Necessary and sufficient condition for the existence of positive solution to singular fractional differential equations

Abstract

In this paper, we discuss the existence of positive solution to singular fractional differential equations involving Caputo fractional derivative. Necessary and sufficient condition for the existence of $C^{2}[0,1]$ positive solution is obtained by means of the fixed point theorems on cones. In addition, the uniqueness results and the iterative sequences of the solution are also given.

1 Introduction

In this paper, we consider the following singular fractional differential equation:

$$ \left \{ \textstyle\begin{array}{l} {}^{c}D^{\alpha}_{0+}u(t)+ f(t,u(t))=0 ,\quad 0< t< 1 , \\ u(0)= u'(1)=u''(0)=0, \end{array}\displaystyle \displaystyle \displaystyle \right . $$

(1.1)

where $2 < \alpha\leq3$ is a real number, ${}^{\mathrm{c}}D^{\alpha}_{0+}$ is the Caputo fractional derivative and f may be singular at $t = 0,1$.

Singular differential equation boundary value problems (BVP for short) arise from many branches of applied mathematics and physics. The theory of singular boundary value problems has become an important area of investigation in recent years. Differential equations of fractional order arise from many engineering and scientific disciplines as the mathematical modeling of systems and processes in the fields of physics, chemistry, control theory, etc.; see [1–4] and the references therein. Recently, much attention has been paid to the existence results of solutions for fractional differential equations, for example [5–17].

In [5], Bai and Qiu considered the existence of positive solution to problem (1.1), where $2 < \alpha\leq3$ is a real number, ${}^{\mathrm{c}}D^{\alpha}_{0+}$ is the Caputo fractional derivative, $f: (0,1]\times[0,\infty)\rightarrow[0,\infty)$ is continuous and singular at $t = 0$. The sufficient conditions for the existence of positive solution to (1.1) were obtained by using the Krasnosel’skii fixed-point theorem and the Leray-Schauder nonlinear alternative.

In [9], the authors investigated the existence of positive solution to the following boundary value problem:

$$ \left \{ \textstyle\begin{array}{l} {}^{\mathrm{c}}D^{\alpha}_{0+}u(t)+\lambda f(t,u(t))=0 , \quad 0< t< 1 , \\ u(0)= u'(1)=u''(0)=0, \end{array}\displaystyle \displaystyle \displaystyle \right . $$

(1.2)

where $2 < \alpha\leq3$ is a real number, ${}^{\mathrm{c}}D^{\alpha}_{0+}$ is the Caputo fractional derivative, λ is a positive parameter, f may change sign and may be singular at $t = 0, 1$.

In recent years, many results dealing with necessary and sufficient conditions for the existence of positive solutions to integer-order differential equations were obtained (for example, [18–23]) with one of the following conditions:

• (A1) $f\in C((0,1)\times[0,\infty),[0,\infty))$, $f(t,e(t))>0$, $t\in (0,1)$, here $e\in C([0,1],[0,\infty))$; there exist constants $0<\lambda_{1}\leq\lambda_{2}<1$ such that for $(t,x)\in (0,1)\times[0,\infty)$,

  $$ c^{\lambda_{2}}f(t,x)\leq f(t,cx)\leq c^{\lambda_{1}}f(t,x),\quad \forall c \in(0,1). $$

  (1.3)

• (A2) $f\in C((0,1)\times[0,\infty),[0,\infty))$; for each fixed $t\in(0,1)$, $f(t,x)$ is increasing in x; there exists $0<\alpha<1$ such that

  $$ f(t,rx)\geq r^{\alpha}f(t,x),\quad \forall0< r< 1, (t,x)\in(0,1)\times [0, \infty ). $$

  (1.4)

• (A3) $f\in C((0,1)\times[0,\infty),[0,\infty))$; for each fixed $t\in(0,1)$, $f(t,x)$ is increasing in x; for all $0< r<1$, there exists $g (r)=m(r^{-\alpha}-1)$ such that

  $$ f(t,rx)\geq r\bigl(1+g (r)\bigr)f(t,x),\quad \forall(t,x)\in(0,1)\times [0, \infty ), 0< m\leq1, 0< \alpha< 1. $$

  (1.5)

While there are a lot of works dealing with necessary and sufficient conditions for integer-order differential equations, the results of fractional differential equations are relatively scarce due to the difficulties caused by the singularity of nonlinearity. In [7], the authors considered the necessary and sufficient condition for the existence of $C^{3}[0, 1]$ positive solution of singular sub-linear boundary value problems for a fractional differential equation with condition (A2).

Inspired by the previous works, in this paper we aim to establish necessary and sufficient condition for the existence of $C^{2}[0, 1]$ positive solutions to BVP (1.1). In this paper, by a $C^{2}[0, 1]$ positive solution to BVP (1.1), we mean a function $u\in C'[0,1]\cap C^{2}[0,1)$ which satisfies $u''(1^{-})$ exists, is positive on $(0,1]$ and satisfies (1.1).

Throughout this paper, we assume that the following condition holds.

