- Research
- Open Access
Positive solutions for singular nonlinear fractional differential equation with integral boundary conditions
- Hongdan Li1,
- Lishan Liu1, 2Email author and
- Yonghong Wu2
Received: 26 August 2015
Accepted: 23 November 2015
Published: 9 December 2015
Abstract
In this article, we study the existence of positive solutions for a class of singular nonlinear fractional differential equations with Riemann-Stieltjes integral boundary conditions. Using the properties of the Green function and the fixed point theory in cones, we obtain some results on the existence of positive solutions. Our results extend and improve many known results including singular and nonsingular cases.
Keywords
- singular fractional differential equations
- Riemann-Stieltjes integral boundary value problem
- positive solution
- multiple positive solutions
- fixed point theorem in cone
MSC
- 34A08
- 34B16
- 34B18
1 Introduction
In this article, we consider the existence of solutions for the following class of singular fractional differential equations: where \(n-1<\alpha\leq n\), \(n\geq2\), \(p, q\in C(0, 1)\), \(p(t)\) and \(q(t)\) are allowed to be singular at \(t=0\) or \(t=1\), \(f, g:[0, 1]\times(0, \infty) \rightarrow[0, \infty)\) are continuous and may be singular at \(x=0\), \(h: (0, 1)\rightarrow[0, \infty)\) is continuous with \(h\in L^{1}(0, 1)\), and \(\int_{0}^{1}h(s)u(s)\,dA(s)\) denotes the Riemann-Stieltjes integral with a signed measure, in which \(A: [0, 1]\rightarrow\mathbb{R}\) is a function of bounded variation.
$$ \left \{ \textstyle\begin{array}{@{}l} D_{0^{+}}^{\alpha}u(t)+p(t)f(t, u(t))+q(t)g(t, u(t))=0,\quad 0< t< 1,\\ u(0)=u'(0)=\cdots=u^{(n-2)}(0)=0, \\ u(1)=\int_{0}^{1}h(s)u(s)\,dA(s), \end{array}\displaystyle \right . $$
(1.1)
Fractional differential equations have attracted more and more attention from the research communities due to their numerous applications in many fields of science and engineering including fluid flow, rheology, diffusive transport akin to diffusion, electrical networks, probability, etc. For details, see [1–6] and the references therein. On the other hand, boundary value problems with integral boundary conditions for ordinary differential equations arise in many fields of applied mathematics and physics such as heat conduction, chemical engineering, underground water flow, thermoelasticity, and plasma physics. The existence and multiplicity of positive solutions for such problems have become an important area of investigation in recent years.
Cabada and Hamdi [7] studied the existence of positive solutions of the following nonlinear fractional differential equation with integral boundary value conditions: where \(2<\alpha\leq3\), \(0<\lambda\), \(\lambda\neq\alpha\), \(D_{0^{+}}^{\alpha} \) is the Riemann-Liouville fractional derivative, and \(f: [0, 1]\times[0, \infty)\rightarrow[0, \infty)\) is a continuous function.
$$\left \{ \textstyle\begin{array}{@{}l} D_{0^{+}}^{\alpha}u(t)+f(t, u(t))=0,\quad 0< t< 1, \\ u(0)=u'(0)=0, \qquad u(1)=\lambda\int_{0}^{1}u(s)\,ds, \end{array}\displaystyle \right . $$
By means of the monotone iteration method, Sun and Zhao [8] investigated the existence of positive solutions for the fractional differential equation with integral boundary conditions where \(2<\alpha\leq3\), \(D_{0^{+}}^{\alpha}\) is the standard Riemann-Liouville derivative of order α, \(f: [0, 1]\times[0, \infty)\rightarrow[0, \infty)\) is continuous, and \(g, q: (0, 1)\rightarrow[0, \infty)\) are also continuous with \(g, q\in L^{1}(0, 1)\).
$$\left \{ \textstyle\begin{array}{@{}l} D_{0^{+}}^{\alpha}u(t)+q(t)f(t, u(t))=0,\quad 0< t< 1 , \\ u(0)=u'(0)=0,\qquad u(1)=\int_{0}^{1}g(s)u(s)\,ds, \end{array}\displaystyle \right . $$
Zhang et al. [9] considered the following nonlinear fractional differential equation with integral boundary conditions: where \(3<\alpha\leq4\), \(0<\eta\leq1\), \(0\leq\frac{\lambda\eta^{\alpha }}{\alpha}<1\), \(D_{0^{+}}^{\alpha} \) is the Riemann-Liouville fractional derivative, \(h: (0, 1)\rightarrow[0, \infty)\) is continuous, and \(f: [0, 1]\times[0, \infty)\rightarrow[0, \infty)\) is also continuous.
$$\left \{ \textstyle\begin{array}{@{}l} D_{0^{+}}^{\alpha}u(t)+h(t)f(t, u(t))=0, \quad 0< t< 1 , \\ u(0)=u'(0)=u''(0)=0,\qquad u(1)=\lambda\int_{0}^{\eta}u(s)\,ds, \end{array}\displaystyle \right . $$
Motivated by the above-mentioned papers, the purpose of this article is to investigate the existence of positive solutions for the more general fractional differential equations BVP (1.1) under suitable conditions on f and g. Obviously, our discussion is different from that in [7–11]. The main new features presented in this paper are as follows. First, the boundary value problem has a more general form, in which p, q are allowed to be singular at \(t=0, 1\) and f, g may be singular at \(x=0\). Second, the existence and multiplicity of positive solutions of BVP (1.1) are obtained. Third, \(\int_{0}^{1}u(s)\,dA(s)\) denotes the Riemann-Stieltjes integral, where A is a function of bounded variation, and dA may be a signed measure. The work includes the multipoint boundary problems and integral boundary problems as special cases.
The rest of the paper is organized as follows. In Section 2, we present some preliminaries and lemmas that are used to prove our main results. We also develop some properties of the Green function. In Section 3, we discuss the existence of a positive solution of BVP (1.1). In Section 4, we discuss the existence of multiple positive solutions of BVP (1.1). In Section 5, we give an example to demonstrate an application of our theoretical results.
2 Preliminaries and lemmas
In this section, for the convenience of reader, we present some notation and lemmas that will be used in the proof of our main results. They can be found in the literature; see [5, 6, 12, 13].
