Research
Open Access
Singular boundary value problems of fractional differential equations with changing sign nonlinearity and parameter
- Yong Mu1,
- Liying Sun2 and
- Zhenlai Han2Email author
Received: 23 October 2015
Accepted: 4 January 2016
Published: 12 January 2016
Abstract
In this paper, we consider singular boundary value problems for the following nonlinear fractional differential equations with delay: where \(2<\alpha\leq3\), \(D^{\alpha}\) denotes the Riemann-Liouville fractional derivative, λ is a positive constant, \(f(t,x)\) may change sign and be singular at \(t=0\), \(t=1\), and \(x=0\). By means of the Guo-Krasnoselskii fixed point theorem, the eigenvalue intervals of the nonlinear fractional functional differential equation boundary value problem are considered, and some positive solutions are obtained, respectively.
$$\left \{ \textstyle\begin{array}{l@{\quad}l} D^{\alpha}x(t)+\lambda f (t,x(t-\tau) )=0, &t\in(0,1)\backslash \{\tau\}, \\ x(t)=\eta(t), &t\in[-\tau,0], \\ x'(1)=x'(0)=0, \end{array}\displaystyle \right . $$
Keywords
fractional functional differential equation delay boundary value problems singular positive solutionsMSC
34A08 34K371 Introduction
Fractional differential equations have been of increasing importance for the past decades due to their diverse applications in science and engineering, we can describe natural phenomena and mathematical models more accurately. Many researchers have shown their interest in fractional differential equations. The motivation for those works stems from both the intensive development of the theory of fractional calculus itself and the applications such as in economics, engineering and other fields. Fractional differential equations have received much attention, the theory and its applications have been greatly developed; see [1–5].
There have been many papers dealing with boundary value problems of fractional differential equations [6–22] and initial value problems of fractional differential equations [23–31].
However, the results focused on the singular boundary value problems of fractional differential equations with delay are relatively scarce [32–37]. It is well known that in practical problems, the behavior of systems not only depends on the status just at the present, but also on the status in the past.
Thus, in many cases, we must study fractional differential equations with delay in order to solve practical problems. Consequently, our aim in the paper is to consider the existence of solutions for singular boundary value problems of fractional differential equations with delay.
In 2004, by means of the fixed point index theorem, Yu et al. [38] investigate the existence of multiple positive solutions for the third-order three-point singular semipositone boundary value problem where \(\frac{1}{2}<\eta<1\), \(f(t,x):(0,1)\times(0,+\infty)\rightarrow (-\infty,+\infty)\) is continuous and may be singular at \(t=0\), \(t=1\), and \(x=0\) and also may be negative for some values of t and x; λ is a positive parameter.
$$\left \{ \textstyle\begin{array}{l@{\quad}l} x'''(t)-\lambda f(t,x)=0, & t\in(0,1), \\ x(0)=x'(\eta)=x''(1)=0, \end{array}\displaystyle \right . $$
In 2011, Zhao et al. [12] studied the existence on multiple positive solutions for the nonlinear fractional differential equation boundary value problem where \(2<\alpha\leq3\), \(D_{0+}^{\alpha}\) is the Riemann-Liouville fractional derivative. By the lower and upper solutions method and the fixed point theorem, they obtained some new existence criteria for singular and nonsingular fractional differential equation boundary value problems.
$$\left \{ \textstyle\begin{array}{l@{\quad}l} D^{\alpha}_{0+}u(t)+f(t,u(t))=0, & t\in(0,1), \\ u(0)=u'(0)=u'(1)=0, \end{array}\displaystyle \right . $$
In 2012, Su [37] studied the boundary value problem for a singular fractional differential equation with delay where \(1<\alpha\leq2\), \(D^{\alpha}\) is the Riemann-Liouville fractional derivative, \(\tau\in(0,1)\), \(f(t,x)\in C ((0,1)\times\mathbb{R}^{+},\mathbb{R} )\) is continuous and may be singular at \(t=0\), \(t=1\), and \(x=0\) and may have negative values, where \(\mathbb{R}^{+}=(0,+\infty)\). By the Guo-Krasnoselskii fixed point theorem, one obtained the existence results for positive solutions.
$$\left \{ \textstyle\begin{array}{l@{\quad}l} D^{\alpha}x(t)+f (t,x(t-\tau) )=0, &t\in(0,1)\backslash\{\tau\}, \\ x(t)=\eta(t), &t\in[-\tau,0], \\ x(1)=0, \end{array}\displaystyle \right . $$
In 2013, Vong [39] considered the fractional differential equation with an integral boundary condition where \(n\geq2\), \(n-1<\alpha< n\), \({}^{\mathrm{c}} D^{\alpha}_{0^{+}}\) is the Caputo fractional derivative, \(\mu(s)\) is a function of bounded variation, f may have a singularity at \(t=1\). The existence of positive solutions is obtained by the method of upper and lower solutions together with the Schauder fixed point theorem.
$${}^{\mathrm{c}} D^{\alpha}_{0^{+}} u(t)+f\bigl(t,u(t)\bigr)=0, \qquad u'(0)=\cdots=u^{n-1}(0)=0,\qquad u(1)= \int^{1}_{0} u(s)\,d\mu(s), $$
Motivated by the work mentioned above, in this paper, we study the existence of positive solutions of singular boundary value problems for nonlinear fractional functional differential equation where \(2<\alpha\leq3\), \(D^{\alpha}\) denote the Riemann-Liouville fractional derivative, λ is a positive parameter, \(\tau\in (0,1)\), \(\eta(t)\in C([-\tau,0])\), and \(\eta(t)>0\) for \(t\in [\tau,0)\), \(\eta(0)=0\), f is a continuous functional defined on \((0,1)\times\mathbb{R}^{+}\) and which may be singular at \(t=0\), \(t=1\), and \(x=0\).
$$ \left \{ \textstyle\begin{array}{l@{\quad}l} D^{\alpha}x(t)+\lambda f (t,x(t-\tau) )=0, &t\in(0,1)\backslash \{\tau\}, \\ x(t)=\eta(t), &t\in[-\tau,0], \\ x'(1)=x'(0)=0, \end{array}\displaystyle \right . $$
(1.1)
When \(\tau=0\) and \(\lambda=1\), problem (1.1) is reduced to the problem of fractional differential equations and has been studied by Zhao et al. [12]. To the best of our knowledge, no one has studied the existence of positive solutions for singular boundary value problem (1.1). Key tools in finding our main results are the Guo-Krasnoselskii fixed point theorem, and our main results of this paper are to extend and supplement some results in [12, 37, 38].
The paper is organized as follows. In Section 2, we shall introduce some definitions and lemmas to prove our main results. In Section 3, we investigate the existence of positive solution for boundary value problem (1.1) by the Guo-Krasnoselskii fixed point theorem.
2 Preliminaries
In the following section, we introduce the definitions and lemmas which are used throughout the paper. This material can be found in [1, 2].
