On solutions of a class of three-point fractional boundary value problems

Abstract

Existence results for the three-point fractional boundary value problem

$$\begin{aligned}& D^{\alpha}x(t)= f \bigl(t, x(t), D^{\alpha-1} x(t) \bigr),\quad 0< t< 1, \\& x(0)=A, \qquad x(\eta)-x(1)=(\eta-1)B, \end{aligned}$$

are presented, where $A, B\in\mathbb{R}$, $0<\eta<1$, $1<\alpha\leq2$. $D^{\alpha}x(t)$ is the conformable fractional derivative, and $f: [0, 1]\times\mathbb{R}^{2}\to\mathbb{R}$ is continuous. The analysis is based on the nonlinear alternative of Leray–Schauder.

1 Introduction

In recent years, due to the wide application in many engineering and scientific disciplines as the mathematical modeling of systems and processes in the fields of physics, chemistry, aerodynamics, electrodynamics of a complex medium, polymer rheology, etc., the fractional differential equations have been widely studied. An extensive literature related to the existence of solutions of boundary value problems for fractional differential equations addressed by the use of various nonlinear functional analysis method. For example, fixed point theory [5, 7, 9–11, 19, 21, 25–28, 32, 33, 39, 50, 52, 56], the Mawhin continuation method [3, 6, 54, 57], the Green function method [5, 44, 45], the integral operator method [4, 8, 13, 14, 17, 22, 30, 31, 35, 36, 38, 49, 51, 53], the upper and lower solution method [12, 15, 18, 29], the numerical method [40–43, 46, 55], and the technique of barrier strips [4, 16, 20, 24, 34, 37].

In [24], Kelevedjiev got the existence of the solution by using the technique of barrier strips. Then some researchers studied the solvability of vary boundary value problems under the barrier strip conditions. For example, in [32], by using a nonlinear alternative of Leray–Schauder, the existence results for the second-order three-point boundary value problem are obtained,

$$\begin{aligned}& x''(t) = f \bigl(t, x(t), x'(t) \bigr),\quad 0< t< 1, \\& x(0)=A,\qquad x(\eta)-x(1)=(\eta-1)B, \end{aligned}$$

where $\eta\in(0, 1)$, $f: [0, 1]\times\mathbb{R}^{2} \to\mathbb{R}$ is continuous, and $A, B\in\mathbb{R}$. After that, the barrier strip technique was used to research the solvability of the difference problem [16] and the time scale problem [34]. Recently, in [20, 37], the author obtained the existence of solutions for the fractional Dirichlet boundary value problem

$$\begin{aligned}& D^{\alpha}x(t)= f \bigl(t, x(t), D^{\alpha-1} x(t) \bigr),\quad 0< t< 1, \\& x(0)=A, \qquad x(1)=B, \end{aligned}$$

under barrier strip conditions, where $1<\alpha\leq2$ is a real number, $D^{\alpha}x(t)$ is the conformable fractional derivative, and $f: [0, 1]\times\mathbb{R}^{2}\to\mathbb{R}$ is continuous.

To the best of the authors’ knowledge, there were few papers discussing the solvability of the multi-point fractional boundary value problems with the technique of barrier strips. Our effort is to use the nonlinear alternative of Leray–Schauder to the unreached areas. In this paper, we consider the following fractional boundary value problem:

$$\begin{aligned}& D^{\alpha}x(t)= f \bigl(t, x(t), D^{\alpha-1} x(t) \bigr),\quad 0< t< 1, \end{aligned}$$

(1.1)

$$\begin{aligned}& x(0)=A,\qquad x(\eta)-x(1)=(\eta-1)B, \end{aligned}$$

(1.2)

where $1<\alpha\leq2$ is a real number, $D^{\alpha}x(t)$ is the conformable fractional derivative, $\eta\in(0, 1)$, $f: [0, 1] \times \mathbb{R}^{2}\to\mathbb{R}$ is continuous, and $A, B\in\mathbb{R}$. We note that if x is a solution of (1.1), (1.2), then there exists $\xi\in(\eta, 1)$, such that $x'(\xi)=B$. Accordingly, the boundary value problem

$$\begin{aligned}& D^{\alpha}x(t)= f \bigl(t, x(t), D^{\alpha-1} x(t) \bigr),\quad 0< t< 1, \end{aligned}$$

(1.3)

$$\begin{aligned}& x(0)=A,\qquad x'(1)=B,\quad 0< \eta< 1, \end{aligned}$$

(1.4)

can be considered as a limiting case of the problem (1.1), (1.2) when $\eta=1$. Consequently, our result for problem (1.1), (1.2) gives an existence result for problem (1.3), (1.4).