Definition 2.1
The Riemann-Liouville fractional integral of order \(\alpha>0\) of a function \(y : (0, \infty) \rightarrow\mathbb{R} \) is given by provided that the right-hand side is pointwise defined on \((0, \infty )\).
$$I_{0^{+}}^{\alpha}y(t)=\frac{1}{\Gamma(\alpha)} \int _{0}^{t}(t-s)^{\alpha-1}y(s)\,ds, $$
Definition 2.2
The Riemann-Liouville fractional derivative of order \(\alpha>0\) of a continuous function \(y : (0, \infty) \rightarrow\mathbb{R}\) is given by where \(n = [\alpha]+1\) with \([\alpha]\) denoting the integer part of a number α, provided that the right-hand side is pointwise defined on \((0, \infty)\).
$$D_{0^{+}}^{\alpha}y(t)=\frac{1}{\Gamma(n-\alpha)} \biggl( \frac{d}{dt} \biggr) ^{n} \int_{0}^{t}\frac{y(s)}{(t-s)^{\alpha-n+1}}\,ds, $$
Lemma 2.1
[6]
Let
\(\alpha>0\). If
\(u \in C(0, 1) \cap L^{1}(0, 1)\), then the fractional differential equation
has
as the unique solution, where
\(N = [\alpha]+1\).
$$D_{0^{+}}^{\alpha}u(t)=0 $$
$$u(t) = C_{1}t^{\alpha-1}+C_{2}t^{\alpha-2}+ \cdots+C_{N}t^{\alpha -N}, \quad C_{i} \in\mathbb{R}, i =1, 2, \dots, N, $$
From the definition of the Riemann-Liouville derivative we can obtain the following statement.
Lemma 2.2
[6]
Assume that
\(u \in C(0, 1) \cap L^{1}(0, 1)\)
with a fractional derivative of order
\(\alpha>0\)
that belongs to
\(C(0, 1) \cap L^{1}(0, 1)\). Then
for some
\(C_{i}\in\mathbb{R} \) (\(i =1, 2, \dots, N\)), where
\(N = [\alpha]+1\).
$$I_{0^{+}}^{\alpha}D_{0^{+}}^{\alpha}u(t)=u(t)+ C_{1}t^{\alpha -1}+C_{2}t^{\alpha-2}+ \cdots+C_{N}t^{\alpha-N} $$
In the following, we present the Green function of the fractional differential equation boundary value problem.
Lemma 2.3
Given
\(y\in C(0, 1) \cap L^{1}(0, 1)\), \(n-1< \alpha\leq n\), the problem
is equivalent to
where
with
$$ \left \{ \textstyle\begin{array}{@{}l} D_{0^{+}}^{\alpha}u(t)+y(t)=0, \quad 0< t< 1, \\ u(0)=u'(0)=\cdots=u^{(n-2)}(0)=0, \\ u(1)=\int_{0}^{1}h(s)u(s)\,dA(s) \end{array}\displaystyle \right . $$
(2.1)
$$u(t)= \int_{0}^{1}G(t, s)y(s)\,ds, $$
$$ G(t, s)=G_{0}(t, s)+\frac{t^{\alpha-1}}{1-\Gamma}g_{A}(s) $$
(2.2)
$$\begin{aligned} &G_{0}(t, s)=\frac{1}{\Gamma(\alpha)} \left \{ \textstyle\begin{array}{@{}l@{\quad}l} {[}t(1-s)]^{\alpha-1}-(t-s)^{\alpha-1}, & 0\leq s\leq t\leq1 , \\ {[}t(1-s)]^{\alpha-1}, & 0\leq t\leq s\leq1 , \end{array}\displaystyle \right .\\ &\Gamma= \int_{0}^{1}t^{\alpha-1}h(t)\,dA(t),\qquad g_{A}(s)= \int_{0}^{1}G_{0}(t, s)h(t)\,dA(t). \end{aligned}$$
Proof
We may apply Lemma 2.2 to reduce (2.1) to the equivalent integral equation for some \(C_{i}\in\mathbb{R} \) (\(i =1, 2, \ldots, n\)). Consequently, the general solution of (2.1) is Since \(u(0)=u'(0)=\cdots=u^{(n-2)}(0)=0\), we get that \(C_{2}=C_{3}=\cdots=C_{n}=0\) by (2.3). Then, we obtain On the other hand, we can combine (2.1) with yielding So, by (2.4) and (2.5) we have Let Then Multiplying both sides of (2.6) by \(h(t)\) and then integrating from 0 to 1, we have Then, by (2.7) we obtain Let From (2.8) and (2.9) we deduce that Substituting (2.10) into (2.6), we obtain The proof is complete. □
$$u(t)=-I_{0^{+}}^{\alpha}y(t)+ C_{1}t^{\alpha-1}+C_{2}t^{\alpha -2}+ \cdots+C_{n}t^{\alpha-n} $$
$$ u(t)=-\frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha -1}y(s)\,ds+C_{1}t^{\alpha-1}+C_{2}t^{\alpha-2}+ \cdots+C_{n}t^{\alpha -n}. $$
(2.3)
$$ u(t)=-\frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha -1}y(s)\,ds+C_{1}t^{\alpha-1}. $$
(2.4)
$$u(1)=-\frac{1}{\Gamma(\alpha)} \int_{0}^{1}(1-s)^{\alpha-1}y(s)\,ds+C_{1}, $$
$$ C_{1}=\frac{1}{\Gamma(\alpha)} \int_{0}^{1}(1-s)^{\alpha-1}y(s)\,ds+ \int _{0}^{1}h(s)u(s)\,dA(s). $$
(2.5)
$$\begin{aligned} u(t) ={}&{-}\frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha-1}y(s)\,ds+ \frac {t^{\alpha-1}}{\Gamma(\alpha)} \int_{0}^{1}(1-s)^{\alpha-1}y(s)\,ds \\ &{}+ t^{\alpha-1} \int_{0}^{1}h(s)u(s)\,dA(s) \\ ={}&\frac{1}{\Gamma(\alpha)} \int_{0}^{t}\bigl[t^{\alpha-1}(1-s)^{\alpha -1}-(t-s)^{\alpha-1} \bigr]y(s)\,ds \\ &{}+\frac{1}{\Gamma(\alpha)} \int_{t}^{1}t^{\alpha-1}(1-s)^{\alpha -1}y(s)\,ds+t^{\alpha-1} \int_{0}^{1}h(s)u(s)\,dA(s). \end{aligned}$$
$$G_{0}(t, s)=\frac{1}{\Gamma(\alpha)} \left \{ \textstyle\begin{array}{@{}l@{\quad}l} {[}t(1-s)]^{\alpha-1}-(t-s)^{\alpha-1}, & 0\leq s\leq t\leq1 , \\ {[}t(1-s)]^{\alpha-1}, & 0\leq t\leq s\leq1 . \end{array}\displaystyle \right . $$
$$ u(t)= \int_{0}^{1}G_{0}(t, s)y(s)\,ds+t^{\alpha-1} \int _{0}^{1}h(s)u(s)\,dA(s). $$
(2.6)
$$\begin{aligned} \int_{0}^{1}h(t)u(t)\,dA(t)={}& \int_{0}^{1}h(t) \int_{0}^{1}G_{0}(t, s)y(s)\,ds\,dA(t) \\ &{}+ \int_{0}^{1}t^{\alpha-1}h(t) \int _{0}^{1}h(s)u(s)\,dA(s)\,dA(t). \end{aligned}$$
(2.7)
$$ \biggl(1- \int_{0}^{1}t^{\alpha-1}h(t)\,dA(t) \biggr) \int _{0}^{1}h(t)u(t)\,dA(t)= \int_{0}^{1} \int_{0}^{1}h(t)G_{0}(t, s)\,dA(t)y(s)\,ds. $$
(2.8)
$$ \Gamma= \int_{0}^{1}t^{\alpha-1}h(t)\,dA(t), \qquad g_{A}(s)= \int _{0}^{1}G_{0}(t, s)h(t)\,dA(t). $$
(2.9)
$$ \int_{0}^{1}h(t)u(t)\,dA(t) =\frac{1}{1-\Gamma} \int _{0}^{1}g_{A}(s)y(s)\,ds. $$
(2.10)
$$u(t)= \int_{0}^{1}G_{0}(t, s)y(s)\,ds+ \frac{t^{\alpha-1}}{1-\Gamma} \int _{0}^{1}g_{A}(s)y(s)\,ds = \int_{0}^{1}G(t, s)y(s)\,ds . $$
Lemma 2.4
Let
\(0\leq\Gamma<1\)
and
\(g_{A}(s)\geq 0\)
for
\(s\in[0, 1]\). Then the Green function
\(G(t, s)\)
defined by (2.2) satisfies
- (1)
\(G:[0, 1]\times[0, 1]\rightarrow[0, \infty)\) is continuous.