The Riemann-Liouville fractional derivative of order α (\(n-1<\alpha<n\)) of a function \(f:(t_{0},+\infty)\to\mathbb{R}\) is given by where n is the smallest integer than or equal to α and \(\Gamma (\cdot)\) is the gamma function, provided that the right side is point wise defined on \((t_{0},+\infty)\).
$$D^{\alpha}_{0^{+}} f(t)=\frac{1}{\Gamma(n-\alpha)}\biggl(\frac{d}{dt} \biggr)^{n} \int^{t}_{t_{0}}\frac {f(s)}{(t-s)^{\alpha-n+1}}\, ds,\quad t>t_{0}, $$
The Riemann-Liouville fractional integral of order α (\(\alpha>0\)) of a function \(f:(t_{0},+\infty)\to\mathbb{R}\) is given by where \(\Gamma(\cdot)\) is the gamma function, provided that the right side is point-wise defined on \((t_{0},+\infty)\).
$$I^{\alpha}f(t)=\frac{1}{\Gamma{(\alpha)}} \int_{t_{0}}^{t}\frac {f(s)}{(t-s)^{1-\alpha}}\, ds, \quad t>t_{0}, $$
From the definition of the Riemann-Liouville derivative, we have the following statements.
Let \(\alpha>0\). If we assume \(u\in C(0,1)\cap L(0,1)\), then the fractional differential equation has \(u(t)=c_{1}t^{\alpha-1}+c_{2}t^{\alpha-2}+\cdots+c_{N} t^{\alpha-N}\), \(c_{i}\in\mathbb{R}\), \(i=1,2,\ldots,N\), as unique solutions, where N is the smallest integer greater than or equal to α.
$$D^{\alpha}_{0^{+}}u(t)=0 $$
Assume \(u\in C(0,1)\cap L(0,1)\) with a fractional derivative of order \(\alpha>0\). Then for some \(c_{i}\in\mathbb{R}\), \(i=1,2,\ldots,n\), where n is the smallest integer greater than or equal to α.
$$I^{\alpha}_{0^{+}}D^{\alpha}_{0^{+}}u(t)=f(t)+c_{1}t^{\alpha-1}+c_{2}t^{\alpha -2}+ \cdots+c_{n}t^{\alpha-n} $$
Next we introduce the Green function of boundary value problems for fractional differential equations.
Lemma 2.1
[12]
Let
\(2<\alpha\leq3\)
and
\(h:[0,1]\)
be continuous. Then the unique solution of the boundary value problem
is
where
$$ D^{\alpha}_{0^{+}}u(t)+\lambda h(t)=0, \quad t\in(0,1), \qquad u(0)=u'(0)=u'(1)=0 $$
(2.1)
$$u(t)= \int_{0}^{1}\lambda G(t,s)h(s)\,ds, \quad t \in[0,1], $$
$$ G(t,s)=\frac{1}{\Gamma(\alpha)} \left \{ \textstyle\begin{array}{l@{\quad}l} t^{\alpha-1}(1-s)^{\alpha-2}-(t-s)^{\alpha-1},& 0\leq s\leq t\leq1, \\ t^{\alpha-1}(1-s)^{\alpha-2},& 0\leq t\leq s\leq1. \end{array}\displaystyle \right . $$
(2.2)
The following properties of the Green function play important roles in this paper.
Lemma 2.2
[12]
The function
\(G(t,s)\)
defined by (2.2) satisfies the following conditions:
- (1)
\(G(t,s)>0\) for \(t,s\in(0,1)\);
- (2)
\(q(t)G(1,s)\leq G(t,s)\leq G(1,s)\) for \(t,s\in(0,1)\), where \(q(t)=t^{\alpha-1}\).
Lemma 2.3
The function
\(G^{*}(t,s):=t^{5-\alpha}G(t,s)\)
satisfies the following conditions:
$$\frac{1}{\Gamma(\alpha)}t^{4}s(1-s)^{\alpha-2}\leq G^{*}(t,s) \leq \frac{1}{\Gamma(\alpha)}t^{5-\alpha}s(1-s)^{\alpha-2} \quad \textit{for } t,s\in(0,1). $$
Proof
The proof can be obtained easily by Lemma 2.2, so we omit it here. □
The following lemma is fundamental in the proofs of our main results.
Lemma 2.4
[40]
Let
E
be a Banach space, and let
\(K\subset E\)
be a cone. Assume
\(\Omega_{1}\), \(\Omega_{2}\)
are open and bounded subset of
E
with
\(0\in\Omega_{1}\), \(\bar{\Omega}_{1}\subset\Omega_{2}\), and let
\(T:K\cap(\bar{\Omega}_{2}\backslash\Omega_{1})\rightarrow K\)
be a completely continuous operator such that
- (i)
\(\|Tu\|\leq\|u\|\), \(u\in K\cap\partial\Omega_{1}\), and \(\|Tu\|\geq\|u\|\), \(u\in K\cap\partial\Omega_{2}\); or
- (ii)
\(\|Tu\|\geq\|u\|\), \(u\in K\cap\partial\Omega_{1}\), and \(\|Tu\|\leq\|u\|\), \(u\in K\cap\partial\Omega_{2}\).
3 Main results
In this section, we discuss the existence of positive solutions for boundary value problem (1.1). For convenience, we give some conditions, which will play roles in this paper.
- (H1):
-
There exists a nonnegative function \(\rho\in C(0,1)\cap L(0,1)\) such thatfor all \((t,x)\in(0,1)\times\mathbb{R}^{+}\), where \(\varphi_{1},\varphi _{2}\in L(0,1)\) are nonnegative for \(t\in(0,1)\), \(h_{1},h_{2}\in C(\mathbb{R}_{0}^{+},\mathbb {R}_{0}^{+})\) are nondecreasing, \(g\in C(\mathbb{R}^{+},\mathbb{R}_{0}^{+})\) is nonincreasing, \(\mathbb{R}_{0}^{+}=[0,+\infty)\), and$$\varphi_{2}(t)h_{2}(x)\leq f\bigl(t,v(t)x\bigr)+\rho(t) \leq\varphi_{1}(t) \bigl(g(x)+h_{1}(x)\bigr) $$$$v(t)= \left \{ \textstyle\begin{array}{l@{\quad}l} 1, & t\in(0,\tau], \\ (t-\tau)^{\alpha-5},& t\in(\tau,1). \end{array}\displaystyle \right . $$
- (H2):
-
and there exists a constant \(k>0\) such that$$0< \int_{0}^{\tau}s(1-s)^{\alpha-2} \varphi_{1}(s)g\bigl(\eta(s-\tau)\bigr)\,ds< +\infty, $$$$\int_{\tau}^{1}s(1-s)^{\alpha-2} \varphi_{1}(s)g\biggl(\frac{k}{2}(s-\tau )^{2}\biggr) \,ds< +\infty. $$
- (H3):
-
There exists a subinterval \([a,b]\subset(\tau,1)\) such that \(\int_{a}^{b}s(1-s)^{\alpha-1}\varphi_{2}(s)\,ds>0\).
