\frac {b_{1}}{a_{1}b_{2}+a_{2}b_{1}+a_{1}a_{2}M}, \\& \gamma_{0}=\min \biggl\{ \frac{G_{m}}{G_{M}},\frac{F_{1m}}{F_{1M}}, \frac {F_{2m}}{F_{2M}} \biggr\} ,\qquad \overline{G}=\frac{b_{1}b_{2}}{\rho}, \\& \overline{F}_{1}=\frac{b_{2}}{\rho}, \qquad \overline{F}_{2}= \frac{1}{\rho} \biggl(b_{1}+a_{1}\lim_{\tau\to{+\infty}} \int_{0}^{\tau}\frac{(\tau-r)^{\alpha -1}}{\Gamma(\alpha)p(r)}\,\mathrm{d}r \biggr). \end{aligned}$$

Lemma 2.3

If (H0) holds, then \(G(t,s)\), \(F_{1}(t)\), and \(F_{2}(t)\) defined in Lemma  2.2 satisfy
  1. (1)

    \(G(t,s)\) is a continuous function and \(G(t,s)>0\) for \((t,s)\in [0,{+\infty})\times[0,{+\infty})\);

     
  2. (2)

    \(F_{1}(t)\), \(F_{2}(t)\) are continuous functions, and \(F_{1}(t), F_{2}(t)\geq0\) for \(t\in[0,{+\infty})\);

     
  3. (3)
    $$\begin{aligned}& G_{m}\leq G(t,s)\leq G_{M} \quad \textit{for }(t,s) \in[0,{+\infty})\times[0,{+\infty}), \\& F_{im}\leq F_{i}(t)\leq F_{iM} \quad \textit{for }t\in[0,{+\infty}), i=1,2; \end{aligned}$$
     
  4. (4)
    there exist constants \(0< l_{1}< l_{2}<{+\infty}\) such that
    $$\begin{aligned}& G(t,s)\geq\gamma_{0} G_{M}\quad \textit{for }(t,s) \in[l_{1},l_{2}]\times[0,+\infty), \\& F_{i}(t)\geq\gamma_{0} F_{iM} \quad \textit{for }t\in[l_{1},l_{2}]\textit{ and }i=1,2; \end{aligned}$$
     
  5. (5)

    for any \(s\in[0,+\infty)\), \(\lim_{t\to{+\infty}}G(t,s)=\overline {G}<{+\infty}\), \(\lim_{t\to{+\infty}}F_{1}(t)=\overline{F}_{1}<{+\infty}\), \(\lim_{t\to{+\infty}}F_{2}(t)=\overline{F}_{2}<{+\infty}\).

     

Proof

(1) For \(0\leq t\leq s\), it is easy to see that \(G(t,s)>0\).

For \(0\leq s< t\), by (H0) and (1.2) we have
$$G(t,s)\geq\frac{1}{\rho} \biggl( b_{1}b_{2}-b_{1}a_{2} \int_{s}^{t}\frac{(t-r)^{\alpha -1}}{\Gamma(\alpha)p(r)}\,\mathrm{d}r \biggr) \geq\frac{b_{1}a_{2}}{\rho} \biggl(\frac{b_{2}}{a_{2}}-M \biggr) > 0. $$
Hence, \(G(t,s)>0\) for \((t,s)\in[0,{+\infty})\times[0,{+\infty})\).

It is easy to see that \(G(t,s)\) is a continuous function.

(2) It follows from (1.2) and (H0) that (2) holds.

(3) By (H0), for \(t,s\in[0,+\infty)\), we have
$$G(t,s)\geq\frac{1}{\rho} \biggl( b_{1}b_{2}-b_{1}a_{2} \int_{s}^{t}\frac{(t-r)^{\alpha -1}}{\Gamma(\alpha)p(r)}\,\mathrm{d}r \biggr) \geq\frac{b_{1}a_{2}}{\rho} \biggl(\frac{b_{2}}{a_{2}}-M \biggr) =G_{m} $$
and
$$G(t,s)\leq\frac{1}{\rho} \biggl(b_{1}+a_{1} \int_{0}^{t}\frac{(t-r)^{\alpha-1}}{ \Gamma(\alpha)p(r)}\,\mathrm{d}r \biggr) \biggl(b_{2}+a_{2}{\lim_{\tau\to{+\infty}}} \int_{s}^{\tau}\frac{(\tau -r)^{\alpha-1}}{\Gamma(\alpha)p(r)}\,\mathrm{d}r \biggr) \leq G_{M}. $$

It is easy to see that \(F_{im}\leq F_{i}(t)\leq F_{iM}\) for \(t\in [0,{+\infty})\), \(i=1,2\).

(4) By (3) there exist constants \(0< l_{1}< l_{2}<{+\infty}\) such that
$$G(t,s)\geq G_{m}=\frac{G_{m}}{G_{M}}\cdot G_{M}\geq \gamma_{0} G_{M}\quad \mbox{for } (t,s)\in[l_{1},l_{2}] \times[0,+\infty). $$
Similarly, we have
$$F_{i}(t)\geq\gamma_{0} F_{iM}\quad \mbox{for } t \in[l_{1},l_{2}] \mbox{ and } i=1,2. $$
(5) By (H0) and \(0<\alpha<1\), for any \(s\in[0,+\infty)\), we can show that
$$\begin{aligned} \lim_{t\to{+\infty}}G(t,s) =&\frac{1}{\rho}\lim _{t\to{+\infty}} \biggl( \biggl(b_{1}+a_{1} \int_{0}^{s}\frac {(t-r)^{\alpha-1}}{ \Gamma(\alpha)p(r)}\,\mathrm{d}r \biggr) \biggl(b_{2}+a_{2}{\lim_{\tau\to {+\infty}}} \int_{s}^{\tau}\frac{(\tau-r)^{\alpha-1}}{\Gamma(\alpha)p(r)} \,\mathrm{d}r \biggr) \\ &{} -b_{1}a_{2} \int_{s}^{t}\frac{(t-r)^{\alpha-1}}{\Gamma(\alpha)p(r)}\,\mathrm{d}r \biggr) \\ =&\frac{1}{\rho} \biggl( \biggl(b_{1}+a_{1} \int_{0}^{s}\lim_{t\to{+\infty}} \frac {(t-r)^{\alpha-1}}{ \Gamma(\alpha)p(r)}\,\mathrm{d}r \biggr) \biggl(b_{2}+a_{2}{ \lim_{t\to{+\infty }}} \int_{s}^{t}\frac{(t-r)^{\alpha-1}}{\Gamma(\alpha)p(r)} \,\mathrm{d}r \biggr) \\ &{} -b_{1}a_{2}\lim_{t\to{+\infty}} \int_{s}^{t}\frac{(t-r)^{\alpha-1}}{\Gamma (\alpha)p(r)}\,\mathrm{d}r \biggr) \\ =&\frac{1}{\rho} \biggl(b_{1} \biggl(b_{2}+a_{2}{ \lim_{t\to{+\infty}}} \int _{s}^{t}\frac{(t-r)^{\alpha-1}}{\Gamma(\alpha)p(r)} \,\mathrm{d}r \biggr)-b_{1}a_{2}\lim_{t\to{+\infty}} \int_{s}^{t}\frac {(t-r)^{\alpha-1}}{\Gamma(\alpha)p(r)}\,\mathrm{d}r \biggr) \\ =&\overline{G}. \end{aligned}$$