- (2)For any \(t, s \in[0, 1]\), we have \(c_{0}(t)\phi(s) \leq G(t, s)\leq\phi(s)\), where$$\begin{aligned}& \phi(s)=\phi_{0}(s)+\frac{g_{A}(s)}{1-\Gamma},\qquad \phi_{0}(s)= \frac {\tau(s)^{\alpha-2}s(1-s)^{\alpha-1}}{\Gamma(\alpha-1)}, \\& \tau(s)=\frac{s}{1-(1-s)^{\frac{\alpha-1}{\alpha-2}}},\qquad c_{0}(t)=\min \biggl\{ \frac{(\alpha-1)^{\alpha-1}t^{\alpha -2}(1-t)}{(\alpha-2)^{\alpha-2}}, t^{\alpha-1} \biggr\} . \end{aligned}$$
- (3)
Then, taking \(\theta\in(0, \frac{1}{2})\), for any \(t\in [\theta, 1-\theta]\) and \(s\in[0, 1]\), we have \(G(t, s)\geq\gamma _{0}\phi(s)\), where \(\gamma_{0}=\min_{\theta\leq t\leq1-\theta }c_{0}(t)\in(0,1)\).
The proof of Lemma 2.4 is similar to that in [14], so we omit it.
Remark 2.1
By Lemma 2.4 we can obtain that \(\frac{\alpha -2}{\alpha-1}\leq\tau(s)\leq1\), \(\phi_{0}(s)\leq\frac{1}{\alpha-1}\). That is, \(\phi(s)\) is bounded, and there exists a constant \(\overline {g}>0\) such that \(\phi(s)\leq\overline{g}\). It is easy to verify that \(\max_{0\leq t\leq1}c_{0}(t)\leq1\).
Let \(E=C[0, 1]\) and \(\|u\|=\sup_{0\leq t\leq1}\mid u(t)\mid\). Then \((E, \|\cdot\|)\) is a Banach space. Let and and let \(K_{r}=\{u\in K: \|u\|< r \}\). It is easy to see that K is a cone in E and \(\overline{K_{R}}\setminus K_{r}\subset K\subset P\).
$$P=\bigl\{ u\in E:u(t)\geq0, t\in[0, 1]\bigr\} $$
$$K=\bigl\{ u\in P:u(t)\geq c_{0}(t)\|u\|, t\in[0, 1]; u(t)\geq \gamma_{0}\| u\|, t\in[\theta, 1-\theta]\bigr\} , $$
Throughout the paper, we need the following conditions:
- (H1):
-
\(A: [0, 1]\rightarrow\mathbb{R}\) is a function of bounded variation, and \(g_{A}(s)\geq0\) for all \(s\in[0, 1]\);
- (H2):
-
\(h\in C(0, 1)\cap L^{1}(0, 1)\), and \(0\leq\Gamma =\int_{0}^{1}t^{\alpha-1}h(s)\,dA(s)<1\);
- (H3):
-
\(p, q:(0, 1)\rightarrow[0, \infty)\) are continuous, \(p(t)\not\equiv0\), \(q(t)\not\equiv0\), \(t\in[0, 1]\), and$$\int_{0}^{1}\phi(s)p(s)\,ds< +\infty,\qquad \int_{0}^{1}\phi(s)q(s)\,ds< +\infty; $$
- (H4):
-
\(f, g:[0, 1]\times(0, \infty)\rightarrow[0, \infty)\) are continuous, and for any \(0< r< R<+\infty\), \(\lim_{m\rightarrow \infty}\sup_{u\in\overline{K_{R}} \setminus K_{r}}\int_{H(m)}(p(s)f(s, u(s))+q(s)g(s, u(s)))\,ds=0\), where \(H(m)=[0, \frac{1}{m}]\cup [\frac{m-1}{m}, 1]\).
In what follows, let us define the nonlinear operator \(T:\overline {K_{R}} \setminus K_{r}\rightarrow K\) by
$$ (Tu) (t)= \int_{0}^{1}G(t, s)\bigl[p(s)f\bigl(s, u(s) \bigr)+q(s)g\bigl(s, u(s)\bigr)\bigr]\,ds,\quad t\in[0, 1]. $$
(2.11)
Lemma 2.5
Suppose that (H1)-(H4) hold. Then \(T:\overline{K_{R}} \setminus K_{r}\rightarrow K\) is a completely continuous operator and the fixed point of T in \(\overline{K_{R}} \setminus K_{r}\) is the positive solution to BVP (1.1).