It is true that the conformable derivative has some controversy. Some researchers think that the conformable derivative does not contribute “new mathematics”. The conformable derivative for differentiable functions is equivalent to a simple change of variable $D^{\alpha}[f(x)] = x^{1-\alpha} f'(x)$. It was noted that a criticism of the conformable derivative is that, although conformable at the limit $\alpha\to1$ ($\lim_{\alpha\to1} D^{\alpha}f =f'$), it is not conformable at the other limit, $\alpha\to0$ ($\lim_{\alpha\to0} D^{\alpha}f \neq f$) because $x^{\alpha}/\alpha$ is undefined at $\alpha=0$.

However, some other researchers think that the conformable derivative and its generalizations can still be interesting and valuable, specially leading to some physical insight with use in the applied settings. We refer the reader to [1, 2, 47, 48] for details as regards the conformable fractional derivative.

The main results of the paper is based on the following nonlinear alternative of Leray–Schauder.

Theorem 1.1

([32] (Nonlinear alternative))

Assume thatUis a relatively open subset of a convex setKin a Banach spaceE. Let$N: \overline{U}\rightarrow K$be a compact map and assume$p\in U$. Then either

• (1) Nhas a fixed point inU̅; or

• (2) there is a$u\in\partial U$and$\lambda\in(0, 1)$such that$u=\lambda Nu+(1-\lambda)p$.

The paper is organized as follows. In Sect. 2, the definitions of the conformable fractional order derivative and integral are given. In Sect. 3, by the use of the technique of nonlinear alternative of Leray–Schauder and barrier strips, the existence of the solution is obtained. In Sect. 4, some examples are presented to illustrate the main results.

2 Conformable fractional order calculus

Definition 2.1

([23])

Suppose $\alpha\in(n, n+1]$, $u:[0, \infty)\rightarrow R$, and u is nth-order differentiable for $t>0$. Then the αth-order fractional derivative of u is defined as

$$\begin{aligned}& D^{\alpha}u(t)=\lim_{\varepsilon\rightarrow0}\frac {u^{(n)}(t+e^{\varepsilon t^{n+1-\alpha}})-u^{(n)}(t)}{\varepsilon} \end{aligned}$$

provided the limit of the right side exists.

If u is αth-order differentiable on $(0, a)$, $a>0$, and $\lim_{t\rightarrow0^{+}}D^{\alpha}u(t)$ exists, then define $D^{\alpha}u(0)=\lim_{t\rightarrow0^{+}}D^{\alpha}u(t)$.

Lemma 2.1

([13])

Let$t>0$, $\alpha\in(n, n+1]$. Function$u(t)$isαth-order differentiable if and only ifuis$(n+1)$th-order differentiable, moreover,

$$\begin{aligned}& D^{\alpha}u(t)=t^{n+1-\alpha}u^{(n+1)}(t). \end{aligned}$$

Definition 2.2

([23])

Let $\alpha\in(n, n+1]$, αth-order fractional integral is defined as

$$\begin{aligned}& J_{0+}^{\alpha}u(t)=I^{n+1} \bigl[t^{\alpha-n-1}u(t) \bigr]=\frac{1}{n!} \int ^{t}_{0}(t-s)^{n}s^{\alpha-n-1}u(s)\,ds, \end{aligned}$$

where $I^{n+1}$ is the $(n+1)$th-order integral.

Remark 2.1

With Lemma 2.1 and Definition 2.2, for $\alpha\in(n, n+1]$, $i=0, 1, \ldots, n$, there hold

$$\begin{aligned} D^{\alpha-i} \bigl[J_{0+}^{\alpha}u(t) \bigr] &=t^{n+1-\alpha}D^{n+1-i} \bigl[I^{n+1} \bigl(t^{\alpha-n-1}u(t) \bigr) \bigr] \\ &=t^{n+1-\alpha}I^{i} \bigl[t^{\alpha-n-1}u(t) \bigr]. \end{aligned}$$

Lemma 2.2

([23])

Let$a \geq0$, $f: [a, b] \to\mathbb{R}$satisfy,

• (i) fis continuous on$[a, b]$,

• (ii) fisαth-order differentiable on$(a, b)$.