• (H) $f\in C((0,1)\times[0,\infty),[0,\infty))$, $f(t,x)$ is increasing in x; there exists a function $\eta: [0,1]\rightarrow [0,+\infty)$ satisfying $\eta(r)>r$ ($0< r<1$) such that

  $$ f(t,rx)\geq\eta(r)f(t,x),\quad \forall0< r< 1, (t,x)\in(0,1)\times [0,\infty ). $$

  (1.6)

Remark 1.1

Inequality (1.6) is equivalent to

$$ f(t,rx)\leq\frac{f(t,x)}{\eta(r^{-1})},\quad \forall r> 1, (t,x)\in(0,1)\times [0,\infty ). $$

(1.7)

Remark 1.2

Condition (H) includes conditions (A1), (A2) and (A3) as special cases.

Remark 1.3

The function η defined in (H) satisfies $\eta(1)= 1$, and $\eta(r) \leq1$, $\forall r \in(0,1)$.

Remark 1.4

If condition (H) holds, then there exists a strictly increasing function φ satisfying $\varphi(r)>r$ ($0< r<1$) such that

$$ f(t,rx)\geq\varphi(r)f(t,x),\quad \forall0< r< 1, (t,x)\in(0,1)\times [0,\infty ), $$

(1.8)

without loss of generality, we may suppose that η is strictly increasing on $(0.1]$.

Proof

If there exist $t_{0}\in (0,1)$, $x_{0}> 0$ such that $f(t_{0},x_{0})=0$. By the monotonicity of f and (1.7), we have $f(t_{0},x)\equiv0$, $\forall x\in[0,+\infty)$. Set

$$ \Omega=\bigl\{ t\in(0,1): \exists x_{1}> 0\mbox{ such that }f(t,x_{1})=0\bigr\} . $$

(1.9)

For any $r\in(0,1)$, denote

$$ D_{r}=\bigl\{ c: f(t,rx)\geq c f(t,x), (t,x)\in\bigl((0,1)\backslash \Omega \bigr)\times(0,\infty)\bigr\} . $$

(1.10)

It is clear that $\sup D_{r}$ exists. Let $\psi(r)=\sup D_{r}$, then

$$ f(t,rx)\geq\psi(r)f(t,x),\quad \forall(t,x)\in(0,1)\times [0,\infty ), $$

(1.11)

and $r<\eta(r)\leq\psi(r)\leq1$. For any $0< r_{2}< r_{1}<1$ and $x\in[0,\infty)$, we have

$$ f(t,r_{1}x)=f \biggl(t,r_{2}\cdot\frac{r_{1}}{r_{2}}x \biggr) \geq\psi (r_{2})f \biggl(t,\frac{r_{1}}{r_{2}}x \biggr)\geq\psi (r_{2})f(t,x). $$

(1.12)

By the definition of ψ, we get $\psi(r_{1})\geq\psi (r_{2})$, therefore ψ is nondecreasing. Let $\varphi (r)=\frac{\psi(r)+r}{2}$. It is clear that φ is strictly increasing on $(0,1)$, satisfies $\varphi(r)>r$ and

$$ f(t,rx)\geq\varphi(r)f(t,x),\quad \forall(t,x)\in(0,1)\times [0,\infty ), r \in(0,1). $$

(1.13)

The proof is completed. □

2 Basic definitions and preliminaries

In this section, we present some preliminaries and lemmas that are useful to the proof of the main results, we also present here some necessary definitions.

Definition 2.1

The Riemann-Liouville fractional integral $I^{\alpha}_{0+}$ and derivative $D^{\alpha}_{0+}$ are defined by

$$ I^{\alpha}_{0+}u(t)=\frac{1}{\Gamma(\alpha)}\int^{t}_{0}(t-s)^{\alpha-1}u(s) \,ds $$

(2.1)

and

$$ D^{\alpha}_{0+}u(t)=\frac{1}{\Gamma(n-\alpha)} \biggl(\frac {d}{dt} \biggr)^{(n)}\int^{t}_{0}(t-s)^{n-\alpha-1}u(s) \,ds, $$

(2.2)

where $n=[\alpha]+1$, $[\alpha]$ denotes the integer part of number α, provided that the right-hand side is defined pointwise on $(0,+\infty)$.

Definition 2.2

(see [2])

The Caputo fractional derivative of order $\alpha> 0$ on $[0,1]$ is defined via the above Riemann-Liouville fractional derivative by

$$ {}^{\mathrm{c}}D^{\alpha}_{0+}u(t)=D^{\alpha}_{0+} \Biggl[u(t)-\sum^{n-1}_{k=0}u^{(k)}(0)t^{k} \Biggr], $$

(2.3)

where $n=[\alpha]+1$.

Remark 2.1

(see Theorem 2.1 of [2])

If $u(t)\in AC^{n}[0,1]$, then the Caputo fractional derivative of order $\alpha> 0$ exists almost everywhere on $[0,1]$ and can be represented by

$$ {}^{\mathrm{c}}D^{\alpha}_{0+}u(t)=\frac{1}{\Gamma(n-\alpha)}\int ^{t}_{0}(t-s)^{n-\alpha-1}u^{(n)}(s) \,ds, $$

(2.4)

where $n=[\alpha]+1$, and

$$AC^{n}[0,1]= \biggl\{ y : [0,1]\rightarrow\mathbb{R}\mbox{ and } \frac{d^{n-1}y}{dt^{n-1}} \mbox{ is absolutely continuous on }[0, 1] \biggr\} . $$

Lemma 2.1

(see Lemma 2.5 of [2])

Let $\alpha> 0$, $u\in L[0,1]$ and $D^{\alpha}_{0+}u\in L[0,1]$, then the following equality holds:

$$I^{\alpha}_{0+}D^{\alpha}_{0+}u(t)=u(t)+c_{1}t^{\alpha -1}+c_{2}t^{\alpha-2}+ \cdots+c_{n}t^{\alpha-n}, $$

where $c_{i}\in R$, $i=1,2,\ldots,n$, $n=[\alpha]+1$.