Proof
It follows from (H4) that there exists a natural number \(m\geq2\) such that It is easy to see that for any \(u\in\overline{K_{R}} \setminus K_{r}\) and \(t\in[\frac{1}{m}, \frac{m-1}{m}]\), we have \(\lambda_{0}r\leq u(t)\leq R\), where \(\lambda_{0}=\min\{c_{0}(t): \frac{1}{m}\leq t\leq \frac{m-1}{m}\}>0\). Let So, by (H3), (H4), and Remark 2.1 we have This implies that the operator T defined by (2.11) is well defined.
$$\sup_{u\in\overline{K_{R}} \setminus K_{r}} \int_{H(m)}p(s)f\bigl(s, u(s)\bigr)+q(s)g\bigl(s, u(s)\bigr)\,ds< 1. $$
$$\begin{aligned}& M_{1}=\max \biggl\{ f(t, x):\frac{1}{m}\leq t\leq \frac{m-1}{m} , \lambda _{0}r\leq x\leq R \biggr\} , \\& M_{2}=\max \biggl\{ g(t, x):\frac{1}{m}\leq t\leq \frac{m-1}{m} , \lambda _{0}r\leq x\leq R \biggr\} . \end{aligned}$$
$$\begin{aligned} & \sup_{u\in\overline{K_{R}} \setminus K_{r}} \int_{0}^{1}G(t, s)\bigl[p(s)f\bigl(s, u(s) \bigr)+q(s)g\bigl(s, u(s)\bigr)\bigr]\,ds \\ &\quad\leq\sup_{u\in\overline{K_{R}} \setminus K_{r}} \int_{0}^{1}\phi (s)\bigl[p(s)f\bigl(s, u(s) \bigr)+q(s)g\bigl(s, u(s)\bigr)\bigr]\,ds \\ &\quad\leq\overline{g}\sup_{u\in\overline{K_{R}} \setminus K_{r}} \int _{H(m)}p(s)f\bigl(s, u(s)\bigr)+q(s)g\bigl(s, u(s)\bigr)\,ds \\ &\qquad{} +\sup_{u\in\overline{K_{R}}\setminus K_{r}} \int_{\frac{1}{m}}^{\frac {m-1}{m}}\phi(s)\bigl[p(s)f\bigl(s, u(s) \bigr)+q(s)g\bigl(s, u(s)\bigr)\bigr]\,ds \\ &\quad\leq\overline{g}+(M_{1}+M_{2}) \int_{\frac{1}{m}}^{\frac{m-1}{m}}\phi (s) \bigl(p(s)+q(s)\bigr)\,ds \\ &\quad\leq\overline{g}+(M_{1}+M_{2}) \int_{0}^{1}\phi(s) \bigl(p(s)+q(s)\bigr)\,ds< +\infty. \end{aligned}$$
Now, we show that \(T:\overline{K_{R}}\setminus K_{r}\subset K\). For any \(u\in\overline{K_{R}}\setminus K_{r}\), \(t\in[0, 1]\), we have Hence, On the other hand, by Lemma 2.4 we have and at the same time, we can get Thus, \(Tu\in K\). Consequently, \(T:\overline{K_{R}}\setminus K_{r}\subset K\). Finally, we prove that T, which maps \(\overline{K_{R}}\setminus K_{r}\) into K, is a completely continuous map. Let \(D\subset\overline {K_{R}}\setminus K_{r}\) be an arbitrary bounded set. Then from the previous proof we know that \(T(D)\) is uniformly bounded.
$$\begin{aligned} (Tu) (t) &= \int_{0}^{1}G(t, s)\bigl[p(s)f\bigl(s, u(s) \bigr)+q(s)g\bigl(s, u(s)\bigr)\bigr]\,ds \\ &\leq \int_{0}^{1}\phi(s)\bigl[p(s)f\bigl(s, u(s) \bigr)+q(s)g\bigl(s, u(s)\bigr)\bigr]\,ds. \end{aligned}$$
$$\|Tu\|\leq \int_{0}^{1}\phi(s)\bigl[p(s)f\bigl(s, u(s) \bigr)+q(s)g\bigl(s, u(s)\bigr)\bigr]\,ds. $$
$$(Tu) (t)\geq c_{0}(t) \int_{0}^{1}\phi(s)\bigl[p(s)f\bigl(s, u(s) \bigr)+q(s)g\bigl(s, u(s)\bigr)\bigr]\,ds \geq c_{0}(t)\|Tu\|, \quad t \in[0, 1], $$
$$(Tu) (t)\geq\gamma_{0}\|Tu\|, \quad t\in[\theta, 1-\theta]. $$
Next, we show \(T(D)\) is equicontinuous. In fact, for any \(\epsilon>0\), there exists a natural number \(m\geq2\) such that Since \(G(t, s)\) is uniformly continuous on \([0, 1]\times[0, 1]\), for the above \(\epsilon>0\), there exists \(\delta>0\) such that for any \(t_{1}, t_{2}\in[0, 1]\), \(|t_{1}-t_{2}|<\delta\), \(s\in[\frac{1}{m}, \frac{m-1}{m}]\), Consequently, for any \(u\in D\), \(t_{1}, t_{2}\in[0, 1]\), \(|t_{1}-t_{2}|<\delta\), we have where This shows that \(T(D)\) is equicontinuous. By the Arzela-Ascoli theorem, \(T:D\rightarrow K\) is compact.