Then there exists $c\in(a, b)$ such that

$$D^{\alpha}f(c)=\frac{f(b)-f(a)}{\frac{1}{\alpha}b^{\alpha}-\frac {1}{\alpha}a^{\alpha}}. $$

Given $\alpha\in(n, n+1]$. Define

$$\begin{aligned} C^{\alpha}[0, 1] =& \bigl\{ u\mid u(t)=J^{\alpha}_{0^{+}}x(t)+C_{n}t^{n}+ \cdots +C_{1} t+C_{0}, \\ & x(t)\in C[0, 1], C_{i}\in\mathbb{R}, i=0, 1, \ldots, n \bigr\} . \end{aligned}$$

By the linearity of integral operator $J^{\alpha}_{0^{+}}$, the space $C^{\alpha}[0, 1]$ is a linear space. For $u\in C^{\alpha}[0, 1]$, according to Remark 2.1, there are $D^{\alpha-i}u(t)\in C[0, 1]$, $i=0, 1, \ldots, n$. Let

$$\Vert u \Vert _{\alpha}= \bigl\Vert D^{\alpha}u \bigr\Vert _{0}+ \bigl\Vert D^{\alpha-1}u \bigr\Vert _{0}+ \cdots+ \bigl\Vert D^{\alpha -n}u \bigr\Vert _{0}+ \Vert u \Vert _{0}, $$

where $\|u\|_{0}=\max_{t\in[0, 1]}{| u(t)|}$. The following lemmas obtained in [13] are fundamental to our main results.

Lemma 2.3

([13])

The space$(C^{\alpha}[0, 1], \| \cdot\|_{\alpha})$is a Banach space.

Lemma 2.4

([13])

The set$F\subset C^{\alpha}[0, 1]$is sequentially compact if and only ifFis uniformly bounded and equicontinuous, i.e., for$\forall\varepsilon>0$, $\exists\delta>0$, s.t. for any$| t_{1} -t_{2}|<\delta$, $\forall u\in F$, ${i=0, 1, \ldots, N-1}$, we have

$$\bigl\vert D^{\alpha-i}u(t_{1} )-D^{\alpha-i}u(t_{2}) \bigr\vert < \varepsilon,\quad\quad \bigl\vert u(t_{1} )-u(t_{2}) \bigr\vert < \varepsilon. $$

Lemma 2.5

([13])

Assume that$u \in C[0, 1]$with a fractional derivative of order$\alpha\in{(n,n+1]}$that belongs to$C(0,1) \cap L(0,1)$. Then

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

for some$c_{k}\in\mathbb{R}$, $k=0, 1, \ldots, n$.

Now, we present the Green function.

Lemma 2.6

Given$y \in C[0,1]$and$1<\alpha\leq2$, $1<\eta<2$, the unique solution of

$$\begin{aligned}& D^{\alpha}w(t)+y(t)=0,\quad 0< t< 1, \end{aligned}$$

(2.1)

$$\begin{aligned}& w(0)=0,\qquad w(\eta)-w(1)=0 \end{aligned}$$

(2.2)

is

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

where

$$ G(t,s)= \textstyle\begin{cases} s^{\alpha-1}, & {0\leq s \leq t \leq1 , s \leq\eta};\\ s^{\alpha-1}-ts^{\alpha-2}-\frac{s^{\alpha-2}t(1-s)}{\eta-1}, & {0 \leq\eta\leq s \leq t \leq1};\\ ts^{\alpha-2}, & {0 \leq t \leq s \leq\eta \leq1};\\ \frac{ts^{\alpha-2}(s-1)}{\eta-1}, & {0\leq t \leq s \leq1, \eta \leq s}. \end{cases} $$

(2.3)

Proof

Applying Lemma 2.5, we reduce Eq. (2.1) to an equivalent integral equation,

$$ \begin{aligned} w(t)&=-I^{\alpha}y(t)+c_{0}+c_{1}t \\ &=- \int^{t}_{0}(t-s)s^{\alpha-2}y(s)\,ds+c_{0}+c_{1}t, \end{aligned} $$

for some $c_{0}, c_{1} \in\mathbb{R}$. By the boundary condition (2.2), we have

$$\begin{aligned}& c_{0}=0, \\& c_{1}= \int^{\eta}_{0}\frac{\eta- s}{\eta-1}s^{\alpha-2}y(s)\,ds - \int ^{1}_{0}\frac{1-s}{\eta-1}s^{\alpha-2}y(s)\,ds. \end{aligned}$$