Lemma 2.2

(see Lemma 2.4 of [2])

If $\alpha> 0$ and $y(t)\in L[0,1]$, then the equality

$$D^{\alpha}_{0+}I^{\alpha}_{0+}y(t)=y(t) $$

holds almost everywhere on $[0,1]$.

Lemma 2.3

(see Property 2.8 of [2])

Let $\alpha>\beta> 0$. If $y(t)\in L[0,1]$, then

$$D^{\beta}_{0+}I^{\alpha}_{0+}y(t)=I^{\alpha-\beta}_{0+}y(t). $$

Lemma 2.4

If $2 < \alpha\leq 3$, $y\in L[0,1]\cap C(0,1)$ and

$$ \lim_{t\rightarrow0+}t^{\alpha-2}\int_{0}^{1}(1-s)^{\alpha -2}y(ts) \,ds=0, $$

(2.5)

then the problem

$$ \left \{ \textstyle\begin{array}{l} {}^{\mathrm{c}}D^{\alpha}_{0+}u(t)+y(t)=0 ,\quad 0< t< 1 , \\ u(0)= u'(1)=u''(0)=0, \end{array}\displaystyle \displaystyle \displaystyle \right . $$

(2.6)

has a unique solution

$$ u(t)= \int_{0}^{1}G(t,s)y(s)\,ds, $$

(2.7)

where

$$ G(t,s)=\frac{1}{\Gamma(\alpha)}\left \{ \textstyle\begin{array}{l@{\quad}l} (\alpha-1)t(1-s)^{\alpha-2}, &0\leq t\leq s\leq1 , \\ (\alpha-1)t(1-s)^{\alpha-2}-(t-s)^{\alpha-1}, &0\leq s\leq t\leq1. \end{array}\displaystyle \right . $$

(2.8)

Proof

Deduced from Lemma 2.1, the solution of (2.6) satisfies

$$u(t)=u'(0)t-I^{\alpha}_{0+}y(t)+c_{1}t^{\alpha-1}+c_{2}t^{\alpha -2}+c_{3}t^{\alpha-3}. $$

By direct calculation of $u(0)$, $u'(0)$ and $u''(0)$, there is $c_{1}=c_{2}=c_{3}=0$. Consequently,

$$ u(t)=u'(0)t-\frac{1}{\Gamma(\alpha)}\int^{t}_{0}(t-s)^{\alpha -1}y(s) \,ds $$

(2.9)

and

$$ u'(t)=u'(0)-\frac{1}{\Gamma(\alpha-1)}\int^{t}_{0}(t-s)^{\alpha -2}y(s) \,ds. $$

(2.10)

By $u'(1)=0$, we have

$$ u'(0)=\frac{1}{\Gamma(\alpha-1)}\int^{1}_{0}(1-s)^{\alpha -2}y(s) \,ds. $$

(2.11)

Therefore,

$$u(t)=\int_{0}^{1}G(t,s)y(s)\,ds. $$

On the other hand, for

$$u(t)=\int_{0}^{1}G(t,s)y(s)\,ds= \frac{t}{\Gamma(\alpha-1)}\int^{1}_{0}(1-s)^{\alpha-2}y(s) \,ds- \frac{1}{\Gamma(\alpha)}\int^{t}_{0}(t-s)^{\alpha-1}y(s) \,ds, $$

we have $u(0)= u'(1)=0$ and $u'(0)=\frac{1}{\Gamma(\alpha-1)}\int^{1}_{0}(1-s)^{\alpha-2}y(s)\,ds$. From (2.5), we get

$$\begin{aligned} u''(0)&=\lim_{t\rightarrow0+}\frac{u'(t)-u'(0)}{t} =\lim_{t\rightarrow 0+}\frac{\int^{t}_{0}(t-s)^{\alpha-2}y(s)\,ds}{\Gamma(\alpha-1)t} \\ &=\lim_{t\rightarrow 0+}\frac{t^{\alpha-2}}{\Gamma(\alpha-1)}\int_{0}^{1}(1-x)^{\alpha -2}y(tx) \,dx=0. \end{aligned}$$

(2.12)

By Definition 2.2 and Lemma 2.2, we have u is a solution of problem (2.6). The proof is completed. □

Remark 2.2

If $\alpha=3$ and $y\in L[0,1]$, then condition (2.5) holds naturally.