Let \(X:=\{x(t):x\in C([-\tau,1],\mathbb{R}),x(t)=0 \text{ for } t\in [-\tau,0],x'(1)=x'(0)=0\}\) be a Banach space with the maximum norm \(\|x\|_{[-\tau,1]}=\max_{-\tau\leq t\leq1}|x(t)|=\max_{0\leq t\leq1}|x(t)|\) for \(x\in X\). Let K be a cone in X defined by
$$K=\bigl\{ x\in X; x(t)\geq0 \text{ for } t\in[-\tau,1]\bigr\} . $$
Define and for any \(x\in K\).
$$\begin{aligned}& \bar{\eta}(t)=\left \{ \textstyle\begin{array}{l@{\quad}l} \eta(t), &t\in[-\tau,0], \\ 0, &t\in(0,1], \end{array}\displaystyle \right . \\& \omega(t)=\left \{ \textstyle\begin{array}{l@{\quad}l} 0, &t\in[-\tau,0], \\ \int_{0}^{1}\lambda G(t,s)\rho(s)\,ds, &t\in(0,1], \end{array}\displaystyle \right . \end{aligned}$$
$$\begin{aligned} x^{*}(t)&=\max\bigl\{ x(t)+\bar{\eta}(t)-\omega(t),0\bigr\} \\ &= \left \{ \textstyle\begin{array}{l@{\quad}l} \eta(t), & t\in[-\tau,0], \\ \max\{x(t)-\omega(t),0\}, & t\in(0,1] \end{array}\displaystyle \right . \end{aligned}$$
It is easy to know that the restriction \(\omega|_{[0,1]}\) of ω on \([0,1]\) is exactly the solution of
$$\left \{ \textstyle\begin{array}{l@{\quad}l} D^{\alpha}x(t)+\lambda\rho(t)=0, & t\in(0,1), \alpha\in (2,3], \\ x'(1)=x'(0)=x(0)=0. \end{array}\displaystyle \right . $$
Since \(f:[0,1]\times C[-\tau,1]\rightarrow\mathbb{R}\) is a continuous function, setting \(f(t,x(t-\tau)):=h(t)\) in Lemma 2.1, we see by Lemma 2.1 that a function x is a solution of boundary value problem (1.1) if and only if it satisfies
$$x(t)=\left \{ \textstyle\begin{array}{l@{\quad}l} \int_{0}^{1}\lambda G(t,s)f(s,x(s-\tau))\,ds, &t\in(0,1), \\ \eta(t), &t\in[-\tau,0]. \end{array}\displaystyle \right . $$
Consider the following operator: Set and Then (3.1) is equivalent to Obviously, if ỹ is a fixed point of operator T in (3.2), then is a fixed point of operator T defined by (3.1). Lemma 2.1 implies that Thus if then Let Then one can verify easily that the function x satisfied boundary value problem (1.1). As a result, in the following we will concentrate our study on finding the fixed points of operator T defined by (3.2).
$$ (Tx) (t)= \left \{ \textstyle\begin{array}{l@{\quad}l} \int_{0}^{1}\lambda G(t,s) (f(s,x^{*}(s-\tau))+\rho(s) )\,ds, &t\in(0,1], \\ 0, &t\in[-\tau,0]. \end{array}\displaystyle \right . $$
(3.1)
$$y(t)=\left \{ \textstyle\begin{array}{l@{\quad}l} t^{5-\alpha}x(t), &t\in(0,1), \\ 0, &t\in[-\tau,0] \end{array}\displaystyle \right . $$
$$y^{*}(t)=\left \{ \textstyle\begin{array}{l@{\quad}l} \max\{t^{\alpha-5}y(t)-\omega(t),0\}, &t\in(0,1], \\ \eta(t), &t\in[-\tau,0]. \end{array}\displaystyle \right . $$
$$ (Ty) (t)= \left \{ \textstyle\begin{array}{l@{\quad}l} \int_{0}^{1}\lambda G^{*}(t,s) (f(s,y^{*}(s-\tau))+\rho(s) )\,ds, &t\in(0,1], \\ 0, &t\in[-\tau,0]. \end{array}\displaystyle \right . $$
(3.2)
$$\tilde{x}(t)=\left \{ \textstyle\begin{array}{l@{\quad}l} t^{\alpha-5}\tilde{y}(t), &t\in(0,1], \\ 0, &t\in[-\tau,0], \end{array}\displaystyle \right . $$
$$\left \{ \textstyle\begin{array}{l@{\quad}l} D^{\alpha}\tilde{x}(t)+\lambda f ((t,\tilde{x}^{*}(t-\tau))+\rho (t) )=0, &t\in(0,1)\backslash\{\tau\}, \\ \tilde{x}(t)=0, &t\in[-\tau,0], \\ \tilde{x}'(1)=\tilde{x}'(0)=0. \end{array}\displaystyle \right . $$
$$ \tilde{x}(t-\tau)+\bar{\eta}(t-\tau)-\omega(t-\tau)\geq0\quad \text{for } t\in[0,1], $$
(3.3)
$$\tilde{x}^{*}(t-\tau)=\tilde{x}(t-\tau)+\bar{\eta}(t-\tau)-\omega(t-\tau). $$
$$x(t)=\tilde{x}(t)+\bar{\eta}(t)-\omega(t). $$
Define the cone and for any \(r_{2}>r_{1}\geq\max\{k,2c\}\), \(R_{2}>R_{1}\geq\max\{k,2c\}\), where and k is the constant in (H2).
$$K_{1}=\bigl\{ y\in K: y(t)\geq t^{2}\|y\| \text{ for } t \in[0,1]\bigr\} $$
$$\begin{aligned}& \Omega_{1}=\bigl\{ y\in K_{1}:\Vert y\Vert < r_{1}\bigr\} , \\& \Omega_{2}=\bigl\{ y\in K_{1}:\Vert y\Vert < r_{2}\bigr\} , \\& \Omega_{3}=\bigl\{ y\in K_{1}:\Vert y\Vert < R_{2}\bigr\} , \\& \Omega_{4}=\bigl\{ y\in K_{1}:\Vert y\Vert < R_{1}\bigr\} \end{aligned}$$
$$ c:=\frac{\lambda}{\Gamma(\alpha)} \int_{0}^{1}s(1-s)^{\alpha-2}\rho (s)\,ds< + \infty $$
(3.4)
Lemma 3.1
Let (H1) and (H2) hold. Then the operator \(T:K_{1}\cap(\bar{\Omega}_{2}\backslash\Omega_{1})\rightarrow K_{1}\) is completely continuous.
Proof
First we show that operator T is well defined on \(K_{1}\cap(\bar{\Omega}_{2} \backslash\Omega_{1})\). For any \(y\in K_{1}\cap(\bar{\Omega}_{2} \backslash\Omega_{1})\), we know that and Then, for \(t\in[0,1]\), we get where c is defined as (3.4). Thus, for \(t\in[0,1]\),
$$r_{1}\leq\|y\|\leq r_{2} $$
$$y(t)\geq t^{2}\|y\|\geq t^{2} r_{1} \quad \text{for } t\in[0,1]. $$
$$\begin{aligned} t^{5-\alpha}\omega(t)&=t^{5-\alpha} \int_{0}^{1}\lambda G(t,s)\rho(s)\,ds \\ &\leq\frac{t^{5-\alpha}}{\Gamma(\alpha)} \int_{0}^{1}\lambda s(1-s)^{\alpha -2}\rho(s)\,ds \\ &\leq\frac{t^{2}}{\Gamma(\alpha)} \int_{0}^{1}\lambda s(1-s)^{\alpha-2}\rho (s) \,ds \\ &\leq t^{2}c, \end{aligned}$$
(3.5)
$$\begin{aligned} \begin{aligned}[b] y(t)-t^{5-\alpha}\omega(t)&\geq t^{2}(r_{1}-c) \\ &\geq\frac{r_{1}}{2}t^{2}. \end{aligned} \end{aligned}$$
(3.6)
In view of (H1), (H2), and Lemma 2.3, we show Hence, T is uniformly bounded and T is well defined.