$$ \sup_{u\in\overline{K_{R}} \setminus K_{r}} \int_{H(m)}p(s)f\bigl(s, u(s)\bigr)+q(s)g\bigl(s, u(s)\bigr)\,ds< \frac{\epsilon}{4\overline{g}}. $$
(2.12)
$$\bigl|G(t_{1}, s)-G(t_{2}, s)\bigr|< \frac{\epsilon}{2(\overline{p}M_{1}+\overline {q}M_{2})}. $$
$$\begin{aligned} &\bigl|(Tu) (t_{1})-(Tu) (t_{2})\bigr| \\ &\quad=\biggl| \int_{0}^{1}\bigl(G(t_{1}, s)-G(t_{2}, s)\bigr) \bigl(p(s)f\bigl(s, u(s)\bigr)+q(s)g\bigl(s, u(s) \bigr)\bigr)\,ds\biggr| \\ &\quad\leq 2 \int_{H(m)}\phi(s)\bigl[p(s)f\bigl(s, u(s)\bigr)+q(s)g\bigl(s, u(s)\bigr)\bigr]\,ds \\ &\qquad{}+ \sup_{u\in D} \int_{\frac{1}{m}}^{\frac{m-1}{m}}\bigl|\bigl(G(t_{1}, s)-G(t_{2}, s)\bigr)\bigr|\bigl(p(s)f\bigl(s, u(s)\bigr)+q(s)g\bigl(s, u(s) \bigr)\bigr)\,ds \\ &\quad< 2\overline{g}\frac{\epsilon}{4\overline{g}}+\frac{\epsilon }{2(\overline{p}M_{1}+\overline{q}M_{2})}(\overline{p}M_{1}+ \overline {q}M_{2}) \\ &\quad=\epsilon, \end{aligned}$$
$$\overline{p}=\max \biggl\{ p(t):\frac{1}{m}\leq t\leq\frac {m-1}{m} \biggr\} , \qquad \overline{q}=\max \biggl\{ q(t):\frac{1}{m}\leq t\leq \frac {m-1}{m} \biggr\} . $$
Finally, we prove that \(T:\overline{K_{R}}\setminus K_{r}\rightarrow K\) is continuous. Assume that \(u_{0}, u_{n}\in\overline{K_{R}}\setminus K_{r}\) (\(n=0,1,2,\dots\)) and \(\|u_{n}-u_{0}\|\rightarrow0\) (\(n\rightarrow \infty\)). Then \(r\leq\|u_{n}\|\leq R\) and \(r\leq\|u_{0}\|\leq R\). For any \(\epsilon>0\), by (H4) there exists a natural number \(m\geq 2\) such that On the other hand, for any \(t\in[\frac{1}{m}, \frac{m-1}{m}]\), we have \(\gamma_{0}r\leq u_{n}(t)\leq R \) (\(n=0, 1, 2,\dots\)). Since \(f(t, u)\), \(g(t, u)\) are uniformly continuous in \([\frac{1}{m}, \frac{m-1}{m}]\times[\gamma_{0}r, R]\), we have that uniformly on \([\frac{1}{m}, \frac{m-1}{m}]\). Then the Lebesgue dominated convergence theorem yields that Thus, for the above \(\epsilon>0\), there exists a natural number N such that for \(n>N\), we have It follows from (2.12), (2.13) that for \(n>N\), This implies that \(T:\overline{K_{R}}\setminus K_{r}\rightarrow K\) is continuous. Thus, \(T:\overline{K_{R}}\setminus K_{r}\rightarrow K\) is completely continuous. It is clear that if u is a fixed point of T in \(\overline{K_{R}}\setminus K_{r}\), then u satisfies (1.1) and is a positive solution of BVP (1.1). □
$$\sup_{u\in\overline{K_{R}} \setminus K_{r}} \int_{H(m)}p(s)f\bigl(s, u(s)\bigr)+q(s)g\bigl(s, u(s)\bigr)\,ds< \frac{\epsilon}{\overline{g}}. $$
$$\begin{aligned}& \lim_{n\rightarrow\infty}\bigl|f\bigl(s, u_{n}(s)\bigr)-f\bigl(s, u_{0}(s)\bigr)\bigr|=0, \\& \lim_{n\rightarrow\infty}\bigl|g\bigl(s, u_{n}(s)\bigr)-g\bigl(s, u_{0}(s)\bigr)\bigr|=0 \end{aligned}$$
$$\begin{aligned} &\int_{\frac{1}{m}}^{\frac{m-1}{m}}\phi(s)\bigl[p(s) \bigl(f\bigl(s, u_{n}(s)\bigr)-f\bigl(s, u_{0}(s)\bigr)\bigr)+q(s) \bigl(g \bigl(s, u_{n}(s)\bigr)-g\bigl(s, u_{0}(s)\bigr)\bigr) \bigr]\,ds\rightarrow0, \\ &\quad n\rightarrow\infty. \end{aligned}$$
$$ \int_{\frac{1}{m}}^{\frac{m-1}{m}}\phi(s)\bigl[p(s) \bigl(f\bigl(s, u_{n}(s)\bigr)-f\bigl(s, u_{0}(s)\bigr)\bigr)+q(s) \bigl(g \bigl(s, u_{n}(s)\bigr)-g\bigl(s, u_{0}(s)\bigr)\bigr) \bigr]\,ds< \frac{\epsilon}{2} . $$
(2.13)
$$\begin{aligned} \|Tu_{n}-Tu_{0}\| \leq{}&\sup_{u\in\overline{K_{R}}\setminus K_{r} } \int_{H(m)}\phi (s)\bigl[p(s) \bigl(f\bigl(s, u_{n}(s) \bigr)-f\bigl(s, u_{0}(s)\bigr)\bigr) \\ &{}+q(s) \bigl(g\bigl(s, u_{n}(s)\bigr)-g\bigl(s, u_{0}(s) \bigr)\bigr)\bigr]\,ds \\ &{}+\sup_{u\in\overline {K_{R}}\setminus K_{r}} \int_{\frac{1}{m}}^{\frac{m-1}{m}}\phi (s)\bigl[p(s) \bigl(f\bigl(s, u_{n}(s)\bigr) -f\bigl(s, u_{0}(s)\bigr)\bigr) \\ &{}+q(s) \bigl(g\bigl(s, u_{n}(s)\bigr)-g\bigl(s, u_{0}(s) \bigr)\bigr)\bigr]\,ds \\ < {}& \overline{g}\frac{\epsilon}{2\overline{g}}+\frac{\epsilon}{2} =\epsilon. \end{aligned}$$
To prove the main results, we need the following well-known fixed point theorem.
Lemma 2.6
[15]
Let
P
be a positive cone in a Banach space
E, \(\Omega_{1}\)
and
\(\Omega_{2}\)
be two bounded open sets in
E
such that
\(\theta\in\Omega_{1}\)
and
\(\overline{\Omega _{1}}\subset\Omega_{2}\), \(A:P\cap(\overline{\Omega_{2}}\setminus\Omega _{1})\rightarrow P\)
be a completely continuous operator, where
θ
denotes the zero element of
E. Suppose that one of the following two conditions holds:
- (i)
\(\|Au\|\leq\|u\|\), \(\forall u\in P\cap\partial\Omega _{1}\); \(\|Au\|\geq\|u\|\), \(\forall u\in P\cap\partial\Omega_{2}\);
- (ii)
\(\|Au\|\geq\|u\|\), \(\forall u\in P\cap\partial\Omega _{1}\); \(\|Au\|\leq\|u\|\), \(\forall u\in P\cap\partial\Omega_{2} \).