Lemma 2.5

(see [9])

The function $G(t,s)$ has the following properties:

• (1) $G(t,s) \leq\frac{1}{\Gamma(\alpha-1)}t(1-s)^{\alpha-2}$, $\forall t,s\in [0,1]$;

• (2) $G(t,s) \leq \frac{1}{\Gamma(\alpha-1)}(\alpha-2+s)(1-s)^{\alpha-2}$, $\forall t,s\in[0,1]$;

• (3) $G(t,s) \geq \frac{1}{\Gamma(\alpha)}(\alpha-2+s)t(1-s)^{\alpha-2}$, $\forall t,s\in[0,1]$.

Lemma 2.6

Suppose that u is a positive solution of BVP (1.1), then there exist $L_{u}, l_{u}> 0$ such that

$$ l_{u}t\leq u(t)\leq L_{u}t,\quad \forall t \in[0,1]. $$

(2.13)

Proof

By Lemma 2.4, u can be expressed by

$$ u(t)= \int_{0}^{1}G(t,s)f\bigl(s,u(s)\bigr) \,ds. $$

(2.14)

From (1) of Lemma 2.5, we have

$$ u(t)\leq \frac{t}{\Gamma(\alpha-1)}\int_{0}^{1}(1-s)^{\alpha -2}f \bigl(s,u(s)\bigr)\,ds. $$

(2.15)

By (2), (3) of Lemma 2.5, we get

$$ u(t)\geq\frac{ t}{\Gamma(\alpha)}\int_{0}^{1}( \alpha-2+s) (1-s)^{\alpha -2}f\bigl(s,u(s)\bigr)\,ds $$

(2.16)

and

$$ u(t)\leq \frac{1}{\Gamma(\alpha-1)}\int_{0}^{1}( \alpha-2+s) (1-s)^{\alpha -2}f\bigl(s,u(s)\bigr)\,ds. $$

(2.17)

Inequalities (2.16) and (2.17) imply $u(t)\geq\frac {t}{\alpha-1}\| u(t)\|$.

Let

$$ l_{u}=\frac{\| u(t)\|}{\alpha-1},\qquad L_{u}= \frac{\int_{0}^{1}(1-s)^{\alpha -2}f(s,u(s))\,ds}{\Gamma(\alpha-1)}. $$

(2.18)

Then (2.13) holds. The proof is completed. □

Lemma 2.7

Assume that $g(x)$, $\{g_{n}(x)\}$, $h(x)$, $\{h_{n}(x)\}$ are Lebesgue integrable on $[0,1]$, satisfy

$$ \bigl\vert g_{n}(x)\bigr\vert \leq h_{n}(x),\qquad \lim_{n\rightarrow\infty}g_{n}(x)=g(x),\qquad \lim _{n\rightarrow\infty}h_{n}(x)=h(x), \quad \textit{a.e. } [0,1] $$

(2.19)

and

$$ \lim_{n\rightarrow\infty}\int_{0}^{1}h_{n}(x) \,dx=\int_{0}^{1}h(x)\,dx, $$

(2.20)

then

$$ \lim_{n\rightarrow\infty}\int_{0}^{1}g_{n}(x) \,dx=\int_{0}^{1}g(x)\,dx. $$

(2.21)

Proof

By $|g_{n}(x)|\leq h_{n}(x)$, a.e. $[0,1]$, we have

$$ \bigl\vert g(x)\bigr\vert \leq h(x), \quad \mbox{a.e. }[0,1]. $$

(2.22)

Set

$$ k_{n}(x)=h_{n}(x)+h(x)-\bigl\vert g_{n}(x)-g(x) \bigr\vert , $$

(2.23)

then $k_{n}(x)\rightarrow2h(x)$ ($n\rightarrow\infty$) a.e. $[0,1]$. By the Fatou lemma, we get

$$\begin{aligned} \int_{0}^{1}2h(x)\,dx&=\int_{0}^{1} \lim_{n\rightarrow \infty}k_{n}(x)\,dx \leq\liminf _{n\rightarrow \infty}\int_{0}^{1}k_{n}(x) \,dx \\ &=2\int_{0}^{1}h(x)\,dx-\limsup _{n\rightarrow\infty} \int_{0}^{1}\bigl\vert g_{n}(x)-g(x)\bigr\vert \,dx, \end{aligned}$$

(2.24)

which implies

$$ \lim_{n\rightarrow \infty}\int_{0}^{1}\bigl\vert g_{n}(x)-g(x)\bigr\vert \,dx=0. $$

(2.25)

Thus

$$ \lim_{n\rightarrow\infty}\int_{0}^{1}g_{n}(x) \,dx=\int_{0}^{1}g(x)\,dx. $$

(2.26)

The proof is completed. □

3 Main result

Theorem 3.1

Suppose that (H) holds. Then the necessary and sufficient condition for BVP (1.1) to have a $C^{2}[0, 1]$ positive solution is

$$\begin{aligned}& 0< \int_{0}^{1}(1-s)^{\alpha-3}f(s,s)\,ds< + \infty, \end{aligned}$$

(3.1a)

$$\begin{aligned}& \lim_{t\rightarrow 0+}t^{\alpha-2}\int_{0}^{1}(1-s)^{\alpha-3}f(ts,ts) \,ds=0, \end{aligned}$$

(3.1b)

$$\begin{aligned}& \lim_{t\rightarrow 1-}\int_{0}^{1}(1-s)^{\alpha-3}f(ts,ts) \,ds=\int_{0}^{1}(1-s)^{\alpha -3}f(s,s) \,ds. \end{aligned}$$

(3.1c)

Proof

(i) Necessity. Assume that u is a $C^{2}[0, 1]$ positive solution of BVP (1.1). In the following, we will divide the rather long proof into three steps.