$$\begin{aligned} (Ty) (t) =& \int_{0}^{\tau}\lambda G^{*}(t,s) \bigl(f\bigl(s,\eta(s- \tau)\bigr)+\rho(s) \bigr)\,ds \\ &{} + \int_{\tau}^{1}\lambda G^{*}(t,s) \bigl(f\bigl(s,(s- \tau)^{\alpha-5}y(s-\tau )-\omega(s-\tau)\bigr)+\rho(s) \bigr)\,ds \\ \leq&\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{0}^{\tau }s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g \bigl(\eta(s-\tau) \bigr)+h_{1} \bigl( \eta(s-\tau) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s) \\ &{}\times\biggl(g \biggl(\frac{r_{1}}{2}(s-\tau)^{2} \biggr)+h_{1} \bigl(y(s-\tau)-(s-\tau )^{5-\alpha}\omega(s-\tau) \bigr) \biggr)\,ds \\ \leq&\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{0}^{\tau }s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g \bigl(\eta(s-\tau) \bigr)+h_{1} \bigl( \eta(s-\tau) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s) \biggl(g \biggl(\frac{r_{1}}{2}(s-\tau)^{2} \biggr)+h_{1} \bigl(y(s-\tau) \bigr) \biggr)\,ds \\ \leq&\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{0}^{\tau }s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g \bigl(\eta(s-\tau) \bigr)+h_{1} \bigl( \eta(s-\tau) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s) \biggl(g \biggl(\frac{k}{2}(s-\tau)^{2} \biggr)+h_{1}(r_{2}) \biggr)\,ds \\ < &+\infty. \end{aligned}$$
In fact, for \(y\in K_{1}\cap(\bar{\Omega}_{2} \backslash\Omega_{1})\), \(t\in[0,1]\), in view of Lemma 2.3, we have and Hence, \(T: K_{1}\cap(\bar{\Omega}_{2}\backslash\Omega_{1})\rightarrow K_{1}\).
$$\begin{aligned} \|Ty\|&\leq\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int _{0}^{1}s(1-s)^{\alpha-2} \bigl(f \bigl(s,y^{*}(s-\tau)\bigr)+\rho(s) \bigr)\,ds \\ &\leq\frac{\lambda t^{2}}{\Gamma(\alpha)} \int_{0}^{1}s(1-s)^{\alpha-2} \bigl(f \bigl(s,y^{*}(s-\tau)\bigr)+\rho(s) \bigr)\,ds \end{aligned}$$
$$\begin{aligned} (Ty) (t)&\geq\frac{\lambda t^{4}}{\Gamma(\alpha)} \int_{0}^{1}s(1-s)^{\alpha -2} \bigl(f \bigl(s,y^{*}(s-\tau)\bigr)+\rho(s) \bigr)\,ds \\ &\geq t^{2}\|Ty\|. \end{aligned}$$
Next we show \(T:K_{1}\cap(\bar{\Omega}_{2}\backslash\Omega_{1})\rightarrow K_{1}\) is continuous and compact. For any \(y_{n}, y\in K_{1}\cap(\bar{\Omega}_{2}\backslash\Omega_{1})\), \(n=1,2,\ldots\) with \(\|y_{n}-y\|_{[-\tau,1]}\rightarrow0\) as \(n\rightarrow\infty\). Since \(r_{1}\leq\|y_{n}\|\leq r_{2}\) and \(r_{1}\leq\| y\|\leq r_{2}\), for \(t\in(0,1)\), we know and Then, for \(t\in[0,1]\), we have This implies that \(\|Ty_{n}-Ty\|_{[-\tau,1]}\rightarrow0\) as \(n\rightarrow\infty\). Hence T is continuous.
$$y_{n}(t)-t^{5-\alpha}\omega(t)\geq\frac{r_{1}}{2}t^{2} $$
$$y(t)-t^{5-\alpha}\omega(t)\geq\frac{r_{1}}{2}t^{2}. $$
$$\begin{aligned} \bigl\vert (Ty_{n}) (t)-(Ty) (t)\bigr\vert =&\biggl\vert \int_{\tau}^{1}\lambda G^{*}(t,s) \bigl(f\bigl(s,(s- \tau)^{\alpha-5}y_{n}(s-\tau )-\omega(s-\tau)\bigr)+\rho(s) \\ &{}-f\bigl(s,(s-\tau)^{\alpha-5}y(s-\tau)-\omega(s-\tau)\bigr)-\rho(s) \bigr)\,ds\biggr\vert \\ \leq&\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{\tau }^{1}s(1-s)^{\alpha-2}\bigl\vert f \bigl(s,(s-\tau)^{\alpha-5}y_{n}(s-\tau)-\omega(s-\tau)\bigr) \\ &{}-f\bigl(s,(s-\tau)^{\alpha-5}y(s-\tau)-\omega(s-\tau)\bigr)\bigr\vert \,ds \\ \leq&\frac{2\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{\tau }^{1}s(1-s)^{\alpha-2} \varphi_{1}(s) \biggl(g\biggl(\frac{k}{2}(s-\tau )^{2} \biggr)+h_{1}(r_{2}) \biggr)\,ds \\ < &+\infty. \end{aligned}$$
Next we prove T is equicontinuous.
Since \(G^{*}\) in uniformly continuous for \(t\in(0,1)\), that is, for any \(\epsilon>0\), there exists \(\xi_{0}>0\), when \(t_{1},t_{2}\in[0,1]\) and \(|t_{1}-t_{2}|<\xi_{0}\), we have Thus, for any \(y\in K_{1}\cap(\bar{\Omega}_{2}\backslash\Omega_{1})\), we get Thus T is equicontinuous. Accordingly to the Ascoli-Arzelà theorem, T is completely continuous. The proof is completed. □
$$\begin{aligned} \bigl\vert G^{*}(t_{1},s)-G^{*}(t_{2},s)\bigr\vert =& \frac{\epsilon}{2} \biggl( \int_{0}^{\tau}\lambda\varphi_{1}(s) \bigl(g \bigl(\eta (s-\tau)\bigr)+h_{1}\bigl(\eta(s-\tau)\bigr) \bigr)\,ds \\ &{}+ \biggl.\biggl.\int_{\tau}^{1}\lambda\varphi_{1}(s) \biggl(g \biggl(\frac{k}{2}(s-\tau )^{2}\biggr)+h_{1}(r_{2}) \biggr)\,ds \biggr)\biggr.^{-1}. \end{aligned}$$
$$\begin{aligned} \bigl\vert (Ty) (t_{1})-(Ty) (t_{2})\bigr\vert \leq& \int_{0}^{\tau}\lambda\bigl\vert G^{*}(t_{1},s)-G^{*}(t_{2},s)\bigr\vert \varphi_{1}(s) \bigl(g\bigl(\eta(s-\tau)\bigr)+h_{1}\bigl(\eta(s-\tau)\bigr) \bigr) \,ds \\ &{} + \int_{\tau}^{1}\lambda\bigl\vert G^{*}(t_{1},s)-G^{*}(t_{2},s) \bigr\vert \varphi_{1}(s) \biggl(g\biggl(\frac{k}{2}(s- \tau)^{2}\biggr)+h_{1}(r_{2}) \biggr)\,ds \\ < &\frac{\epsilon}{2}+\frac{\epsilon}{2} \\ =&\epsilon. \end{aligned}$$
Now we prove the existence of positive solutions for boundary value problem (1.1) by using the Guo-Krasnoselskii fixed point theorem.