3 Existence of a positive solution
In the following, for convenience, we set
$$\begin{aligned}& \overline{f}_{0}=\limsup_{x\rightarrow0}\max _{t\in[0, 1]}\frac{f(t, x)}{x}, \qquad\overline{g}_{0}= \limsup_{x\rightarrow0}\max_{t\in[0, 1]}\frac{g(t, x)}{x}, \\& \underline{f}_{0}=\liminf_{x\rightarrow0}\min _{t\in[\theta, 1-\theta ]}\frac{f(t, x)}{x}, \qquad \underline{g}_{0}= \liminf_{x\rightarrow0}\min_{t\in[\theta, 1-\theta]}\frac{g(t, x)}{x}, \\& \overline{f}_{\infty}=\limsup_{x\rightarrow\infty}\max _{t\in[0, 1]}\frac {f(t, x)}{x}, \qquad\overline{g}_{\infty}= \limsup_{x\rightarrow\infty }\max_{t\in[0, 1]}\frac{g(t, x)}{x}, \\& \underline{f}_{\infty}=\liminf_{x\rightarrow\infty}\min _{t\in[\theta, 1-\theta]}\frac{f(t, x)}{x}, \qquad \underline{g}_{\infty}= \liminf_{x\rightarrow\infty}\min_{t\in[\theta, 1-\theta]}\frac{g(t, x)}{x}, \\& m_{0}= \biggl[ \int_{0}^{1}\phi(s)\bigl[p(s)+q(s)\bigr]\,ds \biggr]^{-1},\qquad m_{1}= \biggl[\gamma_{0} \int_{\theta}^{1-\theta}\phi(s)p(s)\,ds \biggr]^{-1}, \\& m_{2}= \biggl[\gamma_{0} \int_{\theta}^{1-\theta}\phi(s)q(s)\,ds \biggr]^{-1}. \end{aligned}$$
Theorem 3.1
Let \(f, g : [0,1]\times [0, +\infty) \to [0, +\infty)\) be continuous. Assume that (H1)-(H3) and \(\overline{f}_{0}< m_{0}\), \(\overline{g}_{0}< m_{0}\) hold and one of the two conditions \(\underline{f}_{\infty}>m_{1} \) and \(\underline {g}_{\infty}>m_{2}\) is satisfied. Then BVP (1.1) has at least one positive solution.
Proof
First, we consider the case \(\underline {f}_{\infty}>m_{1} \). Since \(\overline{f}_{0}< m_{0}\), \(\overline{g}_{0}< m_{0}\), we can choose \(\epsilon_{0}>0\) such that \(m_{0}-\epsilon_{0}>0\) and \(\overline {f}_{0}< m_{0}-\epsilon_{0}\), \(\overline{g}_{0}< m_{0}-\epsilon_{0}\), and also there exists \(r_{0}>0\) such that for any \(0< x\leq r_{0}\), \(0\leq t \leq1\), we have Set \(\Omega_{0}=\{u\in E:\|u\|< r_{0}\}\). We can see that \(\Omega_{0}\) is a bounded open subset in E, and for any \(u\in\partial\Omega _{0}\cap K\), \(0< u(t)\leq r_{0}\), \(t\in[0, 1]\), from expression (3.1) we deduce the following inequalities: Consequently, On the other hand, since \(\underline{f}_{\infty}>m_{1}\), we can choose \(\epsilon_{1}>0 \) such that \(\underline{f}_{\infty}>(m_{1}+\epsilon _{1})\), and also there exists \(R_{1}>r_{0}>0\) such that for any \(x \geq R_{1}\), \(t\in[\theta, 1-\theta]\), we have Let \(\overline{R_{1}}=\frac{R_{1}}{\gamma_{0}}\) and \(\Omega_{1}=\{u\in E:\|u\|<\overline{R_{1}}\}\). It is easy to see that \(\Omega_{1}\) is a bounded open subset in E, and from the expression (3.3) we have that for any \(u\in\partial\Omega_{1}\cap K\), \(u(t)\geq\gamma_{0}\|u\|= R_{1}\), \(t\in[\theta, 1-\theta]\), Consequently, Applying Lemma 2.6 to (3.2) and (3.4) yields that T has a fixed point \(u^{\ast}\in K\cap(\overline{\Omega_{1}}\setminus\Omega_{0})\), and hence \(u^{\ast}\) is a positive solution of BVP (1.1).
$$ f(t, x)\leq(m_{0}-\epsilon_{0})x, \qquad g(t, x) \leq(m_{0}-\epsilon_{0})x. $$
(3.1)
$$\begin{aligned} \|Tu\| &=\sup_{t\in[0, 1]}\biggl\vert \int_{0}^{1}G(t, s)\bigl[p(s)f\bigl(s, u(s) \bigr)+q(s)g\bigl(s, u(s)\bigr)\bigr]\,ds\biggr\vert \\ &\leq(m_{0}-\epsilon_{0})\sup_{t\in[0, 1]} \biggl\vert \int _{0}^{1}G(t, s)\bigl[p(s)+q(s)\bigr]u(s)\,ds \biggr\vert \\ &\leq(m_{0}-\epsilon_{0})\sup_{t\in[0, 1]} \biggl\vert \int_{0}^{1}G(t, s)\bigl[p(s)+q(s)\bigr]\,ds\biggr\vert \|u\| \\ &\leq m_{0} \int_{0}^{1}\phi(s)\bigl[p(s)+q(s)\bigr]\,ds\|u\| =\|u \| . \end{aligned}$$
$$ \|Tu\|\leq\|u\|, \quad \forall u\in\partial\Omega_{0}\cap K. $$
(3.2)
$$ f(t, x)\geq(m_{1}+\epsilon_{1})x . $$
(3.3)
$$\begin{aligned} \|Tu\| &=\sup_{t\in[0, 1]}\biggl\vert \int_{0}^{1}G(t, s)\bigl[p(s)f\bigl(s, u(s) \bigr)+q(s)g\bigl(s, u(s)\bigr)\bigr]\,ds\biggr\vert \\ &\geq \int_{\theta}^{1-\theta}G(t, s)p(s)f\bigl(s, u(s)\bigr)\,ds \\ &\geq\gamma_{0}(m_{1}+\epsilon_{1}) \int_{\theta}^{1-\theta}\phi (s)p(s)\,ds\|u\| \\ &\geq\gamma_{0}m_{1} \int_{\theta}^{1-\theta}\phi(s)p(s)\,ds\|u\|=\|u\|. \end{aligned}$$
$$ \|Tu\|\geq\|u\|, \quad \forall u\in\partial\Omega_{1}\cap K. $$
(3.4)
In a similar way, when the case that \(\underline{g}_{\infty}>m_{2}\) holds, we can prove that the conclusion of Theorem 3.1 also holds. The proof of Theorem 3.1 is completed. □
Theorem 3.2
Assume that (H1)-(H4) and \(\overline{f}_{\infty}< m_{0}\), \(\overline{g}_{\infty}< m_{0}\) hold. Then BVP (1.1) has at least one positive solution.
The proof is similar to the previous one, so we omit it.
4 Existence of multiple positive solutions
In this section, we discuss the existence of multiple positive solutions of BVP (1.1). We obtain the following existence theorems.