It is obvious that \(\lim_{t\to{+\infty}}F_{1}(t)=\overline{F}_{1}<{+\infty}\) and \(\lim_{t\to{+\infty}}F_{2}(t)=\overline{F}_{2}<{+\infty}\). □

Let
$$ E=\Bigl\{ u\in C [0,+\infty ):\lim_{t\to{+\infty}}u(t)< + \infty\Bigr\} $$
(2.4)
be a Banach space with the norm \(\|u\|=\sup_{t\in[0,+\infty)}|u(t)|\), and
$$P=\Bigl\{ u\in E:u(t)\geq0, t\in [0,+\infty ), \inf_{t\in[l_{1},l_{2}]}u(t) \geq \gamma_{0}\|u\|\Bigr\} $$
be a cone in E.
For \(r>0\), we denote
$$\begin{aligned}& K_{r}=\bigl\{ u\in P: \Vert u\Vert < r\bigr\} , \qquad \partial K_{r}=\bigl\{ u\in P: \Vert u\Vert =r\bigr\} , \\& S_{r}:=\sup\bigl\{ h(u):0\leq u\leq r\bigr\} ,\qquad S_{r}':=\sup\bigl\{ g_{1}(u):0\leq u\leq r\bigr\} , \end{aligned}$$
and
$$S_{r}'':=\sup\bigl\{ g_{2}(u):0 \leq u\leq r\bigr\} . $$
It follows from (H1) that \(S_{r}\), \(S_{r}'\), and \(S_{r}''<+\infty\).
We define the operator \(T: P\rightarrow E\) by
$$\begin{aligned} Tu(t) = & \int_{0}^{+\infty}G(t,s)p(s)f\bigl(s,u(s)\bigr)\, \mathrm{d}s +F_{1}(t) \int_{0}^{+\infty}g_{1}\bigl(u(s)\bigr) \psi_{1}(s)\,\mathrm{d}s \\ &{}+F_{2}(t) \int_{0}^{+\infty}g_{2}\bigl(u(s)\bigr) \psi_{2}(s)\,\mathrm{d}s, \quad t\in[0,+\infty). \end{aligned}$$

We can easily get the following Lemma 2.4 from Lemma 2.2.

Lemma 2.4

If \(u\in P\), then the boundary value problem (1.1) is equivalent to the integral equation
$$\begin{aligned} u(t) = & \int_{0}^{+\infty}G(t,s)p(s)f\bigl(s,u(s)\bigr)\, \mathrm{d}s +F_{1}(t) \int_{0}^{+\infty}g_{1}\bigl(u(s)\bigr) \psi_{1}(s)\,\mathrm{d}s \\ &{}+F_{2}(t) \int_{0}^{+\infty}g_{2}\bigl(u(s)\bigr) \psi_{2}(s)\,\mathrm{d}s, \quad t\in[0,+\infty). \end{aligned}$$

Lemma 2.5

(See [1, 18])

Let E be defined by (2.4), and \(\Omega\subset E\). Then Ω is relatively compact in E if the following conditions hold:
  1. (a)

    Ω is uniformly bounded in E;

     
  2. (b)

    the functions belonging to M are equicontinuous on any compact interval of \([0,+\infty)\);

     
  3. (c)

    the functions from Ω are equiconvergent, that is, for any given \(\varepsilon>0\), there exists \(T(\varepsilon)>0\) such that \(|f(t)-f(+\infty)|<\varepsilon\) for any \(t>T(\varepsilon)\) and \(f\in\Omega\).

     

Lemma 2.6

If (H0) and (H1) hold, then \(T:P\rightarrow P\) is completely continuous.

Proof

We divide the proof into three steps.

Step 1: We show that \(T:P\rightarrow P\) is well defined.