Therefore, the unique solution of problem (2.1), (2.2) is

$$ w(t)=- \int^{t}_{0}(t-s)s^{\alpha-2}y(s)\,ds+ \int^{\eta}_{0}\frac{t(\eta - s)}{\eta-1}s^{\alpha-2}y(s)\,ds - \int^{1}_{0}\frac{t(1-s)}{\eta -1}s^{\alpha-2}y(s)\,ds. $$

For $0 \leq\eta\leq t \leq1$, one has

$$ \begin{aligned} w(t)&=- \biggl( \int^{\eta}_{0}+ \int^{t}_{\eta} \biggr) (t-s)s^{\alpha -2}y(s)\,ds + \int^{\eta}_{0}\frac{t(\eta- s)}{\eta-1}s^{\alpha -2}y(s)\,ds \\ & \quad- \biggl( \int^{\eta}_{0}+ \int^{t}_{\eta}+ \int^{1}_{t} \biggr)\frac {t(1-s)}{\eta-1}s^{\alpha-2}y(s)\,ds \\ &= \int^{\eta}_{0}s^{\alpha-1}y(s)\,ds+ \int^{t}_{\eta} \biggl(s^{\alpha -1}-ts^{\alpha-2}- \frac{s^{\alpha-2}t(1-s)}{\eta-1} \biggr)y(s)\,ds \\ &\quad + \int^{1}_{t}\frac{ts^{\alpha-2}(s-1)}{\eta-1}y(s)\,ds \\ &= \int^{1}_{0}G(t,s)y(s)\,ds. \end{aligned} $$

For $0 \leq t \leq\eta\leq1$, one has

$$ \begin{aligned} w(t)&=- \int^{t}_{0}(t-s)s^{\alpha-2}y(s)\,ds + \biggl( \int^{t}_{0}+ \int ^{\eta}_{t} \biggr)\frac{t(\eta- s)}{\eta-1}s^{\alpha-2}y(s)\,ds \\ &\quad - \biggl( \int^{t}_{0}+ \int^{\eta}_{t}+ \int^{1}_{\eta} \biggr)\frac {t(1-s)}{\eta-1}s^{\alpha-2}y(s)\,ds \\ &= \int^{t}_{0}s^{\alpha-1}y(s)\,ds+ \int^{\eta}_{t}ts^{\alpha -2}y(s)\,ds+ \int^{1}_{\eta}\frac{ts^{\alpha-2}(s-1)}{\eta-1}y(s)\,ds \\ &= \int^{1}_{0}G(t,s)y(s)\,ds. \end{aligned} $$

The proof is complete. □

3 Existence results

Theorem 3.1

Let$f: [0, 1]\times\mathbb{R}^{2} \to\mathbb{R}$be continuous, $A \in\mathbb{R}$, $B \geq0$. Suppose there are constants$L_{2} \leq L_{1}$such that$L_{2} - B < 0 \leq L_{1}$and

• (1) $f(t, x, p)\geq0$, for$(t, x, p)\in[0, 1]\times [A+\frac{L_{2}-B}{\alpha-1}, A+B+\frac{L_{1}}{\alpha-1} ]\times[L_{1}, L_{1} + B]$;

• (2) $f(t, x, p)\leq0$, for$(t, x, p)\in[0, 1]\times [A+\frac{L_{2}-B}{\alpha-1}, A+B+\frac{L_{1}}{\alpha-1} ]\times[L_{2}-B, L_{2}]$;

• (3) $\frac{L_{2}-B}{1-\eta} \leq f(t, x, p)\leq\frac{L_{1}}{1-\eta }$, for$(t, x, p)\in[0, 1]\times [A+\frac{L_{2}-B}{\alpha-1}, A+B+\frac{L_{1}}{\alpha-1} ]\times[L_{2}-B, L_{1} + B]$.