Theorem 4.1
Assume that (H1)-(H4) hold and
\(\underline{f}_{\infty}>m_{1}\)
are satisfied. Further, assume that the following condition holds:
where
\(\phi(s)\leq\overline{g}\), \(\partial\Omega=\{u\in E: \|u\|=1\}\). Then BVP (1.1) has at least two positive solutions.
$$(\ast)\quad \int_{0}^{1}p(s)f\bigl(s, u(s)\bigr)+q(s)g\bigl(s, u(s)\bigr)\,ds< \frac{1}{\overline{g}},\quad \forall u\in\partial\Omega\cap K, $$
Proof
Since \(\underline{g}_{0}=+\infty>m_{2}\), there exist \(\epsilon_{2}>0\), \(0< r_{1}<1\) such that for any \(0< x\leq r_{1}\), \(\theta \leq t\leq1-\theta\), we have Set \(\Omega_{2}=\{u\in E:\|u\|< r_{1}\}\). Then \(\Omega_{2}\) is a bounded open subset in E, and for any \(u\in\partial\Omega_{2}\cap K\) and \(0< u(t)\leq r_{1}\), by expression (4.1) we have Consequently, Since \(\underline{f}_{\infty}>m_{1}\), there exist \(\epsilon_{3}>0\), \(R_{2}>1\) such that for any \(x\geq R_{2}\), \(\theta\leq t\leq1-\theta\), we have Let \(R_{3}=\frac{R_{2}}{\gamma_{0}}\) and \(\Omega_{3}=\{u\in E:\|u\| < R_{3}\}\). Then \(\Omega_{3}\) is a bounded open subset in E, and so for any \(u\in\partial\Omega_{3}\cap K\) and \(u(t)\geq\gamma_{0}\|u\| =R_{2}\), \(t\in[\theta, 1-\theta]\), by expression (4.3) we have Consequently, On the other hand, by (∗) we have for any \(u\in\partial\Omega \cap K\), \(t\in[0, 1]\), Consequently, Therefore, from (4.2), (4.5), and Lemma 2.6 it follows that BVP (1.1) has a positive solution \(u_{0}\in K\) with \(r_{1}\leq\|u\|<1\). In the same way, from (4.4), (4.5), and Lemma 2.6 it follows that BVP (1.1) has a positive solution \(u_{1}\in K\) with \(1<\|u\|\leq R_{3}\). Therefore, BVP (1.1) has at least two positive solutions \(u_{0}, u_{1}\in K\). □
$$ g(t, x)\geq(m_{2}+\epsilon_{2})x . $$
(4.1)
$$\begin{aligned} \|Tu\| &=\sup_{t\in[0, 1]}\biggl\vert \int_{0}^{1}G(t, s)\bigl[p(s)f\bigl(s, u(s) \bigr)+q(s)g\bigl(s, u(s)\bigr)\bigr]\,ds\biggr\vert \\ &\geq \int_{\theta}^{1-\theta}G(t, s)q(s)g\bigl(s, u(s)\bigr)\,ds \\ &\geq\gamma_{0}(m_{2}+\epsilon_{2}) \int_{\theta}^{1-\theta}\phi (s)q(s)\,ds\|u\| \\ &\geq \gamma_{0}m_{2} \int_{\theta}^{1-\theta}\phi(s)q(s)\,ds\|u\|=\|u\|. \end{aligned}$$
$$ \|Tu\|\geq\|u\|,\quad \forall u\in\partial\Omega_{2}\cap K. $$
(4.2)
$$ f(t, x)\geq(m_{1}+\epsilon_{3})x . $$
(4.3)
$$\begin{aligned} \|Tu\| &=\sup_{t\in[0, 1]}\biggl\vert \int_{0}^{1}G(t, s)\bigl[p(s)f\bigl(s, u(s) \bigr)+q(s)g\bigl(s, u(s)\bigr)\bigr]\,ds\biggr\vert \\ &\geq \int_{\theta}^{1-\theta}G(t, s)p(s)f\bigl(s, u(s)\bigr)\,ds \\ &\geq\gamma_{0}(m_{1}+\epsilon_{3}) \int_{\theta}^{1-\theta}\phi (s)p(s)\,ds\|u\| \\ &\geq\gamma_{0} m_{1} \int_{\theta}^{1-\theta}\phi(s)p(s)\,ds\|u\|=\|u\|. \end{aligned}$$
$$ \|Tu\|\geq\|u\|, \quad \forall u\in\partial\Omega_{3}\cap K. $$
(4.4)
$$\begin{aligned} \|Tu\| &=\sup_{t\in[0, 1]}\biggl\vert \int_{0}^{1}G(t, s)\bigl[p(s)f\bigl(s, u(s) \bigr)+q(s)g\bigl(s, u(s)\bigr)\bigr]\,ds\biggr\vert \\ &\leq \int_{0}^{1}\phi(s) \bigl[p(s)f\bigl(s, u(s) \bigr)+q(s)g\bigl(s, u(s)\bigr) \bigr]\,ds \\ &\leq \overline{g} \int_{0}^{1} \bigl[p(s)f\bigl(s, u(s)\bigr)+q(s)g \bigl(s, u(s)\bigr) \bigr]\,ds\leq\overline{g}\frac{1}{\overline{g}}=1=\|u\| . \end{aligned}$$
$$ \|Tu\|\leq\|u\|, \quad \forall u\in\partial\Omega\cap K. $$
(4.5)
Theorem 4.2
Assume that (H1)-(H4) hold and
\(\underline{g}_{\infty}>m_{2}\)
are satisfied. Further, assume that the following condition holds:
where
\(\phi(s)\leq\overline{g}\), \(\partial\Omega=\{u\in E: \|u\|=1\}\). Then BVP (1.1) has at least two positive solutions.
$$\int_{0}^{1}p(s)f\bigl(s, u(s)\bigr)+q(s)g\bigl(s, u(s)\bigr)\,ds< \frac{1}{\overline{g}}, \quad\forall u\in\partial\Omega\cap K, $$
The proof is similar to the previous one, so we omit it.