For \(u\in P\), there exists a constant \(r_{0}>0\) such that \(\|u\|\leq r_{0}\). By (H1) and Lemma 2.3, for \(t,s\in[0,+\infty)\), we have
$$G(t,s)p(s)f\bigl(s,u(s)\bigr)\leq G(t,s)p(s)v(s)h(u)\leq G_{M}p(s)v(s)S_{r_{0}}. $$
Since \(G(t,s)\), \(F_{1}(t)\), \(F_{2}(t)\) are continuous with respect to t, by using the Lebesgue dominated convergence theorem, for \(t_{0}\in[0,+\infty )\), we have
$$\begin{aligned} \lim_{t\to t_{0}}Tu(t) =& \lim_{t\to t_{0}} \int_{0}^{+\infty }G(t,s)p(s)f\bigl(s,u(s)\bigr)\, \mathrm{d}s+\lim_{t\to t_{0}}F_{1}(t) \int_{0}^{+\infty }g_{1}\bigl(u(s)\bigr) \psi_{1}(s)\,\mathrm{d}s \\ &{} +\lim_{t\to t_{0}}F_{2}(t) \int_{0}^{+\infty}g_{2}\bigl(u(s)\bigr) \psi_{2}(s)\,\mathrm{d}s \\ =& \int_{0}^{+\infty}G(t_{0},s)p(s)f\bigl(s,u(s) \bigr)\,\mathrm{d}s +F_{1}(t_{0}) \int_{0}^{+\infty}g_{1}\bigl(u(s)\bigr) \psi_{1}(s)\,\mathrm{d}s \\ &{} +F_{2}(t_{0}) \int_{0}^{+\infty}g_{2}\bigl(u(s)\bigr) \psi_{2}(s)\,\mathrm{d}s. \end{aligned}$$
So, \(Tu\in C[0,+\infty)\), and we get
$$\begin{aligned} \lim_{t\to+\infty}Tu(t) =& \int_{0}^{+\infty}\overline {G}p(s)f\bigl(s,u(s)\bigr)\, \mathrm{d}s +\overline{F}_{1} \int_{0}^{+\infty}g_{1}\bigl(u(s)\bigr) \psi_{1}(s)\,\mathrm{d}s \\ &{}+\overline{F}_{2} \int_{0}^{+\infty}g_{2}\bigl(u(s)\bigr) \psi_{2}(s)\,\mathrm{d}s< +\infty. \end{aligned}$$
It is obvious that \(Tu(t)\geq0\), \(t\in[0,+\infty)\), by Lemma 2.3. Moreover,
$$\begin{aligned} \inf_{t\in[l_{1},l_{2}]}Tu(t) \geq&\inf_{t\in[l_{1},l_{2}]} \int_{0}^{+\infty}G(t,s)p(s)f\bigl(s,u(s)\bigr)\, \mathrm{d}s+\inf_{t\in[l_{1},l_{2}]}F_{1}(t) \int_{0}^{+\infty}g_{1}\bigl(u(s)\bigr)\psi _{1}(s)\,\mathrm{d}s \\ &{}+\inf_{t\in[l_{1},l_{2}]}F_{2}(t) \int_{0}^{+\infty}g_{2}\bigl(u(s)\bigr) \psi_{2}(s)\,\mathrm{d}s. \end{aligned}$$
By Lemma 2.3 (4) we have
$$\begin{aligned} \inf_{t\in[l_{1},l_{2}]}Tu(t) \geq &\gamma_{0}\sup _{t\in[0,+\infty)} \biggl( \int_{0}^{+\infty} G(t,s)p(s)f\bigl(s,u(s)\bigr)\, \mathrm{d}s +F_{1}(t) \int_{0}^{+\infty}g_{1}\bigl(u(s)\bigr) \psi_{1}(s)\,\mathrm{d}s \\ &{}+F_{2}(t) \int_{0}^{+\infty}g_{2}\bigl(u(s)\bigr) \psi_{2}(s)\,\mathrm{d}s \biggr) \\ \geq&\gamma_{0}\|Tu\|. \end{aligned}$$

Hence, \(T:P\rightarrow P\) is well defined.

Step 2: We can verify that \(T:P\rightarrow P\) is continuous.

Let \(u_{n},u\in P\) and \(\|u_{n}-u\|\rightarrow0\) as \(n\rightarrow+\infty\). Then there exists a constant \(r_{1}>0\) such that \(\|u_{n}\|,\|u\|\leq r_{1}\). We have
$$\begin{aligned} 0 \leq& G_{M}p(s)\bigl\vert f\bigl(s,u_{n}(s)\bigr)-f \bigl(s,u(s)\bigr)\bigr\vert +F_{1M}\bigl\vert g_{1} \bigl(u_{n}(s)\bigr)-g_{1}\bigl(u(s)\bigr)\bigr\vert \psi_{1}(s) \\ &{}+F_{2M}\bigl\vert g_{2}\bigl(u_{n}(s) \bigr)-g_{2}\bigl(u(s)\bigr)\bigr\vert \psi_{2}(s) \\ \leq& G_{M}p(s)v(s) \bigl(h\bigl(u_{n}(s)\bigr)+h \bigl(u(s)\bigr)\bigr) +F_{1M}\bigl(g_{1}\bigl(u_{n}(s) \bigr)+g_{1}\bigl(u(s)\bigr)\bigr)\psi_{1}(s) \\ &{}+F_{2M}\bigl(g_{2}\bigl(u_{n}(s) \bigr)+g_{2}\bigl(u(s)\bigr)\bigr)\psi_{2}(s) \\ \leq& 2G_{M}S_{r_{1}}p(s)v(s)+2F_{1M}S_{r_{1}}' \psi_{1}(s) +2F_{2M}S_{r_{1}}'' \psi_{2}(s) \\ \in& L^{1}\bigl( [0,+\infty )\bigr) \end{aligned}$$
and, for \(s\in[0,+\infty)\),
$$\begin{aligned}& f\bigl(s,u_{n}(s)\bigr)-f\bigl(s,u(s)\bigr)\rightarrow0 \quad \mbox{as } n\rightarrow+\infty, \\& g_{i}\bigl(u_{n}(s)\bigr)-g_{i}\bigl(u(s)\bigr) \rightarrow0 \quad \mbox{as } n\rightarrow+\infty, i=1,2. \end{aligned}$$
Then, by the Lebesgue dominated convergence theorem we have
$$\begin{aligned} \|Tu_{n}-Tu\| \leq & \int_{0}^{+\infty}G_{M}p(s) \bigl\vert f \bigl(s,u_{n}(s)\bigr)-f\bigl(s,u(s)\bigr)\bigr\vert \,\mathrm{d}s \\ &{}+F_{1M} \int_{0}^{+\infty}\bigl\vert g_{1} \bigl(u_{n}(s)\bigr)-g_{1}\bigl(u(s)\bigr)\bigr\vert \psi_{1}(s)\,\mathrm{d}s \\ &{}+F_{2M} \int_{0}^{+\infty}\bigl\vert g_{2} \bigl(u_{n}(s)\bigr)-g_{2}\bigl(u(s)\bigr)\bigr\vert \psi_{2}(s)\,\mathrm{d}s \\ \rightarrow&0 \quad \mbox{as } n\rightarrow+\infty. \end{aligned}$$

Therefore, \(T:P\rightarrow P\) is a continuous operator.