Then the problem (1.1), (1.2) has at least one solutionxsuch that

$$L_{2}-B \leq \bigl(D^{\alpha-1}x \bigr) (t)\leq L_{1} +B,\qquad A+\frac{L_{2}-B}{\alpha-1} \le x(t) \le A+B+\frac{L_{1}}{\alpha-1}. $$

Proof

By the use of the Tietze–Uryshon lemma there exists a continuous function $g: \mathbb{R}^{2} \to[-1, 1]$ such that

$$\begin{aligned}& g(x, p)=1, \quad\mbox{on } \biggl[A+\frac{L_{2}-B}{\alpha-1}, A+B+\frac {L_{1}}{\alpha-1} \biggr]\times[L_{1}, L_{1} + B]; \\& g(x, p)=-1, \quad\mbox{on } \biggl[A+\frac{L_{2}-B}{\alpha-1}, A+B+\frac {L_{1}}{\alpha-1} \biggr]\times[L_{2}-B, L_{2}]. \end{aligned}$$

For each integer $n\geq1$, set

$$f_{n}(t, x, p) = f(t, x, p) + \frac{1}{n}g(x, p). $$

Then

$$ f_{n}(t, x, p)>0, $$

(3.1)

for $(t, x, p)\in[0, 1]\times [A+\frac{L_{2}-B}{\alpha-1}, A+B+\frac {L_{1}}{\alpha-1} ]\times[L_{1}, L_{1} + B]$;

$$ f_{n}(t, x, p)< 0, $$

(3.2)

for $(t, x, p)\in[0, 1]\times [A+\frac{L_{2}-B}{\alpha-1}, A+B+\frac {L_{1}}{\alpha-1} ]\times[L_{2}-B, L_{2}]$.

Consider the boundary value problems

$$\begin{aligned}& D^{\alpha}x(t)= f_{n} \bigl(t, x(t), D^{\alpha-1} x(t) \bigr), \quad0< t< 1, \end{aligned}$$

(3.3)

$$\begin{aligned}& x(0)=A, \qquad x(\eta)-x(1)=(\eta-1)B. \end{aligned}$$

(3.4)

Making the change of variables $w(t)=x(t)-\mu(t)$, where $\mu(t)=Bt+A$. It is clear that $x(t)$ is a solution of (3.3), (3.4) if and only if $w(t)$ satisfies

$$\begin{aligned}& D^{\alpha}w(t)= f_{n} \bigl(t, w(t)+\mu(t), D^{\alpha-1} w(t)+D^{\alpha-1} \mu (t) \bigr), \end{aligned}$$

(3.5)

$$\begin{aligned}& w(0)=0,\qquad w(\eta)-w(1)=0. \end{aligned}$$

(3.6)

Define $T_{n}: C^{\alpha}[0, 1] \to C^{\alpha}[0, 1]$ as

$$ (T_{n} w) (t)= \int^{1} _{0}G(t, s)f_{n} \bigl(s, w(s)+ \mu(s), D^{\alpha-1}w(s) + D^{\alpha-1}\mu(s) \bigr)\,ds, $$

(3.7)

where $G(t, s)$ is the Green function defined in Eq. (2.3). The standard arguments show that $T_{n}: C^{\alpha}[0, 1] \to C^{\alpha }[0, 1]$ is completely continuous. Furthermore, the solvability of the problem (3.3), (3.4) is changed as the existence of the fixed point of the operator $T_{n}$.

Now, we are in the position to show that the operator $T_{n}$ has a fixed point $w_{n}$ that satisfies

$$\begin{aligned}& L_{2}-B\leq \bigl(D^{\alpha-1}w_{n} \bigr) (t)\leq L_{1},\quad t\in[0, 1], \end{aligned}$$

(3.8)

$$\begin{aligned}& \frac{L_{2}-B}{\alpha-1} \leq w_{n}(t)\leq\frac{L_{1}}{\alpha-1},\quad t\in[0, 1], \end{aligned}$$

(3.9)

for all $n\in N$. Once this is achieved, then, by combining (3.7), (3.8), (3.9) and Lemmas 2.3, 2.4, the sequence $\{ w_{n}\}$ has a subsequence which converges in $C^{\alpha}$-topology to $w_{0}$, and then $x(t):=w_{0}(t)+\mu(t)$ is a solution of (1.1), (1.2) such that

$$\begin{aligned}& L_{2}-B \leq \bigl(D^{\alpha-1}x \bigr) (t)\leq L_{1} +B, \\& A+\frac{L_{2}-B}{\alpha-1} \le x(t) \le A+B+\frac{L_{1}}{\alpha-1}. \end{aligned}$$

Define U as the open and bounded neighborhood of $0\in C^{\alpha -1}[0, 1]$ such that

$$\begin{aligned}& U= \biggl\{ v\in C^{\alpha-1} [0, 1] \Bigm|\frac{L_{2}-B}{\alpha-1}< v(t) < \frac{L_{1} }{\alpha-1}, L_{2}-B< D^{\alpha-1}v(t)< L_{1} \biggr\} . \end{aligned}$$