For convenience, we denote and there exists a subinterval \([\beta,\gamma]\subset(\tau,1)\),
$$\begin{aligned} \xi_{1} :=&\frac{1}{\Gamma(\alpha)} \biggl( \int_{0}^{\tau}s(1-s)^{\alpha -2} \varphi_{1}(s) \bigl(g\bigl(\eta(s-\tau)\bigr) \\ &{}+h_{1}\bigl(\eta(s-\tau)\bigr) \bigr)\,ds+ \int_{\tau}^{1}s(1-s)^{\alpha-2}\varphi _{1}(s) \biggl( g\biggl(\frac{k}{2}(s-\tau)^{2} \biggr)+\epsilon r_{2} \biggr)\,ds \biggr) \\ >&0, \end{aligned}$$
$$\begin{aligned}& \zeta_{1}:=\min_{t\in[\beta,\gamma]}(t-\tau)^{2}, \qquad \zeta _{2}:=\min_{t\in[\beta,\gamma]}t^{4}, \\& \xi_{2}(t):=\frac{\zeta_{2}}{\Gamma(\alpha)}h_{2}\biggl( \frac{r_{1}\zeta_{1}}{2}\biggr) \int _{\beta}^{\gamma} s(1-s)^{\alpha-2} \varphi_{2}(s)\,ds>0. \end{aligned}$$
Theorem 3.1
Let (H1), (H2), and
\(\xi_{2}^{-1}r_{1}<\xi_{1}^{-1}r_{2}\)
hold. Then the boundary value problem (1.1) has at least one positive solution if
for each
$$ \lim_{y\rightarrow+\infty}\frac{h_{1}(y)}{y}=0 $$
(3.7)
$$\lambda\in\bigl(\xi_{2}^{-1}r_{1}, \xi_{1}^{-1}r_{2}\bigr). $$
Proof
Let \(\epsilon>0\). Then in view of (3.7), there exists a \(M>0\) such that Choose then for \(y\in\partial\Omega_{2}\), like for (3.6), for \(t\in [0,1]\), we obtain Then from (H1), (3.8), (3.9), and Lemma 2.3, we get Therefore, for \(y\in\partial\Omega_{2}\), we have \(\|Ty\|\leq\|y\|\).
$$ h_{1}(y)\leq\epsilon y \quad \text{for } y>M. $$
(3.8)
$$r_{2}\geq\max\{M+1,r_{1}+1\}, $$
$$\begin{aligned} y(t)-t^{5-\alpha}\omega(t)&\geq t^{2}(r_{2}-c) \\ &\geq\frac{r_{2}}{2}t^{2}. \end{aligned}$$
(3.9)
$$\begin{aligned} (Ty) (t) \leq&\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{0}^{\tau }s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g \bigl(\eta(s-\tau) \bigr)+h_{1} \bigl( \eta(s-\tau) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s) \\ &{}\times\biggl(g \biggl(\frac{r_{2}}{2}(s-\tau)^{2} \biggr)+h_{1} \bigl(y(s-\tau)-(s-\tau )^{5-\alpha}\omega(s-\tau) \bigr) \biggr)\,ds \\ \leq&\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{0}^{\tau }s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g \bigl(\eta(s-\tau) \bigr)+h_{1} \bigl( \eta(s-\tau) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s) \biggl(g \biggl(\frac{r_{2}}{2}(s-\tau)^{2} \biggr)+h_{1} \bigl(y(s-\tau) \bigr) \biggr)\,ds \\ \leq&\frac{\lambda}{\Gamma(\alpha)} \int_{0}^{\tau}s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g \bigl(\eta(s-\tau) \bigr)+h_{1} \bigl( \eta(s-\tau) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda}{\Gamma(\alpha)} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s) \biggl(g \biggl(\frac{k}{2}(s-\tau)^{2} \biggr)+h_{1}(r_{2}) \biggr)\,ds \\ \leq&\frac{\lambda}{\Gamma(\alpha)} \int_{0}^{\tau}s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g \bigl(\eta(s-\tau) \bigr)+h_{1} \bigl( \eta(s-\tau) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda}{\Gamma(\alpha)} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s) \biggl(g \biggl(\frac{k}{2}(s-\tau)^{2} \biggr)+\epsilon r_{2} \biggr)\,ds \\ \leq&\lambda\xi_{1}< r_{2}. \end{aligned}$$
On the other hand, for \(y\in\partial\Omega_{1}\), like for (3.6), for \(t\in[0,1]\), we obtain
$$\begin{aligned} y(t)-t^{5-\alpha}\omega(t)&\geq t^{2}(r_{1}-c) \\ &\geq\frac{r_{1}}{2}t^{2}. \end{aligned}$$
(3.10)
Thus from (H1), (3.10), and Lemma 2.3, we get Therefore, for \(y\in\partial\Omega_{1}\), we have \(\|Ty\|\geq\|y\|\). Then T defined by (3.2) has a fixed point \(\tilde{y}\in K_{1}\cap(\bar{\Omega}_{2}\backslash\Omega _{1})\). In view of (3.10), we have It is easy to know (3.3) is satisfied. The proof is completed. □
$$\begin{aligned} \|Ty\| \geq& \int_{\beta}^{\gamma}\lambda\min_{t\in[\beta,\gamma ]}G^{*}(t,s) \bigl(f\bigl(s,(s-\tau)^{\alpha-5}y(s-\tau)-\omega(s-\tau)\bigr)+\rho (s) \bigr)\,ds \\ \geq& \int_{\beta}^{\gamma}\lambda\min_{t\in[\beta,\gamma ]}G^{*}(t,s) \varphi_{2}(s)h_{2} \biggl(\frac{r_{1}}{2}(s- \tau)^{2} \biggr)\,ds \\ \geq&\frac{\lambda\zeta_{2}}{\Gamma(\alpha)}h_{2}\biggl(\frac{r_{1}\zeta _{1}}{2}\biggr) \int_{\beta}^{\gamma}s(1-s)^{\alpha-2} \varphi_{2}(s)\,ds \\ \geq&\lambda\xi_{2}>r_{1}. \end{aligned}$$
$$\begin{aligned} t^{\alpha-5}\tilde{y}(t)-\omega(t)&=t^{\alpha-5} \bigl(\tilde {y}(t)-t^{5-\alpha}\omega(t) \bigr) \\ &\geq\frac{r_{1}}{2}t^{\alpha-3} \\ &>0. \end{aligned}$$
Denote
$$\begin{aligned} \xi_{3} :=&\frac{1}{\Gamma(\alpha)} \biggl( \int_{0}^{\tau}s(1-s)^{\alpha -2} \varphi_{1}(s) \bigl(g\bigl(\eta(s-\tau)\bigr)+h_{1}\bigl(\eta(s-\tau)\bigr) \bigr)\,ds \\ &{}+ \int_{\tau}^{1}s(1-s)^{\alpha-2}\varphi _{1}(s) \biggl(g\biggl(\frac{k}{2}(s-\tau)^{2} \biggr)+h_{1}(R_{1}) \biggr)\,ds \biggr) \\ >&0. \end{aligned}$$
Theorem 3.2
Let (H1)-(H3) and
hold. Then boundary value problem (1.1) has at least one positive solution if
for each
where
\(A:=\min_{t\in[a,b]}(t-\tau)^{2}\), \(M^{*}\)
is a positive constant.