Step 3: We can show that \(T:P\rightarrow P\) is relatively compact.

Let Ω be a bounded subset of P. Then there exists a constant \(r_{2}>0\) such that \(\|u\|\leq r_{2}\) for each \(u\in\Omega\).

By Lemma 2.3 and (H1) we have
$$\begin{aligned} \|Tu\| =& \sup_{t\in [0,+\infty )}\biggl\vert \int_{0}^{+\infty }G(t,s)p(s)f\bigl(s,u(s)\bigr)\, \mathrm{d}s +F_{1}(t) \int_{0}^{+\infty}g_{1}\bigl(u(s)\bigr) \psi_{1}(s)\,\mathrm{d}s \\ &{} +F_{2}(t) \int_{0}^{+\infty}g_{2}\bigl(u(s)\bigr) \psi_{2}(s)\,\mathrm{d}s\biggr\vert \\ \leq& \int_{0}^{+\infty}G_{M}p(s)v(s)h\bigl(u(s)\bigr) \,\mathrm{d}s +F_{1M} \int_{0}^{+\infty}g_{1}\bigl(u(s)\bigr) \psi_{1}(s)\,\mathrm{d}s \\ &{} +F_{2M} \int_{0}^{+\infty}g_{2}\bigl(u(s)\bigr) \psi_{2}(s)\,\mathrm{d}s \\ \leq& G_{M}S_{r_{2}} \int_{0}^{+\infty}p(s)v(s)\,\mathrm{d}s +F_{1M}S_{r_{2}}' \int_{0}^{+\infty}\psi_{1}(s)\,\mathrm{d}s +F_{2M}S_{r_{2}}'' \int_{0}^{+\infty}\psi_{2}(s)\,\mathrm{d}s \\ < & +\infty. \end{aligned}$$
So, \(T(\Omega)\) is uniformly bounded.
For any \(\overline{T}\in(0,+\infty)\), since \(G(t,s)\), \(F_{1}(t)\), and \(F_{2}(t)\) are continuous, we have that G is uniformly continuous on \([0,\overline{T}]\times[0,\overline{T}]\) and \(F_{1}\) and \(F_{1}\) are uniformly continuous on \([0,\overline{T}]\). This implies that, for any \(\varepsilon>0\), there exists \(\delta>0\) such that, when \(t_{1}, t_{2}\in [0,\overline{T}]\), whenever \(|t_{2}-t_{1}|<\delta\) and \(s\in[0,\overline {T}]\), we have
$$\begin{aligned}& \bigl\vert G(t_{2},s)-G(t_{1},s)\bigr\vert < \varepsilon, \\& \bigl\vert F_{i}(t_{2})-F_{i}(t_{1}) \bigr\vert < \varepsilon,\quad i=1,2. \end{aligned}$$
Therefore, for \(t_{1}, t_{2}\in[0,\overline{T}]\), whenever \(|t_{2}-t_{1}|<\delta\) and \(u\in\Omega\), we can show that
$$\begin{aligned}& \bigl\vert Tu(t_{1})-Tu(t_{2})\bigr\vert \\& \quad \leq \int_{0}^{+\infty}\bigl\vert G(t_{1},s)-G(t_{2},s) \bigr\vert p(s)f\bigl(s,u(s)\bigr)\,\mathrm{d}s \\& \qquad {}+\bigl\vert F_{1}(t_{1})-F_{1}(t_{2})\bigr\vert \int_{0}^{+\infty}g_{1}\bigl(u(s)\bigr) \psi_{1}(s)\,\mathrm{d}s \\& \qquad {}+\bigl\vert F_{2}(t_{1})-F_{2}(t_{2}) \bigr\vert \int_{0}^{+\infty}g_{2}\bigl(u(s)\bigr) \psi_{2}(s)\,\mathrm{d}s \\& \quad < \varepsilon\biggl(S_{r_{2}} \int_{0}^{+\infty}p(s)v(s)\,\mathrm{d}s +S_{r_{2}}' \int_{0}^{+\infty}\psi_{1}(s)\,\mathrm{d}s +S_{r_{2}}'' \int_{0}^{+\infty}\psi_{2}(s)\,\mathrm{d}s \biggr). \end{aligned}$$
Hence, \(T(\Omega)\) is locally equicontinuous on \([0,+\infty)\).
Since
$$\begin{aligned} Tu(+\infty) =& \int_{0}^{+\infty}\overline{G}p(s)f\bigl(s,u(s)\bigr)\, \mathrm{d}s +\overline{F}_{1} \int_{0}^{+\infty}g_{1}\bigl(u(s)\bigr) \psi_{1}(s)\,\mathrm{d}s \\ &{} +\overline{F}_{2} \int_{0}^{+\infty}g_{2}\bigl(u(s)\bigr) \psi_{2}(s)\,\mathrm{d}s. \end{aligned}$$
By Lemma 2.3 we conclude that
$$\begin{aligned}& \bigl\vert Tu(t)-Tu(+\infty)\bigr\vert \\& \quad \leq \int_{0}^{+\infty}\bigl\vert G(t,s)-\overline {G}\bigr\vert p(s)f\bigl(s,u(s)\bigr)\,\mathrm{d}s +\bigl\vert F_{1}(t)- \overline{F}_{1}\bigr\vert \int_{0}^{+\infty}g_{1}\bigl(u(s)\bigr) \psi_{1}(s)\,\mathrm{d}s \\& \qquad {} +\bigl\vert F_{2}(t)-\overline{F}_{2}\bigr\vert \int_{0}^{+\infty}g_{2}\bigl(u(s)\bigr) \psi_{2}(s)\,\mathrm{d}s \\& \quad \rightarrow 0 \quad \mbox{as } t\rightarrow+\infty. \end{aligned}$$
Hence, \(T(\Omega)\) is equiconvergent at infinity.