To prove that $T_{n}$ has a solution $w_{n} \in\overline{U}$ such that (3.8) holds, it suffices to verify, in view of Theorem 1.1, that if $w\in\overline{U}$ satisfies Eq. (3.6) such that

$$ w(t) = \lambda(T_{n} w) (t) $$

(3.10)

for some $\lambda\in(0, 1)$, then $w\in U$, i.e., for $0< t<1$,

$$\begin{aligned}& \frac{L_{2}-B}{\alpha-1}< w(t) < \frac{L_{1} }{\alpha-1},\quad \mbox{and}\quad L_{2}-B< \bigl(D^{\alpha-1}w \bigr) (t)< L_{1}. \end{aligned}$$

(3.11)

Now let $w\in\overline{U}$ satisfies Eq. (3.6) for some $\lambda \in(0, 1)$. Since $L_{2}-B\leq(D^{\alpha-1}w_{n})(t)\leq L_{1}$, by Lemma 2.2, there exists $c \in(0,t) \subset(0, 1)$ such that

$$w(t)-w(0)= \bigl(D^{\alpha-1}w \bigr) (c)\cdot\frac{t^{\alpha-1}}{\alpha-1} $$

and

$$(L_{2}-B)\cdot\frac{t^{\alpha-1}}{\alpha-1}\leq w(t)\leq L_{1} \cdot \frac {t^{\alpha-1}}{\alpha-1}. $$

Let $x(t)=w(t)+\mu(t)$, then $x(t)$ satisfies

$$ L_{2} - B \leq \bigl(D^{\alpha-1}x \bigr) (t) \leq L_{1} + B $$

(3.12)

and

$$(L_{2} - B)\cdot\frac{t^{\alpha-1}}{\alpha-1} + Bt + A \leq x(t)\leq L_{1} \cdot\frac{t^{\alpha-1}}{\alpha-1} + Bt + A. $$

In particular

$$ A + \frac{L_{2}-B}{\alpha-1} \leq x(t)\leq A + B + \frac{L_{1}}{\alpha-1}. $$

(3.13)

Suppose that $D^{\alpha-1}w(t_{0})=L_{1}$ for some $t_{0}\in[0, 1]$. We claim that $t_{0}<1$. In fact, due to $w \in C^{\alpha-1}[0, 1]$ and $w(\eta) = w(1)$, by the use of the Lemma 2.2, there exists $\xi\in(\eta, 1)$ such that $D^{\alpha-1}(\xi)=0$. Taking into account the condition (3), integrating Eq. (3.5) from ξ to 1 yields

$$\begin{aligned} D^{\alpha-1}w(1) =& D^{\alpha-1}w(\xi) + \int_{\xi}^{1} \bigl(D^{\alpha}w \bigr) (s)\,ds \\ =& \int_{\xi}^{1} f_{n} \bigl(s, w(s) +\mu(s), D^{\alpha-1}w(s)+D^{\alpha-1}\mu (s) \bigr)\,ds \\ \leq& (1-\xi)\frac{L_{1}}{1-\eta} < L_{1}. \end{aligned}$$

Hence $D^{\alpha}w(t_{0})\leq0$ because $D^{\alpha-1}w(t)$ attains its maximum at $t_{0}$.

On the other hand, by (3.13) we get

$$\begin{aligned} D^{\alpha}w(t_{0}) =&\lambda f_{n} \bigl(t_{0}, w(t_{0})+\mu(t_{0}), D^{\alpha -1}w(t_{0})+D^{\alpha-1}\mu(t_{0}) \bigr) \\ =&\lambda f_{n} \bigl(t_{0}, w(t_{0})+ A+ Bt_{0}, L_{1} +t_{0}^{2-\alpha}B \bigr)>0. \end{aligned}$$

This contradiction proves that $D^{\alpha-1}w(t_{0})< L_{1}$. Analogously, we have $D^{\alpha-1}w(t_{0})>L_{2}-B$. Thus we get

$$\begin{aligned}& L_{2}-B< D^{\alpha-1}w(t)< L_{1}. \end{aligned}$$

(3.14)

Inequality (3.14) together with the relation $w(t)=w(0)+D^{\alpha -1}w(d)\cdot\frac{t^{\alpha-1}}{\alpha-1}$ implies that

$$\begin{aligned}& \frac{L_{2}-B}{\alpha-1}< w(t)< \frac{L_{1} }{\alpha-1}. \end{aligned}$$

(3.15)

This completes the proof. □

Analogously, we can obtain the following result.