$$\frac{2\Gamma(\alpha)}{M^{*}A\zeta_{2}} \biggl( \int_{a}^{b}s(1-s)^{\alpha-2}\varphi _{2}(s)\,ds \biggr)^{-1}< \xi_{3}^{-1}R_{1} $$
$$ \lim_{y\rightarrow+\infty}\frac{h_{2}(y)}{y}=+\infty $$
(3.11)
$$ \lambda\in \biggl(\frac{2\Gamma(\alpha)}{M^{*}A\zeta_{2}} \biggl( \int _{a}^{b}s(1-s)^{\alpha-2} \varphi_{2}(s)\,ds \biggr)^{-1},\xi_{3}^{-1}R_{1} \biggr), $$
(3.12)
Proof
It follows from (3.11) that there exists a \(M^{*}>0\) such that Choose Then for \(y\in\partial\Omega_{3}\), like for (3.6), for \(t\in [0,1]\), we obtain Thus from (H1), (3.13), (3.14), and Lemma 2.3, we get Therefore, for \(y\in\partial\Omega_{3}\), we have \(\|Ty\|\geq\|y\|\).
$$ h_{2}(y)\geq M^{*}y \quad \text{for } y>M^{*}. $$
(3.13)
$$R_{2}\geq\max\biggl\{ R_{1}+1,\frac{2M^{*}}{A}\biggr\} . $$
$$\begin{aligned} y(t)-t^{5-\alpha}\omega(t)&\geq t^{2}(R_{2}-c) \\ &\geq\frac{R_{2}}{2}t^{2}. \end{aligned}$$
(3.14)
$$\begin{aligned} \|Ty\| \geq& \int_{a}^{b}\lambda\min_{t\in[a,b]}G^{*}(t,s) \bigl(f\bigl(s,(s-\tau)^{\alpha-5}y(s-\tau)-\omega(s-\tau)\bigr)+\rho(s) \bigr)\,ds \\ \geq& \int_{a}^{b}\lambda\min_{t\in[a,b]}G^{*}(t,s) \varphi_{2}(s)h_{2} \biggl(\frac{R_{2}}{2}(s- \tau)^{2} \biggr)\,ds \\ \geq& \int_{a}^{b}\lambda\min_{t\in[a,b]}G^{*}(t,s) \varphi _{2}(s)h_{2}\biggl(\frac{R_{2}A}{2}\biggr)\,ds \\ \geq&\frac{M^{*}A\lambda R_{2}}{2} \int_{a}^{b}\min_{t\in [a,b]}G^{*}(t,s) \varphi_{2}(s)\,ds \\ \geq&\frac{\lambda M^{*}AR_{2}\zeta_{2}}{2\Gamma(\alpha)} \int_{a}^{b} s(1-s)^{\alpha-2} \varphi_{2}(s)\,ds \\ \geq& R_{2}. \end{aligned}$$
On the other hand, for \(y\in\partial\Omega_{4}\), like for (3.6), for \(t\in[0,1]\), we obtain
$$\begin{aligned} y(t)-t^{5-\alpha}\omega(t)&\geq t^{2}(R_{1}-c) \\ &\geq\frac{R_{1}}{2}t^{2}. \end{aligned}$$
(3.15)
Thus from (H1), (3.12), and (3.15), we have Therefore, for \(y\in\partial\Omega_{4}\), we have \(\|Ty\|\leq\|y\|\). Arguments similar to those at the end of the proof of Theorem 3.1 show that boundary value problem (1.1) has a positive solution. The proof is completed. □
$$\begin{aligned} (Ty) (t) \leq&\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{0}^{\tau }s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g \bigl(\eta(s-\tau) \bigr)+h_{1} \bigl( \eta(s-\tau) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s) \\ &{}\times\biggl(g \biggl(\frac{R_{1}}{2}(s-\tau)^{2} \biggr)+h_{1} \bigl(y(s-\tau)-(s-\tau )^{5-\alpha}\omega(s-\tau) \bigr) \biggr)\,ds \\ \leq&\frac{\lambda}{\Gamma(\alpha)} \int_{0}^{\tau}s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g \bigl(\eta(s-\tau) \bigr)+h_{1} \bigl( \eta(s-\tau) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda}{\Gamma(\alpha)} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s) \biggl(g \biggl(\frac{k}{2}(s-\tau)^{2} \biggr)+h_{1}(R_{1}) \biggr)\,ds \\ \leq&\lambda\xi_{3}< R_{1}. \end{aligned}$$
Theorem 3.3
Let (H1) and (H2) hold. Furthermore assume that
- (H4):
-
there exists a subinterval \([\beta,\gamma]\subset(\tau,1)\) and a positive constant r such thatwhere k is defined in (H2), c is defined as (3.4) and \(\lambda\in(0,+\infty)\).$$\begin{aligned} r >&\max\biggl\{ k,2c,\frac{\lambda}{\Gamma(\alpha)} \biggl( \int_{0}^{\tau }s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g\bigl(\eta(s-\tau)\bigr)+h_{1}\bigl( \eta(s-\tau)\bigr) \bigr)\,ds \\ &{} + \int_{\tau}^{1}s(1-s)^{\alpha-2} \varphi_{1}(s) \biggl(g\biggl(\frac {k}{2}(s-\tau)^{2} \biggr)+h_{1}(r)\,ds \biggr) \biggr)\biggr\} , \end{aligned}$$
Then boundary value problem (1.1) has at least one positive solution y with \(0<\|y\|<r\).
Proof
In view of (H4), we choose \(n_{0}\in\{1,2,\ldots\}\) such that
$$\begin{aligned} r >&\frac{\lambda}{\Gamma(\alpha)} \biggl( \int_{0}^{\tau}s(1-s)^{\alpha -2} \varphi_{1}(s) \bigl(g\bigl(\eta(s-\tau)\bigr)+h_{1}\bigl( \eta(s-\tau)\bigr)\bigr)\,ds \\ &{} + \int_{\tau}^{1}s(1-s)^{\alpha-2} \varphi_{1}(s) \biggl(g\biggl(\frac{k}{2}(s-\tau )^{2} \biggr)+h_{1}(r)\biggr)\,ds \biggr)+\frac{1}{n_{0}}. \end{aligned}$$
Let \(N_{0}=\{n_{0},n_{0}+1,\ldots\}\). Fix \(n\in N_{0}\) and consider the family of integral equations where \(\kappa\in(0,1)\), We claim that any solution y of (3.16) for any \(\kappa\in (0,1)\) must satisfy \(\|y\|\neq r\). Otherwise, assume that y is a solution of (3.16) for some \(\kappa\in(0,1)\) such that \(\|y\|=r\). Then \(y^{*}(t-\tau)\geq\frac{1}{n}\) for \(t\in(0,1)\). In view of Lemma 2.3, we have Thus, for \(t\in(0,1)\), we have Then like for (3.6), for \(t\in(0,1)\), we have Then from (H1), for \(t\in(0,1)\), \(\kappa\in(0,1)\), we have Hence we obtain This is a contradiction and the claim is proved.