By Lemma 2.5 we obtain that \(T:P\rightarrow P\) is completely continuous. □

Lemma 2.7

(See [19])

Let E be a Banach space, \(P\subseteq E\) be a cone, and \(\Omega_{1}\), \(\Omega_{2}\) be two bounded open subsets of E with \(\theta\in\Omega _{1}\subset\overline{\Omega}_{1}\subset\Omega_{2}\). Suppose that \(T: P\cap(\overline{\Omega}_{2}\setminus\Omega _{1})\rightarrow P \) is a completely continuous operator such that either
  1. (i)

    \(\|Tx\|\leq\|x\|\), \(x\in P\cap\partial\Omega_{1}\) and \(\| Tx\|\geq\|x\|\), \(x\in P\cap\partial\Omega_{2}\), or

     
  2. (ii)

    \(\|Tx\|\geq\|x\|\), \(x\in P\cap\partial\Omega_{1}\), and \(\| Tx\|\leq\|x\|\), \(x\in P\cap\partial\Omega_{2}\),

     
holds. Then the operator T has at least one fixed point in \(P\cap (\overline{\Omega}_{2}\setminus\Omega_{1})\).

3 The existence of positive solutions

For convenience, we give the following notation:
$$\begin{aligned}& h^{\varphi}=\limsup_{u\rightarrow\varphi} \frac{h(u)}{u},\qquad f_{\varphi}=\liminf_{u\rightarrow\varphi}\inf_{t\in[l_{1},l_{2}]} \frac{f(t,u)}{u}, \\& g_{i}^{\varphi}=\limsup_{u\rightarrow\varphi} \frac{g_{i}(u)}{u},\qquad g_{i,\varphi}=\liminf_{u\rightarrow\varphi} \frac{g_{i}(u)}{u}, \end{aligned}$$
where \(\varphi=0 \) or +∞, and \(i=1,2\). We denote
$$\begin{aligned}& A=\max \biggl\{ G_{M} \int_{0}^{+\infty}p(s)v(s)\,\mathrm{d}s, F_{1M} \int _{0}^{+\infty}\psi_{1}(s)\,\mathrm{d}s, F_{2M} \int_{0}^{+\infty}\psi_{2}(s)\,\mathrm{d}s \biggr\} , \\& B=\gamma_{0}\cdot\min \biggl\{ G_{m} \int_{l_{1}}^{l_{2}}p(s)\,\mathrm{d}s, F_{1m} \int _{l_{1}}^{l_{2}}\psi_{1}(s)\,\mathrm{d}s, F_{2m} \int_{l_{1}}^{l_{2}}\psi_{2}(s)\,\mathrm{d}s \biggr\} . \end{aligned}$$

Theorem 3.1

Suppose that (H0) and (H1) hold. If
$$A\bigl(h^{0}+g_{1}^{0}+g_{2}^{0} \bigr)< 1< B(f_{+\infty}+g_{1,+\infty}+g_{2,+\infty}), $$
then the boundary value problem (1.1) has at least one positive solution.

Proof

Since \(A(h^{0}+g_{1}^{0}+g_{2}^{0})<1\), then there exists a constant \(r_{1}>0\) such that, for \(u\leq r_{1}\), we have
$$ h(u)\leq\biggl(h^{0}+\frac{\varepsilon_{1}}{3}\biggr)u,\qquad g_{i}(u)\leq\biggl(g_{i}^{0}+\frac {\varepsilon_{1}}{3} \biggr)u,\quad i=1,2, $$
(3.1)
where \(\varepsilon_{1}\) satisfies \(A(h^{0}+g_{1}^{0}+g_{2}^{0}+\varepsilon_{1})\leq1\).
Therefore, for any \(t\in[0,+\infty)\), \(u\in\partial K_{r_{1}}\), we can get
$$\begin{aligned} \bigl\vert Tu(t)\bigr\vert =& \biggl\vert \int_{0}^{+\infty}G(t,s)p(s)f\bigl(s,u(s)\bigr)\, \mathrm{d}s +F_{1}(t) \int_{0}^{+\infty}g_{1}\bigl(u(s)\bigr) \psi_{1}(s)\,\mathrm{d}s \\ &{} +F_{2}(t) \int_{0}^{+\infty}g_{2}\bigl(u(s)\bigr) \psi_{2}(s)\,\mathrm{d}s\biggr\vert \\ \leq& \int_{0}^{+\infty}G_{M}p(s)v(s) \biggl(h_{0}+\frac{\varepsilon _{1}}{3}\biggr)u(s)\,\mathrm{d}s +F_{1M} \int_{0}^{+\infty}\biggl(g_{1}^{0}+ \frac{\varepsilon_{1}}{3}\biggr)u(s)\psi _{1}(s)\,\mathrm{d}s \\ &{} +F_{2M} \int_{0}^{+\infty}\biggl(g_{2}^{0}+ \frac{\varepsilon_{1}}{3}\biggr)u(s)\psi _{2}(s)\,\mathrm{d}s \\ \leq& A\bigl(h^{0}+g_{1}^{0}+g_{2}^{0}+ \varepsilon_{1}\bigr)\|u\| \\ \leq&\|u\|. \end{aligned}$$
On the other hand, since \(B(f_{+\infty}+g_{1,+\infty}+g_{2,+\infty })>1\), there exist constants \(\overline{r}_{2}>0\) and \(M_{i}\), \(i=0,1,2\), with \(f_{+\infty}>M_{0}>0\), \(g_{1,+\infty}>M_{1}>0\), \(g_{2,+\infty}>M_{2}>0\) such that, for \(t\in[l_{1},l_{2}]\),
$$ f(t,u)\geq\biggl(M_{0}-\frac{\varepsilon_{2}}{3}\biggr)u,\qquad g_{i}(u)\geq\biggl(M_{i}-\frac{\varepsilon_{2}}{3}\biggr)u, \quad i=1,2, $$
(3.2)
where \(\varepsilon_{2}\) satisfies \(B(M_{0}+M_{1}+M_{2}-\varepsilon_{2})\geq1\).
Let \(r_{2}=\max \{r_{1},\frac{\overline{r}_{2}}{\gamma_{0}} \}\). In view of the definition of P,
$$ \inf_{t\in[l_{1},l_{2}]}u(t)\geq\gamma_{0}\|u\|\quad \mbox{for } u\in P. $$
(3.3)
According to Lemma 2.3, for \(u\in\partial K_{r_{2}}\), we have
$$\begin{aligned} \|Tu\| =& \sup_{t\in [0,+\infty )}\biggl\vert \int_{0}^{+\infty }G(t,s)p(s)f\bigl(s,u(s)\bigr)\, \mathrm{d}s +F_{1}(t) \int_{0}^{+\infty}g_{1}\bigl(u(s)\bigr) \psi_{1}(s)\,\mathrm{d}s \\ & +F_{2}(t) \int_{0}^{+\infty}g_{2}\bigl(u(s)\bigr) \psi_{2}(s)\,\mathrm{d}s\biggr\vert \\ \geq& \int_{l_{1}}^{l_{2}}G_{m} p(s)f\bigl(s,u(s) \bigr)\,\mathrm{d}s+F_{1m} \int _{l_{1}}^{l_{2}}g_{1}\bigl(u(s)\bigr) \psi_{1}(s)\,\mathrm{d}s \\ &{} +F_{2m} \int_{l_{1}}^{l_{2}}g_{2}\bigl(u(s)\bigr) \psi_{2}(s)\,\mathrm{d}s. \end{aligned}$$
By (3.2) and (3.3) we have
$$\begin{aligned} \|Tu\| \geq& \int_{l_{1}}^{l_{2}}G_{m} p(s) \biggl(M_{0}-\frac{\varepsilon_{2}}{3}\biggr)u(s)\,\mathrm{d}s +F_{1m} \int_{l_{1}}^{l_{2}}\biggl(M_{1}- \frac{\varepsilon_{2}}{3}\biggr)u(s)\psi _{1}(s)\,\mathrm{d}s \\ &{} +F_{2m} \int_{l_{1}}^{l_{2}}\biggl(M_{2}- \frac{\varepsilon_{2}}{3}\biggr)u(s)\psi _{2}(s)\,\mathrm{d}s \\ \geq& B(M_{0}+M_{1}+M_{2}-\varepsilon_{2}) \|u\| \\ \geq&\|u\|. \end{aligned}$$