Step 1: By Lemma 2.4, u can be expressed by

$$ u(t)= \frac{t}{\Gamma(\alpha-1)}\int_{0}^{1}(1-s)^{\alpha -2}f \bigl(s,u(s)\bigr)\,ds -\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha -1}f \bigl(s,u(s)\bigr)\,ds. $$

(3.2)

For any $t \in(0,1)$, Lemma 2.3 implies

$$ u'(t)= \frac{1}{\Gamma(\alpha-1)}\int_{0}^{1}(1-s)^{\alpha -2}f \bigl(s,u(s)\bigr)\,ds -\frac{1}{\Gamma(\alpha-1)}\int_{0}^{t}(t-s)^{\alpha -2}f \bigl(s,u(s)\bigr)\,ds $$

(3.3)

and

$$\begin{aligned} u''(t)&= -\frac{1}{\Gamma(\alpha-2)}\int_{0}^{t}(t-s)^{\alpha-3}f \bigl(s,u(s)\bigr)\,ds \\ &=-\frac{t^{\alpha-2}}{\Gamma(\alpha-2)}\int_{0}^{1}(1-s)^{\alpha -3}f \bigl(ts,u(ts)\bigr)\,ds. \end{aligned}$$

(3.4)

It is clear that $u'(t)\geq0$, and $u''(t)\leq0$, $\forall t \in(0,1)$.

$\forall\varepsilon\in(0,\frac{1}{2})$, $t \in(0,1)$, we deduce that

$$ \int_{\varepsilon}^{1-\varepsilon}(1-s)^{\alpha-3}f\bigl(ts,u(ts) \bigr)\,ds\leq -\Gamma(\alpha-2)t^{2-\alpha}u''(t). $$

(3.5)

Let $t\rightarrow 1$, noticing (H) and the existence of $u''(1^{-})$, we have

$$ \int_{\varepsilon}^{1-\varepsilon}(1-s)^{\alpha-3}f\bigl(s,u(s) \bigr)\,ds\leq -\Gamma(\alpha-2)u''\bigl(1^{-} \bigr),\quad \forall \varepsilon\in \biggl(0,\frac{1}{2} \biggr). $$

(3.6)

Thus $\int_{0}^{1}(1-s)^{\alpha-3}f(s,u(s))\,ds$ is well defined, that is, $u''(1)$ is well defined. By Lemma 2.6, we have

$$\begin{aligned} \int_{0}^{1}(1-s)^{\alpha-3}f\bigl(s,u(s) \bigr)\,ds &\geq\int_{0}^{1}(1-s)^{\alpha-3}f(s,l_{u}s) \,ds \\ &\geq \int_{0}^{1}(1-s)^{\alpha-3}f \bigl(s,\min\{1, l_{u}\}s \bigr)\,ds \\ &\geq\eta\bigl(\min\{1, l_{u}\}\bigr) \int_{0}^{1}(1-s)^{\alpha-3}f(s,s) \,ds \\ &\geq\min\{1, l_{u}\} \int_{0}^{1}(1-s)^{\alpha-3}f(s,s) \,ds, \end{aligned}$$

(3.7)

which implies

$$ \int_{0}^{1}(1-s)^{\alpha-3}f(s,s)\,ds< + \infty. $$

(3.8)

On the other hand, we have

$$\begin{aligned} u(t)&= \int_{0}^{1}G(t,s)f\bigl(s,u(s)\bigr)\,ds \leq \frac{t}{\Gamma(\alpha-1)}\int_{0}^{1}(1-s)^{\alpha -2}f(s,L_{u}s) \,ds \\ &\leq \frac{t}{\Gamma(\alpha-1)}\int_{0}^{1}(1-s)^{\alpha -2}f \bigl(s,(1+L_{u})s\bigr)\,ds \\ &\leq\frac{t}{\Gamma(\alpha-1)\eta([1+L_{u}]^{-1})} \int_{0}^{1}(1-s)^{\alpha-3}f(s,s) \,ds. \end{aligned}$$

(3.9)

Since u is a positive solution, then

$$ \int_{0}^{1}(1-s)^{\alpha-3}f(s,s)\,ds > 0. $$

(3.10)

Inequalities (3.8) and (3.10) yield (3.1a) holds.