$$ y(t)=\left \{ \textstyle\begin{array}{l@{\quad}l} \kappa\int_{0}^{1}\lambda G^{*}(t,s) (f_{n}(s,y^{*}(s-\tau))+\rho(s) )\,ds+\frac{1}{n}, & t\in(0,1), \\ \frac{1}{n}, & t\in[-\tau,0], \end{array}\displaystyle \right . $$
(3.16)
$$f_{n}\bigl(t,y^{*}(t-\tau)\bigr)+\rho(t)=\left \{ \textstyle\begin{array}{l@{\quad}l} f(t,y^{*}(t-\tau))+\rho(t), & y^{*}(t-\tau)\geq\frac{1}{n}, \\ f(t,\frac{1}{n})+\rho(t), & y^{*}(t-\tau)< \frac{1}{n}. \end{array}\displaystyle \right . $$
$$ \|y\|\leq\frac{\kappa\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int _{0}^{1}s(1-s)^{\alpha-2} \bigl(f_{n}\bigl(s,y^{*}(s-\tau)\bigr)+\rho(s) \bigr)\,ds+ \frac{1}{n}. $$
(3.17)
$$\begin{aligned} y(t) \geq&\frac{1}{n}+\frac{\kappa\lambda t^{4}}{\Gamma(\alpha)} \int _{0}^{1}s(1-s)^{\alpha-2} \bigl(f_{n}\bigl(s,y^{*}(s-\tau)\bigr)+\rho(s) \bigr)\,ds \\ \geq&\frac{1}{n}+t^{\alpha-1}\biggl(\|y\|-\frac{1}{n}\biggr) \\ \geq&\bigl(1-t^{\alpha-1}\bigr)\frac{1}{n}+t^{\alpha-1}\|y\| \\ \geq& t^{\alpha-1}\|y\| \geq t^{2}r. \end{aligned}$$
$$\begin{aligned} y(t)-t^{5-\alpha}\omega(t)&\geq t^{2}(r-c) \\ &\geq\frac{r}{2}t^{2}. \end{aligned}$$
$$\begin{aligned} y(t) =&\frac{1}{n}+\kappa\lambda \int_{0}^{1}G^{*}(t,s) \bigl(f_{n} \bigl(s,y^{*}(s-\tau )\bigr)+\rho(s) \bigr)\,ds \\ =&\frac{1}{n}+\kappa\lambda \int_{0}^{1}G^{*}(t,s) \bigl(f\bigl(s,y^{*}(s-\tau) \bigr)+\rho (s) \bigr)\,ds \\ \leq&\frac{1}{n}+\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int _{0}^{1}s(1-s)^{\alpha-2} \bigl(f \bigl(s,y^{*}(s-\tau)\bigr)+\rho(s) \bigr)\,ds \\ \leq&\frac{1}{n_{0}}+\frac{\lambda}{\Gamma(\alpha)} \int_{0}^{\tau }s(1-s)^{\alpha-2} \bigl( \varphi_{1}(s) \bigl(g\bigl(\eta(s-\tau)\bigr)+h_{1}\bigl( \eta(s-\tau)\bigr) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda}{\Gamma(\alpha)} \int_{\tau}^{1}s(1-s)^{\alpha-2} \\ &{}\times \biggl( \varphi_{1}(s) \biggl(g\biggl(\frac{r}{2}(s-\tau)^{2} \biggr) +h_{1}\bigl(y(s-\tau)-(s-\tau)^{5-\alpha}\omega(s-\tau) \bigr) \biggr) \biggr)\,ds. \end{aligned}$$
$$\begin{aligned} r =&\bigl\Vert y(t)\bigr\Vert \\ \leq&\frac{1}{n_{0}}+\frac{\lambda}{\Gamma(\alpha)} \int_{0}^{\tau }s(1-s)^{\alpha-2} \bigl( \varphi_{1}(s) \bigl(g\bigl(\eta(s-\tau)\bigr)+h_{1}\bigl( \eta(s-\tau)\bigr) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda}{\Gamma(\alpha)} \int_{\tau}^{1}s(1-s)^{\alpha-2} \\ &{}\times \biggl( \varphi_{1}(s) \biggl(g\biggl(\frac{r}{2}(s-\tau)^{2} \biggr) +h_{1}\bigl(y(s-\tau)-(s-\tau)^{5-\alpha}\omega(s-\tau) \bigr) \biggr) \biggr)\,ds. \end{aligned}$$
Now the Leray-Schauder nonlinear alternative theorem guarantees that the equation has a solution \(y_{n}\), in \(\bar{\Omega}_{r}=\{y\in C[0,1]:\|y\|\leq r\}\), for \(t\in(0,1)\).
$$y(t)= \int_{0}^{1}\lambda G^{*}(t,s) \bigl(f_{n} \bigl(s,y^{*}(s-\tau)\bigr)+\rho(s) \bigr)\,ds+\frac{1}{n} $$
Next we claim that \(y_{n}(t)\) has a uniform sharper lower bound. In view of (H1) and \(\|y_{n}(t)\|\leq r\), we obtain Choosing \(\delta(t)=\frac{\lambda t^{4}}{\Gamma(\alpha)}h_{2}(\frac{r\zeta _{1}}{2})\int_{\beta}^{\gamma}s(1-s)^{\alpha-2}\varphi_{2}(s)\,ds\). Then we conclude that there exists a function \(\delta\in C(0,1)\) that is unrelated to n such that \(\delta(t)>0\) for a.e. \(t\in(0,1)\) and for any \(n\in N_{0}\),
$$\begin{aligned} y_{n}(t) =&\frac{1}{n}+\lambda \int_{0}^{1}G^{*}(t,s) \bigl(f_{n} \bigl(s,y^{*}_{n}(s-\tau )\bigr)+\rho(s) \bigr)\,ds \\ \geq&\frac{1}{n}+\lambda \int_{\beta}^{\gamma}G^{*}(t,s) \bigl(f\bigl(s,y^{*}_{n}(s- \tau)\bigr)+\rho(s) \bigr)\,ds \\ \geq&\frac{\lambda t^{4}}{\Gamma(\alpha)} \int_{\beta}^{\gamma} s(1-s)^{\alpha-2} \bigl(f \bigl(s,y^{*}_{n}(s-\tau)\bigr)+\rho(s) \bigr)\,ds \\ \geq&\frac{\lambda t^{4}}{\Gamma(\alpha)} \int_{\beta}^{\gamma }s(1-s)^{\alpha-2} \biggl( \varphi_{2}(s)h_{2}\biggl(\frac{r}{2}(s- \tau)^{2}\biggr) \biggr)\,ds \\ \geq&\frac{\lambda t^{4}}{\Gamma(\alpha)}h_{2}\biggl(\frac{r\zeta_{1}}{2}\biggr) \int _{\beta}^{\gamma}s(1-s)^{\alpha-2} \varphi_{2}(s)\,ds. \end{aligned}$$
$$y_{n}(t)\geq\delta(t). $$
Then we prove \(\{y_{n}\}_{n\in N_{0}}\) is an equicontinuous family on \((0,1)\). Since \(G^{*}\) in uniformly continuous for \(t\in(0,1)\), that is, for any \(\epsilon>0\), there exists \(\zeta _{0}>0\), when \(t_{1},t_{2}\in[0,1]\) and \(|t_{1}-t_{2}|<\zeta_{0}\), we have Thus Therefore, \(\{y_{n}\}_{n\in N_{0}}\) is an equicontinuous family on \((0,1)\). By the Arzelà-Ascoli theorem, there exist a subsequence \(N_{1}\) of \(N_{0}\) and \(y\in C(0,1)\) such that \(\{y_{n}\}_{n\in N_{1}}\) is uniformly convergent to y and y satisfies \(\delta(t)\leq y(t)\leq r\) for any \(t\in(0,1)\). By the Lebesgue dominated convergence theorem, in view of we have Then boundary value problem (1.1) has one positive solution with \(0<\|y\|<r\). The proof is completed. □
$$\begin{aligned} \bigl\vert G^{*}(t_{1},s)-G^{*}(t_{2},s)\bigr\vert =& \frac{\epsilon}{2\lambda} \biggl( \int_{0}^{\tau}\varphi_{1}(s) \bigl(g\bigl( \eta (s-\tau)\bigr)+h_{1}\bigl(\eta(s-\tau)\bigr) \bigr)\,ds \\ &{} +\biggl.\biggl. \int_{\tau}^{1}\varphi_{1}(s) \biggl(g\biggl( \frac{k}{2}(s-\tau)^{2}\biggr)+h_{1}(r) \biggr)\,ds \biggr)\biggr.^{-1}. \end{aligned}$$
$$\begin{aligned} \bigl\vert (y_{n}) (t_{1})-(y_{n}) (t_{2})\bigr\vert \leq&\lambda \int_{0}^{\tau}\bigl\vert G^{*}(t_{1},s)-G^{*}(t_{2},s) \bigr\vert \varphi_{1}(s) \bigl(g\bigl(\eta (s-\tau) \bigr)+h_{1}\bigl(\eta(s-\tau)\bigr) \bigr)\,ds \\ &{} +\lambda \int_{\tau}^{1}\bigl\vert G^{*}(t_{1},s)-G^{*}(t_{2},s) \bigr\vert \varphi_{1}(s) \biggl(g\biggl(\frac{k}{2}(s- \tau)^{2}\biggr)+h_{1}(r) \biggr)\,ds \\ < &\frac{\epsilon}{2}+\frac{\epsilon}{2} \\ =&\epsilon. \end{aligned}$$
$$y_{n}(t)= \int_{0}^{1}\lambda G^{*}(t,s) \bigl(f_{n} \bigl(s,y^{*}_{n}(s-\tau)\bigr)+\rho(s) \bigr)\,ds, $$
$$y(t)= \int_{0}^{1}\lambda G^{*}(t,s) \bigl(f\bigl(s,y^{*}(s- \tau)\bigr)+\rho(s) \bigr)\,ds. $$
Theorem 3.4
Let (H1) and (H2) hold. Assume that there exists a subinterval
\([\beta,\gamma]\subset(\tau,1)\)
satisfying
where
\(\lambda\in(0,+\infty)\). Then boundary value problem (1.1) has at least one positive solution.