Therefore, by (i) of Lemma 2.7 and Lemma 2.3, the boundary value problem (1.1) has at least one positive solution \(u\in\overline{K}_{r_{2}}\setminus K_{r_{1}}\). □

Remark 3.1

It follows from the proof of Theorem 3.1 that the boundary value problem (1.1) has at least one positive solution \(u\in P\) if one of the conditions \(f_{+\infty }=+\infty\), \(g_{1,{+\infty}}=+\infty\), and \(g_{2,{+\infty}}=+\infty\) holds.

Theorem 3.2

Suppose that (H0) and (H1) hold. If
$$A\bigl(h^{+\infty}+g_{1}^{+\infty}+g_{2}^{+\infty} \bigr)< 1< B(f_{0}+g_{1,0}+g_{2,0}), $$
then the boundary value problem (1.1) has at least one positive solution.

Proof

It follows from \(B(f_{0}+g_{1,0}+g_{2,0})>1\) that there exist constants \(r_{3}>0\), \(M'_{i}\), \(i=0,1,2\), with \(f_{0}>M'_{0}>0\), \(g_{1,0}>M'_{1}>0\), \(g_{2,0}>M'_{2}>0\) such that, for \(t\in[l_{1},l_{2}]\) and \(0< u\leq r_{3}\), we have
$$ f(t,u)\geq\biggl(M'_{0}-\frac{\varepsilon_{3}}{3} \biggr)u, \qquad g_{i}(u)\geq\biggl(M'_{i}- \frac {\varepsilon_{3}}{3}\biggr)u, $$
(3.4)
where \(i=1,2\), and \(\varepsilon_{3} \) satisfies \(B(M'_{0}+M'_{1}+M'_{2}-\varepsilon_{3})\geq1\).
Thus, for any \(t\in[0,+\infty)\) and \(u\in\partial K_{r_{3}}\), we have \(\inf_{t\in[l_{1},l_{2}]}u(t)\geq\gamma_{0}\|u\|\) and
$$\begin{aligned} \|Tu\| =& \sup_{t\in [0,+\infty )}\biggl\vert \int_{0}^{+\infty }G(t,s)p(s)f\bigl(s,u(s)\bigr)\, \mathrm{d}s +F_{1}(t) \int_{0}^{+\infty}g_{1}\bigl(u(s)\bigr) \psi_{1}(s)\,\mathrm{d}s \\ &{} +F_{2}(t) \int_{0}^{+\infty}g_{2}\bigl(u(s)\bigr) \psi_{2}(s)\,\mathrm{d}s\biggr\vert \\ \geq& \int_{l_{1}}^{l_{2}}G_{m} p(s)f\bigl(s,u(s) \bigr)\,\mathrm{d}s +F_{1m} \int_{l_{1}}^{l_{2}}g_{1}\bigl(u(s)\bigr) \psi_{1}(s)\,\mathrm{d}s \\ &{}+F_{2m} \int_{l_{1}}^{l_{2}}g_{2}\bigl(u(s)\bigr) \psi_{2}(s)\,\mathrm{d}s. \end{aligned}$$
It follows from (3.4) that
$$\begin{aligned} \|Tu\| \geq& \int_{l_{1}}^{l_{2}}G_{m} p(s) \biggl(M'_{0}-\frac{\varepsilon _{3}}{3}\biggr)u(s)\,\mathrm{d}s +F_{1m} \int_{l_{1}}^{l_{2}}\biggl(M'_{1}- \frac{\varepsilon_{3}}{3}\biggr)u(s)\psi _{1}(s)\,\mathrm{d}s \\ &{} +F_{2m} \int_{l_{1}}^{l_{2}}\biggl(M'_{2}- \frac{\varepsilon_{3}}{3}\biggr)u(s)\psi _{2}(s)\,\mathrm{d}s \\ \geq& B\bigl(M'_{0}+M'_{1}+M'_{2}- \varepsilon_{3}\bigr)\|u\| \\ \geq& \|u\|. \end{aligned}$$
On the other hand, since \(A(h^{+\infty}+g_{1}^{+\infty}+g_{2}^{+\infty })<1\), there exists a constant \(\overline{r}_{4}>0\) such that, for \(u\geq \overline{r}_{4}\), we have
$$h(u)\leq\biggl(h^{+\infty}+\frac{\varepsilon_{4}}{3}\biggr)u, \qquad g_{i}(u)\leq\biggl(g_{i}^{+\infty }+\frac{\varepsilon_{4}}{3} \biggr)u \quad \mbox{for } i=1,2, $$