Theorem 3.2

Let$f: [0, 1]\times\mathbb{R}^{2} \to\mathbb{R}$be continuous, $A \in\mathbb{R}$, $B < 0$. Suppose there are constants$L_{1}$, $L_{2}$such that$L_{2} \leq L_{1} +2 B$, $L_{2} \le B < 0 \leq L_{1}$and

• (1) $f(t, x, p)\geq0$, for$(t, x, p)\in[0, 1]\times [A+B+\frac{L_{2}-B}{\alpha-1}, A+\frac{L_{1}}{\alpha-1} ]\times[L_{1}+B, L_{1}]$;

• (2) $f(t, x, p)\leq0$, for$(t, x, p)\in[0, 1]\times [A+\frac{L_{2}-B}{\alpha-1}, A+B+\frac{L_{1}}{\alpha-1} ]\times[L_{2}, L_{2}-B]$;

• (3) $\frac{L_{2}-B}{1-\eta} \leq f(t, x, p)\leq\frac{L_{1}}{1-\eta }$, for$(t, x, p)\in[0, 1]\times [A+B+\frac{L_{2}-B}{\alpha-1}, A+\frac{L_{1}}{\alpha-1} ]\times[L_{2}, L_{1}]$.

Then the problem (1.1), (1.2) has at least one solutionxsuch that

$$L_{2} \leq \bigl(D^{\alpha-1}x \bigr) (t)\leq L_{1},\qquad A+B+ \frac{L_{2}-B}{\alpha-1} \le x(t) \le A+\frac{L_{1}}{\alpha-1}. $$

Accordingly, we get the following corollaries as consequences of Theorems 3.1 and 3.2 for the boundary value problem (1.3), (1.4).

Corollary 3.1

Let$f: [0, 1]\times\mathbb{R}^{2} \to\mathbb{R}$be continuous, $A \in\mathbb{R}$, $B \geq0$. Suppose there are constants$L_{2} \le L_{1}$such that$L_{2} - B \leq0 \leq L_{1}$and

• (1) $f(t, x, p)\geq0$, for$(t, x, p)\in[0, 1]\times [A+\frac{L_{2}-B}{\alpha-1}, A+B+\frac{L_{1}}{\alpha-1} ]\times[L_{1}, L_{1} + B]$;

• (2) $f(t, x, p)\leq0$, for$(t, x, p)\in[0, 1]\times [A+\frac{L_{2}-B}{\alpha-1}, A+B+\frac{L_{1}}{\alpha-1} ]\times[L_{2}-B, L_{2}]$.

Then the problem (1.3), (1.4) has at least one solutionxsuch that

$$L_{2}-B \leq \bigl(D^{\alpha-1}x \bigr) (t)\leq L_{1} +B,\qquad A+\frac{L_{2}-B}{\alpha-1} \le x(t) \le A+B+\frac{L_{1}}{\alpha-1}. $$

Proof

It suffices to note in the case $\eta=1$ that the boundary condition $x'(1)=B$ implies that $L_{2}-B< w'(1)=0< L_{1}$ which is what is required in applying the condition (3) to show $t_{0}<1$ in the proof of Theorem 3.1. □

Corollary 3.2

Let$f: [0, 1]\times\mathbb{R}^{2} \to\mathbb{R}$be continuous, $A \in\mathbb{R}$, $B < 0$. Suppose there are constants$L_{1}$, $L_{2}$such that$L_{2} \leq L_{1} +2 B$, $L_{2} \le B < 0 \leq L_{1}$and

• (1) $f(t, x, p)\geq0$, for$(t, x, p)\in[0, 1]\times [A+B+\frac{L_{2}-B}{\alpha-1}, A+\frac{L_{1}}{\alpha-1} ]\times[L_{1}+B, L_{1}]$;

• (2) $f(t, x, p)\leq0$, for$(t, x, p)\in[0, 1]\times [A+\frac{L_{2}-B}{\alpha-1}, A+B+\frac{L_{1}}{\alpha-1} ]\times[L_{2}, L_{2}-B]$.