Step 2: From u is a $C^{2}[0, 1]$ positive solution, we get

$$ \lim_{t\rightarrow 0+}t^{\alpha-2}\int_{0}^{1}(1-s)^{\alpha-3}f \bigl(ts,u(ts)\bigr)\,ds=-\Gamma (\alpha-2)u''(0)=0 $$

(3.11)

and

$$ \lim_{t\rightarrow 1-}\int_{0}^{1}(1-s)^{\alpha-3}f \bigl(ts,u(ts)\bigr)\,ds=\int_{0}^{1}(1-s)^{\alpha-3}f \bigl(s,u(s)\bigr)\,ds. $$

(3.12)

Similar to (3.7) and (3.9), we have

$$ f\bigl(s,u(s)\bigr)\geq\min\{1, l_{u}\}f(s,s),\quad \forall s \in (0,1) $$

(3.13)

and

$$ f\bigl(s,u(s)\bigr)\leq \frac{1}{\eta([1+L_{u}]^{-1})} f(s,s),\quad \forall s \in(0,1). $$

(3.14)

Then, for any $t\in(0,1)$, we have

$$ \int_{0}^{1}(1-s)^{\alpha-3}f\bigl(ts,u(ts) \bigr)\,ds\geq\min\{1, l_{u}\}\int_{0}^{1}(1-s)^{\alpha-3}f(ts,ts) \,ds $$

(3.15)

and

$$ \int_{0}^{1}(1-s)^{\alpha-3}f\bigl(ts,u(ts) \bigr)\,ds\leq \frac{1}{\eta([1+L_{u}]^{-1})} \int_{0}^{1}(1-s)^{\alpha-3}f(ts,ts) \,ds. $$

(3.16)

Combining (3.11) with (3.15), we obtain (3.1b) holds.

Step 3: $\forall\{t_{n}\}\subset(0,1)$ satisfies $t_{n}\rightarrow 1$ ($n\rightarrow\infty$). Set

$$ \begin{aligned} &g_{n}(s) =\min\{1, l_{u} \}(1-s)^{\alpha-3}f(t_{n}s,t_{n}s),\qquad g(s)=\min\{1, l_{u}\}(1-s)^{\alpha-3}f(s,s), \\ &h_{n}(s) =(1-s)^{\alpha-3}f\bigl(t_{n}s,u(t_{n}s) \bigr),\qquad h(s)=(1-s)^{\alpha -3}f\bigl(s,u(s)\bigr). \end{aligned} $$

(3.17)

It is clear that $\{g_{n}(s)\}$, $g(s)$, $\{h_{n}(x)\}$, $h(x)$ are Lebesgue integrable on $[0,1]$, and

$$ \lim_{n\rightarrow \infty}g_{n}(x)=g(x), \qquad \lim _{n\rightarrow \infty}h_{n}(x)=h(x),\quad \mbox{a.e. }[0,1]. $$

(3.18)

From (3.13), we get

$$0\leq g_{n}(x)\leq h_{n}(x). $$

Equation (3.12) yields

$$ \lim_{n\rightarrow\infty}\int_{0}^{1}h_{n}(s) \,ds=\int_{0}^{1}h(s)\,ds. $$

(3.19)

By Lemma 2.7, we have

$$ \lim_{n\rightarrow \infty}\int_{0}^{1}(1-s)^{\alpha-3}f(t_{n}s,t_{n}s) \,ds=\int_{0}^{1}(1-s)^{\alpha-3}f(s,s) \,ds. $$

(3.20)

Then (3.1c) holds.

(ii) Sufficiency. Let $P=\{u \in C[0,1]: u\geq0\}$. Clearly P is a normal cone of $C[0,1]$. Denote $e(t)=t$, and

$$P_{e}=\{u \in P: \exists L_{u}, l_{u}>0\mbox{ such that }l_{u}e \leq u\leq L_{u}e\}. $$

Set

$$ Au(t)= \int_{0}^{1}G(t,s)f\bigl(s,u(s)\bigr) \,ds. $$

(3.21)

For any $u\in P_{e}$, by (3.1a), (3.13), (3.14) and Lemma 2.5, we have

$$\begin{aligned} Au(t)&\leq \frac{t}{\Gamma(\alpha-1)}\int_{0}^{1}(1-s)^{\alpha-2}f \bigl(s,u(s)\bigr)\,ds \\ &\leq\frac{t}{\Gamma(\alpha-1)\eta((1+L_{u})^{-1})} \int_{0}^{1}(1-s)^{\alpha-3}f(s,s) \,ds \end{aligned}$$

(3.22)

and

$$\begin{aligned} Au(t)& \geq \frac{t}{\Gamma(\alpha)}\int_{0}^{1}( \alpha-2+s) (1-s)^{\alpha -2}f\bigl(s,u(s)\bigr)\,ds \\ &\geq\frac{t}{\Gamma(\alpha)} \min\{1, l_{u}\}\int_{0}^{1}( \alpha -2) (1-s)^{\alpha-2}f(s,s)\,ds, \end{aligned}$$

(3.23)

which implies $A: P_{e}\rightarrow P_{e}$ is well defined.