where \(\varepsilon_{4}\) with \(A(h^{+\infty}+g_{1}^{+\infty}+g_{2}^{+\infty }+\varepsilon_{4})<1\).
Let
$$r_{4}>\max \biggl\{ r_{3}, \overline{r}_{4}, \frac{G_{M}S_{\overline{r}_{4}}\int_{0}^{+\infty}p(s)v(s)\,\mathrm{d}s +F_{1M}S'_{\overline{r}_{4}}\int_{0}^{+\infty}\psi_{1}(s)\,\mathrm{d}s +F_{2M}S''_{\overline{r}_{4}}\int_{0}^{+\infty}\psi_{2}(s)\,\mathrm{d}s}{ 1-A(h^{+\infty}+g_{1}^{+\infty}+g_{2}^{+\infty}+\varepsilon_{4})} \biggr\} . $$
For any \(t\in[0,+\infty)\), \(u\in\partial K_{r_{4}}\), we denote
$$\begin{aligned}& D_{1}=\bigl\{ t\in [0,+\infty ): u(t)\geq\overline{r}_{4}, u \in\partial K_{r_{4}} \bigr\} , \\& D_{2}=\bigl\{ t\in [0,+\infty ): 0\leq u(t)\leq \overline{r}_{4} u\in\partial K_{r_{4}}\bigr\} . \end{aligned}$$
We have
$$\begin{aligned} \bigl\vert Tu(t)\bigr\vert =& \biggl\vert \int_{0}^{+\infty}G(t,s)p(s)f\bigl(s,u(s)\bigr)\, \mathrm{d}s +F_{1}(t) \int_{0}^{+\infty}g_{1}\bigl(u(s)\bigr) \psi_{1}(s)\,\mathrm{d}s \\ &{}+F_{2}(t) \int_{0}^{+\infty}g_{2}\bigl(u(s)\bigr) \psi_{2}(s)\,\mathrm{d}s\biggr\vert \\ \leq& \int_{D_{1}}G_{M}p(s)v(s)h\bigl(u(s)\bigr)\,\mathrm{d}s +F_{1}(t) \int_{D_{1}}g_{1}\bigl(u(s)\bigr)\psi_{1}(s)\, \mathrm{d}s \\ &{}+F_{2}(t) \int_{D_{1}}g_{2}\bigl(u(s)\bigr)\psi_{2}(s)\, \mathrm{d}s \\ &{}+ \int_{D_{2}}G_{M}p(s)v(s)h\bigl(u(s)\bigr)\,\mathrm{d}s +F_{1}(t) \int_{D_{2}}g_{1}\bigl(u(s)\bigr)\psi_{1}(s)\, \mathrm{d}s \\ &{}+F_{2}(t) \int_{D_{2}}g_{2}\bigl(u(s)\bigr)\psi_{2}(s)\, \mathrm{d}s \\ \leq& A\bigl(h^{+\infty}+g_{1}^{+\infty}+g_{2}^{+\infty}+ \varepsilon_{4}\bigr)\|u\| +G_{M}S_{\overline{r}_{4}} \int_{0}^{+\infty}p(s)v(s)\,\mathrm{d}s \\ &{}+F_{1M}S'_{\overline{r}_{4}} \int_{0}^{+\infty}\psi_{1}(s)\,\mathrm{d}s +F_{2M}S''_{\overline{r}_{4}} \int_{0}^{+\infty}\psi_{2}(s)\,\mathrm{d}s \\ \leq& \|u\|. \end{aligned}$$
Hence, by using (ii) of Lemma 2.7 and Lemma 2.3, the boundary value problem (1.1) has at least one positive solution \(u\in\overline{K}_{r_{4}}\setminus K_{r_{3}}\). □

Remark 3.2

It follows from the proof of Theorem 3.2 that the boundary value problem (1.1) has one positive solution \(u\in P\) if at least one of the conditions \(f_{0}=+\infty\), \(g_{1,0}=+\infty\), and \(g_{2,0}=+\infty\) holds.

Theorem 3.3

Suppose that (H0) and (H1) hold. If
  1. (1)

    \(B(f_{0}+g_{1,0}+g_{2,0})>1\), \(B(f_{+\infty}+g_{1,+\infty }+g_{2,+\infty})>1\) and

     
  2. (2)

    there exists a constant \(c>0\) such that \(\max\{S_{c},S'_{c},S''_{c}\} :=S^{*}< A^{-1}c\),

     
then the boundary value problem (1.1) has at least two positive solutions.