Then the problem (1.3), (1.4) has at least one solutionxsuch that

$$L_{2} \leq \bigl(D^{\alpha-1}x \bigr) (t)\leq L_{1},\qquad A+B+ \frac{L_{2}-B}{\alpha-1} \le x(t) \le A+\frac{L_{1}}{\alpha-1}. $$

4 Some examples

Example 4.1

Let $\alpha=\frac{3}{2}$, $A=0$, $B=\frac{1}{2}$, $\eta=\frac{1}{5}$, consider the following problem:

$$\begin{aligned}& D^{\alpha}x(t)= f \bigl(t, x(t), D^{\alpha-1} x(t) \bigr),\quad 0< t< 1, \end{aligned}$$

(4.1)

$$\begin{aligned}& x(0)=0,\qquad x \biggl(\frac{1}{5} \biggr)-x(1)=- \frac{2}{5}, \end{aligned}$$

(4.2)

where $f(t, x, p) = \frac{t^{2}}{8} \sin(x^{2}+t^{2})+ p^{3}$.

Choose $L_{1} = 1$ and $L_{2} = -\frac{1}{2}$, then $L_{1}+B=\frac{3}{2}$, $L_{2}-B=-1$ and

$$\begin{aligned}& \biggl[A+\frac{L_{2}-B}{\alpha- 1}, A+B+\frac{L_{1}}{\alpha-1} \biggr] = \biggl[-2, \frac{5}{2} \biggr], \\& \biggl[\frac{L_{2}-B}{1-\eta}, \frac{L_{1}}{1-\eta} \biggr] = \biggl[- \frac{5}{4}, \frac{5}{4} \biggr]. \end{aligned}$$

After a simple computation, we have

• (1) $f(t, x, p)\geq1 \ge 0$, for $(t, x, p)\in[0, 1]\times [-2, \frac{5}{2} ] \times [1, \frac{3}{2} ]$,

• (2) $f(t, x, p)\leq0$, for $(t, x, p) \in[0, 1] \times [-2, \frac{5}{2} ] \times [-1, -\frac{1}{2} ]$,

• (3) $-\frac{5}{4} < -1 \leq f(t, x, p) \leq\frac{5}{4}$, for $(t, x, p) \in[0, 1] \times [-2, \frac{5}{2} ] \times [-1, \frac{3}{2} ]$.

That is to say that all the conditions of Theorem 3.1 are satisfied, so the problem (4.1), (4.2) has at least one solution x such that

$$-1 \leq D^{\frac{1}{2}} x(t) \leq\frac{3}{2},\qquad {-}2\leq x(t) \leq \frac {5}{2},\quad \mbox{for } 0 \le t \le1. $$

Example 4.2

Consider the following problem:

$$\begin{aligned}& D^{\alpha}x(t) = f \bigl(t, x(t), D^{\alpha-1} x(t) \bigr),\quad 0< t< 1, \end{aligned}$$

(4.3)

$$\begin{aligned}& x(0)=0,\qquad x'(1)=-1, \end{aligned}$$

(4.4)

where $\alpha=\frac{3}{2}$, $A=0$, $B=-1$, $f(t, x, p) = t^{2} \sin(x^{2} + t^{2}) + p^{3}$.

Choose $L_{1}=2$ and $L_{2}=-2$, then $L_{1}+B=1$, $L_{2}-B=-1$, $L_{2} \leq0=L_{1} +2B$, $L_{2} \le B < 0 \leq L_{1}$ and

$$\begin{aligned}& \biggl[A+B+\frac{L_{2}-B}{\alpha- 1}, A+\frac{L_{1}}{\alpha-1} \biggr] = [-3, 4]; \\& \biggl[A+\frac{L_{2}-B}{\alpha- 1}, A+B+\frac{L_{1}}{\alpha-1} \biggr] = [-2, -3]. \end{aligned}$$

After a simple computation, we have

• (1) $f(t, x, p) \geq0$, for $(t, x, p) \in[0, 1] \times [-3, 4] \times[1, 2]$,

• (2) $f(t, x, p) \leq0$, for $(t, x, p) \in[0,1] \times [-2, -3] \times[-2, -1]$.

All the conditions of Corollary 3.2 are satisfied, so the problem (4.3), (4.4) has at least one solution x such that

$$-2\leq D^{\frac{1}{2}} x(t) \leq2,\qquad {-}3\leq x(t) \leq4, \quad\mbox{for } 0 \le t \le1. $$