It is clear that $e\in P_{e}$, so there exist positive numbers $L_{e}>1> l_{e}>0$ such that $l_{e}e \leq Ae\leq L_{e}e$. Noticing $\eta(r)>r$ on $(0,1)$, we can choose a positive integer m large enough such that

$$ \biggl(\frac{\eta(l_{e})}{l_{e}} \biggr)^{m}>\frac{1}{l_{e}},\qquad \biggl(\frac{\eta (L_{e}^{-1})}{L_{e}^{-1}} \biggr)^{m}>L_{e}. $$

(3.24)

Let

$$ u_{0}=l_{e}^{m}e,\qquad v_{0}=L_{e}^{m}e, \qquad u_{n+1}=Au_{n},\qquad v_{n+1}=Av_{n}, \quad n=0,1,2,\ldots. $$

(3.25)

It is easy to see that

$$ \biggl(\frac{l_{e}}{L_{e}} \biggr)^{m}v_{0}=u_{0} \leq v_{0}= \biggl(\frac{L_{e}}{l_{e}} \biggr)^{m}u_{0} $$

(3.26)

and

$$\begin{aligned} u_{1}&=Au_{0}=\int_{0}^{1}G(t,s)f \bigl(s,l_{e}^{m}e(s)\bigr)\,ds \geq\eta(l_{e}) \int_{0}^{1}G(t,s)f\bigl(s,l_{e}^{m-1}e(s) \bigr)\,ds \\ & \geq \cdots\geq\eta^{m}(l_{e})\int_{0}^{1}G(t,s)f \bigl(s,e(s)\bigr)\,ds =\eta ^{m}(l_{e})Ae\geq \eta^{m}(l_{e})l_{e}e\geq l_{e}^{m}e=u_{0}. \end{aligned}$$

(3.27)

In a similar way, we can get $v_{0}\geq v_{1}$. It follows from the increasing property of A that

$$ u_{0}\leq u_{1}\leq\cdots\leq u_{n}\leq\cdots \leq v_{n}\leq\cdots \leq v_{1}\leq v_{0}. $$

(3.28)

Therefore, $u_{n}\geq u_{0} = (\frac{l_{e}}{L_{e}} )^{m}v_{0}\geq (\frac{l_{e}}{L_{e}} )^{m}v_{n}$. Let

$$ c_{n}=\sup\{c> 0| u_{n}\geq cv_{n}\},\quad n=1,2,\ldots. $$

(3.29)

Then $u_{n}\geq c_{n}v_{n}$. Noticing (3.28), we have $1\geq c_{n+1}\geq c_{n}$. Thus, we can suppose that $\{c_{n}\}$ converges to $c^{\ast}$. It is clear that $0< c^{\ast}\leq1$, we now prove that $c^{\ast}=1$. In fact, if $0< c^{\ast}<1$, then

$$ u_{n+1}=Au_{n}\geq A(c_{n}v_{n}) \geq A \biggl(\frac{c_{n}}{c^{\ast}}c^{\ast}v_{n} \biggr) \geq \eta \biggl(\frac{c_{n}}{c^{\ast}} \biggr)\eta\bigl(c^{\ast}\bigr) A(v_{n})\geq \frac{c_{n}}{c^{\ast}}\eta\bigl(c^{\ast}\bigr) v_{n+1}. $$

(3.30)

Therefore $c_{n+1}\geq \frac{c_{n}}{c^{\ast}}\eta(c^{\ast})$. Let $n\rightarrow\infty$, we have $c^{\ast}\geq\eta(c^{\ast})$, which contradicts (H). Hence $c^{\ast}=1$.

For each natural number p, we have

$$ \begin{aligned} &0\leq u_{n+p}-u_{n}\leq v_{n}-u_{n} \leq v_{n}-c_{n}v_{n}\leq (1-c_{n})v_{0}, \\ &0 \leq v_{n}-v_{n+p}\leq v_{n}-u_{n} \leq (1-c_{n})v_{0}. \end{aligned} $$

(3.31)

Since P is normal, then

$$ \|u_{n+p}-u_{n}\|\rightarrow0,\qquad \|v_{n}-v_{n+p} \|\rightarrow0,\quad n\rightarrow\infty, $$

(3.32)

which implies $\{u_{n}\}$, $\{v_{n}\}$ are Cauchy sequences. There exist $u_{\ast}$, $v_{\ast}$ such that $u_{n}\rightarrow u_{\ast}$, $v_{n}\rightarrow v_{\ast}$. From (3.28), we get $u_{n}\leq u_{\ast}\leq v_{\ast}\leq v_{n}$. By (3.31), we have $\|u_{\ast}-v_{\ast}\|\rightarrow0$. Then $u_{\ast}=v_{\ast}$ is a fixed point of A.

Equations (3.1b) and (3.16) yield

$$ \lim_{t\rightarrow 0+}t^{\alpha-2}\int_{0}^{1}(1-s)^{\alpha-3}f \bigl(ts,u_{\ast }(ts)\bigr)\,ds=0. $$

(3.33)

Then

$$ \lim_{t\rightarrow 0+}t^{\alpha-2}\int_{0}^{1}(1-s)^{\alpha-2}f \bigl(ts,u_{\ast }(ts)\bigr)\,ds=0. $$

(3.34)

By Lemma 2.4, $u_{\ast}$ is a positive solution of BVP (1.1).

Noticing (3.13), (3.14) and (3.1a), it is clear that $u_{\ast}\in C'[0,1]\cap C^{2}(0,1)$. Again from (3.1b) and (3.16), we get $u_{\ast }''(0+)=u_{\ast}''(0)=0$. By (3.1c), (3.14) and Lemma 2.7, we have (3.12) holds, which implies $u_{\ast}''(1^{-})$ exists. Therefore, $u_{\ast}$ is a $C^{2}[0, 1]$ positive solution of BVP (1.1). The proof is completed. □

Theorem 3.2

Suppose that (3.1a), (3.1b), (3.1c) and (H) hold. Then:

• (i) BVP (1.1) has a unique $C^{2}[0, 1]$ positive solution $u_{\ast}\in P_{e}$.