5 Example
Example 5.1
We consider BVP (1.1) with \(\alpha =\frac {7}{2}\), \(p(t)=q(t)=\frac{1}{200\sqrt{1-t^{2}}}\), \(h(t)=t^{-\frac{1}{2}}\), \(f(t, x)=5-t+| \ln x |+x^{\frac{3}{2}}\), \(g(t, x)=2-t+|\ln x|\), and Thus, \(\Gamma=\int_{0}^{1}t^{\frac{5}{2}}t^{-\frac{1}{2}}\,dA(t)=2\times (\frac{1}{2})^{2}=\frac{1}{2}\). Obviously, \(g_{A}(s)\leq1\) and \(\phi(s)=\phi_{0}(s) +\frac {g_{A}(s)}{1-\Gamma}\leq4=\overline{g}\). We define the cone in which For any \(0< r< R<+\infty\) and \(u\in\overline{K_{R}}\backslash K_{r}\), we have \(u(t)\geq c_{0}(t)\|u\|\), \(t\in[0, 1] \). Then \(0\leq r c_{0}(t)\leq u(t)\leq R\), \(t\in[0, 1]\). Since \(|\ln x|\) is decreasing on \((0, 1)\) and is increasing on \((1, +\infty)\), we have It is easy to verify that \(\frac{|\ln c_{0}(t)|}{200\sqrt{1-t^{2}}}\in L^{1}(0,1)\). The absolute continuity of the integral yields that So, by (5.1) and (5.2) we get By (5.2) we know that, as \(m\rightarrow\infty\), the limit of (5.3) is 0. On the other hand, by calculation we have where with \(\gamma_{0}=\min_{\theta\leq t\leq1-\theta}c_{0}(t)\). On the other hand, for any \(u\in\partial\Omega\cap K\), \(t\in[0, 1]\), we have So we obtain Therefore, the assumptions of Theorem 4.1 are satisfied. Thus the problem considered possesses two positive solutions in K.
$$A(t)= \left \{ \textstyle\begin{array}{@{}l@{\quad}l} 0, & t\in [0, \frac{1}{2} ), \\ 2, & t\in [\frac{1}{2}, 1 ]. \end{array}\displaystyle \right . $$
$$K= \bigl\{ u\in C[0, 1]:u(t)\geq c_{0}(t)\|u\|, t\in[0, 1]; u(t)\geq \gamma_{0}\|u\|, t\in[\theta, 1-\theta] \bigr\} , $$
$$c_{0}(t)= \min \biggl\{ \frac{(\frac{5}{2})^{\frac{5}{2}}t^{\frac {3}{2}}(1-t)}{\frac{3}{2}^{\frac{3}{2}}}, t^{\frac{5}{2}} \biggr\} . $$
$$ \bigl|\ln u(x)\bigr|\leq\bigl|\ln rc_{0}(t)\bigr|+|\ln R|\leq|\ln r|+|\ln R|+\bigl|\ln c_{0}(t)\bigr|. $$
(5.1)
$$ \lim_{m\rightarrow\infty} \int_{H(m)}\frac{|\ln c_{0}(s)|}{200\sqrt {1-s^{2}}}\,ds=0. $$
(5.2)
$$\begin{aligned} &\sup_{u\in\overline{K_{R}}\backslash K_{r}} \int_{H(m)}\frac {1}{200\sqrt{1-s^{2}}} \bigl[2-s+\bigl|\ln u(s)\bigr|+5-s+\bigl|\ln u(s)\bigr|+\bigl(u(s)\bigr)^{\frac {3}{2}} \bigr]\,ds \\ &\quad\leq \int_{H(m)}\frac{1}{200\sqrt{1-s^{2}}} \bigl[7-2s+2|\ln r|+2|\ln R|+2\bigl|\ln c_{0}(s)\bigr|+R^{\frac{3}{2}} \bigr]\,ds \\ &\quad\leq \bigl(8+R^{\frac{3}{2}}+4|\ln r|+4|\ln R| \bigr) \int_{H(m)}\frac {1}{200\sqrt{1-s^{2}}}\,ds + \int_{H(m)}\frac{1}{100\sqrt{1-s^{2}}}\bigl|\ln c_{0}(s)\bigr|\,ds \\ &\quad=\frac{8+R^{\frac{3}{2}}+4|\ln r|+4|\ln R|}{200} \biggl[\arcsin\frac {1}{m}+\arcsin1-\arcsin \biggl(1-\frac{1}{m}\biggr) \biggr] \\ &\qquad{}+ \int_{H(m)}\frac{1}{100\sqrt{1-s^{2}}}\bigl|\ln c_{0}(s)\bigr|\,ds. \end{aligned}$$
(5.3)
$$\begin{aligned}& \underline{g}_{0}=\liminf_{x\rightarrow0}\min _{t\in[\theta, 1-\theta ]}\frac{g(t, x)}{x}=\liminf_{x\rightarrow0} \frac{1+\theta+|\ln x|}{x}=\infty>m_{2}, \\& \underline{f}_{\infty}=\liminf_{x\rightarrow\infty}\min _{t\in[\theta, 1-\theta]}\frac{f(t, x)}{x}=\liminf_{x\rightarrow\infty} \frac{4+\theta +|\ln x|}{x}+x^{\frac{1}{2}}=\infty>m_{1}, \end{aligned}$$
$$m_{1}=m_{2}= \biggl[\gamma_{0} \int_{\theta}^{1-\theta}\frac{\phi (s)}{200\sqrt{1-s^{2}}}\,ds \biggr]^{-1} $$
$$\begin{aligned} f\bigl(t, u(t)\bigr)+g\bigl(t, u(t)\bigr) &=7-2t+2\bigl|\ln u(t)\bigr|+u(t)^{\frac{3}{2}} \\ &\leq8+2\bigl|\ln c_{0}(t)\bigr|. \end{aligned}$$
$$\begin{aligned} \int_{0}^{1}p(s)f\bigl(s, u(s)\bigr)+q(s)g\bigl(s, u(s)\bigr)\,ds &= \int_{0}^{1}\frac{1}{200\sqrt{1-s^{2}}}\bigl[f\bigl(s, u(s) \bigr)+g\bigl(s, u(s)\bigr)\bigr]\,ds \\ &\leq \int_{0}^{1}\frac{1}{200\sqrt{1-s^{2}}}\bigl[8+2\bigl|\ln c_{0}(s)\bigr|\bigr]\,ds \\ &= \frac{\pi}{50}+\frac{1}{100} \int_{0}^{1}\frac{|\ln c_{0}(s)|}{\sqrt {1-s^{2}}}\,ds \\ &\leq\frac{\pi}{50}+\frac{1}{10} \\ &< \frac{1}{4}. \end{aligned}$$
Declarations
Acknowledgements
The authors would like to thank the referee for his very important comments that improved the results and the quality of the paper. The authors were supported financially by the National Natural Science Foundation of China (11371221, 11571296), the Specialized Research Foundation for the Doctoral Program of Higher Education of China (20123705110001), and the Program for Scientific Research Innovation Team in Colleges and Universities of Shandong Province.
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)
School of Mathematical Sciences, Qufu Normal University, Qufu, People’s Republic of China
(2)
Department of Mathematics and Statistics, Curtin University, Perth, Australia
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