$$\begin{aligned}& \max g(\cdot)+\max h_{1}(\cdot)\leq\frac{r_{2}}{\frac{\lambda}{\Gamma (\alpha)}\int_{0}^{1}s(1-s)^{\alpha-2}\varphi_{1}(s)\,ds}, \\ & \min_{0\leq y\leq r_{1}} h_{2}(y)\geq\frac{r_{1}}{\frac{\lambda\zeta _{2}}{\Gamma(\alpha)}\int_{\beta}^{\gamma}s(1-s)^{\alpha-2}\varphi_{2}(s)\,ds}, \end{aligned}$$
Proof
In view of Theorem 3.1, for \(y\in\partial\Omega _{2}\), \(t\in[0,1]\), we obtain Then from (H1), (3.18), and Lemma 2.3, we get Therefore, for \(y\in\partial\Omega_{2}\), we have \(\|Ty\|\leq\|y\|\).
$$ y(t)-t^{5-\alpha}\omega(t)\geq\frac{r_{2}}{2}t^{2}. $$
(3.18)
$$\begin{aligned} (Ty) (t) \leq&\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{0}^{\tau }s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g \bigl(\eta(s-\tau) \bigr)+h_{1} \bigl( \eta(s-\tau) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s) \\ &{}\times\biggl(g \biggl(\frac{r_{2}}{2}(s-\tau)^{2} \biggr)+h_{1} \bigl(y(s-\tau)-(s-\tau )^{5-\alpha}\omega(s-\tau) \bigr) \biggr)\,ds \\ \leq&\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{0}^{\tau }s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g \bigl(\eta(s-\tau) \bigr)+h_{1} \bigl( \eta(s-\tau) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda t^{5-\alpha}}{\Gamma(\alpha)} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s) \biggl(g \biggl(\frac{r_{2}}{2}(s-\tau)^{2} \biggr)+h_{1} \bigl(y(s-\tau) \bigr) \biggr)\,ds \\ \leq&\frac{\lambda}{\Gamma(\alpha)} \int_{0}^{\tau}s(1-s)^{\alpha-2} \varphi_{1}(s) \bigl(g \bigl(\eta(s-\tau) \bigr)+h_{1} \bigl( \eta(s-\tau) \bigr) \bigr)\,ds \\ &{} +\frac{\lambda}{\Gamma(\alpha)} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s) \biggl(g \biggl(\frac{k}{2}(s-\tau)^{2} \biggr)+h_{1}(r_{2}) \biggr)\,ds \\ \leq&\frac{\lambda}{\Gamma(\alpha)}\frac{r_{2}}{\frac{\lambda}{\Gamma (\alpha)}\int_{0}^{1}s(1-s)^{\alpha-2}\varphi_{1}(s)\,ds} \int_{0}^{\tau }s(1-s)^{\alpha-2} \varphi_{1}(s)\,ds \\ &{} +\frac{\lambda}{\Gamma(\alpha)}\frac{r_{2}}{\frac{\lambda}{\Gamma (\alpha)}\int_{0}^{1}s(1-s)^{\alpha-2}\varphi_{1}(s)\,ds} \int_{\tau}^{1} s(1-s)^{\alpha-2} \varphi_{1}(s)\,ds \leq r_{2}. \end{aligned}$$
On the other hand, for \(y\in\partial\Omega_{1}\), like for (3.6), for \(t\in[0,1]\), we obtain Thus from (H1), (3.19), and Lemma 2.3, we get Therefore, for \(y\in\partial\Omega_{1}\), we have \(\|Ty\|\geq\|y\|\). Arguments similar to those at the end of the proof of Theorem 3.1 show that boundary value problem (1.1) has a positive solution. The proof is completed. □
$$\begin{aligned} y(t)-t^{5-\alpha}\omega(t)&\geq t^{2}(r_{1}-c) \\ &\geq\frac{r_{1}}{2}t^{2}. \end{aligned}$$
(3.19)
$$\begin{aligned} \|Ty\| \geq& \int_{\beta}^{\gamma}\lambda\min_{t\in[\beta,\gamma ]}G^{*}(t,s) \bigl(f\bigl(s,(s-\tau)^{\alpha-5}y(s-\tau)-\omega(s-\tau)\bigr)+\rho (s) \bigr)\,ds \\ \geq& \int_{\beta}^{\gamma}\lambda\min_{t\in[\beta,\gamma ]}G^{*}(t,s) \varphi_{2}(s)h_{2} \biggl(\frac{r_{1}}{2}(s- \tau)^{2} \biggr)\,ds \\ \geq&\frac{\lambda\zeta_{2}}{\Gamma(\alpha)}\frac{r_{1}}{\frac{\lambda \zeta_{2}}{\Gamma(\alpha)}\int_{\beta}^{\gamma}s(1-s)^{\alpha-2}\varphi_{2}(s)\,ds} \int_{\beta}^{\gamma}s(1-s)^{\alpha-2} \varphi_{2}(s)\,ds \\ \geq& r_{1}. \end{aligned}$$
Declarations
Acknowledgements
This research is supported by the Natural Science Foundation of China (11572205, 61374002, 61374074), and supported by Shandong Provincial Natural Science Foundation (ZR2013AL003).
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)
Department of Basic Courses, Shandong Women’s University
(2)
School of Mathematical Sciences, University of Jinan
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