• (ii) For any initial value $\omega_{0}\in P_{e}$, the sequence of functions defined by

  $$ \omega_{n}=\int_{0}^{1}G(t,s)f \bigl(s,\omega_{n-1}(s)\bigr)\,ds,\quad n=1,2,\ldots $$

  (3.35)

  converges uniformly to $u_{\ast}$ on $[0, 1]$.

Proof

(i) It follows from Theorem 3.1 that BVP (1.1) has a $C^{2}[0, 1]$ positive solution $u_{\ast}\in P_{e}$. Let v be another $C^{2}[0, 1]$ positive solution of BVP (1.1). Lemma 2.6 implies $v\in P_{e}$. So there exist two positive numbers $0< l_{v}<1<L_{v}$ such that

$$ l_{v}e(t)\leq v(t)\leq L_{v}e(t),\quad t \in[0,1]. $$

(3.36)

Let m defined by (3.24) be large enough such that $l_{v}>l_{e}^{m}$ and $L_{v}< L_{e}^{m}$. Then

$$ u_{0}\leq v\leq v_{0}. $$

(3.37)

It is clear that A is an increasing operator and $A v= v$, therefore

$$ u_{n}\leq v\leq v_{n},\quad n=1,2,\ldots. $$

(3.38)

Let $n\rightarrow\infty$, we get $v=u_{\ast}$. So the $C^{2}[0, 1]$ positive solution of BVP (1.1) is unique.

(ii) For any initial value $\omega_{0}\in P_{e}$, there exist two positive numbers $0< l_{\omega_{0}}<1<L_{\omega_{0}}$ such that

$$ l_{\omega_{0}}e(t)\leq\omega_{0}(t)\leq L_{\omega_{0}}e(t), \quad t\in [0,1]. $$

(3.39)

Let m defined by (3.24) be large enough such that $l_{\omega_{0}}>l_{e}^{m}$ and $L_{\omega_{0}}< L_{e}^{m}$. Then

$$ u_{0}\leq\omega_{0}\leq v_{0}. $$

(3.40)

Notice that A is an increasing operator, we have

$$ u_{n}\leq \omega_{n}\leq v_{n},\quad n=1,2, \ldots. $$

(3.41)

Let $n\rightarrow\infty$, then $\omega_{n}=u_{\ast}$. It follows from (3.28), (3.31) and (3.41) that $\omega _{n}$ converges uniformly to the unique positive solution $u_{\ast}$ on $[0, 1]$. The proof is completed. □

4 Example

Example 4.1

Consider the following problem:

$$ \left \{ \textstyle\begin{array}{l} D^{\alpha}_{0+}u(t)+ f(t,u(t))=0,\quad t \in(0,1), 2< \alpha< 3, \\ u(0)= u'(1)=u''(0)=0, \end{array}\displaystyle \displaystyle \displaystyle \right . $$

(4.1)

where

$$f(t,x)=t^{-\sigma}x^{\beta}, \quad \sigma, \beta\in(0,1). $$

Obviously, assumption (H) holds. By Theorem 3.1, we have that the necessary and sufficient condition for the existence of a $C^{2}[0, 1]$ positive solution to BVP (4.1) is

$$\beta-\sigma+\alpha>2. $$

Example 4.2

Consider the following problem:

$$ \left \{ \textstyle\begin{array}{l} D^{\alpha}_{0+}u(t)+ a(t)f(u(t))=0,\quad t \in(0,1), 2< \alpha< 3, \\ u(0)= u'(1)=u''(0)=0, \end{array}\displaystyle \displaystyle \displaystyle \right . $$

(4.2)

where $a\in C((0,1),[0,\infty))$,

$$f(x)=\left \{ \textstyle\begin{array}{l@{\quad}l} x^{\beta}+x, &x \in[0,1], 0< \beta< 1, \\ 2x^{\beta},& x \in(1,+\infty). \end{array}\displaystyle \displaystyle \displaystyle \right . $$

Let

$$\eta(r)=\frac{r^{\beta}+r}{2}, $$

then assumption (H) holds. Noticing

$$x^{\beta}\leq f(x)\leq2x^{\beta}, \quad x\in [0,+\infty ), $$

by Theorem 3.1 and Lemma 2.7, we have that the necessary and sufficient condition for the existence of a $C^{2}[0, 1]$ positive solution of BVP (4.2) is

$$\begin{aligned}& 0< \int_{0}^{1}a(s)s^{\beta}(1-s)^{\alpha-3} \,ds< +\infty, \\& \lim_{t\rightarrow 0+}t^{\alpha+\beta-2}\int_{0}^{1}a(ts)s^{\beta}(1-s)^{\alpha -3} \,ds=0, \\& \lim_{t\rightarrow 1-}\int_{0}^{1}a(ts)s^{\beta}(1-s)^{\alpha-3} \,ds=\int_{0}^{1}a(s)s^{\beta}(1-s)^{\alpha-3} \,ds. \end{aligned}$$