Proof

Since \(B(f_{0}+g_{1,0}+g_{2,0})>1\), similarly to the proof of Theorem 3.2, there exists a constant \(0< r< c\) with
$$\|Tu\|\geq\|u\|, \quad \|u\|=r. $$
Since \(B(f_{+\infty}+g_{1,+\infty}+g_{2,+\infty})>1\), there also exists a constant \(R>c\) such that
$$\|Tu\|\geq\|u\|,\quad \|u\|=R. $$
On the other hand, by condition (2), for any \(u\in\partial K_{c}\),
$$\begin{aligned} \|Tu\| =& \sup_{t\in [0,+\infty )}\biggl\vert \int_{0}^{+\infty }G(t,s)p(s)f\bigl(s,u(s)\bigr)\, \mathrm{d}s +F_{1}(t) \int_{0}^{+\infty}g_{1}\bigl(u(s)\bigr) \psi_{1}(s)\,\mathrm{d}s \\ &{} +F_{2}(t) \int_{0}^{+\infty}g_{2}\bigl(u(s)\bigr) \psi_{2}(s)\,\mathrm{d}s\biggr\vert \\ \leq& \int_{0}^{+\infty}G_{M}p(s)v(s)h\bigl(u(s)\bigr) \,\mathrm{d}s +F_{1M} \int_{0}^{+\infty}g_{1}\bigl(u(s)\bigr) \psi_{1}(s)\,\mathrm{d}s \\ &{}+F_{2M} \int_{0}^{+\infty}g_{2}\bigl(u(s)\bigr) \psi_{2}(s)\,\mathrm{d}s \\ \leq& G_{M}S_{c} \int_{0}^{+\infty}p(s)v(s)\, \mathrm{d}s+F_{1M}S'_{c} \int_{0}^{+\infty}\psi _{1}(s)\,\mathrm{d}s +F_{2M}S''_{c} \int_{0}^{+\infty}\psi_{2}(s)\,\mathrm{d}s \\ \leq& S^{*} \biggl(G_{M} \int_{0}^{+\infty}p(s)v(s)\,\mathrm{d}s +F_{1M} \int_{0}^{+\infty}\psi_{1}(s)\,\mathrm{d}s +F_{2M} \int_{0}^{+\infty}\psi_{2}(s)\,\mathrm{d}s \biggr) \\ < & c. \end{aligned}$$
Namely,
$$\|Tu\|< c=\|u\|, \quad u\in\partial K_{c}. $$
According to Lemma 2.7, the boundary value problem (1.1) has at least two positive solutions \(u_{1}\), \(u_{2}\) with \(0<\|u_{1}\| <c<\|u_{2}\|\). □

Remark 3.3

It follows from the proof of Theorem 3.3 that the boundary value problem (1.1) has at least two positive solutions \(u\in P\) if one of the conditions \(f_{0}=+\infty \), \(g_{1,{0}}=+\infty\), \(g_{2,{0}}=+\infty\), \(f_{+\infty}=+\infty\), \(g_{1,{+\infty}}=+\infty\), and \(g_{2,{+\infty}}=+\infty\) holds.

Theorem 3.4

Suppose that (H0) and (H1) hold. If
  1. (1)

    \(A(h^{0}+g_{1}^{0}+g_{2}^{0})<1\), \(A(h^{+\infty}+g_{1}^{+\infty}+g_{2}^{+\infty })<1\), and

     
  2. (2)
    there exists a constant \(C>0\) such that, for any \(t\in[l_{1},l_{2}]\) and \(u\in[\gamma_{0}C,C]\), we have
    $$\min\bigl\{ f(t,u),g_{1}(u),g_{2}(u)\bigr\} > \gamma_{0}B^{-1}C, $$
     
then the boundary value problem (1.1) has at least two positive solutions.

Proof

The proof is similar to that of Theorem 3.3.

It is easy to get the two positive solutions \(u_{3}\), \(u_{4}\) with \(0<\|u_{3}\| <C<\|u_{4}\|\). □

4 Illustration

Example

We consider the following boundary value problem:
$$ \textstyle\begin{cases} \frac{1}{\mathrm{e}^{t}}(\mathrm{e}^{t} \,{}^{\mathrm{C}} D_{0+}^{\frac{1}{2}}u(t))' +f(t,u(t))=0,\quad t\in(0,{+\infty}), \\ u(0)- \lim_{t\to0^{+}}p(t) \,{}^{\mathrm{C}} D_{0+}^{\alpha}u(t)=\int_{0}^{+\infty} \frac{g_{1}(u(s))}{1+s^{2}}\,\mathrm{d}s, \\ \lim_{t\to{+\infty}}u(t)+2\lim_{t\to{+\infty}}p(t) \,{}^{\mathrm{C}} D_{0+}^{\alpha}u(t)=\int_{0}^{+\infty} \frac{ g_{2}(u(s))}{1+s^{2}}\,\mathrm{d}s, \end{cases} $$
(4.1)
where \(f(t,u)=\frac{\mathrm{e}^{-3t}(1+t)(u+\mathrm{e}^{-u})}{10\sqrt {t}}\), \(a_{1}=1\), \(a_{2}=1\), \(b_{1}=1\), \(b_{2}=2\), \(p(t)=\mathrm{e}^{t}\), \(\psi _{1}(s)=\psi_{2}(s)=\frac{1}{1+s^{2}}\), and
$$g_{1}(u)=g_{2}(u)= \textstyle\begin{cases} \frac{u}{10}, &0\leq u\leq100, \\ 60\sqrt{u}-590, &100< u< +\infty. \end{cases} $$

It is obvious that \(f:(0,{+\infty})\times[0,{+\infty})\rightarrow [0,{+\infty}) \) is a continuous function and singular at \(t=0\).

Let
$$v(t)=\frac{\mathrm{e}^{-3t}(1+t)}{\sqrt{t}},\qquad h(u)=\frac{u+\mathrm{e}^{-u}}{10}. $$
Then we have \(M\approx0.610503\), \(A=2.19551\), \(h^{+\infty}=\frac {1}{10}\), \(g_{1}^{+\infty}=g_{2}^{+\infty}=0\), and \(A(h^{+\infty }+g_{1}^{+\infty}+g_{2}^{+\infty})=0.219551<1\).

On the other hand, let \(l_{1}=\frac{1}{2}\), \(l_{2}=1\). So we have \(B\approx 0.03\), \(f_{0}=+\infty\), \(g_{i,0}=\frac{1}{10}\), \(i=1,2\). Namely, \(B(f_{0}+g_{1,0}+g_{2,0})=+\infty>1\).

By using Theorem 3.2 the boundary value problem (4.1) has at least one positive solution.

Declarations

Acknowledgements

This work is supported by the National Natural Science Foundation of China (No. 11171220) and the Hujiang Foundation of China (B14005).

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
College of Science, University of Shanghai for Science and Technology

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